Mathematics Applied to Continuum Mechanics
SIAM'S, Classics in Applied Mathematics series «rr;ists of books that were previously allowed to Lcr out of- print. These books are republished by SIAM as a professional service because they continue to be important ramrurces for mathematical scientists. F.d.itor-inlChief R+7ben E. t)'Matlle*y, Jr., University of Washington Editorial Board Richard A. 13nGtkiï, University of Wiuorrsin ,Maclist}fl Leah Edelsrcin-Veshct, University of British Columbia Nicholas J. High,-tlsti, University of h1
Problems in the Nation! Sciences
Johan G. F. L3t•linfante and Bernard Koltt>rant, A Survey of Lie Gaups and Lie Algebras with Applications and Currlpttuuional Methods James. M. Chtet,,^a, Numerical Analysis: A Second Course Anthony V Fiaoco and 4iarrh l`: Mc-LÂxrnick, Nonlinear Progrrunrning: Sequential [?nrrrrurrcrinEcl Minimization Techniques E H. Clarke. Optimization and Nonsnuxrth Analysis Chxrr-Ke F. [ :arrier and Ctrl E. Pearson. Ordinary Differential I;quutrrms
Lct, Bteirtkrn, Probability R. J3
G. M. Wing, An Introduction to Irrreirrirnit imbedding
Ahrarlia u u Berman and Hubert J. l'letrHr14 N 16, IVunriegtrtive Matrices in the 2'vfntfit°}nurirr4d SCieliCis C }lvi L. Nlatsk;a3aria► n, Nunlinettr Programming
*Carl Friedrich Gauss, Theory of the Combination of Observations Least Subject to Errors: Part One, Part Two, Supplement. Translated by G. W Stewart ltich.-rrcl
trudltcr:torr tir M{urtx Analysis
U. M. Ascher, R. M. M. Miatdxij, and R. Russell, Numerical Solution of Boundary. lûlue Problems for Ordinary Differential Equations K. E lircnan y,S. L. Campbell, and L. H. Petzolri, Numerical Solution of Irtit.itté-Vuluc Problems in Differentia/Algebraic Equations
Charles L. Lawson and Richard J. Hanson, Sobtirrg Least Squares Problems J. E. l)enttrs, Jr. atu.l Rtrben B. SclSnarlel, Numerical Methods for Urtctnrsrrcù,teci Optitni^urt[»r arid Nonlinear Equations Richard E. Barlow and Frarik Prnsc.itian, Mttthentatiat! Theory of Reliability
C rnelius Lanczos, Linear Differential Opercators Richard Bellnrtn, Introduction to Matrix Analysis, Second Edition F3etesfttircl N. farlett, The Symmetric Fiipenaxtfue Prob+[em
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11
Classics in Ap p li e d Mathematics {ccintinued} Rkh:trd Halxrinsm, Mathematical Models: Mechanical L'ibrurioris, Ii,piilation Dynamics, and Traffic Flow Perer W. M. John, Statis,tietrl Design and Analysis of Experiments Tamer Balar arid Geer Jan Olsder, Dynamic Noncooperative Gaine Theor y, Second Edition Emanuel Parxen, Stochastic Processes Pettar Kokotovit, Hassan K. Khalil, and John C7' Rellly, Singular Pertunccriun Itilezfinds in Conn til: Analysis and Design Jean Dickinson Gibbons, Ingram (Akin, a tx! Milto n Sobel, Selecting and Ordering Populations: A New Statisticril Methodology James A. Murdock, Pt:rtuthartüins: Theory and Methods ivar Ekiklatxl :and Roger Térn:nxt, Convex Ariatiysis err ul litriatia,nul Problems Ivor StaEcgokl, Boundary Value Problems of ll+iatPtematicul Physics, Volumes I rand IP J. M. Ortega and W C. RI7einlxildt. IterYrzi :+e Solution of Nonlinear F.gt u uions in Serenté Variables David Kindcrlehrer a n d Guido Statnfutcciiia, An Inr rud trctiarn nr Variational Inequalities and Their Applications E Nattcrer, The Mathematics of Computerized Tomography Avinash Cr. Kak and Malcolm Slaney, Principles af Camaptateri z ed Tomogrrtiphic Imaging R. Wong, Asymptotic Approximations of Integrals O. Axeissitti a nd V. A. i3arlcer, Finite Element Solution of Boundary Value Problems: Theory and Computation Da vid R. $rillinger, Time Series: Data Analysis and Theory J oe l N. Fr:ttiklin, Methods uf Mathematical Economics: Linear ana Nonlinear Programming, Fixed•Puint Theorems Philip Hartman, Ordinary Diffèrentiarl I:yuutions. Second Edition Pvlichacl D. lntriligator, Mathematical Optunritrrnarn and Economic Theory rr Elliptic Problems Philippe C . Ciarlet, The Finite Element Method Jr Jane K. Cullum and Ralph A. Willoughby, I..c:itc-zus Algorithms /Or Large Symmetric k.igcntaallzr Computations, VOL 1: Theory Vidyasagar, Nonlinear Systems Analysis, Second Edition Robert Mattheij and Ja:tp t+9ulcna:ir, Ordinary Differential EQxacairir; in Thcory and Prcictic:e Shand S . Gupta and S. l'ancliapakes:rn, Mtcltipie Decision P'rDrcclfrt,s- Tlter)r} find Mctfiutie,lurr, of Selecting and Ranking Populations Eugene L. Allgower ari d Kurt Georg, Introduction to Ntariteracal Continuation Merh ❑ aIt Leah E.clvlstein-Keshet, Mauherncuircli Models in Biofriry
Heinz-Otto Krciss arid lens Lorenz, Initial P3014 nLlar'}• Volute Problems and the NaavterStokçs Equal ions J. L. Hodges, Jr. and E. L. Lehmann, Basic Concep ts uf Probability and Statistics, Second Edition George F. Carrier, Max Kraxik, and Carl E . l'e,irso ii, Functions of u Complex Variable; Theory and Technique Friedrich Pululshcitn, Optimal Design of F.xperirnenzs l5rael GANT, Niter Li ncasier, and Leiba Rodman, Incuriurtt Subspaces of Matrices with Applications Lue A. Seg el with G. H. Hand e lm a n, Mathematics Applied to Continuum Mechanics
i ii
A beetle Makin vravcs on a water our niirrlacc. From experience with ship wavIM ane would expect the disturbance to be confined to a V-shaped region behind the object. Here they precede the object; perhaps bredt have a mysterious organ to reverse the natural order of cause and cf e.ct! Or perhaps 'here is more to we propagation +han meets the eye. See the fxneluding ponton of Section 9.1, The photograph appeared on the cover of Science thé (Nev. 14. 1969. Copyright 1969 by for American Association for the Adunnecr-rr.,,i of Science). in connection with an article by V. A. Tudor'. Wave-Making b} lhhirhgLE BeetleslGyrinidrie)." pp- $9799. Reproduction is by permtssion.j
9 P Mathematics Applied to Continuum Mechanics b ti Lee A . Se ge l With additional material on elasticity by (3. H. Handelman
Society for I nd ustrial acid Applied Mathematics Philadelphia
Copyright
2C07 by th e Society for lusltttiirutl and Applied l^l,itlii nt:i[u s
This SIAM edition is an unabridged republication of tl;r work Itrtit published b y iiie Macmillan Publishing t i., New York, NIT I098 7li `i 4 i l I
Ail rii;Itis reserved. Primed ill. the United Slimes itt America. No pan or this hook may I v reproduced, stored. or Ir ;instlliUc`J in any manner wttluvztt the Milli en rierm[syiLiiti clj the publisher- For itifcuinaiiitiit, write li> the Society ittr Industrial and Applied Matlwmaitics, 3600 University City Science (renier, Philadelphia. l'A 1 41104-16M. Library of Congress Cataloging-in-Publication I?a tat Segel, Lee A. Maihcutalits applied t o Coniinittuln mechanics I ice A. S L'i;l'I; wit h :il lt li[iL71l:lI material un elasticity by 0.11. Handelman. cm. -- (Clsthsica tti atph+licxl ut:ttheutattics ; 51) Originally lttihlisixccl; New Yo rk ; M:xinilLiu, c1977, lrtcludrs bibliographical retcrenci,s and index. ISBN- I 0: 0 -H9ii71-620-9 ISBN-13: 978- 0-89871-620 -7 1. ( nutria' mechanics, 2, Mattliclnalics. L Handelman, il. 11. II. Title. QA1308.2.S4 i 2ûe7 5H --dc22
2006052201
T
Slarn
is a registered trademark.
To Rutkr,e/
C ontents FDk1ïW[H;1] TO Ti [F: CLASSICS EI]ITH rh
xvii
PREFACE CONVENTIONS
xih uxïïi
PART A GEOMETRICAL PREREQUISITES FOR THREE -DIMENSI O NAL CONTINUUM MECHA NICS CHAPTER I VECTORS, DETERMINANTS, AND MOTIVATION FOR TENSORS 1.1
3
Vectors in a Cartesian coordinate system, transformation Of vector
components, and the summation convention 3 Prajecnorr 4 f Transforrnurton of coordinates—Geometric approach 5 1 Notattorr t' f Vectors— Alyehruwc point of view y 1 Vectors—Geometric point of view LO f Ret}ici' of linear dependence 12 1.2 Determinants and the permutation symbol 14 The permutation symbol is f Determinants 16 / The" ed" rule 20 1.3 The consistency requirement 23 The consistency of Newton's second law 23 f Physical laws: General statement rJersusparticular numerical i;ersion 24 I Guessing the ren cnr transformation law from the consistency regeurrement 25 .
1.4 The tensor as a linear tianstormation f.lneur transformations 28 f The stress !ensui induces a linear transformation of dïferenttal area into differertfialforce 29 1 A linear transformation of differential area to differential force can he identified with the Stress tensor 31
CHAPTER 2 CARTFSIAN TENSORS 2. I Tensor algebra 33 Definitions and elementary properties 34 / Special results for second order tensors 41 f Isotropic tensors 43 1 T!w vector associated with an unite mmetric tensor 46
2.2 The eigenvalue problem
49
Eigetrralues and eiqenvectors of syn vrretrir sensors
ix
55 f Principal axes 57
Contents
2.3 The calculus of tensor functions
6o
TireorerrZ5 fi1r tlPrltottriYS UfferisCir fields 61 f integral rïteorenrs 64
{
Representation theorems 6,7 APPENDIX
2.1 Some basic equations of continuum mechanics
72
PART B PROBLEMS 1Iti CONTINUUM MECHANICS CHAPTER
3
Viscous 1't,t.iIps 3.1
The Navier Stokes equations
78
Analysis of the local velocity field 7ô f Assumptions that underlie the eyteafixxi !to J Der m artonr of the fuJK.l (-quartons S 2 t Boundary cexr.stiu g5 f Incompressible viscous flow 86 cortl s
3.2 Exact solutions 93 Solutiot I: Ham rc3uerte
flow y; r Solution 1: Plane Poiseuille flow g; f Su/liftoff 3: Rayleigh rinperisWe 'kw 98
3_3
On boundary layers
1o5
Comparative magnitudes of t•isc•ous urul truvse•rel Wrrns 105 ? Re}redfds plumber icr7 / Boundary layer equations for steady.. _flow past a fiat plate leg l3otttadar y conditions l l
f
3.4 Boundary layer flow past a semi - infinite flat plate t 15 Formulation t l6 / Blasius similarity solution i 16 t Defects in the Blasius suiirtioz t 18 f Boundary layer separation 120 , Slightly rïsc'uus uniform floor past streamlined bodies 121 / Slighuh• viscous uniform flow past bluff bodies 122
3.5 \torticity changes in viscous fluid motion t 29 F'ortui{v cctnvertinn and stretching i29 ( Viscous roJrffrr'fy diffusion and boundary genrratinn 13 t f Voracity trr fwer-cltinerrsinnalJ1 oxs 132 f Vwtirify in istY+us flo w s J1;2 j Surrtrnarr of the role of voracity 134 puricla
3.6 Slow viscous now past a small sphere
1 35
tùrrriukriiun 136 i Snllution 137
APPENDIX 3-1
Navier — Stokes equations in cylindrical
coordinates t38 AIEENDIX 3.2 Generation of confidence in the boundary layer equations by construction of a finite difference scheme for their solution 139
xi
Contents CHAPTER 4 FOUNDATIONS OF ELASTICITY
4.1 Analysis of local motion 144 Strain tensor in material coordinates 144 I Geometrical inerpre•tation of strain components 147 J Straw, tensor in spatial coordinates 149 / The rotation rumor 15o j Principal axes of strain 161 f Compatibility ry , uatiom 153 1 Some examples of strain 155
4.2 Hooke's constitutive equation and some exact solutions 1 59 Gr neralized Hooke's laui 159 J Interpretation of the elastic coefficients via exact _solutions 162 / Tension of a çylindrical bar 163 f Shear of a rectangular hat i64 / Compression of a rectangular parallelepiped I66 4.3 Final formulation of the problem of linear elasticity [68 Srrn merry of `general equations, boundary conditions, and minci eDrrditians 169 if Novier's equations 171 1 Beltrami-Mlc! equations 17
4.4 Energy concepts and the principle of virtual work 174 Energy balance 174 J Prinrrple of virtual work 177 1 Uniqueness theorems 178 Potential energy rninurrization in equilibrium 179
4.5 Snrne effects of finite deformation 184 Kinematle s I li4 r Comparison with linear theory for nonlinear elasticity i l g / Simple shear 1g9
187 , A ronsti /dive equation
CHAPTER 5 SOME EXAMPLES OF STATIC PROBLEMS IN ELASTICITY
5.1 Bending o f beams 194 Bending by rc•rrnrnal couples, Formulation 194 j Solution [ g6 f lrr rerprerar ion 1 99 j Jnfrrxhit•tion to the engineering diary of be n din g Basic assumptions zoo f Equations of engineering bending theory 204 / Boundary I orrelrrions 208 J Traveling-wave solutions 2to / Buckling of a f►eatrr 21 z / Variational methods titi elasticity 214
5,2 Si. Venant torsion problem
219
Warping,/renrrion 220 • Stress funrtlon 224 f Further proper:1es of the stress function 226 1 Modified stress function 229 f Rectangular cross section 230 f Elfiptiraf cross section 233 J St. Venant principle 235
5.3 Some plane problems 240 Equations for plane strain 241 f Airy's stress function 241 / Boundary conditions 242 J Polar coordinates 245 / Kirsch pro biens. Stress concentration 248 f Plane stress ego / Generalized plane stress 252 f Concluding remarks 255
/
xü
Consents
CHAPTER 6 INTRODUCTION TO DYNAMIC PROBLEMS IN ELASTICITY
6.1 Elastic waves iii unbounded media
259
Dilatational and rotational wave equations 259 1 Waves via the Helmholtz representation z6c f Plane wave solutions 261
6.2 Propagation of discontinuity surfaces 264 A condition on mild discontinuities 265 f Further jump conditions 266 f Saloir for discontinuity velocities 269 f Orientations of discontinuity sur}aces z70
6.3
Reflection of plane shear waves 271 Formulation 272 f Attempted solution with a reflected wore 273 I Additional reflected wore 274 I interpretation of results 276
6.4 Elastic surface waves 278 Formulation 278 f Solution: Plane waves that decay with depth 279 Î Analysts of the solution z8i f Occurrences of Rayleigh wages 283
6.5 Internal reflection
285
6.6 Love waves 289 Formulation and solution 290 f Examination of the nature of dispersion 293 f Further characteristics of the solution 294 f Concluding remarks 395
PART C WATER WAVES CHAPTER 7 FORMULATION OF THE THEORY OF SURFACE WAVES IN AN INVISCID FLUID
7.I Boundary conditions
301
Kinematic boundary candi lion 3c i f Basic facts about surface tension 343 f Quick derivation of a dynamic boundary condition 304 / Detailed derivation of the dynamic boundary condition for an ins, iscid fluid 3p5 f Forces on a surface element 308 f C'on.segttencrs of local equilibrium 3 i 3
7.2 Formulation and simplifica ti on 319 Equations for two-dimensional wales in an infinitely wide layer of inuiscid Add 320 f Static and dynamic pressure ;2z f Nonaiimensionalization 324 Linearization 326
f
7.3 Order-of-magnitude estimates, nondirnensianalization, and scaling 327 ,Estimating the size of terms tir equations governing water waves 328
f
Scaling 329 j dise of dimensionless scaled variables 331 f Pressure scale 33 2
Ka i
Contents
C HAPTER 8 S OLUTI O N IN THE LINEAR THEORY 8. 1 A solution of the lineari z ed equations 335 Assumption ofa solution of exponential type 135 ` Verification that t311 t-rirrcfiric rrrS 3 36 ;' Return la dimensional notables 339 frrterprt'fulrnrr of art.
are satisJted
solution 340
8.2 Initial value problems: Periodic rases 346 Superposing solutions of exponential type 346 r Another way to write the real part of complex seams 349 r Satisfying .rural rorrditrous 352 8.3 Aperiodic initial values 354 Abandoning dimensionless r ariahlecs 354 , Solution Ha Fourier integrals 35s A yurllitath.e feature of the solution 3s9 i Superposition and the delta Junction 360
APPENDIX 8.1 Bessel functions 365
CHAPTER 9 GROUP SPEED AI1I7 GROUP VELOCITY 9.1 Group speed via the method of stationary phase 3 69 Needier are asymptotic approximation 369 i Stationary-phase approximation 3® Application of die r ppro_ imptïon 370 / Interpretation. Cutup .speed 374 Phase speed 375 Special conditions Hear extrema of group sly 377 I Some applications ter flow pan anodes 378
9.2 Experiments and practical applications 381 Experiments on the Collapse of a rectangular hump of Water 382: Comparison of theort and es rerimem 384 ! Practical application of theory 3 88
9.3 A kinematic approach to group velocity 391 Properties trf sfarf'ft Carving ware trains 392 j The phase junction in regions with an rarrveying number of naves 393 Integral and dir)ererruial expressions of stare rurrserrsJtion 395 i Group and phase refortir 39r5 f Energy propagation 397 I Asymptotic form of the surface: A terse derivation 1913
9.4 Ship, duck. and beetle waves 401 Catlsequerrers of steady* motion 403 / Steady waves induced bi' a point source 404 J Partial drff[^retrtial equation for the phase ji nctiorr 408 A PPENütX 9.1 The method of stationary phas e —An informal discussion 412 Motivation 412 j DrueJ4apnrent of a theorem 414 1 Heuristic derivation of rire
key approximation 415 f
GerrPralftatrort 416
xi v
Cantina
CHAPTER 10
ltii]NLINEAtt E FF EC TS
10.1 Formation of perturbation equations for traveling waves 418 Change of variables 419 / Consequence of traiwliny-wave assumption 420 Sertes solution 421 / Determination of succes_sic.e Sets of equations 422 / Remarks 425
10.2 Traveling finite amplitude waves 426 -
Izwest order equations 426 f Second order equations 428 I Normalization 43o /
Completion of second order calculations 43 r f Third order calculations 432 ir Discussion 434 / Resnnwrt ccue 436 / Near resonance 438 f Sperial resonant
solutions to cantpure with experiment 439 / Recapitulation 441 1 Fuld remarks - Formalism and rigor 442
PART D
V A RIATI O 1ti A L METHODS A N D EXTREMUM P R IItiCIPLES CHAPTER 1 CALCULUS OF VAR lATIQ]ItiS
11.1 Extre m a of a function—Lagrange multipliers 447 Unconstrained extrennplization of afurtction 4r}$ f Consrrained exrrerntalizQtion of a function 450 J inequality corrslramts 455
11.2 Introduction to the calculus of variations 461 The brach istocltrune 461 f A general extremdlizatiwr problem 462 j The Eider equation 463 f Natural boundary conditions 467 / Are solerrtarrs to the Euler eqrtatiorr Fxtrrrnuls? 471
11.3 Calculus of variations—generalizations 476 .4 More deritwtwes 476 f B. More functions 477 / C. More independent variables 478 ! D. Integral constraint 479 ! E. Furxttarraf r:onsrraint 48 r / F. End point free to move nn a gmerr curve 483 -
APPENDIX 11.1
Lernma A
496
A PPENI3IJt i1_2
Variational notation 497
CHAPTER 12
CHARACTERIZATION CF EIGENVAI.UPS AND EQUILIBRIUM STATES AS EXTREMA
12.1 Eigenvalues and stationary points Soo Three stationary value problems 5oa f A parti culier self-adjoint positive problem 5o2
C o n ter: ts
121 EigenvaIues as minima and the Ritz method
504
Motivation 504 / Specification of a linear operator Soli I The lowest eigenw due 5o9 f Higher eigenvahies5to 1 The Ritz method 513 / Validity and utility of the Ritz rNerhod 515 I Generalization 5 1 7 1 Example: Transverse vibrations of a tapered hollow beam 518 f Natural boundary conditions 521
12,3 The Courant maximum -minimum principle 528 A problem in vibration theory 52g l The max-rnin principle Sty f Application to the Ritz meihat1511
12.4 Minimal characterization of linear positive problems
531
Particle equilibrium as a minimum of porenrull energy 534 f The loaded membrane ç37 ! Equivalence of an inhornagE ntou.c equation and a noirrimizarion problem 539 / The Ritz nmerhod applied to the torsion problem 54 t A['PEN
nix l2. 1
Self adjouit operators on vector spaces 546 -
Reif vei for spaces 547 f Scalar products 549 / ,Linear self-adjoin! operators 55I f Eigenuahw problems 552 1 Positive operators 554 / flrlhnnnriial eherrunrs 556 f fnhornngeneaus problems 55g f Th e adjoins 560 f Borsch and Hither; spaces y62 i Completely emiirruous operators 56i
BIBLIOGRAPHY
57[
HINTS AND ANSWERS
579
INDEX
585
Fo r e w or d to the
Cl assi cs E dition
in the 196es, continuum mechanics was undergoing al revolution from :1 "feudal" science-1.4th fiefs of fluid mechanics, elasticity, and esoteric combinations of those sciences with eIectromagnetics —ta a "cosmopolitan' Ibpolit:an' uiencc consisting of a thtsmry of everything continuums. Love's Treatise un tiffe Mathematical Theory of Elasticity and Lamb's Hydrrniynatralics had brought rigor tua these areas, and Physilc articles rigori2ed the more abstract "rational Truesdl'twoHnhicer mechanics' approach. To further complicate matters . , Science was smarting to become computerized, with codes to approximate solutions TO incre Singly cuLnplcX problems. Against this backdrop, Lee Segel set out rra write a book for a course in applied mathematics at Rensselaer. The text for the first semester was Mathematics App! cd to Deterministic Problems in the Natural Sciences, also published by SIAM. The present volume was the text for the second semester of this course. As texts fier such a course, the first enjoyed more success than the second. This is [wanly because mathematics was starting to be applied to as wide spectrum of disciplines in the physical and social sciences and partly because continuum mechanics took on less of a role as a source for interesting and challenging ilia [LIMA kill problems. Still, in the decades since this book's first publication, ccmtinuulma approaches were attempted for the description of a multitude of mechanics prxobleins, ancluLling mixtures, reacting fltaails, heterogeneous solids, multiphase flows, and structureinteractions. Successful treatment of each added crlmplic.uions that required an understanding of the fundamentals of continuum mechanics that went beyond the manipulations required for describing Navrer—Stokes fluids or linear elasticity. This text possesses as quality that nukes it an important scientific tool that can he recommended to anyone interested in understanding tige vagaries of continuum mechanics. It is aimed at explaining the science of continuums mechanics with clarity winning out over rigor, with explanation and motivation cinphasi:ed over manipulation. This is a book from which it is possible to learn continuum mechanics. As such
it shares a niche with many others. both modern and classical. Fiut the student who uses this book to add understanding to a rigorous text will reach as transcendent appreciation of the science of continuum mechanics.
Donald Drew Rensselaer Polytechnic Institute
Preface
1114 istext develops and uses mathematics la analyze cont uum models of
fluid flow and solid deformation. It is intended for upperlevcl undergraduates and graduate students in applied mathematics, science. and engineering: all of the material has been tested in various courses for such students. There is an emphasis on the process of achieving understanding, even at the expense of limiting the topics covered Although applied mathematics has flowered far from its original roots in physics, continuum mechanics remains a core course. One reason is that the concepts that theoreticians use to organize experience are developed most fully in classical subjects such as continuum rriechanics. Subjects of comparable depth -electromagnetism and quantum physics, for example—do not share with continuum mechanics a concentration on relatively familiar natural phenomena and therefore are not so well suited to the rapid development of physical intuition. Moreover, because it deals with problems that are at once deep and widely understandable, continuum mechanics has been for two centuries a major source of significant new applied mathematics. Boundary layer theory provides a paradigm for the development of a concept from one invented for a particular class of p roblems in continuums mechanics to one presently used in a wide variety of applications. Another example is provided by the study of water waves, which continues to stimulate the generation of new techniques for analyzing nonlinear partial differential equations. This volume is a sequel to C. C, Lin and L. A. Segel's Mathematics Applied to Deterministic Problems ïn the Natural Sciences (hereafter referred to as I). It is permeated with the spirit discussed in the preface of That work. In particular, a case-study approach is used, heurism dominates rigor, and brevity is sometimes sacrificed to permit depth of exposition. In I various ideas were introduced in extremely simple contexts, so that their essence could be clearly seen. The danger that these ideas would thus seem rather trivial is countered here by providing more realistic examples. To give one illustration, regular perturbation theory was introduced in I to approximate the solutions of simple algebraic and ordinary differential equations; in the present volume the theory is applied to the system of nonlinear partial differential equations with a "free" boundary that governs water waves. Material in I is a prerequisite for the present work only to the extent that here we assume familiarity with the basic mass and momentum conservation laws of continuum mechanics_ Remarks that refer to I can be skipped. .
xis
Preface
Various themes that appear there are taken up again here, but in a selfcontained fashion. There is no doubt, however, that background from I is useful here, for the totality of material in both volumes is designed to provide a thorough understanding of many topics. As in I, '`we have assumed that the potential reader has had an introductory college course in physics and is familiar with calculus and differential equations. . . . We make considerable use of such topics as directional derivatives, change of variables in multiple integrals, line and surface integrals, and the divergence theorem." Complex analysis is used in a few exercises and in Section 5.3. This volume is divided into four parts: A. Geometrical prerequisites for three-dimensional continuum mechanics_ B. Problems in continuum mechanics_ C. Water waves_ D. Cxtremum principles. The table of contents gives a more derailed outline. Part A features a careful discussion cf how the requirement that natural laws be "essentially the same" in different Cartesian coordinate systems leads to the introduction of Cartesian tensors_ The bulk of Part A is devoted to a fairly formal development of tensor algebra and calculus. Tensors are applied in Part B to derive the constitutive equations for viscous fluids and for linearly elastic solids. (For those who desire it, paths are indicated that reach these equations without the use of tensor machinery.) Boundary layer theory plays the central role in the discussion of viscous fluid motion. A variety of static and dynamic problems are available once the equations of elasticity haw been derived. The treatment of water waves in Part C provides the most detailed discussion in the text, The concept of group velocity plays a central part, and the role of nonlinearity is considered. Experimental verification of the theory is also discussed. In this area, therefore, the text brings the reader to the frontiers of modern research, Moreover, the material on elastic waves in Part B, taken together with that on water waves in Part C, provides at the same time an illustration of the unifying power of a single concept (wave motion), and also the refinements and modifications that are necessary when this concept is applied to different areas. The text concludes with a discussion in Parr I) of extremum principles, considered both as natural frameworks for physical laws and as bases for direct methods of computation. As Part D proceeds, simple concepts of functional analysis are introduced, since they form a natural language for
the discourse. There is considerable independence among the various parts and chapters. Part D, for example, can he begun at once, although knowledge of mechanics will lend urgency to some of the problems posed there. The material on
Preface
xxi
water waves (Part C) has as a prerequisite only the equations of incompressible inviscid fluid motion_ If one is willing to accept the governing equations more or less as axioms, much of the material on elasticity and fluid mechanics in Part B can be made quickly accessible_ Although the motivation for tensors in Chapter 1 requires knowledge of the stress tensor, if one is willing to forego extensive motivation it is possible to acquire the formal methods from Part A, Section 1.1, and Chapter 2. This book grew out of the course "Foundations of Applied Mathematics" introduced by G. 1i. Handelman at Rensselaer Polytechnic Institute around 1957 and taught every year since then, many times by the author. (A precursor of this course was given by A. Schild and Handelman at Carnegie institute of Technology, now part of Carnegie -Mellon University.) Special thanks are due to Dr. Handelman, not only for his general influence, but also for his willingness to write most of the material in the chapters on elasticity. (The principal author is responsible for final editing.) Other writers, teachers, colleagues, and students have all been influential. Joseph plum, Paul Davis, Donald Drew, Eitan Klein, Simon lvloehon, William Siegmann, and Marc Triebitz have been particularly helpful with the present volume_ Many secretaries have given excellent service_ As was the case with I, the publishers, particularly editors Everett Smethurst and Elaine Wetterau, have provided first-rate assistance. Finally, help with the index and other last-minute editorial matters was provided by Joel, Susan, Daniel, and Michael Segel. The author was partially supported in 1968-1969 by a Leave of Absence Grant from Rensselaer. Further support was received during 1971-1972 from National Science Foundation Grant GP33679X to Rensselaer and from a John Simon Guggenheim Foundation Fellowship_ That year was spent as a visitor to the Department of Applied Mathematics, The Weizmann Institute of Science, Rehovot, Israel; the author joined this department in September 1973, but he retains an association with Rensselaer Polytechnic Institute. He is thus formally indebted to both institutions for support during years of on-and-off writing; the support was always generously given and the acknowledgment is correspondingly warm_ L. A_ S.
C onventions
F
ACH chapter is divided into several sections (e.g., Section 5.2 is the second section of Chapter 5). Equations are numbered consecutively within each section_ Figures and tables are numbered consecutively within each chapter. When an equation outside a given section is referred to, the section number precedes the equation number. Thus " Equation (6.3.2)" refers to the second numbered equation of Section 6.3. Hut if this equation were referred to within Section 1 of Chapter 6, then the chapter number would he assumed and the reference would be to "Equation (3.2)," The fourth numbered equation in Appendix 3.1 is denoted by (A3. I.4). A double dagger (I) preceding an exercise, or a part thereof. signifies that a hint or an answer will he found in the back of this volume. The symbol L] signifies that the proof of a theorem has concluded. The preceding volume is referred to as "1." A brief bibliography of useful references (some repeated from E) can be found at the end of this volume. When one of these books is cited, the style "Smith (1970)" is employed_
1%; X1.11.
PA RT A Geometrical Prerequisite s for Three-Dimensional Continuum Mechanics
1 Vectors, Determinants, and Motivation for Tensors CHAP1 ER
T
ENSORS have both computational and conceptual value. We try to illustrate both aspects in Part A. We begin Chapter 1 with a review of vectors which emphasizes that more than direction and magnitude characterize these entities. In a consideration of how vector components transform when different Cartesian coordinate syslerns arc used, we introduce a standard summation convention that makes the required formulas more condensed and hence more manageable. Determinants are developed in a computational fashion with the aid of the alternating symbol. Most readers will be familiar with at least the principal results, but the alternating symbol is intrinsically worth studying, and it provides an approach to determinants that should he new and appealing. We make contact with the central content of the chapter when we turn to a detailed discussion of the fact that physical laws cannot intrinsically depend on the coordinate system used. From this discussion we are able to provide strong motivation for the definition of a tensor. An optional section provides another motivation, based on the connection between tensors and linear transformations. We consider only Cartesian tensors, which means that only Cartesian coordinate systems are allowed. Generalization re f our considerations are possible and useful, but they are beyond the scope of this book.
1.1 Vectors in a Cartesian Coordinate System, Transformation of Vector Components, and the Summation Convention Recall that a Cartesian coordinate system is based on tfsrec mutually perpendicular coordinate planes. These intersect in three mutually perpendicular lines called the x,-, x 2-, and x raxes, A positive direction is assigned to each axis. The coordinate system is called right- (left-)handed if the third finger of the right (left) hand points in the positive x 3 -direction when the thumb and index finger are pointed in the positive x i - and x 1 -directions,
respectively. As shown in Figure 1.1, a point P can be reached by following three displacements GA, AB, and BP parallel to the three coordinate axes. Select a unit of length_ If these displacements (positive or negative) have the 3
4
Vectors, Determinants, and Morivarion for Tensors [Ch. 1 s2
^r u
A
x3
FIGURE 1.1 _ Any paint F ran be reached by rhree displacements that are parallel ta the coordinate axes.
measured values x i , x2 , and x3 , respectively, then the coordinates of the point P are (x i , x2 , x 3). The distance OP is given by (xi + xx + x3)'12. Unit coordinate vectors ell), etas, and et' ) point along the respective axes. The above material wiilI be familiar to the reader, but we wish to emphasize that some fundamental assumptions are implied, Le., that we curt find three mutually perpendicular lines and that the law of Pythagoras holds. By contrast, on a spherical surface great circles play the role of lines. We can find mutually perpendicular great circles through any one point, but the taw of Pythagoras does not hold for distances measured along great circles. Although we can always conceive of a Cartesian coordinate system, the fact that we can use it to describe the space of our common experience is an experimental fact; i.e., we must verify experimentally that Pythagoras's theorem holds. Thus, in Figure 1.2, since OB _ OC and AB = AC, we should have 0A 2 } 0B2 = AB 2 and OC 2 + 0A 2 = AC 2 . If these relationships can be verified, we speak of the space as Euclidean. otherwise as nonEuclidean.* PROJECTION
A fundamental relationship in the Cartesian coordinate system is the rule of projection. if PQ is a displacement with components (u, u, w) and RS is a line with direction at angles (a, fl, y) to the coordinate axes, then the * The Euclidean nature of the space of common expetieucc has primarily been established indirectly by the agreement between observation and multitudes of theoretical predictions whose derivation assumes that space i Euclidean.
Set. 1.
i] vec tors in a Cartesian Coordinate System
5
A
FIGURE 1.2. Measurements on a construction of this sort con Ha principle be used in an effort in frird out whether or not "real" space is Euclidean. lidean.
projection of YÇ on RS is a COS tx f t' cos }.i + iv cos y.* This is a familiar result, but it is not trivial when the lines PQ and RS are skew to each tither. Readers should demonstrate this to their own satisfaction ]Exercise 1). We shall now use the rule of projection to study how components change when new axes are selected. A change of axes may result in a simplification of some problems - this is sufficient reason to study such changes, but there are more fundamental reasons, too, as we shall see. TRANSFORMATION OF COORDINATES—GFOMFTRtt' APPROACH
Consider two sets of Cartesian axes, primed and uriprimeci, with common origin O (Figure 1.3). I_,et a displacement OF have components (A ,, - 2 , A 3 ). Y^
ri ^
ir
e^
FIGURE 1.3. Two sets of Cartesian axes. rotated with respect to one another_ Either can be used to describe the drsplucernent P • We remind the reader that the projection of a point Pon RS is the rntcrsectiort c]t RS with the perpendicular dropped From P crnrp RS. The projection of f 2 is the union cif the prnjeriions of its points.
â
Vectors, Derermitrarrrs, anti Motivari ri
for Tensors ( h_
!
Denote the projection of a Y on the x'c -axis by x', , and so on. By the above rule of projection, xi =x l cos (x r ,x'r )+x i cos (x 2 ,x'r )+x 3 cos (x,,x r ), where (x; , x:) - (xt, x,) is the angle between the positive x ;-direction and the positive 3(i-direction. In general, 3
x; = ^ x; [ os (xi. xi),
j= i,2,3.
( 1 )
i= i
For the inverse transformation, the rule of projection* yields
X; : / x s
cos Wit , x;),
i
= t, 2 , 3.
(2)
This must be compatible with (1). To show that this is so, we substitute (I) into (2), obtaining 3 3 Xi =
EI 14E x k cos (xk , x j) cos (X j, x;).
j-
But l
E cos (xi, x^) cos (x^ , x^) ^ t i}
if i = k 1, k -=1,2,3; if i # k,
(3)
since the sum is merely the projection of a unit displacement along x i upon the xk-axis. By (3), then, we have the identity x ; T x; verified for each i.
NOTATION
The preceding calculations are simple examples of a type of manipulation that is very often encountered_ Experience indicates that the following new notational conventions, once they have been mastered, shorten and clarify the formulas_ iu a given equation, an index (subscript or superscript) is called a dummy if it is used in a summation_ Other indices are called free_ In (3), 1 and k are free subscripts and j is a dummy subscript. Range convention A free index takes on the values 1, 2, 3. The same free indices must appear in each term of an equation. • Equations (I) and (2) can easily be derived akgcbraicatly (Exercise 2).
See. 1.11
Verfors in a Carresian Coordinate System
7
Summation canventioo No index wiérthe written more than twice in a single term.* Suppose that an index, say t, appears twice ill a single term. Then the symbol , ws11 he terni_ understood to precede that From now on we shall use the range and summation eont.enfions unless an expficir stalemerit is made to the contrary. For example, instead of 3
X Tn
=
3
E E A+„ra Y
n = l, 2, 3,
pq,
P"-t Y'-t
we shall writs
X ,.n =
A.,,
Yp4
The same free subscripts, rn and it are used on both sides of the equation. The subscripts p and q are repeated, so that summation is implied_ Two special abbreviations occur repeatedly. The Kronecker delta is defined by ,
f,
r = jw
(4)
}.
The transformation symbol is defined by
r ; , = cos (x ;, x;).
(5)
Using these definitions and the summation convention, we see that (1) to (3) can be written briefly as
By the same reasoning as that used to obtain (8), we also have
(9)
[P,i fj& = ,
Readers should satisfy themselves before going further that because of our various definitions and conventions, the following four equations convey ' Vio &atiort of this requirement is [he biggest single cause of the errors !hat beginners make in manipulations involving the summation convention_
Vectors, Dewr'minants, and Motivation for Tensors 10.
precisely the same information, respectively, as the four equations just written
Readers will also find it helpful in rapidly acquiring familiarity with the conventions if they expand some of the compactly written formulas below_ Note that fix ^ try , in general. To remember the definition (5), memorize: The second subscript goes with the prime. The same phrase makes it easy to remember the transformation laws (6) and (7) and to avoid errors in future formulas. In 0), for example, the dummy subscript i appears an the unprimed quantity xi . Since the second subscript goes with theprirne, the first subscript of must also be i, The free subscript j is associated with the primed components on the left side of (6) and with the second {primed} position in ir.f As our first new calculation using the summation convention, we shall verify that the distance between two points is invariant under a rotation of axes. The distance d between two points P and Q is given ven b . (
d 2 = [ r(Q)
xi(P)]1xi(Q)
v
xr(P)].
In the new coordinate system, we have
l02 = [xiQ) — xi(P}] [x (Q) — x;(p)] = [xi(Q) -- xr1 11](1J[xk(Q) — xi,(P)Y1,By use of (fi) we sec that Lk;(Ql — 'c,{Pn L-xk(Q) — Adfl]r 5a.
( 10)
Only terms when k = r need be retained in (10), since /5 ;t = 0 when k # i, so one can write either = rx,(Q)
xr(P}7
- x.(P)1
The changes that have been made arc as follows - (i) the order of multiplying the real numbers .x ; and r,,, has been reversed, compared with (6): Iii) the free subscript i of (7t' has had its ''name" changed top (a permissible change.sinee it does not violate the rule against writing an index more than twice in a single tcrml; (iiit t he dummy suhstnpt ; of (4t' has had its "name changed to Q la permi+sibtc clhangrh and Iiv) in j9), both the dummy subsctipl j and the free subscripts t and k have been changed in a permissible way. ¶ There is a notation for the transformation symbol with which one does not have so irinembei that the second subscript goes with the prime hecausc the order 01 subscript's is immaterial In this notation cos tx„ xi') _ t r.; '_ +' ir . Because she extra primes are a bit burdensome and because a departure from coiltventionatitycreates 50111e confusion, + 10c shall nut adopt this notation here, but we do recommend it as intrinsically superior -
See_ t-)? Vecrors in a Currestan Coordinate SySIe»r
9
or
(c01 — [xk( 12) — xi,(P)7 kk(Q) xi(P)),
so that (d']^ = d 2 . VECTORS— ALGE B RAIC POINT OF VIE.1+V
Consider the position coordinates x ;c of a particle in motion so that x k = fk(r), where r is the time. In another coordinate system, the position will be given by x; gjt), i-e-, x1
- /k j fk( 1) _ gp{r )-
=
The velocity of the particle is = h(r)
in the xk-system
=
in the x,`.-system.
.
and j{r)
(Here - Or) Thus. from ( l 1), f; — f^}vk •
Multiply by /,) (and sum): — ^ ï ^iv y", =
=
Vf
We see that the transformations linking the velocity components o f and v1 arc the same as the transformations (6) and (7) for the displacement components_ We thus designate as vectors quantities x whose components x ; to (6) and (7). We speak of the displacementandxftrsomcig rector, the velocity vector, and so on Similarly, acceleration is a vector, since
= O r) = ki(r). Thus a vector" is an entity that is associated with three components in any Cartesian coordinate system. The components Li ; and v; in two coordinate systems are linked by the relations ti ;
âl;
t1ït
;,
.
(12)
11v and w arc vectors with components v i and w ; in some coordinate system, their sum v + w is defined as the entity with components ti ; + w 1 . If c is a real number, the scalar multiple or is the entity with components ct7 ; . It is easy to show that the sum of two vectors is a vector, as is the scalar multiple of a vector.
to
Vectors, Deterrrtinonrs, and Motivation for Îensora
[C h .
r
VECTORS--GEOMETRIC POINT OF VIEW
In showing above that 112 (d') 2 , we have verified that if Uk are the components of any vector, then its m ag nitude t , = (r*vk)t"2 is invariant under a rotation of axes. If z 0, then t r /v are the components of a dIrection vector or unit vector, which has unit magnitude . The geometric picture of a vector as an arrow (indicating magnitude and direction) implicitly assumes the validity of the operations used in an algebraic treatment, namely, (1) the decomposition into components along mutually perpendicular coordinate axes, and (ii) the addition of components along the axes to determine the resultant sum of two vectors. The correct geometrical description of vector that makes these assumptions explicit is as follows: .4 vector is a quantity with direction rind magnitude_ The sim of ewes vectors is determined by the parallelogram luR'. It can easily be shown that this geometric description is equivalent to the algebraic definition given above. To show that a vector sum requires the addi ti on of components, one simply projects the parallelogram along the three coordinate axes (Exercise 3). Note that quantities with equal direction and magnitude (which follow the parallelogram law) are regarded as the same vector. Also recall that from the geometric point of view, the scalar multiple cv is a vector with a magnitude I c i times the magnitude of if and a direction the same as (opposite to) that of v if r is positive (negative). The geometric point of view is useful in furnishing a proof of the following important result_* Theorem I. Let IL .j denote the determinant of the transformation matrix L with components f, i . Then )LI = + l if the primed and un1 primed systems are both right-handed or both left-handed, and I LI = if one system is right-handed and the other left handed_ Proof f Let us work in the primed coordinate system, which we assume for the moment to be right-handed. We can regard the kth row of the transformation matrix to be formed of the components of the unit coordinate vector etk', for —
-
et° = ccje&1
implies that
Hence the components
e ,t . eta = t.kieui - et =s = rio it = C. k , .
are just the t k; , for (5) certainly implies that coo. ti', I (13)
Thus by a standard vector theorem I L I = et . em A e{3 ",§ so that I LI has a magnitude equal to the volume of the rectangular parallelepiped spanned by et1), e12}. and e" ), and has a positive or negative sign depending on whether • Another proof is given in Example 1.2.4, 'r An algebraic proof is provided in Example 4 of the next section. 4 We use the symbol A to denote cite vector product_
Sec- . J - f 1
Vectors in a Cartesian Coordinate
Si Stmt
ti i
these vectors form a right- or left-handed system. in the present case this parallelepiped is a unit cube, so that the theorem is proved when the primed coordinate system is right-handed. Completion or the proof is left for Exercise 4. a REMARK. If the transformation between two Cartesian coordinate systems does not involve a reflection, the transformation is called a proper rotation. Otherwise the rotation is impropm, In this terminology. Theorem 1 stales that {
L
—I
for proper- rotations, for improper rotations.
14
Note that evert, vector /ras a direction cout a xntetyrtitrtde but that the converse is not true. For example, the finite rotation of a rigid body about a lixed axis can be specified by a direction and a magnitude. It is natural to represent such a rotation by an arrow with the specified dircx-tton and magnitude. But if we do this. we find t hat the deco mposition into components has no meaning. Consider, for example, the rotation of the unit square OABC through an angle oC (n/2) ,1`2 radians 130 about the GB axis seeFigure 1.4). The "natural" representation of this rotation is an arrow R along 0E with length {rrf2) J2. if R is a vector, its components must be vectors along the x- and y-axes, each of length rr f 2. But these hypothetical component vectors correspond to rotating the square rtj2 radians about OX and then rr/2 radians about O Y. leaving the square perpendicular to the plane. Obviously, the sum of these two components does not represent the same rotation as R. even though their vector sum is R. Thus the concept of component does not apply to a finite rotation; a
{
r1
F LGt; Rf t . 4_ The arro w
a
R provides u natural
represetlrarlQf7 for rotati on of
the squar e ()ABC.' about the fine UR, through an angle (1t{20, The direction (0]12. 8ut R is not a of R u indirrlrerl, and it magnitude should eabvioiesfe vector_
12
Vei tors, Dererrnir vats, oiid Morivariurl for Tenors
[Ch. I
rotation cannot be represented by a vector even though it has a direction and magnitude. See Exercises 5 and 2.6 for further information about finite rotations. KEVIEw OF LINEAR DEPENDENCE
Each vector V satisfies V = Yet l1 + V2 e(21+ 1/3031 where el" represents a unit vector in the direction of the xk-axis. The question may be asked: if we choose three arbitrary vectors AL, A2, and A3, is it possible to find three constants (c t , c^ c 3 ) such that an arbitrary vector V is expressible in the following form? V = ci A ; .
(I5)
Let the components of A ; be denoted by A ir . Since two vectors are equal if and only if their components are equal, the answer to our question depends on the solution to the linear system of equations C1A11 + C2AL2 + C3A13 s Vi , CLA21 + C2 A22 + c3A23 —
(16)
C 1 A 3L + c2A32 + C3A33 ` V3e
with the coefficient matrix
A
= (
AL1 Al2
A 13
An
A23 •
A32
A3i A32 A33
The reader will doubtless recall that if the determinant IA I of the matrix A does not vanish, then (16) has exactly one solution for the uriirnowns c t ,c 2I c3 , and hence there is a unique representation_ Otherwise, this is not the Case. If I AI = O, then thereexists at least oneset ofconstanis (c 1 , c 2 , c j ) # (0, Q. 0) such that c i A r =D
,
(17)
where the zero vector 0 has components (0, 0, 0) in any coordinate system. If (l7) hoIds, the vectors A, arc said to be linearly dependent. If I A # 0, they are said to be linearly independent. Any vector can be represented as a linear combination of three linearly independent vectors. When three vectors arc linearly dependent, we may express one of them as a linear combination of the other two, e.g_, A2 = ^LA t + k 3A3 -
Sec. 1.1]
Vet-tors in a Cartesian Coordinate
Svswm
t3
if one constructs the triangle of Figure 1.5 to represent the sum above, it is evident that the plane containing A, and A.3 must also contain A. . Three linearly dependent vectors are coplanar.
FIGURE 1. 5. A t , A2 , and A ! are linearly dependent, Therefore, flier
are
coplanar.
EXERCISES
:1. Demonstrate the projection rule discussed ai the beginning of the section, t2. [Derive (I) and (2) algebraically, starting with the relation x fee" = 3. Show that the "parallelogram law" of vector addition impliescomponeniwise addition, and conversely. 4. Complete the proof of Theorem I. 5. Let x be transformed into y by a right-handed iotatton through au angle O. The axis of rotation is given by a unit vector n. Show that
y— x =
(1 — cos (J)[x
—
(x • n)nl t sin Oiri n x).
6. (a) Simplify the result of Exercise 5 when [1 is small, to obtain
y- x='O A (b)
wherc6 -fln.
According to parr (a), what is the result of two successive small rotations? What does this imply about the possibility of specifying small rotations by a vector? (c) How small does B have to be for the formula of part (a) to give 10 per cent accuracy? $7. (a) Write down the matrix L for the coordinate transformation obtained by rotating 180C about the x 3 -axis. (See Figure 1.6.) (b) Calculate (and check by using common sense, which makes the answer obvious)the new components of a vector having components at in the old system. $S. 1f A„ A: and A are the components of a vector with respect to three different Cartesian coordinate systems, prove that the values of A are the same whether (ï) one transforms A i into A: directly or (ii) one transforms A i into A, and then A' into A . ,
14
vectors, Determinants, and Motivation for Tensors [Ch. I
Fi Ou RE 1_6. The prtl ►ird cuordenate system is rotated 180- with respect tu rite
r€ 3 -ttxis_
eurpmd,aboth
9. Consider the numbers 25
-9 25
3 5
4 5
B
—16 25
12 25
3 5
12
4
5
(a)
By verifying (8) and (9), show that I. can be considered a matrix describing the transformation between Ca rtesian coordina te systems rotated with respect to each other. (b) Point P has coordinates (0, 1, —1) in the unprimed system. What are its coordinates in the pruned system? (e) Show that 2x6 —Ix2+x3= I and %Ix, t i x 2 — 151 x 3 = 5 describe the same plane. 10. (a) Explain why the usual geometrical definition of a "vector product" of two vectors A and B yields a vector. (b) Explain why one does riot get a vector if the magnitude AB sin f] is replaced by ABeos 6 or AB sin 20. Substantiate your discussion with detailed calculations.
1.2 Determinants and the Permutation Symbol As was illustrated twice in the previous section. a number of mathematical problems reduce to questions wherein determinants play an important role_ Although readers will be familiar with at least the main results from the
Sec. 1.2J
Deae•rrninaarias and aht• Perarruaarrorr Symbol
15
theory of determinants, it is appropriate here to sketch that theory starting with first principles. The mode of presentation, via the permutation symbol rill,(to be defined below), will be new to many. Even leaving aside its value in determinant theory, the permutation symbol must become an object of familiarity, for it is used repeatedly in the remainder of the chapter. in particular, although the proofs of the main properties of determinants (Theorems 2 to 8) may be omitted, the "ed" rule of Theorem 1 l must be mastered. A somewhat abstract approach is now commonly used in introducing determinants, particularly to mathematicians but increasingly to engineers and physicists as well. For example, the important theorem that the determinant of the product of two matricesequals the product of their determinants is arrived at by the use of elementary row and column operations, expressed with the aid of elementary matrice& l lere we take a direct computational approach, as is perhaps fitting for applied mathematicians. The simplicity of notation allowed by the summation convention and the compact storage of information in the permutation symbol, however, give the direct treatment an unexpected elegance. In the remainder of the book we shall only need results for 3 by 3 determinants and the corresponding three-subscripted alternating symbol. We shall restrict our presentation accordingly. Nevertheless, it should be clear that generalization to n by n determinants is almost immediate. TH E PERMUTATION SYA7BO1.
Definition I. (a) A permutation of a list of integers is a listing of these integers in another order. Example; (312) is a permutation of (123) (b) A transposition is a special permutation in which two adjacent integers are interchanged. Example: (132) is a transposition of (312) in which the first two integers are interchanged. (c) A permutation is said to be even (odd) if it can be accomplished by an even (odd) number of transpositions. Example: (312) is an even permutation of (123) because it can be accomplished by two transpositions, as follows: (312) —' (132) —, (123). The trivial permutation (123) — (123) is even, because it can be accomplished by zero interchanges, or two [(123) (213) (I23)]. (d) The cyclic permutations of (123) are (123), (231), and (3E2). Cyclic permutations of (123) can be obtained by picking three consecutive integers from Figure I.7, proceeding in the clockwise direction. The anticyclic permutations of (123), namely (132). (213), and (321), can be obtained by proceeding in the counterclockwise direction. As the reader can easily show, all cyclic permutations are even, all anticyclic pe rmutations are odd. In Definition lc we have implicitly assumed that any permutation can be accomplished by a succession of transpositions and that all such successions which accomplish a given permutation are either even or odd. For proofs that these reasonable assumptions are valid, see texts on algebra such as .
16
Vectors, Determinants, and AMutivazion for Tensors [Ch. 1 p,nlicycii c
FIGURE 1.7. The cyclic and anticyclic
perrnularte3n s
F. W. Ficken, Linear Transformations and Matrices Prentice-Hall, 1967), Chap_ 4, Sec. E.
of th e first three integers -.
(Englewood Cliffs, N.J.:
Definition 2. The permutation symbol or alternator • 1, j, k = l, 2, 3, has 27 values, one for each of the 27 ordered sets of subscripts. These are defined as follows: E;pi
if two of the integers ï, j, k are equal,
-û
—.7
or if all three are equal, 1
(1)
if (i}k) is an even permutation of (123),
—I
if WO is an odd permutation of (123).
The following relations are almost obvious_ Formal proof is left to the reader. = - F:JikT
f; lk} - - F,i1k ,
&ï}k =
(2a, h) (3)
1.14i•
❑ LTFRMINANTS
Throughout this section, A and B will denote the matrices
A„
A 1z A13
A 2L
An A2.i
A3l
A32 A.33
PL [ PL2 PLI
B,
and (
B22 S23 -
B_ll B32 B33
We wish to associate a certain number with a 3 by 3 matrix like A, called its determinant. We shall use the notations det (A) or IA I or sometimes det (A1).
Sec. 1.21 Determinants and the Permutation Symbol Definition 3 IA
I
-E
17
01 4 iO4 2, A 3k -
1
(4)
To make sure that the reader has grasped the meaning of the permutation symbol, we shall write out the nonzero terms_ There are
IAI
t123A11A22A33 + t312A13A21A32 + E231Al2A23A31
+
E321A13 44 22
A ]1
32.411,423 A32 + 6 213Al2A21A33
` A 11 A 22 A 33 + A13A21A32 + Al2A23A]1 — A13422 A 31 —
A11
Az3
Theorem 1
A32 — Al2A21A33•
r
iAl = ^U ■ ArlAl2A k3
(5)
An alternative statement of the theorem is IA TT ] — A I, where A Tr is the transpose of A. Outline Cf proof. Corresponding, say, to the term !;231 A l2A23A31 in (4) is the term e312 A31A 1 z A23 in (5). Both terms consist of a product of the same A's, but what of their sign? The sign depends on the parity (evenness or
oddness) of the subscripts not in their natural order. That this is the same for both terms can be seen by noting that the number of interchanges required is the same as the number of interchanges necessary to change one order of the A's into the other. For example, in the present case. two interchanges are required: A 12A23A31'"' Al2A31A23+•
A31A1zA23.
Reading from left to right, we see that these two interchanges put the second subscripts into natural order: (231) - (213) —' (123). Reading from right to left, we see that two interchanges also put the first subscripts into natural order: (312) --. (132) ' (123). C -
Theorem 2. If B is the same as A except for the interchange of a pair of rows or columns, then IA I=— I B I. Proof. Suppose that B is obtained from A by the interchange of the first two rows so that B2, = A1f,
B11 —
A21, B31 = A31-
Then
EincAliA2JA3k , where in the last step we have interchanged the dummy subscripts i and j. The theorem follows by (2a). The same reasoning holds(Thisoftenul.) in the other circumstances. 0 IBI = EB i.
B B 3J<
E iik
A 2, A 1j A 3k =
18
Vectors, Determinants, and i4lotiixrtion Or Tensors [Ch, !
Theorem 3. If A has two identical rows or two identical columns, then J l = O. Proof. Interchange the identical rows or columns. The resulting matrix is unchanged and therefore has the sar -nc determinant as A, but by Theorem 2 its determinant is also — A. U
Theorem 4. If' B is the same as A except that a row or column of B is a multiple c of the corresponding row or column of B, then 181= r'IAI Proof: Left to the reader. Theorem S. If B the same as A except that the jth row (column) of B jth row (column) of A plus a multiple of the kth row (column) ofequalsth A. k j, then IB1 1AI. Proof Left to the reader. Definition 4. The minor M t of an element A ;i is the determinant of the matrix formed by deleting the ith row and jth column of A. The corresponding cofactor C';, is defined by [.'t, _ (
Theorem 6 (The Cofactor Expansion of a Determinant)_ of this theorem we do not use the summation convention,
in the statement
3 IA I =
Ai C i1 .
t—
1, 2, 3,
J -1
(6a, b)
3
1A1 = ^ A^^^: i= l
j
= I,
2, 1
Outline of Proof. The idea is to write out one of the sums in (4) or (5). For example, writing out the t-summation in (5), we find that
l AI = Ar[ F= i
A j2 Ak3 +
Az1C2jk A }z AA] + A 31 E 3.04 A riA^3
-
The factors of A L ,, A 2 L , and A 3L arc readily recognized as the appropriate for an expansion along the first column. We leave the full proof to cofatrs the reader. 0 Theorem 7. (We again employ the summation convention.) A3C
= IAIbr; ,
A ; t`^; t = I A i vf •
(7a, 13)
For (la), if p = i, we have a restatement of (6a). lip # [, the left zero, for it is the cofactor expansion of the determinant of a sideof(la) matrix whose pth and ith rows are identical. This follows from the fact that the elements of a given row of a matrix never appear in the cofactors of that row, since it is always deleted. h Proof.
Sec _ J_2) Deferminarit.5 arid the Permidtatiun Svnrbn!
4
Theorem S (Cramer's Rule).
Given the linear equations A ii xi = b, with coefficient matrix A, we have
(8)
I A ix; = b,C., k .
(9)
If TA l is nonzero, we can divide by it and obtain a unique solution to (8). If I AI = 0, there is no solution unless b,C kk = O. Note that the same reasoning as was used in Theorem 7 shows that the right side of f9) is the expansion of Dk I, where R k is a matrix that is identical with the coefficient matrix A except that its kth column is composed of the "right-side - elements h 1 , h z . and b } . Proof. We multiply the ith equation of (8) by the cofactor C E, for some fixed k, and sum en is CrkAi i X j
The result follows at once from (7h). [1 Theoremt 9 A A AA r = r oc A, ; AA .A •
F„,I Al
^.
{
1Qa, b)
Proof_ To prove (10a), we Erst note that unless r i s, and t are all different, the right side is the determinant of a matrix with at least two identical columns and so is zero by Theorem 3_ The left side is zero in this case by the definition of the alternating symbol_ If (rsr) — (123), then (10a) is just (5)_ If (rsr) is an even (odd) permutation of (123), the right side is therleternhirslnt of a matrix which differs from A by an even (odd) number of column interchanges and so equals (minus) I AF by Theorem 2. So does the left side. [I Corollary
I ABI = I/1)1Bi. Proof:
Left to the reader,
Illeurem 10
1
,14,
Al
A„
.
Air A la
A jr
A ka
Air
A t
=
t., ik f.
qr I A I .
41 1)
Suppose that at least two of the integers (ijk) are equal, or that at least two of the integers (pqr) arc equal_ Then the determinant on the left of (1 I) has at least two rows which are equal or two columns which are equal_ In such a case, both sides of l 1 I) equal zero. if (ijk) _ (I23), the determinant on the left of (1 l) differs from] A I because of an even (odd) number of column interchanges if (pq) is an even (odd) permutation of (123), The validity of (1 1) in this case follows from Theorem 2, as it does when (ijk) is a permutation of (123. n
Proof.
20
Vectors, Determinants, and Motivation for Tensors (Ch. 1 THE "ED" RULE
Theorem 11
I 8ilk ep4k Proof.
64'6" — 45' 15 jP .
(i2)
From Theorem 10 with r = k and A i, = ai f , 64 5i& eiit e P4k = (5}p bie a fr akp a kq kJ, vip
Expanding the determinant by its third row, we find that
big bik I ei,«epqk
bkp
°It q
Alp alp
6a + a aip a,^ kx ik 6h, b k ^
( 13) ^ bro 6ip 6,01^Je
bip 1 664 1 +3 .o 6k 6.4alp 64 afil
Interchanging the columns of the first determinant in (13) gives the desired result: b;p S;q ^'ke^k = b a lp
= b ;p ô;^
— a ^ bJp . Q
„^
As stated in (3), a cyclic permutation of subscripts does not change the sign of the alternator. Hence we can write (12) in various ways by performing a cyclic permutation of subscripts on either or both alternators: Eilk £kp4
—
Ei.AEpQk — 8kiiEkpq — E1ki4kp — a4 15 l4 — (5 44 6 1p"
(l 4)
We shall term this remarkably useful formula the a rule (pronounced "ed"rule). The eb rule is well worth remembering, but it is better to remember it in a form that is independent of a particular choice of letters for subscripts. To accomplish this, ignore the dummy subscript. [In (12), ignore kJ If "first," "second," "inner," and "outer" subscripts are labeled as on the left side of Figure I.S, then subscripts on A are (first) (second)-(outer) (inner), as on the right side of Figure E.S. Readers should satisfy themselves that this scheme produces the correct formula for all cases of (14). We shall now give some examples that show the efficiency of expressing determinants by means of the alternator and the usefulness of the eb rule. More examples will be given in the section on tensor fields and in the exercises.
Sec. I.2J
Zi
Defermrniaills and the f erniumtton Symbol
tUuicr l
I
I
,
e^}x
^^
#
t
^^ _ 6ip
^
^^
^
SUl{ , i1111
—1
trIrst1
^
(
^ 6119
iL I [tr
Ir tier
`
IS r L unJ!
F I u LI R L
i4fnenlwric diagram for the - ed - rule.
Example 1. let a, t1, and c be vectors with components a,. h ; and
c, relative to a triad of base vectors r [". Express the vector and triple scalar pi oduL Ls using the permutation symbol. Solutier (151
bA C
a' (b
A
C) —
r_,jr 1ei}11 i .
t,161
"fhesc Fr rmtil:is tC}IInii' directly from wetl-known connections between the left - hdrid ^
sides of (l5) and (16) and cleterrninanls. It is often more convenient to use the notation (th component oft = Ft],
(17)
10 Write 1_1) A c], = r:,r hi t,.
(
lS)
Example 2. Prove that (a A b) A e — (a • clti — fh • c)a.
SOlution.
119)
Using (ta) twice. we obtain [[u A h) Ae
]^
— Lm
k[t Rle -
(20)
The f:b rule cannot yet be applied iinrncdialely because the dummy subscript :appears second in i:m but appears first in r..; [compare (14)]. Therefore, we iiarlspose subscripts before applying the rule. , o
[{a Jti b) A clm = —(.f y it Î1R rt, -
-
— (6„,160 -. 6.,.ih,p)arhscr -
u+Rhvr'v t b,„ ur {'r
- -- (h • c )[a]m + (■ • cl[b]#
Example 3. Lei position x depend on the in i tial position A at turne r. Find a compact formula foi the time derivative of the J; eobiari J - el(x i . x z . x j )likei ti , 4 _ , A3)-
22
Vectors,
,Solution.
Dew-minions, and Motivation for Tensors [Ch_ 1
The Jacobian determinant can be written r7x, E iik
—
ex3 r7x 3 —
—
OA , r? A J OA k
(2t)
.
so (using the rule for differentiating products) J
a,, ax, ax3 eA f dA $rl, + ...
= —
^
Here and in the following discussion, we shall not write Oil t two additional terms whose form is obvious by symmetry. Introducing the velocity = k,. we have, by the chain rule,' _ El 3v 1 dxD ^x^ dx^ J ^ N $xD ÔAt aAJ PAk +
l+^ tl t ,J + = f^xp
...^
where we have used Theorem 3. [fence
1-1 = rv'v )IJ.
( 22)
Example 3 provides an elegant proof of formula (13.4.11) of I , which is essential in the study of kinematics.
Example 4. Let ILI denote the determinant of the transformation matrix I. (with components I"). Use the results of the present section to show that I LI — 1 for proper rotations, ELI = —1 for improper rOlationS. Chant of Solution [An alternative proof of (I.14). Also see Exercise 14]. Fut = fo in 010h) and multiply by I t 141I 1." I ^!'Y 17Y/IM = ^7^ ^ R ^ nil Orr S./ 11 Il^^w•
Using (10a) and (1.9), we deduce that ei, 6 ,.,
f
f w = rti^ w ,
so I L ! ' =
But I/.I is a continuous function of the elements of the transformation matrix L. Also. jf = 1 for the identity transformation (wherein e,,— S;;) by direct computation, and therefore for all proper rotations by continuity. Improper rotations have ILA = —1, for they can be obtained by continuous transformation after the particular reflection wherein
0
U
—
1
• As required by the chain rule (compare Appendix 13.1 of I), the vertical line indicates that the sutrsritution r = x(A. t) must be made a fter the partial differentiation_
Sec_ 1_3) Th e Consistency Requirement
23
E XLRCl4F. R
1. Prove that cyclic (anticyclic) permutations are even (odd)-
2. Prove (2) and (3). 3. (a) Prove Theorem I by writing out the terms. (b) Prove Theorem 4, 4. (a) Prove Theorem 5_
(b) (c)
Complete the proof of Theorem 6. Prove the Corollary of Theorem 9. 5. Formalize the continuity arguments that were used to complete the solution of Example 4. $6. Write the result of Exercise 1.5 in the form y. = R i x } . This shows that a nine-component entity is needed to represent a finite rotation. A study of how such components change when new coordinates are introduced will lead us to the concept of a tensor) 7, Lei Y = a A b_ Use the permutation symbol to show that (a) 1 v 1 2 2 IbI 2 sin 2 fi;(b)v•a =0, v• b=0. fa1 Verify that the (f rst)(secon(1)-(outer)(inner) scheme is correct for all eases of(14). 9. Use the Eh rule to determine an expansion of a n (b A e).
10. Show that la A bj 2 = 1aI 2 1b1 2 - (a -11)2. 11- Show that (a n b) • (c A d) _ (a - c)(b • d) — (a - d)(b • e). 12. Find two different expressions equivalent to (a A b) A (c A d). 13. Using the permutation symbol, work out a way to differentiate a determinant. 14. Use results about determinants to deduce I L f 2 = 1 from (1.8).
1.3 The Consistency Requirement We attempt here to provide motivation for the formal definition of a tensor that will he given in the next section_ THE CONSISTENCY OF NEWTON'S SECOND LAW
Versions of the same scientific law in different coordinate systems must be consistent. As an example of what we mean by this, consider Newton's second law. In a certain Cartesian coordinate system, let F ; and a ; denote the components of the force acting on a particle and the components of its acceleration, respectively, Let Fï and ; denote corresponding quantities in another Cartesian coordinate system. In the first system, Newton's second law can be written F; = ma i .
(1)
In the second system the law must have exactly the same form, namely,
Fi
= ma;.
(2)
24
Vectors. Determinants, and MOftuR[rOn for Tensors
I4Ch.
^
One reason fur this similarity can be found in the definitions of the components of vectors like F: The components F; are the projections of F on the fixed mutually perpendicular unit vectors l' (° }. The components 1.-; are the projections of on the fixed mutually perpendicular unit vect ors e 14 . These definitions show that components in the unprimed and primed systems conceptually identical and only rotationally different. 111 (1) is true, are therefore, (2) must be truc, because the correct notational change of adding primes to vector components has been made. Let us check directly that (2) holds if and only if (1) holds_ Since the primed and unprimed components of a vector are linked by (l .6). (2) can be written l vI Fp - ni/ uF or
I pg
- nia) = ü.
(3a b)
Equations(3b) can be regarded as three homogeneous equations for the three unknowns Fp — snap . By Theorem 1.1 the determinant of the coefficients is either t 1 or - 1, NO that 1 3 h) has only the trivial solution Fr
—
rr^t^
—
0,
which is the same as ( l). By reversing the above reasoning, one can show that (I) implies (2). PHYSICAL LAWS: GI';NERAI. STATEMENT VERSUS PARTICULAR NUMERICAL VERSI(7N To he more precise about the meaning of the consistency requirement, let us first slake clear that we are not presently interested in how laws appear to different observers who are moving in relation to each other or to some fixed "inertial - frame of refcrence. Rather, we are concerned with how laws are expressed in the different coordinate systems available to a single observer. After selecting a coordinate system, an observer of a vector obtains a rule for assigning to each point in three-dimensional space at least one triple of real numbers, and a rule for assigning exactly one point in space to each such triple. Since these rules must be perfectly definite, there always exist transformation formulas that allow one to pass hack and forth from triples obtained in one coordinate system to triples obtained in another. We distinguish between the general statement of a law and its particular numerical version. The general statement can be expressed independently of any coordinate system (F — ma) while the particular version relates numbers obtained using a particular coordinate system (Fi = ma.). in general, different versions of a given law, i.e., relations between dif ferent sets of numbers, are obtained if different coordinate systems are used ( ^ = mo o r; = mdai). These different versions are found from the general
Sr(_ 13)
The Consisienc y Rama-Linen'
25
statement of a Jaw by applying the coordinate rules appropriate to different coordinate systems. Consistency* demands that the same result be obtained if u given version of the low is obtained direct!) , from the general statement (Fr, = ma p ) or is obtained indirectly by transformation of another version 1Fp — map follows from F; = mo and the appropriate transformation law (1"6)H. in the previous paragraph, we cited F — txta as an example of a general coordinate-independent statement of a physical law, This example used vectors, entities that have a coordinate-free intepretatiou which invnlves length, direction, and addition by means oft he parallelogram law(Section 1. 1). Combinations of vectors, sometimes involving derivative operators, also have coordinate-free interpretations and therefore can also he used in the general statement of a physical law. To mention the simplest example, the scalar product can be described in a coordinate-free manner, as the product of the lengths of the two vectors involved times the cosine of the angle between them. f But we cannot restrict ourselves to combinations of vectors in formulating physical laws_ More complicated entities called "tensors" are necessary_ We turn now to a discussion that will mntivate the detailed definition of a tensor. GUESSING THE TENSOR TRANSFORMATION LAW FROM THE cON4tRTF'NCV REQUIREMENT
A fundamental result in continuum mechanics is that the stress vector t(x, t, n) is a special function of the unit normal vector n given (using summation notation) by r (x, (, n) = Pi; T, x, t)-
(4 )
(Compare (4.2.17 of L) Here I; is, by definition, the jib component of the stress vector acting on the surface element whose exterior normal points in the direction of the 6th Cartesian base vector O il_ We shall examine (4) in the light ofihe Consistency requirement_ This will lead naturally to the Concept of a second order tensor.§ Consider a new Cartesian coordinate system with mutually perpendicular unit base vectors el` I . Using the same reasoning that we used in proceeding from (l) to (2), we see that in the new system (4) must become t^ = rti T;^.
'
(5)
Cbuarruece or invariance arc alternative [rims fur cnnsislenCy.
t "!ï he divergence of a vector has a coordinate-free inierprerarinn [Exercise 9(a}j. Thus no surprise [hat the mass conservation equation epfct - V • v) = 0 has the same Form in a rotated Cartesian coordinate system (sec Exorcist 14.110f It 4 If desired, rcfercnCc to material outside this chapter can be avoided by using air finite
rotation reIalion requested in Exercise 2"t, instead of (4).
z6
'errors, Derrrrninanis, and Motivation for Tensors !Ch. )
Here 71i is the jth component [the projection on eu v] of the stress vector acting on the surface clement whose exterior normal points in the direction of e". We know the relations between the components of the vectors t and n in the two systems_ Assume that (4) and (5) arc consistent; let us determine what the relations must be between the quantities Tü and the quantities T. Using appropriately modified versions of (1.6), we substitute for r; and n; in (5): (,) t p = / o n, 7- 1 .
We substitute for r p by using p instead of j as a free subscript in (4): I pl ► t i J ip
= (Wi g
Employing a frequently used trick, we rntiltipty both sides of (7) by*
I Il i
Rini;
^
rJ
ri-
/ q,rJg7 f i'
But, by (1.8),
with which (8) becomes
tr; T;, =
tr#f qi n^ T; 1 .
We rewrite (10), changing the dummy subscript on the left and reordering the factors on the right:
nq 9 r = ri q q ^
^^
•1
(1 1)
Now (4) is true for any unit vector n. In particular, it is certainly true if n has the components (1, 0, 0)_ For this n, (111 implies that ^lr
= ta rr ri t .
(12)
Equation (12) plus similar orics corresponding to vectors n with components (0, 1, 0) and (0, 0, 1) can all be written compactly as
T4, =
/ e/r,T^.
(13)
We remind the reader that because of the summation convention, (13) symbolizes nine equations, corresponding to q, r = 1, 2, 3. We have established that i (4) and (5) are consistent, then (13) must hold. But (13) can be verified directly (Exercise l4.2.8 of 1). A straightforward calculation, essentially the reverse of the one in the previous paragraph, then demonstrates the consistency of (4) and (5) (Exercise 5)_
* instead air we could have usod as u subscript any Irurrs except p, q. and t. The Inner three choices wnu[d have violated the prohibition against repeating a subscript more than twice. in a single tam ,
sec. f.3) 17W C omrstrrecr Rrvrrcmenr
27
It is important to noie that our derivation of 11 3) did not use any special properties of the quantities involved in (4) and E5) . We needed only these facts: (a) t and n satisfy the vector transformation law (1.6). (b) We can determine the nine quantities T , in any Cartesian coordinate system. (c) n is arbitrary. It turns out that there are many physical laws of the form (4) whose con-. stituents satisfy (a), (b), and (c) and which are consistent with respect to Cartesian coordinate systems. Sets of nine quantities satisfying the transformation law (1 3) arc therefore common, and consequently they have been given a special name, second order Cartesian tensor (hence the name stress lemur for T). In essence, our discussion of (4) indicates that if a two-subscripted entity appears in a physical law, then that entity must represent the components of a second order tensor. Comparing the tranSforrrlation law for the 3' components of a vector (or first order tensor), = with that for the 3 2 components of a second cider tensor, — /, r /
,
q
there is no difficulty in generalizing to the 3" components of an nth order tensor. And this is no empty generalization, for higher order tensors occur frequently. Having completed sufficient motivation to make our definitions credible and to make their development seem worthwhile, we pass in the next section to a formal study of Cartesian tensors_
tr Xk,R Ct*itü S 1. Show that if (1) and (2) are consistent and if acceleration is a vector, then force is a vector. f ollow the line of reasoning used to obtain (131. 2. Why is it not permitted to cancel rt r from both sides of(' I). even when it is guaranteed that no components of n q are zero, unless (11) holds for arbitrary rig? 3. Explain why the passage from (l I) to (13) can be achieved by using a slight modification of the following result :
r•, rt, = 0 for arbitrary n, implies that t', What is the geometric interpretation of this result? 4. Show that (13) implies that Tqt f q afn =
—
0_
Vectors,
Defermmcpus, and Motivation for Tensors [Ch.
I
5. Show that (13) implies that (4) and (5) are consistent. 6. Specialize (13) to obtain the transformation law for T12 . Write out all nine terms. Show that (15) gives the same result. Unless you are absolutely confident that (13) and (15) are identical, repeat the problem for T23 7. (a) Calculate fii for the pair of Cartesian coordinate systems related by
(b) By what rigid rotation can one coordinate system be carried into the other? (Give the axis of rotation and the amount of rotation.) (c) Show that if T is a second order tensor, then, according to (15), its components T arc related to its components Tp,2 by a transformation of indices symbolized by 1 --3 L',
3',
2
3
Use common sense to check this result_ 8. (a) Same as Exercise 7(a) for the coordinate systems related by
e LLl = e (21
e ra.I
_
_ e {iI
[3i
= e [il
(b) Same question as Exercise 7(b). (c) Show that Tv, = -- T;2, T32 = T 1 . 9. (a) From material in 1, Section 13.4, and/or from the divergence theorem, develop a coordinate-free interpretation of the divergence. (b) Discuss the results of I, Exercises 14.1.9 and 14.1.10, in light of the
ideas presented in the present section.
1.4 The Tensor as a Linear Transformation In this section we shall show that a second order tensor can be identified with a Iinear transformation of a vector space into itself. We shall be led to a natural characterization of tensor components and their transformation law. Knowledge of the connection between the stress vector and the stress tensor will be assumed, as given in Section 14.2 of I_ To be sure that the mathematical prerequisites are secure, we begin with a review of some relevant definitions. LINEAR TRANSFORMATIONS
Definition 1. A iransforrnurion of a vector space* into itself is a rule that associates with any vector v one and °nly one image vector w. We write (1)
w = TM_ Examples of such transformations are T(r) = 4v, T(v) = O, r(v) = q+ v (q a fixed vector). (a, b; * Only ordinary Euclidean 3 space 14LI/
considered
in this stctiun.
c)
See. i 41
The Tensor
as
1-.rm•ur 7rcuts/rnmutiurr
Definition 2. The transformation (1) is linear if it satrslics the following requirement: if C i and e2 are any two scalars and rr'' and vrZF are any two vectors. Then
r.3 0 2 f} = c i TN" r) + r r T(0 21).
T{t r vin
Transfotmarion (2a) is linear, ;rnrr
E'xsantp[^ti.
Tf
(3)
^
w
+ ri 021) = 4(r ^ ►r " + r2Yl2l)
ill
ra4vrriJ
l
i
i -r
4.
Tli(ri) + e 2 f`Wlrl-
5irnitarly,12h1 is Linear Transfnrmatutn (20 is nr ► r luxar, 4mcc Tfr .^ yiit
+
'Y#r} 4 r. 20 :1' 4 (1
(4)
bui c
Tit') i t Thel F l — t r f u iii i q) f_ r -ri(Y '21 i qk
he nuit sides of 14) and (5) arc nor the same tot arhnrary r i and
15)
ez
THE STRESS TENSOR IN11UL'E.S A LINEAR TRAN5FU1iMATloh OF DIFFFRFNTIAt, AREA INTO 13iFF'F:RF.NT1At. FORCE
To begin to see how a second order tensor can he c1mracterired as a linear transformation, recall the relation between the stress vector t and the stress trnsnr T [Equation (14-2-17) of l]= fr. I rr(x, t`
tk, t,
(ti)
for given x and t, lb) provides a transformation t - Tint of the unit normal vector n into 1. This is not a transformation in accord with Definition 1, however. because T is defined only for unit %ectors. We circumvent this difficulty as follows. Lei E be a closed surface and let P be a portion of E containing x. Let dS be the area of P and denote the unit exterior normal to 1 at x by n(x} (See Figure 1,9- Note that dS is not necessarily small.) Define df and dA by -
df - t r^.5
dA
-
n dS
(711.1))
The vector dA is a differential area lis magnitude is the area of P and its direction is that of n(xl, a certain unit normal vector to P. The smaller dS is, the nearer all exterior normals to P are to n(x) (provided that E is smooth and orientable and that dS retains its shape as'i1 shrinks'. The vector df is a differential contact force As in (14.2,4} of 1 the actual contact force fr excrtrd on f satisfies
11m -° -
^c • r, dS
Consequently, for small dS, fj,
dS. dlf
(S1
30
'ertors, Dererminanrs, and Motivation for Tensors [Ch.
!
n(x)
FIGURE 1.9. A patch of the surface wish area dS_ the normal .error a point x in (his parch is shown.
n ai
Sometimes it is useful to say that when dS is" infiniresimal," then the contact force exerted on P is given by d f; and to say Further that dA is a vector whose magnitude is the area of an "infinitesimal" patch containing x and whose direction is that of the exterior normal to this patch. But keep in mind that d f and dA are defined for any d5, large or small. Denote the components ofd f and dA by cif and dA,. Multiplication of (6) by dS then gives
df^
=dA ) T; ; .
(9)
We interpret (9) as a transformation df = T(dA) of the area differential dA into the force differential di'. Our transformation now satisfies Definition 1, because it is defined for any vector dA. Furthermore, the transfuntialion is linear. For if dA and dB are two area differentials with components dA, and (IB„ and a and h arc scalars, then TO dA + b du) has the components (a dA + dB1)T;i = a dA i To + b
so T(4 dA + b dB) = aT (dA) + bT (dB). There is one unsatisfactory aspect of our argument. For any given coordinate system, the transformation T is defined by (9), with the aid of the components Ti . Would the transformation be the same if another coordinate system were used? If not, T is improperly defined. We can rephrase the question this way: Suppose that we use a primed Cartesian coordinate system to compute the components I; and that in this
Sec_ 1.4(
The Tensor as a Linear Tre ►xsjurrnalrora
system the components of df and dA are d[; and del). Is it truc that d[ dA ; Tii ? But this is just the question we answered affirmatively in Section 1.3 when we showed that (3.4) and (3.5) are equivalent if and only if the transformation law (1.6) ho lds_ A LINEAR TRANSFORMATION OF DIFFERENTIAL AREA TO DIFFERENTIAL FORCE CAN BE IDENTIFIED WITH THE STRESS TENSOR
We now explore matters from the opposite point of view. The transformation df T(dA) is linear if and only if T(c dA) = cT (dA),
T (dA + dB) = T(dA) + T (dB).
(l0a, b)
But (10) provides perfectly natural initial assumptions to make about the relation df = T'(dA). Equation (lDa) states that if the differential area is multiplied by a scalar c, i.e.., made c times as large, then the force is increased by a factor of e. Equation (10b) states that the force on the sum of two differential areas is the sum of the forces acting on each. It is therefore natural to see what we can discover about the differential force merely from the knowledge that it is obtained from a linear transformation of the differential area. More generally, what can we discover about a transformation T(v) knowing only that it is a linear transformation of a vector space into itself ? Let el l s, el 11 , and e13 F be a set o f base vector. I. or any vector v, then,
v
view ,
(l 1)
where the t are the components of v_ To find the effect of the transformation T(v), we first observe that the linearity of T implies that T(r) = T(4.40")1= vi T( e ffi)
€ l2)
Now Ile" )) is by definition a vector and socan be written as a linear combination ore' ", el'', and 0 13 . The same holds for T(e (21 ) and T(€131). Thus for some coefficients T -; we have
T(e'') = Ti j et p .
(13)
A combination of 112) and (I3) yields T(v) = /01,6".
(14)
Alternatively, if we denote the image of by w, t hen w = T(w) and rt e = v i To.
(15)
Thus if the components of v are known. the components of its image w are given by (15), provided that the nine quantities Tuare known. These quantities are called the components of T relative to the base vectors O'. As (I3) shows, these components are given in terms of T(et` ) ), the images of the base vectors.
32
Vf`{- rors, Uerfr+nuiurrrs,
and Motivation
jor
Tensors
[Ch . 1
The linear mapping T has a new set of components with respect to a different system of base vectors a 'r_ These are given by the analog of (13): T(etnrr)
(16)
fAA
What is the relation between the f' and the 7';t„? To answer this, we start with the following links between the primed and unprimed base vectors (see the proof of Theorem 1.1): e (nr =
f
.
e i j)
From (13), (17b), and the linearity of T
/ !AT e[^'l
eI ' }
lA
-
(17a, b)
:
Ti e( )) = T(e{' } ) — T(/1,, ")
T(e [mi )-
Transforming the last expression, using (16) and (17a), we find that f I ji
fix, T,„„Few = f i„, f ,
f,r
i11
(18)
For each i, (1 8) is a vector equation. Setting components equal, we obtain T j =„^Ii„l ^A^ '
(19)
Equation (l9) is the transformation law used in the definition of a second order Cartesian tensor. We have therefore shown that if we use (13) to associate components with a linear transformation T of a vector space into itself, then Tis a second order Cartesian tensor. Conversely, we showed at the beginning of this section that if T is a second order Cartesian tensor, then it can be associated with a linear transformation of a vector space into itself. Thus the transformation law (19) between components of Tholds if and only if Tis a linear transformation of a vector space into itself_ Either the transformation law or the linearity property can be taken as the essential ingredient in the definition of a tensor. We choose to define tensors by means of their transformation law, so that we say that tensors are characterized by linearity. LXt•:RCiSES
1. Show that T (dA) is a linear transformation if and only if (LW) holds. 2. Show that if T is a linear transformation, then T(4) = O. 3. The components of the linear transformations (2a) and (2b) a re the same for all coordinate systems. Find them. +4. A transformation T is linear and transforms a set of base vectors eri as follows: TW' F) 2e(2), T(0'1 = e i 1 + et -”, T(e(3}) _ eU (a) Find the associated components 7;,. (b) Find the image of the vector with components (3, 1, 2). 5. Repeat Exercise 3.7(c) but use (13) and (lb) instead of the transformation law (3.15). 6. Repeat Exercise 18(c) using the method described in Exercise 5.
2 C artesian Tensors
CHAPTER
1-[AFTER I provided motivation for the study of Cartesian tensors and for the transformation laws that form the central aspect of their definition. This chapter is concerned with a methodical study of tensor properties. We begin with algebraic considerations, such as definitions of tensor sums, differences, and various kinds of products and "quotients?' Next we consider the eigenvalue problem for second order tensors, a matter of considerable importance but one that to most readers will be only a small extension of familiar material concerning eigenvalues for matrices. We shall largely be concerned with tensors whose components are functions of space and time. 01 great future utility. therefore, will he the study of tensor calculus that concludes this chapter.
2.1 Tensor Algebra This rather formal section presents a number of definitions and theorems concerning Cartesian tensors_ The adjective "Cartesian" may be left out, but "tensor" and "coordinate system" always refer to Cartesian tensors and to Cartesian coordinate systems in ordinary three-dimensional (Euclidean) space.
Our discussion starts with the definition of a tensor that is based on the rules For transformation from one coordinate system to another. We continue by showing how tensors may he added and subtracted. Various kinds of multiplications are defined and discussed, and division of a sort appears in the "quotient rule." We also discuss symmetric and asymmetric tensors and the relation among tensors, vectors, and matrices_ The discourse is largely algebraic, so that we adopt a Formal theorem-proof style. Many of the definitions and theorems we present will be formulated for second order tensors. /The required generalization to nth order tensors is often straightforward_ In such cases we leave the generalization to the reader and thus avoid the cluttered subscripts that are necessary in dealing with nth order tensors_ in Section 1.4 we showed how a second order tensor can be interpreted as a linear transformation of a vector space into itself. This is a coordinate-free description and thus automatically establishes the possibility of using second order tensors in the general statements of physical laws. It is possible to extend this approach and thereby to characterize an nth order tensor as a certain "rnultilincar" transformation. Some authors use this characterization to define tensors. For our purposes the approach of the present section 33
34
Cartesian
Tensors [Ch.
1
is preferable because fewer algebraic preliminaries are necessary and because more useful skill at component manipulation is acquired. DEFIItiIl IONS AND ELEMENTARY PROPERTIES
Definition 1. A zeroth order tensor or scalar s is such that (a) In any Cartesian coordinate system there is a rule for associating s with a real number. (b) This real number is the same no matter what Cartesian ccnrdinatc system is employed. For example, the mass of an object is a scalar_
Definition2.
An nth order tensor (or nth rank tensor) T, n = 1, 2, 3,
-
is such that* (a) In any Cartesian coordinate system there is a rule for associating the tensor T with a unique ordered set of 3" quantities ?; (called the components of T). (h) if i .. ;n and 7' ... J are the components of Tin two different Cartesian coordinate systems, then ^
_ NOTE-
.,
in in
Jr -- •lti -
{l)
For a second order tensor, 11) can be written
T;,.
Tp4 = f
REMARKS-
By Cell =- #s;
we mean that in a certain Cartesian coordinate system, T j is the (ij)th component of T . The components of a second order tensor can be conveniently displayed as a matrix. Example 1. Lei ! be an entity with nine components given by [fl u = baj.
Show that l is a second order tunsot (Oiled the unit timor). Solution. We have a rule giving the conlpuncnts I in tiny coordinate system- (The same set of nine numbers, given by the identity matrix [
t=
ü
1}
D 1 0 , il {} 1
• The relation between the ahstract Cntity ealtcd a tensor and the set of numbers that are its components in a given coordinate system is somewhat analogous to the relation between the abstract idea of a sphere and a particular diawing or model of a sphere.
Sv c.
2.11 T ensor Algebra
35
gives ). he components in all coordinate syslerns.l It remains to prove the validity of the transformation law
But. using 0.1.814 %mr
1e
f+ mr^ ~ %,nI
Example 2. Lei E he an entity with 27 components given by
Show thai E is a third order tensor (called the alternating tensor). prorated that transformations between coordinate systems are restricted 10 proper rotariwns. Suhrivn. The rule for determining cornpcnenis being given, rt remains to prove the transformation law. t,d, =tep % IQ ti R. f;p4o
-
Hy (I-Z.IEIb), the right skie of this equation is ri o. ILl and the result ftrllo s.. since CLf, the determinant of the transformation matrix, is unity for proper rotations, [See Example
1.2.4 or Equation (1.1. I4l.] Note that if reflections were not prohibited, there would be transformations
l , and E would not be a tensor-When we speak of " the for which IL j = tensor E" and when we derive consequences from the tensorial character of E. we shall he implicitly assuming that reflections are prohibited. -
Example 3. I.ei A I , A2. and A ; be vectors. In a given Cartesian coordinate system, deter 111111C a rlumbCr by evaluating the determinant whose ith row consists o f the components of A„ f - 1, 2. 3. Show that this number is a scalar, provided that re fle c t ions are prohibited. Solution. Let A,„„ be the nth component of A. We must prove ghat e' = c. where r
-
1 -0.A1•A2 1 i 34 ,
t' ' = F: py ,.4
' rr
Ap 4 13 r
[hit by the transformation law for vectors,
and the result follows ai once by the transformation law for E that we just proved Example 2.
iii
Definition 3. Let A and B be second order tensors with components A il and B,,. Let c be a scalar. The sum A + B is defined by
[A + 8] = A The difference A
-
+ B.
(2)
A ;; - B i ,-
(3)
.3
B is defined by [A - B]
Cartesian Tensors [Ch. 2
36 The scalar multiple cA is defined by [cAi} = cA,1 .
(4)
In particular, [ — I A l1 = [ - Ali _ — A ;i . Theorem 1. If A and B are nth order tensors and c is a scalar, then (a) A + B is a tensor, (b) A — B is a tensor, and {c) cA is a tensor. Proof of (a). (The proof is very simple, but as it is the first such proof, we shall write it out in detail, for second order tensors. As we continue, proofs wilt become less detailed.) In any Cartesian coordinate system, there is a rule for associating components with A and B. Thus (2) provides a rule for associating components with A + B. To show that A + B is a tensor, it remains to demonstrate that the required transformation law holds. Let A 1 and A I be the components of A and B in one coordinate system and let A;,,,, and B be their components in another coordinate system. By (2), the corresponding components of A + B are (A ; + B, j ) and (A,,,, -# B;,,,,). Since A and B are tensors, the following transformation laws are valid: A. — t,,/ jn A,
(5)
B ij =.„,/ L
(6)
Adding (5) and (6), we obtain (AF, + B, 1) = 4,,i (A; ,In + A:,r}, which is the required transformation law for the components of A + B. Proof of (b) und (c). Left to the reader. O
Definition 4_ The zero. tensor 0 is an entity whose components in any Cartesian coordinate system are all zero. That is, for second order tensors, t_q,• = ^
(7)
Theorem 2,
The zero tensor is a tensor. Proof for second order tensors. The rule for obtaining components is given by (4 The required transformation law, 7 1,
is clearly satisfied if I; =
Tm
„
`
j
11n 11 ,r f rnn ,
= 4 for i, j, m, rt = I, 2, 3. Q
Definition S. Two tensors are equal if their di fference is the zero tensor; i.e.,
A =B-t
-B = -D.
(8)
It follows (Exercise 2) that to prove that tensors are equal, one needs only show that they have the same components in a single coordinate system-
Sec. 2.1) Tensor Algebra
That is, for second order tensors,
(9)
A = B [A] ^R = [ 81).
We emphasize that only tensors of the same rank can be equal (for the subtraction required in Definition 5 is only defined for tensors of equal rank)_ Theorem 3. The following " natural" laws for manipulating tensors hold: Let A, B, and C be nth order tensors and let a and b he scalars. Then
A+(B+ C)— (A+B)+C, A +0= A,
A+(
—
A-+ B B 4 A,
( 10a,h)
A)= O,
(11a, b).
al A + B) - uA + aB,
(12)
(ti + b)A -- uA + bA,
1 13)
(ab)A = LOA),
i A = A_ -
(14a, bl
[Theorem l plus the laws of OW and (t I), show that tensors of the same order form a commutative group under addition. Because (12), (l3), and (l4) also hold, tensors form a vector space under addition and scalar multiplication.] Proof of (IUb). Let ,
[A] ^ 1 = fi,,
[B] 0 = B
,,.
Then, by Definition 3,
[A + B] , 1 - A + B.
[B + A] { = B tu + A, 1 .
Since the addition of real numbers is v:ommutative,
A, f + B
— B, ►
so [A +B]_ [B t A], 1 , -
and the commutative lain (lob) follows, by (9). Proof of the rest of Theorem 3 is left to the reader. fl
Consulting Section 1.1, we observe that a jirsr order tensor is identical with a vector- Indeed, our previous definitions could be thought of as generalizations ofdefinitions familiar to us from vector algebra, We turn to some useful notions that can only be defined when tensors of order 2 or higher are considered. Definition 6. A contraction of an nth order tensor, ,t > 2, is formed by setting two of its indices equal and performing the resulting sum. An example of bow contraction leads to a new tensor whose order is 2 less than the original tensor is provided by the Following theorem.
Car#esiw7 Tensors [Ch. 2
38 Theorem 4. Let T be a fourth order tensor, where
=
L^^+imn
TOR r -
Define f1 by ^ iR
in
—
—
Tim.
(15)
Then Li is a second order tensor, Proof The rule for obtaining the components of U is given by (15). Since T is a fourth order tensor, ^ ^p G j4 eJxr° J,_, TpgFi•
^4^JRR
so, setting
m j,
= But 1 fig
^
1R
(16)
c5-11 yr
as in (1.1.9), so e [pf R7 7;1, 1
Thin
-
By (15), the above equation is e quivalent to FF r which is the required transformation law. p Definition 7. Let A be an nth order tensor and B be an mth order tensor, where [AL i R = A i ,... 4 ,
[8 ],, ^ — Br,
Then the tensor product AB is defined by [ABI,...
D = A; ... ,nBi.,, ,
J...
(I7)
Note that the right side of (17) is the product of two real numbers. Example 4. if v and w are first order tensors with components u ; and w then
[vw],i — IJ i The ÿth component of vw, the product of the ith component of a with the jth component of w, can be found in the ith row and jihî column of the following array: CJ w 1
VI. W2
l} 1 14'3
NW) — 31;44' J
112 .19d2
r..12 w3 -
OE)
V3 WI 113 w1 31 3 W3
Two vectors have a scalar product, a vector product, and a tensor product These products arc, respectively, a scalar, a vector, and a special second order tensor called a dyad (see Exercise 24). NOTE-
S e t. 2.1) Tensor Algebru
39
Theorem 5. Let W = AB, where A is an ,nth order tenser and B is an mth order tensor. Then W is an (n + m)th order tensor. Proof for n I, in = 2. Suppose that A, and BA , and An, and }3 , are the components of A and ft in two different coordinate systems. By definition of W, A r j,
,
Werra
= Am Bp
(19)
Since A and B are tensors, A[
Bjk = /
irnA+n E jp i 4 Bpe .
Therefore, from (19), we obtain the desired transformation law,
Prooffor Me general rase.
Left to the reader. Q
Definition 8. Let A and B be second order tensors with components A 11 and B. Then the contraction product A • B is defined by [A B] — AuBig. The contraction product of tensors of arbitrary order is defined similarly, by summing ever adjacent indices .
Theorem 6. Let A be an !nth order tensor and B be an nth order tensor. Then A • Bis a tensor of order rn + n 2. Proof. A • B a contraction of the (m + n)th order tensor AB. 11 —
Easily proved results (Exercise 16), and useful ones, arc
rod[T - e ^1i^ = T . e M. T. e (1x — 1-
- ,,
(20a, b) (20c)
These equations assign interpretations to each individual tensor component. In Section 1.3 we showed in essence that if t = n • T holds for arbitrary t and n, then T must satisfy the tensor transformation law. it is useful here to restate this result in a general context. To make this section's presentation more complete, we shall repeat the main points of the proof. Theorem 7. Suppose that in any Cartesian coordinate system, there is a rule associating a unique ordered set of nine quantities with T. If, for an arbitrary vector a, a • T is a vector, then T is a second order tensor.* Tilts theorem brings ow
a sect1rtd order censor's characterization as a mapping of a vector
pasx into] itself (compare Section I.4).
s
2
Cartesian Tensors
40 Proof.
Let a •
T = v. We must show the following. Suppose that
(2la, b)
and o; T: ; _ 14,
a, Ti = t?
and suppose further that a and v satisfy the vector transformation law, so = /ad*. and t' 1
(22a. b)
I 11 04.
Then it must be true that Tpq — / ipf
But upon substituting (21a) and (21b) into (22a) and then using (226), we find that — { JP t
F
/ jig T2p-
To employ (16) for the purpose of moving all !'s to one side of the equation, we multiply by f 1,: / 1,/ jr„ a: Tip =
,a;Tlp
Tu.
or -
f.k,] = O.
Since a, is arbitrary, an argument used in Section 1.3 shows that the terms within the square brackets must vanish, which is equivalent to the desired conclusion. Q REMARK.
If a and b are real numbers, b # 0, their quotient alb is
defined by
a
=.c=.a = be.. By analogy, v = a • T could be interpreted as expressing T as the quotient of r and a_ Theorem 7 is called a quotient rule because it can be regarded as asserting that the quotient of two first order tensors is a second order tensor. While suggestive, the "quotient” aspect of tensor equations need not be taken too seriously. The following is another quotient rule of a rather general character. Theorem 8. Suppose that in any coordinate system, there is a rule associating a unique set of 3h quantities with T Suppose that for an arbitrary frith order tensor A, TA is a tensor* of order ni + n. Then T is an nth order tensor. The proof of this theorem and of other quotient rules will be given as
exercises _ • Sink T is opt known to be a knsut, strictly speaking TA is not defined. In any courdinase sysielll, by TA we mean the analog of t 17l.
See. 2. If Tensor Algebra
41
SPECIAL RESULTS FOR SECOND ORDER TENSORS
Definition 9. Let A be a second order tensor with components A. Then the transpose of A is denoted by A T and is defined by [AT%]IJ = A1,.
(23)
Theorem 9. If A is a second order tensor, then A' is a second order tensor. Proof. Left to the reader Definition 10. If
A — A 1 ` or A ll = A
^; ,
then A is said to be symmetric. If
A=
or A,1 = - A l; ,
then A is said to be antisymmetric (or skew-symmetric). For example, in a nonpolar material, the stress tensor is symmetric. A second order tensor may be neither symmetric nor antisymmetric, but it can be written as the sum of a symmetric and an antisyrnmetric tensor, as is shown in the Following theorem_ Theorem 10. Let T be a second order tensor. Then one can write T — S + A,
(24)
where S is a symmetric second order tensor and A is an antisymmetric second cider tensor. This decomposition is unique. Outline of proof T=+{T+ 1.T`)+
1(T —
T T`)-
( 25 )
On the right side of (25), the first term is symmetric and the second is antisymmetric. The uniqueness proof is requested in Exercise 9. G Matrices are an aid in performing certain manipulations with the components of second order tensors_ For example, in a given coordinate system [A], 1 can be written in the ith row and jth column of a 3 by 3 matrix. We shall denote this matrix by (A). Thus A l2
A,
A2z
A23
Ati2
A33
3
Note that A has a meaning independent of any coordinate system, but the matrix (A) is defined with reference to a particular coordinate system. In (18) we displayed the matrix corresponding to the second order tensor vw.
Car#eswn Tensors [Ch_ 2
42
Theorem 11_
Fer second order tensors A and B, (26)
(A - B) = (A)(B).
In words, the matrix of the contraction product of and B is the product of the A and B matrices. Proof.. The component in the pth row and qth column of the matrix (A • B) is A ; Bi . A F,Bk is found by taking the"dot product" of the pth row of (A) and the qth column of (Bl. By definition of matrix multiplication, this is the clement in the plh row and qth column of the matrix product (A)(B). f] Theorem 12. The vector transformation laws U^
Z2^
=
Duli uj,
can be written, using matrix notation, as l^ r
I' 2
(EH = /21
/l2 ^ i3
^i
/22 /23
^2
i 31
/ 33
^3
!^1
/ 32
and / 12 1 ;',
1'2
6) =
/22
(r1
/ 32
respectively As before, lei L be the transformation matrix, and let (v) be the column vector with components 07,, v 2 , t] 3 ). If a prime indicates use of a primed coordinate system and if a superscript (Tr) denotes matrix transpose, then we can w rite (v) • rr = (v) rr L (v) = L(vj', (27a) .
Prou[
Left to the reader.
Theorem 11 The transformation law for second order tensors, A ',„ni
can be written (A) = L(A)1.».
(27b)
Left to the reader. REMARK. Since tensor equations are valid if they are demonstrated in any single set of components, there is a theory for second order tensors that completely parallels the corresponding matrix theory. We have seen this in Theorem 10; the decomposition into the sum of symmetric and antisymmetric Proof.
Sr c _ 2 1]
Tc•usor Algebra
43
tensors is the same as the corresponding matrix decomposition. To give another example, one can term a second order tensor nonsingular if any of its component matrices has a nonzero determinant. If R is a nonsingular tensor. then [Exercise 25(c)J there exists an inverse tensor, which we denote by R with the property
R• R '' = R - ' i R = f_
[28)
ISOTROPIC TENSORS
There are certain materials whose structure singles out one or more directions. Examples are the direction of the grain in a piece of wood, and various symmetry axes in a crystal lattice. Other materials, like water. appear to have no internal preferred direction. The constitutive equations for materials of the latter kind must be expressed in terms of special tensors whose components are always the same. We shall now study this class of" isotropic" tensors. Definition 11_ Tensors, such as I and E, whose components are given by the same set of numbers in all coordinate systems, are called isotropic. ll follows immediately from the definition of zero order tensors that they are isotropic. Are t here any isotropic tensors of order I ?There is at least one, the zero tensor of order 1. Excluding this case ]which we shall term trivia°}, are there any others? There are not, as we shall now show. Theorem it There are no nontrivial isotropic tensors of order 1. Proof (We proceed in a way that makes up in conceptual simplicity what it lacks in elegance. We rule out the possibility that there is a nontrivial vector which has the same components in oil coordinate systems by showing that there is no such vector even if only three special coordinate systems are considered) Suppose that v is an isotropic vector and that its components are y, relative to a certain basis e l n. Consider another set of basis vectors satisfying
e
tx _
^
^z^
^
i zp
=e` , a, e" }- =
(29)
It is clear from Figure 2.1 that the components v: satisfy t•' l -
= fi 3 ,
l'2 ,
Z'3 = VI.
(30)
This cyclic change of subscripts, symbolized by l'
►
2,
2'
3,
3' -' I.
(31)
can also be verified by calculating the appropriatef and applying the vector transformation law (1.1.6). (The calculations are almost the same as those requested in Exercise I,17.)
44
Cartesian
Tensors [Ch_ 2
r fJl 0 cll
FICiURE 2.1, A
rülulron of axes
ecjuivulenf fo a cyclic exchange of subscript.
Since Y 1S isotropic, = rL,
11 2 ,
t' 2
(32)
118 = V3-
Combining (30) and (32), we Find that (33)
=112 =V3 = 1 3 =l1 } • et`l., Consider a third set of basis vectors satisfying V1
cox
y e{21, 1121" =
_
t1}
ear = ea}
(34)
f rom Figure 2.2, = i VI,
but, by isotropy, VZ = v2 .
Thus o2 : — L . This is consistent with (33) if and only if v / = v2 Consequently, the only isotropic vector is the zero vector. U
v3 = O.
Theorem 15. Any isotropic tensor of order 2 can be written 21 for some scalar A. Proof, Suppose that T is an isotropic tensor of second order and suppose that its components are T i relative to a certain basis ems. As in the proof of the previous theorem, consider a new set of basis vectors satisfying (29). By (31),
T12
— T22,
7-22 =
= T23,
T23
= 13
T3Z
- 1132
T 21 = T32
,
T33, ,,
T33` T11,
(35a)
T31 = T12 ,
(35b)
T13 =
T21
.
(35c)
See. 2.11 Tensor Algebra
45 t
[ll
, P
fll
/
F Lc; t I R E 2.2. The cJnubk-prrn:rd "
mtenr Is r o ta t ed 90° about the e' as axis. -
Since T is isotropic, . T IL
T11.
T22 - ^=2 T 33
- T33 -
(36)
From (35a) we deduce that Ti 1 - T22 - T3 3=
(37)
for some number A. Similarly, we can deduce Irons (3Sb) and (35c) that T1 2 =t T2 3 = Ti
,
•
T1
= T32
-
i3-
(38)
New consider the components ❑ f T relative to a third set of basis vectors satisfying (34). Applying the transformation law, one finds, as in Exercise I.3,8, that 1'12 =
—
f21, T31 = T32 .
But, by isotropy, 11 . ^13
M T12, T 31 — T31-
Thus T1 2 - -Tn.,
1-3 l
= T3 z -
(39)
For (38) and (39) to he consistent, we must have 712 =
T23 = ;1 = T21
—
T32
- f13 = O.
(40)
The equations of (37) and (40) show that if T has the same set of components for the three sets of axes related by (29) and (34), then T necessarily has the
Curiesian Tinsors [Ch . Z
46
form 21. But )i is a tensor by Example 1 and part (c) of Theorem 1, and Al certainly has the same set of components for all axes. [] Theorem 16. Any isotropic tensor of order 3 can be written 1.E for some scalar 1. (If reflections are allov.ed, E is not a tensor, so the only isotropic tensor of order 3 is O.) Proof. The argument is entirely analogous to that required for Theorem 15. Details are left to the reader. Theorem 17. If T is an isotropic tensor of order 4, then y qr) [T]pgrs = okr5 Urs ± ;ii(15Fr ya ± ^
+ i { b pr b gS
—
cPs qr)
(41)
for some scalars 1., /4, and Pti. Proof. Again the same ideas suffice, but the details are now somewhat lengthy. The reader is asked to supply a proof in Exercise 19. THE VECTOR ASSOCIATED WITH AN ANTISYMMETRIC TENSOR
To conclude this section, we shall present some formulas involving the alternating tensor that will be seen to be useful in our discussion of kinematics, To motivate our discussion, we start with the fact (see Exercise 21) that ifa rigid material is in uniform rotation about the origin, then the velocity v at the point x is given by (42)
r = ül A x,
where w is the angular velocity vector. in terms of the alternator, (42) can be written e,;k toi xk .
ft;
(43)
Since {43) holds for arbitrary x, the quotient rule implies that the quantities r ft cw} are the components of a second order tensor. Not surprisingly, it proves useful to regard v as given by the contraction product of this tensor and x. As will be seen in the formal development that starts in the next paragraph, it is conventional to deal with a tensor whose components are —E, ik to . With a vector CO we shall associate an antisymmetrk tensor Q defined by ^
=
^ Py r
^^
ir
^ = F_ - w.
or
(44)
Thus 4
(t2) _
— €^ 3 Wg
ca 3
^l 2
0
c), .
— (.1)1
0
(45)
Sec 2.1] T eruur Algebra
47
An explicit formula for w in terms of can be obtained by multiplying both and using the th rule: sides of (44) by
Eip4r/pq = Ei pq € pgr C+J
gipE
çç= f^ (60e5 î, - U9,üi9)(0,. =
3ui —
That is U li
= 8 i ^q ^i ^4 -
(46)
Similarly, (46) implies (44) when 12 is antisymmetric [Exercise 20(b)]. Sometimes (46) is written w = i113 _ (47 ) What we have shown can be summed up as Follows.
Theorem 18.
Corresponding to any antisymmetric tensor 13 there exists a vector to [given by (46)] such that 12 = - w. Here w is called the vector of the anuisymmetric tensor O. Corresponding to any vector w, there exists an antisymmetric tensor 12 [given by (44)] such that (46) holds_ We motivated our present discussion with some remarks concerning rigid body rotation. In this connection we can now draw the following useful
conclusion. —12 • x fct some Theorem l9. If a velocity vector v satisfies v = x • it antisymmetric tensor Q that is independent of the position vector x, then the motion is a uniform rigid body rotation about the origin with angular flpt . velocity w given by ta i = Proof. From Theorem 18, do) is defined as above, then Up =
xg f2gp = Xi? çpr w r —
'eprQ tt: r
X4
re)
A x]p.
Comparing with (42), we see that the velocity is that of a rigid body rotation
with angular velocity w. D REMARK. Later [see (4_1_33)] we use a result analogous to Theorem 19 to show, in essence, that infinitesimal displacements of the form u = S2 • x arc rotations. E XERCISES
1,
Prove Theorem I, parts (b) and (c).
2. Prove (9). 3. Verify (10), (11), (12), (13), and (14)_ is a 4. Prove from first principles that if T is a second order tensor, then scalar. 5. Prove Theorem 5: (a) for n = m = 1; (b) for n 1, rn = 3; (c) in general_ 6. Prove Theorem 8: (a) when m = 1, n = 2; (b) in general. 7. Prove Theorem 9.
Corleswn Tensors [Ch. 2
4^
relative to 8. Let T be a second order tensor with components T. and two different hases. Prove that (a) Td : T, implies that 7'1 = (b) 1; = — 7; ; implies that r ; 9. Show that the decomposition (24) is unique. 10. Prove (a) Theorem 12; (b) Theorem 13. 11. Suppose that for any Cartesian coordinate system there is a rule for associating T with 81 components 'C ity Suppose that for arbitrary vectors v and w, v • T w is a second order tensor. Prove that T is a fourth order tensor. 12. In a certain Cartesian coordinate system, the components of the vectors v and w are (1, 2, 3) and (4, 5, b). Find the components of vw. 13. Give a geometric interpretation of Example 3 in terms of the triple scalar product (see Example I.2.1). In particular, state why reflections must be prohibited. 14. in any Cartesian coordinate system there is a rule for associating T with components Ti such that T ; = Tfl . For arbitrary vectors a, a • T • a is a scalar. Prove that T is a symmetric tensor. 15. Let L be an entity whose components with respect to any unit vectors e[ 'l are given by [L]1 = ut`I• etrr. Here in'L,u[ 21, and u13 l are anorthonormal set of vectors that are regarded as fixed, while components of L are computed with respect to various coordinate systems_ Is L a second order tensor? 16. (a) Demonstrate (20). (b) Specialize (2Db) to the case where T is the stress tensor. Show that the original interpretation (Section 14.2 of i) of the components of T is recovered. 17. Prove Theorem 16. 18. What theorems and definitions show that the right side of (41) is an isotropic tensor'? 19. Prove Theorem 17 in the following manner (Jeffreys and Jeffreys. 1962, Sec. 3.031). (a) By considering special rotations, as in the text, show that ,
14 1111 — U2222 = 11 3333
^
CI.
W1122 = "2211 = 11 4233
11 3322 ! 14 33[1 — "1L33 " t 2+
"2323 = "1313 = "3L31
u2121 = 11 L212. = U 3232 — C3,
"2112 = W2332
1.1 3223 — 11 3113 = 1411331 a c4 ►
"1221
—
so that one can write
[T]
s = C 2 va
rs
+ C 3 e p, b y +
+ (C l — L.2 — C3 —
C 4)L•'pgr,,
where laps = 1 if all subscripts are equal and equals zero otherwise. „
Set. 211
The l:r,yerwakee Problem
49
(b) From the fact that T „x v x4,Xr Xs must be a scalar if x is a vector with components x , (why?), deduce that c, — r 2 — e 3 e4 = D. 20. (a] Verify (45). (b) Show that (46) implies (44) when .0 is antisymmctric. 21. Deduce (42) from Exercise I _ I.6(a). $22. Let To and £k be components of second order tensors_ Suppose that for arbitrary rpm iij
ilkmFkm-
(a) Show that if 7 is symmetric, then CI?,, Cjam. (b) Suppose that the E„ are the components of a symmetric tensor and that the CIA ., are the components of an isotropic tensor. Show that ;.=
2
E0
+ ft kk & i
for certain coefficients 1.4 and A. 23. True or false: If ecri r gives the components of et'k in a primed coordinate system, then (20c) implies that T- = etr}' T ,pg e rr
-r
9
Explain your answer. 24. The tensor product of two vectors is called a dyad_ A sum of such products is called a dyadic. 1(a) Write the dyad ab as the sum of symmetric and asymmetric terms, t(b) Let e( ') be a set of mutually orthogonal unit vectors. Show that (c) Try to state and prove some little theorems involving dyads. (d) Show by construction that every second order tensor is a dyadic. 25. (a) If U is a nonsingular symmetric tensor, show that U - t is symmetric _ (b) Show that (A B} = BT- AT . (c) Prove that a nonsingular second order tensor R possesses an inverse so that (28) holds. (d) Show that (C • D) - 1 = D - ' - C - ' `
2.2 The Eigenvalue Problem In this section we shall be concerned primarily with the problem of determining the eigenvalucs associated with a symmetric second order tensor.* We shall thereby discover a coordinate system in which the components of such a tensor assume an especially simple form. The eigenvalue problem turns out to be one of the most fundamental in all mathematics, but it is at first not
•
Many readers will he familiar with the eigenvalue problem fur square matrices andlor linear transformations. They wits Find tittle new in this section.
Kb, 2
Cartesian rr'nsurs
50
clear why one ought to consider it at ail. We can, however, find some motivation in the sort of investigation that moved Cauchy to begin the study of eigenvalue problems, namely a deeper study of stress. The relation between the stress vector t and the unit normal n is typically that shown in Figure 2.3(a), (We consider a particular point x and time t throughout, so we do not explicitly indicate the dependence oft on x and t.) It is not unreasonable to anticipate that there are directions n such that the corresponding stress vectors t(n) point in the n-direction, as in Figure 2.4(a). In such a case t(n) — do for some scalar A. Because tl — n) _ — t(n), a small •' flake" whose relatively large flat surfaces are perpendicular to n is subjected to a particularly simple set of normal stresses [Figure 2.4(b]]. The stresses simply act to extend the flake, or to compress it if A is negative. Compare Figure 2.4(b) with Figure 2.3(b); in the former there is a shear stress component of t perpendicular to n. This can perhaps be seen more clearly in Figure 2.3(c) in which the stresses have been resolved into components that are respectively normal and tangential to the flake. ti
tal
1
tb}
k)
2.3. (a) Typical refarrorPslirp between th e unit exterior aorr+ta l n and the stress rector t(n). (h) The stress rectors on opposite /ares aj' a small thin "flake" are equal in magnitude and point in opposite directions. (c) Stres.s vectors (dashed arrows) are resolved into components of shear (along the surface of the N1► e) und r.vtension (normal to the flake). FIGiJRe
Pure extension is a relatively simple state of stress. It is worth investigating whether such states generally de exist_ We thus ask ourselves the question: "Are there any directions n such that 1(n) = An for some scalar A?" Using the relation t(n) = n • T we can pose the question in terms of the stress tensor T: Given T, find n such that n - T = An. Indeed, if we can find any vector v satisfying v • T = ;ry
( 1)
we shall have virtually achieved our objective. We merely need to divide both sides of (l) by lid to see that r/Jv) is the desired unit vector n.
Sec. 2.2] The Eigenualue Problem
51
tl+tij - hn
1(n)
1( (J)
F1€;t3RE
2:4. (a)
(b)
Th e 3urftwe element with direction n suffers u purely normal
stress.. (t}) A flake that suffers purely extension, not to shear_
normal Stress will be subject only to
Let us select a particular set of base vectors 0 1 , In ternis of the associated components of v and T, CO becomes
o; T = "iv) or
v; }; _
(2a, l})
,
Fquurioris (1) and (2) are alternative sraterrterirs of the rnathernaticai problem that wewisii todiscriss Note that throughout this section by tensor" we shall understand "second order tensor." Writing (2b) as .
26,j ) = O,
(3)
we see that it can he considered as a set of three equations for the three unknowns r; - These equations clearly have the trivial solution y 1 = O. This solution is unique unless the determinant of coefficients vanishes. For a nontrivial solution to exist, therefore, it is necessary that the determinant of coefficients be iero, i.e., li
fJ^.`t
CI; —
--
A
TL 3
T12
7 :21
T22 -
T3]
T32
A
723
T3 3
;
O.
(4)
A
Expanding the determinant in (4), we find that --J. 3 + 1,1 2 - 1 +1 3 =a,
(5)
where l i — Ti.. z
T11 12 1
(6a)
T1 2 ^22
1 3 = dot (74).
+
Tl 1
T13
431
T33
+
T.2 2
3
T32
T3 3
(6b)
()
Cartesian Tensors [Ch. 2
52
Equation (5) is called the characteristic equation. Its roots (the zeros of the characteristic polynomial) are the eifienvalues or characteristic values of T. We shall denote them by A" ), i = I, 2, 3. The cigenvalues satisfy (1), a coordinate-free equation, so that they must have the same value no matter what coordinate system is used. The coefficients in (5) must therefore be scalars. This can be verified directly. For example, consider the first principal invariaat 1 , also termed the trace and denoted by tr T. This is the contraction of a second order tensor, and hence is a scalar (Exercise I.4). The coefficients of the characteristic equation can be economically characterized. To do this, we need two definitions. The principal dingoes' of (Ti j) consists of T11 , T22, and T33. An mth order principal minor of a third order matrix is an rn by m determinant formed by deleting the row and column of 3 — m terms in the determinant's principal diagonal. Thus the first, second, and third order principal minors of the matrix l 4 7
2 5 8
3 6 —9
are the determinants
^ll = 1, 1 5 1= 5, 1 - 91= —9, 3 7 1 —9 '
IS8 —9 '
1 4
2 5 '
and I 4 7
2
5 8
3 6 —9
respectively.
Theorem 1. The principal invariants I I , 12, and
equal, respectively. the sum of all different principal minors of orders 1, 2, and 3. Proof. We merely verify by inspection that the determinants in (6) are principal minors of the required order. p 13
Corresponding to each eigenvalue An», we can determine a nonzero eigeovector or characteristic vector v that satisfies (2). We denote the eigenvector corresponding to Al°1 by vtrl. With this notation, (2) reads t,4P 7 = xiri vl rybij (no sum on p). vrTii = Ali4v114 or (7a, b) Because our discussion was motivated by means of the stress tensor, we began with the eigenvalue problem (1) REMARK.
v •T= kw,
Sec. 211 The Eigenraahse Problem
53
but frequently there arises the related problem
T•w= .^w. It is easily seen that the characteristic equation corresponding to the latter problem is the same as (4), so one can speak of the eigcnvalucs of T But the equations that give the components of the eigenvectors are not the same in the two cases, so one must speak of Left and right eigenvectors of T. In the important special case wherein T is symmetric, one refers to the eigenvectors, for then the right and left eigenvectors are identical. Exemple I. Find the eigenva Ries and eigenvectors for the symmetric second order tensor T whose components in a certain coordinate system are I
0
0
^ 4
9
0 0
Solution.
-
4
In this case (4) becomes l
0
0
—A Il 0 4
A da 4
t]
4
=0
(8)
9 — ►i 4
or
0—(l —A)(
^
—h)^4 —
—6 1 (l—^l=(i— i^)(2—).1t3 —dl,
SO
Am=1, A''=2, A''-3. (Each individual can choose for himself which eigenvalue he chooses to designate kto, which ,1(21, and which dr31 .) When A _ At') = 1, the three equations of (7) become 0=0,
+ ^^^' t^' =o
.
-
3 r = 0,
^ _ at
(42"
Equations (9h) and (9c) have the unique solution 4" = t` ` = 0, so
1,
for any nonzero constant a
ut
( 9 a, b, c}
Cartesian Tensurs
54
[Ch. 2
When 2. = AI" = 2, (7) becomes
- n^r > _ {),
_
o, (43)42]
2f + (1») = 0. (1 a, b. C)
(-%(2)tl'
j
[The rosiest way to obtain the matrix of coefficients in (10) is to make the stibstiIUtian A -= Al21 in the matrix whose determinant appears in (8).1 Multiplication of (10c) by -
gives (10b), so (t0b) and (10c.) are dependent. We find that vS21 =
=0,
b, ll21 fb,
for any nonzero constant b. Thus 1.{21 = be {21 +
I31
When A = 1131 = 3, (7) becomes - 2 0»
T
U,
-
1 u,] f
4
—e^-)v^]} — (-)0311 = 0.
= 0,
—
which yields ^
{]I
= — V eeS^1 # C4 1]1 ,
for any nonzero cons t ant c. (By deflnition,eigenvectors arc nonzero, so e = O is un accepta bte.} If we want our eigenveclors le be of unit length, we can take V II)
=
e I^1 .
v121 =
^1^1
2
^t
+^
Observe that the unit vectors
2
{31,
^
1]^ - – 23 ^
– —
^^1 + 2 e1!}
(li ^
of (11) are mutually perpendicular: l,t11.
Yo
This is not an accident, but it is connected with the symmetric character of the particular matrix whose eigenvalues and eigcnvectors we just determined. For comparison, here are the results of a second example. Ex*mp!e 2. Consider the second order tensor whose components in a certain coordinate system are 1 t}
0
fl
0
1
t)
–4
U
Determine a set of eigcnvectors and discuss their orthogonality_ Partial solution. As the reader is asked to show [Exercise 3(a)1, the characteristic equation is (1 —2.)(A.' 4 4) = 0.
Sec. 2.2! The Eïye►rralue Problem
55
so ;he cigenvalues are a. 0 = t,
„p i]
2i,
dit3i =
with corresponding right eigenvcctors ue 1L
vr2t
= b&2
f 2jbe''f,
11°1
T
real
—
i^! ZiC+C^ .
and L are any nonzero constants_ Note that there are complex eigenvalucs and that the corresponding eigentectors have complex components. When complex components are possible, one usually takes the scalar product of vectors to be where a, h,
{r, w> - L
,
(12)
where an uverbar denotes complex conjugate_ This definition preserves the property of nonnegativc length: (13) Bul
n ow (w,") = (w. i>.
(14)
(Fer real vectors, v •w = w• v_) Whatever scalar product we use, v (21 and v'" are no longer or ihogonal, i.e., their scalar product is not zero For
v42 '. N 3 ' = 6e• + 4hr. # C; <1,1 n. 03)) =bc-4bc#O. FIGENVALLES AND EIG[NVECTDRS OF SYMMETRIC TENSORS We now present two theorems for symmetric tensors that are suggested by the difference between Example 1 and Example 2. (The importance of symmetric tensors is indicated by the observation that the stress tensor of a nonpolar material is symmetric.) We first show that we can discard the possibility of having complex cigenvalues and eigcnvectors with complex components. Definition I. A tensor is said to be real if its components arc real numbers_ Theorem 2. The cigenvalues of a real symmetric tensor are real numbers. Each eigenvcctor of such a tensor can be chosen so that its components are real. The result on which the proof of Theorem 2 depends is, it turns out, best stated as a separate lemma. Lemma 1. Let u and r be vectors. If T is a symmetric tensor, then u•T•v=v T- u.
Proof
(15)
By the properties of the contraction product, both sides of (l5) are scalars, So whatever coordinate system we choose, we shall obtain the same
Cartesian Tensors [C h. 2
56
quantities for the right and left sides of (l 5). Using the symmetry of T,
u - T;v= uT } v —r^11 v;=aT;u;=Y•T - u_ 0
(16)
Proof uf Theorem 2. Suppose that v-T
,v
(1 7)
v # O.
,
Taking the complex conjugate of (17), and using the fact that T is real, we obtain
V- T =v
(18)
We take the scalar product of (17) with v and subtract the scalar product of (18) with v v- T• k— v• T- v= A v• v= v• v. The k it side of this equation equals zero by the symmetry property (15). Thus (A
—
Inv I) = 0,
( 1 9)
-
since for any vectors v and w (in particular, when w =
v•w=w•v.
(20)
But v•vEIv
U,
since v^O
,
(21)
so (f 9) implies that A = A, Thus A is real. With 1_ = A, (18) becomes
v T- =Ai
(22)
which shows that if v is an eigenvector, so is i, Adding (22) to (17) and multi-
plying by f we find that ,
i(v +
• T = A(v + v) or (Re v) • T = ).[Re v).
Thus Re y is an eigenvec_tor if it is not zero_ Similarly, so is Im v. At least one of these will be a real eigenvector.* 0
Theorem 3. Figenvectors of a real symmetric tensor that correspond to distinct eigen values are orthogonal (perpendicular). Proof Let u and v be eigenvectors corresponding tu the eigenvalucs p and A, / # _. 'Then
u -T
=
v - T = A v.
(23a. h)
• In this discussion, by a vsxtnr we mean a first order tensor whose components may be complex numbers_ If r has components u,, then Rev. by definition • has components Re The reader should vcc ify that Re w is a vector if r is.
Sec - 2.21
The Eigenvalue Problem
57
We take the scalar product of v with (23a) and the scalar product of u with (23b):
u •T•v= pu •v,
v- T• u=2v•u-
(24a, b)
Subtracting (24b) from (24a) and using the symmetry property (15), we obtain
pu•v -
-
iv• u
0,
(25)
or, using (20),
(,u — 1)(u • r) — U. But j # 1., so u v = 0, as required. -
(26)
1[1
Careful examination of the proofs of Theorems 2 and 3 will show that they required only that T generate a linear transformation of vectors w, that T satisfy the symmetry property (15), and that the vectors be subject to a scalar product satisfying certain properties such as (20) and (21). A splendid chapter in mathematics is centered on the demonstration that, by means of shrewdly chosen definitions, the class of linear symmetric operators on vectors with scalar products can be made to include a large number of seemingly unrelated objects, and thus that theorems which are true for the relatively humble linear operator on "arrows" (vectors) remain true in far wider contexts. Appendix 12.1 provides elements of the required generalizations. PRINCIPAL AXES
Consider a real symmetric tensor T with distinct eigenvalues. According to Theorems 2 and 3, such a tensor possesses three real mutually orthogonal eigenvectors. We expect that the components of T will have an especially simple form if we take the eigenvectors of T as a basis for an intrinsic coordinate system. To explore this matter, let 7; ; be the components of T relative to an arbitrary basis e[ °. Choose a new basis Or satisfying Or = Id'', where Y{`" denotes the mutually orthogonal eigenvectors corresponding to the distinct eigenvalues The new components Tp q are found from the transformation law, "
T
(27)
+pig;
But trip
T e [ri . eIrl' -_
(28)
—
where OF) is the t;th component of vim With the characteristic equation (7a) and the orthogonality property, the transformation law (27) thus gives
T pq = viFlvJ } Tg1
Auit.[p}u'y } =
na
(29)
CdrresiQn Tensors [CJ,. 2
58
In matrix form, AO} 0 (no
-
-
A(F) ^
0
t^
n
( 30 )
3
0
M
We sum up as follows. Definition 2. If a second order tensor T possesses mutually orthogonal eigcnvectors, these vectors can be used as a basis for a coordinate system. In such a case, the components of T are said to be determined with respect to principal axes. Theorem 4. Let T be a real symmetric tensor with distinct cigenvalues_ Then, with respect to principal axes the components of T are given by the diagonal matrix of(30). The elements on the principal diagonal are the eigenvalues. It is often the case that once a result has been derived, it can be proved more simply. in the present instance, using (1.20a) we can provide a one line derivation of (30). Indeed, using the cigenveetors of T as a basis, we have Tl = LT . 0 "11 =- {) use a l t = At4(5,, We now illustrate what can happen when the cigcnvalues are not distinct. Example 3.
Ccunsider a second order tensor whose cc}mponertls in a
certain CoOr di-
nate system are
(31) Show that there are only two distinct eigenvectors_ What does this imply about the possibility of choosing a coordinate system in which the components of the tensor Form a diagonal matrix? Pzrriaisofrriaa.
The characteristic equation is (1 - 2)(2 — 2) 2 = 0,
so the ergenvaluc-s are ^[rt
= 1, au]
=
2.
Corresponding to At't, we find the eigenvcctor titi = act"; corresponding to Pt und 2{31. we find only the single eigcnvector v 121 = be [Exercise 3(b)J. With only two eigcnvcctors we cannot determine the three directions necessary for principal axes. At first, one might think of taking the third direction as perpendicular to rt'r and r1 '}, but this is no help_ If we did this, our - new - set of base vectors would turn out to be the old one, with respect to which the components, given by (31), are not diagonal-
See_ 2.2)
The E#yer+tohre Problem
5^#
The matrix (31) is not symmetric. It turns out that real -Sy cr?letrtt - tensors always have three linearly independent eigcnvectors, even when the cigenvalues are not distinct. We now provide a proof al this fact, although certain details are left to Exercise S .
'i ltenrern 5. A real symmetric tensor T always possesses three o r t h o n o rm a l matrix (30) can always be found foreignvctors.Thuadlinte such tensors whether or not the cigenvalues are distinct. Proof. Let T,-, be the components of T with respect to a particular Cartesian coordinate system. If the characteristic polynomial z
-`
— +l
-12)+ 13 — 0
has repeated zeros, certain equations must relate these components. (If zeros are distinct, the result is guaranteed by Theorem 4.) For example. if is a triple root, the polynomial is + 3cP
3(. 2 ; + r 3 ,
so that the equations
f i -- 313 3, 1 2 =
3
must hold. In the erase of repeated roots, choose continuous functions T,{c) in such a way that the equations above do not hold for c # 0, and such that 7;14) = 7,,(*;), 7,0) = T.
(32)
The 7:11c) are components of a real symmetric tensor TO.) with distinct eigenvalues ME) and hence, by Theorem 3. wish orthonnrmal eigenvect€ rs e9r.p eliV) = ii 11 . Taking the limit as c
0, we find that e ln(0) - 01t10) =
Thus the e"u(D) are the desired eigenvectors. If these vectors are taken as new axes, the diagonalization of T follows as in the proof of Theorem 4- LI EXERCISES
I. Verify that 14) implies that #5) and (6). 2. Show that J J 4!T•; T! — Ti, Tt ) and hence verify i hat 1 2 is a scalar. 3. Verify the results riven an the text for ta) Example 2, (hl Exampk 34. t(a) Find the left cigcnvectors of the tensor discussed in Example 2. part (a) guess an orthogonality theorem con(b)Fromthesulf cerning right and left cigenvectois_ Prove the theorem. ,
6o
Curresidn Tensors [Ch. 2
S. t(a)
Find thceigenvalucs andeigenvec:tors fnrthctensorwithcnrnponents
ü
0
ü
D
0
1 _ D
—1
(b) Note that the eigenveclors corresponding to nonzero cigenvalues have the curious properly that v • v = 0. Show that this property holds for all antisymrnetric tensors. 6. Verify (13) and (14) 7. Find the eigenvalues and corresponding eigenvectors of unit length for the tensor T whose components in a certain coordinate system are .
0
--I 3
l\
0
4
0
.
Verify that the eigenvectors are mutually orthogonal. Give a geometric description of the rotation of axes that will put the coordinate matrix of T into diagonal form. S. (a) Write down an explicit set of components T»:) that satisfy (32)in the case where the characteristic polynomial has a triple zero. (b) Repeat (a) in the case where the characteristic polynomial has a double zero and another distinct zero_
2.3 The Calculus of Tensor Functions Until now we have considered tensor cunstants. This was a necessary preliminary to a discussion cf tensor functions whose components are functions of one or more variables_ Exam* I. (a)
f he cyuxtiun of a space curve Is
x^ X(s),
aSs
where the position vector x is a function of the parameter s. (b) The veiociiy of n particle is given by v = v(r[, where the velocity vector y is a function of the time r. (c) By T = 7'(x, r) we denote the stress tensor of a material at position x, time r. As here, when a tensor is a f-unciit}n of position x we sometimes speak of a tensor field. ,
To see clearly how the transformation laws most be modified when we pass from constants to functions, it is perhaps easiest to concentrate first on vector fields. Let r(x) denote such i held, so that (to give a coordinate-free description) a vector v is prescribed at each point X. if a Cartesian coordinate
Sec. 2.3] the Calculus of Tensor Functions
6i
system is selected, the vector field can be described by assigning components at each point: 11; =
VAX', x2, x3}.
In another, primed, system, the description is YY
= I '1
X 1+ x 2 , x3)r1 •
What relation must obtain between the two triples of functions u, and vj2 To see, focus attention on a particular point 3t, with coordinates (1,, in the imprinted system. Let
=
^l..
so that the triple of numbers (xi , +c2, . J are the components in the primed system of the same point x. Thus . 3 ) and [xi, xi , _TO are components of the same vector; they must therefore be related by the usual transformation law. Consequently,
viii.
R29 k"!) _ iii
lc;
where
-
c
❑w
^]
The tildes were introduced to focus attention on a single point. "this point was arbitrarily selected, so that the tildes can be dropped. Furthermore, it is convenient to make the notation more compact hy using the functional relationships x = x(x') and x' � re(x)
(2a, h)
to denote x P - /x9 and x' — x r Pa
(3d, h)
'
With this (1) can be written r,.(x) = /,..0114x) J.
(4a)
CPO = u^Cx(x)^^;^
(4h)
Similarly, we also have
The corresponding transformation laws for second order tensor fields are fjx]
{ Îm f fn ^mN` x'S ii)^
(53)
+ m r^ T xix')].
(511)
and Tm,t[x']
The concepts of limit, continuity, and derivative are defined for tensor functions in exact analogy with their definition for scalar functions. For example, if we denote the partial derivative with respect to time hy r?r , then
T(x,t)= him ai-o
T(x,f +Air)At
T(x '
^ ] .
62
Cartesian Tensors
[Ch. 2
Since the Iii are constants, differentiation of the transformation law shows that the time derivative of a tensor (if it exists) is a tensor of the same order. There arc a number of notations connected with the process of differentiating a tensor field with respect to a position coordinate x 1 . Two frequently used notations for ô I fi:3x p are ap T and Tu ,. Direct notation is achieved by means of the "del" operator V, defined by ;
[V]e du
(6)
x
Thus, by definition, if T isasecond order tensor, f T?
(7)
Jpij — Op 7 7 ;i -
In words, VT is often called "the gradient of T" and is sometimes written "grad T" Divergence and curl are also defined by generalizations of their familiar definitions for vectors: (div TjJ - [V' T1 id,T}!
[curl
rig
-
(8 )
[V A T]ia = r ut, i3J Ti+4
- F
(9)
^JR ^F9•!"
As in the previous section, our discussions will often be confined to second order tensors, but they have obvious extensions to higher order tensors. For example, if T is a third order tensor, VT and V - T are defined by
1 ij k = [V- nik t a drift = Ti". rTheorem 1. If 7' is a tensor field of order ni, then VT is a tensor fjeld of order m + 1. Proof when rn = 2. The rule for determining components is given in (7). Let us write it out in full, using standard notation for partial derivatives: [V ()] A^,
[T,^(x)].
(loa)
We must prove the transformation law,
Employing (IOa), and explicitly noting the role of the independent variable, we can write this law in the form epLeirn^^al
Ô^^
aR
^! j{7^^^ '= ^ ` ^ •
(
1 Oh)
See. 2.3) The' Calculus of Terrsur Functions
[^
' and -1;i and remembering [ hat the / ,, are But using relation (Sb) between "F„„ constant, we have f
C1
x; 7 m„(x') = imi
Tii[x(x')]
The partial derivative on the right side can be determined by use of the chain rule:
c C^]Ck
7 ;;[x(x')] =
G^]CP
Tf(x)
ex
^
(l2a)
,
xix'j OXk
From (3a), however, (12b) We thus obtain the desired transformation law (10) by combining (I L) (12a), and (12b). ,
Theorem 2. If T is a tensor field of order m, then V - T is a tensor field of order rn -- I. in particular, if v is a vector field, then V - v div v is a scalar field. Proof. V- T is a contraction of the (ni + 1)st order tensor VT, se the conclusion follows by a result of the type of Theorem F4. C Theorem 3. If T is a tensor field of order n,, then V A T is also a tensor held of order rr2. In particular, if iv is a vector field, then V A t' is a vector held. Proof_ Left to the reader.
THEOREMS FOR DERIVATIVES OF TENSOR FIELDS
The subscript notation makes it easy to derive useful identities. For example, if and v are differentiable scalar and vector fields, then V-
(bv) = (04 ) .), `
tea + ^^
L
` 4d • 1 + dŸ • v.
Here are some examples of how the alternator can be used in deriving certain useful formulas involving vector fteids. The proofs are rather easy, but we shall write out several of them to provide examples. Theorem 4. Proof.
div curlv =9•0nk yf} ❑ ` ( lI
A Y l = ^ijk ^k . li — ^i ^ pit. ail
(13)
where the last equation assumes that ta k has continuous second partial derivatives so that the order of partial differentiation can be interchanged.
Cartesian Tensors [Ch
64
_
2
We rewrite the first equation in (13), exchanging the dummy subscripts i and j. V • (V A v) = e1;^ i :^ ; _
(14)
^
13ut =
so (14) can be written Y • (Y A 1 }=
—
E l jf
k.
Comparison of (1 b) and (13) gives the desired result. 'theorem S. curl grad J — V n Proof Left to the reader _
(Vbj =
Theorem 6. V A (V A v) V(V• ^?
Proof.
z
V) ---
0
O.
V 2 v, where
= i1m. ;; .
1 Jsing the cb rule,
^,{ &., gn) tj uk A— ia,a,v, = (V(V•v)— V z vl m -
ç ,j [Y^ A V A l'3, — ERnn. a n Erjk ;Vh = ([} rn^" ^
,,
In writing the last expression, we assumed that tin has coritir+uous second partial derivatives so that the order of On and Om could be interchanged_ (In order that the expression V A V A r be defined, v must, of course, possess second partial derivatives, Nit they need not be continuous) U Theorem 7. Proof [Y
V
A
(V
A
v)
IV(x • v} — (r • V)v. 6,,,4S,01Jn r?11.k
A (V AV)] m = ERnaiVnEijk^ j i }k 4(^m 1 ^ n k
=
t;n l^ n.
rn — L' n {^ n Um — ;i0m(ün v n ) — i1 n en t1„'
[IV(v • v) — (r • V }v L .
0
INTE(;RAt. THEOREMS
It is worthwhile recording the results of the divergence theorem and Stokes's theorem* using subscript natation: Divergence theorem n it),- da — jiff i v; ; tdr.
• All vector integral lheorctssm requtre sultictrns smoothness of the functions and boundaries invo [vcd. See,
KeIIog (1924]_
Sec. 2.3] The Calculus of Tensor Functions-
65
Stokes's theorem: I F d5 =
Eiikng ejG'kdff.
