Wolfgang Hackbusch SPRINGER SERIES IN COMPUTATIONAL MATHEMATICS
Elliptic Differential Equations Theory and Numerical Treatment
t2J
Springer
18
Springer Series in Computational Mathematics Editorial Board R. Bank R.L. Graham J. Stoer R. Varga H. Yserentant
18
Wolfgang Hackbusch
Elliptic Differential Equations T heory and Numerical Treatment
Translated by Regine Fadiman and Patrick D.F. Ion
123
Wolfgang Hackbusch MPI für Mathematik in den Naturwissenschaften Inselstr. 22-26 04103 Leipzig, Germany
[email protected]
ISSN 0179-3632 ISBN 978-3-540-54822-5 (hardcover) ISBN 978-3-642-05244-6 (softcover) DOI 10.1007/978-3-642-11490-8
e-ISBN 978-3-642-11490-8
Springer Heidelberg Dordrecht London New York Library of Congress Control Number: 2010923170 Mathematics Subject Classification (2000): 35J20, 35J25, 35J30, 35J35, 35J50, 35J55, 65N06, 65N12, 65N15, 65N25, 65N30 © Springer-Verlag Berlin Heidelberg 1992, First softcover printing 2010 © B.G. Teubner, Stuttgart 1987: W. Hackbusch, Theorie und Numerik elliptischer Differentialgleichungen. Mit Genehmigung des Verlages B.G. Teubner, Stuttgart, veranstaltete, allein autorisierte englische Ü bersetzung der deutschen Originalausgabe. This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer. Violations are liable to prosecution under the German Copyright Law. The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Cover design: SPi Publisher Services Printed on acid-free paper Springer is part of Springer Science+Business Media (www.springer.com)
Foreword
This book has developed from lectures that the author gave for mathematics students at the Ruhr-Universitiit Bochum and the Christian-Albrechts-Universitiit Kiel. This edition is the result of the translation and correction of the German edition entitled Theorie und Numerik elliptischer Differentialgleichungen. The present work is restricted to the theory of partial differential equations of elliptic type, which otherwise tends to be given a treatment which is either too superficial or too extensive. The following sketch shows what the problems are for elliptic differential equations.
A: Theory of elliptic equations
Elliptic boundary value problems
E:Theoryof iteration methods
B: Discretisation: Difference Methods, finite elements, etc.
t--------
-------
C: Numerical analysis convergence, stability
Discrete equations
D: Equation solution: Direct or with iteration methods
The theory of elliptic differential equations (A) is concerned with questions of existence, uniqueness, and properties of solutions. The first problem of
VI
Foreword
numerical treatment is the description of the discretisation procedures (B), which give finite-dimensional equations for approximations to the solutions. The subsequent second part of the numerical treatment is numerical analysis (0) of the procedure in question. In particular it is necessary to find out if, and how fast, the approximation converges to the exact solution. The solution of the finite-dimensional equations (D, E) is in general no simple problem, since from 103 to 106 unknowns can occur. The discussion of this third area of numerical problems is skipped (one may find it, e.g., in Hackbusch (5] and (9]). The descriptions of discretisation procedures and their analyses are closely connected with corresponding chapters of the theory of elliptic equations. In addition, it is not possible to undertake a well-founded numerical analysis without a basic knowledge of elliptic differential equations. Since the latter cannot, in general, be assumed of a reader, it seems to me necessary to present the numerical study along with the theory of elliptic equations. The book is conceived in the first place as an introduction to the treatment of elliptic boundary-value problems. It should, however, serve to lead the reader to further literature on special topics and and applications. It is intentional that certain topics, which are often handled rather summarily, (e.g., eigenvalue problems) are treated here in greater detail. The exposition is strictly limited to linear elliptic equations. Thus a discussion of the Navier-Stokes equations, which are important for fluid mechanics, is excluded; however, one can approach these matters via the Stokes equation, which is thoroughly treated as an example of an elliptic system. In order not to exceed the limits of this book, we have not considered further discretisation methods (collocation methods, volume-element methods, spectral methods) and integral-equation methods (boundary-element meth-
ods). The Exercises that are presented, which may be considered as remarks without proofs, are an integral part of the exposition. If this book is used as the text for a course they can be used as student problems. But the reader too should test his understanding of the subject on the exercises. The author wishes to thank his collaborators G. Hofmann, G. Wittum and J. Burmeister for the help in reading and correcting the manuscript of this book. He thanks Teubner Verlag for their cordial collaboration in producing the first German edition. Kiel, December 1985 W. Hackbusch This translation contains, in addition to the full text of the original edition, a short Section (§3.5) on the integral-equation method. The bibliography has also been expanded. The author wishes to thank the translators, R. Fadiman and P. D. F. Ion, for their pleasant collaboration, and Springer-Verlag for their friendly cooperation. Kiel, March 1992 W. Hackbusch
Table of Contents
Foreword . . . .
v
Table of Contents
VII
Notation
XIII
1
2
3
Partial Differential Equations and Their Classification Into Types . . . . . . . . . . . 1.1 Examples . . . . . . . . . . . . . . . . . . 1.2 Classification of Second-Order Equations into Types 1.3 Type Classification for Systems of First Order 1.4 Characteristic Properties of the Different Types
4 6 7
The Potential Equation
12
2.1 2.2 2.3 2.4
12 14 17 23
Posing the Problem Singularity Function The Mean Value Property and Maximum Principle Continuous Dependence on the Boundary Data
The Poisson Equation 3.1 3.2 3.3 3.4 3.5
4
1 1
........... .
Posing the Problem . . . . . . . . . . . . Representation of the Solution by the Green Function The Green Function for the Ball The Neumann Boundary Value Problem The Integral Equation Method . . . .
27 27 28 34 35 36
Difference Methods for the Poisson Equation
38
4.1 4.2 4.3
38 40
4.4 4.5
Introduction: The One-Dimensional Case The Five-Point Formula . . . . . . . . . M-matrices, Matrix Norms, Positive Definite Matrices . . . . . . . . . Properties of the Matrix Lh Convergence
44 53 59
vrn
Table of Contents 4.6 4.7
4.8
5
General Boundary Value Problems . . . . . . 5.1
5.2
5.3
6
Discretisations of Higher Order The Discretisation of the Neumann Boundary Value Problem . . . . 4.7.1 One-sided Difference for au/an . 4.7.2 Symmetric Difference for au/an 4.7.3 Symmetric Difference for au/an on an Offset Grid . . . . . . . 4.7.4 Proof of the Stability Theorem 7 Discretisation in an Arbitrary Domain . 4.8.1 Shortley-Weller Approximation . 4.8.2 Interpolation at Points near the Boundary Dirichlet Boundary Value Problems for Linear Differential Equations . . 5.1.1 Posing the Problem . . . . . . . . . 5.1.2 Maximum Principle . . . . . . . . . 5.1.3 Uniqueness of the Solution and Continuous Dependence . . . . . . . . . . . . . 5.1.4 Difference Methods for the General Differential Equation of Second Order 5.1.5 Green's Function . . . . . . . . . . . General Boundary Conditions . . . . . . . . 5.2.1 Formulating the Boundary Value Problem 5.2.2 Difference Methods for General Boundary Conditions . . . . . . . . . . . . Boundary Problems of Higher Order . . . . . 5.3.1 The Biharmonic Differential Equation . . 5.3.2 General Linear Differential Equations of Order 2m 5.3.3 Discretisation of the Biharmonic Differential ..... Equation
62 65 65 70 71
72 78 78 83 85 85 85 86 87
90 95 95 95 98 103 103 104 105
'!boIs from Functional Analysis . . .
110
6.1
110 110 111 112 114 115 115 116 119
6.2
Banach Spaces and Hilbert Spaces 6.1.1 Normed Spaces 6.1.2 Operators 6.1.3 Banach Spaces 6.1.4 Hilbert Spaces Sobolev Spaces . . . 6.2.1 L2(il) . . . . 6.2.2 Hk(il) and H3(il) 6.2.3 Fourier Transformation and Hk(m.n )
Table of Contents
6.2.4 HS(n) for Real s ~ 0 6.2.5 Trace and Extension Theorems Dual Spaces . . . . . . . . . . . 6.3.1 Dual Space of a Normed Space 6.3.2 Adjoint Operators 6.3.3 Scales of Hilbert Spaces Compact Operators Bilinear Forms
122 123 130 130 132 133 135 137
Variational Formulation 7.1 Historical Remarks Equations with Homogeneous Dirichlet 7.2 Boundary Conditions . . . . . . . . Inhomogeneous Dirichlet Boundary Conditions 7.3 Natural Boundary Conditions 7.4
144
The Method of Finite Elements
161
8.1 8.2 8.3
161 167 171 171 174 178 180 182 182 185 185 190 193 193 194
6.3
6.4 6.5 7
8
IX
8.4
8.5
8.6 8.7
The Ritz-Galerkin Method .... Error Estimates Finite Elements . . . . . 8.3.1 Introduction: Linear Elements for n = (a, b) 8.3.2 Linear Elements for n c m? . . 8.3.3 Bilinear Elements for n c m.2 8.3.4 Quadratic Elements for n c m.2 8.3.5 Elements for n c m.3 . . . . 8.3.6 Handling of Side Conditions Error Estimates for Finite Element Methods 8.4.1 Hi-Estimates for Linear Elements 8.4.2 L2 and HS Estimates for Linear Elements Generalisations............... 8.5.1 Error Estimates for Other Elements . . . 8.5.2 Finite Elements for Equations of Higher Order 8.5.2.1 Introduction: The One-Dimensional Biharmonic Equation The Two-Dimensional Case 8.5.2.2 8.5.2.3 Estimating Errors . . . . Finite Elements for Non-Polygonal Regions Additional Remarks . . . . . 8.7.1 Non-Conformal Elements
144 145 150 152
194 195 196 196 199 199
X
Table of Contents
8.8 9
10
200 201 201 202 202 203
Regularity . . . . . . . . . . . . 9.1 Solutions of the Boundary Value Problem in H"(n), 8 > m . . . . . . . . . . 9.1.1 The Regularity Problem . . . . 9.1.2 Regularity Theorems for n = m.n 9.1.3 Regularity Theorems for n = m.+.. 9.1.4 Regularity Theorems for General n c m.n 9.1.5 Regularity for Convex Domains and Domains with Comers . . . . . . . . . . . 9.1.6 Regularity in the Interior . . . . . 9.2 Regularity Properties of Difference Equations 9.2.1 Discrete HI-Regularity 9.2.2 Consistency . . . . . 9.2.3 Optimal Error Estimates 9.2.4 m-Regularity
208
Special Differential Equations
244
10.1
244 244 246 247 247 249 251
10.2
11
8.7.2 The Trefftz Method . . . . . . . . . . . . 8.7.3 Finite-Element Methods for Singular Solutions 8.7.4 Adaptive Triangulation 8.7.5 Hierarchical Bases 8.7.6 Superconvergence . . . Properties of the Stiffness Matrix
Differential Equations with Discontinuous Coefficients 10.1.1 Formulation . . . . . . .... . 10.1.2 Discretisation A Singular Perturbation Problem 10.2.1 The Convection-Diffusion Equation 10.2.2 Stable Difference Schemes 10.2.3 Finite Elements
208 208 210 215 219 223 226 226 226 232 238 240
Eigenvalue Problems
253
11.1 11.2
253 254 254 256 260
11.3
Formulation of Eigenvalue Problems Finite Element Discretisation . . . ..... . 11.2.1 Discretisation 11.2.2 Qualitative Convergence Results 11.2.3 Quantitative Convergence Results 11.2.4 Complementary Problems Discretisation by Difference Methods
264
267
Table of Contents
12
Stokes Equations 12.1 12.2
. . . . . . . . . . . . .
XI
275
Systems of Elliptic Differential Equations Variational Formulation . . . . . . . . 12.2.1 Weak Formulation of the Stokes Equations 12.2.2 Saddlepoint Problems . . . . . . . . . 12.2.3 Existence and Uniqueness of the Solution of a Saddlepoint Problem . . . . . . . . . . . 12.2.4 Solvability and Regularity of the Stokes Problem 12.2.5 A Yo-elliptic Variational Formulation of the Stokes Problem . . . . . . . . . . . . . . . . . 12.3 Mixed Finite-Element Method for the Stokes Problem 12.3.1 Finite-Element Discretisation of a Saddlepoint Problem . . . . . . . . . . . . . . . . . 12.3.2 Stability Conditions . . . . . . . . . . . . 12.3.3 Stable Finite-Element Spaces for the Stokes Problem 12.3.3.1 Stability Criterion ......... 12.3.3.2 Finite-Element Discretisations with the Bubble Function . . . . 12.3.3.3 Stable Discretisations with Linear Elements in Vh . . 12.3.3.4 Error Estimates .... Bibliography
275 278 278 279
Index . . . .
307
282 285 289 290 290 291 293 293 294 296 297 300
Notation
Formula Numbers. Equations in Section X.Y are numbered (X.Y.1), (X.Y.2), etc. The equation (3.2.1) is referred to within Section 3.2 simply as (1). In other Sections of Chapter 3 it is called (2.1). Theorem Numbering. All Theorems, Definitions, Lemmata etc. are numbered together. In Section 3.2 Lemma 3.2.1 is referred to as Lemma 1. Special Symbols. The following quantities have fixed meanings:
A,B, ... B,Bj
matrices boundary differential operators (cf. (5.2.1a,b), (5.3.6» the complex numbers HOlder- and Lipschitz-continuously differentiable functions (cf. Definition 3.2.8) k-fold and infinitely continuously differentiable functions cgc>(.fJ) infinitely differentiable functions with compact supports (cf. (6.2.3» distance of the function u from the subspace VN (cf. Theorem 8.2.1) surface differentials in surface integrals dr,drx diagonal matrix with the diagonal elements db d2, ... diag {db d2, ... } a function; often the right-hand side of a differential f equation g(.,.) Green's function (cf. Section 3.2) h step size (cf. Sections 4.1 and 4.2) H"'(.fJ), HS(.fJ), HG(.fJ), Ho(.fJ) Sobolev spaces (cf. Sections 6.2.2 and 6.2.4) KR(Y) open ball about Y with radius R (cf. (2.2.7), Section 6.1.1) I identity or inclusion (cf. Sections 6.1.2, 6.1.3) L a differential operator (cf. (1.2.6» or the operator associated with a bilinear form (cf. (7.2.9')) L the stiffness matrix (cf. Section 8.1) matrix of a discrete system of equations (cf. (4.1.9a»
XIV
Notation
linear space of bounded operators from X to Y (cf. Section 6.1.2) £<Xl(n) space of essentially bounded functions (d. 6.1.3) £2(n) space of square-integrable functions (cf. Section 6.2.1) IN the natural numbers {I, 2, 3, ... } n=n(x) normals (cf. (2.2.3a» the zero matrix o Landau symbol: j(x) = O(g(x» if Ijx)1 ~ const Ig(x)1 0(·) p cf. (8.1.6) a grid function, right-hand side of the discrete equation (4.1.9a) R, R+ the real numbers, the positive real numbers Rh,Rh restrictions (cf. (4.5.2) and (4.5.5b» 8(·,·) the singularity function (d. Section 2.2) supp(f) the support of the function j (cf. Lemma 6.2.2) U = u(x) = u(Xl. ... , x n ) a function, e.g., a solution of a differential equation Uh a grid function (discrete solution; d Section 4) VN, Vh finite-element spaces (cf. (8.1.3) and Section 8.4.1) independent variables x, (x, y), (x, y, z) a vector of independent variables x = (Xl. ... ,xn ) 7l the integers r the boundary of n n the boundary points of a grid (cf. (4.2.1b), (4.8.4» -y(.,.) fundamental solution function p( A) spectral radius of the matrix A n an open set in m.n or a domain (cf. Definition 2.1.1) nh a grid (cf. (4.1.6a), (4.2.1a), (4.8.2» Ll the Laplace operator (cf. (2.1.1a» Llh the five-point difference operator (cf. (4.2.3» V gradient (cf. (2.2.3a» 8+,8-,8°,8t,8t,... differences (cf. (4.1.2a-<:), (4.2.3» 8t, 8; differences in the normal direction (cf. (4.7.4» 8/8n normal derivative (cf. (2.2.3a» (.,.) scalar product (d. (2.2.3c), (4.3.14a» (., ·}xxx' duality form (cf. Section 6.1.3) (-,.) scalar product (d. Section 6.1.4) (., ·)0, (., ·)L2(0), 1·10, 1·IL2(0) scalar product and norms on £2(n) (-, ·)Ie, (-, ·)a, I· lie, 1·la scalar products and norms on Hk(n), resp. HB(.fl) 1. I the Euclidean norm (cf. (2.2.2» II . 112 the Euclidean norm and the spectral norm (cf. Section 4.3) norms equivalent to I· lie, I· Is (cf. (6.2.15), (6.2.16b» maximum norm (cf. (4.3.3», row sum norm (4.3.11), or supremum norm (2.4.1), cf. also Section 6.1.1 n the vector (1,1, ... ) (cf. Section 4.3)
£(X,Y)
1 Partial Differential Equations and Their Classification Into Types
1.1 Examples An ordinary differential equation describes a function which depends on only one variable. Unfortunately, for many problems it is not possible to restrict attention to a single variable. Almost all physical quantities depend on the spatial variables x, y , and z and on time t. The time dependence might be omitted for stationary processes, and one might perhaps save one spatial dimension by special geometric assumptions, but even then there would still remain at least two independent variables. Equations that contain the first partial derivatives u x•
= Ux.(Xl,X2,··· ,xn) = 8U(Xb'"
,Xn)/8Xi
where (1 :5 i :5 n), or even higher partial derivatives u X • Xj , etc., are called partial differential equations. Unlike ordinary differential equations, partial differential equations cannot be analysed all together. Rather, one distinguishes between three types of equations which have different properties and also require different numerical methods. Before the characteristics for the types are defined, let us introduce some examples of partial differential equations. All of the following examples will contain only two independent variables x, y. The first two examples are partial differential equations of first order, since only first partial derivatives occur. Example 1.1.1. Find a solution u(x, y) of uy(x,y)
= o.
(1.1.1)
It is obvious that u(x, y) must be independent of y, Le., the solution has the form u(x, y) = rp(x). Thus u(x, y) = rp(x) for some arbitrary rp is a solution of (1). Equation (1) is a special case of Example 1.1.2. Find a solution u(x, y) of CUx-uy=O
(c constant).
W. Hackbusch, Elliptic Differential Equations: Theory and Numerical Treatment, Springer Series in Computational Mathematics 18, DOI 10.1007/978-3-642-11490-8_1, © Springer-Verlag Berlin Heidelberg 2010
(1.1.2)
1
2
1 Partial Differential Equations and Their Types
e
Let u be a solution. Introduce new coordinates = x + cy, 'f/ = y and define v(e, 'f/) := u(x(e, 'f/), y(e, 'f/» with the aid of x(e, 'f/) = C'T/, y(e, 'f/) = 'f/. Since v'1 = ua:x'1 + u1l Y'1 (chain rule) and X'1 = -c, Y'1 = 1, it follows from (2) that v'1(e,'f/) = O. This equation is analogous to (1), and Example 1 shows that v(e, 'f/) = cp(e)· If one now replaces e, 'f/ by x, y one obtains
e-
u(x, y) = cp(x + cy).
(1.1.3)
Conversely, through (3) one obviously obtains a solution of Equation (2) as long as cp is continuously differentiable. In order to determine uniquely the solution of an ordinary differential equation u' - J(u) = 0 one needs an initial value u(xo) = Uo. The partial differential equation (2) can be augmented by the initial-value function
= ua(x)
u(x, YO)
for x E
m.
(1.1.4)
on the line y = Yo, with Yo a constant. The comparison of Equations (3) and (4) shows that cp(x+cyo) = ua(x). Thus cp is determined by cp(x) = ua(x-cyo). The unique solution of the initial value problem (2) and (4) reads u(x, y)
= ua(x -
c(yO - y».
(1.1.5)
The following three examples involve differential equations of second order. Example 1.1.3. (Potential equation or Laplace equation) Let n be an open subset of m.2 • Find a solution of Ua:a:
+ u 1lS/ = 0
in
n.
(1.1.6)
m.2 with the complex number z = x + iy E GJ, the solutions can be given immediately. The real and imaginary parts of any function J(z) holomorphic in n are solutions of Equation (6). Three examples are Rezo = 1, Rez2 = x 2_y2 and Relog(z-ZO) = log ..j(x - xo)2 + (y - yo)2 if ZO ~ n. To determine the solution uniquely one needs the boundary values u(x,y) = cp(x,y) for all (x,y) on the boundary r = an of n. If one identifies (x,y) E
Example 1.1.4. (Wave equation) All solutions of ua:a: - us/1I
=0
(1.1.7)
are given by u(x, y)
= cp(x + y) + 1/;(x -
y)
(1.1.8)
where cp and 1/; are arbitrary twice continuously differentiable functions. Suitable initial values are, for example, u(x,O)
= uo(x),
uS/(x,O)
= Ul (x),
(x E
m.)
(1.1.9)
where uo and Ul are given functions. If one inserts (8) into (9), one finds uo = cp + 1/;, Ul = cp' _1/;' , where cp' is the derivative of cp, and infers that
1.1 Examples
3
From this one can determine cp and 1/J up to constants of integration. One constant can be chosen arbitrarily, for example, by cp(O) = 0, and the other is detennined by 1.£(0,0) = 1.£0(0) = cp(O) + 1/J(0).
Exercise 1.1.5. Prove that every solution of the wave equation (7) has the form (8). Hint: Use € = x + y and 'f/ = x - y as new variables. Example 1.1.6. (Heat equation) Find the solution of Uw.r:-Uy=O
(1.1.10)
(physical interpretation: 1.£ = temperature, y = time). The separation of variables u(x, y) = v(x)w(y) gives that for every c E IR
u(x, y) = sin(cx) . exp( _c2 y).
(1. 1. 11 a)
Another solution of (10) for y > 0 is
u(x,y) = n=: 1 V"17rY
1
00
uo(€)exp(-(x -€)2/(4y))d€,
(1.1.11b)
-00
where uo{·) is an arbitrary continuous and bounded function. The initial condition matching Equation (10), in contrast to (9), contains only one function:
U(x,O) = uo(x).
(1.1.12)
The solution (l1b), which initially is defined only for y > 0, can be extended continuously to y = 0 and there satisfies the initial value requirement (12).
Exercise 1.1.7. Let uo be bounded in IR and continuous at x. Then prove that the right side of Equation (llb) converges to uo(x) for y -+ O. Hint: First show that u{x, Y) = uo(x) + J~:[uo(€) -uo(x)]exp( -{x _€)2 /(4y)) d€/v'41iY and then decompose the integral into subintegrals over [x - 8, x + 8] and (-oo,x - 8) U (x + 8, (0). As with ordinary differential equations, equations of higher order can be described by systems of first-order equations. In the following we give some examples.
Example 1.1.8. Let the pair (1.£, v) be the solution of the system 1.£:1> + Vy
= 0,
V:l>
+ u y = o.
(1.1.13)
If 1.£ and v are twice differentiable, the differentiation of (13) yields the equations 1.£:1>:1> + v:I>Y = 0, v:I>Y + U yy = 0, which together imply that 1.£:1>:1> - uyy = O. Thus 1.£ is a solution of the wave equation (7). The same can be shown for v.
4
1 Partial Differential Equations and Their Types
Example 1.1.9. (Cauchy-Riemann differential equations) If U and v satisfy the system U:r; + Vy = 0, V:r; - u y = in n em?, (1.1.14)
°
then the same consideration as in Example 8 yields that both the potential equation (6).
U
and v satisfy
Example 1.1.10. If U and v satisfy the system U:r;
+ Vy = 0,
V:r;
+ U = 0,
(1.1.15)
then v solves the heat equation (10). A second-order system of interest in fluid mechanics can be found in
Example 1.1.11. (Stokes equations) In the system
(1.1.16a) V:r;:r;
+ Vyy -
Wy
U:r; +Vy
= 0,
(1. 1. 16b) (1.1.16c)
=0
and v denote the flow velocities in x- and y-directions, while w denotes the pressure.
U
1.2 Classification of Second-Order Equations into Types The general linear differential equation of second order in two variables reads
a(x, y)uzz +2b(x, y)u:J:1I + c(x, y)uyy +d(x, y)uz + e(x, y)u y + f(x, y)u + g(x, y)
(1.2.1)
= o.
Definition 1.2.1. (a) Equation (1) is said to be elliptic at (x, y) if
a(x, y)c(x, y) - b(x, y)2 > o.
(1.2.2a)
(b) Equation (1) is said to be hyperbolic at (x,y), if
a(x, y)c(x, y) - b(x, y)2 < O.
(1.2.2b)
(c) Equation (1) is said to be parabolic at (x,y) if
ac - b2 = 0
and rank
[~ ~
:]
=2
at (x, y).
(d) Equation (1) is said to be elliptic (hyperbolic, parabolic) in is elliptic (hyperbolic, parabolic) at all (x, y) E n.
(1.2.2c)
n c rn.2 if it
1.2 Classification of Second-Order Equations into Types
5
Occasionally the parabolic type is defined only by ac - b2 = O. But one would not want to call the equation Ua:a: (x, y) + Ua: (x, y) = 0 parabolic, nor indeed the purely algebraic equation u(x,y) = O. Example 1.2.2. The potential equation (1.6) is elliptic, the wave equation (1.7) is of hyperbolic type, while the heat equation (1.10) is parabolic. The definition of types can easily be generalised to the case where more than two independent variables occur, say xl, ... ,Xn . The general linear differential equation of second order in n variables x = (Xl,"', Xn) reads n
n
l : ~;(X)Ua:.a:j i,;=l
+ l:ai(X)Ua:. + a(x)u =
f(x).
(1.2.3)
i=l
Since ua:.a:j = ua:ja:. holds for twice continuously differentiable functions, one can assume that, without loss of generality, (1.2.4a) Thus, the coefficients
~;(x)
define a symmetric n x n matrix A(x) = (~;(X»~;=l
(1.2.4b)
which therefore has only real eigenvalues. Definition 1.2.3. (a) Equation (3) is said to be elliptic at x if all n eigenvalues of the matrix A(x) have the same sign (±1) (Le. if A(x) is positive or negative definite). (b) Equation (3) is said to be hyperbolic at x if n - 1 eigenvalues of A(x) have the same sign (±1) and one eigenvalue has the opposite sign. (c) Equation (3) is said to be parabolic at x if one eigenvalue vanishes, the remaining n - 2 eigenvalues have the same sign, and rank(A(x), a(x» = n where a(x) = (al(x),···, ~(x»T. (d) Equation (3) is said to be elliptic in n c m,n if it is elliptic at all x E n. Definition 3 makes it clear that the three types mentioned by no means cover all cases. An unclassified equation occurs, for example, if A(x) has two positive and two negative eigenvalues. In place of (3) one can also write
Lu=f,
(1.2.5)
where (1.2.6) is a linear differential operator of second order. The operator
6
1 Partial Differential Equations and Their Types n
L Q,;,j 8 j 8Xi8xj, 2
i,j
which contains only the highest derivatives of L, is called the principal part of L. Remark 1.2.4. The ellipticity or hyperbolicity of Eq. (3) depends only on the principal part of the differential operator. Exercise 1.2.5. (Invariance of the type under coordinate transformations) Let Eq. (3) be defined for x E il. The transformation 4"': il c lRn -+ il' c lRn is assumed to have a nonsingular Jacobian matrix S = 8{Pj8x E C 1 (il) at x. Prove that Eq. (3) does not change its type at x if it is written in the new coordinates = 4"'(x). Hint: the matrix A = (Q,;,j) becomes SAST after the transformation. Use Remark 4 and Sylvester's inertia theorem (cf. Gantmacher [1, Chapter X-§2]).
e
1.3 Type Classification for Systems of First Order The examples 1.1.8-10 are special cases of the general linear system of first order in two variables. U:J;(X,
y) - A(x, y)Uv(x, y) + B(x, y)u(x, y)
= f(x,y).
(1.3.1)
Here, U = (11.1,"', um)T is a vector function, and A, Bare m xm matrices. In contrast to Section 1.2, A need not be symmetric and can have complex eigenvalues. If the eigenvalues Al, ... , Am are real, and if there exists a decomposition A = S-lDS with D = diag{Alo"', Am}, A is called real-diagonalisable. Definition 1.3.1. (a) System (1) is said to be hyperbolic at (x,y) if A(x,y) is real-diagonalisable. (b) System (1) is said to be elliptic at (x,y) if all the eigenvalues of A(x, y) are not real.
If A is real or possesses m distinct real eigenvalues the system is hyperbolic since these conditions are sufficient for real diagonalisability. A single real equation, in particular, is always hyperbolic. Examples 1.1.1 and 1.1.2 are hyperbolic according to the preceding remark. System (1.13) from Example 1.1.8 has the form (1) with A
=
[0 -1] -1
0
.
It is hyperbolic since A is real-diagonalisable:
11]-1[-10][11] A = [ -1 1 0 1 -1 1 .
1.4 Characteristic Properties of the Different Types
7
The Cauchy-Riemann system (1.14), which is closely connected with the potential equation (1.6), is elliptic since it has the form (1) with A=
[01 -1] 0 '
and A has the eigenvalues ±i. The system (1.15) corresponding to the (parabolic) heat equation can be described as system (1) with
[0 -1]
A= 0
0
.
The eigenvalues (Al = A2 = 0) may be real but A is not diagonalisable. Hence, system (1.15) is neither hyperbolic nor elliptic. A more general system than (1) is Al1lx
+ A2Uy + Bu = f.
(1.3.2)
If Al is invertible then multiplication by All gives the form (1) with A
=
-All A2. Otherwise one has to investigate the generalised eigenvalue problem det(AAl + A2) = O. However, system (2) with singular Al cannot be elliptic, as can be seen from the following (cf. (4) with ~l = 1,~2 = 0). A generalisation of (2) to n independent variables is exhibited in the system (1.3.3) with m x m matrices Ai = Ai(X) = Ai(xl,'" ,xn ) and B = B(x). As a special case of a later definition (cf. Sect. 12.1) we obtain the Definition 1.3.2. The system (3) is said to be elliptic at x if for any vector '~n) E IRn the following holds:
o=f ({l,'"
n
det(LAi(x)~i) =f O.
(1.3.4)
i=l
1.4 Characteristic Properties of the Different Types The distinguishing of different types of partial differential equations would be pointless if each type did not have fundamentally different properties. When discussing the examples in Sect. 1.1 we already mentioned that the solution is uniquely determined if initial values and boundary values are prescribed.
8
1 Partial Differential Equations and Their Types
In Example 1.1.2 the hyperbolic differential equation (1.2) is augmented by the specification (1.4) of u on the line y = const (see Fig. 1a). In the case of the hyperbolic wave equation (1.7), uti must also be prescribed (cf. (1.9» since the equation is of second order.
(a)
"u~IJ (b)
u, uti
u, uti
Figure 1.4.1. (a) Initial conditions and (b) initial boundary conditions for hyperbolic problems
It is also sufficient to give the values u and US/ on a finite interval [Xl. X2] if u is additionally prescribed on the lateral boundaries of the domain n of Fig. lb. This prescription of initial boundary values occurs, for example, in the following physical problem. A vibrating string is described by the lateral deflection u(x, t) at the point x E [Xl. X2] at time t. The function u satisfies the wave equation (1.7) with the coordinate y corresponding to time t. At the initial instant in time, t = to, the deflection u(x,O) and velocity Ut(x, 0) are given for Xl < X < X2. Under the assumption that the string is firmly clamped at the boundary points Xl and X2, one obtains the additional boundary data u(xl, t) = U(X2' t) = 0 for all t. For parabolic equations of second order one can also formulate initial value and initial boundary value problems (cf. Fig. 2). However, as initial value only the function u(x, Yo) = uo(x) may be prescribed. An additional specification of Uy(x, Yo) is not possible, since uS/ex, yo) = ua:a:(x, yo) = u(j(x) is already determined by the differential equation (1.10) and by Uo.
(a)
_·B· (b)
u
u
Figure 1.4.2. (a) Initial conditions and (b) initial boundary conditions for parabolic problems
The heat equation (1.10) with the initial and boundary values
u(x, to) =uo(x) U(Xb t) =!Pl (t),
[Xl, X2], U(X2' t) = !P2(t)
in
fort>to
(1.4.1)
1.4 Characteristic Properties of the Different Types
9
(cf. Figure 2b) describes the temperature u(x, t) of a wire whose ends at x = Xl and x = X2 have the temperatures CPl(t) and CP2(t). The initial temperature distribution at time to is given by uo(x). Aside from the different number of initial data functions in Figure 1 and Figure 2, there also is the following difference between hyperbolic and parabolic equations.
n
Remark 1.4.1. The shaded area in Figure 1 corresponds to t > to, and in Figure 2 to y > Yo. For hyperbolic equations one can solve in the same way initial-value and initial-boundary-value problems in the domain t :5 to, whilst parabolic problems in t < to generally do not have a solution.
°
If one changes the parabolic equation Ut - U xx = to Ut + U xx = 0, the orientation is reversed: solutions exist in general only for t :5 to. For the solution of an elliptic equation, boundary values are prescribed (cf. Example 1.1.3, Figure 3). A specification such as that in Figure 2b would not uniquely determine the solution of an elliptic problem, while the solution of a parabolic problem would be overdetermined by the boundary values of Figure 3a.
(a)
(b)
U
Figure 1.4.3. Boundary value conditions for an elliptic problem
An elliptic problem with specifications such as in Figure 1b in general has no solution. Let us, for example, impose the conditions U(x, 0) = u(O,y) = u(1,y) = and Uy(x,O) = Ul(X) on the solution of the potential equation (1.6), where Ul is not infinitely often differentiable. If a continuous solution U existed in n = [0,1] x [0,1] one could develop U(x, 1) into a sine series and the following exercise shows that Ul would have to be infinitely often differentiable, in contradiction to the assumption.
°
Exercise 1.4.2. Let cP E CO [0, 1] have the Fourier expansion 00
cp(x)
= I>~vsin(v7rx). v=l
Show that: (a) the solution of the potential equation (1.6) in the square n = (0,1) x (0,1) with boundary values u(O,y) = u(x,O) = u(l,y) = and u(x, 1) = cp(x) is given by
°
10
1 Partial Differential Equations and Their Types 00
u(X, y)
= '" .:( ) sin(v7rx) sinh(v7rY). ~sm V7r
(b) For 0 ~ x ~ 1 and 0 ~ y < 1, u(x, y) is differentiable infinitely often. Hint: !(x) = Ef3vsin(V7rx) E 0 00 [0, I], if v-too lim f3vv" = 0 for all k E IN. Conversely it does not make sense to put boundary value constraints as in Fig. 3a on a hyperbolic problem. Consider as an example the wave equation (1.7) in n = [0, I] x [0,1/7r] with the boundary values u(x,O) = u(O, y) = u(l, y) = 0 and u(x,I/7r) = sin(v7rx) for v E IN. The solution reads u(x,y) = sin(v7rx)sin(v7ry)/sinv. Although the boundary values, for all v E IN are bounded by 1, the solution in n may become arbitrarily large since sup{ 1/ sin v: v E IN} = 00. Such a boundary value problem is called "not well posed" (cf. Definition 2.4.1). Exercise 1.4.3. Prove that the set {sin v: V E IN} is dense in [-1,1]. Another distinguishing characteristic is the regularity (smoothness) of the solution. Let u be the solution of the potential equation (1.6) in n c m? As stated in Example 1.1.3, u is the real part of a function holomorphic in n. Since holomorphic functions are infinitely differentiable, this property also holds for u. In the case of the parabolic heat equation (1.10) with boundary values u(x,O) = uo the solution u is given by (1.11b). For y > 0, u is infinitely differentiable. The smoothness of Uo is of no concern here, nor is the smoothness of the boundary values in the case of the potential equation. One finds a completely different result for the hyperbolic wave equation (1.7). The solution reads u(x, y) = !p(x + y) + tP(x - y), where !p and tP result directly from the initial data (1.9). Check that u is k-times differentiable if Uo is k-times and Ul is (k - I)-times differentiable. As already mentioned in this section, in the hyperbolic and parabolic equations (1.1), (1.2), (1.7) and (1.10) the variable y often plays the role of time. Therefore one calls processes described by hyperbolic and parabolic equations nonstationary. Elliptic equations which only contain space coordinates as variables are called stationary. More clearly than in Definitions 1.2.1b,c the Definitions 1.2.3b,c distinguish the role of a single variable (time) corresponding to the eigenvalue oX = 0 in parabolic equations, and to the eigenvalue with opposite sign in hyperbolic equations. The connection between the different types becomes more comprehensible if one relates the elliptic equations in the variables Xl, ... , Xn to the parabolic and hyperbolic equations in the variables Xl,"· ,xn,t. Remark 1.4.4. Let L be a differential operator (2.6) in the variables x = (Xl! ... ,xn ) and let it be of elliptic type. Let L be scaled such that the matrix A(x) in (2.4b) has only negative eigenValues. Then
1.4 Characteristic Properties of the Different Types Ut
is a parabolic equation for u( x, t)
+ Lu =
0
11
(1.4.2)
= u( Xl, ••• , X n , t). In contrast
Utt
+ Lu =
0
(1.4.3)
is of hyperbolic type. Conversely, the nonstationary problems (2) or (3) lead to the elliptic equation Lu = 0 if one seeks solutions of (2) or (3) that are independent of time t. One also obtains elliptic equations if one looks for solutions of (2) or (3) with the aid of a separation of variables u(x, t) = cp(t)v(x). The results are u(x, t) = e-Atv(x) in case (2), (1.4.4) u(x, t) = e±iv'Xtv(x) in case (3), where v(x) is the solution of the elliptic eigen v al ue problem Lv=>..v.