BS
Let us consider the divergence theorem a little further. Suppose that v has only one nonzero component, so that v, --riA5 i, for sonie scalar function rti and for any fixed integer j, j = 1, 2, or 3. Then the divergence theorem gives
n1 el da =ffJ i c' dt
j= 1, 2, or 3.
(l7j
li
dR
Since (17) is truc for any j, it follows that
dR
n()) da l fff d_'
(18)
R
which is Gauss's theorem. This theorem and the divergence theorem itself,
j
i
^ - v dr,
n-vdcr_
can be generalized to make a statement about general tensor fields, as we shall soon show_ First, we give an example that clarifies the theorem to follow. Example Z. Show that the quantity on the left of (20) is a (Constant) third order tensor. Solution. In primed and unpriined systems, let the components of nTdir be given by ' rr
-
nT da OR
P4f
r:p7Qdrs OR
and
nT
n, T r d u _
^
A te
We rriLLSt
iik
PR
prove that
nT da or
air
But the expression within the square brackets vanishes because nT is a third order tensor.
66
Cartesian Tensors [Ch_
2
Theorem 8. Lin is the unit exterior normal to a region R and T is a tensor, then
jEjn.Tdu= JJf v
(19)
OR
JJJVT dr.
n T"da ^R
(20)
R
Proof of (20) when T is a second order tensor. The quantity on the right of (20) is a third order tensor, with components given by
[
Jff
fffiv! k
vTd i = ]
ijk
!t
dT.
R
As for the third order tensor on the left of (20)- for any fixed j and k, can simply be regarded as an ordinal y function of x so, by (17),
Tf(x)
,
{J
I k dtr —
aR
fJJ
c' i 7 & dr.*
R
We have thus proved that the components of the tensors in (20) are equal in one arbitrary coordinate system. As usual, this is enough to ensure that the tensors themselves are eq ual. 0 Trivial modifications make this proof of (20) serve when T is an nth order tensor. Utilization of the same ideas leads at once to a proof of (19) and to a proof of the following generalization of Stokes's theorem. Theorem 9. As in the usual version of Stokes's theorem, consider a surface S with normal n Let OS be the curve bounding S and t be the appropriately oriented unit tangent vector to OS. Let T be a tensor of arbitrary order. Then
Ç
T.tds_-fJV •
A
T.nd7
(21)
s
as
* To remember these theorems, note that a V in a triple integral change; to an n in thc corresponding double integral_ The same mnemonic work with indcz natation, as is shown by the index vcrsinn ni{t9l.
ffn' T, da = ffia: -ri, a r_ ret
Here n, •- ► df . Note, however. that although n, i = Ti rs,, the former order (bumming on adjace nt mnemonic is to be employed. subscripts) must be usai if the
Sec. 2_J] The
Calculus of Trrrsor Fiala iarrs
67
The exercises contain some applications of tensor calculus to physical situations. A number of other applications will appear later_ In particular, Appendix 2.1 lists some fundamental equations of continuum mechanics in subscript notation. We shall devote much further study to these equations. REPRESENTATION THEOREMS
Theorems 4 and 5 state that u=Vny and
w= V
imply that ❑
V•u=0 and
A w--
l1,
respectively. Of great use are the converses of these theorems and their generalizations. We provide some classical results of this ty pe here The first two theorems concern a continuously differentiable vector field connected domain. wdefinovrasmply .
Theorem 10. There exists a single-valued function b such that w — if and only ifVAw=O_ Proof. See Exercise 14.
❑
Theorem 11_ If and only if V- w = 0 does there exist a solenoidal (i.e., divergence-free) vector field w such that
w =V
A
r,
Alternatively, if and only if V • w z- 0 does there exist scalar Fields 1,10 and y such that
w- V Prnof.
A
( OVA -
See Exercise 15 or At-is (1962, Sec. 143).
Theorem 12 (Helmholt-r. Representation Theorem). Let w he finite, continuous, and approach a limit at infinity. Then there exist scalar fields , , and y and a solenoidal vector field v that permit one to write ,
w=Vint+VAv
or w — Vc+VA ifiVy
Proof Given w, we wish to obtain 4' such that V -(w — Vt) = O. Hut V 2 0 = V- w is a Poisson equation whose solution is given by the integral formula (16.1.7) of i. Since w — gc is solenoidal, the result follows by Theorem 11.
n
68
Cariesldn Tensors [Ch. 2
EXERCISES
1. (a) Prove Theorem 1 when T is a fist order tensor; i.e_, show that if v is a vector field, then Vv is a second order tensor field. (b) As an exercise in juggling subscripted subscripts, prove Theorem l in the general case. 12. Prove Theorem 3 when m = 1. 3. Prove 'theorem 5. A full statement of the theorem requires an assumption about the scalar field 0; what is this assumption? 4. State Theorems 7 and 8 using the terms "curl, ,, "grad," and "div." S. Let u, v, and w be vector fields and let and 1,6f be scalar fields. Find alternative expressions for the following: (a) grad (0); (b) div (u A v); (c) V • [u A (v A w)]; (d) V • (V0 A V+ ); (e) V A (0); (1) curl (v n w); $(g) curl (0r). 6. (a) Show that V n(unr)=u(V•V)+(v V)u--v(V - u) — (u•4)v. $(b) Show that V(u-v) = (u- V)v + (v• V)u + u A (V n v) + V A (V n u). 7. When T is a third order tensor, write out a proof (a) of Theorem 8; (b) of Theorem 9. The next few exercises exploit the power of the formalism that we have developed to derive several important results concerning the vorticity (local spin) of fluid motion. Section 3.5 contains a general discussion of the role of these and similar results in understanding fluid flows. S. (a) The vorticity w of a velocity field v is defined by ru = curl v. Show that curl a =
Dt
— w • grad v + w div v,
(22)
where Dt
(b)
4,r+v• V
Dv
and a = — Di
are the substantial derivative and the acceleration, respectively. Deduce the Beltrami equation. w) D(p - , — p` r w• grad r+ p -r curl a. (23a) Dt
(c) By definition, p = p(p) in a barotropic fluid. Consequently, one can V J p - I dp. Take this result for granted here. write p - IV
Sec_ 131 The {.'alcufus of frrr.sur horcriom
t^y
fit is the subject of Exercise 15.13 of E.) Show that in an inviscid barotropic fluid subject to a conservative body force.
D(0. -1) 1o). grad rDi = P -
(23b)
9. Integrate (23 b) as follows: +(a) Define V, W, and S as the velocity, vorticity. and density in material coordinates:
V(A, r) = v[x(A. t), t]. W(A, t) = [x(A, t) tj, .6(A, i) = p[x(A, ^) f]. ,
Write (23b) in indicial notation after having made the substitution x = x(A. t). (h) introduce a function C(A, t) by the equations r _},
W
-
?A ;
For C to be well defined, we must assume that J # (J, where J is the Jacobian 41(x1ax2, x3)
MA I , A2,143) Why? Show that r3C = O. Deduce the rorikity diteoprnenr equation
'W - 6n 'W,, ' J
( 24)
where
Grit
r7x; ,
— ^t^ t o ), ,
W u. _ W(A, r o).
The vorticity development equation was obtarned by Cauchy in 1815. J. Serrin recently gave an entirely different derivation, which is essentially outlined above_ 10. Consider barotropic motion of an inviscici fluid subject to conservative body forces. (a) Show that the vorticity development equation implies that a portion of fluid initially in irrotational motion (co = ü) will remain irrotational motion (Lagrange --Cauchy theorem). if to = Q in V A Y .= 0, then v = V for some Function cP. The existence of has far-reaching consequences, some of which are explored in the material concerning fluid mechanics_ (b) Show that in a two-dimensional flow (w = iVaz -= 0) a fluid particle retains its initial value of p ru].
7°
Cartesian Tensors
[Ch. 2
11. Using subscript notation, show that (14.3.4) of I can be written
JJJ
x
A
(f'
Dr ~pf—V.
T di,
T) dz =
R
(25)
R
where T is the stress tensor and [7s], = ik TA. Assuming conservation of linear momentum, deduce the symmetry of the stress tensor. 12. This problem deals with angular momentum for polar fluids. The starting point is Equation (14.15) of I. (a) Show that the couple stress vector c(x, t, n) satisfies c = n. C for some couple stress tensor C(x, r). (b) Obtain a differential equation form of (143.5) of I. assuming that integrands are continuous. Subtract the vector product of x and (14.2.19) of 1_ Deduce DI
p^t =
g+P - C + Ti -
(26)
In general, then, internal angular momentum is generated by body torques, stress couples, and the antisymmetric part of the stress tensor. 13. The purpose of this exercise is to show that balance of energy and certain invariance requirements lead to balance of linear momentum, balance of angular momentum, and so on. This material appears in a note by A. E. Green and R. S. Rivlin [Z. Angew. Muth. Physik 15, 290 (19fî4)]. (a) Energy balance is postulated in the form + pet Llr — i 1[71
I1
+
pF^ v 1Itit
—h + C i v;] drr,
(27)
dR
where r is heat supply per unit mass and time (perhaps from radioactivity) and h is heat efflux per unit area per unit time. What are the other symbols in (27), and what is the meaning of the other terms? (b) Equation (27) is supposed valid for all velocity fields, in particular for v, + a i , where a ; is an arbitrary vector that is constant in space and time. Suppose that e, r f , F1 , h, and r are the same for two motions that differ only by constant rigid translations, represented by aj . Write (27) when v i is replaced by vi + a i , subtract (27) in its original form, and deduce the integral formulation of the balance of linear momentum.
Sec. 2-3]
(c)
The Cuir rha. of Tenu-or Functions
7I.
Just as in Section 14.2 of 1, one can now derive =
(d)
"Ii ;
(28)
and the differential equation for linear momentum balance. Use these results and (27) to derive 1JJt i
—
pr —
I t J, k] dT
— fri h rla.
(29)
eR (e)
Assume that e, r, T. and h are the same for two motions which differ only by a uniform rigid rotation. Now if u t -# v + uniform rigid rotation, then vi.k ta}.k +
Rho
PO]
where i2 1k is an arbitrary constant antisymmetric tensor. Explain. Make the transformation (30) in (29) and obtain nfk
Jf$
TLJ d = 0.
(31)
R
(g)
Show how (31) implies that TA — Tk ^. (This part requires knowledge of the material at the beginning of Section 3.1.) As in (14.4.20) of I,
h = 101
(32)
for some vector Q1 . The usual constitutive equation that is taken to relate the heat flux hector Q r with the temperature f is Q1 = xB,} (Exercise 3.1.5). Use this and the symmetry of the stress tensor to rewrite (29) in the form of a differential equation involving the rate of deformation tensor. 14. Let VA W-0 in a simply connected domain D, where w is continuously differentiable. Let 0 be a fixed reference point in D, and let P be a point corresponding to x. Define P
¢{x) =
w • t ds, 0
^
where Stokes's theorem permits the line integral to be taken along any path in D (t being the tangent vector). Prove that a0/ 0x, = w ; by applying the definition of partial derivative and by taking as a path from x to x + (4.4et 1 the straight line parallel to the coordinate vector e'`'. 15. Let * be the continuously differentiable solenoidal vector field that is the subject of Theorem IL (a) Let the differential equations dxidt = w ; [xtr)], t a parameter, 1, 2. Using r.,; a have two solutions given implicitly by ,f,(x)
72
Cartesian Ierisnrs
[Ch. 2
geometrical reasoning, show that one can write w= 2 Vf,
Af2
for some scalar function A. lb) Let f3(x) be such that J = del (Of!f lx,) # 0. Regarding the f as coordinates of x, show that row ;
N.
Px, afin
,
and deduce that 1, is a function off, and f2z fonly. (c,1 Let
where f2 is held constant in the integration_ Identify x with fz and v with irvy; then complete the proof of Theorem 1 I.
Appendix 2.1 Sonic Basic Equations of Continuum Mechanics We give here for reference, and in order to show examples of subscript notation, certain basic equations of continuum mechanics. These equations were discussed in l; a summary of the discussion appears in Section 14.5 of I_ We consider a nonpolar single-phase continuous medium that does not exhibit electric, magnetic, or chemical effects. To characterize this medium, we employ the following unknowns:
density p,
pressure p, heat flux vector q ; ,
velocity vector u„
e,
specific internal energy
stress tensor T j .
Regarded as given is the body force per unit mass f,. The field equations arc as follows. Conservation of mass: Op Dt +pL " =
D Dt
a
-+ at
(la) ^'
t1 r 1 .
(lb)
Balance of linear momentum:
Dn^ P^ t
F.^c + ^c. J•
(2)
Appendix 2.11 Some Basic Equations of Continuum Mechanics
73
Balance of moment of momentum:
7 = To .
(3)
Balance of energy: ^^ ^^ra;or
+ e ^=
Pli
-
- T A ].
i);[gi
(4)
There are also important thermodynamic relations (which can be found in Section 14-5 of I), but these will not be given here Examples of constitutive equations, which single out particular types of continuous media, are the following. For an inviscid fluid the stress is purely normal :
(5)
so that the momentum conservation requirement (2)gives the Filler equation, Dv, P ^^ = PI;
- 1^.1 •
The Newton-Fourier law of cooling states that hear flow to temperature gradient: =
(6)
LS
proportional
(7)
PART B Problems in C ontinuum M echanics
CHAPTER 3 Viscous F luid s
P
B develops the constitutive equations for a viscous fluid and a linear elastic solid. Effects of viscosity and elasticity arc explored by solving selected particular problems of mathematical interest This chapter is concerned with illustrating some of the effects of viscosity on fluid motion, and with outlining some of the mathematical ideas that are required in the relevant theoretical investigations. We begin with a derivation of the Navier -Stokes equations, the simplest equations that exhibit viscous effects. By far the largest part of viscous flow theory is based on these equations, for they seem~ to apply in almost all circumstances for common fluids like water and air. Our derivation requires a knowledge of Cartesian tensors, although an alternative derivation not using tensors is outlined in the exercises for Section 3.1_ A sample of the variety of phenomena governed by the Navier-Stokes equations is provided by the investigation of a few exact solutions. We then obtain a simplified version of the equations. This version holds in the "boundary layer" region of rapid variation that is present near solid boundaries in relatively fast, slightly viscous flows. The boundary layer equations are solved for the simplest case, flow past a semi-infinite flat plate. With the background gained, one can obtain a synoptic view of the general problem of fast slightly viscous flow past an obstacle. Another general matter that is considered at this point is the role of voracity in viscous flow. To avoid the impression that relatively slow, highly viscous flows are not of interest, we conclude the chapter with a brief discussion of the important "Stokes flow" past a sphere. The central mathematical idea of the chapter is that of the boundary layer_ The applied mathematician G. F. Carrier has aptly summarized* the importance of Ludwig Prandil's I904 paper in which the ideas of boundary layer theory were first set forth. As Carrier says, its ideas not only permitted Prandtl to solve his problem but have also given rise to a very large body of mathematical formalism commonly referred to as `singular perturbation theory' or `marched asymptotic expansions': more importantly, they also have provided the basic foundations for the heuristically-reasoned, spectacularly successful treatment of many important problems in science and engineering. This success is probably most surprising to rigor - oriented ART
In the paper, "Heuristic Reasoning in Appricd rviaihemalics" [Quart. Appt. Math. 30, I I- IS( 1972)); the coi ire issue dills journal os devoted in symposium, - The FtintrcofApplied Mathcrnarics"
77
Viscous Fluids [Ch . 3
78
mathematicians (or applied mathematicians) when they realize that there still exists no theorem which speaks to the validity or the accuracy of Prandtl's treatment of his boundary layer problem; but seventy years of observational experience leave little doubt of its validity and its value."
3.1
The Navier Stokes Egnations —
Section 15A of I directed attention to the motion of a knife blade through a very viscous fluid (cold molasses). We were led to the hypothesis that shear stresses depend on the relative velocities of adjacent fluid layers. Now, "very viscous fluids" exhibit a relatively large amount of shear stress. For example, air and water do not seem very viscous compared to molasses. We argued, therefore, that it might be reasonable to neglect shear stresses altogether For air and water, and hence to consider purely normal stresses and the accompanying inviscid flow. The ensuing discussions of stratified fluids and compressible flows, for example, led to results that generally appeared in accord with experience, so the neglect of shear stresses seemed reasonable when analyzing those phenomena. But D`Alcmbcrt`s paradox made it evident that the inviscid assumption was nor always a good one. We must develop a constitutive equation that is more widely applicable than the equation I — —pn, which defines an inviscid fluid. ANALYSIS OF THE LOCAL 'VELOCITY FIELD
Since shear stresses depend on relative velocities, let us look more closely at the relative velocities of two nearby points x and y_ We regard x as a fixed reference point and we do not explicitly indicate dependence on the time r_ Then (using the summation convention), a Taylor expansion about x gives L'r(y) — «.(x)=. ^(x)(vs — x3}+ --- , .
(1)
where • - • denotes terms small compared to those retained. More precisely, • - • denotes c„(y s -- x i), where (2 ) if v has continuous second partial derivatives in a domain containing x_ In the language of differentials, if dx°y — x, then
dv, = ur.,(x) dx,.
(3)
Here dv is the first order approximation to dv, where Av = v(y) -- v(x)
w(x • dx) — v(x).
(4)
Ncrwer -Sivkrs Equatio:ra
See. 3 -fj
79
We shall be making the assumption that the stress tensor 7(x) depends on relative fluid velocities only through the velocity gradient tensor v ,4x). En a first attempt to postulate a dependence of stress on relative velocity, it is reasonable to neglect the higher order terms in (1) Should we not obtain results in accord with experience, we can try to formulate new equations that take some of these higher order terms into account At this point it is useful to examine in detail the relative velociticsassfsc,ated With ur . 5(x) in (3), remembering that x is a fixed reference point. To do this, we split up t}, ,5 into its symmetric and antisyrnmetric parts. which we denote by D,„ and 1,. Thus ,
Vr,
s —
v^ here 17 r ^ = D„
Drs 1-
and
— I/sr-
W,
( 5)
Using (2.1.25), we see that the components of D and i•1% are given by /4 rs
(6)
2(r)r.1 — Vs, r)-
'
From (3) and (5), dL' r
= D,a dx s + Wa dx„
(7)
so the relative velocity associated with v i,., can be regarded as the sum of the separate relative velocities associated with D,s and 1;, . By Theorem 2.1.19, the relative velocity described by 14 dx, is a solid body rotation with angular velocity vector —el, where (1)q
i.vi rs(t`'r. s
Hut EWrs'L'r,s = 4
'
,r
'yrs
r,s
s.rs
so
= 4£grs ii .r = AIV A Vlq-
(8)
The relative velocity associated with W is thus equivalent to a solid body rotation with angular velocity vector z curl v, W is called the rotation tensor_ When V A v O (and hence v = Vrk for some scalar function (P), the now is called ïrrotational. We now examine the effect of D by considering relative velocities given by dv w dx • D. Let us write out the components of this equation. To obtain the simplest possible expressions, we choose the principal axes of the symmetric tensor D as basis vectors_ We obtain 4'1Y1
—
v i(x)
-
X,,)
(no
sum on
i),
(9)
where the quantities Pare the eigenvalues of D. Consider an arrow of length L 1 emanating from point x and pointing along the x t -axis (Figure 3.1). Let denote the components of the tip so that, using (9). Y.`" = x ; -t -
u,(y 1 ) — r1(x) = Atr1L1 bi, -
(10a, b)
80
Vs-scram Fluids (C'h 3 xi
x
^
f
l/
x2 _ ^^.,^xl
F1GI;RE 3.1. A rector of length L, aligned
euh, Ike
r r axis_ If the axes are
principal axes, this rector elongates at a rah, per wiry length yiren N. the first
eiyenralue of the deformation tensor. According to (10h), the velocity at the arrow tip differs from the velocity at the arrow hase by an amount d'"L, directed along the arrow. The relative motion of (9) thus causes the arrow to undergo a pure elongation at a rate of P"L L length units per unit time. Dividing by L 1 , we see that »'' represents the rate of elongation per unit length. In like manner, arrows along the x 2 - and x 3 -axes suffer an elongation per unit length per unit time of magnitudes and 13^, respectively. (If one or more of the /1. (1) are negative, the corresponding arrow suffers a negative contraction.) The relative velocity associated with D is thuselongati,r equivalent to a deformation that can be described as the superposition of three elongations in three mutually perpendicular directions given by the cigenvectors of D_ f) is called the deformation tensor or rate-of-strain tensor. Let us summarize the kinematic results we have obtained. It is helpful to rewrite the fundamental equation (7) in the form v(y)=b(x)+(y—x) - (x)+(y—x) - D(x)+--..
(1 I)
Since higher order terms are neglected in ( 1 l ), one often says that (1 I )describes the velocity field in an infinitesimal neighborhood of x. Our discussion has shown that this velocity can be regarded as the superposition of a uniform translation with velocity v(x), a slid body rotation with angular velocity vector I curl i (x), and pure elongations along the three mutually perpendicular principal axes of D(x) at a rate pe r unit length equal to the respective eigenvalues of D(x). ASSUMPTIONS THAT UNDERLIE TFTF CONSTITl.TIVF EQUATION
Six assumptions underlie the relation that we shall ultimately obtain for thesrno: 1. As already stated, we shall assume that the stress tensor T at a point depends only on the velocity gradient tensor at that point. That is, only the first derivatives Vr, r of the velocity components yr will enter. 2. As a result of a change in its makeup, the response of the fluid could be different at different locations in space and time. We rule out consideration of such cases by assuming homogeneity in space and time.
Sec_ 3.1)
T he !V ar ier -Stokes Equarior,s
8i
3. For a given distribution of stress, the response of t: fluid generally caries with its thermody namic state. An example is the fact that molasses becomes less viscous as the temperature increases_ We will have to take this Into account in our constitutive equations. 4_ According to Section 15.1 of I, our constitutive equation must he such that it reduces w the hydrostatic relation Ti = pi5ti when the fluid is motionless_ Putting together the consequences of our discussion thus far, our constitutive relations must take the form 7;f = — pi5 j)+ft
p. 0),
MO. p,0)=ff,
112a,b)
where we have selected the density p and the temperature B to represent the thermodynamic state of the fluid. Equation (12a) contains nine functions fij which are arbitrary except for requirement (12b). N u. The hydrodynamic pressure p rn (12) is generally defined merely as the coefficient of 45 4 . Only in the motionless state can the hydrodynamic pressure be identified with the pressure encountered in equilibrium thermodynamics. 5. We now make the ass:lrrmptiori chat the dependence oaf T on ❑ is linear. This is the simplest extension of inviscid theory. It should he valid if the deformation rates are "small" in some sense, for then products of the derivatives of velocity components will be relatively small With the linearity assumption, (12a) becomes (13)
The coefficients C E ,„ will depend on p and H. By the quotient rule, these coefficients are the components of a fourth order tensor, C. Since 'IJ is symmetric, it must be that (14)
C.' jrrs -
To see this, we interchange i and j in (13) and then use the symmetry of the stress tensor T and the Kronecker delta to obtain = 'l Îj
+
C jirs u,,1.
Upon subtraction of (15) from (13) we find that for arbitrary y r .„ B„rar, s = ü,
where Brs = Cur. —
Cjirl'
In particular, (16a) holds when u 1.1 = 1 and all the remaining cr , are zero. Thus S i , _ O. Proceeding similarly, we seee that all the Br, must be zero, which yields the desired result. 6. As a final simplifying assumption, we restrict our consideration to fluids that are isotropic, by which we mean that there is no preferred direction
82
J./sumo
fluids [f.'li_ 3
in the Muid. To see the consequences of isotropy, consider the constitutive equation (13) in another, primed, Cartesian coordinate system:
— f ", ŸJ +
=
Isotropy demands that the contribution of u„ to 7;1 be identical to the contribution of Cr . , to Teo . Thus 's , _ ['^ # ; i.e., the tensor C is isotropic. In order to obtain a better perception of the maihematical consequences of isotropy, we shall discuss a particular instance. To this end, let us consider a velocity field of the form u e = u i (x 2 ku2 — 1N 3 - O. As illustrated in Figure 3.2(a4,we shadl foots attention on the stress component in the x i direction, which is due to the action of the fluid in the domain x 2 > 4 on the fluid in the domain x 2 < O. This stress component. 72 1 , is the same all akwig the plane s 2 = I]: for definiteness we consider the w1 rests at the origin.
In Figure 3.2(a) this stress is denoted by a heavy arrow. Let us now consider a new flow wherein the velocity field is rotated counterclockwise by 9G [Figure 3.2(b)]. if we perforrn clockwise rotation of the tarok, Figurc 3.2(b) appears identical to Figure 3.2(4 Because the material is assumed to he isotropic, the correspondingly rotated stress [heavy arrow in Figure 3 2(b) I roust be identical with the original stress. The reason is that since there is no preferred orientation in the fluid, one cannot distinguish between an urtrotated look at the vciocity held of Figure 3.2(a) and a rotated look ai the (different} velocity field of Figure 3.2(b). Equivalent to rotating the book, in order to view Figure 3.2(b) sideways, is a 94" clockwise rotation of aptes Thus, using primed axes, the stress component T'2, in Figure 3.2(b) must be identical with the cotrtponent 72 1 in Figure 3.2(a). Similarly, the velocity gradient v'1 , 2 in Figure 3.2(b) must be identical with v 1 _ 2 in Figure 3.2(a). But from the respective constitutive equations for the two flows 1(13) and its primed counterpart] 721 —
T21 — C sI2e 2
C31t1v1.2•
,
(17)
Since we have lust seen that ^21 =
%t,
ut.t =
it follows from (1 7) that • 2112
= C2112.
5incc (13) must hold for arbitrary velocity fields, repetition of the argument we have just given for various special cases gives the result that for all , j, k, and 1,
Cffia =
That is, the tensor - in (13) with components
Cud must be isotropic
DERIVATION OF THE FINAL. EQUATIONS
According to Theorem 2.1.17, since the C,,kt isotropic fourth order tensor, we can write
are the components of an
Mfrs = ^%ij ars + /41:5ar45js + a;s ( jr) + 1 f (6 ir ^ js y 451s 6 jr)
for some scalars it, ti, and
K. It is
(18)
easy to see that the symmetry requirement
(14) holds if and only if K = 0 [Exercise 1(a)]. Now, using the decomposition
The Nai•ter Siokrs F"quarirms
Ser. 3. 1]
83 t2
= ritjt ^j
t2 .
Ob
!
3.2. (a l With the given velocity profile, t h e heavy arrow represents at utiyiR. on lo n e r (shaded) material du e to action or upper iurt,sltaded) material. (b) A 90' cYrunterr•Irki a•zge rotational the vault ypro frle hit ia )yit -es xhe profile here. The effec t cdi th e unshaded region on rire shaded results in a eorrresfJr+RdiruJi}• rotated .stress ( lrerr r ► arrow). This is because the material is Isotropic : orfiervise, resistance to dejerrmarinan would in general depend an errr eerteuton, so that a rotated velocity field would not far rise to a rotated stre ss. FlcURL
Strum
viscous fluids [Ch. 3
84
of Lr into its symmetric and antisymmetric parts Dr_, and — 4-ilrs Drs +
COrs Wr, ,
But with K = Q, (18) shows that C ur, is symmetric in its last two indices and we know that F'rs is antisymmetric — so the second term in the preceding equation is zero [Exercise 1(b)]. Therefore, in its final form the constitutive equation shows that the stress tensor is determined by just the symmetric part D of the velocity gradient tensor. This final form is [Exercise 1(c)]
1 îj
(—p + h11rr)"i1 + 2pDi J •
( 1 9)
The viscosity coefficients ). and p depend on the density p and temperature 0 (or any other convenient pair of thermodynamic variables). In direct notation, since D r, = v, , we have T = [--p
A(V - v)]! + 211)-
(20)
When our constitutive equations are substituted into the equation for linear momentum conservation (A2.l.2), we obtain [Exercise 3(a)] the ltiavier —Stakes equations P
If d1 and
p
Dr = pp — Vp + Di
• v.) + V • (2p]) !
(21)
-
can he regarded as constants, (19) simplifies to Dv
+ PL+ p !)1 _ pf + V[- ` p
v7 +
+',
(22)
where [Exercise 3(b)] ^ EV V/s
^
ü{
2 [^ t •' i
ex.
Dx 2
Pxj
[^;^,
(23)
In the incompressible case, when V • v — 0, we obtain simply p
f
= pf
Vp +
^, ►
(24)
while (20) becomes T = —pl + 2pD.
(25)
It is worth commenting on the fact that our analysis [which follows Batchelor (1967)] has shown that the stress tensor does not depend on the rotation tensor W If a more general relation for the stress tensor were sought (for fluids exhibiting more complex behavior than water and air), then this lack of dependence of W could not be deduced. But W still does not enter the
Sec. 3.11 7he11avier StokesEquemiruu
165
constitutive relations that are traditionally postulated. This is because of the assumptive ui material indifference-- that constitutive equations are the same for all observers. Observers who are rotating with respect to one another would see different solid body rotations but the same deformation, so the assumption of material indifference will not permit dependence of T on U hut will permit dependence on D. Written down in a precise mathematical manner-, the observer invariance' requirement that underlies the material indifference assumption is used to restrict more complicated constitutive equations. See Jaunzcmis (1967) for an exposition of this type of analysis. "Sufficiently small" deformation rates were required for the validity of the linearity assumption (13), but it can be argued that this means merely that strain should he small over the very short time that is characteristic of molecular motion. Thus it is perhaps not surprising• but in any case it certainly is true that there is no reason to doubt the validity of the Navier Stokes equations for common fluids such as water and air except under the most extreme conditions.* aoUNüArcY CONDITIONS
Equation (24) differs from the in viscid hauler equation 02.1 .6), owing to the in the former of the term /W . Addition of this term has the important mathematical consequence of increasing from one to two the highest order of spatial derivatives present. 1f experience with ordinary differential equations is any guide, a new boundary condition is called for (beside the requirement that fluid does not penetrate the boundary[. This view is perhaps reinforced by a feeling that the adjacent viscous fluid adheres to the knife blade as it slices through the cold molasses, Surely, the shear stresses present in a real fluid must somehow restrict the free slip along a boundary allowed by inviscid flow theory. But might not the fluid slip over a very thin stagnant layer next to a fixed wall? Could it be that the tangential velocity is proportional, via a very small constant, to the normal velocity gradient? Maybe molasses does not slip past the knife, but what about a fluid like mercury in contact with a glass wall? (Mercury does not wet glass.) What about violently eddying turbulent flow? What about gases, particularly gases of low density? Such questions have been a subject of research for over 150 years. To name only prominent nineteenth-century scientists, the investigators included Coulomb, mb, Navier, Poisson, Poiseuille, Couette, Stokes, and Maxwell. presence
• It has, in Iaet. recently been shown that an appropr.atc approach to viscous equations through kinctx theory to subtler matter than had previously been supposed) flocs not yield the Neuter Stokes equations in t e strictly two-dimensional raise [l. Opperthcim and T Keyes, Phys. Rev. A7,1384 (1972j. and ë, 100 (1973)]. There are no such unexpected results in three dimensions, however: thus even normal two-dimensional hydrodynamies is "sa4ed." roc such calculations are meant to provide a first appru,imatwn 10 three-dimensional problems x bun venation in one direction is small.
86
Viscous Fluids
[CPI 3
It is now generally accepted that for liquids and gases under "ordinary" conditions it is appropriate to assume that fluid adheres to a rigid wall. To be precise, let x be any point on the boundary of a rigid body immersed in fluid, and let x be a point in the fluid. Then the velocity v(x) will be assumed to satisfy the adherence boundary condition .
]lrn w(x) = v(x w)
,
(26)
where v(x,,) is the known wail velocity. Evidence for acceptance of this boundary condition includes direct microscopic measurements of the extremely small velocity of particles near rigid walls. Unless they are interpreted with some subtlety, however, direct measurements tend to answer questions about the molecular model of fluid flow. What interests us here is not molecular questions, and certainly not the question (assuming it is meaningful) of whether fluid "really" adheres to the boundary. We wish to he assured that the adherence boundary condition is an assumption in accord with the rest of the assumptions involved in the continuum mode! of fluid flow. Such assurance comes indirectly but nonetheless convincingly from the ever-increasing number of instances wherein viscous flow theory, with [he adherence boundary condition, provides results in accord with experiment. Nonetheless, under "extreme" conditions the correct boundary condition remains in doubt.* IIVCOMPRESS1t3I,E VISCOUS FLOW
Except for a few remarks, in our subsequent discussion of viscous fluid flow we shall assume the fluid to be incompressible. (Indeed, we shall usually take the density to be uniform_) For an incompressible fluid of specific volume i. - '
Dn
Dt =
1 Dp Di =
so the term pD(p - ')/Dr does not appear in the fundamental thermodynamical equation (14.4.12) of L Thus the pressure never arises in thermodynamic considerations. Re-salts concerning the pressure which are traditionally deduced from considerations of equilibrium thermodynamics do not necessarily apply to the hydrodynamic pressure. For example, there is no a priori reason to assert that the hydrodynamic pressure in a moving viscous incompressible fluid can be interpreted as the mean normal stress on a stationary wall. Nevertheless, that this interpretation is, in fact, valid * For djsc;ussinn and further references, see Serrin 11959, p.240). An mterestatg account of earlier work can be found ita CïoIdslcin 11938, Vol. 2, pp. 676-80). An argument from kinetic theory that yicids t he no-slip boundary co=idil ion, plus further intalesling comment. is pm, on pp. 83-84 of K. E. Meyer, bur odurrion io Me)hema rim! Fluid Dynamics (New York : ury,1971).
Sert. 3. f ]
7 he Novier-Stokes Equations
87
can be deduced from the constitutive equation (25) and the adherence boundary condition (26) (Exercise 4). Consider a viscous fltaid of uniform density p. Let us divide the Navier-Stokes equation (24) by p and make the definition
v4_
(27)
p The resulting momentum equation, and the appropriate equation of mass conservation, are
Dv
1 f — p gp -
—
D1
V-V
vVY,
(28)
0-
(29)
These equations will form the basis of our discussion of viscous fluid flow. We can regard vas a constant, so we have four equations for the pressure and the three velocity components.` The dimensions of the viscosity p and the kinematic viscosity y are as
follows (Exercise 6): dimensions of p = mass/length-time -,
(30a)
dimensions of v = length L ftime.
(30b)
Typical values for p and y are (at 1 atm pressure and 20 °CJt
:
p (g/cm-s): glycerine = 9, water = 10 -2 , air = 2 x 10 v (cm 2 /s): 2 , air = 15 x 1(3 water = 1
a2
Note that the effect of viscosity is represented by the term proportional to v in (28) and that yis 15 times as large for air as it is for water. The qualitative variation of p and v with temperature is depicted in Figure 3.3 for water and air. (Since the density of water is very nearly l gJcm` over the entire range of temperatures considered, v 12 for water when cgs units • One might be icinplcd to pose as a fundamental mathematical problem: "Given initial values of p and r and given the morion or any solid bvuneur ies, solar US l and 129) subjetl to thr boundary condition ( 26)... A comparable "fundamental problem - of purr rnathrrnai cs would be "characterize the conseslucnt:es of the group axioms," fn both c u.c the incredible richrre55 of phenomena emt]raced by the cr{uatiuns or axioms precludes the solution of such pt claims. In both cases, taste and judgment must be used to select trilerestrng or wolthwhrle phenornrna and to formulate and coke problem, or to prove theorems that will elucidate such phrnom€na. The more "applied" an applied mathematician is, the more he will rely on relevance to a sr.:lei:Idle matter as a criterion for dcterminirt$ which phenomena arc inletcstuil;. t An essential ingredient of the physical "feel required to docffrciive applied mathematic-5 is u knowledge of the rough magnitudes of the various parameters Char appear in the ovations_
68
t'isruus Fluids 1C i. 3
,u on (tags LIMOS.
I
atmosphere pressure'
FIGURE kine»kitit
3 3.
Appro.—tit—tame rYlriutluN 1+•02 temperature rrf viscosity (0) und for air and w a t er- Data from C'v1rL►tNrn (1938}, pp, 5-1,
are used.) There is very little variation with pressure. See Goldstein (1938, Chap. 1) for further details and additional references. At the moment, the viscosity coefficients pc and v may seem to lack a clear connection with our intuitive feeling for viscosity. This connection will be more firmly established below when we indicate how these coeftïcierits can be measured by studying viscous flow in a pipe [compare (23)] or the passage of a small sphere through viscous fluid [compare (6.11)]. Further clarification comes via an exact solution (2.2a, b. c) to the Navier Stokes ouations, a shear flow in which the only nonzero velocity compnnent u is proportional to the nnrrnal direction j':
u = ky,
k
a constant.
In this case the sheaf- force per unit area T} , is proportional to the velocity gradient via the viscosity:
Historically, the viscosity coefficients first appeared when such a rather natural proportionality was assumed. The tensorial Navier-.Stokes assumption (25) came much later, EX ER CI5F:5
L. (a) Show that (14) implies that rt — 0 in (18). (C,,u and l rs -=-14 implies that C ars Ws — O. (b) Show that C# (c) Duce (19) from (13) and (IS), with x = 0_
Sec_
1.11 The !Warier-Stokes Equations
Kq
I Show that (20) implies that the principal axes of T and D coincide. Would you expect this to remain true if the linearity assumption were dropped? If the isotropy assumption were dropped? Back up your answers with physical reasoning. 3. (a) Verify (21). (b) Verify (22). 4. From (25) and the adherence boundary condition, deduce that the magnitude of the normal stress exerted by a moving viscous incompressible fluid on a stationary waIl is equal to the hydrodynamic pressure p_ 5, Let q i denote the ith component of the heat flux vector q. Give reasons for assuming the constitutive equation gf i cii 0 j , where 0 is the temperature and the quantities ri, may depend on thermodynamic variables. From an isotropy requirement deduce the Newton-Fourier law:
rcV[i.
q^
ti, Derive (30a) and (30b) from (2 5) and (27). .
The object of the next set of problems is to deduce certain facts about energy changes, particularly as these changes are affected by viscosity. A knowledge of Cartesian tensors is assumed. 7. From the momentum balance equation (A2.1.2) deduce the kinetic: energy equation (14.4.32) of I:
fff, ,„ d _ JJj'pr. t
R(i)
cit +
-vdd— JJJT:D ctr.
RIO
BO)
where (using the summation convention)
T:D=
f 1:1{ .
Section 14.4 of 1 contains interpretation of this equation and several deductions from it. [We assume that T,, _ TA ; R(r) is a material volume.] S. (a) Modify the equation of Exercise 7 in the case where the body force f is given by f = — VP and the function P is time-inde pendent_ P is called the potential energy per unit mass or specific potential energy. Give a justification for this terminology by interpreting the equation just derived. (b) For an incompressible viscous fluid, show that
T: D = 2nD : D.
(31)
The quantity = 2AiD : D is called the viscous dissipation (per unit volume) (compare Section 14.4 of I). Note that this dissipation is nonnegative.
Viscvi13 Rinds
ye
(['h. 3
$9. The viscous dissipation vanishes if and only if D , U. Show by integrating the resulting partial differentia! equations that D = U if and only if W loY lot certain constants cuj and Ir. T what motion does this velocity field correspond? is it reasonable that it is the only viscous incompressible motion with no dissipation of energy? 10. (a) Prove that
D(V • v) Dr
V-a =
(h)
l
D:D
2 1w l ^ ,
-
(32)
where st E DvfDt and to is the vorticity vector V A v. Use (12) to show that for a viscous incompressible fluid.
fJf
(I)
dr
ff a - n eta_
riwI l dr +
(33)
If the boundary dR is composed of rigid walls on which•v vanishes, deduce the following remarkable relation between the dissipation and the magnitude of the vorticity vector: J]JtI) dT = J.L Jf
(34)
= /21E1)1 2 . LWe have here E. It is not, in general, true that followed Serrin's (1959) presentation of results obtained by Hobyleff and Forsythe toward the end of the nineteenth century.]
N n -I
The next three exercises outline a derivation of the equations of twodimensional viscous flow that makes no explicit use of tensors_ 11. (a)
Verify that the following equations describe a two dimensional velocity field at any point y rear a fixed reference point X. -
2
vAy) — VAX) =
E{Y# - x k )
[^Li
+ higher order terms
2
;
E Ûr„ — xkP ,k t
2 + ^=
I
(}'k
-
-.1q)44fk
+...^
1 — 1, 2,
where 1c? 4^
D ^`
eV,
2 dx ; + r^x r ^'
I^;,
=
t —
^^k
2 ax,
^^i
^ c^x k
S'a'c". â_r J The Nai ler Stokes Equations
9 1.
In the equations above, rau,{Ox k , Dtk , and I^I; at-e Evaluated at the fixed reference point x. (b) Suppose that higher order terms are neglected. Show that if D a, tl, then the velocity field is that of a material that rigidly rotates with constant angular velocity while it simultaneously translates with velocity v(x). (Compare Exercises 9 and 4.1.5.) 12. The object of this exercise is to determine expressions for the components corresponding to the Die, in a primed coordinate system rotated by an angle t} with respect to an original unprimed system. so that
x a =x; cos[ --x r2 sin B, x; =x, cos (a)
0 t- x 2 sin 0,
x 2 =x', sin f1+.tie cos 0; xz— —x, sin t)+x 2 cos O.
If velocity components in the unprimed and primed systems are denoted by (t;,, r: 1 ) and (r;l, r/ 2 ), respectively, verify that
=
u' r.x i
cos 0
(X 1
(e l cos tl +
12 2
sin 0)
+ sin 0- •(--t sin 1 0+r•2 cos 0) 1 —^
J (D 1 1 +D1a)+- -(D rl — D22 } cos 2U+D 13 sin 20.
2
(h) Note that the result in (a) is identical in structure to the transformation law for T1 1 which was derived in Exercise 14.2.8 of I. Show that the identity in structure continues to hold for the transformation laws giving the components of D; 2 = D21 and D. (Section 1.3 reveals the basis for the coincidence" in the formulas.) 13. In an inviscid fluid Ti =_ -p ; - For a viscous fluid, additional terms are needed on the right side to represent a local stress corresponding to a local shearing motion. Exercise 11 shows that the WI, terms in the local description correspond to a rigid rotation, so it is sensible to assume that there is no shear contribution from these terms. As for the remaining terms, the D ; ,, the simplest assumption (presumably appropriate if derivates are small enough) is a linear one. Thus we assume that
7^ r = — p+A1D1r 7;2 —
T22 — —p
A 2 D 13
A ri D12 ,
(35a)
B 1 D I1 + 133D,2 f HiD11,
(35b)
+ C~2Dr2 -i- C73D22.
(35c)
This exercise outlines the consequences of the isotropy assumption, which maintains that in fluids with no preferred direction the relations (35a, b, c) should be the same whatever coordinate system is used.
Viscous Fluids [(_h.
92
3
(a) Using the results of Exercise 12, substitute for D', ,, ❑ ', 3, and D P22 in r12
= B1Di1 + B2 D i e + B 3 D '22
[8,, B2, B 3 as in (35b)]
and deduce that
( 36a)
{D1 s + D22)( 8 1 4 B3) = 0,
1(7'22
—
T11) = iB2(D22 — D11) + (B,
B3 )0 12=
(36b) T12 ' B2 D12 f I(D11 — 1)22)(B1 — B3).
(36c) (b) (c)
Combining (35) and (36), deduce several relations among the A's, B's, and C's. Repeat the above steps, starting with
fi l = - p +
41011 f A2 1712 + A 3 D'22 .
Duce that A2 — 81 = 83 = C2 = 0 ,
and that one can write C 1 = A3 = A,
8 2 =2 1,
C 3 —A2—Â
+ 2p.
(d) Assuming that  and p are constants, use the results just obtained concerning the dependence of the stress on velocity gradients to deduce the Navier-Stokes equations (22) 14. Consider the following steady flow (already investigated to some extent in Exercise 1114 of I). L'1
(a) (b)
=
2x 1 + 3x 2i
V2
=0
Find the rates of deformation and rotation. Determine the principal axes of the deformation tensor and ils diagonal form. Consider a circular disk centered around the origin, of radius 1What becomes of the shape of this disk after a time interval 0.1 unit? Plot the curve of the new boundary. (Assume that the time interval is sufficiently small so that one may calculate the displacements by Ei[ 1 = u, ❑f,
(c)
X1 — Xi,
Ax e = v 2 ❑I.
Consider points on the circle bounding the disk.) Describe the relationship between the results obtained in (a) and (b):
State clearly the deformation and the rotation involved. Verify explicitly for at least one boundary point.
Sec . . 3.21 Exact Solutions
93
15. In a certain Cartesian coordinate system, a two-dimensicjrial steady flow has the velocity components
v, = 2x t + 3x 2 , n2 = x, — x 2 . (a)
Find the components of the following three tensors: velocity gradient, defntmation, and rotation. (b) Consider a set of material points (x,, x 2 ) that at time r = 0 lie on the unit circle x; + x = I. Find the locus of these points at time r =F 0 <E L Proceed as follows. Let (x,. x 2 ) —, (t. ,, 3 .2) in the time interval (0, r:). (1) Justify the assumption ,
y, = x, + (2x, + 3x 2 )e,
y 2 = x, 4 (x1 — x 2 )c.
(ii)
Use perturbation theory to solve for x, and x, in terms of r, and y 2 . Neglect OW) terms. (iii) Impose x; + 4 = I. Neglect OW) terms. (c) Using elementary methods, one can find a new rotated coordinate system in which the result of (iii) becomes revealed as an ellipse in standard form. Write down this standard form Manna fulrcrtcitwurr using part (a). Explain your reasoning.
3.2 Exact Solutions We begin our study of t he Navier-Stokes equations for a viscous fluid of uniform density by displaying three exact solutions (obtained without approximating the equations). We shall briefly discuss the use of such solutions in determining the value of the viscosity p. Since no terms are discarded as "negligible," there is no question that exact solutions represent one possible behavior allowed by the equations. It is natural to be skeptical about whether this behavior is typical. In response to this skepticism, one can only say that experience indicates that exact solutions which can be interpreted as solving "reasonable" physical problems generally illustrate behavior that is broadly typical of other physical situations which cannot be exactly analyzed. SOLUTION t: PLANE COUETTE FLOW
Consider viscous incompressible fluid contained between infinite parallel plates a distance d apart. Suppose that the bottom plate is stationary and that the tip plate is moving with a uniform speed U. Solution of this problem should give us an idea of the effects of the adherence boundary condition and of how tangential viscous stresses drag the fluid along. We choose an x-v-z Cartesian coordinate system. The components of the velocity vector will be denoted by (u, Thus the situation is as depicted
Viscous Fluids ICh. 3
94
Y
=d
F
r
u=
^
4 I
Y
=o =D
t
f z
Fit; v H is 14. The configuration that generates plane Couertefi"ow of u viscous fluid. The fluid is contained between two infinite parallel plates. One of these is stationary and dhe other »ores with speed U.
in Figure 14. Seeking the simplest possible solution, we ignore the effect of body forces and set f 0_ (Gravity, the most important body force, only adds a hydrostatic pressure in any case; see Section 7.2.) Since the plates are infinite, it seems reasonable to look for a solution in which the only nonzero velocity component is u. It is also reasonable to guess that u is a function only of the vertical component y.. With the assumptions u = w = 0 = h(y), mass conservation is automatically satisfied. With f = O the t^ momentum equation (1.28) implies that
_P s -
p+ vhrr= 0,
pr = 0+
A:= 4.
(la, b, c)
From (lb) and (1c) it follows that the pressure can depend only on x and t. Since h is assumed to depend only on y, we deduce from (lb) that p x and her must be constants. [Our reasoning is the same as that used in separation of variables; compare the material following (15.2.3) of I.] We consider a flow due entirely to the effects of tangential viscous stresses and, therefore, look for a solution with pi, = O. Then h,,,, = O. From the adherence boundary conditions
h= 0aty=0,
h=U tit y= d,
we obtain the final exact solution u
Uy = d,
t7
=
= 0,
p — constant.
(2a, b, c)
The linear increase of horizontal speed is shown in Figure 3.5. As in the figure it is customary to position the graph between the bounding walls and to draw inafew velocity vectors to make the flow easier to visualize. Such a diagram is called a velocity profile.
Se a- . 12]
Fxorr S^Jurrons
95 d
L
=0
F t G LP R E 3. 5. The ratted ►• profile for plane (`ouerfe j1uH 1_ et uscomputethe tangential stress exerted by the moving flail on the lower plate, Reverting to subscripts for a moment. we obtain the desired stress, using (1.25) and (1.6),
cr r
T,, = D# 1
c_r 2
r
t•2 --
ex,
In our current notation. the above equation is written ^
CU['t ) -r Cy ex —
Since u = fl} d and r – C. we find that L
(2d)
The stress is directly proportional to i,' and inversely proportional to d. if we could move a plate parallel to another plate at constant speed, then we presumably could measure p b} finding the proportionalil constant. The plates would have to he large enough so that end and side effects would he neglib_ Note that the force on the lower plate increases with p.m conformity wit h our intuitive discussion of the relative fonces required to slice a knife through molasses and through water. SOLUTION 2: PLANE POISEUILLE FLAW, To see how viscosity impedes flow forced by pressure gradients. let us consider horizontal flow between motionless parallel plates at y = ±d forced by a constant horizontal pressure gradient pz
–
C,
C a constant.
(Positive C corresponds to pressure decreasing with increasing x. Accordingly, we expect the fluid to flow in the direction of increasing x when C > û.) (Ti + constant, all With the assumptions u = h( '^, t _ +^ = ü, p = equations are satisfied, provided that –
p -1 C +i'IJ j ^ = 0 or
C -t - Ph„
s ^-
Viscous Fluids ICÏt_ 3
96
Adherence requires that
h =U
at
d,
So
C t1 =
(i} 2 —
(3)
The parabolic velocity profile is depicted in Figure 3.6. ^-^
-
-d
3.6_ /he t rlrrt ir; 1►roifile fur plane Paistti.rilfr flow TTxe jirm• is driven by if { tmsruur pressure gradient that LE dlrrr-rt'd parallel fr, flhr hrteurdin0 fnre F tCï ll s L
pfrunt•.ti_
Of chief physical interest here is the urmRss flow rtde per unit width, i.e.. the mass of fluid per unit time, per unit distance in the 2-direction. that passes a given plane _r = constant. This is given by
f
/}!! d}' _ 1t Cr t d 3. -
(4)
d
The pressure can be measured_ Cnnsequently, comparison of a measured mass flow with (4) could in principle provide an alternative way of measuring viscosity. In fact, viscosity is measured with the aid of solutions similar to but not identical with, the ones that we have just derived_ The first solution we considered (Figure 3.5) is called plane Couette flow. (Ordinary) Couette now is the tangential flow of a viscous hind in the annulus between infinitely long concentric rotating cylinders (Figure 3.7). If the gap between the cylinders is small compared to the mean radius (R, t R2)/2, then plane Couette fl ow should be a good approximation to ordinary Couette flow [Exercise 1(b)]. But the latter is easier to approximate experimentally and can be described in its own right by an exact solution [Exercise 1(a)] _ Indeed, an important method of measuring viscosity is to suspend the inner cylinder from a wire whose twisting is proportional to the torque exerted by stresses transmitted by the viscous fluid located between the inner cylinder and a rotating outer cylinder.
Sec. 3.2] Exert Solutions
97
^-^-^
--
yR -^-Rr
FIGURE 3.7. In Cnuette flow the fluid is confined bemaeetw infinite coaxial
cylinders_ The outer (inner) cyirr<der rotates with angular
velocity n,(12, ).
The exact solution illustrated in Figure 3.6 is called plane Poiseuille flow_ A more useful solution is that for (ordinary) Poiseuille flow in which fluid flows along a circular pipe because of an axial pressure gradient (sec Figure 3.4 The solution is considered in Exercise 2. Instead of (4), the following formula relates mass flow rate through the pipe E with an imposed pressure gradient of magnitude D; E
—8
Dv -1R4.
(5)
A practical method of measuring viscosity consists of determining the pressure gradient D and mass efflux E and solving for y from (5). Plane Couette flow and plane Poiseuille flow show the qualitative features or ordinary Couette and Poiseuille flow_ The former have been studied here so that we would not need to introduce cylindrical conrdinates and thereby divert attention from the issues we wish to emphasize at this point. It might appear that we can sum up by noting that the viscosity p or the kinematic viscosity ' can be determined by measuring the mass efflux in
Fee taR E 3.8. Poiseuille flow in o circular pipe, like plane Poiseuille flow. is driven by, a constant pressure gradient. The parabolic uxwl r ek eity profile is depicted.
Viseurs Fluais (Ch. 3
9$
Poiseuille flow and that the Navier- Stokes equations themselves can be checked by using this viscnsiLy to compare measured and predicted torques in Couette flow. While broadly correct, such a summary conceals certain difficulties. I. Measurements require theories for their interpretation, For example, confident measurement of torque by examining torsion of a wire requires secure theoretical understanding of the relation between torque and torsion (see Section 5.3). As another example, to measure pressure one needs a device and a theory to explain its operation_ In practice, a degree of empiricism is usually necessary. This is well illustrated by the lengthy discussion in Goldstein (1938, pp. 2481t,) concerning a pressure-measuring device called the Pilot tube. 2. Given an understanding of measuring devices, there remains the problem of whether the exact solutions involved are appropriate. Cylinders cannot be made infinitely Icing, nor can they be made perfectly circular. Measuring devices disturb the flow whose properties are being subjected to measurement. Failure to reconcile theory with experiment could be due to the fact that inevitable small deviations from theoretical ideality may mean that the appropriate mathematical solutions are entirely different in character from the simple exact solutions. One ought to examine the stability of the exact solutions to small perturbations (see Section 152 of I and Exercise 4). To illustrate this point, we remark that Stokes derived the Poiseuille flow solution from his equations and found complete disagreement with the experiments available to him. He delayed publication of his solution. Now we know that the experiments involved the violently eddying turbulent flow which replaces Poiseuille flow at sufficiently high imposed pressure gradients. SOLUTION 3:
RAYLEIC:M IMPULSIVE FLOW
We turn La a problem that illustrates how viscous effects develop with time. Consider fluid in the domain y > O which for r < O is motionless and is bounded below by a motionless horizontal plate. Suppose that at time t = 0 the plate instantaneously begins to move, in its own plane, with speed U. Presumably, viscous effects will bring the entire body of fluid into motion. How does this happen'? 0 (sec Figure 3.9). As above, this We assume that u = of y, t), r' requires that p r = ps = 0, p. = constant. We take p. = 0 so that the motion is due entirely to viscous effects. The governing equations then require that —
Ur
= v2 I1.1r,
14
-0
_
f
> 13,
for t
O.
while, by assumption,
(6)
Sec_ 3.2] Exact So luerulis
99 ^
I/I1II/INI/IIfIIiI7II7l7ÎÏÏTT speed = 6 tor r <6 = Ulnrt]Q
lit Ruylc•igh 01IL,uLsitV f/un•. the /lurd iti lr+c•rrneel us rune stay 14 an 'infinite plane. The plane r'ssudrlrrll ti' set into motion purulle1 itself; with speed ['.
FIGURE 3.9
The adherence boundary condition implies that fort > Q: tl(U, rl
U.
(71
A person with considerable insight might realize that in this simple problem the speed U plays only a limited role. All that the solutions will tell us is how u/U passes from its initial value of zero to its ultimate value of unity, at a given spatial position and I -or a fluid of a given viscosity. We thus expect that E4
(8)
= ![t', r,
the point being that l:` does nor! appear on the right side of (8)_ As is frequently the case, mathematics can be used to verity physical insight. With the change of variable it' = Nit/. (61 and (7) become = uu},
LA (}, f) — 1
> U).
(9E41:1
Since these equations do not involve U, re cannot depend on I1, from which (8) follows. The ratio uXU is dimensionless, It must, therefore, depend only on a dimensionless combination of y, t, and v. Since the dimensions of r are length/time, y, t, and w can only occur in the combination y 2 1 - Vv. More precisely, u/U = g(y 2 fvt) for some fu nct ion g. It is cnnvenient to write this last equation in the equivalent form:
u'
-
Lr
w hc°re7 = 2 }.`{vt}
,tz
(10)
for then the final answer will be expressible in terms of a certain tabulated function.
too
Visenu s Raids
tCh. 3
To test the conclusion (10), obtained by physical reasoning,* we substitute into the governing equation (9). By the chain Hile, (9a) implies that —
iv - 1 12t - 3f3 3y' _ v . iv - ] - l^n
or
F" -# 2r1F' = 0. For any fixed t, y --• 0 implies that ri - O. sc (9b) implies that (12)
F(0) = 1.
Note that for fixed y, rl cx:, so (12) requires that at any fixed location Oas r as r of if1 : -• I -• oo. This is a reasonable result_ Equation (11) is of second order, so there must be a boundary condition in addition to (12). We require that at any fixed time the velocity far from the plate approaches zero: Hat u(y, $) ---- 0,
t >0.
(13)
Trouble threatens, however, when further thought indicates that it is just as reasonable to require that at any fixed position the velocity approaches zero as r ,l, 0 (i.e., t > O, r - 0): lini u(y, r
s) -
0,
y > 0.
(14)
,o
It i s reassuring that both (13) and (14) amount to the same condition on yy r namely, } F('(),
11m F(/) = U
(15)
0,2
However arrived
at, the final problem, given by (11), (12), and (15), is a simple one. Regarded as an equation for F', (1 1) is first order. Its solution is
F' = C I exp ( - r7 2 ) T h us
F(rJ) =C, f e ^d^+
C2.
(16)
• Scc Scc`iiun 3,4 and Exercises 3.4.$-tO for other material concerning - similarity' solutions tu partial differentiae equations in which dependent .iambics appear only in certain combination t In [Eats and papers it is customary to vrgaruxe the subject matter its a "straight line" manner in which the mathematical problem is completely formutticd and then is solved. From time to time, however. it is useful to remind [he reader that creative scientific thinking mitre often than nul results in an interaction between the formulation of a problem and its solution-
See. 3.21 Exact Solutions
aoI
t'trrl - err, I
FIGURE 3. W. in tQuyfer`gh in,pulswr flaw, u! (i = Ftg), where ij = bl vt) "' A graph of F is presented in thF,r figurer.
Since the Leal value of Pi is of special interest, we use zero as the lower limit of the indefinite integral in (161. Because of this, (12) simply requires that C2 = 1, whence (IS) requires that
C 1 f e_
:
d + l = 0.
Thedefinit e integral in (17) hay the val tic ec
F(er),
where F (r)
( I 7)
[See Exercise 3.2(d)of 11. Taus = I
—
e
R
In 0
d.
(IS)
The right side of (18) happens to be a tabulated function, called the complementary error function and denoted by erf L (ri). The error /unction itself, erf (rr), is just the integral term in (18), so F(rt) =
err, 01) — l — erf (t,).
(19)
We can ascertain the main features of our solution without resorting 10 tables. Merely from the fact that the solution is a function of q 1}4,11) -- " 1 see that the place where n f U — constant is at a certain value of v(v r) - t, 2.*we Thus the portion of fluid that moves with less than a given per cent of its final velocity U is farther from the plate than a certain plane whose height is proportional to (vt')" Since exp ( x) decreases rapidly with increasing 4', the graph of F must have the qualitative features of Figure 3,10. From a table (ear by numerical —
- constant has a unique * Physical intuition would kid one to believe that the equation sululion, this k confirmed by IN, which shows that Fl(s) is continuous and monotonically
decreases from I to 0 asp/ goes horn 0
Le r[,.
:
o2
Viscous Fluids (Ch. 3
integration) we ascertain that F(17) = 0_005 when r1 _ 2. As a consequence, when 71 > 2 or y > 4(v0" the fluid speed has not reached even half of l per cent of its final value. Since erf, (1) = 0.16, when y > 2(%s)u 2 the fluid speed does not exceed 16 per cent of its final value_ From the highly simplified situation typical of an exact solution, we wish only to draw the most general type of conclusion. Such a conclusion in the present case is that the purely diffusive cruse of setting fluid into motion tykes effect in r region whose thickness is of order (vl)'". This spreading at a rate proportional to r''' is the single most important characteristic of diffusion. (Compare the remarks in 1 at the end of Section 4.1 .)
EXERCISES
In the following exercises, u, u, and w denote the velocity components when Cartesian coordinates x, y, and z are used When cylindrical coordinates r, O. and z are employed (sec Appendix 3.1 for relevant equations), the corresponding velocity components are denoted by Or ) , r.{e) , and w_ 1. (a) Solve for the Couette flow of Figure 3.7 by looking for a solution of the form Gtr} _ w Ü, ut'M = V(r), p = [Sec Exercise 4(b) P(r). for the answer.] (h) Find the stress 7. e exerted by the fluid on the inner cylinder_ Show that if the gap is small compared to the mean radius, then the result is dose to that of (2d). Explain why. 2. Solve for the Poiseuill e flow of Figure 3.8 by looking for a solution of D. Derive (5). the form i( = t = Q, w = f (r), p = = in as much detail as possible the similarities and differences 3. Discuss between (4) and (5). 4. In this problem we outline the beginnings of an investigation into the stability of Couette flow. Knowledge of stability theory, as in Section 15.2 of I, is assumed. (a) Show that the Couette flow of Exercise 1(a) satisfies the inviscid equatiionsof motion, so that viscosity enters only by means of the adherence boundary condition. In a first attempt to find out something about its stability, we shall regard Couette flow as inviscid. (b) Examine the stability of the following exact solution (Couette flow) of the inviscid equations: V(r) = Ar + Er - ^
p = P(r) = i{2 rT L ,
!] ( ri = w =
(2 ntfi A = 2R1 R1— R^ T
0;
8^
(Dt
112)RiM
T Ri`Ri
.
Sec.
.I21 Exarl Solutions
103
Assume disturbances that writing
are
c;[r 1 X, z, l4'
=
1 ,^ '^'t r,
axisymmetric (independent of 0),
Zt^^ _
I, f),
/(r) + L` B1 (r, z, 1),
P = P(r) + P(r,
Z+
I).
Linearize. Write ^,{r}^itoi+
i , z,} t Crl(r
As)
and make analogous assumptions For the forms of 0 19, iv, and p. Eliminate ° ',ti3, and p. To simplify notation, write WI X. Deduce -
dr
(1 ;
a
dr
rX) + 1, 2 (a
— 1)X
0,
X(R I ) — X(R 1 ) —
where
(I3(r) =
r
d
rlr
(r
(c) Note the similarity between this equation and (152.21) in 1. Exploit this similarity and draw conclusions such as Deductions I and 2 of 1, Section 15.2. (The required theorems are still valid, with substituted for dR jdz.) The mathematical similarity evidenced in our investigations of the stability of Couette now and of a stratified fluid is due to the physical similarity of the destabilizing effects of gravitational force and centrifugal force_ 5. In deriving the flow of Figure 3.5,we set px = O. Keep the sameboundary conditions, but now set p = — C, Ca constant. Show that the resulting fl ow is a superposition of the flows of Figures 3.5 and 3.6_ Sketch a typical velocity profile. 6. (a) Verify (11). (b) If u/U — y(y V /vt), find the equation satisfied by q. Recover (I1) by a change of variable. (c) Show that the assumption of (10) would be invalid if (9a) were 24; = vurrs 7. Show that 1„` exp (— e) 8. Instead of (16), take
do
is a convergent improper integral.
F(t7) — CI J ed4 +
C2.
Find Ci and C2 (a) when c = aç (b) for arbitrary finite c_ Verify that the solution is the same as (18). 9. Put a filled teacup on the center of a record turntable and suddenly turn the turntable on to 33 revolutions per minute. About how long does ,
L04
Visci us Fluids ICh 3
it take before the fluid is in solid body motion? Show that the estimate obtained from the diffusion distance (4 02 is much too large. This discrepancy has been explained only recently [see l -1. Greenspan's monograph The Theory taf Rotatin g Fluids New York : Cambridge University Press, 1968)]. 10. if the variable Li is regarded as corresponding to temperature, what problem in heat flow is described by exactly the sanie equations as those appropriate to the Rayleigh impulsive flow? Discuss the physical plausibility of the solution in the context of heat flow. +11. A viscous incompressible fluid fills the region z > O. It is bounded below by an oscillating plane, so :r{x, y, U e) = U exp (iw1),
U and co constants.
.
Find the velocity field. (Take p = constant.) 12. Advantages accrue in certain conditions if portions of an airplane wing are madeof porous material and a small amount of air is continually sucked into the wing. This relatively simple problem gives a first idea of the effect of suction. Consider two - dimensional flow of a viscous incompressible fluid in the half-space z > 0. Let the x-wise velocity component is approach a constant U as z # oc, . Suppose that at the porous plane z — 0 there is a perpendicular suction with speed W, so that at z : 0, w = — W where IV is a positive constant. Find a solution of the form u = u(z),
w = ff ,
u = 0,
p = constant.
14. Consider viscous steady incompressible flow
u = t? = i)
,
iv — f (x,
y)
in a pipe of uniform cross section whose generators are parallel to the z-axis. (a) Show that the axial pressure gradient satisfies ;1_p = — l,
constant, rdr so that the problem reduces to that of finding a solution of the equation ) 0 -? ♦
p
which vanishes on the curve in which the pipe cuts the x, y-plane. (b) The lines — 1 =0,
x+2--y,/3=4,
x+ 2+y
=0
bound an equilateral triangle whose sides have length 20 and whose centroid is the origin. For the pipe whose cross section is
Set_ 3.3] Orr Bnundur}' Lr^}r^rs
1 05
this triangle, show that f (x, y) is a constant multiple of (x — 1)x+2—y )(x+2+
yam)
and find the constant. $(c) Find the tangential force per unit length on the triangular pipe. NOTE. The mathematical problem that arises in viscous flaw down a pipe is identical to a problem that emerges in the study of the torsion of an elastic beam (see Section 5.2)_
3.3 On Boundary Layers* Books on science are full of plausible but noncompelling assumptions that lead to theories whose predictions are eventually confirmed with stunning accuracy. It is refreshing and instructive to contemplate instances where plausible reasoning led to trouble. One such instance conciûrrts the extensive nineteenth century study of the flow of air past a solid body_ There "the trouble had been by no means the lack of a theory, but rather the existence of an almost overwhelmingly large body of theory, constructed by many of the best mathematical physicists of the nineteenth century, according to the most respectable physical principles. This theory gave, for the motion of a wide variety of shapes through the atmosphere, which is treated as `a perfect the fullest information, none of which accorded with the most Rued' elementary observation of the facts. The failure of the theory was particularly disturbing because the only known mechanical properties of the air which had been neglected, namely compressibility and viscosity, could reasonably be supposed to produce small effects in the type of problem considered."t This section is concerned with the steady flow of fluid past obstacles. ft can be shown (as in the work of Lighthill just cited) that compressibility can be neglected in such flows provided that typical speeds are small compared to the speed of sound, that there are no large, imposed temperature differences, and that there are no motions of atmospheric dimensions. It is the aim of this section to explore the subtle effect of small viscosity when such conditions are satisfied. -
COMPARATIVE MAGNITUDES OF VISCOUS AND INVISCID TERMS
Suppose that you run at a speed of 15 kilometers per hour directly into a wind blowing at a speed of 25 km/h. If viscosity is neglected, theory predicts an absence of wind resistance (D'Alembert's paradox). Apparently, viscosity should be retained in equations that describe the air flow. But let us estimate * Appreciation of this Section will be greatly enhanced by ptior study of the maierial in I on sealing (Section 6.3) and on singular perturhalions (Chapter 9). t Quotation from M_ J. Light hill's introductory chapter in Rnsenhead (1963)_
[Ch_ 3
Viscous
106
the relative size of the inviscid and viscous terms. We assume that the wind blows along the x axis. Consider the Navier-Stokes equations for viscous flow ( 1.25), now written with an asterisk to emphasize that the variables are dimensional. With ajar* = 0 (steady flow), these equations are -
du* ^r +^ 0x*ôy*
1 Op* Mu* alai. p c7x* + 1 4-012 + ^Ÿ * ou•d * tr• r7f t^* ^p* 020 ^ 4 1R- e^y +u { # t - : #[^y* OX* [ lx Y e(y*) 2
c7u* (]D* ^^* + a^ ^ *—°'
,
( la. b, c)
A typical inviscid term is u* ate*/dx* and a typical viscous term is v 132 14*/21x*) 2 . For flow disturbed by the runner, it would appear than u* will be of the magnitude U0 (25 + 15) km/h or U0 x cm's. Significant changes in u* would be expected to take place over a distance of the order of some representative body size such as the width of a chest or a leg; say, over a distance of L — 10 cm if so, we would expect that
u
#
e?u* a2u* v *)Z ex* r3lx*)^
UO2 1.- -1 ,,U0 L
-x
Ut, f.
where
U0 L v
^
(4 x 104 crrt/s)(1#1 an) 15 x lU - ^ crn 2 js
10^.
(2)
We have used the value of v for air given at the end of Section 3.1. According to our estimates, the inviscid terms are 100,000 times as large as the viscous. The paradox seems sharper: Neglect of the viscous terms leads to the physically unreasonable result that the runner feels no resistance from the wind, but the viscous term truly seems negligible. Ludwig Yrandtl,at the begin ning of this century, had the insight required to resolve the paradox and point the way toward a procedure for calculating the effect of viscosity in many flows_ The essence of his ideas lies in the recognition that there is an extremely thin region, close to a solid body, where the fluid velocity undergoes great changes. Just outside this boundary layer the fluid "slips" along as predicted by inviscid theory. The velocity changes rapidly as the boundary layer is traversed, and becomes zero on the boundary as required by the boundary conditions for a viscous fluid_ The above estimates are not valid in the thin zone of rapid change, where the viscous terms are important no matter how small the viscosity.
-ei-Nr-
3.31
Chi
REYNOLDS
ro^
l'.r1Ye•fs
rw UMBER
Further progress is facilitated by the use of scaling. To permit some appreciation of this process for those who arc not familiar with it.? we summarize its main elements- The object is to introduce dimensionless variables in such a way that the maximum order of magnitude of the various terms is correctly estimated by the parameters that precede them. Roughly speaking, to do this one must nondimensionalize the dependent variables with constants estimating their maximum values. The dimensional independent variables x* and ys must be nondtntensionalized with lengths that are estimates of distances in which one observes significant changes in the dependent variables. The constants or sr ales that are appropriate in this nondimensionalization may differ from region to region_ The dimensionless group that we computed in (2) is an example of the Reynolds number, Re. The concept of scales is basic to the dehnitiuri of Re: Re — (overall velocity scale)(overall length scale) kinematic viscosity
LW_ t•
(3)
The word "overall" is used in (3) because there can he more than one scale of variation in viscous flow past a body. Thus the Reynolds number is an estimate of the ratio between the convective acceleration terms§ such as rr*clu*/(x* and viscous terms such as y r '- re*fi(x*)' except ni possible - special" regions of "unusual" variation. As we shall see, when the Reynolds n u miter is an order of magnitude or more greater than unity. it ceases to provide meaningful estimates near the solid boundaries, Prandtl recognized that the f'Alernbert prediction "no drag in nonviscous flows" is only slightly wrong in some slightly viscous flows. For flow parr streamlined bodies one can say that there is "slight drag in slightly viscous bows." The reason is that for such flows the itwi cc`td float' panern But when an obstacle is is only altered in the thin boundary layer region. bluff(i.e., blunt), the presence of viscosity in minute amounts alters the entire flow pattern. BOUNDARY LAYER EQUATIONS FOR STEADY FLAM: PAST A FLAT PLATE
We shall return later to some of these general ideas- Now we shall discuss in some detail the effects of a small amount of viscosity on a simple but representative "streamline" flow_ To this end, let us consider the case of a flat plate immersed in some steady two-dimensional flow_ Denote the (dimensional) velocity components by rc*(x*, y*), u*[x*, y*) and the pressure
by p*(x*, y*)- l'he flow could be disturbed by a number of objects, but we di,eussed in Section .63 or Rcrnernber ihat this discussion mals with steady flows_ The term PORN' may provide the
t Scaling is
dominant acccleration firms where the
vclocity at a pnirti varies with lime.
to8
Viscous Fluids (Ch. 3
shall fix attention on the fiat plate, which we imagine to be positioned along the x*-axis. The y*-axis is normal to the plate. We shall assume that the viscous flow does not differ significantly from the inviscid, except near the objects (Figure 3.11). We have already denoted the overall length and velocity scales by L and Zia, respectively. In the mainstream, then, a typical convective acceleration term such as u* Ott*fOx* should be of magnitude U /L. We expect a pressure gradient term like -- p - t 'p*/cox* to be the same magnitude, so we select pUt as a pressure scab:_ Thus the following are appropriate scaled dimensionless variables in the rnainstreamt :
X=
x*
y*
:a*
L. Y=U = U
Y=
v*
P=
o
u
p* P
^, (4)
With (4), the equations of steady viscous motion (1) are
f'L r x
-
UVx +
VU y _ —P x + Re t (Lf xx + Urr), VVy = `Pr ± Re - '
U x +. Vie = 0
(Vxx
( 5a)
4- Vier),
( 5b) ( 5 c)
-
Let U(X, Y), V(X, Y), and P(X, Y) denote the solution to the inviscid problem. These functions satisfy the above equations when the viscous terms are entirely ignored (formally, Re oo). Also satisfied are the boundary conditions of vanishing normal velocity. In particular. V vanishes on the plate, so from (5a) with Re oc we obtain the following result, to be used below: U(X, O)11x(X, 0) -- . F(X , O)
( 5 d)
-
-
Let us reiterate in present terms a point that we have already made. We are considering flows where Re may be 10 5 , 106 , or even larger. Nevertheless, the terms in (5a, b) preceded by Re r cannot be entirely disregarded. The f Here is more detail on the scaling isce also Section 7.3). Lei us use (4) Ow keep the pressure scale. P o unspecified for the moment, so lhal p p'/ Po . The Navicr--Stokes equation (la) becomes —
Uâ t U U^ r'll _ -— L U d } i. v Y
P4
—
op
— pL — ÔX
+ viscous terms.
Since Ua , P. and L are scales, by the definition of - scale" the approximate magnitudes of the three terms that appear explicitly in the preceding equation art ti /L, Uall. and PdpL. respectively. Away from obstacles, when Re > 1 we expect an essentially inviscid fl ow with all nonviscous terms in rough balance. This requires that U6IL = P01pL or Po pUo, determining the pressure scale P. NOTE, From the viewpaini of kinetic theory, pressure can be regarded as molecular momentum transfer per unit area. The pressure is of magnitude pti because momentum per unit volume of magnitude MI D is typically transported by bulk convection at a velocity of magnitude [1, D. ,
-
—
e _ 331 On Sc
Boundary
Lavers
109 Y.
l C[l[7i1:1rlt
& r15 a 3f
rirm5LIr[ =
P.
ja'
-
F ai tJ R E 3.11. A fiat
plate af' length 1.
and rrrher objet r.► émnreraiil tit a r-i-tc'rrWs
fluid.
reason is that the change of variables (4) introduces a scaling which is tint valid near the obstacles placed in the flow. Only away from these obstacles does this change of variables ensure (for example) that UU x and U }- r are approximately of unit magnitude, so that the ratio of Re -- 1 U ri, to UUx is the negligible quantity Re - I. We now turn our attention to the selection of scales that are appropriate within the boundary layer region. We expect the tangential velocity component ce*, the component parallel to the plate, to vary from its inviscid value outside the boundary layer to zero on the plate. Consequently, the maximum value of u* should be of magnitude Li e , the scale of the inviscid component. Variations along the plate have a scale that should be about the same as the scale L of inviscid variation, but the relatively rapid variations normal to the plate will have some scale e5 that is smalt compared to L. What about the normal velocity component v*'P We know that u* vanishes on the boundary. We shall estimate it at a representative distance 6 into the layer. We do this with the aid of the continuity equation ôu*/fix* + (70/4* 0, which shows that the normal derivative 4 t,*ley* has the same magnitude as 490/49x*, namely. U 0/L. Consequently, using Taylor's formula, We find that u*(x*, y*) x t.*fx*, 6)
v*(x*, O) + o * (x*, 0) Y
0
+
-
L
.
(6)
For a moment, let us take the pressure scale to be the unspecified constant Pl. Using the various scales to nondimensionalize, we write
v-
Z1 *
Lf^'
v=
^? *
u*
I.J 0bJL,
p=
Pi
, x r
x• L
,Y =
Y* ^
(7)
L_
110
Vr xc-r)uS Fluids [CA, 3
With (7) the steady Navier-Stokes equations (1) hecorne U,2,
eu
L
{fix
- 14 [
[UP] Ût + ^X
[
+
eu
fil L t' ; b ^-^
P,
ey
Fr ^1^
_
_
L
C^`}^
tip
rvU0
ax +
[
021,11
^ex2
vU o r' 2 ir + ^ 62 ^
oy2, (sa)
Op
(^y +
vU0B ^x v [vt.I.1P2 L3 Dx 2 + 31. (11, Y 2 (8 h)
U 01 iU u U eü a Q #. ['x 4 L [^y
(8c)
The object of our scaling was to choose variables so that the squarebracketed parameters in the preceding equations give the magnitudes of the terms of which they are a factor. 'These magnitudes must he such that the viscous terms are retained in the boundary layer- As we now show, this requirement allows us to determine b an d P 1 . Since ES is small, the largest viscous term in (8a) is proportional to a ufdy 2. This term will be retained, however small v is, only if b is such that the magnitude of this term is the same as the magnitude of the convective acceleration terms. Consequently, we require Uâ/f. = vU0/b2, SO
b = vrf
or
U0
b = Re - " 2 L.
(9 )
Knowing that 5 is proportional to v", we see from (8b) that i p/ra y is proportional to v_ Consequently, the change in p across the boundary layer of thickness b is of order V312 . TO first approximation, certainly, the pressure does not Mary ar~ross the boundary layer. This is an important qualitative conclusion of our analysis. In particular, it means that the pressure scale can he taken to be the same as the inviscid pressure scale: P, = p1426.
(10)
Using the expressions for the boundary layer thickness and the pressure scale given in (9) and (10), we can rewrite the governing equations (8) in the form + rlti y =
-4 rr,, y + Re - 1 m kt, -
pr = Re -1 (^ urn„ — LL J + t ») f- Re -2 uxx,
u + !Jy ^ 0. „
(11a, b, c)
We have assumed that Re > 1 and have introduced certain scaled dimensionless variables. We did this in such a way that in the boundary layer the relative magnitudes of the various terms in (11) should be approximately given by the Reynolds number factors that appear. [Compare the different dimensionless version (5) of the equations, the latter version being scaled
Sec. 33] On Bounrkorl Layer s for the mainstream.] When Re f, then we deduce from ( I I l that to a first approximation the following equations hold in the boundary Iayer:
tux + illy = - P x + u }y.,
py
tl,
u x -1- U}, = (i
(12a. h, e)
Equation (12h) states the previously noted fact that the pressure does not vary across the boundary layer. in (12a) we thus may replace p x(x, y) by its value just outside the boundary layer region. `phis is sometimes determined by measurement and is sometimes calculated by inviscid flow theory_ In the latter case, to first approximation we may use the inviscid equations evaluated right on the plate (rather than a thin boundary layer width away)_ Employing (5d), then, we see that the equations of (t 2) reduce to uu, +r)u
=D£1' +U) ,
fi(x, 0)V x (x, 0)
u+z'. = U, assumed known.
(131
The final steady boundary layer equations (13) have been derived for the flat plate. Nevertheless, it can be shown that they a re essentially valid for boundary layers near arbitrary two dimensional bodies, provided that x and 1' are interpreted as orthogonal coordinates measured along and perpendicularly to the wall, respectively (Rosenheati, 1963, pp. 201f1.). The only restriction is that wall curvature should not be so great that the centrifugal effect of the resulting curved flow requires a nonnegligible normal pressure gradient to balance it (That is, the wall's curvature radius must be large compared to the boundary layer thickness.) We shall restrict ourselves to flow past flat plates, but the reader should keep in mind that most of oui discussion applies much more widely. -
B c, N DAKI' CONDITIONS
We turn now to the question of boundary conditions. Since the most obvious defect of inviscid theory is that it permits fluid to slip along solid boundaries, we surely will want to require that both velocity components vanish at y — O. Next we must somehow link the boundary layer solution with the mainstream_ A way to accomplish this is called patching_ This method is used for a particular flow problem at a particular large Reynolds number. One requires that the boundary layer velocity component u(x, y) join with the corresponding exterior component U(x, Y) at some definite distance from the boundary y = O. Suppose for definiteness that Re 10 6 . Calculations show that the various possible solutions rl(x, y) cease varying appreciably as y increases beyond y = 5 or y = 10, say. Since y!Re" = Y, y = 10 corresponds to Y — 0.01. One can then patch together an approximate solution by using boundary layer flow for Y S 0.01 and inviscid flow for Y 0.01, where the boundary layer solution is chosen to satisfy u(x, 10) — 11'(x, 0.01) (see Figure 112). Since (typically) results do not differ much if patching is applied
112
Viscous Hair's
[Ch 3
Irtwiuid now: CFO. Yl, film, Y}
Patch here {i=re 1-T -
-
-
liaund ar y rayer Ilow ulr. kl. ntr.}'r
TfT TT^ ^ ^TT T^ ^TTTTTT F I U u R +:
3.12. Pinching toyerlrer lxurndurr• laver and !nr •isrid flow.
at y = 20 and Y 0.02, one has confidence that the general nature of one's results is independent of the somewhat arbitrary nature of the manner in which the boundary layer flow is linked to the mainstream_ We shall adopt a somewhat deeper philosophy of linkage called matching.* The idea here is to produce results that are correct in the limit as Re ar . One must drop the notion that the linkage is to be enforced at some tangible value of a normal coordinate. The boundary layer scale ô is proportional to Re - ' r2 , so certainly as Re all the boundary layer variation has occurred far closer to the wall than any fixed (i.e., Re-independent) positive value of the mainstream coordinate Y. On the other hand, any fixed value of the boundary layer coordinate y is at a definite position within the boundary layer and cannot be an entirely suitable place to Link with the mainstream. Therefore, we assert that the inviscid component LF(X Y) at Y = 0 should he equated with the boundary layer component ar(x, y) at y — oz;. From what we have just written. what eke could be appropriate?t The numerical results will not differ much from those obtained in the patching approach, but the matching approach is more readily capable of proof and extension. The previous paragraph still has the spirit of seeking a !oration For matching Better understanding of the boundary condition emerges if one frees oneself as much as possible from the patching frame of mind and seeks an overlap region where boundary layer and inviscid flow merge_ This region should be composed of values that, as Re ; a;, approach the wall, on the one hand, and the far extremities of the boundary layer, on the other- Such a set is given in terms of a function e(Re) by ,
^r
rr
i.e., y
— Re -
we.
(l4)
• We proceed in a manner that is else 311 ial ly self-cnnlained, but the reader who has studied Elementary singular perturbation theory, as in Chapter 9 of t will have a much more Secure understanding of the issLcs inwolrcdt One care also argue that ilk matching of {, at Y = O with a al y = cc Ô the most natural mathematical formulation of the fact that as Re ne, any tixcd 4alus oft lc free-stream variable Y. however small. corresponds to larger and larger 4alues of the boundary layer variable y. ,
Sec. 131 Ott Boundary Layers
113
provided that lim Re -I12 19(Re)
1im Q(Re) = Re
Re +^
^^
Using (14), we substitute for Y and y in terms of
O.
(15a, b)
the intermediate variable ri
and write our matching requirement as dim U [ x, Its-•oc \
JJJ
= dim Re-t
^ r;
i1(x,
R
((6) s)
We stress that ri is held constant in the above limits. Here we work only to the lowest approximation wherein the mainstream equations (5) and the boundary layer equations (11) are simplified by taking them in the limit Re x. The limiting equations contain no trace of Re, so (16) reduces to U(x, 0) = u(x, CO, the requirement previously arrived at. The notion of an overlap region gives this requirement added plausibility. More important, it provides the foundation of an approach that is more easily extended when higher approximations are desired. We have derived matching conditions for the horizontal velocity component 14, but what about the vertical component u? The structure of the equations (13) indicates that u should not he specified as y x : second order y-derivatives appear on u but only first derivatives on a. Thus two boundary conditions are anticipated on u (at y = O and y cc) but only one on u (the vanishing of y at y = 0, which is certainly required by the physical problem). Turning to the streamwise coordinate x, similar considerations indicate that we should require one condition on u and none on u, for (13) contains terms proportional to u„ but u is never differentiated with respect to x. Consequently, if we prescribe the entering horizontal velocity u(x, y) at a certain location (say x = 0), then we expect to be able to compute the flow downstream of that location. Summarizing, we find that the boundary layer problem for flow above* a flat plate of length I. has the following mathematical formulation, to first approximation: ux + vy = 0,
1411, + Vtl y ` U U ' ury ,
O < y< cc, O<x< L. u(x,0) =u(x,0)=0, u(0, y) = uo(y),
0<x< L,
y > O.
lln7 u(x, y) ; U(x, 0), 0< x < L . U(x, 0) and u o(y) are prescribed functions.
'
Flow below is determined by symmetry.
Vis-cous Ftyids
11 4
(C }I. 3
Formulation of the boundary layer problem is hardly a straightforward mailer, and the critical reader may well fed uneasy at this point. Can one trust "physical intuition" when one has no genuine experience of the flow inside the slivery thin boundary layer region? Awareness that a mathematical model is never identical with a physical situation is all the sharper when, as here, rather bold approximations are being attempted. Only further study of partial differential equations and of fluid mechanics can build lull confidence in boundary layer theory_ Nevertheless, the reader may find some reassurance in the discussion of Appendix 3.2, where we show that there appears to be just enough information in (17) to (201 to produce a unique numerical solution_ Moreover, it may be reassuring that a simplified "caricature" of the boundary layer problem seems to have a unique solution (Exercise 5). since most of the The "thickness .' of the boundary layer is of order Re variation across the layer takes place in a multiple or two of the normal length scale = Re - " L. For the runner mentioned at the beginning of this section, L = 10 cm, Re = l0', so the boundary layer thickness is of magnitude 113 mm, It is noteworthy that although the boundary layer is exceedingly thin, yet (as we shall see more clearly below) developments within it have vital and even dominant roles to play in the formation of the entire flow pattern. ' +3 ,
EXERCISES I. Estimate the overall Reynolds number: (a) For a fast-moving automobile. (b) For a hard-thrown baseball or hard-kicked soccer hall. (c) For a large jet airliner at cruising speed, (d) For a dust speck falling in air. le) For the motion of a whiplike cilium that propels a paramecium. (The cilia beat about 30 limes per second with an amplitude that is a sizable fraction of their length of about 10 pm = 10 - 3 cm.) 2. Estimate the boundary layer thickness in cases(a),(b), and (c) in Exercise 1. }3. la) A definite displacenient thickness 6 1 is sometimes used to characterize the extent of the boundary layer. By definition
45,(x)r FI _—^ j
1') dy. a(x LI( X , 01 1
Justify this definition. (b) A momei tum thickness is also useful on occasion. Give an appropriate definition. Explain your answer_ 4. This exercise is concerned with a change of variables, due to R. von Mises, that transforms the boundary layer equations (17) into a single partial differential equation that is well adapted for further analytical
Sri- 3 41 Boundary L
as
i
15
and theoretical work. For further information sec, For example, Rosenhead (1963. p. 212), or the paper by W. Waiter cited at the end of Appendix 3.2. Introduce new independent variables { , i) by = s(x, y) = x,
y)
n=
u x, s) rlw.
(2 I )
Let a new dependent variable w be defined by
ww^b, rï) = u 2[xt , q) f^f4 q)], .
.
where .va. qj = and y(. in are the inverse relations found from [2])_ Show that the problem (17) to (201 transforms into the following: iw d W
[' 'W
1
r
n , UI = 0, 410, ,I)
lim
41L1,
D < < L. f} < rf < r .
4)1.
ql
0< < L
-
Hire (t) — Cî`lt, 0). 5. Confidence that a mathematical problem is well posed can be obtained by showing that a simplified version of the problem has a unique solution. ln this spirit. sketch a solution (using separation of variables) to the following version of the boundary layer equations. wherein the convective velocities have been replaced by constants and the matching boundary condition is applied at a finite boundary layer "edge. ,, f htr r = fix) +
rr =r=flon f'=O.
tr 1 + i•s z
O.
0< x
<
L,
0 < }• < M .
u —jrf_tton} = M.
u=yl}`1 on .x= O.
Here u and b are given constants; J. g, and h are given Functions.
3.4 Boundary Layer Flow Past
a
Semi infinite Flat Plate -
We limit our further work with the boundary layer equations to a discussion of the most important exact solution, that relevant for uniform flow along a fiat plate. In this flow the velocity vector approaches the constant O) far from a plate laid along the x-axis. The inviscid solution is simply t! * (x*, y*) = [lo u * = O. The effect of small viscosity is expected to he to a thin region along the plate and to a wake region behind it. Inconfied ,
order to locus on a single effect, we shall consider the case of a semi-infinite plate, extending along the x-axis from 0 to °. Here, there is no wake_
1 sir
Vi_srnus Fluids ICIL 3
FORMULATION
Introduction of a stream function i*(x*, y*) such that
d1P * _^*
ti *
U*
Y
.00* ax*
-
(1)
automatically guarantees satisfaction of the continuity equation clr4•
ay",
(^x *
icy*
terms of the nondirnertsional stream function kfix, y) , (I) should read
In
u=— u= -- (2) It follows [Exercise 1(a)] that when the stream function is employed, the boundary layer variables are defined by __
!* y (UOLv) 1/2
x=
x* L
y* = (vL/U l,) t !3 ' (5
Y*
(3a)
i.e., x* s {-
i'(x•, Y*)
y* (v'f-1Uo) 11 #
(L!0 Lv)"2
( 3 b)
In terms of ', with U = 1 , the boundary layer problem for uniform flow above a semi-infinite flat plate: is x > f}, y > O.
= X)Y• tfi =(x, 0) _ ,Jr,,(x, 4) °- O,
y > 0.
/)(0, :) = 1, litre rfr)(x, y)
x > 0.
1,
x > 0.
(4a) (4b, c) (4d) (4e)
We expect the flow just at the entrance to the boundary layer to be undisturbed by the plate, so we have imposed the condition u* = U0 at x* = O. The dimensionless form of this condition is u1 =or = 1, because the velocity scale of the inviscid flow is its constant horizontal component U o . BI.ASIUS SIMILARITY SOLUTION
The above problem has a solution of the special similarity type in which the independent variables appear only in a certain combination. Experienced f
Further discussion of the legitimacy and uscfu Irless of the stream fund ion may be found in 15.1.3 and 15.1.4 of I. We have delayed the introduction of the stream function to maximize experience with systems of partial differential equations and to minimizt the appearance of Concepts that are somewhat special. albeit very useful.
Um-61,n
5rr. 3.4] Boundurt• Layer Firm. Prj.sr a .Semi-rrf nice
Flrtr Harr
117
analysis anticipate such solutions in relatively simple problems involving infinite or semi-infinite boundaries, whence determination of [heir precise form is a relatively simple matter [Exercise 2(b)7. Moreover, similarity solutions are the subject of a still-developing theory based on the invariance of problems under continuous groups of transformations (see Exercises 8 to 10). A way to guess the character of the solution front general principles stems from the observation that no geometric length scale L remains in our semi-infinite problem. This suggests that the form of the function 0 in (3b) must be such that I. does not appear. One possibility is iJ(x, p)
(2x)tr2 f(+11.
ry
ü
(2x) -
(5)
For then, from (3), * !Mx * y*) = (U 0 L;')l(3 2x*
-1
t+
= (2110 ux*}t^2 f ^,*
^
^ 2Vx*
2
_.. 3
.,
1; 2
In (5) the constant 2 has been inserted for cnnvenicnce; the alternatives to (5) lead to essentially the same final results [Exercise 2(a)],t Using (5), we find (Exercise ( (b)] ll -
(I?),
c _ (2_ )'
P " f•(ij LL l
— Pia J
d ter)
(7)
It is straightforward to verify that the problem (4) now becomes
1" t fir= t) , f(0) = f '(0) = 0, j '( -r )= I.
8la d y
There is indeed a solution of the form (5). If an x appeared explicitly in (8a), then f could not be a funclinn of r7 nnly, but this did not happen. Both the limits x — ► 0 and y —0 correspond to Ft x. so the two boundary conditions (4d) and (4e) cor respond to the single condition (8d), giving the three boundary conditions expected in a third order differential equation. 'the problem (8j could be solved numerically by assuming f "(0) = A, and computing j at = P1)1, rr = 0, 1, 2, .... Results of various computations can be used to select a good approximation to that value of A which provides a Anoi her way to Ira u ii t he form of
is to reason that in the absence oratternaiis'cs a multiple
of Y• itself must he the c€7rrect Icngih scats i, al a Indeed. if
L
distance sc. downstream from ittc leading edge
.
is replaced by 7r* in (h), then one Aims t
011% ..
4 21-T [j VA * ) " 0
t.^
k 2 . INA* US' .2 )
which is equivalent to tbl. We shall see seticral limes in what follows how the .x' plays the role of the overall length scab_
distance
riB
liiscrr:3s fluids (C7r. 3
solution satisfyingf'(cx;) = I (see Exercise 3). i-i. Blasius, the first to discuss this similarity solution, solved (8) by a series method in his 1908 paper. But the following observations give a more elegant and economical way of proceeding (see Exercise 14 for another approach). 1f F() is a solution of (8a), then so is uF(ori), for any constant a. Choosing F"(Ü) = 1 for convenience, one solves the initial value problem# F' + FF" = O. E(C) ` F'(4) = 0, F10) = L
(9)
if u is chosen to satisfy a
lim
(lo)
then aF(ini) will be the desired solution (Exercise 4). This solution is tahulated, for example, in Rosenhead (1963, p. 224)_ There is good agreement between the predicted velocities and measurements made in a number of experiments_ Of considerable interest is the predicted drag. This is obtained by integrating the wall shearing stress T x . at location x*, where [Exercise 1(c)1 t7u * L x'
_P
}^
*
= f• =
o
" 2 pilo(R x •} 112f10].
(llj
[T he answer is expressed in terms of a local Reynold; number R.. x* iv.] We find that the drag on one side of a unit- width portion of the plate which extends a distance L downstream is 21l2119W,LR,7 (12 f"(0). A customarily defined dimensionless drag coefficient C D is given by ,1 -
2312/ 10)R "
(12)
The numerical solution provides f "(0), with the result C.' = 1.3282I "2
(13)
An important general conclusion is that the viscous drag on a fixed area decreases to zero as the viscosity decreases to zero. DEFECTS IN THE BLAS1US SOLUTION
We note that the predicted shearing stress act ually becomes infinite at x 0 like x - "2. The integral of the stress remains finite but doubt is certainly cast on the validity of the theory for small values of x. Iii pursuing this matter, we recall that the boundary layer approximation is derived for flows whose overall Reynolds number is large compared to unity. For the flat plate, we (ft9) is much easier than chat of (13t, for only a single - march t Numerical initial values is required_ outward from attic given
.S+•r . 3.41 Boundary Layer Ffulr Pair o .Semr-infinite fur Harr
119
have seen that the correct choice for the length scale is the downstream distance x*; i_e., the flow is characterized by the local Reynolds number LF a x*/v. We must thus expect failure of our theory close enough to the Leading edge, so that {l a x*/v < 1, say. [Note that the theoretical defect reveals itself in the discontinuity of Oy at the origin, a discnntinuity required by (4c11 and 044) An incorrect picture of the flow over the short distance vfl a ' should have only a small effect on overall drag predictions, but nonetheless improvement of the theory is of considerable interest_ It can be shown that
1.22(2x)
"
as pc-
cr2 i
(14)
(see Exercise 6). As x ; 0 for fixed y, the vertical velocity component is thus seen to become unbounded. Consequently, the boundary layer solution is defective not only at the leading edge of the plate, but all along the line x : O. Except near y = 0, there seems to he no reason for this defect Indeed, except near the leading edge, the defect disappears if parabolic rather than Cartesian coordinates are employed. Formula (14) also applies for fixed positive x at the" edge" of the boundary layer (y ) and shows that the flow there is directed slightly away from Further, there is erificatioii of our assertion that ex, the plate. unlike 1411x, xij, will be predicted, so that no attempt should he made to match z to the -
inviscid flow. From the very form of the solution (7) we can deduce the qualitative result that the boundary layer thickness increases parabolically with downstream distance. Whatever criterion on u is used to define this thickness, the deduction follows from the fact that u is constant when q is constant, je., when x is a constant multiple of y 2 . Tabulation off' gives quantitative results_ it turns out that f is more than 90 per cent of the way to its ultimate value of unity when ti exceeds 2.4, and more than 99 per cent of the way when a exceeds 16. Thus if boundary thickness is defined as the location ' y , where the horizontal velocity component reaches 90 per cent of its inviscid value, then y2 g — 2.4(2 x * vilie) 1 ". if one employs the displacement thickness 0;,the dimensional version of the quantity defined in Exercise 3.3(a). it is found that = 12(2x. ilLf0) t r In principle the boundary layer approximation should become increasingly accurate as the speed U U is increased, but in practice steady boundary layer flow develops an instability when L!c x*/v exceeds a value of around 600. (See Section 15.2 of I on stability, particularly its concluding cxample.t) Once instability sets in, oscillations and irregularities become more prominent as x* increases and the flow soon exhibits the random eddying that is characteristic of turbulence. Local drag is much larger in turbulent flow than in See also Chapter SoFt:' C. Lin, Theory uf ffOrodynurrar Str3hoiir}' I New York Cambridge University Press, [955), lo supplement i he general stability coFesidrt atinns nt I wlth an account ,
o r rescarcir a=] this pariiCU[ar prvblcm.
120
t'i.sCUUS
M. irl.s
1C:h_ 3
/Laminar (nonturhulent). Roughly speaking, this is because the eddies are more efficient at mixing than molecular diffusion,so that a larger,"effec Live diffusion coefficient" must replace p in the computation of viscous stress, BOUNDARY LAYE R SEPARATION
Another deviation between theory and practice can be illustrated by considering uniform flow past spheres and cylinders. We shall restrict ourselves to two-dimensional flow past cylinders in which the circulation is zero. (Compare Section 15.E of I.) In both situations, therefore, the standard inviscid solutions are symmetric fore and aft_ Pressure decreases from the Forwardrnost point to the corresponding side point or equator, and then increases symmetrically. If boundary layer theory is used to improve on the inviscid results, profound consequences are seen to arise from the increase in pressure over the rear of the body. (The terni adverse pressure gradient is used when pressure increases with downstream distance. See Figure 3.13.) Since the pressure within the boundary layer is the same as that outside (to first approximation), the boundary layer flow is progressively slowed by the adverse pressure gradient. Beyond some point P the boundary
0
F tG t,tit t 3. # 3. Symmetrical inriscid flow pusr cylinder or _spfrerre. S.- stagnation is :OW. Both in the graph giving lrre.ssure und o n the cylinder. point,whervlcy the hoa ry line singles out the region of adverse pressure gradient n,he re in ?Alen > O.
Syr. 3.4]
l3rrurnfur}• Layer
Paw a Semi-infinite Flat Phare
12 I
layer is predicted to contain a region of upstream flow. But in essence this means that upstream fluid is deflected by a zone emaciating from P and extending outward into the flow. The boundary layer is said to have separated from the body at (or near) P (Figure 3.14). Obstacles in uniform flow can he divided into two classes, bluff nr streamlined, depending upon whether nr not significant separation occurs. As we shall now indicate very briefly, the character of the flow and the degree of Our understanding are very different in the two classes.
P
F1auR E 3. 1 4. Successive tangential velocity profil es in a separating boundary Slaw bac'kftr3 ► } . omits in rlae .shaded region, elownsire'urn of the separation layer. point P. SLIGHTLY VISCOUS UNIFORM FLOW PAST STREAMLINED BODIES
Separation is a negligible factor in uniform flow past a slender body with a pointed trailing portion, provided that t he body is not badly misaligned with the flow_ A first approximation to such streamline flows is provided by inviscid theory, and a correction by boundary layer methods_ In principle, a more accurate solution can be obtained by modifying the inviscid outer flow, so that it corresponds to flow past a body with a normal "blowing" given by r.(x, coo) as calculated by boundary layer theory. Passing beyond the first term of boundary layer theory has turned out, however, to require remarkably subtle analysis. This can be illustrated by considering the boundary layer downstream of the trailing edge of - a flat plaie of length L . The stream function must still satisfy (4a), but now the solution must be matched to the Blasius solution of(7) and (S) upstream of the trailing edge. The no-slip conditions on the plate must be replaced by symmetry conditions v = U and dujdy = O. A solution to this problem was published by S_ Goldstein in 1930. This solution had a relatively thin " wake layer" within the ordinary boundary layer. Only in 1969 and 1970, respectively, did K. Stewartsori and A. Messiter find that Goldstein's solution gave a defective description of the pressure. To fix this up, a layer had to be introduced that was thin compared to the
Viscous FJuirii [C'h. 3
J 22
wake and whose length shrank to zero as Re co. This may well seem an academic matter, but it is not. An accurate first correction to the Blasius drag formula (i3) is only obtained if the "triple deck" structure of the boundary layer at the trailing edge is taken into account. (And, indeed, aerodynamicists are aware of the fact that flow details at the trailing edge of wings have a remarkably large effect upon drag. l Thus a presumnably accurate cor rection to the drag coefficient for a flat plate only appeared six decades after Blasius's original calculation! It almost goes without saying that there is no proof that the formal calculations provide an appropriate approximation to the exact solution as Re —k OD. Streamline flows at large Re contain regions of turbulence in both boundary layer and wake.° Although much of value is known about turbulence, and its effect can be estimated in many practical situations, there is to this day no satisfactory foundation on which to base investigations of this phenomenon_ The combined statistical and nonlinear nature of turbulent flow gives rise to theoretical problems that have so far resisted all efforts to construct a theory of generality and power. Si.IGHTI.Y VISCOI:s [:NIFORNi FLOW PAST 81-11_, IF1 BODIES
Boundary layer separation will be present when the conditions cited above for streamline flow are not met. In particular, in the phenomenon of stall separation occurs even for flow past a streamlined airfoil if there is sufficient misalignment with the mainstream. Separation generally brings about an overall flow pattern which is markedly different from that predicted by straightlbrward inviscid theory; its main feature is a large and unsteady wake region. The idea that boundary layer theory can be straightforwardly used to correct a relatively simple inviscid flow must now be abandoned. Yet this theory remains useful for bluff bodies if ii is employed in conjunction with measurements of the mainstream flow. Also, the combination of inviscid and boundary layer theory can give worthwhile results somewhat upstream of the separation point-t Special note must be made of the fact that in spite of separation, the inviscid solution For the cylinder (studied in Section 15.4 of I) is still of great use in calculating genuine flows. With the aid of conformal mapping, this solution can he transformed so that it provides a uniform two-dimensional inviscid flow past an airfoil. Excluding stall, separation effects are negligible in such flows. An outstanding theoretical problem is to calculate the flow pattern around a bluff body, say a sphere or a cylinder, in the limit Re -• a;. Because of Al sufficiently high speeds compressibility effects become important. but these hate been excluded from our considerations here a circular cylinder al Reynolds numbers of order Qf magnitude I0 4 . separation occurs .
t For
ahoul 80° downstream from the forwardrnost tsiagnation) point. At higher tpxd their: is turbulence in the boundary layer forward or the separation pOint.and separation is delayed.
Sec. 3,411 Baum/elf I. Luger Flan Past rt Sears-urr intre flat F`Harr
123
separation, a pronounced lack of symmetry between upstream and downstream flow is anticipated. The limiting inviscrd solution can and certainly will contain discontinuities in velocity between certain adjacent streamlines. The general characteristics of this limiting solution remain unknown, although slow progress seems to be forthcoming from a combination of advances in asymptotic theory, numerical calculation, and experiment. Calculations and experiments cannot provide definitive results because the actual flow becomes unsteady at sufficiently high Re. The limiting solution is still of interest, however. even though it must be unstable to perturbations, The mathematical challenge rrf providing an asymptotic analysis of a prototype nonlinearity is perhaps sufficient motivation for continued study of tlhe problem, but the limiting solution should also be of use in approximating flows where Re is moderately large. For the flat plate, as we have seen, the drag approaches zero as the viscosity decreases. The same is true in general for streamlined bodies. But for bluff bodies the radically different nature of upstream and downstream velocity and pressure fields in large Reynolds number flows is responsible for the significant drag actually found, in contrast to the small drag expected from a slight alteration of D'Alembert's classical "no-drag" result for inviscid flow. The dramatic difference between the drag for streamlined and nonstreamlined bodies is illustrated in Figure 3.15. l
•
streamlined turfed au u l a rtrc•rtlur = 4 x IO ° r L. .{ Bak hrlrrr. I 967. e;rfsrda•r ►t itlr he ►urne dray at speed p.336.) The appeared ert hug.. .a.. Frrlktk'r. t'. M . and Walker, 11b'_ S. Aeronaut. Research Council. R ep . and Mern,, rrrr, l2.41 . Re•drut+ri tt all t"iiiLlltt
3 15 4c•Irt7nutre• crass sct'rie+tr rtJ .
fit
pernis.Ycm,
Inviscid flow, boundary layer, separation, turbulence the development of these concepts and others like them has made it possible to understand the general outlines of results as complicated as the cylinder drag plotted in Figure 3_16_ Moi ivaied by the desire to build better aircraft and ships, the study of uniform flow past obstacles forms one of the most developed parts of fluid mechanics. As we have briefly indicated, much has been accomplished but much remains to be explained. The following survey articles provide recent discussions of orne of the advanced topics that we have mentioned: M. Van Dyke, "Higher Order
Viscous Fluids [Ch, .ï
12 4
1.5
i .o
05
o lD° ee''
?atip
R. —
3. 16. Qrealitatire befutr•ior of the dray per unit axial length as u junction of Reynolds number for a circular cylinder of radius a_ The cross-hatched region represents a drop in dr a y due to bnuexdeary layer tur b elP n ce and corrsegreenr rnot Yrrur n t of rhe .separarir err mind from e i br l ur 80 to 120'. }lurelriny is tutcessary because results depe n d ro some . extent o n th e amount of turbulence in the wind tu ►trtel. (Botrlrekxr, 1 967, p. 34 i , or Ro.senhrad, 1963, p. 106.) FIGURE
Boundary Layer theory," and S. Brown and K. Stewartson " Laminar Separation," Ann. Rev. Fluid Mech. I W. Sears and M. Van Dyke, eds. (Palo Alla, Calif.: Annual Reviews, inc., 1969), pp. 265-92 and 45-72, respectively; P_ A_ Lagerstrom, "Solutions of the Navier-Stokes Equation at Large Reynolds Number," SIAM J. App!. Math_ 28, 202-14 (1975); K. Stewartson, "Multistructured Boundary Layers on Flat Plates and Related Bodies," Advances in Applied McwhaMics 14, 145--239 (1974), also "On the Asymptotic Theory of Separated and Unseparated Fluid Motions," SIAM J. App[. Math. 28,501-18(1975). ,
E X E R CIS E 5
1. (a) If Oi* and W are, respectively, dimensional and dimensionless stream functions, then ;11 = *IC, where C is a constant. This constant must be such that (2) is satisfied. Verify the text's result C U 0 15. Check that C is dimensionally correct. (b) Verify (7) and (8). (c) Verify (11). 2. (a) Suppose that = (2x) :11g(0), 0 = xjy2 . This accomplishes the goal of eliminating dependence on L just as well as assumption (5). Comment.
Sec_ 3.41
Boundary Layer Flow Pusf uSr m i-infinif e flu Haw
l'S
By substituting into (4), (5), and (6), discuss the process of determining whether these equations have a solution of the form = _eig(yx/ for some constants a and b. (c) Modify the differential equation so that the problem does not have a solution of the assumed form_ Modify the boundary conditions to the same end. 3. (a) Set up a difference scheme that could be used to solve the initial value problem consisting of (8a) to (Sc) plus the condition #'"(ü) = A. A a parameter. lb) Let f(x) = NIA); the notation makes explicit the fact that the limiting value of ' depends on the initial condition .4. revise an algorithm that in principle could be used to determine to arbitrary accuracy the value A which ensures f(x•) = 1. 4. (a) Verify that if (l C) holds, then aF(arl) will satisfy the equations of l8), as asserted in the text. Op) (Project) Using the method of (a(, write a computer program and use it to determine the solution to (s). 5. As was shown in Exercise 15.1.7 of I, in steady inviscid incompressible flows without body forces. Bernoulli's equation guarantees that (b)
Ps + ip[(u') 2 + (r"*)2 J = p + where p x is the pressure far from the obstacle. Use this to interpret the definition of the drag coefficient (12). 6. (a) To obtain the behavior of the solution to (8) for large Values of 17. one assumes that . f{.f )
_
-
rr
B + 4(O.
OW small,
B a constant.
Why is the assumption made? (b) Using the above assumption. deduce the approximation
B + A(ri — BY 2 exp [ - 40l — B1 2 ].
A a constant.
[Equation (14) follows at once, using the value of B obtained from the numerical solution.] 7. Separation occurs for a circular cylinder before the region of ads Erse pressure gradient begins, according to Figure 113_ Discuss. In the present section and elsewhere, we have studied problems for partial differential equations that have similarity solutions, wherein the dependent variables essentially appear only in a certain combination. The next three exercises explore this matter further. Further references are G. Birkhof,, Hydrodynamics, A Study in Logit, Fact, and Similitude (Princeton. N.J.: Princeton University Press, 1960) [also issued by Dover Publications] and G. W. Bluman and J. D. Cole, Shniturit y Methods for Constructing Solutions to Ordinary and Partial Differential Equations (New York: Springer-Verlag, 1974).
c a6
4'i.vcnus Fluids
(Ch. 3
8. (a) Consider the transformations {1i Ay'', x + Bx, and y -1 Cy_ Show that the boundary layer problem (4) is invariant under these transformations if and only if C = A and S = AC. (b) if the problem has a unique solution, the solution must also be invariant under the above set of transformations. From this, deduce the form of the similarity solution (5). NOTE. The idea that invariance to a group of stretching transformations requires a solution in which variables our in certain combinations can be regarded as underlying the reduction of parameter number obtained by the use of dimensionless variables. 9. To afford further practice in the use of similarity solutions, consider the fl ow of a thin two dimensional jet into fluid at rest. It can be shown [e.g., Rosenhead (1963, p. 75410 that the flow is governed by the boundary in layer equation (4a) but with the boundary conditions -
0
as y
O„ = 0
at y
► y,
cc, 0,
x>0
(symmetry in x-axis);
x > {]
1l► ^ dy = M, M a constant (a)
(flow dies out far from jet);
(same flow of momentum across each section).
Considering the same class of transformations as in Exercise S, deduce that there is a solution of the form xi3{•(7)
r=
yx- 213
Show that the problem has a solution which is a numerical constant times cc tank arl, where ct is determined by the momentum condition. 10. Consider diffusion of heat in an unbounded homogeneous medium wherein the temperature depends only on the distance [rani a plane, line, and point, respectively. The governing equation for temperature T, (lD)
T
= KV 2 T,
thermal conductivity,
then takes the form
7;= x[T„ +(n—l)r - 'T, where n = i , 2, 3, respectively. (a) We are interested in the diffusion of a fixed quantity of heat, initially at the origin. Show that the corresponding restriction on Tis
fc Jo
Tr" dr = constant.
.Ser. 3.4]
Boundary Layer A ra . I'u.0 a Semi-irlfrrlite flat plurr
127
lb) Show that search for a similarity solution leads to an ordinary differential equation which can be written f' -
fl" + i ► i(f '
f)
0
r 2 /4hr.
for a certain function f f), where =
:AO Solve the equation of (b) subject to the boundary conditions
f
as r
and f ` ,- ► iÏ
+(d) Use superposition (with integrals) to generalize the
solution of
(c)
so that you can deal with the diffusion of a fixed quantity of heat which is initially distributed in an arbitrary fashion. le) What is the solution for a line source at r — 0 (two-dimensional problem) producing heat at a constant rate? The next three exercises give a glimpse of a subject that is the focus of considerable current interest, the interconnectton between chemical reaction and flow. The material for these exercises is taken from Chapter 2 of Levich 11962), 11. Consider boundary layer flow past a semi infinite flat plate. -
(al Show that deep within the boundary layer (t.e. "very" near the plate) the horizontal velocity component u* satisfies ,
u* ^
k}^* I-14 vx • '
(15)
where k is a constant. In fact, k is about (b) Show that (15) is reasonable in view of the following fact. lithe thickness b of th e boundary layer is defined as the distance from the plate then numerical calculations show where u' is 90 per cent of that e5 5.2(vx*/Uoli l 12. Suppose that a chemical is present to a fluid !hawing past an object, and that the chemical reacts very quickly with the material of which the object is composed. It is assumed that the chemical concentration C*(x*, v) is governed by the following problem, where u' and v* are given functions. ,
u* C'* -- ►
la )
Co
r)C* cx*
a.,,.
re. [' c*
,1L'*
+ ^* Cy * =-
D
far from the object.
+
r7(y*)2
f 16a)
C* = O on the object. (16h)
What physical assumptions are implied by (16)'' (b) Suppose that both u• and i.* have a typical magnitude U 0 , and that u* and u* typically undergo significant changes in a distance L. Introduce scaled dimensionless variables rnto (16). Gomment
i28
Viscous Fluids [C'1,. 3
briefly nn the advantages of this procedure. Discuss in general terms the type of solution expected when the Peclet number Pe ï fi e L jD satisfies Pe i> I (a common occurrence). 13. We wish to examine WI) in the case where there is a "concentration boundary layer" close to the semi-infinite flat plate y* O. 0, x* Flow far from the plate satisfies = h 0 ^a* = t]. We expect that if D is small enough, the concentration boundary layer thickness d will be small, even compared with the ordinary viscous boundary layer thickness b. Let us introduce scaled dimensionless variables that are apprnpriate within such a Concentration boundary layer, set up an apprnpriate mathematical problem, and draw a general conclusion without solving. (a) Choose scales for x* and y*. (h) Formula (IS) can be employed for n*. Why? Use it and (a) to obtain a scale for ü*. (c) We assume that Ou*/ex* + De1*f y* = O. (What does this imply?) Consequently, the two terms on the left of (liôa) should have the same magnitude. Why? The scaling for v should now be apparent. (d) Find a formula for the order of magnitude of d in terms of the other parameters. Use it to show that the concentration boundary layer is about one-tenth as thick as the viscous boundary layer in those eases (which occur frequently) when Pe 1000 Re. (e) Set up a dimensionless mathematical problem for determining the leading approximation to the concentration, within the concent ration boundary layer. Remember: u* and u* are regarded as given, u* by (1S) and us by a r:nrresponding formula. (f) As a rough approximation, the rate at which chemical is being absorbed at the boundary, per unit area, is taken to be DCad. Justify this approximation. 14. This exercise is concerned with a methnd for obtaining the Blasius Nat!. Acad. Sci. 27, similarity solution that is due to H. Weyl 578 (1941)]. (a) Let 9(q) = F"(7). w here F is defined by (9). Show that (9) is equivalent to an equation of the form g = (Dig), where .
(Ng) = ex (b) The above equation can be treated by means of the following successive approximations scheme (compare Section 7.2 of I): fa=U, 1,2,....
Show that (z — C) 3 exp (—
g 3(2) = exp 0
^
dt;
-
Sec. 3.5} Vortiriry (7ruriyes in Viscous Fluid Mork*: (c)
(d)
129
Demonstrate that O(g) O, (g) (Dig*) if g s y, Show that all odd approximants (i.e., all g with odd subscripts) are greater than or equal than even approximants. Show further that as n increases, the odd approximants decrease and the even ones increase. Prove that r
g,(z)]
-
,
and thereby deduce the existence of a unique solution q to which the approximants converge.
3.5 Vorticity Changes in Viscous Fluid Motion If a Muid moves with velocity v(x, t], the fortieity trl(x, t) is defined by ru(x. ti = curl v(x, r)
(1)
A major result of Section 3.1 allows us to interpret to as twice the angular velocity vector of the sotid body rotation, which, together with translation and elongation, makes up the local fluid motion. It turns out that great insight into the nature ol- fluid motion is attained by studying the mechanisms by which vorticity changes, in general and in particular instances_ An introduction to these matters is the subject of this sectton-
In a viscous fluid of uniform density, subject to a conservative body force, an equation for the evolution of vorticity can be obtained by taking the curl of the Navier-Stokes equations (1.28). By judicious use of various identities (Exercise l) one arrives at the following result:
r
+ V • VW
w •''v + V
2 (1.1.
(2)
The pressure does not appear in (2), and this is one reason why, it is advantageous to study the development of vorticity. VORTICITY CONVECTION AND STRETCHING
Let us first consider how vorticity changes in the absence of viscosity. With v = D, (2) becomes Po.)
at
^-
v • Vw = w - Vv,
i.e.,
Dw
Dr —
op V v.
(3)
As discussed in Section 13.3 of 1, the term ' Vt.) represents a change in vorticity due to convection, in which voracity is bodily carried from place to place by the fluid motion.
Viscutis Fluids [Ch. 3
As a start toward understanding the fa. Vv terms consider at time r two material points P and Q (points fixed in the fluid). Suppose that P is located at x and Q at x + En, a distance r. along a unit vector n (Figure 3. i 7). A short time Ar later, P will have been displaced by 4'(x, r)Ar + ()UAW] while the corresponding approximate displacement for Q will her(x + nr, r)At O[(Ar) 2 ]. Neglecting 01Arl 1 terms, we see that the original vector in which -t joins P to Q will have been transformed into a new vector given by un + [+.(x + un) - v(x) Via --- un i (n V+)[:Ar + O(t: 2 ). -
(4)
Subtracting the original vector en and dividing by Ar, we conclude that to lowest order in a the quantity tai[ Vi is the rate of change per unit time of a {short) material vector En emanating from x. n
E t. 1.)^
1
[Ie -pYipr
1 ,. - [ Î A t
F u te r 1.11 A short line of parades PQ rs riirrte+f in sink. At [utrf? the line P'Q' by flow with t•elueirr v -
We can write D(Fn)
Dr
= [:n• V r
(as i— O).
(5)
Using (3) and (5), we sex. that if a—cn is initially zero, then it remains zero, for D(m — un) = tl. f)r
(6)
[The result holds if the differential equation has a unique solution, for w — Fla ; 0 is certainly one so lution of (6) that satisfies the initial conditions_ Uniquenesswill obtain if l Vr is bounded-j What we have shown is this. Units can be chosen so that to can initially be identified with a material vector to of small length F. The smaller t is the more nearly it is true that to continues to be identified with, and to move like, the • Our discussion hut! will be intuitive. A rtg,ornus appruaCh Section 2-3 on tensor caluul«s-
ta ou[li=3rd in trie cxEfeLSCS [or
Sec-. 3.5] Ifurtii ihti
n ges in Viscous Changes
Fluid h#otdnn
11
material vector i.n. Thus (31 can be interpreted as describing evolution of the vorticity vector equivalent to that which would recur if that vector were an (infinitesimally long) material vector. Relative motion of the fluid at the two ends of the vector causes it to undergo a rigid rotation (relative velocity normal to +o) and a contraction or elongation (relative velocity along (,))Rotation merely changes the orientation of the vorticity. But velocity changes along u} effectively alter the extent of the spinning region and bring about annihilation or production of vorticity, just as a twirling ice skater can change his rotation speed by laterally extending or pulling in his arms. One can think of vortex lines throughout the fluid passing from one vorticity vector to the "adjacent - one, and vorticity being generated by the stretching of these linesYtsCOUS VORTICITY DIFFUSION AND BOUNDARY GENERATION The introduction of 'viscosity brings into [he picture two new ways for vorticity to change. First, the vino term in (2) is responsible for vorticity diffusion; local spin al one location will, through shear stresses, engender spin at an adjacent location. etc. Second, the necessity for the fluid to adhere at the boundary brings about a shear. Thus in a viscous fluid the boundary is a source (or sink) of vorticity so that an initially irrotational vise iw fluid will not remain irrotational. To obtain formal confirmation ofthe boundary's role as a vorticity source, note from the vorticity equation (2) that the diffusion lean is the negative divergence of a (tensor) vorticity flux — vVw. Consider a fiat plate along z = a: here
so
rrwu=$1.=D
!t
i
=0
c+r rit C'L' Fv =[),--= — - + ,— =0. 0. y' f.: fly i'.t
where we have used the mass conservation condition V-v = O. The %vorticity generated by the plate is given by corrlputing the following expression at z =O: ^^•
iz
= r
l p fi l l) z = vV 2 =- ^ - . i7
p f. y.
We see that vorticity is generated even in the limit r O. It can be shown that there is a unique velocity field v(x, r) that corresponds to a given vorticity field ra(x, t). subject only to the restriction that the normal velocity component of the fluid equals that of an adjacent boundary (with
suitable conditions al infinity). In a short time interval (I, + ,)the vorticity (0(x, t) is convected, stretched, and diffused according to (2) but additional vorticity appears near the boundary (from the boundary source). This provides the unique w(x, r + At) that gives identical normal and tangential fluid and boundary velocities, as required by [he no-slip boundary condition for viscous fluids.
[
Viscous Fluids [Ch_ 3
32
Pressure perturbations propagate with the speed of sound (infinitely fast in an incompressible fluid). Vorticity changes are far slower; this is a Fundamental reason why study of vorticity is so revealing. For further information, start with M. J. Lighthill's article in Rosenhead (1964 In particular, Lighthill discusses how the ideas of the previous paragraph cart be made the basis for a numerical calculation of the flow. VOR7rCITY IN TWO-DIMENSIONAL FLOWS
Great tangling of vortex lines with concomitant vortex stretching is a major feature of turbulence. But such matters are beyond our scope here, and we shall restrict ourselves in what follows to two-dimensional flows [wherein w = (u(x, y), r(x y), 0)]. For such flows the vorticity vector is ,
eu
' ( 0r 0, cu),
^
^u (7)
The stretching term vanishes, since dw
w Vv =
= Q.
(8)
i.I
Vorticity changes only from convection and diffusion, according to the following simplified version of (2): c7rxJ —
+ w • 17r13 — v P r,3.
(9)
at
Integration of (9) over an area
II
Or J 1
w
.
A
bounded by a curve C gives (Exercise 2)
A
-
,
J
u„{ + v
-
-
an
ds,
(l0)
C
where v„ is the component of w in the direction of the exterior normal n, and ira / n is the derivative oral in the direction of n. We see the roles of convection and diffusion again, for the net vorticity within A is increased by the net amount of w convected and diffused across the boundaries. As expected, the rate of diffusion is proportional, via the viscosity y, to the normal derivative of w. Let Us examine how these ideas can be used to interpret the exact solutions of Section 3.2. VOitT'ICITY IN PARTICULAR YISCOL:S FLOWS
From (12.3), the downstream velocity component u in flow between parallel plates a distance d apart is u=
(d 2 - y1 )
(- C = pressure gradient).
Ser. 3.51
VI)rrirrrl. Changes in I['r.uwus Mad Afert?o/rr
^
33
The vorticity
is of maximum magnitude at the walls y = ±d and decreases to zero at the center. Convection ducs not play a role here, as there is no variation along streamlines. There is continual upward flow of vorficity(into the fluid from the lower boundary, through the fluid, and out at the upper boundary) at the Constant rate —tiI!
_
(gyp-
r1Y
=C
.
(I2)
The pressure gradient forces fluid past the boundary 3' = —d, but the fluid must adhere at the wall, so a spin is imparted to the fluid layer adjacent to the wall. This spin sets fluid in the "nex!" Dyer spinning, and so on (vorticity diffuses). Exactly the same process proceeds from the boundary at y d hut the spin is in the opposite direction. At the center of the fluid. opposite spins meet and annihilate each other, giving risc to the absence of vorticity noted above. When thehorizontal plate ofSolution 3,Section 3?, is snatched into motion at speed U, there instantaneously develops a discontinuity in the horizontal velocity component u- Th e horizontal speed is U at the plane and zero just above it. At the initial instant, the vorticity cry can be regarded as infinite at this discontinuity and zero elsewhere. A plane of velocity discontinuities is called a vortex shot here the instantaneous start of the motion from rest creates a vortex sheet, and diffusion of vorticity from this sheet accounts for the entire supply of vorticity in the now thereafter. This picture is affirmed by the exact solution 12.I8. which yields ,
CO
all
Oy
=—
c
^
=
Lf(?rid)
= - }1i1 (4TCV 3 r 1 ) - 1 ''3 exp {
1 ' 2 c°xp(— re ),
— rp 3 J .
r >ü
t
>
[11 = ( 4 l,r) i fz • ^
03) (14)
Upon substituting y = ü into (14), we obtain zero. Thus there is no flux of vorticity through the boundary after the motion starts. But (13) shows that vorticity is relatively appreciable when and only when is of order of magnitude unity or less, that is, when y is no greater in order of magnitude than (wt) I ' 2 . The spieading of an effect across a distance of magnitude (a)U 2 in time r is characteristic of diffusion with diffusivity v. (Compare the end of Section 4.1 in L) Here we see, in addition, an overall decrease in intensity t 312 , as the fixed amount of vorticity is spread over an ever-widening like area
134
Ili semis f kids [Ch_ 3
Suppose that a second plate were parallel to the plate which instantaneously begins to move with speed U. Diffusion would rearrange the fixed amount of vorticity that is brought into being until ultimately the vorticity was evenly spread between the two plates. This is precisely what we see in Solution 1 of Section 3.2, the ultimate steady flow with constant vorticity throughout (see Exercise 3). As a final illustration of the explanatory power of the vorticity concept, consider boundary layer flow. Since downstream speed is of the order of the mainstream speed U S„ a parcel of fluid at downstream location x* will have spent a time of order of magnitude x*1I} o near the plate (which commences at x* = 0). In this time, vorticity will have diffused a distance of order of magnitude (vr) 1 ' 2 = (vx* f U o )" But the region of extensive vorticity is coincident with the boundary layer, so this physical argument gives a boundary layer thickness of order (%x*/[J 0 )' 12 , a result obtained mathematically in Section 3.4. In two dimensional flow past an obstacle, diffusion is responsible for lateral spread of the vorticity generated at the boundary while convection sweeps the vorticity downstream. Bee.ause organized bodily transfer of vorticity by convection is so much more efficient than random diffusive transfer, flow speeds do not have to be very large before vorticity will be confined to a thin boundary layer region near an obstacle. -
SUMMARY OF THE ROLE OF VOMTICITY
As we have seen, vorticity in the interior of an incompressible fluid of uniform density is changed by stretching, convection, and viscous diffusion. None of these mechanisms can create vorticity where it is initially absent_ We have examined how vorticity can be generated at a solid boundary in viscous flow. But no such possibility exists in the absence of viscosity. This leads to the first great fact about vorticity. For a flow that can he idealized as (a) inviscid, (h) of uniform density or having a pressure that depends only on density [sec Exercise 1(h)1, and (c) devoid of body force or having a conservative body force—for such flows, if motion starts from rest or otherwise can be ascertained to be free of vorticity at some time, then vorticity is always zero_ Mathematically, this means that V A r = O, This, in turn (given certain smoothness conditions), guarantees the existence of a function 0. such that r = V4_ Profound consequences result, of which we mention only Bernoulli's theorem (Exercise 15.1.7 of I) and the fact, exploited for example in Section 15.4 01 1, that is harmonic if the flow is incompressible. As for the generation of vorticity, we have examined one major method-the spin imparted by solid boundaries to a viscous fluid. To mention one more, the pressure term drops out of the vorticity equation because it has been assumed to have the form of a gradient and, therefore, to vanish when the curl operator is applied to it. More generally, a contribution from the
Sec. 3.6] Slow Viscous Hoot Past a Strut" Sphere
1
35
pressure term will remain in the vorticity equation, with concomitant thermodynamic generation of vorticity. An important example occurs in shock waves (e.g., aircraft sonic booms). There, discontinuities in thermodynamic variables across the shock are accompanied by a production of vorticity. Expertise in fluid mechanics requires a much deeper study of vorticity than is appropriate here. But we have introduced some of the key concepts that are required, and have given samples of the general principles and their particular implementation.
EXERCISES
1. (a) (b)
Derive (2). Begin by employing the identity found in Exercise 15.1.7 of I or in Theorem 23.7. Show that the text's assumption of constant density is not necessary and that (2) holds for a barotropic fluid subject to a conservative body force.
2. Derive (10). 3. Show that the net amount of vorticity per unit length is independent of time in the exact solution (2.18). Show that this same amount of vorticity is found in the first exact solution of Section 3.2. (This bears out remarks made a few paragraphs above. Note that the vorticity which is instantaneously generated when a plate is impuis ivel ystartcd is the same whet tier or not a second plate is present.) 4. Discuss the distribution of vorticity (a) in the solution of Exercise 2.1; (h) in the solution of Exercise 22 5. Establish a connection between the inviscid version of (9) and the differential equation of Exercise 15.1.4 of I. What is the viscous version of the latter equation?
3.6 Slow Viscous Flow Past a Small Sphere If a small sphere falls steadily in a viscous fluid, the me, all Reynolds number is small. The scientific importance of this one "slow-flow" problem is attested to by the fact that its solution has been used (i) as the basis for accurate determination of fluid viscosities, (ii) as an essential element in Millikan's oil-drop experiment for measuring the charge of the electron, and (iii) as a necessary part of the Einstein-Perrin method for precise determination of Avogadro's number. [Item (iii) is discussed at the end of Section 3.1 of I.] To mention just one application of more general small Reynolds number flows, such flows accompany the motion of flagellae and cilia, the tiny whiplike organs that propel microorganisms through fluid and that also propel fluid through channels in various creatures, including man.
r3 6
Viscoua Fluids [Ch. 3
We shall now outline the calculations that provide a first approximation to slow flow past a sphere. The original work can be found in a famous paper of Stokes published in 1851. Useful formulas for the divergence and curl in spherical coordinates are quoted in passing_ Exercises ask the reader to fill in many of the omitted details. FORMULATION
As we stated below (3.3), the Reynolds number estimates the ratio between (dimensional) convective acceleration terms like u duii?x and viscous terms like r i u2/13x2. When the Reynolds number is small. the convective acceleration terms should he negligible to a first approximation. (it is to be expected that nonlinear terms are negligible in a slow flow.) Theis the pressure-gradient terms should be of the same magnitude as the viscous terms. The dimensional equations governing uniform density low Reynolds number flow therefore are, to lowest approximation, V • v - 0,
V
A
ca — _ fit - 'VP
(La , b)
Here we have used the following relation between the vorticity ea and the
Laplacian of the velocity vector v (given in Theorem 2.3-6)_ V 2111 = V(V • v) — V A [u
[U = V A
,
-
In (1), p is the pressure and p is the viscosity coefficient. From (1 b) we obtain VA
(2)
(Vhw)= Q.
We use spherical coordinates (p. ch, 0), where
x=
p Sin
¢
COS
El,
y
=
p sin ch sin 0,
z = p COS ch.
Unit vectors a {`'', el', and point along the directions p, rp, and U increasing. The corresponding velocity components are t. l", 6}, and 06} . From references such as Rosenhead (1963, p. 133), we obtain the following general formulas, where subscripts denote partial derivatives: p = Pn e" + p i p, e + (p sin O)- pe e'.
V - v = p 1 (p 2 010 + (p sin 4i pV
A v -
(sin 0) - 1(061 sin + [(sin 0 1(of -
(3a)
' (LC'bi sin 0)0 + (p sin 0) - '(v'°')e,
(3b)
— (041) e aervi -
(pu
ro})r] e i E
[(pi's %
—
( »)^etet_
(4)
We take the origin at the center of the sphere and assume that far from the sphere, flow of speed U is directed along the positive z-axis. There should be no swirl, so aim LP) = O. Consequently, from the requirement V • x = O, using (3b), we deduce the presence of a stream function Jr(p, 0) such that t] tPi
(P2 sin /) 14/
v"' _ —(p stn 0)-1 fi r,.
(5)
See. 3.45] 51o11' YisCaus
NOW PUS' a
Snug} Splaerp
137
Only the third component of the vorticity will be nonzero, so the problem can be reduced to
1 2
+
—
P2
c
a a2 2 where c : cos 0,
{'c 2
(6)
plus suitable boundary conditions (Exercise 3) . SOLUTION
It is not difficult to verify that the appropriate solution for flow past a sphere of radius a is ^
2
^^ s'n^
^ - l) (l ^ ü
^[3
^
^,
(7)
^
From (lb), the pressure is therefore given by
(8 )
P = P«L — ip Gra ff 2 cos cfr-
In (8), ? is used to denote density. Remember, p ordinate in this sect lor'_
used for a spherical co-
In general orthogonal coordinates, formula (1.25),
T, 1 = -- {^ii + 2pD ;. remains true, provided that, for example, 7, 2 is now interpreted as the stress vector on the surface element with normal in the direction of the first coordinate vector, projected in the direction of the second coordinate vector. To compute the stress on the sphere, we need the following formulas for selected components of the rate ofdeformation tensor in spherical coordinates : Don _ 2(21'),,
{}^ — ({a sin Or ' (0 ), -4_ p(p
D,m = p(p- rrn n + p -1 (0 nl ) 4
(9)
Using these, alter some calculation one can obtain the final rather simple result that the drag force Fd is given by Fd = 6npLra.
(10)
If the sphere has density 6', then its predicted terminal speed of fall V through a fluid of density (5 in a gravitational field of acceleration g is (Exercise 6) V = a 1g
'(Zr — L5)_
(11)
We note that full understanding of the equations in spherical coordinates requires a tensor calculus which is not restricted to Cartesian coordinates_ This is beyond our scope here, so we have just quoted various needed results. Aris (1962), among others, presents derivations of the required formulas.
L
(Ch 3
vJSC{,us
^^
The low Reynolds number approximation obtained above (Stokes flow) looks considerably more straightforward than its high Reynolds number counterpart, but it should be mentioned that there are actually strong similarities between the two. When improvements are considered, it becomes apparent that the Stokes now is actually an "inner" approximation which matches the "outer" uniform flow. Further progress requires modification of the outer flow. For additional material on these matters, see Rosenhead (1963, Chap. IV).
EXERCISES
I. fa) For what physical reason does the density not appear in the governing equations ( 1)? (b) Verify (2). (c) Verify (8). (d) Deduce from ru (l) a differential equation that in vol ves o n l y the pressure. 2. Justify the introduction of the stream function, as in (5). 3. Derive (6), and provide suitable boundary conditions. 4. Verify that (7) is the appropriate solution to (6). 5. Carry out the somewhat lengthy calculations necessary to obtain (10). 6. Verify (i I )_
Appendix 3.1 Navier--Stokes Equations in Cylindrical Coordinates A study of general tensors, which is beyond the scope of this hook, makes routine the transformation of the Navier-Stokes equations into cylindrical coordinates [or any other coordinates; see Aris (1962)]. We shall only state the results. Thus the principal equations for a viscous fluid of uniform density p are as follows, where subscripts denote partial derivatives. (Body forces are neglected.) Conservation of mass: V - Y = r
1 (^L;^^ M ^ r +
r
-F ri' ` - f).
(l}
Balance of linear momentum: Di]I
Dr D LP' )
D1
►1
- r
^
[Z!
Ld}]^ = —P
+ r - ' ^{ " L1 1ak = —
J — 1 }r + 4'[V ^[r t! r ^
( pr)_
D
Dr
l
pe + 1,[p^L''e' -Jr 2r
= _ p - 1 p,
vV214,.
^ trj t!
— 2r
-2(01)0
2 (r' !
— r
j
^ )e^ +
(2)
2r'Le j ^.
(3)
(4)
.1ftp0lr.6 j?l
et?
r
biro. Li/rrrrrrrue+
AS? rltr.
Here it". 1 ,`rt`, and w are velocity components in the directions r. O. and increasing, i.e., along the unit coordinate vectors eI". et"r and et".
D Dr
V'
^
r'
►
—+at"'_
t'r e12 r
-
r'r
+ r
l 51
f 112 rr'a_; . + r r'r + r'`' r 1)-' + I r'
(6)
The relation between stress and rate of strain is still when p # fir.
1=
P +
(7]
In f7) p and if represent all possi ble combinations of r. q, and k, For example. Tt, is the stress component along e l "' on a surface element whose exterior normal points along 0". The components of the rate-of-strain or deformation tensor are given by 1Drr = (1 ," 1 )r , iDBlr = r I (tnrr 4 r Y". } =
Dv) = r lit' + It `I,, ^ w
-
,
=w
} _r =
1^1
The vurttcfty vector is given by ILO = ❑ AV
t1 .101ld irF +
[ ( O rFf_
e [r tti4'u + r -I [rr [1^ , - (0' 1 0
wr l e
rü^
l`)
Note that the major vector operators have been expressed in polar coordinates: div in (I ), curl in (9), and grad in the first terms on the right of (2k (3), and (4} (the components of p ' Vp).
Appendix 3.2 Generation of Confidence in the Boundary
Layer Equations by Construction of a Finite Difference Scheme for Their Solution Part of the subject of partial differential equations deals with the question of what initial and boundary conditions are appropriate in various situations. It is usually a major task to show that a problem is well posed. i.e., that there is precisely one solution and that small chances in the problem cause small changes in the solution. Here we illustrate on the boundary layer equations a relatively simple way to generate confidence that a mathematical problem makes sense. The idea is to see whether one can establish a finite difference procedure which in principle coutd he used to determine an approximate solution to the problem, and which requires exactly the given boundary and initial conditions. This procedure has independent value as the possible basis of a numerical solution.
Viscous Fluids IClt. 3
: 4n
In contemplating a numerical analysis of the problem (117-20), we first observe that the semi-infinite interval in y trust be replaced by (0, y, ), where y h. is large enough so that no significant variation in the solution occurs. We then limit our calculations to a grid of points, as in Figure 3.18, where u and v. are prescribed as shown. For simplicity in exposition we consider both the horizontal and vertical distances between adjacent points to be the same constant h, but this requirement can easily be dropped.
—
w
e e t
t
ü —r
rp
o a o o o o
O^
R+
D
a D
o o
D
�
O 0 û U
Q
d
�
O
ü
V
O
o
o
d
40 D Q
a
O
C.
# • • •
i
•
O d
O O
• •
•
u - r=4 FIGURE 3. lb. Velocity CoxNporrenls u and r are ta be dererrn inert at the unshaded
grid points. Only u is known at the fruif-shaded points.
We wish to replace derivatives by difference quotients in such a way that we can determine u and a at the various grid po i nts. We shall use the notation r O) " f,. u(th,ih) = ll i , Cf(iit, O) ^Ux(Jh, (1) One possible way of proceeding uses (3.I7b) and the equation obtained by substituting (3.17b) into (3.17a)_ The. a equations are u ; + r+ = Q,
—
rt, , r + r' , = L F i
"
+ 11 .y .
(2a, b)
The first step is to use (2b) to compute :r along x = O. In doing so, we make the approximations vy(i{i, jh) = h -1 (v+! — Jh) —
h -! (sdt .1
(3)
—
fi i (t++,1+! -
1- s.),
— 214
1.j +
141,j- L }.
Upon evaluating (2a) at (x, y) = (ih, jet), using these approximations, and then solving for v,, f , we find that (u1,1
- ^C i j^1 j-1 .
.
— if — h ^(n1 .1 -^ `^ 21414 + ui.i-1)]
(4)
Appe'ndi_v 3.2] t,errrrurrl,+r uj Cilrr/rrierue in doe-
Bind! trklxl'
Le1te'r hyler^frrul ►
141
z r^
F Iii u ti r_ 3.19. firjurrrrrrrrrrar used w obtain a Mk' IYrlree u j r-. Thur sire required twines of u and r• I1re k 1lenrrr at er pone $3 i►rdie used ht' sltudins{ doe Hi r1rra1 rr,ylf halves. resfu'ru!-rh', 4.0 rhe r•Urreivrrefirrrl circle
Ism Figure 3.19). Specializing to i = 0. j = I, we see that t' ;,., is determined by an expression involving the known quantities $1 0 0 , ]f ,+ 8,+0 . 2 (Is is given along x = 0),A, and t' #, p (1 is zero when t --= Ûj_ Repeated use of (4) gives t) all the way along the initial line, up to but not including r e .N . (Note that c , is used in determining t'n.! _ 1 .) r1 The next step is to determine u alongx II. This we do by replacing 12a) by — r1; ) ♦ -11} = 0. h` l { i — ^' i.1
which yields trr^ 1.1 = Ir i.}
i' e
-^
+ r ";
-1
15)
Figure 3.20). Specializing to i — 0, j = I we observe that r1 r , is determined in terms du o. , (given), D o. , just computed). and re.. = 0. Repeated use of (5) allows computation of u along x h up to and including uI N, Since ti t. r , is given, rr is known on all grid points of _v' — h. The use of (5) with i = 0, j = N provides the missing value of r>, r a• N- By repeating the entire calculational process, we can successively determine vertical lines or ti's and 1.1 's. AU the given information has been exploited, and no more seems necessary, so added confidence is derived in the appropriateness of our boundary conditions. In particular, we see that additional prescription of ()along y = hi or x = 0 would be inappropriate, for r at these locations can be computed from the data given. The scheme we have outlined is the general type that is suited for use with an electronic computer. Untouched, and beyond our scope, is the question of convergence as h . [3 of the exact solution of the difference scheme to the actual solution of the differential equations_ Similarly left to other sources is determination of the accuracy of the approximate solution to the difference scheme computed by the machine at fixed h and with inevitable roundoff error. (see
—
Viscous Fluids [Ch 3
42
u
=
?
_
^+
ftu"L'Q E 124. lirformatriu ►t used f[a obtain a new value of it. known information. us ïrr Figure J. I9_
rxldrCYifr's
SJladiary
Of course, other difference schemes can be constructed. For example, one could use (3.17a) rather than (2a) to march rightward in the computation of u. Selection of the scheme that is best in some sense is another question which has to be left to a study of numerical analysis. See, for example, R. Richtmyer and K_ Morton's Difference Meih +ds for Initial Value Problems (New York: Inter'science, 1967) for valuable discussions of the type of proh!erns we have mentioned, We must noie the possibility of a "scheme" that does not require the value of u nn y = O. This seems to cr]ntradrel our statement that ai! given data are necessary. The scheme proceeds as before except that (5) is replaced by h - 1 (u,i l
li .!
J l + ÎI
1 fU
^ 1 — U ..) ]
—
t}
or
By taking = O,) = 0, L 2,_..,u can be computed onx = h star bug with y — 0, without knowledge of u(ii, O) = u I O . But u 1.s _, cannot be computed by this scheme, nor can uz,_ 2 and u 3 _ i when it is repeated, and so on. The whole process of finding evidence that our problem is appropriately forrnulatcd rests on the demonstration that u and tt al each grid point iS uniquely determined. If one or more of the unknowns is arbitrary, the scheme is inadmissible. Now it might be thought that u l - i can be computed from yy I.", I J uo.iv-11 ♦ 12- VD. k -1 - VO.'h'-21 =
h
}
but in using (6) we are indefensibly employing the same expression for t^ , at (0, N -- 2) and at (Q, N — 1), In the limit N —. co, h -6 Q, Nh • } , this means that ra},fO, 1 O, which is generally faise ,
.
Appendix
Generation of Canfid[rrCN in rhi'
BoülrifClfi' 1.-L11 -*V' {:i
M ürNN.'S
1 43
Use ofa finite difference scheme to verify the appropriateness of boundary conditions is not an approach that is free of problems. The total number of unknowns at the various grid points is always far exceeded by the number of equations that could be regarded as approximations to the given set of differential equations. One wishes to devise a set of difference equations that give a unique set of values for all these unknowns, These equations should reduce to the original differential equations in the limit 11 • 0, and no further relationships among the variables should result. Any '*sensible" explicit difference scheme satisfies these requirements, provided that and r grid distances are appropriately related. To prove that a solution to the original problem exists, it remains to show that the sn to the difference equation approaches a limit, and that this limit is a solution to the original differential equations. This may not he so, and proofs in any case are not easy,s But experience teaches that sensible formal procedures are justifiable more often than not"; this is why the ability to construct a sensible difference scheme gives confidence that a solution exists to the original problem_ In the present context there arc serious extra difficulties because of the necessity of passing from the finite domain treated in a difference calculation to the infinite domain specified in the original problem. Indeed, questions of existence and uniqueness for boundary layer problems are definitely not trivial, as is illustrated, for example, in the discussion of classes of similar solutions in Rosenhead (1963, Chap. V, Sec. 21). In some cases, for example, uniqueness is only obtained if decay as r — x is required to be as fast as possible. It is hard to see how the necessity of such extra conditions can be anticipated by the present approach_ But the very difficulty of a rigorous treatment increases the value of a heuristic approach, provided that the limitations of heurism are not forgotten.
* We have pointed nut that the process is not without difficulty, nonetheless, formulating appropriate difference equations and Mien passing to the limit proi.idcsanelIcctiveway to obtain existence and other theorems for partial differential equations. For the wave, heat, and Laplace equations, a classical reference is the paper al- Courant, Friedrichs. and Lewy tired ore p. 12t of 1 [IBM J. i 1. 125 (1967)]. Major works concerned with the existence of solutions to the boundary layer equations (3.I7-20) have employed the method of tines wherein difFerencing with respect to one variable transforms the original prnhlem unto a system of ordinary differential er;4ations_ Q_ A. 01einik [Sager Math. 4, 583 (19631] uses the rrari.sverse line rnethod in which the "limclike" x variable is discrttized_ W. Walter [Arch. Rarional Mech. Areal. 39, 169 (1970)] proves an cxislcnce theorem by means of distretixing the other, spacelike, variable after making a preliminary transformation (see Exercise 3.41
CHAPTER
4
F ound a tions of Ela sticity
I
chapter we shall lay the foundations of the classical threedimensional theory of elasticity. We shall restrict ourselves to the assumption of very small deformations and use as our stress-strain or constitutive relation the mathematical representation of an isotropic, homogeneous, linearly elastic solid. Simple exact solutions will be shown to provide relations between the various coefficients that appear in our linearity assumptions. Considerations of energy and virtual work will provide uniqueness theorems. The chapter concludes with a brief introduction to solid mechanics when the assumption of very small deformations is dropped. It should be noted that the derivation of the equations governing linear elastic solids bears strong similarities with the corresponding derivation (in Section 3. I ) of the Navier Stokes equations for a viscous fluid. N THIS
4.1 Analysis of Local Motion The ftcld equations that apply to all continuous media must be supplemented by constitutive equations that describe the particular nature of a given material_ This section contains a prerequisite to the postulation of constitutive equations in solid mechanics, a local mathematical description of material deformation. "Compatibility equations" are also discussed. Cartesian tensors are used throughout. STRAIN TENSOR IN MATFttIAt. COORDINATES
Consider a particle located at the initial time t o at Po , a point with the coordinates A. The medium, of which this is a generic particle, is deformed. At time t, the particle is located at P, which has the coordinates x(A, 1). The continuous medium is said to be strained when the relative position of two particles is altered. We shall now develop a quantitative measure of such a
strain. We introduce the displacement vector U defined by U=x—A
or U ; =x ; — A.
(la)
With explicit indication of dependent variables. (la) is written L',(A , t) = x{A, s) — A ; .
(lb)
Here we are referring the variables to material coordinates and thus are using the material or I,sgrangian description. 144
See. 4 !]
A nr11f•sis of
Lora! Moti o n
Let us write (lb) in the
1 45
forrni + i^^(A , ti•
x,{A. EP =
(2)
[In words, (2) states that the position x ; of the particle that was initially at A is now (at time t) equal to its initial position plus its displacement from that position.] Let us compute the derivative of x,(A, r) a+ii h respect to A. Using index notation_ we find that
xi.i _ o3i
(3)
Ut.j•
where the argument (A, r) is understood in x,., and L'
. Equation (31 will prove useful in a moment. At time t — 0 consider a vector dA whose initial and terminal coordinates are A, and A; + dA,, respectively. (For the present, regard dA 1 dA 2 , and dA 3 as arbitrary numbers.) If dA is embedded in deforming material, at time t its end points occupy the positions x(A, I) and MA + dA, r) (see F rgure 4 I) ,
,
We can write
x,-(A # dA, 9l — Y,(A. t) = x,, 4(A, t) dA ; + - - • ,
(4)
where - - indicates further terms in a Taylor expansion and the summation convention is used (as it will be throughout our discussion of elasticity). The smaller I dA I is, the closer to zero is the contribution made by these additional terms (assuming that x i. j is continuous in a neighborhood of A). A convenient way of stating this fact is to say that if dA is infinitesimal, and if dx(A. dA, r) is defined by dx ; =x i(A+d A, t)— xpA, t),
(SI
then we have
(6)
dA,_
The ellipses (• • -1 are absent from 16) because (6) is "exact" when dA is infinitesimal.
xrA*dA tl
1 1
f.0
1 ^+iA n
1
^I
t / dA
A
F to u RE 4.1. Defarmarra.3 dx
Jrne
^^f; tA + dA. ti
^
1
f
//
A + dA
changes rhe t nfi nftesrma l l i ne Element dA emu t1 new elmnt
Foundations of E%ISlteit} (Ch. 4
1 46
If we combine (6) and (3), w e obtain dx ;
_ [64 + UJA, r )] dA ;
Or
dx ; = del ; + f.+ i.i(A, t) dA ) .
(7)
Equation (7) has an important geometric interpretation. A straight line is generally deformed into a curve_ By contrast, (7) implies that an infinitesimal straight line (or Fine dement) is deformed into another infinitesimal straight line (Exercise l). Consider two line elements ds„ and ds4 through the point P 0 . After deformation, these become the line elements do and ds' passing through the point P- In component farm, the various line elements may be written as
dsa = dA ; et^M ds = dx ; e" I ,
dtir,
ds' = d)610 1 .
Here el is, as usual, a unit vector pointing out along the x ;-axis. The length of the line element ds will be denoted by ds and a corresponding notation will be used for the other line elements. We are now ready to introduce a measure of local deformation. It is not obvious how this measure should be chosen; indeed, more than one choice can be made-Trialand error indicate that one appropriate measure is the change in the scalar product ds„ • ds4 due to deformation. Thus we consider ds • ds' — ds„ - dsQ = ds ds' cos 0 — ds„ ds' o cos 0 c ,
(8)
where the angles are shown in Figure 4.2. Employing (3), we may write this difference in scalar products as dx ; dx; --
dA, dA; = (hi, + 1r1 }(o,k + U4 . 0 dA, dAx — dA 4 LC, = dA i dA: + (1i ,, + l`l j , k + U,,,Ui.x) dA } tfA k — dA i dA t (9) (U'k,i + I;F J k + [1;. }Ui, k j dA) dAi. ,
1
We designate the quantity in parentheses as twice the material [or Langian] strain tensor,* which we denote by 2 thi . Thus F
k1 = ^ k ^
j. k +
Ui.
(1U)
4 k-
Note that the strain tensor is symmetric: rl = difference in scalar products can now be written as ,
k
ds • ds' — ds o • ds#, = 2;4 1 dA ' The word "tcosur" is appropriate since ^ is Farmed by addition tives of a vector.
(11.) and contraction
ofideriva-
sec _ ¢_11 Analysis of Loc•ai Motion
1 47
A,
5
dst,
« A2 fil)
A1
FI C;URE 4.2. r]c}{rrrrrrarirsn changes the
angle between two line dements from
Igo ro A. Frequently, the strain is so small that I fi r J ! 4 1. In such cases it is natural to replace the strain tensor by its first nrder approximation, writing rhh i — (j +
;ark
(12)
When the strain tensor is given by the linear relation ( 12), we shall speak ref
infinitesimal strain . GEOMETRICAL. INTERPRETATION OF STRAIN COMPONF:NISti
now show huw changes in ienjths, angles, and volumes can bu described in terms of the components rh) . We consider first the special case when the two line elements ds o and ds4, coincide and lie along e": d , = dA; =ds o ,dA 2= d4 =dA 3 — dA — 0. In this case dx ; =dx; and [from We
(11)]
rfs 2 --- dsô = 21 1 , dsb.
(13)
Let us define p 1 by the relation dw = dsa( l 4 - m i )
-
(14)
Thus
1.11 =
ds — ds" ds n
(15)
in other words, p is the change in length per unit original Length parallel to the A s -axis. Substituting (14) into (13), we find 2a
1
= 2 fc, +
(16)
i48
Foundations of Fla_sririty ICh. 4
The solutions of this quadratic equation in j1 are Iii =
±11
+ 2/11) 112 — 1 -
Only the plus sign yields a meaningful root, since p 1 must vanish whenever 11 = O. Thus ra (17) p, — (1 + 2 7i 1 )' — 1 and (18) ds =, ds o(l + 2 ) `/2 1, and (17) implies that
For infinitesimal straits, I m i l
p 1 = 1 + 3.(27 11 ) + 0( 71i1) — 1
Thus ii +,
or
t7t
^
^
Nt•
(L9)
12 2 , and +Iaa represent the change in length per unit original length
el), e'2 t, and em, respectively_ Let us consider a second example. Here we examine the deformation of line elements which are initially orthogonal, so that
parallel to
dA t ds d , d4 2 —dA 3 = 0,
dAï=dif 3 - 0,
dA; — dso.
From (8) and (11) we find that ds ds' cos i} = 21 1 r ds L dso _ ,
(20)
Let y 2 = n — O. Thus y . 2 represents the decrement in the right angle that was orig i nally formed by elements in the directions of c' and ec2. We have ,
cos $= cos (-n — y t2 ) i sin y i 2, ds = ds (,(1 4 2ri l I )" and ds` ds'a(1 + 211 22 ) 1 ' 2 so that, from (20),
sin Yi
2/ 12 _ (1 + 410 1/2 (1 + 2g22)" 2.
(21)
For infinitesimal strains, (21) reduces to Y12
2 1 12
(22)
'
when i # j, rl rt represents one-half the decrement in the right angle that was originally formed by elements parallel to &' 1 and e1r1. A further interesting interpretation is found by considering the Jacobian which represents the ratio of a volume element after deformation to a 1, volume element before deformation (see Section 13.4 of I). The Jacobian is given by
Thus we see that
fur infinitesimal strains
=
Nx1, x#* xA
00 ] , #1.# r A3)
_ det
r At ♦ (id aAj
see ¢- f( Analysis
Loral MraliUrl
1 49
Expanding this determinant for the case of infiniteNtrrtaII 3t rains, we find (Exercise 3) that
J^ 1irhi .
(23)
In this ease, then, the trace 1.5f r),,, namely q„. represents the change in iwrlunie
after delimiiution, per unit original whittle. At this point it is worthwhile looking hack at what we have done in this section, in older to emphasize that there are two independent levels of approximation involved in arriving at the infinitesimal strain tensor (l2). Using the word loosely, "strain - at point 1' may be characterized by examining the deformation of various lines that pass through P. As is customary, we confined our attention to the fate of such lines in the limit as t h eir length approaches Zero. It can be said that in characterizing deformation we only examined tenpoints that are "nearest neighbors" of P. This restricted the precision or our characterization but was clearly the appropriate first step_ Having decided to limit our attention to the deformation of anfinitcstmal straight lines, we derived an expression (9) that can he used to provide information such as the increase in length of such lines, the change in angle between a pair of them, eic. Equation 19) provides this information even in cases n f the most extensive distortion. For a large class of important problems, however, we can limit ourselves to very small distortions. It is this rel,rlively simple but very useful linear theory. wherein )10) is simplified to (14 that concerns us almost exclusively in this volume. Rut Section 43 gives a taste of the theory when the strain is chararterifed by arbitrary distortions of infinitesimal line elements, STRAIT TENSOR IN SPATIAl. ; OORI}INA"iES
An argument entirely similar to that just used can be applied to develop a measure of strain in spatial coordinates For this purpose, we write - A,(1i, 1l Note the following connections between velocities r and V and displacements u and 11 in spatial and material coordinates. respectively: V-
six s1!
_
r''1A + U)
-
r'!
r' ll
Du
ci
Dr
We can develop the spatial (or F ulerianl strain tensor, c l ((x, [j, defined by t:r! —2(4r. i
-I
( 25 )
where. ; = i1j0x,. The individual compiments can he studied through examples similar to those used in the previous section (Exercise 4) For small strains. WC find that E..1 ` 1(14r. r -
I
ii ) , ,).
I
(2
i)
Foundations of EÏustirity
J5
[Ch. 4
For t — j, tt is the change in length per unit length in the deformed state parallel to the x ;-axis. For i j i: o is one-half the decrement in the angle between two directions that end up in the deformed state parallel to c'i and est, respectively. Comparing (12) and (26), we see that for infinitesimal strains the two strain tensors have the same form except that a material description is used in the former case and the differentiation is with respect to the material variables - while a spatial description is used in the latter case, and the differentiation is carried out with respect to the spatial variables. It should be noted that the difference between differentiating with respect to spatial variables rather than material variables is of second order. To be explicit, ira vertical line denotes a substitution 01- material variables for spatial, we have, by the chain rule, that
_
BfJ k _ k ex,,, Ûut C^(A,,, OA p = ax. 8A p cx^ DA ^U
&k
x ,n
^ Frrz
4-
^^^
t^F 16U„,
ôxp
ax,„
a
(27)
p
Neglecting second order terms, we find that t) _ ô14(x, i)
cOA ,
exp
- A. ffi
(2R)
and also
duk(x, [) Oxp
,au ,,(A, 1) P
(29) Mk.
o
Thus the difference between the two derivatives can be neglected when we are dealing with infinitesimal strains In this and the next three sections w e shall restrict ourselves to infinitesimal strains. In many classical texts, the infinitesimal strain tensor is denoted by t o , although each component is interpreted as if one were considering the material strain tensor_ Also, no distinction is made between :d ; and U 1 . Although this is not strictly consistent with the notation used up to now, we shall continue the practice of failing to make a sharp distinction between material and spatial coordinates in the remainder of the chapter except for the Final section (4.5) on nonlinear elasticity. THE ROTATION TENSOR
We shall now employ the strain tensor to show that an infinitesimal piece of material, alter deformation, is translated and rotated (a rigid motion) and also elongated along three mutually perpendicular axes. We begin by noting that (30) d u ; = u r {A + dA) - N ; (A) = ui . f(A) dAi.
Sec. 4. ! l
Analysis Of Laced MorioR
151
Thus points near A are displaced a common amount 1i ; (A) and arc subject to a dA i . According to Theorem 2. I.10, Ili 1 can further alteration given by u1. or be written as the sum a symmetric tensor and an antisymmetric tensor,
namely, (31)
coo ,
where €ii = ^( 1^c + cir1. i ) and wu =
l.j
.^
— ^i)
[32a, b]
The antisyrnmetric tensor w 1f is known as the rotation tensor. As has been shown at the end of Section 2.1, we may associate with such an antisymmetric tensor the rotadon vector fit , where [analogously to the fluid mechanical case in (3.1 .8)]
flk
""
ifL' n u)k •
ij (2) if
(33a)
As in (2.1.44), Cmr.i[ 1-2k
— WmA •
(33b)
Now [using (7)], dx ; — dA, _c = dA f +w dA i ,
and [using (33b)], we obtain co if dA i = Efl f2k dA = [dA Iti III.
Thus cow dA i is a small rotation through the angle I5li, about an axis parallel to Sk through the tail of dA. 1f eti 0, then the whole deformation reduces to a small rigid body translation and rotation.* We note that the components of the strain tensor c ;; for i = j arc known as the normal strains and the components for i # j are known as the shearing strains_ We have shown that normal strains represent elongations [see (19)] and shearing strains represent angle changes [see (22), and remember the equivalence of e ii and 11 i; in the case of infinitesimal strain], PRINCIPAL AXES OF STRAIN
Since the strain tensor €, 1 is symmetric, it has three mutually orthogonal characteristic vectors which are known as the principal axes of strain. The characteristic values are called the principal strains and are frequently denoted by, , c 2 , and r, . When referred to principal axes, the strain tensor takes the form t 1f
=
Ei
❑
❑
et
0
❑
❑
0.
• This result can also be shown cilrecLly (Exercise 5) ,,
t5z
Foundations of Elasticity
ICI?. 4
It follows that eki dA j corresponds to a stretching, by a percentage equal to the principal strains, along the principal axes of strain. If we take as our system of axes the principal axes of strain, a rectangular parallelepiped with edges d r , dA 2, and fill deforms into the parallelepiped shown in Figure 4.3. It is seen from Figure 4.3 that the change in volume is given by (1 + E 1 )(1 4 €2)(I + € 3 )
dA r dA 2 dA 3 — dA, dA 2 dA 3 .
Recalling that we have confined ourselves to small strains, we find that the change in volume is given by
(c, + t: 2 + E 3 ) dA, dA
Cl/13-
Asis Section 13.4 of 1, the dilatation A is defined as the change in volume per unit original volume We see from the above equation that -
(34a)
d=e: l +f: } + E,.
We note that this expression is the first invariant of the strain tensor evaluated in a coordinate system that coincides with the principal axes. Thus we can conclude that A = e.;; _
(34b)
This result coincides with the analysis of the Jacobian found above_ It can be shown (see Exercise 11.1-20) that the planes on which the maximum shearing strain will occur arc those planes that cut the principal planes (i.e., the planes normal to the principal axes) at 45°. The maximum shear strain will be the maximum of the three quantities Itl — t:1 , ^
E{
I
2 1t- 2
— 1.
3,
(35)
EF3 — ErI-
drt x
FIGURE 4.3. Deformation changes one infinitesimal rectartyulur
parallel-
epiped to another, provided that the original parallelepiped is oriented along the principal axes of straw. (For clarity of presentation, the defèrrtuitxur pictured is much larger than it should he
SET_ 41] Analysis of Loral Milian
1
53
COMPATIBILITY EQUATIONS Suppose that we are given the nine components of the strain tensor, t:,3(x, z), co . Can we guarantee that there will be a displacement vector, u,(x. i), that will satisfy the following equation? =
'PP i l ,
+
(36)
j, i)'
That is, are there three (single-valued) displacement functions u i , u z , and u that satisfy the six distinct equations provided by (36) when c, j is a given symmetric tensor? As we shall see, sometimes one can determine an expression for the strains in a given situation, but the integrability question we have posed must be answered before one is assured that this expression is appropriate. To be more precise, some problems in elasticity are naturally posed entirely in terms of the stresses: The stresses are given on the boundary of an object and are to be determined in its interior. We shall see that once the stresses are known, it is a simple matter to determine the strains. Knowing the strains, can one now find the displacements? (This question does not arise in fluid mechanics, for it is of virtually no interest to consider situations where stress is prescribed on the boundary of a fluid region) The integrability problem can be restated more conveniently in the following equivalent form. Given a strain tensor ar ; and an antisyrnmetric rotation tensor 0.) ;i , under what circumstances can we guarantee that there is a displacement vector u, satisfying the following equation? (37)
u.^ — e,) + cu .
The two forms (36) and (37) are equivalent because the heart of the matter is to relate deformation and displacement ; any required solid body rotation can easily be added. (A formal proof is requested in Exercise 12.) ' Because the problem has been changed from (36) to (37). we can regard it (for fixed i) as a question to which we can apply Theorem 2.3_10. This states that if a continuously differentiable vector field B; = Box} is defined over a simply connected region in space, then there is a single-valued function 4 _ ¢(x) such that
B;0,,
081
if and only if 41-, Bi.
(39)
0_
For our particular problem, we are seeking three functions such that ea,. = + a. As stated in (33h), we may write
[I,, 142,
and u
w J = ZIA i 4 .
We observe (Exercise 6) that nk, k — 1
(1)r1.
_ 0.
(0}
Found/J tiorls nf EIra.slrri 1• tC"h_ 4
154
We can thus formulate our problem once more as follows: Given the symmetric tensor E; and the vector 11. 1, with 1-4 .1, O, find necessary and sufficient conditions for the existence of a vector u ; such that + c;ik il_*
(41)
We may nnw apply our integrability theorem to each component of given in (41). Consequently, necessary and sufficient conditions are esri(Et!
+
€ig, n ).,, =
as
0.
Using the "ed" rule of Theorem 1.2.11,
( 15s145ph — skiFif' k P SI.1
si
+
r25 ^.
The integrabilility condition now reduces to —
t
sr^
c si. p
(42)
We seek a condition that is expressed solely in terms of the c o . Because of the introduction of II, this is easily accomplished by another application of the integrability theorem, this time to (42). The necessary and sufficient conditions for the existence of a displacement vector become Qrs = Û
where
Qrs
(43)
rini Espj E ll. -
The integrability conditions (43) are known as compatibility conditions, since they are the requirements that a given strain tensor be compatible with a sing Ze-valued displacement vector_ Since Q,# = s, [Exercise 7(a)], there are six independent equations. These can be taken to be the following [Exercise 7(b) or Exercise 14(a)] : ,
1.23 = ( - 23.1 + £31.2 + 6 12.3),1, 122.31 = ( — '6 31.2 + £12.3 + E23, X1
4 33,12 = ( -0 12.3
12, 12
+ E23+l
+'6 31.2).3*
(44)
1 11.22 + 1122, i I*
2E23, 23 = 6. 22.33 + C 33.223
2 &31.31
F33. t 1 + 6 11.33'
A multiply connected region M can be made simply connected as required by the integrability theorem by making appropriate cuts. if displacements are to be single-valued in Ai, they must certainly have this property in the cut region. Thus the conditions of (44j are still necessary, even in multiply • It is conventional lu use similar notation for the alternator symbol p, fl. Tbc dangcr of confusion should no be great_ c1, __ L i
E iji,
and the strain
Sir_
4_ Ij
Anufi'ses
Lut
at Morion
155
connected regions. in addition, one must require that the same displacements result at every point on each cut, no matter how these points are approached. (The additional condition has to be verified only on any nne set of cuts that makes M simply connected.) SOME EXAMPLES OF STRAIN
order to illustrate in concrete terms some of the general ideas just discussed, we present three elementary worked examples_ In each case the reader should be sure that he agrees with the given information; (a)a description or the strain, (b) the principal axes of strain and the principal strains, (c) the dilatation, and (d) the maximum shearing stress. In
Example 1. Consider uniform dilatation in which the sirain tensor is given by
Solution. in the particular coordinate system chosen, the strain tensor represents a change in length per unit length of the amount a in each of the x,-dire Lions_ The s hear strains vanish in this coordinate system_ On the other hand, we note that the tensor is isotropic; hence the strain tensor will be the same for every coordinate system_ The principal strains are e; every coordinate system is a set of principal axes; and the maximum shearing strain is zero. The cubic dilation is given by =t = 3e.
or a=
Example 2. Consider simple extension given by the strain tensor e, = e,
i = j = I.
t;r = 0 forail other i and j .
S lulion. In this state cf strai n, the material undergoes a change nie per unit length in the x i -direction only. Since the tensor is already in diagonal form, the given sel. or coordinates form a system of principal axes and the principal strates arc e. 0, and O. The dilatation equals e_ The maximum shearlrtg strain is le and planes of maximum shearing strain are shown in Figure 4.4. x
rr
^
XI (j}
(bl
F! c: u R E 4.4. Planes of maximum sfrenrinrg .srrain in sim ple exrensrnn along r!u' x, -{ixis_
156
Fouxulrarir3rl.s of Elastic- iry [Ch_ 4
Example 3. Consider simple shear in which the sale deformation is a change in the angle between the x,- and x,-directions of 2s. The strain tensor For this case is U
=
s
0
s ü C:). 0 0
(4s) []
The before and after pictures are shown in Figure 4.5, for a section that was a unit square before deformation_ The reader shuuld use this figure to check his results. N o r E. Rotation is not specified, since is not given. ln Figure 4_5(b), sufficient rotation is assumed to align the bottom of ltie figure with its position prior to deformation_ x,
= x,
!h! F tci u K r 4.5 Srrrrin
sfef'r1r. (a ) Be frrre deformation. ( h) After difor-
rrrfrrian
Solution. It can be shc'vn [Exercise 9(a)] that the principal axes of strain are the x 3 -axis and two mutually perpendicular vectors that make angles of 45 0 to the x ! - and x2 -axes. The principal strains are s, - s, and O. Figut 4.5 enables us to confirm these facts frorri the geometry_ It is apparent that the diagonals l'o and undergo the greatest changes in length. 1 fence they must lie along the principal axes of strain, which we thus sec make angles of 45° to the x,- and x 2 -coordinates. The principal strains can be computed from the geometry of the deformed parallelepiped_ We note that t = Io = - According to the law of cosines, the diagonal C is given by 2 ['
= [ I - 1 -- 2 cos (inn — 2s)]" 2 = (2 — 2 sin 2s)' ` 2 — 44112 23130
where advantage has been taken of the s mall strains to linearize the equations. Thus one principal strain is given by
^ ^n = -s. —
lU
Similarly,
I" = [I + 1 — 2 cos (in + 2s)]+r2 = 2 i + 2 0. -I- sj_ Consequently, the other principal strain is
r fu
set-. 4.11 Arscrlj•,ws fit Local Motion
157
The orientation assumed in drawing Figure 4.51bj is assured if u 2 (0, t), 1 ) = u3(0, t7, 1 ) = 0.
ia ;(ü, D, oj = 0,
(46)
Under these assumptions one can show [Exercise 9(b)] that thedisplacerrrents are given by H 1 - 2sx2,
11 = ar = O.
(47)
,
W e conclude with a mention of'the important case of plane strain_ Here the displacement vector is given by
u # - u Q(x t , x 2 ),
or ,
u s - 0,
-
I, 2.
The strain tensor is given by Evd
=
t: o {x
i: e3 = 0,
i
,, X 2 } — 2(u
3.^
a,
+ z^ ,) ^
— l, 2;
1,2,3. EXERCISES
1. Strictly speaking, (7) deals only with the end points of the vector dA. Show that our discussion implies that all points on dA are "deformed'" into a new straight line. 2. Use the quotient rule to show that the Lagrangian strain tensor is indeed a tensor_ 3. Verify (23), and deduce the text's interpretation of the trace of the strain tensor, for small strains_ 4. Carry out the development of the Fuleriari strain tensor, including the interpretation of the components for both general strains and infinitesimal strains. 5. If u; + tr ; - D for all i and j, show by direct integration that the deformation can only be a small rigid rotation plus a rigid body translation. (This exercise is very similar to Exercise 11.}.) 6. Verify (40). 7. (a) Show that (43) implies that Q,. Qom' (h) Show that (44) follows from (43). 8. Sketch the following displacement fields. Calculate the extensions, shears, dilatation, and rotation. (a)
u,
(h) ex 1 , ii i = u 3 Here e is a constant.
D-
9. (a) Find, by algebraic calculation, the principal axes of strain and the principal strains for simple shear_ (b)Demonstrate that Figure 4.5(b) is correctly drawn. 10. A cylindrical bar (not necessarily circular) with generators parallel to the x 3 axis undergoes the displacement -
u1
- Ox2x31 I4 = OXIx3
,
4i 3 =
xi x2), ,
ftxurda rior,s of nawiriry [C h . 4
l 5b
where H is a small angle. Clive a geometric description of the displacement. Find the strain tensor, the principal strains, and maximum shear strains. 1 L Let w = w(x 1* x 3 )and let Greek subscripts range over the values 1 and 2. The displacement vector is given by IJa =
UV a
• ü3 = w } -- — -
,
2(1 — 2v)
X 3 w a^ ti
where v is constant. Find the strain tensor. Evaluate T33
+v C33 ]
^1'
1—
^ik
1 2_ (a) Show that if u satisfies (37), then it satisfies (36). f.(b) Show that if a function u can be found that satisfies (36), then one can always find a (perhaps different) function u that satisfies (37) as well. 113. The compatibility conditions (43) represent six conditions on the six different components of the symmetric strain tensor. There is thus only a single tensor that can satisfy these conditions." Discuss. 14. (a) With the aid of (1.2.11), show that the compatibility conditions (43) can be written
Q rs
—
O.
, = tes.iii
— rr,.,I =
0.
(48)
$(b) Show that (43) is identically satisfied if €; = ^; + 14i.; for a smooth function n_ Why is this result not surprising? 15. Let V be a simply connected region with boundary V. Consider the incompatibility tensor Q defined in (43)_ (a) Show that Q„ . ,= O. $(b) Suppose that Qr,4 0in V r#s; ,
121
r =
Q2I = Q33
^
{]
on ôV.
Show that Q„= Oin V Thus if three of the compatibility equations hold in general and three others hold only on the boundary, then in fact all six independent equations hold generally.' * This exercise is adapted fro m K. Wa hixu , "A NINE on the Coniiiiiuris of Compatibility,” Mai, ph ys , 36, 306-312 (1957). Comments on the significance of the result are grade an the end of Section 43.
Se c 4_2] Nno4t•`:1- f ûnslifutrt•e Equation and Sot»r Exact .awtfitlotLs
t
59
4.2 Hooke's Constitutive Equation and Sonic Exact Solutions In this section we supplement the field ,equations, which apply to any continuum, by a suitably generalized version of Ilooke's "stress is proportional to strain" assumption for cylindrical bars under tension. The generalization involves an isotropic tensor of order 4, and hence two coefficients. (The third coefficient, normally found in the expression for a general fourth order isotropic tensor, is zero by symmetry.) Certain exact solutions will he shown to relate these "Lame coefficients" to the more down-to-earth elastic constants such as Young's modulus. This modulus was introduced in Chapter 12 of I, where a one-dimensional theory of elasticity was presented. The discussion here is independent of the earlier one, but of course bears many similarities to it GENERALIZED HOOKE'S LAW
Let us now state the basic equations governing the motion of a continuous medium. Anticipating a fuller discussion at the beginning of the next section, we exploit the assumed smallness of the displacement and its derivatives to replace the (material) velocity by the partial derivative with respect to time of the displacement: au, Du, (Jul aJ1 i - n t^ i ^ Di ^t ^
Similarly, we shall make the approximation 3 ;n; {Fl 1
Derivations of the basic equations can be found in Chapter 14 of l; the equations are listed in Appendix 2.1. With the above approximations the mass-conservation requirement takes the form
Dp
_ 0.
(1)
The three equations derived from the balance of linear momentum become
P t
! cox t-
{Tii ) +
fi .
(2)
Consonant with its conventional use in elasticity theory, we use ; ro denote the body force per unit volume. The notation pf, is used for this quantity in I, and in the present volumeexcepl for the chapters on elasticity.) Balance of angular momentum gives (3) = Energy equations will not be required in
our discussion.
I643
Foundations of £tusrit- 4y [Ch. 4
The body force per unit volume.. .6(x, r), is normally considered to be known. Thus (1) and (2) comprise four equations; and there are 10 unknowns - the density, the 3 components of the displacement vector (used in preference to velocity in elasticity), and the 6 components of the symmetric stress tensor. 1f we were to introduce the strains, the discrepancy between the number of equations and unknowns would not be altered, since we would he adding 6 components of the (symmetric) strain tensor and 6 relations between the strains and the displacement vector_ Constitutive equations, the stress - strain relations still remain to be postulated. In Chapter 12 of I. on the one-dimensional motion of a bar, we used as our stress- strain relation l llooke's law, which states that stress is proportional to strain. Let us consider this case in somewhat more detail_ Suppose that we apply a tension test to a cylindrical har of structural metal, such as steel. That is, suppose we measure the elongation of the bar as a function of an applied axial force. The. results turn out to be of the form shown in Figure 4.6_ The abscissa, e, is the uniform axial strain, i.e., the change in length per unit original length. The ordinate is the stress T computed by dividing the applied force F by the original cross-sectional area a.
T
T
u tr
__
f ! J ! I
r t 1
1 {}
k' t c u k E 4.6. SfrPs.s T eErso- strain e in a tension test on a cylindrical b u r
We find that up to the point P, the proportional limit, the stress T is proportional to the strain e. Between P and Y, the yield point, the relation is no longer proportional_ lithe load is released, however, the unloading curve will duplicate the loading curve; and, when the load is reduced to zero, the strain will likewise vanish. Beyond Y, the material becomes plastic and unloading is indicated by the dashed line. We observe in this case that when the load is reduced to zero, there is a permanent strain left in the specimen. if the material is subsequently reloaded, the stress strain curve will follow the unloading tine, which is approximately parallel to OP. After the reloading portion meets the original loading curve, the stress-strain diagram continues as though no unloading - reloading cycle had taken place. The stress corre-
Sec. 4_2j Hookr's Consfivaire E.quainori and Soin Lxucs Solutions
rbr
spending to the point U, a local maximum of the graph, is usually referred to as the ultimate load ; point B represents rupture of the specimen. Depending on the particular material, portions of this stress. strain curve may he missing or somewhat altered_ In some materials, the stress-strain curve beyond the yield point may he almost fl at. Moreover, the unloading reloading cycle may look something like a hysteresis loop. One should also realize that the apparent drop in stress after U is somewhat illusory. We recall that the stress plotted was defined as applied force divided by the original cross-sectional area At the point U. the area of the specimen has contracted. If- one were to calculate the stress by dividing the applied force by the instantaneous area, the curve would not drop. For low-carbon steel, P is about 35,000 pounds per square inch (psi) and 11 is about 60,000 psi. For iron, P is about 25,000 psi and Y is approximately 30,000 psi. The initial slope of the stress strain diagram is known as Young's modulus and designated by E. For steel, E 30 x 106 psi, and for aluE = 10 x 10' psi_ (I psi = 69,870 dynesjcrn minum, Classical elasticity is concerned with the linear portion of the stress-strain curve from O to P. It should be remembered that not all materials display this behavior_ For example, rubberlike materials usually have only a curved stress strain law that is approximated by a straight line through the origin only for extremely small stresses. We shall assume that the material is initially stress free, so that all components of stress vanish when the components of strain are zero. Furthermore, extrapolating the tension test results, we assume that the stresses are linear functions of the strains. Consequently, we may write -)
(4)
Ci1hmEkrn-
The quotient rule shows that the 81 coefficients C oin, form the components of a fourth order tensor C_ For most of our applications, we shall assume that the material is homogeneous, which means that the components of C11 ,, are constants independent of both x and t. There will be a few cases in which we shall want wallow these coefficients to be functions of x, but this will be particularly noted at the appropriate time. The components of C_'^ can be found by removing a representative sample from the material and applying a series of tests wherein both the stress and strain components can be measured_ A wide class 01 materials in which we shall be interested are devoid of "grain" or crystal orientation. They thus have the property that the values of the coefficients, so determined, will be independent of the orientation of the sample tested. Such materials are called isotropic. Consider (4) and its counterpart in another Cartesian coordinate system ^i 1 — ^, ijkm Ek
•
t62
Frrundarrrxrrs of F laxslirïr}.
lC'!1. 4
In an isotropic material the contribution, say, of £ 2.3 try T 2 , must be the same as the contribution of E3 3 to T, 2 . Th115 C 1223 =
0 1223
or, in general, C'ukra That is, C must be a fourth order isotropic tensor. The general form of such a tensor is given in (2-1.40. The scalar found there must be zero, however, since Tu is invariant to an interchange of i and ), but the terms multiplied by K are not. [See Exercise 2.1.22 or the derivation of (31.49).] Consequently, (4) takes the form` T,-; = 4E1:4 + 11-41‘ 15 i;-
(5)
Equations (5) are known as the generalized Hooke's law, and the constants . and p are known as the Lamé constants. INTERPRETATION OF THE ELASTIC COEFFICIENTS VIA EXACT SOLUTIONS
We shall consider a number of simple, "armchair" experiments that will give us some physical insight into the two Lam& constants, as well as certain important combinations ct the two. To do so, we must first solve the linear equations (5) for the strains as functions of the stresses. In (5), let us put i and j equal to a common value k. Performing the indicated sum, we find that Tkk -
(2p + 3,110: kk
or f.k k
= --
k
-_
21.i + 3A7
(6)
Thus we obtain the strain-stress equations
2u
2p(2p + 3,1) '
(7)
• The saint final result is obtained From the seemingly more general assumption Ti r To see this, one employs the decomposition ^1.^.
= i.F,. +
i 06w
of (1,31). Then one can make the assumption that an observer rotation will not affect the constitutive equation (principle of material indifference). Alternatively, one can prove that t ic contribution associated with w. in fact, vanishes. - flic discussion is precisely parallel to that for the corresponding case that arises in fluid mechanics. as presented above (3.1.19) and also bckrw {3.1.25].
Sec_ 4_1] Hoake'.s
Curpsrrlurree•
&yurclrrxi und Satin, Exact Su}urrm.ti.
t63
We shall now consider a sequence of situations that are so simple that exact solutions to the governing equations can be guessed. In all these situations the (symmetric) stress tensor is identically constant, there are no body forces, and there are no variations with time, so the linear momentum equation (2) and the angular momentum equation (3) are identically satisfied. The density p is constant, so the mass-conservation requirement (1) is also identically satisfied. The various situation s a re specified byprescribingstresses on theboundarics of bodies having given shapes. From the guessed form of 71 1 we compute the strains from (7). In all our examples the strain tensor is constant, so the compatibility conditions (1.43) are certainly satisfied. TENSION OF A CYLINDRICAL BAR
As our first example, we consider a cylindrical bar under tension, as illustrated in Figure 4.7. The magnitude of the force per unit area is prescribed as N at the ends of the bar, and zero along the sides.
FIGURE 4.7. A c 'li ufrrt ul brrr Wider rrrnwlr
The stress tensor in this case is given by
f) {l 7;o — 0 U 0 , 0UU N
for then both the end and side forces have their assigned values (Exercise 1). The example under study is precisely that of the tension test_ Therefore, we expect the axial stress and the axial strain to be in a ratio given by Young's modulus E:
N Elt
(R)
r64
Foundarrons of F.Ja.siiriry
1C.h. 4
Furthermore, it has king been known that lateral contraction occurs in the tension test. The ratio of lateral contraction to longitudinal extension is designated by Poisson's ratio v. That is, we expect that —
f:22
^1L
_
— ^3 3
=
(9)
V.
^li
From the known values of I we compute the t: i f from (7) and find accord with (8) and (9), provided [Exercise 2(a)] that ^
=^(2^+3^. ) + $.1'
ti:
2(..1-^ ^)
.
(10a, h)
Solving for the Lamé constants, we find [Exercise 2(b)] that
A
_ E l,), ^ ^ 2(1 +
Ev
(1 + v)(1 — 2 ► j '
(lla, b}
and that
v_ E
d
(12)
2,u{31 + 2}^j
The strain-stress equations ofMean now he written very simply in terms of the "engineering constants," Young's modulus and Poisson's ratio: = (1 + ►+)7; f - u 3rkk
^
.
(13)
These equations have a rather simple interpretation_ We note that (i 1 —
TrLr r2 2T33 (14)
and similarly for the other axial strains. Equation (14) states that the tensile strain in the x 1 '-direction is the elongation produced by the tensile stress in that direction minus the contraction produced in that direction by the tensile stresses in the x 2- and x 3 -directions. One could, in fact, develop the stress strain laws by starting with this fact as a hypothesis, together with the assumption that the principal axes of stress and strain coincide. From this it follows that the stress strain Laws in principal axes are given by (14) with corresponding expressions for the other two components. Since the constitutive relations must be tensorial, they must be given by (13) in any other Cartesian coordinate system. -
SHEAR OF A RECTANGULAR BAR
The second case we shall examine is that of pure shear stress. Imagine an infinite cylindrical bar of rectangular cross section that stretches along the x3 -axis. Subject the lateral faces of the bar to shear stress having magnitude s
See 4_4 Huokr's Cotrsatufirr Equation and Same FA ucv Sr, fuftr,ns
165
-r7 -
5
+
^
I
?
i
/ ^
/ ^ / ^
Fi[;t7 sE 4.8. !lie arrows represent Shear farces applied rrl u rectangular hur
.
The miss section delimits into Me dashed parallelogram
and orientation as in Figure 4.8, Then (Exercise 4) the stress tensor is given by F tiJ
s
7:i — s 0 {] .
(15)
t]
0 0
We see from (7) or (13) that the only nonvanishing strain component will be ' l 2 _ If we write E, 2 = ' , then (according to the interpretation of the nffdiagonal strain components found in Section 4.1), the cross section deforms into a parallelogram, as shown in Figure 4.8. The stress-strain law (5)yields T1 — 21"F 2
(i6a)
or, alternatively, my.
(16h)
will be proIn linear elasticity the (correct) expectation that the angle has led engineers to introduce a proportionto the applied shear s portional ality constant called the shear modulus G:
From (I fib) and (I lb) we sec that
E We expect that, for reasonable materials, G > 0 and E > O. Thus I +y>O
or w> —1.
We expect, additionally, that longitudinal extension is accompanied by lateral contraction. Consequently, the inequality stated above will be replaced by the assumption (17) v >p,
Foun dations of Elasticity [Ch. 4
166
COMPRESSION OF A RECTANGULAR PARALLELEPIPED
The final case that we shall consider involves pure compression. Consider a rectangular parallelepiped with faces normal to the coordinate axes. Impose on each face an inward normal stress of magnitude p, p > 0_ Then p =
0
—
0
0
p
U. p
0 0
—
The shear strains vanish and the normal strains satisfy
ECE l _ EE22 = 1;€.33 = (2); — 1)p.
The dilatation, A, is given
by
FA = E kk = — 3(1 — 2v)p. (18) We expect that the dilatation will be proportional to p. Thus the bulk modulus or compression, k, is defined by
= —
(19)
A
(The minus sign is present because dilatation is negative in compression.) From (18) and (19), k
_ 3(1 — 2v)
.(20)
In terms of the Lame constants, (20) can easily be shown (Exercise 5) to become k = À + p.
(21)
It is expected that material will not expand under the influence of a purely compressive stress. For this reason we assume that k 0, i.e., [from (20)] that y< To summarize the inequalities concerning Poisson's ratio, expectations that elongation will not lead to lateral expansion and that compression forces will not cause expansion give rise to the assumption .
4-
(22)
An incompressibility assumption gives v = . Typical values of v fall between 1 and -I. For example, v = 0.25 for cast iron and glass; v — 0.29 for carbon steels. EXERCISES
1. (a)
Consider the cylindrical bar under tension, as in Figure 4.7. Show that if the stress tensor is as given in the text, then the end stresses are N and the side stresses are zero.
Sec. 47] Hrxake's C un.srtfuter'e Lquuttora rind
2. 3.
4.
5.
Sonie E.*tic'! Oknolls
lb
(b) Find the displacements, choosing the arbitrary rigid motion to make these as simple as possible. (a) Verify (10); (b) verify (11) and (12). Show that the principal axes of stress and strain coincide for an isotropic material. Prove the somewhat more general result that if A,, is a symmetric tensor and u is a scalar, then A ii and A rp + c^ have the same principal axes_ What is the relation between the principal values of the two tensors? Show that the stresses in Figure 4.8 are indeed associated with the stress tensor that is given in (15). Compute the strain tensor that corresponds to this stress tensor. Establish (21) as well as the following relations among the elastic constants: 3k v 3k E _ E 3k(1 2v) ,
1 +r
— 9k — -
-
-
—
6. Consider the following displacement for an elastic problem with vanishing body forces: u3(x, r) ï u3(x, t) = 0.
e+1(x, r) = U(xt, r),
Find the corresponding strains and stresses. Find the equation that u(x t , r) satisfies. How does this compare with the displacement equation developed for the one-dimensional motion of a bar in Section 12.1 of 1? 7. Ii is asserted that measurements on the strain of a certain motionless solid can be fitted if the following assumption is made: y 2 + Y 2 xZ
X2
0
(ce}) = y 2 + 2 2
xZ
X
x .
(23)
21
(a) What body forces are implied by the above assumption and (2)? (h) Show that even though the equilibrium equations 12) can be satisfied, (23) nonetheless cannot bean appropriate deduction from the experimental evidence (assuming that the material is not torn). 8. (a) Deduce from (1.4S) the following form of the compatibility conditions for a 13ookean material [one that satisfies (5)]. V 2 Ts + e ,3 T,
.
=
(24a) (c5„v 20 + 0. „).
TR
(24b)
s^^
,
Here 0
-
(b) Using the equilibrium equations (3.8), deduce v I ! V-27;,+
V^
rs
l -4- ^' ^r* ^^[) ^
{
1r
s
^ ^- s . ^ = 0.
(25)
16s
Foundations of E1as1icùr L('h-
4
(c) Show that 1 — Ir ,r1
^3r:.
'
■^
and hence that
V xO=^ + ^(7-f). (d) Obtain the following alternative form of the compatibility equations in terms of the stresses V
V2
7; ♦ ® .ri + ' 1+w l -
b,(V - f) + Ii
Jr.F
+
J â.r
O. (26)
9. Show (using the results of the previous exercise) that when the body Force is constant, then e._=_ Tkk and en are harmonic, and that the stress and strain components are biharmonic. 10. This exercise considers the equations governing an imaginary, " quasiclastic," incompressible material that is indifferent to deformation but that resists rotations relative to absolute space. See the end of Section 6.3 for remarks connecting this material with the "ether_" (a) Lei the constitutive equation be T = kw, where Tis the stress tensor and w is the rotation tensor of (4.1.321). Show that the antisymmetry of implies the antikymmetry of T, and find a relation between the vectors (denoted by 12 and r) that are associated with to and T (b) Provide a physical interpretation of the constitutive equation in terms of moments and a "twist modulus" k by considering the special case 0 12 = -0 21 O, - 13 = WA = W23 = W 32 = O. (e)
Use
linearized momentum equations and neglect body force. Derive Ov
yt =-
k
2
V A i?.
an , t^t
1 =^ (Vnv ) ,
V•v =O, V - f2 = O .
(These equations have the form of Maxwell's equations for empty space!)
4.3 Final Formulation of the Problem of Linear Elasticity In line with our use of the linearized stress-strain law and the linearized relation between the strain tensor and the displacement vector, we wish now fully to linearize the governing equations of elasticity. A major step has already been accomplished—replacing the velocity v by the time derivative of
ei. 4.3] Final Format: imp of l ►1e Pr,,hJeni
eq
Linear nastway
169
the displacement, by ignoring the relatively small nonlinear terms in the material derivative. An additional step is necessary. We exploit the assumed smallness of the displacement u to approximate the density at the spatial point x and the time t by the density of the particle that was initially at x. (This particle will now, at time r, generally be quite close to x, since displacements arc small.) Material and spatial descriptinns of density have been denoted by (5. and p, respectively: r(A,
t) =
p[x(A. r), t],
rMA(x, t
t] = p(x, r)_
Thus our assumption can be written
p(x, t)
c5{x, 0).
(1)
Another way of stating this assumption requires introduction of the deviation p' from the initial density^(x=
r) =
plx,
t)
—
b{x, i)),
I iI C L
(2)
Substituting p(x, f) Mx, 0) + p(x I) into the rxlasS-conservation equation {2.1), we obtain, to first approximation [ Exercise 2(a)], —
cep
{fit
,
(`t1,(x, 0) + --+ r)(x, 0)
ei
C
r^tr
r^ x,
îor
— 0_
(3)
As will be clear immediately, p does not enter any of the remaining linearized equations_ -I - hus the displacement u ; can be computed separately and (3) can then he used to calculate the density correction p. This is seldom done, as this correction is normally of little interest. Still, a careful worker should at least use ( 3) to estimate the magnitude of li , in order to he sure that he has been consistent in assuming that this magnitude is small. We shall use the notation p(x) - b(x, t)). Thus pix) is the known density of the material that was initially located at x, and p(x, t) p(x)_ From now on we shall usually not bother to make the distinction between p(x, z) and p{x); p +mill h • regarded as the known Mina) devrsit y. SUMMARY OF CENERAL EQUATIONS. BOUNDARY CONDITIONS, AND INITIAL CONDITIONS
Let us now summarize the linearized field equations that hold in the interior of an elastic body R. The density p(x) and the body force per unit volume f,{x, t) are considered to be known. The linear momentum equations (2.2) are p (x) u ,
=
Î
c.'
Orr
(4)
Fâundafion.s of EiastiCily
170
[Ch. 4
The strain-displacement relations (1.32a) are
(5)
= ^1^i . 1 + ii j i) ^
and the stress-strain laws (2.5) are To. = 2pic ;i + 1.Ekk ^i i j .
(6a)
Alternatively, one can use the strain-stress equations (2.13) Ftij
— {l + '}T 1 w v7k
(6h)
^ . }
The conservation of mass equations need not be considered here since the density variation can be determined by{3)once the displacement n i(x, t) has been ascertained. We also note that the compatibility equations (1.43) will automatically be satisfied if the rig are defined as in (5). The boxed relations comprise 15 equations in 15 unknowns—the 3 displacement components u ; , the 6 independent components To of the symmetric stress tensor, and the 6 independent components r ; , of the symmetric strain tensor. In addition to these field and constitutive equations that hold in the interior R, boundary conditions must be specified over the boundary R. There are a number of important classes of boundary conditions. For example, over part of the surface, c?R 1 , the displacement vector u i may hespecified. Over another portion of the surface, OR 2, the stress vector r ; may he specified. Since = T", a linear combination of the components of the stress tensor is specified over the surface OR t _ There are ether situations that arise when only certain components of the stress and certain components of the displace-
ment are specified on the boundary. Complete formulation of the problem also requires knowledge of initial conditions_ Typically, u and au/Or are prescribed at some initial instant To summarize, a general initial boundary value problem of dynamic elasticity is to find z (x, i), £ 04DX, t), and Tutx, t), so that the boxed equations (4), (5), and (6a) are satisfied for x in a region R, r > O. Also to be satisfied are the following conditions:
0) =
uFox(x),
I x),
(x, Or t
ui(x} r) _ te i (x, r1,
X
1f(x, r)
x
XE R.
(7a, b)
r 7 U. E
aR 1 ,
U.
Here le', u;'', z , tr and also ff, are given functions. Furthermore, OR
OR
R.
Sec - . 4.3)
Filial formulurrulI i3 f Mt" Pro,'lem r,f
Linear L`'tu n( a's.
171
A general boundary value problem in static elasticity results when p (12 2r ; j[lt 3 is deleted from (4) and when the initial conditions (7) are also deleted_ In particular, for the stresses we have the equilibrium equations +j, =0.
(8)
NAVIEWS tŸ4)LlATLDNS
If boundary conditions involve only [he displacement vector u, it is probably best to have the governing equations expressed entirely in terms of this vector. To accomplish this, we write (6a) as = 1( uu1,, + u ,. r) 4- ).ir $,_ & ô,j _
Substituting for the stress tensor in (4) and collecting like ternis, we find
P
t,
LI,
u ^. J , + (p
[^f
+
1)ru3 ,^ +
.
(ria]
Equation (9a) may be written in vector notatinn as u
p
- 2 =
^
i^t
u + (p
+ ^.j grad div u 4 -f.
( 9 h)
In terms of the engineering constants, (9a) may also be expressed as
2(1 + v) us •;; + 1 — 2rr+'"
P r
+ It
(9c)
Equations (9a), (9b), or (9c) are known as Navier's equations. BE LïMAM1—%1rCHE[.t. EQUATIONS
Consider Navier's equations in the static case, for which (9c) can be written (using different subscripts)
+ v)
l
u4'"+ 1
-
2v^ i '^+
- ^ ^k
=
^
(10
)
As we shall now see, in this case we can eliminate the displacement and write the equations entirely in terms of the stresses. To do this we differentiate (10) and change the order of differentiation in the first term, obtaining
kk
2(1 + v) = ^} t+ k . , = V.
1 .
Ili
1 1
2v +. rk f -
The interchange of k and j yields 1
+1
2v u'''fk
+
2(1
+ v) E
fj. i = O.
(12)
172
l burui'urraas
of Elasticity
[Ch.
4
We next add (11) and (12), utilize (5), and observe that differentiation with respect to x i and xk can he reversed. We find, upon multiplying by E12, that F,B - • +
F —
1A + 2 ► ' Fii
(13)
(1 + u) (^k f ).k)= {]_
We can now use (6b) in (13) together with the fact [a consequence of (6b)] that (1 — 2074.
By this means we find that ç y+ 1 + /{ vTr p.nnL1jk + i iLk + (1 1.)(J*J +
(1 +
— O.
(14)
This result can be further simplified by first observing the result of setting j = k in (14):
+ y
f
y ^^,*.
Inserting this back into (14), we finally find the Beltrami Michell equations that express the governing relations of static elasticity exclusively in terms of the stresses:
1 + I
v
v c. ji+ j - _ v , .i
+.Îk.,+Î;. =a
(15)
There are six distinct equations, for the six distinct components of the symmetric stress tensor. Extension of these results to the dynamic case is requested in Exercise 2(b)_ The equations of (1 5) are identical with those of (2.26), which were obtained by manipulating the compatibility equations (1.43). Thus the BeltramiMichell equations themselves can be regarded as compatibility equations. When formulating the governing equations in terms of stresses one can, in fact, use the equilibrium equations* (8)and the compatibility equations in any of the forms (2.25), (2.26), or (15). Six independent equations constitute the compatibility conditions, and three more comprise the equilibrium conditions. It may seem that there is a danger of inconsistency, for only the six independent components of the symmetric stress tensor are unknown. in fact, we see here the danger of guessing at consistency by a naïv e counting of equations—for the reasonableness of the physical requirements suggests that it is not inconsistent to demand satisfaction of the nine prescribed equations for the six unknowns. Closer examination of the situation reveals more accord with "common sense" than first appears. There are nine partial differential equations for the six unknown components of T, but only three boundary conditions are • We consider here the equilibrium equations for simplicity. but our analysis can easily be extended 10 the time-dependent case.
sec. d_3]
Email Fo rrrrauttrnrr of Me. Problem r+} Linear Vanua)
1 73
obtained by prescription of the stress ii • T. It seems that there are three too many equations and three too few boundary conditions. But Exercise 1.15 shows that three: of the compatibility equations can be required to hold only on the boundary, without changing the basic structure of the problem. With this requirement, the problem takes the "sensible form" of six partial differential equations and six boundary conditions for six unknown 1unctinns_ (See Exercise 3 for an analogous case from fluid mechanics.)
E!iERCItiF-ti
1. Show that the stress vector t, acting on a surface with unit normal n, rs given by %,
t
2. (a)
(h) 3. (a)
ri ^. ^ t I— 1
. x+
2(1 + t•)
'
^
r,
Verify (3). Find Beltrami Michell equations for dynamic elasticity, in which the term is retained in (9). Let v(x)denote the velocity field in the steady motion of an irrotatronaI incompressible fluid contained in a region R. Justify the following equations: ❑ nV' - U.
v•
n —};
^^Y•
❑ •# =
n r[r = U.
l1-
tn. R. on r'R
(
( f gi ► i'n}_
l6a. h)
{ 1 tx. d)
a2R
(b)
(c)
The equations of 110 consutute a problem for three unknown functions t} r , u2, and t'3 with four equations and only two boundary conditions. Nonetheless, show that it has a unique solution by quoting an appropriate result from potential theory. . O. Now assume that Q 3 = f]2 = Û Let Q, = r1 ;c t•, t . Show that in R, Q3 = it on i'R. Prove that Q3 = 0 in R. (We now see that the problem can be put into the more conventional form with three equatrcjns
Qr =
0, Q2=0. V -r— Q,
in R;
and three boundary conditions
173 = 0 v - n = f'
Y- n do = O.
,
i
on
N
Note that the last condition can be regarded as sort of a 'compatibility boundary condition" for incompressibility.)
L 74
Frruadurir3as
LAUD-CIO'
[Ch.
4
4.4 Energy Concepts and the Principle of Virtual Work In Chapter 12 of I, on the longitudinal motion of a bar, we developed several energy concepts. A similar program can be carried out in the general case of linear elasticity, It can he seen that the basic ideas which we shall employ here are exactly the same as those employed in the one-dimensional case. The chief differences between the two programs arise frcrn the technical difficulties of working in three dimensions. We also extend the ideas of I by showing that in a certain sense the equilibrium configuration provides a minimum potential energy. E N E RGY BALANCE
We consider an elastic body with interior R and hounding surface OR. Recall that the rate of working is given by the scalar product of force and velocity. Thus, in the time interval O < t < T, the work of the external forces WE will be given by
dr. di ^- J
^Ÿf -
0
❑T
fJ.1 ti^j dr dt
(
I)
.
r7R
To compute the work of the internal forces for 0 ç t Ç I; we consider an arbitrary portion of R, which will be designated by D, and the corresponding boundary D. The ith component of the total force acting on D is given by
da ` D
OD
f dr + t^
1),N da -
JJJIJ D
ïD
Since D is arbitrary, we can conclude that the internal force per unit volume Thus the work of the internal forces, W r , is given by equals f; +
Pvz- Efif
ll i + i^tl üT
üt.
(2)
As in Exercise 1(a) and l(h), it can be shown that
ffl
7 i j ui
dT
=
d er -
Tii oi dr,
(3a)
''
rt
âR
and that Tj,
^ rj .
(3h)
Using the stress-strain law (2.5), we find [Exercise. 1(c)] that Taco =
-
Tr ( 2 etjii1 + dleki a/j).
(3e)
Sec, 4.41 Energy Concepts and the Principle of Virtual Mirk
175
We conclude [Exercise 1(d)] that Y■ 1 J
fT 0
(
1 dz + 5ff J R
1 ûi des
dt —
in 1( 21.m ,j F.k1 + xtkrFji)IÔ di R
ôA
(3d)
It is anticipated that the work by external forces will be transformed into work by internal forces plus a stored potential energy of strain. Accordingly, we define the change in strain energy as the difference between the work of the external forces and that or the internal forces. From (l) and (3d), we see that change in strain energy = fJI(2PEijEJ + A',c i,o») dtIQ,
(4)
Consequently, the strain energy, E, , satisfies EE(T) _ fir (2 .icii t,; + IÏtu riJ)Iô dr + constant. We set the arbitrary constant equal to zero, thereby identifying the arbitrary zero of strain energy with the strain-free state (E,; 0). Thus
j'j
f
'
1
Ar, k t: 1J) di.
(5)
In terms of the engineering constants,
E, 2(1 E+ v)
JJf(
aki ev + 1 k 2v tttii di.
(6)
R
By (2.22), 0 v 5 /, so that Er is nonnegative, as is at once evident from (5). [Note that co, cii = (402 .1 That any deviationfrom the unst rat ned stare Iead.N to a positive strain energy is an important fact: as we shall see shortly, it permits us to deduce uniqueness theorems for linear elasticity. To obtain a useful "mixed" expression for E, we use the stress-strain relations to write 244flJ ci! + kekA tii = Maki + AEkk 61 f)e;i --=
This gives T i t o dz_
(7a)
R
In principal axes, with principal stresses and strains Ti and r; , we have E, _ 5ff TA dr.
(lb)
176
Foundations of ' Rusticity (C'h_ 4
To interpret [7h), recall the elementary physics problem of determining the potential energy stored in a spring. Let the spring possess a Ilooke's constant k, so that the force F exerted by a spring stretched a distance x is given by F(x) = kx_ The work required to stretch the spring a total of s units is W(5) =
J
F dx = j kx dx = iks i isf•'(s).
s1
o
v
The correct result may be obtained by multiplying the total extension s by the average of the initial and final forces 0 and F(s). (See also Exercise 12.4.5 of I.) it appears that a time integration, as in (1) and succeeding equations, is the most convincing way to obtain the factor that must appear in (7). it is now manifest that the local potential energy stored in a strained elastic body, EE , can be regarded as being due to the stretching of three 1-lookean springs aligned along the mutually perpendicular principal axes_ This is no surprise, for we have seen that local deformation can be viewed as stretching along such axes. it is sometimes useful to define a stress energy E,,, which is merely the strain energy written in terms of the stresses. By applying the strain stress relations (2.13), we find from (7a) that
E
0
2E
v}T,•; T1 — vTi; T;i]
fir
dr.
(8)
R
Since Ei ^ Û Ed O. Let us return to the momentum equation (3.4), multiply through by Ai , ,
and integrate over the volume R, After a little manipulation [Exercise 2(a)], we obtain
= ffi idi7+
f11
itR
(9a)
where the kinetic energy K(i) of the material in R is defined by
fC _ i ffi pfû
,
(9b)
dr_
R
itut {Fxercise 2(h)]
fJf T,i 1ot =
(10)
R
We thus arrive at the energy-llence relation
dr^
dE + di
,^,^,^ R
^; ci
dr
+
t; ti; d o . au
(II
)
Sec 4.41 Energy Ci.»
pis and Hie Prr+u ip( u► ! i"irru al ^b'r.rl►
1
77
Thai is, the rate of change of total energy (kinetic plus strain) equals the rate of working of body , forces and surface stresses. A one-dimensional version of this relation (in the absence of body forces) is given in (12.4.8) of 1. PRINCIPLE OF VIRTUAL WORK
We now derive a useful relation for s[rni problem that is essentially a variant of the divergence theorem, Consider any set of stress components T, that satisfy the equations (3.8) of static equilibrium
+
f
—t]
in R.
(12)
No form of compatibility condition need hold, but the usual relation tj = n - T^ is assumed to link the stress tensor and the stress vector_ Let u# be any set of virtual displacements. These can he any smooth functions defined in R. In particular, the virtual displacements need not satisfy the Javier equations. We define a corresponding set of virtual strains, ej, • by the relations
= Yu'1 + rr
}
( 1 3)
j.
By standard manipulations
fff( ^; u*)^ H
t, u,* d o- _ ?H
r`R
= ff'fi;ur dr +
(l4)
? r u* r rfT ,
But using the symmetry of the stress tensor and ( 12), we can write
{[J J•
1
dr =
f tr^` t^r = — ff
f A u*
[1T
.
Interchanging dummy subscripts, we see that (14) implies that
Jfj i» dr + ff) tu* d o — ff.{ 7j; ^
/I
di
= JjJ(I,.0 + TWIN dT
F[
= iff T, i(u*
f + üj, i) d
-r.
Employing (l3), we obtain our final result,
JJJ x
fiut ch
4- ffil ur da =T'tri dt. OR
R
(16)
L 7li
Foundations fit f:7a.sticiry
!Ch. 4
Equation { l6) is known as the principle of virtual work.* When multiplied by 1, Equation (16) has the interpretation that the work of the body forces on the virtual displacements, plus the work of the surface stresses on the virtual surface displacements, equals the work of the stresses on the virtual strains. The factor of I is necessary for reasons discussed in the interpretation of (7l). UNIQUENESS THEOREMS
The energy-balance relation and principal of virtual work are convenient tools for establishing the uniqueness of the solutions to problems of elasticity. Let us first confine our attention to the general boundary value problem in static elasticity. Recall from Section 3 that in this problem u, is prescribed on M IL and t; on CR2, where L R , -- r1R 2 OR. We assert that the solution of this problem is unique. To prove this assertion, let us assume that there are two solutions, u;, Et. ) ; and 4. 771, c;} . Define "difference" displacements, strains, and stresses by It
u; =,
=
Tr,
= Tri — T7j .
(17)
From the linearity of the equations, we see that for x on r:'R
u; = 0
t ; = To ni = 0 for x can eR i ,
1- urthermore,
Tt = D
for x in R,
so the difference stress components T o satisfy an equilibrium equation in which body forces on absent. We can now apply the principle of virtual work to these displacements and stresses. Using (7a) we find that
2F,
11 1.
ToE, dr
t;11;
da +
[,u, da + fJjf,u, dr = O.
tut
sand u ; can only be a rigid body motion (see Exercise 1.5). Since t4, {] on ôR 2, the rigid body motion also vanishes and the solution must be unique, (If rR 2 is empty, the solution is unique only up to a rigid body displacement.) Now let us consider the general initial boundary value problem of dynamic elasticity that was described in Section 4.3. Again, let us assume that there are two solutions, er; and 147, and define difference displacements, strains, and stresses as in (17). Thus Since E( is norinegative definite, it follows from (5) that
= 0 on OR,
E IS
and t; _ Tl;ni = 0 on t1R2,
• We sec from (1e1l'isiificalion for the use of the word "virtual" in the Sense "in effect but not in fart." Equation [ I6J involves functions i1* and Ef., chat in general have no physical antcrpretation, but these functions have the effect of displacements and strains in a work balance_
Sec- 4.4] Energy Concepts an d the Principle
of virtual mirk
1.79
and
Pnï
`
To in R.
(The difference solution again satisfies a momentum equation with vanishing body force.) Applying the energy-balance relation to these: displacements and stresses, we find that dK rf t
dG r + =Û c[
[
c^ r
K + E t = constant. Now at time t — 0, x1,(x, 0) = i (x, 0) = to*, Oj = O. Thus the sum of the kinetic and strain energies must vanish, but, since each is positive definite, it follows that both K and E, vanish separately. From the vanishing of K we deduce that ri ; 0; thus the problem described by the difference in the two solutions is static. As in the previous case, the vanishing of E^ implies that the solution is tiniquePOTENTIAL ENERGY MINIMIZATION IN EQUILIBRIUM
The principle of virtual work (16) points the way toward important energy minimization properties possessed by elastic media in equilibrium. In our study of these properties we shall be considering minima among a For the remainder of the sec lion, certain class of virtual displacements these will be smooth functionst that are prescribed on a pari OR, of the boundary and that correspond to prescribed stresses [through (15) and (3-6a)] on the remaining part of the boundary r R 2 We associate a potential energy i'(5) with i as follows
t • u da.
VV(U) = Eau) — 55f f • L [fir --
(1 8 )
OR
Here the strain energy EA) is given by (5) with 1: ;1 = 1-(1-4i4 + Consider virtual displacements of the form iü = u + u''
(19)
where u is the actual equilibrium displacement and ü is a constant. Using (5) and (3.60, Tii
f AERk 15i»,
t As usual, We assume sufficient smouthc]css of 1unLLIL}ns and regions to carry nui required
manipulations.
Possession
on the displacements.
or three conisnunus duriv.sn ►'CS tur ns oui
to be a Sufficient candinori
^13fl
Foundations of Elasticity [Ch. 4
one can show (Exercise 6) that
(7, rîl9 Edu + Ou*) a^o
= fir 1 c* dz. T R
(20)
I fence
()
+ u*) =
firTij Eti dr — U
if
*
}'n*
d o- . (21)
J^
R
According to the principle of virtual work (16), the right side of the above equation vanishes. We have thus proved that the potential energy V(û) has a stationary value corresponding to the equilibrium displacement.f In fact V is a minimum at equilibrium, as we now demonstrate. We need not restrict ourselves in what follows to "near-equilibrium „ displacements. Thus we consider virtual displacements u of the form u + u* where u is the equilibrium displacement as before_ An essentially straightforward calculation [Exercise 8(a)] shows that E (u u*) — EL(u) =1ff
^
r.
di + Q(u *)
(22)
where ou*) - fiJ[p(Li:lj2 +
P(41)2 1 df.
(23)
R
Using the divergence theorem and the fact that To satisfies the equilibrium equations, we find [Exercise 8(b)] that
F L(u + u *) , E L(u) =
filfi ; it* drr + fif f i Lt* dz + 12(u*). iR
(24)
rH
It follows at once from the definition of V that V(u i- u*) — V(u) =
Q[u • )-
( 2 5)
Hut Q(u') is nonnegative and only vanishes when the virtual strain is zero, i.e., when u* corresponds to a rigid body motion. With no real loss of generality we can assume that the strains are prescribed at at least one point, so that in fact u'" = 0 when Q(u*) = 0. We have proved the following result. 'Theorem I. Of all virtual displacements, the true displacements (which satisfy the equilibrium equations) provide an absolute minimum to the potential energy. t Jh srarronariry of V can also be swdied by the Techniques
Eie rcisc 1J_3.26.
of variaiiurral calculus—see
Set. 4.4] Eflery}' Concepts [cnd the Principle of Virtual Work,
I hi
We now indicate the proof to a useful converse of Theorem Theorem 2. if a particular virtual displacement minimizes the potential energy, then it is the true displacement (i.e., it satisfies the equilibrium equations). Proof Suppose that for some fixed virtual displacement u, the set of virtual displacements u t u* has the property that fnr all u'
V(u + u*)
V(u).
(26)
It follows by another use of the divergence theorem [Exercise 9(a)1 that
—
'.;
+ put dr +
7;1 Ra i — Out d o + Q(us)
0.
(27)
The surface integral vanishes because of the boundary conditions. Thus, for arbitrary virtual displacements u*, we have +
dr + Q(u*) > 0.
(28)
Therefore, it must be that the equilibrium equations are satisfied: 1.1+
f—
(29)
to the contrary, that for sonie i
1.70 at a point and therefore* by continuity, in a sphere about this point_ (Reversal of the above inequality is handled similarly.) Pick a smooth virtual displacement, proportional to a factors, that is positive within thissphereand vanishes outside it. A contradiction can now easily be generated from (28). The reason is that the first term is proportional to E. but the quadratic second term is proportional to £ 2 and thus cannot cancel the first term if t is small enough. CI There are a number of other minimum principles in the theory of elasticity ., for example, the true stress minimizes a certain "complementary energy_" (See Sokolnikoff 1956.) Extensions can be made to dynamic problems by a version of Hamilton's principle (see Exercises I l .3.2 and 1 1.3.2U). Aside from their general theoretical interest, such principles provide practical means of calculation. We shall see examples of this in our study of variational methods in Part D. We note in conclusion that minimum principles play a vital role in the field of elastic stability_ A glimpse of the issues involved is provided at the end of Section 5.1.
I.
Fourzdrsriont of Elasticity [Ch. 4
82
ExEitCI5E5
1. (a) Derive (3a). (b) Derive (3b) by writing 441 as the sum of a symmetric and an antisymmetric tensor. (c) Show further that a(ct,E0) _ ^ t?[ ^t
J.
(d) Use the above results to demonstrate (3d). 2. (a) Derive (9a). (b) Derive (l0). 3. A cylindrical bar of length L, with generators parallel to the x 3 -axis, is oluriiform density p. Gravity acts in the direction of the negative x 3-axis. The bar is supported by a uniform vertical stress on the top surface x 3 = L and all other sin faces are free of stress. (a) Formulate an appropriate mathematical problem that should yield the appropriate stress tensor under equilibrium conditions. (h) Show that the stress tensor given by 0
0
0
U
0
(]
0 0
(e)
p9x 3
satisfies all conditions of (a). Find the corresponding strains and show that the displacements a re given by —
p9x3xr E
_ pg(xl
—ppx 3 x 2 11 2
^- y'x r
+
2E
}'xi
— L2
)3
There will be some arbitrariness in the answer to (c). Why does this not contradict the uniqueness theorem? Remove the arbitrariness by assuming that there is neither displacement nor rotation at {0.0, L). (e) To what shape is the originally flat top surface warped under the action of gravity? 4. Consider two equilibrium states of an elastic body R. The body forces h and surface stresses z, produce displacements u ; . The body forces f and surface stresses t' prcxluce the displacements u. (d)
Si e'_ 4.4]
blend Caneeps und Me l'rini'ipre nf Virruur WOrk
1E3
(a) Prove the reciprocal theorem of Betti and Rayleigh:
fIf' tittdc JJJJu
d
T=
dû-ieR
f51 f?w
d
(h) ff T, and Eli are the stresses and strains corresponding (of, and t, and Tr, and arehe stresses and strains corresponding Wirt and rr, , develop the following alternative form of the reciprocal theorem: fir dr = ffJT 'r° j€} x ^
ij dr.
5. The reciprocal theorem can serve as a computational device in a number of cases. Let us consider one such example. Let
f * = 0 and (a) (h)
1-fi
bir .
Find the corresponding surface stresses r*. Show, from the reciprocal theorem, that the total change in volume AV
- JiikidT
is given by
1 ^ 2v
AV ^ =
x
7k►
d=.
(c) Conclude that I 2v ❑ V = -- ^
,x, du f
—
OR
fIl
fx;d2
6. Derive (20) and (21). 7. By following the lines of the argument given in the text, but proceeding from first principles, show that the equilibrium displacement of a simple Nookean spring provides an absolute minimum to the potential energy. N. (a) Verify (22). (b) Verily (24) and (25)_ 9. (a) Derive (27). (h) Following the line of argument given in the text, provide a format proof that (28) implies (29) In particular, provide an explicit formula for a thrice continuously differentiable virtual displacement that vanishes outside the sphere. DO. Show that if body forces vanish and if displacements are prescribed over the whole boundary, then it is the strain energy itself that is minimized at equilibrium. -
FoUrtdarions of Elasticity [Ch.
I H4
4
4.5 Some Effects of Finite Deformation In this section we take a brief look at some effects that were exeludcd from consideration heretofore, owing to the neglect of nonlinear terms in our characterization of deformation_ We begin with another examination of kinematics, based on the algebraic "polar decomposition theorem." Our deeper understanding of kinematics is used in formulating a rather general constitutive equation. We use this to reinvestigate simple shear. Although consideration of this one elementary case obviously can only scratch the surface of the subject, nonetheless we can discern some important effects of finite deformation and we have laid the groundwork for further study. Our approach is based on the book of Jaunzemis (1967), which is a good source for some of the reasoning that we shall omit, and which provides a useful bridge to modern research in finite deformation theory. KINEMATICS
We showed in Section 4.1 that in the linearized case local deformation can he regarded as a superposition of a solid body rotation and stretching along mutually perpendicular axes. We now show that the same type of result is true in the nonlinear case_ Later we demonstrate that the linear result can be recovered as a special case of the nonlinear theory. We begin with (1.6) dx• = x ; JA) d A ; .
(I)
which we write in the lams dx, = f {A) dA,
or
dx = F- dA.
(2)
Here F is a common notation for the position-gradient tensor Vx. We assume that F is nonsingular, so that each initial line element dA is transformed into a unique line element dx, and conversely_ tinder these circumstances the transformation is characterized by the following theorem. Polar Decomposition Theorem. if F is nonsingular, then it has a unique right decomposition of the form
F = R •tI -
(3)
R'r = R ' ;
(4)
1 Pere R is orthogonal,
and U is symmetric and positive. (A symmetric tensor is termed positive if all its eigenvalues are positive-) There is a unique left decomposition of the form F= V •R,
(5)
where R is the same tensor that appears in (3) and if is positive and symmetric.
Sec. 4.5J Srarrre Etr•Crs of FtntiY iJefnrmarron
185
A transformation associated with an orthogonal matrix' such as R preserves length, for y t = R rj x j implies that
= RraR^k^xr = x k x
(6)
since Rrjpfk =
(R1.R)
Thus the transformation induced by R is a rotation and R is called a rotation tensor, Moreover. (I and V induce stretching (or contraction) along their three mutually perpendicular eigenvectors and are termed the right and left stretch tensors, respectively. Proof of the polar decomposition theorem. Define w by
iv;—F
.1 1.
W e k n o w that ►
= 0 if and only if Y = 0,
f 7)
since F is nonsingular. Thus the right side of 14'î 11 r
t, I r ik t: 1 l k
(8)
is a positive definite quadratic form, since (7) guarantees that the form vanishes if and only if L. — v t = r 3, = O. Associated with this form is a (symmetric) matrix C:
i.e., C = F T `F•
(F r ' )k
(9)
Since the form is positive definite, C has positive eigenvalues. In principal axes, C is thus represented by a diagonal matrix with positive entries. Let U be the tensor whose representation with respect to these same axes is a diagonal matrix consisting of the positive square routs of the corresponding entries for C. In these axes, and therefore in all axes. U has the properties
U T ' = 11.
(10a, El
u• I) =C.
Since U has positive eigenvalues, it is nonsingular. We can thus define a tensor R by
(I 1l
1,1 = F-
Clearly_ 13) is satisfied. Also R t::
R lr,R T (1) Y
}
i
r-
urthcug❑ nal.
for [using Exercise ? t 251 h11
FTF-F-Ü t =
,- fi As here, we Swatch frçcl} bciwcCn Ytrtso
U
,u (12
and the inalricci that represcnl them in s[io -ne
coordinate system. Thus R is a main. that gi c Lorrmpo11 r of R. and (H T ' R,, is t he fifth component of the matrix product formed by the transpose of R and R uself Carte55an
i'•iaurrdafions nf EfusriRtily
lC'h. 4
Suppose that there were two right decompositions,
F = R• U
F R- U.
and
(13)
We have FTr •
F = i1 i r
• R Tr •
R. lrl = U • U and similarly F i r- F = iJ - U.
By considering the equal symmetric tensors U. U and U F in principal axes, we see from the uniqueness of the positive square root that U = V. Now R = R follows from (13). Thus the right decomposition is unique. Suppose that F — V - P provides a left decomposition. Since P• P T` = 1, we have
= P • (p Tr
J'
.
I/
Fs)
( 14)
-
Because of the uniqueness of the right decomposition F = R • [:, (14) that imples
P IT• V - P
P=R,
ii.
(15)
Thus there is a decomposition of the form (5), and the required properties of V now follow at once from the known properties of i1 [Exercise I (a)]. L C is called the right Cauchy-Grew deformation tensor_ Comparison with
(1.10) shows that it is related to the material or Lagrangian strain tensor if as follows [Exercise 1(b)] : =
1(C — 1.
(1 b)
An analogous development can easily be carried out using spatial coordinates. We write
dx , where f-1 = A .,. ,
,
(17)
A tensor e is defined by _ { T Lr 1 cik = (1 1 )jk = JLl
{ ik
(IS)
-
This tensor is related to the spatial or Lulerian strain tensor z by E—
4I
I
-
C)-
(19)
But, using the chain rule, we find that
_ fixa
.2
_
cox ;
G * xLAt A —A <:I
rAjx) X^
= } f^x )• i
A= A410
Consequently,
F[A(x)] • f(x) = I,
ie.. f(x) = F `[A(x)]-
1201
lf we make the definition B(x) = F[A(x)] - F'r[A(x)],
(21a)
See - . 4.51
S wnw F/fects of Finite Deformation
LA7
it follows that fi r
•f =1F 'l'`• F ' = B
-
t? l b l
In terms of the left Cauchy—Green deformation tensor B, then, (19) becomes
f??1 From the kit decomposition F = V • R, one now can deduce [Exercise ?(a)] that the analogous formula to (leb) is B — V • V or, mere precisely,
B(x} — V [ A(x)] • !IAN)"
(23)
COMPARISON WITH I.I;4FAR THEORY
We now show that use of the polar decomposition theorem leads to a characterization of strain that, as it should, reduces to the results in Section 4.1 when strains are small. Let us briefly recapitulate our earlier results, being careful now to distinguish between spatial and material coordinates. We began with
d[l, = U;, r dA,
or
dt: = VU•dA
(24)
`which is 11.x(7), taking care to use material variables] Decomposing [lie displacement gradient tensor V U into the sum of a symmetric and an asymmetric tensor, we saw that small deformations could be regarded as the sum of a rigid translation and rotation plus a stretching along the three mutually perpendicular eigenvectors of the symmetric tensor with components (25)
)=, Comparing (2),
dx, = and (1.7), dx, = dA + Cl ; ï dA = {bk, + ti c )dA i , ,
we see that [he position gradient tensor F and the displacement gradien[ tensor VII are related by =à ;1 +V t.) or F= f+ VU.
1261
We write the tensors of the polar decomposition f = R - U in the following way_
R =i + fi<,
U=L+LF.'
(27)
Let us assume that the rotation described by R and the stretching described by [' are "small." We interpret this to mean that R and t) arc "nearly" the identity transformation, so that products of k and (J` can be neglected Do rust confuse the displacement vector U and the stretch tensor t1_
R urrdarirN1s of Ffrlstrriry [Ch_ 4
188
in calculations such as the following: ^=
R r rfL
= ^^
* R T' ^^i
Fo
+ Te `
+
R+
(28)
.
From (28) we see that in a linearized theory when corrections are entirely neglected,
kTr+k - U,
(29)
so that k is antisyrarnetric. Moreover, the symmetry of O follows from the symmetry of f7_ Thus since
F=R•fi= (1+R)•[f4. )=1+
+t1
(30)
+• ,
comparison with (26) shows that in linearized theory
Vu = k+Cf.
(31)
Decomposition of a tensor into the sum of a symmetric and asymmetric tensor is unique (Theorem 2.1.10), so it must be that R1J
i
(^} ,
.r — L r^,i 4 )
V =
(32a, U)
l ^ f lJ^ .t . ^
In the limit of small deformations, the two approaches thus yield exactly the saine result. We remark that the fundamental way to decompose a transformation is to regard it as equivalent to a succession et- simpler transformations. This is reflected in our discovery (for example) that F can he written as the product of the simpler tensors R and U. In the Linear theory of elasticity, however, it is sufficient to write a deformation as the Shim of simpler deformations. This is because one can analyze each deformation separately, after which (because the governing equations are linear) one can add the results to obtain an analysis of the full deformation. A CONSTITUTIVE EQUATION FOR NONLINEAR t'LASTICLT1r
We have introduced several related tensors-4 it, U L B, C each of which characterizes strain_ A constitutive law is obtained by postulating that the stress tensor T is a function of one of these strain tensors_ We are at liberty to select the characterization of strain that will lead to the simplest equations ,
—
for the particular class of problems under consideration. We shall confine ourselves to isotropic materials. Tu begin with, let us assume merely that T is some function of the position gradient tensor F. Because of our isotropy assumption, it can now be shown (Jaunzernis, 1967, p. 429) that T can be regarded as a function of B; not surprisingly, the rotational part of the deformation does not enter at all. The stretching is directly characterized by V but we work in terms of R = Itt • V The reason is that the components of B can be simply expressed in terms of the velocity gradient tensor, while this is not the case for its square root V
Sec. 4.5] Some Efidiyis of finite De'ft.rt aiksi
1{9
Making some restriction in generality, hut still retaining wide scope, we limit further consideration to polynomial equations of the form h
►
E c„ 8'".
`
(33)
n=e
One expects that for an isotropic material the scalar coefficients r- ; can only be functions of the principal invariants B, B 2 , and 23 of 8. This can be demonstrated (Jaunaemis, 1967. pp. 430 -32). Considerable further simplification can be obtained by invoking the Cayley-Hamilton theorem to write 8" as a linear combination of t, 8, and 82 . With this we see that f33) is equivalent to an assumption of the form ,
T=C'l+C .,f3+CLB z ,
(34)
For some constants co , C"1 , and C'1 that depend on B 1 , B2, and B. For what follows it is somewhat more convenient if we write powers of B as linear combinations of B - t , I, and B, and therefore assume that
T=
048
+ 01 - y,8-1
(35)
SIMILE SFIi*AR
To get some inkling of the possible effects of nonlineartty,* let us reconsider the case of simple shear, whose linear theory we discussed in Section 4.1. In an inverse approach to the problem, let us assume a standard shearing displacement of the form x 1 = A l + 25,4 2 , x 2 = A 2 .
)C3 - A 3
,
ti
a constant.
(36)
The displacements are [in correspondence with formula I l 47) of our linear analysis]
U 1 =2sA s
(^',
,
-
L' 3-p-
(37)
The displacements (36) follow from (37), but it remains to he shown that the governing equations are satisfied by (36). The components of the position gradient tensor corresponding to (36) are given by F; = cox ; fr A ) , i.e., by the matrix 1
F
-
0 0
2s
U
I 0 . (1 I
(38)
The components are all constants, so we have a case of homogeneous deformaiioil in which all measures of strain are independent of posit ion_ In terms of the component matrix T that represents T in the given coordinate system, the 4
Nonlinear effects of a diPfereni character aie discussed al the cnr4 of Seclron 5.1.
19-0
Foundations of Elastic* [Ch- 4
constitutive equation (35) thus becomes T =aB
that is, 1 + 4s 2 2s T^ 7 I 2.1
0
0
0
1 0 0 1 — 2s L] Q-4- f3 0 1 0+ y — 2s 1 + 4s 2 0 . 1 0 0 l Q 0 l
(39)
From (39) we see that the coefficients are functions of s r : a=
li =
a(x 2 ),
j(52),
Y = }'(5 2 )-
(40)
We also observe that the equilibrium equations in the absence of a body force are satisfied ( Ti . i = 0), since 1 is constant. In addition, we note that the components of the Lulerian strain tensor c = 1(1 — B - ') are given by Q
s
0
(41)
c , s —2s 2 u).
0
0 0
In linear elasticity, the Fundamental result concerning simple shear is that the shear stress is proportional to the shear strain through a constant elastic modulus. This is expressed in (2.16a) by Ti
(42)
= 2148 t2 -
Here we have, by (39), Tip = 2(z — y)s,
i.e., T12 = 2 sx le1z *
where gs 2 ) ; cc(s 2 ) — 16 2 )-
(43)
The nonlinear shear modulus At(s 1 ) depends on the square of a measure of shear. Since
{44}
{s') = {O) 4 O(. 2 ),
the fact that the modulus is nut strict ly constant is not expected to be noticeable until rather large shear is imposed. This expectation is in accord with observations. In contrast with linear elasticity, more than shear stresses are required to sustain simple shear_ We find from (39) that {A 3 =
a + ! + Y, T22
133 4 44E2,
T 1 3 = T31 ï Tz 3
=
= T33 + 4125 2 , Tll —
7 'i 2 = [ } -
It follows that 1711 — T22 = 2-ïri 2
.
(45)
Sec. 4.3]
Sam e f f}erts r ► j Firrite
Deforrrtution
191
This is a particularly "robust"' deduction, as it is independent of the response functions a, fi, and y. The normal stresses T, z and 7 clearly must be present (if s # Q). Moreover, they cannot be equal. Suppose that the material was originally a cylinder with a square cross section in the (A 1 , AA-plane. As shown in Figure 4.5(b), the cross section distorts into a parallelogram- If the sirCSSes on the deformed domain are as designated n Figure 4-9, one finds i Exercise 2(c)] t hat the shear stress S and the no rural stress N are given by _
1 1 + 4s
N _ 1.23 —
2N T — -
12
+45 2
S(46}
Thus normal as well as shear stresses must be imposed on the boundary to obtain a pure shear deformation. If normal stresses are not supplied, one expects some (positive or negative) dilatation of the material to result from attempts to shear it Such Kelvin effects are indeed observed.
^v
r, t I-. i e, u R E 4 9
11r'r,rnrui
and shear stresses cm tlrt< lures raf u sfieetred sy14eAre• he1 ir
.
For fixed thermodynamic variables, Iinear elasticity requires specification of only two constant moduli ). and . One of these can be found in principle from a single comparison between shear stress and shear deformation- By contrast, cur constitutive equation of nonlinear elasticity (35) requires determination of three functions (a, /3, and }') of three variables (B1, B2, and B 3 ). Because B 1 = B2 and B3 = 1 in simple shear, these functions can be determined only to a limited degree by a study of this prohicmEXERCISES
1. (a) Use (15) and known properties of U to show that Vis symmetric and positive. (b) Verify (16) and (19).
192
Foundations of Eiasririty - ICh. 4
2. (a) Verify (23). (h) Demonstrate (32b) using (16). (c) Verify (46). 3. Give full justification of the passage from (33) to (34). 4. Consider simple extension in which Xi
=
VA I,
x2 ; 0111A ,
x3
= VA3.
Here V is the imposed stretch in the A 3 - direction, and). characterizes the lateral contraction. (a) Find B and its invariants. Use this to write equations for the stress components. (h) Suppose that the material can be regarded as incompressible. Show that this implies that B3 = 1. Assume incompressibility and T22 = Q. Show that the applied stress impose the conditions T1 T31 and the stretch V are then related by —
T33 = (V 2 — V '){a — }'V
1 ),
where a and ' are functions of V - '(2 + V 3 ) and 11(2 4 V - ] ). 5. Show that the same general conclusions follow if B is replaced by C in (33).
CHAPTER 5 Some Examples of Static Problems i n Elasticity
I
4 we set up the basic equations governing the classical theory of linear elasticity and we examined a few of [he general properties enjoyed by these equations. It is our feeling that one cannot acquire any ideas about the import of these general field equations without experience in seeing solutions to specific problems such as those considered in the next two chapters. These represent hut a minute portion of the extensive literature devoted to the study of specific solutions of the equations of elasticity. Many problems of classical elasticity are far from easy and they have continuing interest both from the mathematical and the technical point of view. Our chief concern throughout the text has been to illustrate the making of mathematical models of problems that arise from other sources. It might possibly appear that the previous chapter has completed the task within the confines of the linear, elastic world. Hut we shall illustrate that such a view is illusory. Onc must examine each case to decide which dependent and which independent variables are actually in contention and how the boundary conditions, stated in physical terms, are to be translated into requirements on the displacements or surface stresses. The first section of this chapter is devoted to the theory of bending beams or bars. We begin by seeking an exact solution that will illumine the principal features of bending, albeit subject to some rather special assumptions. From this special solution we derive a relation between the applied bending moment and the resultant center-line curvature and we use this relation as the basis of an approximate "engineering theory of b ending" under rather general conditions_ Toward the conclusion of Section 1 we give some indication of the errors involved in our approximation. (It is quite typical of the difficult subject of elasticity, however, that full justification of various well-known and largely reliable approximate solutions such as engineering beam theory can be found only in the advanced literature, and perhaps not even there.) The section closes with a discussion of the buckling of a beam under axial compression. We turn in Section 2 to an analysis of the deformation of a cylindrical bar subject to torsion. Here our approach is to make certain guesses as to the general form of the solution, and then to solve for the remaining undetermined functions. This "semi-inverse" method only works for somewhat to CHAPTER
193
194
Some Examples of Statie Pr f kms in Etusricrty [Ch_ 5
special boundary conditions. As we shall see, however, by virtue of modern formulations of the classical "St_ Venant principle," it can he shown in some circumstances that the resulting error is negligible. Section 3 dells with classes of elastic problems that can be regarded as two-dimensional. An illustration of an averaging approximation is provided • this allows the assurnption of two-dimensionality for certain problems concerning thin plates. Solution of a particular problem illustrates the technically important fact that stresses are apt to reach surprisingly large valu es inthegbordfls,
5.1 Bending of Beams RENDING BY TERMINAI. Cpl PLES; FORMULATION
As our first example we consider the bending of a prismatic bar or beam by couples applied at each end. Suppose that the bar is of length L and crosssectional area A, and that a couple M = Mj is applied at the end x 3 = L and an equilibrating couple —M is applied at the opposite end x 3 = O. Let the X , axis be situated paral lel to the generators of the cylinder so that it passes through the centroid of the cross section. Let the x,- and x 2 -axes he along the principal axes of inertia of the cross section [Figure 5.1(a)]. -
1
■■■ ,■
•••••
••.
lb}
Fit; U a E 5.1 _ (a). Chaire of u x-rs bor investigation «f Mailing of bar of
ler+yfJ+ The linear distribution ol'erid trur -tro1rs Mat rlrurcu•teri.srrr L by couple M. (b) of ptire bending-
Se t, 3.1J
Br li dirrr; of Beams
L•]S
We digress to mention a kw matters connected with the rotation of a rigid body. Primarily, we wish to make sure that such terms as principal tries of inertia are understood. The terms we need can be defined in a purely formal manner. hut it is preferable to introduce them in their proper context. Recall that for planar regions A,
A
= loc i dx 2 and !vf, = A
x^f► +#r^
J..
A
give the (firs!) moments of area and mass, respectively, about the Jr{ axtx: ff 3
fi
—
1, 2. The centroid or center of mass has coordinates {.ti ,. M JJPdA
l2}
determined by
i 4I i.
I
Here = 1tiXr. r j j is the density, so M is the mass of A Suppose that a rigid body moves wa tt one point (the originj fixed. The inertial tensor N is defined as follows:
JJ [(x xX!
N—
xx .br+da.
A
Thus the components of N are given by POr
^P JJ
[(x
+ Xl I5 au — YrX ilk dkr f1X ^.
T
A In particular,
^ZI, f1X i efX z
i° A
is the second moment of mass about the -r r -axis. It is also called the moment of inertia about this axis. Mlorce]vçr. f Y t2
=
JJvi
l:iP i'L. i d1L •
A
is the negative of one of the products of inertia_ Since N is symmetric, principal axesof inertia can be chosen will) respect to which the products of inertia vanish. The components of N then form a diagonal matrix consisting of the principal moments of inertia The main applications of these ideas stem from the fact that the angular momentum of a rotating raid heidy is given by N • to. where at is the arigeilar velocity vector. See, for example, Jcffrc y% and Jeffreys (1962. Chap. 3).
We attempt to formulate a simple problem that will have an exact solution. Trial and error shows that this can be accomplished if the couple M is produced by a linear distribution of surface tractions,* as shown in Figure • Truc•rion is a syrurnym kir
stress
that is nflen used in elasticity theory
196
Some
5
Examples of Stark Problems irr Elvstieiry [Cit.
The stress boundary conditions thus are t t = 0,
on the lateral surface;
t 1 = t 2 =0, t 3 =ax 1 , r 1 = t 2 = 0,
(la)
on x 3 = L;
r3 = -ax 1 ,
Oil x3 = 0-
(lb) (lc)
The moment M about the origin that is produced by the tractions on the end surface x 3 = L must have the components M 1 = M3 = Q and M 2 = M. But
M1 _ J•f EUk r)Ik d o- , A
where and
r 1 = x 1 , r 2 = x2 ,
r 3 E L.
Consequently, since products of inertia vanish in the chosen coordinate system, M1 =
M3
= 0,
x2ax I
f O.
d
M2 = ff
X
do
— al -
A
Here fis the second moment of area about the x 2 -axis. To obtain we require that a= —
M
M2 =
M,
(2)
—.
I
Thus we see From (lb) and (lc) that at x 3 = L, 13 = I:11 s = {3 ^
J4flx 1 ^
I
3a)
and at x3 — 4, t3 = — T33 =
Mx 1 I
( 3 b)
SOLUTION
We now show that all conditions can be satisfied if the stress tensor takes the simple form 0 0 Q Ti _ [} o 0 0 0 —Mx 1 Jf
(4)
Sec. S. I J Bending of Beams
197
The boundary conditions on x 3 = 0 and L are clearly met, as are the equations of equilibrium (4.3.8) (in the absence of a body farce) O On the lateral surface n 3 = O. If we restrict Greek subscripts to the values I and 2. we see that the stress vector acting on the lateral surface is given by r, a
Tfn} = Te no .
This vanishes as required by ( la). The strains can be obtained from Hooke's law (4.2.13).
Ex,; = I l + ►')T,-, — 1. 7-4 05,
,.
We find that vMx, EC s-- -, EC22
i'Mx —
r
,
E1:33
—— f
=4, i#j.
(5)
Thus the compatibility equations (4.1.43) are satisfied [Exercise l(a)1. The displacement vector, p,, is then found by solving the first order partial differential equations (4.1.36) that express the strain components in terms of the displacements:
I
0
Mx1
vMxi
vMx, nt. —
U2. 2 =
El
1. 2 1" U 2. 1 = 11l,
El
11 3.3
•
142, 3+ 14 3. 2 = 0■
=
143 .
— T E!
1 + ti
^.
3 — O.
(6a, b, e)
I6d, e, f)
Integrating (6a), (6b), and (6c), we see that VMX
Id' =
142 =
AEI
+
vMxix2 El
1.02 ,x3),
+ jl(x t, X3),
(7)
x'' 3C2). u^ = — M Elx3 + M
To begin determination of the "functions of integration," u. v, and w we apply (6f), obtaining u 3 = ^.3(x2•x3) _ — u3.^ =
Mx 3
F.1
— K , Ix^, x 2)•
It then follows [Exercise l(b)] that w = xi
f(x2) + 9(x2)-
(8)
198
Some Examples
❑f Static Problems in Elasticity [Ch. 5
From (6c) we find that —
,,+f xifr(x2) — g(x2) = v.3(x1 , x3).
Hence, by a separation of variables argument,
f(x2) — Ax 2 + B and g(x 2 ) — Cx 2 + D. Combining
(7),
(9)
(8), and (9), we obtain ^ E^x3
u3
+ Ax ] x 2 + Bx ] + Cx 2 + D.
(10)
Now we can return to (6) and (7) to find
Mx3 u 1.3 !
r..i, 3(x2, x3) —
El
Ax 2 — B.
Thus
v11r1x i MA 2E1 + 2E1
Ax 2 x 3 T$x 3 } h {x 2 ).
We have a new function, li(x 2 ), to determine; but it depends upon but one variable. Similarly, utilizing (6e), we can proceed as above to obtain the following counterpart of (11): uz =
vhfx ] x2 El
Ax 1 x 3
—
Cx3 + k(x 1 ).
(12)
Finally [Exercise 1(b)]. appl ication of the same ideas, using (6d), yields U1
=
M(xl+ vxi-vx1) 2E1
14 2
=
wMx]xI —C.x 3 +Fx 14 +K,
t2 3
—
—
&x j --Fx2 +H,
El
_
Mxyx3
(13)
+Bx t +Cx 2 + D.
Only the constants B, F, H, C, K, and D are undetermined. We note that the linear terms that contain these constants represent a rigid body motion (Exercise 2). The constants can be determined, therefore, by adding the requirement that the displacement and rotation at the point x = 0 vanish. The first condition, u = a at x O, yields
Sec. 5.1 J Bending nj Bea n.s
1 94
The vanishing of the rotation W u at the origin gives rise to three scalar equations, 11 1. 2 — 11 2.1 = fi . u1,3 — 1 3,2 + = L} ,
114 3,1
— ^l =
0,
which, in turn, imply that
F=CH
o
.
Under these additional requirements, the final displacements (13) become u
M(xi + ux; — v4) 2E1
1
—
uMx 1 xx u3 =
u3
El
(14)
•
iM1x 1 x 3 = — El -
This important exact solution is termed pare bending_ LN"1"ERYRETA [ION
Let us examine the displacements (14) in some detail. The lines originally parallel to the axis of the beam, the x 3 -axis, will be called filaments. (This notion will soon prove useful in an approximation wherein the beam is replaced by a bundle of one-dimensional elastic Fibers that are capable of resisting only extension or contraction) Filaments Chat lie on the central plane, x = 0, undergo neither extension nor contraction. On the other hand, the filaments that lie on the x, < 0 side of the central plane will he extended (u 3 › 0); those on the x, > 0 side will be contracted (u 3 < a). We designate the coordinates of a point after displacement by x;, ,
x;—x,+ u,i(x)The x 3 -axis (x 1 = x2 = Ü) will be called the neutral axis. After displacement, we see from (14) that points on the neutral axis will occupy the positions x^
Mxi =
xz = f},
2E{
(15)
x r3 = x3.
Since the strain component E3 = u 3.3 vanishes when x 1 = L} we see that (according to our linear theory) the neutral axis is neither elongated nor compressed. Equation (15) shows that it is bent into a parabola that remains in the {x 1 , x 1 ) plane. Along the neutral axis, we also have ,
-
dxi _ Mx3
dx3
El
(16)
Some Exarnpfe s raf Static- Problems ïa Elasticity [Ch. 3
zoo
-
and
d 2x . dx 3
_
M
(17)
Eaf
From {14) and (16) we see that u3
_
--.xi
8xi(O,0,X3)
ex
(IR)
Throughout our whole discussion, it has been assumed that the displacements are sufficiently small that linear theory can he used. In this approximation the second derivative d 2x',{dx3 equals the curvature h of the neutral axis_ When it is desired to extend our ideas to include nonlinear effects, it tarns out that (17) can still be employed, provided that it is expressed in the form M Ï
1 R
(19)
where R is the radius of curvature. This is the Eider--Bernoulli rule for the bending of beams. (Note that the introduction aril allows this rule to beexpressed in acoordinate-fre,ernanner-) In the form M = EIR - ,the Euler -Bernoulli rule states that the applied moment M is directly proportional to the curvature R - ,with proportionality constant the flexural rigidity 'f. As expected, the rigidity increases with the stiffness (measured by Young's modulus F). The dependence on the moment of area I is explained by the observation that a bar will be more rigid if its material is concentrated tar from the center line, where the strain is greatest. LNTRC1Ut.[:TL(]N Ti] THE: ENGINEERING TLiE.URY O F BENDING— RAMC ASSUMPTIONS
We are now in a position to derive an approximate theory of the bending of beams in which the applied forces that produce the deformation are not purely terminal couples. We shall examine bending by lateral forces. (Our approach is in the spirit of the discussion of the lateral motion of a bar found in I Chapter 12.) A key assumption in the development is the use of the Euler-Bernoulli bending law (19). We recognize that this is a hypothesis that is somehow more restrictive than those already used in 0115- development of three-dimensional linear elasticity, since it has been derived for bending by couples rather than the more general forces considered here. A more complete theory of bending, which is technically known as flexure, is considerably more difficult mathematically. We shall not consider this more complicated problem but rather refer the reader to Sokolnikoff (1956). It should be noted, however, that one approach to flexure leads to the Dirichlet problem in two dimensions. This is essentially the same mathematical problem that arises in the next section when we analyze torsion. ,
See. 5. 1 1 Bending of Beams
2)L
We shall consider a uniform prismatic bar with the x 3 -axis passing through [he centraid of each cross section. The x 1 and r 2 axes are directed along the axes of principal moments of inertia of the cross section as shown in Figure 5.2, where ara I-beam section has been chosen For definiteness. For simplicity, we shall take the top surface to be horizontal, hut this is not necessary for our argument_ The cross-section area will be designated by A, the mass per unit volume by cr, and the mass per unit length, d 4, by p_ .
FIGURE
5.2. Di irihution a/ Alive cm the tap q an 1./war o
The problems that we have chosen to emphasize in this chapter conic from the literature of elastic equilibrium; that is ; they are independent of time. We shall depart from this program for the present example and consider time dependence. The analysis is not made much more difficult by this extension, interesting wave-propagation problems can then be considered, and the static equations will, of course, arise as a special case. We suppose [hat to the upper surface is applied a force that acts in the vertical direction only and varies only along the length of the beam. This force may then be specified in terms of its distribution per unit ierigrh, fix, r), where
fix,
= f (x 3 , I^[.
(20a)
The internal forces will be specified in terms of resultant forces and resultant moments acting on a section of the beam. Consider a portion of the
Some Examples of Static Problems in Elasticity [Ch. 5
202
f
.
1-
MO; r, +}
tit
r1l17. f+ r FIGURE 5.3. Force', and bending truurlrtttt arum; un arhitrur# +t•ynre+ft of a lu■rattt. 4surrt•iatexl uptll tln ttrtlsHle7iI rxrr she k i t (right} rnd of the bur ii v dut (rrrx.ti.ti) rr•/tre:sertriny head (mil) off to: rrrrlHl -. trr-r-rrrditict tr+ fllr rrgltrhuurl ride Am rr< trrr repres;krnarrr,.l u { raf+tmYlt ►
beam between x 3 � a and x 3 = b. as shown in Figure 5.3. The forces acting on the ends of the bar can be resolved into vertical and horizontal components. It then becomes clear that these forces sum up to a shear force Q, a bending moment M. and a tangential Force P. We assume that the latter is tern as we are restricting consideration to bending only. The shear force Q is defined in the following manner: Q(b, 1; + ) is the total vertical force exerted by the material on the side x 3 > h on the material on the side x3 < h across the cross section x3 = b;Q(a. t; --) is the total vertical force exerted by the material on the side x ^ < a on the material on the side x 3 >a across the cross section x 3 — a. Since vertical force acts in the x r direction, we may write (in terms caf the vertical unit vector ï shown in figure 5.2) -
l_'tlbt
Qih.r. +)— Ob. t)i_•
It can be shown IF.xercisc 7) that. for arbitrary b, Q(h. r: —1 = —QV?, r: +) _ —Q[h, In similar fashion we can define the bending ftputreeni M as follows: M(h, t; •1 ) is the total moment exerted by material x 3 > h (In the material x3 < h across the cross section x 3 -= b; M(a, t: ) is the total moment exerted by material x 3 a on the material x 3 > a across the cross section x 3 a. All the internal and external forces are assumed to act in the (x r , x3)- plane. —
• The (*uanlily Q(b, t) introduced in (206} is the Magnrtudc al the force (IOC tO the stress ttx 1 , .x 3 , b, i;
i2(h. r) _
JJ T1(xi.x.k iidx 1 di. .4
net
or resealwii vrrItcal
s .rl 8rnr1"fry
13rurna
201
Cooseyuently, we may write ; + j = M(6,
taj.
(1a)
Moreover (Exercise 7). m(/), r; - )
- Mih. t; +)
^bl(h tij.
(21b)
The various shear forces and bending nieinents are shown in Figure 5.1 We shall new assume that the displacements fulfill the following assumptions. (al The center line moves only vertically with the displacement n 3 , r (ln particular, u 3 — Q for points on the center line.) (b) All vertical displacements are the same as that of the center line_ (c)There is no displacement in the x ; -direction. (d) Plane sections that are normal to the conic' line before deformation remain plane and normal to the deformed center line. We shall shortly provide a rationale for these assumptions based on what we have previously learned about pure bending by terminal couples. For the moment, lei us observe that the first assumption is a consequence of the pure 'bending result WI. Assumptions (h) and (c) arise from the "thin" geometry of the heanu since its cross section dimm:nsions are small compared with its length, lateral contractions produced by longitudinal extension (Poisson ratio effect) can be neglected. Assumption (d) can be interpreted as saying that the deformed surface which corresponds to the cross section x 3 = constant is replaced by its tangent plane at the point of intersection with the deformed center line. It is left as an exercise to show that in pure bending the normal to this tangent plane is parallel to the tangent vector of the deformed center line (Exercise 5). The nature of the first two displacement components can immediately be determined from assumptions (a), (b). and (c): -
u,
]'(x3 • r),
it s = O.
Let us turn to the slightly more difficult task of characterizing u 3 . To this end, consider a point Q on the undeformed fiber BQ with equation x i = x ? . jSce Figure 5-4. Note that since the x,-axis points downward, the fiber depicted must be associated with a negative value of 4'".) The displacement of Q can be described as Follows_ (ii) By assumption lb), Q suffers a vertical displacement QS that is equal in magnitude to PP From this we deduce a result that we shall need in a moment : The length of SP' is — x;°j. (ii) The horizontal component of displacement, SQ', must be such as to transfer Q to a point Q' on the straight line PQ' that is the image of PQ [ by assumption (d)]. If the angle at P' of the triangle SP'Q' is denoted by AO, it follows that u 3 (.ti 1 . x 3 . r) = — x, tan All .
20 4
Some Exam ples of Static Problerns rn Elurticity [C'#+.
5
i+ j(Q)
—
x a)
Flu u K r 5 .4. fle}rirmatrrur of a small srymen r of u hnarrr ac•cr ► rdir# to the —
engineering erürrJ rheorti•_ ••
The minus sign arises in a situation wherein we have chosen the xi -axis to point downward and have drawn the deformed center line with positive curvature. Other choice4, however, consistently analyzed, will lead to the same result [Exercise 6(a)]. Since AP'' IS a right angle [assumption (d)], we see that tan Ail Summarizing our
ay(X, , t) ^
rx3
information concerning the displacement vector, we write
= y[k, , r),
11 3 =
tr 3 = — Y 1
r1yfx3 if
.
t1
(22
)
Now we can give further support to assumptions (a) to td). We argue that the qualitative Features of pure bending should apply locally in thin bars, for here the detailed distribution al forces should not matter. And indeed, equations (22) are the same as those which would be obtained by assuming the displacements of pure bending (14) and setting the Poisson ratio ratio v' equal to zero [Exercise 6(b)]. EQUATIONS OF ENGINEERING BENDING
THE a
Having described the displacements and the stress resultants, we are in a. position to write down the equations of motion for a portion of the beam taken arbitrarily between x 3 = u and x 3 = b, as shown in Figure 5.3_ Imposing linear momentum balance, we have t;
^
) + 11(xj, r) dx 3 + Q(h, r; +) =
^
rff
dx, dxi dxxQ(a. r3r2
A
(23)
Sec. 5.11
Benriin,y ui #rrurrrs
205
From (22) we see that u 2 vanishes while u 3 is an odd function of x i , Thus the choice of the x 1 - and x 2 -axes passing through the centraid leads to 2u
3
ylx 3. r1
dx z dx 2 = A
1r
, u.
124a)
A
Consequently, using 120). we find that
f (x 3 , r) rix a
(.4,
+^, ^( I)
QUA, 1) = a ri
r^ 2 }d-^ ^, 1) #ry r
.0
v
dx 3
t
(24b)
This can be rewritten as
dQ(x 3 . r) . /'[x^,r)+ rx
r12 ).lx,
p
l} â
{).
3
w
Since the section between x 3 = a and x # = b has been arbitrarily chosen, we may apply the IDubois Reyrnond lemma* to obtain the partial differential equation describing Ow balance of linear momentum (70.x
r)
{x t) = p
3,
xa
(J t JAX3, rl f,+2
(2i)
Next we write down the conscqucncesof the balance of angular momentum, where the moments are taken about the point x 1 = x2 = 0 x 3 - u- In our calculations, we shall designate by h the width in the x,-direction of the flat, top ; udace on which the external load fix 3 fi is applicd.t Equating the rate of change ill- angular momentum to the torques acting, we thus have. ,
,
f
M(u, r; —) +-
4
(bk
NO, r; f ) f
[lk } A
2
lx - ak) A h
Q(b, r: +- ) =
(x
dx 3 dx 2
ak) n rr
[^U
u
I e^x t dx 2 [!Ji 3 . 1
A
{27) W e now have to calculate the v ector products in (27) and simplify the integrals. W e find that L^ df (x -[dc) n
f^
l!
try
_ ^ ^ x^ r^t ^ r^xs
d 3y ^ 03yy0 ) 3 + x r r7t 2 Px + (x } y u dt' ^
02, — kx2 ^ -
• In the prrSencc
of dascon ITrlulitcs, ;he procedures must he altered. Soc per unir kngxh, so that
t Remember that I wds defined iu be then force
area_
(28) Exercise 12 the force per
Wilt
Some Examples of Static Problems il Elasticity (Ch. 5
2o6
Since the x,- and x 2-axes pass through the centroid of the cross section and are directed along the principal axes of inertia,
fJ
f
J'
x i xi dx, dx 2 = 5x 2 dx, dx 2 = 0,
Jx dx, dx 2 = 1.
A A One can then show that only the j component in (28) does not vanish [Exercise $(b)], and we obtain
— M(a, t ) + M(b, r) + f(x 3 — a) f (x 3 , t) dx3 + (b — a)Q(b, 1) a) 1 at ^^x+A(x3— x3
t
(29)
2 dx3-
As in the development of (25), (29) can be rewritten in terms of a single integral: [(
Q(x3,
x3
r)]
y —d 1
de e x
3
+A(x3—a) -
dx 3 =0.
(30)
The Dubois- Reymond lemma may be applied again, with the result that
â49Q (x3 — a)f(x3 , t) + (x3 --a) + Q(x3, t) 3 =a1 r
x +P(x3—a) d3
(31)
Multiplying (26) by (x 3 — a) and subtracting the result from (31), one obtains DM ^x
pl
^3y
+ Q(x3^ r) — A &2 ôx 3^
(32)
Among other consequences, (32) shows that the shear force Q can be found from the bending moment M and the deflection y. The shear force can be eliminated completely by differentiating (32) with respect to x 3 and subtracting (26), with the result
em
P
2 a4Y 1Or 2 - PK dt 2 3X 2—_ ex2+
49 2y
(33)
Here x is the radius of gyration defined by x2 - 1/A. Our problem has now been reduced to a single partial differential equation (33) connecting the bending moment and the deflection, i.e., one equation with two unknowns.
Ser' 5.1 j
f3e•ndinr) e}J
Beano
207
As in the case of the lateral motion of a bar, we are missing a constitutive equation connecting t h e stresses with the strains. We shall assume for this relation the linearized Euler Bernoulli law (17) developed for the problemil of' pure bending. For small displacements and slopes, we thus impose the condition -
hi = ^1 { ^ PA
134)
Substituting (341 into (33), we obt ekrr the final fourth order partial difterential equation for the displacement y(x 5 , r)
E
(35)
We can obtain the equation for static deflection of a beam by suppressing the time dependence in (35). For this case the displacement itx j ) depends on the force distribution f (x3) according to the equation
y El
(3)
ds 43 = J:
The second term in (35), due to Lord Rayleigh, represents the effect of rotary inertia, as can be traced from the derivation [Exercise Hlcl]. In many applications, its effect is small and it is usually neglected. Under these circumstances, (35) is replaced by f^d }'
^] 4 + C^X3
t•
y
(37)
p ^' — — ' 1. i i
Looking back, we can obtain more insight into cur approximation procedure_ From the footnote to (20b) we see that A ()(k, t) can be regarded as the average vertical force acting at the section x. Similarly, A - 1 M can be regarded as the average j-component of the moment vector. Nonvertical stress resultants and non-j components of moment resultants are completely . ignored. We wish to utilize only the average quantities Q and M. Thus in (23) and (27) we can impose linear and angular momentum balance only on arbitrary vertical slices cf the bar, not on completely arbitrary regions as required by an "exact" theory. It should be a good approximation to replace a function that does riot vary appreciably by its average, so the use of Q (fur example) should be appropriate if stresses do not vary significantly across a section of the beam. Such arguments again lead to the confusion that our approximations should provide useful answers for beams whose cross-sectional dimensions are sufficiently small. 'L
208
SOFTre f.xurr2ples uif S[u1 w
Problems in Elasticity [Ch. 5
BOUNDARY CONDITIONS
If the beam is of finite length, we must append approximate boundary conditions to the differential equation (37). Since we have reduced the geometrical freedom to the deflection of the center Tine only, and since the precise stress distribution is replaced by stress resultants, we cannot expect to formulate boundary conditions by prescribing components of displacement or components of traction at points of the boundary surface. In place of these, our boundary conditions are given in terms of the deflection or the stress resultants. Because the governing equation contains fourth partial Jeri •atives. we shall need two conditions at each end of the bears. We shall consider several of the standard boundary conditions that might occur at the right end of the bears, x 3 = L. It will be clear that in most cases exactly the same boundary conditions hold at the left-hand end of the bar. Possible exceptions are the subject of Exercise 13(a). The end x 3 — L is said to he built in lilt is so restrained that there is no deflection and the slope remains zero (Figure 5.5). Thus the boundary conditions become
built-in :
y(L, r)
ay (Z ' f)
r FIGURE
,
r,'x a
l^.
( 3 8)
.
A buifr -in encl.
A simply supported end is one in which both the deflection and bending moment vanish, as would occur in the case of a hinge [Figure 5.6(a)]. In terms of the deflection, this boundary condition is described by simple support: y(L, r)
d2
t) '
= 0.
(39)
[The vanishing of the second derivative follows from the Euler Bernoulli law (34).] It should he noted that both boundary conditions (38) and (39) remain the same whether the problem is static or dynamic. The diagram depicting the simple support is frequently simplified to the form shown in Figure 5.6(b). There are situations that arise in practice in which the applied loads are not transverse to the beam and hence have an axial component. -
Sec. 5.!)
Bending n/ beams
209
^.x
r^^S^^^r^rr^ #d)
FIGURE i,6. Various typ es of simply step forre1! end (a) .4 hinged end. Fihereui and bending moment r-ettti.sf+. th) A stntpfrftecf representation rif the hinged ord. (C) A support f1ie'tr ran srefrfiih• an axial reaction. (d) support Mai freely permits r1-vial mutton deffec•tiem
The simple support may be capable of opposing an axial force, as indicated in Figure 5. 6(c), or nor, as shown in Figure 5.6((1.)_ The third important boundary condition is the free end. In this case, the bending moment and the shear force vanish, or
PAL, r) = Q(L, t)
—
U.
From (32) and (34),
3'
t) __
N
13 y
A ôr2
f
x3
D'y Y^
(gai
Thus the conditions foi a free end may be written in terms of the displacement t) as -
y(x ; ,
c2y(L, :i)= 03C1
t,
a3 t) A Di ox 3 p
E
Oa, t)
(41)
r7x3 — U.
Boundary conditions (41) take into account the effect of rotary inertia. if this is neglected and (37) is used as the governing partial differential equation, then the corresponding boundary conditions, replacing (41), would be (12 y(L, t) = û , dx1
0 3 y(L, t)
it should be noted that boundary conditions
0.
(42)
(42) also represent a free end in
the case of static deflection. Another useful boundary condition is the one that corresponds to a concentrated force F at the end x 3 = L (Figure 5.7). ln this case the bending moment will vanish and the shear resultant at the end mist be in equilibrium
Some Examples of Static Problems in Elasticity [Ch. 5
210
i
{
FIG U RE 5.7_ .4 free end sichfeet to a concentrated force
F.
with the applied force. More precisely, by examining the forces on a slice of the bar between L — A and L and noting that momentum and external load terms are negligible in the limit A --> O. we see [Exercise 9(a)] that the force boundary condition requires that (43)
Q(L,t;—)+ F =O.
Since F = Fi and Q(L, r ; —) _ --42(L, r)i, (43) can be replaced by the scalar equation
Q(L, r) = F.
(44)
In terms of the displacement. the force and moment boundary conditions are
fil y(L, t_) ^ A [^1 2 ex 3
a3 AL, r}
F
e2AL' r^ ,
eC3
=
^.
(45)
If either the effect of ro tary inertia is ignored or the problem is independent of time, the boundary conditions are simplified to 2 e3 t, t) (46) — El - >L" t) = F. = Q, ex3 ex; Another way to obtain (44) is as follows. Consider (24b) for the right end section, for which b = L. We must set Q(L, t) = O, for there is no material to the right of x = L. On the other hand, we must add the concentrated force term F to the left side of the equation. Upon taking the limit a L, we obtain (44). In like manner, if there were a concentrated force G at x 3 = K, the left end of the beam, the corresponding boundary condition would be Q(K, t) = —G. The condition of vanishing moment similarly follows from (29). TRAVELING-WAVE
SOLUTIONS
We can gain some insight into the nature of the beam motion by examining a simple harmonic wave solution to the governing equation (35), for the homogenous case that occurs when f is set equal to zero. Following the standard procedure (as in 1, pp. 37881), we consider y(x3, t) = exp
`(x3 — cr)].
(47)
Sec 5.1 .1 Besdt,k of Beams
2i1
Here c• is the velocity of propagation of the wave and A is the wave length. Note that the wave propagates along the axis tithe bar, just as in the case al longitudinal motion. On the other hand, the displacement is in the i- direction normal to the axis of p ro pagation. This wave is said to he transverse (in contrast to the longitudinal wave of I, Chapter 12). Substituting (47} into the equation of motion (351, with f set equal to zero, WE note that there is a Cornrnon exponential factor in every terni. 11 this factor is suppressed, the following algebraic equation remains: 4rt 2 EIJ -; — pn 1 47*. 2
2 c-2
--
p; 3 .
is
08)
Solving for the speed c, we find that
4rr2E! c•2
p( J. 2 + 47( 2
(49) }
Equation (49) includes the effect of rotary inertia. In the event that this is neglected, the wave speed is given by
r =
2n^:'! A p
—.
Observe that the influence of rotary inertia depends upon the ratio of the radius of gyration of the cross section to the wavelength of the transverse wave. Unless this ratio is large, rotary inertia is of little significance in determining the wave speed_ On the other hand. rotary inertia must be included for short wares whose length is comparable to the radius ofgyration or smaller. Lt should be pointed out that in those situations where the Rayleigh rotary inertia term is of some significance, there is ail additional term that has not been included which is of the same order of magnitude. The further term takes into account the fact that in flexure there is a distribution of shear stress across the cross section; the assumption of plane sections remaining plane must be changed. This effect has been analyzed by Timoshenko, who found that it can be represented by a modification of the coefficient in the rotary inertia term together with an additional term in the differential equation proportional to 0 .1 .4 /01 4 , See S. Timoshenko. Vihruta'n Problems in €nqureeririy, 3rd ed. (New York: Van Nostrand. 1955. pp. 32e1 31. 334 35), A final observation applies whether or not rotary inertia is considered. We recall from I. p. 377, that the speed of propagation of longitudinal waves depended on the physical constants of the bar only- In the case of transverse waves, we see that the spedd of propagation r depends not only on the properties of the beam but also on the wavelength of the particular wave, 1. Such dependence of the wave speed on wavelength is called dispersion. To examine dispersion more closely, we differentiate e 2 in l49i with respect
12
Some
Examples of Static Probkms in Elasticity [Ch.
5
to A and find
dc 2
— 4rr 2 E}
dA $
p(^ $ + 47z 1 rc 1 ) 2-
(51)
Th u s
and the dispersion is anomalous in that long waves travel slower than short waves. The effects of dispersion are given detailed treatment in Part C. Calculations are largely carried out for water waves, but sec Exercise 9.1.8. Additional interesting solutions to the equations we have derived are the subject of exercises. BUCKLING O F A BEAM
If a long, straight elastic beam is gradually subjected to a greater and greater axial load P. a critical load P = P1. will arise at which the column suddenly will deform into a bowed state. This phenomenon is a prototype of elastic buckling. One expects that the governing equations of the problem will always be satisfied by an equilibrium statecOrresponding to a straight, slightly shortened, column. For P . P1 , however, it is anticipated that this state will become unstable to any small perturbation, resulting in the appearance of the bowed equilibrium state. It is therefore the case that buckling cannot be described by the classical linear theory of elasticity; in that theory, as was shown in Section 4.4, there is only a single possible state of the column at equilibrium. To give some idea of the issues involved, we now give one formulation of the problem that gives rise to a prediction of the critical load Pr at which buckling ensues_ For a far more comprehensive introduction to the important topic of elastic stability, the reader is referred to C. L_ Dym, "Stability Theory and Its Applications to Structural Mechanics" (Leyden: Noordhoff International Publishing, 1974). Although we have proved it (in Section 4.4) only for linear elasticity, we shall take as the basis of our presentation the assumption that equilibrium is characterized by an extremum of potential energy. In our analysis we shall continue to employ the approximate theory of bending that we have developed in this section, As mentioned under (22), our approximate theory gives results that are identical with those that would be obtained if Poisson's ratio is set equal to zero. With such an assumption, equation (4.4.6) for the strain energy gives
E. =
J)Lj
f f
:
d
i
.
(52)
Sec. 5.1]
blending oJ
2t3
We suppose that the buckling is associated with bending in the x i -direction. Using the expressions in (22) for displacemen ts, we thus find that the only nonzero strain component isr — —x i s". Consequently,
E, = 1E1
(}"j' ra?.x 3
(53)
c3
where, as before, I is the second area moment of the cross section about the x2 -axis. Suppose that there is a normal compressive force P at each end of the beam but that the sides are stress-free- Body forces are neglected. Then. according to 14.4.18), to obtain the total potential energy we. need only supplement the strain energy by the term r .,. Pr+ 3^
- ff rtr da =
P V
s= ^ i
^ k3
I.
( 54 )
R
Here we use the facts that
t(i i ,
=
 ^
t.(x i ,x 3 ,L,i1= — ^i ,
where A is the area of the cross section. We now utilize a nonlinear result to rewrite (54) in terms ay. In doing so, we make an additional constitutive assumption, namely, that the column is of such a material that it suffers no change in length during the deformations under consideration. Our variational principle concerns a given substance, so we use material coordinates and obtain from (4.1-10) that on the centerline 2 /13 3 — , 1.e.„ 2f13. 3 + (U 3.3 )2 ± (iLI s 3 ) = Û. (55)
❑
3y 3 , we obtain to first
Solving for the small quantity
b3.3
— i(^Ial; _
approxinisition
-102 -
(56)
Note that we are indeed taking nonlinear effects into account here, for if nonlinear terms were neglected entirely in (55) we would obtain U3 O. In the purely linear case, then, the contribution (54) from the compressive force to the potential energy is negligible. Employing (53), (54), and (56), we find the following expression for the potential energy V:
V(y) =
J
r.
[EI(y,.)3
—
P(y)1 dx 3 .
( 5 7
For definiteness, let us also assume that the column is pinned at both ends, so that
},{0} — Y ,}( 0) = JAL) = y.,( f,) = f1-
(58)
214
Some Examples of Static Problems in Elastievly [Ch. S
One can then show (Exercise 4 or Exercise 11.3.1) that the condition of extrerial potential energy leads to
El
^*
_ ^ 0.
(59)
The eigenvalue problem formed by (59) and boundary conditions (58) has only the trivial solution unless P=P
„
-
n'Pc ,
Pc = ir 2 EIL - '
(60)
in which case w
(tut);
== Cr sin - — , c.„ arbitrary. f
(61)
It appears that a natural candidate for the critical buckling load is P 1 Pc , for the first sign of nonuniqueness appea.rs here_ In considering how to proceed further, one might conjecture by analogy with point mechanics that equilibria correspond with statienary points of potential energy, but only energy minima correspond to stable equilibria. One might also conjecture that the arbitrariness in amplitude found in the buckled solution at P I would disappear in a more accurate theory. These conjectures are correct, as can be seen by consulting p]ym (op. cit.) or the literature cited therein. Some idea of the issues involved can be gleaned from the beginning of Section 12.4. Here arguments are given to demonstrate that, for a single particle subject to a potential force, an equilibrium point that corresponds to a minimum in potential energy is stable in the sense of l.iapunov. YAR1ATlUNAt.
METHODS IN ELASTICITY
In Chapter 11 we introduce the calculus of variations, a subject concerned primarily with the extrcmalization of integrals under various constraints. By applying variational techniques, one can show that the virtual displacement which minimizes the potential energy satisfies the correct equilibrium equations and boundary conditions (Exercise 11.3.26). Moreover, the same techniques can be used (Exercise 11.3.2 7) to demonstrate a generalization of Hamilton's principle of particle mechanics: The virtual displacement that cxtremalizcs the time integral of the difference between the kinetic and potential energy in fact satisfies the differential equations (4.3.4) of dynamic elasticity. The variational formulation of elasticity provides a useful method for obtaining appropriately simplified problems in special cases. This is illustrated in Section I 2.4 where the differential equation governing a flexible membrane is derived by assuming a sensibly approximated form of the potential energy_ Variational formulations of elasticity find another important application as a computational tool. If the correct displacements are known to minimize
Svc_ 5
- (
I
/knifing
of Bums
215
some integral, then a finite linear combination of suitable functions can be inserted into the integral and the constants of combination can then besought that provide the smallest possible value to the integral over the class of functions under consideration. The problem is thus reduced to an algebraic one that is well-suited to modern computers. This approach is illustrated at the end of Section 12.4, in connection with the torsion problem. For further discussions of variational methods in elasticity, see Weinstock (1952, Chapter 10) or Sokolnik off (1956, Chapter 7). EXERCISES
1. (a) Show that the strain tensor given by (5) satisfies the compatibility equations. (b) Carry out the detailed calculations that lead from (7) to ( 13)2. It was remarked in the text, following (13), that the linear terms in these equations represent a rigid body motion. Show this to be the case. 3. Consider a beam of square cross section bent by terminal couples_ Utilizingthe displacements given by (14), sketch the deformed middle surface, x , ï 0, Also sketch the boundary of the deformed cross section given by 3 = constant. 4. In (57), let y = yo + Ey, satisfy the boundary conditions (58) for all L. Show, using integration by parts, that the condition
d
V(yo -4- Ey
z
implies that y o must satisfy (59). t5. In our discussion of the geometrical approximation used in the approximate theory of bending, we stated that in pure bending by terminal couples, the normal to the deformed cross section at the neutral axis was parallel to the deformed neutral axis. Prove this statement. 6. (a) Suppose that the deformed center line has a curvature opposite to that shown in Figure 5.4. Show that it continues to be true that rr j = , (1 34:3, ilat- 3,(h) Prove the statement made after (22) that the displacements given there are the same as those for pure bending if the Poisson ratio is set equal to zero in the latter case. 7. Establish (as in I, p. 361, or from properties of the stress tensor) that for arbitrary b, Qlh.z: +) _ — Q(b,r; $. (a) Show the validity
(b) (c)
—
),
M(b,i; +)
—
-- M(b,i;
—
}-
of (24a). Complete the calculations needed to derive (29) from (27). By tracing its origin, show that the second term in (35) represents the effect of rotary inertia.
2115
Same Lxa»rpits of Static Problems in Elostrkiry
[Ch. 5
9. (a) Derive boundary condition (43) by the first method suggested in the text_ (see F igure 5.8). (b) A beam carries a heavy particle of mass in at x,
—L
.t 3
L FI[;I:RI.
5_8 A n end carrying a heavy lrurfir'1r r+ f rrraas en.
Show that the boundary conditions at x 3 = L are given by ni (C)
t
=
El
x3
—
ph.2
c•
Ô
+ In
2 x3
(62)
How would (62) change if we had to take into account the moment of inertia of [he dumbbell-shaped end mass shown in Figure 5.9?
F i t:UIeE 5.9. An earl carrying ar Jrmrhfletl - .siipprsrf nurss, r {•yuirrng crur ► rde ration erf die moment 4 iitPYrrrr_
10. Find the static deflection y(x 3 ) of a beam (as a Figure 5.10) that is clamped at x 3 = i, free at x 3 = L, and subject to the vertical loading
f _ (x 3 — L)F 0
.
x;
FI GU R t 5.10. Acantilever beam subject to a frneark decreasing toad-
Sec 5.1 I ldendiny uj Beams
217
11. Determine the natural frequencies of vibration of a beam (as in Figure 5.11) that is simply supported at x 3 = 0 and x3 = L. To carry out this consider solutions written in the form progam, ytx 3 , r) = ri(x 3 ) exp Irtt) of differential equation (35), with f = 0, and boundary conditions 139). -
I
r3
^
o
t, FIGURE 5. 1 1 . A beam simply setpporred tit bo th ends
(a)
Show
that u(x 3 } must satisfy d4u + ptc 22 2 dx1 dx 3 3 2 du(0) = d2 Lj 14(0) = u(L) = dx3 3
E!
p
ate
=
0,
—
(b)
— O.
Show that the only solution of this boundary value problem is, to within a multiplicative constant, u(x3) = sin acx 3 .
How a re a and , related? What are the permissible values of a and hence the natural frequencies or eigenvalues 1.? I2. We shall study the static deflection of a simply supported beam of length L carrying a concentrated unit load at an interior point .x 3 — 5 (see Figure 5.12). Thus the deflection y(x 3 ) will satisfy (c)
dxa d4 -2 Y
_ = 0, 0<x 3 < ,
<x3 <
L,
(63a)
3
Y"
o
Fi uuRE 5.12. X3 =
2y^o) = y(L) = d2 A3 ) = 0. ddx 3
t
c
A simply supported beam subject to a concentrated loud at
Some Examples of Stark Problems in ErQsflCli}' lC'h 5
2 i t3
The mathematical problem is not well posed until the conditions at x 3 = are stated. We assume that y(x 3 ) is continuous at x 3 =
dy
dx3
(63b)
(63c)
i s continuous at x 3 =
[!n essence, two bars are "built in" to each other al x 3 = ç, compare (38)-J +(a) Give arguments to support the conditions = 0*
xâ
(63d)
S
and da y
_ (63e)
A3 . = El
(b)
In order to indicate the dependence un both the deflection as
-
1.
3
and
, we write
y(x3) = g(x3 , 4), where y(x 3 , ç) is known as the influence function or Greee's function_ it represents physically the deflection at x 3 produced by a unit transverse load at . Further define
g(x3,
[x3}, Y> (x3),
X3 < <x3.
Derive 6Ely.,(x3) = x 31, - - L)(x3 — 21 L +
3)
6Ely fi(x3) = 4(x 3 - L)( 3 — 2x 3 L + utilizing (63a) to (63e). Note that g(.. 3 , } = gi , x 3 ): i.e.. the Green's function is symmetric. In physical terms, this is another version of the BettiRayleigh reciprocal theorem (see Exercise 4.4.41 and states that the deflection at x 3 due to a unit transverse force at is equal to the deflection at S produced by a unit force at .v 3 . Argue physically that the deflection produced on the beam by a general force distribution f (x 3 ), i.e., the deflection satisfying fÎ 4 .,
El — = [(x3)• dx 3 x z^ ) = d^^^^} — y(L) = ^ 0 y( ^) — 0, 3 ^
^
E
vc stands fur afl ejump atx 3 = tr;i.e.,Q fl : =
+ l — fIt — l•
Set.
5.21 Sr. Versa -to Tur.tiior+ Prr,hlrm
21y
is g iven by Ax3) = f r glx3, ^ 1^15) d . , JJ 4 .
(64 )
Id) Verify that the conditions of the problem are satisfied by (64). lei Find the position where the beam deflection is a maximum. Discuss its position with respect to the center of the heam. Determine the maximum deflection where = IL. al Graph the shear stress and the bending mornerlt as functions of x 3 . Discuss in detail. (g) Repeat the exercise for the case where the left end of the hem is hinged rather than simply supported. 13. #(a) What changes, if any, should be made in (4l), (45), and (62) if they are to apply at the lefl-hand end of a bar. (h) Obtain (63e) from the second boundary condition in (46) and from the corresponding condition for the left end of a bar. 14. Derive (26) directly under the hypothesis that only vertical accelerations are relevant and that these are due to a vertical body force/ . and a vertical stress Q.