(1.4.5)
2 The Potential Equation
2.1 Posing the Problem The potential equa.tion from Example 1.1.3 reads Llu=O innclRn ,
(2. 1.1 a)
where Ll = 8 2 18x~ + ... + 82 18x;" is the Laplace operator. In physics, Equation (la) describes the potentials; for example the electric potential when contains no electric charges, the magnetic potential for vanishing current density, the velocity potential, etc. Equation (la) is also called Laplace's equation since it was described by P. S. Laplace in his five-volume work "Mecanique Celeste" (1799-1825). However, it was L. Euler who first mentioned the potential equation in 1752. The connection between the potential equation for n = 2 and function theory has already been pointed out in Example 1.1.3. Not only is the Laplace operator an example of an elliptic differential operator, but it actually is the prototype (the so-called normal form). By using a transformation of variables, any elliptic differential operator of second order can be transformed so that its principal part becomes the Laplace operator (cf. Hellwig [1, 1I-§1.5]).
n
In the following
n will always be a domain.
Definition 2.1.1. The region connected. 1
n c lRn is called a domain if n is open and
The existence of a second derivative of u is required only in the boundary
n,
not on
r=8n of n. For a prescription of boundary values u=
onr
to be meaningful, one has to assume continuity of u on considerations lead to 1
(2.1.1b)
n = n u r. These
n is called connected if for any x,y E n there is a continuous curve within n that connects x and y, that is, a -y: S E [0,1] ...... "Y(s) E n continuously with
"Y(O) = x, "Y(1) = y.
W. Hackbusch, Elliptic Differential Equations: Theory and Numerical Treatment, Springer Series in Computational Mathematics 18, DOI 10.1007/978-3-642-11490-8_2, © Springer-Verlag Berlin Heidelberg 2010
12
2.1 Posing the Problem
13
Definition 2.1.2. The function u is said to be harmonic in fl if u belongs to <J2{fl) n CO{fi) and satisfies the potential equation (1a). Here CO{D) [Ck{D), COO (D)] denotes the set of continuous [k-fold continuously differentiable, infinitely often differentiable] functions on D. In general one should not expect that the solution of (1a,b) lies in C2(fi), as we show in the following example. Example 2.1.3. Let fl = (O, 1) x (0,1) (cf. Figure 1.4.3a). Let the boundary A solution of the boundary values be given by cp(x,y) = x 2 for (x,y) E value problem exists but does not belong to C2(fi).
r.
PROOF. The existence of a solution u will be discussed in Theorem 7.3.7. If u E <J2(1i), then it follows that u..,..,(x,O) = cp..,..,(x,O) = 2 for x E [0,1]; in particular u..,..,{O, 0) = 2. From the analogous result ullll (O,O) = CP1l1l(0, 0) = 0 one may conclude ..1u(O,O) = 2 in contradiction to ..1u = 0 in fl. _ In the case under discussion one can also show u E C 1 (fi). That this statement is generally false, is shown by Example 2.1.4. In the domain fl = (-!,!) x Figure 1) introduce the polar coordinates
x
= rcoscp,
n
y
(-!,!) \ [O,!) x [O,!) (cf.
= r sin cp.
(2.1.2)
ro ro
Figure 2.1.1. An L-shaped region
The function u(r, cp) = r2/3 sin«2cp-7r)/3) is the solution of the potential equation (1a) and has smooth boundary values on r (in particular u = 0 holds on ro c r) although the first derivatives in r = 0 are unbounded, i.e. u ¢ Cl(fi).
14
2 The Potential Equation
This follows from the fact that, along with U x and u y , the radial derivative Ur
= U x cos
for r
-+
o. In order to check that Llu = 0, use the following.
Exercise 2.1.5. Show that (a) In terms of the polar coordinates (2) in the Laplace operator takes the form
EP
Ll
10 Or
= 8r2 + ;
m.2,
1 02
+ r2 0
(2.1.3)
(b) In terms of the three-dimensional polar coordinates x
= r cos
= rsin
y
z
= rcostjJ
the Laplace operator is given by 02
2 0 1 [ 1 02 + r2 sin2 tjJ 0
Ll = 8r2 + ; or
0 02 ]
+ cot tjJ otjJ + OtjJ2
.
(2.1.4)
Note. In the n-dimensional case the transformation to polar coordinates leads to 02 n -1 0 1 (2.1.5) L l = 2- + - - - + - B 8r r or r 2 ' where the Beltrami operator B contains only derivatives with respect to the angle variables.
2.2 Singularity Function The singularity function is defined by s(x,y) :=
{
- ';2 log Ix -
yl
Ix _ yl2-n (n - 2)wn
for n
= 2,
for n
> 2.
(2.2.1a)
for X,y E IRn , where Wn
= 2r(1/2t /r(n/2)
,
W2
= 27r,
W3
= 47r,
(2.2.1b)
with r the Gamma function, is the surface of the n-dimensional unit sphere. The Euclidean norm of x in IRn is denoted by ..
Ixl
=
1/2
(LX;) .
(2.2.2)
i=1
Lemma 2.2.1. For fixed y E IRn the potential equation in IRn\{y} is solved by s(x,y).
2.2 Singularity Function
15
The proof can be carried out directly. It is simplest to introduce polar coordinates with Y as origin and to use (1.5), since s(x, y) depends only on r= Ix-yl. For the next theorem we need to introduce the normal derivative 8/ 8n. Let nbe a domain with smooth boundary r. Let n(x) E m.n denote the outer normal direction at x E r, i.e. n is a unit vector perpendicular to the tangential hyperplane at x and points outwards. The normal derivative of u at x E r is defined as
8u(x)/8n = (n(x), Vu(x»,
(2.2.3a)
where (2.2.3c) is the gradient of u and n
(x,y)
= 2:XiYi
(2.2.3c)
i=l
is the scalar product in m.n . In the case of the sphere KR(y) (cf. (7» the normal direction is radial, and Bu/8n becomes 8u/8r with respect to r = Ix - yl, if one uses polar coordinates with the origin at y. It follows from 8s(x, y)/8r = -Ix - yll-n /wn , that
8s(x,y)/8n = _R1-
n
/w n
for x E 8KR (y).
(2.2.4)
The first Green formula reads (cf. Green [1])
r uL1vdx = - lnr (Vu, Vv) + lan r u 88vn dr
ln
(2.2.5a)
and holds for u E Clem, v E C2(Q) if the domain n satisfies suitable conditions. Here Ian' .. dr denotes the surface integral. Domains for which Equation (5a) holds are called normal domains. To see sufficient conditions for this refer to Kellogg [1, Chapter IV) and Hellwig [1, 1-1.2]. Functions u, v E C 2 (Q) in a normal domain n satisfy the second Green formula (2.2.5b)
Theorem 2.2.2. Let n be a normal domain, and let u there. Then u(y)
E
C2(Q) be harmonic
= far [s(X, y) :n u(x) - u(x) 8~ s(x, y)] drx
for all yEn. Here 8/8nx and drx refer to the variable x.
(2.2.6)
16
2 The Potential Equation
PROOF. By
(2.2.7) we denote the ball with centre y and radius r. Since the singularity function s(·,y) is not differentiable on x = y, Green's formula (5b) is not directly applicable. Let n" := n\K,,(y), with € so small that K,,(y) en. Since n" is again a normal domain, it follows from Llu = Lls = 0 (cf. Lemma 1) and (5b) with v = s that
r
Jan"
[u(X)aa s(x,y) - s(X'Y)aa U(X)] drx = nx n
o.
(2.2.8a)
We have an" = anu aKE(y). However, at x E aKE(y), the normal directions of an" and of aK,,(y) differ in their signs. The same holds for the normal derivatives, so that the integral in (8a) can be decomposed into fan • ... = Jan' •. - J8K.(y) . ". The assertion of the theorem would be proved if we could show that J8K.(y)··· - -u(y) for € - O. The normal derivative au/an is bounded on aK,,(y), and J8K.(Y) s(x,y) dr converges like O(€llog€1) or O(€) towards zero, as can be seen from (1) and f8K.(y) dr = €n-1 Wn . Thus, we obtain
r
s(X'Y)aa u(x)drx- O
JOK.(y)
From J8K.(t/) dr
= €n-1 Wn
r
(2.2.8b)
n
and (4) one infers
u(y)aa s(x,y)drx = -u(y).
nx
JOK.(y)
(2.2.8c)
The continuity of u in y yields
r
J8K.(y)
(u(x)-u(y»aa s(x,y)drx :$ max lu(x) - u(y)l- 0 as € - O. xE8K.(y) nx (2.2.8d)
Equations 8b,c,d show that
J8K.(y)
ruins - sinuJdr -
that (8a) proves the theorem.
-u(y) (€ - 0), so •
Any function of the form
1'(x, y) = s(x, y) + ~(x, y)
(2.2.9)
is called a fundamental solution, of the potential equation, in n if for fixed yEn the function ~(.,y) is harmonic in n and belongs to C2(il).
Corollary 2.2.3. Under the conditions of Theorem 2 for every fundamental solution in
n the following holds:
2.3 The Mean Value Property and Maximum Principle
u(y) = Ian ['Y(X, y) :n u(x) - u(x)
8=x 'Y(x, y)] drx.
PROOF. (5b) implies fan[iP8uj8n - u8iPj8n] dr
17
(2.2.10)
= o.
•
For a possible weakening of the condition iP = 'Y - s E C 2 (ll) for iP(·, y) E
Cl(ll) n C 2 (il) refer to Hellwig [1, 1-1.4]. Exercise 2.2.4. Let il = KR(Y)' Define 'Y(x, e)
={
(nJ)wn [IX - el 2- n - (KJPlx - (,I)
-2~
[log Ix - el-Iog
(~.rllx -
2-n]
ell)]
for n > 2 for n
(2.2.lla)
=2
with x, e E il, e' = Y + R 2 1e - yl- 2(e - y) and show: (a) 'Y is a fundamental solution in il, (b) 'Y(x, e) = 'Y(e, x), (c) on the surface r = 8KR (y) the following holds:
~'Y(x,e) = ~'Y(e,x) = 8n{
8n{
__I_(R2 -Ix - yl2)lx _ €I-n
Rwn
(e E r).
(2.2.llb)
2.3 The Mean Value Property and Maximum Principle Definition 2.3.1. A function u has the mean value property in il if u E CO(.a) and if for all x Eiland all R > 0 with KR(X) c il the following equation holds
(2.3.1) Since f8KR(X) dr = wn~-l the right side in (1) is the mean value of u taken over the surface of the sphere. An equivalent characterisation results if one averages over the sphere KR(X).
u E Co(ll) has the second mean value property in il if fKR(X) u(e) d€ for all x E il, R> 0 with KR(X) C il. Show that
Exercise 2.3.2.
u(x)
= R,!lwn
this mean value property is equivalent to the mean value property (1). Hint:
fKR(X) u(e) d€ = fOR faKr(x) u(e)dr{ dr. Functions with the mean value property satisfy a maximum principle, as is known from the function theory for holomorphic functions:
18
2 The Potential Equation
Theorem 2.3.3. (Maximum-minimum principle) Let n be a domain and let U E CO(n) be a nonconstant function which has the mean value property. Then u takes on neither a maximum nor a minimum in n.
PROOF. (a) It suffices to investigate the case of a maximum since a minimum of u is a maximum of -u, and -u also has the mean value property. (b) For an indirect proof we assume that there exists a maximum in yEn: u(y)
=M
~
u(x)
for all x E n.
In (c) we will show u(y') = M for arbitrary y' En, i.e. u == M in contrast to the assumption u ¢ const. (c) Proof of u(y') = M. Let y' E n. Since il is connected, there exists a path connecting y and y' running through n, i.e. there exists a continuous cp: (0, 1]- n with cp(O) = y, cp(l) = y'. We set
I := {s
E
[0, l]:u(cp(t»
=M
for all 0 ~ t ~ s}.
I contains at least 0, and is closed since u and cp are continuous. Thus there exists s* = max{s E I}, and the definition of I shows that I = [0, s*]. In (4) it is proved that s* = 1 so that y' = cp(l) E I and hence u(y') = M follows. (d) Proof of s* = 1. The opposite assumption s* < 1 can be made and shown to be contradicted by proving that u(x) = M in a neighbourhood ofx* := cp(s*). Since x* E il, there exists R > 0 with KR(X*) C n. Evidently, it follows that u = M in KR(X)* if it is shown that u = M on aKr(x)* for all 0 < r ~ R. (e) Proof of u = M on aKr(x*). Equation (1) in x* reads M
= u(x*) =
r
u(~) drd(wnrn-l).
JaKr(x*)
In general we have u(~) ~ M . If one had u(~') < M for~' E aKr(x*) and thus also u < M in a neighbourhood of ~', one would have on the right side a mean value smaller than M. Therefore, u = M on aKr(x*) has been proved. • Simple deductions from Theorem 3 are contained in Corollary 2.3.4. Let n be bounded. (a) A function with the mean value property takes its maximum and its minimum on an. (b) If two functions with the mean value property coincide on the boundary an, they are identical.
n
PROOF. (a) The extrema are assumed on the compact set = n U an. According to Theorem 3, the extremum cannot be in n if u is not constant on a connected component of n. But in this case the assertion is also obvious. (b) If u and v with u = v satisfy the mean value property on an then the latter is also satisfied for w := u - v. Since w = 0 on an, part (a) indicates that maxij w = mini) W = O. Thus u = v in n. •
2.3 The Mean Value Property and Maximum Principle
19
Lemma 2.3.5. Harmonic functions have the mean value property.
PROOF. Let u be hannonic in nand Y E KR(y) c n. We apply the representation (2.6) for n = KR(Y). The value s(x,y) is constant on8KR(Y): let it be denoted by a(R). Because of (2.4), Equation (2.6) becomes
The equation agrees with (1) if the first integral vanishes. The latter follows from Lemma 2.3.6. Let u E 0 2 (0) be harmonic in a normal domain we have 8u -8 dr=o. ail n
i
n.
Then
(2.3.2)
PROOF. In Green's formula (2.5a) substitute 1 and u for u and v respectively.
•
Lemma 5, Theorem 3, and Corollary 4 together imply Theorems 7 and 8:
Theorem 2.3.7. (The maximum-minimum principle for harmonic functions) Let u be harmonic in the domain n and nonconstant. There exists no maximum and no minimum in n. Theorem 2.3.8. (Uniqueness). Let n be bounded. A function harmonic in n assumes its maximum and its minimum on 8n and is uniquely determined by its values on 8n. The representation (1) of u(y) by the values on 8KR(Y) is a special case of the following formula which will be proved at the end of this section and which provides Equation (1) for x = y.
Theorem 2.3.9. (Poisson's integral formula) Assume we have
E
(2.3.3) is given by the function
u(x)
= R2 -Ix RMln
YI2
r laKR(Y)
Ix - ~In
which belongs to OOO(KR(y)) n CO(KR(Y) ).
dre for x E KR(y),
(2.3.4)
20
2 The Potential Equation
The mean value property only assumes 1.£ E COCa), while harmonic functions belong to C2(n)nCO(.a). This makes the following assertion surprising: Theorem 2.3.10. A function is harmonic in n if and only if it has the mean value property there.
PROOF. Because of Lemma 5 it remains to be shown that a function v with the mean value property is harmonic. Let x E KR(X) C n be given arbitrarily. According to Theorem 9 there exists a function 1.£ harmonic in K R(X) with
According to Lemma 5, U has the mean value property, as does v, and Corollary 4b proves that 1.£ = v in KR(X), i.e. v is harmonic in KR(X), Since KR(X) C n is arbitrary, v is harmonic in n. • An important application of Theorem 10 is Theorem 2.3.11. (Harnack) Let 1.£1. 1.£2, ••• be a sequence of functions harmonic in n and converging uniformly in li. Then u = limk-+oo Uk is harmonic inn.
PROOF. The limit function is continuous: 1.£ E CO(li). The limit process, applied to Uk(X) = I8KR(x) Uk (e) drd(wnR!'-l) yields Equation (1) for Uj i.e. 1.£ has the mean value property. According to Theorem 10, 1.£ is also harmonic inn.
•
Theorems 3 and 7 on the maximum-minimum principle pertain to global extrema. The proof of Theorem 3 does not yet exclude local extrema in the interior. It merely shows that 1.£ is then always constant in a circle KR(Y) C n. As is known from function theory one can derive from this u(x) = M in :0, if 1.£ is analytic, i. e., there is a convergent power series expansion in a neighbourhood of any x En. Indeed, the following theorem holds whose proof can be found, for example, in Hellwig [1,III-§1.5]: Theorem 2.3.12. A function harmonic in
n
is analytic there.
The proof of the Poisson formula still needs to be carried out. (a) First we must show that u in (4) is a function harmonic in KR(y), i. e., it satisfies ..11.£ = O. Since the integrand is twice continuously differentiable and the domain of integration r = 8KR(Y) is compact, the Laplace operator commutes with the integral sign:
Llu(x)
= (Rwn)-l
1
el-n] dre
for x E KR(Y)'
(2.3.5)
2.3 The Mean Value Property and Maximwn Principle
21
According to Exercise 2.2.4 there exists a fundamental solution 'Y(x, e) such that
eE r,x E KR(Y)· (2.3.6) From Llx~ = ~Llx'Y(x, e) = 0 and (5), one infers that Llu = O. (b) The expression (4) defines u(x) at first only for x E KR(Y)' It still needs to be shown that u has a continuous extension on KR(Y) = KR(Y) u r (Le., u E CO(KR(Y))) and that the continuously extended values agree with the boundary values cp:
lim
u(x)
x-+z
= cp(z)
for
Z E
r.
(2.3.7)
xEKR(Y)
By Equation (6), putting u R2
Let
Z E
r
-Ix _Y12 f
== 1 in Corollary 2.2.3 gives the identity
Jr Ix -el-ndr~ = 1
Rwn
for all
x E KR(Y)'
(2.3.8)
be arbitrary. Due to Equation (8) one can then write: (2.3.9a)
R
Figure 2.3.1. Construction of ra, n
We define ro = r n Kp(z), n = r\ro (see Figure 1) and split the expression (9a) into u(x) - cp(z) = 10 + 11. where 1. = R2 ,
-Ix - YI2 f cp(e) - cp(z) dJ', Rwn Jr. Ix-eln ~,
i=O,l.
2 The Potential Equation
22
Since
Iirof
dr~1 :S max l
re~ l
dr~
iro Ix - el n
~~In,
it follows from Equation (8) that (2.3.9b) Because of the continuity of
> 0 such that for given 10 > 0 (2.3.9c)
10 :S 10/2. Set C
m~Er
l
1 zi :S 0(10) := 210 (p)n 2 4G.
and Ix - zl :S p/2. The last inequality implies Ix - el ~ ~
for
eE Tt
(see Figure 1).
Together with R2_lx_yI 2 = (R+lx-yl)(R-lx-yl) :S 2R(R-lx-yl) :S 2Rlz - xl :S 2RO(€) one obtains
I
f
= R2 -Ix &In
yI 2
irl
II11:S
10
2·
(2.3.9d)
Thus for every 10 > 0 there exists a 0(10) > 0 such that Ix - zl :s:; oCt:) implies the estimate Iu(x) -
r
2.4 Continuous Dependence on the Boundary Data
23
2.4 Continuous Dependence on the Boundary Data Definition 2.4.1. An abstract prdblem of the form A(x)
= y,
x E X, Y E Y,
is said to be well-posed if for all y E Y it has a unique solution x E X and if the latter depends continuously on y. It is important to recognise whether a mathematical problem is well-posed since otherwise essential difficulties may occur in its numerical solution. In the case of the boundary value problem (1.1a,b) Xc C 2 (n) nCO(n) is the space of functions harmonic in nand Y = CO (r) is the set of continuous boundary data on r = an. The topologies of X and Yare given by the supremum norms:
lIulioo := sup lu(x)1 xED
and
1I
(2.4.1)
xEr
The question of the existence of a solution of (1.1a,b) will have to be postponed (see Section 7). The uniqueness, however, has been confirmed already in Theorem 2.3.8, if n is bounded. That the boundedness of n cannot be dropped without further ado is shown in
Example 2.4.2. The functions U(X1.X2) = Xl in n = (O,oo) x m, u(X1. X2) = log{x~ + x~) in n = m,2\Kl{O) U{X1.X2) = sin{xl)sinh(x2) in n = (O,7r) X (O,oo)
(2.4.2a) (2.4.2b) (2.4.2c)
and also the trivial u = 0, are solutions of the boundary value problem Llu = 0 in n, u = 0 on r = an. For bounded n the harmonic functions {solutions of (1.1a,b)) depend not only continuously but also Lipschitz-continuously on the boundary data:
Theorem 2.4.3. Let
n be bounded. If uI and uH
are solutions of
then
(2.4.3) PROOF. v := u I - u ll is a solution of Llv = 0 in n and v =
24
2 The Potential Equation
-
The definition (1) of 11·1100 implies (3).
If lI
3 shows that the associated solutions satisfy lIu n - ulloo -
Theorem 2.4.4. Let D be bounded. Let
PROOF. Since according to Theorem 3 lIun - umll oo ~ lI
b)
c)
Figure 2.4.1. Approximation of a by a' and
a"
Let iln be a sequence of domains with rn = aDn. We say that rn - r if dist(rn • r) - O. Here, dist(rn , r) := sup{dist(x, r) : x Ern}, dist(x, r) := inf{lx - yl: Y E r}. Further, one must specify when
0, there exist numbers N(E) and O(E) > 0 such that the following implication holds: n ~ N(E),
x E r,
Y Ern,
Ix-YI ~ O(E) => l
f.
The sequence Un E CO(Dn) converges uniformly to u E CO(D) if
(2.4.4a)
2.4 Continuous Dependence on the Boundary Data
lim sup{lun(x) - u(x)l: x E iln n il}
n-+oo
= O.
25
(2.4.4b)
Remark 2.4.5. (a) Let K be a set which is compact (Le. complete and bounded) with r C K, and rn C K for all n. Let 4>n E COCK) converge uniformly on K to 4>. If CPn = 4>n on rn and cP = 4> on r then (4a) is satisfied. (b) Let iln C il hold for all n and let Un be the following (not continuous) continuation of Un onto fi: Un = Un on fin, Un = u on fi\li n . Then (4b) is aquivalent to uniform convergence Un -+ u on fi in the usual sense. Theorem 2.4.6. Let fln C il, with fl bounded, and let rn functions Un, which are harmonic in iln , be solutions of Llun = 0
in iln ,
Un
= CPn
-+
on rn.
r. Let the (2.4.5a)
Let CPn E CO(rn ) converge uniformly in the sense of (4a) to cP E CO(r). Then the following assertions hold: (a) if there exists a solution u E C2(fl) n cO(fi) of Llu
=0
in il,
u
= cP
on r
(2.4.5b)
then Un -+ U holds in the sense of (4b). (b) if conversely Un -+ U E co(n) is satisfied in the sense of (4b), then u is the solution of (5b).
PROOF. (a) Let the continuation Un be defined as in Remark 5b. Since u is uniformly continuous on n, there exists liu (f) > 0 for all f > 0 such that (2.4.6a) Set Ii* (f) := min {liu (~) , Ii (~)} with Ii from (4a). Because rn -+ r there exists Nr(f), so that dist(rn , r) S 8*(f) for n ~ Nr(f). For n ~ N*(f) := max{Nr(f), N(f/2)} (N from (4a» we want to show that Iun(x) -u(x)1 S f, for x E For x E the estimate is trivial because Un(x) = u(x). For all x E fin C il, however, there holds
n.
n\nn
IUn(x) - u(x)1
= lun(x) - u(x)1 S
IIJ,~ IU n - ul
(2.4.6b)
(cf. Theorem 2.3.8), because Un -u is harmonic in iln . It remains to estimate IUn(x) -u(x)1 for x Ern. For x Ern with n ~ N*(f) there exists y E r with
Ix - yl S li(f/2). Thus we obtain lun(x) - u(x)1
= ICPn(x) -
u(x)1 S ICPn(x) - cp(y)1 + Icp(y) - u(x)1 S
S ICPn(x) - cp(y)1 + lu(y) - u(x)1
s 2f + 2f = f
from (4a) and (6a). Since x E rn is arbitrary it follows that IUn -ul Sf on rn, and (6b) proves the uniform convergence Un -+ u on fi. Hence, from Remark 5b it follows that (4b) is satisfied.
26
2 The Potential Equation
(b) Let K C Q be a compact set. Since r .. --+ r there exists a N(K) such that K C Q.. for n ~ N(K). Thus, the sequence {u .. : n ~ N(K)} converges uniformly in the usual sense on K to u so that one can apply Theorem 2.3.11: consequently u is harmonic in K. Since K C Q may be chosen arbitrarily, it follows that u E C2(Q). By assumption, we already have u E c°(.?i). That the boundary value u =
r.. =
=
'Pn
= 1 on 8K1/ .. (0),
=1
satisfy the condition
Remark 2.4.7. In Q = K1(0)\{0} c m.2 the potential equation has no solution u E C 2 (Q) n CO(n) which assumes the boundary values u(x) = 0 on 8K1 (0) and u(x) = 1 in x = O.
3 The Poisson Equation
3.1 Posing the Problem The Poisson equation reads Llu
= f in il
(3.1.la)
with given f E CO(il). In the physical interpretation f is the source term [for example, the charge density in the case of an electrical potential u). To determine the solution uniquely one needs a boundary value specification, for example, the Dirichlet condition u = cp on F.
(3.1.lb)
Definition 3.1.1. The function u is called the classical solution of the boundary value problem (la,b) if u E C2(il) n CO(D) satisfies the equations (la,b) pointwise. Until we introduce weak solutions in Section 7, "solution" will always mean "classical solution" . The solution of the boundary value problem (la,b) will in general no longer satisfy the mean value property and the maximum principle. But these properties still hold for the differences of two solutions Ul and U2 of the Poisson equation, since Ll(Ul -U2) = f - f = o. Thus the uniqueness of the solution of problem (la,b) immediately follows and Theorem 2.4.3 can be brought over: Theorem 3.1.2. Let il be bounded. (a) The solution of (la,b) is uniquely determined. (b) If u I and u II are solutions of the Poisson equation for boundary values cpI and cpII, then we have (3.1.2)
PROOF. (b) The proof of Theorem 2.4.3 can be repeated verbatim here. (a) If u I and u II are two solutions of (la,b) then (2) shows that \luI -uIIlloo ::; IIcp - cplloo = o. Thus u I = u II . • Theorems 2.4.4 and 2.4.6 can be transferred likewise.
W. Hackbusch, Elliptic Differential Equations: Theory and Numerical Treatment, Springer Series in Computational Mathematics 18, DOI 10.1007/978-3-642-11490-8_3, © Springer-Verlag Berlin Heidelberg 2010
27
28
3 The Poisson Equation
3.2 Representation of the Solution by the Green Function Lemma 3.2.1. Let the solution of (1.1a,b) belong to C2(n), where normal domain. Then u may be represented as
u(x) =
-In
1(e,x)f(e) de +
l
n
is a
[1(e,X) :n u(e) - u(e) a!e 1(e,X)] dre (3.2.1)
for every fundamental solution 1 in (2.2.9). The proof is the same as in Theorem 2.2.2 or in Corollary 2.2.3. The term = n\KE(x) becomes IOe 1f £leo Since the singularity of 1(e, x) is integrable at e = x, IO e 1 f de converges to 10 1/ £le as € -+ o.
IOE1..:1ude with n E
Exercise 3.2.2. (a) Let n c lRn be bounded, Xo E n, f E CO(n\{xo}), and I/(x)l :::; Clx - XoI- s for s < n. Show that 10 /(x) dx exists as an improper integral. (b) Let n c rn.n be bounded and let Xo(e) E n depend continuously on ED, with D compact. Let f(x, e) be continuous in (x, e) E .0 x D with x -=I xo(e) and let 1/(x,e)l:::; Clx-Xo(e)I-S, s < n. Show that F(e):= 10 f(x,e) dx is continuous: FE CO(D).
e
In the boundary integral in (1) one may replace u(e) bY'P(e) (cf. (l.1b». The function auj an on r , however, is unknown and cannot be specified arbitrarily either, since the boundary values (1.1b) already determine the solution uniquely (cf. Theorem 3.1.2). To make Ir18ujande vanish one must select the fundamental solution so that 1(e, x) = 0 for e E r, x E n. Definition 3.2.3. A fundamental solution 9 in (2.2.9) is called a Green function (of the first kind) if g(e,x) = 0 for all E r, x En.
e
The existence of a Green function is closely related to the solvability of the boundary value problem for the potential equation: Remark 3.2.4. The Green function exists if and only if for all x E n the boundary value problem .14> = 0 in nand 4> = -s(·, x) on has a solution 4> E C 2 (n).
r
The above consideration results in Theorem 3.2.5. Let n be a normal domain. Let the boundary value problem (1.1a,b) have a solution u E 02(.0). Assume the existence of a Green function of the first kind. Then one can express u explicitly by
3.2 Representation of the Solution by the Green Function
u(x)
=-
r g(e, x)/(e) d(e) -1rune cp(e) !.l0 g(e, x) dre.
in
29
(3.2.2)
In the following we reverse the implication. Let the existence of the Green function be assumed. Then, does function u defined by Equation (2) represent the classical solution of the boundary value problem (1.1a,b)? Here it must be proved, in particular, that u E C2(il) and Llu = I. Firstly, it is not even clear yet whether the function u(x) defined by Equation (2) depends continuously on x since the definition of a fundamental solution 'Y(e, x) does not require continuity with respect to the second argument x. Despite that, the Green function is g(e, x) with respect to x in C2(il\ {e})' as the following result shows (cf. Leis [1, p. 67)). Exercise 3.2.6. Let il be a normal domain. Let the Green function exist for il, and for fixed y E il let g(.,y) E C2(il\{y}) (weaker conditions are possible I). Now prove that g(x,y)
= g(y,x)
(3.2.3)
for X,y E il.
Hint: Apply the Green formula (2.2.5b) with ilE = il\[KE(x') U KE(X")], x',x" E il, u(x) := g(x,x'), v(x) := g(x, x"), and use (2.2.10). If one tries to reverse the assertion of Theorem 5, one encounters the surprising difficulty of having to set precise conditions on the source term I. The natural requirement 1 E C°(il) is necessary for u E C2(ll), but it is not sufficient, as the following theorem, whose proof will be appended at the end of this section, shows.
Theorem 3.2.7. Even il the boundary r and the boundary values cp are sufficiently smooth and il the Green function exists, there are lunctions 1 E C°(11) to which no solutions u E C2(ll) correspond. Theorem 7 shows that Equation (2) need not represent a classical solution for 1 E CoO?). However, a sufficient condition for 1 to do so is Holder continuity. Definition 3.2.8. 1 E CO(ll) is said to be Holder continuous in II with the exponent>. E (0,1) if there exists a constant C = CU) such that I/(x) - l(y)1 ~ Clx - yl>'
for all x, YEll.
(3.2.4a)
We write 1 E C>'(Q) and define the norm II/lIc>.(.Q) as the smallest constant C which satisfies (4a):
_._
II/lIc>.(o) .- sup
{I/(X) - l(y)l. Ix _ yl>'
. x, Y E il, x
}
:f y .
(3.2.4b)
30
3 The Poisson Equation
The function 1 E Ck(n) is said to be k-fold Holder continuously differentiable in n (with the exponent 'x), if DV 1 E CA(n) for alllvi ~ k. Here v
= (Vb' •• ,vn ) with Vi
E
Z,
Vi
~
0,
Ivi
= Vl + ... + Vn
(3.2.5a)
is a multi-index and (3.2.5b) a lvi-fold partial derivative operator. The k-fold HOlder continuously differentiable functions form the linear space ck+A(n) with the norm (3.2.4c)
= k + ,x
one also writes C 8 (n) for ck+A(n). The k-fold Lipschitz continuously differentiable functions 1 E Ck,l(n) are the result of the choice ,x = 1 in (4a, b). For reasons of completeness let us add that
If s
(3.2.4d) is the norm in Ck(n) for integer k ~ O. Exercise 3.2.9. (a) 1 is said to be locally Holder continuous in il iffor each x E n there exists a neighborhood Ke(x) such that 1 E CA(Ke(x) nil). Prove that if n is compact then 1 E CA(n) follows from the local Holder continuity in n. Formulate and prove corresponding statements for ck+A(n) and Ck,l(n). (b) Let s > O. Show that IxI8 E C 8 (KR (0», if s ¢ IN, otherwise IxI8 E C 8 - l ,l(KR (0». Hint: 1 - t 8 ~ (1 - t)B for 0 ~ t ~ 1, s ~ O. The function u from Equation (2) can be decomposed into Ul +U2 where Ul = - J(J gl cJe and U2 = - Jr <pagl an dr; U is the solution of the boundary value problem (1.1a,b) if we are able to show that Ul and U2 are solutions of .dUl
= 1 in
n,
Ul
= 0 on r,
.du2
= 0 in il,
U2
=
Theorem 3.2.10. If the Green function exists and satisfies suitable conditions then U(x)
= - f
is a classical solution of .du = 0 in il, with U
(3.2.6)
=
The proof goes in principle just as for Theorem 2.3.9 (cf. Leis [1, p.69]).
3.2 Representation of the Solution by the Green Function
31
Theorem 3.2.11. Suppose the Greenfunctiong(·, x) E C2(n\{x}) for x E
n
exists, and let it be f E c>.(n). Then u(x) =
-10 f(t;)g(t;, x) df.
(3.2.7)
is a classical solution of Llu = f in n, with u = 0 on r. PROOF. The boundary condition u(x) = 0 for x E r follows easily from g(x, t;) = 0 and (3). The property u E C 1 (n) and the representation uz.(x) = - fn!(t;)gz.(t;,x)df. result from the
n n:
m.
n be bounded and A := {(t;, x) E Exercise 3.2.12. Let n c x t; =I- x}. For the derivatives of f with respect to x assume D~f E CO(A) and ID~f(t;,x)1 ~ Clx - t;1-s with s < n for any Ivl ~ k. Prove that then F(x) := fn f(t;, x) df. E Ck(!]) and DV F(x) = fn D~f(t;, x) df., Ivl ~ k.
To prove u E C2(n) this step cannot be repeated since gz.Zj (t;, x) = O(It; - xl- n ) has a singularity which is not integrable. We write the derivative in the form
uz.(x) = -
fa
[f(t;) - !(x)]gz.(t;, x) df. - f(x)
fa
gz.(t;, x) df..
(3.2.8a)
Let 8j F(x) be the difference quotient (F(xE ) - F(x»/f., with xj = Xj + f. and x~ = Xi for i =I- j. The product rule 8j (FG)(x) = G(x)8j F(x) + F(xE )8j G(x) applied to Equation (8a) gives
8j u z• (x) = -
kf [f(t;) -
f(x E)]8j gz• (t;, x) df. - f(x E)8j 88
f
~Jn
g(t;, x) df..
Since [f(t;) - f(x)]gz,Zj (t;, x) = O(It; - xl>.-n) is integrable, the limit f. -+ 0 results in the formula
uz.zj(x)=-
kf l!(t;)-f(x)]gz.Zj(t;,x)df.-f(X)8~8;~kf g(t;,x)df.. (3.2.8b)
Equation (8b) implies
Llu =
10 [f(t;) - f(x)]Llg df. - f(x)Ll 10
9 df. = - f(x)Ll
10 g(t;, x) df.,
so that it remains only to show that Llfngdf. = -1. Choose KR(Z) so that x E KR(Z) c n. The Green function has the form (2.2.9): 9 = s+!P. The first two terms in
f g(t;, x) df. = f g(t;,x)df.+ f !p(t;,x)df.+ f s(t;,x)df. Jn Jn\KR{Z) JKR{Z) JKR(Z) are harmonic in KR(Z), so that Ll fKR(Z) s(t;,x) df. = -1 is what has to be proved.
32
3 The Poisson Equation Let a(r) be defined by
s(~,x)
=
a(l~
- xl) (cf. (2.2.1)). For fixed r > 0
set
1n _ 1 f s(~, z) dre. (3.2.8c) wnr 18Kr(z) For all x fj 8KR(Z) (Le., Ix-zl =I r) v is harmonic, since s(~, x) is nonsingular on 8Kr (z) and satisfies .dxs(~,x) = O. Since s(·,x) is harmonic in Kr(z) for r < Iz - xl, the mean-value property (2.3.1) holds, which can now be written vex) :=
vex)
= s(z, x) = a(lz -
xl)
for Iz - xl > r.
(3.2.8d)
Using Exercise 3.2.2b we see that vex) is continuous in IRn , so that we also have (3.2.8e) vex) = a(r) for Iz - xl = r. Thus v is harmonic in Kr(x) with the constant boundary values (8e). The unique solution is therefore
vex)
= a(r)
for Iz - xl ~ r.