5.2 St.
Venant Torsion Problem
Inn the preceding section we locked at two basic approaches to solving problems in elasticity. In one case, the bending of a beam under terminal couples, we were fortunate to be able to choose a geometry and boundary conditions that permitted us to obtain a simple exact solution, in closed form, to the field equations and the boundary conditions. A second case, the bending of a beam under transverse loading, is far more complicated when viewed from the exact theory of elasticity We chose to attack this problem by making physical assumptions on the nature of the displacements, strains, and stresses. As a result, we did not obtain solutions to the exact field equations nor to the exact boundary conditions hut rather to equations and boundary conditions that were developed in the same spirit as the exact ones and, hopefully, represent some form of averaged approximation. In this section we shall look at another type of approximation, known as the St. Venant semi-inverse method. In particular, we shall examine one important problem from this point of view, the cylindrical torsion problem. The semi-inverse approach consists in partially prescribing the form of the displacements with one or more functions to be determined. These are found by requiring that [he field equations he satisfied. Except in unusual cases, such restricted forms of solution will not he sufficiently flexible to allow arbitrarily prescribed boundary conditions at each point of the
St Arne
22Q
Examples of Static Problems in Elasticity [Ck!_
5
boundary; rather, the boundary conditions will be satisfied in some global sense. This leads to the interesting question as to how close such a solution will be to an exact solution of the problem. We shall comment further on this at a later stage. WARPING F U NCTION
A cylinder is said to be under iursion when it is subject to a twist_ As we shall see, in general the twist causes a change in cross-section shape and a concomitant axial distortion or "warp" of the cross section. We shall, now apply the semi-inverse approach to the torsion problem, which is prescribed as follows. Consider a prismatic bar of length L, not necessarily of circular cross section (see Figure 5.13)_ The lateral surface S is free of tractions. Stresses are applied to the top and bottom surfaces, x 3 (] and x 3 = L. that give rise to a torque M3 about x 3 -axis (and no net force)_ The torques applied at the top and bottom surfaces are, of course, taken te be equal in magnitude and Oppositely directed (otherwise no static solution is possible). The bar will twist under the action of these torques. X
1 z 1t2
5.11 A prismatic• bur subject to equa l and opposit e Wrques applied at its ends. FIGURE
One of our ultimate goals is to determine the amount of twist (suitably defined) as a function of the applied torque, the geometry of the cross section, and the elastic properties of the material. Since the lateral surface is free of tractions, we have t,(x) — 0
for x on S.
(1)
Sec_ 5.2] .Si'_ Yenanr Torsion Problem.
:21
In terms of the stress tensor, (1) can be written as
T o rr, = D on 5,
i?)
where ni is the outward normal to the lateral surface S. Since the bar is prismatic, the normal to the lateral surface will only have components in the A i and x 2 directions. if the Greek subscripts are restricted to the values I and 2, (2) can then be written To lu* =Don S.
(31
We now assume that the motion consists of two parts*: (a) A small rotation about the x 3 -axis of constant magnitude 0 radians per unit length. (h) A displacement in the x 3 -direction, known as v.arping, that is the
same for each cross section. Ry performing an appropriate solid body rotation, we can impose the condition that there is no rotation at x 3 D. Let us calculate the displacements that result from these assumptions (see Exercise 4.1.10). Consider a generic point (x t , x,) in an arbitrary cross section, as shown in Figure 5.14. In terms of polar coordinates r and i, xi
r cos
i1,
x2
r sin 0.
Ftc i UR r; 5. t4 A ,rrc lrrrrr ► ti mum from Mc l4cr,tir rg iJu• cylinder
cur milk at 3
Milli CI
rhrcuO
.
• The M,mi-inverse approach is introduced at this point. We purrrulf r prescribe sonic features of the solution and see what this entails_ In a plrreiy inverse appiudch, we would rumpled'. prrsr:.ribe the 501MiCKIS and then work hod; ward to see what pr«hJem Itas been solved. The possibility that assurnptton 1st} and Ib) are appropnals might well occur to an experienced engineer. R eaders should perhaps suspend judgment until they see that indeed i torso assumfuror provide a solution to a torsion problem. Then they might try in conviritv thr rnsckes rt poMrripri that with experience they too would have made a correct guess concerrang the form of the solution.
222
Sume Examples of Static Problems in £lasturty (Ch_ 5
After the rotation described in assumption (a), the point will have the coordinates x i = r cos (P f Ox 3 ),
x 2 = r sin (ç1 + (]x 3 ),
where x 3 is the vertical coordinate of the particular cross section. Hence the displacement vector has the components 11 1 = r cos OPf Ox3) -- r cos if
t1 2. = r sin (Of + Ox3) -- r sirs .
,
(4)
L'xpanding the trigonometric functions in (4), taking into account the assumption that the total angle of twist Hx 3 is small, and returning to rectangular coordinates, we obtain (Exercise 1)
63( 2 X 3.
t#2
= fÎJ{ i x 3 .
(5)
By virtue of assumption (b) we must add to these two displacement components a displacement in the x 3 -direction that will be a function of x 1 and x2 only. We shall designate this by O0(x i , x 2 ), where is known as the warping function and an expected proportionality to 0 has been explicitly indicated_ It is also convenient to use the two-dimensional alternating tensor,* 413, defined by I, or= 1,/5=2; co _ —I, ac = 2, = 1; Q, 2=
^3.
With this, the components of the displacement vector can be written a 3 — 00(x. I , x2).
14. — — eppOxpx3,
f (6)
Our problem is now reduced to finding the equation satisfied by the warping function in the region A, a generic cross section, as well as the boundary conditions on B, the bounding curve of A (see Figure 5.15)_ One effect of our particular assumption on the form of the displacement vector can be seen at this stage. The displacements of the end surfaces at x 3 - 0 and x 3 = L cannot be arbitrarily prescribed but must be taken from (6). The end surfaces will warp just as any other cross section. The equation satisfied by 4 (x 1 , x 3 ) can be found by applying Navier's equation (43.9a) in the static case with vanishing body force: + = û.
,uuj.1i 4. tit+ A
(7)
From (6), one can conclude (Exercise 1) that
=
(8)
^
and u3. 1J =
• We use e rather sirasn tensor.
than
.sa
.
(91
c (as in (he threc-dimcnstonal alternatntl;n avoid confusion with the
Sec_ 5 2
Sr
lrenunt Torsio n Problem
223
l',
FIGURE 5.IS. [.rait ttrraye7rr recur t and tam rrornrul r ector n fu r ht hue^r^d^iiy curie B r+/ rlrr r -r4.).E+ se^r rirur.
In addition, i.i
—
`e4f(x,x3)
—e p fiesi k ,x 3 — 0
With all this. (7) reduces to — 0
(10
in A.
)
Thus we see that the warping function satisfies Laplace's equation in two dimensions Next, we detcrmine the boundary conditions that must be satisfied by ç(x i , x 2 ) by considering the implications of (3), the condition that the lateral surface is free of traction. Pnom the stress- strain relations, we have .
=
Au"
+u;F1+
Using the displacements given by (6), we find [Exercise 2(a)] that
{çti a — esf x p ) = T3 .
1;3 =
(11)
All other Try are equal to zero. Note that 1-, 3 is independent of x 3 , so that the stress exerted on a cross section x 3 = constant by material above it is the same for all cross sections. We can now calculate 13) in terms of the warping function, with the result charts — ex n— 0
on
or
=
ea ,x 01 z
o n B.
Therefore, we see that the warping function satisfies a second boundary value problem of potential theory-
(12) Neumann problem
or
Some Exa,rrpfes of Srarir Problems in Elasiirily
224
[Ch. 5
STRESS FUNCT1f1N
We now shall derive an alternative formulation of the problem by concentrating our attention en stress. By (11), the only nonvanishing components of the stress tensor are 70 3 . Thus the equations of equilibrium reduce to a single equation, a
;, a = 0.
(13)
As the reader is requested to demonstrate in Exercise 2(b), (13) is a sufficient condition 1er the existence cf a scalar functions(x 1 , x i ). called the strm function, with the property (14)
Ta3=ea/34'J.
(The stress function is analogous to the stream function used in fluid mechanics. See Exercises 15.1.3 and 15.1.4 of 1_] From (3), we have on the boundary eat,
.ig
?la
= 0_
The last relation can be simplified by introducing the unit tangent vector T. From Figure 5.15 or from an algebraic argument [Exercise 2(c)], we see that (1 5)
= Thus the boundary condition becomes
T j t p = 0, or, upon introducing the arc length s along the boundary
dt ils
B, (16)
Equation (16) may be integrated at once along the boundary, resulting in the final condition '
= constant
on B.
(17)
We note From (14) that the values of the stress tensor are independent or the choice of the constant. In the case of a simply connected cross section, this constant may be taken to he zero. The situation is not so simple, however, in the case of a multiply connected cross section associated with a prismatic bar having cylindrical holes (Figure 5.16). The constant may be taken to he zero on one of the boundaries, B for example; but the values on each of the Bp, must be determined from the additional other boundaries, B 1 , B 2 condition that the resulting strains integrate into displacements which are single - valued in the multiply connected domain. The development of the appropriate Formulas can he found in Sokolnikofl'(1956, pp. 169f .). The equation that the stress function `l' satisfies in the region A can be found by observing that the stresses calculated from P will be equal to those ,
..
,
Set . 5.2}
Si. :'er:un[ Tar.erprr Prahicm
F io LIRE 5.16.
Cross srrfiurr
2 -'5
of rr prr_srneurr bar Hrth c ►•linrlrkul iruk.^
determined from the warping function 0 in (11 ). Thus
ems ' —
— eox0i).
The partial derivatives of P can be written explicitly by observing (Exercise 3) that T .19
I
.6014)440 — T .r -
We can then conclude that .r =
¢ — x e ).
(18)
Differentiating (1$) with respect to x we find [ Exercise 4(a)J that in A.
(19)
Assuming that the cross section is simply connected, we have, in addition, the boundary condition ^
= il
on
a
( 24)
We thus se e that the stress fu ncti on satisfies Poisson's equation with vanishing boundary conditions. We expect to have to verify that the compatibility conditions arc satisfied when a problem is formulated in terms cf stresses las it is when the stress function is employed). This verification can be carried cut [Exercise fi(c)i The result is a foregone conclusion, however, since the existence of singlevalued stresses is assured by the first steps taken in our approach to torsion-
Some Examples of Stark Prnhlems rrr ElusdiriRy
226
[Cli. S
FURTHER PROPERTIES OF' THE S -i RFSS F- UN('TIUN
Prandtl* observed that the stress function can be obtained experimentally through the following analogy. The small deflection w(x,, x 3 ) of a thin membrane under uniform tension 1; fixed to a boundary B, and subject to a net uniform pressure p, satistiest 4V
...
y
^-+
—
'
w =
oli B.
Here w is taken positive "upward _ in the direction of increasing x 3 , and p is the excess of upward pressure over downward pressure. The mathematical problem for the deflection w is identical with (19) and (20), the problem for the stress function `f'. (There is also an analogy, although not as useful a one, with viscous flow down a cylindrical pipe. See Exercise 12. WI Taking advantage of the membrane analogy, we can find '' by cutting out an area from the top of a fiat box in the shape of A and fastening ai thin membrane around the edge of the hole. Upon application of an appropriately chosen pressure in the box, the membrane will deflect by an amount that is proportional co the corresponding value of the stress function. This technique can be used to find the solution to the torsion problem for complicated cross sections, fur which analytic methods prove difficult_ Moreover, we shall find additional properties of `I' that will make the membrane analogy even more informative, We Leave to Exercise 13 verification of our assumption that the resultant force on the end sections is zero. Let us calculate the moment supplied by the tractions to the top surface. (This must reduce to the applied moment with the single nonvanishing component M 3 .j The fitly component of the moment, M i , is given by
JJ
Mi _
;xfi k dxidx2.
(21)
A
The normal to the top surface x 3 = L has n l : r^ = 0, n3 = 1, so that the traction there is given by tk
—
Tikri —
(22)
T3k•
Inserting (22) into (21), we obtain M;—
JJ
c.x J T3k d.xJ dx
?..
A
• L. Yranail, Phys. Z.. a it 9031. ^ A simple dcrivailiotr fif lhc onc-dïmEnEional version of ibis equalinri preCedcs (7.1./31. Th€ two-dimensionst version can be nbtaincd by linearizing (7.t.35).
Se` 4-. 3.11
Sr 6`r•nanr 1'r► rsro+r Problem
2.27
Si rice 7 '33 = Û. we lind [Exercise 4i b)] that o n the top surface M 1 = -- L JJeT p d, 1
dxx =
L
J$
A
o 'V dx , dx;.
(23a)
a
Green's theorem may he applied to 123a1 with the result that =L 1tIJ,i j d s .
(23b)
.
f
Since vanishes on the boundary of the simply connected cross section, we see that 14 ^
0.
(24)
Equation (24) states, as we originally proposed, that only the x 1 component of the moment al the e n d of the beam diners from zero. For some comments abnut the multiply connected case, sec Exercise 5. Ile third component of the moment. M I , is given by -
ti4 3 =
JJ .
eX 1 I:.c1X L (IX F-
A
hi terms of the stress function, this becomes M
3 = f^ ez ^e^ ; x Q
`f' rl .x ,
(25)
el).:,.
A
By Exercise 3, we may write M3
+c,
— 54rXr'Pi dJ( i dx 2 — — A
rl.i i dx 2 ,
( 26 )
A
or
M 3 — — ff (x 2 1-1') dx 1 dx 2 f 2 II`
dx 1 r.x
A
We now apply the divergence theorem with the result that M3 =
jxii 'k' ds + 2 TIT dx, rtx
^.
A
Since the stress function vanishes on the boundary B. we finally find that M 3 — 2 fig' dx dx 2 A
(27)
228
5
Some Examples of Static Problems in Elasticity [Ch.
On x 3 _ 0, a minus sign must precede TA, so that we obtain the required oppositely directed moment on this face_ Equation (27) yields a significant global quantity, the applied torque_ This, in turn, is directly proportional to the angle of twist per unit length, 0, as is readily seen from (19) and (20) [Exercise 6(a)J. In terms of the warping function , one finds [Exercise 6(b)] that M1 =
where
D,
DO,
(28)
the torsional rigidity, is given by
1J ex
D = 4.1 +
dx i dx 2
.
(29)
Here J stands for the polar moment of area of the cross section: 3 = ff x„x„ dx, dx 2 . A
The torsional rigidity is important to engineers, since it plays the same role in the technical theory of torsional vibration of rods as dues the flexural rigidity El in the transverse vibration of beams- [Compare (29) with the Euler-Bernoulli law (1.19): M = (E1)R -1 .] Noting from (10) and (12) that ck depends only on the geometry of the cross section, we sec that (29) clearly spells out the roles played by the geometry and the material modulus p in the determination of the torsional rigidity, It can he shown f Exercise 9(d)j that of all cross sections with the same polar moment of inertia, the circular one has the largest torsional rigidity. Equation (27) tells us that the membrane analogy yields the applied torque, fer it is nothing mere than twice the volume enclosed by the deflected membrane, We can also find some other interesting features of the stress function, To this end, let us consider the Level curves of this function; i_c_ the lines q' = constant. The normal to these lines is given by the gradient o Since the T* 3 are the only nonvanishing components of the stress tensor, we see that the stress on a cross section plane x 3 constant is purely a shear stress with two components T 13 and 7 -23 . The component of this shear stress vector in the direct ion of the normal to the constant stress function lines is proportional to ,
.
-
1";3 tiJ .o
rap .# T a = 0.
(30)
Thus the shear stress vector TT 3 is tangent to the constant stress function lines. Furthermore, if 7' represents the magnitude of the shear stress vector, then T2 = Tas Ta 3 = eat t.v e, T. 1' = +. .9'
T 2 = !grad cY 1 2 .
(31)
Sec. 5.21
Si. Venom. Torsion Problem
The maximum value of the magnitude will then occur at the maximum of 'grad 'Pl. In view of the fact that the gradient is directed :long the path of steepest ascent of the surface represented by 4'(x,, x 2 ), we see that the maximum shear stress will occur where the constant stress function lines are most densely packed. In particular, the maximum shear stress can be determined from membrane experiments by searching for the steepest portion of the membrane. Additional information can he determined by calculating the Laplacian of T'; (4'_ 0 4-1 0 jam„ = 2`!E'# 24'.0, 91.0, ? 0_
nA
+ 241,0 `P.vma
Functions such as T 2 that have a nonnegative.. Laplacian are termed subharmonic. They share a number o1 important properties with harmonic functions (the subset of suhharmonic functions whose Laplacian vanishes). In particular, one can show (Exercise 7) that a suhharmonic function is either identicallyacnrtstant or else it has its maximum valueon the boundary. Consequently, since T L is suhharmonic, the maximum value of the shear stress will occur on the boundary. This result simplifies the determination of the maximum shear stress for a specific problem. We finally recall that the behavior of DNA molecules can he illuminated by a determination of the stress function (I, pp. 106- I08). MODIFIED STRESS FUNCTION
Equation (19) states that `l:' satisfies a Poisson equation with constant right-hand side_ We can gain some additional techniques for solution by introducing a modified stress function that satisfies Laplace's equation. To this end, we observe that br,,Ic pti = (x ]C a, 1 ) ` 2(.+€ AP ) .0 — 2x .13.0 .= 4. We thus define a modified stress function I(x 1 , x r ) by fJ
_ 'F + ; µ0x,x ii .
(32)
From (19) and (20), we see that 2. must satisfy. O =xx,
in A, on B;
(33) (34)
i.e., E is the solution of a Dirichlet problem. The relationship between E and the warping function rfi can he determined by writing alternative expressions for the stresses T, 3 . [The constant pfd was inserted on the right side of (32) te make this relationship particularly meaningful.] f ^y , PO4,0 (Y` Tai' 0 — e af x tr ) = e. e `P, — ) 11k.0[ 1 .0 — 4-(x ? 0.01 (35)
230
Some Examples of Static Problems in Elasticity [Ch- 5
Consequently, 0•a — 00.xp ` e4E.P — eaPxY 6 7o
or
0.0 = eu0 L.0 t i.e., 0.1 = E,2+
^3=-E1
(36)
-
Thus the warping function. it and the modified stress function, E, satisfy the Cauchy-Riernann equations; the modified stress function is the harmonic conjugate of the warping function. Furthermore, introducing the complex variable z=X, + 1x2 , we see that the complex.-valued function
f(z) = f(x, -4- ix 2 ) _ (x,, x 2 ) + iE(xi, x 2 )
(37)
is an analytic function of the complex variable z. Thus the methods of potential theory (introduced in I, Chapter 16), as well as the power of analytic function theory, are available as techniques for solving the torsion problem for specific cross sections. The examples that we shall consider in the following sections are rather simple and can be handled with elementary methods. It is not our purpose in this chapter to consider the more elaborate mathematical techniques required by more complicated problems. An elegant and readable account can be found in Soknlnikoff (195b, pp. 134-65)RECTANGULAR CROSS SECTION
The rectangular cross section can be studied very conveniently through separation of variables and double Fourier series techniques. Let us consider such a section of width a and depth h as shown in Figure 5.17. We assume for definiteness that b > a. The determination of the pertinent quantities will proceed from finding the modified stress function E(x t , x 2 ). We note from (33) that E must be a solution of Laplace's equation in the rectangular area. From (34), E satisfies on the boundary a, x =
Ç
+ ,
Ex
+ b _
+
-
$
(38a, b)
Rather than calculate E(x,, x 2) directly, we introduce a new harmonic function f(x i , x-' ) by means of
f = E.]i + l =
- 22 + 1. ,
(39)
Differentiating (38a) with respect to x 2 and (38b) with respect to x 1 , we sec that ±lb) = 1. E.22(± -ia. xa) = 1,
Sec.
3.2.1 St.
Vermin Torsion Problem x
231
^
1 ^ 1is41 R E
5 17. Constant srrc'.ss
jirrJrfNfn lines
m
ü r1'r rr.tujufur 111iH4,41
wider
torsion..
. 1'hus, from (39), our new
function f satisfies the boundary conditions
f(± la, x2)=O.
f(x1, ± b1= 2 .
(40a, b)
Consequently, the introduction of the new harmonic function provides the advantage of yielding a problem with simpler boundary conditions. Of equal importance is the fact that the stress function and (he stresses can be determined by integrating f (x i , x 2 ). Since f will be found as a series expansion, the convergence of the series for the stresses can expected to he improved over that For f. Lei us assume an infinite series of product solutions in the Form* u.
}-oc„ x 2) =
L,
rnxn(x71Yn(x2y
(41 )
n
Hach term of the series 141) is expected to satisfy Laplace's equation. Consequently,
X; Y,,
4 X.
Y = Q,
where the primes refer to differentiation with respect to the appropriate variable. Since X, is a function of x, only, and Y. is a function of x 2 only, we have by the usual arguments lL A
X,,
^
A
• We tike the liberty of skipping a few steps in the separation of variables mch hod-See 1. pp. 123 27 or 553 55, fora fuller rreaimerti. See also ExCralse 14. -
-
Sc one Examples of Static Problems in lEdasriiriry [Ch. 5
232
where k n is a constant to be determined. Thus the factors in the product solution are x
"
_
cos
kry x i ,
siri k„x i ,
— {
cosh k„x z , s1A h ,^ n x z .
Since the boundary conditions are even in both x, and x 2 , we seek solutions with the same property. Thus we take as a purported solution,
f (x„ x2) =
c,1 cos k r x, cosh k„x 2 -
(42)
r=6
Because of boundary condition (40a), we require that cos 4Rk n =-0
nr J
=
(2n + 1)n a
,
ér
0, I, ...
(43)
Boundary condition (40b) now requires that Q,
j ( x , +4-h) — 2 = ,
E c„ cosh
Ibk„ cos k r,x . ,
(44)
p -4
a< By virtue of (43), the cosine functions are orthogonal in the interval x, la; thus, multiplying both sidesof(44) by cos k",x, and integrating over this interval, we obtain _ Cm
8(—
(45)
(2m + 1)n cosh (ibkm)
Consequently, fix ' , x 2 ) =
8
(--- 1j " cosh E — - - - - krix2 — ccTh kn x 1 .
7[ R , 4
2n + 1 cosh i-bk n
(46)
One can readily show [Exercise 1O(a)] that the series (46) converges uniformly x i and x 1 for 1x21 < {b , and hence may be integrated term by term. tri Noting from our expression for T 3 in (35) that T13 =
(E,2 — x21,
T23 = p9(— , + x1),
we may use (39) and (46) to find that
}1 3
T23
—
=
°`'
(- I r
sinh k„x z cos kn x,, 7r 2 "p (2n+ 1) 2 cosh k„bJ2
$apo
pi) 2 x 1 —
845
`x
( — 1 ^R
CC}Sh ^.„^
x
n 2 ,X0(2n+ 1) 2 cosh k„b(2
Sill khx1 -
(47)
(48)
See.
5,1] Si.
Yer:aror Torsion ProFifem
233
There are actually "constants of integration," g(x,) and h(x i ), that at first sight must be added to (47) and (4R), respectively. But g and h can he shown to vanish [Exercise l0(a)] by applying the equation of equilibrium 3, I
}
(49)
23.2 = 0,
1
and the boundary conditions x 2 ) -- T23(x1, !b) = O. As we have previously proved, the maxtrnum shear stress will occur on the boundary. Whether from direct calculation or from the constant stress function lines sketched in Figure 5.17, we see that the maximum must occur either at (±Q/2, 0)1 or at (O, ± bj2). It can he shown [Exercise KO)] that, in fact, the maximum occurs at the first pair of points. Thus
,„ , = T23(2 a. 0 = a .r41f l — I
ech } ^,I „l i^ - n ^
l
(t1 1
1
The convergence caf the series appearing in (50) is extreme]). rapid. As a result, the maximum shear stress can be approximated by taking only one term: f rnan
[7^lri t — ^; sech
^^
21.4)]_^-
The error is less than 0.0019(80u/70) (Exercise I l). ELLIPTICAL CROSS SECTION As we have previously noted, our solutions to the torsion problem are of semi - inverse type in the sense that part of the displacement vector is specified. We now consider an example of a solution that can best be described as being purely inverse. The procedure consists in selecting a specific modified
stress function with, perhaps, some yet-to-be determined constants and then determining the boundary on which the prescribed boundary conditions will be met. This is hardly a systematic method of solution and whether the resulting boundary represents an interesting cross section or not will depend on how skillful we were in our original selection. The pure inverse approach is certainly one of educated guessing, but when it is successful, it represents an extremely simple technique of solution. For the example in question, we look at the very elementary analytic function Z 2 = (X1 + Jx^ ) ? = Xi — x2 ♦ ^Ïx^x3.
We may use either the real part or the imaginary part as a possible stress function. Experimentation shows that the real part offers a useful solution. We take advantage of the fact that addition of a constant does not change a
234
Some Examples of Siwk Problems in Elasticity [Ch.
5
function's harmonic character, nor does multiplication by a constant. Thus we study the harmonic function k 2,
-- x3)
=
where the constants e 2 and k 2 are at our disposal. From boundary condition (34),
C 2 (xÿ
—
x 2i ) +
k2 _ Ix; +
x^
on the boundary B. Thus we see that the points un the boundary 13 must satisfy —
e2 i + (4 t
= k 2.
(51)
Equation (51 ) states that the boundary 13 is the ellipse X22
1
o f Figure 5.18(a). Here ti
= #
k2 —
c ^'
^^
z
k2 =
2+ r 2;
2 Cc =
u^ — b 2
k2
2(al # b1),
=
a^b^
(52)
a2 + b^
Direct computation (Exercise 12) shows that the stress function is given by
a3b 2
x2 xi
Thus the shear stress lines are ellipses. The twisting moment is
found
(Exercise 12) to be rrp061 3 b 3
(54)
Ail 3—Q2 +bz-
Y2
^
(b) F, IG u k E 5. G$, (a ) Elliptical boundary obraim'd from rhe inverse merhod ni suhetinm. (h) tines al constant warping in on elliptical cylinder UndPr f UFSi Off
XI
Se . 5.1
Si.
'Vent ri
Torso
Problem
235
Since the modified stress function E is conjugate to the warping function 4 , , we see that
4 = —2t xt _iv 2
(55)
' 2
The lines of constant warping are rectangular hyperbolas, These are shown schematically in Figure 5.18(h), where the dashed lines represent warping in the negative x 3 -direction and the solid lines in the positive direction. Note that when a = h and the ellipse is a circle, then e-2 z- u 2 — W = il and there is im u'arpIngS i. VENANT PRINCIPLE
We remarked when we began our discussion of the St. Venant torsion problem that the approach is of a semi-inverse nature, in that part of the displacement vector is specified a priori. The price that must be paid for this simplification is the inability to specify the exact distributio n of stress across die bases of the cylinder; only the resultant torque can prcscribed. The distribution of end tractions is a consequence of the solution to the problem, and this may or may not agree with the actual applied tractions_ Justification of the procedure is twofold, First, it is extremely difficult in practice to determine the actual distribution of traction, although the resultant torque can he measured accurately, Second, one invokes the St. Venant principle. This hypothesis refers to cylinders that are long compared with their cross-sectional dimensions. It states, roughly speaking, that the difference between the stresses and displacements produced by two different equipollent* distributions of end traction will be only a small perturbation everywhere except near the ends of the bar. The precise meaning of the St. Venant hypothesis and its justification have been the subject of many studies, almost from the time of the original St. Venant paper in 1855. in addition, a number of these studies have also considered whether the principle can be extended to other cases. Reference to some of the early investigations of the question will be found in Love (1944, pp. 327 28) In recent years. important steps toward clarifying St. Venant's principle have hem' made.t Purely verbal statements of a "principle" that was actually a conjecture have been replaced by rigorous estimates. To illustrate wh at has been accomplished, we shall state some basic resultsdue to R. Tou pin. (These results will be given for a standard f-lookeatt material, although . ] oupin treats a more general case.) Consider two problems involving the deformation of a semi-infinite cylinder of cress section Ai fx,, xal E A ,
0 S x3 <
-
Disirihuiions of stress arc equipollent if they lead to the same net force d+u! mermen( t t-Dr references sec pp. 190 207 of M E_ G un in's survey, 'The Lopez r 1 h e ory uI EtastC in Vol- (ha of the Liw;rlupediu u) Physics (Kew York! Springer-Veriag. 1973)
•
Some Examples of Static Problems in Ela sticity [C k 5
236
Suppose that the loading applied to the exterior of the cylinders is identical in the two cases, except at the end x 3 — O. On this end the loads are different hut are equipti lent. We expect that the solutions will not differ appreciably
at points far from x 3 0 (see Figure 5.19).
iq
FIGURE 5.1e3. A semi-infinite cylinder or cross section R, Suppose that the stress distributions acting on two such cyttrrders tux, equipollent on the end section at .r s = 0 arrel identical mi the rest of the buurrckrrt' Then St Yeyrnemr`.c principle asserts that the deformations should !xr nearly equal when _Y ; is sufficiently large.
In uniqueness proofs (as we have seen in Section 4.4, for example), rather than show that two possibly different solutions are in fact identical, it is easier to show that the difference between these solutions is zero, Here, too, we shall consider the difference of the two problems mentioned in the previous paragraph. We therefore must demonstrate that solutions are "small" far from a section at x , = b which is loaded with stresses that give rise to a zero net force and moment_ We are given that the remainder of the boundary is free of stress (and that stress on the section x 3 = ±c approaches zero as 3e,
Let EAR) demote the strain encrgy in the bar to the right of the section xx = x`. Recall from its definition in (4.4.5) and from the succeeding discussion that the strain energy is a nonnegatwe definite quantity which vanishes only when the deformation is a rigid body motion. The notion that the solution is "small" far from x 3 = 0 is made precise in terms of an inequality showing that the strain energy decreases exponentially with x3_ This inequality states that for any positive constant L Eix 3) S EAD) exp — x3
—
L
(56)
The decay length d(L) is specified as follows. Consider solutions u(x, r) to Navier's equations (4.3.9} that are of the form
s x, 1) = fax) exp (mg),
x e B.
(57)
If the boundary of the region B is free of traction, such solutions represent free vibrations of constant frequency ca.* The frequencies of vibration are • Vibrations and we motions arc discussed in Scciion 6.1,
SOL 5.2]
Sr.
Venrrrrf
Tprs+en Prrr{rtr ► ri
237
determined by a certain eigertvaluc problem. Let too(L) denote the lowest (nonzero) vibration frequency of a cylinder having cross section A and length L (composed of the same homogeneous material as the original cylinder). Then
AL) =
2 g(l
l58)
2p
In Chapter 12 we discuss the characterization of eigenvalues as minima of certain quotients of integrals, The inequality furnished by such a characterization of w0(L) plays an important role in the proof of Toupin's theorem. The length L can be chosen arbitrarily, it is obviously preferable to select it so that the exponential in (56) decays as quickly as possible. The bound (56) is global. in that it guarantees the decay of an overall integrated property of the solution. (In the language of functional analysis, the nonnegative definite functional E, is ri suitable norm with which to measure the solution.) One would like to be able to estimate the decay of local quantities, such as the displacement at various points. This, too, Toupin provides, in terms of an estimate for the magnitude of the strain tensor at any point x in terms of the strain energy stored in a sphere of volume V centered at x. If the strain energy within such a sphere is less than some quantity L , then ciptk.,„(x) U V -', (59) where K is a constant that depends on the elastic moduli_ Note that the estimate is large if a given point can only be surrounded by a small sphere. (The sphere is, of course, required to lie within the elastic body.) This is quite satisfactory, for strains are expected to he large at points that are near slits or are within thin panels_ As was aptly remarked by 3aunzemts (1967, p. 352} "the sensitivity of solutions to changes of boundary data has important ramifications in all branches of physics. It is of particular concern in elasticity theory because here the mathematical boundary conditions are quite complex and require information that is often practically unobtainable." E X ERCI S ES
1. Establish equations (5), (8), and (9). 2. (a) Show that To is given by (11) and that all other I;, $(b) Show that (14) is a justified assumption. (c) Use f4.0 to deduce (15) from the fact that t = k A FL 3. Shaw (without merely enumerating possibilities) t hat eao e, 0 s 45j' .
4. (a)
Complete the calculations that yield (18) and (19). (b) Verify (23a) and (23b).
—
O.
238
Some Exampl e s or Sunk Problems in Elasticity [ Ch_ 5
5. We concluded in (24) that M a 0 for a simply connected cross section, by using the fact that 'V _ 0 on the boundary B. In the case of a multiply connected cross section, the boundary values on L are given by
`N =0 on B,
q= k i on B1 ,
where k i is a constant and i = 1, ... , n, as indicated in Figure 5.16. l4fodify the proof of (24) for the multiply connected boundary conditions given above. b. (a) Deduce from (19) and (20) that 'F is proportional to B. (b) Develop the formula for the torsional rigidity given in (29). (c) Check that the compatibility conditions are satisfied when the torsion problem is formulated in terms of the stress function. 7. We have noted the we can prove that the maximum shear stress occurs on the boundary from the maximum principle for subharmonic functions. This exercise outlines a proof similar to that used in 1, pp. 570-72. (a) Start from the identity .
1f(tiV v 2
07 2 14) da =
^ ^
(14 R
lr
—
ds
,
R
where OR is the boundary of the region R and a is the outer normal. Let (. 1 , L: 2 ) be a point in R and r 2 = (x - o)(xo - fia] In particular, lct u = 0, where 0, > 0 and = - In r. Show that
2
(h)
fji(I
z} -
In '4), dc f
j' I (. 0 ,l OR
f^
ln r ï^ - - in r i c rt ds. an
By taking R to be a circle of radius p centered at (. ^ l , 2 ). prove the mean value inequality 4( 1 ,X71s
1
2n pJeRç
ds.
(c) To prove that there is no interior maximum unless is constant, 2 ) = 1 , x assume the contrary and Iet Cx M, the maximum value, where (x i , x 2 ) is an interior point of R. As shown in Figure 5.20, there is another interior point (y i , y 2 ) in R, where 0(y r , y 2 ) < M. Let (z 1 , z 2 ) be the first point on a curve C in R going from (x 1 , x 2 ) to (y i , y,), where iP(z r , z 7 ) = M. Let r be a circle centered at (z,, z 2 ) lying in R. Show that there must be an arc of F on which < M. Hence show that there can he no interior maximum_ Show that the only cross section that will admit a warping function which vanishes identically is a circle.
Set . . 5.2]
Sr. Venom.
Tar NUM
Problem
239
F [tir1RF 5.20, Construction s!?unvrrm Mar u cunrrudirirnrr results trim? the drutytolita on inferior maximum rrt ( -► i . x :).
ussur?aP r
9. We shall develop here a very useful alternative Formula for [he applied moment, using Green's theorem in the plane. (a) Starting from (2b), show that
L (b)
r 4 x a n ii ç{i d5,
—
r'V
Show that =J
(c)
where ^'t
A
- f 43 ` -- if s_
an
Finally, obtain the Dice-Weinsrein formula,
M3 = J -
J5
ctii fix ] clx 2 -
(am)
Using (60), prove that of all cross sections having the same polar moments of area .1, the Circle has the largest torsional rigidity. 10. (a) Provide all details omitted in the text's separation c1l` variables approach to finding a harmonic function f that satisfies 140) Also verify (47) and (48). (b) Prove that for a rectangular cross section the maximum shear stress is at (±u p) and verify (50). 11I. This exercise develops an approximate formula for the maximum shear stress on the boundary of a rectangle by starting with (50) written in the (d)
.
,
form
T=Ba 1
8
—
^2 sech
1 rib
sech
^^ + ^ — R =^
-
+ 1)t^i3j2a)] • 2 (2r^ + 1)
240
Some Exumptes of S7nrie Problems iM Elasticity
[Ch. S
(al Show that the series appearing in this expression can be bounded above by
9 (b)
sech (Zr+II
^^
.
^=i
Show that this expression cari he further bounded above by -n
x
e?4f+ —^ ^ n=
(c)
-
5
exp nil 2 ü
Sum this series and obtain a numerical estimate for the worst case,
which occurs when b = a. 12. For the elliptical cross section, find [he stress function `!', the lins of constant shearing stress, and the tordue angle relationship (54). 13. Show that the text's formulation of the torsion problem does indeed provide a zero net force on each of the two ends of the cylinder. (a) lise [he stress function. (b) lise the warping function.
5.3 Some Mane Problems An extensive literature has been developed on [he solution of static elasticity problems that are essentially two-dimensional in formulation. Concern with two dimensional static elasticity arises from two main motives: (I) there are many important technical problems that are two-dimensional in nature; tii) there is independent interest in the powerful mathematical methods that have been developed for the solution of the biharmonic equation, which plays a central role in the theory. It should also be noted that the experimental methods of photoelasticity are particularly well adapted to such two-dimensional problems, so that many experimental verifications of the mathematical theory have been made. In our brief introductory discussion of these matters,wc shall first consider the problem of plane strain and show how this can be formulated as a boundary value problem involving the biharmenic equation. A garticular example, the Kirsch problem, will be solved_ Although the mathematical techniques required are elementary, the solution is important as an introduction to the technically important problem of stress concentrations. We shall also consider the problem of plane stress, which is of somewhat limited applicability, and a more useful approximate theory of generalized plane stress. The resulting boundary value problems will be shown to be the same as those occurring in plane strain, and thus solutions obtained for one case can readily be translated into those for the other_ -
Set'_ 5.3)
Some Plane Prohlemt..
24 1
EQUATIONS FOR PLANE STRAIN
Let us consider a cylindrical body whose axis is parallel to the x 3-direction. We include multiply connected regions as well as infinite regions bounded by cylindrical holes. The surface tractions and body forces, if any, are assumed to be independent of the coordinate x 3 . The basic assumption of plane strain is that there is no displacement in the x 3 direction and the ot her two displacement components are independent of x 3 . As usual, Greek subscripts will take on only the values I and 2, whereas Latin subscripts range from I to 3. With this, the assumption of plane strain may be stated as -
Jig
= 4.1C1,x2 j,
u3 = O.
(1
)
The strain-displacement relations become =
^.
0
+
Up, g),
E;y
= O.
From Hooke's law,
T1 _ ?let,1 + .ich x ki a we have Te
T
T,r 3=0,
T0(Xt• Y2),
(2a)
T33 =
Carrying out a contraction in Hooke's law, we see that 1-33+
T33 =
or T33
=
(2p + 31)c,
'+' f
(2b)
.
Thus (2a) and (2b) show that the complete stress system can be determined onceTe is known. To simplify our discussion, we shall also assume that the body forces vanish.' The equations of equilibrium, To.) 0, become, for our stress system, —
Tt 1.t +Tt2.2 = 0,
T21.1 +
T22.2 =
(3a, b)
^•
AIRY'S STRESS FUNCTION
Equations (3a) and (313) can be regarded as integrability conditions. We can thus deduce that there exist functions V 1 (x i , x 2 ) and V2(x t , x2 ) with the properties Tt l = V1,2 , T12 =
—
V1,1, T21 = V2,2 , T22 =
-
V2.1•
• In the important cases where body forces are derivable from a potential, the resulting complications are not too serious. We shall not consider this case, however.
242
Some Examples u/ Slade Problems in £lastïrirr
[Ch. 5
Since the stress tensor is symmetric, we must have
V1.1 + F2. 2 =
(4)
^-
Equation (4) is again an integrability condition, and yields the conclusion that there exists a function U(x 1 , x 2 ) with the property II, = U , 2 and 1,2 — ff . We see that the components of the stress tensor can be found from U by ,
7', r = (1 32 , Ti2 =
U.129
1 a2
—
(5)
.11-
In terms of the two-dimensional alternating tensor eo we have (6) tiQr Cre U, fP' The function ti(x r , x 2 ) is known as Airy's stress function The next task is the determination of the partial differential equation and the boundary conditions that must be satisfied by this function. We recall that in view of (6), the stresses will satisfy the equations of equilibrium. In order that these stresses lead to compatible strains, the Beltrami-Michell equations (4.3.15) must be satisfied. For our purposes it is necessary and sufficient to use just one of these equations, namely {in the absence of body forces)
7;.. pp = U. In terms of the stress function, this becomes A= e., e,r, il mg* ` ï v 1) ,. n66 or
t 'r ,au00
(7)
Thus the stress function satisfies the hiharmonic equation in the region D of the Ix , x 2 )-plane occupied by the cross section of the cylindrical elastic body. ,
BOUNDARY CONDITIONS
Let us now turn to calculating tractions on a cylindrical surface in terms or the stress function. The trace of such a surface in the (x 1 , xj plane will be denoted by C. as shown in Figure 5.21(a). The normal to the surface will have ri, U. In addition, we shall make use of rp , the unit tangent vector to C. The traction t ; satisfies -
Therefore, t 3 = 0, and
In terms of It by (6)
Ser.
5.3]
Some Plum. Prvfr fe m.%
2
43
^rU
la I
FIG ufil: 5.21. (al Crrr C re frre,sivrrt parr erf a cylindrical boundary ime•r:u•e•red hr the (.r, ,x 2 )-plaire, which is perpendirrdur w the directrices s }f the cylinder. Depic•red also are the normal te ^{rrl. the rurryent (ro ), and die srress ree•rvr (ii ). t hl Normal (t `) and Irnr,ycwriuJ (il ) components of the stress recto; (0. ]-
Since the unit tangent vector and unit normal vector obey —
n. = e'ae z$3,
we may write r t — e#4,1 L or r y .
(10)
Thus we see that if the stresses are given along all or part of a bounding curve C, the derivatives of Li must satisfy (l0) along C'_ In the classical formulation of boundary value problems governed by the hiharmonic equation, the value of the Function and its normal derivative are specified along the boundary. We shall show that this is the appropriate formulation here when the stresses are given along the bounding curve C. To do so, we must investigate (10) somewhat more closely.
2
44
Some Examples r.J Static Problems in Elasticity [Ch_ 5
Let us rewrite (10) in terms of the arc length s along C as t = etfu
au ^s
or
0} =
rj.2
t2{s) —
'
au ,
Integrating with respect to s, we see that the first partial derivatives of U are determined on the boundary by
fr2(s}ds + c t =_ F,(s) + c
#.l,i(s] 11 .2(s) =
J
1 (s)ds :
+ c x - Fx(s) + C 2 .
Furthermore, I-'.1n + U, 2 1i 2 —
0 t1 [^fr
= (F 1 + cj }n r + (F2 + r'Anx
and
c? I1 _ U.rtz + 1.). 2 r 2 = Os — (F r -+- c s }t 1 4- (F 2 -I-c 2 )t 2
.
Integrating once more with respect to s, we have, for some function ii, )(s) = H(s, cr, CO 4 r 3 he traction is known along a boundary curve C, U and O i t iefi We see that Kale are also known along C and the resulting boundary value problem is of the classical type. It should be noted that this formulation introduces three constants or integration. Just as in the torsion problem, these can be neglected for simply connected regions since the stresses will satisfy compatibility conditions guaranteeing single valued strains and displacements. In the case of multiply connected regions, additional conditions to guarantee global singlevalued behavior must be added, and these will suffice to determine the unknown constants. These conditions were first developed in a paper by J. H. Michell_• For future purposes, it is convenient to consider a decomposition of the traction t into a component tangent to a curve C and a component normal to C. The former will be denoted by IT and the latter by e N [see Figure 5.21(b)]. Since -
au t ^ ^ ^^^ ^s - • On the Direct Determination of Str esa in an Elastic Solid, with Application inrheThcory of Plates." Proc.
Land. Morfl. Sur. 3(1, tOO 24 L 1899)
Ser 5 3] -
-
Some Plane Problems
2 45
we have r
r ^
r P rr _~°N, i, as•P
DU ix., I T = — n -- r! as .
^
In similar fashion,
r^ = rfinP _ ePpU.p^ t, na = G.ow t;,to =
._ rp-
( 12 )
In the event that the plane strain problem is stated in terms of stress boundary conditions, we see that the mathematical problem requires solving the biharrnonic equation for U with the boundary conditions either given in the form (10), or (1 l) and (12), or, after preliminary integration, given in terms of U and on the boundary_ Solutions can be found in a number of ways. For appropriately shaped boundaries, separation of variables provides the simplest and most direct approach. A much more powerful approach is possible, using complex variables. Discussion of these methods would take us somewhat beyond the scope of this chapter_ The interested reader will find a very good account in Sokolnikollf; (1956, pp_ 262 324 PfÎl.Ali COORDINATES
Ire order to consider a certain example, which is naturally handled with polar coordinates, we have to make some preliminary calculations. We shall see that these rather straightforward results have some interesting features in themselves. From (15.4.4) of I or other sources, the biharrnonic V 2 V 2 ü = 0 can be written in polar coordinates as a2 U
t 011
I 02U
ir e + rS• +-r (10 2 are t r Pr + r3 002 =°
(1 3)
If we look for product solutions of the Form 14r, Û) = el sin aU or
U(r, 0) = r' cos 20,
we find (Exercise I) that such solutions fall into the following groups: to o +b o Inr + ca r i +da r2 Inr,
(14)
(ea + Ïo In r + gür2 -t h o r 2 In r•)1J,
115)
(a i r + b i r - i + c i r ln r + d i r 3 ) sin L,
( 1f )
0,
(17)
(e z r + fir- i + gir In
r + h i d) cas
(A i r + B i r - i + C i r ln
r +
D i r ; )0 sin 0,
(l Ii}
(L i :- -4- F i r - ' + G i r ln
r +- If i r 3 }0 cos 0.
(19)
(a.
+
-
b q r ' + cr' +2 + dr - x •2 ) sin
(e.ra +J f
20,
r -' • 9.e+2 + ir o r' 42 ) COS `JCfr
Here a 0 0, 1, arid, for the moment, need not be an integer.
(20) (21)
246
Some Examples of Static Problems in Elasticity [Ch_ 5
Let us assume that the elastic region contains at least a portion which covers the full range in 0,0 <. O 27r. There is no need for the stress function U to be single-valued in this region, but the stresses themselves certainly must have this property. We shall examine the implications of this requirement by calculating the traction_ To resolve the traction into polar coordinates, we shall use the results found in (11) and (12). Let us first consider C to be an arc of the circle, r = Constant_
In this instance, n 1 = cos 0, n 2 = sin @, t 1 = — sin 6.
72 = cos fl,
as seen in Figure 5.22. For this particular case, we shall denote f s by T, r and r by TA. The notation is chosen to indicate that the normal to the curve is in the direction of increasing r, the normal component is in the same direction, whereas the tangential component is in the direction of increasing O. From (12) we have i
= Tr =
.D x r ;u U,11 suie 0 — 2 U.12 sin H cos Q + UI.22 cos t OE
(22)
Similarly,
(T =
T— ^.
Uartvnp = —U,, I sin Ei cos 0 + U.13(cos2 0 — sin 2 0) +
i},22
sin @ cos B. (23)
We now convert (22) and (23) into expressions with derivatives expressed in polar coordinates. A straightforward calculation (Exercise 2) yields
^ l}
U . , cos El + U ,z sin H, —U .1 r sin
+ U . a r cos
^,
U .I , c0s 2 0 + 2U . , 2 sin 0 cos (i + U . 22 sin e (24)
—
U .11 r sin 0 cos 0 + U .22 r sin V cos 0
+U,rzr{cos2 — sin e z r
0)+
— U I t r 3 srR^f^ — ^U 1 2 ar2 sin
r 1
DU ,
-
^$
cosB +
U ,22 r 2 cos 2 Û
—
r
^ .
— U r
Se c. 531
Some Plane Prubkrrs
FIGURE
247
5, 22. Nnrmu! (a,) und
(To
)
reciors
Of
ap o»rr on a circular
ar c .
Returning to (22) and (23), we find that 1 0 2 U I DU T" 172" + r fY'
(25) (26)
Next, we consider the curve C to be a radial line, 0 = constant. as seen in Figure 5.23. In this case, + N is the same as Tim, and r as i . From (11) and (12) we find that ,
= f,' ,, cos 2 0+ 2Lr . , 2 sin 0 cos B+ U , zr cos 2 and tT = der = — U . i , sin 0 cos 0 + U.1 x(cos 2 0 — sin 2 0) + U .22 sin 0 cos 0. Again, we may use (24) to show that
t-
Tna f
ar
( 2} 1
, 1 (1.0
(28)
iïr TN
z FIGURE
5.23_
Monza( (n,) und lanye'rrfral
(ri„)
o -rt -rors at u paw
c,tt u radia!
24N
Some Examples of Slade Problems in Elasricity iCh.
5
(26) and (28), we see that T. _ Ter . We should note that T,., J. and 6 are actually the components or the stress tensor in cylindrical coordinates. In order to take advantage of separation of variable techniques for other geometries, it is desirable to develop tensor analysis in orthogonal curvilinear coordinates and, in fact, in general curvilinear coordinates. We can now return to the particular solutions given by (14) through (21) and determine which of these will lead to periodic stress components. By applying (25) to (27), it is seen (Exercise 3) that a must he an integer n, and the most general stress function that will lead to periodic stress components is Comparing
=u u +bfl lnr
c o r' + dc,r 2 ln r+e4 i9
+ ([1 1 r- + b r ' } c i r In r + d 1 r 3 + A 1 rD) sin Û ,
+(e, r+ f, r - Y +
1 r In r + Jr , r 3 + E j rfl) cos
"f 1y [(are + b"r " + c„e+1 -4- dyr1 sin n(^
f "
=2
f (er r" i jr a
+ y R rn* 3
+
KIRSCH PROBLEM: STRESS C[)ItiCF:fti
Î2r r - "+ 2 ) cos nil.
( 29 )
TRA F- 1(1Iti
An important, although elementary, application of these results is the problem first analyzed by Kirsch.* Consider an infinite elastic body with a cylindrical cavity bounded by
r2
= x
i
a2 ,
as shown in Figure 5.24. The cavity is stress-free and a constant traction Tit = S is applied at infinity. The boundary conditions for this plane strain problem are thus 1 3 =1'-=
at r=
a
(3O)
r= cx1.
(31)
and TY 1 = '.22 = '6;
at
The boundary condition (31) can be expressed directly in terms of the stress function as r G^
I 2 gx 2
s^
at r = 00
or
U --- ir 2 (1 — cos 20)5
at r = oc.
•"Dxe 7 heorie der Etastziias und (Ire Bedilrtnisse der Fcsligkcristehre," Y. Dew Acher lrryrn.
42, 797-8C17 1 1fi9B1.
Sec. 5.3] Soma
FIGURE
apposite
plane problems
2 4y
5.24. An infinite body with a circular hole. sublec . r ± x tKirsrh problem). 'torero, stresses at A'
to evurrl und
We may summarize the boundary conditions in terms of the stress function as follows: !im r-.x
I 0 1 Er a 2 ate
U
rt
(32)
= -SCI — ccs 20),
4
I r'd!
+-
a ar
^
at r = a,
(33) =
--ü
atr= a.
(34)
and
! OU
I ûz U OF)
In view of the condition (32) at infinity, we try for a solution in the form consisting of only two sets of terms from (29j, namely, ['(r. fl) = b o In r + t 0 r2 + cf0 r 2 ln r + (e,r
+ J3 r - 2
+
y 1 r 4 + i1,) cos 20. ( 35 )
It should be noted that the constant term a o has been omitted, since (33) and (34) only involve derivatives. Applying (32), we see that d o =-
= 0,
c o = iS,
,e2 =
1
The remaining constants are found from I33} and (34]. From (33) we find that -4 (36) ba = — 4a3S and }S — 6f2 a — 4hx u -2 = 0. Boundary condition (34) yields
+6fz a +2h z a -2 ^D.
(37)
Sum[' Examples t4 Skin( Problems in Elasticity [Ch. S
25.0
Solving (36) and (37), we finally find that li e = ia 2 S and 1-2 = - 10 ¢ S. Thus the desired stress function is
Cl ( r, 0) — —1So 2 In r is?
+ (- 45r 2 -- 1St? r— 2 + 4Sa 2) cos 20.
1381
The stresses can now be readily calculated from (38)_ We recall that at the boundary of the cavity, T. and Tr s will vanish_ On the other hand, the stress component Toe , which is known as the hoop stress, need not vanish. From (27), utilizing (38), we find that 4a(11,
U) = S(1
—
2 cos 20).
(39)
Relation (39) has some important consequences. We note that the minimum value, - S will our at t} = 0, n, i.e., in the direction cif the applied stress. On the other hand, the maximum value will occur in a perpendicular direction, i.e., t} art,la. The maximum is 3S. Thus the effect of the circular hole has been to magnify the effect of the applied stress to triple its value. This is a classical example of stress concentration produced by cutouts and curves in the boundary_ The multiplicative factor 3 is known as the stresscoiicenrration facrrrr. One of the important problems of applied elasticity is the calculation of stress-concentration factors for particular boundary configurations and loading programs in order to ensure that the maximum stresses throughout the body stay within design limits. Unduly high stressconcentration factors require modifications in design that will reduce the stress below specified limits. One finds, for example, that sharp corners normally produce a higher stress concentration than rounded ones; the consequences of this observation can be seen in the shapes of common structural elements ,
—
.
PLANE STRESS
In our discussion of plane strain, we considered an elastic cylinder t hat was infinite in extent in the x 3 direction. Assuming that the applied loads were such that the displacement was independent of x 3 , we found that the resulting problem was plume in that the solution depended only on x t and x 2 . We now consider another possible plane problem in which the body is essentially a plate that is thin in the x 3 direction and can carry only stresses parallel to this plane. Under these assumptions, the stress components To vanish. If there are no body forces, the equations of equilibrium arc 7 = 0_ same argument used in a following the rialyiing plane conThus, strain, we clude that there is a stress function U with the property -
-
lab = e.i
.VP •
(40)
See.. 5.3] Some Pkunt' Problems
251
Rather than pursue the argument through a further analysts of the sires tensor, it is somewhat more convenient to consider the differential equations governing the displacement vector. We recall Hooke's law (4.2.5), — 2ND + i cks, L J . ,!
Setting T33 = Q, we find that
2f.la: 33 + At: k A = 0, which leads to C33
}
+
2)/
"os
Returning to liooke's law, we can now write the non'.arlishing components 1 of the stress function in terms of the displacement components ^^ 1 and 11 2 only. We find that Tap
—
t1 li 1 . 0 +
2 0. up .i t
t` . : ^^c
.
(41)
The equations of equilihrium, including a body-force term, (] TQ3.7 j; — 0,
can now be written in terms of the displacements as Deus .,,p[1 + 2).(J. + 2p) 1 1 + IIUe
+ fp = 0.
(42)
We thus observe that it is possible to write fold equations for plane stress that include only those components of the displacement which are in the plane of the stress field. It is of interest to compare (42) with similar equations governing the case of plane strain. Although we have developed the latter by working exclusively with the appropriate stress equations, displacement relations can be Found simply by using Navier's equations (4.3.9) with the observations that
u 3 - O.
u, = u a(x l , x2), We then find that, in the equilibrium case,
►gyp.=a + (&+ } a.a +10
-
0.
(43 )
Comparing (43) and (42), one might he tempted to draw the conclusion that the problems of plane strain and plane stress are mathematically equivalent, provided one introduces a new Lamé constant, 1, in the plane stress problem defined by — 2p).
^l
+
2p
252
Sony Examples of Static Probfems in Elasrrciry [C`h. 3
But dependent variables may depend on the third coordinate, x 3 , since we have so far only exhibited those field equations that show the dependence on x 1 and x 2 . Therefore, we are left with a question: is plane stress a plane problem? Is there, in fact, a dependence on x',? The answer, unfortunately, turns out that in general all three of the spatial variables are present. This can be explicitly shown by considering the structure of the solution. The detailed calculations are somewhat far afield hut can be found in Love (1944, pp. 206-207). Love shows that the stress function U(x 1 , x 2 , x3) is given by
U nix , x 2 ) + U 1 (x r , x2)x3 — 4-J.(3A + 2.t)- lOexjHere T.0 = 0e + f is a constant, i t,(x t , x2 ) is plane harmonic,
00 and Ste' L.i
=
Similar expressionsshowing the dependence of the displacement vector on x J are available. We note that save for exceptional cases, the stress function, the stress components, and the displacement vector all will depend on x 3 . Plane stress is therefore not a plane problem in the sense that all three spatial variables enter into the solution. It should also be observed that the dependence on x3 is of a specific form. Consequently, the boundary conditions, whether stated in terms of stresses or displacements, must be given in the corresponding form as a function of x 3 , or else the whole plane stress analysis is not applicable_ GENERALIZED PLANE. STRESS
The question still remains as to whether one can develop a meaningful two-dimensional theory for stresses on a thin plate. We now show that this can be answered affirmatively with what is known as generalized plane stress, a theory that is based on an appropriate averaging of the basic quantities across the thickness of the plate. We begin by stating the assumptions that lie behind our approximation. We consider an elastic body in the Form of a cylinder with generators parallel to the x 3 -axis. The top and bottom surfaces are located at x 3 -- +h, as shown in Figure 5.25- The middle surface is, of course, x 3 = O. It is assumed that the half thickness, h, is small compared with the other dimensions. It is further assumed that the surface tractions are applied only to the cylindrical surface and these are distributed symmetrically with respect to x j the middle surface. In addition, we suppose that the body force f hasabout no component in the x 3 direction, and the remining components fa are also symmetrically distributed with respect to x 3 . As a consequence, the displacement component 143 in the x 1 -direction will be an odd function of x 3 and -
Sec. 5.31
2 53
Surin' Plume Problems
---
)- _
^
""1 Jr
^/
-1
-
h
y
------'" 77 ` ^^_^^^^. kt ^^ ^ FIC 1,Rr 5.25. C'Ylindrrrltf lrkrr[' of riieckurss
lateral
2h ritur r% +rrrcrll ttrrrrliurrd 1,, rrti
drrr7enSrCUrS.
1130C1, x2, D) =
O. Furthermore, 7;3(x i , x,, ± h) — 0,
since the tractions vanish on both the top and bottom surfaces. We next define stress and displacement averages by xl, Y2) —
J ^ 1(x 1 ,x 2 ,x)dx 3 h
and b fipc i . x 2 ) = 2h Jt1x21x3) dx3. Since u 3 is an odd function of x 3 ,
ù 3(x,, x2 1 = O. From the equilibrium equations + fr
T^
=
Q
we see for the case i — 3 that
T3., =
Q
TA 3, 3 =
0.
or Ts3.2
♦
(44)
Let us now apply (44) to the top and bottom surfaces, x 3 = 4-ii Since Tz3(x,• x2, ±h) = D, T33.3`x1, x2, ±h) = Q-
(45)
Some EA -ampresof Static Probkrru in Elasticity ECG 5
2 54
Recalling [hat (46)
T.3 3(-‘1 , X2+ ±h) = 0,
we see from (45) and (46) that if T33 were expanded in a Taylor series in x3 about ±h, it would be 0(h 2 ). We now make use of our assumption that the plate is thin, i.e., h is small, to neglect T33 throughout the whole thickness of the plate. Having exploited the third of the equilibrium equations, we now return to the first two ►
^^ ,
i
+ ^ = 0-
Computing averages across the thickness, we see that {
1 ^s.# + 2!i
'' ^
+T
T3a.3 ^^ 3 + Je
_ m
0,
-8
where h _fafx 1 x3l`2hf xi f,( ,x 2 ,x 3 )rix 3 . ^
1
Now, h
Jr."
Tii
0 I•
x2.
x 3J AI
^ T3a(xt. ]c2, X3)
=
O-
- h
-a
Thus our averaged equilibrium equations become T0n. +
= Q.
(47)
These are of the same form as the actual equilibrium equations of plane stress, It now remains to find the relations between the averaged stress and displacement components, to substitute these into (47), and thus to determine averaged Navier equations. Since we have assumed T33 = D, the corresponding stress-strain relation is = 21441 . 1 f Î411, Q t 14 3. 3lC onsequently, 14 3.3
R —{ + 2g1)- 114x.¢.
Hooke's law for the components of the stress components in the plane is aA = pi( 14 g.p t 14 0..) + .au k kba0 . Replacing 14 3.3 from the relation found above, we have }^^ = ü[uo. s + 14 0. l + 241(A + 214 1 U f. âQ0
Sec. 3-3]
Some Heine Problems
255
By integrating across the thickness of the plate. we obtain the Following stress-average displacement-average relations tip.2 ) +
roe =
(48)
where
2=
2Ap(2+2p) -
'.
Substituting (48) into the averaged equilibrium equations (47), we f nd that ç or + )A F.6^ + ,r, = 0.
(49)
Comparing (43) and (49), we see that the equations for the averaged displacements in generalized plane stress are exactly the sono: as those for the dis -placemntsi r,povdethLamcnsirepldby Consequently, the mathematical problems for the two cases are exactly the same, and results obtained in one case cari be interpreted directly in terms of the other. Thus we cari write down at once the solution of the Kirsch problem for generalized plane stress. At the heart of all of [hese problems of the plane type is a boundary value problem involving the biharmunic equation. CONCLUDING REMARKS As noted at the outset of this chapter, our purpose
has been to introduce the reader to several classical problems in static elasticity_ We have delineated a number of concepts that are of importance to engineering design, such as flexural rigidity, torsional rigidity, and stress concentration_ In addition, two of the very important partial differential equations of mathematical physics have appeared, Laplace's equation and the biltarrnonic equation. We have, in addition, discussed a number of approaches to handling the rather difficult problem of static elasticity_ An elementary exact solution has been exhibited in the case of the bending of a beam by terminal couples. A much more complex exact solution appeared in the problem of plane strain_ Several different types of approximate analyses have been considered. In the engineering theory of bending, for example, we started with a global result derived from a much simpler case., and have assumed that it would be approximately applicable in a more complicated situation. Rather than deal with stresses and strains, we developed a new theory of resultants, with the Euler-Bernoulli law serving as a stress ..strain relationship. This approach has been employed successfully in a wide variety of applications. The St. Venant torsion problem represents a different type of approximation, in which part of the unknown displacement is specified and part is determined from the field equations. This form of semi inverse method has Certain deficiencies in that one cannot expect to find a solution of the field -
2513
Some Fxurrrpfes of Static Problems En Elasticity !Ch. 5
eq ual ions and the boundary cond t ions except under fort uito us ci rcumst ances. In our case we were not able to specify the boundary conditions at will, and the success or failure of the method depends how significant the change to the boundary conditions turns out to be. Generalized plane stress represents another form of approximation that can also be applied in other situations. Here one takes ad vantage of a small geometric quantity to neglect certain of the unknown functions and uses averages of the other quantities over this small dimension. In sum, the examples considered here not only should provide an introduction to classical static elasticity but also should help the reader gain greater proficiency in the art of approximation.
F.?CERi'ttii::S
i. The purpose of this exercise is to develop the product solutions given in (14) to (21). Since the dependance on r in the hiharmonie operator written in polar coordinates is similar to that of the Eau equation in ordinary differential equations. it is slightly more convenient to introduce the change of variables r
^
exp t.
(a) Show that the resulting partial differential equation has only constant coefficients. (h) Consider product solutions of the form
[.I(t, 0 ) = T( r)6) (11), where T =
exp Ai
and
0 cos all or e — sin ae.
Show that this choice of functions leads to the tndicial equation (# -° al )[(A — 2) 2 — x 2 ] = O.
(c l
1501
This serves as the connecting link between the constants 1 and aa. Equation (SO) implies that a= A, —d, A-- 2,2—:t and
A—x —a cx+ 2,--a- 2. ,
,
Show that (20) and (21) then follow from the above. W) The remaining equations require a careful analysis of the double roots that can occur for either a or Perform this analysis. (e) Rewrite the solutions in terms of r and [l rather than ; and O. 2. Derive (24).
Sec. 5 .7] Srume P lan e Problems
257
3. By adding the requirement that the stress components must be periodic in 0, show that the most general solution of [he biharmonic of the product form in polar coordinates is given by (29). 14. (This exercise and the following make use of the solution of the Kirsch problem given in the text to find the solutions of addi[ioral questions through simple modifications.) Consider the situation shown in Figure 5.26, in which the stresses at infinity are Tiz(cc)=D
an d
T22(cc) =
Find the stress function and the shear concentration factor_
P
FIG U R E
Th e Kirsch problem irith sfres,r applied or
i= — +
x.
5. From Exercise 4, find the stress function and the stress concentration factor for the infinite solid with a circular cylindrical hole under biaxial tension, wherein T11(cc) = T,1(co) = P, 1.12 (co) = O. #6. Consider the same solid as ire the previous exercise, subject to pure shear, i.e. (Figure 5.27), 11 z fx)= T2i(x)= S.
T, i (oc.)= T22(3c) =a
Find the stress function, the stress concentration factor, and the location on the boundary of the maximum hoop stress.
Some Examples al Stark Prahlem-s nn Elasticity lCh_ 5
25 H
.ti
i ^
l[; l' T.1 f:
irr frrarr
5 27 An rrrl`rraire J , a,e11 +i air ca ciir rrlcl+
.uhlec r
it par
slrrar r rf
^
7. Consider the same body as in the previous exercises subject to internal pressure in the cylindrical cavity and zero traction at infinity. i.e. (Figure 5.28), ^y + + O. = ` !' , ^ d — D, 15 ^ ( ^ !y = T. 2 ( 2 2 ( x. Why is the negative sign taken for the traction on the hole? Find the stress function and the stress concentration factor.
F il RE 5.28. .9rr rra{rrrrrrr body with a circular kirk. SOW, rri irNel7tal{rrr.+sure tige
r rlirreatruerl curio- and
:ell)
traction [Jr rrrfrriirry.
itr
Note that this problem can be solved by the superposition of two stress systems_ The first is biaxial tension (Exercise 5) and the second is a state of constant pressure, Tz a (r, 1)) = T22fr. 0) = P, Ti2 (r. O) = 0. (51) Show that the stress function L', corresponding to (5 l) is given by rl = —zPr2. —
6 Introduction to Dynamic Problems in Elasticity
CHAPTER
W
: HAVE just considered some examples of elasticity problems that were independent of time. We now turn our attention to dynamics (i.e., timedependent problems). As one might surmise from our experience with statics, time-dependent boundary value problems can present serious mathematical difficulties. Consequently, our discussion here will be confined to extremely simple geometries. The resulting mathematical questions will prove rather elementary; nevertheless, it wIl] be possible to see a number of interesting physical phenomena. Initially, we shall examine the displacement equations and show that they are wavelike in character. Taking our clue from the analysis of one-dimensional discontinuity surfaces examined in (,Chapter 12, we shall make parallel calculations for the three-dimensional case. Finally, we shall consider the effect of boundaries in some very simple cases. A good reference for elastic wave motion, which contains comparison between theory and experiment, is Graft - 11975).
6.1 Elastic
Waves
in Unbounded Media
We shall conne our attention to an elastic medium which is homogeneous, that the elastic moduli are constant througheut the body_ ln addition, We shall assume that the body force vanishes, se that the field equations are also 5❑
homogeneous.•
Navier's equations î4.3.9c) for homogeneous media in the absence of body forces are t4r 1 ar,, ► ^ + ^ — 2r^ u 2P — .,
11)
where -
49"
ULLATATIONAI. AND ROTA -IIOl AL WAVE EQUATIONS
Differentiating (I) with respect to _x r (and summing) and interchanging the order of partial differentiation, we find that p(1
+
v)(1 — 2i) .. 12)
E(1 — v) • hfl eihulk are avai Lab
for taking into accouns the presence rif body loves. with the result
chas the essence. of she mathrmattral, prnhkem is k:nnsained to the homogeneouscm,e. z5y
26c
Introduction to Dynamic Problems
in Elasticity [Ch. b
We recall that the cubic dilatation, A, is given by Cif
;.
Thus (2) may be written entirely in terms of the cubic dilatation as (3)
C; 2A,
.i1
where ^ —
^ °—
14 1 —r) p(1 v) (1 , 2x' )
+2p
(4)
p
We see from (3) that the cubic dilatation ❑ satisfies a ware equation, known as the dilatational wave equation. The speed of propagation of such dilatational waves, CD, is known as the dilatational wave speed. Dilatational waves are frequently called longitudinal waves, or irrotational waves, or, in seismology, P-waives (where P stands for pressure). Navier's equations (1) also admit another wave-type solution. Differentiation of (1) with respect to x i, yields
+ (1 —
2v) 1 uj,1ik ^ 2p{1 + OE -
(5)
Interchanging the role of the subscripts t and k in (5), we obtain ^k. 11J
+ (1 — 2v) t o fkï
2p(1 + v)E -
(6)
We recall from (4.I.32b) that the rotation tensor W ik is given by
Thus subtraction of (6) from (5) yields ^ik.J1
_ Cr ^..
rt
(7)
We see that the rotation tensor satisfies a wave equation, the rotational wave egnation. The speed of propagation Cr is given by
Cr
2p(1+ >) — P
(8)
These waves are also known as eqnivolurninal waves, or transverse waves, or shear waves, or S-waves. The speed of propagation CT is frequently referred to as the shear speed. WAVES VIA THE HELMHHOLTZ REPRESENTATION
These results can also be found in an alternative way. We start with the Helmholtz representation of an arbitrary vector field, Theorem 2.112. Recall that according to this theorem the vector field u i may be written as
Se['. 6.1]
Elastic Wrn`r.S in #hrfiuu ►xir[1 Media
2ô1
where and w, — Cijk kirk. j
;
V1 =
and k.k
—
U.
In words, this representation states that the vector field can be written as the sum of the gradient of a scalar field and the curl of a divergence-free vector field. Direct calculation (compare Theorems 2 . 14. and 23.5) will also show that }^; t
= C} and
c;1k Vi. 1 = Q;
in other words, u , has heen decomposed into the sum
of an irrotational vector
v, and a solenoidal vector w ; . Let us consider first the case where u, is solenoidal, i.e., v, _ 0 and id, =- w, . Navier's equations (1) then reduce to /ILL;
2p{ I + v} E
i+;,
which is the wave equation governing equivoluminal waves. Next, let us look at the case where the displacement is irrotational, i.e., ►c'; = D and u ; = v,. This can be best analyzed (Exercise I ) by rewriting (I) in the equivalent form 211 — v) ( x+ . 11 + 1 — 2v 2ti
2p1I + i 1 . ipq“:ysrtrr.sl,p
-—
L
f91
In the case being considered, however, Eer = 4, so that i9l becomes l ,(1
u I ,;J
^
— 2r)(I + r) .. u„
E(1 — v}
which is the dilatational wave equation. Consequently, an arbitrary displacement can he regarded as the sum of equivoluminal and dilatational
waves. PLANE: WAVE SOLUTIONS
Among the many particular solutions of the wave equation, one of the simplest and most useful is the plane wave solution. p} a plane wave we mean a scalar or vector function of space and time that is constant on a plane in space which moves at a constant speed normal to itself. Such a plane is given by atx ; — et =
where GA moving plane.
constant,
1 and r is a constant. The vector at ; is the unit normal to the
11as664
iiofro d:1040ff iv 0)7W/ilk - Prohir+ms
262
[Ch. 6
Let us consider a displacement vector of the form Q k x{ — CI- -
11;
Certain formulas are slightly simpler if we use Navies s equations written in terms of Lam& constants, i.e., ;tut t1 F ( 1 4
t
-
(10)
=
For our particular form of displacement, (10) becomes
(+
4- p a aj te;.
p)a i
Rearranging the terms in (11), we
=
u;
d dC
',
obtain
CO. +a t a; + (i
—
pc 2 )5 ; f]u; = 0.
(12a)
The succeeding calculations can be somewhat simplified by letting A
+
^
=
Pc x — ^•
te;
- 1; .
(13a, b, c)
this (12a) becomes (A70Œ1 — 06,j)V1 , 0-
(12b)
Equations (12b) are a standard algebraic eigenvalue prohietn_ The require-
ment for a nontrivial solution is del (Axra 1 — Œô; ) 3 0-
(14)
Once the eigenvalue a has been found from (14), the values of the speed of propagation e of the associated wave will be determined by (13b). Since [here are at most three distinct values of the eigenvalue. there will be at most three speeds at which an elastic plane wave can propagate. The eigenvalues for this particular problem can be found rather simply by returning to the vector equation (12b), written as
Aa;a;
= rrl;.
(15)
We observe that any vector in the plane perpendicular to the unit normal a.;
will satisfy a lr =0-
Such
will therefore be an eigenvector of (15). corresponding to the eigenvalue a = 0 Since there are two mutually independent vectors in the plane, a — 0 is an eigenvalue of multiplicity 2. Now the sum of the eigenvalues is equal to the trace of the matrix (Theorem 2.2.1 or Exercise 4). Therefore, a vector
Are ; 12 1.= A=0 +0+a. Thus the three
eigenvalues
are a = A, O. 0.
Ser
Wares in
L'rrboulzded M e diu
2 63
1.et us now interpret these results in terms of the physical variables of the problem- Corresponding to rr = D, it follows from (13b) that (16)
C2 =
which is the square of the speed of a shear wave. Since the cigenvector lies in the wave front plane
a; xi - et = constant, the corresponding displacement vector is parallel to the wave front. The is polarized in the plane of the front, The remaining eigenvaltle rr , A, yields
wave
,
C
2 —
+ 2p P
Returning to (15), we find
ct i ci j
= j;.
(17)
Equation (17) states that [! is equal to its projection along the unit normal ai; i.e., Vi is parallel to the normal to the wave front. Consequently, there is a second possible elastic wave that is propagated at the speed of a dilatational wave and represents a displacement perpendicular to the plane of the wave. These two waves are the only plane waves possible in an clastic body.
EXF:RCISFS
1. Derive (9)2. Using the form ^i = CP.e+
equations (1) can he put into the form
show that Navier's
2(1 - v) 1-V
p41
.Jl
+ %)(l - 2v) (1
-F
-4-
cijk Ok.Fa
_
2p(1 -+- ] ,
- O. 1
i
(Thus we can conclude that the displacement h i satisfies Navier's equations if it is decomposed into a scalar potential which satisfies the dilatational wave equation and a vector potential tif k which satisfies the rotational wave equation.) 3. (a) Consider a solution of Navier's equations of the form u l -= f (x i
Find the value of e.
-
ce).
u Z = O,
u, = f}.
horro€!uctipp to Dynpinir Problems i n Elasticity
26 4
[Ch. 6
(h) Make a similar calculation for a displacement of the form u, = u 3
u2 = QLXI — Cr)
Q,
Explain how these examples justify the use of the words" longitudinal" and "transverse." 4. For the 3 by 3 matrix A consider the eigenvaluc problem given by (c)
F(a) = det If the
cigenvalues
are
Cf,, ô
and
(Ail --
03,
ab ;) = O.
then
F(a) = (rr - a)(0. 2 - (c)(0 — ,
4
Compute F"(0) to show that the trace of the matrix is equal to the sum of the eigeIivalues_
6.2 Propagation of Discontinuity Surfaces Section 12.3 of i provided an analysis of the one-dimensional motion of a discontinuity surface. A discussion given by Keller• will form the basis of a generalization to be given here. Let us denote the moving discontintaty surface by 0(x , t) = 001 , x 2 , x 3 , r) � 0.
(1)
Equation (I) can either he represented as a moving surface in three-dimensional x i-space or as a fixed surface in four-dimensional (x, r)-space_ Let us consider the first interpretation for the moment. The trace of ¢ T D in 3space at two different tunes r i and! : i 2 is illustrated in Figure 6.1. From (1) we have
=il
+
-
( 2)
On the other hand, the unit normal n, to the surface f❑ r fixed r is given by 1
Nick
k 0,k
Utilizing this result in (2), we find that ( 3) ,Î.k 0.k This useful expression for the normal component of the velocity of a point on a moving surface will he used again in Section 7.1. * 1-I. B. Keller. "PropagatiUn of Stress Discuntinuitiesin Homoge.rcuus Elastic. Materials," SIAM Rcr. d, 356 ,8 I ( 1964) .
Sec. 6.2]
Propagation o,/' Discontinuity Surfa ces
0=
265
0
I
r=
t=tt FIGURE
6.1.
Two .vuccrssït•e positions of a discontinuity surface.
A CONDITION ON MILD DISCONTINUITIES
Now let us consider (1) as represented by a fixed surface F in four dimensional (x, 1)-space. We shall denote the four-dimensional unit normal to the -
surface F by (vi, v0). if ;, r) represents a point on F, the jump in a Function f(x, r), denoted by If (x, t)i is defined by ! (x,
llm [ f (x ; — Si'
f - fa%) - for,
!t0
F.i"
r + c 'u)]
Just as in the one-dimensional case, we shall restrict our attention to mild discontinuities in the displacement vector, By this we mean that (i) the displacement vector is continuous across and (ii) any derivative of the displacement taken tangential to I - iscontincous across I Assumption (i)implies
r,
that
[ ,1
=L}
on F.
(4 )
If(a i , a d } is any direction tangential to the surface, fl; l', 4- a o r d = Q. The continuity of the tangential derivative. astiumptiun (ii 1. can then be expressed as rr;^llr.r'
f c^ o^tr ^ _ 0. ,
( 5)
Relation (5) states that if we consider, for fixed i, the four-dimensional vector ([ u,.fl , coo, ti will be perpendicular to every tangential direction along F. Therefore, this vector will be parallel to the normal to F. Thus there is a parameter };, corresponding to the fixed value of i, with the properties = kip .; and [fir ; ➢ Eliminating the parameter b„ we finally conclude that
—
o + ^• 1
(6)
l+rrrr,durfiorr fo Dynamic Problems in Efasficrr4' [C'h. h
266
FURTHER JUMP CONDITIONS
The equations of motion will now be examined to determine further conditions on the jumps - Recall that continuity of the integrand is always required to pass from postulated balance laws, involving integrals over arbitrary regions, to differential equations_ We are specifically interested in situations where continuity does not obtain, so that we begin with the momentum balance equation in its integral form, ti
rlr
^^rpriid dr =
d ^;; rr, du
Jr'
+
^1{Ilj
^tirl
,
^^^ -
I, di.
(
7)
RI l I
This is the fundamental Cauchy equation [i.e.. Equation (14.2.51 of I] with the following modifications. (i) As is always done in linear elasticity, we approximate vi by ts. (ii) We have expressed the stress vector in terms of a symmetricstress tensor. (iii) Wehave followed thecon ention used in elasticity thatf represents the body force per unit tro fume. We shall integrate Mover an arbitrary time interval Tgivenbyt l < f tz. Our assumption of small deformation implies that the spatial region R varies only slightly in time (provided, perhaps. that a rigid body motion is appropriately applied--but this will not change the force balance under consideration)- Thus, our region of integration G in four dimensional (x, ti-space will be a slight perturbation of a hypereylinder with directrices located parallel to the r-axis. A region of this type is depicted in Figure 6,2, but with a restriction for clarity to two space dimensions. In set notation, the integration region G is the Cartesian product of R and T, i_c., G = R x The boundary of G will be denoted by r"G, the element of volume in G by cle and the element of surface area on c?G by dcrThe essential step in our derivation of the dynamic jump conditions is placing the possibly discontinuous terms of the momentum equation into a surface integral over the boundary r3G of G, To this end, for a fixed subscript i we define a 4-vector t Iil by -
,
(
**t^ )
t ct+^ s =t
r
u b^s ^1 * b2 , ^3 ) — 4 - 4011, , T,l
7r13
T. 1);
that is, t+1
On the "fiat" surfaces cf G we find the following. On the top of G (when r t 2 ): —
vu = 1; v, — ❑, j # 4; and ViIv j = — pa ;
(8a)
On the bottom of G (when r = r, ): v„
-- 1; v } = L), j it Q, and
ar y
l=
(8h)
Se{
ab7
_6.21 P► operyeuNrrr u ( !)r,+c rarrrrrrrir r SWIM CAG
bay . l'i u1'3
I;u
-4.
12
Y^
Flcit+kr•. b.?. .4rr whims/3 -t -l uhürre ilrr'flriY'
G
_ fi r
G, l#n hl rr cfr,tiiruurhlfl
^
On the nearly cylindrical sides of G. it is consistent with linear theory to approximate by a vector that is parallel to the top and bottom surfaces of G_ Thus, on the sides of G, Vp = 0; r1 = si, j
0
and ,i''1' J
T1r.
(#Sc)
With (S) it is quite easy to verify (Exercise l) that integration of CO with respect to r, from r, to I , . yields .
jjfe)v1d5 + . 0.6
riff
.f. tit = 0
or (9)
268
fnrrodurtion to Dynamic Problems in Ekstitrty
1C11. 6
We now consider our arbitrary region G cut by the fixed discontinuity surface F. The surfacedivides G into two volumes G, and and the boundary 13G into two hypersurfaces I- , and r as shown in Figure 6.2. As is also shown in this figure, we select the normal v that points into G2 as the unit normal on the discontinuity surface F. Note that (: 2
G,u G2=G,
r, 06 2
,=r t +
F, cG4
r,+ri .
Equation (9) will now be applied to the region consisting of all G. to the region consisting of G, alone, and to the region consisting of G1 alone. This yields
+ ffiCVvi — fill f. di, JjJ' r, ffiEpv, drs { ^^^ ^^, l' ^r(s = — fill fi GI
T- 2
iiG=
r,
T
GI
r
r
G2
Note that the sign preceding the second integral on the left-hand side of the last equation is negative, since the normal No has been chosen directed into G 2 , whereas the normal used in Green's theorem is an outward normal. Note also r are evaluated on F that in the second and third equations the quantities as this surface is approached from within G, and G2 , respectively_ Now we subtract the second and third equations from the first. This yields namely an integral involving jumps taken over a portion of the surface
F.
— p^r:^^4} drr ^ 0.
(10)
Since the portion of the surface l' a ppearing in (10) is an arbitrary section of 0(x, t) = 0, the DuBois-Reymond lemma may be applied to yield the followlug equatio n, involving the jumps in the stress components and the velocity on
(t l) The unit four-dimensional normal vector (v.,, v a ) can also be written in terms of4i(x, r) as ( v) , v01 "
{o
.k'f'.k
+ ( , ) 1 ] - 1 rx(^^* 431
-
Therefore, (11) becomes f^^i rid4b-
(12)
Sec_ 6.2J Propaguriorr uf fJr-tic-aririmrirl' Surfel[ e's
-{,g
Our object is tu obtain a relation involving the displacements through the use of (I2). From Hooke's law, = ^(^rii + ir 1.i ) }- ).u k . k -5 0 ,
we find from (12) that A p .^ + ^trr.. t t .r = a c^
fGi
QT411^0.f = ft Qtr, ;j
üui. cl
(13)
1 he jumps in the spatial derivatives can also be replaced by jumps in the velocity by means of (G). This yields
-
Efi and
06
u^fi , k ^
1110. k
^
Equation (13) can thus be reduced [o
"1
"
+
+ P)Efij ,I4. , P^tri
(4
This can be rcwrit[en as
pQii;N) 2
(i + /.t0.346 .1^.• — l^Q6i 1 + — — ^ ^ ^ j,
^.^ F.j
We further reduce
(14)
( l4) to _
{ ^f
+
)21.&1+ P ^.(019
_^,^^,, ^.k ^J4 ^^1^ = Q,
(15)
SOLVING FOR DISCONTINUITY VELOCITIES
Equations (1 5) are a set of linear, algebraic, homogeneous equations in the jump in the velocity [1î ; 1j. With the identifications
3 € —
(W - -
ai =
,
r
,
this system is essentially the same as (1.12a), which was previously considered in our discussion of plane waves- That is, the unknowns in the first case are the tr; and in the second case the C Li J ^, but [he coefficient matrices are identical. Thus we can write down the eigen values and the eigenvectors at once, with the result
( (b) 2 _ _ü 0.k 0.k y If
latroduc•tion to Dynamic Problems in Elasticity
2jo
ICh. 6
or
(ii)) x
(17)
^}
0,01).k
^
We observe from (3) that the left sides of (16) and (1 7) provide the square of the speed of the discontinuity surface normal to itself. Consequently, the dis one of two speeds, just as the case for the-contiuysrfaemovnlt plane wave. These are the dilatational speed,
Cr, _ and the shear speed,
Returning to (16) and (I 7). we see that the discontinuity surface must satisfy the partial differential equation (W) 2
or
(62 Following the nomenclature of geometrical optics, (18a) and (18h) are known as eiconal equations_ In contrast to the electromagnetic case, there are two equations corresponding to the two distinct wave speeds. ORIENTATIONS OF DISCONTINUITY SURFACES
Useful information can also be found concerning the cigenvector. [û ; L the jump in velocity across the discontinuity surface- If the surface propagates at the shear speed, C r , (16) holds and (15) becomes (
j.(0.1,(13.1,1 - 'DM =
+P
or = 0.
(19)
Equation (19) states that when the discontinuity surface propagates at the shear speed, the jump in velocity is perpendicular to the normal to the surface in 3-space. In other words, the jump in the velocity is tangential to the moving discontinuity surface. Alternatively, if the discontinuity surface propagates at the dilatational speed, then (15) becomes
,,}^,, ; s►" .k
.^
Q 6 1 ^ a.,[ 6,i. ^
Srr. e5, 3J
271
ReJlecrrurr r), Plant Shear Kid rws
Recalling that the unit normal, re,, to [Ire moving discontinuity surface Sd t isfies 6 .1e 0 0.e► 04
It
,I
!
-
.
we sec that the jump in the velocity will satisfy Il rl
= rr t nAû li.
120)
Equation (20) states that when the discontinuity surface propagates at the dilatational speed, the jump in the velocity is directed along the normal to the moving surface. This coincides with the result previously «hraincd for plane wave propagation.
t:xEttc IwF:
I. Provide all omitted details for the passage from (7) to (9)-
6.3
Reflection of Plane Shear Waves
Our previous sections have dealt with the behavior of elastic waves in unbounded spaces. Our discussion here will he concerned with the effect (if a boundary. In particular, we shall consider an elastic half-space occupying the region x 2 2 0, as shown in Figure 6.3. l he surlace x 1 = li is assumed free of applied tractions.
%7ff77777ff/77777J/7!!J/7!/l177s
a„ IIY
a F Kr u K t
61
A plane thew Kate
This wut •r rs reflected ai angle (1j, Khasi, direction a makes un angle
rrae-rdrrrr ai angle tJ an u free unlace In ue.idr ur►re. a 1r,rrjltiurlrnul "tint upprars
FA
rrh rive 1 rrrrrul
fnfrodurriolr tri Dynamic Prnhtums +n Elasticity
272
[Ch. $
FORMULATION
Since the normal to the free surface is (0, — 1, 0), the boundary condition on the free surface is f;
- T3 = 0
on x 2 = C.
(1)
We shall now consider the following special problem subject to the boundary conditions (1). An incident plane shear wave, whose displacement vector lies in the (v i , x2 )-plane, is propagated toward the free boundary x2 = a. We wish to determine the waves that result so that both Navier's equations and the boundary conditions (1) are satisfied. One can think of the physical situation as being set up in the following way. At time r = D a disturbance is activated that thenceforth continuously propagates a plane shear wave toward the free boundary of a half-space. The question is: After a time long enough so that initial transients have died away, what kind of elastic displacements will have been generated by the incoming wave? The total displacement field can he regarded as the sum of the given incident wave plus a remaining scattered wave.* The latter is due to the presence of the boundary, for in the absence of a boundary all conditions will be satisfied by the incident wave itself. The scattering problem under consideration here has, as we shall see, the unique distinction Elf being a major implement in shattering the theory of the elastic ether. The scattering of a plane wave by a free surface is the simplest prototype of a large class of problems that are important in geophysics. In this context, the free surface is an idealization of the earth's surface in which gross curvature effects have been assumed negligible, as have local inhomogcneities. The free surface is thus naturally thought of as horizontal, and the plane to which the displacement is confined (plane of polarization) would then be regarded as vertical. The waves in question are consequently sometimes called SV-waves, "S" for shear and' . V" for vertical plane of polarization. Horizontally polarized shear waves are called Sl-1 waves, while pressure waves (i.e., compression waves) are called P waves_ The letters P and S are also regarded as standing for primary and secondary, for earthquake detectors first register the faster-moving pressure waves and then the relatively slow shear waves. The plane SV wave under consideration here will be associated with a fixed normal direction n lying in the (x i , x 2 )-plane and making an angle f with the positivex 1 -axis_ Let Idenote a unit vector in the (x x 2 )-plane that is parallel to the trace of the wave front in this plane. Their the ensuing displacement vector must be parallel to r to ensure that the wave is a shear wave. We shall further assume that this wave is harmonic in time, since more general plane waves could then be constructed by Fourier integral methods. A
une- dimCrlsional scattering prohlem was cunsxlenYl
,3r
the enc4 of Section I2.2 of 1.
Sec_ OJ) Reflearrrn of Plane shear Weii.es
273
Referring to figure 6.3, we see that n and T are given by ri
i Sic] l— jeos 0,
T= i cos (i-+-j sin 0,
or, in terms of the Cartesian notation, n, = sin U, n 2 —
—
cos B, t t =
cas
U, T 2 = sin O.
Thus the time-harmonic incident wave of amplitude A is given by
ur"t = to exp i(&n • x -+- tur) or tlsil
k X ^ + ron *
— ra o exp
It should be recalled from our work on one dimensional waves that ki2rt is the wave number and cv/27r is the frequency. Since ue has been constructed d s a transverse or shear wave, it follows that k and w are related by -
(4) where CI —
as in (I.ti).
A -i TF;114I EU SO I. rED N WITH A R1r'FLECTED WAVE
A straightforward calculation, similar to one that we shall make shortly, will indicate that the stresses calculated from eea`' will not satisfy the boundary conditions (1). We attempt to remedy this situation by adding to the solution another plane shear wave, one that is reflected from the free boundary. As indicated in Figure 6.3, the normal ni of the reflected wave has the components sin OR and cos OR. The corresponding unit tangent vector t in the (x t , A2)plane has the components cos Ux, — sin OR. The resulting transverse plane wave i4," of unknown amplitude B is given by.
ir(»
Iq B exp i(ktmr t.x ) -+- ter)-
(5)
We now attempt to determine the constants B and UR in (5) so that the total displacement
ut: ) will yield stresses that satisfy the boundary conditions ( I). Since the displacement represents a shear wave, the cubic dilatation vanishes, and the stress-strain relations become
=
L^{te
;.x
+ u ► . ,) I+v
(6)
uk. a<
(7)
We again remind the reader that Greek subscripts such as re and ; takç nn Lhe values L and 2 only. In (3). the letter i stands for "Incident" when used as a supersCrip ; ii is used again as ihr
imaginary unit.
ftarndurlrv ►i ta DJvrarrrrr Problems Elasticity [Ch.
274
6
Returning to the boundary conditions (1), we note that T32. = Q by the twodimensional character of the displacement vector. The boundary condition 122 U implies that 2. x = 0 on x 2 = 0. This may be written explicitly through (3h), (5), and (6) as —
1. 2 ikn 2 A exp i(kn r x ! + cur) + f2 ikm 2 B exp ijkm l x i + ctr[) = O
(8)
Since (8) must hold identically for all values of x i and I, the exponential factors in each term must be identical. Consequently, n, = m,, i.e.. sin = sin OR. Furthermore, since the normals to the incident and reflected waves make acute angles with the positive x,-axis. F] = O R . This means that !]
Tr = I 1, T 2 = —f2
and n 2
Thus (8) implies that A = — B. We see that we have only needed one of the two nontrivial bocindaryconditions to determine all the unknown parameters in the reflected wave. The remaining boundary condition, T1 2 = 0, requires that when x 2 = a, 14
1.2 + 14 2.1 = O.
Making use of (6) and dropping the common exponential factor, we see that this implies that f
,ikri,A + fitkni 2 B + r 2 ikriA + f ikrri 1 B — O.
Further simplification can be obtained by means of the relations between the components of T i„ tQ, na , and m previously found, with the result that
l*in2 + T 1 n t )A - (T rn2 + T 2 n r )B = t} Consequently, A = B, which, combined with our previous result, implies that A = B = O. Thus the only solution with the structure (6) is the trivial solution consisting of no incident wave and no reflected wave. AUDI IIOhïAL itEFI.EC t Et) WAVE
To the reader who has some familiarity with the corresponding problem in acoustics or electromagnetics, this result should he somewhat surprising, since the corresponding problems in either of these other areas are solved by precisely the method we have just employed. The underlying difficulty lies in the fact that the elastic problem requires the satisfaction of two nontrivial boundary conditions, in contrast to the other cases, which have essentially one such requirement We have, however, one additional fact in elasticity that we have not used: Longitudinal waves, as well as shear waves, are possible. It will now be shown that through the addition of a reflected longitudinal plane wave al", the boundary value problem can he solved_ We consider then a displacement of the form t1Q ^
14(: ) 1- ri (¢`) + ui"r
5'ur. 6.3] ROW . lion of Ham. Sissur Warc•.s
2 75
Ie, = r,A exp i(kn 1xy + am) + TaB exp r(krr^x. + wr) 4- Q f exp r(k L arx, + tor).
(9)
Hereea R is the unit normal to the reflected longitudinal wave front, as shown in Figure 6.3. It has the components
a2 =
0, = sin 4.
Cos 4).
Since the displacement represents a longitudinal wave, it is directed along the normal or,,, as is seen in (9). Again, from our analysis of longitudinal plane waves, we may write [using (4)] kr
1 =
ca
k r _ C- ï
SO —
CD.
-
k y
(10)
'p'
where C r) is the velocity of longitudinal or dilatational waves. Furthermore, since the dilatational speed exceeds the shear speed, k I_ < k. As we have seen in the previous analysis, the exponetitials in each of the O. Therefore, three terms of 19) must be the same on the free surface x2 k L rr,
(ll)
k sin fi. = k t sin 0.
412)
kn I = km t or k SLIT t)
Consequently, the angle that the normal to the reflected wave front makes with the x 2 -axis is given by s i n 0 =— sin O.
sin CI) =
(13)
7
Referring to (9), one observes that (12) can be interpreted as stating that the speed of each of the three terms which make up the displacement along the free surface x 2 = O must be the same The nonvanishing stress components generated by the longitudinal wave along x2 = 0 are given by -I-
v U
E 12 —
2(1 + i)
1-
2v r.r
4':2+ I ^ l
(i
-
2 :I
If exponential terms are suppressed, this implies that T
^,
Tit') --
l+ v
E
Ctk r ai+
— CIk^^t02.
I+r
v
4 1—
-
6
Introduction to Dynamic Problems in Elasticity [Ch.
276
(Here " " denotes proportionality.) We now enforce the requirement that T22 = O at x 2 = 0 by adding the contribution of the longitudinal wave to (8). This yields ^~kL^cr2
v t1
-2;
t kt 2 ►4 2 A+ kr‘gm 2 B , 0_ ^
Because t2 = — T
111 = m111,
,
m2 = —112,
T =
r; .
we may simplify the last relation to + kt 2 n 2(A + B) = Q_
Ck z 0 +
(14)
In similar fashion. the boundary condition T 12. — 0 at x 2 : C becomes 2Ck L a i a 2 + k(-r 1 11 2 + r2n1)(A — B) = 0. (15) The components of the normals and tangents may be put in trigonometric form, following which ( 14) and (15) become v
0,
(lba)
k(A — B)(sin 2 S — cos 2 0) + 2Ck i_ sin rti cos 0 — O.
(16b)
MA t B)( — sin 0 cos 0) + CIL c os 2 0 +
—
1 — 2v
The remaining important quantities to be determined are the reflection coefficients CIA and B/A. These are the ratios of the amplitudes of the reflected longitudinal and transverse waves to the incident transverse wave. They can be found by solving (16a) and (16b) simultaneously, with the results {Exercise 3) (k4 jk) sin 40
A cos' 20 + (k Lik)2 sin 20 sin 20
117a)
and
B
_ cos 2 20 — (k,jk) 2 sin 20 sin 20
A
cos' 20 + (k 1Jk) 2 sin 20 sin 20
(17h)
INTERPRETATION OF RESULTS
Some observations about the solution are in order. On the basis of (4) and (10), it is seen that the reflection coefficients depend only on the Poisson ratio +° and not on Young's modulus E or on the density p. The speed of propagation along the free surface is independent of the frequency; i.e., there is no dispersion. Note that a reflection coefficient could in principle be complex, corresponding to a phase shift in one or both of the reflected waves las can be seen from (9) by writing, for example, B = (BI exp i(arg B).1
Sec.
6.31 Reflection oJ Mire Shear FSrure_k
2 77
Since kjk i. > 1, {13) shows that there is a critical angle of incidence ] r sin -1 (k ifk) for which sin r — I. For O > $; there is no real angle ¢ that fulfills OM; i.e., the assumed forrrr of solution is not appropriate. The case 0 > û; is taken up in Section 5_ Our calculations indicate, and further investigations substantiate, that both transverse and longitudinal waves appear when a wave of either type is scattered by any nonhurnogeneity, such as a force-free surface, a crack,or an inclusion of a different material. This fact has particularly striking relevance to nineteenth-century ideas that optical waves are carried by an elastic ether. It became established experimentally that all optical waves are polarizable, which means that they must be transverse. This is incompatible with an elastic propagation medium, which invariably contains a longitudinal component in a scattered transverse wave. The difficulty disappears, however, if the ether is regarded as a material insensitive to deformation but responsive to rotations relative to some fixed "absolute - coordinate system. As shown in Exercise 4210, the resulting equations of motion for small displacements in the absence of body forces are formally identical to Maxwell's equations! See Snmmerfeld (1950, Sec. IS), For further remarks on this fascinating, albeit eventually abandoned, line of investigation in theoretical physics.
EX FLR C, t S t S
1. For the case y = A, determine the critical angle_ Mut the reflection coefficients f3/A and C/A as a function of the angle of incidence O. 2. A re there any values of the angle of incidence 6 for which the amplitude C of the reflected longitudinal wave vanishes? If [here are, calculate the corresponding value of the amplitude of the reflected shear wave, B. :3. Complete the solutions for B/A and C/A given in (17a) and (174 In particular, show that
lc 2 ki
2( l - - t^) ^
1
—
(18)
^v
4. Consider a solution of Na,vier's equations without body forces of the form tl,
=
x2 , [)•
Show that the resulting displacement consists of a general elastic wave polarized in the (x 1 , x,)-plane and a shear wave polarized in the x 3 -direction. 5. Carry out an analysis similar to that of this section for an incident longitudinal wave making an angle et with the normal. Is there internal reflection in this case?
2r
Introduction to Uynaniir. Problems in Elastic* EC1i. 6
6.4 Ela stic S urface Wa v es We shall consider another type of elastic wave solution for the half-space x 2 4 with a free surface. This type is not orate plane wave form considered in the previous section, but rather exhibits a dependence on x 2 that causes disturbances to die out as x 2 becomes large. The displacement is assumed to be polarized in the {x 1 , x 2 )-plane and to be independent of x 3 ; i_e., L!¢
r! 3
= i (x1, X2, r),
—
O.
The decay with x 2 that is characteristic of the waves under discussion means that the effect of the waves is confined to the vicinity of the free surface. Such surface waves arc also known as Rayleigh waves, in honor of the first person to analyze their nature.. Rayleigh was particularly interested in these waves as possibly forming an important component of disturbances from earthquakes as well as the collision of elastic solids. Further investigations, both theoretical and observational, have shown that this is indeed the case. We shall have more to say about the importance of Rayleigh waves at the end of this section_ FORMULATION
Turning to the details of the required calculation, we decompose the displacement into an irrotational part and a dilatation-free part, writing tea --- t:Q +
VV.,
where V1.2 — V3 1
from irrotationality and ^.Q
—
from the equivoluminal requirement. From the calculations of Section 6], it follows that va and we satisfy (^
+ ^^)t trA — P^" a
and Pwa. pp = ^1Ü.
We recall from {1.4) and {1.8) that n=
+
C
p
•
1—— p
The boundary conditions are the same as those of the previous section,
T2; — 0
fo r x x = 0.
* See Lord Raytoigh. On Wavcs Propagated Pray. Londr. t Math, Sac.. 17, 4 1. I (1E85). -
(3)
Along the Plane Surface of an EIaslic Solid."
Sec. 6.41
Elastic Surface Wcnres
2
79
In addition, we require that a s approaches zero when x 2 becomes positively infinite. By means of the stress-strain relations, the boundary conditions of (3) can be written in terms of the displacements as
D
(4a)
+ , )u2, 2 + u ,.1 = 0,
(4b)
U2 . 1 +uI . 2=
and
(2
at x 2 = 0. The third boundary condition, for T23, is identically satisfied. SOLUTION: PLANE WAVES THAT DECAY WITH DEPTH
We look for displacements of the form 14a =Aexp[— ax 2 - i{sx l rio = B. exp [ bx 2 f i(sx1 — wt)]. —
The coefficients a and b are taken to be positive, to ensure that the displacement will go to zero as x 2 becomes large. In addition, the wave number s/tit and the frequency to/2n arc taken to be the same in both ternis of the displacement ; the complex exponentials will thus match en the boundary x 1 = O. The speed of propagation along the free surface is w{s_ Equations (l) and (2) will be satisfied if
2,u)(b! — s 2 ) _ — pw x
(5a)
and
[p2
—
s2) =
(5b)
Equations (5a) and (5b) can be further simplified by introducing
h2 J Pf)2 . A +2/1
kz
—
pm` ,
1-1
with the result that
b 3 —s 2 + h 7 =Q,
a 2 —s' + k 2
=
(6a, b)
Since wo. = O, ^
isAt
— A2
a r D,
which may be rewritten as A 1 = —iaA,
fi t R A5.
Similarly, the irrotationality condition v i. 2 B 1 = — isB,
—
B2 —
v2.1 =
bB.
(7)
is satisfied by (g)
The proposed form of the solution for the displacement still has three undetermined constants, the amplitudes A and B and the wave number s. if the value of s can be found, then the speed of propagation c along the free
Introduction E a Dynamic Problems in Elasticity [Ch. 6
28a
surface can be determined. To find these constants, we have at our disposal the two boundary conditions (4a) and (4b)_ Before making detailed calculations, we can see that these boundary conditions will provide a pair of linear, simultaneous, homogenous equations in A and B. To ensure a nontrivial solution. we should find a determinant condition that will determine the third parameter s. If the search is successful, the am plitudes A and R will then only be known to within a multiplicative constant_ This should have been anticipated, however, since the original boundary value problem is homogeneous and any displacement that satisfies the conditions can only be fixed to within a constant multiplier_ The calculation of the boundary conditions can be sirnpliled by using real solutions rather than complex exponentials, and thus rewriting the two parts of the displacement vector as w2 = A. exp ( - lc2)[ees (Sx t - air) + i sin (sx t = B. exp (- 1).C 2 )(cos (sx i - (oar) + r sin (sx r t
-
-
tot)], oit)].
Employing (7) and {8), we see that real components of the displacement take the form
, , = Aa exp (- ux 2 ) sin (sx r - we),
w 2 - As exp
V 1 = as exp (- bx 2 ) sin (sx y - oat),
1,2
►
( - nx2)
cos (xx, — c:xr
Pb exp (- hx 2 ) cos (sx - cot). ,
Boundary condition (4b) is most conveniently written as f
(t a)
+ 2 p( 0 2. 2 + 2.2) - Û
at x 2 = O. This implies that 2pAsa + [20 2 +
- s 1 )]B = 0,
(9)
where the common trigonometric term cos (sx, - rig) has been suppressed. In the same fashion, boundary condition (4a) yields
(10)
A(a l + 5 2 ) + 21 sB = 0,
where the common factor sin (sx, - tot) has been dropped from each term. The simultaneous linear homogeneous equations will possess a nontrivial solution only if the determinant of the coefficients vanishes, i.e., only if 4µubs 2
-
[Â(b 2
-
s 2 ) + 2p 1) 2](a 2 + s 2 )
a
(1l)
With the aid of (5) and (6). (t 1) can he reduced Further [Exercise 1(al], to the form ^x k2 . 2 1 f^ l - z 1 --1)2 ` ^ (12) S
5
S
Solutions of the algebraic equation (12) for s 2 will enable us to determine speed of propagation c, the amplitude ratio A/B, and the decay coefficients a and b.
Sec 6.1
Elasrir Surface Wares
281
ANALYSIS OF THE SOLUTION
Further analysis of (12) will produce some simplifications as well as some interesting physical results. Recalling that the speed of propagation c in the horizontal direction is given by cots, and keeping in mind the definitions of h 2 and k 2 , w e see that Ca
k2 2
C Cr^
and
h2
, 5
c2
(13a)
^
n
Furthermore,
k2 C 3 (13b)
c2r
h2 `
From (6b) we infer that k 2 < s 2 . t loreover, it follows from equations (1.4) and (1.8) that c2 < ('r< cra-
(14)
In addition, the material properties of the problem enter (12) only through CT and Ca. In particular, the frequency CD is absent. Thus if (12) is regarded as an equation for c 3 , the roots will depend only on the density and the elastic constants_ We therefore conclude that the speed of propagation is less than or equal to the speed of propagation of shear waves and dilatational waves; and, in addition, the waves are not dispersive_ The quartic equation (12) can be simplified through the use of (I3); one finds [Exercise I (b)] that it reduces to a cubic in the variable € 3 namely,
—
8e4 + (24 — 16V)c 2 — I6(1 --
) _ 0,
(15)
where c2 —
c
2
CT
2
[
r
a
2
= ^i k2 .
We must find a solution e2 of (15) that lies in the range 0 S ii' S L Although this generally must be done numerically, it is possible to see rather simply that there is such a solution. One need only observe that for e2 4, the lefthand side has the value — 1 6- 1 — 1 ), which is negative. On the other hand, when e2 = I., the left hand side of (15) has the value 1. Thus, by continuity, there is a solution in the desired range. Specific cases are described in Exercise 2. Once ti has been obtained, the ratio of the amplitudes AJB can be found from (10). Another important aspect of this problem is the determination of the rate of decay of the wave with increasing depth_ As is the case with water waves, the decay is exponential [compare (8.2.2) and (8.2.3)]. Since a < b, the slowest decaying constituent of the displacement is the transverse wave component. The decay can be measured by asking for the depth at which the surface -
Introduction to Dynamic. Problems in EPustrrrl}•
282
(Ch_ b
displacement is 1/e times its value at the free surface_ ln other words, we seek the value of x 2 for which
e - ux 1
=e
or (16) It is more instructive, however, La measure the decay depth in a somewhat different form. We recall that sj2rr is the wave number. The wavelength  is given by JR
2rE — s
.
We shall measure x 2 in units of wavelengths, by writing
X2 =
^.
Equation (16) can then be put into the form i
u. Using (6b), we find that
2 ra'
•^I 2
fI1? = s s 2 -- -^T
or (17)
27ro — ( I — e2)-112,
Another aspect of the decaying nature of the surface wave may be seen by returning to the formulas for the displacement, u i = [Aix exp (--ax e ) + Bs exp (—bx 2 )] sin (sx r — our), u 2 = [A s exp (— ax 2 ) + Bb exp ( — bx 2 )] cos (sx j — wr).
{ 18)
From (10), B
.4{a 2
^
s2)
2bs
and, suppressing the trigonometric terms,
2b
x
--
exp {—ux2j(a2
"12 + s2) — exp (—ax 2 2s
+
s2
exp
—(b — u)x z } ,
{19)
— (b -
(20)
2s2 —
a^ + s ^
exp
Sec_ 6.41 Elastic 5rrr/iree Wares
2133
Since a < h, the exponentials appearing within the brackets are decaying. is it possible to find a depth x 2 For which the amplitude of either u 1 or a 3 will vanish According to (19), there will he such a depth, x 2 , provided that 20h L!^ + s 2 From (6a) and (fib) it
can
be shown that
ab —
ri
^
+s
(21)
— (ii 215 22)]—1` 2 El — 024201,2
— (22)
(k2fs2)
^
On the other hand, s must satisfy the eigenvalue equation (12), with the result
that
2ttb u# s
if ?
2
k^ -
^ s
_
1 2
(2
—
2 c' ;.
(23)
Since()< < 1, 21 )does hold. Thus there is a depth at which the amplitude of the horizontal displacement will vanish. Below this depth, the amplitude will change sign. In similar fashion, the possibility of a sign change in the vertical displacement will depend on the magnitude of s2 /(a 2 + s 2 ). An argument similar to the one above will yield 2s 2 7 a - + s)
2 2 — (k^ls^)
> I.
(24)
Therefore, the amplitude of the vertical displacement will not change sign as x 2 increases. OCCURRENCES OF RAYLEICIi W A VE S
We have mentioned that Rayleigh was correct in thinking that surface waves will be excited by earthquakes. Indeed, seismograms show that Longitudinal waves appear first, followed by shear waves, and then Rayleigh waves, in accord with the ordering of speeds given in (14). in hindsight, this ordering is to be expected. The reaction of the material to the volume change in compression should be more vigorous than that to orientation change in shear, while the free surface permits a still easier distortion and hence a still lower speed. Rayleigh waves, in fact, form the principal part of the seismic signal. The other waves are attenuated throughout a three-dimensional volume, while Rayleigh waves only spread their energy in a two-dimensional region near the earth's surface. The calculations presented in this section can be regarded as the simplest of ever-more sophisticated analyses that are being
2S4
Inrrodrrrriorr in Dynamic Problems irr Edrlsriri1 y
[Ch. 6
made of seismic signals to reveal details of the earth's structure and the nature of earthquakes_* Modern technology employs Rayleigh waves of 1 billion hertz or more to store and recognize electronic signals_ Delay lines, for example, are advantageously fabricated from such ultrasonic acoustic waves because their speed is typically 10 - as small as the spewed of light. Typically, most of the acoustic energy is concentrated a few hundredths of a millimeter from a hard crystal surface, so the waves can be generated and detected with relative ease. Piezoelectric effects con vert electric signals to pressure signals. and vice versa. A recent article ) concludes that "methods of processing acoustic and visual information and displaying the information with the help of acoustic techniques are only in their infancy." As is so often the case, a classical phenomenon has found new uses that will doubtless press theoreticians to far-reaching extensions of earlier results. EXERCISES
1. (a) Derive (12) from (11)_ (b) Derive (I5) from (12) using the relations among h, k, C. CT, and C D _ 2. (a) For the Poisson ratio 4• = A. show that the cubic (15) can be split into a linear and quadratic factor. leading to the roots
= 4,
2♦ 4,
2—
.
Only the third root is meaningful (why?), leading to the result ë = 0.9194. (b) For an incompressible material, v = 1,derive the solution i-F = 0.9553_ 3. (a) By means of sketches, convey a picture of Rayleigh waves when as in Exercise 2(al. Depict to scale such features as the decay of amplitude with depth and the depth at which the horizontal displace-
(b) (c)
ment changes sign. Repeat (a) for an incompressible material, as treated in Exercise 2(b). Compare the above results with those for water waves, e.g., in (8.22)
and (8.2.3). 4. Given a specific value of the Poisson ratio u determination of the Rayleigh speed c requires the solution of a cubic equation, which can prove tedious. On the other hand, this exercise shows that the problem becomes somewhat simpler if one asks for all of the roots 4 < e < I corresponding to all values of the Poisson ratio in the range 4 Ç v Ç I. (a) Show that 2e — 1 . (25) v 2(0 — 1) ,
• A classical rctsrcnce is H. Jeffreys, The Earth. 4th cd (New York: Camtm-rdgc Umversny Pr ess). 1962_ # p_ S. Kinoand i_ Shaw. - Acoustic Surtrace Waves, ,, Set. Amer., SI - 68 (pct. 1972).
Sei -. 6.51 internal Reflection
285
(b) Considering (15) in the form c•h
8('a +24r2 -16
y
^
16(1' 2 — 1)
use (251 to compute the Poisson ratio as a function of the Rayleigh speed and plot your results. i It should be observed that the full range 0 < ez < I may not he allowable, since c 1 and 0 < 5. Verify (221. (23), and (24). 6. Generalize the discussion of this section so that it applies when x = 0 is the interface between two elastic media. Set up the relevant equations for the amplitudes in dimensionless form, but do not aitcrnpi to solve.
63
Internal Reflection
At the conclusion of the discussion in Section 6.3 concerning an incident shear wave on a free surface, il was noted that if the angle of incidence 0 was sufficiently large so that 0 > 0, , ac.,
kL
s inN > I or
!' sin t)> I,
I 11
CT
the solution given would not apply. The analysis of surface waves in Section 6.4 gives a clue to another form of solution that may answer theyuesl ion. The basic idea consists in replacing the reflected plane longitudinal wave by a longitudinal surface wave.* This short section provides the required calculations. We see that the incoming shear wave is reflected, but with a phase difference and the conconemitant generation of a surface wave_ The case when the critical angle is exceeded is thus called internal reflection. We take as our trial solution tt„
11 T
-^
,
wher e Tit = T,
is the
(2)
4 imp i(ku r .t,
incident sheaf usive written in complex
foi tit- h
ere, as in (3.2l and (3.4), f^1
T
^
= COS 0, T2= sin
U. n l t Sin
0, 11 2 =- — i:115 fl,
k =
-, C T
• The same idea can he arrived al lnrnriIly by using the tact that trigonninetrh fue^- uuns M complex srgunrenls call be expressed to term~ id hyperbolic funs:nods. Tins leads ire an r 1 dependence ere ilie form of a linear coinhi ealinn of sinh ht i alai cosh br a if the soiuunn is io be hounded as K i Y then the only possibility is the clpossent all ciependencz assumed in { 41 below .
2g6
Iorrodurriori iu Dynamic Problems in Elosiieiry (C.'h- b
and the angle of incidence fi satisfies (1)- The reflected shear wave is designated by ua' and given by
u^r =
rÿ13
exp i(k ►n y x + wt),
(3)
where, on the basis of the analysis of Section 3 'following (3.8)],
z 4 = cos 0,
r` _
—
sin 0,
m, = sin 0,
m 2 = cos 0_
The horizontally moving longitudinal surface wave is denoted by given by
+ wt)],
ua° = C¢ c,=xp [ — bx,
u411 and is (4)
where
k L = k sin B
or
co sin
k,
0
Cr
Our previous analyses show that the exponential terms will balance properly on the free surface, x 2 =- O. Since ua" is a longitudinal wave, (4.8) with k L identified with s shows that the amplitude components must satisfy
C1 = - IC,
C2 =
(
)c.
'
I5a)
F urthermore, (4.6a) and (4.13) give 1,1 2 =k1-11 =ki
^
C
^
.^
(5b)
-
D
The amplitudes of the second and third waves need not be real; we write them in the form B=II(cosb
—
i
sin s
),
C=I<(cost —i sin
4
(6)
The analysis can be handled more conveniently by considering real solutions. Taking the real parts of each of the three waves, we find that u
i
= A cos 0 cos
ell = H cos O[cos u— —
,
b cos
u I — A sin Q cos Or - , -1- sin b sin {(r*] ►
H sin O[cos b cos ip ` + sin 6 sire 411,
(7a, h) (8a)
(8b)
where
tfr i - ic(x i sin 0 ± xp cos 0) +
(8c)
Also u!^ — K exp( u
—
bx
— sinEcos ]( i- cos t sin xi,
°t = ( fGb/k z ) exp ( — bx z ) [cos E cos X + sin e sin x],
(9a) (9b)
SFC. 15.5]
fnrrrna! fleet- Min
287
where X - kLx i
_
(9c)
In (8) the pair of trigonometric terms could be written as the single term cos (i' — 6)_ Thus the presence of a complex amplitude in (3) is equivalent to a phase shift of —d when written in rial terms. Similarly. the presence of in (9) is equivalent to a phase shift of —E in the surface wave. From (4.4a) and (4.4b), the boundary conditions on x 2 = 0 are U41.2
=
O,
Gila)
—(7.1 l —2v 14 r. r
(nib) Ob)
+ U2 .1
v 14 2.2
We shall now examine whether(10a)and (10b)can be satisfied by appropriate choices of the unknown constants H, K, 6, and E. Substituting (7), (8), and (9) into ()Oa), we obtain 0 — kA sin y(cos 2 0 — sin 2 fi] + Hk cers` fl(sin h cos y — cos Fi sin y-) 4- Hk sin 2 Oleos d sin y — sin cosy) + 2K bfsin r: cos y — cos L sin y), since when x 2 = 0, rjr# = +4x i sin 0 + lot — k i.x 1 + cor ; y. ln order for this equation to hold for all values of x a , the coefficients of sin y and cos y must vanish separately_ Consequently. 0 = kA cos 20 — Hk cos 20 cos b
—
2Kb cos c,
0 — Hk cos 20 sin S + 2Kb sin t.
(l 1) (12)
Carrying out the same program in (10b), we find that O = kA sin 20 -1- Hk sin 20 cos 6 + ?Ka sin r:,
(13a)
0 = —H% sin 2(1 sin (5 + 2Ka cos
(13h)
r.,
where ^
vk r, — 0 2 j4}( l -- -1 — 2v
(14)
Equations (1 I4, (12), (13), and (14) constitute a system of four equations for the two amplitudes H and K and the two phase angles r: and .5_ Multiplying 4l l) by sin c and (12) by cos r and adding, we find that Ak cos 20 sin E. —
HI(
cas
2O(cos 6 sin r — sin (5 cos r) = O
or
A sin r:— H sin
(r: —
=0.
(75)
lrrlrodtr['fi4n lo Dynamic Prohfems Eipslieif}
28$
Similarly, multiplying I l 3a) by cos £ and 113b) by sin F. and subtracting, we obtain .4 cos -- ff cos ( .c — b) - 0.
(16)
Thus (15) and (16) are a system of linear, algebraic, homogeneous equations in A and H which must possess a nontrivial solution- The required condition is
sin (5 -- 2s) = O. We thus obtain the important relation between the phase angles.
(17)
b= 2c. In addition, (15) then yields
H
—A.
(IS)
Returning to (I I)and (12), we multiply the first by sin b and the second by cos b and add, with the result that
Ak cos 20 sin 6 — 2Kb sin (6 — — 0. Since b
2c, we obtain for the unknown amplitude K the relation
K = 1 Ak c o s 20 cos c.
(1 9)
Summarizing our results ai this point, we see that we have obtained the phase angle 6 in terms of € in (17), the amplitude if in (18}, and the amplitude K in terms of E in (19). All that remains to he done is to calculate the remaining phase angle c. This is accomplished thi ough (13) and (14). Multiplication of (13a) by sin (5 and (13b) by cos 6 and subsequent addition yields (since 6 4 2t.) Ak sin 20 sin E+ Kcr-0.
(20)
Equations (19) and (20) form a pair of linear homogeneous equations in A and K that must have nontrivial solutions. The determinant of the coefficients must therefore vanish; i.e., ak
cos 2F1 cos E +bk sin 20 sin r —
(-
Consequently, the phase angle rés given by tan c ` —
a/b t an 20
(21)
Equations (17), (18), (19), and (21) complete the determination of the unknown constants that appear in the displacements given in (7), (8), and (9). We finally need only show that the parameters k, k t , and b which appear in these equations depend only on 0 and the Poisson ratio v. It should be noted
See. 6.61 Lore Wares
289
from (19) and (21) that these parameters enter only in the ratios of r}/b and k/b. From (14), (1- 2v )^-v
b
v)^-.
—(!
(22)
On the other hand,
kL = k sin 0; and the question is reduced to finding the ratio h/k_ From (5b) and (4.I3b),
k
b
CT
Using (3.18), we can write finally b'` k
sin
D
1-- 2v —
?(1 — v)
(23)
Equation (23) then allows us to calculate the amplitudes H and K and the phase angles F and 6 in terms of the incident amplitude A, the i rtr:ident angle 0, and the Poisson rouie v. This determines the reflected shear wave and longitudinal surface wave_ Eï7iLRCISE
1. ln Exercise 11 the reflection coefficients 131A and CIA for the case v = 3 were to be calculated as a function of the angle of incidence [I, 0 < (a) Now carry out the computations for the angle of incidence ranging between the critical angle and rri2. (b) Plot H/A, K/A, and £ as functions of O. 2. Discuss possible muitiplc-valued solutions of the equations for 6 and ^,
6.6 Love Waves It has been noted in Exercise 3.A that if the displacement vector u is independent of x is then u is a combination of a general elastic wave polarized in the vertical (x t , x 2 )-plane, and a horizontal shearwave (SH-wave) polarized in the x 3 -direction. Should a given boundary value problem require only SH waves for its solution, the resulting equations are essentially scalar, rather than vector. In such cases one can study somewhat more complicated geometries than those previously considered, A felicitous exploitation of the mathematical simplicity of SH waves to permit study of a geometrically complicated and physically important problem occurs in A. E I-_ Love's Adams Prize Essay of 1910_* The problem ` A F. t 1^{3ve, Some Problems in C;eodynurerics (New York. UEver, J967}, pp. 160 ti3. Originally published by Cambridge University Press 0914
inirodursion ru bynamic Problems in Elrisiiriey [CA, 4
290
was to explain the transverse movement that takes place in the early pha ses of the large waves in the earth as observed by seismological measurements. It had been proposed that the movement was a manifestation of waves that travel through the crust without penetrating very far into the interior. We shall present Love's analysis or this proposal. We model the situation as a horizontal layer, parallel to the (x,, x # }-plane and of finite depth h, overlaying a semi-infinite region. The density of the overlaying "crust" will be denoted by p and its shear modulus by i. The corresponding quantities for the semi-infinite region, which represents the remainder of the earth, will be denoted by p* and p*. The resulting configuration is shown in Figure 6.4. We ask whether it is possible to find an SH wave that travels through the layer and only penetrates a small distance into the underlying substrate.
ü
P. p
77/227727///7.///////7/
fJ///f//J///J/// ///!/!J
I
.P
`
xt FIGURE 6.4, A<•rasi at thickness h U1 •erJa pirirj o senn-kjFniir suhsiraw wïi^i ûi^{/ownr damn- prupernes
FUR M tJLATtr]M AitiU SOLUTION
The displacement in the layer will be denoted by u 3(x r , X. t) and that of the substrate by u3(x 1 x s , 1). The wave speed in the layer is given by cT = and in the semi-infinite space by c'2 = p*/p*. We look for displacements that are harmonic both in time and in the horizontal direction x i . Thus we take ,
i4 3 ` l (x 21 exp [t5x 1 — C!}j and
}
-
Le = y(x z ) exp i(sx, — u_,t).
Here e Ws is the horizontal speed of both waves.
Se•r. 6.6]
Lure Wares
291
The displacements must satisfy Navier's equations in the form U y,^
_
^2u3
2 df2
(3)
r
and Us _
I t^2u3 [ -
(4)
.
11
Substituting the displacement given by (I) into (3), we find that f(x 2 )
must satisfy
d2f 2 w2 - 2 - s - 2J= O.
iiX 2
CT
Two cases arise, if 51 > w 2 /c 2 then f = A exp(hx 2 ) + B exp(
-
6x,),
(5)
where
c,3 2)112
b=
—
? .
C T,
On the other hand, if c2) e /4 3 s 2, then f = M sin
(axa
+ D cos (ax 2 j,
(G)
where 02r ii 52 a= (.7 r Equation (4) yields a differential equation for the unknown function g(x 2) in the displacement (2). On the other hand, we are seeking displacements in the semi-infinite substrate that do not penetrate far into the region. Thus the only physically possible solution is easily seen to he g = E exp(--kx 2 ),
(7)
where (0 2
k=
si—
112
*z ^r
Boundary conditions must be given on the free surface x 2 = 0 and at the interface x 2 = h. The traction vanishes on the free surface. Thus t i =T2 =0 The only nonvanishing component of 32 , so that Hooke's law yields
at x 2 = O.
To foi the given set of displacements is
u 3. 2(x1, t], () = O.
(8)
29 2
introduction to Dynamic Problems in Elasticity
Kit_ 45
It is assumed that the displacement as well as the traction is continuous at the interface. Consequently, u3(x1 , h , r) T u31 1, h, r)
(9)
and 1 32(x1, h, r) = T32(x1+ h, 1)
ar pi3,2(x1+ h , t) — p *4 3,2(x1, h, t).
(1O)
Suppose that cu e/ei < s 2 . Using (1), (5), and (#l), we find that A = B, and the full displacement can be written in the form u 3 = .el cosh bx 1 exp i(sx 1 — cot).
(11)
Substituting (2). (7). and Ill) into boundary conditions (9) and (10), we find ,sat cosh bh = E exp (—kb), pbsit sink bh = — kte*E exp (—kh). (12) In order for (12) to give nontrivial solutions for sif and E, the determinant of the coefficients must vanish. This yields I.01 — —
tanh bh.
( 1 3)
Since bJ, is positive, tanh bh is positive. Then the right-hand side of (13) is negative, whereas the Left-hand side is positive, which is impossible. It must be that
w2
> s=, 2 cT with the amplitude f (x2) given by (6). This requirement together with the decay requirement in the semi-infinite space implies that
co * C3
<s<
co — C 1-
(14)
In terms of the horizontal wave speed c = cots, this inequality may be written in the form cT < c < (15) Inequality (1 5) not only supplies a comparison between the wave speed and the shear speeds in each of the media, but also shows that no solution of the type we are seeking can exist unless the transverse speed in the substrate exceeds that of the layer. 1f (15) is not satisfied, we must give up the requirement of exponential decay in the half space. From (1), (6), and boundary condition (li), we find that M = 0 and the displacement in the layer is given by
u 3 = D cos ax e exp ï(sx i — wt).
(16)
Ser. 6 6
Lore Wares
2
93
Boundary conditions (91) and (10) then yield D cos ah — E exp (—kh)
—
0, pan sin ah
—
N*kI- exp (—kh) = () (17a)
IF (17a) is to provide nontrivial solutions for D and E, then it must be that tan ah.
(17b)
F.1
Equation (17b) can he put into more instructive form by noting that the variables k and a are functions of the wave speed c and the wavelength A = 2rrfs. In particular. we note that a=
22r (C2 — 1)1'2
(18a)
A
and k=
j (I — ('*2 )rr2
(I#lb}
,
where
C = -, C* _ — . CT er
(L $c}
"l -hus (17b) can be written as p*(1 — [*2)1f2 =
F4C 2 —
l}112
tan ^(C2 —
(19)
EXAMINATION OF THE NATURE OF DISPERSION
Equation (19) provides a relationship between the wave speed c - and the wavelength , It is interesting to note that not only the shear speeds c . and CI, enter (19), but the shear moduli p and p.' enter explicitly as well. (Compare 4.15, the dispersion relation for Rayleigh waves.) In view of the fact that ci = pfp and cT2 = v•/p*, (19) can be written in terms of p *fp, namely,
r 2 112 c 2 (I — C* ) (C 2 — I in — tan p c *
2)h
(C=
(20)
Two useful results can be seen at once from (19).. As the wavelength becomes long (A -. co), the right-hand side approaches zero and c must then approach rf, On the other hand, if r approaches its lower limit C r . then the fact that the right-hand side of (19) approaches a finite limit as C —, I requires that 2nh (c2
(2n 4-
Consequently, the wavelength A must approach zero.
(21)
Irrrrodueriarr lu Dynamic ProbiYms in Etu.srir+ry [Ch. fi
2 94
In order to examine the dispersion in more detail, it is convenient to write the dispersion relation explicitly in terms old- and cr 2 . eliminating explicit dependence on c. Using (18a), we see that
cz
= c2
(22 )
+1,
and from (18c) and (19) it follows that 4
tan aft =
[-
iF
I-
CT
-
C T2
CT 1 12 T3
(23)
Equation (23) is somewhat more useful if it is put in the final form
2rt 17•1.
,4 2
c^
_
c F — c^ T [ *2
-I-
*^
' ^ *z c ^z T
x
tan'
2 h {(aA)- }r
CT
Differentiating (22) with respect to )., we find - LIZ (..?')' 1 ch. D ^ 47r 2 c 4 ï + 1
(24)
^.
(a).)
d(u).)
).- -
Since u is positives we observe that dcfdA has the same sign as d(a)fdA_ Differentiating (24) with respect to A and solving for d(u).)fdl., we obtain
Kh - + --)
d(aA) 2i
Kh
WA) —
(2 5)
where
c
tan {(a01)( il/À)}
pc" CT S - C72 sec' (WARW.)} The requirement c*r > c r implies that K > O. Since all of the coefficients in (25) are positive, it follows that d(aX)/dd is positive, and hence dc Ida is positive. It has thus been demonstrated that the motion is dispersive, and the dis -persionmal,.gerwvspoatdhigersp than shorter waves. FURTHER CHARACTERISTICS OF THE SOLUTION
Returning to (2) and (7), we sec that the displacement in the substrate is given by — E exp (- kx 2) exp i(sx j - cot).
The depth of penetration can be measured by considering the ratio of the amplitude at a depth of one wavelength to the amplitude at the interface. This quantity is simply exp ( U), which can be written entirely in terms of the speed of propagation c
exp
exp [ -_2ir(1
(26)
Sec. h.h] Lore Wares
2 95
Since c increases with increasing wavelength J, we see from (26) that longer waves penetrate further than shorter ones. It should be noted that the wavelength a, need not be a single-valued function of the speed r_ This follows from the fact that the tangent function which appears in (17h) and (19) is multibranched.1 ne same values for tan ah can be obtained by a sequence (i i ) of a's, where O
a t h< n , Irt < u h< xr ....
To these different values of u, there corresponds the same value of the speed c. Since a 2 h > `-n, reference to (I 6) shows that there will be nodal planes in
the layer for these higher modes. Denoting the wavelength corresponding to a; by ).;, it is seen from (22) that since the speed of transmission is the same in both cases, ap:ir = az
^^
Thus
It has been found in seismological problems that the most important case is the one without nodal planes, since the energy transmitted by the longer waves is greater than that associated with the shorter waves. CONCLUDING REMARKS
the general discussion as well as the special examples considered in this chapter, the important fact that there are two speeds of propagation in an elastic medium has been emphasized. We have noted that the basic field equations can be written as scalar and vector wave equations, each with a different speed of propagation. in many cases, however, we find that the boundary conditions require that both types of waves be present. This is one of the major reasons why elastic wave propagation problems are frequently more difficult mathematically than the corresponding problems in either In
acoustic or electromagnetic wave propagation_ We also observe that in our study of propagation in infinite and semiinfinite media, no dispersion took place_ (For example, the speed of Rayleigh waves is independent of their length.) This is in contrast with the results that we developed in the preceding chapter on the engineering theory of transverse wave motion in beams_ We can recognize a duc to the origin of dispersion from its appearance in Love waves, in the presence of boundaries a finite distance apart. indeed, dimensional reasoning shows that there cannot be dispersion in the absence of a length scale (such as the width of the dust or the beam), for otherwise how could the wave "know" how long it was and hence how fast it should travel? It seems that dispersion should be regarded as a normal state of affairs,. but dispersion cannot be present in problems lacking an intrinsic length scale_ Drspersrvc waLrür waves arc diuusscd al Icnglh ,n what Follows. slatting in Section 8.1.
296
hir+awr#rciior: ro
Dynamic Problems in Elasticity (Ch . 6
EXERCISES
Study Love waves for the case in which the substrate is infinitely rigid, ii-e-, u 3 = 0 for x 2 = h. lithe motion is dispersive, find the relation between the speed and the wavelength. 2. Study Love waves for a layer that has free surfaces both at x 2 0 and x 3 4 h. 3. The variation of h/A as a function of cir e- can be found in the following manner. Choose a value of Or e- in the range !.
I
^
• CT ^ _ CT CT
(27)
^
Calculate u; and solve for h/. from l7bl written in the form
(9(k)= tan
[u^]
1 .
Carry out the calculation for various valuesofc/c T in (27) for p*/p = I and the case te fir = 2, and 3. In addition, calculate the corresponding values of exp ( kA). (This can be done without undue effort using a programmable pocket calculator.) 4. We have seen that solutions of the form (1) and (2), with the restriction that the displacement dies out in the substrate, are only possible if c r < cr. Suppose that this inequality is reversed and the restriction of decay in the substrate is replaced by boundedness. Examine what can be said about the solution to such a problem. S. Are there alternatives to the text's assumption that in Love waves the displacement and the traction must both be continuous at the interface? Discuss. ,
—
PART C Water Waves
CH APTER 7 For m uIation of the Theory of S urface W aves in an Inviscid F luid
T
IE REAMER will surely grant that the wave concept is one of the most fruitful in theoretical science. Our senses continually record the arrival of sound and light ; we learn in elementary physics courses that these phenomena are due to pressure and electrodynamic waves. Quantum mechanics tells us that the objects around us are "really" light wave packets of probability densities. Other examples abound. As the theoretical physicist Sommerfeld has written, "Ever since waves were studied, water waves have served the natural scientist as a model for wave theory in general: . Part C is concerned with waves on a water surface. These are a paradigm for wave phenomena because of their simultaneous generality and accessibility to observation. In referring to generality, we have in mind the fact that the appearance of water waves is usually influenced by dispersion (dependence of wave speed on wavelength) and nonlinear effects. in special cases. water waves are essentially nondispersive (long waves) or linear (small-amplitude waves)_ Now, "ordinary „ sound waves are nondispersive (so that we may speak of the" speed of sound) and little affected by nonlincarities, but nonlinear effects in "strong" sound waves cause shocks or sonic booms. Light waves are dispersive (hence the rainbow) but are not ordinarily affected by nonlinearities_ With the advent of lasers, however, the new field of nonlinear optics was born. Thus an expert in the theory of nonlinear dispersive water wave is prepared for complications that may emerge in other wave systems. Special instruments are necessary to record even rudimentary facts ..
concerning most wave systems, but visual inspection suffices to reveal many rather subtle aspects of water waves. This accessibility to c'bçervatio i means that readers will possess the basic familiarity with the phenomenon that is a prerequisite to theoretical analysis, Elastic waves were discussed in the previous chapter_ The reader who becomes familiar with both elastic and Fluid waves will find the contrast enlightening. There exist many similarities, for example, the presence of dispersion except in the absence of a length scale. Another similarity resides in the important effects of density stratification. We considered one prototype problem of this type in elasticity when we treated hove, waves in Section 6.6. There are some basic results for the fluid mechanics case in Section 15.2 290
'oc
Formulation of the Theory
of Surface Wares arr hiri rid Fluid [C h. 7
of I, particularly in Exercise 8. See also C. S. Yih, D of Nonhortiogeneotis Fluids (New York : Macmillan, 1965). Understanding of density effects in one
medium is, of course, helpful when such effects are to be studied in another medium. Although similar in many respects, elastic waves also differ importantly from water waves. Perhaps the primary difference is that the a cisterice in elastic media of both transverse and longitudinal waves brings about interactions not found in fluidlike media that do not support shear stresses. By contrast, ar adequate study of water waves must concern itself with nonlinear effects, while these effects (at least for the usual applications) are normally negligible for elastic systems. In I we have derived field equations that suffice for an initial study of water waves, namely, the equations for an inviscid fluid. Complete mathematical characterization of the problem requires more thought, however, for it is necessary to seek appropriate boundary conditions at the air-water interface. A careful derivation of these boundary conditions is the subject of Section 7.1 In Section 7.2 we Formulate a mathematical problem that can serve as a model for a wide variety of situations involving water waves. This problem is rather difficult. so we conclude Section 7.2 by writing down a simplified problem, wherein nonlinear terms are deleted and boundary conditions are applied at the mean position of the interface. A justification of this simplification is the subject of Section 7.3. The linear problem is studied intensively in Chapters 8 and 9. Part C concludes with Chapter 10, an introduction to a study of' nonlinear effects. NoTt: to INSTRUCTORS. The standard treatment of water waves is based on Laplace's equation for the velocity potential. with appropriate boundary conditions obtained by the use of Bernoulli's equation. Here we proceed directly from the Fuler equations, thereby skirting certain strictly hydrodynamical topics such as Bernoulli's equation and the justification of the irrolationality assumption. The d isadvantage of slightly more complicated calculations is more than offset, we believe, by the advantage of giving experience in dealing with systems of partial differential equations. Moreover. in a viscous motion starting from rest, nonlinear effects lead to a second order vorticity throughout the fluid. There is a resulting mass transport, even in the limit of vanishing viscosity.* Weak turbulence initiated and maintained by the waves, or stemming from the wind, is responsible for the appearance of vorticity that affects the nonlinear properties of the waves.t It thus is likely that much future research on water waves will not employ the irrotationality assumption. • tier M S. Longuei-tiwir,s, "Masb TransPurs ru tksr liaunddry Layer ata FrceGscrttatmg Surface," J Flu i d .4fech. 8. 293 3O5 119b0) t O. M. P h illips. "A Nate an the Turbuicrsex Generated by Gravity Waves," J. Geaphys. .
R e s, â6, 2t389-21s93 t tS ► fi!
) .
Sec. 7,1 ] 8oeundar.1 C'cuWlrtrrrtvi
3a I
7.1 Boundary Conditions In the first part of this secuun we shall derive the "kinematie boundary conditions- a mathematical expression of the fact that the surface F(x, t) = 0 is the boundary between two immiscible substances. No knowledge of the nature of the substances will be necessary the discussion is entirely geometric. Logically, then, this topic should perhaps have been presented with the kinematic material of Section 13.2 in 1. li is presented here because there would seem to be higher motivation for studying the kinematic boundary condition in a familiar and tangible selling, that of water waves. Since our treatment olfthis condition is entirely independent of the succeeding treatment of water waves. it can be read al any 1 ime. In the latter part of this section we define and use the notion of surface tension in deriving a forte balance condition that holds at the interface between immiscible inviscid fluids,
kINL!1f1A'ftC - Bt)UNl)Aiiii ('[1Nili7iO N li is an idealization to say that a surface Fix, t1 = O separates two immiscible substances. According to the molecular point of view, the separation is not abrupt. but rather there is a mixed transition zone joining two regions of almost pure composition. Idealizing this transition zone by a surface is permissible and sensible when one is interested in phenomena which vary signilicantly only over distanc:.es that are large compared with the width of the transition zone. Suppose, for definiteness, that a surface Fix, r) = O is deemed to separate water and air. (Replacing a transition zone by a surface iscertainly permissible here, for even centimeter-long surface ripples have a scale that is perhaps 1 million times greater than typical transition widths.) We assume that all the water remains on one side of the surface and all the air on the other, i.e., that the bounding surface is impermeable. If nu fluid is to pass through fix. tI = 0, then ur any point x„ N this .+rrrfue -e the crmrtpoilettt of the /ljtiti reiu( it ►' v(x,,. 11 than is tnsrutauneuasl ' normal to F = f3 at x e must equal i (x c, , t), the normal speed t / F U ar x{,. The italicized requirement just stated, while roughly correct, must be modified because the fluid velocity al a boundary point may not be defined. The reason is that the tangential velocity component may suffer a discontinuity across the boundary (since the boundary is impermeable). Suppose, therefore, that the impermeable surface l-ix, r} - O separates one substance in region l2 from another in region R 2 . Let n be a unit normal to at x 0 : ,
VF'Ix ü , t)
n = IVF(xe,, tl^ ^
11)
302
Formulation of the theory of Swjw a Naves in an lrrriscid Fluid [Ch. 7
Then the boundary condition that we shall impose is line v{x , i) - n = VR(x 0 , r) ^
(2)
—x w
;Er,
for x€, on F=O;i— 1, 2. We now seek an expression for i.„ in terms of F.* Let x = %At) describe the trajectory of a point P on the surface F = O. Then F'Lxdt), t] = O.