(3.2.8f)
The equations (8c,d,f) yield
f
s(~, x) drt; = wnrn-1a(max{r, Iz -
xl})
18Kr(z)
and then, since 0 < Iz - xl < R,
s({,x)~=
f 1 KR(Z)
Wn
l
iz-xi
fR f
10
s({,x) dre dr
18Kr (z)
rn-1a(/z-xl)dr+wn
lR
o
rn-1a(r)dr
Iz-xl
xln
lR
rn IR -Wn rn a(lz-xD+wn-a(r) -a'(r)dr n n Iz-xl Iz-xl n Wn Rna(R) _ Wn fR rn (_ r 1- n ) dr = Wn Rna(R) + fR ::.. dr n liz-xi n Wn n liz-xi n =Wn
Iz
-
= Wn Rna(R) + R2 _ Iz - x1 2 . n 2n 2n From this we see that, independently of R, n, and z, there results
.d
f
s(~,x)~=-l.
lKR(z)
From Theorems 10 and 11 follows
•
Theorem 3.2.13. Under the same assumptions as for Theorems 10 and 11 Equation (2) gives a representation for the classical solution of the boundary value problem (l.la,b).
3.2 Representation of the Solution by the Green Function
33
Finally we put forward two inequalities for the Green function as exercises: Exercise 3.2.14. In il, and ill C il2, respectively, let the Green functions 9, 91 and 92 exist. Show (a) 0:59(X,Y):5s(x,y) forx,yEilCIRn , n~3 (3.2.9) What is the inequality for n=2? (b) 91(X,y):5 92(X,y) for X,y E!h C il2 (3.2.10) Hint: Exercise 2.3.14. Supplement: Proof of Theorem 7. If we make use of a later theorem (Theorem 6.1.13) then Theorem 7 follows from Theorem 3.2.15. The solution u does not depend continuously on I, il the supremum norm (2.4.1) is used as the norm in CO(n), and that from (4d) is used as the norm in C 2 (n). PROOF. Let {} = Kl(O) C IR2 und cp = o. The disk {} is a normal region for which the Green function is known (cf. Theorem 3.3.1). By Theorem 11 there exist solutions un E Q2(n) of Llun = In in il, un = 0 on r for the functions 2 _x21Pn(r), r:= lxi, In(x) = x 2 r2
I rl-
Pn(r):= min ( n· r, log '2
1)
'
which belong to CO(n) and are uniformly bounded: II/nlloo = 1jlog2. By Theorem 11 we have un(x) = 9(~, x)ln(~) de. Since I/n(~)I :5 nl~l, it follows from Exercise 12 that
In
U:1Z1 =
-fa
9z1zJn de =
-fa
ipzlzJn de
-
fa
sZlzJn de at x = 0,
where 9 = ip+s. The first integral is bounded since ip E C 2 (n). The derivative of the singularity function is SZlZl (~, 0) = (~~ - ~i)jl~14. The special choice of In gives for x = 0 U:1Z1 (0) =
-fa
ipzlzl (~, O)/n(~) de + fa[~~
The surface integral K := IaKr(O)[~~
-~~]21~1-6Pn(I~l)de.
- ~iI21~1-5 d~
> 0 does not depend on
I:
r E (0,1], so that the second integral takes on the form In := K r-1pn(r) dr. Since t[rllog(rj2)1)-1 dr diverges as f -+ 0, we deduce In -+ 00 as n -+ 00. Since lIunllc2(.a) ~ IU:1Z1 (0)1, it follows that the map I f-+ U is not bounded, and thus not continuous. •
34
3 The Poisson Equation
3.3 The Green Function for the Ball Theorem 3.3.1. The Green function for the ball KR(y) is given by the function in (2.2.1180). For f E C>'(ll) and cP E C2+>'(F) with 0 < .A < 1, the representation formula (2.2) defines a solution u E C2+>'(ll) of the boundary value problem ~u = f in n, u = cp on F. The proof of the theorem follows from a result of Schauder, which is cited in Theorem 9.1.20. In the case n = 2 the plane IR2 can be identified with GJ by the correspondence (:c, y) +-+ z = :c + iy. The following considerations are based on Exercise 3.3.2. Let the map cp: z =:c + iy En ....... be holomorphic. Show
C= e+ i11 = p(z)
E n'
(3.3.1)
Equation (1) shows, in particular, that a holomorphic transformation of coordinates maps harmonic functions into harmonic functions. An arbitrary simply connected region with at least two boundary points can, by the Riemann mapping theorem, be mapped by a conformal mapping Pzo: zEn ....... P.zo (z) E Kl (0) onto the unit disk such that PZo (zo) = 0 for any given ZO E n. Let g(C,C /) be the Green function for Kl(O). One may check that G(z, zo) := g(Pzo (z), 0) is again a fundamental solution. Now z E an implies pzo(z) E aKl(O), i.e., G(z,zo) = O. Thus G(z,zo) is the Green function in n. This proves Theorem 3.3.3. Let n c IR2 be simply connected with at least two boundary points. Then there exists a Green function of the first kind for n. The explicit forms of various Green functions can be found, for example, in the book by Wloka. [I, Exercises 21.1-21.8]. Of numerical interest might be the fact that with conformal mapping one may remove comers which are disturbing (e.g., reentrant comers) (cf. Gladwell-Wait [I, p.70]). Example 3.3.4. Let n be the L-shaped region in Example 2.1.4. Choose P = z2/3: n --+ n'. Then P is conformal in n. The sides of the angle Fo C an (cf. Fig. 2.1.1) are mapped into a single line segment, so that n' has no more comers sticking in. The Poisson equation ~u = f in n corresponds to the equation ~v(C) = ~IClf(C) in n'.
3.4 The Neumann Boundary Value Problem
35
3.4 The Neumann Boundary Value Problem In (1.1b) and in (2.1.1b) the boundary values u =
o
on u(x) =
onr.
(3.4.1)
In physics this second boundary condition, as it is also called, occurs more frequently than the Dirichlet condition. For example, if u is the velocity potential of a gas, then ~ = 0 means that the gas can only move tangentially at the boundary r. Except in some unusual cases, the boundary value problem Lu = f in nand ou/on =
n
Theorem 3.4.1. Let be a normal region. The Poisson equation Llu with the Neumann boundary condition (1) is only solvable if
£
cp(e) dr~ =
In
=f
(3.4.2)
f(x) d:x..
If a solution u does exist, then u + c, with c any constant, is also a solution.
PROOF. (1) One may repeat the proof of Lemma 2.3.6 for Llu = f. (2) Obviously u + c satisfies the same equation.
•
Later, in Example 7.4.8, we will show that the Neumann boundary value problem for the Poisson equation has a solution if and only if (2) is satisfied, and that two solutions can differ only by a constant. In the representation (2.1) both the values u(e), for E r, and the normal derivative ou/on occur. The Green function of the first kind was chosen in such a way that g(e, x) = 0 for E In the case of the second boundary conditions (1) one makes the assumption that o'Y(e,x)/on~ = c (c: constant), i.e. MJ(e,x)/on~ = c - os(e,x)/on~
e
e r.
for ifJ
= 'Y -
s. The Corollary 2.2.3 with u == 1 and 'Y
= s gives
L:=
£dr.
Since ifJ must be harmonic (Le., f := LlifJ = 0), from Equation (2) we see that cL + 1 = 0 is a necessary condition for the existence of ifJ. Thus the condition on the Green Function of the Second Kind for the potential equation is
o'Y(x,e)/onx = -1/
£dr.
36
3 The Poisson Equation
Thus the term fr u8"{/8ndr in (2.1) becomes const· frudr. Since u is only determined up to a constant (cf. Theorem 1), one can fix this constant with the additional condition fr u dr = o. This gives the following result, if we write 9 for "(:
u(x) =
-1 f(~)g(~,
x) ~ +
1'P(~)g(~,
x) dr~.
The Green function of the second kind for the ball KR(O) C IR3 can be found in Leis [1, p. 79].
3.5 The Integral Equation Method In the representation (2.1) of the Poisson solution the singularity function s can, in particular, be chosen to be "(. If in addition one imposes the given Neumann data (4.1), one obtains
u(x)
=
£
k(x, ~)u(~) dr~ + g(x) for x
with the kernel function k(x,·) :=
-8s(~,x)/n~
E
n,
(3.5.1)
and the functions
g(x) := gl(X) + g2(X), gl(X):= g2(X) :=
£s(~,x)'P(~)dr~
1s(~, x)f(~) dr~.
The right-hand side in Equation (1) with the unknown boundary value r, can be used as an ansatz solution:
u(~),
~ E
4>(x) =
£
k(x, ~)u(~) dr~ + g(x).
(3.5.2)
The first summand on the right of (2) is called the double-layer potential (dipole potential); gl is the single-layer potential, while g2 is a volume potential. For each u E CO(r) the 4> in (2) is a solution of the Poisson equation (1.1a) in n. However, 4> is also defined for an argument x in the exterior domain m.n\n. A closer look at the kernel function k(x,e) shows that it is in fact only weakly singular for the case of smooth boundaries r. Thus 4> is also defined for x E r. The function 4> which is now defined on all IRn is not continuous at points of the boundary r. At Xo E r there exists both an interior limit 4>_(Xo) for x -+ Xo, x E n and an exterior limit 4>+(Xo) for x -+ Xo, x E IRn\n. In addition we have the third function value 4>(Xo) of (2). Their connection is given by the following jump discontinuity relation (cf. Hackbusch [7, Satz 8.2.8]):
3.5 The Integral Equation Method
4'+(Xo) - 4'_(XO) 4'+(Xo) + 4'_(xo)
= 2u(xo), = 24'(xo),
37
(3.5.3a) for Xo E r.
(3.5.3b)
In order that the ansatz (2) does indeed give the solution u in (1), the boundary value 4'_, continued from the interior, must agree with the function u which is put in the integral: 4'_ = u. Now one can solve Equation (3a,b) for !P_: 4'_(Xo) = 4'(Xo) - u(Xo). The equation 4'_ = u thus leads to
u(Xo)
= 44'(Xo) =
l k(xo,e)u(~") dr{ +
g(Xo) for Xo
E
r.
(3.5.4)
Equation (4) is called a Fredholm integral equation of the second kind for the unknown function u E CO(r). The original Neumann boundaryvalue problem (l.la), (4.1) and the integral eqation (4) are equivalent in the following sense: (a) If u is the solution of the Neumann boundary-value problem, then the boundary values, u(e), E r, satisfy the integral equation (4). (b) If u E CO (r) is a solution of the integral equation (4), then the expression (1) gives a solution of (l.la), (4.1) in the entire domain n. The transformation of a boundary-value problem into an integral equation, and the subsequent solution of the integral equation is referred to as the integral equation method. It allows, for example, a new approach to existence statements, in that one shows the solvability of (4). The integral equation (4) can also be attacked numerically. If methods similar to the finiteelement method described in Section 8 are used, then the result is called the boundary-element method (BEM). One can find references to the integral equation method in, e.g., Hackbusch [7, §§7-9] and Kress[l].
e
4 Difference Methods for the Poisson Equation
4.1 Introduction: The One-Dimensional Case Before developing difference methods for the partial differential Poisson equation, let us first recall the discretisation of ordinary differential equations. The equation a(x)u"(x) + b(x)u'(x) + c(x)u(x) = !(x) can be supplemented with initial conditions U(Xl) = Ul. U'(Xl) = u~ or with boundary conditions U(Xl) = Ul, U(X2) = U2. The ordinary initial value problems correspond to the hyperbolic and parabolic initial value problems, while an ordinary boundary value problem may be viewed as an elliptic boundary value problem in one variable. In particular one can view - u"(x)
u(O)
= lex)
= 'Po,
for x E (0,1), u(l) = CPI
(4.1.1a) (4.1.1b)
n
= ! in the domain = (0,1) with Dirichlet conditions on the boundary r = {O,l}. Difference methods are characterised by the fact that derivatives are replaced by difference quotients (divided differences), in the following called, for short, "differences" . The first derivative u'(x) can be approximated by several (so-called "first") differences, for example, by the forward or right difference: as the one-dimensional Poisson equation -Llu
(8+u)(x) := [u(x + h) - u(x)Jlh, the backward or left difference
(4.1.2a)
(8-u)(x) := [u(x) - u(x - h)Jlh, or the symmetric difference
(4.1.2b)
(OOu)(x) := [u(x + h) - u(x - h)l/(2h), (4.1.2c) where h > 0 is called the step size. An obvious second difference for ul/(x)
is (-8-8+u)(x) := [u(x + h) - 2u(x) +u(x - h))/h 2 •
(4.1.3)
ao,
One also calls 8+ , 8-, and 8-8+ index/ D7 / emphdifference operators. The product 8-8+ may be viewed as 8- 08+ or as a+ oa-, Le .. (a+ a- )u( x) = 8+(8-u(x».
W. Hackbusch, Elliptic Differential Equations: Theory and Numerical Treatment, Springer Series in Computational Mathematics 18, DOI 10.1007/978-3-642-11490-8_4, © Springer-Verlag Berlin Heidelberg 2010
38
4.1 Introduction: The One-Dimensional Case
Lemma 4.1.1. Let[x - h, x
+ h] en.
39
Then
a±u(x) = u'(x) + hR with
IRI ~ 4I1uIlC2(Si)
if 1.1. E C 2 (n)
(4.1.4a)
aou(x) = u'(x) + h 2 R with
IRI ~ ~lIuIlC3(.a)
if 1.1. E C 3(n)
(4.1.4b)
a-a+u(x)
= u"(x) + h 2 R with IRI ~ 112"uIC4(0) ifuEC4 (n).(4.1.4c)
PROOF. We give the proof only for (4c). If one applies Taylor's formula u(x ± h)
= u(x) ± hu'(x) + h 2 u"(x)/2 ± h3u"'(x)/6 + h4R 4,
R4 = h- 4
l
z ±h
z
u""(e)(x ± h - e)3/3!dI;. = u""(x ± ~h)/4!,
with ~ E (0,1), to Equation (3), the result is (4c) because R u""(x - ~2h)]/24.
• • • • • • •
(4.1.5a) (4.1.5b)
= [u""(x+~lh)+
•
Figure 4.1.1
• · • • · · · · · li h We replace il = (0,1) and ilh
Grid for h = 1/8
n = [0,1] by the grids
= {h, 2h,···, (n -l)h = 1- h},
(4.1.6a)
nh = {O, h, 2h, ... , 1 -
(4.1.6b)
h, I}
of step size h = l/n. For x E ilh, a-a+u(x) only contains the values of 1.1. at x, x ± h E h • Under the assumption that the solution 1.1. of Equations (la,b) belongs to C 4 (n), (4c) yields the equations
n
(4.1.7) If one neglects the remainder term O(h2) in Equation (7), one obtains - a-a+Uh(X)
== h-2 [-Uh(X - h) + 2Uh(X) - Uh(X + h)] = f(x) for x E ilh . (4.1.8a)
These are n-1 equations in n+1 unknowns {un(x),x E equations are supplied by boundary conditions (lb):
nh }. The two missing (4.1.8b)
Uh is a grid function defined on Uh
nh. Its restriction to ilh yields the vector
= (uh(h),uh(2h), ... ,Uh(1-h))T.
40
4 Difference Methods for the Poisson Equation
If in (Sa) one eliminates the components Uh(O) and uh(l) with the aid of
Equation (Sb), one gets the system of equations (4.1.9a) with 2 -1
-1
2
-1 2
-1
-1
-1
(4.1.9b)
2 -1
-1 2
qh = (f(h) + h- 2<po, f(2h), f(3h), .. " f(I- 2h), f(I- h) + h- 2
4.2 The Five-Point Formula First we select the unit square
n={(x,y):O<x
nh := {(x,y) E n:x/h, y/h E Z} for step size h
=
(4.2.1a)
~ (n E IN). The discrete boundary points form the set
n:= {(x,y) E r:x/h,y/h E Z}.
(4.2.Ib)
nh := nh U rh = {(x,y) E nh:x/h, y/h E Z}.
(4.2.Ic)
As in (1.Ib) we set
y
Figure 4.2.1 A two-dimensional grid
0
0
0
• • • • • • • • • • • • • • •
0
0
• • • • • • • • • •
0 0 0
.
:
Points of n h
0:
Points of rh
0 0 0
x
4.2 The Five-Point Formula
41
In the Poisson equation -Llu == -U",,,, - ulIlI U
IP on
=
= f in il,
r,
(4.2.2a) (4.2.2b)
the second derivatives u"'''' and UuIl can each be replaced by the respective differences (l.4c) in the x and y directions: (LlhU)(X,y):= (8;8;% +8;8t)u(x,y) = h-2[u(x - h,y) +u(x + h, y) + u(x,y - h) + u(x,y + h) - 4u(x,y)]. (4.2.3) Since on the right side of (3) the function u is evaluated at five points, Llh is also called the Five-Point Formula. The discretisation of the boundary
value problem (2a,b) using Llh leads to the difference equations -LlhUh(X) Uh{X)
= f(x)
= IP{x)
for x E ilh,
(4.2.4a)
for x E n.
(4.2.4b)
Through (4a,b) one obtains one equation per grid point x = (x,y) E il h , and hence one equation per component of the grid function Uh = (Uh(X»xEGh. Except in the one-dimensional case, there exists no natural arrangement of grid points, thus one cannot immediately obtain a matrix representation as in (1.9b). The only natural indexing of Un is that through x E ilh or the pair (i,j) E 71? with x = (x,y) = (ih,jh). Let the matrix elements be given by Lxe:=
{
4h-2 if x = ~ E il h , _h- 2 ifxEilh,~Eilh'x-~=
o
(:h) orx-~= (±t),
otherwise.
(4.2.5) For x = ~, Lxx is a diagonal element; in the second case, Lxe = _h-2, we say that x and ~ are neighbours. If one eliminates the components Uh(X), x En, with the aid of Equation (4b), then Equation (4a) assumes the following form: (4.2.6a) LxeUh(e) = %(x) for x E ilh,
I:
eE!2h
where qh
:= A
+ IPh,
A(x):= f(x),
IPh(X):= -
I: LxeIP(~)·
(4.2.6b)
eErh For the proof split the sum -LlhUh(X)
=
I: LxeUh(~) eEGh
into L:eEnh ... and L:eErh .... The second partial sum is -IPh and is moved to the right side of the equation.
42
4 Difference Methods for the Poisson Equation
Remark 4.2.1. fh is the restriction of f to the grid ilh . For all points far from the boundary we have 'Ph(X) = 0; here x E ilh is said to be far from the boundary if all neighbours x± (0, h), x± (h, 0) belong to ilh. In the case of homogeneous boundary values 'P = 0 we have % = fh. The system of equations (6a) can be expressed in the form (1.9a):
where the matrix Lh = (L~)x,~En" and the grid functions Uh = (Uh(X))xEn" and % = (qh(X))xEO" are described by their components (cf. (5), (6b)). Strictly speaking Lh is not a matrix in the usual sense but a linear mapping since the indices x E ilh are not ordered. Exercise 4.2.2. (a) An N x N matrix P is called a permutation matrix if w := Pv, for all v E IRN, has the coefficients Wi = v1f(i), 1 ~ i ~ N, where 7r is a permutation of indices {I, .. ·, N}. Show that P is unitary, Le., p-l = pT. (b) Let I be an index set with N elements. Let the "matrix" coefficients Aa ,8, 01, f3 E I, be given. To the arrangement OIl. ... ,OIN of indices corresponds the N x N matrix A = (aij )i,j=l, ...• N with Qjj = Aa,a;' Let A be the matrix which belongs to a second arrangement all"', aN of I. Prove that there exists a permutation matrix P such that A = PAPT . (c) Let Aa,8 = A,8a for all 01, f3 E I. For each arrangement of the indices the corresponding matrix is symmetric. A possible linear enumeration of indices x E ilh is lexicographical ordering (h, h), (h,2h),
(2h, h), (2h,2h),
(3h, h), (3h,2h),
(1- h,h), (1 - h,2h),
(h,1 - h),
(2h, 1 - h),
(3h,1 - h),
(1 - h, I - h).
(4.2.7)
Generally, the point x = (Xli"', Xd) precedes the point y = (Yl,"" Yd) in lexicographical order, if for a j E {I, ... , d} the conditions Xi = Yi (i > j) and Xj < Yj hold. Each line in (7) corresponds to a so-called x-row in the grid th. A vector Uh whose (n - 1)2 components are enumerated in the series (7) thus separates into n - I blocks (so-called x-blocks). The block decomposition of the vectors generates a block decomposition of the matrix Lh which is given in (8). Exercise 4.2.3. (a) With the lexicographical numbering of grid points the matrix Lh has the form
IT=[~I ~I I 4.2 The Five-Point Formula
Lh
=h
-2 [
T -1
-1 T
-1 .... -1
..
T -1
i '
.
-I
-1
43
~1 ~1
'
(4.2.8) where T is an (n - 1) x (n - 1) matrix and Lh contains (n - 1)2 blocks. 1 is the (n - 1) x (n - 1) identity matrix. (b) Let il be the rectangle il
= (0, a)
x (0, b)
= {(x,y):O < x < a, 0 < y < b}.
Let the step size h satisfy the conditions a = nh and b = mho Show that the discretisation (4a,b) in the corresponding grid ilh leads to a matrix which also has the form (8). But here Lh contains (m - 1)2 blocks of the size (n - 1) x (n -1). Another frequently used arrangement is the chequer-board ordering (or red-black ordering). To this end one chessboard pattern divides ilh into "red" and "black" fields: ilhr := {(x, y) E ilh : (x ilhb := {(x,y) E ilh: (x
+ y)jh odd}, + y)jh even}.
(4.2.9)
First one numbers the red squares (x,y) E ilh r lexicographically, and then those of ilh b • The partition (9) induces a partition of vectors into 2 blocks and a partition of the matrix Lh into 2 . 2 = 4 blocks.
Exercise 4.2.4. With respect to the chequer-board ordering, the matrix Lh assumes the form 4
A 4
(4.2.10)
4
4
where, in general, A is a rectangular block matrix because for n even, ilh r and ilh b contain a different number of points. The complete (n - 1)2 x (n - 1)2 matrix Lh in (8) or (10) is needed neither for the theoretical investigation of the system of equations Lh Uh = qh nor for its numerical solution. All properties of Lh considered in the following are invariant with respect to re-numbering of the grid points. Even though numerical methods for the solution of LhUh = qh implicitly use an arrangement of grid points (with the exception of special algorithms for parallel computers), they never employ the complete (n - 1)2 x (n - 1)2 matrix L h . Every usable
44
4 Difference Methods for the Poisson Equation
algorithm must take into account that Lh is sparse, Le., it has substantially more zero than nonzero elements. In the following we again return to indexing by x E nh. Nevertheless we will continue to refer to the Lh defined by (5) as a matrix. The difference operator Llh is also described by the star
<1.
~h
-2
[1 ~4 1]
(4.2.11)
The general definition of a difference star (with variable coefficients) reads
C-l,l(X,y) C-l,O(X, y) C-l,-l(X,y)
CO,l(X,y) CO,o(x, y) CO,-l(X,y)
Cl,l(X,y) Cl,O(X, y) Cl,-l(X,y)
00
=
L
t;j(x,y)uh(x+ih,y+jh),
(4.2.12)
',j=-oo
in which the zero coefficients have not been written out. Attention. The star (11) does not represent a submatrix of Lh! The coefficients of the star appear in each row of Lh.
Remark 4.2.5. Note that the difference operator -Llh cannot be equated with the matrix Lh since Llh does not contain information on the type or place of the boundary conditions. We shall say the matrix Lh belongs to a difference star (12) if the system of equations LhUh = Qh results from the difference equations in x E nh after elimination of the Dirichlet boundary values Uh(X) = cp(x), x E n. Even if the vector Uh in LhUh = qh contains only components Uh(X), x E nh , one occasionally equates Uh with the grid function on which assumes the prescribed boundary values (4b) on n.
nh
4.3 M-matrices, Matrix Norms, Positive Definite Matrices The elements of the matrix A are denoted by aa{3, Ot, f3 E I. Here A and the index set I assume the places of Lh and nh . We write
A
~
B if aa{3
and define analogously A byO.
~
B, A
~ ba {3
for allOt, f3 E l,
> B, A <
B. The zero matrix is denoted
4.3 M-matrices, Matrix Norms, Positive Definite Matrices
45
Definition 4.3.1. A is called an M-matrix if aaa
>0
for all a E I,
aa{3::;
0
for all a
i:- {3,
(4.3.1a)
A nonsingular and A -1 ~ O.
(4.3.1b)
The inequalities (la) can immediately be proved for Lk (cf. (2.5)). However we still need criteria and auxiliary results to prove (lb). The index a E I is said to be directly connected with {3 E I if aa{3 i:- O. We say that a E I is connected with {3 E I if there exists a "connection" (chain of direct connections) (4.3.2)
The index set I together with the direct connections form the graph of A (cf. Fig. 1). Frequently A has a symmetrical structure, i.e., aa{3 i:- 0 holds if and only if a{3a i:- O. In this case a is (directly) connected with {3 if and only if {3 is (directly) connected with a. Definition 4.3.2. A matrix A is said to be irreduci ble if every a E I is connected with every {3 E I.
CD, @, CD: Indices _
: Connections
Figure 4.3.1. Graph of the irreducible matrix A
In the case of the matrix A = Lk, two indices x, y E ilk are connected if and only if y = x or if y is a neighbour of x. Arbitrary x, y E ilk can evidently be connected by a chain x = x(O) , x(l), ... ,x(k) = y of neighbouring points. Thus Lk is irreducible. Exercise 4.3.3. Prove that A is irreducible if and only if there is no ordering of the indices such that the resulting matrix has the form A
=
[All
o
Al2] , A22
where All and A22 are respectively square nl x nl and (nl ~ 1, n2 ~ 1), and A12 is an nl x n2 submatrix.
n2
x
n2
matrices
46
4 Difference Methods for the Poisson Equation
The important question as to whether A = Lh is nonsingular can be treated as a special case of the following statement. 4.3.~ershgorin) Let Kr{z) denote the open disk {( E CD: Izand let Kr{z) := {( E CD: Iz - (I ~ r} denote the closed disk. (a) All eigenvalues of A lie in
Criterion
(I < r},
UKrQ(aaa)
with ra =
aEI
L
laa,8l·
~¢Q ~El
(b) If A is irreducible, the eigenvalues even lie in
aEI
aEI
PROOF. (a) Let A be an eigenvalue of A and u a corresponding eigenvector which, without loss of generality, satisfies /lu/l co = 1, where
(4.3.3) is the maximum norm. There exists (at least) one index 'Y E I with lu-yl
= 1.
Assertion 1. lu-y I = 1 implies
From (*) follows A E K r ., (an) and hence the statement. To prove the assertion use the equation from Au = AU associated to the index 7:
Av.-y =
L a-y,8u,8, ~I
that is
(A - an )u-y =
L a-y,8u,8' ~~
From lu-yl = 1 follows IA - a-y-yl = I(A - a-y-y)u-yl ~ II:,8¥-y a-y,8u,8l. By taking the modulus into the sum and by using IU,81 ~ /luI/co = 1, (*) follows. (b) Let A be irreducible and let A be an arbitrary eigenvalue of A with associated eigenvector u which in tum is again normalised by /luI/co = 1. The case A E UaE/ Kr,,(aaa) immediately leads to the statement. Therefore let A ¢ UaEIKrQ(aaa) be assumed.
f:. 0, i.e., 7 is directly connected with {3; then lu-yl implies IU,81 = 1 and IA - a,8,81 = r,8'
Assertion 2. Let aa,8 and IA - a-y-yl
= r-y
=1
Part (a) proves the existence of a 'Y E I with Iv.-yl = 1 and IA - ani ~ r-y. According to the assumption, IA - ani = r-y must hold so that assertion 2 is
4.3 M-matrices, Matrix Norms, Positive Definite Matrices
47
applicable to "I. Since A is irreducible, for an arbitrary {3 E f there exists a connection (2) of "I with {3: ao = "I, a1. ... ,ak = {3, aCto_lao f; O. Assertion 2 shows lua,1 = 1 and l,x - aa,a, I = ra, for all i = 0"" ,k; in particular, ,x E 8Kr~(af3f3) for {3 = ak. Since {3 was chosen arbitrarily, it follows that ,x E naEl 8Kr",(aaa), and the statement is proved. Proof of Assertion 2. Besides the inequality chain (*) there also holds l,x an I = r -y so that all the inequalities in (*) become equations. In particular
L: 1a-yf3ll uf31 = L: 1a-yf31 f3~-y
f3~-y
must hold. Since lu.al ~ lIuli oo = 1, the identity la-y.al lu.al = la-y.al must be satisfied for each summand. Hence a-y.a =I- 0 implies lu.a I = 1. The application of Assertion 1 to {3 yields l,x - a.a.al ~ r.a. The assumption ,x
Definition 4.3.6. (a) A is said to be diagonally dominant if
L: laa.al < laaal
(4.3.4a)
.a~a
for all a E f. (b) A is said to be irreducibly diagonally dominant if A is irreducible, the inequality (4a) holds for at least one index a E f and
L: laa.al ~ laaal
for all a E f.
(4.3.4b)
~¢",
~EI
Note that while an irreducible and diagonally dominant matrix is irreducibly diagonally dominant, the reverse need not hold. The matrix Lh from Section 4.2, while not diagonally dominant, is irreducibly diagonally dominant, for Lh is irreducible and satisfies (4b). At all points near the boundary-Le., those x E ilh, which have a boundary point yEn as a neighbour-however, (4a) holds: L~¢'" laa.al ~ 3h- 2 < 4h- 2 = ~EI aaa· The spectral radius p(A) of a matrix A is given by the eigenvalue that is largest in modulus: p(A) := max{I,xI:,x eigenvalue of A}.
(4.3.5)
48
4 Difference Methods for the Poisson Equation
In the following we split A into
A
=D -
B,
D
= diag
{a",,,,: a E I},
(4.3.6a)
where D is the diagonal part of A: d",,,,
= a",,,,,
d"'f3
=0
for a
f- {3.
(4.3.6b)
B := D - A is the off-diagonal part: (4.3.6c) Criterion 4.3.1. Let (6a-c) hold. Sufficient conditions for
p(D-1B) < 1
(4.3.7)
are the diagonal dominance or the irreducible diagonal dominance of A.
PROOF. (a) The coefficients of C := D-IB read caf3
= -aaf3 / aaa (a f- {3), COla = O.
From the diagonal dominance (4a) follows that ra := Lf3¥'" IC"'f31 < 1 for all a E I. By the Gershgorin Criterion 4a all eigenValues oX of C lie in UaEI KrQ(caa ) = UaEI KrQ(O) so that P,I ::; maxr", < 1 and hence p(C) = p(D-1B) < 1 also follows. (b) If A is irreducibly diagonally dominant then rf3 ::; 1 for all {3 E I and r", < 1 for at least one a. According to Criterion 4b all eigenvalues of C lie in Uf3E I K r ,8 (0) U (n f3E I 8Kr ,8 (0)) . This set lies in K1 (0) if n pEl 8Kr,8 (0) C K1(0). At first let us assume that all rf3 agree: rf3 = r for all {:J. Since ra < 1 for one a E I, it follows that r < 1 and nf3 8Kr,8(0) = 8Kr (0) C Kl(O). But if all rf3 are not equal then nf38Kr,8(0) is empty. Thus in both cases oX E K1(0) holds and (7) is proved. • Exercise 4.3.8. (a) Weaken irreducible diagonal dominance as follows: Let A satisfy the inequalities (4b) and for all {3 E I let the connection (2) exist for an index a E I for which the strict inequality (4a) holds. Prove that even under this assumption p(D-IB) < 1 holds. Hint. Use Exercise 5. (b) Show that the geometric series S = L~o C" converges if and only if p( C) < 1. Then the following holds: S = (I - C) -1. Hint: Represent C in the form QRQ T (Q a unitary, and R an upper triangular matrix) and show IiC"lioo ::; K[p(C))". (c) Let u be a vector. We define lui as the vector (!) with the entries lui a := lu",l. For two vectors one writes v ::; w if Va ::; w'" (a E I). Show that: (1) AB 2: 0, if A 2: 0, B 2: 0; AB > 0, if A > 0, B > 0; (2) AD> 0 if A> 0, and D 2: 0 is a nonsingular diagonal matrix; (3) Av ::; Aw if A 2: 0 and v ::; w; IIvll oo IIwll oo if 0 S v S w;
s
4.3 M-matrices, Matrix Norms, Positive Definite Matrices
49
(4) Au ~ IAul ~ Alul if A ~ O. The importance of inequality (7) results from Lemma 4.3.9. Let A satisfy (la). Let D and B be defined by (6a-c). A is an M-matrix if and only if p(D-1B) < 1. PROOF. (a) Let C := D-1B satisfy p(C) < 1. Then the geometric series S := E:'o Cv converges (cf. Exercise 8b). From D-1 ~ 0 and B ~ 0 one infers C ~ 0, C v ~ 0, and S ~ O. Since I = S(I - C) = SD-1 D - B = (SD)-1 A, A has the inverse A -1 = SD- 1. D-1 ~ 0 and S ~ 0 result in A -1 ~ O. From this (lb) also results, i.e., A is an M-matrix. (b) Let A be an M -matrix. For an eigenvalue>. of D-1 B select an eigenvector u =I- O. According to Exercise 8c we have 1>'1 lui = I>,ul = ID-1Bul ~ D- 1Blul·
Because A -lD ~ 0 (cf. (la,b)) one obtains -A -lDD-1Blul ~ -A -lDI>'llul so that lui = A -l(D - B)lul = A -1D(I - D- 1B)lul ~ A -1Dlul- A -1 D I>'llul = (1 -I>'DA -1Dlul
follows. For 1>'1 ~ 1 we would get the inequality, lui ~ 0, i.e., u = 0, in contradiction to the assumption u =I- O. From this follows 1>'1 < 1 for every • eigenvalue of C = D- 1B, thus p(D-1B) < 1. Criterion 7 and Lemma 9 imply Criterion 4.3.10. If a matrix A with the property (la) is diagonally dominant or irreducibly diagonally dominant, then A is an M -matrix. Theorem 4.3.11. An irreducible M -matrix A has an element-wise positive inverse: A -1 > O. PROOF. Let a,/3 E I be selected arbitrarily. There exists a connection (2): a = ao, at,···, ak = /3. Set C:= D-1B. Since ca'_la, > 0, it follows that
"n,"','Yk-IEI
According to Lemma 9, p(C) < 1 holds, so that S := E:'o CV converges. Since Sa{3 ~ (C k )a{3 > 0 and a,p E I are arbitrary, S > 0 is proved. The assertion results from A -1 = SD -1 > 0 (cf. proof of Lemma 9). • In the following we derive norm estimates for A -1.
50
4 Difference Methods for the Poisson Equation
Definition 4.3.12. Let V be a linear space (vector space) over the field of real numbers (K := JR) or complex numbers (K := GJ). The functional 11·11 is called a norm in V if lIuli = 0 only for u = 0, lIu + vII ~ lIull + IIvll for all u, v E V, IIAull = IAI lIuli for all A E K, u E V.
(4.3.8a) (4.3.8b) (4.3.8c)
Example. Let V = JR#1 where #1:= is the number of elements of the index set I. The maximum norm defined in (3) satisfies the norm axioms (8a-c). If one views the elements u E V as vectors, 11·11 is called a vector norm. But the matrices also form a linear space. In the latter case one calls II . II a matrix norm. A special class of matrix norms is contained in Definition 4.3.13. Let V be the vector space with vector norm 11·11. Then one calls (4.3.9) IIIAIiI := sup{IIAull/liull: 0 ::J u E V} the matrix norm associated with the vector norm
11·11.
Exercise 4.3.14. Let 111·111 be defined by (9). Show that: (a) (b) the following holds:
111·111 is a norm;
IIIABIlI ~ IIIAIIIIIIBIII,
(4.3.10a) (4.3.10b) (4.3.10c) (4.3.lOa)
1111111 = 1 (I: unit matrix),
IIAull ~ IIIAlllliull, IIIAIiI ~ peA).
Example. The matrix norm associated with the maximum norm II ·1100 (cf. (3» is called the row sum norm and is also denoted by II . 1100. It has the explicit representation (4.3.11) Exercise 4.3.15. (a) Prove (11). (b) For matrices 0
~
B
~
C there holds
IIBlloo ~ IIClloo. In the next theorem we denote by 1 the vector having only ones as components: :ao< = 1 for all a E I. For the notation v
~
w see Exercise 8c.
Theorem 4.3.16. Let A be an M -matrix and let a vector w exist with Aw ~ 1. Then IIA-11l 00 ~ IIwll oo .
4.3 M -matrices, Matrix Norms, Positive Definite Matrices
51
PROOF. As in the proof of Lemma 9, let lui be the vector with the components luo,!. For each u we have lui ~ lIull oo l ~ IlullooAw. Since A-1 ~ 0, we obtain
lA-lUi ~ A-llul ~ IlullooA -1 Aw = Ilulloow (cf. Exercise 8c) and IIA-lulloo/liulloo ~ Ilwll oo . Definition 13 implies IIA -11100 ~ Ilwlloo.
that
•
How to estimate with the aid of a majorising matrix is shown in Theorem 4.3.17. following holds
Let
A
and
A'
be M-matrices with
A'
~
A.
Then the
(4.3.12)
PROOF. A-1 ~ 0,
A'-l ~ A-1 follows from A-1 - A,-l = A-l(A' - A)A,-l and A' - A ~ 0, A,-l ~ 0. The remainder follows from Exercise 15b.
•
Exercise 4.3.18. Prove (12) under the following weaker assumptions: A is an M-matrix, A' satisfies (la) and A' ~ A. Hint: Repeat the considerations from the first part of the proof of Lemma 9 with the matrices D' and B' associated to A'. Exercise 4.3.19. Let B a principal submatrix of A, that is, there exists a subset I' c I such that B is given by the entries bal' = aal', a, f3 E I'. Prove that if A is an M-matrix, then so is Band 0 ~ (B-1 )al' ~ (A -1 )al' holds for all a, f3 E I'. Hint: Apply Exercise 18 to the following matrix A': a~1' = aal' (a, f3 E I'), a~a = aaa (a E IV'), a~1' = 0 otherwise. Another well-known vector norm is the Euclidean norm
lIull2 := [c L IUa12]l/2
(4.3.13)
aEI
with fixed scaling constant c > 0 (for example, the choice c = h 2 in connection with the grid functions from Section 4.2 results in the fact that c LaEI represents an approximation to the integration J(1). The matrix norm associated to 11·112 is independent of the factor c. It is called the spectral norm and is also denoted by II . 112. The name derives from the following characterisation. Exercise 4.3.20. Prove: (a) for symmetric matrices there holds peA) (cf. (5)). (b) For each real matrix holds:
(c)
IIAI12 = [p(AT AW/ 2= [maximal eigenvalue of AT Aj1/2. For each matrix holds IIAII~ ~ IIAlbIlATI12. Hint: (b) and (lOd).
IIAII2
=
52
4 Difference Methods for the Poisson Equation
For the proof in the exercise use the scalar product (u, v) := C
L u",v",
(4.3.14a)
"'EI
(c as in (13)) and its properties
(u, u)
= lIull~, (Au, v) = (u, AT v), I(u, v)1
~ lIulbllvll2.
(4.3.14b)
Here the case K = lR is always used as basis., i.e., all matrices and vectors are real. Definition 4.3.21. symmetric and
A matrix A is said to be positive definite if it is
(Au, u) > 0 for all u
# o.
(4.3.15)
Exercise 4.3.22. Prove that (a) a symmetric matrix is positive definite if and only if all eigenvalues are positive. (b) All principal submatrices of a positive definite matrix are positive definite (cf. Exercise 19). (c) The diagonal elements a",,,, of a positive definite matrix are positive. (d) A is called positive semi-definite if the inequality (15) holds with "~" instead of ">". A positive semidefinite matrix A has a unique positive semidefinite square root B = A 1/2, which has the property B2 = A. If A is positive definite, then so is A 1/2. A corollary to Exercise 22a is Lemma 4.3.23. A positive definite matrix A is nonsingular and has a positive definite inverse. The property "A -1 is positive definite" is neither necessary nor sufficient to ensure the property "A -1 ~ 0" of an M-matrix. In both cases, however, (irreducible) diagonal dominance is a sufficient criterion (cf. Criterion 10). Criterion 4.3.24. If a symmetric matrix with positive diagonal entries is diagonally dominant or irreducibly diagonally dominant then it is positive definite. PROOF. Since r", < a"'Cl<> resp. r", ~ a",,,,, the Gershgorin circles which occur in Criterion 4 do not intersect the semi-axis ( -00,0], so that all the eigenvalues must be positive. By Exercise 22a then A is positive definite. • Lemma 4.3.25. Let Amin and Amax be respectively the smallest and largest eigenvalues of a positive definite matrix A. Then there holds
IIA -1112 = 1/ Amin.
(4.3.16)
4.4 Properties of the Matrix Lh
53
PROOF. Exercise 20a shows that IIAII2 = p(A) and IIA -1112 = p(A -1). From (5) then result p(A) = Amax and p(A -1) = l/A min, since Amin > O. •
4.4 Properties of the Matrix Lh Theorem 4.4.1. The matrix Lh (five-point formula) defined in (2.5) has the following properties: Lh is an M-matrix, Lh is positive definite,
IILhlloo ~ 8h- 2 ,
II Lhl12
(4.4.1a) (4.4.1b) (4.4.1c) (4.4.1d)
IILh"ll1oo ~ 1/8, < 8h- 2 ,
~ 8h- 2 cos2(7rh/2)
II Lh"1112 ~ ~h2 sin- 2
t 2h )
=
2!2
+ O(h2) ~ 1~'
(4.4.1e)
PROOF. (a) In Section 4.3 we already noticed that Lh is irreducibly diagonally dominant and satisfies the inequality (3.1a). By Criterion 4.3.10 then Lh is an M-matrix. (b) Since Lh is symmetric and irreducibly diagonally dominant, (lb) follows from Criterion 4.3.24. (c) That IILhlloo ~ 8h- 2 can be read from (2.5) and (3.11). To estimate Lh"l one uses Theorem 4.3.16 with w(x, y) = x(l - x)/2. Then we have Lhw ~ 1 (even that (LhW)(X, y) = 1 unless y = h and y = 1 - h) and IIwlloo ~ w(1/2, y) = 1/8. (d) The inequalities (ld,e) result from Lemma 4.3.25 and
Lemma 4.4.2. The (n _1)2 eigenvectors of Lh are u"'" (1 ~ V,j.L ~ n -1):
u""'(X, y) = sin(v7rx) sin (J-L7rY )
(4.4.2a)
The corresponding eigenvalues are l~v,j.L~n-1.
(4.4.2b)
PROOF. Let n~D be the one-dimensional grid (1.6a) and let u"(x) .sin(v7rx). For each x E nkD there holds 8-8+u"(x) = h-2 [sin(v7r(x - h)) + sin(v7r(x + h)) - 2sin(v7rx)] = 2h- 2 sin(v7rx)[cos(v7rh) -1]
since sin(v7r(x ± h)) = sin(v7rx)cos(v7rh) ± cos(v7rx)sin(v7rh). The identity 1 - cos~ = 2sin2(~/2) then implies
54
4 Difference Methods for the Poisson Equation (4.4.2c)
Let L~D be the matrix (1.9b). Note that (a-a+u)(h) also involves the boundary value u(O) which (L~Du)(h) does not; similarly (a-a+u)(1 - h) depends on u(1). However, since u(O) = sin(O) = 0 and u(1) = sin(v7r) = 0 we have L~Duv = -a-a+uv, and (2c) can be brought over:
1$v$n-1.
(4.4.2c')
The two-dimensional grid function uVP- in (2a) can be written as the (tensor) product UV(x)uP-(y). Now we have that (LhUVP-)(X, y) is equal to the sum uP-(y)(L1DuV)(x) +uV(x)(L1DuP-)(y), so that (2b) follows from (2c'). • In the sequel we want to show the analogies between the properties of the Poisson equation (2.2) and the discrete five-point formula (2.4a,b). The analogue of the mean-value property (2.3.1) is the equation 1 Uh(X,y) = '4[u h(x-h,y)+u h(x+h,y)+uh(x,y-h)+uh(x,y+h)]. (4.4.3)
From (2.3) and (2.4a) with
f
= 0 we obtain the
Remark 4.4.3. The solution Uh of the discrete potential equation (2.4a) with f = 0 satisfies Equation (3) at all grid points (x,y) E {lh. As in the continuous case the mean-value property (3) implies the maximum-minimum principle.
Remark 4.4.4. Let Uh be a non-constant solution of the discrete potential equation (2.4a) with f = O. The extrema max{ Uh(X): x E nh} and min{uh(x):x E nh} are assumed not on n h but on n.
PROOF. If Uh were maximal in (x, y) E nh, then because of Equation (3), all neighbouring points (x± h, y) and (x, y± h) would have to carry the same values. Since every pair of points can be linked by a chain of neighbouring • points, it follows that Uh = const, in contradiction to the assumption. The last proof indirectly uses the fact that Lh is irreducible. The irreducibility of Lh corresponds to the assumption in Theorem 2.3.7 that n is a domain, i.e., connected. The result of carrying over Theorems 2.4.3 and 3.1.2 reads as follows:
Theorem 4.4.5. Let ul and u~ be two solutions of (2.4a): -L1hu~ = f for different boundary values u~ = !pi (i = 1,2). Then the following holds: (4.4.4a) (4.4.4b)
4.4 Properties of the Matrix Lh
55
PROOF. Let wh:= u~ -uk. (1) In the case that O. (2) Let M be the right side of (4a). -M ::; Wh ::; M on implies the • inequalities -M ::; Wh ::; M on rh.
n
The discrete analogue of the Green function 9(X,e) is h- 2 L-,;l. Let oe be the scaled unit vector
_{h-
oe(x) -
2
0
if x if x
=e
of e
(4.4.5a)
The column of the matrix h- 2 L-,;l with index
eE ilh is given by (4.4.5b)
For e E ilh fixed, 9h(·,~) is a grid function defined on ilh. The domain of definition is extended to ilh x nh: (4.4.5c) The 9h(X, e) are entries of h- 2 L-,;l: 9h(X,~) Lh implies Remark 4.4.6. Uh(X,e)
= h- 2 (L-,;l)xe. The symmetry of
= Uh(e, x) for all x,e E nh (cf. (3.2.3)).
The representation (3.2.7) is recalled by: Remark 4.4.7. The solution Uh of the system of equations (2.4a) with boundary values
Uh(X) = h2
L
Uh(e,x)i(e),
x E ilh.
(4.4.6)
eEah
Equation (6) is the component-wise representation of the equation Uh = L-,;l ih. The factor h 2 compensates for h- 2 in (5a). It was introduced so that the summation h2 I: in (6) approximates the integral fa. Since in Section 3.2 we consider the Poisson equation L1u = i, but here the equation -L1u = i, the right-hand sides in (3.2.7) and (6) differ in their signs. The discrete Green function is positive also (cf. (3.2.9)): Remark 4.4.8. 0 < 9h(X, e) ::; h- 2/8 for x,e E ilh . PROOF. The upper bound follows from
56
4 Difference Methods for the Poisson Equation
lIo{lIoo = h- 2, and
(lc). gh > 0 can be inferred from L-;;l > 0 (cf. Theorem
4.3.11).
•
The bound gh(X,~) :::; h- 2 /8 is too pessimistic and can be improved considerably.
Remark 4.4.8'. 0< h(X~) < 9
,
! _ 10g(lx -
- 4
~12 + 2h2) < !log hi.
log 16
-
log 4
(4.4.7)
This estimate via 0(1 log hI) reflects the logarithmic singularity of the singularity ftmction s(x,~) = -log(lx - ~12)/41T. Exercise 4.4.9. For the proof of inequality (7) define Sh(X,~) := 1/4 -log(lx - ~12 + 2h2)/ log 16,
'Uh(X):= Sh(X, 0)
and carry out the following steps: (a) Sh(X,~) ~ 0 for all x E h, ~ E n h, (b) -Llh'Uh(O) = h- 2 , (c) -Llh'Uh(X) ~ 0 for all x E }R2 (longer calculation I), (d) -Llh(Sh(-,~) - gh(·,e» ~ 0 for fixed ~ E nh, (e) 9h(X,e):::; Sh(X,e).
n
Let
L
9h(~,X)
X
E n h·
{En"
Since
r h:= {~ E n: ~ is not a corner point
(0,0), (1,0), (0, 1), or (1, I)}
instead of over the points near the boundary, then the definition of
= -h L
8;;gh(~' x)
x E ilh,
(4.4.8)
{Er;.
where 8;; is the backward difference with respect to the direction of the normal n:
4.4 Properties of the Matrix Lh
57
8;;Uh(', x) = h- 1IUh(', x) - Uh(' - hn, x)] (note that Uh(',X) = 0 for' E r~ points next to the boundary.
c
rh). The variable, - hn ranges over all
Remark 4.4.10. Equation (8) corresponds to the representation in Theorem 3.2.10. The summation h EeEr~ approximates the integral frFinally we want to take a closer look at the estimate for the solution Uh = L-;;l% through lIuhl/oo ::; I/L-;;ll1ool/%IIoo ::; IIqh1l00/8 (cf. (lc)). According to (2.6b), qh = Ih +
n
Theorem 4.4.11. According to (2.6b), let % =!h +
Iluhlloo ::; II L-;; 11/00 max I/(x)1 + max l
~~
~~
~~
(4.4.9)
PROOF. Set Uk := L-;;l!h and u h := L-;;l
Here, h2Enh and fn,hLrf. and fr correspond to each other. Theorem 4.4.12. Under the assumptions of Theorem 11 there holds
II u hll2,nh ::; IIL-;;1I\211/112,fl"
+ ~1I
::; 11611/112,iJh
+ ~II
PROOF. (1) It suffices to consider the case of the potential equation (Le., f = 0). Let the restriction of
58
4 Difference Methods for the Poisson Equation
for x E r{.. Let the mapping <.Ph 1-+ Uh = L-,;lcph be given by the rectangular matrix A: Uh = Acph. According to Equation (8) the entries of A read
= -h8;;Yh(e, x) = 9h(e - hn,x) = 9h(X,e - hn), x E ilh, eE r~. Since A ~ 0 (cf. Remark 8), one obtains the row sum norm IIAlloo := maxx"E~ a~ as IIAcphlloo for the choice of CPh(X) = 1 in all x E r~. The ax~
solution Vh
= ACPh
then reads Vh
= 1 (why?), so it follows that
(2) The column sums of A are s(e) := "ExEah ~ = "ExEah 9h(X,e - hn) for E r~. The grid function Vh := h-2Lh"ln at the points hn near the boundary has the values
e
e-
s(e)
= Vh(e - hn),
eE r~,
n normal direction,
e
as is implied by Remark 7. Let = (6, ~2) E r~ be a point of the left or right = 0 or 1). As mentioned in the proof of (lc), LhWh ~ 1 boundary (i.e., holds for Wh(X,y) := x(l- x)/2. Since L-,;1 ~ 0, then so is Wh 2:: L-,;ln and hence Vh :$ h-2wh. In particular we have the following estimate
el
s(e)
= Vh(e -
hn) :$ h-2wh(e - hn)
= h- 2h(1 -
e-
h)/2 :$ h- 1/2
e
at the point near the boundary hn = (h,6) or (1 - h, e2) . For a point from the upper or from the lower boundary one obtains the same estimate if one uses Wh(X, y) := y(l- y)/2. Since the column sums s(e) are the row sums of AT, we have proved
IIATlloo = max{s(e):e E r~}:$ h- 1/2. (3) We have IIATAII2 = p(ATA) :$ IIATA1100 :$ IIATIloolIAlloo :$ h- 1 /2 (cf. Exercise 4.3.20a, (3.lOd), (3.10a)) so that the solution Uh = Lh"lcph = Aih satisfies the following estimate
= h2
L <.Ph (x)(A A<.Ph)(x) T
xEr~
Since IIcpll2,r~
= l14>h Ib,r~, the assertion follows.
•
4.5 Convergence
59
4.5 Convergence Let Uk be the vector space of the grid functions on n k . The discrete solution e Uk and the continuous solution U e CO(n) cannot be compared directly because of the different domains of definition. In difference methods it is customary to compare both functions on the grid nk. To this end one must map the solution U by means of a "restriction"
Uk
U 1-+ Rku
e Uk
(u: continuous solution)
(4.5.1)
to Uk. In the following we choose Rk as restriction on nk: (4.5.2) The limit h -+ 0 is made precise as follows. Let H C m+ be a subset with accumulation point zero: 0 e H. For example, the step sizes considered so far form the set H = {1/n:n e IN}. For each he H let Uk be equipped with the norm II . Ilk. Definition 4.5.1. (Convergence) The discrete solutions Uk (with respect to the system of norms II· Ilk, hE H) to u if
e Uk converge (4.5.3a)
We have convergence of order k if (4.5.3b) The proof of convergence is usually carried out with the aid of the concepts of "stability" and "consistency". The discretisation {Lh: h e H} is said to be stable with respect to 11·1100 if sup
IILhllioo < 00.
(4.5.4)
kEH
For the discretisation defined in Section 4.2, the stability has been proved in (4.1c) with respect to the row sum norm, and in (4.1d) with respect to the spectral norm. The grid function Ih in -L1huh = Ih is the restriction Ih
= Rhi
(4.5.5a)
of I, where in (2.6b) Rh was chosen as the restriction to ilh: (Rhf)(X)
= I(x)
for all x
e nh .
(4.5.5b)
The formulation of consistency becomes simpler if instead of the matrix Lh one considers the difference operator -L1h on which it is based (cf. Remark 4.2.5). Let Dh be a general difference operator, and let -DhUh = Ih discretise -Llu = I. The discretisation described by the triple (Dh, R h, Rh) is said to be
60
4 Difference Methods for the Poisson Equation
consistent of order k with respect to operator), if
11·1100
(consistent with the Laplace (4.5.6)
for all U E Ck+2(O). Here K is independent of hand u. Remark 4.5.2. Let Rh and Rh be given by (2) and (5b). The five-point formula L1h is consistent of order 2: estimate (6) holds with k = 2 and K = 1/6. PROOF. The expansion (l.4c) can be applied in the x and y directions and yields L1hRhU(X, y)
= L1u(x, y) + h2(Rr,; + R,,)
with IR:J;I, IR"I ~ (1/12)lIuII C4(i1)' (4.5.7)
•
Theorem 4.5.3. Let the discretisation (Dh' R h , R h ) be consistent of order k. Let the matrix Lh associated to the difference opemtor Dh (cf. Remark 4.2.5) be stable with respect to II ·1100' Then the method is converyent of order k if U E Ck+2Ci'i).
PROOF. Wh := Uh - RhU satisfies the difference equations -DhWh
=
-DhUh
+ DhRhU = !h + DhRhu
= Rhf + DhRhU = DhRhU -
RhLlu
in flh.
Because of the homogeneous boundary condition Wh = 0 on rh we have that = L,/(DhRhu - RhL1U) (cf. Remark 4.2.1). (4) and (6) imply (3b). •
Wh
If one substitutes the concrete values IILh"llloo ~ 1/8 and K obtains
= 1/6 one
Corollary 4.5.4. Let the continuous solution of the boundary value problem (2.2a,b) belong to C4(ii). Let Uh be the discrete solution defined in (2.4a,b). Then the converyence of Uh to
U
is of second order:
h2 IIUh - RhUlioo ~ 48I1uIIC4(i1).
(4.5.8)
The assumption U E C4(1i) can be weakened to Corollary 4.5.5. Under the condition U E C 3 ,1(1i) we also have IIUh - RhUlioo ~ (h 2/48) Ilull c 3.1 (ii)" PROOF. The remainder term R4 in (1.5b) may also be written as
(4.5.8')
4.5 Convergence
61
The Lipschitz estimate lu"'(e) - u"'(x)l ::; Ie - xlliull c 3.1Cfi) implies R.t ::; Ilullc3.1(?i/4! so that in (7), (6), and (8) the norm Ilullc4(?i) can be replaced • by lIullc3.1Cfi)'
If, however, one further weakens u E e 3 •1 (n) to one obtains a weaker order of convergence. Corollary 4.5.6. Under the condition of order s - 2:
U
U
E
eB(n), 2 < s < 4,
E eB(n), 2 < s ::; 4, Uh converges (4.5.8")
where Ks := 1/[2s(s - 1)] for 2 ::; s ::; 3 and KB := 1/[2s(8 - 1)(8 - 2)] for 3 ::; 8 ::; 4.
The proof results from
Even though the proofs of convergence are simple, the results remain unsatisfactory. AB can be seen from Example 2.1.3, the continuous solution of the boundary value problem (2.2a,b) generally does not even satisfy U E e 2 (li), although one needs at least U E e s (n) with s > 2 in Corollary 6 for convergence. Stronger results can be obtained by an analysis which will be discussed in Section 9.2. That errors of the order of magnitude of O(h2) occur even under weaker conditions, is shown in Example 4.5.8. If one solves the difference equation (2.4a, b) for -Llu = 1 in [l = (0,1) X (0,1) and u = 0 on r, one obtains at the centre x = y = 1/2 the values Uh(!, i), which are shown in the first column of Table 1. The exact solution u(!,!) = 0.0736713· ., results from a representation that one can find in Example 8.1.11. The quotients €h/€2h of the errors €h = u(!,!) - Uh(!,!) approximate 1/4. This proves Uh(!,!) = u(!, !)+O(h2), although U ¢ e 2(n). At (!, !) Uk has furthermore the asymptotic expansion Uh = u(!, + h2 + O(h4). The error term e(!,!) independent of h, is eliminated by using the Richardson extrapolation
e(1.1)
The extrapolated values are already very accurate for h column of Table 1).
(!, !)
= 1/16
!)
(cf. the last
62
4 Difference Methods for the Poisson Equation
'!able 4.5.1 Difference solutions for Example 8 h
Uh(!,!)
1/8 1/16 1/32 1/64
0.0727826 0.07344576 0.0736147373 0.0736571855
Uh,2h(!,
!)
0.0736668 0.07367106 0.07367133
Uh(!,!) Quo-Uh(!, tient
fh =
!)
8.89xlO- 4 2.26xlO- 4 5.66xlO- 5 1.41 x 10-5
0.250 0.251 0.249
fh,2h
Quotient
4.5xlO- 6 0.06 2.7xlO-7
4.6 Discretisations of Higher Order The five-point formula (2.11) is of second order. Even if the solution U belongs to C 8 with s > 4, no better bound for ..1hRhu - Rh..1u than O(h2) would result. An obvious method for constructing difference methods of higher order is the following. As an ansatz for the discretisation of the one-dimensional equation U" = f choose
cm
k
(DhUh)(X)
= h- 2
'2: CIIUh(X + vh). II=-k
The Taylor expansion provides ~
(DhRhU)(X)
= '2: a"h,,-2u" (x) +O(h2k-l),
k
a"
=
,,=0
('2: cllv")/j.LL II=-k
The 2k + 1 equations ao = al = a3 = a4 = ... = a2k = 0 and a2 = 1 form a linear system for the 2k + 1 unknown coefficients CII. For k = lone obtains the usual difference formula (l.4c); for k = 2 a difference of fourth order results: 2
-1
4
5
h (DhUh)(X) = 12[uh(x-2h)+uh(x+2h)]+3[uh(x-h)+uh(x+h)!-2uh(x).
If one applies this approximation for U" to the x and y coordinates, one obtains for -..1 the difference star
2 -h12
[
1 -16
-1~
60 -16 1
-16
1
1
(4.6.1)
(cf. (2.12». The difference scheme (1) is of fourth order, but presents difficulties at points near the boundary. To set up the difference formula at
4.6 Discretisations of Higher Order
63
(h, h) e ilh, for example, one needs the values Uh( -h, h) and uh(h, -h) outside nh (cf. Figure 1). One possibility would be to use scheme (1) only at points far from the boundary and to use the five-point formula (2.11) at points near the boundary. The above complications do not occur if one limits oneself to compact nine-point formulae; by this one means difference methods (2.12) which are characterised by "ca {3 # 0 only for -1 S a,/3 S 1" (cf. Fig. 1). An ansatz with the nine free parameters ca {3, -1 S a, /3 S 1, leads, however, to a negative result: there is no compact nine-point formula with (DhU)(X) = -Llu(x) + O(h3 ). In this sense the five-point formula is indeed optimal. Nevertheless, nine-point procedures of fourth order can be obtained if one also selects the right side fh of the system of equations (2.6a,b) in a suitable manner.
..
.
II
•
•
•••
II
•
X
•••
.. : Difference Scheme (1) .: Compact nine-point formula
• Figure 4.6.1. Scheme (1) and the compact nine-point formula If one applies the compact nine-point scheme
Dh := _
h~2 [=! ;~ -1
to
U
-4
=! ]
(4.6.2)
-1
e eS(n), the Taylor expansion results in
[8
8
~
4 4 h2 h4 -Dh'U = -Llu - 12Ll2u - 360 8x4 + 4 8x2 8y2
]
+ 8y4 Llu + O{h6 ). (4.6.3)
Here it is crucial that the error term can be expressed by - Llu and hence by
f·
For the special choice of the restriction Rh via 1/2 4
1/2
]
I,
(4.6.4)
1/2 i.e.,
fh(X,y) =(Rhf)(x,y) 1 :=12 If(x - h,y) + I(x + h,y)+ f(x, y - h)
+ I{x, y + h) + 8/(x, y)], (4.6.4')
64
4 Difference Methods for the Poisson Equation
one obtains the expansion (4.6.5) which, because 1 = -Llu, agrees with (3) up to O(h4). The matrix Lh of the system of equations which results after the elimination of the boundary values Uh(X) = cp(x), x E r h, has the entries L
~
= h- 2 {
e, e e
20/6 if x = -1/6 if x - = (±h, ±h) or (±h, =Fh), -4/6 ifx - = (±h,O) or (O,±h), o otherwise.
The right side of the system of equations LhUh
(4.6.6)
= qh is
qh := Ih+CPh, Ih := ~I according to (4'), CPh(X) := -
L
L~cp(e). (4.6.7)
~Er,.
The discretisation (Dh , Rh, Rh) with Dh from (2), Rh from (5.2), Rh from (4') is called the mehrstellen method (cf. Collatz [1]).
Exercise 4.6.1. Let Dh and Lh be defined respectively by (2) and (6). Prove that (a) Lh is an M-matrix; (b) IILh'llloo ~ 1/8 (stability), IILhiloo :5 Wh- 2 /3; (c) IIDhRhu - RhLlulioo :5 ifCo·llullc&(o), if u E CS(Ji) (consistency).
Theorem 4.6.2. Let ilh be defined as in Section 4.2. Let Uh be the solution provided by the mehrstellen method LhUh = % with Lh from (6), and qh from (7). Let the solution u of the boundary value problem (2.2a,b) belong to C6(lJ) or C5.1(.G). Then the following error estimates hold in the respective cases: llh4 IIUh - RhUlioo :5 14401Iullc5.1(o)· (4.6.8a) In the case 01 the potential equation, i.e. 1 = 0, we even have resp.
IIUh - RhUlioo ~ Kh6I1ullc8(O)'
IIUh - RhUIloo ~ Kh6I1ullc7.1(O)' (4.6.8b)
ilu E CS(n), resp. u E c.,·l(n). PROOF. As in Theorem 4.5.3, (8a) results from Exercise (lb,c). In the case that 1 = 0 the O(h4 )-term also vanishes in (3). •
4.7 The Discretisation of the Neumann Boundary Value Problem
65
4.7 The Discretisation of the Neumann Boundary Value Problem The Dirichlet boundary condition u{x) = !p{x) was used directly in the difference method; a discretisation was not necessary. A different situation arises for the Neumann boundary value problem
-.1u = finn = (O, 1) X
au/an =!p
(0,1),
on
r.
(4.7.1)
The normal derivative, which reads explicitly
au -=-u an y au an = -Ua;
au -=U an a; au -=u an y
for X= (x, 0) E r, for x={O,Y)Er,
for x = (x, 1) E r, for x = (l,y) E
r,
(4.7.2)
like the Laplace operator, must be replaced by a difference. We will investigate three different discretisations.
4.7.1 One-sided Difference for
Bu/on
The Poisson equation leads to the (n _1)2
= (l/h -
1)2 equations
(4.7.3) which require the values of Uh(X) for all x
n~ := nh U r~,
r~:=
E
n~, where
n \ {(O, 0), (0,1), (1,0), (1, I)}.
To obtain a further 4(n - 1) equations for {Uh(X): x E rh}, we replace, at all x E r~, the normal derivative au/an =!p by the backward difference
(4.7.4)
-.- .. .. I
• -
Discretisation of the Neumann condition Discretisation of the Poisson equation
-i-- · Figure 4.7.1.
If one inserts the corresponding normal directions n for the four sides of the square one obtains
66
4 Difference Methods for the Poisson Equation
h- 1 (Uh(X,0) - Uh(X, h») =
} for x
h- 1 [Uh(0, y) - uh(h, y)) =
} for y = h, 2h, ... , 1 - h.
= h, 2h, ... , 1 - h,
(4.7.4') Equations (3) and (4) yield (n + 1)2 - 4 equations for as many unknowns.
Exercise 4.7.1. After a rescaling of equations (4) to h- 18;;Uh(X) E r~, these equations, together with (3), form a system Lhuh that Lh is symmetric and satisfies (3.1a).
x
= h- 1
As in the Dirichlet problem the variables Uh(X), x E r~, can be eliminated with the aid of (4) in (3). At the point (h, y) near the boundary, for example, Equation (3) becomes h-2(3uh(h, y) - uh(h, Y - h) -uh(h, y+ h) -uh(2h, y») The star h- 2
[ -1 -1
4 -1
h- 2
-1] thus becomes
0
-1 3 -1
[-1
0 3 -1
h-2 [
= f(h, y) + h- 1
-1] ,
h-2 [
-1] ,
h-2 [
-1
-1 3 -1
0].
-1
-1 3 0
-1] ,
near, respectively, the left, right, upper and lower boundaries. At the comer points one even has to replace two boundary values, so that, for example, at
(h, h) E fl. the star reads h-2 [0 -
~
-1
l
Except for the special case
h = 1/2, one obtains for the (n - 1)2 values Uh(X), x E ilh, the system of equations
with
(4.7.5a) if x E ilh is far from the boundary, if x = (h, h), (h, 1 - h), (1 - h, h), (1 - h, 1 - h), otherwise (4.7.5b) if x, E are neighbours, otherwise for x =1= (4.7.5c) fh(x) = (Rhf)(X) := f(x),
e nh
and
e
4.7 The Discretisation of the Newnann Boundary Value Problem
67
Remark 4.7.2. (a) Lh is symmetric and satisfies the sign condition (3.1a). (b) With lexicographical arrangement of the grid points of ilh, Lh has the form T-I -I
Lh =h -2
[
-I
T-I ..
.... -I
The matrix Lh is singular because the system LhUh = %, like the continuous boundary value problem (1), is, in general, not solvable. The analogue of Theorem 3.4.1 reads as follows.
Theorem 4.7.3. The system of equations (5a) is solvable if and only if _h2
L
f(x) = h
L
(4.7.6)
xEr~
xES'h
Any two solutions of (5a) can only differ by a constant: uJ. - u~
= ell, c E IR.
PROOF. Evidently, Lhll = 0 holds, i.e., 11 E kernel(Lh)' Furthermore, Theorem 4 will then imply dim(kernel (Lh)) = 1. This proves
kernel(L h ) = {ell: e E IR}
(4.7.7)
and thus the second part of the assertion. (5a) is solvable if and only if the scalar product (v, qh) vanishes for all v E kernel (LI) = range (Lh).L. Because = Lh and (7) of
LI
(ll,%) = 0
or
L
%(x) = 0
(4.7.6')
XEnh
is sufficient and necessary. According to Definition (5c), (6) and (6') agree. • Let condition (6) be satisfied. System (5a) can be solved as follows. Select an arbitrary Xo E ilh and normalise the solution Uh (determined except for one constant) by
(4.7.8) Let Uh be the vector Uh without the component Uh(Xo). Let Lh be the submatrix of Lh in which the row and column with index Xo have been left out. Let Q.h be constructed likewise. Then
68
4 Difference Methods for the Poisson Equation £hU = qh
(4.7.9)
is a system with (n - 1)2 - 1 equations and unknowns. Theorem 4.7.4. The system of equations (9) is solvable; in particular, £h is a symmetric M -matrix. Under condition (6), Uh = L;:lqh' supplemented by (8), yields the solution Uh of system (5a).
PROOF. (a) As a principal submatrix of Lh, £h is symmetric. In {}h \{Xo} any two grid points can be connected by a chain of neighbouring points so that Lh is irreducible. For all x E ilh \ {Xo} there holds (3Ab); at neighbouring points ofXo we even have (3Aa), so that £h is irreducibly diagonally dominant. According to Criterion 4.3.10, Lh is an M-matrix, thus nonsingular. (b) If Uh is the solution of (5a), one can assume (8) without loss of generality, so that Uh restricted to nh \ {xo} also solves Equation (9) and has to agree • with the unique solution Uh.
As a corollary of Theorem 4 one obtains that rank(Lh) ~ rank(Lh) = (n - 1)2 -1, i.e., dim(kemel (L h = 1 and hence (7) holds. Another possibility for the solution of Equation (5a) is to pass to an extended system of equations
»
(4.7.lOa)
with where
(T
(4.7. lOb)
can be prescribed arbitrarily.
Theorem 4.7.5. Equation (lOa) is always solvable. If for the last component of the solution Uh we have A = 0, then condition (6) is satisfied and Uh represents the solution of system (5a) which is normalised by nTUh = EXE S1h Uh(X) = (T. However, if A f. 0 holds one can interpret Uh as solution of Lhuh = ilh where ilh = qh - An belongs to j(x) := I(x) - A and j and cp satisfy condition (6).
PROO F. 11 is linearly independent of the columns of Lh so that rank IL h , n] = rank (L h ) + 1 = (n _1)2. Likewise, (R T , 0) is linearly independent of the rows of ILh, R] so that rank (L h ) = (n - 1)2 + 1, i.e., Lh is nonsingular. The other statements can be read from (lOb). • Attention. One should either use Equation (lOa,b) or Equation (9), after first replacing % by ilh := % - (IT %/IT n)l. As a justification of this recommendation note that the condition for solvability of the continuous problem is JS1 I dx + Jr cpdT = 0 and that this does not at all imply the discrete solvability condition (6). For smooth functions f and cp Equation (6) can be shown to hold up to a remainder of order O(h). Thus it is generally unavoidable to
4.7 The Discretisation of the Neumann Boundary Value Problem
69
replace ih and qh by ih -All and qh -All (A = llT %/IT 1). In the case of Equation (lOa,b) this correction is carried out implicitly. If, however, (9) is used without any correction, the resulting solution can be interpreted as a solution of LhUh = qh with qh(X) = %(x) for x '" Xo and qh(Xo) := - Ex#Xo %(x). That is, here too, an implicit correction of qh is carried out, with the difference that the correction is not distributed over all components as before, but is concentrated on qh(XO). If Equation (6) is satisfied up to order O(h), then qh(XO) and qh(XO) differ by O(h-l). Therefore the solution Uh of Equation (9) contains a singularity at the point Xo.
Theorem 4.7.6. (Convergence) Let u E C3.1(1i) be the solution of the Neumann problem (1). Let Uh
= (u;) be the solution of Equation
(lOa).
Then there exists acE IR and constants C, C' independent of u and h such that IAI ~ C'hllullco.1(i'i)'
IIUh - Rhu - dll oo
~ C[hllulb,l(i'i) + h21Iullc3,1(i'i))' (4.7.11)
PROOF. (a) We have A=
nT qh/nTn = (h2 LJ(x) + h L
Because f fl f dx + f r 'P dr = 0, the first factor consists only of the quadrature error O(hllullco,l(i'i»' (b) According to Theorem 5, Uh is the solution of LhUh = qh := qh - All. This corresponds to the difference equations -LlhUh
= ih :=!h -
Al
in ilh,
a;;Uh
=
on r~.
The difference Wh := Uh - RhU satisfies -L1hWh
= -LlhUh + LlhRhu = LlhRhu + ih - Al = L1hRhu - RhLlu - XI =: Ch in ilh,
au a;;Wh = a;;Uh-a;; RhU =
(4.7.12a)
on r{ (4.7.12b)
The errors of consistency are
IIChiioo ~ ~h2I1ullc3'1(i'i) 1
It/J(x) I ~ '2hllulb,l(O) + IAI
(cf. Remark 4.5.2), (cf. Lemma 4.1.1).
Since the solution Wh exists, condition (6) is satisfied with Ch and t/J:
70
4 Difference Methods for the Poisson Equation
Then Wh := Wh - cl with c = nTWh/nT n is the solution of (12a,b) normalised by nTWh = O. The application of the following Theorem 4.7.7 to Equation (12a,b) provides the inequality (11). • Theorem 4.7.7. (Stability) Let condition (6) be satisfied. Let the solution Uh of (3), (4) be normalised by IT Uh = O. Then there exist constants Ct, C 2 independent of U and h such that (4.7.13)
The proof of this theorem, which corresponds to Theorem 4.4.11, will be supplied in Sect. 4.7.4. 4.7.2 Symmetric Difference for
au/an
AB can be seen from Theorem 6, the one-sided difference a;; causes the error term O(h). It seems obvious to replace a;; by a symmetric difference. To this end the five-point discretisation is set up at all points x E h = [Jh u n (cf. (2.1c)): -LlhUh = fh := Rhf in'lih , (4.7.14a) where Rh is the restriction to 'li h . The symmetric difference &,! is defined by
n
(~Uh)(X) := (2h)-1[Uh(X + hn) - Uh(X - hn)] = !f1(x) in x E
n.
(4.7.14b)
Here, we assign two normal directions to the corner points, so that for each of the corner points two equations of the form (14b) can be set up. In the comer x = (0,0) one has, for example, the normals
The corresponding equations (14b) also contain different (!) values IP(O+, 0) = limz -+o!f1(x,O) and !f1(0,0+) = limy-+o!f1(O,y). For x E rh the difference formula (14a) needs the values in points x + hn outside of 'li h • These can be eliminated with the aid of (14b) so that a system of equations Lhuh = qh remains for the (n + 1)2 components Uh(X), x E h .
n
Exercise 4.7.8. (a) If the grid points of'li h are arranged in lexicographical order, Lh has the form
Lh = h- 2
[
T
-21
-/
T
-/ -/
'. I'
T -/ -21 T
T
=
[_i -;
-1..
-1
4
-1
-2
4
I.
4.7 The Discretisation of the Neumann Boundary Value Problem (b) Lh is not synunetric, but DhLh with Dh = diag {d(x)d(y)}, d(O) = d(l) 1/2, d(·) = 1 otherwise, is symmetric.
71
=
The analogue of Theorem 3 reads: Theorem 4.7.9. Equation (14a,b) is solvable if and only if nT Dhqh = 0 for Dh in Exercise 8b. Any two solutions may differ by only a constant. The formulation of nT DMh = 0 with the aid of f and cp reads: _h2
2:
d(x)d(y)f(x, y) = 2h
2:
(4.7.15)
d(x)d(y)cp(x, y),
where the summand for the corner points occurs twice in the second sum, and both the different limits for cp are taken into account.