(3)
Differentiating (3) with res pe ct to r, we obtain
dx f.
F,
O,
(4)
But dx Pfrlr is just the velocity of point l'. To obtain the normal speed V, we project this velocity vector on the unit normal n of I!). Thus, from (4). ` VF IVFI
•d
F,
x^ !
dr
(5)
G VF
Employing (5), we see that boundary condition (2) becomes limv(x, r). VF(x 0 , r) = — F,(xo. r).
(6)
11 -*So
ti*
This is usually
R,
written
F, + v -F — O.
(7a)
Equivalently, the kinematic boundary condition is l] J-
Dr
= 0
( l b)
with the understanding that the condition holds as the surface F = 0 is approached from either side, When the interface equation is written z = ((x, y, r) and v = (u, y. w), then (7) takes the form w =Cr + u t; x + ttç as the reader can show [Exercise l(b)]
on z=C( x, y, t),
(8)
-
• This wrss obn inc l in (6.2.3)- We repast the brref required argument hezc Ioz cornpletcness and because we wish tarts C and [] to be accessible wrthout the knowledge of Cartesian tensors that was dcvclnped in part A and applied Iii Part B.
Set - . 7.1]
Boundary C onuliii oris
303
BASIC F A C TS ABOUT SURFACE TENSION
As one learns in elementary physics courses, observations of the properties of a liquid-air interface suggest the need to take into account an elasticlike properly of the surface. One revealing experiment involves a soap film on a wire ring. A loop of thread in the film will quickly assume a circular shape if its interior is punctured (Figure Ti). Presumably, in Figure 7.l(ak each element of thread is motionless because of a balance between forces resulting from the presence of surface film on either side of the element. When the interior film is destroyed, surface forces act only on one side of each dcment of the string, pulling it out. Note that in the final state of Figure 7.1(b), symmetry shows that the forge associated with a portion of surface acts normally to the bounding string.
()
W)
7.1. (ai A wire rira! .► upporiini a soup film ni which rests u loup of thread. (b) The confiquraiton afier the film in the interior of the loop is punctured FIGURE
ln another common experiment, fluid in a thin tube rises until its weight is balanced by surface forces (Figure 7.2). `l o explain this, one postulates a surface tension force per unit length of magnitude T that acts at each point P on the bounding curve of the surface, in a direction that is normal to this curve and in the tangent plane to the surface at P. The horizontal components of this force add to zero. A vertical force balance on the column gives (in the notation of Figure 7.2)
2ttrT cos a = rrr 2 hpg,
rigig i.e., 7' _ — - 2 cos a
The surface tension coefficient T is thus expressed in terms of readily measurable quantities. On a molecular level, surface tension can be understood as a manifestation of the significant attractive forces that exist between fluid molecules. The molecular viewpoint is outside our scope here, but good discussions can be found in most basic physics texts.
lai
Formulation of the Theury of Surface I ailles in an frniisrid Muid [Ch 7
F I G u k E 7.2. A tlrrn rube of against yrar+r r r magnitude' y J.
u/ °W #rr I p r.w held up h 4 swr;lfue•e rerr irurr
QUICK DERIVATION OF A DYNAM1( BOUNDARY CONDITION
We now present a rapid derivation of a form of the dynamic boundary condition that is sufficiently general to suffice for nearl y all future needs in the present work. Then we shall retrace our steps more carefully. We shall restrict consideration to inviscid fluids. Figure 7.3 presents an edge C of a (cylindrical) surface demerit that does not vary in the y-direction, and that departs only slightly from the plane z = constant_ It can be shown that body and inertia forces on the element arc of higher order, so that the surface fore must add to zero. (As will be shown in the following discussion, the argument is similar to that used in Section 14.2 of 1 to demonstrate the principle of local stress equilibrium.) We impose the condition that the net upward forces, due to surface tension and to a pressure
a'
.
rlal
r
I
^ 01 I i3 ^ r
I k
I
izi i
I
1vi .
uKIE 7 -3. S►rrJure fernier: rir ► rrrrrr
r r acting on ri
^^
+ ❑x
rrr•arl.r planar cylindrical .surfrare •
Net- . 7.1 )
Boundary ['urrdi r i^ns
30 5
difference across the surface, add to zero.* the net upward pressure force (per unit distance in the y-direction) is approximately
(PB — Max + O(Ax) 2 .
(9)
Here p a is the pressure acting on the bottom of the surface and p r is the pressure acting on the top. The O(Ax)' error term in (9) arises from estimating the length of C try Ax. Let 1,1 denote the angle that the tangent to C makes with the positive x-axis, O S tfi < i (figure 7.3). Assuming a constant tangential tension of magnitude T, we obtain the following expression for the upward force contribution from this source: -
T stn [rr — fi(x)] + T sin Jr(x -+ ❑ .).
(IU)
Since the surface is nearly flat, rr — i/i(x) is small, so that sin [rr—+k(x)]- tan [rr
—
Vi(x)]_
tan (x)_
(Lk) --
^
Because sin tti(x + Ax) x tan tÿ(x f Ax), (JO) can be replaced by
—
z T C^.K
dz (x + ^x) + T dx
Expanding the second term in (1 l), we obtain the following expression {evaluated at v}:
— T dz + dx
2z dz T €t A.x IL- O(Ak) 3 = T ^ Ax T f dx` dx dx 2 ^
,
0(11vj 1 _ ( l2}
Equating the sum of (9) and (12) to zero, dividing by Ax, and then tak irtg the limit Ax # 0, we finally obtain
d2 .7
PT — F ►s = f dx z •
(13)
This is the desired dynamic boundary condition for a surface that varies only in one dimension. It is a linearized condition, restricted to surfaces that deviate only Slightly from a plane_ DETAILED DERIVATION OF THE DYNAMIC BOUNDARY CONDITION FOR AN tNYI4{'ill FLUID
We shall now reconsider the dynamic boundary condition, proceeding in a much more careful manner than before. The r€.triction to nearly flat boundaries will be dropped. Our discussion will contain some detailed arguments, which do not make for rapid reading. Time spent in careful `
Our more dead ed dtscussron iatertri this scctron shows that the x-components of the various
forces add to zero if we assume that T is cnnstant.
306
Fornrukrer un nj the Theory eery
of Surkier Wares in on Inriscid Fluid jCh_ 7
perusal of these arguments, however, will help the reader toward a genuine understanding of the iSSLtcs involved. h turns out to be helpful to regard an interface vii as a special state of matter, confined to a surface. For an ordinary material region, the influence of adjacent exterior substance is represented by a certain force per unit urea given by the stress t, a vector defined an t he boundary of the region in question. Analogously, we hypothesize that the influence of outside adjacent rnatenal on a patch of surface material P can be regarded as equivalent to the influence of a force per unit length I acting around the curve C that forms the boundary of P (Figure 7.4).* The total force of this kind would thus be given by ( 14 )
where dN is an element of arc length_
ds
FIGURE 7 4. A force p er emit letu y th 1 is hy7rrrlhesi_er.f to ere -i nri the boundary
porch p
raJ
surface Mai ,seppreale.v two rrollmiuis,
of'ex
Uwe consider a material region R(r) that intersects the interface .0 between two different substances, then the influence of outside material will require a body-force integral and a surface-stress integral, as in Section I4.2 of 1, plies a line integral such as (14). Here C will be the curve formed by the intersection of .0 and the houridary of R(r} (Figure 7-5). Thus instead of (14.2.5) of 1, we have the momentum balance equation
dr,,, #t
pv
=
fff
p
dr+
Jf t
pig
du+
f
i d-5.
(15)
(,i
There is an appealing symmetry in (15). Linear momentum can be represented as changing due to distributed sources per unit volume given by pF, • A full analogy would requirc irltroduciion of an " intcrfacc body force," hut these seems no compelling physical mason ro inteOdticr such grim-alit).-
Sec. 7.11
Boundary Co rirfil rorts
307
Ft ci.' R E 7_5. A rtrctrrrat region R(r) rhat mraddhs she
iMrerfac•e .1 be tw een
two d'rlieremt materials. sources per unit area (acting on the boundary) given by t, and sources per unit length (acting on a certain boundary curve) given by 1. Although we shall not pursue it. one can derive consequences of (15) which are analogous te the consequences of (14.2.5) that were developed in Section 14.2 of I [sec Aris (1962}1 What we wish to emphasize is that (15) shows that the fundamental equation (14.2.5) of I a generalization F = ma for continua, cannot be "obviously correct," because sometimes (when surface tension is present) it is not correct at all We shall at once adopt the simplest reasonable hypothesis concerning the nature of 1. Let n(x, r) be a unit normal to a small piece P of the interface at point x and time I. Let ulx, r) be a unit tangent to the boundary C at point x, time r. Choose n and u so that the unit vector N = u A n points outward from P (Figure 7.6). We shall assume the consri:wive equation ,
I(x, r) = T(x
r)N(x, r)
for x on C.
(1 6) Equation (16) states that the "rnernbranelike" force per unit length i acts in the direction of N with a magnitude that is independent of N; T, the magnitude of the membrane stress, is the surface tension.
>"• !CURE
,
7.6. Tangent cold rtorrrtp l rectors at a perirri on
surface P.
the boundary C
308
Formulation of thp
Afar),
of Surface. Wa v es in an ltaaisrid Fluid [Ch. 7
What justification is there for the constitutive assumption that surface effects are representable by a stress acting in the direction of N? First of all, it is the simplest sensible assumption and one that at least explains the static phenomena of Figures 7.1 and 7.2_ It is reasonable to adopt this assumption and sec if it leads to sensible answers. Moreover, there is an analogy between tension in the surface phase and stress in the bulk phase (the interior of the fluid)_ It is frequently useful to postulate that a fluid has zero viscosity; in this case the internal stress acts normally. One hopes, not ir1 vain as we shall see, that it will similarly be useful to ignore the possible presence of"surface viscosity" and thus to postulate normal-acting stresses in an interface as well as in the bulk phase. It should be pointed out that there is considerable interest in physicochemical problems where the standard surface tension postulate is inappropriate. Ark ( 19ô2) discusses some of the theoretical generalizations that are required when "surface viscosities" are considered. Nonetheless, it is clear that the hypothesis of normally acting surface tension rs at least as important for an analysis of surface effects as is the {i nviscidl hypothesis of normally acting stress in bulk phase fluid mechanics. FORCES ON A SURFACE ELEMENT
Having postulated a certain kind of interfacial stress, we shall derive the desired dynamic boundary condition for uni invirc•id fiuidt by requiring conservation of momentum to hold for a fluid element "straddling" the in terface_ Consider a portion of the interface cut out by planes x — constant and y = constant, where each pair of planes is a small d istance ki apart (Figure 7.71. Lei us locus our attention on a material element R whose midsurface is this portion of the interface and whose thickness has the uniform value r. . We shall apply the conservation of linear momentum requirement (15) at the instant t. It is not hard to show that the i nertia term and the body force term will be hounded by a multiple of r.j. . For example, suppose that the mid surface has the equation z = Mix, y) for (x, y) in the square S,,. 11 l' denotes the volume of R, then
JfJPF
di
!+
ax (pf ]
VR }
where the maximum is taken over some fixed domain containing R. But k,t • Surface viscosity seems to
haVC
no detectable cficx:t in pure liquids, but may be imponanl in
solutions_ t Effects of viscosity are usually secundary in water wave problems [sec Bachelor (1457, 370A-
Sec. 7.1] Boundary Conditions
309
I
'
I
I
^ I
I ^
+ I I t
I I I ^
-- — —'— — —
^
—r ^^
^r
u
Fm
u sc a=_ 7.7 A
.sr1r}urr " flake -
R.
can be found by multiplying the area of the rnidsurface by the thickness cp . Thus lfI< = efa f f(1 4 Ms
)"' dx dy Ç
-
F1.e' max if I t M .
-
4 M )" 2
sY
Similarly, the forces acting on the vertical sides of R will be bounded by the product of ell' and a constant independent of f: and p. Since x can be made arbitrarily small, the remaining terms of (15) have the property
Zi
rn 2 1f
, i]
.1;
+ f5tar + ^Ir ds
O.
(L 7)
i
Here C is the intersection of the interface .a and the boundary OR of R, and . is the portion of ,i that is interior to C. A plus or minus superscript is affixed to .^ depending on whether the surface is approached from the top (z > O) or the bottom (z < O). Equation (17) is a local equilibrium principle of the type derived in Chapter 14 of 1 - ln the present case the stress and surface tension forces on a surface element .o are in equilibrium. We now restrict ourselves. for simplicity, to a two-dimensional situation in which there is no variation in the y-direction. In particular, the interface height does not vary with y, so the interface is a cylindrical surface with generators parallel to the y - axis (Figure 7.8). The surface tension forces acting on the front edge of the interface and its back (curves cut by planes y = constant) are exactly equal and opposite and so add to zero. The surface tension force per unit length that :acts on each of the Iwo straight lines
310
Formulation of she Theory of Surface Waves in an #nviuid Fluid (Ch. T T{si.j N(s L j
th mugs 7.$. Edge stresses on a surJur fr. The surfil €e fs o cylinder wiih arr equation of 'he form f ( x, 2J = ^.
forming the side boundaries of the cylindrical interface does not vary along one such line. These force per unir length vectors are denoted by T(sJN(s L) and T(s n )N(sR ), respectively (Figure 7.8). The subscripts L and f refer to the left and right for one looking in the direction y increasing. The reason for the use of the letter s will appear in a moment. The surface tension_ forces acting on the left and right sides of the interface, each side having length , are pT(sL)N(sj and pT(sR )N(sR ). A "head-on" view of the from edge of the interface, as seen by someone looking in the direction of increasing y, appears in Figure 7.9. We designate by e the curve making up this front edge, The total length of r will be denoted by A__ Its arc length, taken increasing as x increases, will be denoted by s. At the left
Y EOuKF 7.9. The frorrr edge of t he interface dPpirre,d in tï:qure 7.$. fop and t ^ bottom ^7rYS]71 rf} magnitudes Cire {) r f S) and Sll , respectively.
Sec,
7.11
Boundary Conditions
3 11
TABLE 7.1 Forces on the iriterfaee. Upward
Rightward
tin directio n of z increasing)
(in dircction of s increasing)
Top: Bottom: Left:
Right:
< 5 '1( 5 R ,
1
= I,
3,4-
and right ends of c, s will be given the values st_ and s R , so — By 0(s) we shall mean the positive acute angle between the x-axis and the tangent to c at s We indicate in Figure 7.9 typical pressures p ris) and Nis) acting normally to the top and bottom surfaces. Note that in this view Ms ') and N(s,) are tangent to c at its left and right ends. We are now in a position to obtain the information needed for determining the consequences of the local equilibrium equation (17). In evaluating the line integral, we need only consider the contributions from the left and right edges of the interface, since the contributions from the front and back edges add to zero. We list the required information in Table 7. I . In the table, we have employed for c the parametric equations x = x(s), z =
z(s); s i < s < sx,
( 18 )
where the dependence on t has not been explicitly shown. We have also used the relations dz
Sin 0= ±—, fI' ï
-
dx
c[ts$ —
(19)
d5
The "Wngen[ angle" ti is defined somewhat differently from its counscrpan that was
used above, Which unambiguous do inition one adopts is essentrilly irrirrratereal. Of importance is the care with signs that we emphasize here and that we glossad over earlier by employing the .
'usual" formula tan
kir dzidx.
312
rF o rmulal ion of tlrp Theory of Surface Wares in an litriscid fluid
[Ch. 7
r
4x FIGURE'
7.10. Diagram used to rabtain the relationships ( C 9) between Me
rariyrnrx:l anyJr° El pfird dLrirorrres kith respect to arc feriylh s.
it,i2; therefore, the signs must be (sec Figure 7.10). By definition, 0 5 il chosen to make sin nonnegative. Let us carefully derive the first entry in the first column of the table Tri begin with, in Figure 7.9 the arrow marked p i correctly denotes the stress exerted by the top fluid on the inter face .0 since, because we are regarding the Fluid as inviscir1. R stress of triagnitude p r. acts in the direction of the exterior normal (which points out of .0 into the top fluid). Equivalently, a stress of magnitude +pr aCts in the[lirection of the interior normal (as shown). The corresponding force associated with an element of c having length ds is p 3 ds. The "top" vertical force upward on .0 is thus -
—
-
J
sR
r
pt(s) cos O(s)]p ds = —
fl_
FA p1{s) ds ds (s)p s^
dx pr(5 r)
( 01. ;
ds
(e haw used the integral mean value theorem in obtaining the above equation.) Let us turn to the third entry of column one The force upward due to the surface tension acting on the left-hand end of c is
oi5t) s i n vi s it } .
= J^^l^
)
f
dz —
Ts [tiL)l
L
The negative sign appears since, as can be seen from Figure 7.9, for s near S L , z decreases as s increases; so dzjds is negative. But suppose that the Figure were di awn differently; would the result be the same? consider Figure 7.11.
Here, since N(s L ) points downward, the upward force is
—117'(a L ) in 0 (sL) =
—
reT(S,<.)
ds
(s t),
because z now increases with s. The final result is indeed the same. Derivation of the remaining entries will be left as exercises for the reader.
See. 7.11 6owdary Conditions
7.tt. Diagram
of a rsptx'orr.dorrmrard
3
a^
[o n6,nm ih.• sur/are tension Jorre
tira
eAte
cylindrical swr/nee_
CONSEQUENCES OF LOCAL EQ(JiLi8R1UM
The sum of the upward forces is the vertical component of the bracketed term in (17). This sum Can be written dx (Si } + d
Pa(s } dx [s^) — Pr L^S
ds
!s
T
dz S t C Sy C S R .
[s a=a s
(We used the mean value theorem on the last two entries of the first column of Table 7.11 We apply the limit called for in (17) in such a way that c always contains a certain fixed point (X, 4 In this case, the points characterized by s r , s 2 , and s s all approach (x, z). Using the fact that the ratio of the arc length of curve c to its chord length ü approaches a constant [Exercise 7(b)], we find that
ds [I
ds^
Ap^x=U,
(20)
—
for each point (x, z) on the interface. Here (21)
AP =- Pr — PR-
Similar manipulation of the sum of the rightward forces (Exercise 8) gives d
r
TAL I ds
+ 4P ds
0_
(22)
-
Having derived (20) and (22), it is natural to spend a little time combining them in various ways in an attempt to find, if possible, more striking forms of these boundary conditions. In particular, one expects to be able to relate dp to some local geometric properties of the surface_ This expectation is borne out, as we now show. On writing out their first terms, (20) and (22) become
Apx, = 0,
(23)
7^ x, + Tx L + Apz , = 0,
(24)
7szs + Tz. —
Formulation of the Amu of Surface Waves in an inviscid Fluid [Ch_ 7
314
where we employ a subscript to denote differentiation. We now multiply (23) by rF and (24) by x,, for the purpose of eliminating 6p. Upon adding, we obtain 714 + 4) + T^2r ^ + X3 X65,) = 0.
BUt xi + zâ so (upon differentiating with respect to
(25a)
1,
s)
x,x„+z zu =0.
(25b)
,
Hence T = O. In two-dimensional motion of an inviscid fluid. tic , surface tension must be constant (in space). It is now easy to demonstrate that both (20) and (22) become 7',c =
Here
K
4.
(26)
is the curvature of c, defined as usual by d = ds
_
angle of tangent vector with positive x-axis
=
tan
(z.) x,
(27)
Thus Zi1
JC,s
Xs
2s
K ü 2 y ^ ?L a — Xs$ Z,
(28)
as the reader can show [Exercise 9(a)]. Suppose that the interface equation 2 = (x, t). Then [Exercise 9b)] K + C - 312 `
2)
c has the (29)
The elegant formula (26) is the final form of our dynamic boundary condition_ (Keep in mind the constitutive assumptions that we have made= inviscid fluid, normally acting surface tension.) We see that the pressure suffers a discontinuity 4 which is proportional to the curvature u, with proportionality constant equal to the surface tension f_ 1f x is positive, then the interface is concave upward as in Figure 7.1a. But from (26), positive K implies positive 4 = p r — pa . This gives a sign check, because the surface > p g . Negative K implies a picture like bulges as one would expect with Figure 7.11; PT < p g , which is again as one would expect. EXERCISES
The first four exercises involve two-dimensional flows, sa that the interface reduces to a curve in the xy-plane_
Sec. 7.1 ] Boundary C ündisians
315
1. (a) Show that any curve xy = Constant can be though[ of as an interface dividing two immiscible materials whose motion is described by the velocity vector v = xi — yi(b) Derive (8). 2. (a) In a two-dimensional steady motion any streamline can be considered as a boundary between immiscible materials. Why? (b) Verify the statement of Exercise 2(a) for the particular example worked out in Section 112 of 1. In that section the velocity vector is given in Equation (17). the streamlines in Equation 12 I). 3. (a) In two-dimensional unsteady motion a streamline generally cannot be considered as the boundary between immiscible materials. Why not? (h) Show that the streamlines of Exercise 13,2.4 of 1 cannot be considered as boundaries separating immiscible materials moving with the give velocity field. +4. h can he shown that the following two equations Carl he considered as giving particle paths for the flow of a fluid of uniform density: x = a + e6 sin la -f tj,
y =5
—
cos la
r).
00) (31)
Various values of (u, b) correspond to various fluid particles. Thus the spatial velocity field is i^
= exp fb(x, y, 11 cos [a(x, y, z) + r], = exp [b(x,y, t)] sin [a(x, y. r) -I- r], ]
where a(x, y, r] and b(x, y, t) are given implicitly by (30) and (31). Show that the kinematic houndarycondition is satisfied along the curve derived from (3Oland (31) by setting h - constant and regarding casa parameter. This is GerSrtier's rrochuitial wave, the single known exact solution to the problem of wave motion in a semi - infinite medium with no surface tension (that we formulate below) [see Milne - Thompson (1968)]. 1-5. Show [hat if the surface F(x, r) = 0 separates water and air, then it always consists of the same fluid particles. }6. Static problems involving surface tension are expressed in terms of the surface tension coefficient T. the gravitational constant g, and the density of the medium p. What conclusions can be drawn using dimensional analysis? 7. (a) Derive those entries in Table 7 that are not derived in the text_ (b) In the limit used to obtain (20), to what quantity does )/ i tend? 8. Derive (22)9. $(a) Derive (28). (h) Derive 129).
316
Formulation of the Theory of Surface Waves hi [ar lniiseid Fluid [Ch. 7
10. Derive the dynamic boundary condition without the restriction 0/0y = O.
Proceed by the following steps. *(a) Consider the interface element .fir of Figure 7.12. with equation = C(x, y). whose projection on the xy-plane is the square whose z corners are
(ye, yo -4 J1)J (A + /1, }'o + Fe)Write down an integral relationship requiring that the pressure forces and surface forces add to zero. (xo, yo), (xo + Jz, yo),
I
FIGURE 7.12. An interface element .0 whose projection on the -.. Y-plarrr square.
Is
a
(b) Combining the integrals and using the mean value theorem and the integral mean value theorem. deduce the following relationships:
+- C ÿ) I n
ôy [TCxC,,ll + —
^ [} • (1 # ^ ){1 + Ÿ
C.
ApC }, W — ^ [ T{1 -i- c!}[1 + + ^^ [
A P = 6 [ l c,(1 + +
^x
[Tc x(1 -3-
ÿ)
Cr) - ;], "
+
J ,zi
+
1,2
J.
(32)
+ cr) +
7- c. c,(1 +
+- ^
''/]-
(33)
Sec.
7.J1
Boundary Condilions
317
(c) Use 134) to eliminate Ap from (32). carry out the differentiations. and ohtain T. = O. Similarly, show that T,. = 0- Hence derive from (34) Ap ° T[
I + C) — 2Cz4,C., + C yy(l + CŸ)1(1 + ; + rÿ21
- ]" 2 (35)
The coefficient of Tmust have a geometric interpretation that can be described without reference to a particular coordinate system_ One such description is ic 1 + k:2, where !Ki and lc 1 are obtained as follows. Construct the tangent plane 2: to the surface 5 at point P_ Constructtwo planes r, and rt 3 perpendicular to each other and to I (Figure 7.13). Define c 1 as the cu rvature al P of the plane curve r ; formed by the intersection of ri and .0, 1 = 1, 2. The planes n, and n, are determined by the above description only up tri a rotation about their line of intcrsectior]. But it can he shown that xi f- rc2 has the same value however the planes are rotated; 3{rc ti # li 2 ) is called the mean curvature- For further details, see accounts r3f differential geometry. such as that in Chapter 2 of 1J. Struik, Di prentiai Geometry (Reading, Mass.: Addison-Wesley, 195U). REMARK.
FIG [1 R E
7- I 3. Diagram us ed to ex plain Ihe -
er,rrr eRr -
of
orrura ['urralure
11_ Improve the derivation of the dynamic boundary condition (26). This could be the subject of a brief essay. Here are some suggestions. (a) Give stronger physical reasons For assuming the constitutive equation (16). (h) Justify the statements above (17) concerning the forces on the vertical sides of R. (c) List precise smoothness conditions that guarantee the validity of the various mathematical steps. tZ. (This exercise requires knowledge of regular perturbation theory; see Chapter 7 of 1.) $(a) Let : = /ix) be the interface between fluid and air. Let the fluid be motionless, inviscid, and subject to constant atmospheric
318
Formula iton of the Theory of Surface Wares if! un Irwi..ccW Fluid
10i.
7
pressure and the force of gravity (acting in the direction of —z}. Let p be the density, assumed constant. Show that the dynamic boundary condition becomes —T - p9' + ((1 + CD-3r2 = C,
C a constant. (36)
(h) Figure 7.14 represents a fluid hounded on one side by a vertical plane wall. Show that with the coordinate system given in this figure, C must be zero_
FIGURE 7_ 1 4. 5raiiortary ntittiscus Own a fluid with sur farE• re•nsrun is in c•orriur•r
With a y e,riicuf u411.
Make the change of variables rÿ= h ,x= xh cot O.
(37)
Then show that (36), with C = 0, can be written =1 2 C[1 -#
'- d
')T",
dR
(38) '
where 2=
^9f?s
E—
r 7. ,
cat 3
O
(39a, b)
.
Show that appropriate boundary conditions are C(0) — 1,
(
0
as
-c -. oc
,
(40)
and also C '(0)
-1
.
(41)
If you are familiar with scaling procedures (Section 6.3 of i),
justify (37).
See. 7.2] Formul
t(d)
lion
319
and Simplification
For certain liquids. r. is small compared with unity. Therefore. we look for a series solution (►r)_Ço( )+zrg)+cÇÇ(7+ ... to (38) and 140). We regard the parameter r i* as fixed when we calculate this solution. Later we shall impose 141) and find a value for 7 2 . Show that (38) and 14C) imply that —
'i
-r/C0
= 0;
CD(0) = 1,
z i CL = ilr 2Ca(C1) 2 ;
Co(cc) = f-
C i (o} — C 1 (cc} = U.
‘_.
Find the equations satisfied by *(e) Show that i_ o = exp (— M. and find findr_ 2 . 1(1) Show that (41) implies that
L,.
if you are courageous.
1 =- rl + r:r; + O(r: 3 ). Assume that 7 2 = I + ar. + 0(r 2 ) and find a. [Note that although there is an f: in the denominator of (39a). nevertheless. r 2 = 0(1). as was implicitly assumed above.] ff your courage persists. find the O(c) Correction to -r. +(g) Consider fluid contained between two parallel vertical walls_ About how far apart would the walls have be before we could use the Le estimate the height of the meniscus? (h) This problem can be solved exactly, See for example. Landau and Lifschita (1959. p. 2351. For sufficiently small r:. show [ha' the expansion given above agrees with an expansion of the exact answer (assuming convergence of the various series).
to
result of lfl
,
7.2 Formulation and Simplification We shall formulate a problem involving surface waves of an inviscid fluid (which we shall call "water") whose density can be regarded as uniform. "Air," a relatively light fluid whose influence is negligible. will be assumed to occupy the space above the water surface. Although a number of simplifying assumptions will be made, the mathematical problem that emerges will still be quite difficult. Thus the section concludes with a further simplification, neglect of nonlinear terms. We use an asterisk (*) to show that variables are dimensional. As usual, subscripts denote partial derivatives.
320
Formulation of
the
Theory of Surface Waves in an lrmisrid Fluid [Ch. 7
EQUATIONS FOR TWO-DIMENSIONAL WAVES IN AN INFINITELY WIDE. LAYER OF JN VISCID FWD
We shall restrict ourselves to two-dimensional motions (u* = e3fe3y* The body force will be that corresponding to a uniform vertical gravitational acceleration of constant magnitude g_ The lower boundary of the Fluid will be assumed to be an impermeable horizontal plane (therefore, we cannot cQstsider such phenomena as the breaking of waves on a sloping beach). We choose a coordinate system so that the lower boundary has the equation z* — — H (H a constant) anti so that gravity acts in the direction °I - decreasing z* (Figure 7.15).
FIGURE
7_ i S. Natation
for urresiryae+art ol water NY]r e'.s in fluid al mean dept h H.
We first write down the requirement of linear momentum balance. From (A2.1.6) with f — gk, taking into account definition (A2 1_1b) of the material derivative, we have u* -t If*wx. + V0* u'. — — pi
w + u*w:. -1- x•*w=. = —p
x. ,
t p r: . —
(l )
g.
(2)
Here te* and I4 are velocity components in the directions x* Increasing and z * increasing, respectively_ The constant p is the density and p* denotes the pressure1r or a fluid of uniform density the mass-eonservatiorr requirement (A.2.1.1) is u =. + ti4 1. = o. (3) The condition that boundary condition
nn fluid permeates the bottom gives rise to the
at z' =
—
H,
w* _ 0.
bottom (4)
We shall assume that "side" boundaries are so far apart that their influence can be Neglected_ Consequently, the horizontal variable x• has the domain — cx < x* < co. Once such an infinite domain is allowed, experience teaches that there may be solutions which behave badly aslx* I —, c:. There is — really" rio such thing as an infinite domain in nature, so it is not a straightforward
Sec- 7.2] For w/ariun and Srtrplrfieution
32]
matter to assign boundary conditions 'al infinity." !Nonetheless, there are some types of behavior (e.g., exponential "blowup") that are almost certainly intolerable in a mathematical mode!. We deal with the situation for the present by imposing the following somewhat vague side boundary condition: solution "well behaved" as lx' l
(51
oc.
— C*(x*, r*]_ When Iet the water-air interface have the equation have, from (111), the surface kinematic boundary condition at r* = C * (x * , t#),
'' = Cr. + u*4*..
i,
•=0 (6}
For reasons that will be discussed shortly, we shall assume that [he pressure above the water surface is a constant p . With this assurruption s (' _26) and (1.29) give the surface dynamic boundary condition T(*.iotl •+ (f *•)3]
at z* = C • (x', l'1,
312 M
pA - p*.
(71
Here Tis the surface [ension. As is the case even with the simplest mechanical problems, initial conditions must be imposed to obtain a fully specified situation_ For, all other things being equal, the initial state of the water determines its subsequent history. Later we shall give examples of appropriate conditions; now it will suffice to make a formal statement that such conditions need in be imposed. We shall start measuring time from the initial instant, which will thus be designated r* = O. Hence we require the initial condition at t• = 0,
suitable conditions prescribed_
(g)
We have been thinking of a water layer having "depth" H , but we must make precise what we mean by this. The depth varies when surface irregularities are present. We can, however, speak ofa layer whose mean depth is H. Since the layer bottom is located ël ÿ* = —11, this means that 2* = O must be the mean position ni - the interface_f Hence the function Mx*, two must haven spatial mean value of zero. if * has period L, we therefore impose the mean Might requirement ([or a periodic interface) 1
i-
1, ^o
c.v.,t *) dx* = O.
(9a)
if C* is not periodic, we must make the more general requirement
Urn
L
J-14*(x*, t*}dx* = O.
(9b)
# Although the definition ]c prohahl}- known to most readers. for rnmpteteness she mean value of a function is defined in Exercise 1.
322
Formulation of the Theor y of Surface Waves in an h+riseulflu rd (Ch. 7
Possible changes in mean height could be permitted by imposing (9) only al t* = O. The mathematical problem posed by (1) to (9) seems likely to have a solution. lr, (1) to (3) we have three equations for the three unknowns u*, w', and p*. Also, en each part of the boundary equations (4) to (7) provide a N o T E.
boundary condition on a*, w*, and p*. [We are temporarily interpreting (6) and (7) as a sine implicit condition on p* and w*. This can be obtained in theory" by eliminating L'_J Initial conditions are provided by (8). while (9) is an extra normalization condition that pins down the average location of the bounding surface. Even if we could prove that the system of equations (I) to (9) has a unique solution, it would be quite another matter to characterize this solution in any detail. The system is nonlinear, and some of the boundary conditions are to be applied onacurve z' _ *(x', t* ) whose location will only be known when the problem is solved. These features cause considerable technical difficulty, so we shall soon restrict our attention to a simplified system. Before simplifying the equations, however, it is advantageous to reformulate there somewhat, by employing a slightly different form of the pressure which arises when a possible mot ionless state is "subtracted out:" STATIC AND DYNAMIC PRESSURE
A steady-state (Nat* = 0), exact solution to (1) to (9), cur responding to a motionless layer, can be found at once. The surface bounding a motionless layer should be plane, so we look for a solution wherein
u*= IA* —ÿ* = Equations (3) to (6) are identically satisfied, while (1) shows that p* is independent of x'_ Equations (2) and (7) require that p* = po(z*). where
pa(z*) _ — Paz * + pa ,
— H S z* < 0.
(1%
Initial condition (S) is irrelevant for our steady-state problem. Finally the mean height condition (9) is certainly satisfied. 11 is not difficult 10 discern the physical meaning of the solution just found_ The static pressurept, a distance — z* units below the surface of a motionless layer is due 10 the weight of a fluid column —z* unfits in height and one unit in cross section, plus the weight of the atmosphere. Returning now to the original problem ( I) to (9), we find that it is convenient to introduce the difference between the total pressure p' and the static pressure p" rather than the pressure itself. We accordingly define the dynamic pressure
p* by p * (X *, I*. t * ) = p* x * , z*. V') (
no(z').
(1 l)
Sec. 7.21
Fornrulariocr und Simplification
323
In terms of the dynamic pressure rather than the, total pressure, the governing equations [(1) to (9)] become
+ w*W# = —p 'i^•
,
wt*, + u*wx. + w*w= = — pp= ,
u*. + w= = 0. At x• _ — N,
x,* _ O.
As I x*
solution well behaved.
cc,
(15)
(16)
w* = Cr. + u*Cr..
At z* = 4*(x*. r*)
^^^•f •tl + (C1•)2]-312 = — P* + p9‘*-
is
= 0.
(13) (14)
At z* _ C*(x*, r*),
At
( 12)
(17)
suitable initial conditions prescribed.
(18)
(19)
For waves o f period L in x *,
^
f
L
x*. r*} dx* =
a
U,
(20a)
and for an aperiodic interface
1 i:, 2K
urn —
x
_
C*(x#, t*) dx* - 0_
(20a)
,^
Comparison of (12) to (20) with (1) to (9) shows that only two essential changes occur as a result of introducing the dynamic pressure i•)*. (i) The body force terms g is dropped in passing from (2) to (13), (ii) In passing from 7) to (18), we drop the constant term pa and add the term pgC . If there were no free surface [and hence no dynamic boundary condition (18)), the equations involving the dynamic pressure p * would thus be exactly the same as the equal ions involving the total pressure p* except that the gravity term would be absent in the former. Consequently, for an inviscid fluid of uniform density, in the absence of free surfaces one can always assume that gravity is nor present. The pressure obtained as part of the solution will be the dynamic pressure. The effect of gravity can be taken into account merely by adding the static pressure and the dynamic pressure to determine the total pressure. When free surfaces are present. the static pressure might better be termed the mean static pressure, as it gives the force on a unit area due to the (temporal) average weight of a column of fluid whose actual height fluctuates because of surface deflection. Consequently, as can be seen from the dynamic boundary condition ( le), in the presence of a free surface gravity must be taken into account throughout the calculations. [
324
Formulation of th e Theory of Surface Wares in an fm ist id Fluid [Ch.)
Note that the constant atmospheric pressure does not appear anywhere in the governing equations (12) TO (20). This is because motions are driven by pressure derences, so that a uniform pressure has no dynamical effect. (The situation would be different if thermodynamic effects entered, but recall that we are considering a fluid of uniform density.) In particular, the motion of the hounding surface is entirely unaffected by a constant atmospheric pressure_ The surface moves just as if it were entirely free of a constraining pressure force, and is thus often called a free surface_ In restricting consideration to fluids bounded by a free surface, we are tacitly putting to one side the important problem of wave generation by wind—for here air-pressure variations are all important. Subtler effects are
also being neglected. For any motion of the water will drive a motion of the adjacent air, producing air-pressure variations. These, in turn, will affect the motion of the water. Since the density of air is much less than that of water, the effect of the latter motion will be negligible except when it continues for too long a time. The exception must be noted, because small effects can accumulate to noticeable levels if they act over long periods. NON D1M ENSIONALIZATION
We shall now introduce dimensionless variables to reduce t he number of parameters on which the solution explicitly depends_ (Compare Section 61 of I_) For the time being we shall restrict our attention to "fairly regular" wave trains, which have the property that the distance between one crest and the next does not vary much. Any one of these distances can be chosen as a typical wavelength L (Figure 7.16). "Fairly regular" wave trains also have the property that the maximum heights of the individual waves do not differ appreciably; consequently, any one of these heights can be chosen as a typical amplitude A. The time it takes any selected wave crest m to travel a distance L will be called a typical period P. (In the special case of an exactly periodic wave train that propagates at constant speed C without changing shape, P will be the temporal period of the waves and C = LIP.) 1 n dealing with fairly regular waves, the quotient of a typical wavelength L and a typical period Pis sometimes referred tc3 as a typical wave speed. We now
1
„,
L and typical umpliructe R. !he ware cr est m 1rai5els a distance approximately equal to L in the typical per i od P. Ffoi1KE 7.I6. A
"
fairly regular" wave train of typical wavelength
Sec. 7,21 Formulation and Simplification
introduce new (unstarred) dimensionless variables, in terms of L, the density p:
la
x`
P,
x — L
w* w= ^^^,,
u*
u= A ^ , /
p
,
^
z
=
A,
325
P, and
Z^
f.
p* pLY^3
,
2'
^ — C` . A
(21)
in terms of these variables, (12) to (2C) can he written Exercise 6)
x + sm.} =
!1, + +
14*
+ rtx
At
z_
(24) ( 25)
w = CF 4- EuC Y .
At z — ES
p
L
031
+ w= =
At z = EC,
For waves of period
= — p_+
solution well hehaved_
As 1x1 i ac',
At t = 0,
(221
w = 0.
— ^J ,
,
— pY .
(26) (27)
BC„(1 + v2 (x ) -3t z ).
suitable initial conditions prescribed.
(28)
(29)
in x* (and hence of period unity in x),
f
C(x, t) dx = O.
(30a)
For aperiodic waves, lim A^
Here
t—
A L'
ce
1 21L
^
_} , H
‘, (x,
t } dx — O.
(30b)
- k
l
T p L. 2 '
qP 2 I
(3 l )
are dimensionless combinations of parameters that appear in the dimensionless equations (22) to (30)_ We see that the severe dimensional parameters A, L, H T, p, y, and P can appear only in the four dimensionless combinations of (31). The first two parameters of (31) are geometric and provide, respectively, the ratios of typical wave amplitude and mean water height in typical wavelength. The parameter B, sometimes called a Bond number provides a measure of the importance of surface tension. The Froude number Frrieasures the importance of gravity. It is thesquare of the ratio of two times, Pond WO". The former is a typical wave period and the latter is the trmc it takes a particle, starting from rest, to fall a distance equal to half a typical wavelength.
326
Fcar m ufari lm of fhe Th eory of Surface Waves
i1ltHS['id fluid [Ch. 7
LINEARIZATION
The mathematical problem posed by (22) to (30) is difficult_ There arc two main reasons for this. First, the equations contain many nonlinear terms, and there is no general theory for nonlinear partial differential equations. Second, the boundary conditions (27) and (28) must be imposed on the curve z = but {(x, t) is one of the unknown variables of the problem. If we restrict our considerations for the time being to waves of small amplitude, then both difficulties disappear. We expect that such waves accompany flows wherein velocities, pressures, surface deflections, and their rates of changes are small—so that all nonlinear terms can be neglected to a first approximation. The assumed smallness of surface deflection has another consequence: surface boundary conditions can safely be evaluated at the mean position of the interface. With nonlinear terms neglected and surface boundary conditions imposed at the mean water level z = 0, (22) to (30) simplify to the following sels of equations [with a superscript (I) denoting a first approximation]: =
_
f
w it {
t) x
(32)
(JJ
=
(33)
O. (34)
Ate — — h, w4 = D As 1x1 At
a= 0,
At z = 0, At
i
as,
(35)
solution well behaved.
wO,
P(
iJ
=_ ^+ , ^
(36)
(37)
^ F(Cl t t — 8CV2)-
— 0, suitable initial conditions prescribed.
(38)
(39)
For waves of unit period ,
J
^{ ^ ^[x,
^}
âx = 0.
(40a)
0
For aperiodic waves, li ril .0
I 2K
II: ^t IX, t} dx = O. _a
(40h)
The two succeeding chapters are devoted to a study of the linearized equations (32) to (40). The reader can begin this study at once if he wishes. But sooner or later he should peruse the next section , where the linearization process is examined more carefully. Let us mention here the most important point covered by this reexamination. Equations (32) to (mil) can be obtained
Sec.
7 3 Order-af Magniiiide Esiimares, Narjdimeresichalization, and Scaling
327
from (22) to (30) by setting c equal to zero. but we have provided no firm justi fication for such a formal step. EXERCISES
1. Let f be a continuous function defined on the interval [i . b]. Let x o = a, x1, x„ = b divide [a, bl into n equally spaced intervals of width Ax. Write down an expression for the average value of the pi + I quantities f (xi), i = 0, I, 2, ... , n. Multiply numerator and denominator by Ax and pass to the limit, thereby motivating the following definition: average value off (x)
1 -
b
f (x) dx.
2. Show that for a function of period L, (9a) holds if and only if (9h) holds. 3. (a) Formula tea mathematical problem that is relevant to wavegeneration by wind. (h) How would the Formulation in the text be altered if the bottom were irregular? In particular, how would (9) be changed? 4. In Section 15.4 of 1, gravity was neglected in finding the force on a body immersed in a uniform stream. Show that the presence of gravity adds a buoyancy force. 5. True or false : Even though a free surface is present, if surface-tension effects are dominant, then the effect of gravity can be determined in the same way as recommended in the text for situations in which free surfaces are absent. Explain your answer. 6. (a) Verify that the variables of (21) and the parameters of (31) are dimensionless_ (b) Verily (22) La (30). 7. If you are familiar with the concept of stability (see Sections 11. l and 152 of I), show that (32) to (40) would he obtained in analyzing the stability of a motionless layer of fluid to small perturbations.
7.3 Order-of-Magnitude Estimates, Nondimensionalimation, and Scaling In Section 2 we passed in a rather formal way from the dimensional set of equations (2.12 -20) to the linearized dimensionless set (132 -40). We shall now try to make this passage more convincing. For full appreciation of the ensuing discussion, the reader should be familiar with the content of Sections 6i and 6.3 of I on dimensional analysis and scaling, but the present section can be read independently. It will suffice to know that a rough definition of the concept of the niter- of magnitude of a function in some region is "an estimate of the function's maximum absolute value, for independent variables in the given region_"
328
Formulation of the Theory of Surface Waves or an }txviscirf Fluid (Ch. 7
ESTIMATING THE SIZE OF TERMS IN EQUATIONS GOVERNING WATER WAVES
In Section 2 we decided to neglect the nonlinear terms in (2.22) and (2.23). Using language introduced in Section 13.3 of 1, we are evidently assuming that the convective acceleration terms (preceded by r) are negligible compared with the local acceleration terms w and w. To gain more insight into this matter, let us estimate the order of magnitude of the acceleration terms for a train of "fairly regular" waves of typical wavelength L, amplitude A, and temporal period P (see Figure 7.16). The first step is to determine the order of magnitude of the velocity v* possessed by the fluid composing the wave train of Figure 7.16. Is it AL, or A/P or LJP, or an entirely different quantity? The first possibility can be ruled out on dimensional grounds, since AL has dimensions (length) 2 . Such quantities as A/P, LJP, (LIP) - (AIL)" all have the correct dimensions, so the requirement of correct dimensionality is of limited usefulness. More helpful is a knowledge of the general features of water waves, which yields the fact that during u period P, particles near the surface of the water travel a distance roughly equal to the amplitude A. The farther below the surface a particle is, the less extcnsive is its motion. Thus, since it is maximum values of Ive1 that we arc concerned with, v* has order of magnitude A/P. There are at least three ways that a person could acquire the knowledge to estimate v*. (i) He could have observed water waves.f Watching a piece of seaweed tossed about by ocean waves, he would see that the seaweed moves hack and forth over a distance approximately equal to the amplitude A, not to the wavelength L, in a period P. (ii) He could have drawn upon a knowledge of wave motion obtained from a study, either theoretical or experimental, of wave motion in other media- (iii) He could have solved the linearized problem (2.32 40) and computed the path of a particle (see Exercise 8.1.5). This last way might seem dishonest. We are attempting to provide convincing motivation for an approximation to be used in solving a problem. We there turn around and say that lithe problem is salved, then we can supply the motivation sought_ This is a typical and legitimate procedure, however Uncertain of t he validity of what one is doing, one makes an approximation for a formal mathematical reason. One solves the approximate problem, and hopes that the physical insight thereby gained will clarify the physical significance of the original approximation. We have argued that v* is of order of magnitude A/P. What about the local acceleration vt? To determine its magnitude. we begin by recalling that to a first approximation v* has period Pin t * . In a time Pf2. then, Iv* I rises from a ,
t Theoretical instruction 1n fluid mechanics. can b.e enhanced Ihruugh Trie use of films precd by the 1J.5. National C'onuncücc for 1-lurd Mechanics Films e nd disrrrbutrd by Encyiopacdia Eirilairniea Edllcalionik Corporation, 425 Miehigarti Ave.,
Chicago. Ill
-
6&61I- here
portions of A. l3ryson's him " waves in F1uicis" could certainly be shown to advantage. See the book of film notes, éf ssrrorrd Experiments ut Fluid Mechanics (C'amhtrdgc, Mass.: MIT Press,
1972).
Se c. 1.31
O rder - of-Magrrrfude Essinmtes, Nondlr ►rensio+rafizariori, and Scaling
32 9
value of zero, more or less levels off around its maximum value, and then falls back to zero. If we assign equal time intervals to each of these three behaviors,# we are led to the conclusion that 1 v* rises from zero to its maximum value of A/P in about P/6 time units. Suppose that I v* I = 0 at time r0 . As an estimate for I rh.*Jot * j, we have
r * (ic * P 6)1 ^v [ a l A/P – 0 6A
11
P/6
-
– – P/6
px
(l]
Similarly, keeping in mind that to a first approximation v* has a spatial period of L, we find that
j at,* j
AjP
(2)
ex";
We further assert that derivatives with respect to z* also reduce I v* I by a factor roughly equal to L/6. (This, too, requires some advance knowledge of water waves, such as direct observation that – unless the water is very shallow velocities seem to change no more rapidly in the horizontal direction than in the vertical) Thus it seems legitimate to assume that all spatial derivatives multiply orders of magnitude by a factor that is roughly equal to 6/L.§ As a first application of our order-of-magnitude estimates, we educe that
* ` Vsid
is of order of magnitude
Y*Jc''i *
(AM kl6ïL) A/Pl _ A IA/Pk(6 P) L
—
[1
Thus our neglect of the nonlinear comer ire acceleration terms means that
we ►viith their length ., are considering wades whose urnpilitade is smart compared i.e., AIL C L There is more that we can accomplish, now that we have an idea of the magnitudes of the various terms- As will be iliustratrd al once, we can choose dimensionless variables in such a way that each term is preceded by a combination of parameters that explicitly reveals the term's magnitude. Appropriate approximations are thereby suggested. SCALING
We could define a nondimensional horizontal variable x in a number of ways, such as x = x*{L or x = "A or x = Y' L) or x = x* /6A. The quantities L, A, A 2/ L, and 6A are called intrinsic reference lengths, since they t The three inter vats may well not be equal, so we Cannot be confident Ilya' our estimates are at ail precise. But ii is rough orders of rnrtgnilude that we seek, for these are usuafIystill:cornt for the purposes of comparing terrn sIn making order cif magnitude estimates, we have nni considered the efeci of the bottom. This is appropriate unless the water layer is shallow Thc kcy ED a proper scaling foi shallow water waves is i he understanding That the depth is, in fact. I he appropriate length scale to use in farmin g the scaled vertical independcni variable and velocity compnneni lser Stoker 11957. -
Sec. 2.4)1.
-
330
Formulation
of the Theory of Surface Waves un an lmiacid Mud [CIS_ 7
are standards of length that are expressed in terms of parameters appearing in the particular problem under consideration. Intrinsic reference velocities, intrinsic reference times. and intrinsic reference pressures are defined by analogy with intrinsic reference lengths_ They are specific velocities, times, and pressures that are formed from the parameters that appear in the formulation of a problem, and to which the velocities, times, and pressures of the problem can be compared in defining dimensionless variables. Examples of possible intrinsic reference velocities for our water wave problem are AIP and LIP. As we have seen, the choice of an intrinsic reference quantity is not unique_ We recommend that intrinsic reference quantities be chosen so that when they are used in forming dimensionless variables, then each term in she equations governing a problem is the product of a constant factor giving 'hose terms' order of magnitude and a function of unit order of magnitude_ The variables of that problem will then be termed scaled. The particular intrinsic reference quantity associated with each dependent and indcpcndunt variable will be called the scale of that variable. As we shall see, when variables are scaled (generally not an easy task), then semiautomatic perturbation procedures will often simplify the mathematical problem_ The concept of scale needs some discussion before its meaning becomes clear. To give one example, we assert that asa velocity scale in the water wave problem we have been discussing we should lake AIP, the order of magnitude of v* We would then define a dimensionless velocity v by (4)
so that A v* _ (-13)v_
(5)
We further assert that a reasonable choice for a length scale is L/6, so that dimensionless horizontal and vertical variables are defined by x*
x —
116
,
_*
a _ L{6
(6)
We have
0 0x*
d
6 L 8x'
ax*
60 L
( 7 a, b)
Consequently, using (4), v+-V
*Y*
— rriC} %i y ,
PLP
Vv.
(8)
Srr. 7.31 Order-oj-Maanin d
E_sJimmies.
dime+rsroncii 3 ion,
and Sca/i, i
33 I
Our choice of length scale has automatically led to results obtained above. That is.17) is equivaleiit to the assertion in the paragraph Following (21 that "all spatial derivatives multiply orders of magnitude by a factor that is roughly equal to b/L" Moreover, (Pam omat ically gives the numerator of the quotient in (3) as the order of magnitude of Or- V*v*_ The discussion of the preceding paragraph illustrates the following point. In scaling independent var iables one chooses those particular reference quantities which ensure that upon nondimensionaliairrg. terms involving derivatives appear preceded by a factor giving their order of magnitude f give another example, an appropriate time scale is P/6, for if
t* P/6'
(9)
then
ev* 11 4
A / P av P16 Ot
6A e?v 132 el
(IO1
Again, the correct order of magnitude appears expliciily on the right side [compare (1 )] while the dimensionless factor iv/r t is of unit order of magnitude. The surface deflection * has not yet been scaled, but this presents no difficulty. By definition, A is the maximum value of so that the scaled dimensionless surface deflection 5 should be defined as follows:
—— • A USE:
lll}
OF DtMENS1()NLES5 SCALED VARIABLES
A number of poilus of interest will emerge if we insert the dimensionless variables defined in (4), (6), (9), and (l l) into the kinematic boundary condition (2.61 We obtain the following:
at z = 6EC,
tit = 65, + 6^u
(12)
Y,
where
A
(13)
L
With the emergence of the parameter z, we are led naturally to the study of the subclass of water waves wherein I c 14 1. Such waves will be said to have small amplitudes, with the implicit understanding that amplitudes are being compared with wavelengths. For small-amplitude waves, a natural first approximation to the boundary condition (12) is obtained by retaining 0(1) tcrmst and by neglecting 0(e) t
Set Appendix 3 i ,
of 1
for definitions concerned wish the ••oh
natation used here.
f ïxmularian of Theory of Surface Wares in an &wsCid Fluid Irh. 7
33 2
terms. We obtain at z =ü w= ,
her .
"Automatically" we have emerged with a linearized boundary condition, applied at the mean position of the interface. Moreover, power-series expansions in c naturally suggest themselves as a means to improve upon the first approximation_ It now appears wise to drop the "6" that appears in (6) and (9), the definitions of scaled independent variables. We have seen that it is relative orders of magnitude O() compared with 0(l) that matter in our discussion. These are not affected by the "6." Moreover, scaling proceeds via order-of-magnitude es:imaies, and the appearance of precisely the number "6" might lull us into the illusion of an accuracy that we cannot legitimately claim. Keeping in mind that we can use a somewhat more accurate scaling if sharp numerical guesses are ever rLuired, we shall therefore employ the following dimensionless variables: t ;
is
^, x =
x` ^
,
__
z* ^,
u
u• ^^^,.
w•
a'^ = ^, {,
^' Ciis
These are the same variables that we used i n (2.24 With them, (12) is replaced by the following
at z =t:C. w—Sr - + This is identical, as it should be, with the corresponding equation (2.27). PRESSURE SCALE
We have avoided for as long as we could the question of the pressure scale. As often seems to be the case, this scale is the most difficult one to select. Perhaps the difficulty occurs because pressure measurements—in contrast to measurements of lengths, times, velocities, etc.—require instruments not possessed by the casual observer. In a first approach to the scaling we can argue as follows. We have seen above that when IE.I C I the convective acceleration is small compared with the local acceleration_ Then the acceleration of a fluid particle is of the magnitude of eviler*, i-e-, AP -2 . Consequently, the inertia of a unit volume of fluid has magnitude pAP -2 . How many such blocks of fluid are pushing in a coordinated fashion on a given unit area? As the scale of velocity changes is L, we would expect fluid in a column of length roughly comparable with L to be moving in the same direction. Consequently, the dynamic pressure, which is due to the inertia of moving fluid, should have magnitude pA P -2 L. This is the pressure scale used in (2.2!), so the justification of the choice of variables in Section 2 can be regarded as reasonably complete.
sec_ 7.3I Order-of Marirtiterdr Esti+rrot!'.s,
NondimeAsjorializiitnn, wid Scaling
333
Since the pressure scaling is subtle, it is worth looking at it from another point of view. The approach that follows is rather lengthy. but il is typical of the sort of argument that is sometimes necessary. For the moment, let us denote the unknown pressure scale by rt; the dimensionless (dynamic) pressure p is therefore defined by
A
(15)
In
The dynamic boundary condition (2.18) is the single must charauertstic feature of our water wave problem. so that it is perhaps not surprising that this condition turns out (after trial and error) to provide the key to scaling the pressure. In terms of our scaled dimensionless variables, (2_lli) can he written z = F.ÿ, Tfsr - 2 Cx,1 (I + F:2- C!) 3r] : —1rp + pgAly. 116) Pressure, as a force per unii area, has dimensions (mass)) length) (time) - 1 . The only parameter that has a mass dimension is the density. Consequently, it is convenient to write '
pressure scale n = (densï'ty)(length) 2 (tinie) r _
(17)
It is a ratio of lengths that forms the small parameter in our problem. so let
us concentrate on selecting the correct lengths in (17). Suppose that n V. If so, since the other terms in (16) are 0(A ) or smaller, the leading approximation to (16) For small I; = AIL. is p = O. This is a contradiction. We assume that the pressure is order unity: then we deduce that the order-unity term is zero, set that the magnitude of the pressure is small compared with unity. Suppose that in ^ A l . Then [he leading approximation ru (16) would not contain the pressure, being -
TAI -2 r,, = = pgAC.
(18)
From ( 18), when is positive, iâ xx is positive and the curve z = 1:(x, t) is concave upward. When 1Ç is negative, this curve is concave downward. Such concavity relationships are exactly opposite to the ones that roust occur for wavelike solutions (Figure 7_l7)_ We therefore roust reject the possibility that Concave down ward !
Concave
upward
i' luuxc 1117_
Ce+reial•rrr refaximshipl ferr o llpt;d
wart- mole
334
Formulaliom of !he Theory of Surface Wares in an a*nviscrd fluid (Ch. 7
the dynamic pressure is of smaller order than the 0(A) velocities and surface deflection. The other possibilities having been shown to lead to contradictions, it must be that the pressure is of order A. That is, from (17), Zr = pA[(length}(time] - }.
(19)
The crucial part of the scaling is now finished, for the leading approximation to the dynamic boundary condition has been established to be, in general, a balance among a pressure terra, a surface tension. term proportional to Ç , and a gravitational term proportional to C_ It remains to specify the square-bracketed contribution to (19). This contribution must be 0(1) if p is to be O(A). Both g and TJpL have the right dimensions, but each has the possibility of failing to be 0(1) if, respectively, gravity and surface tension are relatively unimportant. Moreover, it is L and P that have determined our scales thus far, so we anticipate that these
parameters will also enter here. Thus as a final version of (19) we lake it
t pALP -2
.
This is in agreement with our first approach to the scaling and hence with the scaling used in (2.21). Having struggled to understand the implications of the linearized water wave problem given at the end of the previous section, we are now ready to try to solve the problem. We begin a discussion of the solution in the next chapter, EXERCISE*
1. Six methods for obtaining order-of-magnitude estimates were listed at the end of Section 6.3 off_ Which of these were used at the beginning of the present section?
" Also sec Chapter 6 of 1. which cm-Rains exercises on dimensional analysis and scaling
C H APTE R 8
Solution in the Linear Theory
I
7 we formulated a system of linear partial differential equaaccompanying initial and boundary conditions, to describe the tions, with motion of small-amplitude water waves. In the present chapter we solve these equations. Section 8.l exhibits particular solutions composed of sinusoidal functions. in Section 8.2 it is shown that an infinite series of these functions will provide a solution that satisfies given spatially periodic initial conditions_ Section 83 treats the general initial value problem with a "smeared-out" integral superposition of the sinusoidal solutions. Our treatment of initial value problems illustrates the use or Fourier series (Section 8,2) and Fourier integral (Section 8.31. These topics have already been discussed in Chapters 4 anti 5 of f. [In I. Fourier analysis was used to solve the partial differential equation that describes diffusion, Here, a .s}•su ni of partial differential equations is being studied.) No knowledge of the material in I is assumed in the present chapter, but those who are familiar with this material cari rapidly skim part of the discussion in Sections 8.2 and 8.3. N CHAPTER
8.1 A Solution of the Linearized Equations In this section we shall find a particular periodic solution to the linearized equations for small- amplitude water waves_ in the next section we shall see that this solution is not so special as it may appear at first, since general periodic waves can be synthesized from it and solutions very similar to il, In the text we shall discuss the simplified case h — H}f, ac, which is applicable when the typical wavelength L is small compared with the mean depth H. The reader is asked to carry through the more complicated calculations required when this simplification is not made (Exercise 4). A,SSuMPTtrtN OF
A SOLUTION OF EXPONEN`i
IAL TYPE
The equations to he considered are those given at the end of Section 7.2, with it aC
= O,
+ As
z
-.
—
Ask! At z = 0,
oo cc,
,
wr 1 '
+pcir= 0, a » #
= O.
-' 0.
(2) (3)
solution well behaved.
w in = City*
(la, b,c)
p
ut = f .• IC"'
—
BrSS[]•
Solution periodic in space, with zero mean value. 335
(4a, b) (5)
Soiuiionire the Linear Theory ICS. 8
336
To find a solution that satisfies these conditions, we employ the following generalization of the well-known result concerning the exponential nature of solutions to ordinary differential equations with constant coefficients, Systems of linear homogeneous partial differential equations and !wundciry conditions with constant coefficients Lech as II) to (4)] laemr yo /ions in which each dependent variable is a different constant multiple of the same product of exponential factors, one factor for each independent variable.* We shall now consider a solution of this type and seek special choices For the various constants so that all required conditions are satisfied_ In particular, we take ut 't(x, z, r)
û
a'tt ttx, z, 1) i F ^ '(x, r. tJ#
-- Re
i„.
exp (cwt} exp (earx) exp (kz),
(6a)
h
and
5r 1 (x, t) = Re ÿ exp hew) exp where ü, +;.,
(tax).
(6h)
1i, C, w, a, and k ate constants_
R I:MARKS. (1) There is. of course. no factor exp (k:) in f6h) because the surface height tÿr" does not involve z. (ill Since the solution to our physical problem cannot be imaginary, we have prefixed the right side of (6) by "Re," indi sting our intention to take the Feat part of the final answer. (iii) We do not at first restrict w, a, and k to be real, so the presence or absence of i in the exponents involves no loss of generality. But the choice of exponents does reflect art anticipation of the final form of the solution. VERIFICATION THAT ALL CONDITIONS ARE SATISFIED
We note that a must be real. Otherwise, the solutions will become exponentially large either as _v —# * or._ or as x — — ac , which would violate 13) (Fxercise 2). Since a is real. wit satisfies the condition 151 Substitution of (6a) into (l) yields the following, on cancellation of the exponential factors that are common tea each term:
arrUw eafe
—f,
(7a)
k ji = 0,
(7h)
= O.
(7cl
+it
ic+ r
♦ i4tt•
Thesystemofhomogeneotisequationshasthetrivialsolutionü = t4 — p _ 0, corresponding to the motionless state_ For a nontrivial solution to exist, the determinant of the coefficients in (7) must vanish. so that W(2 — k 2 ) — ü. • [.-oarpyrc Stttion 15.2 of I, whetfi thrt abrrit appro ach is uscd.
(8)
. J]
Sec 8-
A Solution of Me Linearized F.gaarion.s
337
Suppose that ua = Q Then the only nontrivial solution to (I) to (5) of the form (6) is
_ 0,
= Fr r r =
C13
1
=
0'
1/ I J
=
h exp (h 2),
where Ce is arbitrary [Exercise 1(a)]. In this flow the horizontal speeds of adjacent horizontal planes of fluid are related by a'r' Y û exp (kz). But one can verify by inspection that (L } to (5) is satisfied by ti ll) = 1(4
w tI J
=
pri]
criJ
=
0.
Here f is an arbitrary function; it need not even be continuous. Such a general shearing motion is possible because planes of inviscid fluid can move over one another without hindrance_ Were ire to include the nontrivial solutions corresponding to w — 0, we would therefore be considering a combination of waves and at somewhat artificial shear flow. We wish to consider waves in the simplest context, so Met shall uhruti's ignore possible solotions a frrre°,cfxfridrng tb w — O.* For the remainder of t h is section we a swine that a > 0- The case of nega ( i vc a will be studied in Section 82) Since to # O. (X) implies that k = # a. Boundary condition (2] cannot he satisfied if < O. so
Since k is positive, the solutions (6) decay as z t_ The determinant of the coefficients in the homogeneous equations (7) is zero, so their solution will not he unique. In face, there turns out to be a one parameter family of Solutions_ Anticiipal ing this, we denote this parameter by a anis write —} —
-
=
(LO)
a,
where a is a complex constant, nonzero (to avoid the trivial solution). but otherwise arbitrary. From boundary condition (4a) plus Equations (7b) and (7c), we find —
fi — r,a 2 a - `a, ii = =utc (a it 0).
(l la, b, cJ
Equation (7a) is a consequence of (lb) and (7c) and provides no new information. [- the only nontrivial solutions corresponding to a T {) have w = 0 The full nonlinear equations (7.2.22-28) ate satisfied fur arbitrary 1 by u — fir1, k' l' C - D. This leads to the realiL1tion that even if rnilial cundiiruns arc prescribed. uniqueness requires a more prrcic condition ihan t71161 as r-Z 1 — x _ For example, if the initial disturbance is confined to a bounded region. Il seems approprraie to demand that their be no
inward
flow of energy ai sufficiently large values of 1x1 Another possibility is to prescribe some' inward new " as' x and to rei uire an ourward flow of energy as x —0 x,_ Such a " radiation — —
condition" was discussed at the end of Section 12.2 of l in connection with a problem involving clastic waves.
SoJurion in rhe Linear Thelon; [Ch. t3
338
(Exercise lb) and will be ignored, as stated above.] The remaining boundary condition (4b) implies that r.4a2a - ' = 1.(1 + Hal } or
ro 2 = F(a + Ba 3 )
(12)
e2 ; F(tx i ^ Bah
(13)
or
where t'=
—
a
(14)
.
IEt E M A R K. Since R is nnnnegative, ill is real and the solution (6) is oscillatory in time_ From (12). given a. there are two possible values of w. They are equal in magnitude and opposite in sign. 1t might avoid a little confusion if we write the two possibilities as ru = w, and to = — eo 1 . ri t > 0; but we choose to simplify the notation by considering the two roots to be ru and — w. where by vi we now mein the posiitve root of (12). With this change to notation. till is valid for solutions proportional to exp Innl), lout for solutions proportional to exp (— iieit) we most use
=
rrvu,
p=
iw z a - ' a .
oat?.
(15)
Using (l 11 and 04), we see that one set of solutions of the type IN is as follows: 11 4111 X. 2, r) r*"1it(x, z, t)
p`tt(x, 2,1)
—
=
Re
t^
ir4i 0a 7 a
exp (az)
exp [r'a( ( + ri}]
'
;11 1x, t) = R e exp [ia{ x +
This set represents leftward-traveling waves with dimensionless spatial wave number a and dimensionless temporal frequency w_• The dimensionless spatial and temporal periods are 2n/a and 2n/co. respectively, and the dimensionless wave speed is c. Equation (i 3) gives r as a function of a and the dimensionless parameters F and B. • Those unfamiliar with the basa facts concernsng traveling pcnodwt waves can consult Section 12.2 of I.
Sec_ 8.1] A Sohiiion of they Li eearezed Equations
339
Using (15), we see that the - negative w" set of solutions is 14" )(x, }(x, z. r)
w - in)
wt' qx, z, t) = Re
exp (az) exp [t 4x - (t)].
to 2 ,2 - l
plrl (x, r, r) CI] (x, r} =
Re exp 1icc(x - ci)].
This set represents rightward-traveling waves of wave number a and frequency r1). RETURN TO DIMENSIONA1. VARIABLES
Let us reintroduce dimensional variables. Then the Factor giving the periodic dependence cif nur scilution becomes exp
ax*
±
f * = exp [f W(xe ± Cr')],
where a
W
j and
rL
C
=
^.
From (16a) we see that the dimensional s pa tial and temporal periods are 2ni_jcc and 2KPicu. Equations (16h) and (I 6c) relate the dimensional wave speed C and the dimensional wave number W t o their dime nsionless counterparts c and ;x. Let us consider a class of waves of various dimensional wavelengths i = 1, 4.... We wish to compare the properties of these waves, so we choose a single representative length scale L and time scale P to serve for all. We have just seen that the spatial period (i.e., the wavelength) is given in general by 2nLja, where a is the dimensionless wave number. Consequently, the dimensionless wave number r ; of the ith wave will satisfy
2ati_ -- Lt .
(17)
a;
Recall that
9L
F
PilL
2
(18)
Upon substitution of (16e) and (18) into the dimensionless " speed equation" (13), we see that the ith wave has a dimensional speed C, given by
C?
^
T
-
ur
1
C1 , y1^ `' -1- TK ; . J ,
(19a)
Satdion
34e
in Me Lamar Theory (Ch. 8
Here
r- T
and K ; - a`= 2
L
(19b,
Li
where we have used (17) in obtaining the second equation of (19c). There are two limiting types of water waves in which, respectively, gravity and surface tension effects dominate. To bring this out, we write (19a) in the form
CC;°r]= + [C;``]'.
(20a)
where
[ cry
!
2rt f '
[
^"
= 2rtr ]
^
(20b, c)
INTERPRETATION OF THE SOLUTION
We see from (19a) that the speed C ; of a wave of length L4 = 27111(1 depends nra irs 1nglh. Such waves are called dispersive. As we shall see, a great deal about the wave's qualitative behavior can be deduced from dispersion relations* like (l9a), which provide the quantitative dependence of speed 2n Î; f.; then, from (20), C? [Cr]'. Thus when upon length. If gLif 2n gravitational effects dominate, waves move with a speed approximately equal to Coy, the speed of gravity waves of length L i _ if gL ;12nt 2n17L1 . then surface terisic,n, effects dominate and the waves move with a speed approximately equal to 0", the speed of capillary waves of length L i . As expected, surface tension effects dominate when waves have relatively small lengths. For then the surface possesses relatively high curvature and, as seen from the dynamic boundary condition (7.1.26), high curvature is a concomitant of large surface tension effects. Dispersion of gravity waves is said to be iwmull because longer waves move faster, just as in the case of light waves moving through air, where the longer, red waves move faster than the shorter. blue waves, Dispersion of capillary waves is called artumalous because shorter waves move faster. Figure 8.1 contains a graph of Cl when both capillary and gravitational effects are important. The graph's most interesting feature is a minimum wave speed er . For waves on water,
p = 1 gjcm a . T = 72 g/s 2 .
g = 980 cm/s 2 ,
sa [Exercise 3(b )]
e; = 23 emirs,
Lc = 1.7 cm.
(21)
In water, no waves move at a speed slower than 23 cm/s_ At speeds higher than this, two types of wave are possible: relatively long waves dominated by • A case of dispersive waves in elasiicily is discussed in Section 6.6.
Ser. s.lj A Solution of the Linearized Equations
341
SquBre or WE'VE speed
L,
^
Fra1 kE 8.1, Graph yn•hy l dimensional) wcu•r speed C, Karr u! lrrrettir Speeds Cr atul Cr lesr pure ,yrar-et y and rmrflur.r wares are also shown.
L
-
.
Note the rtisiettre (3:1 a mammon ware speed.
gravitational effects, and relatively short waves dominated by surface 1ensinn effects. To see one physical consequence of the existence ofa minim urn wage speed, consider the steady flow of water past a fixed obstacle say, mo emertt of a gentle stream past a fishing limo_ Lei the speed of the fluid far from the obstacle be a constant Y One anticipates that the presence of the obstacle will result in the formation of a wave system that is stationary when viewed from the obstacle.* Equivalently, such a wave system must move along with an obstacle that passes al speed V through stationary water. But our results tell IS that a train of stationary waves cannot, in fact, appear unless V is greater than 23 cm/s. The same cnnsidcral ions apply to the steady motion of a ship, but the speed ofa ship would of course normally be far greater than 23 cm/s. An important physical result to be gleaned from our solution is that fluid speed and pressure decay rapidly with depth. When = i — r/cc (at a depth of half a wavelength. speed and dynamic pressure are reduced by a factor of e - x 0.04. That is, motions half a wavelength below the surface are about 4 per cent as fast as they are at the surface_ At one wavelength below the surface, the velocities and dynamic pressures have about one-fifth of I per cent of their surface magnitudes. These results are in accord with the common experience that one has to swim only a surprisingly small distance underwater to avoid the effect of surface waves. EXERCISES
1. (a)
Fill in the details omitted under (8) by
examining the consequences
of satisfying 48) with w = 0. (b) Find all nontrivial solutions to (l l to 15) of the form (6) with
x = O.
• Such wave systems are discussed rather fully in Section 9 4 and more briefly al the cnd Section 4.1_
of
Solution is
34z
the L.bwar Theory [Ch. 8
2. Show that (3) is violated if Im (a) # D in (6). 3. (a) (consider solutions proportional to exp l— ja i x)exp (iwt) exp (ix i z)Show that the relationship between frequency and wavelength is still given by (12). (b) Verify (21). $4. Find periodic solutions to (7.2.32-40) without the simplifications h = co. In particular, find the dimensional wave speed_ 5. (a) Show that, to a first approximation, particle paths for gravitydriven periodic waves in a fluid of infinite depth are circles whose radius near the surface is nearly the wave amplitude A. Note that the radius decreases rapidly with depth. The particle path equation dxfdt = v(x, tj is, in general, nonlinear and difficult to salve. Approximate as follows. Let a be a fixed point near the initial particle position x 0 . No malter what the particle path is, for a sufficiently short time all points on it will be near a (Figure $.2). For this length of time we can solve the (simple) approximate equation (kids = v(a, I), x(0) = x4 . For our water waves, the resulting particle paths are such that a particle which starts near xa will always remain near xp (a result that ceases to be correct after a "long" tinge)
^ t
r
a
FIGURE 8.2.
The pad; elf o panicle Meat %fairs at same pwni x
o near
a.
The
solid portion of the path is near sr. so thor the particle path equation can he appropriately sitnplf%d.
(h) In Section 7.3 we assumed that particles move a distance A, approximately, in time P, so that the velocity scale is AJP. Is this assumption verified? 6. Show that for waves in a fluid of finite depth the particle paths are ellipses, to a first approximation. Note that the interface] distances are independent of depth.
Se,['. 8,11 A Solution vi the Linearized Equations
343
7. The a im of this exercise is to sh ow that the basic " separa t io n of variables" solution for radially symmetric linear waves on a fluid of infinite depth is
u
— iwJ 1(ar)
P' n
-.
C1
iwJ 0 (ar) Off-' J0(ar)
exp (leur) exp (az^
iw1 t (ar) + C2 iwJ0(crr) exp ( (i)2Œ - 'J0(ar) —
—
iwt ) exp (az),
C = C r Ja(cxr] exp (iwt) + Cr Mari exp (
—
it a),
where al and a are still related by (12). Here r (x 2 + y 2) 112 and u is the radial velocity (in the direction r increasing). (a) Show that the linearized equations tl, + f ryr -=0 ,
W^ _ 4 Pr
O.
(ru }Jr
± rw, = 0,
and the dynamic boundary condition is rr P = F — m],r+r- 'Cri).
(For material on polar coordinates, see Appendix 3.1 and Appendix 15.2 of l.) (h) Since r and r appear only as derivatives, assume that i and z dependence is exponential and look For a solution of the form u w= p
ù(r) w( r) exp (+ iwr) exp (az) 15(r)
C = C(r) exp (i irnt). Show that the problem reduces to the following equation for w(r}.
r(W }' + rac2 tiv = a. (c)
Complete the calculations_ Possibly of help is the material on Reesel functions in Appendix R.I.
The next three problems introduce king small-amplitude waves, which turn out to be governed by the classical wave equation. Tidal waves (i.e.. tsunamis) are an important example of such waves_ 8. Imagine a channel along the x-axis with the z axis pointing upward. Consider long (compared with the fluid depth), two-dimensional (no y-variation), small-amplitude (ignore all nonlinear terms) waves in an inviscid incompressible fluid under the action of gravity. Since the waves -
Solution in the Linear Theory [Ch. 8
344
are long, it is plausible that the vertical velocity terms are negligible. ignore surface tension and assume that the pressure at the free surface must take the constant value p a (a) Deduce
(z• _ C#)
.
pc•
r^?u*
X
(
22
)
^
from the requirement that linear momentum be conserved [see (721) and (7.2.2)]. Start by deleting terms containing w' and its derivatives in (7.2.2). Then integrate with respect to z* and obtain (23)
p* = PA + P9(C' — z*).
which stales that the pressure at a point is due entirely to the weight of the water above that point. [Our neglect of vertical acceleration has led us to (23) - It might be more convincing to begin with (23)1 (h) Suppose that the mean depth of the channel has the constant value H and the channel has constant width b (Figure 8.3). By using the fact that influx of mass into a given fixed volume must be compensated for by an increase in fluid height, derive the continuity equation
(bN+brâ*)
+a
(24)
#*) = 0,
where small nonlinear terms are again ignored. Side View
z
'
End View
-
0
z • — —fi
FIGURE
(c)
II ^6 - --i
8.3. l'lms uf awar r in a M [inr+pi.
Combining (22) and (24), derive (25)
City — Oa.* = 0, In the situation described by Exercise 8, consider an isolated wave
f(x —
/i ii);
f(x)=—Q when x< A,x>B.
=
(We drop asterisks for the remainder of these exercises.) Suppose that this wave strikes a rigid vertical wall which forms a cross section of the channel.
Sec.
g.11 A Solution GI the Liarired Equations
345
(a)
Show that the wave is reflected without change of shape_ (b) Show that during the impact of the wave on the wall, the water rises to twice the normal height of the isolated wave. (c) Briefly compare the results of (b) with the related discussion in Section 12.2 of I. 10. Show that Che derivation of Exercise 8 can readily be extended Co channels for which the depth H is variable: H H(x). 11. Suppose that the depths of a channel for x < 0 and x > [l are if, and H 2. A progressive wave =
ci
Cos cox — e1 1)
,
c 1 = gH 1 ,
travels along the portion of depth H 1 . Determine the amplitudes of the reflected and transmitted waves, as multiples of the amplitude of the incident wave. To obtain conditions to be applied at the discontinuity x D, first integrate (22) and (24) [without asterisks] with respect to x from x u Y e Co x = }. g , where 1. 1 < 0,1. 2 > O. Then let y,, 3 —, O. Show that must be continuous at x = 0 while s(x, 1) must suffer a prescribed discontinuity there. (What is the physical meaning of the latter condition?) Use (22) to prescribe the discontinuity in the free-surface slope (x . Finally, assume that the entire solution is the sum of the incident wave (displacement Ç i ) plus a reflected wave and a transmitted wave_ Show that the incident and reflected waves must have the same wavelength but that the wavelength of the transmitted wave differs From this value. Discuss your results. Do they depend in a sensible way on the parameters of the problem'? If these parameters are assigned typical values, are the results reasonable? Compare your results with those for the analogous elastic problem, as discussed in Sec. 12.2 of I. 112. Expand in powers of ; h the expression for waive speed that was obtained in Exercise 4. Thereby show that for a depth If satisfying H = 3T7pg, the speed is nearly constant for waves whose length is moderate or large compared with H r,. What is He for water? [Such nearly nondispersive water waves can he used to visualize the development and interaction of (nondispersiv e) SOU nd waves in air. See "Shallow Water Wave Generation by Unsteady Flow,"J. F_ Ffuwcs-Williams and
D. Hawkings, J. Fluid Mech. 31, 779 (1968)] I1 (a) Defend the following equations as a model for tides in large flat basins: 1711
^t (
— ^f1)2J =
0 Jx 4 l4 — ^}, ^ ^x
^t
eV al
+2Wtr = — y
ax ou)-- $y (ha].
d
0y
l^ — a
346
Sofurion in
the 1.irrear Theory (Ch . 8
Here a and y are velocity components, C the tidal elevation, and h the height of the basin (assumed TO be constant in what follows)_ What are and w? (b) Assume a time factor of exp (kit), solve for u and ❑ in terms of t:, and derive an equation of the form VC + k 2 C = O. Show that vanishing of the normal velocity component at a vertical boundary implies that
°C-- ïr^0, where the derivatives of C are, respectively, normal and tangential. N O T E: Solution of this problem is the subject of Exercises 12.2-16-20-
8.2 Initial Vaine Problems: Periodic Cases In the previous section we ignored initial conditions in the process of exhibiting some particular solutions to the linearized water wave equations. Now we begin our study of the initial value problem. Since the position and the velocity of the free surface is more readily observable than interior velocities and pressures, we shall consider conditions on the surface, namely, C( ` i(x,O)_ f(x),
al ix, O)
9(x),
( la, b)
where f and g are given functions. In this section we shall consider the important special case where f and g are periodic. The aperiodic case will be discussed in the next section. SUPERPOSING SOLUTIONS OF EXPONENTIAL TYPE
In mathematical terms, we are faced with an initial boundary value problem for a system of linear partial differential equations with constant coefficients. Our procedure is an illustration of a general approach to such problems that is of wide applicability. A fundamental fact upon which this procedure is based was illustrated in Section 8.1: Systems of linear differential equations and boundary conditions with constant coefficients always have solutions of exponential type. A second basis is the lemma that a linear combination of solutions to a linear homogeneous problem is also a solution. (See, for example, the concluding portion of Section 2.1 of I.) Therefore, one can attempt to satisfy the initial conditions by taking a suitable linear combination of solutions to the equations and boundary conditions. 'lb do this superposition successfully, it is necessary (but not always sufficient) to have available all linearly independent exponential-type solutions of the given equations aril boundary conditions (Exercise 12). in dealing with the initial value problem for small-amplitude water waves, then, our first task is to find all these solutions for Equations (IA 5). -
S„. S.l]
)r- rnm
lnirial
Vcrlue Prnhle n,s:
Periodic eases
347
(1.10) and (1.1 1 ), one exponential solution for
pin
will .0
ci1 1^
is the real part of -- L!k
Cxp (22)
1{.0
exp (2z) exp (a^)
0,12^
exp ( iacx) exp ( i t^f ) -. J<: j ,
(2)
1 where w, is the positive root of (1.12)_ From ([ 10) and (1.15), another solution is the real part of exp (2z). — ia►
exp (cz)
uu 2 a r
e.x1? (az)
exp (lax) ex
( -iwt) - F' 2
(3)
It will prove convenient to have the time dependence given by the real factors sin wr and cogs wt. To accomplish this. instead of the linearly independent solutions EF and Ez of (2) and (31 we consider the linear combinations (E t + EA/2 and (E, E2)f2i• The first of these gives -
sin cot exp (2:.) -it) sin t exp (22) exp (iwrx). rr Ct " I cos um exp ( z) cos (Or - ito
(4)
The second linear combination gives iv)
cos wt
exp (2z)
co cos wt _r a eo ar sin an
exp(agi) exp (a:)
exp (iced.
(5)
sin wt We assert that all real solutions of exponential type ca,r be rxbtiined by i, rq the real parr (fa complex linear combination of(4)and (5). For simplicii y r1k we show this only for 4-,411. but treatment of ru t), w' 1 I, and p{ n is essentially the same_ By - the real part of a complex linear combination of (4) arid (5)" we mean
[for CIrf} (6)
Re [A cos alt exp (iax) + B sin cftr exp where A and B are arbitrary rx rnplcx constants. if
A 4 A rrJ + ,A'
,
B - BIrl -F ri7^
^ 1y
Sohiiinn in the LFrear Theory [Ch. 8
3 48 then (6) can be written as
A '} cos an COS ax — AOl cos an sin ax
+ Or) sin wt cos
sin all sin ax_
(7)
By "real solutions of exponential type „ we mean real linear combinations* of solutions having the form exp (± iwt) exp ( ± iax).
(8)
Here r > 0 as discussed above (1.15). Also, in the factor exp (± ianc) the presence of the + symbol means that all possibilities are covered if a is restricted to positive values. [As noted under (1.11), the case a = U can be ignored.] Solutions proportional to both exp(iwt) and exp(—in.t) appear explicitly in (2) and (3) and implicit ly in (4) and (5). It remains only to consider solutions proportional to exp (—iax). The dependence of (o ors a is the sanie for solutions proportional to exp (ia.x) as it is for solutions proportional to exp 1—iax) (Exercise 1.3). Therefore, a complex linear combination of solutions proportional to exp (— lax) is C cos (Jr exp ( — fax) + D sin wt exp (— fax),
where C and D are arbitrary complex constants. Taking the real part, we obtain Ct' cos tot cos ax + C.
cos an sin ax
+ D"1 sin ira cos GLx + Dr'' sin we sin ax.
(9)
Expressions (7) and (9) are linear combinations of the same four functions, so if (7) is included in a list of solutions LC be superposed there is rio need to include (9). In the remainder of this section We shall consider situations in which the initial position C and "velocity" C r of the interface are periodic. We shall use as a length scale the spatial period that we require the solutions to have. Consequently, the initial functions of (1) have the periodicity property f(x + I) = f(x),
y(x + I) = g(x)-
(10)
The solutions of (4) and (5) will initially have a unit spatial period if and only if they always have such a period. For a given term to repeat itself in
• Since the cxpresSionS
in (8)
by means of complex constants.
are complex. real linear combinations of course arc composed
SEC.
8.21
Value Problems: Periodic Cases
349
grit distance, that term must have a least spatial period of l/n, n stn integer_ We can thus only accept solutions for which 2rrlct = l/rt or a = 2rt1, rr = 1, 2, 3, .... We shall denote the corresponding frequencies by w,,. From the dispersion relation (1-12), the W are given by the following formula: ,
F[2nn + Bl2rrrle],
at!
r1
= l, 2, 3, ..
I l 1)
--
In the climactic step of the separation of variables method, we attempt to satisfy the initial conditions by superposing all linearly independent solutions cal the equations and boundary conditions. Therefore, we assume that ![o,,
l ^I
=- Re
.^^
^
L1i n l{ ^ Ii l'C
h=I
^
^Ir1 5
sin ion ('
exp (2nrrzl
sin to,/
exp (2nrrz)
cos (Uh l
CxFi 12111iz)
exp 12:Ixi.r)
COS tt^ n f
-
+ Re ^
A°n
-
n=
1
cos t o „ t
exp
cos it),, t
exp (:_117iY )
s in rr1 r
exp (2tltrz)
(2rIR.7 )
exp (2>*reix).
(12)
sin frJ^r
ANOTHER WAY TO WRITS: THE REAL PART OF
COMPLEX slims
We shall soon want to multiply both sides of equations derived from (I 2) by a complex constant, hut [he operations of multiplication by a complex constant and taking the real part do not commute. Fortunately, we need not abandon compact complex notation, since difficulties caused by the lack of commulativity can be circumvented if we rewrite ( 12) with the aid of the relation Re.:=
ÿ *1,
(13)
where * denotes complex conjugate. Using I 0), we can write the first sum in (12) as sin --
) L n -1
tr7„
r43rt —fin
rt}„
t exp (2iln21
sin a.)„1 r:xp (2r1r})
„
['0 d n
exp (2rrrit)
exp (? ► rrixl
cos to„ r exp (2/177.z) [•W„ exp 12nitiix) COS
rJn r
exp ( —2nrli-r}]'
exp i
(
—
?•lrtit)]
.i1,* exp ( — 2rlrrix)]
[a n exp (2r17ir.C) + .4^* exp (
711n1 l )1
.
(14) There is a similar expression for the second sum.
Solution ira the Linear Theory
35 0
[Ch. 8
We can write each line of (14) as a single sum if we redefine the constants so that the second terms have the same form as the first terms, except that n appears instead of n. This notational symmetry is attained if we make the definitions —
[13-11
A,r
1st —
,,, A
- „
(15)
we.
i- , B„ —
—
fa,,. B-„=
. (16a,h,c,d)
(We have also inserted a factor j so that this factor will disappear in the final form of the sum) We must examine in detail how (15) and (16) lead to an elegant notation. First, let us consider the surface displacement. We see from the last line of (14) that the assumed form for the cos w,t contribution to VI becomes A„ cos w„ r exp (2rlrrix) f n=1
yA
,
cos co_„ t exp (
2ntzix).
„=
This can be compactly written as A„ cos w„ t exp (2inzix),
provided that we make the definition Ala
=û.
(16e)
We next turn our attention to the horizontal velocity component_ Using (15) and (16), we see from the first line of (l4) that the assumed form for the sin w„t terms in ur 11 is sin w„t exp (2rtrtz) exp(2rtrzlx) w =1
+ X KO „A_, in r^_t exp (2nnz) exp (-2rinlx). -
(17)
=t
Further juggling is required to achieve notational symmetry_ We write exp (2nttz) as exp (I 2 ► trr 1 z) in the first term and as exp (I — 2r,tz I x) in the second term. To take care of the opposing signs of the first and second sums, we use the signum function sgn where sgn x — 1, x > 0;
sgn x = —1, x < O.
We can now write (17) ast
stt rr = E -irr^ „ sgn N
t sgn (2nn)
Section 8..
(2rrtt)A„ exp (I2r171Izj exp arirtix) sin w„ f _
— CC
sgn (n), but uh.c of sgn (inn) permits readier generalization of our results rn
Sec.
8.21 Indio! Value Problems: Periodic Cares
35 1
Additional manipulations of the kind just performed (Exercise 1 ) allow one to transform (12) into the final assumed form,
- it „ sgn (2rnn) —
(A„ sin cot — B. cos tf„t) exp (I2rin
(A„
w„
sin w„ r
—
IZ)
B,, cos û,,r) exp (1 2r7a'tIa) exp (2rlrcix)
(A„ cos cu.: -+ $„ in r „r) exp (I2i1ar[0
bu I
A. cos L11 t f B„ sin r.iJ.n l ,
(IS) Having arrived at the solution (1 ), we no longer need concern ourselves with the constants .d, and at, that were used earlier. We must, however, adhere to the restriction
AA
A_,,,
B* = B_,,.
(19).
This follows from the definitions (1t) of A„ and BA in terms of ,nt. and At.. Without (l9}, the sums in (l8) will not represent real quantities. Example 1. As a check, show directly that w` i ' Is real!.
SFvfruoinm. (w ill
•=
- c0„(A:
Tn make the last factor look right, we j - —iI, Then
p.,,r19* _ y _ w
ra„!}exp(I2mtlZ)cxp( - 2n rr
sin Lud —
define a new dummy summailor, index
sin ())._ 4 : — B'_ cosa, t r1exp(I2inl-)expl2jnrx}. ,
j- a
l'rDm (19),
A# i = AT* = A} and
B*_ f = B,.
Using this and (151, we obtain
y —[op ; sinrrl f t - fi r COS ay) cxpf121^I-1cxp (2:12ix )
[+nsrr]* = 1 ^
--^
d 11
w
Y^.
j by
351
xiokaian in the Linear 77reary [Cif. 8
Example 2.
As a check, show formally that the continuity equation (i -Ic) is satis-
fied by (J 8).
Sofrafiun. Since our calculations are to be formal, we assume that tiffe order of drfïerentiation and summation cati he interchanged, so that wvl
s
E{^rlJIG1n sgn [2110
I2nrrIcu,](14, sin w,r
J3,cos eu,r]
..-^
cxp ll 201n lz) imp f2nnixj n sgn
(tr) ;
^
Q
!PI .
SATISFYING INITIAL CONDITIONS
We now substitute fit'', as given in (18), into the initial conditions (1). Assuming that term-by-term differentiation with respect to t is permissible, we obtain the requirements a
f {x) _
A„ exp (2rirrtx),
9(x) =
t.1„B„ exp (2nitix)- 120a, h)
The right sides of (20a) and (20b) are complex } ourier series. To determine A,,, we proud in a manner analogous to I he procedure used to determine coefficients in a real Fourier series (sec Section 41 of I). We multiply both sides of 120a) by exp (— 2knix), k an integer, and integrate over the interval [D, 11 Since
exp [2rti(rt — kJ dx
r+ rx=
^°`
1,
0
k,
(21)
we find that ^k
—
{ J (x} C7[^j +
2 ^i ^r i %]
C^x
,
Ifî = ^ 1 ^2 . . . . ,
^^ ^^G)
Si rr^ i larly, wk
Bk = f g(x) imp ( — 2itit[1x) dx,
k
= ± l . + 2. . . , .
(23)
Equations (18), (22), and (23) provide a formal solution to the initial endue problem. (i) Asa check, formulas (22) and ( 23 ) do satisfy (19). (ii) A REMARKS. considerable advantage of writing Fourier series in complex form is that the integration (2 C), unlike the corresponding integral for Fourier series involving sines and cosines, does not require manipulation of trigonometric identities. Although we shall not go into details, we also remark that the complex notation simplifies any nonlinear manipulations that one might choose to do. (iii) The coefficients A k involve only [he initial surface displacement f (x) and
Sec. 8.2] Iniliol Value Problems: Periodic Cases
353
the coefficients B* involve only the initial surface velocity g(x)_ Had we used exponential rather than trigonometric forms for the time dependence, the relation among f, g. and the various undetermined constants would have been more involved (Exercise 5). (iv) if we had used the solution (l2) instead of (20a), we would have obtained f(x) —
y .s^
Re w
=
exp (2rrrrex),
,r^ = sir +
t
or
(.. cos 2nlrx — .S.i1,1f stn 2nnx)_
} (x) =
(24)
w, l
The Fourier coefficients .s. „' and .c1 can the advantage of complex notation is lost.
he obtained in the usual
way, but
EXERCISES
L Complete the calculations, begun in the text, that are necessary to obtain (18). 2. Complete the verification, started in Example I, that the sums in (la) represent real quanti ties_ 3. Complete the Formal verification. started in Example 2, that the alleged solutions of (18) actually satisfy (1.1-5). 4. Extend t he fo rm u las of this section to the case of fi n i to H (see Exercise L4). 5. Verify remark (iii) following (23) by repeating the calculations without making the shift from (2) and (3) to (4) and (5). In particular, show that the final answer is the sank as that obtained in the text. 6. Replace (15) by w„ = w and find the counterpart of (18)_ Which convention For defining w,, n < O do you prefer? 7. Satisfy initial conditions starting from (24) rather than from (2Clal. Verily that the final answers are identical. 8. As a check, verify that formulas (22) and (23) do provide a solution that satisfies (l). +9. Find Ak and BIG when the water surface is initially motionless and initially has the shape pictured in Figure 8.4. —
It i I I.
^
i r—
!
x
^
I
] —
^
E - ,
^
I I
11 ia
I'
r^
5
FIG LIItE 8.4. Initial displacement needed /Or Exercise 9. Only one period is 5I1nK-n via Ant-lion ► i•irfl period unify.
Solution the Lin ear The ory
354
[Ch. 11
10. (a) Solve the initial valu: problem for the following special case of (l) :
Ox, ü) = cos
(2Trx
C,(x,
0) = O.
)
The resulting motion is given the name
srariding wait's. Why is the
name appropriate? (b) Find the particle paths and sketch some typical ones. I L la) Consider PE, the potential energy of gravity waves (B - 0) in one spatial perry]. Take z = 0 as The zero point of potential energy. If, as in this section, the spatial period is used as a length scale show that
PE = iP9A 1 L
f 1 {2 dx.
lb) Show that KE, the kinetic energy par spatial period, is given by (, KE _ "piA31.F (cl
-
'
j
J Iii + i4. 1 )r1k d:_ o
Using (18), find expressions for K E and PE. Show that
KE + PE = ifr9A 21.. ^ ^^A^ 1 + n= — aS
Note that energy is conserved, as expected. i12. Prove the validity of the italicized statement precuisng (2)
.
83 Aperiodic Initial Values In this section we shall consider the initia conditions Cm p) = f(x).
Lix,0l = g(x),
(lab)
as in the previous section, but we shall no longer require f and g to be periodic. This means that solution of the initial value problem involves Fourier integrals rather than the Fourier series that are relevant for the periodic case. We shall provide a formal treatment of the problem. As a first deduction from the mathematical manipulation, we shall then show that a symmetric initially motionless surface hump splits into congruent left- and rightmoving segments_ The section concludes with a perspective of the initial value problem in terms of the delta function_ ABANDONING DIMENSIONLESS VARIABLES
Two rativenitons. valid kir the remainder of our discussion of the linear
theory of water waves, will be made before we proceed. (al We shall drop the superscripts on utli tv"°, prat, and ts'r'
.+tprriodir India! Values
Sec. 83] (b)
We shall take A
355
L. = I cm and P = Is.By(7.2.21)
a* = Pi, x* Lx, i' =tLP - r v, C*=ELC,
p* =
0P-2p.
where c = AIL Thus no notational difference remains between dimensionless and dimensional distances and tunes, hut the latter have the units centimeters and seconds. Similarly, dimensional velocities and displacements have units of cm/s and cm. The dimensional dynamic pressure 15* divided by the density p has the units crn 2 /s 2 . Convention (b) can be viewed as using the centimeter is a reference length and the second as a reference time. We need no longer continue to employ carefully chosen length and time scales because our previous effort in This direction has served its purpose of helping us formulate a simplified version of our original problem. In many instances, even at this stage, careful choice of reference quantities makes inessential but nonetheless welcome simplifications in the answer to a given problem. We make convention (b) lx:cause we shall be cnnsidering -a wide class of problems and do not wish to keep changing reference quantities. SOLUTION VIA FOURIER INTEGRALS
Solution of the general initial value problem proceeds in a manner that is closely analogous to the procedure used in solving the problem when the initial surface displacement and velocity were periodic. Again we supLrpose all possible solutions of exponential type. Before, the periodicity condition allowed only those solutions whose wave number ar satisfied a = 2nn, r' an integer Without a periodicity requirement, a can be any real number. Since we have a continuum of wave numbers to "sum over,,, we shall superpose by means of an integral. Just as in Section 8.2, it is advantageous to write the solutions with their x-depdence expressed as a complex exponential. We therefore superpose the solutions of (2.18) by replacing 2nn by w„ by w(a), A n by A(a), and B„ by B(a). Integrating over a, we are led to the assumptions —
r^ sgn a[A(a) sin on — H(a) i^
cc3s
r u] exp (1 a 12 ± iax) da,
^i
(2a)
.a.i[A(a} sin cm' — R(a) cos rar] exp ( Ia I z+ iccx} r^a,
142(x, 2, 1) =
(2b)
^
G7 2 ( a l'tA(a:} cos cur + B(a) sin (in] exp (1 a 1 a+ iacx)
p[x, x, r}__ a
(2c)
ax, 1}
f
[A (a) cos lug -+
B(a) sin rllt] exp (fax) d a.
(2d)
Autun in the Linear Theory [Ch. 8
356
ln these equations. w is a function of a defined for nonnegative values o r as the nonnegative root of the dispersion relation (1.12). With L = P = 1, we have from (7.2.31) that B TIp9. F = g. Thus (1.12) becomes
0.4ct)
T ° T f p, a
= 9a + î'a 3,
0.
(2e)
For negative values of a, w is defined by (4 — cf) = ca(a),
(3a)
-
in analogy with the relation between m and co_ given in (2.15); A(a) and Bloc) are unknown functions that satisfy
A#(a) = A(
—
a),
Bs(a) = B(
—
a),
(3b, c)
in analogy with (2.19). Example 1. Asa partial check on our work so far, show that C is real. Safruiarr.
From (2d),
*(x, [)
[A' la) ces c a)t +
=
B•(a) in w(a)r] exp (— lax)
da.
B y (3b, c).
tA(—a) cos w(a)t + G1(
Z*(x,t) =
With the chan g e of variahlc a= ^ * (x, t) =
^
^
—
—
a)sincv(a)1]exp(
—
iax)da-
{^ this becomes ,
[ AV) cos { o( — P t + 1:3 O) sin (01-13)i] exp (ifix)1 - d^}.
or, using (3a), 4•(x, l) ;
sin L^(^^r] exp {r^ix) d{s = ^(x, t). 1 l^J^{^[Cos+-u^^^t + BO ^
h)]. E xnmpk 2. As a check, show formally that Hi +- P_ — 0 [Equation 11. ft)]. Sohuiwr. Differentiating under integral signs, from (2h) and (2e). w, + Pi — f {— tU s + cil' NA cos wt + B sin cm)
ex p{ E a l z+ iax) da. = O.
REMARK. The reader shouki be confident that the quantities
given in (2) are
formal solutions of the governing equations (1.I). To obtain (2) we superposed a continuum of solutions (and did some harmless but valuable notational juggling). The basic idea of the superposition is as follows. Symbolically, if a vector s(x, I. ml is a solution to a set of linear homogeneous equations for any real value of a. then the linear combination
r{x, t • a,jd.; i- 1
(da; - ar — oer - r 1
ffperirhfir India) Values
S. 831
357
with q
tr = f^ •
° h.
is also a solution. Assuming that +t exists. we cannot be sure that the limn ^
>w
lim rr
y s(x, t aoA2, _^ s{x
r, a) d a
I
is a solution only because we cannot be sure !hat ditfcren tiaiion under the integral sign permissible. nut (2) is a formal solution because in format c:alcuta[ions we assume that is such limit inicrchanges are possible. Moreover,
is a formal solution if it exists Formai eaiculatuons are meaningful because they are correct if appropriate restrichomi arc made on the entities involved. i is often appropriate ta defer the task of determining a sufficient set of restriciiois.
Our task now is to obtain an expression for the unknown functions A(a) and B(ar) from the integral equations resulting from substitution of (2) into the 'initial conditions 1 I) A(^j exp (irfxj dix
f (x) =
g(x) = f 23(^)r^^(^) exp (i^cx) da. (4a, ^)
,
cc
To find Alla), we mimic the derivation of (2.22) as closely as possible* and write
r
^
lirn Jexp( -
(x)dx = lim f exp ( - !p_rl
i.
r Ala} exp (iaxJ da dx.
r
Interchanging the order of integration. we Fuld that
exp(
Jim
i lrx)f(x)dx =
r-.cJJ r ^ r
A(a) I ^ c;xp [^(a - tr}.t] d. Jr1 a x j r
lirn
,•^JJJJ
-
flic integral in the square brackets is easily evaluated. When a # , I J
^
r
exp
[i(^
[I(bC - Rt)i] - cxp [- i(ü il jf^ — ox] d x = QX^ — — -- --
_
i(a -
2 2—
- sin (a -
so lirn r-^ '
f
I
a
exp( —i}tx)f(x)dx = r
lim ^ 2A(a) 1,^;
sin (a a—
p)i
da.
l- hc unknown functions A and B can also be dercrmtned from (5.3.1) and (5.3 8] of I.
(5}
Sokiioti iii the Linear Theory [Ch. 8
358
The right side of(5) has no readily apparent formal limit as it stands but with the change of variable (a — p)i = v it becomes* lim 2 r oc
A(^ + p
^
-
sin ^ ^
riv _=
where
C
u -1 sin v dr.
-
(6)
-m
We can determine the numerical constant C to any desired degree of accuracy by numerical methods, but in fact it can be shown by various analytic methods that C — n (Exercises 2 and 3). Putting all our results together, we see that (5) becomes l
2n
^
exp (
iux)f (x) dx
A(p)•
(7)
^ .3D
Similarly,
g(x) dx = ^)^(N^ 2Irr ^ eXP ( — iJ^)
(8)
In an important special case, the free surface is initially motionless and its initial position is an even function. That is, f(x)
f(
x),
(9a, b)
9(x) = O.
—
With (9a), equation (7) can be written A(^ll
2
cos px f (x) dx n
fo cas px f (x} dx,
(10)
since
f^
s ill f!X f (x) [SX = 0, ^
•
Equation (10) shows that A is an even function. Since B
r) f
A(a) COS wt cos ax da = 2 f4(a)ccswi!
0, (2d) becomes
cosaxdcc. (1 2)
^
where we have used the relation w(--a) = w(a).
•
We have assumed Thal
r-ao
A{ l
+ + tJ= A(2l. 11
This is certainly true for fixed finite v if A is continuous at FA., but w ranges from — oc to + ec. Min integrals con verge, however, contributions for Targe M must be small. Th is gives confidence that whit we have donc is valid- For proof. see Section 5,3 of 1_
Sec. 8.3]
.periodic Initroi Vuhies
359
Exampk 3. Suppose that the surface deflection is initially given by a rectangular bump.
(1x. 0)= 0,
1x1 >R;i
CÇx.0)
— Q. xi
R:
C,{x.g1= 0
(13)
(sec Figure 8.5). Find the surface deflection for later times. L
FIGURE 5 5. Surface deflection ui time - r= D. in the form taf a rerIOngulur hump_ ti,rrnglrr (15) giees the del: ecir`vn at later +rmes-
alutian. li is not dear that the given initial displacement will result in a motion whose convective acceleration is negligible as assumed by our linearized ihcory Only if such a motion is predicted will the results of the theory be consistent with the assumptions made ilt deriving it. Keeping this in mind, and noting that (9) holds, we use (tUM to find
I
ALA = -
t`
ccas
^ ü
Q sin pH
dx -- -
ln`
(14)
horn (12). 2Q
;(x, r) = -
-
sin all? I- — cos to(a)1 cos cue da.
( 15 )
G
However complicated a function at is of a, the integral could be evaluated numerically. The job would be difficult, as the interval of integration is infinite and the integrand is oscillatory. As we shall see in the next chapter, however, it will be possible ED simplify (15) so that no complicated calculations need be made. Note from (15) that as could have been anticipated, the surface deflection and its derivatives will indeed be small lithe initial height Q is small. A QUALITATIVE FEATURE OF
THE SOLUTION
We cannot emphasize loo strongly that although in a sense we have solved a wide class of problems, our work is almost worthless unless we can proceed further. Our purpose is to acquire a better understanding of water waves, but this does not yet emerge from the "solution" provided by (2), (7), and (S). Further effort is necessary, A discussion of our results forms the subject matter of the next chapter. Here we prescrit the derivation of but one general conclusion, when the free
Sviutia ►r in the Linear Theory [Ch. 8
360
surface is initially symmetric and motionless. In this case, we have by a slight modification of (12),
‘(x, r) = 2 Re
f
A(a) cds
CI1t exl7 (lx)
dm
Writing cos oil in terms of exponentials, we obtain
+sx(x, t) + tâ L(x, t},
(16)
where C R (x, t) — Re E J A(a) exp Max — wt)] (17)
Ci( x, r) = Re
fa
A(a)exp [i(ax +
r^t)] da.
We have CA— lc, t) = Re ^A(a)exp [i(— ax — oil)] da -- a
= Re
.a
A *(a) exp [i(ax + tot)] da,
because Re z = Re z. But A is real [by (10)], so that
Thus an originally motionless surface deflection splits into two parts, a portion ICR traveling to the right, and a symmetrically placed congruent portion CL traveling to the left. Since there is no distinguished direction in .a homogeneous fluid layer of infinite extent, this result could have been anticipated. Consider nondiypersiue wrwes for which the phase speed wfa is a constant c (independent of a). For such waves we deduce from (17) that (Exercise 8) CR(x, =
(x — ct, )
,
1( , t) — R(x + et, C),
(18)
where r = wfa. The original motionless deflection splits into a pair of congruent disturbances, of half the initial amplitude, that travel to right and left unaltered in shape with a constant speed c. Fly contrast, since their constituent sinusoidal waves of different lengths travel at different speeds, dispersive waves generally change their shape as they move. SUPERPOSITION AND THE DELTA FUNCTION The following considerations give a better understanding of how complicated linear phenomena can be regarded as the superposition of the effects of simple "causes." Example 3 can be regarded as giving the effect (surface displacement) caused by an initially motionless, rectangular surface bump,
Srr_ 13.31 Aperiadir lrririuf irufuea
;6e
centered at the origin, of width 2R and height Q_ In the solution given in (15), x is the distance from the original center of the bump_ If the bump were centered at x = 4„ then (15) would give the correct later "effect, - provided fi Thus tut initial motionless hump of width that x were replaced by x centered at 4; , would give a surface displacement of height f and —
❑
,
2
.I
Jo cc
sir[
Ilk
2
cos co(a)1 cos cr(x —
da.
( 19 )
Because the governing equations are linear, an initial motionless "step-bump" like that of Figure K.b(a) would cause a surface displacement which was the sum of the surface displacements due to the bumps of Figure 8.6(b) to (d). We should be able, therefore, to find the effect of a general motionless initial displacement ax, û) = f (x) by approximating f(x) by a step function, using (19). and taking the appropriate limit_ Multiplication of the numerator and
°tl
I
r I r I I
I 1 I I I
i
1
I
Ji
b
h
1 I I I I
(2}
i
❑tl
Rh]
d( 3
[r)
t) 6 t1
R h!
(d)
E} FIGURE 8.6. Th e three re*rfuagles of (:t) are separately depiered i,r tb), Cc). and id), in the !incur theory of water wa v es, the surftx'f deflection din' IC an initial configuration (a) inn b e ob tai ne d by summing the rlefieerions du e 10 irriti[e! cvrrfigurafiuns of the form (b), (c l, and {d).
362
Solution in the Linear Theory [Ch_ 8
the denominator by At la facilitates determination of the limit. We obtain C(x, r) —
f
2 iirn x
ef, -0 r a
f(
d2) cos (loos cos a(x ti) Air der 2 at6 2
aJr^ ( r
or
r) —
f() cos co(a)t cos ac(x — rÿ) da f:M
^
(20)
^ a
where we have used the well-known result Sin a
Inn
a
a-•0
=
1.
As it should he, (20) is the same as one form of the original solution (Exercise 13), A similar way to obtain the same final answer begins by considering (15) in the limit Il Oy 212R . The result is ^ sin aR cos cu(a)f cos ax ria o aR
2QH
lirn
C(x, [) =
Ti
R^o
WA - I
Or C(x, t) — -
l^
cos tot C(7s ax da _
(2l)
a
^
Here g(x, t) can be regarded as the displacement at time t due to a unit volume of water (per unit distance in the y direction) initially arranged in a tower of infinite height above the point x = O. That is, the original displacement is —
C(x, 0) = b[xj, where the delta function 6(x) satisfies (formally) ,51x ) _ 0, x
Cl;
I)dx — 1 ^.
(22)
Suppose that the motionless tower were initially located above x = e; and con tained f() units of wafer, i.e.,
«(x, 0)—Z(x — )I (0, Then the resultant displacement would be 1
-7r
CE
if ( a j cos cog CPIs ac(x — Oda, 0
Se e. 8J1
Aperrrxlfr.' lniriui Values
163
, as before_ Finally, it is plausible to where we have replaced x by x regard a motionless initial displacement as a composite of delta functions. ;led to obtain the resultant equation for by Integration. This yields -
1 jr(x r) -
. f(f,.
cos tat eels o(x — ) do! kit, a
123)
which indeed agrees with (20). R r M A R K. When NI) = 6(0, then (23) must reduce to (21). This is implied by the fermi] property
glo3(0 4 = ,((}). a property that is slightly more palatable than (22), and, in fact, is the basis of a well-founded definition of the delta function. Note that the integral in (211 does not exist in the conventional SLnSC_ Our recent manipulations have been circular, in that we started with a general solution, we specialized to the case of an initial rectangular bump, then we recovered our original result by superposing infinitely tall and thin initial bumps. Such formal experience with the delta function is a worthwhile preliminary to later study (not possible within the confines of this volume) of the delta function in its proper setting within the theory of "distributions" or "generalized functions. -
E7► 6RCI
S
1. Show that the improper integral in (6) converges. 2. (a) if you are familiar with contour integration, use that method to show that the constant (' defined in (6) has the value i. (b) Use a computing machine to evaluate C In three significant figures. 3. Certain formal manipulations with the Laplace transform yield C = 7E, where C is defined in (6)_ The Laplace transform F{s} of a function [(t). sometimes denoted by L( {tr)]. is defined by L[fltl] = 1 (s) = -
J,a
C]{p (
-
titlf
eh
if the integral converges. (a) Show that formal interchange of the order of integration gives F(r) de =
erg (- stji - ' f (OW.
+(b} Formally set ti = U in the above equation and use the result to obtain C. 4. Verify that u, w, and p, as defined in (2), are real.
Soh aia ►r in the Linear Theory [Ch. 8
3454
S. Show formally that the governing equations (1.1-4) are satisfied by the solution defined in (2). 6. Show that if f(x) is odd [f ( x) _ f(xi], then –
A( 43 .4) sin µx dµ f(x)1
–
provided that A(p) _ f f(x) sin ,ux dx_ rf p
7. Show that the formulas for A and B in (7) and (8) satisfy the conditions on A and B given in (3h, c). S. Verify (18). 9. Specialize the formulas of this section to the case when the initial conditions are C(x, 0) – (bfrr b 2 + x 2 ). Cr(x, 0} = O, where b is a positive constant. Consider infinitely deep water and assume that surface tension effects dominate gravity effects. Do either (a) or (h). (a) Work out A(a) from (7) by contour integration. (h) If you are unfamiliar with contour integration. verify that (4a) is satisfied if A(a) (2n) - ` exp (–bled). 10. Suppose that water waves were nondispersive (like waves in a string). Sketch the appearance of the free surface which is initially deformed as in Figure 8.5. Is this in accord with your physical expectations concerning water waves? (If so, revise your expectations.) 11. Extend the results of this section to the case of finite II. 12. Generalize the remarks about left and right moving waxes that were made after (17), when the restriction (9) is not assumed 13. Using(2d)and (7), show that when Cr(x, 0) = 0, C(x, t) can be written .
^
C(x, t) =
nI
^
^
^a
cos a(x ^ ^)f (+;) cos c.u(a)r d« 4g.
14. Find C(x, t) when
C(x, 0 ) _ exp ^
I
I— )
,
(x, 0 ) = 0-
R
15. Find C(x, t) when
C(x, D) = Q sin (.0x, Ix I
R;
((x,1)) =
-
0
, 1x1 > R,
1,(x, O) = 0.
Discuss the limiting case R ac. Obtain (19) by using the result of Exercise 13 when 16.
((x, O) = fad' , ix –
C IN , : p)=0.
C(x, (]) = 0 otherwise,
17. Make precise and verify the statement that the velocity, pressure, and surface displacement resulting from the initial conditions (1) can be
Appendix 8.11
Besse) Flint-lions
365
determined by adding the solutions corresponding to initia! conditions Clx, Oj =
f
Lk, D} = 0;
C(x, O) O,
(x O) = g(x).
18. Generalize Exercise 2_11 to show that. assuming convergence of the integrals, the sum of kinetic energy (KE) and potential energy (FE) for the waves given by (2) is KE + PE = >rpgA'
JT
I 4 1J1 2 +
1 B(P)I 2 ] d#.
(24)
Do this by writing down expressions for dx, etc.
r
and then using the following version of 14a) and 171'
exp [rxfx — p)lAla1 tfx dot — 2rAW).
• f.
(25)
^
19. +(a) Show that (2d) can he written = Re y, where
yfx, r) =
F(a) exp
—
F(a) _ 4(2) -
^x
so that, from (24), KE + PE
ra j,y 1'
f.,
F(012 da
126)
(b) Use (25) to show that KE + PE = .pyA'
127(
^
21). With the aid of several graphs. characterize the behavior of the transform A(p) Found in Example 3 as a function of Q and R. Given A(p), what can be ascertained concerning the parameters Q and R that characterize the
initial disturbance?
Appendix 8.1 Bessel Functions The following equation often arises, particularly when cylindrical wordinates are employed:
r 2 F"(r) 1- rF'(r) + (a ) r 2
—
p2 )F(r) -
Here p and a are parameters; the degenerate case
from the discussion.
f_
(1)
= 0 is often excluded
Solution in the Linear Theory [Ch. 8
366
Solutions of (I) are called Bessel functions. Asa result of over a century of analytical and numerical work, there exists a most extensive body of knowledge about these functions. [The classical reference, which only covers work up to 1922, is G. N, Watson, Treatise on the 'Theory of Besse! Functions (New York: Cambridge University Press, 1966).] Work on the Bessel equation is representative of the subject special functions, a large part of which can be regarded as a study of functions defined as solutions to certain important ordinary differential equations. Since so many of their properties are known, it is almost as meaningful to say that the solution of sonic problem is a Bessel function or a Legendre function, For example, as it is to say that it is a trigonometric function. (Indeed, sines and cosines can profitably be defined as solutions to the differential equation d2p/Jix 2 + y = U.) It cannot be denied, however, that mastering the properties of typical special functions is a Far more extensive task than the comparable task for trigonometric functions. Here we wish only to list some of the most essential properties of Besse! functions. A kw more results are given in Exercise 3.1.4 of I. The Bessel equation (I) has a regular singular point at the origin. The method of Frobenius provides the following Maclaurin expansion of one solution to the equation, the Bessel function of the first kind, of order p:
^r^ ^ `^ ( - 1 j"'{r/2)^ k2 )„,z_-0 m { r(p + m + 1 )'
J ^r)
r
U.
(2 )
lip is not zero or an integer, J_(x) is a second linearly independent solution. Otherwise, we have the "exceptional case" of the Frobenius method [see, e.g., Boyce and DiPrima (1969), or E. Kreyszig, Advanced Engineering Mathematics (New York: Wiley, 1967)]. Then the appropriate solution is Y (x], the Bessel function of the second kind:
Yr(r)
2 -.l
r 2 +
^
+
l
p^ ( (;)
^
y)
1r
r^
n
m =a
a
'(
p
^
-
I^" 1 (h. + hm+ p)(+'12e^ m !(rn+ 1^) !
tr^ —
—
r > O,p = O. t ,2, ----
in?
(3)
Here h c =Ü
,
h= ^ i !-
I
1
y = lir n '
— lii m) = 0.577 ....
m— Cr.
For small r, the first terms of the above series provide good approximations to alp and Y. To obtain approximations for larger,. we must recognize that r = co is an irregular singular point of (1)- Appropriate methods (Coddington and Levinson, 1955) yield series expansions for 1 F and Y. which are asympto-
Appendix 8_1] Besse/ f wrrricros
367
tic ally valid as r i cc. The first terms of these expansions provide the following approximations -
lAr) ^' x 1;(r)--
f cc's
Jii sm
r — ^ nr — ^ px ^ ,
( 4 a)
(r —4 ,- 2 RR).
(lb)
These expressions are exact when p = - . Although not at all evident from the Frobenius expansions (2) and (3). qualitative behavior of Besse! functions (oscillation with decreasing magnitude as r increases) is evident fmm (4), a single term of the expansions about the irregular singular point at r = x _ This is perhaps to be expected. for more should emerge from an analysis if we confront the function head on at its worst singularity. From (2) and (3) one can derive the following derivative formulas, where Fp stands for either Jp n r Yr : dr ^r New)]^ err"F_
dr
Aar),
i] _ —ar FF p + „W.
[r -PFp(
(5a)
(5b)
Using (5), we can deduce
d; [1.. P(ar)j
+ 1~ p(ar)
! ^IF p - r(ar) — pr —
af' F+ r (ar) f pr - rF p{ar) {
=-- ^af^_
Jar)
Ep + I(ar)]
(6a) 16h1 56c1
EXERCISE
1. (a) Show that formal differentiation of(2) and (3) yields (5a) and (5h). (b) Derive (6a) and (6h)_
C HAPTER 9 Group Speed and G roup V elocity
I
N THE previous chapter we exhibited explicit formulas for the solution of rather general water wave problems. It might seem that this would be the occasion For considerable satisfaction. but this is not the case. We seek understanding of natural phenomena; no such understanding emerges from passive contemplation of complicated formulas_ In the present context, and a great many others, insight does appear when solution formulas are simplified by means of asymptotic approximation methods. Roughly speaking, this is because "understanding" a complicated matter requires discernment of its essential aspects and relegation of details into the background_ AsyrupWlic approximations do just this. for they generally ignore a function's regular points and even its regular singularities and concentrate on revealing behavior in the neighborhood of an essential singularity.* For water waves, the asymptotic approximation is obtained by the method of stationary phase," which can be regarded as a variant of the more general method of steepest descent_ The former method allows one to simplify the Fourier integrals so that the behavior of the solution can he well characterized by introducing the notion of wimp .spred.t Group speed is a central unifying concept in wave propagation. It arose from observations on water waves by keen observers such as Scott Russell {cited on p. 31SO of Lamb (1932)]. The mathematical basis for the phenomenon was established by the great nineteenth-century British applied mathematicians, Stokes, Kelvin, and Rayleigh§ ; contributions of great value have also been made by a number of contemporary applied mathematicians. In this chapter we apply t he method of stationary phase to water waves and then compare some of our predictions with experiment. We also briefly. of the theory of water waves, particularly surveyompactil ns in connection with "tidal waves." We next turn to a kinematic approach to group selncity. which more clearly reveal; the physical basis of the • This poor is developed in L. A. Segel. "The importance olAsyimptour Analysis in Applied Mathematics." Amer. Malt. Morirhli 73.7 -14 (1966). t Group speed hears the same relation to group veloemy as ordinary speed does to oidinary
vciociiy. Many authors do not deem it wurthwhiit is make this distinction cxpthcrtiy and always refer to "group velocity_ .. The Irish mathematical physicist (or applied Enathrrnatician) Hamilton artiliriparcd some of lhcworl; on group speed during hm mrd-nirctcenth-t. mur' Ntudy of optics_ See Chapter 9 of Lamb (4932) for htsiorical notes_
368
Sec_ 9.1] Group Speed ria the Method of Siorionnr}• Phose
3t9
results and has the further advantage of being capable of extension to wave systems for which slight inhomogeneities or nonlinearilies must he taken Into account. Using this approach. we predict the striking wave pat tern associated with a ship—or a water beetle.
9.1
Group Speed via the Method of Stationary Phase
Pursuing the gnat of deriving physical understanding of water waves from thesolution formufas(8.3.21, (8.3.7). and (8.181. we obtain an approximation to these formulas for large values of x and r that permits a clear physical description of our rultx. NEEID FOR AN ASYMPTOTIC
APPROXIMATION
We assert that it should be informative to approximate our formulas in the limit r CC. Simplification would also he expected from approximations valid near r = 0, but the results should he of less interest than approximations valid for large t. One reason is that the behavior at small time is expected to be dominated by the particular initial conditions imposed, while for large times it is likely that effects arising from the intrinsic nature of water waves
will be prominent. Those versed in the theory of complex functions can apprcciale another reason. It appears from (8.3.2) that for fixed x the surface deflection C(x, r) [as well as u, w, and p] is analytic at t - 0, or any other finite value of t, but has an essential singularity as (We arc temporarily regarding r as a complex variable_ It is a central result of function theory that this is necessary to understand the behavior of C, even though we are ultimately interested only in real values of the time r.) One expects that a better understanding of the nature of iÿ will be gained by examining near its - worst " point, al the essential singularity r = x , rather than near points of good behavior Such as r = t) Turning to approximations based on the spatial variable, we assert that simplifications tinder die assumption of large x should be particularly informative. Cierierally, the initial disturbance is to some extent concentrated near x =- O. Thus approximations for small x, like approximations for small t, should be strongly influenced by the initial disturbance. Approximations valid for large x . , however, like approximations valid for large 1, should reveal more about the intrinsic nature of water waves. In addition, for fixed t the surface deflection appears to be analytic at x = 0 but appears to have an essential singularity at x = x. As above, therefore. complex function theory prompts us to examine the solution for large x. STATtONARY-t'FtASF. APPROXIMATION
Once the idea of looking at approximate behavior for large x and t has occurred to us, we are faced with the technical problem of finding suitable approximation formulas. Kelvin faced just this problem in his work on
Group Sp ee d and Grua!) Velocity (Cu.I 4
water waves; he solved it by initiating the method of stationary phase. The basic problem solved by this method is the approximation for Targe real A of the integral
1(a, b, 1.) =
r_ ^ U_
f(a ) ex p[if g(a)] (la,
(
1
)
^^
Suppose T hat g'(a) =0 fora ac < h if and only if a — cc ; , and that Oaf) 0, I = 1,..- MIN 1), a < a; < b. Under suitable conditions the following basic stationary phrase theorem can be demonstrated: ,
f(u, h,ï.}
^•
rr 2 { (a i)
1
i= r
t
Cxp [i Â9(Œ) +- ^ rn sgrl g'.(7^) ^-
l
(2) No
E.
The results arc still valid, with obvious slight changes, if a =
—
or h = ay. Since the error decreases as A — cc, the approximation (2) is
asymp to t ic_ A generalized stationary phase result, slightly vague but very useful, is this. Ifexp [(Mall oscillates rapidly compared withf (a), we have the approximation -
^
-
f (a) exp [ih {a)]
da
_ ,^2n >1: I it"(a !) I ; -
"2s f^a}
exp [i!I(a lj
sgn h"(a))], (3)
^
where
hlac)=C] if and only if* =ay. h'
. j).)-P6
D,
}= 1,---. N, N> I-
We discuss the derivation of (2) and (3) in Appendix 9.1. APPLICATION üF TIM APPROXIMATION
Consider the right-moving wave of (8.3.17): .,
R OC,
r) = R e
f A (a) 1.!xp [+rtX — iciga]r } da.
(4)
We can put (4) into the form (1) by moving from the origin at a fixed speed V so that x can be replaced by Vt. This gives
A(a) exp {r![a V -- co(a)1} da,
cÿx = Re
V =+,
(5)
a
By regarding Vas fixed and lettings —# co, we can now use the basic st at t ionaryphase theorem (2) to find an approximation to (S) that is valid for large t. Hut this theorem guarantees the validity of the approximation only for the single
see. 4 f 1 Group Speed ria the Merhod of Stationary
Phase
371
-
value of x- satisfying x = Vr. We desire results that hold for a range of values, so w e turn to the generalized version of the theorem, The exponential factor in the iritegrand of (4) should vary rapidly compared with A(a), as required, rovided that either x or r has sufficiently large values_ Comparing the integrals in (4) and (3), we set h(a) = ax — cel(a)i.
Points of stationary phase satisfy Ji'(a) .= 0 or —
aa'(tr)t — [}-
(6)
We anticipate that (6) will have a unique solution to the examples with ,t,iic.h we shall he dealing. We denote this solution by a(x, rj, so that x
—
Olt = 0-
(7)
This notation explicitly indicates the dependence of the stationary - phase wave number on x and r (which are parameters as far as the integration with respect to a is concerned). Equation uatian (7) can he regarded as an implicit definition of DIX, t}, the wave number giving the principal contribution to the surface deflection at (x, r)_ (We shall discuss this last poin t in more detail shortly.) In (3) we must substitute the value of h(s) at the stationary - phase point. This is simply h[a(x, 1)] i a{x, r}x — w[rx(x, ills
(g)
Other portions of (3) are handled similarly. In taking the real part of the integral as required by (4), we must bear in mind that A is real [by (8.3.10)]. With all this, application of (3) to (4) gives
isdx, t} ^
f fi, (x, *
A(x, r) cos [a(x, i)x —
(Aix, i )r — ()sgn u^ "(x,
r)], (9a}
where' fl(x, r)
—
A[a(x, f}] f^(x, r) .
°to[ot(x, t}], w"(x, t) =
d2 co(a) j 2 ^a
alx.+l
(4h, c, d) ncprnding on whether one nr two indepcndcnt variables arc displayed. the 5amc letters for different functions on opposite sidrs of the rçuatit ns in (9b- c. d). For example, alx.il is the valut al and t for thr function '1(, 1. whilr 44Œ) is the valut al a of the different (unahion .1i k
arc
used
Group Speed arid Group Velocity [Ch. 9
3 72
Example 1. Apply the method of stationary phase to simplify the expression for the surface deflection, due to an initial rectangular bump. that was Found in Example 8.3.3. For simplicity. ignore the effects of Finite depth and of surface tension. Sabrrïon. Using (8.3.2e), when depth and surface tension effects are ignored we have w=(
„al"].
w — ^^il 2 a -1l2 .
,
^
=
rl3
`
- 3ra
110a, ", C}
From (8.3.14), the function A(a) appropriate to this problem is Q site aR an
A(a)
( l Od)
The contribution of the right-moving wave is given by (9) where, from (1). a(x, t) is determined by
This equation has a solution if and only ifx O. (Similarly. the left-moving wave will make a nonncgligible contribution to the asymptotic expression we sack if and only if x < O. This is to be expected.) For x > 0, then, using gi 4x2 ^
u{x, r)
.
we find from (9) that for sufficiently large values of x and r the nght-moving wave is given by Cp
1 os 4x- c 4x 1
41:) ix ng
sin
(l2)
4
In Section 9.2 we shall show that (12) compares favorably with experiment.
Recall that the essence of the stationary-phase approximation lies in (6), which states that at point x and time t the surface deflection is mainly due to those wave numbers Œ(x, t), which are such that (13) For values of x and r satisfying x = VI, V a constant, (13) is independent of x and t and yields (assuming a unique solution) a = av ,
where cola,) = V,
a r a constant.
(14)
In more ponderous but explicit notation, we can write 4x, 1)1r Ye -
w'(x,
( Vt,
(15a)
r) = û y .
of x(x, toix=v, = '(ay)
V•
(15b)
5er. 9.1]
Group Speed um the M e thod of Siationary Ph ase
373
If the "wave number a(x, r) that explicitly appears in (9) is evaluated when Vt, then, by (15a), the constant U y is obtained. It thus seems that ifyou start from the origin and move with constant speed I", then at your location x = Vi there always appears the same wave number a,.. t he one given by (15a). This is not an entirely satisfactory statement, however, for the wave number (or spatial period) ar u polar has no obvious geometrical meaning- It turns ou t, nonetheless, that our basic idea about a v is correct. To clarify the situation, let us examine the appearance of the waves at time t near the point x which satisfies x = Vt. This requires a Taylor expansion doll constituents of (9). We start by considering the amplitude AIx, rl. Near x = Vr, Taylor's formula gives x
A(x, ii = 111V I, r) i- A,(Vr, t)(x
Vr) - 4 ;AYL(Vt. t)(x — V
-
+-
(16)
From the definition (9b), A(x, r) we find, by the chain rule, that Ax =
We can determine ax by implicit differentiation of (13): — ua"oc = 0,
so a x(x, t) = [rw"(x, r)] ' .
(l7a)
Using (15a), we see that the first two terms of (1 b) give A(x, r)
A(a ,) +- 'gay) (x — V;) + - - • . rm"far)
(17b)
Equation (17b) indicates that if x is restricted to be within a fixed distance of Vt, then we can replace the function A(x, t) by the constant A(a i.) if we are willing to make an error proportional to t We next expand
`.
h(x, :) - a(x, r)x — o{x, r)r about x = Vr. Since hx(x, t) =
4x
—
W"t)oc, + ü.
we find, using ( 15h), that
h, Vt, I)
= ay .
Differentiation of (19( gives. with the aid of (17a).
h, Vt, t) = [ruy"'(izr.)] -
374
Group Speed and Group Velocity [Ch. 9
Therefore, it appears that by allowing another 00 -i ) error we can replace h(x, t) by the first two terms in its Taylor series, i.e. , by h(VI,r)+h,(Vt
,
r)(x- Vt)=a t.Vr —(4a jt+a v(x— Vr) av x -
Similarly, offx, t) can be replaced by w"(a v ). We finally have for x near Vr, C.
^
^ (arj cos [ax — w[a Y)t — 4 n sgn m +'(°^r) - (20)
^
Example 2. For large t, what does the deflection C i, of Example 1 look like near x = Vi, V a given constant? Solution. From (I l }and (10), =
2,
e4aYl = 40wl ' 12 V 2
, S113
to°Iciw1 = -i4 - 'I/ 3 ,
ayR
(21)
sn 1 4
r sin I ^ gR V -') cos ( - gl^ - ^x - ^ g ^t +
CR . 4Q1 9
1
4
(22)
INTERPRETATfON: CROUP SPEED
Equation (20) follows immediately From (9) if the formal substitution V: is made in a, cry, and of. This substitution would be unconvincing without an argument such as the one just given, however. In the term ct(x, r)x, for example, why should one be able to replace x by Vt in the factor a(x, t) but not in the factor x? Such a replacement is legitimate because a(x, t) is a slowly varying function of x —a feature that was not evident before we made the calculations that led to (20). [The slow variation of a with x for large times is explicitly shown in (l 7a).] Equation (9) is an approximation valid for all sufficiently large values of x and s. Equation (20) is a fiirTher approximurion of (9) that is valid for x near Vt when f is large enough to justify our neglect of O(t - ' j terms. In order to cover" all large values ofx with (20), one must take all permissible values of the Constant I' Personification of the formulas may help. Equation (20) provides the surface deflection seen at large time by a man moving with uniform speed V, starting at the origin at time zero, in a magic boat that does not disturb the water surface as it moves. To get a complete picture of the surface deflection at large time, one must initially send out many men in many boats moving at various speeds. The appearance of (9) seems difficult to characterize generally. Only in (20). a special case of (9), do we see a function having manifest wavelike character.
See_ Q.!]
Group 5peerl via tleP MI hod ❑f 5ialronary Phase
375
This is illustrated by the results of Examples l and 2. The asymptotic approximation ( l 2) does not have the form of a slowly varying wave. hut later, when (22) can be used. this form emerges. Since(20) describes the surface deflection near x — Vt, for sufficiently large time an observer who is moving at speed V (starting at the origin at t = 0) will continually see waves of wave number « Y . This can only happen if waves of wave number a. move with this speed. The term group speed is applied to V. From (14), the group speed corresponding to the wave number at, is given by the formula V = '(cc v)- We conclude that whatever its initial
shape, al large times an initial disturbance sorts itself out into a wave Train consisting of oscillations of slowly varying wavelength_ A section of the wave train where the oscillation has a wavelength approximately equal to 1 y - 2n fa y, mores at a speed approximately equal to the group speed V = ri (a v ). The approximation improves as time increases. PHASE SPEED Recall that a purely sinusoidal wave, proportional to sin (cce x -- cap, r), say, moves with the phase speed wda e . We are considering a wave train formed of a continuous superposition of such waves. The above-mentioned role of phase speed is therefore inaccessible to direct observation, because without preliminary Fourier analysis it is impossible to isolate one of the continuum of harmonics that make up the wave train. As we shall now show, the phase speed can be observed by focusing attention on a point of the free surface at a certain fixed distance d above the mean height. The progress of such a point is given by the function x x(t) that is defined implicitly by a solution of C R(x, t) = d. Implicit differentiation will give x the speed of the point. Once again it is the approximation 120) that allows us to obtain a clear-rut result_ Using (20 ), implicit differentiation of bR = d gives ,
— it - 311 cos H - I -
"(
sin #! H = O,
(23)
where H
cofar)t — *7I sgn
W "(a Fr).
For large r, the first term on the left side of (23) is negligible compared to the second, so that the speed of the point x can be estimated from Ham O or ay x —(.u(a,,)=0.
Thus for large time points near x 4 Vt that are at a fixed distance S above the mean height move with speed w{av}fav, the local phase spy_ In particular, points at the mean height (d = 0) move with this speed. Further, it can he shown that maxima, minima, and points of inflection also move with the local phase speed (Exercise 3).
Group Speed and Group Velocity [Ch. 9
n6
Example 3. Discuss phase and group speed for water waves, ignoring surface tension and depth effects. Solution.
The frequency (Ai is given in terms of the wave number a by rat = 'Jga-
The phase speed is wfa = gJa. The group speed is LAO = 1 gJa. The phase speed is twice the group speed. Both speeds are decreasing functions of the wave number, so that both are increasing functions of wavelength.
F iG tr R t 9.1. successive "snapFirols" Al a,rloa -ly raw Ow ware tre7in. The lines RR' enclose wares of approximate leng th The speed of such a ware group is less than that of the point of tY ►mmon phase labeled P I . The lines QQ' enclose a ,group of longer wares, whir* more ..faster than those associated w ith RK The local maximum 1 3 mores with as phasr , speed toward the n ight of the ,
QQ . grouP.
The roles of phase and group speeds are shown schematically in Figure 9.1 wherein the free surface is depicted at the equally spaced instants t o , t p -f ❑f, r a + 2/it, .... We consider for definiteness a situation, like that of Example 3, in which both group and phase speed increase with wavelength and the phase speed at a given wavelength exceeds the group speed at that wavelength. On the left of the figure, between the pair of lines marked RR', are waves of approximate wavelength A t .* On the right, between the pair of lines marked A,. There is a transition region QQ', are waves of longer wavelength There is an intrinsic imprecision in the nolion of wavelength. since the distance b€Iwœn successive crests (or troughs) is slowly varying_
Sec. 9.1)
Group Speed via ihr Method fJJ Sialionar}' Phase
377
between the two sets of waves. The transition region is actually long, so that there is a continual gradual change of wavelength. We have shortened this region in order to he able to depict more clearly groups of waves moving at different speeds. The speed of the waves of length A L is given by the slope of the line RR', The longer waves of length 1.2 move with the greater speed given by the slope of QV. (To compute the slope in the conventional manner, one must look at the figures sideways so that the independent variable r increases to one's right. Although units are not given on our schematic diagram, we assume that they have been chosen so that the slopes are numerically equal to the relevant speeds.) A point where the free surface always has the mean height has been labeled P 1 . This point moves with the local phase speed, that is, with the phase speed associated with waves of length A L . As this is larger than the local group speed, P 1 moves forward relative to the group of waves having wavelength J L. Eventually, the point P I will be more accuntidy associated with waves of wavelength larger than A i and hence with a larger phase speed, Also depicted is a local maximum P 2 moving with the phase speed, larger than that of P 1 , associated with waves of length )r. SPECIAL. CUl*lt]ITIUNS NEAR EXTREMA O F GROUP SPEED When applying the method of stationary phase in (3). we replaced Pita) in the neighborhood of a stationary-phase point a, by what we took to be its first two nonvanishing terms. h(;) + }h"(Œ1)(Œ
—
ai)2 .