Remark 4.7.10. The sums in (15) represent summed trapezoidal formulae. fdx+ cpdr = 0 follows Equation (15), except for a remainder Thus, from O( h2l1ullcl.l (0)'
In
Ir
Theorems 4 and 5 can be transferred without difficulty. The convergence theorem, Theorem 6, becomes Theorem 4.7.11. Let u E the solution of LhUh
=
c3 •1 (n)
(D~% )
be a solution of (1). Let Uh
with Lh
= [D~.fh ~].
=
("f: ) be
We have conver-
gence of second order:
The proof is essentially the same as for Theorem 6. An additional technical difficulty is the fact that the consistency error LlhRh u - Rh Llu must also be Instead of treating the determined in x E n although u is only defined in difference equation and the boundary discretisation separately as in (12a,b) one should directly analyse the equations LhUh = % from which the values Uh(X + hn) outside of li have already been eliminated.
n.
4.7.3 Symmetric Difference for
au/an on an Offset
Grid
If we offset the above grid by h/2 in the x and y directions, we obtain the grid
in Figure 2. The points near the boundary of r. We set
flh
are at a distance h/2 from
72
4 Difference Methods for the Poisson Equation
n:= {(X,y) E r: ~ - ~ E Z or
*-~
E
z}
(cf. Figure 2). To each point near the boundary x-hn/2 (x E rh) corresponds an outlying neighbour x + hn/2. o
0
0
0
0,r--"--"'--•••• +0 ot • • • • ,0
0: Points in ilh
ot •••• *0 0, ...••••-to
.. : Points in rh
I
I
I
I
I
I
~
o
-~-.-
0
0
. 0
Figure 4.7.2. Offset grid
The discretisation of the Poisson equation (1) is -LlhUh(X) = f(x)
in x E ilh,
h- 1[Uh(X + hn/2) - Uh(X - hn/2») = cp(x)
in x
(4.7.17a)
En.
(4.7.17b)
The difference (17b) is symmetric with respect to the boundary point x and nevertheless agrees with the backward difference 8;; at the grid point x+hn/2.
Remark 4.1.12. After eliminating the values Uh(X + hn/2), x E n, we obtain a system of equations LhUh = Qh, where Remark 2 is also valid for this matrix L h . In contrast to Section 4.7.1, Lh is of size n 2 x n 2. The Theorems 3, 4, 5, and 7 hold analogously. Theorem 6 holds with the inequality (16) instead of (11).
4.1.4 Proof of the Stability Theorem 1 In the case of Dirichlet boundary values the stability statement of Theorem 4.4.11 follows immediately from the maximum principle and the bound for L-,;1. The corresponding statement of Theorem 7 for Neumann boundary values, however, cannot be proved that easily. In the literature one can only find weaker estimates which on the right-hand side of Equation (13) contain an additional factor Ilog hi. However, this factor is not to be avoided if one uses Equation (9), LhUh = qh, without condition (6) being satisfied. Let the discrete Green function 9h(X,€), x E ilh Un, € E ilh be defined by -Llh9h(X, €) = 6(x, €) :=
{h- 2 o
e} _{h- /(4 - 4h)
for x = otherwise
1
0
(4.7.18a) } if x is near the boundary if x is far from the boundary ,
4.7 The Discretisation of the Neumann Boundary Value Problem
73
(4.7.18b) where Llh and a;; act on x. 9h exists since LXE!h 8(x,~) solvability condition (6').
Lemma 4.1.13. For arbitrary % with :o.T% Uh(X) := h2
L
= 0 (c/.
=
0 proves the
(6')),
qh(~)9h(X'~)
(4.7.19)
{EO"
represents a solution
0/ LhUh = %.
PROOF. At points far from the boundary X E ilh, (LhUh)(X) = -Llhuh(x) = qh(X), At points near the boundary x E ilh, from a;;Uh = 0 and (6'), one has the identity (LhUh)(X)
= (-Llhuh)(x) = Qh(x) -
h- 1 4 _ 4h
L
Qh(~)
= qh(X),
•
{EO"
Equation (18a,b) determine 9h up to a constant.
Theorem 4.1.14. The Green function
9h(X,~)
can be so selected that
(4.7.20) This inequality corresponds to the bound (4.7) in the Dirichlet case. Before Theorem 14 is proved, we want to show that Theorem 7 follows from it. Analogously to Jo[l + jlog Ix-~111 dx ~ Kl and Jr[l + liog Ix-~1I1 drx ~ K2 we obtain h2
L [1 + jlog(lx - ~I + h) 11 ~ K~, h L [1 + Ilog(lx - ~I + h)1l ~ K~. xEr:.
From (19), (20), and qh
nT
= ih +
If Uh := Uh - nTuh/1T 1 is the solution of LhUh = qh normalised by Uh = 0, one can see that
so that the inequality (11) and Theorem 7 are proved. It remains to prove Theorem 14.
•
74
4 Difference Methods for the Poisson Equation
e
Lemma 4.7.15. Let e be one of the unit vectors (1,0) or (0,1). Let E nh be such that + he E nh also. For all these e and let there exist a function Gh(X) = Gh(X; e) with the properties
e
e
e,
(4. 7. 21a) a;;Gh
=0
on r~,
(4. 7. 21 b)
IGh(x)1 ~ hG'/(lx - el + h), where G' does not depend on e and Theorem 7.
(4. 7. 21 c)
e. Then Theorem 14 holds, and thus also
PROOF. For e,e' E nh let 9h(X,e,e') be defined as a solution of a;;gh(·,e,e') =0 on
r~,
-Llh9h(X,e,e')=h- 2 {_\
o
!~~:~,}
in Dh.
otherwise
For e' = e+he, gh(·,e,e') agrees with Gh in Lemma 15. For arbitrary e, e', one finds a connection e = e 1 , ... ,e' = e' with ek+l - e k = ±hek , e k = (1,0) or (0,1). Since 9h(·,e,e') = 'E~=19h(·,ek-l,ek), (21c) implies the estimate
eo,
,
Igh(X,e,e')I ~
hI: G'/Ox -
e k- 11+ h).
k=l If one considers first the case Xl = el = e~, X2 ~ e2 < e2 and one applies 'E~kl 11k ~ const . (1 + log(k2h) -log(klh», one obtains Igh(X, e,e')1 ~ G"[1
+ pog(lx -
el + h)l + pog(lx - e'l + h)ll·
In the general case, this bound also comes out, but one must choose the connection e k such that Ix - ekl ~ min{lx - el, Ix - e'I}. Since we know that h 'Ee/Er~ Ilog(lx - e'l + h)l ~ const for all x E Dh, gh(X,e):= 4.!:4h
I: gh(X,e,e') e/Er~
satisfies the inequality (20). Because 'Er 1 = (4-4h)jh the equations (18a,b) h are also satisfied, i.e., gh(X, e) is the Green function. • We shall construct the function Gh needed in Lemma 15 explicitly. The basic building block is the discrete singularity function Sh(X,e) on the grid in m.2 Qh:= {(x,y) E m.2 :xlh,ylh E Z}. Lemma 4.7.16. and
The singularity function defined by Sh(X, e) := ah(x - e)
4.7 The Discretisation of the Newnann Boundary Value Problem
75
j1f j1f +a:2112)/h - 1 d'T]l d'T/2 2 sin ('T]d2) + sin2('T/2/2) = __1_ j1f j1f sin 2«xl'T]l + x2'T/2)/(2h)) d'T]l d'T/2 2 1
ei(a:1111
O"h(X) := - 1611"2
-1f
-1f
811"2
-1f
-1f
Sin ('T]1/2) + sin2('T/2/2)
for all x, eE Qh has the property -Llhsh(X, e) = h- 2 for x = o otherwise.
PROOF. Let e(x, 'T]) := exp(i(xl'T]l
eand -Llhsh =
+ x2'T/2)/h). Note that
-Llh(e(x, 'T]) - 1) = 4h- 2e(x, 'T])[sin2('T]d2) + sin2('T/2/2)]
1:1:
and
e(x,'T])d'T] =
•
{4~2 ~~~~;~}.
For multi-indices a = (aba2) E 7l.2 with al ~ 0, a2 ~ 0 one defines the partial difference operators of order lal = al + a2 by (4.7.22) Starting with the representation in Lemma 16, Thomee [1] proves
For the construction of the function Gh in Lemma 15 we select, without loss of generality, e = (1,0) and keep E ilh with + he E ilh . The function
e
e
(4.7.23) satisfies (21a) but not the boundary condition (21b). Let the symmetrisation operators Sa: and Sy be defined by 1
1
Sa:Uh(X,y):= '2[Uh(X,y)+Uh(h-x,y)], SlIUh(x,y):= '2[Uh(X,y)+Uh(X,h-y)] for (x,y) E Qh. The function (4.7.24) is symmetric with respect to the axes y = h/2 and x = h/2. Thus we have (21b) on the left and lower boundary: a;G~(O,y) = a;;G~(x,O) = 0
(4.7.25)
Furthermore, G h satisfies condition (21a) just as Gh does. For each f3 E 7l.2 we define the operator P{3 by
76
4 Difference Methods for the Poisson Equation
41 [Uh(X + f31L, y + f32L) + Uh(X +Uh(X + f31L, y - f32L) + Uh(X - f31L, y -
(P,8Uh)(X, y) =
f31L,y + f32L) f32L)],
where L = 2 - 2h, and set G~(x) := (P,8G~)(x) - (P,8G~)(O).
Lemma 4.7.18. For f3 = (f31.f32) with 18QG~(x)1 ~ hK/If31 3 Here, K is independent of the
(4.7.26)
1If3l1oo ~ 2 there holds
allial ~ 2, x E ilh. choice of the ~,~ + he E ilh.
(4.7.27)
for
PROOF. According to the definition we have G~(O,O)
= O.
(4.7.28a)
The operator P,8 preserves the symmetry, i.e., Uh = SzUh implies P,8Uh SzP,8Uh. Thus we have that G~ = SzG~ = SlIG~ and consequently
=
(4.7.28b) G~(x) is the linear combination of h8;ah(X-e) for differentl with Ix-el+h ~ K'If3I, if x E ilh . According to Lemma 17,
(4.7.28c) so that (27) follows for form
lal = 2. Let lal = 1. 8QGh(X) can be written in the
aaG~(O)
I
+ L[8QG~(Xk) -
aaG~(xk-l)]
k=l
with x O = x, x' = x, IIxk - Xk-1ll oo = h. Each summand has the form ±h8'YG~ with hi = 2 so that (28b,c) lead to the estimate (28d) (4.7.28d) Likewise one infers from (28a) and (28d) the inequality (27) for proves Lemma 18. Since
L:,8EZ2\{O}
lal =
0, which •
1f31-3 < 00, Gh(x) :=
L
G~(x)
(4.7.29)
,8EZ2
exists. Since -LlhG~(X) = 0 in ilh for all f3 #- (0,0), Gh also satisfies Equation (21a). As already mentioned in the proof of (28b), Gh = SzG h = SliGh
4.7 The Discretisation of the Newnann Boundary Value Problem
77
holds so that G h also satisfies Equation (25): 8;;Gh = 0 at the left and lower boundary. The proof of 8;;Gh = 0 at the lower boundaries requires Lemma 4.7.19. Gh is L-periodic: Gh(X, y) = Gh(X + L, y) = Gh(X, y + L).
PROOF. Let G~(x+f31L, y+f32L)-G~(f31L, f32L) be abbreviated to 'Y{3(x, y). Definition (29) says that (4.7.30a) Now'Y{3 can be written as a sum over the differences h8t G'h (X+f31 L, f32L+J-th) from J-th = 0 to p.h = x, and over h8tG'h(f31L + vh,f32L) from vh = 0 to vh = y. For 1f311 ~ 2 the distance between the arguments of ~ and ~ + he is always::; (1f311 - 1)L. Each summand is thus, by Lemma 17, bounded by O(h2j(lf311-1)2) = O(h2jf3f). As a sum of such terms then 'Y{3 can be estimated by (4.7.30b) 1'Y/31 ::; O(hjf3~) for 1f311 ~ 2. We want now to show that the following equation (30c) also holds:
Gh
= k-+oo lim
"L.J "'Y{3 = k-+oo lim "L.J "L.J 'Y{3. L.J 1{31-11::;k 1/3219 1-k::;.Bl 9+k 113219
(4.7.30e)
The sums (30a) and (30c) differ by Dh := L:1{321::;kl'Y(1+k,{32) - 'Y(-k,{32l The values 1 + k and -k that can be given to f31 satisfy 1f311 ::; 2k. The absolute value of the sum Dh is then bounded, from (30b), by L:1{321::;k O(hj k 2 ) = (2k+ 1)O(hjk2) = O(hjk), so that the limits of the double sums in (30a) and (30e) are the same. Changing the variable f31 to f31 -1 then transforms (30e) into
(4.7.3Od) The first part of the identity k
L
[G'h(f31 L, f32L) - G~(f31L + L, f32L)]
/31,f32=-k k
=
L
[G~(-kL,f32L)-G~(kL+L,f32L)]
f32=-k
k
=
L
[G'h(h + kL, f32L) - G'h(kL + L, f32L)]
(4.7.30e)
(32=-k
is elementary; the second results from the symmetry G~ = S",G~. As above, the summands of the last sum can be bounded by O(hjk2) so that (30e) vanishes for k ..... 00. Together with (3Od) one obtains
78
4 Difference Methods for the Poisson Equation
= k->oo lim The proof of Gh{x, y) Since L
=2-
Gh(l, y)
'Yf3{x + L, y)
"L..J
= Gh{X + L, y).
-k~f31.f329
= Gh{x, y + L) is analogous.
2h, the symmetry Gh
= Gh(h -
1, y)
= 811:Gh and the periodicity yield
= Gh(h -
1 + L, y)
= Gh(l -
•
h, y),
i.e., 8;;Gh{1,y) = 0 on the upper boundary (l,y) E rho Likewise we show that 8;;Gh{X, 1) = 0 on the right boundary. Thus (21b) is also proved. It remains to show (21c): IGh{X)1 ~ hC/{lx - ~I + h) in nh . The G~ for 11.81100 ~ 1 are linear combinations of h8; O"h(X for = and other ~ nh, which are generated by 811:, 8 11 , and Pf3. For all there holds Ix - el ~ Ix - el, so that from Lemma 17 follows
-e)e e e
e
On the other hand, Lemma 18 shows that
I
L
G~{x)1 ~ h
1If3l1co~2
L
K/I.813 = hK' ~ hK"/(lx - ~I + h).
1If3l1oo~2
The assumptions of Lemma 15 are hence satisfied: Gh with (21a-c) exists. Thus theorems 14 and 7 are proved. •
4.8 Discretisation in an Arbitrary Domain 4.8.1 Shortley-Weller Approximation Let the boundary value problem
-.L1u =
f in
be given for an arbitrary domain over n one obtains
n,
u = cp on
r
(4.8.1)
n. If one places a square grid of step size h
flh := {(x,y) E fl: x/h E 71.
and y/h E 7l.}
(4.8.2)
as the set of grid points. By contrast, the set r h of boundary points must be defined differently from the case of the square. The left neighbour point of (x, y) E flh reads (x - h, y). If the connecting segment {(x - t9h, y): 19 E {O,l]} does not lie completely in fl, there exists a boundary point (x - sh, y) E
r
with (x, y) E flh' S E (O, 1), (x - t9h, y) En for all 19 E [0, s), (4.8.3a)
4.8 Discretisation in an Arbitrary Domain
79
which now, instead of (x - h, y), is called the left neighbour point of (x, y) (cf. Figure 1). Likewise the right, lower, and upper neighbour points can be boundary points of the following form: (x + sh, y) E r
with (x, y) E {}h, s E (0,1], (x + 'I9h, y) E {} for all '19 E [0, s), (4.8.3b) (x, y - sh) E r with (x, y) E {}h, S E (0,1], (x, y - 'I9h) E {} for all '19 E [0, s), (4.8.3c) (x, y + sh) E r with (x, y) E {}h, S E (0,1], (x, y + 'I9h) E {} for all '19 E [0, s). (4.8.3d) We set
n
:= {boundary points which satisfy (3a), (3b), (3c), or (3d)}.
(4.8.4)
A grid point (x, y) E {}h possessing a neighbour from n, is said to be near the boundary. All other points of {}h are said to be far from the boundary. As can be seen in Figure 1, a point may be near the boundary although (x ± h, y) and (x, y ± h) belong to {}h (namely if not all connecting segments lie completely in il). • E .fJh near the boundary • E {}h far from boundary DEn on boundary
Figure 4.8.1. [Jh and rh
Exercise 4.8.1. For a convex domain .fJ show that
n
= {(x, y) E
r:
x/h E 71. or y/h E 7l.}.
If one wishes to approximate the second derivative u"(x) with the aid of the values of u at x, < x < x", one can use Newton's divided differences: u"(x)
= 2[u(x") -
u(x) _ u(x) - u(xl)]/(x" _ x') x"-x x-x'
+ Rem.
(4.8.5)
Exercise 4.8.2. Show that (a) the Taylor expansion implies
IRem I ~
1 (x" - x)2 + (x - x ' )2 3 x" _ x' lIu llc3([z',zll])
max{x" - x x - x'} ~ 3 ' lI u llcs([z',zll)) for the remainder in Equation (5) if u E C 3 ([x' , x"]).
(4.8.6)
80
4 Difference Methods for the Poisson Equation
(b) In Equation (6) one can replace the norm of C 3 ([x /, x"]) by that of C2,1 ([x', x"]). (c) If x" = x + h and x' = x - h the difference in (5) agrees with the usual second difference a-a+u(x). To set up the difference equation for -Llu = fat (x,y) E ilh we use the four neighbouring points (x - Sth, y), (x + Srh, y), (x, y - suh), (x, y + soh) E n h U rh, 0
< s.
$ 1,
as defined above. The subscripts to the S are suggest respectively left, right, under and over. For points far from the boundary s. = 1 holds; for points near the boundary (x, y) at least one neighbour lies on r h and the corresponding distance s.h may be smaller than h. Equation (5) with x' = x - Sth and x" = x + srh provides an approximation for u..,..,. Analogously one can replace Uyy by a divided difference. One obtains the difference scheme of Shortley and Weller [1]:
-DhU(X,Y):=h-2{(~+_2_)u(x,y)(2 )u(x+srh,y) StSr SoSu Sr Sr + St (
2
St Sr + St
_ Su
)u(x-sth,y)-
(
2
So So + Su
)u(x,y+soh)
(2 ) u(x, y - su h )} . So + Su (4.8.7)
Remark 4.8.3. If St = Sr =
Su
= So = 1, Dh agrees with Ll h .
The discrete boundary value problem assumes the form -DhUh =!h:= Rhf on n h with (Rhf)(X) = f(x) for x E ilh , Uh =!P on rho
(4.8.8a) (4.8.8b)
The five coefficients on the right side of Equation (7) define the matrix elements Lxe for = x and for the four neighbours of x. Otherwise we set Lxe = o. The right side of the system of equations
e
e
(4.8.9a)
in which Uh is interpreted as a grid function on nh (not nh Un) is given by %
= !h + !Ph,
!Ph(X):= -
L
Lx~!p(e)·
~Er,.
Again CPh(X) = 0 holds for points x E ilh far from the boundary.
(4.8.9b)
4.8 Discretisation in an Arbitrary Domain
81
Theorem 4.8.4. Let n be bounded and contained in a strip (xo, xo+d) x IR or IR x (Yo, Yo +d) of width d. For the matrix L,. belonging to the Shortley- Weller discretisation the following holds: (a) L,. is genemlly not symmetric; (b) L,. is M -matrix with (4.8.10) PROOF. (a) Let x = (x,y) E n,. be near the boundary, but its neighbour x, = (x + h, y) E n,. be far from the boundary. Then it holds that LXXi = -2h- 2/( 1 + Bt) f. _h- 2 = LXiX' if Bt < 1. Other than in Exercise 4.7.1, in general no scaling can be found so that D,.Lh (Dh diagonal) becomes symmetric. (b) Lh need not necessarily be irreducible and hence irreducibly diagonally dominant. But the weaker condition in Exercise 4.3.8 is satisfied and proves the M -matrix property. (c) For the proof of (10) we use Theorem 4.3.16. If the domain nlies in the strip (xo, Xo + d) x IR we select Wh(X, y) := Rhw, W := (x - xo)(xo + d - x)/2. The remainder in Equation (5) contains only third derivatives that vanish for w. Thus D,.w,. agrees with Llw = -1:
The corresponding system of equations reads L,.w,. = q,. := J,. +
Exercise 4.8.5. Prove the analogue of the derivative (4.4.9):
With (10) stability has been proved. The order of consistency however is only 1, for at points near the boundary eh := DhR,.u - il,.Llu
is of the order of magnitUde O(hl). Here, Rh is the restriction to n,. u n. il,. has been defined in (Sa). We want to show that nevertheless there is convergence of order 2. The difference w,. := u,. - R,.u between the discrete solution u,. = Lh"1q,. and the solution u E C 3•1(a) of (1) satisfies -D,.w,. = - D,.u,. + D,.RhU = ilhf + DhRhU (4.8.11a) (4.8.11b)
82
4 Difference Methods for the Poisson Equation
so that Wh = L/:1ch follows. Ch can be written as c'K + cK, where c'K (resp. cK) is the error of discretisation of the x difference (resp. y difference). In turn, c'K . spl't' -z = C2:,1 + C2:,2 : IS 1 Into (.;h h h
C2:,2(X,y):= {c'K(x,y) h 0 Analogously one defines
if SI = ~r = otherwIse
1},
C",,1 ' - J: h . - Ch
_
c",,2
h'
cK,l and cK,2 and sets
+ Ch, _11,1
Ch1..= C"',1 h
The errors ~ are described by (5.7):
IIw~lIoo S IIL/:1I1oollc~lIoo, Ilc~lIoo S ~h2I1ullc,s.1(n).
(4.8.12a)
With K := ih3I1ullc2.1(n) define
Uh = Kl in ilh,
Uh
=0
For x E ilh far from the boundary Ch(X) however, we have
Ch(X)
=K
L
L~
~En"
= -K L
on rh,
= 0; for x
L~
Ch:= LhUh. E ilh near the boundary,
(x E ilh near boundary).
~Er"
Consider, for example, the case x = (x,y) E ilh near the boundary with (x - SI h, y) En. According to Exercise (2a, b) the x difference has the error
e=
Ic2:,1{x)1 < h
-
2+ 2 h 2+ 2 Sr St lIull 21 - = h- 2Sr St K < K 3 Sr + St C • (0) 8 r + St -
2h-2 St{Sr
+ St)
= -KL
~
.
The analogous estimate for cK,1{x) gives Ic~(x)l S Ch{X). Because c~(x) = Ch{X) = 0 at the x E ilh far from the boundary one has -Ch S c~ S Ch. Since L/:1 ~ 0 it follows that -Uh S wl S Uh, i.e., (4.8.12b) Equations (12a,b) together with wl
+ w~ = Wh = Uh -
Rhu prove
n
Theorem 4.8.6. Let satisfy the assumption of Theorem 4. The conue1Yence for the Shonley- Weller method is second order: IIUh - RhUlioo S
S if u E C 3,1(i'i).
~h3I1ullc2.1(n) + IIL/:1I1oo~h2I1ullcs.l(n) ( 1 h3
'3
d 2 2) + 48 h lI u ll c3.1(i'i)'
(4.8.13)
4.8 Discretisation in an Arbitrary Domain
83
Exercise 4.8.1. Show that if one uses the Shortley-Weller discretisation for all grid points near the boundary, but the mehrstellen method from Section 4.6 for all points far from the boundary, we obtain a method with convergence of third order: IIUh - RhUlloo = O(h3 ).
4.8.2 Interpolation at Points near the Boundary
Instead of discretising the Poisson equation at grid points x E ilh near the boundary, one could also try to determine Uh(X) by interpolation from the neighbouring points. If, for example, x = (x,y) E ilh is near the boundary, (x - slh, y) E n and (x + srh, y) E ilh Un, then linear interpolation yields Uh(X, y)
=
[SlUh(X + srh, y)
+ SrUh(X -
slh, y)JI(sr
+ s!).
Thus, at the point x we set up the equation (4.8.14a)
e
= (x - slh, y) should be a boundary point, one can replace Uh(e) by cp(e). If, however, (x, y + soh) or (x, y - suh) is a boundary point, we choose interpolation in the y direction:
Since
(4.8. 14b) At any x E
nh
far from the boundary the five-point formula (2.3) is used:
-(..1 hUh)(X) = fh(x) = Rhf(x) = f(x)
(x far from boundary). (4.8.14c)
Theorem 4.8.8. Let il satisfy the assumptions of Theorem 4. Let the discretisation be given by (14a-c) with the choice between (14a) and (14b) being made in such a way that always (at least) one boundary point is used for the interpolation. Let the system of equations resulting after the elimination of the be Lhuh = %. Lh is an (in geneml unboundary values Uh(e) = cp(e), E symmetric) M -matrix which satisfies the estimate (10). The discrete solutions Uh converge with the orner 2, if U E C 3,l(n):
e n
(4.8.15)
PROOF. The M-matrix property and (10) are proved as in Theorem 4.2. Let x E ilh be a point near the boundary at which (14a) is used. The error of interpolation is ck(x,y):= (Sl
+ Sr)RhU(X,y) -
slRhU(X + srh,y) - SrRhU(X - Slh,y),
1 I ~ '2SrSl(Sl 1 ICh(x,y)
+ sr)h2 IIUIICl,l(ii)'
84
4 Difference Methods for the Poisson Equation
With this c~ and K := h2I1ulb.l(ii)' one can essentially just repeat the proof of Theorem 6. • By rescaling and adding the equations (14a,b), one obtains
su)
h-2 {(s/ + Sr + So + Uh(X, y) _ ..!.Uh(X _ s,h, y) S'Sr SOSU s, 1 1 - -Uh(X + srh, y) - -Uh(X, y + soh) Sr So
--.!...Uh(X,y - Suh)} = O.
(4.8.16)
Su
Using this device one can obtain a symmetric matrix Lh, even at arbitrary n. Exercise 4.8.9. Show that the discretisation (14c), (16) leads to a symmetric M-matrix. The estimate of convergence reads
In (14a,b) linear interpolation was chosen because the values at the neighbouring points were sufficient for it. Constant interpolation by
U(X, y) = u(x - s,h, y) = rp(x - s,h, y),
if (x, y) E n h, (x - s,h, y) E
n,
is less desirable since it only provides first-order convergence: IIUh -Rhulloo = O(h). By contrast, interpolation of higher order is very applicable indeed (cf. Pereyra-Proskurowski-Widlund [1]). However, it is described by an equation which also contains points at an distance of ~ 2h. Higher boundary approximations are in particular then necessary if one wants to apply extrapolation methods (cf. Marchuk-Shaidurov [1, p. 162 ff.]).
5 General Boundary Value Problems
5.1 Dirichlet Boundary Value Problems for Linear Differential Equations 5.1.1 Posing the Problem In Section 1.2 we have already formulated the general linear differential equation of second order: (5.1.1a) Lu=f in
n
with (5.1.1b) We mentioned that, without loss of generality, one can assume tlij(X)
= aji(x)
(5.1.1c)
so that the matrix (5.1.1d)
A(x) := (aij(x))i,j=l .... ,n
is symmetric. A differential operator apparently more general than (1 b) is
~[I
a2
i,j=l
'3
L = LJ aij ax.ax'
III]
ana a2 ~[Ia a II] + ax' aij ax' + ax.ax. aij +LJ ai ax' + ax' ai +a. 3
,
i=l
'3
'
•
(5.1.2) But since, for example, (aU u) = aUu"',"'j + (aau/aXj)u"", the operator (2) can be described in the form (lb), provided the coefficients are sufficiently often differentiable. According to Definition 1.2.3, Equation (1) is elliptic in n if all eigenvalues of A(x) have the same sign. One can assume without loss of generality that all eigenvalues are positive so that A(x) is positive definite (cf. Exercise 4.3.22a). Thus, L is elliptic in n if
a!j
a!,
n
L: tlij(X)ei~j > 0
for all x E
n,
0
#- e E m,n.
(5.1.3a)
i,j=l
W. Hackbusch, Elliptic Differential Equations: Theory and Numerical Treatment, Springer Series in Computational Mathematics 18, DOI 10.1007/978-3-642-11490-8_5, © Springer-Verlag Berlin Heidelberg 2010
85
86
5 General BOWldary Value Problems
For any x E n there exists c(x) := min{aij(x)eiej : lei = I} and it must be positive (c(x) is the smallest eigenvalue of A(x)!). Hence one can also write (3a) in the form (3a'): n
L: aij(X)eiej ~ c(x)leI
c(x) > 0 for all x E
2,
n,
e E JRn .
(5.1.3a')
i,j=l
Definition 5.1.1. The equation (la), or the operator L, is defined to be uniformly elliptic in if
n
inf{c(x):x En} On
r
=
> 0 (c(x) from (3a')).
(5.1.3b)
an we impose the following Dirichlet boundary value condition: 11.
=
on
r.
(5.1.4)
5.1.2 Maximum Principle In general, the maximum principle does not hold for the equation Lu = f, nor is the solution of the boundary value problem (la), (4) uniquely determined. Example 5.1.2. Let n = (0,11") x (0,11"),
L atj(x*)ux•x; (x*) + a(x*)u(x*), i,j
(5.1.5a)
5.1 Dirichlet Boundary Value Problems for Linear Differential Equations
87
and the Hesse matrix B := (u xiXj (X)*))i,j=l, ... ,n must be positive semidefinite (here, a matrix B is said to be positive semidefinite if it is synunetric and satisfies 2:: 0 for all E IRn). From Exercise 4 follows
<e, Be)
e
n
tr(A(x*)B)
=
L aij(x*)UxiXj(x*) 2:: o.
(5.1.5b)
i,j=l
If the minimum u(x*) were negative, one would obtain from (5a,b) the result I(x*) 2:: a(x*)u(x*) > 0 in contradiction to the assumption I::; o. Part (b) is proved analogously. • Exercise 5.1.4. The trace is defined by tr(A) := I:~1 au. Prove that: (a) ~i 2:: 0 and tr(A) 2:: 0 if A is positive semidefinite; (b) tr(AB) = tr(BA) = I:~j=l ~jbji; (c) tr(AB) 2:: 0, if A and B are positive semidefinite. Hint for (c): Bl/2AB1/2 is positive semidefinite; Exercise 4.3.22d . .AI:, in the case of the potential equation, from Theorem 3 follows the
n
Corollary 5.1.5. Let be bounded (not necessarily a domain); furthermore, assume that the conditions of Theorem 3 hold. If I::; 0 in n [I 2:: 0 in n] and il there exists a negative minimum [positive maximum] of u in it must lie on the boundary an.
n
Remark 5.1.6. The continuity of the coefficients ~j,~, and a of L in Theorem 3 and Corollary 5 can be replaced by the assumption a < 0 in n or by the stipulation: In every compact set Ken let ~j, ~ and a be bounded and let L be uniformly elliptic.
5.1.3 Uniqueness of the Solution and Continuous Dependence
Lemma 5.1.7. Let n be bounded, let the coefficients of L be continuous and let (3a) be valid and a ::; 0 in n. Let Ul, U2 E e 2(n) n eO(n) be solutions of the boundary value problems LUi
If It 2::
= Ii
h in nand
in
n,
::; IP2 on
Ui
r,
=
(5.1.6)
then also Ul ::; U2 in D.
n
PROOF. v := U2 - Ul satisfies the equations Lv = h - It ::; 0 in and v =
88
5 General Boundary Value Problems
Theorem 5.1.8. (Uniqueness) Under the conditions of Lemma 7 the solution of the boundary value problem Lu = f in il and u = cP on r is uniquely determined. PROOF. Let Ut, U2 be two solutions. Lemma 7 with 11 = cP shows that Ul $ U2 as well as U2 $ Ul. Thus, Ul
on
f
= U2.
12 = f,
CPl =
CP2
= •
The next theorem states that the solution depends Lipschitz-continuously and cpo
Theorem 5.1.9. Let L be uniformly elliptic in il. Under the conditions of Lemma 7 the following holds: (5.1.7)
for solutions Ul and U2 of (6). In this inequality the number M depends only on K := sup{lllij(x)l, lai(x)l, la(x)1 :x E il}, on the ellipticity constant defined by m := inf{c(x):x E il} > 0 (cf. (3b)) and on the diameter of the domain il.
PROOF. Let il C KR(Z). For all x E il we have Zl - R $ Xl $ Zl select a
~
+ R.
We
0 so that
ma2
-
K(a + 1) ~ 1
and we define
We compare w with the solution v := Ul-U2 of Lv = 11 on From the choice of KR(Z) we have
r.
w(x) ~ v(x)
for x E
12 in il, v =
CPl-CP2
r.
Furthermore, the selection made of a ensures: (Lw)(x)
=
,a(x)llcpl...-
CP21100, +
$0 {ae2Ra - ~a(Xl~zl+R)lal1(x) a 2+al(x) a+a(x)lHII1 - 121100 '-v-'
'-v-'
~
~
1 ~ m ~ -K ~-K $ K(a + 1)}1I11 - 121100 $ -1111 - 121100 $ l1(x) - h(x) = (Lv)(x). $ 0
~
-{ma2 -
If one uses Lemma 7 with Ul
= v, U2 = w,
the result is v $ w in
n, i.e.,
5.1 Dirichlet Boundary Value Problems for Linear Differential Equations
89
where M := e2Ra • Analogously one proves -w $ v so that (7) follows.
•
Exercise 5.1.10. (a) Let {l be bounded; let the coefficients of L be continuous in {l. Now show that L is uniformly elliptic in (l if and only if L is elliptic inn. (b) Theorem 9 holds for the special case !1 = h without the assumption of uniform ellipticity. (c) Let a strip in rn.n be described by
s = {x E rn.n:O $
("I, x - x*) $ S},
where x* E rn.n is a boundary point of s, "I E rn.n with 1"11 = 1 is a unit vector, and S is the strip width. Show that inequality (7) holds with M := e6a , if n c S and a are as in the proof of Theorem 9. (d) Under the conditions of Theorem 9 show that (5.1.8)
n
Exercise 5.1.11. Let the coordinate transformation P: x E 1--+ ~ E {l' and the inverse p-1: {l' -+ be continuously differentiable. To the operator L (in the x coordinates) let L' correspond in the ~ coordinates. Show that if L satisfies the conditions of Lemma 7, or of Theorem 9, then so does L'.
n
The right side f and the boundary values cp are not the only parameters on which the solution u depends. We next investigate how the solution depends on the coefficients Uij, Ui, and a of the differential operator L. Theorem 5.1.12. Let the coefficients of LI and LII be a!j, aL ai, resp. aU, a~I, all. Let u I and u ll be solutions of
LIu I = LII,uIl = f
in
(l,
u l = u ll = cp on
r.
Let LI satisfy the conditions of Theorem 9 and let u II belong to C 2 (n). With M from (7) we then have
lIu I - uIIlloo $
(.t lIa!j - aU 1100) lIuIl II + (~lIa~ - a: Ill oo) lI lb(i)
M{
C 2(ii)
',3=1
uIl
+ lIaI -
all 1100
II ItJ a~. '3 = aH '3' the condition u
then just
u ll E
(5.1.9)
lIuII 1100 }.
E C1(n) CO(n) n C2({l) will do.
n C2({l)
is sulhcient· =. a~I : U ' , zf also a~,
90
5 General Boundary Value Problems
PROOF. Set f' := L1u II . Then 11/'-11100 = II(LI-LII)uIIlloo can be bounded by the right side of (9) without the factor M. Theorem 9 applied to L1u II = I' and LIul = I implies lIul - ulIlioo =::; Mill' - 11100' •
5.1.4 Difference Methods for the General Differential Equation of Second Order For notational reasons we limit ourselves to the two-dimensional case n = 2. E m.2 require special discretisations at the boundary as General domains explained in Section 4.8. Here we only want to discuss the difference formulae at points far from the boundary. Therefore it will suffice to base our comments on the unit square n= (0,1) x (0,1).
n
Let L be given by (lb). If one wishes to obtain a difference method of consistency order 2, the following choice suggests itself, which will later be modified for reasons of stability.
au (x, y )8: 8; + 2a12(X, y)~~ + a22(x, y)8t 8;+ al(x,y)~ + ~(x,y)8~ + a(x,y) =h- 2 [
+
-a12 (X,y)/2 all(x,y) a12(x,y)/2
(5.1.10)
a22(x,y) -2[au(x,y)+a22(x,y)] a22(x,y)
a12 (X,y)/2] au(x,y) -a12(x,y)/2
~h-l [-al~X'Y) ~(~,y) al(~'Y)] + [~ o
-a2(x,y)
0
0
a(:y) 0
~].
0
Remark 5.1.13. The difference method (10) is a nin&-point scheme of consistencyorder 2. Let Lh be the matrix of the system of difference equations in ated to (10) (cf. Remark 4.2.5):
nh
associ(5.1.11)
The solvability of Equation (11), for arbitrary qh, is equivalent to uniqueness. In the continuous case, uniqueness was essentially given by the condition a=::;O and the ellipticity condition alla22 > a~2 (cf. Theorem 8). These conditions are in general not sufficient to guarantee the solvability of Equation (11). We thus replace (10) with another discretisation.
Theorem 5.1.14. For the coefficients 01 L let (3a,b) hold, and assume that (5.1.12)
5.1 Dirichlet Boundary Value Problems for Linear Differential Equations
91
and a(x, y) ~ 0 in n. Then there exists for L a seven-point difference method of consistency order 1 such that the associated matrix Lh is an M -matrix except for the sign. In particular, the resulting Equation (11) is solvable.
= a22 > o.
Note that Condition (12) follows from (3a) only if all
PROOF. At the grid point (x, y) E nh (cf. (4.2.1a)) we abbreviate the coefficients ilij(X,y), ili(X,y), and a(x,y) to O-ij, ili, and a. The principal part ,,£aijE)2/8xi8Xj (Xl = X,X2 = y) is discretised by the following difference stars:
~ h- 2 8xr 82
8Xl8x2
[
0 1 -2 0
!h-
2
~h-2
1] ,
[ -:
[-1 ~
-1 2 -1 1 -2 1
~
[
-1].
if al2 ;::: 0,
8xfh-2
q,
0
1 -2 1
0]. (5.1.13)
if al2 < o.
-1
For al2 ;::: 0, we thus obtain
a22 - a12 2(al2 - all - a22) a22 - a12 If one introduces at2 := max{al2'0} and a12 := min{al2'0} then both for al2 ;::: 0 and for a12 < 0 we get the seven-point star
(5.1.13')
Exercise 5.1.15. Let Lh be the matrix belonging to the difference formula (13') (note that the coefficients in (13') depend on the spatial coordinates). Assume (3) and (12) hold. Prove that -Lh satisfies the conditions of Exercise 4.3.8a. The first derivative terms ili8/8x. are replaced by the forward and backward differences, respectively, ~8;' if ai ~ 0:
tili .=1
8~i + a: h- l [-a1 -,ali~ la21 -a 2
at'
at]
+ [0
~0 a
0],
(5.1.14)
is defined in the same way as ar2. Because ~ 0, the diagonal where element is negative, the others nonnegative. If one adds these terms to (13'),
92
5 General Boundary Value Problems
the resulting matrix -Lh satisfies the conditions of Exercise 4.3.8a. Therefore -Lh is an M-matrix. It is easy to see that the difference formula (13') is of second order of consistency; (14), however, contains the on&-sided differences so that the total discretisation is only of first order. • The condition (12) can be avoided if one allows larger difference stars, which also contain the values U{X1 ± vh, X2 ± p.h) for fixed V,p. E 71.. (but in general Ivl,Ip.1 > 1) (cf. Brambl&-Hubbard [1]). Layton-Morley [1) point out that with weaker conditions than (12) one may still obtain a matrix Lh, which, though not an M-matrix, does have a positive inverse. To obtain a method of consistency order 2, One must discretise E CLi{}/ {}x;. as in (10). The following corollary shows that -Lh is also an M-matrix when hCLi is sufficiently small. Corollary 5.1.16.
In addition to the assumptions of Theorem 14 let the
following hold:
(5.1.15) Then the discretisation of'ECLi{}/{}x, from (10) together with (13') leads to a seven-point difference method of second order of consistency Buch that -Lh is an M -matrix.
PROOF. -Lh satisfies (4.3.1a) and is irreducibly diagonally-dominant.
•
Exercise 5.1.17. The condition CLi;. ~ la121 + hiCLil/2 instead of (15) is not sufficient. Construct a counterexample with au = ~2 = 1, a12 = 0, h = 1/3, a;, (x, y) variable, and lasl = 6 so that Lh is singular. Considerably weaker conditions for the nonsingularity of Lh than in Theorem 14 and Corollary 16 are needed in Section 9.2 (cf. Exercise 9.2.6, Corollary 11.3.5). In general, Lh is not a symmetric matrix. Symmetry of Lh is to be expected only if L is also symmetric: L = L'. Here the formally adjoint differential operator L' which is associated to L in (2), is defined by L
,=
~
[III {}2
L....t a;'j {} .{} . ',j x, x J
{} II {}
{}2 I] - L....t ~ [ II {} {} I] a, ~ + ~ai +a.
+ ~a,ja-: + {}x,.{}X J.aij x, xJ
,=1
X.
x,
(5.1.16) It is easy to see that a symmetric operator can always be written in the form n
L=
{}
{}
L ~a;'j{x)a-:x + a(x), i,j=l x,
asj =
aji·
(5.1.17)
J
A difference method for this is given for the case fiv&-point star
n
= 2 and a12 = 0 by the
5.1 Dirichlet Boundary Value Problems for Linear Differential Equations
+ [ 00
0 0] a(x,y) 0 . 000
93
(5.1.18)
Theorem 5.1.1B. Let a1l, a22 E 0 2.1(0). The difference method (18) is consistent of order 2. The associated matrix L", is symmetric. If aii > 0 (ellipticity) and a:::; 0, then -L", is a positive definite M-matrix. PROOF. (a) For the proof of consistency, expand
v(x + h/2) := a1l(x + h/2,y)[u(x + h,y) - u(x,y)]
(y fixed)
around x + h/2, and then expand v(x + h/2) - v(x - h/2) around x. (b) The symmetry results from L:c~ = L~:c, for E il", (cf. Exercise 4.2.2c). (c) L", is irreducibly diagonally dominant so that Criteria 4.3.10 and 4.3.24 are applicable. •
x,e
The mixed term ~a12(x'Y):1I 1 [-a12 (XA) - a12(xB) 2h-2 2a12(XA)
o
+ /ya12(x,y)/x
a12(xc) + a12(xB) -2[a12(xA) + a12(xc)] adxA) + a12(xD)
(x - ~,y), XB = (x - ~,y (x+ ~,y - h). Another way of writing (19) is:
with
XA
=
+ h),
Xc
=
can be discretised by 0 ] 2a12(XC) -a12(xc) - a12(xD) (5.1.19) (x + ~,y), and XD =
4[a:b(. -~, .)a; + atb(. -~, .)a;; + a;;b(. +~, .)at + a;b(. +~, .)a:] , (5.1.19') where b := a12. Exercise 5.1.19. (a) For the coefficients of Lin (17) let the following hold: au > 0, a22 > 0, a12 = a2l :::; 0, au + a12 > 0, and also 2a22 (x, y + h/2) + a12(x - h/2, y + h) + a12(x + h/2, y) > 0, and a :::; O. Show that the difference scheme which is described by (18) and (19) has consistency order 2 and that the associated matrix - L", is a symmetric, irreducibly diagonally-dominant and positive definite M-matrix. (b) What is the suitable discretisation for the case a12 ~ 0 ? Exercise 5.1.20. The difference formula from Theorem 14 for the operator L reads au at a; + a12 (at a: + a; a;;) + a22a: a;; + a1 at + a2a: + a when
94
5 General Boundary Value Problems
a12 ~ 0, a1 ~ 0, a2 ~ o. Let the associated matrix be L h . Then prove that the transposed matrix LX describes a difference method for the adjoint operator L' and also possesses consistency order 1.
In general it is possible to show for regular difference methods that L~ is a discretisation of L'. The role of regularity is demonstrated in Example 5.1.21. Let Lu := u" + au' in n = (-1,1) with a(x) ~ 0 for x ~ 0 and a(x) ~ 0 for x > o. According to (14) au' is discretised for x ~ 0 by a(x)a+u(x) and for x> 0 by a(x)a-u(x). Let the associated matrix be Lh = L h,2 + L h,1t where L h,2 and L h,1 correspond to the terms u" and au' respectively. According to the above, 1Vh should be a discretisation of -(av)'. But the differences Lr,1Vh at x = 0 ~nd x = h are
Lr
h-1[a( -h)Vh(-h) - a(O)vh(O) - a(h)vh(h)] = -(av)' - h- 1a(0)vh(0) + O(h), h-l[a(O)vh(O) + a(h)vh(h) - a(2h)vh(2h)]
= -(av)' + h-1a(0)vh(0) + O(h);
LI
thus they are not consistent. Nevertheless, is a possible discretisation of L'v = v" - (av)', for it can be shown that the error Vh - RhV has order of magnitude O(h). To prove stability one has to show IIL;;11100 ~ const. Obviously it is sufficient to prove this inequality for sufficiently small h. In the proof of Theorem 9 we used the fact that -Lw
~
1 in
n,
W ~
0 on
r
for w(x) := exp(2Ra) - exp(a(xl - %1 + R)). Let DhUh(X) be the difference equations from which LhUh results after elimination of the boundary values. We set Wh := 2RhW; i. e., Wh(X) = 2w(x) for x E li h . The following holds: -DhWh
= -2(DhRh -
RhL)w - 2RhLw ~ 2 - 2(DhRh - RhL)W.
Each consistent difference method satisfies II (DhRh -RhL)wlloo ~ O. For sufficiently small ho we thus have
This inequality agrees for x E nh far from the boundary with -LhWh(X) ?: 1. For x E ilh close to the boundary, -DhWh(X) also contains the sum - E~Erh Lx~Wh(e), which is not contained in -LhWh(X). For the discretisations from Theorem 14, Corollary 16, Theorem 18, and Exercise 19, however, L~ ?: 0 (x E ilh, E n) holds, so that because Wh ?: 0 on n,
e
-LhWh ?: -DhWh ?:
n
(h ~
holds for all grid points. Theorem 4.3.16 yields
ho)
5.2 General Boundary Conditions
95
Theorem 5.1.22. The discretisations from Theorem 14, Corollary 16, Theorem 18, and Exercise 19 are stable under the conditions posed there, i.e., IILh"1I1oo :$ const for all h E H = {l/n: n E IN}. According to Theorem 4.5.3 the methods converge. The order of convergence agrees with the corresponding order of consistency.
5.1.5 Green's Function The idea of representing the solution by the Green function can be repeated for the general differential equation (la,b). The Green function (of the first kind) g(€, x) is singular at = x and satisfies
e
Lxg(€,x)
Leg(e,x)
= 0,
= 0
for x,€ E il, x =I-
ej
g(€,x)=O forxEror€Er. Here, L' is the adjoint differential operator (16). If L =I- L', then 9 is no longer symmetric: g(x,€) =I- g(€, x). Under suitable conditions the solution of (la-c), (4) can be represented as
u(x) =
-In
g(e, x)f(€) de
-£
x) dr~
at
where B = B~ = 2:~j=1 nj Uij is a boundary differential operator (nj are the components of the normal vector n = n(e), € E r). Only when the principal part of L agrees with .d is B the normal derivative. In the discrete case the inverse Lh"1 again corresponds to the Green function g(., .).
5.2 General Boundary Conditions 5.2.1 Formulating the Boundary Value Problem Let the differential equation be given by (l.la,b). The Dirichlet boundary condition (1.4) can be written in the form
Bu
=
on
r
(5.2.1a)
where B is the identity (to be precise: the trace on r). In more general settings B can be an operator - a so-called boundary differential operator of order 1: n B =
L b (x)8/8x + bo(x), i
i
x
E
r.
(5.2.1b)
i=1
If one introduces the vector b(x) the form
= (b 1 (x),··· ,bn(x»T,
Bu can be written in
96
5 General Boundary Value Problems
Bu = (b(x), Vu(x)} + bo(x)u(x) or B = bTV + boo
(5.2.1b')
Example 5.2.1. (a) From b = 0, bo(x) =I- 0 there results what is known as the Dirichlet condition u =
')'3:u
=
';(1- x)/2
il:Llu =0 (a)
(b)
Figure 15.2.1. (a) Boundary value problem with changing boundary condition type (b) Dirichlet problem in a disk with a cut
The normal derivative B = a/an (Le. b = n) is important in connection with L = -Ll. For the general operator L in (1.1b) the so-called conormal derivative B with
b
= An (A = (a.jkj=l, ... ,n, a.j as in (Ub»
is of greater importance, as we will see in Section 7.4. Statements about existence and uniqueness of the solution always depend on L and B. We have seen already that for L = -Ll, B = I (Dirichlet condition) uniqueness is guaranteed (cf. Theorem 3.1.2), while the problem associated to L = -Ll, B = nTV = a/an is, in general, not solvable (cf. Theorem 3.4.1).
5.2 General Boundary Conditions
97
The coefficients of B depend on position. Of course, B(x) = 0, Le., b(x) = must not occur for any x E r. But it is possible that =1= 0) in 'Y C r and b(x) =1= 0 in r\'Y. Then there is a Dirichlet condition u =
o and bo(x) = 0, b(x) = 0 (and bo
Example 5.2.3. Let n be the upper semicircle around x = y = 0 with radius 1. Let the differential equation and boundary conditions be given as in Figure la. The boundary condition changes its order at x = y = O. The solution in polar coordinates reads: u = r l / 2 sin(
The same singularity as in Example 3 occurs in the problem described in Figure lb; the solution is also r1/2 sin(
Figure 5.2.2. A domain symmetric with respect to 'Y
Example 5.2.4. Let n = nl U !"h U 'Y be as in Figure 2: let the reflection of n1 in 'Y result in n2. If one seeks a solution of Lu = f in n, Bu =
While the boundary condition Bu =
98
5 General Boundary Value Problems
Then, in addition to BtL boundary condition
=
r\bl u 'Y2), we can require the periodic
(5.2.2) on 'Yl and 'Y2. The solution is periodically continuable in the x-direction (with period X2 - xt}. The origin of periodic boundary conditions is discussed in Example 5. Example 5.2.5. (a) Let il' be an annulus which is described by the polar coordinates r E (rb r2),
5.2.2 Difference Methods for General Boundary Conditions The boundary conditions posing the least difficulties are the periodic boundary conditions. Define ilh as the set of grid points in il and on 'Yl (but not on 'Y2; cf. Figure 3). 'Yl
•
o
Figure 5.2.3. Discretisation for periodic boundary conditions
5.2 General Boundary Conditions
99
The difference equation at a grid point (Xl> y) E 1'1 has a left neighbour h,y) outside Replace Uh(X1 - h,y) in the difference equation by Uh(X2 - h, y). By doing so we have transferred the periodicity onto the difference solution without explicitly discretising Equation (2). Of course, the step size must be chosen so that X2 - Xl is a multiple of h. (Xl -
n.
Exercise 5.2.6. For periodic boundary conditions the structure of the matrix Lh changes. Let L = -.1 in the square n = (0,1) x (0,1). Assume Dirichlet conditions for the upper and lower boundaries, and for the lateral boundaries the periodicity condition (2). In analogy to (4.2.8) exhibit the form of the matrix Lh for a lexicographical arrangement of the grid points. In Section 4.7 we described the discretisation of the boundary condition B = n TV' = a/an for the case that n is a rectangle and r coincides with the grid. In the same situation one can discretise the general boundary condition (1a,b) as follows. outer
r
... ......
inner ---, (x+h,y+h) L·lne ... ~.I< ......h ...-) b2 (x - x) - b1 (y - y) ... ,. \x+ ,y • (x + h, y) (.::) X,y I
=
°
---. I I
Figure 5.2.4. Discretisation of Bu =
Let (x, y) E n = nh n r be a grid point on r. The point of intersection (x + h, y) drawn in Figure 4 is given by y = y + h~(x, y)/bl (x, Y). Note that b1 1= 0, since b TV is not to be a tangential derivative. For u E Q2(D) the directional derivative bTV'u = (b, V'u) in (x,y) is approximated by
[u(x,y) -u(x+ h,Y)]/Vh2 + (y - ij)2 = (bTVu)(x,y) + O(h).
(5.2.3a)
On the other hand, one obtains u(x + h, y) under the same conditions by interpolation up to order O(h2 ):
100
5 General Boundary Value Problems
The combination of (3a) and (3b) leads to the difference formula
(BhUh)(X, y) :
{Uh(X, y) - y+~-tiuh(X + h, y) - ?Uh(X + h, y + h) }
.jh2 + (ii _ y)2
+ bo(x, Y)Uh(X, y)
(5.2.3c)
=
Lemma 5.2.7. Let Rh be the restriction to the grid points U E C 2(n) one has
nh . For a/unction
Bh(Uh - RhU)(X, y) =
0/ the boundary condition has consistency order 1.
Equation (3c) has the form
(BhUh) (x, y) = COUh(X, y) - CIUh(X + h, y) - C2Uh(X + h, y
+ h).
(5.2.3c')
We assume that the difference equations L(En" Lx(uh(~) = fh(x) (before the elimination of the values Uh(~), ~ E n) satisfy the inequalities Lx( $ 0
(x 1: ~),
Lxx
=-
LLx(
(x E nh,~ E
nh ).
(5.2.4a)
(;o!x
Here Lxx > 0 and Lx( $ 0 correspond to the sign condition (4.3.1a). The second inequality in (4a) agrees with (4.3.4b). The corresponding inequalities for (3c') read: CI ~ 0, C2 ~ 0, CO ~ CI + C2. (5.2.4b) Let the difference equations in x E ilh and the boundary equations (3c') given for (x, y) E rh be combined into AhUh = gh. (4a,b) implies ~ $ 0, axx ~ - L(;o!x~(X,~ E nh). Therefore, Ah is an M-matrix if Ah is irreducible and (4.3.4a) holds for at least one x E nh . The above considerations explain why the boundary discretisation should satisfy the conditions (4b). (4b) holds for the case bo = 0 if and only if b2 /bl E [0,1). If ~/bl E [-1,0], (4b) can be satisfied if one interpplates between y and y - h (instead of y and y + h). If, however, I~I > Ibll, i.e., if the tangential component is greater, one could interpolate between (x + h, y) and (x + h, y ± kh), where Ib 2/b l l $ k. For the case b2 /b 1 ~ 1, however, it is more practical to interpolate at the point (x, y + h) between (x, y + h) and (x + h, Y + h). Generally one should choose as interpolation point the point of intersection of the line with the dotted straight line in Figure 4. In the preceding discussion we started with the case il = (0,1) x (0,1). If n is a general region, two discretisation techniques offer themselves (cr. Sections 4.8.1 and 4.8.2).
5.2 General Boundary Conditions
101
(x - 81h,y) = (x, 17). ...
• : Points in f1h .: Points in
n
Figure 5.2.5. First boundary discretisation
First option for discretisation: Let f1h and n be chosen as in Section 4.8.1. At near-boundary points the differential equation is approximated by a difference scheme which in the case of L = -..d corresponds to the Shortley-Weller method. For this one needs the values of Uh at the boundary points E As in Figure 5 let (x, y) = (x - 8th, y) E be a boundary point. Again we can set up an equation analogous to (3a-c). In Figure 5 (x, y) has the same position as (x + h, y) in Figure 4. In general one has to use the point of intersection of the line which is also presented by (x, y) +tb (t E IR), and the dotted straight line in Figure
e n.
n
5. Second option for discretisation: Let ilh be the grid f1h used above. Now let ilh consist of all points of ilh that are far from the boundary. For all (x, Y) E ilh difference equations (with equidistant step size) are declared which involve {Uh(e):e E ilh}. For each point (x, y) E n := ilh\ilh a boundary discretisation must be found (cf. Figure 6). Equation (3a) can be set up with (x, y) and (x + h, y) instead of (x, y) and (x + h, y). The arguments of the coefficients b1 and b2 are (x, y). This point results implicitly from ~(x, y)(x
- x)
= b1(x, y)(y - y),
(x, y) E r.
(5.2.5)
Remark 5.2.8. (a) At the grid points adjacent to the boundary, the first discretisation requires a difference scheme for Lu = f with nonequidistant stepsizes. The second discretisation requires an approximation of the nonlinear problem (5). From a programming viewpoint both procedures are undesirable because of the case distinctions. (b) If the vector b from B approaches the tangential direction, both methods fail since the straight line no longer intersects the dotted straight line from Figures 5 and 6. Here, the second discretisation fails earlier than the first one.
102
5 General Boundary Value Problems
r .. - - -. " ,
(X,Y~
,,
,
Line ' (Ax,yA) +tb
,,*,I (x + h, ii)
,.(X,y).I (x+h,y)
.--- .. I I
• Points in
nh
• Points in n
Figure 5.2.6. Second boundary discretisation
Another option for avoiding the difficulties described above consists in using variational difference equations, at least near the boundary (cf. Remark 8.6.1). Occasionally it is also possible to simplify the boundary conditions by using coordinate transformations. Let Bu = cp be prescribed on 1 cr. If one finds a transformation (x, y) +-+ (e,1]) such that the equations = 0 and b2X( = blY( are satisfied on 1, one obtains for the transformed problem Bu == u8u/8n + 1101.1. = cp on a vertical boundary piece (8/8n = -alae). The discretisation may be carried out as described in Section 4.7. The same reasoning as in the preceding sections proves s tab iii t y and convergence:
e
Remark 5.2.9. Let the difference method DhUh = fh satisfy the conditions of Theorem 5.1.22 so that -DhWh ;::: 1 holds for Wh := K1 + Wh (K > o is constant; Wh is as in the proof of Theorem 5.1.22). Let the boundary discretisation (3c') satisfy CO - Cl - C2
= bo(x) ;::: € > 0 for all x E rho
For sufficiently large K, one then obtains (Bhwh)(x) ;::: 1 for x En. Hence it follows that AhWh ;::: 1, where Ah is the M-matrix defined immediately after Equation (4b). Since IIAh"ll1oo ~ const (cf. Theorem 4.3.16), stability has been proved. The order of convergence is the minimum of 1 (order of consistency of Bh ; cf. Lemma 7) and the order of consistency of DhUh = fh. In general it is not possible to approximate b TV by symmetric differences. To construct a discretisation of Bu = cp with consistency order 2 despite this fact, set: 3
Bhuh(x, Y) := COUh(X, y) -
L CiUh(Xi, yd = g(x, y), i=l
(x, y) E n,
5.3 Bowtdary Problems of Higher Order
103
where (Xi, y,) are three points in nh. If, in order to remove the first two tenns in the Taylor expansion, one uses the differential equation Lu = f in (x,17) and the tangential derivative of Bu =
5.3 Boundary Problems of Higher Order 5.3.1 The Biharmonic Differential Equation In elastomechanics the free vibration of rods leads to (ordinary) differential equations of second order if longitudinal vibrations (= compression waves) or torsional vibrations are involved. By contrast, transversal vibrations ( = bending waves) result in an equation of fourth order. Correspondingly, the bending vibration of a plate leads to a partial differential equation of fourth order. This is the biharmonic equation (plate equation) L12u
= f in n
(5.3.1)
where Ll2 = a 4/ ax4 + 2a 4/ ax2ay2 + a 4/ ay4. Here u describes the deflection of the plate perpendicular to the surface. If the plate is firmly clamped at the edge, one obtains the boundary conditions u
=
au/an
=
(5.3.2)
with
=
and
Llu
= I{J2 on r
(5.3.3)
(simply supported plate). Other examples can be found in (7.4.12b,e). Exercise 5.3.1. Show that if one solves the Poisson equations L1v = f in n, r and, subsequently, L1u = v in nand u =
v =
Remark 5.3.2. The solutions of L1 2 u = 0 do not satisfy a maximumminimum principle (counterexample: u = x2 + y2 in n = KR(O».
104
5 General Boundary Value Problems
5.3.2 General Linear Differential Equations of Order 2m The partial derivative Dcz (a E zn : multi-index) of order lal = al + ... + an is defined in (3.2.5). A differential operator of order 2m has the form
L
=
L
acz(x)DCZ
(x E
.a)
(5.3.4a)
Iczl$2m and defines the differential equation of order 2m:
Lv. = f
(5.3.4b)
in.o.
Ellipticity has been explained thus far only for equations of second order (cf. Definition 1.2.3). Definition 5.3.3. The differential operator L (with real-valued coefficients acz) is said to be elliptic (of order 2m) at x E .a if
L
(5.3.5a)
acz(x)eCZ '" 0 for all 0 '" e E IRn.
Iczl=2m
..
Here, ecz is an abbreviation for the polynomial ef 1 e~2 ··e~n of degree lal. We set P(x, e) := E 1czl=2m acz(x)ecz. Evidently (5a) is equivalent to P(x, e) '" o for all e E IRn with lei = 1. For reasons of continuity either P(x, e) > 0 or P(x, e) < 0 must hold. Without loss of generality, we may assume that p(x,e) > 0; otherwise we scale with the factor -1 (changing from Lv.(x) = f(x) to -Lv.(x) = - f(x)). Since the set{e E IRn : lei = I} is compact it follows that c(x) := min{P(x,e): lei = I} > 0 and this justifies the formulation of (5a) as
L
acz(x)eCZ ~ c(x)leI 2m
for all e E IRn with c(x)
> O.
(5.3.5b)
Iczl=2m Definition 5.3.4. The differential operator L is said to be uniformly elliptic in .a if inf{c(x):x E .a} > 0 for c(x) from (5b). Exercise 5.3.5. (a) Translate L from (1.1b) into the notation of (4a). What are the coefficients acz for L = Ll2? (b) Prove that the biharmonic operator Ll2 is uniformly elliptic. (c) Let acz be real-valued. Why are there no elliptic operators L of odd order? (d) If the coefficients acz are sufficiently smooth one can write L from (4a) in the form L = E1czlSm EI.8ISm( -1)lfJIDfJaczfJ(x)DCZ (cf. (1.1b) and (1.2)). For m = 1 (equation of order 2) we have used one boundary condition; for the biharmonic equation (m = 2) two boundary conditions occur. In general one needs m boundary conditions
5.3 Boundary Problems of Higher Order Bju:=
I:
bjaDau =
(j
= 1,2,· .. ,m) on r
105 (5.3.6)
lal:=;m;;
with boundary differential operators B j of order 0 ::; mj < 2m. Remark 5.3.6. The boundary operators B j cannot be chosen arbitrarily, but must be independent of each other (they must form a so-called "normal system"; cf. Lions-Magenes (1, p. 113] and Wloka (1, Definition 14.1]). In particular the orders mj must differ pairwise. We have a Dirichlet boundary condi tion if B j = 8 j - 1/8n j - 1 = = 1,···, m (cf. (2)). The representation of the solution using the Green function will not be discussed at this point. However, it is remarkable that the Green function (and in particular the singularity function) for L is continuous whenever 2m > n. For L = ..12 in rn?, for example, the singularity function is 8~ Ix - Yl2log Ix - yl (cf. Wloka (1, Exercise 21.9]). ({)/8n)j-1 for j
5.3.3 Discretisation of the Biharmonic Differential Equation The simplest difference formula for L method
Dh = ..1~ =
= ..12 is the 13-point difference
2 [ h- 4 1 -8 2
. 1 -8 20 -8 1
2 -8 2
(5.3.7)
which can be represented as the square of the five-point star ..1h from (4.2.3). Let the grid ilh C il and the boundaries rh C r, "th C rn?\!1 be defined as in Figure 1. The difference equation (5.3.8a)
n
requires the values of Uh in h := ilh U rh U"th. These are determined by the boundary conditions (cf. (2)) for x E r, 8~Uh(X) := (Uh(X + hn) - Uh(X - hn)I/(2h) =
Uh(X) =
(5.3.8b) (5.3.8c)
106
5 General Boundary Value Problems 0
0
0
0
0
0
0
0
IJ
IJ
C
C
IJ
C
C
0
0
• •
0
IJ
0
c
c c c c
0
IJ
• • • • • • • • • • • • • • • • • • • •
0
0
c c
•
IJ
0
0
C
IJ
C
C
IJ
C
C
0
0
0
0
0
0
0
0
• •
0
0 0
Figure 5.S.1. Grid for the biharmonic equation in
a=
(0,1) x (0,1)
As in Equation (4.7.14b) we assign to the comer points two nonnal directions each. Correspondingly, to each comer point belong two equations (Be). Note that two (m = 2) boundary layers occur, r h and "rh, and that two boundary conditions are given.
Remark 5.3.7. If with the aid of (8b,c) one eliminates in Equation (8a) the U"rh}, one obtains a system of equations LhUh = % unknowns {Uh(e):e for {Uh(e): E nh}. Lh is not an M-matrix since the sign condition (4.3.1a) is violated. But L,/ also is in general not ~ 0 (cf. (4.3.1b)).
e
En
Thus the methods of the proof in Section 4 cannot be used here. Instead we will prove the analogue of Theorem 4.4.12.
Theorem 5.3.B. Let Uh be the discrete solution of the biharmonic equation in the square n = (0, 1) x (0, 1) defined by (8a-c). The matrix Lh described in Remark 7 is symmetric and positive definite. It satisfies
(5.3.9a) If IP1
=0
on n (cf. (8b)), then Uh can be estimated by
lI u hlln"
1
:5 25611fhlln"
h- 1/ 2
+ 16.j2IIP2Ir",
(5.3.9b)
where the norms are defined as follows:
Let the sum EXEr" in Il{J2lr" contain the corner points twice, where, for example, the value IP2(X) at x = (0,0) should be interpreted once as IP2(0, 0+), and once aSI{J2(O,O+) (cf. comment on (Be)).
5.3 BOlmdary Problems of Higher Order
107
PROOF. We limit ourselves to the most important part, the inequality (9b). In the infinite grid Qh:= {(x,y) E JR2:x/h,y/h E Z} we set Uh(X) := Uh(X)
for x E ilh,
Uh(X):= 0
for x E Qh \ilh.
After partial summation one obtains
L
uh(x)lh(x)
=
XEah
L
uh(x)Ll~Uh(X) =
XEQh
L
(LlhUh)(X)(LlhUh)(X)
XEQh
(First consider the identities Ev(x)a+a-u(x) = - E(a-v)(x)(a-u)(x) = E(a+a-v)(x)u(x) in the one-dimensional case.) From the definition of Uh it follows that LlUh = 0 on Qh \(ilh un). On one has Uh = Uh = 0, and
n
Uh(X - hn) = Uh(X - hn), Uh(X + hn) = 0, Uh(X + hn) = Uh(X - h(n) + 2h
for x E n. If one sets a';Uh(X) := [Uh(X) - Uh(X - hn)]/h = -Uh(X - hn)/h at x E rh , then
for x E
rh, implies that
(uh,LhUh)ah
= (uh,lh)ah = IILlhuhll~h +2h- 1 Ia;Uhl}h -2h- 1(a;Uh,
where h ·)ah and (., ·)rh are the scalar products belonging to the norms (9c). According to (5.3.10) one estimates (a;Uh,
+ 1
Thus one obtains
IILlhuhll~h
+ 2h- 1 Ia;Uhl}h
::; lIuhllahllfhllah
::; (Uh' fh)rh
+ 2h- 1(a;Uh, CP2)rh
(5.3.11)
+ 2h- 1 Ia;Uhl}h + h- 11
Theorem 4.4.12 shows that Iluhllah ::; IILlhuhllnh/16. Thus for
=
0
(5.3.12a)
= 0, (11) implies the inequality IILlhuhll~h ::; (2h)-1Icp21}h' therefore
lI uhllah
1 1 ::; 1611Llhuhllah ::; 16(2h)-1/21
(5.3.12b)
In the general case we write Uh as uk +ukI , where
108
5 General Boundary Value Problems
Theorem 5.3.9. (Convergence) Let the solution of the biharmonic boundary value problem (1), (2) in il = (0,1) x (0,1) satisfy u E c5,1(a). Then the discrete solution of equations (8a-c) is converyent (at least) of order
i:
Iluh - RhUlioh ::; Clh2I1ullc5.1(ii)
+ C2h3/2I1ullc2.1(ii)'
(5.3.13)
where Rh is the restriction to ilh U rho PROOF. The Taylor expansion shows that
h2 LlUhRhU = Llu + 41 (uz,n:z
1 IRI::; 360 Ilull c 5.1 (ii)in ilh. (5.3.14) Outside ilh one can specify RhU arbitrarily since these values do not appear in (13). In 'Yh we set
+ UgSlSlSl) + h4 R,
RhU(X + hn) := u(x - hn)
h3
+ 2h~2(X) + "'3Unnn -
h4 41 Unnnn
2h5 8
+ T!(8n)
5
u(x),
x En·
The choice is made so that (14) also holds for x E rho Applying Llh to LlhRhU yields 2 2 2 . LlhRhU = DhRhU = RhLl U + O(h lIull c 5.1(ii) m ilh
(Rh is the restriction to ilh). The difference Wh := Uh - RhU (defined in satisfies
nh)
DhWh = DhUh - DhRhU
= Rh(f Wh(X) ~Wh(X)
.&u) + O( h2 I1 u ll c 5.1(ii» =: 9h,
(5.3.15a)
= Uh(X) -
(RhU)(X) = ~l(X) - u(x) = 0 for x E rh,
(5.3.15b)
= 8~Uh(X) - 8~RhU(X) = ~2(X) - [~2(X) + O(h21Iullc2.1(ii»] =: "ph in x En.
(5.3.15c)
If one sets 9h = O(h2I1ullc5.1(ii» and"ph = O(h2I1ullc2.1(ii» instead of fh and ~2 in Theorem 8 (note that ~1 = Or) then (9b) implies assertion (13). • By using more complicated methods one can replace the factor h- 1/ 2 in (13) by C(f)h- E (f > 0 arbitrary) so that the order of convergence is O(h2-E), i.e., almost 2.
Remark 5.3.10. Inequality (9a) shows cond2(Lh) := IILh1l211Lh"1112 = O(h-4). In general a difference method for a differential equation of order 2m leads to the condition (5.3.16)
5.3 Boundary Problems of Higher Order
109
This indicates greater sensitivity to roundoff errors at higher orders 2m (cf. §4.4 in Stoer[l] and Bulirsch-Stoer[l]). Difference methods for boundary value problems of fourth order with variable coefficients and general domains n are discussed in an paper by Zlamal [1]. There too convergence of order O(h3/2) is shown. By contrast, O(h2)convergence was proved by Bramble [1] for the 13-point star (7) in a general domain n if the boundary conditions are suitably discretised.
6 Tools from Functional Analysis
To enable readers without previous knowledge of functional analysis to follow the next chapters we summarise here all definitions and results that will be needed later on.
6.1 Banach Spaces and Hilbert Spaces 6.1.1 Normed Spaces Let X be a linear space (alternative term: vector space) over K where K = R or K = C. In the following the normal case K = R is always intended. K = C occurs only in connection with Fourier transforms. The notion of a norm II·II:X -+ [0,00) is explained in Definition 4.3.12. The linear space X equipped with a norm is called a normed space and is denoted by the pair (X, 11·11). Whenever it is clear which norm belongs to X, this norm is called 1I·lIx and one writes X instead of (X, 1I·lIx). Example 6.1.1. (a) The Euclidean norm 1·1 from (4.3.13) and the maximum norm (4.3.3) are norms on Rn. (b) The continuous functions form the (infinite-dimensional) space CO(n). If n is bounded, all u e CO(n) are bounded so that the supremum norm (2.4.1) is defined and satisfies the norm axioms. If is unbounded, the bounded, continuous functions form a proper subset (CO(n), 1I'lI c o(il» of CO(n). Instead of II 'lIco(o)' we use the traditional notation 11·1100' (c) The HOlder-continuous functions introduced in Definition 3.2.8 form the normed space (CB(n), 1I·lIco co».
n
The norm defines a topology on X: A C X is open if for all x e A there exists an € > 0 so that the "ball" Ke(x) = {y e X: IIx - yll < €} is contained in A. We write Xn -+
x
or x
= n-+oo lim X n ,
if
IIxn - xII -+ O.
Example 6.1.2. Let In. I e CO(n). The limit process In -+ I (with respect to 1I·1l00) denotes the uniform convergence known from analysis.
W. Hackbusch, Elliptic Differential Equations: Theory and Numerical Treatment, Springer Series in Computational Mathematics 18, DOI 10.1007/978-3-642-11490-8_6, © Springer-Verlag Berlin Heidelberg 2010
110
6.1 Banach Spaces and Hilbert Spaces
Exercise 6.1.3. The nonn 11·11: X reversed triangle inequality holds
-+
111
[0,00) is continuous; in particular the
IlIxll -IIYIII :s; IIx - yll
(6.1.1)
for x, y E X.
As Example 1a shows, several norms can be defined on X. Two norms are said to be equivalent if there exists a number
II ·11 and III . III on X o < C < 00 such that 1
C
IIxll :s; IIlxlll :s; Cllxll
(6.1.2)
for all x E X.
Exercise 6.1.4. Equivalent nonns lead to the same topology on X.
6.1.2 Operators Let X and Y be normed spaces with the norms II . IIx resp. II . lIy. A linear mapping T: X -+ Y is called an operator. If the operator norm IITlly<-x := sup{IITxlly/llxllx: O f x E X}
(6.1.3)
is finite, T is said to be bounded (cf. (4.3.9)). Exercise 6.1.5. (a) T is bounded if and only if it is continuous. (b) 11·lly.....x is a nonn. With the addition (Tl + T2)X = T1X + T2X and the multiplication by scalars, the bounded operators fonn a linear space which is denoted by L(X, Y). (L(X, Y), II . lIy<-x) is a nonned space. With the additional multiplication (T1T2)x = T1(T2X), L(X,X) even forms an algebra with unit I, since IIIlIx .....x = 1. I always denotes the identity: Ix = x. Exercise 6.1.6. Prove that IITxlly
:s; IITlly<-xllxllx for all x
E X, T E L(X, Y).
(6.1.4a)
If Tl E L(Y, Z) and T2 E L(X, Y), then T1T2 E L(X, Z) and (6.1.4b)
We write Tn -+ T for T, Tn E L(X, Y) if liT - Tnlly .....x the operator nonn).
-+
0 (convergence in
112
6 Tools from Functional Analysis
6.1.3 Banach Spaces A sequence {xn E X: n ~ I} is said to be Cauchy convergent, or is called a Cauchy sequence, if sup{llxn - Xm IIx: n, m ~ k} -+ 0 for k -+ 00. A space X is said to be complete if any Cauchy-convergent sequence converges to an x EX. A Banach space is a complete normed space.