Our approximation is invalid when h"(a) D and will be poor when h"(a ;) is nearly zero. For waves, hi) = ax — coop. so [as is clear From (9a)] the usual stationary-phase approximation wilt fail al wave numbers where the graph of group speed (Oa) versus wave number a possesses a horizontal Ionyeni. For water waves. provided that the depth is not too small, the group speed has a local minimum V,,, at a certain wave number a,,, [Exercise 5(a)]. Consider an observer located at point VI at time f. If V < V„,, the observer should report a negligible surface displacement after a time. since C .:: f) according to the stationary - phase approximation_ If V is slightly larger than I/,„, on the other hand, the observer should report waves of wave number approximately equal to a n,. If V = V,,, wc can use the results of Exercise A9.1.5, but to describe the transition between the smooth-water region reported by the first observer and the waves reported by the second, it is necessary to use an approximation that is valid, for large time, uniformly for VL V 5 V 2 where Vi < V < V2. Such a uniformly valid approximation can he found in terms of the Airy function Ai, the simplest function whose behavior alters from monotonic to oscillatory as its argument changes. For details, see Jeffreys and Jeffreys (1962, pp. 517-18). These authors modify the
37 8
Group Speed and Group Vorocrry ((k 9
usual stationary-phase result by a clever device. A more straightforward approach can be made using heavier machinery.* A similar uniformly valid approximation must be used to obtain an adequate description of the waves near x ` t H. The reason is that (even in the presence of depth and surface tension effects) the group speed also has an extremum at V y yH [Exercise 5{b)]. It turns out (Jeffreys and Jeffreys, 1962, p. 51 7) that the disturbance becomes mere and more prominent near
x
;r. There are capillary waves ahead of this point, but they die out
relatively rapidly [Exercise 4(a)]. Moreover, these short waves, with their accompanying large velocity gradients. are more strongly damped by the viscous effects not considered in our analysis. Except for shallow layers that will not concern Lis here, 9H > V !, and the discussion of this paragraph applies to the leading edge of the wave train_ To summarize, the principal surface disturbance is a slowly varying wave train whci ,e "front" and "back „ boundaries move with speeds 1 1,„ and
gH, respectively_
SOME: APPLICATIONS TO FLOW PAST OBSTACLES
We shall now apply some of our conclusions to the understanding of flow past obstacles. In doing so, we anticipate to some extent more detailed discussions to follow. By making some applications of our conclusions now, however, we indicate their power to those who cannot at the moment pursue the subject further_ Moreover, the tentative nature ci f some of our arguments provides good motivation for further study. We have seen that lines of constant phase, notably crests and troughs, travel with the phase speed co/a. Sets of waves having roughly the same length, by contrast, have been shown to travel with the group speed drr*fda. 11 can also be shown (see Section 9.3) [hat energy travels with group speed, for waves of a given approximate length retain their energy as they move. The conclusions we have reached have been in the context of an initial value problem, hut they will he shown its Section 9.3 to hold whenever a slowly varying wave train is present_ In particular, the description in the previous paragraphs of the roles played by phase and group velocity holds in the case of steady flow past an obstacle. This should not he surprising, for the effect of an obstacle in a steady linear flow is the same as the superposed effect of appropriately chosen initial disturbances. The obstacle can be regarded as sending out identical disturbances at each instant. At any moment of time, the effect of these disturbance can be obtained by adding the effects of each individual disturbance, with appropriate delays to account for the varying times at which each disturbance has occurred. Sec C. Chester. B. Friedman, and F Utsdl "An Extension of the Method of Steepest Descents," Prat_ Cambridge Phil. Soc. 53, 599-îi l (1957). Sec also our Figure 9.5 and the refermez cited u> >ts caption. ,
Sly. 9_1]
(irotep Spud
rua ihe Myrhod of 51a1ionary Phase
379
Since energy travels with group speed, there is a significant difference between situations wherein group speed exceeds phase speed and situations wherein the opposite situation obtains. In the former (latter) case, wave groups and energy travel forward (backward) with respect to the patte rn of crests and troughs. In this connection it is noteworthy that group speed exceeds phase speed if and only if the latter is an increasing function of wave dumber. Thus the tvavellerigrh (if any} at which the difference between group speed and phase speed changes sign is the some us the Waveleriyth (ij' any) ut which phase speed is a minimum (Exercise 2). For water waves when depth effects can he neglected, this wavelength is 1.7 cm see 18.1.21)]. Solutions of numerous special problems bear out one's intuition that when the surface of flowing water is disturbed by some object (or when the object moves through the water). most of the wave energy will generally be found in waves whose length is of the same magnitude as a typical obstacle dimension.' It is no surprise, therefore, that the wave pattern forms a wake behind a moving ship. To be stationary with respect to the ship, a crest must have a (phase) velocity that is balanced by the component el- stream velocity (as se en from the ship) ahai is normal to the given crest (compare Exercise 4.211. All significant wave groups are gravity waves, with lengths considerably longer than 1..7 cm. Consequently, energy moves backward with respect to the pattern. Each wave group disperses. and this spreading causes adecrease in amplitude and a gradual disappearance of pattern with increasing distance aft of the ship. Water beetles, by contrast, are typically less than 1 cm long, so much of the wave energy they generate will be in the form of capillary waves (lengths less than 1.7 cm'. ]f the beetles move steadily at less than the minimum phase speed of about 23 cmis, no wave pattern can be produced (for no set of crests and troughs can move slowly enough to maintain their position with respect to the beetle). In this case one expects only a highly localized disturbance of the water surface, which in fact is jusi what is observed Beetles can move as fast as 50 to l&) cmis, so there should be instances in which wave patterns appear. Indeed, such a pattern is shown in the frontispiece. Although there a dead beetle is held stationary in a smoothly flowing stream, the insect is properly oriented and the waves are virtually the same as in nature - except that the beetles in question normally move in a jerky manner and so produce a superposition of the patterns obtained for steady motion. Striking in the frontispiece is the fact that all the noticeable waves precede the beetle Our present understanding of the situation would lead us to anticipate this, • In aIJ rases, short waves may be an irnporiant source of energy disswarron. t Thr tengrh of hie brute is in the inrrrrner}rete runtçc. where both capillary and gravity effecis are expected to he impnrtarti. Thus it is not surprising that gravity waves Can a [so be sccn behind beetles. although their relatively great length and srnatl arnphrude make detection difficult_
380
Group Speed and Group
[Ch. 9
however: Since group velocity exceeds phase velocity in capillary waves, the energy moves forward with respect to the pattern. Beetles seem to have an intuitive knowledge of the subtleties of group velocity, for there is some evidence that they use reflections of the waves they generate to warn them of objects ahead of them. (See the article by Tucker cited in the frontispiece caption, and the references listed therein.) Physicists, too, have reason to be concerned with deeper aspects of group velocity and related matters; otherwise paradoxes arise in connection with the velocity of light. An excellent reference here is L. Brillouin, Ware Propagation and Group Velocity (New York: Academic Press, 1960). The principal mathematical issue in Brillouin's book is the asymptotic evaluation of integrals, a topic that we have touched upon briefly in our own development of the group velocity concept. Further development of this concept in the present volume (Section 9.3) is based an an entirely different kinematic approach that is fruitful even in contexts where it seems impossible to express resu lts in the fern3 of integrals. EXERCISES
L Find asymptotic results, as in this section, when there is no initial displacement of the free surface but there is an initial velocity. 1. (a) Show that in general the difference between phase and group speed changes sign precisely at the wave number at which the phase velocity is a minimum. (h) For water of infinite depth, find (as a function of wave number) the ratio of group speed to phase speed. Sketch the graph. Note particularly the result for pu re capillary waves (or, equivalently, the infinite wave-number limit). (c) Find the ratio of group speed to phase speed for water of depth I!, ignoring surface tension. Graph. Do your results suggest any modifications for the concludèngdiscussion of the section in the case of waves that are long compared to water depth? 3. Show that according to {0), maxima, minima, and points of inflection move with the local phase speed. This involves defining x as a function of t by the equation dC R /r?x = [i and then finding ac. In determining x, show that only fi(x. t) need be considered time-dependent, since time derivatives of other terms are small_ 4. (a) When capillary effects dominate, (8_31e) gives (for fluid of infinite depth) 0=
`43;
T_
?
p
,
a > U.
(24)
Find the counterpart of (12). Deduce that capillary wave amplitude, in contrast to gravity waves,decreases relatively rapidly as x increases.
See_ 4_l] F.xperirnenrs and Prorl,rcrt Applications
3b
^
(b) If you keep your eye on a wave of a given length, will a maximumheight point for that wave move forward or backward relative to the wave as a whole? At a given instant, are the longer or the shorter waves nearer to the point of the initial disturbance? Explain. 5. When depth and surface tension effects are both included. the dimensional dispersion relationship is (see Exercise 8.1.4)
ro' — (ga + T c3 ) tanh fl
.
(25)
where H is the depth of the layer, 7 - = Tlp, etc. la) Set tanh xH = I in (25). (This is appropriate except for waves that are long compared to the depth 1/.1 Shot, that the group speed V12) has a minimum 4;,, when x — m„,. where x, w = {1.4ti y T. Find V,, and the corresponding wavelength for water waves [7- = 72 g s 2 . p — 1 g/cm 3 . ti = 980 ems']_ +(b) For small a.
=Wi[a—
w 3 a 3 +••-]
Find co 3 lc) Under what circumstances are the regions of validity of the calculations in (a) and (b) expected to overlap'' For water waves. about how large does f! have to be to ensure this? Sketch the graph tif l'i in this case. Show that a graphical construction for group velocity can be obtained as 6. follows_ Plot phase velocity e verses wavelength .. Construct the tangent line to this curve at the wavelength d o of interest. The dinireti group velocity is the intersection of the tangent line with the vertical axis. 7. Using (25), show that the phase velocity no longer has an interior minimum cm for water waves What if H ^ I7 £ , 17g - t. Show that modifications are suggested to the concluding discussion of this section? 8. (a) Find the relation between group and phase velocity in the anomalous dispersion case provided by transverse elastic waves in a beam [use the dispersion relations (5.I.50)1 Discuss the significance of your results. (b) Repeat part (a) using (5.1.49) and thereby determine sonic effects of rotary dispersion.
f!, 3
H, _
-
9.2 Experiments and Practical Applications A good theoretician should be familiar with all relevant experimental results. Although it is not appropriate here to present an extensive account of water wave experiments, to forbear entirely from mentioning them might foster an inappropriate disregard for reality. Thus we shall compare our theory to a certain set of careful experimental measurements. Moreover, we
Group Speed and Group Velocity [Ch. 9
382
shall provide a brief discussion of practical application of lite theory, particularly to rsummis (tidal waves). EXPERIMENTS ON THE COLLAPSE OF A RECTANGULAR DUMP OF WATER
J. E. Prins* performed experiments to test the validity of the theoretical results we have been discussing. He set up the equivalent (sec Exercise I) of an initial rectangular bump or depression of width 2R and height Q on a motionless channel of water about 18 m long and 30 cm wide (60 ft by 1 ft)# (see Figure 9.2). The depth H was typically 67 cm (2.3 ft), although shallower
FtG u lr
9.2_ Experimental setup used by J. Prins_ The -• bump" of water
un the kit is released w time t = O. Waves propagate to the right and ore prosily absorbed by maternal p(uced on the sloping heath, Thus there is negligible reflection. and merJs -urr ►rtrrits carrt.spund hell to the idealized ruse of an un-
bounded fluid layer.
depths were also used. Prins measured the vertical movement of the free surface at five places along the channel and found the oscillatory waves predicted by our theory when QM < 0.2 and R/H < 1. Our theory does not predict phenomena found by Prins at larger values of Q/ f and R/11, such as "a leading wave being a single wave with solitary wave§ characteristics, separated from the dispersive wave pattern by a more or less flat part at the still wafer level." This is not surprising, as the larger are (2/H and RIB, the larger is the initial bump and the more likely it is that our assumption of small-amplitude waves is inappropriate. -
• "Characteristics of Waves Generated by a Local Disturbance," Trues. Arne r. Geopiiys Union 39, 865-74 (t95). t This was done by placing an airtight box, apcii side down, into the water at the head of the channel_ (The box WAS the same width a5 the cha nnel.) The air in the box was par l ial ly evacuated to obiain a rectangular elevation, in other experiments, compre -ssed air was introduced to obtain an initial recta ngular depression_ At the initial imtant the downstream side of the box was quickly removed. k A solitary mare is long compared with the depth H otthew undisturbed fluid layer. Its maximum height above the undisturbed level, A, need not he small_ It moves without deforming, ai speed % f! i A) (sec Sicker (195711
▪
Sec. 1.2]
Experiments und Practical Applications
383
In Figure 93{a) and (c) we display some of Prins's measurements of the time interval P between successive relative maxima, due to an initial rectangular devation. These are plotted versus r, the time at which the intervening. relative minimum passed one of the fixed locations, x — 4 r . at which measurements were made. ln Figure 9_3(b) and (d), P refers to the time 3.0
1
{1
{1
11 tx ^ I.
r=a ^
I0
r^
^
g
It Ili
HP-1 )^^ r1.%%
E^ r
^
^
^
4
•
?4
115
1f
4
(A)
30
.0 l
0
Idl
tcl
F t (i Li x E 9, 3. Time interval P tzs function of tame z, as defined ln the schematic ware recordings depleted in (a) and (h). In (a) and (c), elet ►ation—Q = 0.3 f! (9 cm). fer lb} and (di, depression-1Q ` —0.3 fr. fer lai and (b(, hump extent R ; fi. In (c) and (d), R = I ft. Depth H ; 23 fr in all cases. Each graph contains four curries, giving successively larger takers of P. Experimentaipoints A. x. 0, are recorded at slalioris x = 5, 15, 25. 35 ft, resper 41474y (5 fi ft 150 con I _ Dashed laies. infinite depth theory [Equation (1) of Solid lines: lines: Kranzer-Keller finite depth theory_ f Redrawn with perrrrisiion from Fig. 3 of J. E. Prins, Trans. Amer. Geophys. Uni art 39, 865 874 (I 958). copyrighted by American Geophysical [brion.] .
,
Group Speed
384
and Group
Velocity [Ch- 9
between succeeding minima, since [in contrast to 9.3(a) and (c)], the waves were caused by a depression. COMPARISON OF THEORY AND EXPERIMENT To obtain a theoretical prediction of P, consider the asymptotic formula (1.22). Choose V = x ; /r so that the successive maxima in question are near x ; - VT when t is close to T. According to (1.22), successive maxima should occur in a time interval ofduration 2rr/(gV t). An observer at x x ; should therefore record a value of P given by —
P=
4rtx i gt •
(1)
As Figure 9-3 shows, agreement between experiment and theory (dashed lines) stems good when f' is less than about I but is only fair for larger values of P. Many factors might he responsible for the partial discrepancy between theory and experiment. The discrepancy occurs for relatively small values of r, so one possibility is that measurements were made before enough time had elapsed to allow use of (1.22). It turns out, however, that most of the discrepancy is due to the effect of finite depth. One can easily provide indications that when P begins to exceed a quantity with magnitude unity, then the waves begin to be long enough to he affected by the bottom's presence (Exercise 2). Further as Prins shows- see the solid line of Figure 9.3), predictions of depth theory arc in gond agreement wish all measurements made_ fi+^[r Figure 9.4 shows traces of the surface deflection when Q = —6 cm, R 10 cm. A noteworthy feature is the continual increase in surface deflection maxima for the first portion of the record obtained at each station. This increase is nearly linear as the reader can check with a straightedge. Just such a linear behavior at fixed x might be expected from (1.12) when s IR
gt z R — 2 eh
gt2R 3. 4x
IZ}
In this case t' R is the product ofa sinusoidal function and a linear function of t
fi)cosr —
CR t
.
(3)
Indeed (Exercise 5), except for a relatively short initiai period, the cosine Factor in (3) varies rapidly compared to t, and the maxima Of Ci (as functions of time) occur if t = t,,, where
gr H
^
^
4x
^
n
r2n,
(),
1, 2,
-•-•
(4)
When (2) holds, then, the maximum amplitudes do, in fact, increase linearly: max CR
x
!
X
/) _r<x
,ri
' 0,1, 2, •
()
Time, r(seç1 14
15
25
FIGURE 9.4. 1aPc heights as a function of rime, recorded at various fixed slolions. Q = — fi, R = } fr, H =re 2.3 ft. 'Redrawn with permission from Fig. 5 of J. E. Prins, Trans. Amer. Geophys. Union 39, 865'874 (1958). copyrighted by American Geophysical Union.]
Group Speed und Group I{frlorir y[C'h.
10
40
r
4
5o
^
rt
hr depth H) versus dimensionless time r ,yfhf, Corrrpurixem of 14ran:r'r Keller (op eft ) ewfrerlareor'rs with measured values of J. Prins," Water Wares Due to a Local Disturbance," F i C+ tr RlE 9.5. Values of the ware hetirthr ti (dirided
Proc. 6th Conc. Coastal Eng. 1Eng. Council Warr Res.), 1958. (f!
R Jt, N = 2} fit, observation at x = 35 ft.! [Redrawn from R. Wiekrel er nl,, "Warrr Wares Generated hy landslides in Recserru}r'rs.° ilnirersiry oJ California ff rdreur#e• Bnrtirrerrtrrrt Laboratory. Berkeley. Calif.. Wave Res. Project Tech. Rcpt. HEL 19 -1, Apr. 196c, Fig. 10: with permission,'
The approximation (2) is valid while r i is less than about 4x 2My ?. Using R 10 cm, q = 10 i cm/s 2 : for the first observation post (x = 150 ern) this condition is t <
10 s. Indeed, at this post the surface deflection maxima 10 s aller the start of the motion. The cease to increase linearly around other posts arty, respectively, three, five, and seven times as far from the origin and the linear increase ceases after around 3 v/10, 5.110, and 7 10 s, as predicted. Caution is necessary in using (1.22)10 predict the wave deflection at the very leadingnbost portion of the wave (see below). Nevertheless, Figure 9.4 shows that this portion does move very nearly at the speed ‘4:11, in accord with the remarks made ai the end of Section 9.1. Figure 9.5 {from a report by R. Wiegel. E. Noda. E. Kuba, D. Gee, and Tornbergi gives a detailed comparison between certain observations of G. Prins and a finite depth calculation by H. C. Kramer and J. B. K elicit (see
• "VYai er WOWS C'rencrarcd by LandsUdes in Reservoirs. - tlrxiuersiiy of Ca1ifomut Hycl raui ic Engineering Lahoraiory, Berkeley, Calif.. Irate Res Proper: Tech. Repi i1£L 19-1. Apr 1966 t "Watcr Waves Produced by Exptoxtons," J_ Appt, Phi;;ee s 3I1. 398 4071i959).
5er. 9.2J
Experiments ravi
Prrrr -1rra1
,ipplrr -rrtrrrn_ti
387
Exercise 44. It is seen that there is excellent agreement between experiment and the simplest Kranzer Keller theory except near the leading portion of the
wave (small values ois ). As t iOf drops below 20 units, t he sircnplest stationary-phase approximation becomes inappropriate, as anticipated in Section 9.1, because 0)"(a) is approaching zero. When o fl, the stationary-phase calculations can he modified without much difficulty (Exercise A9.1-5). and the resulting point is shown as a large circle on the figure_ The points marked "Ursa] theory" involve the Airy function approximations mentioned at the end of Section 9.1 as necessary to describe the transition between the oscillatory and smooth portionsof the freesurface_ There isclosecorrespondence between experiment and the appropriate theory. The good agreement between theory and experiment marks the end of a long road. Had we known that our theoretical work would have required as many simplifications as it did, we might have despaired before we started. Even leaving aside our treatment of water as a continuum and our positing of a stress tensor, there were a number of later and less basic assumptions of challengeable validity. These included our suppositions that the fluid could he treated as inviscid, that the motion could be treated as two-dimcnsional,and that a long enough time had elapsed since the release of the rectangular bump to permit use of formulas simplified by the method of stationary phase. More than superficial discussion of why the neglect of viscosity is appropriate draws on experience of the extensive and continuing work of this century on the subtle effects of small viscosity. As discussed iri Chapter 3, a profound difference between an inviscid fluid and a viscous one is that only the former can slip freely past solid boundaries_ Fluids of small viscosity usually adjust to the "no -slip" condition in a narrow region near such boundaries. Water waves suffer relatively small viscous effects because most of the motion takes place near the surface, where there is no constraining solid boundary (although there is a boundary layer). The bottom boundary has relatively little effect because motion decreases so rapidly with depth. Hecause adjustment to no-slip conditions is confined to thin "boundary layers," side-wall effects are negligible unless the walls are quite close together. In particular. retardation of the flow at the side walls is slight enough so that the initial two-dimensional character of the motion is scarcely altered as the waves travel down the channel. One should always remember, however, that small effects can give rise to a large cumulative influence if they act over a long time. In this instance, no matter how wide the channel. viscous retardation caused by its side walls will eventually have an important effect if the channel is long enough- t detailed discussion of the influence of viscosity on water waves due to an initial disturbance can be found in a paper by I Miles.*) Moreover. certain interesting features of the
• "The Cauchy- Poisson Problem for
Vtscqus
J Fluid lvlerh_ 34.159 70 (19651.
366
Group Speed arrd Group Vrk riry [Ch.
9
flow (such as mass transport) are second order; these may he strongly influenced by a small amount of viscosity. It may seem surprising that agreement was obtained between our asymptotic results and observations made at a station merely 1.5 rn from theplane of symmetry, just two or three seconds after the motion started. But to see whether this agreement should have been anticipated, we must measure distance and time with respect to intrinsic scales (see Section 6.3 of l). In the present case. an appropriate intrinsic length is R and an appropriate intrinsic time is Q{g. Typically_ R 0.1 s Thus our asymptotic 10 cm, v results are accurate for measurements that are made about 15 intrinsic length units and 25 intrinsic time units after the start of the motion. In retrospect at least, this is not surprising. The detailed initial conditions have ceased to dominate in a region that is many disturbance widths from the initial bump. Also much time has clasped compared with the time that it takes a freely falling particle to move from rest a distance equal to the initial bump height. -
PRACTICAI. APPLICATION OF THEORY
In the present context, our investigation of water waves finds its major justification in the theoretical development and experimental verification of general concepts and points of view (linearization, superposition, stationary phase. group speed) whose applicability extends far beyond the particular problem that sve have analyzed. It Should also be mentioned. however. that the study of water Waves is being actively pursued for its own sake because of its applicability to a number of scientific and engineering problems. To give some examples, Stoker (1957, Chap. I I)discusses the numerical prediction of the passage of a flood wave down the Ohio River and through its junction with the Mississippi_ G_ Gadd. reviews the power and limitations of existing water wave theory as applied to the problem of ship design_ W_ Van Dornt uses theoretical considerations very similar to those we have been discussing to supply evidence for his suggestion that gravity waves in a iiquidlike layer may be responsible for certain striking annular mountain rings on the moon. Van Dorn also shows how water wave theory is being used to help understand the often destructive oceanic gravity wave systems caused by earthquakes or violent volcanic eruptions_iï (Such waves are commonly called • "Understanding Ship Resistance Mathematically," 1. last- Muth. Appi. 4, 43-57 (1968). " Tsunamis on the Moon," Nature 220. 1102-1I07 (1968). § "Tsunamis," Contemporary Physics 9, 145-455 (1.968). ¶ It illustrates the subtleties involved in taring w direct one's scientific work according to sociuL criteria that the same theoretical considerations have been used to study the effects of water waves caused by nuclear explosions. (See W_ Van Dorn, "Soule Characteristics of Surface Gravity Waves in the Seat produced by Nuclear Explosions." 1. Geophys. Res. 66, 3645--3862 (1961).
Sec. 9.2J Experiments and Proctwal Applications
389
"tidal waves," but the Japanese word tsunami is now preferred because it avoids incorrect associations with ordinary tides.) Even a sketchy treatment of these applications would require more space than is available, but since the tsunami problem is closely allied in many respects to the Prins experiments, we can give something of its flavor in a few sentences. [Wiegel et al. (op. cit.) show that Prins's experiments and the associated theory are also relevant to waves due to landslides in reservoirs, as in the 1963 Vaiont Valley disaster in Italy] It would be advantageous to be able to correlate the observed tsunami wave form with the seismic disturbance that produced it. Some day this might allow prediction of pessibl4 harmful wave motion from a knowledge of the nature of the earthquak e Now, as often as not, one works backward from observed wave forms to supplement our inadequate understanding of earthquake mechanisms, It is easier to obtain data on the waws than on the initiating disturbance. so there have been several attempts to predict features of an initial water surface displacement that would give rise to existing records of wave passage_ The mathematical problem is thus the inverse of the one needed to analyze the experiments of Prins_ The inverse problem presents no difficulty because of the symmetric relation between functions and their Fourier transforms, as in (8.3.4a) and (83.7). A rough estimate of the source size can readily be obtained by ascertaining which frequencies are dominant in the Fourier transform of the initial amplitude (sec Exercise 8.3.20). As soon as detailed information of any kind is sought, however, complications ensue. Among them are the following: L Radial symmetry is a more appropriate assumption that onedimensionality. The required extension of the theory can be found in the paper by Kranzer and Keller (op_ cif). 2. Tsunamis are caused by deformation of the bottom, not displacement of the free surface. Furthermore, the bottom deforms ever a finite length of time. These and similar matters are dealt with in a comprehensive paper by K. Kajiura.' 3. Typically, waves caused by an Alaskan earthquake are pleasured at Wake Island, almost 4000 kilometers away in the Pacific. Thus the curvature of the earth must be taken into account.' 4. There are effects of depth nonuniformity. To deal with these, it is necessary to take account of the fact that in two-dimensional dispersion not onlydo phasesand wavelengths travel at different speeds, but they move along • "The Leading Wave of a Tsunami, - 8ulf. Earthquake Res_ Iris! 41, 535-71(1963). # See W. Van Dorn, . The source Motion rat the Tsrts+amr of March 9, 1957. as Deduced Crum Wave Measurements al Wake Island,'' Iniern. Union Geodesy Geeophys. Prvr- Tsunami Meetirirrs. Tenth Pacific Science Congress, Univ. Hawaii. 39 48(1963) -
-
Group Speed end Group Velocity ECh.
39D
9
different paths. It is fortunate that the main features of the earliest waves turn out to be insensitive to details of the source that produced them.* 5. Even in the open ocean, the tsunami wave spectrum is overlaid with the spectrum of other waves, such as the tides and the wind-generated swell radiating from storm centers. (Fortunately, the energy peak of the tsunami is at a frequency intermediate to those of tides and swell.) 6. Wave-recording instruments are located near islands whose presence causes diffraction of the tsunami waves and a generation of near-shore waves due to interaction between shore and swell (surfbeat). In summary, earthquakes often induce rapid subsidence of submerged masses whose wave-generating effect is similar to that induced by the surface depression used by Prins to generate waves in the laboratory. This fact links our theoretical considerations and both controlled experiment and natural phenomena. Theory and experiment provide a basis for understanding the natural phenomena, but much remains to be done before the situation can be regarded as satisfactory.
EXERCISES
$1. Show that the "semi-infinite" setup of Figure 9.2 will give the same right-moving wave as would be obtained from the release ofa rectangular bump of height Q and width 2R which was initially placed in an "infinite" channel. 2. (a) Derive a relation between the length of sinusoidal water waves, their temporal period P, and the mean water depth H. (Ignore surface tension.) (b) Show that in the Prins experiments discussed here, finite depth effects are expected w become important if the value of P is greater than a value of magnitude unity. 3. $(a) The insets in Figure 9.3(a) and (b) show typical amplitude traces for an initial elevation and for an initial depression of the same small magnitude_ What is the salient feature of the results? Show that it is in accord with theory (b) lise Figure 9.4 to provide further quantitative checks of (!_ l2). $4. Derive the following result of Kranzer and Keller for waves in a fluid of mean depth H, when xis large but less than r, ytH (C x 0for x Ignore surface tension.
=
V 04 (Jo E{!G) sin (w# kx— xP,
i gH).
^
*W. Van Dorn, "Explosion-Generated Wives in Water of Variable Depth ‘ " J. Marine Res. 22, 123--41 (1964) .
Sec.
9.3] A
Kinematic Approach to Group Velocity
Here k is given in terms of x and expressed by
39 1
by x Vl where the group speed V is
t
,
V = ^^ I ^, 2kH ^
sinh 2kH
and the phase speed C is given by C2 = k
tank kH.
E(k) is the Fourier cosine transform of the initial displacement: E(k) = --
t2 4
.
C(x, 0) cos kx dx.
Show that the approximation (5) is appropriate for the circumstances in which it was used.
9.3 A Kinematic Approach to Croup Velocity Group speed arose in Section 9.1 from an interpretation of asymptotic expansions of the Fourier integrals that give the Surface displacement in water waves. Using the method of stationary phase, we were able toshow that after a sufficiently long time a localized initial disturbance of the water surface will become transformed into a slowly varying wave train_ The concept of group speeds' emerged from our efforts to provide a careful description of the wave train's behavior. It is often the case that after a striking result appears, an investigator is able to demonstrate how that result can be obtained by arguments considerably simpler than those originally used to derive it. Such demonstrations are of great value, for in revealing the essential aspects of a result they deepen one's understanding of it and hence show the way to its wider application. In this section we shall outline a simplified approach to group velocity (a generalization of group speed) based closely on two papers of G. Whitliam.* [The basic idea appears in Landau and Lifschitz (1959).) This approach is termed kinematic because it stems almost entirely from the assumption that the surface is a slowly varying wave train_ For several wavelengths, such a wave train appears to be composed of waves having a certain wave number and frequency. It is natural to assume that the same dispersion relation holds between this local (approximate) wave number and frequency as holds for plane waves having exactly this wave number and frequency. Dynamical • "A Note un Group Velocity,"
J. Fluid Mech. 9,
Energy Propagation sur Three-Din cnsionat Waves," (1961)_ ALternativciy, see Whitharn 0974).
347-57 (!96,O); "Group Vtiocity end Pure Appl Mr1t1]. 14, 675-91
commuer
Gro up Speed and Group Yelnciiy (Ch.
392
9
considerations never explicitly enter, therefore, because results are expressed in terms of an unspecified functional relation between wave number and frequency. The precise relation required by the dynamics of a particular physical context will be seen merely to alter certain details. Moreover, it will be evident that our results are not at all restricted to water waves, hut rather apply generally to dispersive wave systems. PROPERTIES OF SLOWLY VARYING WAVE TRAINS
We wish to make certain general observations about possible forms of the free surface deflection 1(x, t). In the present context, C can depend only on the two horizontal spatial coordinates x i and x 1 ; i.e-, x (x 1 , x x ). It will hecorrre clear, however, that our remarks are readily generalizable to situations (e.g., involving electromagnetic or elastic waves) where the relevant dependent variables depend on all three coordinates. It is convenient to write C as the real part of a complex function and to write the complex function in polar form with argument c
z)
R
e A(x, t)
exp [0(x,
r)] _ A(x, t)
cos [0(x, r]].
(l)
Here A and 4 are real functions. Suppose that A is a constant, a, and 4 is a linear function. Then for some vector constant k and some scalar constants riS and 4 0 , we can write = acos(k x - tut - Oa.
(2)
-
By now familiar reasoning, one can show that the C of (I ) is a plane wave with wave-number vector k and frequency to. The wave propagates in the direction ofkatspeed cnfor, whererx = JItIWe shall now demonstrate that if A is" nearly" a constant and 0 is"nearly" a linear function, then C is nearly a plane wave (which is no surprise) and we shall find formulas for the wave number and frequency of this wave, To accomplish this, we expand the functions in (2) about a fixed point x 1°1 at a fixed time t' ° ': A(x, t) = A(x t », t tof l + - - - , fix, r) = ÿ}(x'°', r t") f y4,( x iv/. t tai } . (x _ x toi} +
4p
tnl ` „olio — ea)) j- . -.-
Suppose that the omitted terms in the above equations are negligible_ Then C : Re r4(xt ° ', r'° ') exp i[V4,(xt °,, rt"'}- x + 4,(x j° ', i' ° ')r + constant]-
(3)
Referring to (2), we see that the solution near (x' ° ', rich ') can be regarded as virtually a plane wave with wave number vector V lx ;d_ r ich ') and frequency 0,(x 10', t"'") For this interpretation to be meaningful. the approximation (3) must he valid over a distance of the order of the local spatial period and a time of the order of the local temporal period. That is. 13) must hold for
Sec- 9.3] A Kinematic Approach ro Group Velocity
roughly the following range of x and 2iE
— x(D)1
ix
i
(xt°l, tto
t I^
t,
393
at least:
2a
I t _ t tor ^ <
(4)
M001)1, 11(») I
We now wish to consider [he function C of (1) in those cases wherein the omitted terms in the expansions are negligible for at least [he range of x, t indicated in {4), and for ail x o and to in the domain of interest_ (hi the present context, the previous sentence provides a reasonably precise characterization of a "nearly constant" function A and a "nearly linear" function gyp.) Let us define a vector function k(x, t) and a scalar function w(x, t)as follows: k(x, r) = V (x,
t),
r(x, t J = — ,(x, t).
Under the conditienS cited, our discussion shows that k can be regarded as the local wave number of a slowly varying wave train, and c:a can be regarded as the local frequency. That is, at around the time t and near the point x, the function C looks approximately like a wave of wave number k(x, r) and frequency w(x, I)_ Since C x U cos 0, the loci of adjacent crests (maxima) are approximately given by 4(x, t) 2rrn, ri = 0, ± 1, +2, .. - - Troughs (minima) lie approximately along ¢(x, t) = if + 2an, n = p, ± 1, + 2, .. - . Crests and [roughs are the extreme phase-s of a wave; thus 4 is called the phase function.• Bear in mind that a unit change in phase corresponds to passage through a complete wavelength. THE PHASE FUNCTION IN REGIONS WITH AN UNVARYING NUMBER OF WAVES
We now present an alternative way to introduce the phase function. To this end, in the domain R under consideration suppose that wave crests lie along certain smoothly varying curves, and that these curves do not terminate within the domain, nor do new crests appear or disappear duringthe course of time. At a given instant t, let us denote the loci of the crests by 45(x, tl = 2nn, n = 0, + I, - - - - Moreover, let us denote the trough loci by (x, t) = tt + 21crt, 'ï= C,±1, have sufficiently small curvature We assume that the curves 0 _ 0, + It, so that intermediate values of 0 can be smoothly tilled in by erecting normals to 0 = nn and interpolating. (see Figure 9-6). We further assume that the two-term Taylor approximation ...
0(x,
r)
0(x 0 , r) + V (x t,, t)• (x — x o )
is a good one, at least when x is within a wavelength of x o . By this second assumption, the curves -= constant are nearly straight lines- Indeed, our * 1 i tnuy }ae convenient to regard the solution as being composed of the super paslir011 of more than unc set of waves, each set having its awn phase function Ship wares provide an example -set Section 9.4 -
Group Speed and Group Velocity [Cll. 9
394
mom ^^^
.
-
■
-
-
^= '_a{Ir ^
=2rin-
4
-
^}
2 r 4 n - 11
Ftc u tt r 9.6. Curves rtOt. t) = constant denote crests (heal y Tines) and
troughs (light lines). The phase function 4i differs by 2rr between adjacent Crests and between adjacent troughs- Intermediate MATS 0J- 4 are obtained by erecting normals to crests and interpolating according to the distance to the nearest trough. Shown are points on the dashed curve Ib = 2 n (rt - j . these points are halfway between the crest cti -- 2 nn and the trough 42 =Nato Wang normals ro 4 = 2nn. —
},
assumptions are clearly fulfilled by a regular wave train with equally spaced straight crests and troughs— they will also be fulfilled if the crestsand troughs are "nearly" equally spaced and "nearly " straight. There is a degree of arbitrariness in defining between crests and troughs. In fact, "fuzziness" is present throughout the discussion in this section. Imprecision in statement and lack of an obvious path to rigor is the price we pay for quick and physically significant results. Of course, we have already seen that the method of stationary phase provides one way to obtain these results that is more satisfactory from a strictly mathematical point of view. Let us consider a point x 0 on = 2rui, and another point x, that is located where the normal to 4 = 27rn at x 0 intersects ¢ = 2rr(n + 1) (Figure 9.7). Using the two-term Taylor approximation, we can write
❑ 4(x,,, ta-(x, — x o) = IV4)(xo , r)IIx,
4)(x„ r) — (xra, r}
But 0(x L , t) — 40 0 . 1) = 2tt(rl + I)
—
2irn = 27r.
Also lx, — x o ^ the distance between adjacent crests, can be identified with the length L of the wave. Thus I Vi I 274. Of course, Ÿ4)(x u , t) is perpendicular to the curve 4)(x 0 , t) = ri. We can therefore write ,
(5a)
Prtl(x, t) = k(x, tj,
-'wjn + t 1
-'arrr
r^ FIGUttt 9-2- Tuxt points an the normal in points are rJ watelertgth apart
th e crest iti{x, s) = 2rin.
These
Group :''Fi[lr it}
9.3]
I
r
3g5
4 -- ?Rirr + I
t=tct^r^r 9.8_ Norrrull.s V0 to efiurr.x
= constant these i -e( tors pooh! ,A way rnutiraaf os Mr direction of Vo. with suCCes.sirrll• owner retluer. rrJ 0 will pan MI' 1r+.1.41
die direction r oe increasing
then (Teal ct_ssuc7Qtt'd po int Igo .
T'Irir.►•.
rl
where k(.r, t) is the Jocai wave number rector in that its direction is normal to the wave crests and 12nc) - times its magnitude approximates the number of waves per unit length. Suppose that one records values of ç at some fixed point x o . Each passage of a complete wave over x o is associated with a change of 2ir in 0. Since ❑e points in the direction of increasing 0, a decrease of 2n in et) corresponds to the passage of waves moving in the direction of V4 (Figure 9.8). Thus -
[O[xi,,ta) — 004. 1 01'2n
is the average rate, over the time interval (t t , t , ) at which waves pass over x0 in the direction of k V_ Consequently, we nay write .
-
cu(x. 1) =
—
Or(x, f),
(5b1
where (eix, r) is [he local frequency in the sense that (2 1- tw gives the instantaneous rate at w hich wave crests are crossing a given point x at time t in the direction of k(x, r). IrYTEG R AI. AND DIFFERENTIAI. EXPRESSIONS OE W AV E CONSERVATION
From (5a1 it fellows that if f` is any smooth curve that joins x o to x,, then
J'
&cx. t)- dr =0lx,, r)
0(x o , r}.
(6)
Group Speed and Group Velocity [Ch. 9
396
This is another expression of wave conservation, for it implies that the number of waves encountered in passing from x o to x, at time t, regardless of the path traversed, is c(x t , r) — ç (x u , r) (also see Exercise 2). Assuming that k(x, r) and w(x, r) possess continuous partial derivatives, we have, from (5a) and (5b), that
^k + Vrn = 0,
(7)
which is a differential equation expressing wave conservation_ For if we integrate (7) along any path C passing from x0 to x i , we have r^r
ck dr = — co(x1) +
x o).
Thus no waves are created between x 0 and x r , because the rate of increase of the number of waves between these points is equal to the net rate at which existing waves cross inward across the ends of any curve joining the points. GROUP AND PHAS E VELOCITY
We are considering problems wherein the governing equations admit exact plane wave solutions whose wave number and frequency satisfy the dispersion relation (8) w = ctr{k). We shall assume that (8) can be used to relate the local frequency ut(x, r) to the local wave number k(x, rl_ Consequently. we can infer from 17) (in two spatial dimensions) that
v x
dr +
}
—
0,
where
V
ak y
{i,}—• 1,2).
(9)
1~ rom (5a), curl k=0
8k1 8k ; or -= ,
ï= = 1,2.
Consequently, (9) can be put into the form c'k
+
(V • V}k = 0,
where V = (kt
, 1
;2).
( 1 0)
We wish to interpret (IQ)_' To this end, consider an observer moving on a path x = x(r) with speed V, so that
The interprcrarion is w•irtua&Ly immediate for those who recognize BD) as staring that the material derivative of k vanishes.
See. 9.3] 4 Kinematic Approach w Croup Velocity
397
Let us compute how k changes with time for such an observer. We have dk[x(r), r] =
dt
Vk
. dx cfr
ck cl'
(12)
It follows from (10) that
ri'k _ D (13)
dt
if (1 1) holds. Accordingly, an observer moving with the group velocity V will continually observe waves of wave number k and frequency 1.90). That is the wave number k and frequency c11(k) propagate with the constant group velocity V(k) - Vw(k). This relatively simple line of reasoning not only has rederived a principal result of Section 9.1 but has extended it to waves in more than one spatial dimension, (See Exercise 4 for extension to slightly nonuniform media_) We turn to a higher dimensional generalization of the concept of "phase speed" that we discussed in Section 9.1_ We have seen that curves of constant phase are given by rP(x, 1) = constant_ Such curves move normal to themselves with the phase velocity a given by -
c^
----o-r^. I^4l
(1 4 )
where k is the unit vector kla. [Equation (7.1.5) gives the speed of a surface normal to itself. Recall that tc = (kl.] Lines everywhere orthogonal to curves of constant phase are called rays. Unlike the group lines given by (II). rays are generally not straight Furthermore, the ray through a point generally dries not have the same direction as the group line through that point (Exercise 3L Note that introduction of (5a) and l5h) into (14) gives e = (e), yik. Keeping in mind the remark under (Sa) that 7j2rr provides the local number of waves per unit length. we observe that this last formula is completely in accord with our original notion of phase speed. ENERGY PROPACATVON
Keeping these general considerations in mind, we turn again to water waves whose amplitude is a function of single spatial variable v. Formula (8.12d) for the surface displacement C can be written compactly as = Re 1. where (Exercise 8_119)*
F(g) exp [[(ax — tut)) d a. .J
F(u)
A(u) + Fi-3(a).
(15)
a
* We return 3o Dur earlier use of r. bu i in une- dirnrnsional prohLc-ms this it €onsi(teni with our Later definition x - Ik
e_
Group Speed and Group Velocity [Cil
398
,
9
The complex amplitude function F(a) is determined from the initial conditions by (8.3.7) and (8.3.8)_ From Exercise 8.119, alternative formulas for the dimensionless energy in the waves (total energy divided by pq 4 2 , where A is a typical amplitude) are
a
^
. Ix(x, 01 1 c^x
(16a)
and
n f JF(a)1 2 da.
(16b)
Equation ( 16b) shows that the energy associated with wave numbers between a and a + da has the constant value 70F(a)1 2 du as time increases_ After a time, localized initial disturbances break up into a train of slowly varying waves in which each value of a can instantaneously be associated with a point in space; i_e_, a< = a(x, r). We have seen that wave numbers propagate with the group velocity; we now see that energy density propagates wirh the group ielaci y, because each wave number is associated with a rased energy density_ The casual observer of water waves keeps his eye on a crest or trough— as we have seen, these features travel with the phase velocity. But the spread of energy is generally more fundamental than the propagation of superficial features. For example, it is the rate at which wave energy flows through a control surface that determines the drag of a ship. It is the speed of arrival of energy in waves of certain intervals of lengths that determines the damage done by certain explosions or earthquakes. In addition, the "radiation condition" for water waves is that the "energy velocity," Le, the group velocity, should always be directed away from the source.• Thus considerations of energy propagation generate many applications of the group velocity concept. ASYMPTOTIC FORM OF THE SURFACE: A TERSE DERIVATION
We now show that relatively simple general considerations yield essentially the same basic asymptotic formula that was obtained earlier by applying the method of stationary phase to a Fourier integral representation of the solution. Consider a segment of a wave train whose wave numbers vary between a and a + Aa. The position of the ends of this segment will be given by V(ar}r + x a and x + Ax _ Via + Aar}1 + x i ,
(17)
where Ax{r) is the length of the segment and the constants x o and x L are chosen so that x(I Q ) and x(r0 ) + Ax(t o ) coincide with the ends of the segment at some time t o after the wave train has come into being. The quantity The radiuIion condition is discussed in Sections 12.2 and I6_3 of 1. Application of this condition in terms of group velocity is rSdc in Seliun 9.4. '
Sec. 9.31
A K incmprir Approach to Grow Velocity
399
V(a) w'(a) is of course the group speed associated with wave number a. If Ax and ❑rx are small enough so that jyI and IF( are roughly constant in the intervals (x, x -+- Ax) and (a. a + Au), respectively then, from (16) and (17), the dimensionless energy in the wave segment under consideration can be given either by I](x. t)I'Ax or by it IF(ar)J'Aa. Equating these expressions, we find that
Ix( = I^ij2zr
^^ -
But, from (17), ❑
V'(a)i(&a) + x i -- x o .
-x
The constant x i — x o is negligible for large time_ Hence, combining the two preceding equations, we find that
Ixi ;r-
211
1 181
—icig
Now C = Re x = I x( cos (arg XISince waves in the segment under consideration have wave number and frequency approximately given by a and wia), respectively, we can write (for right-moving waves) cos (arg x) = cos (arx — rnr - 0 0 1,
00 a constant.
Using (18), we see that for such waves C
w"(or}t
f F(rx)l cos (ax -- cur + O n ).
This coincides with the results obtained in Section 9.1 using the method of stationary phase, except that an explicit expression for the phase 00 was obtained in the earlier derivation.* Our results that energy travels with group velocity and that the disturbance modulus obeys (I8) for large time are not restricted to wafer wanes. They hold for any complex wave function x given by a Fourier integral like (15), if the "energy" density is identified with the square of the amplitude Ix(x, f) I x. The arguments of this section are far-reaching and relatively simple. Consequently, the reader may feel that it was not worth his while to go through the more complicated earlier discussion involving the method of stationary phase. We believe that those discussions are worthwhile, however. • in
Section 9.1 we considered the case B = O. so In becomes I A 1- The absolute-value Sign can be removed [rum A. to give caaCLLy our earlier result (1-20)—fora passible minus sign can be absorbed into 00.
400
Group Speed and Group Velocity
[Ch. 9
and not only because the methods used have many other applications. The arguments involving the stationary-phase method are better able to hold up under critical scrutiny than those just presented. Indeed, the reasoning of this section would not be easily credible without the background of the explicit stationary-phase calculations. On the other hand, the present arguments are not restricted to problems for which standard Fourier analysis is appropriate. For example, Exercise 4 indicates how these arguments apply to inhomogeneous media. We remark, in addition, that direct use of the point of view taken here—that the asymptotic solution takes the form of waves of slowly varying amplitude and phase—is the starting point of much modern work on nonlinear wave theory. EXERCISES
I. #(a) Derive a, + cox = {l the one-dimensional version of (7), directly from a ware-conservation requirement. (b) Derive (7) by appropriate extension of the argument in (a). (c) In the equation of (a), introduce an expression involving phase velocity instead of ci. Point out the strong analogy with ordinary conservation laws_ 12. Let k(x, t) be the vector wave number of two-dimensional waves. Show that if waves are conserved, then there exists a function 4 such that k = VOL 3. 1.(a) Show that the direction of the group velocity V is the same as that of the phase velocity c if w depends only on the magnitude of k, not its direction. (h) Show the converse. Proceed as follows (Whitham, 1960, op. cit.}: Assume that ,
C. to
ek ;
- y(k,, k 3 , k 3 )k ; ,
i - 1.2.3.
(19)
Show that for all , and f. t # j,
a 20) dkt ek
_
ag ck, k '
r7^ ^
so
ak, k "
eg _
a^
a (k?} ! c?(k1) '
Thus if
x
+k2 +kj- a l,
y
,
show that
Y
n ,
soy is a function only of a l . Now integrate (19).
r = k3 ,
Sec. 9.4)
4.
(a)
Ship. Duck, and Belle Wares
401
For slightly inhomogeneaus media, the argument of the [ext can
be adapted by assuming that the relation between local wave number and frequency depends on position: W = c k x). ,
Deduce from (7) and the irrotationality of k that
, e^ + .-- =— i^x ^
Oki at
(20 )
'
and thus k can in principle be determined since its rate of change is known for a point traveling with group velocity. Show that (?al is equivalent to Hamilton's equations dk,
dx; eW
41(Ll r,x+
Lit
,
di
c'k '
t(b) Equation (20 shows that the wave number no longer travels with the group velocity in slightly inhomogeneous media. Show that the frequency still travels wish group velocity_ S, For waves in n spatial dimensions, (15) generalizes to x(x, r) =
J Fk :
explkx
—
ru ) dk,
where dk (kind dx below) denotes an xi-dimensional tiOIurne element Show that dx = (270"
SI j! (k)1' u^k
27[ ar 2
xW
and that (IS) is replaced by
Ix1
= IFI
--
del
Ces ['ii1
9.4 Ship, Duck, and Beetle Waves Ship waves provide an excellent opportunity [o illustrate the clarifying power of the concepts we have been discussing. Sommerfeld's (1950) description of the phenomenon is characteristically elegant. "The wave pattern that is left behind by a ship at sea consists of a system of waves that envelops the hull lengthwise and is interwoven with a system of cross waves. The two systems advance with the boat so as to be stationary relative to it The beauty of the pattern is most impressive when viewed from an airplane or from the top of a high cliff, but the same phenomenon on a more modest scale develops behind a duck swimming in a pond." Indeed, we shall also
Group Speed and Group Velocity [Ch_ V
402
(J_(, i
YZ
^p
« x^
• "J
obi
irrexre s ysrFrrts. { a] Shure based coordinates. where F r G 4r fit II 9.9 Two coordinate Al and wa r es more kith unÿerrrxt speed L1 in the rlrrre•riort of the nte!gs!lrr4• axis. (b) Ship bused coordinates, where strip and wares are stationary and distant water mores in the direction of rhe positive x l axrs with speed U. -
-
-
have reason to give brief consideration to the surface wave patterns generated by aquatic creatures. The classical treatment of ship Waves considers the continuous emission of radial waves caused by a point impulse moving with constant velocity over the surface of the water. The method of stationary phase is used to simplify the appropriate Fourier-Bessel integral, much as we have used this method in Section 9.1. (In fact, toward the end of the nineteenth century, Kelvin developed the method of stationary phase to deal with this very problem.) For full accounts of the classical approach, see Lamb (1932), Sommerfeld (1950), or Sicker (1957)_ We shall provide a kinematic treatment of the problem that utilizes the general considerations of the first half of the preceding section.• Although perhaps less firmly based, [he kinematic treatment is considerably simpler than the classical one, and it more clearly exposes the physical ideas involved. Let x" ' denote [he coordinates of a point P measured with respect to a coordinate system fixed on the "shore," Suppose that a ship moves into still water with uniform velocity U in the direction of the negative x; r-axis [Figure 9.9(a)]. Let x denote the coordinates of P measured with respect to another coordinate system fixed with respect to the ship [Figure 9.9(b)]. Then x=
4- UI-
(I)
Consider plane waves that, when observed in the shore-based x {° " system, appear to have wave number k, frequency w10 ", and hence phase velocity c" ° " J [w ° rfa]k. [As in previous sections, a = Jk I, and k is Che unit vector kfa.] Let the amplitude C of these waves he
C(x, t) = Re b exp (ik. x" ° r * Our exposition follows that of Whitharn (1%1, O. Col and E_ ll.lrscll. "Steady Wave Yaattcrns on a Nonuniform Steady Flutd Flow,.. J Fluid Minch. 9,333 46 (1960). -
Sec. 9.4] Ship, Duck. and Beetle Waves
403
Application of the transformation (I) shows that in the x system C(x, t) = Re
b exp Iik • (x — Ur ) — rcu'"'r]
Re h exp [ik• x — qr./Pc" + k• UP"
(2)
Consequently, viewed from the ship the waves appear to he of a "Dopplershifted" frequency w given by Uit 0 " } k -
(3)
-
Viewed from the ship the phase velocity c = c•k is thus given by e= a - i colot+ k.U)k= en" +(k, Lj)
14)
Using (3), we can find the relationship between the group velocity V viewed from the ship and its shore-based counterpart Vt ° s: acust"
Ok i =
ak * u..
so V — V"
+ U.
(5)
r
Neglecting the effects of surface tension and of depth, we know from earlier work (Section 9-1, Example 3) that
c fü^ _ k 9^^ ,
V ^vs ^ } clos
(6a, h, e)
CONSEQUENCES OE STEADY MOTION
The fact that motion is steady when viewed from the ship has a number of important consequences. First of all, the wave height must be independent of time, so that, from (2),
cotas + k' R.1 = 0 or ie roM
4 Ul k = - 0_ -
( 7 )
Formally, (1) is equivalent to setting at = 0 in (3). We know that wave groups with a wave number of .ti k travel through x — x — r space with group velocity V. Under conditions of steady motion, an unvarying pattern of waves will be seen in the (x t . x2)-plane. What does the role of group velocity imply for this pattern? To answer this question, it is advantageous to bring out the role of k. and & z . the components of k, by introducing slightly different notation. Thus lei us denote the dependence of frequency ai on wave number k as follows [sec (3)] ;
= f (k,, k ).
(S}
The components of the group velocity vector are then given by akt .
1 = 1, 2.
(9)
404
Group Speed and Group Velocity [Ch_
9
Let [x 1 (f), x 2 ()J denote the coordinates at lime t of a point associated with the local wave number k. Such a point moves with constant velocity V(k), so that
dx r
t?f
dx 3
af
dt
ak r '
dt
c?k2 -
(IC)
Thus wave numbers k always lie on lines in the (x r , x 2 )-plane with slopes dx 2
Of irr k 2
dx,
f)f
#ak 1
As we have stated, in steady motion we must set ro 0. Thus the relation f (k1 , k2)= 0
(12)
connects the components of k. If we consider an arbitrary k 2 , (12) determines the corresponding first component k t ; k r = kt(kr)• *
(13)
From (12) we deduce further that
df dk ,
ak 1 A i
ref + ek, — a.
Consequently, (11) implies that
dx, dx r
dk, dk
(t4)
There is thus a one parameter family of straight characteristic lines, with slope given by (14), on each of which the wave-number vector has the same Pudue [k
,(k 2 ), k 2 ].
STEADY WAVES INDUCED HY A POINT $QLRCE
We shall idealize the ship as a point source of waves that is fixed in an exterior flow of velocity U. Such an idealization should give reasonable results far from the ship. Moreover, since the problem is linear, appropriate superpositions can in principle be used to obtain more accurate results_ We suppose that the point source is steadily generating waves of the whole spectrum of wave numbers and that there are no other sources of waves. Locate the origin of coordinates at the source_ Then there will be a steady wave pattern that is associated with the totality of possible characteristic lines passing through the origin, The field of k vectors can be found as described above. Once this is done, the curves of constant phase 0 = constant can be determined: since k = V , these are everywhere orthogonal to the k's. (See Figure 9.10. It might be conceptually helpful to conceive of the * Equation (L21 lnay not have a unique sululion (I 3). For an example. see below_
See. Vd)
Ship, Durk . und Bi'erfe Wares
405
FIG tr R +,1, l;Q. .g I`nrra i7 characteristic line, the teal e-rilrnrher r'errcrr k rs runt. lhrr3uyh a p r+irrt is cJrtlluyorrui A curi e of ccrnrstü►ii !-.'[ for k awkward ^+ irfr lhur point,
required calculations this way: Find the k associated with each point in the plane. construct ati "infinitesimal" u vector that is perpendicular to each k, then "link " up "adjacent" tits into smooth curves of constant phase.' To find the characteristic lines in the case of ship motion, the appropriate special case of (12) can easily be found from (6a) and (7) to be ply
;
+ k3
-
k i ll=0.
(15)
Here axes have been aftgried, so that U = (U, 0).
As the form of (15) suggests, it turns out to be convenient to proceed by introducing polar coordinates (and to reason geometrically)_ We shall see that there is a lack of uniqueness; part of the (x 1 . _x 3 ) plane is covered by two sets of waves. Now to the details. We introduce polar coordinates to describe the wavenumber vector: -
cos y)j. k—k r i+ k 1 ] = (—a sin x)i+(x Thus ' is the angle between k and the vertical (F igure9. I I ). Now e4" lies along k [by (6b)] and c{0' + U is perpendicular to k [by 17)]. Recalling that the x i -axis is in the direction of U, we are led at once to Figure 9.11 and hence to the observation that
.r a s Sin y =
(17)
f1
With (6b), (17}provides the following explicit formula for the relation between the magnitude a and argument x of a wave - number vector:
U sin x
—
9frr•
I
I $)
Viewed from the ship, wave numbers propagate in the direction of the group velocity V, so this is the direction of the characteristic lines [compare
Group Speed and Group Velocity [ Clt. 4
406
U
FIGURE 9.11- Relations among the wave-number vector
It, the phase velocity
vector ct4 ", and the ship velocity U.
(11)]. Recall from (5) that V = U + Vt°t and from (6c) that VW = 10'1. With this we can modify Figure 9.11 to obtain Figure 9.12. W e deduce that the characteristic lines make an angle of y -- 6 with the x i -axis, where tail
6-
(19)
# tan y.
^
FIG i iu 912. Figure 9.11. with characteristic line added.
Equation (P91 can also be obtained analytically. as Follows. By (14)
the slope of the
characteristic lines is given by l dx,dk _ dx, dk 3
dk,fda d1c2{da
(20)
Using t h e definition of x in (t61 we obtain
dx i* cal
tan x + ax'(a)
I
—
af(a) tan a
•
(21)
Thus if we define b by
tan 6 = -- a 'la),
(22)
Sec. 9.4]
Ship, Durk, and Beetle
Wanes
407
['urtir
L.
[Jr uons[an c
phass
Charrclerouic lone
FIGURE
9_ 13. Po la r coordinates tr. 0) fo r t he rurre n f crr„f[url phase
then dx z l dx
= tan (y — b),
123i
The desired result (19) now follows easily [Exercie 3a)}_ Pursuing the goal of deriving equations for the curves of constant phase, we note that if ri a) and 0(a) are the polar coordinates of a point Q on such a curve (Figure 9.13), then, as we have just seen, 0(a)
a( (a) — 6(a),
where 6
tan - 1 (1 tan y).
(24)
The angle y is given implicitly in terms of a by (18). Hut a standard formula from the theory of polar coordinates states [Exercise 3(b)] that dr dg
r tan â .
(25)
From this one can deduce [Exercise 3(c)] that r—
constant a si n 6
(26)
In (24) and (26) we have polar equations for the lines of constant phase. It is best to regard the curves as parametrized by x, using (18) to express a in terms of y. The principal feature of the curves is deduced from the fact that f}(a) has a maximum f„, when y = y am , where [Exercise 4(a)]
= tan -
= 54.7 ° ,
ti — tan
= 19.5° .
(27)
As y increases from 0 to n12, 9 increases to 6,,, and then decreases to zero_ The constant in (26) changes by 2rt from crest to crest (Section 9.3). When 9 increases and then decreases, two families of curves are swept out, as shown qualitatively in Figure 9.14. Along 6 ï 0,y, the two families meet in a cusp and our approach can no longer be used, since near the cusp it is no longer true that the wave system is slowly varying. There is a breakdown, too, in the usual stationary-phase calculation, but a modified procedure gives the detailed asymptotic behavior of the waves near [i = 6,,.
408
Group Speed arid Group Velocity [Ch. 9
Flo uttE 9.14_ Two sers of wan, crests yerrerakd by a moving [^irrt^lm to Land] (1932), p. 4341.
drsfurbance
That a ship moving with constant velocity gives rise to substantial waves only within a wedge-shaped region could perhaps have been predicted. but it may seem surprising that the wedge angle is independent of speed. Certainly, the situation is quite different for sound waves (Exercise 1). This difference should have been anticipated, however, on dimensional grounds (see I, Section 6.2). There is no length scale connected with a point source moving over infinitely deep water. Thus no dimensionless parameter can be formed in the gravity wave problem (from the relevant parameters gravitational acceleration g, density p. and speed U). It is otherwise with sound waves, for example, where the dimensionless Mach number can be formed from the ratio of U to sound speed. One thus expects the wedge angle to depend on Mach number for sound waves (Exercise I } but to he constant for gravity waves in deep water. Consider situations in which both gravity and capillarity are important. Now a dimensionless parameter can be formed, for (9T f p) " 4 has the dimensions of velocity—indeed, .v/2 times this quantity is the minimum phase speed r,„. Thus Li/c,„ is a dimensionless parameter that is expected to play an important rate in the problem_ Indeed, we have already mentioned the fact that there will he no wave pattern if this parameter is less than unity. If (#I/e.) > 1, on the other hand, there is a wedge with half-angle sin -1 (c,,JU) with the property that capillary waves appear only inside the wedge and gravity waves only outside (Exercise 10). Figure 9.15 provides a splendid illustration of a capillary-gravity wave complex, and thereby illustrates a number of the features that we have been discussing. PARTIAL nIFFERENTiAL EQ[JAIIOry FOR THE PHASE
FLNCT1O
About a decade after the papers of Ursell and Whitham, C. Hunter presented a further refinement of ship wave calculations.* We outline this approach briefly here, for it provides an elegant and useful application of the •"Un thc Calculation uf Wavc Patterns," J. Fluid lblrch, .iJ, 637-35 (1972).
Sec. 9.4]
Ship, Duck. and &I tie 4Fûi .t-s
Vic UR E 9. t3
409
A tHi{j in a_stre•ran , with capillary wares ahead of it and gruriry wares behind it. IFrom V_ A . Tucker. "14/ares an d Winer Beetles." Physics reacher 9. 10 14. 19 119711. Fig_ 3 (Copyright 197L hr American Assoc. Phys. Teachers): with pe rmissron] .
-
theory of first order nonlinear partial differential equations. We ranfinc our exposition to the ship wave problem, but it should be fairly clear that considerable generalization is possible. See Hunter's paper for a fuller exposition. Local wave number components k 1 and k t for steady deep water gravity waves are related to each other by( I 5). As in (3.5a), however. these components are connected to the phase function by the equations
k, =
chip
k2 =
eifi ax e
(2H)
u ^^ = o.
( 29 )
Upon substitution of (28) into (15) we obtain z ^
^^
^ + +(")'
This is a first order, nonlinear partial differential equation. According to the classical theory [Garabedian (1964). Sec_ 21] the partial differential equation F ^ L . ^2. ^, ^ Y`
^
„ ^
=0
(30)
Group Speed and Gray; Velocity [Ch.
410
9
has a solution that can be regarded as composed of infinitesimal characrerisfrc strips. These are space curves bearing infinitesimal surfaces with normal vectors (p i , { 2 ), where p ; 434fOx1 . In terms of a parameter s, the strips are represented by
dx; :
ds
F11 ,
T
as !• t `
p lra1} —
pi = Fx^ + p-F6 ds
t
1, 2. (31a, b, c)
[From an analytic point of view, the original equation (30) is equivalent to the system (31).] In the present case, F is independent of x i , x 2 , and 0. F From (31c)we consequently recover the result that p i = k4 is constant along characteristics. If we employ the phase 0 as a parameter, the remaining equations (31a, b) can be written [Exercise 7(a)] dx U+
3)I12
!
dx2 -3 tft ikr(9a )
d(i) Uk ; +
(32)
Integration is a trivial task, since the denominators are all constants_ The solution curves x, = x 1 (44, x 2 - x2(0), are straight lines, the characteristic lines. As before, we ideaii2e the ship as a steady point emitter of waves, so that we build up the solution from the family of characteristics that pass through the origin. To conform with standard practice, it is convenient here to introduce polar coordinates in a different way from (16) by
k 1 _ — a cos 40, k = — a sin i;
so that U cos 0 _
77. (33a, b, c)
With this, upon eliminating a from (32) one obtains [Exercise 7(b)] 4g x 1 = — U 2 ç (5 cos
►
— cos 31/i),
4gx 2 = U 2 f(sin i + sin 30). (34)
Parametric equations for the phase curves 0 = constant have thus been found. A time t can usefully be introduced, namely, the time that it takes wave number k to reach a given value of 0 as it travels with the group velocity oiawcird• along a characteristic. If ç = 0 at the source, we find [Exercise 7(c)] that =
at cos 0.
(35)
Thus only negative values of 0 are to be considered in (34). Since cos 0 > 0 by (33c), no further matters of principle are involved in deducing the wave pattern. The final result is again Figure 9.14, as was obtained by the previous approach. • The ship musa prcpagaIe energy outward (radiuIion condition),
Sec_ 4_4]
sh ip,
Du ck, and 8r•ealr• WrWFs
41
ExFttCtS E S
1. Suppose that a point moving in a straight line with speed U continually emits disturbances (e.g., sound waves) which travel with uniform speed e. That is, the effect of a disturbance at a point at time i y 0 is confined at later times to a sphere of radius ci centered at that point. If c < U ("supersonic travel"), use a simple geometric argument to show that the disturbance is confined to a cone (Mach cone) and find the half-angle at the apex (Mach angi!e)_ 2. Show that the relation between U and cI ° ' in Figure 9.11 can be deduced from the requirement that in a steady flow (viewed from the "shore"). the phase velocity at which a crest travels mast be balanced by the cornponent of stream velocity normal w the crest. 3. +(a) Show that ([9) follows from (ill) and (22). (b) Derive (25). (c) Derive (26)_ 4. (a) Derive (27). (b) Show that Figure 9.14 gives the correct qualitative picture of the wave crests. 5. Show that (24) and (26) are equivalent to (34). 6. Use a computer to plot curves of constant phase and thereby to obtain an accurate Version of Figure 9.14_ 7_ (a) Verify (32). (b) Derive (33c)and (34). (c) Derive (35). 8. la) Use (34) to deduce that waves are confined to a wedge shaped -
region downstream of the ship with vertex angle 2 tan (h) Work out further details of the pattern from (34), perhaps using a computer 9. (a) Consider waves on the surface of deep water that are sufficiently short to permit the neglect of gravitational effects_ Starr with an appropriate replacement for 129) and recover the following p. 469) for curves of constant parametric equations of Lamb (1932, phase when only the surface tension term, is retained in the dispersion equation: .
xl =
ra see { 1 — 2 tan g Lr),
x ] = 3a sec
tan k.
Here a is a negative constant and the parameter RP is a measure of the inclination of the wave-number vector defined by. k,
— 2cos t,
k2 —
- -xsinle.
If a computer with good plotting facilities is available, use it to graph a few constant-phase curves_ 10. Verify the statement [found three paragraphs following (271] that if (1.1/c„,) > 1 _ .. , there is a wedge with half-angle sin -1 (r,,,iUI with the (b)
4 32
Group Speed and Group Velocity [Ch. 9
property that capillary waves appear only inside the wedge and gravity waves only outside? 11. According to the original caption of Figure 9.15, the ratio of the wavelengths of the two types of waves where their crests are perpendicular to the direction of water flow uniquely determines wavelengths and water speed. Since the gravity waves are seventeen times as long as the capillary waves, the wavelengths of the capillary and gravity waves must be 0.042 and 0.71 m, respectively, and the water speed must he 034 m/sec. ' Verify these statements,
Appendix 9.1 The Method of Stationary Phase—An Informal Discussion In discussing wave dispersion, and in many other contexts, one is led Co seek approximations to integrals of the form 1(a, b, ).) = If(x) ex p [i4-1(x)) dx
(I)
The approximations should be valid for large values of the real parameter ;.. MOTIVATION
Instead of (1), it is easier at first to consider its real part, j f (4 cos [(x)] dx.
(2)
Ill. is very large, then cos ).g(x) will be a very rapidly oscillating function. The situation will be like that depicted in Figure 9.16: f (x) will change slowly
FIG Li R E 9.16. As shown, if ;. is large, ens P .g(x)] wilt in genFral eQry rapidly compared ro fix).
App endix 9.1]
The Mt^s h ud rtf Sturronurl- Phase An lnfnrrrro! Drsc -u.ssron
413
Kit
xt 1.u1 l
I
I
I I
I I I
I 11
^e
^i
Ir
I^
FIGURE 9_17. A function ,y with a ronsiunr rryrort No mazer h n x faro' i i s,
there
will be nr>'oscillation of cos [i.g(x)] lor
.x o < r < x I
compared with the change of cos Ag(x). Because of this, the contributions to the integral from intervals where cos Ag(x) is positive nearly cancel the contributions where cos Ag(x) is negative. On the other hand, suppose there is an interval where g(x) is constant. In that interval cos :tg(x) would also be constant, so that there would be no oscillations and no cancellation. It is thus plausible that, for large ;t, the contribution to (2) from an interval where g(x) is constant far outweighs contributions from intervals where g(x) varies. In Figure 9.17, for example, we expect the contribution from 1 3 to outweigh the contributions from 1 1 and 1 3 . Moreover, for very large •1 we expect that the !s contribution will dominate, even though I t is very narrow. In the limit A o) it is plausible that if g(x) has a horizontal tangent at x; (figure 9.18), then the contribution to the integral from the interval [x ; — d, x ; + 6], however small b is, dominates contributions from intervals in which y does not contain a horizontal tangent.
go1
I I t a
FIGURE
9.18. For sufficiently large
arise when x is
rc^^nt.
s,
i. the major contribution M ,
in a narrow region about x , the location of the ,
( 2) will
lrurrcrnirnl
Group Speed cowl Group Velocity [fit.
4 14
9
In the present context, points w here g'(x ; ) = 0 are cal led points of stationary phase. We assert that for large A the principal contributions to (1) do indeed come from these points, and that this fact allows (1) to be approximated by a relatively simple expression. DEVELOPMENT O F A THEOREM
In the first part of the demonstration of our assertion, we suppose that the closed interval [a, f] does not contain a point of stationary phase. Then, since g'(x) 0 for a S x S 13, we can write f(a P, a.) = f (x) exp [a9(x)] dx ` f .
t^g'
dx [exp
dx.
Integrating by parts, we find that i^.!(a, 13,) = _ exp (i).g)
. ' 4l)3Î9„ exp (ia.g) dx.
--
(3)
a
We next employ two forms of the triangle inequality, Ix + 'I
ru
and
Ix1 + IYI
F(X}dX S ^IF(x)I dx. o
I
(4)
Using (4) and the fact that I exp (iA.rg)I ° 1, we see from (3) that I Al(a,
P. ^) I ^ 10) + I (a) s'(1)
+
f'g' — fe" dx ^1
(9') 3
We can conclude that if there is no stationary-phase point in [a, fl], then certainly I(a. fi. AIM = O(A_ ').as A --• x, for IAI(afi. APB is bounded by a constant which is independent of di. We slate this result in the form of a lemma, which contains conditions guaranteeing that the various integrals exist. Lena.
If f (x) and y(x) have, respectively, one and two continuous derivatives for a < x S b, and if y`(x) # C for a < x < b. then I(x) exp [0.9(x)] dx
)
as A.
(5)
We turn now to the case where the interval [a, /3] does contain a point of stationary phase. Call such a point x„ so that g'(x ; } = C_ Suppose further that g"(x;) # O. Let 6 be such that x, is the only zero of g'(x) for I x — x i I ç S. We assert that J +a !^ = f (x) exp [ikg(x)] dx
^ Y a i
(6)
Appendix 9.1] The Method of Swrrnnary Phase—An lnformpf Discussion
can be approximated for S i = ( r) '
Lis
large positive A ) I — 112 f(xi)
4lS
by S i , where
exp [f 4(x,) + ;7Zi sgn g"(xi)]_*
(7)
To be more precise, we assert that under appropriate hypotheses =Si +O(). -
').
(8)
In considering a general case of the integral l(r^ b, A) of (1), let us suppose that there are N points of stationary phase in (a, b), N 1, and that all such points occur in the interior of the interval of integration. Then we can split up [a, b] into subintervals that contain stationary-phase points, where we can use (8), and into other subintervals that do not contain such points and hence, by (3), that make negligible OP, - I ) contribution to I. Such reasoning leads to the basic stationary-phase theorem (12). If one or both of the end points a and b are points of stationary phase; i.e_, C or = 0, then there are corresponding contributions to the if g`(a) approximating sum of 41.2). Not surprisingly, it can be shown that these contributions are precisely half the contribution of an interior point of stationary phase. If g"(x,) — 0 but q "(x i) V- 0, the corresponding contribution to (i.2) is a relatively large 0(AA " I ) term, and the first term neglected now is 0(A -2 °). See Exercise 5_ ,
HEURISTIC DERIVATION OF THE KEY APPROXIMATION
It remains to give evidence for our assertion that (6) can be approximated by (7). We shall give only a heuristic derivation of this result. For a rigorous discussion, see A. Erdelyi, Asymptotic Expansions (N.Y.: Dover Publications Inc., 1954) and also A. Wintner, "Remarks on the Method of Stationary Phases," J. Muth. Phys. 24, 127 30 (1945). In (6) we can make as small as we wish. t# is reasonable to suppose that in the narrow interval [x i --- 5, x i + 6], we can approximate f (x) by f (xi). We must be more accurate in considering g(x), since it is multiplied by the large quantity A. Remembering that g'(x,) = C, we approximate Mx) by the first two nonvanishing terms in its Taylor series about x i . Thus -
x, +d
.,
}l .I (Yi}exp (i+^Cg(xi) - f- I9 (xi)( j/!^ ` ^i) 2 J! dx
Q (` )
An - 6
or
, l exp [l4(x i)1
+
a exp [rrl,L(x — x i ) 1 ] dx,
2f(xi)
s,
• Here sgn is Itic siynung (r ^nr -rron: sgn x
x
=
1xl
^
1.
1.
>0, x c 0. ^
(10)
4 16
Group Speed and Group Velocity [Ch. 9
where
( II)
A — iy"(x ; ). If A > 0, we make the change of variable
^lrl(x — .x
= 42 ,
(I2)
with which the integral in (10) becomes 64.14 1 112
(d1,q)- 112
f0
exp
(13)
For large a., (13) can be approximated by ^«
exp (g2) fig = 0
cos (el d a
- i
J (2 n
}d .
(14)
Numerical methods could be used to evaluate (14), but those having some familiarity with Contour integration should have no difficulty in completing Exercise 4, which shows that its value is it'". Exercise 2 deals with the case A z 0. Putting our results together, we obtain the desired formula (7). CENERAI.IZA'r ION
The presence of the parameter A in (9)ensures that there will be a situation when A is large) wherein one factor of the integrant] performs many oscillations about a zero mean value while the other factor changes only slightly. Whenever such a situation occurs, the same heuristic reasoning that we have just presented suggests that we can approximate the integral by considering only those portions of the interval of integration where the oscillatory factor is stationary. Thus when exp [iii(x)1 oscillates rapidly compared with jr (x), we expect the generalized stationary-phase result of (I.3). What our hypotheses for this result lack in precision, they gain in emphasizing the essential
point. The condition h(x) 4(x), . co, is sufficient but not necessary to guarantee the relatively rapid oscillation of h, and hence to ensure that (1.3) is a good approximation. Often a theorem is proved in the literature in a form different from that desired for a particular application, and the sufficient conditions used for the proof may not be met in practice. This is illustrated in Section 9.1, where we apply the generalized stationary-phase approximation (1.3) without reducing our integrals to the clearly stated form of (1.2). Having some confidence that we could produce the required proofs if necessary, but feeling that the main issues do not lie in that direction, we are content to rely on our intuition that the necessary extensions of the basic theorem can be obtained.
Appendix 9.1]
The Method of Stationary Phase- An Informal Discussion
417
EX [FtC IS ES
1. To acquire some notion of how the stationary-phase concept is relevant even though no large parameter explicitly appears, show that if there is no stationary phase point in [a, l3], then under suitable hypotheses on f and h (which you should state) -
f
f(x) exp cih(x)] dx I < k max p
h ()
,
k is a constant.
2. Obtain (7) when A < O, i.e., when g"(x + ) < O. 3. (Project) If you have access to a digital computer, use it to evaluate the integrals of (14) to three significant figures. By considering the integral of exp (iz 9 over the contour depicted in Figure 4. 9.19, show that the evaluation of 114) reduces to the evaluation of exp ( — t9 dr. The final result can then be obtained with the aid of Exercise 3.2.2(d) of 1.
FIGURE
9.19. A contour in the :-plane
5. Consider 1(a, bd.) of (1). Suppose that in addition to the ordinary stationary-phase points x ; , there are also M points y, satisfying g'(yi) = 0,
g"1,y,) = Q, 9"1v1)
a;
a < yt < b, 1i = 1, 2,... , M.
Show formally that the series in (1.2) must be augmented by l'(,,) (0)1 !3 ,1-
1/3
L i
Ig (y)I - 1/3f( a)exp f
[idlAy i)]•
-
Under suitable hypotheses it can he shown (Stoker, 1957) that the error is now OP.-2/3).
C HAPTER 10 Nonlinear Effects
IJ
this point, our treatment of water waves has concentrated on the extensive classical developments of the linear theory. Now we turn our attention to nonlinear effects. Grappling with nonlinearity is one of the major tasks of the modern applied mathematician, so this is certainly not misplaced effort. Moreover, much understanding of nonlinear effects has emerged in the last few years from studies of prototype problems in water waves. The rather classical material that we present in some detail is of considerable value in itself, and also provides a necessary Foundation for more recent developments. Our discussion of recent results is brief, yet it will afford an appreciation of current research_ NTIL
10.1 Formation of Perturbation Equations for Traveling Waves in Chapters 8 and 9 we studied water waves under the assumption that wave amplitudes were so small that nonlinear effects could be entirely ignored. We first obtained sinusoidal traveling wave solutions to the governing equations. These solutions, although special, exhibited important properties--particularly the dependence of wave speed on wavelength. Moreover, the sinusoidal solutions could be superposed to solve a general initial value problem. Now we shall investigate how a sinusoidal traveling wave solution to the linearized equations is altered if nonlinear effects are taken into account An important new property will arise dependence of wave speed on amplitude. But superposition is no longer possible, so that, in contrast with linear theory, the obtaining of near-sinusoidal solutions is not a dominant step toward full understanding of the problem. "Adding" nearly sinusoidal nonlinear wavesgi%es rise to new effects that we can only describe briefly here. Useful insights into nonlinearity are nevertheless afforded. Our calculations are an example of regular perturbation theory. This subject was introduced in Chapter 7 of I, mostly in the context of a single ordinary differential equation. A system of two ordinary differential equations was discussed in Chapter 8 of I. Here we deal with a system of three prarririi differential equations_ There are two further complications: Boundary conditions are applied on an unknown curve and a series expansion of a certain parameter in the problem is necessary. Although knowledge equivalent to the elements of regular perturbation theory, as discussed in Chapter 4 [8
Sec. 10.11
Formation of Perzurharron Equnl(uns for Traueltmy Waves
419
7 of 1, would doubtless give much more confidence in our procedures, no such knowledge is required here. The reader need only accept the following idea: If various dependent variables and a parameter are known to be functions of a certain other parameter l:, it is reasonable to hope that expansions of all quantities in powers ofc will provide valid results when I k I is sufficiently small. in this section we shall first introduce a coordinate system that "moves with the wave_" Then we assume expansions in the amplitude parameter r.. The original nonlinear system is thereby reduced to a sequence of linear systems. Solutions of these systems are discussed in Section 10.2_ CtiA NICE OF VARIABLES
Consider small - amplitude water waves that retain their shape as they move with constant speed C. We look for waves t hat are spatially periadicand hence, since they do not deform, are temporally periodic_ {Later we shall restrict ourselves to "nearly sinusoidal waves," but now our discussion is more general.) We shall take 27rL to he the spatial period (wavelength) of these waves, 27rP to be their temporal period, and A to be their amplitude_ In terms of the dimensionless variables x = x*/L and r = t*/P, both spatial and temporal periods have the value 27t.f Recall that the pattern rnoves a wavelength 2 rL in a period 27zP, so that
L
_
2nL —
2711)
or C.
_ LP .
We shall look for right -moving waves, so that all variables will be assumed to be functions of x* — Ct* rather than x* and t* separately. Dependence on the second horizontal variable y* will be assumed to be negligible. Using the scaling of (7.2.21), we therefore assume dimensionless dependent variables '"
It .s
u ! A/f' ,
w
* {'
=
— Co A
pALP 2' -
A la d)
that are functions only of the dimensionless independent variables
x —
x*
—
Ct*
-*
r = — L
and
.
(
3e, f}
(Note that x can be regarded as the horizontal variable in coordinates that "move with the wave.") if ti* = Li*
(x* --
L
Cr* z* .
, L
t This is a hetter choice than unify for the dimensionless pct pd - il avoids the appearance of tir in Many later formulas_
Nonlinear Effects [Ch. JO
42
we have
au *_
c'e^•
*
€7x*
.
Analogous relations hold for r*-derivatives of rte* and (; thus the governing dimensional equations (7.2.12-15), (7.2.17-18), and (7.2.20) become -Eu x + E 2 (IA , + w14,) = —Ew1
(21
+ ex (u ' + ww,) — — ep,. Eu
At z = —
— epz,
(3)
+Ew,=0.
(4)
Ew — 0.
(5)
Atz= EC,
r.w = —EC„+E 2 riCz .
At z T Er,
Fp = F[E4
f
(6)
BECz„(1 + F 2 !) -3/']
(7)
x^ c^(x, t) dx = 0_
(8)
n
Instead of boundary condition (7.2.16) and initial condition (7.2.19), we impose a periodicity condition. We seek functions of period 27EI_ in x* and 2mrP = Dr1,VC in is. From (le) this implies that all functions have period 2r in x.
(9)
For a reason that will be explained shortly, we have chosen not to perform a division by E that would simplify the appearance of the equations. CONSEQUENCE OF TRAVELING-WAVE ASSUMPTION
that we have assumed that the nonlinear equations have solutions in the form of waves that travel at a constant speed C without It must be emphasized
changing shape. (Mathematically, we have assumed that all dependent variables are functions of x* and r* only in the combination x• — CO.) The validity of this assumption is yet to be established. Four parameters appear in the dimensionless governing equations (2) to (9)_ As in (7.2.31) these parameters are E
A L•
_ L
_ pg L2 .
rL C2
We have seen that according to linear water wave theory, periodic waves travel with a speed C that depends upon the wavelength 2rtL, the surface tension 7, and the depth H . Only the parameters h. B, and F enter in linear theory, and C occurs only in the Froude number F. Thus we would expect
&c. an Farmaiion of Perturbarion Equations for Traveling Wares
42
this dependence to be expressible in the form F = F(B, h).
(10)
Indeed [Exercise 1(a)], the dispersion relation (9. I .25) can be written as F
[(I + B) tanh h]
(11)
When nonlinear effects are taken into account, we expect that [10) will be generalized to F — F(B, k F).
(12)
In other words, traveling waves are expected to he possible only if the dimensionless speed parameter F depends in a specific way upon the other parameters of the problem. SERIES SOLUTION
We have seen that analysis of water waves can be considerably simplified if it is assumed that the amplitude-to-wavelength ratio i isa small parameter. The most drastic simplification is to retain only terms of lowest order to F. This results in a linearized theory that has been extensively discussed in Chapters 8 and 9. A more comprehensive approach is to assume that the dependent variables can be expressed as power series in e. Since it is reasonable to hope that such series will converge when z is small enough, we shall proceed with the formal calculations. We assume power-series expansions for all dependent kariables. Thus for the horizontal velocity component we posit the expansion
tu(x, z, t, E) = Ew'"(x, z, rj + elu` kx, z, r) - r. 3 ui 3 (x, z, r) +
- - - -
(13)
Note that the O(E) left side of (13) is matched by an OC') dominant term on the right. Superscripts match the power of e; and the lowest superscript is unity, in conformity with the notations of earlier chapters. (This last desirable feature emerges due to the fact that we did not divide the equations by r..) For the other variables we write (no longer indicating independent variables)
+ E3w(3' + - - - ,
Ew = t:w"'
+ E3p{3' f- . . . ,
( 15)
F- t1 + r. 2 cm + f;3 {31 + ...,
( 16)
t:p = Ep"i Ec, —
(14)
-}- E 2 /PP
F(B, h, i<) = F' °}(B, h) + .:F''''(B h) -I,
F" )(B, h ) + - •
(17)
The first term on the right side of (17) is assumed to he 0(1) because we know from linear theory that small-amplitude waves travel at a speed that approaches an 0(1) value as the amplitude tends to zero- Indeed, we expect Ft' B, h) to be given by the right side of (11).
422
[ch- W
Nonlinear Effects
DETERMINATION OF SUCCESSIYE SETS OF EQt1ATICINS
The substitution o f (13) to (17) into (2) yields — cu
t 1 1 — ^3^t21 f ^€
—
3^(3) x
+ « • -
f +(u4e11 + s214(2i + . • .)(euxla + E2ux^1 4 • .
+(0.0 11 + s2w(21 + .. )(E1ei1° E2u=21 + = c 111 _ _ E^ 1 31 +..— P^ 0 1)1.2) P:
.)
- •) (18)
or, collecting powers of c, 111 + EZ ^ re x121+ fa 12 ] s^ _ 111 u^ + p^ x 1 + 1+ t11 eex{ 1 1+ w tl ^ tl] ^ : ♦ E 3[ 14,3 ] + p1.21 + til'=i,lŸ 1 + 1^ 12}u; 11 + 1Nw/.;x1 + -
+
D. (19}
It is well known from the elementary theory of power series that if Cr]
E crie _ 0 for
Ep70,
< eo ,
1
and some constantsc then c = D, n t, 2, . . Since precisely thesame result is true when the Ch are functions independent of e, we equate to zero the coefficient of each individual power of s , which yields: —
,,
0(c): _
0(c2 ): 0(c3 ):
uY,1
(20)
+ Pr11 — 0,
+ px4 —
- -
u=31 +
px31
us1J14*1/ ^ w{7}1îi11
= —
—
once)
— 14
(21) 1,k 11 p up , w121ü =11 .
t21E1111
,
(
22
)
Equations (3), (4), and (5) present no new problems. To treat (6), we first write it in the form
Mix, E C, r) =
t} ± E lrA(x,
( 23)
t]CA, r),
where we have explicitly carried out the required boundary substitution z — El, Now if some function f (c) has a power series expansion about € = a (Maclaurin expansion), then
.f
.f (0) } 41c) + 1E 21 "10 + - - -
Similarly, if we regard w(x, c() as a function of x, then
w(x, ^^ ` ^x, Q) +
E[ ^
,,{
x, eC)
=a
x + ^I
a LLL
w{x, EC4+ +a
. ' .
▪
!see_ 10.1] Formation
of Perturbation EquufroMN for Trar.elirly wave,
4 23
where we use partial-derivative notation to emphasize the fact that only variation with respect to e is being considered: x and ( are deemed constants during the e-ditereniiation. By the chain rule, ^
- w{x, ^^} =[ ^z ,,,(., z) a (Ec), ^E = - ^ ^E SO
[w(x.(4)
j
a-4
Using this, we obtain
w(x,
w(x, Q) + s;w r (x, D) + is2CiwZr(x, 0) + - • - .
(24)
We emphasize that all terms on the right-hand side cf (24) are evaluated at z-0. For the quantities w, w., and ( on the right side of (24). we must utilize expansions (14) and (16). An analogous procedure must be used on u(x, EC). With all this, (6) becomes ci4,0y
11,2 14,(2)
+ I,3 w t3y ^ - •
+ {€ V ! )
t c 2 1^1
+.
. .ww.=4 11
^ E z x,=zlF { . . -}
+ tiFC` 11 + • • .) 2 (iEw* 12 f- • . .) ^ ^^t21 -. ^a^t3y x x E2u13 1 t , 1 + . . - t. f1 2c11 1I f] I I + [F.lf t 1 -
'
li + 2 Vx^ 1 f
+ . . .^ ^ . . ,1 ( 25 )
...),
where all quantities are evaluated at z = O. Upon equating successive powers of a~ to zero, we find that (25) yields a set of boundary conditions to be applied at the mean surface position z = C: wo) } SLL x i
—
w 1 2 1 + r{ 2 1 =
x
11` 131
+ C .3 ) =
(26)
Q S
I! wlll + ^tl y^ill
(27)
=.
: Viy w =21 _ Ixtl,t,l; lt — lg[
+
u11210.11 + 141110cIy
L i)3wxs1
+ ti ( I VI 1J's 1
( 2 g)
• O. Joseph emphasizes the fact that functions initially defined only to the fluid (2 5 aç) are expanded its terms of functions defined under the mean water level 2 - Q. Mapping of the original domain into a layer provides a clear-am way around this apparent difficulty See D. Joseph, "Domain Perturbations: The Higher Order Theory of lnfinitc imal Waves." Arch- Rational Meth. Anal. 51, 295-303 (1973)-
Nonlinear Afftefs [Ch. 10
4 24
Treatment of (7) requires one step not previously used, application of the binomial theorem to obtain {I
+ E2c1)- 3f2
ie 2 cx + .. .
I —
We have carried out the above computations in detail to provide a fairly complete example of nontrivial perturbation calculations. The calculations are straightforward, but as higher order equations are sought, the degree of complication and the resulting possibility of error rapidly increase. The final results [at O() and 0(E 2 )] that we obtain upon expanding the various equations are as follows: l^(E)
At z— 0,
0,
=
0,
(30)
=
0.
(31 )
(29)
0.
—1,,
At z
=
w[ I 1
At z ^ Q , pl I)
(32)
VYL J^
(33)
— P° 11 C11
0.
(34)
^ 2x
CIIJ(X} l
f^Y =
0.
(35)
n
,.raI 21
x
H%I3I s
^.
I 2 i= Px ff77 121 r x
^
L1 s 12r + w 1; 2I =
At z = —h,
_ u lll u Ili _ l wllJ Il
^
ll1^I
L1
: •
^,Il]^ il
=
(36)
l
0.
(37)
( 3 g)
14, l 2l — C.
(39)
At z — 0,
W{2 I + (r'i -- —x1.11 w;I ! + u1lV".
At z = 0,
}^ p121 — 0° '(C 12) ^ B^xxl) r -- C"V"+ F {l-[C1l ` — B4O„zl)•
j
fI,^J( x a
)_
0
(40)
(41) (42)
Some of the O(r) equations have already been given in (22) and (28). The complete set of 0(E 3 ) equations appears in Section 10.2. Keep in mind that periodicity requirement (9) must be satisfied for all dependent variables.
Ser_ 10_11
Formation of Yewrbcuion Equations for Traveling Wanes
425
REMARKS
ow)
Note that the superscripts in each term of an equation always add to r^. For example, three of the terms in the O43 ) equation (28) are 13 l y
_ ci 1 owl!).
tl M} 11 j
SŸt 1
In each case, the sum of the superscripts is three. There is no magic in this; if follows from the fact that in our original series (14) to (17) each term having superscript n is of order rte. This observation about the superscripts is useful. as it provides a check on our calculations. It is appropriate here to build confidence in our approach by showing in general terms how the sets of equations we have derived can be solved successively. We shall find that the O(E) equations have a nontrivial solution only if FtO) bears a certain relation to B and k The left sides of the O(r 2 ) equations are identical with the left sides of the O(E) equations. All the functions on the right sides of the O( z) equations have superscript unity and hence are known once the O(f) equations are solved. An unknown constant F{ " also appears on the right side of the O(c 1 ) equations, but we shall sec that this must he related in a certain way to 1 and h if the O(C- 1 equations are to have a solution. In general, at O(e") we have a set of inhomogeneous equations_ The functions appearing on the right sides of these equations are known, since they are combinations of previously obtained solutions of lower order equations. A new unknown constant P" appears, but this can be determined by the requirement that the 0V) equations have a solution_ From { l 7), successive determinations of the F{"' providesuccessiely more accurate determinations of F and hence, since F = gLJC2, of the speed C of a wave that propagates without change of shape. We shall carry out the details of the solution in the next section. Note that there will be no difficulty with unknown boundary location, because at all orders surface boundary conditions are applied at the mean height z = O. The effect of surface deflection makes itself felt by the appearance of extra terms, e.g., the —0 1) 4' term in (4[1. [The other term in (4[7) that does not appear in its linearized counterpart (33) arises because the unapproximated boundary condition is nonlinear.]
EXERCISES
I. (a) Verify ( 11 ). (b) Complete the derivation of (36) to (42). $(c) Suppose for the sake of argument that Et) was O(i: 2 1. Would the assumption (15) have led to a contradiction? 12. Find the O(r 3 ) equations "that follow from the assumptions (14) to (I7).-
Nonlinear Affects [Ch. 10
426
10.2 Traveling Finite-Amplitude Waves We shall now solve the successive sets of perturbation equations derived in the previous section. This permits us to discern some of the qualitative effects introduced by nonlinearities. It turns out that our solution breaks down for a certain exceptional set of wavelengths; we shall indicate in general terms how to deal with the "nonlinear resonances' among the waves having these exceptional lengths. The details are beyond the scrape of this book. Indeed, our final discussion brings us close to the frontiers of current research. LOWEST ORDER EQUATIONS
Al lowest order the equations to be solved are (l_29 35). No nonlinear terms appear in these equations and boundary conditions are to be applied at the mean height z = O. Indeed, the equations are precisely those of linear theory, in a "wave-bound " coordinate system. We shall take h =riu (infinite depth) to simplify the calculations. We proceed just as in the original linear analysis of Section 8.1 by assuming exponential solutions: -
um yy l
û
LI
;
Re
w eXp (iax) exp {kz},
Pc l l
(1)
15
r(L I
= Re 4" exp (lax).
The periodicity condition ( 1.35) requires that a be restricted to integer values n_ The remaining conditions are satisfied [Exercise 1(a)] if k = n and
n F-1Q) _ = I}nzB "
FI °t= F^ ° '.
(2)
The corresponding solutions can be written as follows: E
FT
= Re
i— n
a„ exp (inx) exp (nz),
(3)
n
c tl, = n
Re u„ exp (irtx)-
Here the arbitrary constants u„ represent the amplitude of the wave with spatial period 27t/n; n = 1, 2, - - - . It is instructive to regard the solution set (3) as eigenfunctions of the governing linearized problem (1.29-35). This problem requires 27E-periodic solutions of various equations and boundary conditions_ Trivial (identically zero) solutions are always possible. The nontrivial eigenfunction solutions
See. I0.211 Traveling Fin[re-Amplinale Wares (3) occur only for a certain discrete set of cigcnvalues Fri given in (21. (in physical terms, this reflects the fact that an infinitesimal wave of length 2ir/n can travel only at one specific speed.) Now consider the problem of finding 2n-periodic solutions of the full nonlinear equations i I.2-8). The trivial solution is again a possibility And again we expect nontrivial solutions only for certain values F. of F. n = 1, 2, .... We are faced with a nonlinear eigen+ralue problem. For "nearly linear" problems, where the ratio of nonlinear to linear terms has some small magnitude F, it is natural to expect that eigenvalues and eigenfunctions have expansions in powers* of E. (Physically, nonlinearity should cause a slight distortion of the speed and shape of the waves predicted by linear theory.) In mathematical terms, then, our object in this section is to find series expansions for certain nonlinear eigenvalues and eigenfunctions. For linear problems, general solutions can be synthesized by a superposition of eigenfunctions. Such superposition is not possible in nonlinear problems, so determination of eigenfunctions is correspondingly less valuable. The delineation of solutions to general initial value problems for slightly nonlinear equations is an object of current research; we cannot go into these matters here. Nonetheless. it is certainly of interest to examine softie solutions of nonlinear problems. Furthermore, nonlinear eigenfunetions turn out to play an important general role when nonlinearities are small. For simplicity, we shall only examine corrections to the lowest eigenvalue, for which n = t. In principle, there is no difficulty whatever in handling a general n (Exercise 7)_ There is just further clutter in an already complicated sequence of calculations. We shall restrict u to real values_ This implies no loss of generality; it just amounts to a special choice of the coordinate origin. Dropping the subscript unity, we thus focus our attention on determining nonlinear corrections to the following solution of the lowest order equations (1.29- 351 (with h = 3c): ut t t = a cos x exp z,
(4a)
WI t' = a sin x exp z,
(4b)
p t t t = a cos x exp ..
(4c)
a cos Y ,
(4d)
=
At lowest order the dimensionless speed (or eigenvalue) F is given by (2) with n set equal to unity: (4e) P°1 I I + B) * Since the parameter t. appears anatyticalty in the equations, there is reason to hope that the solutions are analytic functions of L.
428
Nonlinear Effects [Ch. 10
SECOND ORDER EQUATIONS
We now turn to the 12E 2 ) equations 11.36-42) with h = co. Substituting from (4) into the right sides of these equations, we obtain (Exercise 2(a)] 0 —1412) + ^t2^ x = -- we' -t- iJ^` u=; ' + wi2
—
(5a) a 2 exp 2z,
(Sb)
O.
(Sc)
As -a —no, W 121 _40. At z
0, w121+
At z
plZ'
(5d) (5e)
a' sin 2x.
F't° '(0 3 ' -- BCx') , —rJ 2 (1 + cos 2x) + F" } (41 + B) cos x. ^ rt 2 '(x) dx = 0_ J °2k
(51) ( 5 g)
In order to obtain a solution to the differential equations (5a), (Sb), and (5c), we first eliminate id" ) and w(21, to obtain a single equation for p'3 '. To this end we differentiate (5a) with respect to x, differentiate (5b) with respect to z, and add the resulting equations. By design, d 2' and IN" ) appear only in the expression —0421 + w»)x and thus cancel [see (Sc)]. Consequently. we obtain ^ r7 2 V 2 p42 ' = — 2a' exp 2z, 5^' .E ^
^
ex
^z
(6)
13y inspection, (6) has the general solution pl y ' — —
exp 2z + ff{x, z),
(7)
where H is any function that satisfies Laplace's equation: ' 2 H 0.
(g)
Substituting back into (Sa), (Sb), and (5c), we X 121
= 1Y10, x+ x2) — N s ^
find w"' a — — If f'
(9)
It remains to determine the harmonic function H. We furnish the required boundary conditions by substituting (7) and (9) into (5e) and (5f), after first differentiating the former with respect to x. This gives the following [Exercise 2(b)] : At z = D, At z = 0, H
If
+ at) _
Ft ° '(C 2 ' -- BC° 1 ) =
—
—Za' cos 2x,
Ice cos 2x +
Ft "a(1 +
(10) B) cos x. (1l)
See. 10.1) Traveling Finite-Amplitude Wares
419
Now p(2), and hence H, must have period 2n in x. By a straightforward application or separation of variables, we are thus led to consider the following periodic solution of (8) that decays to zero as z -i — cc [as required by 5(d)] : H = Re
E h„ exp (nz) cxp (ni x ).
(12)
n=L
Here the hi, are (possibly) complex constants. The second order surface deflection ( 12 "(x) must also have period 2nt, so, if sufficiently well behaved, it, too, will have a Fourier expansion. We write C' 2)(x) -= Re X C„ exp (mx),
(13)
l
where the C„ are (possibly) complex constants. No x-independent term appears in (13), since Ca
0,
(14)
by (Sg). Substituting (12) and (13) into (1O) and (1 1), we find (nh„ — F1^^ } exp (lFlt'1 = - 2a 1 cos 2x,
Re
„
(15)
L
Re
y ;h„
— FEn,fiâ„(1 + f3ri 2)]) exp. lam)
M- 1
= -- ra t cos 2x + F{ ' a(1 + B) cos x.
(1 b)
Since the right sides of these equations do not contain sine terms, h„ and C„ must be real [Exercise 2(c)]_ We now solve the pairs of equations obtained from (15) and (16) by successively considering the coefficients of cos 'ix for n = 1. 2..... When n = 1. we obtain fi l
—
Cl
h, — FL°lC,ll + B) = t^[ (l "a(1 + B).
— 0,
(I7a, h)
Since P'(1 + B) = I by (4e), (17) implies that F' l "G(1 + B) = O. Buta #O and
B >O,so lt^ =
0.
(18)
This is our first result. There is no second order correc:ion to the wave speed. Using (18), we see that (17) is satisfied if
I
f —
^
I.
h, = a, '
(19)
where rt l is an arbitrary constant_ That is, C = [ü + €a i + O(E 3 )] cos x +
(20)
Nonlinear Effects [Ch. 10
430 NURibiALIZATIUN
Two arbitrary constants, a and a l , have now appeared, and a little thought will show that new arbitrary constants will appear as successively higher powers of £ are considered_ To clarify the situation, we write (20) in dimensional variables, obtaining ^* = A(a -}- ^ i + . . .) c o s
x•
_ c[*
+--,
14
c - —_
We are now reminded that we have not specified the amplitude scale A. Two of the many ways that we could do this are the following. Possibility 1: Choose A to be the maximum value el+ÿ•. Possibility II: Choose A to be the amplitude of the cos [(x* — Ct•)JL] term in the Fourier expansion of Ç*. These possibilities are equivalent when only linearized terms are considered. but differences emerge when higher order terms are taken into account. Higher approximations to Ç* reveal terms proportional to 2(x* — C1*)
cos
, cos
3(x* — C:te) ,
Increasingly complicated computations would thus have to be made at each stage if possibility I were adopted, Possibility II, on the other hand, can be expressed simply as I ^ ^^^-
x• --- Cr*
-
-
-
rt n
cos
(
x* -- f't*
--
d
x* ^ ^:r*
-
= 1.
L
In dimensionless variables moving with the wave, the above condition becomes 1 ^ 2A (21) ax) COS x dx = 1. ft 0 From this it follows that 2.R I
JJ ü
f
1111(x) c.vs x
dx ` rr,
Çl^^(x) cos x dx = 0,
(22a) f = 2, 3, . . . . (22b)
o
The calculations required by these conditions are not onerous. Consequently, we shall impose (21) or, equivalently, (22). Equation (21) is called a normalization condition. Such conditions make a definite convention as to the meaning of certain symbols used in a problem. Another normalization condition is (2.2.9), whereby z = 0 is specified to be the average water level.
Sec- 10.2] Traveling Finite-Amplitude Wares
43 t
Because possibilities I and II are equivalent when only linearized terms are considered, we can write cA„ = eA 1 + c 2 €2 Ai + c 3 E 3 Ai + - -
(23)
,
where the subscripts I and II distinguish the amplitudes corresponding to the two possibilities and the c; are constants. A change from possibility Ito possibility II can be effected by insertion of (23); this amounts to a rearrangement of the series for the wave amplitude, velocities, and pressure [Exercises 8 and 9). The appearance of innumerable arbitrary constants, which originally troubled us, is now seen to be a manifestation of the fact that there are innumerable ways to select the amplitude scale A. Once a selection is made. all constants are determined, as we shall now illustrate. COMPLETION OF SECOND ORDER CALCULATIONS
Imposing (22), one finds easily [Exercise
2(e)] that
a, =+;,=h,=0.
a=l,
( 24 )
Hence (20) becomes
C=
cos x + --
-
or, in dimensional variables,
x'
1* = A cos
—
CO)
L
+ •-
This is in accord with our choice of possibility II. When n = 2, (15) and (16) give 2h 2 — 4s2 = — 2
h2 — P91 + 417g 2 = —
,
with solution fusing (4e)].
_ 3B
1f+ B .
: = 21 — 2B
h
B
1 -- 2B'
-
2
(25)
There is no solution if B = . This case will shortly be discussed at greater length. When n > 2, (15) and (16) simply become
h„ — nC„=0,
h„ 1-141 + Bn 2 )(:„ =0,
(26)
n> 2;
(27 )
n>2,
(28)
These equations have the solution C„= h„= 0
if
C„—a„,h„= na„
if B= n ' ,
#n
',
where the a„ are arbitrary constants {Exercise 2(f)].
Nonlinear Effects
43 2
[Ch. 10
When B ri`', n an integer, the OW) equations (with the normalization condition (21)] thus have a unique solution given by 2) =
B
+ cos 2x, Cz cos 2x — 2111— 2B
(29a)
— 2 e'2 + 1 e2 cos 2x,
(29b)
Ue21 sin 2x,
(29c)
Ue' cos 2x,
(29d)
3B 1 -213
(29e)
[In obtaining(29c)and (29d), we have used (9) and (5c).] We shall temporarily disregard the cases when B -= n - '.
THIRD ORDER (_AI..(_Ul.A•iI(]NS Since E 1 1 turned out to vanish, we must consider the 0fr 3 ) equations if we wish to see the effect of nonlinearity on wave speed. The appropriate equations are (Exercise 1.2) ^1^1 X —.14,(2)
1l1
At
z = .
3t
}
14
;
(x2 {1 )14 11 x
:
t!t^111 '21 .s
—
171,.3; ^ ^ u l k 1 i,vx1.21 , !!{x I w Ÿk 1
w
(rl^lz) r
,
^ w k2 ^ u{kr I
L k Iw111
—
^21[ I ), (30b)
(30e)
} Wit 3; = O.
w{; ' +
^s3 j — _ cill wlxl _ 7^kiak} _
f tkF) ? w 111
+ u 1t11`S2k + u=klck11ç1!+ E {2j crII• z ^ C, pk
il — FkOi^ Ck^1 , B^Sx311 — }
F ( "(V2)
Ba))
—
C"Ip
F
{2l
z 21
—
^k^I^k11
(30d)
_^
l1IkI )?^ IIk
l(^ 11j — IC I)) (30e)
+ iBF`01(^x" )2Cxx'•
As z -. - co, wc 31
(30a)
(300
0,
f
Z■ 0 3)(x) dx = O.
(30g)
o
Also, from (22h), fa
cos x dx = 0.
(30h)
See. 1021
Trrireling Frnile-Amp!ia de Wars
433
Upon substitution of the known lower order solutions of (4), (18), and (29),
Equations (30a) to (30e) become [Exercise 31a0 -ux31 + p4,31 = U sin x exp (34
(31a)
-3t + psi' = -3U cos x exp (3z),
(31b)
W1, 1 = 0.
0,31
(31c)
At z = 0,
w43 ' + 1 131 = S i sin x + S 3 sin 3x.
(31d)
At z = 0,
p13 ' - FtsOt(l 3) - Ball ) = C 1 cos x + C3 cos 3x.
(31e)
Here S i , C 1 , S3, and
are certain nonzero constants. In particular, it turns out [Exercise 3(h)] that C3
S 1 = - - 1U - -1C2. C 1 = F42 '(1 +
B)
-
(32a)
iF1° 'B + - U # 2
(32b)
.
The solution procedure is entirely analogous to that used on the U(c z) equations. Differentiating (31a) with respect to x and (alb) with respect to z, we obtain, upon adding the resulting equations and employing (3Ic), V2p13) = —SU cos x exp (3z)_ The general solution is f
p"' — —1) cos x
exp (3.1 + Re
E k A exp (nz) exp (ïnx),
{33)
„=1
for some (possibly complex) constants k,,. Paralleling (13), we write
Ct3'
= Re
E +„ exp (inx).
(34)
n=1
where the s; „ are (possibly) complex constants. To convert (31d) into a condition on p03) and C{ 3 ', we differentiate with respect to x and use (31 b), which gives p{"
+ C,32 = ( S i — 3U) cos x + 35 3 cos 3x.
(35)
Substituting (33) and (34) into (35) and (31e), we obtain [Exercise 4 a)1 (
(nk„ - n 2(;„) exp (inx) = S 1 cos x + 353 cos 3x,
Re
(36)
„=1
Re
E tk„ — Ft o1 [4„(1 + Bn 3 )]) exp (inx) = (C1 +
(1) cos x
+
C3
cos 3x.
„-1
(37)
As in the case of the corresponding equations (15) and (16), the coefficients k„ and e„ must be real. The most informative results come from equating to
Nadi/war Efferls [Ch_ 10
434
zero the coefficients of cos x: lc, —
k k — Fi° ( 1 + Bg k = C , + U.
= SI,
(38)
Since FI ° '(1 + B) — 1, for these equations to he consistenit it is necessary that Si =C' t + U. From 132) we thus find (when B # 1 that L
F121 =
1 1B1 1 — 28
28' + B+ 8 8(1 + 8) 2 (1 — 2B) '
—
313
+ B 8(1 + B)
1
2 (39)
When B # lin, the solutions of (36) and (37) are (Exercise 5) =- il, ^3
=
k,
(I +B)(C3 S s) 2(1 — 3B) —
k„= 0,
!=
42 = k 2 =
S
1 [3(1 2( I — 3$)
k3
-
f B)C 3 -
—
(l ,+ 9B1S 3 j. -
4,5,....
(40)
t)ISCUSSiON Let
us combine our principal results fer clarity. Since
F
L Z
= ^
= FiO}-
cFl ll +£ 2 F121 +... *
from (4e), (18), and (39) we have the following expression for the wave speed C: fa
e 2(2B 2- + B+ 8)
1
-
C2 l
+
B
1
B-
8(1 + B)( 1 — 2B) +
é pgL2 .
(41)
A combination of (4d), (29a), (34), and (40) yields t: 3 (1 + B)(C 3 Si) 2(1 — 3B) cos 3x -+- - •
F 2 (1 + B} = f: cos x + 2(1 —8} cos 2x -
—
( 42 )
or, in dimensional form, A
!_
^ cas +
+
x*
—
L
CO)
A 1+ B
^
L2 2( 1 — 28)
A 3 ( I + i6)(C 3 — 5 3)
L3
2(1 _ 3131
cos 2
x* — Ct* ros3 - L
x* — CO) T
L
+ •••,
(43)
where
B
1, 1.,
.
.
.
t t i is typical that the successive terms in the expansion of a parameter such a F are determined by a consistency condition, Le.. a condition l ma must held if ceriain cquatiOns arc to have a solution_ Compare the argument ihat led to the dcicrminulion of F"I in (Is).
Ser IO.l} Traveling Finite-ArnpiillidC
Wares
435
We shall return later to the exceptional cases that we have just had to exclude to avoid zero denominators in (43). For the moment, we concentrate on the contrasting effects of amplitude as given in the dominant correction terms in (41) and (42), depending on whether B . } or B < h is best to cast our results in terms of the wavelength parameter L. There is a critical value of L given by
T B ; ^Lx
l -
2'
i.e., L ^ Î. where L ,
=
^T -
PO
(44)
Î_
(For clean water at room temperature. I. is approximately 2,4 cm.l For L < when surface tension effects are relatively strong, the phase speed Cdecrcascs as the amplitude increases. For L > L, when the situation is presumably dominated by gravitational forces, just the opposite dependence of speed on amplitude is predicted_ The term amplitude dispersion is used when the speed of a wave depends on its height_ The lack of such dependence according to linear theory is a consequence of examining the phenomenon in the limit of vanishing amplitude: When their amplitudes a re very small all waves of a given length move at very nearly the same speed. NOTE In the remainder of Our discussion of water waves. we shall speak of capillary waves or gravity waves, depending on whether L less or gicater than 'L_ Recall that the classification was somewhat different when the effect of length on phase speed was the criterion. A natural critical wavelength according to this criterion is 1.7 cm, corresponding [according to (8.1_21)] to the minimum phase speed. The considerable significance of this speed for linear effects was discussed at the end of Section 9, f. Another important feature of small-amplitude nonlinear problems is illustrated by 1421 A solution that is sinusoidal at lowest order containssuccessively higher harmonics at successively higher orders. This feature appeared in Equations (7.1.24) and (11.2.24) of 1, in connection with discussions of nonlinear effects on the motion of a pendulum. Indeed, harmonic generation is of very frequent occurrence in slightly nonlinear problems for reasons that we shall now outline_ We expect physical phenomena in homogeneous media to he governed by equations wherein the spatial variables appear only as derivatives. Also, unless the phenomenon is being forced by an outside influence, there is nothing to distinguish an origin of time, so temporal variables should appear only as derivatives. Linearized equations will thus have constant coefficients and will admit sinusoidal solutions. The second stage in a perturbation procedure will - force" the solution with harmonics of the linear solution, just as (51) is forced by a cos 2x term. But if a linear constant-coefficient system is exposed to a (nonresonant) sinusoidal forcing, it will generate a sinusoidal response ,
of the same period_ We see from (42) or (43) that the first harmonic (with half the period of the fundamental lowest order cosine term) has a different sign for capillary and
Nonlinear
436
Effects [Ch . 10
to)
(h)
FIGURE 10.1 Representative nonlinear warn prnfrlks_ (a) Graflay weave sharpened crests, broadened troughs. (b) Capillary wives: broadened cress, sharpened troughs.
gravity waves. Consequently, it turns out that nonlinearity sharpens the crests and flattens the troughs of gravity waves (Exercise 6)_ By contrast, in the case of capillary waves nonlinearity broadens the crests and sharpens the troughs, to give a characteristic " dimple-down " appearance that is markedly different from gravity waves (Figure 10.1). RESONANT CASE We see from (41) and (42) that our results "blow up" when f3 = 4, i.e_, when the wavelength equals the critical value L. The trouble can be traced to the fact that the simultaneous equations (38) are inconsistent when B = 4. The structure of mathematics is not shaken by this inconsistency. Trouble arises because we have assumed that a problem has a solution ofa certain form, and, under some circumstances at least, evidently no such solution exists (compare Exercise 14). In the present instance, we have assumed that the water wave equations have a solution that is sinusoidal at lowest order and that travels without deforming. Can we modify this assumption in order to obtain a solution when B = 4? Actually, we need to modify our assumption when B = 1 fez n = 2, 3, 4, ..., for in all these cases the solution will eventually break down_ [Trouble when B = is already evident in (42).1 A clue to the nature of our difficulties is offered by our earlier remarks concerning higher harmonics being " forced" by the sinusoids that appear on the right-hand sides of the successive nonhomogeneous linear equations that we have to solve. The reader will be familiar from ordinary differential equations with the exceptional resonant cases when the forcing is at a natural frequency of t he homogeneous system. In the present case it might be expected ,
Sec. 10.21
Traveling Finite-Atrrplrlurle Waixs
437
that the "resonance" occurs when one of the higher harmonics cos rrx is associated with the same Linearized wave speed as the (fundamental) cos x. Now, from our earlier linear results or from (41) with t = 0, we know that a wave of length 2nL moves with speed C given by C l = gL 1 +
(45)
^x
A harmonic of wavelength 1 Jn times as large moves at the same speed if gL
n
^ n7' 1 + 2 = yI. 1 + i or pt^ 7a?9^-
T P9I 2 = +t'
This condition, B = n ', is indeed the one that holds when our original perturbation calculations fail_ Consider further the case B r 1; i.e., f_ = L. According Co linear theory, the periodic wave
a cos x + h cos (2x + 0)
(46)
will travel without change of shape, whatever the values of the amplitudes a and b and the phase factor O. The reason is of course that each of the two constituent waves travels at the same speed (about 24 ern is), in spite of the fact that one is twice as long as the other. This suggests that (46) is the proper solution to assume at lowest order when B = . Such an assumption was made in a paper by I_ R. Wilton in 1915, and later, in a slightly modified fashion, in an article by W. J. Pierson and P. Fife.' The solution process can be carried forward without contradiction. The results for the surface deflection as given by Pierson and File is (in our notation)
EC = E4cos x + { + + cb 2 }Cos 2x] + ie 2 cos Ix Ic2 cos 4x 4 - - - .
(47}
The solution correThere are two different possible solutions when B = The sponding to the upper (lower) signs is of the gravity (capillary) type, with "dimples" up (down). The coefficient b 2 is not determined al this order- As pointed out by L_ F- McGoldrick, however,t constants like b 2 a re ultimately determined by initial conditions, so they can be chosen at will to improve the agreement with any given experimental data. The dimensional wave speed corresponding to (47) is given by
C
04l ±^+ -
(4$) }
Noteworthy is the presence of a first order correction terns • "Same Nonlinear Properties of Lung-Cresied Periodic Waves with t.rngihs Near 114 Ceniimelers."!. Geophys. Res. dd. 163 79{196t t "On Wilson's Ripples: A Special Case of Rrsonani Inlerac[ions," .l fliaJ Meek. 42. L93-200 (#970). )-
Nonlinear Effects [Ch. 10
438 NEA R RESONANCE
It may appear that all is now in order, yet this is not the case. Formulas like those in (41) and (42) become infinite when B = and we have "cured" this difficulty by finding a special traveling-wave solution corresponding to this value of B. But there is also difficulty when B is sufficiently near -. Our expansions are based on the supposition that the magnitudes of the various terms are given by the various factors 01£ that explicitly appear. However. to give one example. the "second order correction " term in 142) (the term proportional to cos 2x) will, in fact. be 0(c) like the fundamental if B — = 0(4 i.e., if L = Î + 0(c). The proper way to handle the situation when L . E only became well understood during the 1960s. We shall sketch the results here. Focus attention on a situation wherein two waves of lengths close to 2nL and rtL are undergoing a resonant interaction. It turns out that the solution has the general structure of (47), except that the amplitudes and the phases of the two waves undergo slow variation in space and time. This can be motivated by the observation that the near-resonant wave pair proportional to exp i(1 + ae).x and exp 21(1 + be)x can be written in the form ,
[cap (iacx)] exp (ix)
and [exp (2ibex)] exp (2îx).
in this form, the wave pair has the form of the resonant waves exp (ix) and exp (2ix), but with slowly varying amplitudes. One thus looks for a solution with such amplitudes A i , and phases 0 i , of the dimensionless form EC EA 1 (X, T) cos [x + 01(X, T)] - CA 2 (X T) cos [2x + 02(X, Tjl + O(E 1 );
X = ex, 7' = Ft.
(49)
Determining the unknown functions in (49) requires an application of the method of multiple scales (l, Section 11.2). By a fairly straightforward, albeit lengthy calculation, the following equations for A 1 , A2,0 and 02 can be obtained :
r77'
GI
;
f U eX)
U2 }
A1`
OX A2 `
—2 Q sin 0A 1 A 2 ,
(50a)
Q sin fi~ A;,
(50b)
A 1 ^^^, + Ui
1 cos 0A 1 A Z , dX^ ' , — 2 2
(SOc)
A z ieT + U2
cos 0 A;. ê]^ z i —Q
(50d)
Here Q is a certain constant, B - 20 1 — 61 2, and U L and U 2 are the group speeds associated with waves of length 2rtL and rrî, respectively. Of particular
Sec. 10.21 Traveling Finite-Amplitude Waves
439
interest, as we shall see, a re steady solutions 01 (5.0) (ija T = O), without phase O}_ The latter condition requires that cos 0 = O. changes {r3,O t f3X = Taking advantage of arbitrariness in the definition of 0 1 (corresponding to axis translation), we satisfy (5Uc) and (50d) by taking 0, = O, 0 2 — — l sr} U =- E/2. Equations (50a) and (Sob) reduce to
aeseX =
aA, U'
— ax
—2QA,A 2 ,
1.12 1321 2 _
QAt'
(51)
As is typical of modern methods, the problem is now reduced to a system of ordinary differential equations in which nonitnearityis retained bta in afar more tractable form. SPECIAL RESONANT SOLUTIONS TU COMPARE VaTIi EXPERIMENT
Experiments due to L. F. McGoldrick• require (as we shall see) solutions to (51) wherein the harmonic vanishes at a some point x û , so that A i(xa)
—
A.
A 2 (x u ) = (]_
(52)
These solutions can be found in terms of hyperbolic functions; they are graphed in Figure 10.2 (a qualitative version of McGoldrick, op. cit., Fig. 1).
xp
FR:. u>z E 10.2. Amplitudes of rhf resonant flirxdrrmental (A 1 ) and its harmonic 3 ) according to rtivrsctd theory. Although only the fundamental is excited (A by the wav e maker a t position xo , rronlirfeor effects transfer nearly all the energy to the harmonic.
McGoldrick set up apparatus wherein a wavemaker could be tuned to oscillate with any (reasonable) desired frequency. His program was to oscillate the wavemaker at x 0 with amplitude A and with frequency corresponding to the fundamental resonant wavelength 271. According to Figure 10.2, a harmonic of amplitude A 3 should be generated, while the fundamental amplitude A L should decrease to zero with distance from the wavemaker. • An Expel-inicnl on Second-Order CapiLlary Gravity Resonant Wave lntcracLions." J. Fluid Meek 40, 251-71 (1970)_ See also "Weak Quadratic interactions Fu•a-Dimensional Waves," by Y. Kim and T. Hunrutly, l_ F7aaid Mech. 50, LU 7-32 (1970.
Nonlinear F.,ftris [Ch. 10
44°
Viscosity has been omitted from consideration thus far. As resonance is such a delicate matter, it may legitimately be doubted whether the predicted phenomena will show themselves at all in practice. However, McGoldrick argues convincingly that in the situation under investigation the effects of viscosity can be represented by linear damping terms, which tend to bring the individual amplitudes to zero exponentially with distance, Thus the amplitude equations (51) are altered to
^^# i ^` ax
— AI A,,
U2
`—
A2 A z ,
( 53 )
where A, and A 3 are constants. Closed solutions seem no longer possible, but phase plane methods (I, Suction 11.3) give the qualitative behavior of possible relations between A L and A 2 . The results (Exercise 12) are shown in Figure 10.3 La qualitative version of McGoldrick (op. cit., Fig. 2)]. Let us start with a given A 1 and A2 = C as in the inviscid case. We see that the harmonic will still be excited by the resonant interaction, but eventually both the fundamental and the harmonic decay to zero, owing to the effects of viscosity. McGoldrick adjusted the apparatus to oscillate with the frequency = 2703 that should correspond to the resonant wavelength 2rrL_ Since (Exercise 13) 9pg ! L
8 7'
(54)
determination of the proper frequency merely required measurement of the surface tension T. This was done with a duNuoy tensiometer, an instrument for measuring T under static conditions. When the experiment was run, McGoldrick found that "the details of the ensuing wave propagation were utterly at variance with the resonance predictions.„ The trouble was traced to the fact that "the interface is not the ideal scrupulously clean one supposed in the simplest of interfacial models, but is in fact contaminated.” Indeed, the characteristics of the surface seemed to change so fast that measurements could not be completed before conditions altered appreciably_ However. it appeared that a condition of "quasiequilibrium dirtiness" arose after about three days_ This lasted several more days, during which repeatable results could be obtained. Direct measuremauls of harmonic decay gave the logarithmic decrement d? [compare (53)] that is associated with the dirty surface. Measurements of an experimentally determined resonant frequency led, via (54), to an average dynamic surface tension that was less than the static value by as much as 30 per cent. The decrement A, of the fundamental could not be directly measured (because the harmonic is always excited). Instead, A L was chosen so that the theoretical solutions of(53) matched observations. In the end good qualitative agreement with theory was obtained, and adequate quantitative agreement.
Ser. 10.21
Traveling Finite-Amplitude Wares
441
Ai
F t G U R E 10.3, Phase plane for the resonant Ji,nda+nen1 i1 (A ,) and harmonic (A,) amplitudes tchen viscous effects un taken into accourn. An initial condition containing just the fundamental (henry dot) results in a trajectory (heavy line) wherein the harmonic grows because of nonlinear interaction. Ultimately. both amplitudes decoy to zero because of viscous dissipation of energy.
RECAF1TULATt[}1V
Let us recapitulate the findings in our study of nonlinear effects. First, consider nonexceptional values of the surface tension parameter B(B # 1, #^ For each such value we were able to find. by an explicit albeit somewhat lengthy calculation, the first terms in series expansions for nonlinear waves. For near-sinusoidal waves of a given length, the problem of determining wave shape can be regarded as a nonlinear eigenvalue problem. We found the "lowest" eigenvalue F as a series in the amplitude parameter c. In the limit of vanishingly small amplitude (c 1 û) this eigenvalue approaches a constant (which can be found by solving an ordinary linear eigenvalue problem). In more physical terms, we found that waves of the type investigated can travel at an amplitude-dependent speed given by (41). In the limit of vanishingly small amplitude, this speed approaches the constant found by the classical calculations of linear theory (Exercise l5). For the resonant case lB = we pointed out that use of the same assumptions as in the general case would lead to a contradiction. Thus in this case it is necessary to abandon the supposition that the steadily traveling nonlinear wave differs only slightly from its linear counterpart The presence of a harmonic having the same amplitude as the fundamental must he allowed. )
,
442
Nonlinear Effects
[Ch. 10
Although we presented only the final results in this case,once the correct form of the solution is postulated, the calculations present no new difficulties in principle. We saw that near-resonant cases such as B ; also lead to inconsistent results if straightforward methods are used, although the inconsistencies are somewhat more difficult to spot. Here the analysis becomes rather complicated, so we could only sketch the results. Nonetheless, the basic mathematical ingredients should be familiar to readers of I and the present volume— multiple scaling, phase plane, etc. Noteworthy from a physical point of view in the fine theoretical experimental attack on the problem by McGoldrick is the inadequacy of elementary ideas concerning surface tension. Indeed, such inadequacies have recently stimulated a much deeper study of the "'surface phase"; there are profound implications in the basic theory of continuum mechanics and in applications ranging from biology to chemical engineering. The reader is referred to Aris (1962) for a formulation of the fundamental equations. The failure of straightforward perturbation expansions due to "resonance" is a common feature of slightly nonlinear problems. As in the particular case investigated here, it is typical that when nonlinear resonance is carefully investigated, it turns Out that no drastic "blowing up" occurs, hut rather there is a particularly strong energy interchange among the resonating modes. It is possible that the existence of resonant energy interchange is important in the generating of waves by wind (McGoldrick, 1970, op. cit.). Certainly analogous nonlinear "resonant" interactions are important in many other contexts, for example in the generation of regular patterns by instability. [For a survey of some modern work on the formation of flow patterns, see " Nonlinear Stability Theory," by J. T. Stuart, Ann. Rev. Fluid Mech. 3, 347-70 (1971).1 FINA1. REMARKS--FORMALISM AND RIGOR
Only some of the various perturbation methods that we have been discussing have been put on a rigorous footing; the rest are formal. Thus in 1847 Stokes published the first few terms in a formal perturbation expansion for small nonlinear gravity waves (B - 0) in fluid of infinite depth. After a time, considerable controversy developed as to whether the Stokes waves ever really exist. Existence was established in 1925 when Levi-Civita proved the convergence of the Stokes expansions for sufficiently small amplitude-towavelength ratio. (Actually, Nekrassov published a similar proof in 1921, but as his paper appeared in a relatively obscure Russian journal, it was neglected until recently.) In 1926 Struik extended Levi-Civita's results so that they applied to a layer of finite depth. Only in 1960-1961 was the existence question completely settled by Kraskovskii's proof that Stokes waves exist for all amplitudes less than the extreme one at which the waves have a sharp crest. But in 1967, 120 years after Stokes's original paper, Brooke Benjamin and Feir
Sec. WW.!] Traveling fini e- Ampliwde Wares
443
showed formally that Stokes waves would never he observed on sufficiently deep water because they are unstable to certain kinds of small perturbations.* Roughly speaking, the analysis of Benjamin and Feir treats the interaction of the fundamental Stokes wave with "sideband" perturbations having wavelengths respectively slightly less and slightly greater than the fundamental. Calculations of the slowly varying amplitude type which we have discussed show that, even if they are not originally present, the sidebands will soon appear much as Al does when B = Z_ The original periodic wave train gradually degenerates into a collection of wave groups. Although a significant step forward, Benjamin and Feir's paper is but part of an impressive new body of knowledge concerning dispersive waves. A recent reference that lends unity to many of the new results is A. Davey and K. Stewartson's article, "On Three- Dimensional Packets of Surface Waves," Prex•_ Roy. Soc. London A338. 101- 10 .11974) -. also see the hook by W'hitham 11974). It is noteworthy that it took more than a century to justify Stokes' insight in its entirety. and that shortly thereafter his work was significantly extended by formal methods whose rigorous demonstration remains to he accomplished.
EXERCISES
I. (ii) Verify (2). (hi All the solutions of (3) have (dimensionless) spatial period ?r[, but their lit period is 27r/n_ There is a unique relationship between the (dimensional) length of a wave and its speed. Show that this fact is consistent with (2). 2. (a) Verify (5). (b) Verify (10) and (11). (c) Verify (15), (16), and the statement following (16). (d) Show that (22a) and (22b) are indeed the correct formulations of the normalization condition stated in possibility tl. (e) Verify (24). (1) Verify (25) to (28). 3. (a) Verify (31). (h) Verify (32). (c) Determine the constants Ss and C3 that appear in (3 Id) and (31e). 4. (a) Derive (36) and (37). (b) Equate to zero the coefficients of cos rrx in (36) and (37), n i- 1, and solve the resulting equations. In particular, show that ill = I, then there is no solution to the equations for k 3 and
For more information and references eonuersring this and other work on nonlinear water waves, sec an issue of ate Proceedings dings o{ rlle• Ro nil Sorierr(299.. No 1456. 1967) dcvotcd cot trcly 10 "A Di cusmon tin Nonlinear Theory of Wave Propagation in Dispersnc Systems.-'1
444
Nonlinear
Efferta
[Cfr. 10
(c) Give a convincing argument to back up the assertion that if B = 1 jm, ni a positive integer, then the perturbation procedure will eventually yield a pair of equations that have no solution. 5. Complete the calculations necessary to find pt 31, ({ 3 3, u t3}, and wt;) when B # n - ', n an integer. 6. By means of sketches, compare the effect of the cos 2x term in (42) when B > to the case when ii < f. 7. The text treats nonlinear corrections to the solutions (2) and (3) only when n = 1. Modify the calculations so that they apply for general n. 8. Make the necessary changes in the text's calculations to conform with a normalization selected according to possibility I: (a) Through second order. (h) Through third order. 9. Compare the results of Exercise 8 with the results found in the text, as follows. (a) Find e2 [and c 3 if you did Exercise 7(b)) in series (23) relating A l and A te _ (b) Point out explicitly how different normalizations result in a rearrangement of the series for the wave height C and for the pressure p. 10. (Project) Derive (47) and (48). 11. (Project) Derive (50). 12. Use phase-plane analysis to derive the qualitative features of Figure 10.3. 13. Verify (54). 14. (a) Consider the problem
cly dr
Y
,
y(0)
1.
Look for a series solution of the form y = 1 + a 2 / 2 + r2 3 [; + ... + ap e' + • • and show that a contradiction results. (b) Make up a less trivial example than that of part (a) which also illustrates the appearance of a contradiction when a solution of incorrect form is assumed. 15. Show that the lowest order results of (1 ) to 13) are in accord with the corresponding linear theory solutions of Sec. $.1 [found in ( 1.13) and under (1.15)].
PART D Variational Methods and Extremum Principles
11 C al c ul us of V ar i ati ons CH AFTER
I
concluding part of this volume we show how to solve classes of problems wherein a function is sought that extremalizes* a certain integral_ (Such problems are traditionally called varicJrional, for they are often treated by an approach that studies the effect of s mall variations in the integrand.) In addition to showing how to solve variational problems when they arise (Chapter ! 1), we demonstrate in Chapter 12 that it is often advantageous to cast certain other problems into variational form. N THE
Mechanics is replete with variational problems. For example, a thrown ball moves on a path that minimizes the difference between possible kinetic and potential energies (Exercise 3.2). The deflection of a compressed rod can be obtained by requiring that the rod take up a shape that minimizes its potential energy (Exercise 3./). The shape of water waves is governed by the necessity to extremalize a certain energy function (Exercise 3.21). Determining the desired paths and shapes necessitates finding functions that extremali're certain integrals. For example, the rod problem requires that one find a function y(x), subject to certain houndary conditions, that extremalizes the integral
[14fi(y"} x — N(}'`) x ] dx;
L, M N constants. ,
D
This chapter is devoted to developing solution methods for such extremalization problems. We begin in Section 11.1 with Lagrange multipliers, a topic that runs through extremalization problems of all kind. Section 11.2 discusses the simplest class of problems in the calculus of variations. Section 11.3
generalizes earlier considerations so that problems such as the three mentioned above can be treated. Our discussion will be informal; outlines of various proofs wiii be given, but details will usually be omitted.
11,1
Extrema of a Function—Lagrange Multipliers
We begin this section by briefly reviewing the elementary definitions and results concerning the extremalization of a function. We use Taylor's formula to derive necessary and sufficient conditions for the existence of maxima and minima. This approach requires more smoothness than competitive approaches, but the required reasoning is simple and generalizable. • We leave until rarer a careful distinction between extrema and staiiunary values
447
Calculus of Variations [Ch. f 1
448
The section continues with a review of the use of Lagrange multipliers in finding the extremals of functions when the independent variables are constrained. Several of the results will be used in the following discussion.
UNCONSTRAINED EXTREibiALIZATION O F A FUNCTION Let us recall some basic definitions. We say that! has a relative (or local) maximum at x o if and only if there exists a positive number EL such that
f(x 0
whenever 0 <
F) — f(xo) <0
< c,.
(1)
If a ç x u ç b, then f has an absolute (or global) maximum in [a, h] at x0 if and only if
f(x) — f (x o ) ç 0
whenever a ç x ç b.*
(2)
Moreover, f(x) has a relative minimum or an absolute minimum at x o if the inequalities are reversed in (1) and (2), respectively. (For brevity, we shall sometimes drop the adjectives " relative" and "absolute" when no confusion will result.) Suppose that f {x) exists and is continuous in a closed interval [a, N. Here is one way to obtain necessary and sufficient conditions for the existence of a relative maximum at x0 , where ra < xi) < b. From the Taylor formula with remainder, we have
f(x 0 + t:) — f(x0) , 4 - '(xn) + is 2f "(x0 ) + KE ; ,
(3)
where K is x constant given by
KT f""(xu+ f1£} 3!
O
Suppose that f 'Ox o) # O. Then, for sufficiently small I £ I, the sign of f (xo + e) — f( x0 ) is determined by the tf"(x o ) term in (3). 1f 1 '(x 0 ) is positive, then f (x 0 + E) — f (x o ) is positive when L is positive. if f "(x o ) is negative, then f (x o +] — f (x o ) is positive when c is negative. Both of these possibilities contradict (1). Thus a necessary condition for a relative maximum of f (x)
at x0 is that f'(x 0 ) 0. With AO = 0 we see from (3) that the sign of f(x ü + s) — f(x o ) is —
determined for small F by the i£ 2f"(x 0 ) term, if the latter is nonzero. Consequently, a sufficient condition for a relative maximum is f(x 0 ) = 0, f "(x 0 ) < O. Similarly, for a relative minimum a necessary condition is f '(x û ) = 0; and a sufficient condition is f '(xo ) 0, f "(x o ) > O. If f"(x 0 ) = 0, further sufficient conditions can be obtained by considering more terms in the Taylor expansion. Continuing the litany or elementary results, we recall that an extremum is either a maximum or a minimum. Using subtler tools than Taylor's theorem
_
• 111(x)
See_ )!.!] E-rrrema a Function
Lagrange MulripllQrs
449
with remainder, it can be shown (without the necessity of assuming anything about the existence of derivatives) that if x 0 is a relative extremum off(x), then it is necessary that either f'(x 0 ) does not exist or f'(x o) _ O. All points x where f(x) = [] are called stationary points of f. In the one-dimensional case that we have been considering, stationary points are either extrema or points of inflection. (When f' _ 0, the graph off has a horizontal tangent-i.e., the curve "levels off"—so the adjective "stationary' . is an appropriate description of the function's behavior.) Suppose that f is a function of n spatial variables (xi, x x . .., x„i ! x. The definitions of relative and absolute maximum and minimum are obvious generalizations of the corresponding definitions for a function of a single variable. The generalization of Taylor's formula (3) is (Kaplan, 3952, Chap. 6)
f (xt°' } E) - f (0i)—
F ^- (x j°e ) + i =1 Gx I
1r
,
i.j.A -
1
L r+r
O2f
1 [
Xie
ixt°li
(4)
Kijt,EiE c.#[+
Here x {01 = CO() ) .01)1
x(n))
c
—
(CI E
and the K ip,are certain constants involving the values of the third derivatives of f at some point between xt ° l and xt °4 + E. By reasoning exactly analogous to that used above for el function of a single variable, we find that
. L (xto = B'
i = 1, 2,...
► r,
(5)
is a necessary condition for the existence of a relative maximum or a relative 0, where V is an f-dimensional minimum at x a . Sometimes (5) is written ❑f gradient operator. To formulate sufficient conditions for the existence of extrema we recall from algebra the definition of a definite matrix_ I_et Q he a real symmetric n by n matrix with elements Q 13 . This matrix is said to be positive (negative) definite it.
E Qait! is positive (negative)
( 6)
for all nonzero n-tuples ( I ,^ , ... „ „) of real numbers.* Again by paralleling the reasoning used for functions of a single variable one can easily show that a • Necessary and sufficient conditions for positive dehnitcness are (i) that all the eigenvelues of Q are positive, or [ii) that the determinants formed from the by r square array of upper left elements (principal minors) are positive. r = I, 2.. - - . n. The latter condition forms a practical test- Those riot familiar with these criteria can fired thcm proved in algebra texts, particularly those speciaLi2ing in matrices. See, for exanipLe. P. Lancaster, Theory of Muin i'a [New York: Academic Press, 1969), pp. 93ff.
Caknilt,rs ni Variations
450
[Ch, HI
sufficient condition that a (three-times continuously differentiable) function f (x) have a relative maximum at xt0 is Of
Dx; (x
t°i ) = 0,
^ a , (x(15 ') negative definite. Oxi
(7a)
A sufficient condition that xt 0 t be a relative minimum for f (x) is that
_we) ) = [1 da[#
22f ,
r x i r7x i
(x'° i) positive definite.
(7b)
CONSTRAMNEI) E7{TREMALIZAT1UN 4)F A FUNCTION
Frequently, one must find an extremum of f (x i , x 2 , _X3) = f(x) subject to the condition that g(x) = O. Geometrically, this condition means that the point x is constrained to lie on the surface g O. As an example, tet f (k) be the average temperature at any point x within the earth. Lct glx) = O be the equation of the interface between two different rock strata. To Find the hottest point on the interfacc,one must maximize f among all points satisfyingg = O. Because ofthe constraint g = O. the variables lx 1 , x2,x3)are not independent, so the usual necessary condition Vf O is usually riot satisfied.* A possible way around the difficulty is to solve y(x) — 0 for one of the variables in terms of the other two. Upon substituting the result into f (x), one will obtain an ordinary extrerrtalization problem fir a function of two independent variables. We shall give an alternative approach that permits a geometric interpretation of the cond i tions to be derived and that preserves the symmetry among the variables which is lost when one is solved for in terms of the other two The geometric derivation is convincing, but those who prefer analytic proofs can attempt to show that the same answers would result if one variable is eliminated. Exercise 15 guides this attempt in a simple but representative case, For definiteness, we shall now discuss the minimization of but the more general case where f is extremalized is handled by entirely similar arguments. Let the point Q, with coordinates x,,, be a local minimum for j on the surface g = O. First consider the exceptional case where xQ in fact provides the smallest value for f(x) among ct11 points x near x (2 . Here we have Vf (x (2) = 0 by standard arguments. If xQ is not a local minimum off, the level surfaces/= c; in the neighborhood of xy have the general appearance depicted in Figure 1 I . I , where the c ; are chosen so that c g < r 2 < c3 - - - - (By definition, f is constant on its level surfaces. In our example, these would be isotherms.)
f
* Refciring in the above example. VI will gcncIJlly vanish ai the hottest point within the carlh. But this potni wilt floc usuaify be found on the intert`acc,
Set -. 11.11 Extrema of a Fu ►rrtirnr Leitlrwryar Multipliers J=t2
1=r1 1-tar
45 L 1=[-i
I t.I. The tines I"ubeled f= c rEprt'ss°rlr surfaces of constant The whir of f increases as one passes from a level surface to th e one on its right. On the surface g = 0, the smallest value of f is attained at th e poinr Q where the fever .surface f= c is rArtyerrt tri flit" constraint surface y ={]FIGURE
The constraint surface g = 0 is also represented in Figure 11.1. As a point moves along g = 0 from P to (j to R, the function f decreases from the value e s at P to the value c' 3 at Q and then increases to the value e s xi R. For points on y = 0, f has ra relative minimum at Q ssrhesre the level surface f — c 3 its ianyent trey = O. Let x (2 be the position vector at Q. Because the two surfaces are tangent at Q. a normal to f = c 3 at Q. Vf(x (A must he collinear with a normal to g = Oat Q, Vg(xQ ). This assumes that the tiormals exist, which they will if the partial derivatives are continuous* and the gradients are nonzero. Leaving aside the possible nonexistence of normals, we can thus infer the existence of a constant r. such that VINO = .4Vgix (2 )
-
(
)
if x¢ is an unconstrained local minimum of f, then, as we have seen, 01 holds, with ). — O. If Vg(x 42) = 0, then no scalar ). will make (8) valid, unless by chance Vf(x Q ) = 0. • Many resuLis involving functions of several lariabtci. including ihriNe reyuircd io prove that Vf is normal lo the surface] = constant, stern frira the furmula
}ix +
- J{xl = WOO - Ox + 1:• eix.
where km €- O.