Example 6.1.7. (a) m.n is complete, but 0 as well as (Ck.l(D), 1I·lb.,lCD» are complete, thus Banach spaces. PROOF of (b). If {Un} is Cauchy convergent then there exists a limit u* (x) := limn--+ooUn(x) for all xED. Since Un converges uniformly to u·, u· must be continuous, i.e., u* E CO(D). •
Denote by Loo(D) the set of functions that are bounded and locally integrable on D. Here we do not distinguish between functions which agree almost everywhere. In this case the supremum norm is defined by IIuliLOOCD) := inf{sup{lu(X) I: X E D\A}:A set of measure zero}}
(Loo(D), II 'IILooCD» is a Banach space. Exercise 6.1.8. Let X be a normed space and Y a Banach space. Show that L(X, Y) is a Banach space. Exercise 6.1.9. Let X be a Banach space and Z C X a closed subspace. The quotient space XI Z has as elements the classes x = {x + z: z E Z}. Show that X/Z with the norm IIxil = inf{IIx + zllx: z E Z} is a Banach space.
A set A is said to be dense in (X, II· IIx) if A c X and A = X, i.e., every x E X is the limit of a sequence au E A. If (X, II· IIx) is a normed, but not complete, space, one calls (X, 1I·lIx) the completion of X, if X is dense in X and IIxllx = IIxlix for all x EX. The completion is uniquely determined up to isomorphism, and can be constructed in the same way as one constructs m. as the completion of
6.1 Banach Spaces and Hilbert Spaces
113
Theorem 6.1.11. Let Xo be a dense subspace (or just a dense subset) of the normed space X. Let Y be a Banach space. (a) An opemtor To E L(Xo, Y) defined on the subspace Xo with IITolly<-xo = sup{IIToxllylllxllx:O =1= x E Xo} has a unique continuation T E L(X, Y), i.e., Tx = Tox for all x E Xo. (b) For Xn -+ x (xn E X o, x E X) holds Tx = lim,.-+oo Toxn. (c) IITlly+-x
=
IIToIIT+-xo·
PROOF. For x E Xo, T is defined by Tx = Tox, while for x E X\Xo there exists a sequence Xn -+ x, Xn E Xo and Tx = limn -+oo TXn = lim,.-+oo Toxn. It remains to show that lim Toxn exists and is independent of the choice of the sequence Xn E Xo. Since IIToxn - TOxmlly ~ IITolly+-xllxn - xmllx, the sequence Toxn is Cauchy convergent. Because of the completeness of Y there exists y E Y with Toxn -+ y. Similarly, for a second sequence x~ E Xo with x~ -+ x, there exists for the same reason a y' E Y such that Tox~ -+ y'. Since y' - y = lim(Tox~ - Toxn) and IITox~ - TOxnlly ~ IITo(x~ - xn)lIy ~ IITolly<-xollx~ - xnllx -+ 0, Tx is well defined by (b). Part (c) follows from IITxliy IlIxllx = lim IIToxnl/y\lIxnllx for Xn -+ x, Xn E Xo. •
Exercise 6.1.12. Let Xo be dense in (X, 11·11). Let 111·111 be a second norm on Xo, that is equivalent with 11·11 on Xo. Show that the completion of Xo with respect to 111·111 results in (X, III· liD and 11·11 and 111·111 are also equivalent onX. A corollary of the "open mapping theorem" (cf. Yosida [1, §II.5]) is the following, important result:
Theorem 6.1.13. Let X, Y be Banach spaces. Let T E L(X, Y) be injective and surjective. Then T-l E L(Y,X) also holds.
Exercise 6.1.14. Let X, Y be Banach spaces, and T E L(X, Y) be injective. Show that: (a) Yo := range(T) := {Tx: x E X} with 1I·lly as norm is a Banach subspace ofY. (b) T- 1 exists on Yo and is bounded: T- 1 E L(Yo, X). Let X and Y be Banach spaces with X C Y. Obviously the inclusion denoted by I : x E X 1-+ X E Y is a linear mapping. If it is bounded, i.e.,
IE L(X, Y)
that is, IIxlly ~ Cllxllx
for all x E X,
(6.1.5)
then X is said to be continuously embedded in Y. If furthermore X is dense in (Y, II . lIy), then X is said to be densely and con tin uously embedded in Y.
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6 Tools from Functional Analysis
Exercise 6.1.15. Let X eYe Z be Banach spaces. Show that if X is [densely and] continuously embedded in Y and Y is [densely and] continuously embedded in Z, then X is [densely and] continuously embedded in Z.
6.1.4 Hilbert Spaces
A mapping (., .): X x X -+ K (K = IR or K = CIJ) is called a scalar product on X if (x, x)
> 0 for all 0 ::J. x
E
X,
(6.1.6a)
(Ax +y,z) = A(X,Z) + (y,z)
(6.1.6b)
(x,y) = (y,x)
(6.1.6c)
for all A E K,x,y,z E X, for all x,y E X,
where ~ denotes the complex conjugate of A. Exercise 6.1.16. Show that (a) proof use (b) the Schwarz inequality:
Ilxll
:= (x,x)1/2 is a norm on X. For the
l(x,y)1 ~ IIxlillyll for all x,y E X. Hint: consider IIx - Ayll ~ 0 for A= IIxll 2 /(x, y), if(x, y) ::J. o.
(6.1.7)
(c) (-, .): X x X -+ K is continuous.
A Banach space X is called a Hilbert space if there exists a scalar product (., ·)x on X such that IIxlix = (x,x)lt 2 for all x EX. x E X and y E X are said to be orthogonal (x .1 y) if (x, y)x = o. If A c X is a subset of the Hilbert space X then the orthogonal space A1. := {x EX: (x, a)x = 0 for all a E A} defines a closed subspace of X.
Lemma 6.1.17. Let U be a closed subspace of a Hilbert space X. Then X can be decomposed into the direct sum X = U $ U 1., i. e., any x E X has a unique decomposition x = u + v with u E U, v E U 1.. Purthermore, IIxll~ = lIull~ + Ilvll~· Exercise 6.1.18. Let A be a subset of the Hilbert space X. Prove the equivalence of the following statements: (a) A is dense in X; (b) A1. = {a}; (c) x E X and (a,x)x = 0 for all a E A imply x = 0; (d) for every O::J. x E X there exists an a E A with (a,x)x ::J. O.
6.2 Sobolev Spaces
115
6.2 Sobolev Spaces In the following il is always an open subset of JRn .
L2(il) consists of all Lebesgue-measurable functions whose squares on il are Lebesgue-integrable. Two functions u, v E L2(il) are considered to be equal (u = v) if u(x) = v(x) for almost all x E il, i.e., for all x E n\A, where the exceptional set A has Lebesgue measure J.t(A) = o. Theorem 6.2.1. L2(n) forms a Hilbert space with the scalar product (u,v)o:= (U,V)L2(S1):=
L
u(x)v(x)dx
(6.2.1)
and the norm
(6.2.2)
Lemma 6.2.2. The spaces COO(il)nL2(il) and C8"(il) are dense in L2(il). Here C[f(il) := {u E COO(il): supp(u) compact, supp(u) cc n}.
(6.2.3)
supp(u) := {x E il:u(x) # O} denotes the support ofu. The double inclusion il' cc il indicates that il' lies in the interior of il (i.e., il' C il and TI' n
an = 0).
Let Da be the partial derivative operator (3.2.5b). In the following we need so-called weak derivatives Dau which are defined in a nonclassical way. Definition 6.2.3. u E L2(il) has a (weak) derivative v := Dau E L2(il) if for the latter v E L2(il) holds: (w,v)o
= (-l)la l(D aw,u)o for all w E C8"(il).
(6.2.4)
Exercise 6.2.4. Show the following: (a) Let u have a weak derivative Dau E L2(il). If the classical derivative nau exists in il' C il, it coincides there (almost everywhere) with the weak one. (b) If u has the weak derivative Va = Dau E L2(n), and if Va has the weak derivative v a+J3 = DJ3va E L2(il) then v a+J3 is also the weak Da+J3-derivative of u; i.e., Da+J3u = DJ3(Dau ). (c) Let il C JRn be bounded and 0 E il. u(x) := Ixl 17 (0" E JR) has weak first derivatives in L2(il) if 0" = 0 or 20" + n > 2.
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6 Tools from Functional Analysis
(d) For u" E COO(n) let u" -+ L2(n) nonn. Then v = nau.
U
E L2(n) and Da u" -+ v E L2(n) in the
For further applications we consider -
N-
Example 6.2.5. Let n = Ui=l ni, where the bounded subdomains n, are disjoint and have piecewise smooth boundaries. Let k E IN. A function u E Ck-l(a) with restrictions u In.E ck(lii) (1 ~ i ~ N) has a (weak) kth derivative Va = nau E L2(n), lal ~ k, which coincides with the classical one in i.
un
PROOF. The (k - 1)st derivatives exist as classical derivatives so that the assertion need only be shown for k = 1. Let Da = a/ax;, W E C{f(n). The assertion is obtained via integration by parts:
-(Daw,u)o
= -(wz.pu)o = -
In wziudx
I:ln, wzjudx = I: [rIn, wuzi dx - lan. r wun~i) dr] = I:l dx = I: r wvadx = r wvadx = (w,va)o, n, In. ln
=-
i
i
WU Zi
i
i
since every x E ani \ an also belongs to ank \ an for some k =f. i with opposite • nonnal direction n(k) = _n(i). The inequality I(u,v)ol ~ lulolvlo (cf. (1.7)) reads explicitly I
For v I
In
u(x)v(x) dxl
~
[In
lu(x)12 dx
In
Iv(x)12 dx]1/2.
(6.2.5a)
= 1 the result is
In
u(x) dxl
~ J1.(n)1/2[1n lu(x)l2 dx]1/2
(J1.(n): measure of n).
(6.2.5b)
For a E LOO(n),u,v E L2(n) one now has (6.2.5c)
6.2.2 Hk(n) and HG(n) Let k E IN U {o}. Let Hk(n) c L2(n) be the set of all functions having weak derivatives Da u E L2(n) for lal ~ k:
6.2 Sobolev Spaces
117
The Sobolev space denoted here by Hk(il) is denoted by W;(il) in some other places. Usually, there Hk(il) is a space which only coincides for sufficiently smoothly bounded il with W;(il). Theorem 6.2.6. Hk(il) forms a Hilbert space with the scalar product (U,V)k:= (U,V)Hk(n) :=
L
(Dau, D a V)£2(n)
(6.2.6)
lal9
and the (Sobolev) norm lulk := lIuIlHIo(n) :=
L
II Da u IlL2(n)"
(6.2.7)
lal~k
Lemma 6.2.7. O=(il)
n Hk(il)
lies densely in Hk(il).
Lemma 7, whose proof can be found, for example, in Wloka [1, Theorem 3.5] permits a second definition of the Sobolev space Hk(il): Remark 6.2.8. Let Xo := {u E 0= (il): lulle < co}. The completion of Xo in £2(il) with respect to norm (7) results in Hk(il). Definition 6.2.9. The completion of 08"(il) in £2(il) with respect to the norm (7) is denoted by HG(il). Theorem 6.2.10. The Hilbert space HG(il) is a subspace of Hk(il) with the same scalar product (6) and same norm (7). 08"(il) is dense in HG(il). For k = 0 the1l~ holds (6.2.8)
PROOF. (a) 08"(il) C Xo (cf. Remark 8) implies HG(il) C Hk(il) C £2(il). (b) According to the definition 08"(il) is dense in HG(il). (c) For k = 0 the norms (2) and (7) coincide. Lemma 2 proves Hg(il) = £2(il). Because of HG(il) C Hk(il) C £2(il) (k arbitrary) (8) follows. • Lemma 6.2.11. For il bounded,
II· IIH"(n)
and (6.2.9)
are equivalent norms in HC(il).
PROOF. (a) Evidently,
1·lk,o ~ 1I·IIHk(n).
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6 Tools from Functional Analysis
(b) Let u E Coca). For an a with lal = k -1 set v := DOtu E Coca). There exists an R with a c KR(O). For each x E a, Xl E [-R, Rj; thus
(cf. (5b». Integration over x E a yields Ivl~ ~ 4R2Iv:C11~ ~ 4R21ul~ 0' since V:Cl is k-fold derivative. Summation over all a with lal = k-l yields luiLl,O ~ Ck-llul~,o. Now luIJ-l,O ~ Cj-liullo follows likewise for alII ~ j ~ k, and
thus lullo ~ CjCHl· .... Ck-dul~,o· Since lul~ = 2:~=0 lullo ~ Clul~,o' for all u E Coca), the statement follows from Exercise 6.1.12. •
Exercise 6.2.12. Show that (a) for bounded a and k 2:: 1, Hk(a) and H3(n) are different. Hint: Consider the constant function u(x) = 1 and use Lemma 11. (b) Lemma 11 holds even if n is bounded in one direction, i.e., if II lies on a strip {x E lRn : IXkl < R} (k E {I,···, n}). Theorem 6.2.13. Let m 2:: 1. There exist constants C 7](10, m) such that
= C(m)
lulk ~ Clul~mlul~m-k)/m for all 0 ~ k ~ m, u E Ho(ll),
lulk
~ flulm + 7](E)lulo for all E > 0, 0 ~ k
PROOF. (a) Partial integration for
< m, u E HO(ll).
and 7](10)
=
(6.2.lOa) (6.2. lOb)
lal = 1 shows that
IDOtul~ = (DOtu, DOtu)o = _(D 2Ot u, u)o ~ ID 20t ui0lui0 ~ lul2lulo.
Since also lul~ ~ lul2lulo, it follows that lul~ ~ (n + 1)luI2Iulo. If one replaces u by DP u with 1.81 = I, one obtains in the same way lul,2 ~ Clull+llub-l
(1 ~ l
< m; u
E
Ho(ll».
(6.2.lOc)
(b) Let u =I 0 be fixed. Set VI := log lub, WI := [lv m + {m -1)vo1/m and ZI := V, - W,. (lOc) leads to 2z1 - ZI-l - ZI+l ~ e:= loge, where Zo = Zm = O. For z = {Zl.··., zm_d T one obtains Az ~ en. Here A is the M-matrix (4.1.9b) for h = 1. A -1 2:: 0 proves that z ~ c:= eA- 1 n. VI = ZI+WI ~ CI+WI (I ~ 1 < m) implies lull = exp(vl) ~ exp(ci + WI), i.e., (lOa) with C:= exp(ellA -111100)' (c) Elementary calculation shows that for each E > 0, 0 ~ ~ eo < 1 there exists an 7]{ 10, 8 0 ) such that
e
aeb l - e ~ Ea+7](E)b for all a,b 2:: O. Formula (lOb) is a corollary of (lOa,d).
(6.2.1Od)
•
6.2 Sobolev Spaces
119
By similar means, together with (5.3.10), one proves
{DOtu,DPu)o :;; €Iul~ + {4€)-1Iul~ (€ l> 0, lal :;; m, I\PII:;; k:;; m, '11. E Hm{n)), {DOtu,DPu)o :;; €Iul~ +ll{€)lul~ (€ > 0, lal:;; m, IPI < m, '11. E H[{'(n)).
Remark 6.2.14. The set {u E COO{n):supp{u) compact, 1'11.1", in H"'{il). .
(6.2.10e) (6.2.lOf)
< co} is dense
PROOF. Let '11. E H"'(n), € > O. According to Lemma 7 there exists a function COO{n) with 1'11. - uEI", :;; €/2. There exists a E coo{m,n) with a{x) = 1 for Ixl :;; 1, a{x) = 0 for Ixl ~ 2. For sufficiently large R, one also has luE{x)-a{x/ R)uE{x)l", :;; €/2. Thus there exists v{x) = a{x/R)uE{x) E c8"{n)
UEE
with 1'11. -
vi", :;; €.
•
Since "supp(u) compact" already implies "supp(u)
cc m,n"
we obtain
The Leibniz rule for derivatives of products proves
Theorem 6.2.16. E H"'(n).
l\auI\Hk(SJ):;; C",l\al\ck(.i'i)lIuIlHk(SJ) for all a
E
C"'(Q),
'11.
Theorem 16 together with the substitution rule for volume integrals shows
Theorem 6.2.17. (Transformation theorem). LetT:n -+ n ' be a oneto-one mapping onto n ' with T E cmax(""l)(n) and IdetdT/dxl ~ 0 > 0 in n. We write v = '11. 0 T for vex) = u(T(x)). Then '11. E H"'(n' ) ['11. E H~(n')] also implies '11. 0 T E H"'{n) [E H~(n)l and (6.2.11)
6.2.3 Fourier 'Iransformation and H"'{m,n) For
'11.
E C8"(rn.n ) one defines the Fourier-transformed function
u by (6.2.12)
Note that bounded.
u is
described by a proper integral since the support of
'11.
is
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6 Tools from Functional Analysis
Lemma 6.2.18. Let
U
E Co:'(ffin). For R
~ 00
converyes uniformly to u(y) on supp(u).
PROOF. It suffices to discuss the case n tion with respect to { results in
IR(u; y)
= (I/,K)
= 1 (by Fubini's theorem). Integra-
L
(x - y)-l sin(R(x - y»u(x) dx.
Then IR(I;y) = (1/'n") IRt-1 sin(t) dt = 1 for all R > O. Since U E coo(ffin), then also w(x, y) := [u(x)-u(y»)/(x-y) E coo(ffi2n). The estimates w(x, y) = O(I/lxl) and wa:(x, y) = 0(I/x2 ) hold uniformly for y E supp(u). Partial integration yields
IR(u(,) - u(y);y) = IR«' - y)w(·, y); y) = (1/7r) fIR sin(R(x - y»w(x,y) dx
=
-(I/7r)
L
cos(R(x -y»Wa:(x,y)dx/R = O(I/R).
The statement follows from
IR(u; y)
= U(y)IR(I;y) + IR(u(,) -
u(y);y)
= u(y) + O(I/R).
•
PROOF. Lemma 18 shows
Lemma 6.2.20. The inverse Fourier transformation 9="- l u = u is defined for u E Co:'(ffin) by (13):
(9="-lu)(x) := (27r)-n/2 {
1IR"
ei({,x)u({)d{
:= lim (27r)-n/2 [ R-+oo
11{loo~R
ei({,x)u({)d{.
(6.2.13)
6.2 Sobolev Spaces
121
PROOF. We have from Lemma 18
f
(21T)-n/2
f
ei({,x)u(e)£le = (21T)-n
J 1{loo5,.R
f
ei({.x-y)u(y) dy£le
J 1{loo5,.R JJRR
= u(x) +O(1/R).
•
Theorem 6.2.21. !1',!1'-1 E L(L2(IRn), L2(IRn» with 1I!1'1I£2(JRR)t-L2(JRR) = 1I!1'-1IlL2(IRR)t-L2(RR) = 1, i.e., !1' is an isometric mapping of L2(IRn) onto itself The scalar product satisfies (u,v)o = (u,v)o for all u,v E L2(IRn). PROOF. Since Cg:'(IRn) is dense in L2(IRn) (cf. Lemma 2), !1' can be continued to!1': L2(IRn) ..... L2(IRn) (cf. Theorem 6.1.11). The norm estimate follows from Lemma 19. The roles of !1' and :r- 1 are interchangeable (cf. (12) and (13»; thus !1'-1 E L(L2(IRn), L2(IRn» also holds. The second statement results from
(u, v)o =
~(Iu + vl5 -lul5 -lvI5) = ~(Iu + vl5 -lul5 -lvI5) = (u, v)o . •
Exercise 6.2.22. Prove that: (a) With e a
= ell .. ·e~R, there holds (6.2.14)
(b) There exists C for all e E IRn.
= C(k) such that iJ(1+leI 2)k ~ Llal91eal2
~ C(1+leI 2)k
Lemma 6.2.23. (a) lulk = h/LaI9IeaI2u(e)lo holds for all u E Hk(IRn). (b) A norm on Hk(IRn) equivalent to 1·lk is (6.2.15)
PROOF. (a) It suffices to show the statement for u E Cg:'(IRn) (cf. Corollary 15, Theorem 6.1.11): 2
lul~ =
E lal9
IDau l5 =
E lal9
I!1'D au15 =
E
leau (e)15
lal9
(b) The statement follows from Exercise 22b.
=
E lal9
lea I2u(e)
.
o
•
Lemma 6.2.24. Let 8h.j be the difference operator8h•j u(x) := [u(x+~ej) u(x- ~ej)}fh, where ej is jth unit vector. Ifu E Hk(IRn) and 18h,julk ~ C for all h > 0, 1 ~ j ~ n, then u E Hk+1(IRn) holds. Conversely, 18h•j ulk ~ lulk+l holds for all u E Hk+1(IRn), h > o.
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6 Tools from Functional Analysis
PROOF. (a) From !f(u(· + 8ej))(e) = e-i6~iu(e) follows !f(8h ,ju)(e) = sin(ejhI2)u(e). Hence, since 4h- 2sin2(ejhI2) ~ ej for h ~ 1/1el, follows 1!f(8h ,ju)(e)1 2 = ejlu(e)l2 for lei ~ 1/h. Summation over j = 1, ... , n and integration over then gives
¥
e
r .. (1 + leI 2)k+1Iu(e)1 2de 1m.
(lul~+d2 =
~ ~
r ... + (lul~)2 + 11~19/h { (1 + lel 2)klel 2lul 2de
11~1?1/h
1
1~I?l/h
~ (
11~1?1/h
The inteliral over
A2 A2 ... + (Iulk) +~ .LJ(18h ,j u lk) j=l
'" + Cklul~ + nC2 •
lei ~ 1/h vanishes for
(b) For the converse use (8h ,ju)(x)
=
h -+ 0 so that the statement follows. f~~~2u:l)i(x+thej)dt. •
n
Let 8 ~ O. For = m.n one can define the following scalar product (16a) and the Sobolev norm (16b) for all u e C8"(m.n): (6.2.16a) (6.2.16b) The completion in L2(m.n) also defines the Sobolev space HB(m.n) for noninteger order 8. On the basis of Lemma 23 and Exercise 6.1.12 the newly defined HB(m.n ) for 8 e IN U {O} agree with the Sobolev spaces used until this point. Let c m.n . The number 8 ~ 0 can be decomposed as s = k + A with k e INu {O} and 0 < A < 1. We define
n
(u,v)s:=
~ lal::>k
+
lul s
Jlxn
[1
Dau(x)Dav(x)dx
n
[Dau(x) -
D~:(~)~[I~:~(X) -
:= IIUIiHaCS1) := J(u,u)s
(8 = k +A, 0
Dav(y)) dxdY] , (6.2.17a)
< 8 < 1).
(6.2.17b)
The norm I·IB is called the Sobolev-Slobodeckir norm. One can define the Hilbert spaces H8(il) and H3(n) in the same way as in the case 8 = k E IN. The properties of these spaces are summarised in the following theorem (cf. Adams [1), Wloka [1]):
6.2 Sobolev Spaces
123
Theorem 6.2.25. Let s ~ O. (a) For n = IRn the norms (I6b) and (I7b) are equivalent, i.e., both norms define the same space HB(IRn) = H3(IRn). (b) {u E COO(n): supp(u) compact, lulB < oo} is dense in HB(n). (c) CO"(n) is dense in H3(n). (d) aDa(bu) E Hs-Ial(n), if lal ~ s, u E HB(n), a E ct-ial(n), bE Ct(n), where t = s E lN U {O} or t > s. (e) HB(n) c Ht(n), H&(n) c H&(n) for s ~ t. (f) In (IOa,b) k and m can be real. (g) In (11) replace IITllc"Co) by IITllc"co)' with t > k if k rt IN. Exercise 6.2.26. Check, using the norm (I6b), as well as (I7b), that the characteristic function u(x) = 1 in [-1,1], u(x) = 0 for Ixl > 1 belongs to HB(IR) if and only if 0 ~ s < 1/2.
6.2.5 Trace and Extension Theorems
The nature of boundary value problems requires that one can form boundary values ul r (restriction of u to r = an, or trace of u on r = an) in a meaningful way. As can be seen easily, a Holder-continuous function u E CS (n) has a restriction ul r E CB(r) if only r is sufficiently smooth. But, from u E HS(r) does not necessarily follow ul r E HS(r). Since the equality u = v on HS (.0) only means that u(x) = vex) almost everywhere in .0, and r is a set of measure zero, u(x) =f vex) may hold everywhere on r. Also, the boundary value u(x) (x E r) cannot be defined by a continuous extension either since, for example, u E Hl(n) need not be continuous (cf. Exercise 4c). The inverse problem for the definition of ul r is extension: does there exist, for a given boundary value r.p on r, a function u E HS(n) such that r.p = ul r? If the answer is negative there exists no solution u E HB (.0) to the Dirichlet boundary value problem. First we study these problems on the half-space
n=
IR~ := {(Xl.··· ,xn ) E IRn:x n
Functions u E H8(IR~) restricted to r.
r = an =
IRn- 1 x {O}. (6.2.18) will first be continued to it E HB(IRn) , and then it
> O} with
Theorem 6.2.27. Let s ~ O. There exists an extension operator ifJB E L(HS(IR~), HS(IRn» such that for all u E HS(IR~) the extension it = ifJsu and u coincide on IR~. PROOF. For s = 0, set it = u on IR~ and it = 0 otherwise. Since lIuIlL2(Rn) = lIuIlL2(R+), the mapping defined by ifJou := it is bounded: lIifJolIL2(Rn)<-L2(R+) = 1. For s ~ 1 define ifJsu = it by
124
6 Tools from Functional Analysis
(Le., by reflection on r). For s = lone obtains, for example, luh = J2lu/t, i.e., ~s E L(H8(m.~), HB(m.n)). For larger s one uses higher interpolation formulae for u( ... , -xn ) (cf. Exercise 9.1.13, cf. Wloka [1, p. 101)). • In the following the restriction ul r is written in the form 'YU. At first, 'Y is defined only on CO'(m.n ): 'Y: CO'(m.n ) -
cO'(r) c L 2 (m.n -
1 ),
hu)(x):= u(x)
We write x = (x', xn) with x' = (Xl."·, xn-t) E m.n - 1 : HB(r) = HB(m.n- 1).
m.n- 1 . r
for all (x) E r. (6.2.19) is identified with
Theorem 6.2.28. Let s > 1/2. Then '"Y from (19) can be continued to 'Y E L(HB(m.n), Hs-l/2(m.n- 1 )). Thus we have in particular !'YUl s -l/2 S cslul s for u E H8(m.n ). In the case n = 1, i.e., 'Yu = u(O), we have hul S Cslul B.
PROOF. It suffices to show l'Y'ILI:-l/2 s C~I'UI: for u E CO'(m.n ) (cf. Theorem 6.1.11 and Theorem 25a). Let the Fourier transforms of 1£ E CO'(m.n ) and W := 'YU E CO(m.n - 1) be u = !fnu and w = !fn-lw (!flc:k-dimensional Fourier transform). !fn can be written as product !flo !fn-1 where !fn-1 acts on x' E m.n - 1 and !f1 on Xn . Therefore W("x n) := !fn-1U("Xn) has the properties u(e',,) = !flW(e',·) and w = W(-,O). According to Lemma 20, w(e') = W(e', 0) = [!f11u(e', ')I\~n=O has the representation
w(e')
= (271')-1/2
L
u(e',en)den
for all e' E
m.n- 1.
(6.2.20)
Since u E HS(m.n ), O(e', en) := (1 + 1e'12 + e~)B/2u(e', en) lies in L2(m.n) (cf. Lemma 23b). Inequality (5a) yields
The first factor may be computed as K B (l + le'1 2)1/2-B with KB = fIR(! + x2)-s dx < 00, since 8 > 1/2. The second is U(e') := 1I0(e', ')IIL2(IR) E L2(Rn-l) because IIUIIL2(Rn-1) = liOIIL2(Rn) (Fubini's theorem). Together we have:
Integration over e' E
m.n- 1 results in
6.2 Sobolev Spaces
125
Thus Iwl:-l/2 ::; C~lul: is proved with C~ = (Ks/(27r))1/2 (cf. (16b)). In the case n = 1, w(e') already represents 'Yu = u(O), and the integration over e' is not required. • Theorem 28 describes the restriction u(·, 0) = 'YU to Xn = o. Evidently, similarly we have lu(.,xn )ls-l/2 ::; Cslul s for any other Xn e m., with the same constant Cs. The mapping Xn 1-+ u(·,x n ) is continuous [resp. Holdercontinuous] in the following sense.
Theorem 6.2.29. For s
> 1/2 the following
statements hold:
lim lIu(·,x n ) - u(·,Yn)IIH.-l/2(lRn-1) = 0 for all Xn
'IIn--t:r:n
e IR,u e HS(IRn), (6.2.21a) (6.2.21b)
IIU(·,X n ) -U(-,Yn)llH8-1/2->'(lRn-1)::; Ks.~lxn -Ynl~IIUIlH8(lRn) for all u e HS(m.n ), 0::; >. < 1, >.::; s -1/2 (and for>. = 1 if s > 3/2).
PROOF. (a) Let Uv e C~(IRn) be a sequence with Uv -+ U e HS(IRn) and set IPv(x):= lIuv(-,x)IIHo-l/2(lRn-1). The function IPv is continuous in m. and converges uniformly to lIu(.,x)IIH.-l/2(lRn-1) since luv(-,x) - u(.,X)ls-l/2 ::; Cslu v - uls for all x e IlL Thus (21a) follows. (b) ue(·,xn ) := u(·,xn + €) - u(·,xn ) has the Fourier transform ue(e) = ue(e',en) = [exp(ien€) - l]u(e) so that lue (e)1 2 = 4sin2(en€/2)lu(e)1 2 . As in the proof for Theorem 28 set W(.,x n ) := :7n-1Ue(·,X n ), W = W(.,O). The first integral in the estimate of 27rlw(OI2 now reads
1m.(1 + 1e'12 + e;)-S sin2(en€/2) den = (1 + le'1 2) 1/2-s fIR (1 + e)-S sin2(1]t) dt with 1] = ~(1 + le'1 2)1/2. The decomposition of the last integral into subintegrals over It I ::; 1/1] and It I ~ 1/1] shows flR(1 + t 2)-S sin2 (1]t) dt ::; Cs.~1]2~. The remainder of the argument follows the same lines as in the proof of The• orem 28. Up to this point we have obtained HS(IRn) by completion of CQ'(IRn) in L2(m.n). The next theorem shows that for sufficiently large s one can also complete in CO(IRn) n L2(IRn) so that HS(IRn) contains only classical functions (Le. continuous, HOlder-continuous, [HOlder-] continuously differentiable functions).
Theorem 6.2.30. (Sobolev's lemma) Hs(m.n) c Ck(IRn) holds for k IN U {O}, s > k + n/2 and HB(IRn) c ct(m.n) for 0 < t ¢. IN, s ~ t + n/2.
e
PROOF. (a) Let s ~ t + n/2, 0 < t < 1, u e HS(IRn). For given x, y e IRn we want to show lu(x) - u(y)1 ::; Clx - ylt with C independent of x,y. The coordinates of the m.n can be rotated so that x = (Xl! 0, .. ·,0), y = (Yl. 0, ... ,0). An (n-l)-fold application of Theorem 28 to u( .), u(-, 0), u(·, 0,0)
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6 Tools from Functional Analysis
, etc., results in 1£(·,0,0, .. ·,0) E HB-(n-l)/2(1R). Theorem 29 provides the desired estimate with C = KB-(n-l)/2,tlU(" 0,···, 0)IB-(n-l)/2 :5 C'luls ; thus 1£ E ct(lRn) and II 'IICt(Rn) :5 CII ·IIHo(Rn). Then 1£ E CO(lRn) follows from Ct(lRn) c CO(lRn). (b) Let s ~ t + n/2, 1 < t < 2, 1£ E HB(lRn). Part (a) is applicable to Datu E Hs-l(lRn) (cf. Theorem 25d,e) with lal :5 1: Datu E Ct-l(lRn ) for lal :5 1. Thus 1£ E Ct(lRn), etc. • We return to the statements of Theorems 27 and 28. For all 1£ E C8"(IR+') the restriction 'Yu = 1£(·,0) agrees with 'Y~BU. Completion in HS(IR+') yields
This proves the following corollary: Corollary 6.2.31. Let s > 1/2. For the restriction 'Yu := 1£(.,0) we have 'Y E L(H"(IR+.), H,,-1/2(lRn- 1
».
°
With the restriction to Xn = one evidently loses half an order of differentiability. Conversely one gains half an order if one continues w E Hs- 1/ 2 (lRn - 1 ) suitably in IRn. Theorem 6.2.32. Let s > 1/2, wE Hs-l/ 2(lRn - l ). There exists a function 1£ E H"(lRn ) [1£ E H"(IR+.») such that luis :5 C"lwl,,-1/2 and 'YU = w, i.e., W = 1£(.,0).
PROOF. Let u = :1'nu and w= :1'n-lw be the Fourier transforms. 'YU = w is equivalent to (20). For
u(e) = u(e', en) := w(e')(1 + le'1 2),,-1/2(1 + le'I 2 + e;)-" / K", K,,:=
L+ (1
t 2 )-"dt,
one checks that (20) and lui: = K;l/2Iwl:_l/2 hold. Restriction of u E HB(lRn) to IR+. proves the parenthetical addition. • If one replaces IR+. with a general domain n c IRn , then IRn- 1 ~ IRn- 1 x {O} = aIR+. becomes r = an, and the necessity arises of defining the Sobolev space H" (r). We begin with the
°
Definition 6.2.33. Let < t E IRU{ oo} [resp. k E 1NU{O}]. We write n E C t [n E Ck,l], if for every x E r := an there exists a neighbourhood U C IRn such that there exists a bijective mapping
6.2 Sobolev Spaces cp-l E Ct(Kl(O» [cp E Ck.l(U), cp-l E Ck.l(Kl(O))),
127
cP E Ct(U),
cp(U n r) = {e E Kl(O):en = O}, cp(U nil) = {e E Kl (0): en > O}, cp(U n (llln\il» = {e E Kl(O): en
(6.2.22a) (6.2.22b) (6.2.22c) (6.2.22d)
< O}.
Here Kl(O) is a circle (ball) if 1·1 is the Euclidean norm. For the maximum norm I . 100, Kl(O) is a square (cube). Likewise, Kl(O) can be replaced by another circle KR(z) or any rectangle (xl,x1) x··· x (x~,x~).
Figure 6.2.1. Covering neighbourhoods of rand
[J
Example 6.2.34. Let il be the circle Kl(O) C lll2. A neighbourhood ofx* = (1,0) is Ul from Fig. 1. The mapping x E Ul 1-+ cp(x) := e E (-1,1) x (-1,1) with Xl = (1 - e2/2)cos(7rel/2), X2 = (1 - e2/2)sin(7ret/2) is bijective and satisfies (22a-d) with t = 00. The same holds for any x E r. Thus il E Coo. Exercise 6.2.35. (a) The rectangle il = (xl, x1) x (x2, x~) and the L-domain from Example 2.1.4 are domains il E CO· l . (b) The cut circle in Figure 5.2.1b does not belong to CO· l . Lemma 6.2.36. Let il E C t [il E C k •l ] be a bounded domain. Then there exists N E lN, Ut (O:5 i :5 N), Ut , ~ (1 :5 i :5 N) with N
U i open, bounded (0 :5 i :5 N),
U Ui :::> 'ii,
UO CC il,
(6.2.23a)
i=O
N
Ui ;=
ui n r
(1:5 i :5 N),
UU = r i
(6.2.23b)
= 1, ... , N,
(6.2.23c)
i=l
at: Ui ~
0
-4
ai(Ui ) C llln-l bijective for i
ajl E Ct(aj(Ui n Uj
»[resp. at
0
ajl E Ck.l(aj(Ui n Uj))]. (6.2.23d)
On Ui (1 :5 i :5 N) are defined mappings CPi with the properties (22a-d).
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6 Tools from Functional Analysis
PROOF. For every x E r there exist U = U(x) and l/i} CC n. Let UxErU(x)UU V' be an open covering of the compact set Therefore there exists a finite covering through U' := U(x.) (1 ~ i ~ N) and at most one Vj which is denoted by UO. If one sets Ui := U' n r, 0:; = ax.,
n.
A set of pairs {(U., 0:;): 1 ~ i ~ N} fulfilling the conditions (23b-d) is called a C t _ [resp. C k,l]_ coordinate system for r. In Example 34 one has N = 4. The map all is given by al 1 (€1) = (cos(1ret/2), sin(1r6/2» E U1j and likewise a,41(€1) = 008«3 + 6)1r/2), sin«3 + e1)1r/2) E U\ where in each case -1 < ei < 1. On a1(Ul n U4) one obtains a4(al 1(€1» = €1 + 1. Lemma 6.2.37. (Partition of unity) Let {U':O ~ i ~ N} satisfy (23a). There exist functions O'i E Og>(m.n ), 0 ~ i ~ N, with N
supp(O") CUi,
LO'l(x) = 1
for all x E
n.
(6.2.24)
.=0 The general construction of the O'i can be found, for example, in Wloka [I, §1.2]. In the special case of Figure lone may proceed as follows. Let O'(t) := 0 for It I ~ 1 and O'(t) := exp(1/(t2 - 1» for t E (-1,1). Then 0' E C.r(m.) and supp(O') = [-1,1]. In U' from Figure lone defines, for example, tPo(x) := 0'(9IxI2/4),
tP1{X):= 0'(2r - 2)0'(2
vI:
for x = r
(c~
etc.