0.1•11- 6 This requires that the pallid derivatives exist and are continuous in e domain conldining x.
45 2
CQ1[-irlus of Variations [Ch. 11
The symbol A is called a Lagrange multiplier. We have indicated the truth of the following statement concerning the use of Lagrange multipliers in finding extremals subject to constraints. Theorem 1. Suppose that NI gives an extremum of f(x) among all points x which satisfy g(x) = 0. If f and g have continuous partial derivatives in a domain containing X Q , then either
g(x(2) = 0 and
Vg(x Q) = 0,
(9)
or there exists a constant .1 such that
9[a) = D and Vf (x0) — AVg(xQ) = 0.
(10)
Candidates for extreme values of f (x) satisfying g(x) = 0 are, therefore, Points where either f or g fail to have continuous partial derivatives. (i) (ii) Points satisfying (9): g = 0 and Vg = ü. (iii) Points satisfying (10): g = 0 and V( f — ,fig) = 0. li the domain ofg is limited (explicitly, or implicitly, e.g.. by the requirement that a square - root function must have a nonnegative argument), then, as usual, points on the boundary must be given separate consideration. One way to remember the conditions of (]0) is to pretend that the problem is to make an unconstrained extremalization on the function of four variables x3).
f(x1+x2, x3) —
Then we would have at x (2 i F
aF of r7 F ax, — ax 3 = aX 3
•
The 2F /'x ; give the components of V( f — )4) while OFfeA is proportional to g. Note that the results would be the same for constraints of the form g = constant. Sufficient conditions that a "candidate" is actually an extremalizing point can be formulated using Lagrange multipliers but are somewhat complicated.* It is usually no more difficult to verify directly that a "candidate does maximize or minimize f.
Example 1. Find points on the cLIipsaid 9(x, y.
X2 y 2 2 2 zl- ^2 + = {}, b'+c'— l
where
>b2>cz, which are closest to and farthest from the origin. a2
(12)
• For example. sec pp- 431 f. in Vol. 1 of V. L Smirnov. A Course of Higher Marhemaiks (Readins, Mass_: Addison-Wesley, 1964)_
See. 11 _ 1]
Fxrre ►na
Furnitrin -- layrarrye Jllultiplif•rs
453
Sohrtian. As usual, it is easier to cxtrernalize the square of the distance. To exy2 + Z 2 subject to (i l}, treinalixe f(x, y, z) = 12 we use (9) and 1itid candidates by simultaneously solving { I I) and ,
x2
yr
/x 2 +
t2
+ 2 + r .2 - )] -a 11 1
y2 +z2 -a.l
y
(13)
^.
Equation (I 3) gives 2x - 2.+ -2x = 0.
23 , - 0.
2y -
2: - 2Ae - :z = 1]_
(14)
These three linear equations for x, y, and z have nontrivial solutions only for certain values of d- The SO lurions are
â
▪ x=e t . 1'=0. _ • x = 0. y T c2 .
^•
0:
(LSa)
r = 0;
(l 5h)
c3;
( t SCl
.;, x = O, l' = 0, z
where c,, c 2 . and
-
c l are arbitary so far- Substitution of(IS 5i mue U I) gives ± b , c^ = ±C-
e t = # a. r- 2
(16)
The functions f and g have continuous derivatives and Vg is nonzero for points on y = 0. z) given by (15) and (1.6)_• (These points he a1 so the only candidates are the points the ends of the axes of the ellipsoids.) To determine the nature of the extremum provided by the candidates (a, 0, 0) and - u, 0, 0), we wril e = Fa+x',
y
(
z
the primed variables measure the extent of the departure of a typical point (^ y, rj from a candidate point. Subtracting the square of the distance in the origin of the candidate points (± a, 0, 0) kern the corresponding quantity for a typical point, and using the fact th;li trie latter point must lie on the ellipsoid, we find the following: ,
f (x, j, z)
11-1-d2,0,0) = (±a ±
But
t'1 ? +
(ty` + (c'} ? — a2
tijx + ( h ^ z + l rz2 s 1
}a ^ Or
{^Q
♦ ^ ) 2 a2
=
-
b2
(ÿ { }^
uz {r 'lz,
so fix.
a= y, z) - fl±a, U, 0) = (y') 2(II - - ) + (z') 2(1 -
6^
[
ur
(17(
2
y' = z' = 0. so both candidate By the inequalities 112j, this difference is negative unless points yield the same abswlure maximum distance from the origin_ Testing of the other two pairs of candidate points is left no Exercise I.
• Note that the num}wrs1in(15) g ivethe vaLuc of + y2 -4r' for the ascocinjied candid a[cs. The permitted values of t he Lagrange multiplier frequently have some such Ecumenic or physical interpretation_
Calculus of Variations [ CJr . 11
454
To test whether the stationary point x 103 is an extremurn of f, we have tried to determine whether the difference f(x"" 4 x') f(x 1°) ) is positive (minimum)or negative (maximum) for all x' which are such that x = xI°t + x' satisfies the constraints of the problem. For local maxima and minima, this -
—
difference will be of one sign only if Ix; I IxzI, and Jx3I are sufficiently small, and this can be taken into account by neglecting all but the largest terms in the difference. Because xt °1 is a stationary point, the difference will not contain any linear terms. Neglect of higher order terms will usually yield a quadratic expression [as in (17) and Exercise 2] whose character can be determined by various methods such as those noted in the footnote to the line following (6). ,
Example 2. {This example shows the importance of the exceptional case (9).1 Find the minimum distance from 10, 0, - I) to the surface A(x. }', z) = x z +
(18)
— z s = O.
From Figure 11-2 the answer is clearly unity. But let us see how our machinery operates_ r
FIGURE 11.2. The analytic determination of the minimum distance from (0, {) — I) ro the surface depicted tuas an exceptional character herause the surface has a cusp at the closest point ro (0, Q , — Ij .
Solution. To extremalire f(x, y, 2) = xt + y2 + (z + I )i subject to (I 8), we attempt to find candidate* by simultaneously solving (I8) arid F[xe +y2 + (z + 1) 2 —
+y2 —z 3 )]=4
or
2x(1 — d) = 0,
2y(I -- d) = D,
2(z + 1) + 5 rl.z` = ü-
(19a, b, c)
If A = 1, r + 1 S O from (19c), which contradicts z Q, a consequence of (18). If û contradicts (19c). The four # 1, (19a, h) and (18) give x = y == z = D. but z i
Sec. !!.!]
ExtrPrrra
of a Function—Lagrange Multipliers
455
equations (1$) and 119) therefore have no real solution, so no "successful candidates" have yet emerged, even though f and g have continuous derivatives everywhere. But Fg = Oat (O. 0, 0) and q(0. 0, 0) — 0. so the origin is the "candidate"for which we were looking_
Exercise 20 provides an application of Lagrange multipliers that is at once more physically significant and more subtle than the examples treated so far. INEQUALITY CONSTRAINTS
Lagrange multipliers can also be employed in extremalixation problems where the constraints are inequalities. Suppose, for example, that it is required to maximize a smooth function f(x 1 , )(,) subject to the constraint h(x 1 , x 2 , x 3 ) 5 fl_ The maximizing point is thus restricted to be in the closed region R bounded by the surface h = 0 (Figure 11.3). (Think of the problem of determining the maximum temperature_f among all points of R, whether intenor points or boundary points.)
The problem. is to maximize a funrtion f utttony all point's in the region R defined by h O. Suppose that the maximizing point is at x12y. on the boundary b= 0 of It Since rhe gradient points in the direction of maximum kJcrease, ;be normal ŸJr(x{ painrs away from negative txelueS of h and toward positive values. Hence, tf x rs any point interior to R and sufficiently close to 121 ) is x 421, then the angle 9 between x -- x^ 1 and the exterior normal C h(x obtuse. This fact is used in the derivation of 125)_ FIGURE 11.3_
Calculus of Variations
4515
[Ch. II
We can treat the present problem by combining our previous approaches. In the open set of points x satisfying h(x) < 0, we search for local maxima of where Vf = O. Possible maximizing points on the boundary are subject to the constraint h = 0, so a Lagrange multiplier can be employed_ But a maximum attained at x(2) on h 0 will not be a true local maximum for this problem unless f (x12) )is larger than the values of f at neighboring points in R. (Of course, this complication was not present when we were dealing solely with the constraint h = 0.) We shall find an additional necessary condition ensuring that a candidate point on the boundary will, in fact, be a iota) minimum off in the region H. The various conditions can be neatly combined, as the following detailed discussion shows. _f a maximizing point x;° is in the interior of R, then of course Vf (x" I) = O. In this case the usual condition Vf (0°) = ).Vh(xt 1)) holds for 1 = 0. If a maximizing point x(2 ° is on the boundary of R, then Mx" )) = 0,50 by Theorem 1 [if the exceptional case (9) is ruled out], Vf (x 121 ) -= AVh(x 12 '),
/!(x 121) = 0-
(20)
Since x# r} is by hypothesis a local maximum off, then (21)
_f ix) - f(x"l) < 0,
fer any point x that is in R and that is near x(21. But, to a first approximation [by Taylor's formula (4)].
fix) Since Vf
—
f( x(21) = [1(x 121 )] - (x
—
x121 ).
AWE, by (20), for boundary maxima.the above equation gives f (x)
—
f(x 14) _ A[Vh(x1")] . ( x
—
x 12 ').
(22)
Let U be the angle between x — x12F and Vh(xW W W). This angle is obtuse, so cos e < C. (See Figure 11.3--the convexity of R at x (21 is not necessary.) But (22) can he written
{ ix) — ( x i2)) — % I
VJ!(x(2I)I I x — x [21 I cos e,
which is consistent with (21) and a negative cos ( if and only if di is positive. If x12 ' were a minimum, ,f would have to be negative. Let us sum up. Assuming smooth functions, to extremalize f (x) subject to the constraint h(x) 0, we search for points x and constants ) which satisfy
Vf = Wh.
(23)
Solutions to (23) must further satisfy either h
(24)
(and the usual sufficient conditions for unconstrained maxima and minima) or
h= 0 and
A >0{?.<0)
(25)
Sec. 1 1 .11 Extrema of a Function—Lagrange Rlultiptiers
457
when a relative maximum (minimum) is being sought. It can be shown that exactly the same conditions (23) to (25)hold when x is an n-dimensional vector. The same formalism, generalized in are obvious way, applies to the case of p equality constraints and q inequality constraints. Thus suppose [hat it is desired to extremalize f(x) subject to the constraints
i = 1,... p;
FIB{x) =
hi(x)
j = 1,..,
We then search for points x and constants A t i = 1, .. ,
. ,
(26a, b)
p, and p i , j = l . .. _ , c1,
that satisfy
VI
=
é
y
-1
1= t
E Ugh + E
(27)
The A, are determined by the requirements (26a) while the Pi must either satisfy p i 0 if h-<0 or be such that h ; =0 and pi >O (p; < 0) when a relative maximum (minimum) is being sought. The theory of Lagrange multipliers for the general situation in which there are a number of equality and inequality constraints is treated rigorously by Hestenes (1966). His book contains proofs of the results that we have stated here. Exercise 15 shows in a relatively simple case how a proof can be constructed by legitimizing the process of elimination of variables. Indeed, such a process can always he used "in principle" to dispense with Lagrange multipliers. This would not be wise, however, for the multiplier method is not only pleasingly symmetrical but also generally requires less calculation (Exercise 21). -
EXERCISES
1. Complete Example 1 by determining whether the "candidates" (0, + b, 0) and (0, 0, +c) give a maximum, a minimum, or neither_ 2. This exercise is concerned with the determination and classification of the stationary points of
By' + z 3 — 3z f(x,y,z) = Ax 2 3
(A # B)
among all paints (x, y, z) on the cylinder x 2 + y 2 = 1. (a) Show that the stationary points are as follows
(0, 1, 1), (0, 1. 1), (0, 1, 1), and (0, — 1, l) (1, 0, 1), (-1,0, 1), (1, 0, —1), and (-1.0, —1) —
(b)
—
—
when A = B; when, =A.
To determine the nature of the first stationary point, write X
=x',
y = 1 + y',
z =
and show that among all points on the cylinder
.Ï(x', 1 + ÿ', I + z') — f(0, 1, 1) _ (A — $)(x') 2 + 3(z +) 2 + (z')3.
Calculus of Variations
458
iC h. ! f
Sufficiently near (0, 1, l)„ the deviation z' is small compared to 1, so (z')3 can be neglected compared to (z') 2 . With this, it is at once clear that the point in question provides a relative minimum if A > B, and is neither a maximum nor a minimum otherwise. (c) Determine the nature of (1, 0, 1). (The nature of the remaining stationary points follows from symmetry considerations.) Using Lagrange multipliers, find and classify the extrema of x + y on [he sphere x 2 + y2 + z 2 _ 2. Use geometrical reasoning to check your calculations. 4. lise the method of Lagrange multipliers to help find and classify the points that extremalize 1 (x 2 + y) and satisfy y = nix + b•m and b constants. With the aid of sketches, show that your answer is sensible. S. consider the problem of extremalizing f (x, y, a) subject to the constraints g(x, y, z) 0, h(x, y, z) = C. Modify the geometric argument given in the text to show that Vf must be in the plane determined by Vg and Vh, and thereby obtain a Lagrange multiplier formula for this problem. 6. Using Lagrange multipliers, find the points on the curve —
z
=
x 2 +y 2
,
x +y +z = 1.
that are nearest to and farthest from the origin_ Use geometry to determine the nature of the "candidate" points. 7. Use the method of Lagrange multipliers to find the points on or inside the circle xz -f- (y' — u)x = I that extrcmalize 1 — (x 2 + A. Treat all nonnegative values of the paramet er a. With the aid of sketches, show that your answer is sensible.
8. Repeat Exercise 4: (a) When the constraint is y < nix + b. (b) When the constraint is y mx + b. 9. Find the necessary and sufficient conditions for f (x) to have a relative maximum at x a when f "(-x ri ) ~ 0_ lise the Taylor formula approach. H). Fill in the details required for the derivations of (5), (6), and (7). 11. Solve Example 1 by eliminating z. Only one pair of candidates are determined by setting the partial derivatives with respect to x and y equal to zero. How are the other candidates obtained? Compare the calculations needed here with those of Example I. 12. Solve Example 2 by eliminating x 3 + y 2 . Compare this approach with that of the text. 13. (Project) lay solving various examples in the two different ways, try to determine the comparative merits of the Lagrange multiplier method and the method of elimination. 14. Construct an example which illustrates the fact that boundary points of the domain off may give the solution to the problem of extremalizingf subject to the constraint g = C.
See. Mil Ertrrm e,l afiolrrriiu2
La y■anye Mulriplirn
459
15. This problem outlines a proof of the Lagrange multiplier method in a simple case- Let (xo, ye) give a relative extremum of fçx, y} among all points satisfying 9(x, y) = O. Suppose that f a nd g have continuous partial derivatives in a domain containing (x0. ye). +(a) if condition Y" is satisfied, a famous theorem guarantees the existence of a function [t(x) such that
fi [xo • c(xol] + 4 ?(xo)fi[xo , 4(xn)I
= 0.
gs [YO,'(xoln + "(xo,)92(xa,Oro)] = 0, where a subscript r denotes a partial derivative with respect to the ith argument. What is condition Y. and what is the theorem? Quote all theorems necessary for the justification of the preceding equations. (b) Consider the first of the preceding equations minus ). times the second. Shov that if condition Y h olds. then uric can definer b . vo)Iya(x . t'a) and thereby deduce the two-variable counterpart of (el Complete the proof by considering what happens if condition }' fails to hold. in particular, show the possibility of (9). (d) The final result could have been obtained by noting from the equations of part (a) that the ratios i t y i any! f ..cy, have a common value which can be denoted by — Â. Why was the less direct procedure of part (b) used instead? 16. Extend the above proof so that it applies to Theorem L 17. Extend the above proof so that it applies to the situation of Exercise S. 18. Students often assert, without proof, that of Iwo stationary points which they have found, the one that gives the larger value of the function in question provides a maximum and the other a minimum. Show the dangers of this procedure by discussing the extremalization of /(xi in the case where x, and x, are the only values at which J'(x) — 0, and f(Yil < f (x2). 19. Let x be an n-dimensional vector and consider the problem of extremalizing f(x)subject to the constraint c tx) = 0_ Lei x'" } bean extrernalizing point. Using the language of differentials, if x + dx is a point near >K, then it must be that
4f = Vf - dx = D for all dx satisfying (a) (b)
(c)
dg = Vg dx = D.
Why? Show that demonstration of the Lagrange multiplier condition reduces to the proof of a theorem in geometry (or a theorem about rt-dimensional vector spaces with inner products) and prove the theorem. Generalize to the case where there are p constraints rl ; = O,
Cufc•ua#us of Yariarions [CJr
4 60
.
11
20. This exercise uses Lagrange multipliers to prove a theorem of Coulomb (plane case, 1776) and Hopkins (general case, 1847). The approach is that found in J. L. Erickson's "Tensor Fields," Handbuch der Physik, Vol. 3/1, S. Flùgge, ed. (New York: Springer-Verlag, 1960) p. 845. (a) Let f = (ft , fz, 1.3) and g (9 L' 92, 93) be mutually perpendicular unit vectors_ Cite authority for the statement that if the components of the stress tensor are denoted by Tip then ar = L f 7+1y1 gives the shear stress on an element with unit exterior normal f.4 (b) We wish to maximize the shear stress. Justify the assertion that central in doing this is consideration of the function 3
y
i. f= 1
(c)
(d)
3
f i191 +
(
i = J
an +
+ yii9J)•
Provide a complete description of what must be done with this and other functions. Set appropriate partial derivatives equal to zero. Take scalar products of f and g with the resulting equations, use the fact that f and g are orthogonal unit vectors, and dedu ce that a = 13. Now show that g - f is an eigenvector of T with eigen va lue 2a y and that g f is an eigenvevtor with eigenvalue 2a r. y, Call these eigenvalues 1, and J. 2 ; show that the extremal shear stress equals -
—
—
— A2)• Conclude that the maximum shear stress is equal in magnitude to half the maximum of the absolute values of the differences between principal stresses. Show that the plane of maximum shearing stress bisects the angle between the principal planes of stress. Sketch. N O T E. In general the stress tensor, and hence the maximum shearing stress, varies with space and time Since many materials Pail because ofexcessshear, the location and magnitude of maximum shear stresses is a matter of con sidera ble practical importance. 21. Consider the problem of finding points on the quadric surface (e)
3
E Ax x 1
I
(A
j
A bu
(28)
i.1 =1
that extremalizes the squared distance to the origin. Without using Lagrange multipliers. show that the problem reduces to the determination of the eigenvectors of the symmetric matrix A il , with lengths chosen to satisfy (28). Proceed by regarding (28) as an implicit equation for x 3 and
differentiating implicitly. (Note that more effort is necessary in this approach than in the two line calculation at the beginning of Section 12.1.) -
• We do not use the summation convention in the last two chapters. in order that they can be read independently of earlier miitcrial-
See. 11.21 Irtrrndtrc•rtrm in the C àlcuhr.s of Variations
46 t
11.2 Introduction to the Calculus of Variations Children know that the shortest distance between two points is a straight Tine," but it takes a considerable amount of mathernatal knowledge to prove it, particularly in a generalizable way_ As a start, let us restrict consideration to plane curves y = y(x) joining the two points in question. Take the origin of coordinates at one of the points and denote the other point by (U, b). Recall that
ds2
dx 1 + d}'^
or
d =1+
ax
1L
dx.
(I )
These relations sum up a discussion of arc length ds that is familiar from calculus. They remind us that the length of the curve y = y(x) for x between 0 and a is given by
f
ri 4
(y')
112 dx_
(2)
We would like to show that the integral of(2) (when it exists) is smallest when This y = (b/a)x among all Curves y = y(x)starting at(O,O)and end ingal (a, is perhaps the simplest problem of the calculus of variations, a subject that deals with the determination of functions which extremalize integrals. THE BRACHISTUCtIROr4F
A slightly more difficult calculus-of-variations problem, whose answer is not obvious, is the famous brachistochronc problem of classical rnechanics.* Suppose that the curve of the previous paragraph is a "frictionless" wire and that a constant gravitational force acts in the direction of y increasing (see Figure 11.4). For what curve will a point mass that is initially at rest slide from (O, O) to (n, b) in the least time? Energy is conserved. Let us prescribe the potential energy to be zero at (D, 0). Equating the kinetic plus potential energy to a constant, and taking this constant to be the total energy at the origin, we find that rnc 2
+ mg( —
= O,
so t; = (2g},)"
(3)
Here L is the speed of the particle and m is its mass. Now y = dsjdi, where s denotes arc length. Let t o and r# be the times at which the particle is at (0, Oj and (a, b), respectively. Then the time to be minimized is r, — t o = j di. Since di = (is»), we seek to minimize
t,a L1 + (y) 21"(29y ,1" dx
(4)
* Greek: brarriys =Shoot chrunns = time. Publication ur ;his problem by John Bernoulli in 1696 caused discussion and argument [Kai gave birth to the calculus of variariuns as a welldcfined subject. .
Calculus a# Variations [{' k If
462
-
x
Sliding point rnAss
F URE I L.4. in the bruchisrarhrone
prnblem, one is faced with the Risk of the shape of rr frfcttarrfess wire y = fixj so Mai an i+riiially ionar}' particle will slide from CO. 01 to (u, 6j irr mfnitnum time.
c'ho3et,sir+g
among smooth functions y that satisfy the boundary conditions 140)= 0,
yt)= h.
To minimize the total distance traveled, the wire should, of course, be tautly stretched between the two points, On the other hand, an initial period of vertical fail will build up speed most quickly, albeit with a distance penalty. We seek a compromise minimizing the time of travel_ A !GENERAL IxAL E3CTREMALIZATI4T! PROBLEM The problems just mentioned, although interesting, barely indicate the importance of the calculus or variations. Nevertheless, they provide sufficient motivation to proceed by formulating a problem general enough to include
minimization of(2) and (4) plus a number of other integrals of interest. Problem A xtrernalize
(5) among all sufficiently smooth functions Ÿ that satisfy Act) = A,
}'(19 ) = B.
(6)
"Sufficient smoothness" means sufficient continuous derivatives so that the integral in (5) exists and that the manipulations we shall perform are valid. We shall be more specific later. In (5), 1 is an example of what is called a functional. Recall that a (realval ued)f enction y can be thought of as a rule which assigns a single real number y(x) to every number x in its domain of definit ion. A real-valued f motional I, like (5), is a rule that assigns a single real number f() to every function ], in its domain of definition,
5er- - ,1!-2J
Introduction to zlre°
C'alr•ulusof Variations
463
For definiteness we shall first seek a function y(x) that maximizes the functional (5). We denote the maximizing function by y, assuming that it exists. What we shall find is a differential equation that y must necessarily satisfy. The solution of this equation that satisfies the boundary conditions (6) will presumably give us the function we seek. Just as a simple analyticcondition (vanishing derivative) holds at a relative maximum of a fartctkm, not necessarily at an absolute maximum, so we expect a simple analytic Condition at a relative maximum of a functional, not necessarily at an absolute maximum. We must try to deduce something from the assumption that among all sufficiently smooth functions 9 which are near y and which satisfy (6), f(P) is largest where F = y. But when is p "near" y? Since both the function and its first derivative appear in the iniegrand of(5),we shall adopt the following natural (but not inevitable) definition of"nearness": x < fi
{7}
(ÿ'(x) — 3/(x)1 small for a ç x S f.
(g)
I y(x) — y(x)' small for a and
Furthermore, for reasons that will be stated later, by a "smooth" function we shall mean one which has continuous second derivatives in [cc, M. (Onesided derivatives are required at the end points.) We thus seek a condition necessarily satisfied by a function y which maximizes the integral (5) among all functions that (i} are twice continuously differentiable in [a, #] (ii) satisfy the boundary conditions, and (iii) satisfy the "nearness requirements" (7) and (8). Functions satisfying(i), (ii), and (iii) are said to be admissible, for they are admitted to the minimizing competition that is won, by definition, by y. ,
THE E:llr.tvR EQUATION
It would obviously be helpful if we could somehow transform the question of maximizing a functional to a question of maximizing a function. The following observation makes this possible. Let s be any function that is twice differentiable in Ea, f] and satisfies the
boundary conditions s{ac} = D,
AM =
Then if we define 9 by
y(x) = y(x) +
ES(x),
it must be that
1( + es) < !(y)
for < _ It:I < Ei
(11)
if e- is sufficiently small. The reason is that p, as defined by (IC), is admissible and y is the function that minimizes f among all admissible functions_ Requirement (1) is satisfied because of the postulated smoothness of s, {ii) holds because
46 4
Calculus rrf Variuriarl.s
[C'h. I
of (9), and (iii) will be valid provided a is taken to be sufficiently small, as it is in (11). What has been accomplished is the generation of a class of admissible functions which are parameterized by a in such a way that the intcgrand is maximized when E = 0_ if we focus our attention on E. we have afunction of E with a maximum at E = 0: the standard methods of calculus can now be employed. To this end, we regard s as fixed and consider I(}° + Es) to be a function of e, which we denote by .0(e). From (11), .0(4 has a relative maximum at E = O. A necessary requirement is that .J(e) have a vanishing derivative at E = 0, that is, (i.
_o
dE
= O.
(12)
From the definition of f in (5), assuming that we can differentiate under the integral sign, we find that f d dE F(x'
dE
y + E+, + Es'} dx.
(13)
Using the chain rule and remembering that only E is considered to be a variable, we have dr
F'(x, y + €s, y
'
ds Fix, c, )
+ es) =
OF(x, C, ^) -
^y1
+cs re
z-ytcs
d
d ^+
E5)
G
+ Of(x, ^ ^) ^= y +cs a=y +rs"
^^
d
de (/
+ Es'}.
We can write this result in the form
d dt:
F(x, y
F2(x, y + Es,
+ Es, }''+E_ti')
y'+ ES1.S + F3 (x,y+ Es, y' +E `}s',
where the subscripts 2 and 3 indicate partial derivatives of the function ) with respect to its second and third arguments, re prctively_* Fl Thus requirement (12) becomes ,
d3 dE
^
(14)
j" LF 2 (x, }', y')s + F 3(x, y, }r)a`] d.r = 0. r
=o
* T o makc surc that the notation is cicar, we cnyat~asixc 'hail in F,{x. y + ^^ y' + cs") for eaaL7lple, the substitution of y # ts loi the seco nd variable is made aft e r the differential loin with respect it) the third variable_ [C'omparc iht no t ation f1{3) for 41(x)/dx ai x = ü } Thus if Fix, ri) xlnq thcn ,
,
.
,
FAN.
}' + fS, j•' + "
fS'1
[^ + rs) sin x ^
9+
^
Sec. 11.2] aRnlrode.rtrnn in th e C'rdrulu.s of Variations
465
Equation (14) holds for any smooth function s that satisfies the boundary conditions (9). The arbitrariness of s can be exploited if we integrate the second term in (14) by parts: Fs(x. y . _Ile dx
F3(x, y, y`]s I
;-
^
¢
x
r
d
—
dx F3(x, y, y') dx. ^s
(15)
The "end-point contributions" from x = ct and x = ( vanish because of the boundary conditions (9). Requirement (14) can therefore be written F2(xYY') — f
F(x,^', y^^ 5(x)dx = 0.
(16)
[
Sinces is almost entirely arbitrary, one would guess that the square-bracketed terms would have to vanish. This is true, provided that these terms are continuous (Lemma A, Appendix 11.1). We have thus derived the desired necessary condition
d
3F2(x,yY y, y') + ) ! dx Fs(x,
(17)
Equation (17) is called the Euler equation (or sometimes the EulerLagrange equation) of the variational problem_* This equation should be remembered in the form (17), if at ail; but the indicated differentiation can be carried out to obtain
F2 (x, y, y') , F3 1 (x, y, y') —
y _v')y' T F 31(x, Y, Ay" = 0, ,
(1 8)
where the double subscripts denote the appropriate second partial derivatives. Written in the form (18) the Euler equation is seen to be a second order non-
linear ordinary differential equation for y. For the left side of (18) to be continuous, as is assumed in Lemma A, F must have continuous second partial derivatives and y (and hence s) must be twice continuously differentiable. The first of these conditions limits the class of integrals f that we can maximize. As we have seen, the second limits the class of udrraissible ftertcrivns from which the maximizing function can he selected. With small changes, our argument applies to minimizing Ï. Using slightly different notation, we can sum up our discussion thus far in a theorem. * "Typically, Euler would discover the germ of an idea in the course of solving a particular problem. He would then go on to other matters and Lagrange would take up the idea, generalize it, show how it could be used for a large class of problems, and exhibit its relation to other ideas. Lagrange was not just an imitator who followed the paths Euler had laid out. He added so much that he must be rated as Euler's equal. but he is most distinguished by his desire to unify and generalize where. Euler was content with the fragmentary and the particular_" From p. L63 of R. Friedherg,.4ri Adoenreurer's Guide to Nirrrzher Theory (New fork: McGraw-HiLl. 19681_
466
C'uhulas of Variations [['h. ! I
Theorem 1.
Consider lfÇ F(xi
i
Y
dx.
( 1 9)
where F has continuous second partial derivatives (in a domain containing the arguments of interest). We regard as admissible all functions j' that are twice continuously differentiable and satisfy the boundary conditions
y(a) = A ,
(20)
y(13) — B.
If y gives a local extremum of ig., ) among all admissible functions, then y necessarily satisfies the second order Euler equation
Fy(x, y,
d
y) - dx
f
y (x, y, y)
=0
(21)
and the two boundary conditions (20). Make sure that the notation in (21) is understood. For example, F is the partial derivative of F(x, y, y) with respect toy', regard ingx arid as constants. Equation (21) is, of course, the same as (17). The notation used in (17) is more convenient in the process of derivation, where the arguments of the functions keep changing. The notation of(21) has the advantage that one can omit the arguments and still retain intelligibility, writing F. - (dfdx)F,.. .= 0. The reader should he familiar with both notations. Example 1. If y y(x) is the curve minimizing the distance between (4, 01 and {a, b), we have, from (2),
F(x, }', }' ) = [1 + (y''jVrs,
y(ül = U.
Since F has no explicit dependence on y [Le, F,(x, y, y') obtaining F,. = constant or, for- this problem, [1 +(3911112=l [1+
g
Ÿta)
b
122a, b, c)
ü ] we can integrate (21) once, .
er2=
'
w h ere
-
[.
C is a conslani. Squaring and simplify ing. we obtain (Al
K,,
wltereK; = C.' =(1 - C 2 I '.
Thus yr = +X,.% + K2 and K z = 0, +I< = bla. from 22b. £1 The final answer is the expected straight Line y -
In the above example we have used the obvious fact that the Euler equation can he integrated once if F happens to be independent of y. An integration can also be performed immediately if F is explicitly independent of x. This is
less obvious, and more useful.
f!-2] lnlrodur•rrofr Set - -
Theorem I. If
F
to the Calculus of Variations
46 7
depends only on y and y' , then (21) implies that
f— y'F = C,
C a constant.
( 23 )
Proof
(F^y'f)^F ,y'+ Fir y" ^},"^' } . — ^' x(F .•) dx d ,
,
y. = dx (Fy) = l}. —^ l T ^
D
In reducing the problem From (21) to (23), an extraneous solution is introduced; the above equation shows that (23) implies either (21) or = O. But the curve y = constant will generally not provide an extremal- y' (See Exercise 10.) As is shown in Exercise 5, the hrachistnchrnne problem, wherein F provides an example of the usefulness of Theorem 2. [1 + y') 1 )" 2 [2yy] Always check to see whether F is independent of either x or y, for then, as we have seen, an immediate simplification is afforded. NOT F;_
NATURAL. BOUNDARY CONDITIONS
Suppose that an integral from x = a to x fi is to be minimized without any restrictions as to the behavior at its right-hand boundary. Although no constraint is imposed there, nonetheless the minimizing curve must satisfy some condition at this boundary. We now turn to the matter of determining such a natural boundary condition. A brachistochrone problem provides a good example of a situation that involves a natural boundary condition. Suppose, as before, that the particle starts from rest at (0, 0), but now let us require only that the wire end on the line x = a (Figure 11.5). What curve now minimizes the time of travel? In particular, how does the minimizing curve behave at x = a? More generally, consider the following problem.
Problem B Extremalize !(y) T â F(x, ji satisfying
P')dx among ail sufficiently smooth functions j'(ar} = A.
(24)
We proceed as nearly as possible in our earlier manner. We again consider admissible functions of the form y(x) + es(x), but now we drop the boundary condition s(f3) = O. If y is the extremalizing function, then /O. + s) must have an extreme value at E = O regardless of the value of a) as long as s is smooth and sac} = 0. As before, we require that the derivative with respect to E vanishes at E = O. Because of the dropped boundary condition, one of the
468
C'alculus of - Variations [Ch. 11
Ÿ r G U R E 11.5. The solid curve gives the (schematic) shape f a wire along which frictionles s passage oj a mass point minimizes the time of r ravel from (0, U) ttr (a, b). Along which of the dashed curves will a parttcle lear:rng (0.0) cr oss the iirie x r a in minimum rime? 1r is not enough ttodetermine the differential eguatiwr jor such a curve. }n addir ion to rh e Crrndi r ion _0( 0) = O. w e must somehow determine ano ther naturally occurring boundary condition ar x= v.
end-point contributions is nonzero, so that after parts integration, instead of (16) we have , y, y.) — dx t 3(x, y yds(x) dx + F3111, AM, y'(fillsifi) = O. (25) ,
^Fa(x a
Now (25) holds reyardles.s rd the value of s(). It certainly holds for the particular case s(P) = O. [It also holds when s(fl) = I or s(j5") = }, but these possibilities are not of present interest.] When 5(13) = 0, the second term of (25) vanishes and we deduce, exactly as before, that Fa.(x, y, y') T
dx
F3(x, y, y') _ 0.
(26)
Equation (26) is, of course, a condition on y(x) and is entirely independent of s(x). Substitution of(26) into (25) yields
FAA Y{$), y`()] , } = 0,
Sec. If .2) introduction to the C'rlh•Idhts uj Variations
469
for any value ofsf . Taking s(ft) ; a, a # 0, we find
F3[13, y(fl), y'()] = 4,
i.e., F r = O at .x
)3.
(27)
Thus if y(x) is the solution of Problem B, ii must sati.fy the Euler equation
(26), the prescribed boundary condition y(a) = A of (241 and the r:atural boundary condition (27). For many students, the derivation of (27) seems fallacious or difficult to understand. Underlying the difficulty seems to be the (incorrect) feeling that there is something contradictory in first taking s($) to be zero and then taking it to be nonzero. Once again, (25) holds for any value of s{ (i). In particular, (25) holds (i) when s(f3) = O and (ii) when sUili= a. a D. From case (1) we derive (2e), and then from case (ii) we derive (27k The reader might find it helpful to consider the following very simpte prohtem which is salved by reasoning analogous to that used in obtaining (261 and (27) . Problem Given that for att real numbers x A sin 2 x
B CCas 2 x
L v 0,
A and B cansl:inl5;
(28)
find A arid B. Solution. Since (28) is truc for all .r in particular it is true for _r — O. Since sin 0 = 0, cos 0 = I, we roust have B = 1. Equation (28) is also truc when x = ni4. Thus ,
;A+;
—
I
—
O,so A = 1.
The following [able shows the analogy: Trigonometric identity
Na[ural boundary condition problem
(25) holds for arbitrary smooth s(x)
(28) holds for arbitrary real
satisfying s(a) = O. F,(x, y, y') --
of s(x)
d fix
f^(x, r. y') is independent
A and B are independent nix.
,
Since (25) holds when s(f) = 0, (26) must hold. Since (25) holds when sift) = I. then. using (26), (27) must hotd,
Since (28) holds when x = O. B must equal unity. Since (2e) holds when x = a/d, then, since B = I. A must equal unity.
Example 2.
Consider the brachistochrone problem mentioned at the beginning of our discussion of natural boundary conditions_ From the definition (4) of the integral to be minimized, Fr - = y1 29.4 L + (,t' ) 2 }!
L +2 ,
Ca tevlrrs of Variations [Ch. H
47 0
.
so the natural boundary condition at x = O is y = O. Is it physically reasonable that the extremalizing path has a hori -rrrntf:I tangent at its intersection with x — 13? Yes, because at the end of the path it is not productive to invest time in downward, accelerating motion--available speed should be used in heading straight for the goal.
Let us now examine the situation in which natural boundary conditions must be derived at both ends of the interval. In essence, only a slight extension of the previous argument is required, but we shall use the opportunity to cast a little more light on the logic of the argument. Thus, let us consider the problem of exiremalizing Jâ F(x, j<+ j-+') dx without any restriction on the values y at a and 0. Now, after integration by parts, we find that if y T } + Es, then ,
F 34xr y^r ^^'^ ^
d dx F3(xs y, y ) six) dx + F3[fit J'161+ Y +Oils(Â) '
a
Yla)] (a) 2- 0, (29)
F
for arbitrary smooth functions y_ [Compare (25}.1 Schematically, let us denote this set E of smooth functions by the interior of the circle in F igure 11.6. If (29) holds for all smooth functions s, in particular it holds for those functions s that vanish at both a and fi (subset I in Figure 11.6). From the statement just made - "(29) holds for smooth functions s that vanish at a and fl"—we derive d Fz(xi y, Y'} — dx F3(xt }', it) 4 0
,
Ir) FIG Li R E 11.6. F.guatioi (29) holds fora set I of smooth fw crions
s. This collection afficherions is sy,t*bolized by the pouts within the circle. Equation (29) holds a fortiori for the subsets of designated I, I], and Ill_ One is, ref course, permitted to draw deductions from (29) using the special character of the functions that belong to the subsets_
(30)
S. 11.2]
lulrodurtium
10
the C'afrulus p{ Cfarialiwxti
47L
just as in our discussion of Problem A. From (29) and (30) we find that F3E1,YOU), Y'(fl)] ) — F3[Œ, Yla), y'(Œ)]s(cr) — 0, (31) for arbitrary smooth functions s. In particular, (31) holds for smooth functions s which vanish at x = a and have the value 2 at x = $ (subset II in Figure 11.6). If this is the case, it must be that F3 [fl.
y(fl), Y'(fJ)] = 0.
(32)
From (31) and (32) we find that F3[Œ, y Œ), Yr(a)ijs1) = 0,
(33)
for arbitrary smooth functions s. Since (33) holds for functions s which have the value 74 at x = a (subset III in Figure 11.6), it must be that F3[a. Y(Œ1, Y1Œ)] = 0.
(34)
Equations (32) and (34) are the two natural boundary conditions that will be satisfied by any solution y of the problem of extremalizing F(xJ, j')dx without restriction (other than smoothness) on the behavior of ' either at a or at fi .
ARE SOLUTIONS
TO
THE EUI.ER EQUATION EXTHEMALS?
We have seen that if the function y is to provide an extremum for the functional iCv) among all sufficiently smooth functions $, then it is necessary that y satisfy the Euler differential equation (21). This is a second order equation. Two boundary conditions, whether imposed or "natural, ,, must also be satisfied. It therefore appears that we have enough information to determine the solution to our problem, at least in principle. It turns out, however, that the calculus of variations is a subject that is rich in subtle surprises and hidden difficulties. To show this, we now list some of the possible relations between the solution to the minimizing problem and the solution to the Euler equation and its accompanying boundary conditions. I. The Euler equation and boundary conditions may have a unique solution that is rather easily seen to provide the desired answer. An illustration is furnished by the problem of minimizing the distance between two points (see Exercise 4 2. The solution may be unique, and it may provide the desired answer, but this may be difficult to prove. The brachistochrone problem furnishes an example here. it is not difficult to solve the Euler equations and thereby to obtain certain curves (cycloids) as candidates (Exercise 5). It can be shown, but it is not easy to do so, that precisely one such cycloid joins a given initial point and a given (lower) final point and that this curve provides an absolute minimum descent time. See Exercise 11.3.24 for a uniqueness proof in a similar situation.
C àk uius of Varinliums
472
-
[Ch. if
3. The integral to be minimized may have the same value for every function p. That is, I may be path-independent, so there is certainly no path that gives a minimum value to !. It is not hard to show that this happens if and only if the Euler equation is identically satisfied for all admissible functions (Exercise 45). 4. There may be no solution to the problem, and this may be indicated by the fact that the Euler equation has no solution satisfying the boundary conditions. For an illustration, see Exercise 7. The richness of possibilities is aptly illustrated by the problem of finding the curve joining two points (x i , y r ) and (x2+ y 2 ) that provides the minimum surface area upon rotation about the x-axis. Here, there may be zero, one, or two solutions to the Euler equation and its boundary conditions. The classification of these various curves is somewhat involved. Suffice it to say that when there are two solutions, one provides a relative minimum and one does not. In this case, and the others, the curve that supplies an absolute minimum may be different altogether, if one admits to the competition curves that do not have a tangent at certain points [see Pars (1962, expecially pp. 146fî`.)]. Some consideration of this problem appears in Exercise 8. We have just illustrated that, in general, it is difficult to show that a solution to the Euler equation satisfies given boundary conditions and that it actually provides an extremal. But, by a sensible definition. we shall say that any solution of the Euler equation renders the corresponding integral stationary. To give a little more idea of the depth of the theory required for a complete analysis even of relatively simple problems, we return to the definition of nearness" found in (7) and (8). These equations implied t hat the extrema we sought were relative to functions whose values and derivatives were close. Another possible definition of nearness would omit the requirement on the derivatives. With the former definition we speak of extrema relative to weak variafis (competitors to the minimizing curve must have both functional values and values of derivatives close to those of the minimizing curve). With the latter we speak of extrema relative to strong variations (only function values are required to be close). As one might expect, a curve can furnish an extremum relative to weak variations but not relative to strong. To derive a sufficient condition that a solution furnishes a relative minimum of a certain kind, a natural path to follow would be to generalize the second derivative conditions of calculus. It turns out that more or less straightforward generalizations of this kind can furnish further necessary conditions (Exercise 15). Sufficient conditions require deeper analysis, as discussed, for example, by Pars (1962, pp. 1201ff). Note, however, that sufficiency conditions are not required for many applications. This is because various physical requirements are equivalent to the postulation that some integral is stationary, perhaps when subject to certain constraints. "
-
-
See.
1121 h,rroductiiou to rhe• Calculus
14.
Variations
473
EXERCISES
I. if F(x, y, y') = y sin (xy`), find (a) Fx, (b) F,,, (c) T , (d) (dldx)F,,• 2. If F(q, r, s) = gr 2 s 3, find (a) F k (w, r, s), (b) F 2 (x, y, z). (c) Show that Fa(3, 2, 1) = 36. (d) Find F12 and F21. (They should he equal!) *3. Find the Euler equation [from (17) or (21)J y In lxy'). (a) When F[x, Y y') (b) When F(x, y, y') = y exP (y')4, Show that in the shortest-distance problem of Example 1, it entails no loss of generality to consider curves joining (0, 0) to (rt, 0). Then show directly from the integral (without using the calculus of variations) that the straight line provides an absolute minimum. 5. Solve the Euler equation for the brachistochrone problem. Start by deducing —10 + (y`) 2 = K - L 1 y K a constant. Do you introduce an extraneous root when you square this equation? If so, what happens to it? Note that the substitution y = K 2 sin e t in the final integral puts the particle at the origin when t = 0, and is thus more convenient than ,
,
y = K 2 cos' r. Remember that x 2 = x I. - .. Another approach altogether is to regard y as the independent variable and so to change the functional to be minimized to an integral with respect to y. $6. Show that the Euler equation (1 7) is identically satisfied for all admissible functions y if and only if there exists a function G such that F(x, y, y') = (d/dx)G(x,y), so that f v 5 h,, F(x, y, y')dx depends only en y(u) and y(b). 17. Consider the problem of minimizing .! = I[, xiy'}' dx for functions satisfying y^(-1) = — 1, y(1) = 1, Show that if the admissible functions must be twice continuously differentiable, then the problem has no solution. !)o this by solving the Euler equation. Show that the problem still does not have a solution if continuous functions with piecewise continuous derivatives a re admitted_ Do this by considering the function y wherein y(x) — — 1 for x < —ï, y(x)=xfefor I çE, 34x}= 1 for x> E. > 0• y2 > 0. xi c x28. Curves = }r(x), y ? 0, join (x 1 , y ^) to (x 2 , Y2). The curves are rotated about the x-axis. We wish to find that twice continuously differentiable curve which minimizes the area. Find and solve the relevant Euler equation to obtain the "candidate" curves (a)
(b)
y
C L cosh {C 1 - `(x
—
C r )],
C E and
C2 constants.
9. (Project) We refer here to Exercise 8. (a) Consider the family of curves passing through (x i , y t )- Show that these curves have an envelope (Franklin 1940, pp- 525OE). Use this fact and a sketch to deduce that there are two, one, or zero members of the family which pass through (x 2 , y 2 ), depending on the location of (x2 , Yi)-
C'Qfrulus
474
rit
Variations [CJ:, ! I
(b) Suppose that the conditions for admissibility are relaxed so that curves need not be expressed in the form y = y(x). Suppose further that continuous piecewise differentiable curves are admissible. There is now a curve composed of three line segments that always provides a relative minimum and often "provides" an absolute minimum. Can you guess what h is? 10. Consider spherical Coordinates, wherein x
_ p sal tb cos U,
y
= p sin 4) sin 0,
z=
p cos O .
Let curves on the sphere p M2 constant be given in the form 4 = f(0), #(a} Set up the calculus-of-variations problem whose solution is a curve of the above form that cxtrcmalizes the distance between two points on the sphere. (b) Use Theorem 2 tc reduce the problem to a first order differential equation. Snow that if the note after Theorem 2 is disregarded, one will be led to the erroneous conclusion that small circles on the sphere are extrenializing curves [Pars (1960. p. 43]. 11. Derive (23) by writing dx ' (h ` dy F x y. — dx = G x, y, - )dy^. dx — J.() dy } ,
(a)
Verify that dx (.;(x, ^x, y, ^
]'
= F x, y,
dx ^ Y -
r dx . dy
(b) Use the Euler equation d Gp — — G3 = i)
dy
and the hypothesis that F(x, y, y') is independent of x. (This formal derivation is more natural than the one given in the text, but its justification requires more restrictive hypotheses.) 12. In finding the shortest distance between (U, U) and (n, b), one obtains the right answer by trying to minimize S [I -- (y') 2 ] dx rather than rl + (y`) 3 ] " 2 dx. Is this trick always possible? Formulate and prove a theorem. 13. Give the natural boundary conditions for the following: (a) For (2) (comment on the answer). 1(b) For the F of Exercise I. (c) For the F of Exercise 3(a). $(d) For the F of Exercise 3(b).
See. l' 1_21 introduction rn riae f ïzkrrlu.x
!r(trruliorrA
475
Further exercises concerning natural boundary conditions, and more interesting ones, will be found in the next section 14. (a) Formally derive the Euler-Lagrange equation ( I ?) by choosing }•n, _P h to minimize N O T E.
}5'3, - - -
n ^} J j +1 ^a J E F ^* ff11 • 7^^ t^ (^ç+ l i=0 xj+i —
x;
-
where x u = a < x, ‹ x 2 • -
-tex„ + 1 -= b,
and
j(a) =
î
5^(x 1 ),
a. 11b) = P.
Then take the limit as tor; x ;# — xi -s Q.* N T E. In the limit, it is immaterial whether }}, t or}y,is used as the second argument in F. But the choice made here simplifies the calculations, (b) In (a) the function values at the end points x U a and x.. 1 = b were fixed. Allow arbitrary function values at these points and thereby derive natural boundary conditions_ 15. This problem yields Legendre's necessary condition for a relative minimum (maximum). that F33 = Fy. must he nonnegative (nonpositive). (a) Carry one step further the passage from (13) to (14) and derive
d 2fir cle 2 (b)
i
=
j"
fi
f 1: 22 52 -t- 2}•2I55i
+ F33(sr )2 ]
a
Suppose that F 33 (x, y, y' } is negative when x = . for some point in the interval (a, 11). Assuming that the second partials of F are continuous in a domain containing all points of interest, show that y could net provide a relative minimum to 1. Do this by considering an appropriate particular function s. for example, s(x) = C, i -x — I ? 6;
s(x) ^ I —
- 'fix— I
,
Ix —f
6; 6
C 1.
16. James and Johannes Bernoulli solved the brachistoehrone problem in 1697. Johannes wrote: "With justice we may admire Huygens because he first discovered that a heavy particle falls on a cycloid in the same time always, no matter what thestarting point may he. But you will he petrified with astonishment when I say that exactly this same cycloid, the tautochrone of lluygens, is the brachistochrone we are seeking" [quoted in E T. Bell's, Men of Matheniatica (London: Penguin, 1953), p_ l46. show that the cycloid is a tautochrone. • This is the approach used by luter, Lagrange -5 procedure is employed in the cexl_
Calculus of Variations [Ch. 11
476
11.3 Calculus of Variations—Generalizations Experience soon makes evident the need to extrerna]ize more general functionals than those treated in the previous section. If the material that has been presented so far is thoroughly understood, howeNer, its extension should present few difficulties_ A relatively terse presentation of the calculations required to obtain generalized Euler equations is the business of this section. We shall usually denote partial derivatives by numerical subscripts during the course of the derivation, and by appropriate subscript letters in writing the final answer. In part F it is more convenient to use the latter notation throughout. The first six exercises below are, in order, examples of problems requiring the six generalizations that we shall discuss. For proper motivation, these exercises- should be read over before proceeding. A.
MOR E I)ERtVA'rIVES
We wish to determine a function y that extremalizes F(x,!'. ft,
fLY)
n fix
(1
)
among all sufficiently smooth functions .f.J that satisfy _0(a)
Ao,
V(a) = A L , P(P) = B o , l''(f) = B1,
(2)
for given constants AD, AL, Bo , and B 1 . We introduce a smooth function s that satisfies
s(a)=
s'(a )=
49)= 5 '0)—
^.
Considering the admissible functions y + cs, we require that the following function 5(c) have an extremum at e O.
5(e) = fF(x , y + cs, }' +
rs, y" + r") dx.
= 0 now implies that
j:(F a s + F3 s' + F{ s ") dx -= Q _ Upon integration by parts and use of the boundary conditions satisfied by s, we obtain [Exercise 11(a)]
r.
(F2 —
d
dx
a F3 +
dx
; F4 c!x = D.
(3)
e _ 11 _3] Sc
['ufrulas of Variations--GPrterra&2orions
477
From (3) and Lemma A (Appendix 1 1.i) we deduce that it is necessary that the extremalizing function y satisfy the fourth order ordinary differential equation Fr
d dd2 F.+ -F^r a dx ^ dx#
(4)
and the four houndary conditions (2). Generalization to functions F involving the first n derivatives of y is straightforward. B - MORE FUNCTIONS
We wish to find two functions y, and y 2 that extremalize
rF(x,y1.54,y2,y2)dx
/Uri, Ÿs) among all sufficiently
(5)
smooth functions p i and $ that satisfy
.171(a) = AI , .f 1(P) _ R 1 ,
(h)
2(v) = A2, j32(() = Hz,
for given constants A ] , B 1 , A 2 , and B2- We consider admissible functions
k,
of the form ŸA ) !1x) + z;5.dx),
( 7)
where the si are smooth functions that satisfy the boundary conditions
st(a) - si(ii) = 0,
i=1,2
We make the definition
li .5 t , Z ) _F(x. yi +
J
r. tst , y;
+ r: is', , Yz + tz'z,y'3 +t z s'z )dx.
For fixed s i (x) and s i (x), .1(F 1* c 2 ) must be extremai when ri Necessary conditions are
afi ac,
when L I = O , f: 2
GE 2
=
r2
= 0.
= p-
These lead by a short familiar path [Exercise 11(h)] to the simultaneous ordinary differential equations d
F,,, - - dx
F = 0,
1,
2-
The generalization to n functions should be evident.
(8)
CRk uh1S üJ Variations
47S C.
[Ch. I 1
MO RE INDEPENDENT VARIABLES
We wish to determine a function z(x, y) that extremalizes F(x, y, f, 2,,, 2,) dx
}(f)
dy
(9)
R
among all sufficiently smooth functions of two variables 7 which are prescribed on S, the boundary of R. Consider admissible functions f of the form s = z + Es, where s is sufficiently smooth and s =0
on S.
( 10)
Our usual argument requires that .51"(c) vanish at r = C, where (^}
1
fF (x, y, z + cs, z i + rs, z + E.5,) di(dy.
(li)
Thus it must be that
1J
5F3
+ sF 4 + ,F 3)dx dy 0.
(12)
Transferral of the derivatives on s to derivatives on F, and F 3 is required if we are to proceed as before, To accomplish this, we define a vector function V by V = Fo i + F j.
(13)
Using V, the terms of interest in (12) can be written Jjs F4 + s y F 3 )dx dy = jf(V.s.V)dxdy. R
(14)
f^
But because of the identity [Exercise Mc)), V•(sV) = Vs• V + sV V,
(15)
we find, using the divergence theorem, that
ff
(Vs • V)dx dy =
—
f(s0 - V)dxdy +
(n sV)da, -
(16)
R
where n is the unit exterior normal to R. The line integral in (16) vanishes because s 0 on S. With all this, (12) becomes
Jfs[F3 ^
aF s
,
-
^
-
^ [ix dy
=
D.
(17)
Ser. f 1.3]
[ ahuJas of Variations —Generalizations
479
The two-dimensional version of Lemma A shows that the square-bracketed terms in (17) must vanish (if they are continuous), so that—switching to letter subscripts— d —F = — =0 . (18) Ox FY ^F zti The boundary condition (10) must also be satisfied. [It should he clear that FZM denotes the partial derivative of F(x, y z, z x , z y ) w ith respect to z. keeping x, y,,z, and z, constant. In taking a/ex the variable y is held constant. Compare Exercise 8.] Conditions sufficient for the validity of the formal calculations leading to (18) not only limit the class of admissible functions and functions F, but also require the boundary of R to be smooth enough so that all the integrals exist and may be suitably manipulated. If Jl is an integral of a function z of three variables x,, x 2 , x 3 and its first partials z,, z r , z 3 , then [Exercise 12(b}] (18) generalizes to ,
3
F#
a {
r= ` x^
F
,
}
=0
,
zf =
^
c7 z +dx
D. 1NTFC,RAt. CONSTRAINT
We wish to extremalize f
(v} _
^a
F(x, y, 37) dx
(19}
among all sufficiently smooth functions j) that satisfy the houndary conditions !'(Œ( = A,
Yî^^) = B,
(20)
and the constraint -
(l') _
a
4 (x, 37, j7) fix = C,
Ca given constant.
(21)
Denote the extremalizing function by y_ We can no longer proceed by considering admissible functions of the form y + Es, for if s is a fixed function we cannot expect y + Es to satisfy My + Es) = C for an interval of E va lues containing O. Only when such an interval exists can we proceed as before to require a vanishing derivative with respect to ti at E = 0_ Instead, we write y(x) + E l s,(x) ♦ E2 (x), where the s,{x) are smooth and satisfy s,(a) _ $0) 0, i = 1, 2. Regarding s, and s2 as fixed, our problem is to write a condition implied by the fact that .^(E I ,E Z ) _
j F(x,1' + E
o t + E2 S2 .y . + F t S, + F2 s3 )dx
Calculus of Variations [Ch. II
4 8O
has an extreme value at 1 i = E 2 = 0 among all nearby (s,, e2) satisfying j(E,, s z ) = C, where
=G(x, y +
+ Ea sa, Y' + c,si + s2sfd dx.
f
Using a Lagrange multiplier A, the procedure of Section 11,1 leads to 0
0 (.I — Ai) = (.o — Ai) = 0 as t -s 3
at t 1 =-
E2
= 0,
(22)
unless j,,( 0, 0) = /rr(O 0) = 0. ,
(23)
From (22) we can deduce in the usual way (Exercise 11(d)] that
f [
a
^ F. dx
F2
^ Gs
--
—
--
^d G3 x
,^jx) dx = 0,
i
= l, 2-
(24)
We must now be a little careful Equation (24) was derived under the supposition that s i and s2 were fixed. A may thus depend on s i and s 2 . But (24) with I = I yields F2 —
d dx
F3 , (x) dx
G2
d dx
fr3 t(x) dx•
so A is independent of s 2 . (The possibility of a vanishing denominator in the above equation is essentially disposed of in Exercise 13.) Equation (24) with I = 2 is now seen to be of a form allowing application of Lemma A. Thus [Exercise 11(400 y(x) and a. must satisfy
Fy --
dF
^ . — dj G ,, —
d^ G ,
= 0
(25)
plus the boundary conditions (20) and the constraint (21). The exceptional case (23) is the subject of Exercise 13. Note that (25) is appropriate to the unconstrained extremalization of I (F — ).G) dx. By straightforward generalization of the argument used here (Exercise 14), one can show that if Jâ F(x, , ÿ) dx is to be extremalixed among smooth functions 5 satisfying the boundary conditions (20) and the N constraints I:0 ` 1(x, }y, ÿ') dx — 0,
i = 1, 2, ... , 1'If, (26a)
then 1y and the N Lagrange multipliers A; must satisfy Fy —
dx
F—
^- 3
^ C;'^ — d G;) = 0 dx
(26b)
Set. 11.31 Calculus of Variations—Generalizations
q8 i
plus the constraints (26a) and the given boundary conditions y(13) = B.
y(a) = A,
(26c)
Moreover, by combining previous arguments, we may readily generalize to integrands that contain more derivatives, more functions, or depend on more independent variables. E. FUNCTIONAL CONSTRAINT*
We seek to extremalize
FO. xi, x3,x3, x t,z2,xs] dt
(27)
^o
among smooth functions that have prescribed boundary values xAto) = a1+
x4(11 ) =
= 1, 2, 3,
(28)
and satisfy the constraint G(t,
(29)
X1, X2, X31= 0.
Because the constraint does not involve integration, our previous line of attack is inappropriate, for it will not reduce the problem to one involving the extremalization of an ordinary function [Exercise 14(c)]_ We expect a result involving a Lagrange multiplier, and we obtain it by the approach often used in the derivation of such results, use of the constraint to eliminate a variable. (Compare Exercise 11.1.15.) Assuming that the hypotheses of the relevant implicit function theorem are satisfied, in particular that c GJr3x 3 # D, we employ (29) to express x 3 in terms of the other variables: (30)
x3 = f(r, x,, x 2 ),
Using this, our problem becomes that of determining the unconstrained extrema of f,
fl
0(tx1, x2 , xI , x 2)di,
(31 )
u
where it)(t, x 1 , x 2 ,
12) �
F t , X 1, X2+
1 (^, xl, x2>^ xt+ ^2t ^ i ( i ,
Xls
X2)
•
From Part B we know that the x ; must necessarily satisfy the Euler equations é
= 1, 2.
(32)
In this case it clutters the notation unduly to make explicit the distinction between a general admissible function and a particular exurentalizing function. `
482
Calculus of Variations
[Ch_ i
In computing (,b Y,, we must remember that d dt
+
f(t, x t , x2)
When i = 1, the Euler equation (32) gives, in terms of F,
F= , + F f + Fx,
d (Fr. + F= 1 [ 1 ) .=0 — dr Ox t dt
or d d f.. (F„, -- Fx , -}- f Y , — F;^, dr dt
= O.
(33)
But, using (30), we have that G[t, x i , x 2 , f{t, x t , x 2 )] = 0 identically in t, x t , and x 2 . Taking the partial derivative with respect to x i , we find
G„, + G I , f= , = O.
(34)
Consider (33) minus times (34):
di
, ti - AG xti + F x, ` dt x , — D. F x' T AG
(35)
Using our assumption that c'G fr?x 3 # 0, we can define a fe ncrion A(r) by
F — (dirdr)F i ,
(36a)
Equation (35) now implies that
F.
de
F^ — ,
l^Gf
(36b)
x,
'The equation
F.
d dt
F x2 — AG, L3 = 0
(36e)
can be obtained similarly, by considering {32) when i = 2 [Exercise 11(e)]_ Combining (36a), (36b), and (364 we can summarize our results by stating that necessary conditions on the extrernalizing functions and on the Lagrange multiplier function ,i(r) are the Euler equations Fx ,
d F'x, —C x,, di
i
— 1 , 2, 3,
(37)
plus the constraint (29) and the boundary conditions (28). Since Gx - 0, these results can be expressed in a manner that we have corne to expect' the equations (36) are appropriate to the unconstrained extrernalization of F — W. But remember that here A is a function, not a constant.
Sec. J 1 .31 Calculus of F'arrarions
Generalizations
48 3
The above derivation assumed that 060x 3 A C. if this were not the case, but if either r G/r x i 0 or r3G/ x x # Q, we could still obtain the same results by eliminating x i or x 2+ respectively. Thus an alternative to (37) is the possibility that
i = 1, 2, 3. Note that (38) implies that raG/i t
(38)
0 on the extremalizing curve since
G[r, x 1 (1), x 2 (t), x 3 (t}] — 0 implies that
. G r — 0_
+
F. END POINT FREE TO MOVE. ON A GIVEN CURVE
We wish to find a curve extremalizing a certain functional when that curve is required to begin at (a, b) and to terminate on a given curve G(x I , x2) = 0. Let all smooth curves satisfying these requirements be given parametrically by
xi =
x, r), (
c
1, 2, 0 < r
1.
Given that
G[. 1(l), x 2 (l}0 = 0,
Xi(0) — a, x 2 (0) = b,
(39)
consider the problem of extremalizing
x2) _
FEE, x1(t), ,gi(t), xx(t). rx(t)i dr,
where '
:1
di
(40)
REMARK. There is no loss of generality in assuming that the parameter r has the special values 0 and 1 at the beginning and end, respectively, of each admissible curve (Exercise 16). Denote the functions extremalizing (40) by x i and x 2 . Let s and 5 2 be smooth functions defined on [0, I] that satisfy
5,(0) = 0,
(41)
1 = 1, 2.
Then for any such functions s i and s 2 , regarded as fixed, it must be that
e2 ) _
J
F[t, x i
+ e i s,, x 'i + c,s xa +
has an extremum at E1 = E2 = 0 among all
El
E=sx, x 3
+ & zsi] di (42)
and E2 that satisfy the constraint
J(E t , E 2 ) r G[x.(1) + f i s t (1), x 2 (1) + E2 S2 (1)) = 0.
If (0,
0) — jr ,,(d, 0 ) ° 0 ,
(43)
Cîikulus of Variations [03_ i i
484
that is, if Gf:[x1(1), x 2(1)) _ 0,
G=.[xi( 1 ), x2(1)1 = 0,
(44a, b)
we have seen that the usual Lagrange multiplier scheme does not work. Assuming for the moment that neither equation of (44) holds, we have
r?€i
[- (EL, r2) — J(r^
,
e2
)J = 0 at E I = £ 2 = 0,
i=
1, 2.
(45)
On integrating by parts in the usual manner [Exercise 11(f)], we deduce that
^
= f st(f)(F
f,
!
O
d
Fxi di ♦ 3i(1 )1* si
LÎx i
— ).G,Sxt(I), x 2 (1)1s,(1),
1= L
t = 1, 2.
(4.6)
We proceed as in our earlier discussion of natural boundary conditions. We know that (46) holds for any of the class of smooth functions s i and s 2 which satisfy the boundary conditions (41). We investigate, in turn, the consequences of (46) for particular subclasses of such functions, Employing an argument of Part D, we first note that, (46) with i = 1 shows that À is equal to a ratio that is independent of s 2 . Equation (46) with i = 2 certainly holds for those functions s 2 which satisfy s 2 (1) = 0, and from this we deduce that d Fx2
dx z
F — 0.
(47)
Equation (46} with i = 2 also holds for functions s 2 that do not vanish at i — 1, so [using (47)1 we infer that at ! = 1,
Fxi — AG =2 = 0.
(48)
Similarly,
F
d
dx Fr = 0,
(49)
L
and at t = 1, F L..— ÀGx = 0.
(54)
If Gs , and G,r , do not vanish air = 1, we can combine (48) and (50) to obtain Fx, G.
FL . att=1. Gx,
(51)
The two unknown functions x i and x 2 must therefore satisfy the two second order equations (47) and (49) and the four boundary conditions (39) and (51).
Sec. 1 1.3] Calculus of Voriariotls --Generalizations
485
The transversaüty condition (51), in relating x t and x 2 to the components of VG, fixes the direction in which the extremalizing curve approaches the given terminal curve G = O. (In other words, the exact nature of the transverse approach is specified.) This condition was derived under the supposition that neither (44a) nor (44b) is valid. If just one of these equations holds, however, we know that at the end of the extremalizing curve, the end-point curve G = 0 may be, respectively, horizontal or vertical_ If both hold at a point (or if G does not have continuous partial derivatives), G may not have a well defined direction when it is met by the extremal. In all these exceptional cases the location of the possible meeting point is known. -
Example_ For the brachistochrone problem (where now - = didt) F = (if + .rÎ1 1 1 (2Y12) 1 2
so at t = 1, Gx ,
F,u
.ft
VG =
or
Gx
^ ix t • tea)
02 0 ü).
(521
Thus when the particle reaches G = 0 its direction is normal to that curve. in agreement with the remarks made in Example 22 in connection with the natural boundary condition for this problem. EXERCISES
+I. According to the theory of elasticity, a thin homogeneous cylindrical rod compressed by a constant longitudinal force P al both ends is in equilibrium if its potential energy is an extremum. Suppose that the rod lies along the x-axis for 0 < .x S L and let y(x) denote lateral displacement (see Figure 11.7). It can then be shown, as at the end of Section 5.1, that the potential energy corresponding to a deflection ÿ(x) is L
[E!0")2 - P02 ] dx,
(53)
where E is Young's modulus for the material of the rod (a stiffness factor) and ! is the moment of inertia of a cross section. In this problem
L
^
FIGURE 1 L7. A rod under compression.
486
Caicufus af Variations [Ch_ 1 1
we shall assume that L and I are constant and will use K as an abbreviation for P f F#_ (a) Find the Euler equation necessarily satisfied by a function y which extremalizes (53) subject to the imposed boundary conditions y(L) = y'(D) = y'(L) = 0.
y(0)
(54)
[Conditions (54) are appropriate to a bar with ends "built in" so they cannot move or bend. Note that the slight compression due to P is ignored in the boundary conditions] (b) Suppose that both ends of the bar are hinged, so that although their position is fixed [y(0) = y(L) = 0], they arc free to bend. Find the appropriate natural boundary conditions. Solve the resulting problem. Show that as K increases, a critical value is reached at which a nontrivial solution of the problem is possible. (The connection of such critical values with the phenomenon of buckling is discussed in Section 5.1.) (c) Suppose that the end of the bar at x = L is completely free_ What are the appropriate natural boundary conditions? 2, This problem concerns the motion of a particle of mass m due• to a conservative force F(x 1 , x 2 , x i ). We posit the existence of a potentialenergy function V(x 1 , x2 , x3) such that the ith component of F satisfies F; = Let T denote the kinetic energy of a motion with position components x i at time t, so that T`
?TOE,2,
d dt
The difference between the kinetic and potential energies is called the Lagrangian L: f r L(x 1 , x 2 , x 3 , x 1 , x 2 x3) = T(xl, X2 x3) ` V(x l, x2, x3). ,
Let xj = ,(r) describe the trajectory of a point that leaves a at time t i
an d arrives at p when t = t 3 so that x(t, ) = txi 41 2 ) = /30 Hamilton's ,
principle asserts that the particle will move on that path which renders stationary the quantity
Jr:r, ^(xl^
g2,173,
:Xi, X-2, k3)
[^t .
That is, from all conceivable trajectories which start at a when t = t 1 and end at p when t 1 2 , nature selects the trajectory x i = xi(t) which renders stationary the integrated excess of kinetic over potential energy_ (a) Show that Hamilton's principle implies Newton's laws of motion. (b) Generalize to a system involving many particles. (c) Show the converse of (a),
Sec_ H .3] Calculus of tfariunof,—Gen ralizalions
4E7
3. (a) Suppose that a fluid of constant density p is confined to a region R. and that a velocity potential 0 exists, so that the velocity v satisfies v = V4. Suppose further that the potential 0 is prescribed on the boundary S(R). It is desired to determine 01 in such a manner that the net kinetic energy f f l R pv • V - dr is minimized. Show that V 2 ç = 0 is a necessary condition. +(b) Suppose that the kinetic energy is to be minimized without restriction of 4) on the boundary_ Find the natural boundary condition and give a physical interpretation of your results_ $4. A flexible cable of length Land constant mass per unit length p hangs so as to minimize its potential energy. Let the ends of the cable be fixed at (O, O) and (a, b), where a > O. We must minimize
J
o
p9^l+(^`)^^
tl3 dx
among all smooth functions lv that satisfy 5'(0) = C,
jlu)
= b.
and
f [1 + ^ÿ,Yr2 dx —
L
Here g is the acceleration due to gravity and y = fix} is the equation of the cable. Use a Lagrange multiplier to find the shape of the cable. That is solve the relevant differential equation but do not attempt to determine the arbitrary constants. That matter is treated in Exercise 24. $5. Given two points on a surface, the geodesics joining the points are curves whose lengths are extrernal compared to nearby surface curves joining the points. Thus the geodesics of the surface G(x i , x 1 , x 3 } = 0 are found by extremalizing ,
[xi(t) + .q(i) + x3(t)]iri dt, subject to the constraint G(x t, and to the boundary conditions xi(tc) — a-,
x ; (i t) = b„
E = 1, 2, 3.
Show that the principal normal to the geodesic curve at a point coincides with the normal to the surface at that point. (Recall From differential geometry that the principal normal n satisfied n = + ' aids, where dxfds is the unit K is the curvature. s denotes arc length, and t tangent vector.)
Calraus of Varial io ns
4 86
[C IL 1 I
6. (a) Show that the shortest plane curve starting at (0, O) and ending on the given curve Glx, y) = 0 intersects G(x. y) = O at right angles. (b) Solve the problem completely when G(x, y) xy 1_ Sketch. $7. Write out (4) a (a) When F(x, (b) When F(x, y, y' yu ) = (y")2 m x(y') .2 + y 3 . $5l, (a) Given that —
,
F(tz, f3, y, ^ ^:) = ally + exp (ô) + s 2 , ,
if z is a function of x and y, find F3(x, y, z, z+ z)
a
F4(x, y, z, zx* $ y ) —
a
f Fs4x, y, z, Tx,
zy).
[Compare (17).] (b) Write out (18) when F(x, y, r, zx , z,,) = xz sin (zx). 9. Generalize part A to the case where the function F of ( l) depends on and (a) its first three derivatives, (b) its first n derivatives. 10. Generalize part B to the case where instead of (5), one has
f(}',.}2 Y3)_
fa
F(x, }i.}LIYi..92,Y2.yi.}'u, A. }+3}dx.
11. Firl in the details necessary to obtain the following equations: (a) (3); (b) (8); (c) (15); (d) (24) and (25); (e) (36c); (1) (46). 12. Generalize part C to the case where the integrand of (9) depends: (a) On z(x, y) and all its first and second partial derivatives. (b) On a function of three variables and its first partial derivatives. $13. Show that (23) implies G y — (dfdx)G„. _ O. Discuss. 14. (a) Generalize the results of part D to obtain the conditions (26). (b) Generalize the results of part D to the case where the integrals depend on a function of two variables and its first derivatives. (c) Show that previous methods will not work in part E by writing the analog of (22) and attempting to obtain the analog of (24). 15. What are the appropriate natural boundary conditions $(1)) In part D? +(a) In part 13? (c) In part E? +16 l .et x,— x,t), octe t
fi,
i = 1,2,3,
be the parametric equations of a curve. Find a new parameterization that preserves all differentiability properties and for which the parameter runs between zero and one. 17. Recall (Exercise 3) that Laplace's equation V20=0
Sec.. !!.3] C â1adiu.s of Variations— Grnerufizariu ►rs
489
is the Euler equation corresponding to the extremalization of the Dirichlet integral, F
5J
dA —
t
^V^^^ dA.
A
A
Write Laplace's equation in polar coordinates by making the change of variables in the integral and computing the new Euler equation. X18. Show directly that the harmonic function of Exercise 17 actually minimizes the Dirichlet integral. 19. Consider a particle performing planar motion under the influence of a central force field. Lei the particle's position at time t be given by the polar coordinates r(t) and 0(t). (a) Show that the Lagrangian of Exercise 2 is now L = m(^z + r 2192 ) — W(r).
(b)
20. (a)
Show that Hamilton's principle (Exercise 2) yields Euler equations which represent the radial equation of planetary motion plus Kepler's second law that the radius vector sweeps out equal areas in equal times. Use Hamilton's principle (Exercise 2) to determine the equation governing u(x, r), the perpendicular deflection of a point x at time t, of a string of constant linear density p vibrating between Fixed end points at x = 0 and x = L. Here the potential energy V is assumed to be proportional (with proportionality constant p, the tension) to the increase in length compared with the length at rest. For small deflections ( f u,, I C 1, p x constant) show that V — 4,0
^
u Y dx.
^
Find T and derive Purr
—
=
(b) Generalize to the vibrating membrane where V is proportional to the change in area 21. This problem is concerned with a variational principle that yields the differential equation and boundary conditions for the two-dimensional irrotationai fl ow of an inviscid incompressible fluid above a rigid plane at y = 0 and below the free surface y — 10, t). [See C, Luke, J. Fluid Mech. 27, 395-97 (1967).1 I1 x, y, t) is the velocity potential of the two-dimensional flow, we wish to extremalize ^.
Cafct,fus of Variations [ Ch. 1
490 v
FiuuRt 1.1.8. The two-dimensional mariun of a fluid can be described by a rleloeiry potential 4)(x, y, r). The upper boundary y ; 14x, r) isdepicted here in a space time plot.
among all smooth functions and Ii, where has prescribed values when x = x r , x = x 2 , i = t i , and t = 1 2 . Here g is a constant, the gravitational acceleration (see Figure 11.8). (a) Suppose that and h are the extremaiizing functions. Consider J40 + 01, h + s 2 H], where $ and H are such that + tï 1 $ and h + E2 H are admissible. Deduce g,^/ ,j, ^ (^^_ + 3Wy + 4`r + 9^ )y =h r2
x
= (x j
x,fh,J U
0,
(58)
d3- dx dt = 0.
(59)
H dx
di =
l
(4x^, + cr +()
Explain your procedure. (b)
To modify (59), use the formula
JJJv. VeD dr = —fff(V•r}0dr + JJain - y ckr, R
R
OR
where
V ^i ^^+j^t f k ^
Y
(60)
sE'r. 1 1.3] Ca !r•ulus of Vuriutir++u --CrNnerali_ari:+ns
49 1
and R is the three-dimensional region depicted_ (Observe that x, t, and y play the roles usually played by x, y, and z; otherwise (60) is just a three-dimensional version of (16).] Deduce from (59) xs
^ -fr ,
—
J
r1.3^,
(C.0)3,= D
fix di 4
^,
r.
Jx
[v( — ^i : ^ r "
,} ^y y hr + W1W JS h (IA dt
^
ra
r, ^si
^
^^
(^__ + ^Yr^ d 3° fix (If = 0,
ti^ ^ (1 + h!,+ h? ) (61)
(c) From (58) and (61), derive a differential equation and three boundary conditions_ Explicitly and carefully state your reasoning_ 22. +(a) Show that of all plane closed curves of given perimeter, the circle encloses the greatest area. (I'or our purposes, it will he sufficient to show that the circle extremalires the area and to argue geometrically that this extremum must be a maximum.) If A is the area of a curve bounded by C. Green's formula
c(^ _
^^ Q dy) _
fc(P
cY 1r1x-+ ^idrl
c x ay f
A
can easily he specialized to give the result A ï ;4(x dy -- y dx). Use the parametric form of this equation_ (b) Using (a), consider any smooth closed curve of length L which encloses an area A. Deduce the isoperimetric inequality
A<
L2 4J
23. According to Fermat's principle, light travels from (a, a) to ({i, 61 on a path y = ji(x). that minimizes ^
=
f 1 ± }r^ ^)^i^ fÎ^L (c(x, ^)
where
Am) = a,
y(WI)
—
h.
I gives the time taken to travel from (a, u) to ( {i, b) m a medium for which the speed of light at (x, y) is c(x, y). Suppose that the curve Q with equation y = 1-(x) is an interface between two different media so that c suffers a discontinuity as Q is crossed. Let c (x0 ) and e' (x 0) denote the values of c as x approaches a point x 0 on Q from the left and right, respectively. Let 0 - and 0' denote the angles made with the normal to Q by the left and right tangents to the light path (see Figure 1 [9).
Calculus of Variraions
492
[Ch. l
Aimi iblc path_ y — }..(x)
Ex rrzmptizing path: y = y(x)
Ÿ IG uRr 11.9. A n gles V - and fl+ made by a r a y of light crossing udiscontinuity_ Snell's law (62) relates these angles to a sp eed rrtriu_
Derive Sneiar's taw: sin B sin 0+
c-
(62)
c+
Proceed by considering the slightly more general prohlern of extrcmaiizing
Ja
• (xJ , J r ) dx +
J G(x, 5%1.1 dx, ^
where F and G are the functions appropriate to the left and right sides of Q, respectively. 13 is the x-coordinate of the intersection of y = f(x) and y = }1(x). Write ÿ(x) = y(x) +
Es(x),
s(CL) =
s(fi) = O.
Note that
Y{r) + u-siD = Î(D
(63)
implies (formally) that = Ÿ(c) and 13(0) _ 1), where y is the x-coordinate of the intersection of y = f (x) and the extremaiizing curve y = y(x). (a) Compute {e)
L^E
F(x. y + es, y' ♦ es') dx +
fr1 and impose the usual necessary condition. a
W) dd
5rr•. 11.31
ialcuhf_f f1#
^RrfQ^t{l^_L' — ^tf'Jl!' rt^fr 'crlrf}1i ç
493
Substitute for (0) by implicitly differentiating (63). Take into account the fact that y' will be discontinuous at x = y. (c) Show that the usual Euler-Lagrange equations hold on F and G for x < y and x > y, respectively, while Snell's law and the value of y emerge from natural boundary conditions. (d) Compare the advantages and disadvantages of the derivation of Snell's law presented here with that presented in elementary physics texts. fe) Consider the simpler problem of extremalizing J when • is a given fixed constant_ Derive a condition on the discontinuity of y'(x4. This is called the Weierstrass Erdrnwin vertex condition. 24. Exercise 4 reduces the hanging-cable problem to the following question: Is there a unique member of the three-parameter family of catenaries (b)
y — C' 1 =C_; cosh
—
C 3C'
)
that passes through (0, 0) and (a, b) and has length L. L u 2 +-62 ? Following Section 6.7 of Pars (1962)* we outline a proof that the answer is affirmative. We ask the reader to fill in the details. (a) Consider chords of the curve yt = cosh _x that make an angle = tan - I (b/u) with the x 1 -axis. Denote the length of a chord and the length of the arc below it, by 2P and S. respectively. Let R = S/21'_ Show that R varies monotonically from one to infinity as P varies from zero to infinity by deriving the following four equations (the notation is supplied by Figure 11.10)_
w•i = cosh
FIG u R E 1 1.10. The pant (4,g) bisects the chord of the cuterlrlry y t = coshx i •
CQ1rWfus of Varia
494
(1) (ii) (iii)
i f P sin a = cosh ( + P cos a). P sin a sinh sinh (P cos a). S = sinh ( + P cos a) — sinh ( — P S y a ---=sin + 2P
(iv)
r
(ORS
[Ch. 11
cas et).
sinh (P cos al s cos 2 a. f cos a
(h) By (a) there is a single chord for which R has the correct valut L/(u 2 + b 2 ) 1 '2. Show that this implies that one can then solve the problem by a change of scale and a translation. 25. (a) Consider the problem of extremalizing
_
_
fifE
^t
^t Ni r^tiz
u r , Lï^, u^, x1 ^ , dx2 , ^x3,
^û3^
xl ,.. ., r^x3 axis
(LX ]
dx2
dx3
A
among all smooth functions i L û 2 ie 3 having given values on the boundary of R. By slightly generalizing Part C show that the extremalizing funct ions u i necessarily satisfy ,
,
3
P., -- y (' )• = 0, •= 1 x J (b)
where
OU;
u
"
^
^x}
Suppose that u 1 , u 2 , and u 3 are prescribed on a part OR of the boundary of R but are not prescribed on the remaining part aR 2 Suppose further that one wishes to extremalize the integral of part (a) minus
1{4 uk da, t'R
where t is given on the boundary. Show that the equation of part (a) still holds, but in addition there is the natural boundary condition 3
tr =
}=L
Eu„ rti on OR 2
26. Consider the function u that extremalizes the potential energy [given in (4.4.18)1: 3 V(E) V ()=
^
R
moi dx1 dxa dx
3 }^r/ + ^3 (^ ^^ rr2 ^
I.
L
where e rj(u) — -'{u r
+ ai. d,
f and t are given.
3
dcr, y liur,
set. 11.31 Calculus of Variations—Generali=urio,ts
495
Suppose that u is prescribed on a portion OR ] of the boundary. lise the results of Exercise 25 to derive the equations of elastic equilibrium
3 dT,
ox;
L
(
wher
and the boundary conditions
a
e Tif -—^;^ ^ + .^^;^
^
^^)
t=1
3
EI Tej— t j on ÔR 2 .
u, given on aR 1 ,
j=
27. The aim of this exercise is to show that Hamilton's principle for particles (see Exercise 2) extends to continuous elastic media To this end, consider the extrernalication of
$12
i (u) Y{u)l df,
,
where the potential energy V is given in (64) and the kinetic energy T is defined by 3
T(û) = l p fff
3 rtx , dx 2 dx 3
auk)
R
(p = density, a given constant). Using the same boundary conditions as in Exercise 26, show that the same results are obtained except that the equilibrium equations are replaced by the dynamic equations 0214 ; 3 aT . + 2 — 1 ôt ax, P f-
28. Let A ; and y; denote eigen' alues and corresponding eigenfunctions of 28. the Sturm- Liouville problem, Ly—,y,
10) = .013) = O.
Here
L= --d p —d +g dx dx p and q are positive and smooth, +l t
Jn
F(x, y, y') dx,
)2
,
).3 - • • . Let y minimize
where F = p15702 + q$5 .
0
subject to the constraints
IFI6 G{x, j , $5 ) dx = 1 and '
a
J
G,{x, K $') dx
O
,
i = 1,... , N — 1,
Calculus u/ !luriwio ►ns Ef'h_
496
r
^
where G' and G, = YYr (a) (b)
Using Lagrange multipliers show that y is an eigenfunction of the Sturm Liouville problem. Employ the fact that eigenfunctions are mutually orthogonal (I, Chapter 5) to show that y is the Nth eigenfunction. (The variational characterization of eigenvalues is extensively considered in Chapter 12.)
Appendix 11.1 Lemma A Lemma A. Suppose that G =
(1)
G (x)s(x) dx 0
!
P
^
for all functions s possessing n continuous derivatives (n any positive integer) and satisfying
(a) = 0,
s(0) = O.
(2)
If G(x) is continuous for i < x < 13, then G(x) M O. Proof. If G(x) # U for a ' x z j , there exists a point x 0 , Œ ç xu 5 13, such that COO # O. Suppose that G(x 0 ) > O. Then (with slight modifications if x 0 is an endpoint of the interval) if G is continuous, there exists a positive number p such that G(x) >0
fer Ix —
< p.
As we shall show in a moment, it is not hard to construct a function s* which has rt continuous derivatives, satisfies ( 2 ), is identically zero for l x — x 0 1 p, and is positive for rx — x o l < p. But then [taking p small enough so that (— p, p) is contained in (a, e)t, dx
^ G(x)s * (x.) dx > 0,
J xt - p
contradicting (1). The possibility G(x e ) < Q is handled in the same way, Thus G(x) 0_ A suitable function s* is given by s*(x) s'(x)
0,
I x— Xi] f o p [133 — (x -- x 0 ) 2 _11' ,
Ix — x0 1 < p.
If m > n, at least n derivatives of s* vanish at x = x e ± p, ensuring the required smoothness_ G
Appendix 11.21
Variational ,Natation
497
The lemma and the proof generalize in an obviousway to many dimensions. Note the similarity between Lemma A and Dubois-Reyniond lemma that was used so often in I. Lemma A is also associated with the name of DuboisReymond—but here, such attribution would obviously make for ambiguity of reference.
Appendix 11.2 Variational Nutatian In common use is a notation that emphasizes the analogy between extrernalizing a function and extremalizing a functional- First consider a function y. Recall that the differential dy(x o , dx) is by definition the first order change in y(x) at x o due to an arbitrary change dx in x, i.e., dy y(x 0 ) dx. Thus a necessary condition that y have an extremum at x o is 0 at x - x 0 for arbitrary dx. dy(x, dx) To extremalize the functional I(y) - jâ F(x, y, y') rlx cf (11.2. 5), we make a change in notation and introduce the twice continuously differentiable function by instead of the function s of (11-2.9 j and (11.2-10), where by -= s.
(1)
so by(a) = bu(1) 0.
as the lowest order approximation
We define the first variation bl(y, by) to A/, where
AI : I(y + by) - I(y). "Lowest order" means that only terms linear in by and its derivatives are retained. But using by = s and regarding ❑ I as a function of E for fixed s, we have
AI= I(y-}- Es) -I^):
E
df(y + rs) de
=0
The required first derivative is precisely what was computed in our original approach to Problem A of Section 112 Using the result (11.2.14), we therefore obtain ^
I(y + t:s) - 1(y) =
E
(Fr s + F,. s') dx +
-
Thai is
bI
y r)by + F, .(x, y, Y + 1hy'] dx. .
= ^a
( 2)
Necessary for 1 to have an extremum at y yo is the condition bl y, by) = O at y = yo for arbitrary by; and the Euler equation (11.2.21) follows from this condition as in Section 11.2. Hence if one makes a correspondence between
Calculus of rariu+iuns IC h. 11 dx and y, and between dy and 5i, then extremalizing a functional is seen to be closely analogous to extremalizing a function. Formal manipulations are easy with variational notation, because of the close link with ordinary differentials. For example. if f , and I 2 are functionals, then the following rules hold: ± 1 2 ) = ô#, + 61 3 ,
pi)_
6(1 1 1 2 ) — l i I a + 12 61 1 ,
l 3 bU t t 1051 2
(3a,b,c)
Iz
f2 To prove (3b), we define
ff', ', } r (F.} = { [LY OE) + Es)/ 20)+ Fes'} — ) I(Y) 1 A(Y) and note that to lowest order ^ (E) ^ F 1 1(Y) ^
t
Go '{0)
(E)
120 + FS}
s- 0
or
+ } z(y)
dE
1^(Y + a:-s}
Proofs of the remaining formulas are left to the exercises. As another example of the natural character of the formal ruses, we observe that from (2) we can deduce such formulas as
t+ f
y" dx =
a
bf
%1yR - 1 by dx, Q
fi
sin (y') dx = f cos (y') b y' dx.
a
(4a, b)
Q
Example J. F ind a necessary condition that A he a stationary value of f i1ir2, where
ii=
fPly'3 1 +
=
dx.
rr r}' 3 dx.
lsl
Here p, q, and r are given functions caf x, r > O. The extremalizing function p must not be identically zero hut must equal zero at x = a and x Q. Solutiurr. The First variation must vanish. Thus, since .i = i,tj'.,'11i 1t .), we have
r
f ____ — 161 2 )
r 61 1 —
Since I, 34 0, we infer that r5i , — ,lfll = [1. Hence
j {2pyy'
4
20,4
—
22r OA dy = 0
L^l
for arbitrary srnoni h functions by that vanish at a and fl. Performing an integration by parts on the first term, we deduce in the usual manner that —(py'l' + lq
—
.lrjy = [l,
i(a) = ytjf1 -• O.
l7l
Appendix f }.2]
Variational Aü,rurton
499
Note that the eigenvalue problem (7) has been shown to be equivalent to the problem of rendering stationary the quotient of integrals] 1 tf 2 . This provides an example of a matter to he extensively discussed below. It is useful to know something about the variational notation, since it is used by many authors. On the other hand, the formal elegance of this notation can seduce the unwary into an unfounded belief that they understand the elements of the calculus of variations when, in fact, they possess only manipulative skill. We therefore believe that the more basic approach of Section 11.2 should be used at least until the subject is thoroughly understood.
EXERCISES
1. Prove (3b) and (3c) 2. (a) Verify (4a) and (4h). (b) Find c f! exp 3. (Chandrasekhar, 1961, Chap. 2.) Let F - (D 2 — kIG z (D 2 — k 2 )2 I1 where k is a constant and D ; d/dz. Show using variational notation that (D 2 — k2)F = — Rk 2 W
is the Euler equation which provides the required condition if R is to be a stationary value of 1 i /u 2 1 2 , where ,/,( W)_ Jr i_(DF) 2 + k 2 F2 1 dz, o
1 2(W) =
Je G
2 d2.
The function W that renders R stationary is required to satisfy these boundary conditions: at z = 0 and at z = 1, W = 0, F = D, either DW — 0 or D2 W = O. 4. Show that the first variation of the integral I defined in (2.5) is unchanged if a total derivative of the form d
dx Ox,
^]
is added to the integrand_ [This nonuniqueness of the functional whose extremalizati ❑ n leads to a given Euler equation can sometimes be used to advantage. See, for example, p. 9 and elsewhere in L. Landau and E Lifshitz, Mechanics (Oxford: Pergamon Press, 1960).]
CHAPT ER 12
Characterization of Eigenvalues an d Eq uilibrium S tate s a s Extrema
I
N CHAPTER l 1 we showed how variational problems could be reduced to the solution of boundary value problems in differential equations. In this chapter we proceed in exactly the opposite direction by showing that there are often advantages to casting problems in differential equations into variational form. A similar duality is present in the relation between differential and difference equations: Numerical analysts make their living by approximating differential equations by difference equations, but we saw in Chapter 3 of I that it can be useful to approximate difference equations by differential equations. For certain classes of variational problems, determining the EulerLagrange equations leads to a solution, since these equations can be readily solved, either analytically or numerically. In other cases the reverse procedure is useful, for an extremalizalion problem can be handled directly, without resort to a translation to differential equations_ After an introductory section the famous Ritz method for direct solutions to variational problems is thus described in Section 12.2. A variational formulation also permits ready deduction of certain qualitative conclusions. To illustrate this, the Courant max rein principle is used in Section 12.3 to demonstrate several qualitative features of vibration problems (as well as to permit quick proof of the essential feature of the Ritz method)_ Section 12.4 shows how variational formulation of certain positive problems in partial differential equations leads to simple uniqueness and stability proofs as well as to efficient calculation methods. The chapter concludes with a lengthy appendix on certain basic ideas from functional analysis that are useful, although not essential, in pursuing the topics under investigation here. -
12.1
Eigenvalues and Stationary Points
THREE STATIONARY VALUE PROBLEMS
To begin our considerations, it will he worthwhile to consider three instances in which stationary value problems lead to eigenvalue equations. (By definition, functions suffer no first order change al stationary points, i.e., their first derivative vanishes.) In the next section we show that for an impo rtant class of positive problems, the eigenvalues can profitably be characterized as minima, not just stationary points. 500
Sec. 17.1] Er9envalues und Sraiwnury
Puinrs
501
Problem A Determination of points on the quadric surface 3
E Ark r ac ; = 1,
r.,-
where A u = A i
(la. h)
I
that render stationary the squared distance to the origin i: 21 } x-- 2z + V-3 _ (Compare the particular case treated in Example 11.1.1_) Employing a Lagrange multiplier A 1 , we find that at the stationary point (x i , x 2 , x 3 ), we must have 3
0X t
—A
'LA; = p,
'
i= I
k
1, 2, 3.
r.j - t
( 2)
If we use the symmetry condition (lb) and divide by 2, we obtain 3
3 ^
;=1
A,,, xf = ;xk
y A dl x# x 1 =
Also,
j.j =
I
,
k = 1, 2, 3.
(3)
t
This is the much studied problem of determining the eigenvalues of the symmetric matrix with components A r (see Section 2.2), The eigcnvectors of A id , with lengths chosen to satisfy (lak give the stationary points. Problem B Determination of the unit vectors n = (n 1 , n 2 , n 3 ) giving stationary values to n • On), the normal component or the stress I. In terms of the symmetric stress tensor* with components To we must render stationary 3
X I1 t TJ fi,
i. j=
Oa)
l
subject to the constraint
(4b1 Upon introducing a Lagrange multiplier )1., we see that (4) requires finding components rii that satisfy a
3
drz k i, i=E1
3 \
i1r TfF} — EF; = 4 i= I
or 3
E Tioni = A n.% and -1
E
= 1.
(5)
4= 1
Readers who are unfamiliar with the stress tensor can just observe that (4) linplics (Sj and (b)-
502
ChQracterizalion oj Eiqenr:atues
and Equiitbrieoh Stales as Extrema
[Ch. 12
This is the already studied problem of obtaining the principal axes of the stress tensor (Section 2.2). Formally it is, of course, virtually the same as Problem A. Note that (as in Example 11.1.1) the Lagrange multiplier A is numerically equal to the desired stationery nunnal stress, for 3
E
3
nk
ni = Â
f,f°1
= Â.
(6)
1.4
Because symmetric matrices have mutually orthogonal eigenvectors, the stationary values are associated with three orthonormal vectors n. Problem C Determination of a function p(x) that renders stationary
V - J[EI(xHjY) 2 — o
fn23 dx.
(7)
We seek here stationary values of the potential energy V appropriate to the buckling problem for the bending of a bar that is hinged at both ends. [The expression for V can be determined from Exercise 113.1 by replacing the deflection slope y' used there by p and by allowing the moment of inertia I to vary. Of course, I(x) > O.] We are faced with an elementary problem in the calculus of variations. It is easily seen (Exercise 1) that it is necessary that a stationary function p satisfies the following Euler equation and natural boundary conditions:
—d
!{x) dx = Ap,
p'(0) = 0,
p'(L) = O.
(8)
Here A has been used as an abbreviation for the ratio of the compressing force f to Young's constant E. The lowest value of A for which (8) has a nontrivial solution is proportional to the critical buckling load. Mathematically speaking, (8) is an eigenvalue problem for a differential equation of a rather general Sturm-!_.iouville type (compare Section 5.2 of I). One can regard Problems A, B, and C, as illustrating what at timesalmost seems to be an unreasonable importance of eigenvalues. Three rather different stationary value problems lead to the necessity of computing eigenvalues. But here we emphasize a complementary point of view which regards Problems A, B, and C as evidence that eigenvalues can be characterized as stationary t;alnes of some quantity in a variety of ways. We shall explore this matter, whose importance is evidenced by the many times that eigenvalue problems have been encountered in this volume and in I. A PARTICULAR SEt.F-ADJOINT POSITIVF. PROBLEM
The material that we wish to present lends itself well to an abstract formulation. In the context of self-adjoint operators on vector spaces, the basic
Sec. 12_1l
Eigrnratues and Stationary Points
503
simplicity and generality of the ideas is clearly revealed, and an excellent opportunity is provided to illustrate some useful ideas of functional analysis_ Theabstract prerequisites are presented in Appendix 12.1_ The reader who wishes to obtain rapidly the main ideas on eigenvalues and extremalization can, however, omit this appendix and read the succeeding sections with the following particular case in mind_ Consider the eigenvalue problem
— dx
[
px1 dx
+ q(x)u(x) _ ,1u(x
u(a) = u(b) = 0.
),
(9)
We prefer to regard this problem as seeking certain particular functions (or generalized " vectors") that are members of the set S of all twice continuously differentiable, real-valued functions which vanish at a and b.* We take the set S as the domain of the linear differential operator d d L _— dx p
q
(10)
and look for nonzero elements of S ("cigenvectors") which satisfy Lu (The zero element of S is the function z, where z(x) = 0_ Although Lz z is as usual specifically excluded from being an eigcnfunction.) We discussed the eigenvalue problem (9) in Chapter 5 of I (where we included an additional bit of generality that is not needed at present). There, by an integration by parts, we showed the self-sdjointness property
Az,
(Lzi, = (u, Lv)
when u E S
and
G) e S,
where
(f, g)
f (x)9(x) d
—
(12)
^-
r
^
Equation (12) provides an illustration of the generalized "scalar product" ( , ), which will be repeatedly referred to. Furthermore, integration by parts shows that L also provides an example of a positive operator in the sense that if u E S, then
(Lu, u) ? 0, (Lu, = 0
(13)
if and only if u = O.
Those who wish to confine themselves to the special case we have just outlined should continually specialize the abbreviated notation we use to this case. For example, we shall show that the lowest eigenvalue is the greatest lower bound over S of (Lu, u)f(u, u).f In the present case, using the definition The results also hold for other boundary conditions. such as the conditions
kilo) -
ii(b) =C
cf (8).
t We have already shown, in Example I of Appendix 1 1.2 (with r = I), that eigenvalues of (9) render (Lu. ul{(u, u) stationary.
50 4
Charar -rerixarion of Et .yenrahres and Equilibrium Srorrs as Exrrernu [ eh_ 1 2
of ( ,
and an integration by parts,
(Lu, (u, u)
[—(pue)' +
(lulu dx
_ ^ô CP{u1 2
A iJ^ dx
te' dx
7a
qu 2) dx
EXERCISE
1. Verify that the conditions of (8) are appropriate. 12.2 Eigenvalues as Minima and the Ritz Method Using material introduced in the previous Section as motivation, we now characterize eigenvalues as the minima of certain functionats, i.e., of certain scalar-valued operations whose arguments arc elements of an appropriate vector spate of functions. It is then natural to obtain upper bounds to eigenvalues by "doing the best you can" to minimize the functionals by inserting the partial sum of a series that could converge to the desired eigenvector, and choosing the coefficients to make the functional as small as possible. This is the Ritz method,' a technique of considerable practical importance whose presentation will be the subject of the latter part of this section. MOTIVATION
motivate our characterization of eigenvalues, we focus attention on Problem 13 in Section 12.1. Leaving aside its interpretation in terms of stresses, the mathematical problem is to find stationary values among all vectors of unit length, of To
3
given that T1) = 7j. / . =t As we have seen, the stationary values are the eigenvalues 2 defined by T(x) = 71x 1 , x 2 , x a} _—
3
E
E x ; di x; ,
gl F x, — ;0c,,
t = 1 , 2 , 3.
(I)
= 1 where the stationary values are The points on the unit sphere E i3= taken ont are given by the three mutually perpendicular eigenvectors of T 1 . • The method is also called the Rayleigh Ritz method. In his 1877 bank The Theory of Sound (in Sec , 89. for example) Lord Rayleigh characterized as minima certain frequencies that roc in his study of vibrations. He then obtained useful estimates of these frequencies by guessing simple forms of the corresponding eiget+function, perhaps minimizing with respect W a single pa rameter. Although Rayleigh doubtless knew its potentialities, it is to the papers by the German scientist W. Ritz, beginning in 1908. that we owe the systematic exploitation of the method_ The "modern." somewhat abstract version is presented well by Mikhlie (1964), a Soviet contributor to the study and application of variational methods. t A function is said to rake on a given value if the function has this value for some point in the domain under consideration.
Sec. 12.2] Figenvedues -s Mjnanu rued the Ritz Method
505
For reasons that will become clear, it is of particular interest to study the positive definite case where 3
E x t Ti xj > Û Here the level surfaces
E xi To x1 =
K.
Li-I
K a positive constant,
( 3)
are ellipsoids. To fix ideas, let us make a particular selection of constants so that
L = , f2
—
I
1.
A3
(4)
Choosing principal axes, we have 3
Tj1 x,x^ = X^ + -tx3 + x3. i,
}= L
The unit-length eigenvectors associated with the Ai x t11 4 (1, 0, 0), x{2M - (0, 1,
arc ±xt' where
0), x 43} - (0, Q, 1).
Figure 12.1 shows sections by the coordinate planes of the level surface ellipsoids that correspond to stationary values of T, namely
x^ + txx + x
dt
i = 1, 2, 3.
Let us examine the geometric interpretation of the algebraic results. The lowest eigenvalue A l corresponds to the smallest ellipsoid that just touches the unit sphere. This is, x
i
+ 4x
+
9x; - 1,
and it touches the sphere at i xli As can be seen from Figure 12.1(a) and (h), all points on the sphere that are near +x { " can only be touched by larger level-surface ellipsoids of the given class (3). Among points on the
sphere, therefore, T has local and absolute minima at + xl L ^. As shown in Figure 12.1(d) and (f), the ellipsoid (3) for A _ 02.2 touches the unit sphere at + x {2}. In the (x i , x 2 )-plane, nearby points on the unit sphere correspond to smaller ellipsoids, but in the (x 2 , x 3 )-plane, nearby points Correspond to larger ellipsoids. The stationary points +x (21 are thus neither local minima nor local maxima of T(x). The same type of considerations shows that the third pair of stationary points ± x 111 provides local and absolute maxima for T1x) among all nearby
points on the unit sphere. It would he advantageous if the eigenvalues could all be characterized as minima, or as maxima, not merely stationary points. 1f the former were true,
gatr
Characterization of Eigenralues and Equilibrium Stares as Exrremu
[Ch. 12
Yr
t+r
ry
3
{i)
tri
{dM r^
X
sy
sl
rd!
01 i
J~
r G t; R E 12.1. Points on the unit sphere are sought that give stationary values ro T(x) -x + #;x2 + x}. Since level surfaces 7'(x1 = constant are ellipsoids. the problem is equivalent tu finding which of this family of ellipsoids is iwiger+r ro rire sphere. Depicted here as heuty crows are secrions (in one quadrant) mode by the coordinate planes through each of the three tangem ellipses_ Points of rangerrcy, the stationary points. are indicated by heavy dots.
for example, the approximate methods that did "as good a job as possible" in finding a minimum would give an upper bound to the desired eigenvalue. In the simple example just considered, however, no such uniform characterization has emerged from our naive considerations. But we shall now show that a slightly different point of view allows us to characterize all eigenvalues in our example as absolute minima of T over suitably chosen sets. Since T1 is a positive definite matrix, T(x) > 0 when Ix I = I_ Since 1(x) is bounded below (by zero) for x on a closed bounded set (the unit sphere), we know that T(x) has a greatest lower bound b and that there exists a point on the unit sphere where T b. An absolute minimum is a fortiori a stationary point, so the lowest eigenvalue A(t) of 7; in fact, provides an absolute minimum to T(x) among points on the unit sphere. Indeed, T(+ x {, }i = =
.r
b,
and Ttx) takes on its greatest lower bound at ± xu }. Thus the smallest stationary value provides an absolute minimum for T on I x I = 1, as we have already seen geometrically.
See- 12.2] Eiqene•afrrrs us Minima a n d the Ritz Method
507
Because Ti is symmetric, it has mutually orthogonal eigenvectors. Consequently, the remaining stationary points must lie on the great circle x3 -i x3 =1, in the plane x i = 0 perpendicular to x"/ These stationary points of T(x,, X2, X3) necessarily are stationary points of 710, x2„ x3). There are only two pairs of the latter [determined by a two-dimensional eigcnvalue problem that is obtained by setting x 3 = f} in (1)1. Thus the two stationary values of 71x 1 , x2 , 0) on the unit circle x; + x= = 1 will provide the two remaining eigenvalues. But by the same argument as in the three - dimensional case, one of these stationary values is, in fact, the greatest lower bound of the nonnegative function T(x t , x 2 , û):
= min T(x) for x such that x x — 1,x - x'" = O.
(5)
As shown in Figure 12.1(d) and (f), this bound is taken on ai the points ± x' 3 } Repetition of the above argument shows that the third eigenvalue is the greatest lower bound of T(x i x 2 , x 3) among points on the unit sphere that are orthogonal both to xir' and to x(2 . The only such points are s x(3 ', so that this greatest lower bound can only be 71 4 x 13) ) ,
— min T1x)
for x such that x • x'"
—
x x"' = U
(6)
-
To summarize, we knew that the various eigenvalues were smar ionary values of 7'(x) for x on the unit sphere. We demonstrated that, in fact, the eigenvalues are all minima of T1x) for x on subsets of the sphere that are selected in conformity with the fact that successive eigenvectors must be orthogonal to all previous eigenvectors. Our demonstration relied upon the symmetry of T o , to ensure the orthogonality. It also relied on the knowledge that (x) was bounded below (because of its positive definiteness), to ensure that T indeed took on a minimum when x was restricted to the various subsets of the unit sphere. It is certainly not evident from the discussion just given that the characterization we have made fora particular eigenvalue problem is of general utility. Nevertheless, experience with a number of particular problems-- such as Problems A, B, and C of Section 12.1-- gradually brought mathematicians to the realization of an underlying similarity, To gather the fruits of this Au, experience, we write the original eigenvalue problem (1 ) in the form Lu written [where the normalized eigenvectors are denoted by el u = l^r a^ u]^ Y = ,
,
,
(vi,
[t 3 ),
then v — Lu
l o ut
= J-1
The operator L is self-adjoint and positive if the scalar product a lu, Yti r I a i Li(
(7)
508
Characterization of Er genralue► and Equilibrium States as Exfre+Tms [(.'h. 12 ;
is employed (Appendix 12.1). In this notation our previous results can be written [where the normalized eigenvectors are denoted by nil = min (Lu, u), ). r = (Ln11), n u) ), (n`", nu)) — 1;
). 2 =
min (Lu, u), 2 2 = {Ln[2 ', um), {n12 , nu} ) = 0 (n12 ', n (2)) = 1; c..0= ,
^
(8) and so forth. In point of fact, it proves convenient to alter slightly the above characterizations of the eigenvalues. To do this, we recall that a scalar multiple of an eigenvector is also an eigenvector. If (u, u) = 1 and if w = cue, # 0, then (w, w) = (au, au) = a 2 (u, u) = x 2 ,
(9)
so (Lu, u) = (L[a - 1 w], a - r w)
fit)
a - 2 {Lw, w,) (
i
w)
(1 0) J
Thus the unnorrnalized eigenvector corresponding to the smallest eigenvalue can be thought of as providing a minimum of (Lw, w)/(w, w). The condition (u, u) = 1 merely provides the information (w, w) a 2 and need not be taken into account during the search for the minimum. To obtain the second eigenvalue, the search must be restricted to vectors that are orthogonal to the first eigenvector, etc. We turn now to the characterization of eigenvalues in a general class of problems wherein a reliable guide is provided by the results that we have found for a particular case. SPECIFICATION OF A LINEAR OPERATOR
The succeeding theorems and discussions of this section all involve a linear self-adjoint operator L with domain S and range T, i.e., L : S T We assume that Tis a real vector space with a scalar product ( , ) and that S is a subspace of 1 -(in which case the same scalar product will serve for 51. Finally, we assume that if t e T and (t, s) = 0 for arbitrary s e S,
then t = z,
(11)
where z is the zero vector in T. In other words, only the zero vector is orthogonal to all other vectors in T. in the important special case when S = T (L maps the vector space into itself), assumption (11) obviously holds. The reason is that we can take s wand apply the last of the scalar product axioms to deduce: = z from {t, 0=-0. Suppose that T is the space of continuous functions and S is the subspace of functions that vanish on the boundary of the region under consideration and that possess n continuous derivatives. Then (11) is nothing more than Lemma
ser. f 2.2]
Eiyent•aa'ues as Minima an d the Ritz Method
509
A (Appendix 11.1. Finally, it can be shown (Exercise 3) that (11) holds IFS is dense in T This property requires that
if t t= T, then s„ —. r fora sequence {s„}, s n c S n = 1, 2, .. . (12) The approach of s„ to t indicated by "s„ -* t” is specified in terms of the norm ,
.
IL If (a generalized length) defined by II I(
(s, s 1 ' R .
By definiti on, sn
^
if and only if
t
liln II S n —
(131
ill = 0-•
THE LOWEST EIGFNVAL(JE
Theorem 1. Suppose that the set of real numbers (Lu, u)/(u, u) has a greatest lower bound (glb) ). 1 and that this minimum is actually attaincdt when u — u t : gib
(Lu, u)
uE5 0i, II)
J. 1 —
(Lit ' , u1)
^
u^ t S.
,
(14a, b)
W3,140
Then ;., is the lowest eigenvalue of L and u 1 is the corresponding eigenvet;tor. Proof Let s be an element of S and let F. be a real number- Regarding s as fixed and E as varying, we can assert that the following function "f(1:) has a minimum al r: = 0: M(E) —
it/4u t + Es], t + i s) (u t + ES, u, -}- ES)
(15)
But straightforward calculation shows that ME !M = (Lu
i , u1) + 2i(Lu 1 , s) + c2(Ls, s)
`
`(e) (16)
(u1, u1) + 241, s] + $ 2 S,
s)
c)
Since (E) is minimal when e — 0, it must be that Afé`(o) = 0 or
0=
D( 0 ) N'(0) — N(0)D'(0)
[D(0)] 2
"
(a) — , 1 D`(0) U(0)
(17)
We have used the result A l N(0)/D(D), which is an alternative way of writing (14b). Using (16), we deduce from (17) that
(Lu1 — ,^ 1 u 1 s) = O. ,
It can easily be shown [Exercise
(18)
3(d)] that S is dense in T if and only if every clement of
I has an element of S arbitrarily close to (t. t If L is a positive operator. as in the example discusscd at the beginning of this section. then (Cu. ti)/(u, u) is bounded below by uro, and therefore possesses a (nnnnegative) greatest lower hound, Whether or not the bound is attained is a rather deep question that is beyond Our present scope-
510
[. Iraracrerirariun oj L{genvatues and Equilibrium States as Exrrenur [C'h. 12
Since s is an arbitrary member of S, (1 1) i mplies that Lu1 -- AO' = z or Lu =
(19)
The above equation shows that / I is an eigenvalue. To show that it is the lowest, note that if u is any eigenvector with corresponding eigenvalue A, then Lu
).u,
so (Lu, u) _ Mu, u) or 2 =
(Lu, (u, u)
(20)
From (14a), 2. HIGHER EIGFlNVALUES
Theorem 2. L, so that
Let fir , i = 1, 2, ... N — 1 be the first N — 1 eigenvalues of ,
2 1 X 1 2 < .,. ç Let the corresponding orthonormal eigenvectors be denoted by u ; , i 2, ... , N — 1. Thus
(ur,
1 ;)
—
i,J —
•
t
, 2
,-.-, N — L
1, (21)
[We know from Appendix 11 1 that eigenvectors can be chosen so that (21) holds.] Suppose that there exists a real number J.N and an element of S, uN , such that (u N , u ;) = 4 and .
(Lu, u) — --(u, u)
^ N = gla
.fs
fu, NJ-
i — 1, . . . , N — 1,
(Lu s , aN)
(22a)
(22b)
(uh. uN)
is the Nth eigenvalue of L, and uN is the corresponding eigenvector. Proof. Let s be any member of S, the domain of L, and define r by
Then
;.N
h- t f
X ( s. u ïlui ^= 1
(23 )
rr— I (S) u) — E (s, umif = 4,
(24)
=
s
33
Note that by (21) ( I , u)
4
r =1
so t is admissible to the "competition" of minimizing (Lu, u)/(u, u). Thus the
following function (E) is minimal when r
61(z) _
0:
(L[re,i + ^ , UN + ]
Li)
( U N + et, uN + e t )
(25)
Sec. 12.2]
Eigen{salues as Mrrttm ❑ and the Rirr Method
Slt
We deduce (Ltii , — 1r,u. , r) = D
(26)
,
in exactly the same way as we deduced (18). We cannot reach the desired conclusion at once, since r is not arbitrary.* But from the definition oft, IV -I (LAIN
r)
—
=
(Lu x 2NUr+, s) — t
E -
—
ÂNUU,
i
(s, uj4
( 27 )
Using the self adjointness of L and various properties of the scalar product, we see that the ith term of the sum in the above equation equals (S, u1)(Luu, le i)
(s, ur)14ur ur)
(s. Uj)(uri , Lu i ) 0 • (5, OWN, 2rui ) 1 - (s, ui1^4luhi, u 1 ) — O.
Since the sum in (27) is zero, as before we can
(Lu re — Xup 5) = 0, ,
—
deduce that
■ so UN
Thus i, is an eigenvalue. Suppose that 2 is any eigenvalue distinct from A 1 , a.3 , .. _, and 4,1 _ I . Since eigenvectors corresponding to distinct eigenvalues are orthogonal, the eigenveetor ti corresponding to 2 must satisfy (y,
i — 1,2, ... , N — 1_
-=
(28)
As in (20), 2 w (Lu, Ujj(u, c), so u is an element of the set of numbers of which _ . is the glh. Therefore, ). ? ;. N . d R E M A R K. Scalar multiplication of u by a nonzero constant does not change the value of the ratio (L14, &)/(u, er). instead of evaluating this ratio for u, then, we can use r = of II u li. As (t, r} — 1, we can write
---
gib (Lar, z)),
( 29 )
ui!
!p.m = I
with a corresponding statement for the higher eigenvalues. The converse of this remark, aS in the discussion of (9) and (10), shows the equivalence of the Iwo characterizations of the eigenvalues, with or without the auxiliary condition (r, r) = 1. Extension of the above results to the problem
Lu
=Ar#ee,
M positive and self- adjoint,
(30)
A little knowledge of vector space propertiieS permits a more elegant argument [hark that given in the main teat. Equation (26) holds for an arbrtrary vector in the subspace P composed of vectors that are orthogonal to the fixed vectors u,. r = I, ... , N — I. It is not hard to show that Lap,. — f,. "rid e P. Hence.. as before. L& . equals the zero of A but the afro of a subspace is identical wish the zero of the original space. —
5 12
t1
Charuc+rrizulivrt of E+;aernuah ►esartd Equilibrium Mates us Extrema [ Ch.
is straightforward (Exercise 4). For example, the lowest eigenvalue is now characterized by (La, u) r = gib f3 ^
(31)
YcS ^u, t1^
where, as in (Al2.1.24), (u, u) = (Mu, a)_ Example. As in Exercise 11.3.20, one can show that the defection i]fx, y, t) of a
vibrating membrane satisfies pv2 ^^ =-
2^ ^
(32)
^i
where the constant p denotes tension and p is the density of the membrane, which is assumed to vary with spatial position (x, y). We assume that p is continuous in R, the domain covered by the membrane. Suppose that the membrane is fixed along its boundary B. There are time-harmonic Solutions of the form U = cos tor u(x, y) or L' = sin ait u(x, y)
(33)
if pal } u = puI 2 u ,
u = Don 11.
(34)
The problem of (34) can be regarded as the search for elements 14 of the set S = (w lw has two continuous partial derivatives with respect [al and y , ^+ = 0 o B)
(35)
that satisfy Lu =
(36)
wher e Lu =_
❑2 u, Mu
-
-
pu, A =
w'
.
(37)
Both L and M are self- adjoint under the scalar product 91 =
(
J
ffc.})(xy)dcT.
(38i
R
Furthermore, Land M are positive operators, so all cigenvalues are positive_ in particular, using (3l) we see that the lowest eigeuvatue ; t satisfies
at
wV r w du i[ f gib — js wts
]J* p w do
(39)
Using Green's theorem to modify the numerator, and using the idea expressed in (29), we Can also write .l i — glb fJivwi 2IVH.I da,
(40)
where admissible functions must be elements of S and must also satisfy I = {w,w) =px''da.
(4 1)
Sec_ 12.2] Eigenvafues as Minima und she Rirz Meihod
5 13
THE RITZ METHOD
We have expressed eigenvalues as minima of certain functionals. The elements that are possible arguments of these functionals must be selected from a certain well-defined set S, the domain of L. A practical procedure for obtaining approximations to eigenvalues consists of selecting a few elements from S and making the functional as small as possible by using a linear combination of these trial elements. To illustrate the procedure, consider the eigenvalue defined by where u a S, (ee, u) = 1.
=- gib (Lu, ie),
(42)
Let w i , w 2 , ... • w* he a set of trial elements selected from the vector space S_ It is convenient to select trial elements that satisfy the orth ❑ normality condition (w1,
wi) _ c5 ;j•
(43)
We now consider the particular element of S given by the linear combination 14 =
E ïR; w;
( 4)
t=1
and we choose the real numbers a ; to make (Lee, u) as small as possible_ Of course, the /i must be such that the constraint (u, u) 1 is satisfied. This means that h'
1
=
E i= L
N
N
j= L
^j ^ !
N
^ ^ ^i^ J^^i a x°) } = L j' i. 1 = 1 i-L
(45)
where we have employed (43). But N
(Lu, u) =(L[
N
!M
y cL, w; , y x i
i =1
11 ^
= ^ m; a; a i,j
1- 1
=
,
(46)
1
Here
(Lw i , w,).
(47)
Note that = (Lw j , w1 ) = (w) , Lw ; ) = (Lw; , wi ) — m il , by the self-adjointness of L. Consequently, among all 2; satisfying ^
i
EL =
we must choose those that minimize N
2
1
,
(48)
C'haraclerizoiiun r f Eigenvalues a n d Equilibrium Stales as Exrrema
514
[Ch. 12
Using the approach of Section 11.1, we introduce a Lagrange multiplier A and deduce that the minimizing 2, must necessarily satisfy ^
N
y mjlat a i —
Oa{
N
A ^ at =-- 0.
i,1-1
(49)
i= l
Upon carrying out the differentiation and dividing by 2, we find from the above equation that
Er
+77 i O 4 — Act r - 0 or ic= l
i — 1, . .
— AS,k)cck
. ,
N.
( 50)
h
To obtain nontrivial solutions to the eigenvalue problem (50), we require that 0 = determinant
— A A ) _ (— 1)r`A N + - --
,
(51)
an Nth order characteristic equation for A. We have icduced our problem to the solution of the algebraic equation (51), even though the original problem certainly need not be algebraic. Now we must establish a connection between the roots of { 51) and the eigenvalues
of the operator L. To this end it will prove useful also to pose the algebraic eigenvalue problem (50) in the framework of linear self-adjoint operators. {This problem is merely a slight generalization of an illustrative example treated in the previous introductory section.) Consider the Euclidean vector space R N of N-tuplcs a — ([,, al# , _ , j, with inner product - -
N
a• d =
-
where d
x ;; ,
, hti).
(52)
1
The eigenvalue problem (50) can now he written Ma = Aa. The linear operator defined by
d
=
Ma ifb r
(53) ;-t
is self-adjoint, as shown in (Al2.1.13). By Theorem Al2.1.1 the eigenvalues are real. We denote the ith lowest eigenvalue by A ; , so that A, < Ax
...
AN,
( 54)
The corresponding eigenvectors irrr', i = 1, ... N, are determined only up to an arbitrary constant by the requirement that Ma Aa, but this constant must be selected so that the constraint (45) is satisfied. That is, in the present ,
notation, a{'M • ar"t = 1.
Using (55), we see that Mar ie
/4,o° implies that au) • Ma" ) = A i .
(55)
Sec. 12,211 Eigenvalues as Minim and the
Ritz Method
5t5
Now the extreme values of the quantiy (46), whose gib we seek, are
Ece N
F.4 =
{,)^ atri = ^n . p A4
1
4
Mau)
. -
Thus the lowest eigenvalue A t should be the desired approximation to the lowest eigenvalue 1 1 of L. As we have minimized (Lu, u) only over a restricted set of functions, our approximation ought to be an upper bound. Indeed, it is a natural conjecture that A, We shall prove in the
next
1,2 ... N ,
(56)
,
section that this conjecture is correct.
VALIDITY AND C]T- 1LJTY
OF THE RITZ
MET 110f)
We shall briefly discuss conditions for the convergence cf the Ritz method. Then we shall touch on some practical matters connected with the use of this method. As more and more trial elements are taken, the lowest eigenvalue cannot increase. Lei us assume that L is bounded below in the sense a(u, u),
(Lu,
2 a scalar.
(57)
Then the various approximations to the lowest eigenvalue are certainly bounded below by a, so they must converge to some value- It is natural to hope that this value is the correct answer, ,1 1 . To discuss sufflcientconditions for the justification of this hope, we first note that we can assume with no loss of generality that a is positive (and hence L is positives). For if L is not positive, we observe that if L i u + fin. /1 a scalar, then A + is an eigenvalue of L, if and only if À is an cigen slue of L. (57) and {1 > Eat , then J. is bounded below by a positive con- IfLsatie stant. (Lu, y). if L is positive (and self-adjoint) we make the definition Lu, t'] As in Exercise A 12.1.6(c), this provides a functional of two elements [ that satisfies all the axioms of a scalar product. Now a set of elements w i is said to be complete in S if for any u F S, ,
Lim E a i w; = u
for some set of scalars a l , ar e ,
...
]
.
The limit is in the sense of (13); i.e., the norm of the difference between the partial sum and u must approach zero_ if the elements w l arc complete in S when the norm is formed from [ , ] (so that Dug 2 = [u, u]). then convergence * 1[(57) ho lds for rr > O, then 1964).
L
positive 6ul the convcrse l5 not necessarily true (Mikhlin, ,
516
Charactert_atrün of Eigeni alues and Equ[libriurn
Blares ac EYrrerna [Ch. 11
of the Ritz method can be demonstrated for the lowest eigenvalue, Convergence for higher eigenvalues can also be demonstrated under certain conditions (Mikhlin, 1964, Secs. 32 and 78). Proofs of convergence are obviously of high theoretical interest. For practical purposes, however, computations often become formidable when N becomes even moderately large. Typically, this is because calculation of the ni d involves complicated integrals and because it is not easy to determine the eigenvalues of a large matrix. In many cases what one wishes to know is whether a good approximation can be simply obtained. And indeed the Ritz method often provides an excellent approximation to At even if only one or two trial elements are employed. One illustration of this is found in the calculation of I, Section 12.4, where a one-term approximation gives an upper bound to the fundamental frequency of a longitudinally vibrating wedge-shaped bar that differs from the exact value by less than 3 per cent. A further illustration of the method will be given below. There are three main reasons for the outstanding practical usefulness of the Ritz method. 1. The method is most powerful for approximating the lowest eigenvalue, but it is just this quantity that is often of highest importance. To give some examples, the lowest frequency determines the character of a vibration more than the higher frequency overtones, the lowest eigenvalue usually corresponds to the most unstable perturbation (see Section 15.2 of I), the lowest eigenvalue of the Schrôdinger equation* gives the ground state, etc. 2 Because an eigenvalue is the stationary value of a certain functional, a first order error in the eigetttaector will give °lily a second order error in the eigenvalue (see Figure 12.2). 3. In applications, one often has some idea of the general "shape" of solutions. This can be used to guide the selection of trial elements. A common "rule of thumb" in using the Ritz procedure is to compare the eigenvalue estimated when, say, one trial element is used with that obtained with two trial elements. if the answers are close, one feels satisfied with one's approximation. But it could he that convergence is to the second eigenvalue (say), not the first, because both trial elements are virtually orthogonal to the lowest eigenvect or. A theoretical landmark WAS achieved with the aid of a Ritz calculation by Hyllcraas alï t#ie ionization potential of h€Buret 1t is worth quoting in emenso from the account on p. 347 of E. U. Condon and G. H. Shortley's The Theory of Atomic Specira {New York: Cambridge University Press, 19571. "An eighth approximation led to 1,80749tthc = 198322 cm t for the iortiie.ation potential as compared with an experimental value al! 1 48298 ± 6 cm - L. It will be noticed that the theoretical value txeeeds the experimental value by 24 cm - ' , which appears to contradict our statement that the Ritz method always gives Too high an energy value. The discrepancy is duc to the neglect of the finite mass of the helium nucleus and to relativity effects. When these are included the theoretical value becomes 198307 cm ', which is in agreement with the experimental value_ This is an important accomplishment of quantum mechanics since it is known that the older quantized-orbit theories led definitely to the wrong value."
Sec_ 12.2J Eigenualue.e as Minima and the Ritz Methad
517
f
Y
F1 i; ri RE 1 2.2. A schematic illustration of why an D{E] error in rhe eigrrt:.recrar, represented by x a , leads to only an 042 ) error in the eigFnrralue A. The quantity R(A) is minimal when r = xQ, Mid 1 =— R(x o l. Since R'(x,j = O. if we nusr.stimure xa to obttrin x U + E, using R(x 0 4 El win give [r , a good approximation fo A since .
.
R(x r, -f r) — A = R(x U + Fj — Rim ^ R"{xoiE r-
GENERALIZATION
The Ritz method can be used foi - the eigenvalue problem Lu = ;Mu, where M is a positive self-adjoint operator. In this case, as in (31), if (u, y) i (Mgr, ti), (Lu, u]
A, = glb -- -- or urs (u, U}
A, =
gib (Lu, u),
(58a, b)
I,cs
while for higher eigenvalues the minimization is subject to orthogonality conditions of the form
E
i=
Then (Exercise 5) (Lu ) Li) =
P. k =
.4ik?'iYk:
A ,;1
= (LWr, n' }) =
aPk Ys }k •
BIk
(k ►'i, wk .> —
r, + ►'+ I;
(60a, h)
A' Oil u> =
E r,k
1
^^T 1 y
ivt),
(61a, b)
-1
so the lowest eigenvalue is the smallest root of det (A ij — AB ;) = O.
(62)
Note that the coefficient of A N is (— 1 r det (B). This determinant is the nonzero Gram deterrninant associated with the linearly independent trial elements w; and the positive self- adjoint operator M (Exercise 6).
518
( 7rarercarrfxatius rofE -fJlranialurs arid Equilibrium Sitars as Extrema
[Ch. 12
EXAMPLE: 'TRANSVERSE VIBRATIONS OF A TAPERED HOLLOW BEAM
Engineers need to calculate the frequencies at which structural elements freely vibrate (natural frequencies). The reason is that another pant of the system may force it to vibrate ut a frequency near one of ils natural frequencies. if so, "resonance" brings about a large amplification of the forcing amplitude with un attendant likelihood of failure. In practice, usually only the lower natural frequencies are of importance because high frequency forcings in most cases are "shaken out" in the system, and so are too weak to cause harm. Following Mikhlin (1964, Sec. 69), we show how the Ritz method can be used to estimate the frequencies at which a beam can undergo transverse (i.e., bending) vibrations. To provide a true picture of this nontrivial example, we shall quote rather extensively from Mikhlin's presentation. The reader need not concern himself unduly with the details; the main point is to see how the Ritz method can provide a fruitful approach to a practical problem. The transverse displacement u* of a beam performing bending oscillations with frequency ww satisfies the rquutiont E
r)2 * ^^
(.(XS) d
[
; * }a
^l^
d(x
] —
irA • (x')L uu*
—
0.
(63)
Here the beam lies along 'ilie x*-axis from 0 to D, 1*(x*) denotes crosssectional moment of inertia and A•(e) denotes cross-sectional urea. E and p are Young's modulus of elusticity and the density, respectively ; both of these will be assumed constant. To reduce the number of parameters, we introduce the dimensionless coordinate x, where x x* f {compare Section 6.2 of I). We consider a beam that is fixed at x* z 0 and free ut x* = D. Define the dimensionless area and moment of inertia by A(.a) = D 'ANA)), 1(x) = 1) 4 1*(x n. Then our problem is r1(x)e"(x)1" = AA(0e,
corD'pE
u '(o} = u") L) =
ti•(0)
1,
= O.
This problem is of the form 1.0 = AMu fur u E S, where S consists of functions that are defined und four times continuously differentiable on 10, t] and that satisfy (64b). M is trivially self- adjoint and positive (since A O). So is L (since J > 0) for if is E S, v e S. integration by parts shows that (u,
u
La) ^ a
^
[IL,"]," dx ¢ jr Li" le" dx
(a). Lu),
(65)
a
t Equation (63) can be readily ublatncd from (5.1,33) and (5.134) by ncgkctingeffccta of rotary inertia and body form (Kocp is mind that p was used to denote mass per unit kngth in Section 5.1.) The boundary conditions or (Mil) are those of (5_1 .3f1) and (5.I.42).
Sec. 12.4
Eiyenvolues as Minima and the Ritz Method
519
and
Lu) -= f 1(u") 2 dx
O.
[If {u, Lu) = 0, then if __ 0, so n O by (64b).3 The lowest eigenvalue will be given by (58). As trial functions we select the solutions w(C) ❑ f da— w 1 1 A
—
!4 ;
v^
= w,, ' _ 1 = w"
—
_ , ,•„ - =
2
Q_ (66)
Four reasons for this selection can be identified. L The problem (66) is similar to the given problem (64)—In fact, (66) is appropriate to the uniform beam_ Consequently, one can hope that even Just the lowest eigenfunction of (66) will provide a decent approximation of the lowest eigenfunction of (64), provided that 1 and A do not vary markedly with x. 2. The trial functions are eigenfunctions of a Sturm Liouville problem, so their completeness is assured. 3. The trial functions satisfy the correct boundary conditions_ This should be helpful, although it is not necessary, since the free boundary conditions will occur "naturally." 4. The eigenvalues of (66) have been tabulated by V. Faddeyeva [Trudy L11PS No. 6 (1935)]. It is to conform with Faddeyeva's use of the interval L- i, f] that we have introduced the variable c in (66), but the results can be adapted simply by making the transformation x = + ,
(67)
w(x) = x -- ).
The required eigenfunctions of (66) are [Exercise 8(a)] sin xk k(} = sin (2^2)
cosh xk cosh (5,j2) 1
k odd, (68)
cos ak
sinh
cos (4/ 2) Binh (0 012) ,
k even.
The eigenvalues are solutions of the transcendental equation
cos ak cosh cek = 1,
k = 1,2, .._,
(69)
and it is these numbers that have been tabulated by F - addayeva. The functions is fx) = w {(x — 1) are to be substituted into the Ritz equations (60b), (61 b), and (62). Mikhlin considers a bar that has the shape of a hollow truncated cone, as pictured in Figure 12.3. Then
1(x)
A(x) 74 4 6. ^ a }, ^
rz(y s ! a2 ),
where y
= R (R
r)x.
(70)
530
Churarrerixarion tit Eigenvatues and Equilibrium Stairs as Extrrrna
2R L
(Ch. 11
2r L
Fi[tUx 12.1 d tapered lirrf lu K• hewn isa solidly . rural ion whose cross section is rlrpicied here, The cro y s section ïs rotated around the dashed axis. In this case 1 a ^^t _ ^ (l^ r — r x }j^rit + ^3 { 1t x_ 2 r s ^1^+ ) = _ tb (R + rre — u l^+
4 ^
2(12 2 TE
Ha =
^#
r 2 ){R — fi}
(R } rj x — u 1
'
1
1t' 1 (R — r)4 f^^^,
(71)
1.1^ — 1 (le — r 2 )J11 1 + (R
—
r)2.11k r,
(72)
where f^l
Jy V
y j"
^ tx 1r 2
j [1x
44)71)4V) 170+ (^^ 0 d^} r(0%v+(0
0, 1, 2, 3, 4;
n
(73)
(74)
n = 0, 1, 2-
in 1Niklilin's book 2i pages of formulas follow for the f3i', such as the following for i = k. k o dd: 1 i^ ^ --
= at^ ..
^^ at„ Cot a
2
^ °c±rx coi ^+ ^ —OE -11 ^ cat x 12 + ^ 2 4
^ 1 1'±i = a^ ca^ = ^ - 4a# cot a- - + 4 , 4fj,;' ^
+
barkcat ^ —a t cot x ^-
^.
(75)
Sec. 12.2] Etgenruleres cu Mrrirmu und ;he Rir2 Method
52 r
Also J1V
== 03a,
f^x
J1k } = f`tl
144 -
^,
(76)
We present the final results for the case
R _o, r = 215, a =
.
From the formulas given here it is not too much work (Exercise 10) to obtain the first order equation* 0.8526 -- 11.98A = 0. Thus the first approximation to the lowest eigenvalue is = 0.07117.
(77)
Upon finding the lowest roots of the quadratic and cubic obtained, respectively, in the second and third approximations, livfikhlin obtains — 0.06762,
A(13) = 0.06674.
As expected, Aul t > A (12) > AV ). Note that (77) can be used as a first approximation to the lowest root of the characteristic equation for a higher order approximation, to be improved by some perturbation or iteration scheme. From the second-lowest roots, Mikhlin obtains the following two approximations to the corresponding exact eigerivalue A` } = 1.8605,
/VP
= 1.7718.
It appears that accurate results are being achieved. A computer program could be set up to supply results for any given value of R, r, and a. RATURAI. BOUNDARY
CONDITIONS
Successful abstract theories reveal the unifying concepts that underlie seemingly disparate problems. At the same time, the very generality of these theories sometimes makes them incapable of dealing with important special features of a particular case_ An illustration of this fact in the present context stems from the observation that for differential operators, but not for matrix operators or integral operators, boundary conditions are usually a part of the definition of the domain over which the operators act. Consequently, the general theories cannot be expected to bring out the subtleties connected with the fact that under certain circumstances it is advantageous to omit specification of Some of the boundary conditions in describing the domain of a differential operator. We close the present section with an illustration of this point. Miktilin did nut introduce dimensionless variables, so we have had ID carry our a translation
of his resulrs,
Iltrle
522
Characterization cif Eigenvalues and Equilibrium Slates as Exart'tna [C'h. 12
Consider the eigenvalue problem (pu')' + qu = Ani,
u(o} — O,
(78a) (78b, c)
p(b)u"fh) + Ou(b) — O.
We regard this problem as an instance of L14 = Pilo. Here k is a positive constant,. As usual we assume that p, q and r are positive, that pis continuously differentiable on [4, 1], and that q and r are continuous on [0, 1]. A special case of this problem appears in 12.4.10 of I, where equations are given that describe longitudinal vibrations of a bar. The bar is secured at x = h by a spring, and k2 is the relevant spring constant. 13y Exercise 13(a) the operator L is self- adjoint and positive over a space whose elements must satisfy the boundary conditions (78h) and (784 Consequently, the eigenvalue problem can be treated by the methods that we have already presented. If it is, however, Ritz trial functions must satisfy the complicated boundary condition (78c). An alternative approach, as we now demonstrate, avoids this complication. We observe that on integration by parts, ,
j7,
1 — (pie')'
+
+ qu]G dx --= —pu'r ,
(pu'd 4 que) dx.
^
(79)
u
G
Let us define a vector space T consisting of twice continuously differentiable functions that vanish at a. ff u and y are elements of T, then (79} can be written (80a) where now (SÛb)
and (80c) Since p and q are positive, r product ( Exercise l3(b)J. Let
A L = gib R 1 (14), uc T
,
1
possesses all the attributes of a scalar
R 1 (uj
[14,
u]
k2u2(b)
(81)
r/+ L^, u)
where (u, re} i (Mu, u) as above. There is reason to expect that . exists, for R 1 (u) > Û; nonetheless, the lower bound may not be achieved for a function in T. ,
Theorem 3. Suppose that there exists an element u 1 of T such that 2 1 = R02 1 ). Then 2 is the lowest eigenvalue of (7 8 ) and u, is the corresponding eigenvector.
Sec _ 1 2_21 '
Eir#enr•afr i es Qs Minimu and alir R I t- ME' Mimi
Outline of Proof s E T and define
523
As in the proof of Theorem 1 we select an arbitrary 4 3 (0.mDi
= 1{ 1 (u 1 + r..$),
The requirement jr(0) = Q again yields N'i (0) — ). 1 19'1 (0) = O or [Exercise 13( c)]
2[u 1 ,
s]
-- 22 1 (u 1 , .$) + 21ï 2 I11(hh(h) = D.
We now employ (80) and the definition of <
>. Canceling a 2, v,c lind that
(Cll r — £ 1 Mu r . s) -}- [p(b)rri(10 + k 2 u r (h)]ssh) The conclusions that Lit t = ï i Mtt i ,
((h)ra r (h) +
1i r ((t) = b,
u.(h) = ()_
and that f. r is the smallest eigenvalue, now follow by the standard line of reasoning used in problems involving natural boundary conditions [Exercise 13(c)]. d 4
EX
FRCISES
1. Using geometric reasoning, the text provides two different characterizations of the stationary values in Problem B (for all admissible: x, on the one hand, and for increasingly restricted sets of x values. on the other)_ Provide such a discussion for Problem A. 2. Show algebraically that i., and are correctly glen by (5) and (6). (Take the additional constraints into account by means of additional Lagrange multipliers.) 3. #(a) From the identity .
(ii„,.
G„) ` (u, U) = (14, — te, V M — r) + ( ii —
prove that an
fortiori, (fa, um)
al.
r) +
it, zn y implies that (ir,,. 1.) (es. r). (tr ,,. i') (a. ri.
trr.
(u, r) and a
Prove that (I I) holds if S is dense in T. Show that with the inner product (7), l'Nil gives the usual length of of a vector in 3-space. (d) Show that S is dense in Tin the sense of (12) if and only if for every positive c and for every element t of T, there exists an element s of S such that Its — < 4. Extend Theorems 1 and 2 to the problem Lea = AMu. where M is positive and self-adjoint. 5. Extend the Ritz method to the problem LIE = i_Mu by verifying (6O), (61), (b) (c)
and (62).
514
Chwrueterisatiatt of E{genn•atues and Equilibrium Stairs as Extrema (Ch.
f
6. Let L be a positive self-adjoint operator with domain a vector space S possessing the scalar product ( , ). Let v i , ... , v, be elements of S. Similarly to (Al2.1.24), the definition [i i , (Lv i , vi ) introduces [ ], which satisfies the scalar product axioms. (a) Prove that if the r i are linearly dependent, then the following G ram determinant vanishes; ,
[t., 11 v i ] "' D., + vn] [^„, ^ :1 [ v., v„] ::: '
^
(b)
Show that if the Gram determinant vanishes, then there is a vector v = = t a1 eat with not all y = 0, which satisfies [u, u ; ] = 0, i = t, . . n. Deduce [v, u] = 0 and thereby prove the linear dependence of the v i . 7. (a) Verify that (64a) is the dimensionless version of (63). (b) Fill in the details necessary to obtain (65). H. Verify {a) (68); (b) (69); {c) (70); (d) (71); (e) (72). 9. (a)-(e) Verily the five formulas of {75); (f) verify (76). 10. (Project) Find at least a rough approximation to the lowest root of (69) and use it to check (77). 11. (Project) Find fairly accurate approximations to the first two roots of (69) using hand techniques. 12. {Project) (a) Outline a computer program that will successively find the roots of (69). (b) Write and run the program_ (c) Write a discussion of the issues involved in preparing a computer program to obtain the lowest three eigenvalues for the beam of Figure 12.3. Any sensible value of R, r, and a should be accepted by the program. 13. (a) Show the self-adjointness and positivity of the operator described in (78) and the sentences that follow (78). (b) Verify that (80b) defines a scalar product_ (c) Fill in the details required in the proof of Theorem 3_ 14. To afford practice with the Ritz method, consider ,
d2n dx2 (a)
(b)
(c)
+ i_u=0,
u(-1)=4(1)=0.
If 1 0 denotes the lowest eigenvalue, show that 4.4 0 ft2. Find an upper bound for 4/1 0 by evaluating (Lw, w)j(w, w) when w — 1 — x 2 . Compare with x 2 9.87, a result correct to two decimal places. To improve on the approximation alb), use the fact that the lowest eigenfunction is even. This permits the integrals to be evaluated between zero and 1, and suggests as an improved trial function w = 1 — x2 + a(1 — x 2 )x'. Use this to obtain an approximation
Se(' . 12.2] Eigenvalhes as Minima and the Ritz Aleihod
52 5
is
to 410 that is accurate to about one-tenth of 1 per cent. [It probably quickest to employ (62).] (d) If (1 - x 2 ) o a„x' is used as a trial function, to what number do you expect the second lowest root of the Ritz eigenvalue equation to converge as N ac.? 15. The bending oscillations of a plate that is rigidly fixed at its boundary B (a plane curve) are governed by the following problem. which is a two-dimensional version of the beam vibration equation (63) and the "built-in" boundary conditions (5.1.381 '
Vo w f k w - 4,
w=
Crt
O on
B.
182)
Here w is the vertical deflection, 1. is a dimensionless frequency of oscillation, and denotes a normal derivative. Use the Ritz method to find an approximation for the lowest eigenvalue when B is an isosceles right triangle. (a) Reduce the problem to the evaluation of definite integrate_ Defend your choice of trial functions. (b) Complete the calculation (project),
The following five exercises concern an eigenvalue problem that is the subject of "Free tidal oscillations in rotating fiat basins of the form of rectangles and of sectors of- circles,- by A. Pnueli and C. L. Pekeris, Phil. Trans RGy. Sr c. (London) A 263, 149-71 (1968)_ This paper seeks to gain insight into the effect of jagged coastline on tides. The wavelength cf tidal oscillations is so large compared to the depth of the region which contains the water that vertical variations can be ignored. Thus the wave height only depends on the horizontal coordinates x and and the time t. For wave heights of the form ax, y) exp (it7t) it can be shown (Exercise 8. t _13) that
N2 4 k 2 ) _ 0 for (t, y) in R,
en
- it
ids
= 0 for (_x, y) on B.
(83)
Here B is the boundary of the region R that contains the water; ltix (17 2 - 4w 2 )/gh - (o 2 fgh)(1 - r 2 ), where w is the constant angular speed at which R rotates, g is the (constant) gravitational acceleration. h is the (constant) average depth of the water: ?/rtt denotes differentiation along the outward normal to B, and c?/r's denotes differentiation along B_ Note that C is a complex - valued function. One purpose of this series of exercises is to afford practice in manipulating the ideas that we have presented by asking for the extensions required in the complex domain.` (No • It mensions to vector -valued functions arc also useful. For an example. see "A computertrnplerocr3led vector variational solution of loaded rectangular waveguides," by W. J. English ISh,iM 1. Appi, Math. 21, 461-68 ( 1971 }].
526
Characterization el - Eigen ahres and Equilibrium Slates as Ewsemna KA
U
true complex analysis is necessary for this, only an understanding of complex numbers.) 16. First consider the case of no rotation, where t = O. Show that if iÿ makes the integral
1_
J
J(VZ. Ÿ2 — k 2 Z2) dx dy,
E complex conjugate,
stationary among all complex-valued functions Zfx, y) which are twice continuously differentiable, then is an eigenfunction. Use the results of Exercise 17_ 17. (a) The formula
J
fv4 . VII' dx dy =
e n ds ff0 / 4,1.0 dx dy O
R
R
is well known when 4) and 1/ are real - valued: Prove that it remains true when and arc complex-valued functions of x and y. (b) Suppose that Re JJ f (x. y)g(x, y) dx dy = 0,
"Hey "real part of,"
for arbitrary twice continuously differentiable complex-valued functions f. Prove that if g is continuous, then g = O. NOTE. Exercise 16 also requires a one-dimensional version of the above lemma, but the proof of this is essentially the same as that for the
two-dimensional version. +18. Now let rotation be present ('r 0). Show that a correct variational formulation results if one adds the line integral I' to the integral 1 of Exercise 16, where =
c?2 — d5$ F^5
Also show that P is real. 19. One possible way to obtain an approximate solution to the problem of extremalizing 1 -4- 1' begins by assuming
Z=
ay complex constants, p=1
where ^^
Cp
+ kp ^e = 0,
^^
= 0 on B,
k; s
-
• . -
sec. 12.21 Eifiestralues as Minima and the Ric: Merlod
527
(As trial functions, this approach uses "no-rotation" eigenfunctions. It should therefore be useful for sufficiently small values of r) Demonstrate that the above assumption for Z leads to the problem +
r
1
—0)Niai =Q,
=
1, 2,
N.
m-t
/^ Nrm _
N? =JJ? dx dy,
— [ S! , ds = PMrg
L'S
R
Part of the demonstration requires extension of the Ritz method to the complex case.
For small z, a perturbation method is useful in obtaining the approximate eigenvalues k. Nothing further is required here, but the reader may wish to glance at Exercise Al2.1.7 to obtain an idea of how such calculations proceed. 20. Pnueli and Pekeris found that, except for small values of r, the following method was more successful than that of Exercise 19. Following a suggestion of Trefftz, one can consider trial functions of the form NOTE.
1a„ „(x, y; k) + i^i 7,1(x, y; k)], app and fl„ real constants, n-
1
where (V' + k e g„ =(V 2 + k l kl„ - 0 and the real functions and ri„ are complete on the boundary. (a) Show that surface integrals drop out so that one obtains
n
-
fH Langn D! +
1
Lg.) + ifin( ` +1 n ^^^ + tir Drin)] da = 0,
L ^ [1^^^^^7t - g1 DO + 13„(qaD►h + rledring ds =
n=1
8
where
p
=
- it i^5
t^i1
(b) Show that these equations are equivalent to
E qtr a + rP
ds = 0,
n=l •
a
°^s
1I
^ ^
6.
f("a‘n
=1 •
as
+
# ^^ âds = o. n
0,
52$
Chararterizalion of Fryrn, whirs mud Equilibrium Susses as F. .renia loir.
12
(The lutter" represent simultaneous equations for the determination of the coefficients a„ and f1„ and the vanishing of the determinant of these equations yields the cigenvalues r,, for a given k." To furnish a bolier understanding of the method, we note that for a hasin wherein kI a and Kyi s 1x, the following are suitable trial functions fer waves that arc symmetric with respect to the center of the rectangle: = CUh I. ti A Li)ti
]^ n
-
sin it x sin ^i y
-
I)ivl s iiili by !rn ensures that rf„ is real n:YCti when if. is imaginary.)
12.3 The Courant Maximum—Minimum Principle A PRont.l'. M1i 1N ► 'IBRA l ION ttlt,ï11t1'
Consider the longitudinal vibrations of itihotnogeneous bars, i.e.. bars whose composition and shape may vary along their length. Suppose that two bars are identical in every resp e i except that Ilse lirsi (surperscript I) is tihide of a slilfet material limit the second fsuperscript 2j. to other words, the respective Young's moduli satisfy 1.1I 1 0. > L12100. )
where f) is the length of the bar. What can one say about 11w frequency of vibrations? Thai the stiller bar possesses a higher fundamental frequency is casily shown from the variational charactcrir,.ition of the lowcsi eigenvalue. From Section 12.4 of 1. for a bar that is fixed at both ends this is Evi.,1'i02 rf x 10 (w).
1(H')= J
J
t = 1,2.
The set S consists of functions that have certain smoothness properties and that vanish at ii and D. For every admissible function, the qunttetii i is larger for the first bar than the second, ln particular, it nl" provides 111C iiiiliiiituui of #x"`lu•1, ibeii
lin = Rwitu 3 ")
min W 2 110 _
^t]
we* s
> r.}`' = Dut this approach sloes not lead to the desired statement the functions th:it must be tested to obtain the minimum 2, 3 For are orthogonal lo a i ` white those for 411 must be orthogonal to 0 1 Tliese arc different sets of fonctions, so we cannot carry through our previous arguaient. A characterization that does allow us to provide the desired result foc a wide class of problems, and that is also of considerable theoretical interest
n
Sec. il_j]
The Courant' Maximum --Minirrrtrrn Prirrc - rpk•
519
and elegance, is the Caurwu* maximum--minimum pri ncipl e . To this we now turn. The reader will not be surprised to find that we motivate the general principle by turning to the example that we discussed before, the determination of points on the unit sphere that minimize L - t 7;x i xf Previously, We characterized the second eigenvalue of the positive definite and symmetric matrix with components 1; ; by restricling attention Co the plane through the origin that is perpendicular to the first rigenvector, Now we wish to avoid reference to the first eigen vector and still formulaic a problem that gives the second eigenvalue. As Courant proved, this can be done by first considering an arbitrary plane through the origin. Such a plane cuts the unit sphere in a circle of course, and cuts a given level surface in au ellipsc t As before, try to find the ellipse of minimum area that just touches the Circle. Now try to choose the plane so as to maximize this minimum area If your geometric intuition is good, you will see that the plane should be rotated until it is perpendicular to the first eigenvector, so that the maximum minimum characterization leads to the same second eigenvalue as before. In any case, this result follows from the general formulation of the maximum minimum principle, which follows, . 1
-
.
THE MAX—MIN PRINCIPLE
Theorem 1. Consider a real linear self-adjoint operator L with domain S. as described in the material just preceding 12-111. Let the eigenvalues of L be f 1 < ). - • • with correspondmng orth ❑ n ❑ rural a igenveçtors to I . ii ... . I arbitrary vectors r, as follows§: Define a function lit of p —
rnilïv,._.- , rp_ tl =
min RI(irl, we s
f = t,-.-, p — 1,
(2)
where
Mu) • Richard Courant was
head ache
— (L^^_ u) (u, re)
(3)
Mathematical Institute m Gottingen when it waS a world
center For mathematical research. Driven our by the Nazis, he started all over again in the United Stairs and built up at New York University the intcrnaii wliy renowned tnatNUte tttdt bears his name.
through as center. a closed curve obviously results. it is apparent from the form of the attrehraic suhstttutrotn required obtain chip curve that its equation contains only linear and quadratic terse- Corlsequcrrtly, to the curve is an ellipse, t The level surfaces are ellipsoids. When such a surface is cut by a plane
The argurnents arc disirr to follow if we use the more intuitive -- rnux"' and "min -- in preference td lub and gib. To make sure that ihtr notation is understood, we remind the l eader that in (2) m denotes the minimum value of R(rrl among all elements of S that arc orthogonal to the vectors r,. i = I. _ _ _. p t_ As usual, we canon as well minimize (La, 4,11 if we add the additional restriction (di. rrI = I. in conformity with previous praciice. we are assuming the —
existence of all minima and mamma.
Characterization of EiyetrraJ:ws and Equilibrium States a-s Extrema [['h- 12
53D
Then max m = m(u t ,..., Ur— t) =
(4 )
hJ,I
Proof
According to Theorem 2-2, r(ut, u2, - - • u - t ) _ ). . so a fortiori .
,
,
max m> (r11 1 -o complete the proof, we show that the last inequality is also truc in reverse by demonstrating that for arbitrary ^; m(uy, .... v p _ t ) S Ar To provide this demonstration, we exhibit a vector u in S which is orthogonal to the v, and which is such that R(u) ç ). p . We write ,
-.
p
P
u =
E b; ti ; and require (u, u) = 1,
i.e.,
y bi2 — 1.
(Sa, b!
There is at least one vector u in the p-dimensional subspace spanned by that is orthogonal to L t, - • • , _ I and that is of unit length as required by (5b).* But for this vector u ❑F
A
P
R(u) _ (Lu, u) _ Xb ; I.u ; , s= 1
^
[,i =1
F
^^^► ^l^+i
P
—
},b?
Ç
^.
(6)
J1 _ . e =t
t= t
Because R(u) < 2 certainly mir t , _ < the ti's are arbitrary.
bjuj) i,j =I
J
P —
A
hj u; =
- -
, up 1 )
i! p , and the max of rn is also
Example. The pth eigenvalue of the self-adjotnt problem Mc = Ac that arasc in method can he chanicterized using the max mitt principle as Follows:
the Rit?
c Me pic) = C^c •
ii(a tlr ,
. ai'-CI] --
min PM'. i6 R"
A,,=
max p.
(7)
ia 1 Qt
p
Using the max-min principle, one can readily show that all frequencies of vibration are raised in a longitudinally vibrating beam if the beam is made stiffer in the sense of W. For, as in our argument concerning the lowest eigenvalue, Nii21(l}I,..
• In concrete terms, thcp
vp-
-,
t'y— l )
(8)
I linear homogeneous equations
y Nui. t'i = o, —
Cl}Il itlu l+
j — I•- - •P —
I-
1
have al least a one-parameter fancily of solutions for h1 , _ . , b p , dru} a particular solution cats be selected se that (5h ) is satisfied-
Sec. 123] The Coures zt Maximum-Minimum Principle
Sat
Lei the maximum value of the rrzi'' be provided by I, 2.
Then J. A2t _ mt2)(ui2is ..-.
uF} i l C ^tti{ut2w ._.. ^
1)
max ^Tr^^'1U • -
rt. I) ; 4".
^.
ia•, I
(9 )
APPl.1CATIDN TO THE RITZ !METHOD
Let us now prove our earlier assertion that eigenvalues obtained by the Ritz method are bounded below by the exa ct eigenvalues- This is done in an elegant manner by using the max-min principle to characterize the eigenvalues of both the original problem and the problem generated by the Ritz method. The latter case has been dealt with in the example above, and we shall employ the notation used there. Theorem 2. The pth smallest root of the algebraic equation (2-511 obtained in the Ritz method is an upper bound for the pth eigenvaluc of the original problem. Proof Let u 1 , - . - , up - 1 denote p — I arbitrary elements of S. As before, we are considering as trial functions a linear combination du orthonorrnal elements of S, namely, w 1 , . - i . Define p - - 1 n-tuplesa'"; i = I , p — 1; by f Rai t+x where alp = (n•,, v ;). a4') =lt,•^ sulk .
,
,
..
,
Note that the general trial element
C
!! = ;=
t
i
satisfies n
E ^`^w1^ ^} =
(u, vi) = y
Fl
i E c1a lp = e' a' ', 1=
i = t,
, p — 1.
( 12 )
I
Furthermore, using (7), we see that RN)
_
(1-14. u)
c • - ` R —
(u, u) T c • e
^(c)
(13)
Equation (13)just shows that R(u) reduces to p(c) when the trial function (l 1) is used. But we view (13) as an identity between the expressions whose maximum- minimum over suitable sets, respectively, characterizethe original eigenvalue problem and the Ritz approximation to it. Lxp]icitly noting the dependence on the a; '' of (1O), we fix our attention on the vector c = c(a { ", a ( v - 't that minimizes p among all vec[ors which are orthogonal to these si t '', i = 1, ... , p -- 1. Consider the function u of i l 1) with the components c; of this particular c as coefficients- By (12), since ...
}
Characterization of Eigetwafues and Equilibrium
532
Suites as Extrema [Ch. 12
= 0, this u is orthogonal to the ta; . Consequently,
e•
min p(c) = R(u)
for a certain u such that (u„ i4) = 0.
cc
(14)
But the left side of (14) is /t(tt{ tr, ... ,alp L) ) in the notation of (7), so that AR (
Ly
min
. ► air - I );
R(q) = rn(U1, - - -, up-1}-
ii
i4.w} =i
•Then max p(a t oo i
{ L},
} a tr- Lt
m(v i • , vr-
,^
In other words, by (7), A
?
m
(v
1,
- • • Lp
-
L).
But since the i: are arbitrary, Ap
max m(VL , ..-.ep-t) = J.p ,
which is what we wished to prove.
El
E X FRCISES
1. (a) Use the max-min principle to prove a result concerning the qualitative effect of increasing the density of a longitudinally vibrating inhomogencous beam. (b) What is the effect of increasing the moment of inertia I upon transverse beam oscillations? (See Exercise 12.4.4 of [.) 2. Using (12.4.10) of I, discuss the effect of the spring constant k on the longitudinal vibration frequency of a bar_ In particular, consider the limiting cases k = 0 and k = rx . 3. (a) Show that preventing a portion of a vibrating membrane from moving cannot lower the frequency of vibration, while slitting the membrane cannot raise the frequency. t(b) What will be the effect on the frequency of freeing part of the boundary of a vibrating membrane? t4. Regarding the buckling of a bar, consider three situations in which, respectively, the boundary conditions at an end of the bar are
(i) y=y'=O(ii) y =y"= O.
" + K y " T Oy t (iii) 1 _ 0Imagine the other end of the bar to be "built in" (y J y' = 0) in all cases. Referring to Exercise 11.3.1 for notation and results, discuss the relative magnitudes of longitudinal force that are required for buckling in the three situations.
sec. 12.41
Minima! Clrururrerizurrrgr
of linear
positive >prphtt•r11t
533
12.4 Minimal Characterization of Linear Positive Problems A surprisingly large proportion of natural laws can be viewed from either of two perspectives. In one, equilibrium configurations or motions are determined by imposing a necessary balance between various vector quantities. The other formulates the laws as extrenial principles. Since a single world of phenomena is under scrutiny, both points of view must. cf course, he equivalent. Sometimes the formulation of a natural law is bound up with extremum principles from the very beginning of theoretical work. An example can be found in optics- Willebrord Snell (1591- [6261 experimentally discovered the law of refraction that bears his name. Later, in 1657, Pierre de Fermat asserted that "nature always acts by the straightest course"; in particular, light travels between two points so as to minimize* the time of travel_ Assuming a speed of light propagation that varied from material to material, Fermat deduced Snell's law, More often than not, the formulation of a natural law as an extremum principle followed its original enunciation by some time. (An example is provided by Hamilton's variational formulation of Newtonian mechanics, the far-reaching and precise culmination of Maupertuis's grand but ill-supported proposition that nature always acts to minimize sonic "action.") Thus it seems that riot only should a theoretician be capable o f solving extremalization problems that arise "naturally" in the course of his work, but he also should be aware of the fact that natural (or social) behavior often, and with advantage, ran be regarded as being the consequences of an extremalization principle_ Possible advantages arc suggestive theoretical elegance and computational convenience. An illustration of the first advantage lies in the unification of optics and mechanics through the principles of Fermat and Hamilton- this was influential in forming the foundations of quantum mechanics. The second advantage was illustrated in the previous section, where we showed how a variational formulation of certain classes of eigenvalue problems led to the powerful Ritz. met hod of approximate computation_ In classical fields of "hard" science, al least, these days most problems are formulated as a set of equations or inequalities: most often differential but also integral, difference, delay-differential, and so on. One must make an extra effort not to lose sight of the possible advantages of an extrental formulation. In this section we wish to illustrate the possibility of reformulating as minimization problems certain equations that involve positive operators. We begin with a discussion of the fact that in particle mechanics stable equilibrium can he characterized as the state that minimizes potential energy. • It is now known rhai only m a restrictive formulation of
Fermai's
iruc
rosay
tl^xl the time of travel is minirnixcd. In general. all rifle Can asscrl is that the travel trine along the actual path
is stationary with respect to a suitable class of admissible path% Sue pp- 127fr ot' M
Born and E.
-
Wolf's classic,
Principk.s n} ()pries (London: Pergamon Press. 1959)
534
Characterization
EfgOnrrnlues a nd Equilibrium States as Extrema
(Ch. 12
An attempted generalization of this idea to the problem of a weighted membrane leads to a conjectured equivalence between the Poisson problem in partial differential equations and a certain minimization problem. The equivalence is proved. Indeed, a whole class of such equivalences is demonstrated by using functional analysis. (This approach emphasizes a structural similarity that was ultimately perceived to unify a number of particular problems) Again, a Ritz approximation procedure is formulated. Its strengths and weaknesses are illustrated in the context of the classical torsion problem of linear elasticity. PARTICLE EQUILIBRIUM AS A MINIMUM OF PDTENT1Al. ENERGY
The ideas we wish to present can be illustrated by considering the motion along the x-axis of a particle of unit mass that is subject to a force component f (along the x-axis). Let there be a force potential 0, so that f (x) = — ç'(x). Then x =— . (l) '
Upon multiplying by x and integrating with respect to time t, we find that
+00 2 + =
C,
C a constant,
(2)
the well - known expression for energy conservation in this conservative system. By definition, at equilibrium the particle is at rest (dx/dt = ü}. From (1) this means that if x = .x 0 is an equilibrium solution then 4 1(x0) = O. Thus at equilibrium the potential energy is a stationary function of position. We now show that the equilibrium is stable if the potenliui energy is u minimum. As was mentioned in Section 11.3 of I, the precise meaning of "stability" varies with context. This is not due to a mere lack of a standardized terminology but reflects the fact that the appropriate stability question genuinely differs from situation to situation. Thus the downward-hanging equilibrium of a damped pendulum was shown in Section 11.3 of I to be asymptotically stable in that perturbations from equilibrium eventually die out completely. There is no damping in the conservative system under present consideration. All one can hope for is that departure From equilibrium can be made as small as desired by keeping the level of disturbance sufficiently low. More precisely, one wishes that for all time the state of the system will differ by an arbitrarily small amount from the equilibrium state, provided that the initial state departs from equilibrium by a sufficiently small amount. This is the requirement of liapltnor Stability. * In other words, Liapunov stability requires that the state remains close to equilibrium if it starts close enough. * The Russian. scientist Liapunov introduced this definition in a paper published in 1893liapunov's early and profound contributions to siability theory were the forerunners of a Russian exptrtise in this field that Continues strongly to this by
Sec. 12.4] Mirrrrnut C'hararrerrzarir3rr clot Lirrear Posirit•e Problem
535
FIGURE 124 Liapttnov stability of the origin in the phase line. Given
«)
circle of radius E, there fnusl be a corresponding circle of radius point which starts in the lamer circle never mores our of rte farmer.
arl}'
such that a
Asymptotic stability, on the other hand, requires that the state gradually return to equilibrium, not just remain close to il. In the present case, the state of the system is given by the position and speed of the particle_ As we have found in several instances, it is convenient to measure displacement in terms of departure from equilibrium. Thus we introduce x—x0 ,
+x 0 ).
O()
In terms of the potential () is stationary at the equilibrium point i± = 4_ The Liapunov stability requirement is as follows (see Figure 12.4)_ The equilibrium point = Ois Liapunov-stable if for any given positive E, ,
[OW
+
whenever, for some positive
42 ( , )]nr 2 C 6 = (5(0,
for
7
0
[CM + 4 2 ( 0)]' < b.
(3)
(4)
That is one must be able to arrange that a state point will stay within any given circle about the equilibrium poinl (in the phase plane). This must he done by starting the slate point sufficiently close to equilibrium. We shall provide a demonstration of Liapunov stability under the assumptions that d"(0) > 0 and "'i) is continuous when I I is sufficiently small. Taylor's theorem with remainder then implies that
rrld =
(NO)
+ tat' + Oie],
where a = (110}.
(5)
Since 1 has a minimum when = 0, the constant a is positive. Keeping only the two lowest order terms in , we see that the energy-conservation requirement (2) yields (in terms of )
42+ ad z=2[G
-
0)1.
( harurreriru:iun of Eigcnralues and Equilibrium
536
Suie.`
as Euremu [(h- 12
Since the left side of the above equation is nonnegative, we can write
= K',
}a'
K a positive constant.
Thus the state point Os), (t)) moves in the 4 - 4 plane (the phase plane) in trajectories that have the form of ellipses near the equilibrium point O. U). The particular elliptical path is determined by the initial conditions. By ascertaining the major axis of the ellipses, we deduce tirai Ka - ]!2 [ n t}
1-(1)1 112
{
(6)
K,
Thus, as is geometrically obvious, the point ( , 4) will remain close to the origin if K is small enough. But
_
+ a42(0)
(7)
Now if 2
452,
(0) +
e5 > i},
we have
V(02 2 ^
herefore,
from
6 2,
1_4(0)] 2 < tl' K2 - 4"(0) -},
ai;'(0)
+ (4 ).
(6) and (7), L42 (t) + ^(t)]
:^^
^ E5g(a),
(8)
where
g(u) =
/() + R p,xo - rrz, (t + a)ia2
f2
< 1,
u ? 1.
We see, then, that the "nearness" requirement (3) holds if o = k/g(a). We have finished the prerequisites for the remainder of this section. We point out, however, that our demonstration of stability was not complete. Further effort is required to justify the neglect of the 0(tÿ 3 ) term in (5) (Exercise 1). We also point out that the ideas presented here can be generalized to form the second or direct method of Liapunon. To generalize to systems of high order is relatively simple: the ellipses of constant energy become hyperellipsoids, since the energy in the neighborhood of stable equilibrium will be given by a positive definite quadratic form. A more important generalization involves equations for which a positive definite energylike Liapunov function can be found that can be shown to decrease when evaluated at the coordinates of the moving state point. Then the equilibrium point can be shown to he asymptotically stable. For further reading see, for example, a treatise by W. Hahn, Stability of Motion (New York: Springer-Verlag, 1967) or a more elementary paperback, by D. Sanchez, Ordinary D erentictl Equations and Stability Theory (San Francisco: W. H. Freeman, 1 . 968),
Sec. 12.411 Minimal Characterization of Linear Pastore Prohkms
537
THE LOADED MEMBRANE
An important scientific idea, one of those which lay at the foundations of the simultaneous rise of calculus and mechanics, is that definitions and results for discrete systems can be extended to continuous systems by a process of "chopping up," summation, and passage to limiting integrals. With this now-familiar idea in mind, it is natural to conjecture that for continuous conservative systems, too, stable equilibrium is marked by a minimum of potential energy.* Rather than attempt to verify this conjecture, let us assume that it is true and see what some of the consequences are. We choose to study a problem involving two spatial variables so that a partial differential equation will emerge. Experience indicates that it is of more interest than may at first be apparent to select the problem of calculating the deflection of a loaded membrane. imagine, then, a drumhead that is loaded with a weight per unit area described by a continuous function f (x, y). Here x and y are Cartesian coordinates, and the undeflected drum will be regarded as being fixed to a curve B bounding a region A in the xy plane. Assume that the drum is stretched tautly, so that a small deflection w(x, y) will result from the loading. We assign w a positive value if it is downward (Figure 12.5). We proceed under the assumption (validated in Section 4.4) that the actual deflection will minimize the potential energy among all smooth functions that vanish on the boundary. A deflection will be regarded as conceivable (or admissible in the terminology we have been using) if it is smooth and if it vanishes on the boundary B where the membrane is fixed_
FIG U ttE
12,5. Deflec[iurr of a loaded membrane _
By definition, a membrane is a stretchable material which is so flexible that negligible force is required to bend it. Deflecting a membrane by a displaccment ►► (x, y), then, results in a storage of potential energy due solely to the stretching. One can proceed to calculate the potential energy from the standard linear constitutive assumption. However, we shall adopt a fresh (but equivalent) approach. We shall assume that for small deflections at least, the potential energy of stretching is a function of the difference between the
}s
• We have tried to make the chapters on variational methods as self-contained as possible, so we merely point out here that material in Section 4,4 has already shown that classes of Continuous elastic systems do indeed extremalize or minimize a suitable energy functional at equilibrium,
Characterization
538
of Eigentafue ► and Equilibrium Stoles as Exirema [Ch. 12
deformed and undeformed areas. This difference D is
JJ
+2,)`' 2 da—
D = 11(1 +
{9)
dc.
A
A
For sufficiently small deflections,
wx w^
l,
-}
so D :
if-1
w; + ++F^) d a ,
(10)
A
where we have used the approximation (1 + c)' 12 x i + ic. We shall assign a potential energy of zero to the undeforrncd membrane, so that 1 1,(0) = O. For small D fin a familiar process) we keep the first nonvanishing term in a Maclaurin expansion of K(D) about D — O. e thus have for some constant T that
TO
or
Vs 4 T JJ(
+ ► ÿ) da.
(11)
A
An additional contribution to potential energy, V. results from the sag due
to the weight V„,
placed on the membrane. We have
— (work necessary to lift membrane back to a flat position)
Of (x i y) d er . A
Since total potential energy V is the sum of V, and VV,, a conjecture that the deflection minimizes potential energy leads to the following characterization of w. Consider the sel of smooth functions w which vanish on B_ 'The actual deflection w is that member of the set which solves the problem minimize IT if _ + wÿ} der — A
We know, however, that to these conditions' on w: —
T Ÿ2w
---
f,
J
fx1)f(x. y) da.
(12)
A
a
force-balance approach to our problem leads
w—
0 on 8 (F 2
3 -
exz
+ dyz
.
(13)
Here T is the tension in the membrane. * A detailed derivation can be obtained by a slight modification of the derivation of the surface tension boundary condition, found in Section 7. I - In that section, as here, there is no resistaace to bending and a tension resists stretching- In water waves an interface is subject to a net pressure p T p, wihich here should be replaced by the weight density J. Since we are —
.
limiting ourselves to small deflections, the surface curvature can be represented by its leading term- 1 his was ❑ ' in Section 7-1. Since we take deflection to be positive downward, it is —y e w' here.
Sec. 12.4] Minima! Characterization of Linear Pasture Problerni
539
EQUIVALENCE OF AN INHOMOCENEOUS EQUATION AND A MINIMIZATION PROBLEM
We could proceed by showing that, as in particle mechanics so in appropriate problems of continuum mechanics, the minimum potential-energy characterization of stable equilibrium is equivalent to that obtained from a balance of forces. it is of more interest here, however, to provide a direct demonstration of the equivalence of (12) and (13). For the main purpose of this section is to show the equivalence of ce rt ain problems in differential equations and certain minimization problems. Some rearrangement and introduction of new notation will allow us to obtain results of considerable generality with little extra effort. The "rearrangement" is actually an application of Green's theorem, which shows that
J5
(wx f w)
do = JJi
Jiv
vwi da _ J
A
A
vi w do,
(14)
A
since w vanishes on B. We introduce the notation (w, f ) ^`
f
f wx. y) f (x, y) dc.
(15)
We define two sets of functions as follows:
R = {ii l>ti is continuous in AL
(16)
S = fw r is twice continuously differentiable in A, K• = 0 on B}.
(17)
Finally, we make the abbreviation
L-
TV2 .
118)
Note that L takes functions in S into functions in R:
(l9)
L : S —' R.
With all this, our task becomes that of proving the equivalence of the following two problems. Problem A Find an element w e S that satisfies Lw =
f
Problem B
Minimize F(R)= (Lii, ►t•) -- 2(i . f) among smooth functions tit- E S. [ln obtaining Problem B we have doubled the quantity to be minimized in (12), and have employed (141 In proving the equivalence of the two problems, we shall use the following properties of the entities we have defined.
54 0
Characrérizarion of Eigcntalucs and Equilibrium Slates
as Exert ma [Ch_ )2 -
(1)
S and R are vector spaces under the "usual" definitions of addition and scalar multiplication, as in the two-dimensional version of Example IV, Appendix 12.1. (ii) The expression { , ) of (15) defines a real-valued function on pairs of elements of S and R, and this function satisfies all the axioms of a scalar product. {Compare Appendix 12.1, Definition 2.) L is a linear operator {with domain S and range RI and it satisfies the self-adjointness property (Lu, (iv)
if u, r E S.
(u, Lr)
(20)
L is a positive operator, in that if w E S, (Lw, w)
0,
(Lw, w) = 0 if and only if w is the zero element of S. (21a, b)
This follows from (14) and the vanishing of w on B. NOTE. Both (iii) and {iv) are special cases of Exercise Al2. 1.10. (v) If it is true that
(NJ) = 0 for every s in S. then if f E R,f is the zero element. (This isa restatement of Lemma A. Appendix f 1.1.) The remaining results of this section hold whenever the properties (i) to (v) are applicable. This is the case for the original problem that motivated us. but it is also true in many ether instances. (Some of these are the subject of exercises.) The reader who seeks no added generality can simply verify that the various properties which we shall use actually hold for the particular definitions of ( , ), R, S, and L given in (15) to (1$). As in exercise A 12.1.ôc, it can be shown that (iv) implies that if [u, e}
(LRu, v);
u,
€ S;
(22)
then [ , J has the properties of a scalar product. In particular, if ( 23 )
then II J can be regarded as a generalized length or norm in the sense of (Al2.1.8). Theorem 1. if Problem A has a solution, the solution is unique. Proof. if Lw, = f and Lw 2 = f, then Lk, — w 2 ] = z (z being the Zero element of S). Consequently, by either (2 lb) or (Al2.1.8a), ^Iwl
— WI N
(L[x'i — w1]. w, — w 2 ) = 0, so w 1 = w 2 . 0
(24)
Theorem 2. If w is a solution of Problem A, it is a solution of Problem B. and conversely.
See. 1 2 .41 Prooaf.
Min imul Characterization of Linear Positive ProhJerf! w -
i4 i
Suppose that w is a solution of Problem A. so that L w — f. Then w} - 2[^r titi'] + [w, Iv] - [ ► {', iv] = EH' - Si' ^^ [w, +S'^ ,
.
or fliv)
Il çi
' w II 3 - W +• it -
125)
To solve Problem B, we must choose an clement of that minimizes Fits ). From (25) it is obvious that the choice is IT _ w. Note that Min F(i) = - 1I4 2 ras
{26)
To prove the converse, suppose that w minimizes Fi+'). Let s be an arbitrary clement al S. Then if Az) = F(w -}. cs), the condition .i'(0) = 0 is necessary. But (r.) = (Lk. + as]. w + rs) - 2(w + 4s, fl
(Lw, w)
—
2(w. f) + r. (Ls. w) + (Ln. s) - 2(a,1)] + 0(1: 2 ).
Using the self-adjointness property, we find .410) = (L.s. tr) + (Lw, }) - 2(+. !1 so that Lw = f by (r).
2(l.ws - f. ,)
(271
11
By demonstrating the equivalence of Problems A and B, we have in particular proved the equivalence of the minimization pro blcm(12) and the Poisson problem 04 In terms of the membrane deflection interpretation we have shown the equivalence of a force-balance approach and an approach that characterizes equilibrium as the state that minimizes potential energy. Note that for the membrane (Lit.. is•) is the potential energy of stretching. Thus the the positive definite character of L can be regarded as expressing the fact that any deflection requires work to accomplish or (equivalently in this conservative system) Stores energy in the membrane. Many other minimum energy characterizations of mechanical problems can be fitted into the framework of our general theory, Particularl) for problems in elasticity, one expects repeated occurrence of positive definite operators to express the fact that a positive amount of work is required to distort a system from its unstressed "natural - state. -
THE RITZ METHOD APPLIED Ti] THE TORSION PROBt.EM1
To obtain an approximate solution of Problem B (and hence Problem A). we make F(0) as small as possible by choosing appropriate constants 2 in the trial function 1►
=
E 7C j t1•, .
f281
54 2
C.ltaracierrzafion of Eagentalues and Equilibrium Suites as Exirema [Ch. 12
In (28) the w ; are preselected in the hope that a linear combination of them
ca n prov ide a good approximation to the solution w. This is the Ritz procedure again, and it proceeds much as it did in the eigenvalue application of Section 122 With (28) we have N
F(w) (Lw, w)
2(w, f ) =
y ^; ^ 1 rt; ►
M ,
— 2 1 a ; b; ,
4.;= t
(29)
r=
where
(Lw ; , w a),
fa = (w';. f )•
(30)
To find how to choose the coefficients a ; in order to minimize F, we differentiate with respect to ot s and obtain
y "la ak
bf .
(31)
A sufficient condition for theexistence of a unique solution to (31) is the linear independence of the w, and the positivity of L (Exercise 12.2.6). Because it shows the strengths and weaknesses of the Ritz method in an important but relatively easy problem, we now turn our attention to the torsion problem of linear elasticity. This concerns the twisting by stresses on its ends of a cylindrical bar whose generators we imagine to be parallel to the z-axis. Let the cross section A that is intersected by the xy-plane be bounded by the closed smooth curve B, as in Figure 12.6_ As we have shown in Section 5.2, there is a solution in which the cross section located a distance z from the xy - plane rotates through the angle Oz, where 0 is the constant rotation per unit length. The tangential stresses T_ and T) on any section z = constant are independent of z, and can be written
7
^
Ow =^ y . ^=
t?w
(32)
where the stress function w(x, y) satisfies
— V 2 w = 2GO,
w = Q on B.*
(33)
(Here G is the shear modulus, a constant that characterizes the material's resistance to shear.) The sides of the bar are free of stress and the resultant of the forces distributed over any section is zero. Rut these forces provide a
• Equation (33) occurs in several other physical contexts such as viscous fluid flow down a cylindrical pipe with cross section B (Exercise 3.2.l4) and random walk in a region bcundr l bY 811. f quation (14 20)].
Sec. 12.4] Minimal Characterization of - Linear Positive Pfnhieme
543
F l C uRr. 1 2.6. 'Iarsion of a cylindrical beam by u couple of R ►ktyniftKle
couple whose magnitude Al is given by
M T JJ(x T:,
(3 4)
^
so [ Exercise 3(b)3 Ilf = 2 flit} c .
( 1 S)
d
Equation (33) can be put into the form of Problem A merely by making the identification f 2Gfi. Let us test the Ritz principle when A is the square lying inside the lines x — ± 1, y = # 1. From symmetry considerations. the simplest sensible trial function is = otti
(x
x — l )(y 2 — 1).
(36)
From (3l) one finds [Exercise 4(a)] that ;" = Using this value of xi", one u', can substitute the approximation (36) into (35) and find that M
544
Uh uracierimrian of Eigeni lues and Equilibrium Srates
as Extre na In_ 12
where Mt" = 2.22G0. To improve the approximation, it is natural to try 1)[ iz, + a21(x2 ws^i = (x 2 T 1)(y2 + y 2)1 (37) —
.
Here and below, using results given, for example, by Mikhlirt (1964, Sec 65), the results turn out to be = (35)(31)
(8j(277j
GO, ^ {2ai ^
(5^(lU5) GOM (8)(554) >
:
A1,112 /
where 14!12 ' = 2.24660.
(38)
The close agreement between the two approximations Mt" and M"i encourages us to think that our results are accurate_ Indeed, M is actually a bit smaller than 2.250, so the error in MI ' is less than one-fifth of 1 per cent_ is it a coincidence that the approximate values of decreased as more terms were taken, and that both lay above the true value? No, it is not, as is shown in Exercise 6. Of principal interest in deciding whether or not the beam will fail when subject In torsion is the maximum magnitude of the shear stress_ We thus seek to approximate f
= ma?[ ( T'=„ + T}) oz Lx. Yjf A
(39)
The maximum occurs on the boundary of the square, in the middle of each side [Exercise 5(a)]. Thus its value can be determined from (40) r = Iw y( 0. 1 )1. Using the two trial functions IV° and yti.^2', we obtain approximate values = 1.25G0 and 1.41 GO, while the true value can be shown to be 1.35GO. We see that in contrast to the situation for the moment hi, here the Ritz method does not provide a sequence of nondecreasing upper bounds for the correct answer. Moreover, the approximations are not nearly as accurate. For these facts we can give the following reasons. The moment M requires evaluation of an integral. As is usual in such instances, small errors in the approximation to the integrand 1 should not have too large an effect_ Indeed, Exercise 6 shows that M itself can be characterized by an extremal property, so that first order errors in estimating should give rise only to second order errors in M. By contrast, t 1 " and rt e ' were obtained by evaluating, at a particular point, a derivative of an approximate function. This is a notoriously inaccurate procedure: A spurious twitch in the approximation can give rise to a large error. Whaiever method is used, reliable approximations to T arc difficult to obtain. In concluding, we point out that the torsion problem (33) is formally identical to the membrane problem (1 3) when the weight (or pressure) func-
Sc. e 12_4]
Almond Characterization af Linear Posirrioe Pruddcros
545
lion is the constant 2Gf7JT. In one case w is the stress function, in the other w is the deflection_ This membrane analogy was pointed out by Prandtl in 1903. The analogy furnishes a practical procedure for determining the shear stresses in a twisted beam of complicated section. One examines the deflection of a soap film that is fixed to a curve having the same shape as the boundary of the beam, and that is subjected to a uniform pressure. [Sec Sections 93 and 99 of Timoshenko and Goodier (1970). These sections are part of a chapter devoted entirely to the torsion problem.] Of several other scientific questions that reduce to (33), we make particular mention of the random-walk problem discussed in Section 3.4 of 1. There we pointed out how a probabilistic question that arose in a biological context was answered quantitatively by referring to classical calculations on the torsion problem.
EXERCISES L (Project) Justify the neglect of the O( e) term in (5)_ 2. Generalize to three dimensions the text's discussion of Liapunov instability at a minimum of potential energy_ 3. Verify: (a) equations (29) and (31); (b) equation 135). 4, Verify the text's results lai for x; 1 ' [as defined in (36)] and M 11 : (b) for oi412 j, ott22 °* and M i 2 I . 5. (a) Verify (40I. (b) Verify the text's values for r 0 ' and rte'_ 6. (a) Use (26) to characterize the couple M as a minimum, (b) Use (31) to show that the Ritz method provides an upper bound for M. 7. Referring to the example at the beginning of this section, show that equilibrium is unstable if the potential energyis(a) maximum, (b)stationary, but neither a maximum nor a minimum. As part of the problem, formulate an appropriate definition of instability. 8. By sketching relevant circles and ellipses, give an entirely geometric demonstration of Liapunov stability for the example considered in the text. Start from 42 + a a = K 2 . 9. Discuss the stability of equilibrium for the undamped pendulum using the Liapunov approach. (Compare Section 11.3 of I_) 10. The deflection w of a bent plate can be shown to he governed by the equation Vow = qjD, where D is a combination of elastic constants and q(x, y}) is the density of loading_ if the plate is rigidly fixed along its edge B, both w and its normal derivative must vanish at S. [The problem is a two-dimensional, time-independent version of (5.1.37) and (5.1.38).] (a) Discuss the variational characterization of this problem. (Proofs of self-adjoin tness a n d positivity are reg uestcd in the Exercise Al2.1.13.) (b) When q is a constant. find a one-term approximation to w. Assume that the plate is square, and use the Ritz method.
546
Characterization of Eiiernwdurs and Equilibrium 5;ores as
E_Yrrmw
[Ch. 12
11. Show that a change of variable reveals an analogy between planar membranes subjected to a varying loading and unloaded membranes fixed to a nonplanar boundary. 12. The electrostatic potential w[x, y, 2) exterior to a conductor bounded by a closed bounded surface A satisfies for x (x, y, z) exterior to A ; w = wo far x lim lx [wand lir ix I bounded.
E
A;
Here w0 is a constant. (a) Set the problem up to conform with the abstract formulation discussed in this section, proving the requisite self-adjointness and positivity. Begin by considering a region exterior to A and interior to a sphere of large radius. (b) Discuss in general terms the Ritz method and its use in determining the capacity C, where
A
In this definition, a/On denotes the exterior normal derivative.
Appendix 12.1 Self-adjoint Operators on Vector Spaces in Chapter 5 of I we showed that eigenvalues of Sturm--Liouville problems for second order ordinary differential equations were real, and that eigenfunctions corresponding to distinct eigenvalues were orthogonal in a certain defined sense. In Section 2.2 we showed that eigenvalues of real symmetric tensors were real, and that cigenvectors corresponding to distinct eigenvalues were orthogonal. In this appendix we formulate an abstract symmetric eigenvalue problem that embraces the two cases just mentioned and many nacre_ Indeed, we go further with our abstraction. We list axioms that define an abstract "vector" in such a way that operations of addition and scalar multiplication such as one carries out for directed line segments are equally valid, for example, when carried out on certain sets of functions. We also present an abstract version of the usual scalar or "dot" product. This is needed to define sell adjoint and positive linear operators and to derive properties of equations that involve such operators. A few more concepts are mentioned in what is essentially a brief introduction to functional analysis. Some prefer to learn powerful unifying ideas first, and then have them illustrated in a number of concrete cases. They can read this appendix early_ Possibly, an example or two will have to be omitted, but the material presented here is essentially self-contained. Indeed, although the material is
Appendix 12_ f ] Self-udjorrr; Operators on Vector Spaces
547
important, we have placed it in an appendix to emphasize the fact that it can be inserted at any of several points. It is more in the spirit of This book, however, to observe how the abstract grows out of the concrete, If two or three particular concrete instances a re presented first, at the price of some "inefficiency" one gains powerful motivation for studying the appropriate abstraction_ Often the abstract procedure can he obtained by selecting what comparison shows to be the common features of the particular arguments. Applied mathematicians must be vigilant about keeping in mind the concrete manifestations of the abstract ideas. More than that, they must keep in mind that in any particular instance, more can often be accomplished than is encompassable by a general theory. Yet one cannot deny the elegance of the abstract ideas nor the utility of separating the particular aspects of a situation from the common structure that underlies a number of related situations. Our plan in this appendix is to present a number of abstract definitions and a few important theorems. Several concrete examples of the abstractions are given, most of which are discussed elsewhere in ibis book. Many readers will be familiar with at least the beginning of the succeeding discussions, and it would he beyond our scope to convey a detailed picture to those who are not. Thus many proofs will largely be left to the exercises, particularly those that require a more-or-less routine verification of certain properties.
REAL VECTOR SPACES Definition 1. Let S be a set on which is defined the following two operations: (1) addition, which associates an entity !; + tv with any two elements z and w of S; and (ii) (real) scalar multiplication, which associates an entity au with every real number a and element y of S. If the following rules are satisfied for any elements u, y, and w of S, and any real numbers ar and then S is called a vector space over the real numbers or just a real vector space.• (a) r + w is an element of S. (h) {u+ :a)+ w=u+(a+w). (c) There exists an clement z of S (zero element) such that c + z = u for every u in S. -I(d) To every v in S there corresponds an element of S, denoted by -- e, such that u + ( u) -- z. [Notation: By r~i - w is meant to + (—w}1
?,
—
• 11 in the LO rector-space postulates taj to 011, Greek letters can be complex numbers, then we have a rector space over floe complex numbers, Abstraction of the reyutccd properties of real and compkx numbers is shown in algebra texts to give rise to the concept of a field. and thence to the corresponding notion of a vector space over an arbitrary Field. f Many authors denote the zero element by'' but we employ '`z" to emphasize that this element 15 not necessarily a number_
[haracrerizaRiun ufEiyent;afues and Equilibrium Stales as Exirema
548
[Ch_ 1 2
(e)
u } y = y + 14_ [Properties (a) to (e) mean that S is a commutative group under addition.] (f) arty is an element of S. +cow. (g) a(v + w) (h) (a+/3)u (i) (Nfi)v = OEM). ke = V. (j) The following sets S; , with accompanying definitions of addition and scalar multiplication, are examples of real vector spaces. Verification is left to the reader. Example 1. S t is the set of directed line segmen ts in ordinary titrer-dimcnsi nual space. Scalar multiplication and addition are defined in the usual geometrical way (prolongation of the segment, parallelogram law). . a„), where the a are Example 2. S2 is the sel of ordered n-tuplcs a (a 02 . real numbers. if and b =_ (b c , b 3 , _ _ _ hAj are elements of S3 and a is real number, by definition, a + h is the n-tuplc whose ith element is a ! 4 b.; as is the n-tuple whose ith element is aa1 . Also . .
,
z is the n-tuple whose ith element is zero; — a is the n-tuple whose ith clement is — ai .
Example 3. Si is the set of nth order Cartesian tensors. Definitions of addition and scalar multiplication are given in Section 2.!,asare elements of the proofthat the tensors form a vector space with these definitions. Example 4. S4 is the sel of real-valued functions defined on an interval [a, h] and possessing n continuous derivatives. if f and g are elements of S. by definition
f + p is the function whose value ai x isf (x) f
g(x);
of is the function whose value at x is of (xj.
Example S. Ss is the subset of S4 wherein the functions vanish at a and at b. The definitions of addition and scalar multiplication are "inherited" from Example 4_ [A theorem of algebra, applicable here, states that to show that a subspace ofa vector space is itself a vector space one only has to verify the closure properties (a) and (f) of Definition 1 .] A s an illustration of vector space* manipulation we prove that t
= z fur every i) in S.
(1)
(Recall that r is the zero element of S.) This follows from our postulates. since IL , can be "canceled" from both sides of the following equation! in +Da= (I I-(1)u= I
•
Vector spaces arc also called linear spans.
+ r.
Appendix 12_1] Neff-ttdjoint Operutor_ti un Vector Spo c r.►
549
SCAI.AR PRODUCTS
Definition 2. Let S he a vector space over the reals. Suppose that with every pair of elements n and w of S there is associated a real number (u, w) with
the following properties: (a) (air, w) = a(u, w), ce any real number. (L + t), w) = (u w) -F (y, w), where a is also an element of S. (b) ,
(ci (w. e) = (e, w). (d) (v, v) k 0 and (e, e) = û if and only if e = z. Then (ta, w) is called the (real) scalar product (or sometimes inner product) of V and w. 1f (ti, w) = 0, then y and w are termed orthogonal. As the reader can verify, the following are examples (Areal scalar products. Examples 6 to 9 refer to scalar products defined on the vector spaces that were given as Examples I to 4 , respectively. Example 6. For directed line segments r and w, the ordinary "clot" product it formed by multiplying the rnagniLudcsofi a.id Lt byihecosineof the angle hetweenthem. In the present context, the dot product is also correctly called a scalar product. Example 7. if a and hare ti-tuples, a scalar product is defined by (A,b) =
Er u1 b ;
(2)
-
^ -
Example S. If a and b are first order Cartesian tensors, a scalar product is formed by their contraction. Example 9. Lei j and y he elements of the vcaor space of functions defined and continuous on the interval [u. bi. A possible scalar product is given by (j. 9) =
fl.x)g(x)rixl dx,
1.3)
where r is a (fixed) continuous function Chas satisfies rlx) > 0 for u < x b. An important special case is r -= 1. tote for instance t hat under the scalar product
(f. = f f(x)g(. ) dx,
(4)
if f is even and g is odd, then f and g are orthogonal if a = b. As an illustration of manipulation with scalar products, we prove that = ti for any r
tri S
,
(5)
This follows by means of postulate (h) For scalar products. since (t,
tr,„
—(v.v) — tu,y)= 0
(^f
When performing manipulations with a scalar product, one roust he sure that all the appropriate axioms are saiislied for every possible element that can appear in the product In what follows, we shall always assume ;hut this is the case, without further explicit -
indications.
5S0
Charrscter[zuium of Eiyenvalues and Equilibrium Stales as Extrenua lCh. 12
Just as with "ordinary" vectors, elements v, of a vector space S (i called linearly independent provided that
• y a, v, = z
I, ... , n) arc
only if ere = O.
(7)
I
Example 10. Using (1) and (5), show that mutually orthogonal nonzero vectors are linearly independent _ Salado& If the nonzero vectors v, are mutually orthogonal, then (v ; , v1) = Ab u for some constants p,. The constants must be positive. (Why?) But upon taking the solar product of (7) with vj , we find that
E abb. u3) = C or y ct c p, 6 4 = ü.
i^1
r^l
Since
pi > 0, a l = 0. 1 = 1. -..,n. Kali a, are zero. then (I) implies (7). Vectors u in spaces with an inner product can be assigned a generalized length or norm Ilvi by means of the following definition:
u)" 2 .
411 =
[Here and below the nonnegative root is implied by ( ) I J 2 .1 It can be verified [Exercise 22] that the following properties hold:
I vII II III —
Ilvii = 0 if and only if :a
0,
(8a)
z.
(Rb)
IaIIIvII-
I(U, w)I s IIvII IIAII
if + +III s II v II + II + Il
(Schwarz inequality)_
(8c)
(triangle inequality).
(8d)
Example 11. Consider inequality (8c)—which is attributed by various authors to one or more of the mathematicians Cauchy. Schwarz, and Bunyakovsky For continuous functions un Erg b] subject to the scalar product .
,
f
f (x)g(x) dx.
a
this gives
Ç fxx
dx
r^2 Sfxdx^ixdx+r2 ■
■
while the triangle inequality (8d) implies that 1r2 1f2 fu(x) + gx)] dx 5f 3 (x)dx + Çy(x) dx • f •
L,2
Appendix 12.]]
55i
Self-adjoin, Operators on Vector Spaces
The degree to which the directions of ordinary vectorsr 1 and r 2 coincide can be measured by the angle between them_ The cosine of the angle, or (VI ' v211f ► 1 I1bs l) is perhaps an even better measure, for it ranges from —1 (when the vectors point in the opposite direction) to zero (when the vectors are orthogonal) to 1 (when the vectors have the same direction). The Schwarz inequality (8c) permits the same measure 0i- coincidence or correlation to be used for functions, for irx ffixJix ) dx^ f i(x) dx r1 f^(x ) dx
will also range from — I to 1. Such a measure of correlation is frequently used. in the theory of fluid turbulence, for example, the intensity of randomness might be gauged by measuring the correlation between pressure traces taken over a period of time at two different points. LINEAR SELF-ADJOINT OPERATORS Definition 3. Suppose that there is a rule L which associates with every element y in a (real) vector space S precisely one element w in a vector space T Then L is an uperazr r with domain S and range T(or a mapping fro n S info T). We write L : S —, T or w = L[v] or iv = Le. L is a linear operator if and only if for any elements n and li of S and any real numbers a and /3, we have L[oty + flw] = aL[r] + (r.[, J
(9)
or, equivalently [Exercise 3(h)], L[rxv] = rlL[4],
L[o + w] = L[r] + L[151
(14)
Examples 12 to 15 refer to the vector spaces of Examples 1 to 4, respectively. (Proofs are largely left to the reader.) Example 12. ifv is a directed line segment. L(u) = 5i defines a linear operator, for
L[cw + Pw] = 5(au /Ivy) = alSrl t {5w) = irL[r] + fJL[i ]
.
Exttntple 13. Let a be an n-tuple and M a fixed matrix with elements m,,. Let
c = L[a] be defined by c3
E
M4/. x .
j'• 1
Then L is a linear operator. Of a is regarded as a column vector, we can use matrix notation to write L[a] = Mai Example 14. If T is a Cartesian tensor of order n, and Q is a fixed Cartesian tensor of order n, then
L[T] = T • Q defines a linear operator mapping nth order tensors into tensors of order n — 2.
(1 2)
Characterization of Eir enesalurs an d Equilibrium Stairs as Extremes ([_h. 12
55 2
Example 15. Let if- be an element of the vector space of twice continuously differentiable functions_ Then L[ f] = d ifidxr defines a linear operator mapping this vector space into the vector space of continuous functions_ The operators
Ne(f) = f 2 and Ni(f) —
d
dx + 7
are not linear because
N 1 (af + jig) :ü hf 2
+ 2ixffy+ (11 g 1 but
aN 1 (f)
f
l!iN101 = af' +
while
N,(a^ + ^) =
+ ^ ^^ + 7
but
at iV= 1 f ) + ^i1^` 1 ^4)
T
a
df
+ ^i ^x + 14.
Definition 4. Let L be defined on a real vector space S with inner product. L is termed self-adjloint (or se/far/joint ewer S) if (Lu, y) = (u. Lu) for every 14 and u in S. Example 16. If L is a second order differential operator (dfdx)(pd/dx) and S is the vector space of twice continuously diflkrentiable functions that vanish at x = a and x =. b, then the self adjointness of L is asserted in i, Equation (5.2.7)_ The inner product is that of (4).
Example 17. Linear operators associated with symmetric second order tensors are proved to be self adjoint in Lemma 1 of Section 2.2. The " vectors " here are first order tensors, and the inner product is formed by contraction.
Example 18. Let S he the vector space of rr-tuplcs with scalar product (2) and let the linear operator L be defined as in (I l)- if the coefficients rt+ t1 satisfy the symmetry condition M ; j = mg. then L is self- adjoint_ The proof of this statement is virtually identical to that of Lemma I. Section 2.2. One uses definitions, subscript interchange, and the hypothesized symmetry; as follows: R
(b, La)
=
11
X hi 1=
rfJiJa l
1 1- 1
y
b,rtJi,ut
j - J i•
ai E rn„b — (a' Lb). ^
(13)
EIGF.NVAI.IlF. PROBLEMS Let u be a nonzero element of a
vector space S. and let Al be an operator (not
necessarily linear) whose domain is S_ If Mu = for some scalar A, then di is M, called an eigenvttlue of and u is the corresponding eigenveclor.
Appendix 12.1] Seff-adjr ►inr Operators on Vector Spaces
553
Example 19. Consider the vector space of functions that are defined and twice continuously differentiable on ftl, rt] and that vanish at the end points of the interval_ Then the linear operator
L
d^ —
(14
-^ x#
)
has as eigenvalues the squares of the positive integers (i.e.. the nth smallest eigenvalue equals rily The corresponding eigenvectors are any constant multiple of sin ax. „
The question arises: Are the eigenvalues real if the problem "looks" real? The answer in general is negative. (An example is given in Section 2.2— there the complex character of the eigenvalues LS a consequence of the fact that a real polynomial may have complex zeros.) As we now show, however. excursions into the complex domain can be avoided in a restricted but important class of eigenvalue problems_ Consider linear operators on column matrices defined via multiplication by a square matrix with real coefficients, or linear operators on functions defined by differential operators with real coefficients. We have in these cases L[u + iv] = L[u] + iL[r]
,
L [u ] real.
(15a)
Here we are expanding our paint of view to consider a new vector space C with elements u + it where u and :a are in the original vector space S. In general, if a linear operator satisfies (15a),wecall it real. Note that (15a) implies L[u + iv] = L[u] + :L[u] = L[u] — iL[t.] = L[14 + rv].
( I 5b)
We now rule out the possibility of complex eigenvalues and eigenfunctions for real self adjoin: operators L. That is. we show that there cannot be a number 2 2(r) 4- #O. and some u+i A 0 such that
L[u + i c] = J.{u + i v )_
(16)
Theorem 1. Eigenvalues of real self-adjoint linear operators are real. The corresponding eigenvector can always be taken to be real. Proof With ). _ {'' +tit'' (16) is equivalent to tr} L!1 =- ;t u — Lu = /4tr`t,' -F- P'ie. (17a, b) ,
Using (l7) and the self-adjointness of L, we have
0 = (ü, Li)) — (Lu, t)) = (u, A mu -F- A" u) — {. {r1u — = P'[IlL4 11 2 4 110 21
u) ( 1 8)
Since u + it; is an eigenvector of L, it must be nonzero. We can therefore
conclude from (18) that At' ) = 0, so the eigenvalue 1 is real_ Moreover, Lu = 1. {'fu and = J.{"U, so either u or L. cart serve as a real eigenvector associated with the real eigen value 2 = A'''. d
554
characterization of Elyerrt+alues and Equilibrium Stages as 6xlre+rru
Pe. 12
Theorem I means that we lose no eigenvalues or eigenveetors by remaining in the real domain. This is guaranteed to be the case, let us emphasize, only when L is Self-adjoins. We have seen special instances of this result for real self-adjoint Sturm-Liouville operators in Section 5.2 of l and for real symmetric second order tensors in Section 22. We now state the abstract version of the orthogonality results that were proved in these instances. Theorem 2. Eigenvectors corresponding to different eigenvalues of a real self-adjoins operator are orthogonal. if n linearly independent eigenvectors correspond to a single eigen value, n > 1, then one can find n orr hogon l eigenvectors corresponding to that eigenvalue. Proof. Suppose that
Lu= 2.0
Lta
and
—
ttu,
2#
Then, on taking the scalar product of the first equation with u and the second with u and subtracting, we find that (,
Lu) — (u, Lc) — ){t
,
u) — µ{u, u) _ (• — 1.4(u,
(19)
u).
(We have used the symmetry property of the scalar product_) But the selfadjointness of L implies that the left side of (19) is zero_ Since À # the orthogonality relation (u, a) = D is established. To construct mutually orthogonal eigenvectors from a number of linearly independent eigenvectors corresponding to one eigenvalue, one employs the Gram-Schmidt process—just as in Section 5.2 of I. 0 ,
Example N. The self-adjointness proved in (13) implies that the eigenvalrtes of a symmetric n by n matrix are real, and that the eigerivecturs can he regarded as
real
mutually orthogonal.
POSiT1VE OPERATORS
Let M be defined on a vector space S with a scalar product An operator WI is called positive (or positive over S) if (Mu, u)
0 for u in S,
(Mu, u) _ 0 if and only if u
z.
(20)
Example 2l. let She the space of functions that are defined and twice contiguously differentiable on [0, 1] and that vanish at O and L Then L = —d 2 /dx' is positive over S, using the "usual" scalar product (4). For if u E S, integration by parts yields t^
(u, L.u) = ^ Li( — rd") n
fix
^
JP dx o
(1-
(21)
Equality holds if and only in' = 0, or u = constant; and the constant must be zero because of the boundary conditions.
Pierwr Spaces
Appendix f 2.11 Sr}I-rit!'uinr Operutr ► rs
555
A useful generalization of the above material on eigenvalue problems concerns ;Mir,
Lit
(22
where L and M are self- adjoint, and M is positive. By following the line of proof used in Theorem I, one can show that the eigenvalues 2 are real [Exercise 6la)]. if we attempt to repeal the calculations used in Theorem 2 [Exercise 6(b)], we are led immediately to the relation lie', — p)(v, Mu) _- 0 or tr, M u ) = 0,
123a, b)
where tr and r are cigenvectors corresponding to distinct eigenvalues and pt. To interpret (23b) as an orthogonality ri u1t, we define a new scalar product < , ) by
(24)
This is an acceptable definition [Exercise b(c)] because the four postulates of Definition 2 are satisfied by ( ). In particular, the scalar product of a nonzero vector with itself is positive, because 1bf is positive_ Hence we ha ► c the following result. Theorem 3. Ei[.;envectors of the problem Lu AMu (where L and M arc real, linear, and self- adjoint and M is positive) that correspond to distinct eigenvalues arc orthogonal in the scalar product ( • )• E xample 22. Consider the Sturm Lionville eigentialue problem —
ri dry p r i r cix
i_+•
+q
M ; }'(bF
=Q
(25)
where p is continuously differentiable;
q and r are continuous; p, y, r are
positive • toc it 5 x < h.
1261
Here we would define S as the vector space of twice continuously d illeretxtial+ functions satisfying the boundary conditions_ Regarding the equation of [25) as an irts#ance of L}. = AMi , and using the scalar product
110g(x)f#x,
1.1.g1 =
1271
^
we see that. for hi
E
S,
1114tt.ttl= (14u) = j rnJ. ^ 0, After an integration by parts. roe u c S we find
(Lu, =
J
[')'
(28)
that
+
gir l]
0.
(29)
55 6
f`haracverizarion r ►j El-gem-aloes and Equilibrium States as Exirerrra
[C76, 12
Thus both L and M are positive operators. By Theorem 3. eigenvectors y, and y of (25), which correspond to distinct eigenvectors, are orthogonal in the sense that (y'f.
f
or
j',) = U
ulAyAxb-ix)dx = U.
DRTHDItiDItMAI. ELEMENTS
Example 22 again proves a result that was demonstrated in Chapter 5 of I, but it shows that result embedded in a general framework. The SturmLiouville problem in differential equations is seen to provide just one illustration of a self-adjoint operator that generates orthogonal eigenvectors. But it is a most important example, as is indicated in Table 12.1. In Chapter 4 of I we obtained various results for ordinary Fourier series and then generalized to sets of orthonormal functions- it will be illuminating to rederive these results here so that the reader will recognize that they hold for any orthonormal elements in a vector space and will see how experience with particular cases can be summed up in a simple abstract theory. I ouriei s work with particular heat flow problems has been generalized into the abstract theory of Fourier series. which in turn can be specialized to give results for problems that only at a deeper level are seen to be related to the problems with which Fourier originally grappled. We deal with orthonormal elements tr of a vector space with scalar product < , >, so that -
{Hl, 14.>
i j = 1, 2
=à
(30)
,
Definition 5. The orthonormal sequence 04„) is complete in S i f f L S implies that for some scalars a,r , rY
lim
f —
„_
r
Df„LA n
=D
ill
II = (
•
>1'2).
(31a)
By convention we write
f=
E x„ra n
(31 b)
n= I
if (31a) holds. Upon formally taking the scalar product of both sides of (3I bl with ui , we find that an<14n. n =i
i/
'
E Pi = -i
5o
(32) If (32) is satisfied, we say that the a, are (generalized) Fourier coefficients off
TA BLE 12.1 Various important special cases of orthogonal functions of
Name associated with functions
Ox I
rtvi
Interval
I
[ — Tr. n]
x
[0. I]
Fourier
I
0
Besse'
x
kix-1
I _ x2
Legendre tal ---3
MN)
a Sturm—Liouville problem, with equation (25).` Boundary condiiions
Eigenv-atues
_An) = .ti — rr) An) = y1 —1r1
0. n2
y(0) finite
= 1. 2. ...
14 :Jk(ii„) = 0
Nonnormatized eigenfunciions I. sin nx. cos ax
A(Ii.x)
y(1) = 0 0
1
[ — I .. I]
0
( I — .‘ 21 I 2
[ —1,1]
polynomials
Tchebycheff polynomials
(I _ x.2)1.2
Hermite polynomials
eXp (—N 2 )
0
exp ( — x 2 J
( — x. x.)
Laguerre polynomials
x exp ( — x)
0
exp 1 — ..t)
[0, oc. I
y( — I) and All finite y(-1)and it I)
/An + I)
P,,(x)
ii 2
T(x)
0, 2n
H o(x), H(x)
n
L,,(x )
finite AsIx1-4, Y-... y = 0(x'). for some k > 0 y(0) finie: t as X -* x., I. = 0(e) for some k > 0
• See Courant Hither'. Vol. I 11953. pp. 32419:). DRision by appropriate constants gives orthonormal sets of functions. Explicit expressions for these constants can be found, for example. in Vol. 11 of A. Erdelyi et al.. eds.. ilignrr Trunscendensal FuncnionA (Nem York. McGraw-Hill. 1953). -
55$
Clxmae-seri:asrun of Einenrafues and Equilibrium Srares as Emma [Ch. 12
Suppose that one wishes to examine a finite sum
SN
(33)
=
and to choose the coefficients y r so as to minimize the norm of the difference between f and SN. or equivalently, minimize 11 f
Spa =
SN, f SN) -`
(34)
Using the orthonorrnality condition (30) and various properties of the scalar product, we find that N
11 f — S,4, 11 2 ---
N
f ^ ^Yru;,. f— ^ y1i, i-1 1-t N
N
NN •
f)
y yryl Cut, u,)
f) - L yJ
i=1
+ Y ^^ •
<.f, .f) — 2 Y
1=1
i =1
Here the a1 are the Fourier coefficients of (32). Completing squares,
{ + Ji f — X 11 2 — (f. . !) f
N
rw
i=1
1= l
y (ac; - x' 1) -
a; .
(35)
To minimize, we should pick the y1 So that they equal the Fourier coefficients a1 . Thus a Fourier expansion minimizes the "distance" between the clement land ils approximation SN. Putting y ; = a ; in (35), we obtain Bessel's inequality,
= IIfft 2
1 =1
eq. i.e., ffl11 2 ?
If the u, are complete, then we can use (35) with T i to deduce Parseval's Theorem,
i ,e., 11.f 11 2
11 11I 2 =
(36)
a i and lir,_ x
= 0,
( 37 )
1^t
INL-{OHtOGF.NF:O[Js PROBLEMS
In Chapter 5 of I we discussed the fact that Sturm-Liouville problems generate complact orthonormal sets of eigenfunctions. Here we show that such a set, or indeed any complete set of orthonermal eigenvectors of a real The square of the norm differs only by a constant from the pearl-square error that minimized in the parallel discussion of Chapter 4oi1t "Complete" means complete in S for an unspecified space S.
was
Spaces
Appendix 12.1 ] Sefj-ndjuinl Operators on Ÿerlur
554
linear self- adjoint operator L, can be used to solve the problem Lo =f,
(38)
where f is a nonzero element in the range of L. Consider a problem that is different from (31) but related to it, the eigenvalue problem Lu = Au- We know that the eigenvectors u, can be orthonormalized: 1 21 = ;Li u;,
(u ; ,
t
;
i•j = 1 . 2 ,--•- *
(39)
Suppose that the u i are also complete in the domain and range of f._ Then it is natural to try to expand the unknown solution, as well as the known righthand side, in terms of these eigenvectors:
E ^
,
_
(4{]a, h)
•
f = .=i
As in (32), the respective Fourier coefficients are given by = (y, u 1 )
and fl _ (f, u i).
(4 la, b)
inserting the series of (40) into the equation Lv = f we obtain, formally, f
y a i Lu i = ^ a ^r 1e; ,
= Lv _
(42)
,
^
=
—
1
1
Using this and (40b), r = f — I-r ={{i — a A jtt; , -1 ,
(43)
,
Since the zero element has zero coefficients in its Fourier expansion, we find that I^i
(44)
^i f r
Provided that none of the r ; are zero, (44) gives us the desired (unique) solution (45) i
=
Suppose that L does have a zero eigenvalue. 1.et associated linearly independent eigenvectors be denoted by LG., :t , ^, • - - u . ^ - For each of these, (44) implies that the corresponding coefficients g„,, {gy ... ^ + must be zero if Lu = f is to have a solution. In other words, by (41b), if 1 R, = • - • _ ;t,„{ = 0, the problem Lu = f has a solution only if ,,
,
,,
,
(f, Um ) _ ( - r^, 14 n2+1) = --• = ( f, u^^^ =
[].
(46)
Since we no longer are concerned with Lv = rhfu we return 10 the notation ( , I For scalar product_ '
,
560
Charucneri_awion of Eiyenralses and Equilibrium Stares at Ertrema IOr. i 2
[The solution is non unique,' since, from (44), the coefficients a,,,, ... , a,,++ are arbitrary.] We emphasize the fact that she self-adjoins irthomogeneous equation Lu f has u solution only iff is orthogonal to all solutions of the related ltorrwgeneous equation Lu = r. [Compare (46).] As is illustrated in Exercise 7, this necessary condition arises frequently in applied mathematics, especially in perturbation
calculations. THE ADJOINT
Use of the term "self-adjoint" implies that there are linear operators which have something other than themselves as an adjoint. We now demonstrate that as the term "adjoint" is defined, this is indeed the case. Needless to say,. the importance of this demonstration does not lie in its justification of our terminology. The fact is that the increased generality allows far wider application of many of the ideas we have presented. Some of the required generalizations will be outlined in the following paragraphs and in the exercises, but for full treatment the reader must be referred to more advanced works, such as those by Friedman {1956) and Stakgold (1967). Example 23, Let S be the vector space of functions ii that are defined and thrice continuously differentiable on [0, l] and that satisfy
Li(0) = u( 1) = u`(0) = O.
(47)
Let Li
u'".
(48)
Using the usual scalar product, consider (u, Lu) and integrate by parts: , (
v
.
f.t+) =
1
f ti(x)!r°,(x)dx
— [¢;ir " - t;'1t ' + V' u]^ + f (—U"l1/ dx.
(49)
u
o
Suppose that y satisfies v(0) _
01) = 0.
(SO)
Employing (47) and (50), we see that all the boundary terms in (49) vanish so that we can wri re4% Lu) (L u, u) for u c S, u er T, where L't —or_ Wert T is the vector space of
functions a that are defined and thrice continuously differentiable on [0, 1] and that satisfy (50). We emphasise that S and T arc different spaces, because the last boundary condition in (48) differs from the last boundary condition in (50). In general, if
(t;, Lu) =(L'u, u)
for u E S,
(51)
u E T;
• NOW that ihr Existence or a zero tigcnvaluc shows that the linear operator L is not El FAOdated with a one-to-one transformation, for a nonzero victor (as well as zero) is mapped into zero. Nonuniyuencss of any solutions to Lu = f is certainly to be expected .
Append ix 12.11 Srf{-vdjorni Operators
on Vector
Spores
we say that L* is the adjoint of 1. (and vice versa). The spaces S and T arc called dual. [f L + = L and S = T, then we invoke Definition 4 and term L self-adjoint. HE' = L but S # T, then L is called formally self adjoint. NOTE. (i) Eigenvalues of non-self-adjoin! operators cannot he expected to be real, so in (51) and elsewhere one generally employs a complex-valued scalar product wherein part (c) of Definition 2 is modified to -
(u, u) = (v, n).
(52)
Here the bar denotes complex conjugate We shall assume (52) for the remainder of this appendix. (ii) Do not forget the assumption, made explicit under (6), that the inner product is always well defined. Example 24. Consider C", the vector space of complex n-tuples such as zl" ,
7r''). [..et Lz"' be obtained by prertmultiplying rt'' by a fixed n by n matrix M with
complex components rn, . An appropriate scalar product on C ,
(Zlil, =111)
L ^I^I=121 r ^ • c=
defined by (53)
1
With this A
(J.2(1l,
1t 2 l 1 e 11
^ ^ + ^ ^tk^f=!
+1=
_ ^ ^^
I.+=
^
{
11^^1 i
^
If we make the definition
m+ ► ^ +tr1„ we find that (L1"),ztr,] =
zifim141^^ai — (i4",L .2^r^j. ^.
;= i
Thus the adjoint operator in this cast can be obtained by premultiplying by the transposed complex conjugate
or the original matrix M.
To relate eigenvalues of L with those for I. + , we proceed at first as in the proof of Theorem 2. Let u ; and y1 be eigenvectors of L and L + with correThus This sponding eigenvalues f and
Lie; = ;u„
L 't 1 = 1 v1 ,
(54)
SO
(t) t ,
Lur) — {L' u), ui) _ (y1, +cur) —
The left side is zero by the definition of L+, while the right side can be altered to obtain 0
— fe)(G u;).
(55)
362
Characterization of Fr±yc•n, -alue.s and Equilibrium SrQrr %
LIN
Et1re►na l(h J?
Thus, if the eigenvalue associated with ui is distinct from the conjugate of the eigenvalue associated with v j , then tee is orthogonal to u1 . One says that a biorthogonality property holds between the eigenvectors of L and L+. An additional statement can be made. Suppose that the eigenvectors 1r ; OIL are complete in the sense that any eigenvector of I,+ can he expressed as a linear combination of the u i . if (u, u i) = Q for every i, then vj must be the zero vector, for DJ =
E a; ui implies that (v i , v) =
otAvi . ur) = 0.
(56)
This contradicts the assumptionthat uj is an eigenvector. Consequently, (u) , u ; ) # O for some i, whereupon (55) implies that pi = 7t f for this i. Thus the stated conditions are su €- tens to permis a proof that every eigenvuiue of L.' is the complex conjugate cf an eige ► tuulue t3f L. If the cigenvaiues of L are also complete, we can pair off eigenvectors u i of I. and of of L * so that corresponding eigenvalues are conjugates of each other. Coefficients in an eigenvector expansion can now be determined, for if if
f
(57)
Y aiTA' 1- L
then by the results we have demonstrated
{
f, u})
v,)
rx1(u ; ,
A= i
SC
(f, u1)
(58)
(u;, ui)
One application of the above material is this. In certain problems rather than approximate a desired eigenvalue of an operator I. directly, it is easier to approximate the corresponding (conjugate) eigenvalue of L + SAItiACH AND H1LBFltT - 4PACF.S
Functional analysis, an important area of mathematical research in the twentieth century, stems in large part from abstractions of the kind we have been making in this appendix_ Powerful theorems have been developed that can bear directly on problems of great interest in applications (And, as usual, the general theorems often fail to give results of the desired strength in important particular cases.) A few more definitions will allow the reader to understand the statements (at least) of a number of additional theorems in functional analysis. The remaining material is not used elsewhere in the present book. -
• See W. H. Reid, "7he stability of parallel Flo ws... Bask Deretoprnenrs in Arid D iti#rrrrs, M- Holt, ed., Vol. I (New York: Academic Press, 19651. pp. 249--30?_
Appendix 12.11 Sery=adjornr Operators on Vector Spares
563
A metric space M is composed of (i) a set of elements x i , and (ii) a realvalued function d(x t , x i) that is defined for each pair of elements and has the properties d(x ; , x ; ): Ax ; , x i) = 0
ii(x,, xi) ^ 0, (59a, b, c)
if and only if x, = x ; ;
d(x„ x ;) + d(x,, xk)
(triangle inequality)-
A convergent sequence (x ; }, x ; e M, has the property that for every positive e there exists an integer n such that, for some x e M,
d(x ; , x} < e
whenever i > n.
( 60 )
A Cauchy sequence {xj, x ; M, Las the property that for every positive e, there exists an integer n such that d(x; , x2 ) < c
whenever t > n
and
j > n.
(61)
A metric space is complete if every Cauchy sequence is a Convergent sequence. (Do not confuse the concepts of a complete metric space and a set of elements that are complete in a given vector space.) The rational numbers under the usual absolute value metric are an example of a metric space that is not complete, for a Cauchy sequence of rational numbers can, of course, converge to an irrational number A normed vector space (or normed linear space) is a vector space with elements x ; on which is defined a real-valued function II I with the properties
Ilx;ll
0, 11;11 = û if and only if x i is the zero element; ( 62 a, b, c) Ilax;i1 = 1( I IIx ; II; + 11)0. (Ixt + x3 11 ?
A Wormed vector space can always be regarded as a metric space if rune empl oys thenatural metric generated by the norm and writesd{u, tr) = 11 14 — 1'11 ABanchspeiormdvctspaehiomltnsaur metric. A scalar product space is a vector space with a scalar product defined as in Definition 2. Such a space has a natural norm x i II (x ; , x )` r z and hence a natural metric. It is a normed vector space, since equations (8a, b, d) of Definition 2 are the same as equations (62a, h, c). A Hilbert space is a scalar product space that is complete in its natural metric_ A Hilbert space, with its scalar product and norm, has more structure than a Banach space. A Hilbert space is a Banach space, but the converse is not necessarily true_ ,
,
COMPLETELY CONTINUOUS OPERATORS
The abstractions of functional analysis establish useful analogies between structured classes of functions and ordinary geometry_ Proofs in Hilbert space theory, for example, can often be constructed by drawing appropriate
564
C. Iwuruclerizurion of Eigrnrulue.s and EguilibritJrrt Surres as Extrema
[Ch. 12
diagrams and then translating the geometric steps into manipulations of the abstract lengths, scalar products, and angles that are defined by the axioms. The analogy between Hilbert space and ordinary three-dimensional Euclidean space breaks down in connection with compactness. A set of Hilbert space elements is bounded if each has a norm that is less than some constant; the set is compact 'revery sequence of elements chosen from it contains a convergent subsequence. In three (or n-}dimensional Euclidean space, the Bolrano-Weierstrass theorem shows that a closed bounded set is compact. But this connection does not hold generally in Hilbert spaces, as is shown by the Following example. Example mple 25. Consider the closed bounded set S of Hilbert space elements f that satisfy fI f Il 5 I. Let f, be a sequence of orthonorrnal elements in S:
f)
=
No subsequence converges, for the "distance" between any pair of elements is
. as is
shown by the following calculation:
6
-- fJ11 3 - 1Jf - I1 , ^ fjl f IL) ( - ^^++r 1 +{ + IIJ^ +^f^. = lfi• fJ^ = 1+ 6 +I -2. -
Again and again we have encountered examples of Sturm-Lionville problems. These involve differential operators, which "unsmooth," in contrast to "smoothing" integral operators. Thus an advantageous reformulation can be obtained by inverting the Sturm-Liouville problem Lu Au by means of the inverse operator L' (assuming that it exists)_ We have L14 =
Au
implies that
14 =
AL - 'u or L - 'u = 1 - ' il.
We see that the nonzero eigenvalues of L- t and the nonzero eigenvalues of L are mutually reciprocal, and the corresponding eigenvectors of the two operators are identical. For Sturm-Liouville problems, Green's functions can be used to invert the differential operators and thereby to obtain corresponding eigenvalue problems involving integral operators_ The self-adjointness property is retained by the inverse operator (Exercise 27). If we write the inverted problem in the form Au = pie, we expect that the operator A has a property that reflects its smoothing character. Such is the case: The sequence Ax, turns out to be compact whenever x„ is bounded. Operators with this property are said to be completely continuous. Theorems I and 2 of Section 12.2 characterize the eigenvalues of a Iinear self adjoint operator L under the assumption that (Lu, u)f(u, u) has a greatest lower bound and that this bound is attained for an element in the domain of L. The inversion that is necessary to permit the development of Sturm-
Appendix 12_11 Sr4f adjarni Operators on Vector Spaces
565
Liouville theory in terms of completely continuous operators prepares us for the fact that the eigenvalues of completely continuous self-adjoint operators arc characterized by a raximutn property. Let A be a completely continuous self-adjoint operator whose domain is a Hilbert space S. One can prove the existence of a solution to the problem of maximizing I (Au, 01/(u. 4 The maximum gives the modulus of a nonzero eigenvalue, and the maximizing vector is an eigenvector. A sequence of maximum problems with orthogonality constraints (as in Theorem 12.2.2) yields all nonzero eigenvalues and the correspondingcigenveclors. Moreover, if A does not have a zero eigenvalue, the eigenvectors are complete in S. It is fitting to illustrate the theorems of functional analysis with results that show the existence of the complete sets ofeigenvectors that we have been manipulating. Yet it can be argued that we have just changed the nature of the problem, so that the task is now one of proving complete continuity (or some other sufficient condition). There is some truth in this, but the functional analytic approach has the advantage of breaking the problem into a portion where detailed analysis is required to establish the characterization of operators and spaces, and another portion where one derives strong and clean geometric characterizations of these operators and spaces_ Proofs of the detailed assertions that we have made, and an introduction to the literature, will be found in such references as Friedman (1956h Stakgold (1967), and Riesz-Sz.-Nagy (1955).
EXERCISES
I. Verify that illustrations of Definition 1 are provided by (a) Example l ; (h) Example 2; (c) Example 3; (d) Example 4_ L Verify that illustrations of Definition 2 are provided by (a) Example 6; (b) Example 7; (c) Example 8; (d) Example 9. 3. (a) Verify that the L of Example 13 is linear. (h) Show the equivalence of (9) and (10). (c) Verify that illustrations of Definition 3 are provided by Examples 13 and 14. 4. A complex vector space C is introduced in the discussion of eigenvalue problems. Fill out the definition of C and verify that it is indeed a vector space. S. Verify that (15a) is satisfied in the instances mentioned in the sentence preceding this equation. 6. (a) Show that the eigenvalues of {22) are real. (h) Verify (23a, b). (c) Show that < , > as defined in (24) fulfills the conditions of Definition 2.
C hararrerizalion of Eiyertralues and hodarbwr, S1ure.s as Exrrerara (Ch _ 12
566
7. The object of this exercise is to solve via power series the eigenvalue problem
£L lt1(u) = Au, u E vector space S, I c l
Ll° l(u) -
We assume that a solution is known when e = 0. Here L,t °' and i. 111 are self- adjoint linear operators mapping S into R. We assume that the known eigenvalues of Lt ° E are simple; that is, no more than one linearly independent eigenvector corresponds to each eigenvalue. Let u} °w be the orthonorrnal set of eigenvectors corresponding to the eigenvalues of L 1°1 . Assume that this set is complete in S. Let us look for solutions in the form of power series in c: ^ r
(a)
(b)
=u;°)
+Elf
(')
+ 2 ui 2 '+-..,
d1.j = AID) +r.t1' ) +•
.
From equating the OW) terms to zero, find an inhomogeneous equation for le. Determine API by the orthogonality condition necessarily satisfied if a solution to such an equation exists. [See the paragraph following (46)J Assume that [4 1 ° has a series expansion in terms of the u1" } and find the coefficients. To obtain a unique solution, require that te satisfy the normalization condition ( ut,) , uln) f 1.
Carry the calculations one step further, and find Al21 and an expansion for 42). NOTE The formal calculations give the perturbation of any particular simple eigenvalue 2; ° ' even if some or all or the remaining eigenvalues are not simple. (It must be assumed that the unperturbed eigenvectors have been selected so as to form an orthnnormal set_) Calculations giving the perturbation of a multiple eigenvalue are somewhat more complicated. See Courant-Hilbert, Vol. 1 (1953, p. 346). 8. Use the approach of Exercise 7 to approximate the lowest eigenvalue of (c)
z dx 1
+ Exu =
Au,
u'(0) = 1410= O.
9. (a) Show that
d[ du p(x) rx + q(x )u dx
'
!i(a) -}-
nl1'(u)
— 0,
u(b) MAN = 0, can be regarded as an eigenvalue problem involving a self-adjoint operator. • Fora particular closely related problem. see Exercise 7.2.1 I of
Append 12.11 Se11-witurru Opercutors
Prutor Spaces
567
(b) Give conditions sufficient to guarantee that the operator is positive as well as self-adjoint. 10. Prove that
a -); Ox
3
E_
L=
Lit=
= i,t2
x 3)i
is a positive self-adjoint operator if ,4 04x) is positive definite for all x in R. Let the domain of L be the space of twice continuously differentiable functions that vanish on the boundary of R and use the "natural" scalar product. 11. Let L act on functions defined and continuous on a bounded threedimensional region R. Here if v = Lu, then s Y MY)
1ff
142 4'3,
where a given kernel IC(x, y) is defined and continuous in R. Suppose that K(x, y) K(y, x). Prove that L is a linear self- adjoint operator. 12. Suppose that L and L2 map a vector space S into itself. Let L I and L2 be self-aidjoint. Is their product (i.e., their composition) self-adjoint? 13. Show that if V 4 = VW) acts on smooth functions in R that vanish on the boundary B of R. together with their normal derivative, then V' is self-adjoint and positive, 14. Lei the domain of L be the set of functions f that are defined and fourtimes continuously differentiable on [a, bj, where f ) f f(b)
0. 11 L
d d2 d2 p(x) dx 2 dx 2 dx
)
d d—x
ri
prove that L is self-adjoint. Use the scalar product (11 1 f2)
=
J f,(x)f2
(x)dx
and assume that p, q, and r are sufficiently. smooth. 15. (a) Suppose that
d2 L = a 2(x)-dx 2
z (x) dx
adx).
Show that the equation Lu = Au can be put into the form considered in Exercise 9. (b) Show that an eigenvalue problem for a general fourth order ordinary differential operator can be put into the self-adjoint form of Exercise 14 only under special circumstances.
568
Chararreri arion of Eisuennyitue ► and Equilibrium Slates a.s E.rirema ICh. 12
16. Consider (38) in the case where L requires multiplication of an n-tuple by a given symmetric matrix. Compare the text's statements with standard results concerning solutions of linear equations. [Use the scalar product (2).] 17. If L, L 1 . and L2 are linear operators, find alternative expressions for the following: (a) (L+} + ; (b) (L k + L) F c) (L ,L 2 )+ 1$. (a) Verify that (53) is an appropriate complex scalar product. (h) Show that if the conjugate is omitted in (53), there are nonzero elements a(r) and d 2 i whose scalar product vanishes. 19. Extend the text's discussion of (38) to the case where L is not self-adjoint. ; (
.
In the next two exercises L will be a linear operator over a space S composed of real-valued functions that are defined and twice continuously differentiable over [u, b]. Further conditions on S will be stated_ The scalar product used will be that of Exercise 14.
20. if functions in S must satisfy u(0) = u(1) = 0
,
and L =
[Ix
(Fix) —d
x)
dx + q(
.
find L+ and the dual space T 21. Suppose that functions in S must satisfy u(1) = an(0), u1(1) _ (3u'(0),
x and fi constants_ Let Lu = u". Show that L' is formally self-adjoint but is not selfadjoint unless a special relationship exists between a and ji. 22. Consider the norm defined by IILII = (I, O w. (a) Verify (8a) and (8h). (b) Prove (8c), starting with the fact that 0 < (r + 2w, L + 2w) for any real number 2. (c) Use the Schwarz inequality to establish (8d). (d) When does equality hold in (Sc)? in (8d)? 23, Let the elements x d belong to a metric space M. Use the triangle inequality to show that if {x i } is a convergent sequence, then it is a Cauchy sequence. 24. Let {u i } be a complete orthonormal set in the scalar product space S and let u be an element of S_ z (the zero element). (a) Prove that if (u, ui) = 0, i = 1, 2, ... , then u (b) Suppose that S is a Hilbert space. Prove that E7=' , (u, u 1)u j must converge to an element of S. (This is part of the Riesz-Fischer theorem. in the proof, use the convergence guaranteed by Parsevai's theorem to show that the partial sums form a Cauchy sequence.) (c) Use (b) to prove the converse of (a) -.i.e., show that if only the zero element is orthogonal to each of an orthonormal set, then that set is complete.
Appendix 12.11 Self-adfnlrri ()pertnurs on Vector Spat -PA
5
25. Show that the Sturm- Liouville operator — d/dx(pd jdx) + q is selfadjoint when p, q, and the boundary conditions are appropriate to the following equations of Table 12.1: (a) Bessel; (h) Legendre; (c) Tchebycheff; (d) Hermite; (e) Laguerre_ NOTE. Either the equations have a regular singular point at one or both of the end points, or the intervals over which the equations hold are unbounded. The complete theory for these singular Sturm-Liauville equations is rather complex, but it can be found in advanced treatments of differential equations, for example, Coddington and Levinson (1955). 26. Let S be the vector space of functions defined and continuous on [0, 11 The linear operator D = dfdx is defined only on a subspace of S, the set of functions with continuous derivatives. Show, by contrast, that if a linear operator is defined on a single element of ordinary threedimensional Euclidean space, then it is defined on the whsle space. REMARK. The fact that linear operators are often only defined on subspaces causes difficulties in functional analysis, compared to the theory of finite-dimensional spaces_ Marry of these difficulties can be overcome, however, if (as is very often the case) the subspace is dense in the full space. 27. Let a linear operator L provide a one-to-one mapping of the whole vector space S onto itself. Show that L is self-adjoint if and only if L is self-adjoins
Bibliography General Applied Mathematics R., and D. HILBERT. (1953) Methods of Mathemcrrrcof Physics. Vol. 1. New York: Interscience Publishers. a division of John Wiley & Sons, Inc. 561 pp. "Mathematical methods originating in problems of physics arc developed and the attempt is made to shape results into unified mathematical theories." (Volume 2 does not have the classical style of Vol. 1 hut is a compendium DI - important results in the theory of partial differential equations.) JEFFREYS, H., and B. J FFREYS. (1962) Methods of Aiiithematicol Physics_ New York: Cambridge University Press_ 716 pp_ "This book is intended to provide an account of those parts of pure mathematics that arc most frequently needed in physics . _ .. Abundant applications to special problems are given as illustrations_" COURANT,
Ad v anc ed Calculus FRANKLIN, P. (1 940
A Treatise on Advanced Calculus_ New York.: John Wiley & Sons. Inc. Reprinted by Dover Publications Inc_, New York, 1964. 595 pp. 1i11DEBR AND. F. B. (1962) Arkunced Calculus Ji)r Applications. F:elglewood Cliffs. N.J. 7 Prentice-Hall, Inc- 646 pp. KAPtAN, W. (1952) Advanced Cakuhis, Reading, Mass. Addison-Wesley Publishing Company. Inc. 678 pp.
Differential Equations BOYCE, W„ and R. ()iliumA_ (1969) Elementary .r irrential
Equations crud Boundary
Value Problems, 2nd ed. New York: John Wiley & Sons, hic. 583 pp. "This book is written from the . _ viewpoint ttl the appliedr r^dttternattctan whose interests in differential equations may at the same time be quite theoretical as well as intensely practical." COI)DINCiT❑ N. F_, and N_ LEv1NSON. (1955) neon' of Ordinary Diflerenriaf Equations, New York : McGraw-Hill Book Company. 429 pp. Authoritative advanced work. GARAEïFI]IAN_ P_ (1964) Partial Dt_ ere„tïal Equations. New York; John Wiley & Sons, Inc. 672 pp. A graduate text "written for engineers and physicists as well as mathematicians. — Ordinary Difierentiuf Equations. Essex. England: Longman Group Ltd. INcE,.(1927) Reprinted by Dover Publications, Inc., New York. 1956. 558 pp. An excellent source for the classical theory_ ,
5,7 1.
572
Bibliography
KELLOGG, i,
0. (1929) Faurrda[ions of Potential Theory. Berlin: J. Springer. Reprinted by Dover Publications, Inc_, New York, 1953. 384 pp. "It is inherent in the nature of the subject that physical intuition and illustration be appealed to freely, and this has been done. However, in order that the book may present sound ideals to the student, and also serve the mathematician, both for purposes of reference and as a basis for further developments. the proofs have been given by rigorous methods." TYCKONOV, A., and A. SAMuutSKl_ (1964) Partial Differential Equations of Muthernrtfrcaf Physics, Vol_ L San Francisco: Holden-Day, Inc. 390 pp. Vol, ❑ 1967. 250 pp. Representative of a large class of books, this is relatively brief and has good .
illustrations of how the mathematics is used.
Tensor s ABRAM, J. (1965) Tensor Calculus Through Diffrrenriaf Geometry. metry. London: Butterworth & Company Ltd. 170 pp. About 25 per cent of the book is devoted to applications its dynamics and continuum mechanics.
ARts. R. (1962) Fectors, Teri.sors, acrd tiro flask- &guotions of Fluid Mechanics. Englewood
Cliffs, N.J.: Prentice-Hall, hic. 286 pp. "Sets out to show that the calculus of t ensors is the languagc most appropriate to the rational examination of physical field theories," The argument is illustrated by detailed consideration of fluid mechanical equations. KARAMCRETI, K. (1967) Vector Analysis and Cartesian Tensors with Selected Applications. San Francisco: Holden-Day, lac. 255 pp. Begins with vector algebra and calculus, and therefore provides u particularly good reference for readers who hick experienix with these tools_
G. (1960) Cartesian Tensors. London: Methuen & Company 1.td_ R9 pp. Tensors are defined as multilinear functions of direction in this introduction to the subject for applied mathematicians and physicists.
TEMPLE,
Continuum Mechanics HooGE, P. G. Jr. (1970) Continuum Mechanics Ncw York: McGraw-Hill Hook Company. 251 pp. The essentially modern innovation consists in viewing the subject o fall continuous media as a whole and concentrating on those properties and concepts which arc common to all of them. The ever-increasing body of knowledge which must be mustered by the engineer, and, in particular, the use of increasingly complex materials in modern technology has made this new approach necessary." Au introductory graduate text for engineers, this Ilk provides an alternative reference for the derivation of the Navier and Navier Stokes equations from first principles. JAUNZIEMis, W_ (1967) Coritmi um Mechanics_ New York: Macmillan Publishing
Company, Inc. 604 pp. A graduate teat that rakes a "modern" rigorous approach to the subject_ Emphasizes elasticity.
Bihla'cr$raphr
573
SEDüv, L. I. (1971) A Course in ConrinuNm Mechanics (translation from Russian edited by J. R. M. Raduk). Groningen, The Netherlands: Wolters-Noordhoff Publishing, four volumes of 242, 309, 340, and 305 pp. "In recent years, the need for the introduction of a course iii continuum mechanics at more advanced levels at universities and technical high schools has become clear; it is needed as a general base for the theoretical development of thermodynamics, electromagnetism, hydrodynamics, gas dynamics, elasticity. plasticity, creep and of many other groups of phenomena of physics and mechanics, The generality and, at the same time, inseparability of these branches, which at a first glance may appear to be autonomous subdivisions of mechanics and physics, bring about the need to consider them] as a single unit." SOMMERFELD, A. ( 1950)_ "Mechanics of Defarmable Bodies." Vol_ 11 of Lectures urn Theoretical Physic.. New York: Academic Press, Inc. 396 pp, (Paperback edition: 1964 "My aim is to give the reader a vivid picture of the vast and varied material that comes within the scope of theory when a reasonably elevated vantage point is chosen. " Unlike the other references on continuum mechanics, this requires no knowledge of tensors. TRtJESO]ELL, C_, and W. Not.'. (1965) ' . The Non-linear Field Theories of Mechanics," Encyclopedia of Phy. irx, S. F kigge, ed., Vol. 3/3, New York : Springer-Verlag New
York, inc. 602 pp. " The maximum mathematical generality consistent with concrete, definile physical interpretation is sought." TRIJESUI:<[ L, C, and R. TnuPIN. (1960) . ' The Classical Field Theories, - in Encyclopedic' of Physic-5,S_ Fliigge, ed., Vol. 3/1, pp- 226-790. New York: Springer-Vcrlag New York, Inc. "We present the common fnund.rtion of the field viewpoint, We aim to provide the
reader with a full panoply of tools of research, whereby he himself, put into possession of the latest discoveries but also of the profound but all too often forgotten notly achievements of previous generations, may set to work as a theorist. This treatise is intended for the specialist, not the beginner."
Flu i d Mechanics BATCHELOR, G. K. (1967) An Introduction ru Fluid f]rnarnie-s. New York_ Cambridge University Press, 615 pp. "A textbook which can be used by students ❑ f applied mathematics and which incorporates the physical understanding and information provided by past research_" CHANDRASEKHAR, S. (1961) Hydrodynamic and f>iwdrnnrayrtelic Slability. New York: Oxford University Press, Inc. 654 pp. A compendium of results on the linear stability theory of stratified fluids, of rotating fluids, and of many related flows_ CatJRANT, R., and K. 0_ FRIEDRICHS_ ( 1948) Supersonic Fiait and Shock]eru!s New York: Academic Press, Inc. 464 pp.
The classical authority on the theory of compressible flow_ "The book has been written by mathematicians seeking to understand in a rational
Bibliography
574
way a fascinating field of physical reality, and willing to accept compromise with empirical approach_" G O.n5rr_irr. S. (ed.). (1938) Modern Develapments is Fluid Dynamics_ New York: Oxford University Press, Inc. 702 pp. The first four chapters (187 pages) still form an exccllenl introduction to viscous flow. HAPrEI., J., and H_ BRENNER. (1%5) Low Reynolds Number Hydrodynamics. PrenticeHall, Englewood Cliffs. N.J. 553 pp. "The subject matter is largely confined to a development of the macroscopic properties of fluid-particle systems from first principles." KINSMAN, B. (1965) Wind Waves. Englewood Cliffs. N.J.: Prentice-Hall, Inc. 676 pp. "This is a very personal book. It is riot she ;rush about waves. It is an effort to communicate to you what I have so far managed to understand about waves." Full of interesting theory, and with magnificent pictures lo remind the reader that water waves are wet_ LAMtt, li. 0932) Hydrodynamics, 6th ed_ New York: Cambridge University Press_ 738 pp. Also available from Dover Publications, Inc., New York. An encyclopedic collection of classical results. LANlu ta, L. D.. and F.. H. 1.1FSr-UI1rt. ( 1959) F7uid Mechanics. Elmsford, N.Y. Pergarnun Press, Inc_ 536 pp. Translated from the Russian by J. R Sykes and W. }I_ Reid. "The nature of the book is largely determined by the fact that it describes fluid mechanics as a branch of theoretical physics." LEVICH, V. G (1962) Physicochemical Hydrodynamics. Prentice-Ball, Englewood Cliffs, N.J. 700 pp. "Hy physicochemical hydrodynamics we understand the aggregate of problems dealing with the effect of fluid flow on chemical or physicochemical transformations as well as the effect of physicochemical factors on fluid flow" (V. Levich)_ "Higher mathematics is employed throughout the text, neither as an end in itself nor as an elegant dress for the subject matter, but as a potent implement for sharpening physical understanding" (L. Scriven, Editor's foreword) Mu.NE-THOMPSON. L. M. (1968) Theoretical Hydrodynamics, 5th ed. London: Macmillan & Company Ltd_ 743 pp_ A text stressing inviscid fluids and complex variable met hods_ New York ; PRA NOTL, L., and O. G. TIETJENS. (1934) Applied ff,i dro- andAero-mechanics, McGraw-Hill hook Company_ 270 pp. Also available from Dover Publications, inc., New York, 1957, as Fundamentals of Hydro and Aero-Mechanics_ Brief and inexpensive, with considerable physical insight_ RosENtiEen, I.., ed. (1963) Laminar Boundary Layers. New York: Oxford University Press. Inc. 687 pp. " An account of the development. structure, and stability of laminar boundary layers in incompressible fluids together with a description of the associated experimental techniques. SEEd Ett, R. J., and G. TEMPLE (eds.). (1965) Research frontiers to Raid Dynamics. New. York: Iriterscience Publishers, a division of John Wiley & Sons, Inc. 788 pp_ A collection of useful survey articles. SERRIN, J. B. (1959) . 'Mathematical Principles of Classical Fluid Mechanics," in Esuyciopedia of Physics, S_ FlOgge, ed., Vol. 8/1, pp_ 125-263. New York: SpringerVeriag New York, [Tic_ -
"
Bibliography
575
"Our intent ... is to present in a mathematically correct way, in concise form, and with more than passing attention to the foundations, the principles of classical fluid mechanics." STOKER, J_ J. (1957) Winer Waves. New York: Interscience Publishers_ 567 pp_ 'l -he classical theoretical book on the subject. WFtrrt;iM. G. R_ (1974) Linear and Nonlinear Waves. New York: John Wiley & Sons. Inc_ 636 pp. "This book is an account of the underlying mathematical theory with emphasis on the unifying ideas and the main points that illuminate the behavior of waves The emphasis is on the conceptually more difficult nonlinear theory_" Most, but by no means all. applications are taken from fluid mechanics. 'I -he two principal journals are Journal of Fluid Medu nres and Physics of F7ut s. Beginning in 1969, Annual Reviews of Fluid Mechanics, a collection of authoritative surnmarks of research progress, has been published by Annual Reviews, Inc., Palo Alto, Calif_
Elasticity K. F. (1975) Wave Marion in Elastic Solids_ Oxford: Clarendon Press, 649 pp. "The purpose of this book is to present, in one place and in a fairly comprehensive manner, air intermediate-level coverage of nearly all of the major topics of elastic wave propagation in solids . - - - Throughout the book, emphasis has been placed on showing results ... from both theoretical and experimental studies." GREEN, A. E., and W. ZERNA_ (1954) Theoretical Elasticity. New York Oxford University Press, inc. 442 pp. This book is mainly concerned with three aspects of elasticity theory which have GRAFF,
„
atrcednio tyars.fnielcdomats,plexvrib methods for two dimensional problems ... and shell theory-" LOVE, A. E. H. (1944) A Treatise ran the Alatlrerrmtiral Theory of Elusnriry, 4th ed. New York: Dover Publications. Inc. 643 pp. "ft is hoped . - . to present a fair picture of the subject in its various aspects, as a mathematical theory, having important relations to general physics. and valuable applications to engineering." Although not easy to read, this is the classical reference work PRESCOTT, J. (1924) Applied Elasticity. Essex, England: Longman Group Ltd. 666 pp. Reissued by Dover Publications, Inc„ New York, 1946_ "In writing this book 1 have tried to sec the subject from the point of view of the engineer rather [than] from that of the mathematician." SOKOLVLKOF , L S. (1956) Mathematical Theory of Elasticity. New York : McGraw-Hill Book Company. 476 pp. Perhaps the best single reference. "This book represents an attempt to present several aspects of the theory of elasticity from a unified point of view and to indicate, along with the familiar methods of solution of the field equations of elasticity, some newer general methods of solution of the two-dimensional problemsTIMOSHENitO, S_ P., and I_ N_ Goonim. (1970) Theory of Elasticity, 3rd ed. New York: McGraw-Hill Book Company. 567 pp_
576
Bibliography
"The primary intention [is] to provide for engineers. in as simple a form as the subject allows. the essential fundamental knowledge of the theory of elasticity together with a compilation of solutions of special problems that are important in engineering practice and design." A leading journal is the Journal of Applied Mechanics, More than 2000 pages of up-to-date information on Solid mechanics can be found in Encyclopedia of Physics, S. Flagge, ed., Vols. 6a/I-3; .Mechanics of Solids f-111, C. Truesdell, ed. Berlin' Springer-Veriag, 1471
Variational Methods AKHIEZak, N. I. (l%2) Hue Calculus of Variations. New York: Blaisdell Publishing Company. 247 pp. Survey of main problems and methods. Buss, G. A_ (1944 Lectures on the Calculus of Variations. Chicago: University of Chicago Press. 292 pp. A more advanced work with emphasis on such matters as higher dimensionality and variable end points. See also Bliss's treatise Calculus of - Vtwiaiion.s. Mathematical Association of America. 1944, COURANT. R. (1957) Calcidus of Variations_ New York University Lecture Notes, revised and amended by J. Moser. 280 pp. (typewritten). These excellent notes are available in many libraries. COURANT, R.. and D. HILBERT. (1953) Methods of Mathematical Physics, Vol. i. New York Interscience Publishers, a division of John Wiley & Sons, Inc. 577 pp. Chapter 1V packs a great deal of material into l 10 pages. Seventy pages of applications to eigenvalue problems comprise Chapter VI. FORSYTH, A. It (1960) Calculus of Variations. New York: Dover Publications, inc. Paperback edition of 1927 treatise. 656 pp. "Though it does not purport to he a history, the gradual growth of the successive tests has governed the arrangement." GELFANU, I. M_. and S. V. Foulthl- (1963) Calculus of Variritions, revised and translated by R. A. Silverman. Englewood Cliffs, NJ.: Prentice-Hall, Inc. 232 pp. A modern introduction with considerable material on applications_ HEST , M. R. (196114 L'ult UJus uj` Variartorts and Optimal Control dory. New York : John Wiley & Sons. Inc. 405 pp. A unified rigorous approach to a large class of problems in calculus of variations and control theory_ LANCZOB, C_ (1949) The Variational Principles of Mechanics_ University of Toronto Press, Toronto. 307 pp. The variational principles of mechanics are firmly rooted in the soil of that great century of liberalism which starts with Descartes and ends with the French Revolution and which has witnessed the lives of Leibniz, Spinoza, Goe the, and Johann Sebastian Bach. It is the only period of cosmic thinking in the entire history of Europe since the time of the Greeks. 11- the author has succeeded in conveying an inkling of that cosmic spirit, his effort will be amply rewarded" (from the Preface).
Bibliography
577
MIIC}rt_LN, S. G. (1964) Variational Methods in Marhesnurical Physics. Elmsford, N_Y_ Pergamon Press, Inc. 5R2 pp. Advanced and comprehensive treatment centering on direct me thods. PARS, L. A. (1962) An Introduction so she Calculus of Variations. New York : John Wiley & Sons, Inc. 350 pp. "My aim has been to provide a clear and rigorous exposition of the fundamental theory, without losing sight of the practical applications •- in geometry and dynamics and physics." WEINSTOCK, R. (1952) Calculus of VariotiO ns. New York: McGraw-Hill Book Company. 326 pp. An elementary introduction with an extensive treatment of applications. WoormOUSE, R. (1964) A History of the Calculus of Variarinns in she Eighteenth Ceniw y New York: Chelsea Publishing Company, Inc. 154 pp_ Originally published in 1810; contains very interesting accounts of t he pioneering work of the Bernoullis, Taylor, Euler, etc. FR1F1]MAp.i, k3. 0956) Principles and Techniques of Applivd Murhernrirics. New York: John Wiley & Sons, Inc. 315 pp. A studyofabstract linear spaces, operators defined on such spaces, and their use in systematizing the methods and techniques for solving problems in applied mathematics_ .
STAKGOLD, 1. (1967) Boundary Vrdue Problems of Mathematical Physics. New York: Macmillan Publishing Company, Inc. Two volumes of 340 and 40g pp_ Eigenfunction expansions and Green's functions are the principal tools developed for the solution to linear boundary value problems.
F., and B, SZ. -NpcJY. (1955) Functional Analysis. New York Frederick Ungar Publishing Company, Inc. 467 pp. A classic introduction to the study of linear operators_ Applied mathematicians may find this book more readable than many successors. for it seems to use the organizing force thane an end in itself. abstrcionme
Hi nts a n d Answer s SUCTION 1_ I
L Hint: itetord the given displacement as et imposed of three mutually perpendicular displacements along the axes and project the whole figure_ 2. Hinz' Take scalar products with Orr and ERN, respectively.
— I o U -I D D
7.L =
L}
U 1
£' il _ felr1. e4}1• /e t► 1`
S. Hint:
SEC 11oN
6. R, 1 = (cos 6)6 o f (
I — cos
triM ► r; e, j
(sin
1.2
14:, ►r n r,-
S EC'l'1ON 1.4
4. (si)
(} 2
U
l
0
I
I
U t)
(bp
(a, 6, I I
SFCrI{7N 2.1 22. The results requested here are derived in Section U. 24. (a) ab r(eb i ha) + 1(ab bal. IN Hint: Choose the unit vectors as coordinate vectors. -
SECTION
2.2
4. (a) C'orrespi tiding to the eigenvalues 1. 2i, and —21, the required eigensectors are multiples of (1, 0, U), (0, 2i. 1), and (0, —2i, 2), respectively. 5 (a) Eigenvalues are 0, t, — t. Corresponding eigcnvectors are proportional to e'" £121 + ie131 „ £I l I — 1£131 SEt'11oN 23
2, Hint: Use Theorem 1.2.9 with A, = f, i . = mol l+ + grad ¢ A v. 5. (g) curl r 1 6. [bl Hint: [^ A u] p = 4 t pik(^u^1^ xr ^t+;1[^xrj. 9. (a) DO - t NjPr = $- W,fev,lDxjl.- .uw.,1• (e) Hrnf: In your calculations the term $za,AA r shoulid appear. Modify i t by means o f the chain rule. ^
SECTION
9. Hint :
3.1
Employ an argument used in the method of separation of variables. 579
Nr,trs
5 80
and Answers
SECTION 12
Hint: Try tr=u(z,r),u =w= O. 14. (c) Hint : Take advantage of symmetry; find the force on the side along x = 1 and 11.
triple it. SECTION 3.3
3. (a) The product of Lr(x, O)and the integral gives the decrease, caused by the slowing of flow in the boundary layer, of the volume flux along the boundary. To first approximation, the same decrease would be caused by an outward displacement of the inviscid flow by a distance 6 1 .
(b) 15 2(x) _ fo ❑u(x.(x,Y)0) 1 ^
y) u (x' dy U(x, 0)
SECTION 3.4 -Ar3 10. (C) T is proportional la t exp ( —r 1 /41a). (d) Hint: Suppose that the heat were initially concentrated at r = re . SECTION 4-L
6. flint: Use (32b). 12. (b) Hint: If a particular displacement field can be found SO that (36) is satisfied, (36) is still satisfied if the field is augmented by an arbitrary rigid body rotation. 13. flint: Contemplate (39). Also see the end of Section 4,314, (b) Hint: Show that each of the two terrils vanishes separately. 15. (b) Hint: QLL.1 —
Q1,3 — Q13.3
x1 t<
SECTION
111. Him: Consider V(u +
4.4
u`) — SECTION 5,1
5. Hint: The coordinates of the deformed neutral axis are given by (15). Write down the coordinates x: of the deformed cross section corresponding 10 x 3 T a, and find the normal to this surface corresponding to the point x r = x2 = O, x3 = u, 12. (a) Hint: One cannot pass from (24h) to (25) because iQ/0x 3 will not exist at points where Q is discontinuous. One can proceed by applying (24b) to the intervals a, 1, [1 , bI, and [a. b]. (In the latter case, the presence of the force concentrated at C must be accounted for-) A suitahle combination of the resulting three equations leads to an expression For the jump in Q (compare 1, pp. 360-36l). A sirnilar argument gives the jump in M. (c) Him! Difficulties arise owing to the fact that d 3q/dx] is discontinuous at 1 3 = C. Avoid these by considering the sum of integrals from O to x 3 and 1 3 to L. 13. (a) flint: No change is needed in (41). Minus signs must be introduced preceding F in (45) and (46) and preceding ►r2 in (62). SEC -rtoN 5.2 Equation ( 14) can be regarded as giving a vector field as the gradient of a scalar field. 8. Hint: Consider the modified stress function. 2. (b) Rini. :
581
Hints and Answers S ECTION 5.3
S. Hint: 6. Hire:
Rotate coordinates. The problem can be reduced to previous considerations by introducing principal axes. For a similar situation, see Exercise 4.1.9. SEC 1ION 6.3
3. Hint:
Use (1g) to eliminate v in (l ba p. Also use l l 31. SECTION 7.1
4. Hinz:
One could consider this curve to be given by —uly.r) — et' sr[uly.r)+t]= O,
where aly. t) is implicitly defined by 1311. 5. Hint: Let the material description of the motion be given by x — x(A, r). Define F[x(A, r), r]. What significance has the surfacef lA, r } = D? Show that the velocf(A, r) ity of this surface normal to itself is zero. Surface tension effects are probably negligible for quiescent fluid
6. Partial answer:
T/ . surfaces whose dimensions are large compared with 9. (a) Hint: Use (25). 111. (1) Hint: Do not forget to project elements of surface and arc onto the xy-plane. Answer:
r
L L
pf Y
...
+ ( r ! — k)dxrly
.
.,,
T1i + Cl Jx so . N —f T(1 + + +
u}[
_k sY I
+ 01) + Ky]l,b+,, dx
u
Al + a ) + kC y]l ro dx
=o
a
u ^
■
i
(1 + C.! +C;.) -112 [1(1 + r41' ^;ia4 y + ^^w^lao f y ^)
m
^
r.
fp Tu+
a + 13)-u2[41 + Sÿ) -gxb, +kx]IL,dy=o,
where Ap =PT —pa. 12. (a) Him: The pressure is hydrostatic. (c) c-scale = ICL. = b. x-scak = I C I^,/I^I,,,.s
(d) ^ — T2C: ^ I 2 ^Z) ({"o =1[ + ^Q^o^^ +
(e)
CA
16
^_s^
— e^
(1)a= -- 8 B.
(g)
About 2h or 3h apart.
= h/cot U.
r^ $ C^^o);.
Hin ts and Answers
582
SECTION 8.1 4. Answer for left-moving waves of wave number at : w aw l cos (a,x + w e t) s i nh a4(z + h), u = ---ora l sin (oci x + cur t) cosh arts + h), p =cup ai sin (ai x + wi t) cosh at(z + h). = a sin (Ri x + u,i t). Nondimensi❑ nal wave speed c, satisfies 4 = F(a 1 + Ba e ) tanh ce,h. Dimensional wave speed C ; satisfies Ci = (gK I + TK () tanh K t H, where K 1 . a, L '. As it should, this reduces to (191 when H/L, — oa. 12. !lIan for water.
_—
SECTION
8.2
9. Ak 11k -i n - t sin (}krtj, Bk = O. 11. (Ai Hint: Use the dispersion relation. 12. Him: If It solution has been "left out," then one can exhibit an initial value problem that cannot be solved by superposition. SECTION
8.3
3. (b) Hint: Integrate by parts twice to obtain Wind. 19. (a) Hint: Write the trigonometric functions in terms of exponentials and thereby show that I = +
SECTION 91 5. (b) w 3 -- H z j6 -- T/29.
1. Hint:
SECTION 9.2 Show that in the latter case WI, z, 1) , i} (where x = [1 corresponds to the
center of the bump), so that a vertical wall at x = i} will have no effect. 3. (a) One trace is the mirror image of the other- ÇThis is not the ease for disturbances of large amplitude)
Krarlxcr and Keller's notation differs somewhat from ours, but the results are in essence those of Section 9_i, using the dispersion relationship (1-25) appropriate to water of finite depth. [Set T = O in (I25), since surface tension is to be neglected.] 4. !lint:
SEC I10N9.3 t. (a) !lint:
Show that if no waves are created or destroyed,
ix. w(x, r) — w(x + fix, t) = 2.. Hint'
dt
oy[x, t) dx.
The net flux of waves into the interior of a closed curve must varnish.
3. (s) Clint: Find V if =W(a).
4. (6)
#=IkI
Form the scalar product of V and (7).
!lint:
SECTION 9.4 3. (tt) Hint' tan S = tan (x — 0). If the constant-phase curve is given in the form x = 49) cos ti y = r(B) sin B, then tan x — (d y/d6)/(dx/dO). ,
flints and Answers
583
SECT1nN 10
,
1. (c) No; WI would just have turned out to be identically zero. Z. For the answer, sec 12.30).
SECTION 11.1 15. (s) The theorem is the implicit Function Theorem.
SECruoN 11.2 (b) sin (xy'). (c) xy cos (xy'). 1. (a) y'y cos (xy') (d) y cos (x11) + xy . cos (xy'i — xy(xyr" + y'i sin (xy'). 1 (a) ln (xÿ) — 1 + ;•(y')- 2ÿ' = 0. (b) I — y y+'- y'= 0. 6. Hint: When starting from the fact that the Euler equation is identically satisfied, reason that the coefficient of y" must be identically equal to zero, so F = 4(x, y)y' + r(x, y) for some functions <13 and ÿr. 7. (a) Hinr: F} - O. (b) The greatest lower bound for I is zero, but This is not attained for any admissibley. 10. (a) The integral to be minimized has as integrand -
Rdfi^x +
sin3
f ^' rr
13. (b) xy cos (xy) - 6 at the boundary.
(d) y = Oat the boundary _ SECTION 11.3 1. (a)
+
=
y" = 0 at x _ 0, L. (c) y" = U and y ~ + Ky = 0 at x = L. 3. (b) The natural boundary condition V n = O is to be expected, for this is the same as v n = U. and prohibition of flow through the boundary should indeed be the best one can do (at the boundary) to keep down kinetic energy. 4. Hint: Use the tact that the integratids do not depend explicitly on x. Answer y = —(A/pg) — (Afpg)cosh [A—' ya4(x — 13)1, where A, 8, and 41 are determined from (56) and (57). 5. Hint: Show that the Filler equation implies a proportionality between d dx ;
dr d
and
iG cox,
7. (s) 2yt"ry + 4y" y t 307 11 O. ., +y,. + y 0. (b)y -Fxy & (R) .1[y — exP (fx ) — 3 : 35 . (b) xlzxx sin (2x ) — (a[I + z) COS (2x ) + X 51n (z x ) O. 13. 11 G - (d/dx)G,; 2. O, then y renders stationary the integral of (20, and this requirement alone—phis the boundary conditions—will often specify y uniquely. If 3 # C for ,
,
this value of y, then the problem has no solution. 15. (a) F O (at any end point where yi is not prescribed). (h) F,. T AG E = O.
Minis a nd A+r.swers
584
16. Hint: Use a new parameter l'. which is a linear function of E. 18, Hint : Let i he smooth and vanish on the boundary. Show that !(0 + ri) > l(/) by using an appropriate form of Green's theorem_ 21. (a) Him! Use the parametric representation of a curve. SECTION 12.2 3. (a) Him: Use 012.1.8c). 18. Hint: If functions arc single-valued and paths of integration are closed curves, then there are no end-point contributions after integration by parts. SECTION 12.E
3. (h) The frequency cannot be raised, because the minimizing function can be chosen from a larger class, since functions are now admissible even if they do not vanish on the "freed" portion of the boundary. 4. There are zero, one, and two natural boundary conditions in cases (i), (ii), and (iii), respectively, so the classes of admissible functions become wider and the eigenvalues cannot increase.
Authors Cited Ciia[ions from the bibliography are not indexed.
Bell. E. T., 475 Birkhofl', G., 125 Bluman, G. W., 125 Born, M., 533 Brillouin, L., 380 Brown, S., 124 Bryson, A., 328 Carrier, G. F., 77
Chester, C., 378 Cole, J. D., 125 Condon, E. U., 516 Courant, R., 143 Davey, A., 443 Dym,C.L,212 English, W. J., 526 Erddyi;, A., 415 Erickson, J. L.., 460 Faddeycra. V., 519 Howes-Williams, J. F, 345 Fife, P., 437 Fliigge, S., 460 Friedberg, R_, 465 Friedman, B., 378 Friedrichs, K., 143
Jeffreys, H., 284 Joseph, D., 423
Kellum, K.. 389 Keller, H. B., 264 K.ellt:r, J. B., 386 Keyes, T., 85 Kim, Y., 439 Kino, G. S., 284 Kranztr, H. C., 386 Krtysrig, E., 366 Kuhn, F.., 386 Lagerstrorn, F. Ay 124
Landau, L., 499
Hahn, W., 536 Hanratty, T., 439 Hawtings. D., 345 Hunter, C. 408
Rayleigh, Lord, 27B Reid, W. H. , 562 Richrrnyer, R_, 142 Sanchez, D., 536 Sears, W., 124 Segel, L, A., 368 Shaw, J., 284 Shortie'', G. 1-1., 516 Sin irnow. V. 1., 452
Stewartson, K., 124, 443 Struik, D., 317 Stuart, J. T., 442
Lewy. H., 143 LJfshitz, E., 499 Ligfithill, M. J., 105 Lin, C. C., 119 Longuet- Higgins, M. S., 300
Tirnnshlnko, S., 211 Tornber6, G_, 386
Luke, J. C.. 489
Ur sell. F.. 378, 402
McGoldrick, L F., 437, 439 Meyer, R. E., 86 Michell, J. 1-1., 244 Miles. J. W., 387 Morton, K. , 142
Van
Love, A. E. H., 289
(add, G., 388
Gee, D., 386 Greenspan, H., 104 Gurtin, M. E., 235
Pnueli, A., 525 Prandtl, L. 226 Prins, J. E., 382
Nods, E., 386
Oleinik, O. A., 143 Oppenheim, 1., 85 Pekeris, C. L, 525
Philips, O. M., 3ù0 Pierson, W. r., 437
585
Dorn, W., 388. 389, 390
Van Dyke, M.. 123. 124
Walter, W.. 143 Washizu, K.. 15B Watson, G. N., 366 Weyl, H., 128 Whitham, G., 391 Wiegel, R., 386 Winner. A., 415 Wolf, E, 533 Yih, C. S., 300
Subject I ndex Adherence boundary condition, 86
Adjoint, 56{1--61 Adverse pressure gradient, 121) Airy function, 377 Airy stress function, 242-45 Alternating tensor, 35, 222 Alternator, 16 Amplitude dispersion, 435 Anti»yrnmctric tensor, 41 vector of, 46-47 Banach space, 563 Barotrnpic fluid, 68 Beam, vibrating, 518-21, 530-31 Beetles, wave pattern from, 379-80 Beitrami equation. 68 Bdtrami-Mithell equations. 171-72 Bending, 194-215 moment, 202 pure, 199 Besse! functions, 365-67 inequality, 558 Betel -Rayleigh reciprocal theorem, 183 Biharmonic equation, 242, 245 Biurthogonality, 562 Blasius similarity solution. 116-19 Bluff body, 121 Body force, 159 Bond number, 325 Boundary conditions boundary layer, 112-13 elastic, 208-211), 216 natural; srr Natural boundary conditions viscous, 85-86 water wave, 31)1-14 Boundary layer. 71,105-14 chemical reaction in, 127-28 separation of, 120, 122-24 Boundary layer equations, 112 difference scheme for, 139-43 Mises transformation for, 114-15 Boundary operator. 515 Bounded set, 564 Drachistochrone, 461-62, 469, 470, 473, 47 5, 485 Buckling, 212 14, 486 l3uilt-in end, 208 Bulk modulus, 166 phase, 308 Cable, hanging, 487, 493 CaIcuius of variations, 461 Capillary waves, 340, 380, 435 Cartesian tensor, 33: also see Tensor Catenary, 493 Cauchy sequence, 563 Cauchy-Green deformation tensor, 186-87
Center of mass, 195 Centroid, #95
ChM-BM-ASlic equation, 52,514
line, 404 polynomial, 52 value, 52 Vector, 52 Chemical reaction, in boundary layer, 127-28 Cofactor, 18 Compact set, 564 Compatibility conditions, 153-55, 167-68 Complementary error function, 101 Complete, 556 set, 315 space, 563 Completely continuous operator, 564 Complex sums 349-52 Compression, 165-66' Concentrated load, 217-18 Conformal mapping. 122 Consistency condition, 434 requirement, 23-27 Constant coefficients, linear partial differential equations with, 336 Constitutive equation for heat flow, 89 for linear elasticity, 1611` 62 for nonlinear elasticity, 188-89 for surface tension, 307 for viscous fluids, 80 85 Constraint, 450-57, 479 85 Continuum mechanics, basic equations of, 72 Contour integration, 364 Contraction, of a tensor, 37 Conrraction product, 39 matrix of, 42 Convergent, sequence, 563 Coordinates, 3-4 Couette flow, 93-95, 96 stability of, t02-103 Couple stress tensor, 70 Courant max-min principle, 528-32 Covarianicc, 25 Cramer's rule, 19 Critical angle, 277 Curl, 62 Curvature, 314,487 mean, 317 -
-
D'Alembert's paradox, 105 Deformation tensor, 79, 186 Del operator, 62
Delay Unes, 284 Delta function, 36x63 Dense set, 509 I)cterrntnants, 16-22 Diagonalization, of a tensor, 59
s $8
Subject Index
581
Diac-Weinstein Formula. 239 Differentials, 29, 78 Diffusion, 102, 126-27 Dilatation, 152, 155 speed, 2717 ways, 260 Direction vector, 10 Dirich lei integral, 489 Dirtiness. quasi equilibrium, 440 Discontinuity surfaces, propagation of, 264-7! Dispersion, 211--11 243-94, 340, 381 anomalous, 212, 340 nonlinear, 435 normal, 340 Displacement, 144 Displacement thickness, 114 Dissipation, viscous, 89-90 Divergence. 62 -
theorem, 64, 66 Domain perturbations, 423 Doppler shift. 403 Drag coefficient, 118 on a sph'rr, 117 Dual spaces, 561 Dubois-Remond lemma, 497 Dyad, 38.49 Dyadic, 49
Dynamic boundary condition, 304-14, 316 Ed rule. 20 Eiconal equations, 270 Eigenvalues, 49-59, 552 as minima, 505-507 nonlinear, 427 perturbed, 566 simple. 566 as stationary values, 504 505 Eigenvectors, 52-53 Elastic constants, relations between, 167 Elrcirostalic potential, 546 -
End built in, HE free, 209 simply supported, 208 Energy balance, 7€ -71, 174-76 conservation, 354 kinetic, 89, 176, 487 potential, 89, 179 -81, 214, 494-95, 534-36 propagation, 397-98 or wave system, 365 Equilibrium equations in elasticity, 171 local, 309 and potential energy, 214 Equipollent stress, 235 Equivoluminal waves, 260 Error function, 10I Ether, elastic, 168, 277 Euclidean Space, 4 Euler equation. 463-67. 471-73, 475, 499 Eu1er-Bernoulli rule, 200 Euler-Lagrange equation; see Euler equation
Exact solutions to elasticity equations, 162-66 to viscous flow equations, 93-102 Experiment, comparison with, in water waves. 382-88 Fxterrsian, simple, 155, t92 Extrema]izal ion of a function, 448-57 of potential energy in elasticity, 494-95 Exiternum, 448 Fermat's principle, 491 Filaments, 199 Finite deformation, 184 91 First variation, 497 Fishing line, fl ow past, 341 Flexural rigidify, 200 Flexure, 200 Fourier coefficients, 556 integrals, 355-59 raies, 352-53 Free end, 209 Frequency, local, 393 Froude number. 325 Functional, 462, 504 Fundamental, 435 -
Gauss's theorem, 65, 66 Geodesics, 487
Gerstner's trochoidal wave, 315 Gradient, 62 Gram determinant, 524 G ravity waves, 340, 435 Greens function, 218 Group lines. 397 speed, 3613, 374$4
velocity, 391-400 Gyration, radius of, 206 Harmonic generation, 435 Hamilton's equations, 401 Hamilhon's principle. 486, 495 Heat diffusion, 126- 27 flux vector, 71, 89 Helium ionir.atioo potential, and Ritz method, 516 Helmholtz representation Lhenrcrn, 67 Hilbert space, 563 Homogeneous deformarinn, 189 Homogeneous material, 161 Hooka s law, generalized, 162 Hoop Stress, 250 Identity tensor, 44 Impermeable boundary, 301 Incompatibility tensor, 158 Index dummy, 6 Free, 6 inequality constraints, 455-57 Inertia moments of, 195
888
Subject Index
Inertia from.] principal axes of, 195 products o1, 195 rotary, 21)7 tensor, 195 Infinitesimal strain, 147,150 Influence function, 218 inhamogeneous equation, 500 lnhornogenrous media, waves in 401 Inner product, 549 Integrahtlity, 153 Intermediate variable, 113 Intrinsic reference quantities, 330 Invariance, 25 and energy balance, 70-71 Inverse method, 233 lrrotational flow, 79 lrrotatianal waves, 260 Isoperimetric inequality, 491 Isoterm, 450 Isotropic, 82 material, 161 tensor, 43-46 Jacobian, 148 49 derivative of, 21 22 Jump conditions, 218,266-71 kelvin elicit, 191 Kinematic boundary condition, 301-302 Kinematic viscosity, ë7 Kinetic energy, 89, 176,487 Kirsch probkrn, 248-50, 2$7-58 Kronecker delta, 7 Lagrange multiplier. 452-57, 480--82, 3111-5112 Lagrange-Cauchy theorem, 69 Lagrangian, 486 description. 144 Lamé constants. 162 Laplace transform, 363 Legendr ee necessary condition, 475 Level surface, 450 Line element, 146 Linear dependence, 12 elasticity. 169-70. 187-88 independence, 12, 350 operator. 551 transformations. 28 32 t.inearization, 326 Local Reynolds number. 1l8 Local velocity field. 78-80 Long waves, 343-45 Longitudinal waves, 260 Love waves, 289-95 -
Mach angle, 411 cone. 411 Matching. 112-13 Material coordinates, 144 Material indifference, 85 Matrices, and tensors, 41-43 Matrix. definite, 449
Maximum, 448 Maximum-minimum principle. 528-32 Maxwell's equetiosls, 168, 277 Mean height requirement, 321 Membrane, 489, 512, 337-38 analogy, 226, 545 Metric, 563 Minimization, of potential energy, 179-81 Minimum, 448 Minor, 18 Mises transformation, 114-15 Moment, 195 rtia7 -Weinstein formula for, 239 of area, polar, 228 Momentum thickness, 114 .
Natural boundary conditions, 467-71, 493, 521-23 Natural laws, 533 Navier equations, 171 Navier-Stokes equations, 84, 90-91, 93-102. 138-39 Neutral axis, 199 Newton's second law, 23-25 Non-Euclidean space, 4 Norm, 509 natural, 563 Normal principal, 487 Normal stress, 50 Nonlinear eigcnvalue problem, 427 Nonlinear traveling waves, 418-43 Normalization condition, 430 Numbed vector space, 563 Observer invariance, 85 Orthogonal eigcnvcctors, 56-57 functions, 557 vectors, 549 Overlap region, 112 Parallelogram law, 13 Particle paths, in gravity waves, 342 Patching. I I I-12 perks number, 128 Permutation. 13 symbol. 16 Perturbation theory for eigenvalu s, 566 for fluid-air interface, 317-19 for nonlinear water waves, 418-43 Phase function, 393,408-10 speed, 375-80 velocity, 396 97 Physical law. general statement vs. particular version, 24-25 Pipe, viscous flow in, 104-105 Plane strain, 157, 241-50 stress, 250-52, 255 stress. generalized, 252-56 Plastic material, 160 Plata boundary layer flow past, 107-11, 115-19 -
Subject Index
S8g
Plate €vol.] elastic, 525. 545 Poiseuille flow, 95-98 Poisson's ratio, 164 Polar decomposition theorem, 184-86 Polar fluids, 70 Polarization, plant of, 272 Positive operator, 539-41, 554 Potential energy, 89, 179-81, 214 exrremalization of, 494-95 minimum of, 534-36 Pressure, 322 Principal axes, 57, 151-52, 195 Principal diagonal, 32 Principal minor, 52 Principle; see Fermas principle Projection, 4- 5 Proportional limit, 160 P-waves, 2611 Pythagoras, law of, 4 [
Quasi-elastic material. 168 Quotient rule. 40
Radiation Condition, 33 -i, 398, 410 Range convention, 6 Rate-of-strain tensor, 19 Rayleigh impulsive flow, 98 102 waves, 278 Rayleigh Ritz method; see Ritz method Reaction chemical, in boundary layer, Ill28 Real operator, 553 Reciprocal theorem, Belli- Rayleigh, 183 Rectangular bar, shear of, 164--65 -
Reflection coefficients, 276 internal, 285 - 89 Resonance, nonlinear, 436-411 Resonant energy interchange, 442 Rr:SUltant, 2 02 Reynolds number, 1.07 Rigid body rotation, 47, 79, 195 Ritz method, 504, 513-21, 531-32, 542-44 Rotation Finite, 13 improper, I l proper, IC 1 rigid body, 47, 79, 195
small, 13 tensor, 79. 15041, 185
vector, 151 Si_ Venant principle, 235-37 Scalar. 34 multiple, 9 product, 549 Scaling, 107, 329-34 Scattering, 272 Schwas inequality, 550 Second order tensors, 41 Self-adjointness, formal, 561 Self-adjoint operator, 507, 546-62, 566-67 Semi-inverse method, 219-20
Separation of flow, 12U of variables, 23]-33, 343 Sequences, 563 Shear force, 202 modulus, 165 of rectangular bar, 164-65
simple, 156, 189-91 speed. 270 strain, 151-52 stress, 50 maximum of, 229, 238, 460, 144 waves, 260 Sideband perturbations, 443 Signum function, 351 Similarity solutions. 116-19, 125-27
Simply supported end, 206 Skew-symmetric tenser, 41 Slow visçuus flow, 135 3b
Snell's law, 492- 93 Solenoidal vector field, 67 Solid body rotation, 47, 79, 195 Solitary. wave, 382 Sphere slow flaw past, 135-38 Stokes drag (ormula fur, 137 Stabilii}
asymptotic, 134
Liapunnv, 534
S1,3II, 122 Si ationary integral, 472 phase, method of, 370, 412-16
point, 449 value problems, 500-503 Stokes's theorem, 65,66 Strain, 144-50 energy, 175 infinitesimal, 147
normal. 15 plane, 157, 241.-50 principal, 151
shearing, 151 tensor, 146, 149-50 Strain sires equations, 162 -
Streamlined body, 121 Stress, 50
concentration, 250 energy, 176 function, 224-25,242-45 modified, 229 plane, 250-52 LensDr, 25 -26.29- 32 Sty css-strain relations. 160-62 Stretch tensor, 185 String, 489 Sturm-Liouville problem, 557 Subhiirrnronic function, 229 Suction, 104 Summation convention, 7 Superposition, 346 Surface Free, 324 phase, 308
590
Subject index
Surface leant 1 t ension, 303 viscosity, 308 S-wavca, 260 Symmetric tensor, 41, 55-59 Tautochront, 475 Tension, of a cylindrical bar, 163 64 Tensor, 34; uitso sec Deformation tensor, Rotation tensor antisymmetric, 41 calculus, 60-67 components, 31-32 contraction, 37, 39 derivatives, hi di tgona1i ation, 59 difference, 35 equality, 36 -
field, 60 inertia, 195 and matrix, 41-43 nonsingular, 43 product. 38 quotient rule, 40 real, 55 scalar multiple, 36 second order, 41-43 stress. 25-26 sum, 35 symmetric, 41,55
Variations first, 497 weak and st(ong, 472 Vector, 9-13, 488 identities, 23 product. 465 space, 547-48 Velocity gradient tensor, 79 potential, 512-13, 558 profile, 94 scale, 214 Virtual displacements, 177, 181 Virtual strains, 177 Virtual work, 177-79 'Viscosity, 84, 87 effect on water waves, 440 Vortcx sheet, 133 Vorticity, 68, 129 35 boundary generation Of, 131 convection off, 129 -
development equation, 69
diffusion of, 131 stretching ai, 130 31 -
Warping 221
function, 223 Water bttmp, collapse ci, 38148 Wave conservation. 395-96 equation. 377-78,542 function, 396 length, 380 number, 380, 393, 395
transformation law, 25, 34, 61 transpose, 41 Tidal oscillations, 525-28 Tidal waves, 389 Tides
number vector, 522 reflection, 271 trains, 392, 399
equations foe, 345-46 Torsion, 219-35,541-45 Torsionalrigidity, 228 Trace, 52 Traction, 195
Waves capillary, 435 capillary-gravity, 408 dimple-down, 436 elastic, 211, 259-63, 276-84
Transformation
coordinate, 5 6 -
matrix, 10, 22 symbol, 7 Transpose, tensor, 41 Transposition, 15 Transversatity condition, 485 Transverse waves, 260 Trial elements, 513
Triangle inequality, 414, 550 Triple deck, 122 Trip le scalar product, 21 Tsunami, 389 Twist modulus, 168
gravity, 435 in inhoroogerprous media, 401 Love, 289-95 nonlinear, water, 418-43 plane, 261 polarized, 263 ship, 4112-10 small amplitude, 331 SV , SH PV , 272 trochoidal, 315 ultrasonic, 284 water, 299-444, 489-91 wedge angle of, 408 Weierstrass-Erdmann condition, 493 -
Ultimate load, 161 Uniqueness theorems, in elasticity, 178-79 Unit coordinate vectors, 4 tensor, 34 vector, 10 Variational methods, in elasticity, 214 15 Variational notation, 497-99 Variational problems, 447 -
- ,
-
Yield point, 160 Young's modulus, 161 Zero element, 547 tensor, 36 Zeroth order tensor, 34