These O'i{X) := tPi{X)/ tPl(x) satisfy (24). A function u on r can be written in the form ~ O'lu. Each summand O'lu is parametrisable over o:;(Ui ) C m.n - 1: (O'lu) 0 a; : a.(U.) cc m.n - 1 -+ IR.. This in turn makes possible
Definition 6.2.38. Let n E C t [E Ck,1]. Assume (U., a.) and 0'. satisfy (23b-d) and (24). Let s ~ t E IN [s ~ k+ 1] or s < t ¢ IN, t > 1. The Sobolev space HB(r) is defined as the set of all functions u: r -+ m. such that (O'iU) 0 a;l E H&(m.n - 1) (1 ~ i ~ N). Theorem 6.2.39. (a) HB(r) is a Hilbert space with the scalar product N
(U,V)B:= (U,V)HO(r):= L«UO'i) o a; 1, (vO'.)oa;1)HO(lRn -
1) •
• =1
(b) If {(Ui , a.): 1 ~ i ~ N} is another C t _ [Ok,1:] coordinate system of rand {Ui} another partition of unity, then the space HB(r) defined by this is equal to HB{r) as a set. The n017nS of H8(r) and fI8(r) are equivalent.
6.2 Sobolev Spaces
129
PROOF. For (b): Transformation Theorem 17 [resp. 25g] ; for DE CO,l, cf. Wloka [1, Lemma 4.5]. • The trace and extension theorems (Corollary 31 and Theorem 32) can be extended to any domain with a sufficiently smooth boundary. "f now denotes the restriction to r = an: "fU = u Ir. Theorem 6.2.40. Let n E C t with 1/2 < s :::; t E IN or 1/2 < s < t [resp. DE e k ,l, 1/2 < s = k + 1 E IN]. (a) The restriction "fU of u E HII(n) belongs to Hs- 1/ 2(r): "f E L(HS(D), HS-l/2(r)). (b) For each W E Hs-l/2(r) there exists an extension u E HS(D) with W = "fU, luis:::; e s lwls-l/2. (c) For each W E HS(n) there exists a continuation Ew E H8(ntn) with E E L(H8 (n), Hs(ntn)). PROOF. The proofs follow the same pattern. Let U i , Ut and
:::; Cslul s .
•
Remark 6.2.41. (a) Under the conditions of Theorem 40 and the additional condition s > lal + 1/2 there exists a restriction "fDau E Hs-la l-1/2(r) of the derivative of u E HS(D). (b) For each u E Hg(n) with s < 1+ 1/2 one has "fDau = 0 if lal :::; l. Theorem 6.2.42. For n E CO,l holds HJ(D)
= {u E Hl(D): ulr = O}.
PROOF. (a) u E HJ(D) is the limit of UII E CO>(D), all of which satisfy ulIJr = O. Since ulli r -+ ul r with respect to Hl/2(r), it follows that u E H (n) and ul r = o.
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6 Tools from Functional Analysis
(b) Conversely, let u E H1(n) and ull' = O. The proof that u E HJ(n) can be divided up as follows: (ba) By using the partition of unity over {Ui: 0:::; i :::; N} (cf. Lenuna 37), the statement reduces to the case n = ffi.+. (bb) Without loss of generality one can assume n = 1: n = H4. (be) According to Remark 14, for each E > 0 there exists a iiE E COO(ffi.+) with finite support such that lu - iieh :::; rJ/2. For each rJ > 0 there exists CPfj E COO(ffi.+) with CPfj(x)
=0
for x:::; rJ/2,
Icp~(x)1 :::; 3/rJ,
cPfj(x)
=1
for x 2:: rJ,
0:::; cPfj(x) :::; 1.
The function U E := iie - (1 - cpt}iiE(O) satisfies U E E COO(H4), uE(O) = 0, supp(uE) finite, and lu - uE/t :::; CEo Thus X := {v E COO(ffi.+):v(O) = 0, supp(v) finite} is dense in {u E Hl(ffi.+):u(O) = a}. (bd) The statement would be proved if Cgo(ffi.+) were also dense in X with respect to I· /t. Let vEX. Evidently, vfj := CPfjv E Cgo(ffi.+) holds for all rJ > O. Set 8(rJ) := IIv'IIL3(o.'1) and note that 8(rJ) -+ 0 for rJ -+ O. Since vex) = v'l(x) for x 2:: rJ, it remains to estimate the variables IIv - v'lIlL3(o.'1) and IIv' - v~IIL3(o.fj)' Because of v~ - v' = cP~v + (cp'l - l)v', one obtains IIv~ - v'IIL3(o.'1) :::; IIcp~IILoo(o.'1)lIvIlL3(o.'1)
:::; (3/rJ)lI v IlL3(o.'1)
+ IIcp'l -
lIlLOO(o.'1)lI v 'IIL2(o.'1)
+ 8(rJ)·
J;
Since vex) = v'(e) cIe, (5b) implies the estimate Iv(x)1 :::; .jij8(rJ) for all o :::; x :::; rJ, and hence IIvIlL3(o.'1) :::; rJ8(rJ). Then the statement follows from Iv - v'll~ :::; IIv - V'llli3(o.'1) + IIv' - v~lIi3(o.'1) :5 C8 2 (rJ) -+ O. • When n E C1, the normal direction n exists at all boundary points. Analogously to Theorem 42 one proves Corollary 6.2.43. For n E C1 and k E IN holds H~(n)
= {u E Hk(n):o'u/on'lr = 0 forO:::; 1:5 k -I} = {u E Hk(Il):DQul r = 0 forO:::; lal:5 k -I}.
6.3 Dual Spaces 6.3.1 Dual Space of a Normed Space Let X be a normed, linear space over ffi.. As a dual space, X' denotes the space of all bounded, linear mappings of X onto ffi.: X':= L(X,ffi.).
6.3 Dual Spaces
131
According to Exercise 6.1.8, X' is a Banach space with the norm (dual norm)
Ilx'lIxl ;= IIx'IIR<-X = sup{lx'(x)lIlIxllx:o =I x EX}.
(6.3.1)
The elements x' E X' are called linear functionals on X. Instead of x'(x) (application of x' to x) one also writes (x,X')XXXI, or (x',x)x1xx, and calls (-, ')XXXI the dual form on X x X':
(x,X')XXXI = (x',x)x1xx = x'(x).
Lemma 6.3.1. Let the Banach space X be embedded densely and continuously in the Banach space Y. Then Y' is continuously embedded in X'.
PROOF. (a) Let y' E Y'. Because Xc Y, y' is defined on X. (b) Since X is a dense subspace ofY, according to Theorem 6.1.11 and (1.5), for each y' E Y' holds:
lIy'IIYI = sup ly'(x)lfllxlly xEX
~ c1 suply'(x)lfllxllx = C1l1y'lIxl, x
•
Le., Y' is embedded continuously in X'.
To the transposed matrix in the finite-dimensional case corresponds the dual mapping (or the dual operator).
Lemma 6.3.2. y' E Y'
Let X and Y be normed and let T E L(X, Y). For each
(Tx,y')yXY ' = (x,X')XXXI
for all x E X
(6.3.2)
defines a unique x' E X'. The linear mapping y' 1-+ x' defines the dual operator -+ X' with T' y' = x'. The following holds: T' E L(Y', X') and
T': Y'
(6.3.3)
PROOF. If one writes (2) in the form y'(Tx) = x'(x), one can see that x' = y' 0 T. (3) follows from the definitions of the norms;
IIT'lIxl<-YI = sup IIT'y'IIx1flly'IIYI = y'#O
sup
x#O,y'#O
l(x,T'y')lxxxlf[llxllxlly'IIYI]
= sup I(Tx, y')YxYllf[llxllxlly'lIy] = sup IITxllY IlIxlix = IITlIy<-x. x#O,y'#O
x#O
• Example 6.3.3. Let n = (0,1), X = (COCi?), II . 1100), and x E n. The mapping cx: u E c°(li) 1-+ u(x) E IR is a functional: Cx E CO(D)' (the socalled delta function). The Laplace operator .d belongs to L(C 2 (D), CO(D».
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6 Tools from Functional Analysis
The dual mapping Ll' E L(oo(O)', Q2(ll)') is applicable to 6x : Ll'6x is characterised by (Ll'6x )u = Llu(x) for all u E C 2 (ll). Exercise 6.3.4. Let S E L(X, Y), T E L(Y, Z). Show that (TS)'
= S'T'.
Exercise 6.3.5. Show that if T E L(X, Y) is surjective, then T' is injective.
6.3.2 Adjoint Operators
Let X be a Hilbert space (over 1R). Every y E X defined by III(x) := (x, y)x is a linear functional III E X, with 1I/lIlixi (cr. Yosida [1, §III.6]).
= lIyllx. The converse also holds
Theorem 6.3.6. (Riesz representation theorem) Let X be a Hilbert space and I E X'. There exists a unique Y/ E X such that I(x)
= (x,y/)x
for all x E X
and IIflixl
= lIy/lix.
Corollary 6.3.1. Let X be a Hilbert space. (a) There exists a one-to-one correspondence (the Riesz isomorphism) J x E L(X,X') with JXY = Ill' J;l I = y/ that presenJes the norm, i.e., IIJxllxl+-x = IIJ;lllx+-xl = 1. (b) X' is a Hilbert space with the scalar product (x', y')xl = (J;lx',J;ly')x. The dual norm IIx'lIxl from (1) agrees with the norm induced by (x',x,)~2. (c) One always identifies X with X" because x(x') := x'(x). Prom this follows JX' = J Xl , Jx = J Til = T lor T E L(X, Y) if Y = Y" is also a Hilbert space. (d) One can identify X and X': X = X', Jx = I.
x,
Let X, Y be Hilbert spaces and T E L(X, Y). The mapping defined by T* := J;lT' Jy E L(Y, X) is called the operator adjoint to T and satisfies (Tx, y)1I = (x, T*y)x for all x E X, Y E Y;
IITlly+-x = IIT*lIx+-y. (6.3.4) The adjoint and the dual operator only coincide (i.e., T* = T') if X' is identified with X and Y' with Y. TE L(X,X) is said to be selfadjoint (or symmetric) ifT= T*. T E L(X,X) is called a projection if T2 = T. It is an orthogonal projection if furthermore T is selfadjoint. Remark 6.3.B. Let Xo be a closed subspace of the Hilbert space X. An orthogonal projection is given by Tx := y E Xo with
6.3 Dual Spaces
IIx - Yllx
= inf{lIx - Ellx:E E X o}.
133
(6.3.5).
If conversely T E L(X, X) is an orthogonal projection with the range Xo := {Tx: x E X} one has (5) for y = Tx. An orthogonal projection always has the norm IITIIT+-x :s; 1.
PROOF. (a) x can be decomposed uniquely into x = y+z (y E Xo, z E Xl) (cf. Lemma 6.1.17). y is the unique solution of (5). x E Xo implies y = x, thus T2 = T. The analogous decomposition x' = y' + z' shows (x, T*x') = (Tx, x') = (y, x') = (y, y' +z') = (y, V') = (y+z, V') = (x, Tx'), hence T = T*. (b) Let T be an orthogonal projection with range Xo. Let x = y + z be split as above. T2 = T shows Ty = y. For each y' E Xo holds (Tz, V') = (z, T*y') = (z, Ty') = (z, V') = 0, thus Tz E xl. Together with Tz E Xo follows Tz = 0 so that Tx = Ty + Tz = y. (c) Tx = y and Ilxll~ = lIylli + IIzll~ ~ lIyll~ prove IITllx+-x :s; 1. •
6.S.S Scales of Hilbert Spaces We assume:
vcU
are Hilbert spaces with a continuous and dense embedding. (6.3.6)
Lemma 6.S.9. densely in V'.
Under assumption (6) U' is embedded continuously and
PROOF. The continuity of the embedding U' C V' was established in Lemma 1. For proof that U' is dense in V', we use Exercise 6.1.18d (A:= U',X := V'). Let 0 =I- v' E V' be arbitrary and '1.1. := J y 1v' EVe U. According to the definition, '1.1.' := Juu E U' C V' is characterised by u'(x) = (x, u)u for all x E U. For x:= '1.1. = Jy1v' E V follows
(v', u')v1
= (Jy 1V', Jy1u')v = ('1.1., Jy1u')v = '1.1.'('1.1.) = ('1.1., u)u > O.
•
According to Corollary 7d , U and U' can be identified. By this one obtains the Gelfand triple V cUe V'
(V
c
U continuously and densely embedded).
(6.3.7)
Corollary 6.S.10. In a Gelfand triple (7) V and U are also continuously and densely embedded in V'. PROOF. For U C V' see Lemma 9, for V C V' see Exercise 6.2.25.
•
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6 Tools from Functional Analysis
Attention. Likewise one could identify V with V' and one would obtain = V CU. But it is not possible to identify U with U' and V with V' simultaneously. In the first case one interprets x(y) = (y,x}uxul for X,y E U as (y,x)u (in particular for x, y EVe U), in the second case as (y,x)v. U ' C V'
Exercise 6.3.11.
Let (7) be true. Set W := {Jv1u:u E U} and define (x, y)w := (Jvx, JVY)u as the scalar product on W. Show that (a) (b) (c) (d)
W is a Hilbert space; W C V is a continuous and dense embedding; (v, w)v = (v, Jvw)u for all v E V, wE W; l(x,Y)vl :5l1xllullYllw for all X,y E W.
Because U = U' the scalar product (x, y)u can also be written in the form y(x) = (x,y}uxu1. If x E V, then y(x) = (x,y}vxvi also holds. That means that (x,y)u = (x,y}vxv1 for all x E V,y E U C V'. Likewise one obtains (x,y)u = (x,y}v'xv for all x E U and y E V. The dense and continuous embedding U C V' proves
Remark 6.3.12. Let V cUe V' be a Gelfand triple. The continuous extension of the scalar product (-, ·)u to V x V' [V' x V) results in the dual form (', '}VXVI [(" '}v1xv), Therefore the following notation is practical: (x, y}vxvi = (x, y)u (x,y}v1xv = (x,y)u
for x E V, Y E V', for x E V',y E V.
In connection with Sobolev spaces one always chooses U := L2(O) so that the embeddings read as follows:
c L2(f1) c (Hg(o»' HII(ll) c L2(ll) c (HB(ll»'
Hg(O)
(8 ~ 0),
(6.3.8a)
(8 ~ 0).
(6.3.8b)
Exercise 6.3.13. Show that (8a) and (8b) are Gelfand triples. The dual space of H8(ll) is also denoted by H-S(ll) or HOB(ll): HolI(ll) := H- 8 (ll) := (Hg(ll»,
(8
~
0).
The norm of H-S(ll) according to (1) reads: lul-s := sup{l(u,v)L3(.a)l/lv ls :O
-I v E Ho(ll)},
for
8
~0
where (U,V)L3(.a) is the dual form on Hg(ll) x H-II(O) (cf. Remark 12).
Remark 6.3.14. (a) Let II = m,n. The norm dual to
1'1:,
lul~8 := sup{l(u, v)ol/lvl:: 0 -I v E Hs(m,n)},
6.4 Compact Operators
135
is equivalent to I·I-B and has the representation (2.16b) with -s instead of s. (b) The Fourier transform shows Da E L(HB(m.n), Hs-1al(m.n )) for all s Em.. (c) au E HB(n), if U E HB(J1), a E oteri), where t = lsi E IN U {O} or t > lsi.
6.4 Compact Operators Definition 6.4.1. A subset K of a Banach space is said to be precompact [compact] if each sequence Xi E K (i E IN) contains a convergent subsequence X'1e [and limk-HX) X'1e E K]. Another definition of compactness reads: Each open covering of K already contains a finite covering of K. Both definitions are equivalent in metric spaces (cf. Dieudonne [1, (3.16.1)]). For the terms "relatively compact" and "precompact" see also Dieudonne [1, (3.17.5)J.
Remark 6.4.2. (a) K c m.n is precompact [compact] if and only if K is bounded [and complete]. (b) Let X be a Banach space. The unit sphere {x EX: IIxll ::; I} is compact if and only if dim(X) < 00. Definition 6.4.3. Let X and Y be Banach spaces. The mapping T E L(X, Y) is said to be compact if {Tx: x EX, Ilxllx ::; I}, the image of the unit sphere in X, is precompact in Y. Exercise 6.4.4. When is the identity I E L(X, X) compact? Lemma 6.4.5. Let X, Y, and Z be Banach spaces. (a) Let one of the mappings Ti E L(X, Y), T2 E L(Y, Z) be compact. Then T2Ti E L(X, Z) is also compact. (b) T E L(X, Y) is compact if and only ifT' E L(Y', X') is compact. PROOF. (a) Let Ki := {x E X: IIxllx ::; I}. If Ti is compact, i.e., Ti(Kt} is precompact, then T2(Ti(Kt}) is also precompact and thus T2Tl is compact. If, however, T2 is compact, one proves the assertion as follows. Since scaling does not change compactness, IITl lly+-x ::; 1 can be assumed without loss of generality. Hence T i (Kl) is a subset of the unit sphere in Y and therefore T2(Tl(Kt}) is precompact. (b) cf. Yosida [1, §X]. • A special type of compact mapping is a compact embedding: Definition 6.4.6. Let X c Y be a continuous embedding. X is said to be compactly embedded in Y if the inclusion I E L(X, Y), Ix = x is compact.
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6 Tools from Functional Analysis
Together with Definitions 1 and 3 one obtains: X C Y is compactly embedded if every sequence Xl E X with IIXilix :::;; 1 contains a subsequence convergent in Y.
Example 6.4.7. Let nbe bounded. OB(n) C CO(n) is a compact embedding for s > O. PROOF. Functions Ui E oB(Ii) with IIUillc'(Q) :::;; 1 are equi-continuous and uniformly bounded. The assertion is based on the theorem of Arzela-Ascoli (cf. Yosida [1, §III.3]). •
Analogous results can be obtained for Sobolev spaces (cf. Adams [1, p. 1441, Wloka [1, §7J):
Theorem 6.4.8. Let n c Rfl be open and bounded. (a) The embeddings Hg(n) c H&(n) (s, t E R, s > t) are compact. (b) FUrther, let n E CO,l. The embeddings Hk(n) c H'(n) (k, lEN U {O}, k > 1) are compact. (c) Let 0 :::;; t < sand n E or (r > t, r > 1) or n E Ok,l (k + 1 > t). Then the embedding H8(n) c Ht(n) is compact. Remark 6.4.9. In Theorem 8b one can replace n E 0°,1 by the "uniform cone property" (cf. Wloka [1, §2.1]). To ensure n E 0°,1 it is sufficient that the boundary an is piecewise smooth and the inside angles of possible comers are smaller than 211'. Indented comers (cf. Figure 2.1.1) are thus permitted while a cut domain (cf. Figure 5.2.1b) is excluded. In Section 6.S the following situation will arise:
v cUe V'
Gelfand triple,
T E L(V', V).
(6.4.1)
Because of the continuous embeddings, T also belongs to L(V', V'), L(U, U), L(V, V), and L(U, V).
Theorem 6.4.10. Let (1) hold. Let V CUbe a compact embedding. Then T E L(V', V'), T E L(U, U), T E L(V, V), T E L(V', U), and T E L(U, V) are compact. PROOF. As an example, let us do T E L(U, V). Since the inclusion I E L(V, U) is compact we see I E L(U, V') is also compact (cf. Lemma Sb). T E L(U, V), as the product of the compact mapping I E L(U, V') with • T E L(V', V), is compact (cf. Lemma 5a).
Exercise 6.4.11. Let dim X compact.
< 00 or dim Y < 00. Show that T
E L(X, Y) is
6.5 Bilinear Forms
137
The significance of compact operators T E L(X, X) lies in the fact that the equation Tx - AX = y (with X, y EX, Y given, X sought) has properties analogous to the finite-dimensional case. Theorem 6.4.12. (Riesz-Schauder theory) Let X be a Banach spacej let T E L(X, X) be compact. (a) For each A E ClJ\{O} one of the following alternatives holds:
(i) (T - AI)-l E L(X, X)
or
(ii) A is an eigenvalue.
In case (i) the equation Tx - AX = Y has a unique solution X E X for all Y E X. In case (ii) there exists a finite-dimensional eigenspace E(A, T) := kernel (T - AI) # {O}. All X E E(A,T)\{O} solve the eigenvalue problem Tx=Ax, x#O. (b) The spectrum q(T) ofT consists by definition of all eigenvalues and, if not T-l E L(X, X), A = O. There exist at most countably many eigenvalues which can only accumulate at zero. A E a(T) if and only if'). E a(T'). Furthermore,
dim(E(A, T» = dim(E(X, T'» < 00. (c) For A E q(T)\{O}, Tx - Ax = y has at least one solution x only if (y, X'}XXXI = 0 for all x' E E(5., T').
E
X if and
In Lemma 6.5.18 we need Lemma 6.4.13. Let Xc Y c Z be continuously embedded Banach spaces and let X C Y be compactly embedded. Then for every f > 0 there exists a GE such that (6.4.2) IIxlly ~ fllxllx + GEllxllz for all x E X. PROOF. Let f > 0 be fixed. The negation of (2) reads: There exists Xi E X with (lixilly - flixillx)/lIxiliz -+ 00. For Yi := (fllxillx)-lxi E X we thus have (IlYilly -1)/IlYillz -+ 00. From this one infers IIYillz -+ 0 and IIYilly > 1 for sufficiently large i. Since IIYilix ~ 1/f. and Xc Y is a compact embedding, a subsequence Yik converges to y* E Y. Now, IIYilly > 1 implies IIY*lIy ~ 1, i.e., y* # O. On the other hand, Yik also converges in Z to y* since Y c Z is continuously embedded. IIYiliz -+ 0 gives the contradiction sought: y* = o.
•
6.5 Bilinear Forms In the following let us assume that V is a Hilbert space. The mapping a(., .): V x V -+ IR is called a bilinear form if a(x, y + AZ)
= a(x, y) + Aa(x, z),
a(x + AY, z)
= a(x, z) + Aa(y, z)
138
6 Tools from Functional Analysis
for (x, y, Z E V, A E IR) (in the complex case one speaks of sesquilinear forms: a(x,Ay) = ~a(x,y)). a(·,·) is said to be continuous (or bounded) ifthere exists a C s such that la(x, y)l ~ Csllxllvllyllv
for all x, y E V.
(6.5.1)
Lemma 6.5.1. (a) To a continuous bilinear form one can assign a unique operator A E L(V, V') such that a(x, y)
= (Ax, y)v'xv
for all x, y E V,
IIAllvl<-v ~ cs.
(6.5.2)
(b) Let VI and V2 be dense in V. Let a(·,·) be defined on VI x V2 and satisfy (1) with '~E Vl,y E V2" instead of '~,y E V". Then a(·,·) can be extended uniquely to V x V so that (1) holds with the same Cs for all X,y E V.
PROOF. (a) Let x E V be fixed.
= sup{la(x, y)l: x, Y E V, IIxliv = lIyllv = 1}.
(6.5.3)
A is called the operator that is associated to a(·, .). The bilinear form a*(-,·) adjoint to a(·,·) is given by a*(x,y):= a(y,x) (x,y E V). The bilinear form is said to be symmetric if a(·,.) = a*c-, .). Exercise 6.5.2. Show that (a) If A belongs to a(·, .), then A' belongs to a*(·, .). (b) If a(·,·) is symmetric, then A = A'.
Lemma 6.5.8. Let A E L(V, V') be the operator associated to a continuous bilinear form a(·,·). Then the following statements (i), (ii), (iii) are equivalent: (i) A-I E L(V', V) exists; (ii) 10,10' > 0 exist such that inf{sup{la(x, y)l: Y E V, IIYllv
= 1}: x
= 1} = 10 > 0,
(6.5.4a)
inf{sup{la(x, y)l:x E V,lIxliv
= 1}:y E V,lIyllv = 1} =10' >0;
(6.5.4b)
E
V, IIxliv
(iii) the inequalities (4a) and (4c) hold:
sup{la(x,y)l:x E V, IIxliv If one of the statements (i)-(iii) holds, then
= 1} > O.
(6.5.4c)
6.5 Bilinear Forms
f
= f' = 1/IIA- 1 I1v+-v'
(f,f' from (4a,b».
139
(6.5.4d)
From (4a) follows inf{sup{la(x, v)l: V E V,lIvllv = 1}:x E
v, IIxliv = 1} ~ f > O.
(6.5.4e)
Conversely, (4a) follows from (4e) with a possibly laryer f > O. (4e) and (4c) are also called the Babus ka conditions. (4e) is equivalent to sup{la(x, y)l: y E V, IIYllv = 1} ~ fllxllv
for all x E V,
(6.5.4e')
because (4e) is equal to (4e') for all x E V, IIxliv = 1. The scaling condition IIxliv = 1 can evidently be dropped. The left-hand side in (4e') agrees with the definition of the dual norm of Ax so that (4e) and (4e') are also equivalent to (4e"): (6.5.4e") IIAxll v ' ~ fllxllv for all x E V. PROOF. (a) "(i) ::::} (ii)": Let A-l E L(V', V) exist. Then (4a) follows from . f{ III
} ...
. f =
la(x, y)1 . f I(Ax, y)1 sup II x II V II y II V = ..III eY sup lIeY II X II V II Y II V
..III eY lIeY
.. #0 11#0
=
0:#0 11#0
inf sup I(AA- 1 x', y)1 IIA-1x'lIvllyliv
.. 'eY' lIeY
.. '#0
=
11#0
inf IIA-1x'lIv1llx'lIv'
aI/ev' .. '#0
=
1/IIA- 1 I1v+-v'
=
inf IIA- 1x'II- 1 sup I(x', y)1 ..'eY' V lIeY IIYllv .. '#0
= 1/[
11#0
sup IIA-1x'llv/llx'IIV'] z/EV'
.. '#0
=:
f.
In the same way one shows (4b) with f' = 1/IIA'-1Ilv+-v'. Because A'-l = (A-l)" (3.3), and V" = V, it follows that f = f'. (b) "(ii) ::::} (iii)": (4c) is a weakening of (4b). (c) "(iii) ::::} (i)": f > 0 in (4a) proves that A is injective. We wish to show that the image W := {Ax: x E V} c V' is closed. For a sequence {w v } with IIw· - wvllv' -4 0 we must therefore show that w· E W. According to the definition of W there exists Xv E V with Axv = wv . From (4a) one infers via (4e) and (4e") (with x := Xv - xI') that IIxv - xJ'llv :::; IIwv - wJ'lIv' If. Since {wv } is Cauchy convergent, this property carries over to {xv}. There exists an x* E V with Xv --+ x* in V. The continuity of A E L(V, V') proves Wv = Axv --+ Ax* so that w* = Ax* E W. According to Lemma 6.1.17 one can decompose V' into WEBW.L.1f A were not surjective (thus W i V'), there would exist awE W.L with w i O. Then y:= Jv'w = JV1 E V would satisfy y i 0 (cf. Theorem 6.3.6, Corollary 6.3.7). Since a(x,y) = (Ax,Y)v'xv = (Ax,w)v' = 0 for all x E V, a contradiction to (4c) would result. Therefore, A is also surjective, and Theorem 6.1.13 proves A-l E L(V', V). (d) Statement (4d) has already resulted from Part a of the proof. •
140
6 Tools from Functional Analysis
It will be shown that for interesting cases conditions (4a) and (4b) are equivalent (cf. Lemma 17). A particularly simple case is in Exercise 6.5.4. Show that if dim V (4b) with e' = e and conversely.
< 00, then (4a) implies the statement
Definition 6.5.5. A bilinear form is said to be V-elliptic if it is continuous on V x V and there is a constant CE such that a(x, x) 2: CEllxll~
for all x
E
(6.5.5)
V.
Exercise 6.5.6. Show that: (a) If W C V is a Hilbert subspace with the norm equal (or equivalent) to V, then a V-elliptic bilinear form is also W-elliptic. (b) Let a(·, .): V x V - m. be continuous. If a(x, x) 2: CEllxll~ holds for all x E Vo where Vo is dense in V, then (5) follows with the same CEo (c) Let a(., .): V x V - m. be continuous, symmetric, nonnegative (a( v, v) 2: 0) and satisfy (4a,c). Then a(.,.) is V-elliptic with CE ~ e2/Cs (e from (4a,b), Cs from (1». Hint: First prove that la(u,v)1 S [a(u,u)a(v,v)J1/2 (cf. Exercise 6.1.16).
Lemma 6.5.7. V-ellipticity (5) implies (1) and (4a,b) with e = e' PROOF. Let x E V,lIxliv = 1. sup{la(x,y)l:y E V,lIyllv CE proves (4a) with e ~ CEo (4b) follows analogously.
=
~
CEo
I} ~ la(x,x)l2:
•
The combination of Lemmata 1, 3, 7 together with IIA'-lllv+--vl IIA-lllv+--vl (cf. Lemma 6.3.2) proves
=
Theorem 6.5.B. Let the bilinear form be V-elliptic [or satisfy (1), (4a,c)]. Then the corresponding operator A satisfies the conditions
= IIAllvl+--v S CS,
A E L(V, V'),
IIA'lIvl+--v
A- l
IIA'-lllv+--vl
E L(V', V),
with Cs from (1) and C'
= I/CE
= IIA-lllv+--vl S c' [resp. C'
(6.5.6)
= l/e = l/e'].
With the help of the bilinear form a(·,·) and a functional f E V' one can formulate the following problem: Find x E V
with
a(x,y)
= fey) for all y E V.
(6.5.7)
According to Lemma lone can write (7) in the form (Ax- f, y}vlxv = 0 for all Y E V, i.e., Ax = f in V'. The equation Ax = f is solvable for f E V' if and only if A-I E L(V', V). Hence one obtains Theorem 6.5.9. Let the bilinear form be continuous (cf. (1» and satisfy the stability condition (4a,c) [it is sufficient that a(·, .) is V -elliptic]. Then Problem
6.5 Bilinear Forms
(7) has exactly one solution x := A-l f. This satisfies C = lie = lie' [resp. C = liCE].
IIxliv
141
~ Cllfllv' with
Corollary 6.5.10. Under the assumptions of Theorem 9 the analogous statement holds with the same estimate for the adjoint problem Find x* E V
PROOF. IIA-lllv+-v'
with a*(x*, y)
= f(y)
= IIA,-lllv+-vl
(cf. Exercise 2a, Theorem 8).
(6.5.8)
for all y E V.
•
Exercise 6.5.11. Let a(·, .): Vx V - m. be continuous. Let Vo be dense in V. Show that the solution x E V of problem (7) is already uniquely determined by "a(x, y) = f(y) for all y E Vo". The same holds for (8). Problem (7) may be equivalent to a variational problem:
Theorem 6.5.12. f E V'. Then
Let a(·,·) be V-elliptic and symmetric; furthermore, let J(x) := a(x,x) - 2f(x)
(6.5.9)
(x E V)
assumes its unique minimum for the solution x of Equation (7). PROOF. Let x be the solution of (7). For arbitrary z E V set y := z - x. From J(z)
= J(x + y) = a(x + y,x + y) - 2f(x + y) = a(x, x) + a(x, y) + a(y, x) + a(y, y) - 2f(x) = J(x) + a(y, y) + 2[a(x, y) - f(y)] = J(x)
+ a(y, y) ~ J(x) + CEllyll} =
one can read J(z) > J(x) for all z
J(x)
2f(y)
+ CEliz - xII},
•
=f x.
The term "V-elliptic" seems to indicate that to elliptic boundary value problems correspond V -elliptic bilinear forms. In general this is not the case. Rather, V-coercive forms will be assigned to the elliptic boundary value problems. Their definition necessitates the introduction of a Gelfand triple (cf. (6.3.7)): V cUe V'
(U = U', V C U continuous and densely embedded).
Definition 6.5.13. Let V cUe V' be a Gelfand triple. A bilinear form a(·, .) is said to be V-coerci ve if it is continuous and if there exists CK E m. and CE > 0 such that
(6.5.10)
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6 Tools from Functional Analysis
Exercise 6.5.14. Set a(x,y):= a(x,y) +CK(x,Y)u with C K from (10). Let I: V --+ V' be the inclusion. Show that (a) the coercivity condition (10) is equivalent to the V -ellipticity of a. (b) If A E L(V, V') is associated to a(·, .), then so is A := A + C,.I to a{·, .). Why does A E L(V, V') hold?
The results of Riesz-Schauder theory (Theorem 6.4.12) transfer to A as soon as the embedding V C U is not only continuous but also compact. Theorem 6.5.15. Let V cUe V' be a Gelfand triple with compact embedding V cU. Let the bilinear form a{., .) be V -coercive with corresponding operator A. Let I: V --+ V' be the inclusion. (a) For each AE CU one of the following alternatives holds:
(A - AI)-l E L{V' , V) (ii) A is an eigenvalue. (i)
and (A' - Xl)-l E L{V', V),
In case (i) Ax - Ax = f and A'x· - 'Xx. = f are uniquely solvable for all f E V' (i.e., a(x, y) - A(X, y)u = fey) and a·(x·,y) -X(x., y)u = fey) for all y E V). In case (li) there exist finite-dimensional eigenspaces {O} =f E{A) := kernel (A - AI) and {O} =f E'(A) := kernel (A' - XI) such that Ax = Ax A'x·
= X*
= A{X, y)u
for x E E{A),
i.e.,
a{x, y)
for x* E E'(X),
i.e.,
a·(x*,y)
= X(x*,y)u
for all y E V, (6.5.11a) for all y E V. (6.5.11b)
(b) The spectrum u(A) of A consists of at most countably many eigenvalues which cannot accumulate in CU. A E u(A) if and only if X E U(A'). Pu.rthermore dimE(A) = dimE'(X) < 00. (c) For A E u{A), Ax - Ax = f E V' has at least one solution x E V if and only f..L E'{A), i.e., (f,x*}v'xv = (f,x*)u = 0 for all x· E E'{A).
PROOF. With V c U, V C V'is also a compact embedding, i.e., the inclusion I: V --+ V' is compact. A + CKI with CK from (10) satisfies A + CKI E L(V, V'), (A + CKI)-l E L(V', V) (cf. Exercise 14 a). Lemma 6.5.4 shows that K := (A + CK 1)-1 I: V --+ V is compact. Hence the RieszSchauder theory is applicable to K - 1'1. Since K -1'1= -1'(1 -1'-lK)
= -1'(A+CKI)-l{A+CKI -1'-lI} = -I'(A + CKI)-l(A - AI)
with A = 1/1' - CK, the statements of Theorem 6.4.12 transfer via K - 1'1 to A - AI = -1'-1 (A + CKI) (K - 1'1). •
6.5 Bilinear Forms
143
Remark 6.5.16. The spectrum u(A) has measure zero so that under the conditions of Theorem 15 the solvability of Ax - AX = f is guaranteed for almost all A. Problem (7) is solvable if not "accidentally" 0 E u(A). Lemma 6.5.17. Under the conditions of Theorem 15 the inequalities (4a,b) are equivalent.
PROOF. (4a) proves that A is injective, Le., 0 f/. u(A). Theorem 15a shows A-l E L(V', V) so that (4b) follows from Lemma 3. • Evidently a(·, .) remains V-coercive if one adds a multiple of (., ·)u. Generally, there holds
Lemma 6.5.18. Let a(·,·) be V-coercive where V cUe V'. Then a(·,·) + b(., .) is also V -coercive if the bilinear form b(., .) satisfies one of the following conditions: (a) for every € > 0 exists Ce such that (6.5.12a)
(b) Let the embeddings V c X and V C Y be continuous, with at least one of them compact. Let the following hold: Ib(x,x)1 ~ CBllxllxllxlly
for all X E V.
(6.5.12b)
(c) Let the embeddings V eX, V c Y be continuous. Let (12b) hold. For II . IIx or II . lIy assume that for every f. > 0 there exists a C~ such that
IIxlix ~ €lIxliv + C~llxliu
or
IIxlly ~
€lIxllv + C~lIxllu
for x E V. (6.5.12c)
= CE/2 with CE from (10). Then a(·,·) +b(.,.) satisfies the V-coercivity condition with C E /2 > 0 and C K + Ce instead of C E and CK. (b) Lemma 6.4.13 proves (12c). (c) Let the first inequality from (12c) hold, for example. Since the embedding V c Y is continuous, Cy exists with IIxllY ~ Cyllxllv. Choose €' = €/(2CB Cy) in (12c): PROOF. (a) Select €
Ib(x,x)1
~ CB(€'lIxliv +C~,"xllu)CYllxllv ~ ~lIxll~ +Kllxllvllxllu
with K = CBCyC~,. Since Kllxllvllxllu ~ ~"xll~ (5.3.10)), (12a) follows.
+ !K2€-1I1xll& (cf. •
7 Variational Formulation
7.1 Historical Remarks In the preceding chapters it was not possible to establish even for the Dirichlet problem of the potential equation (2.1.1a,b) whether, or under what conditions, a classical solution 1.£ E C 2 (fl) n CO(fl) exists. Green [I] took the view that his Green's function, described in 1828, always exists and that it provides the solution explicitly. This is not the case. Lebesgue proved in 1913 that for certain domains the Green function does not exist. Thomson (1847), Kelvin (1847), and Dirichlet offered a different line of reasoning. The Dirichlet integral 1(1.£):=
f
Ja
IVu(x)1 2 dx =
f tu~.(x)dx
Ja i=1
(7.1.1)
describes the energy in physics. With boundary values 1.£ = II' on r given, one seeks to minimise 1(1.£). This variational problem is equivalent to
1(1.£, v):=
fa
(Vu(x), Vv(x»dx = 0 for all v with v
= 0 on r.
(7.1.2)
The proof of the equivalence results from 1(1.£+ v) = 1(u)+21(u, v)+l(v) and l(v) ~ 0 for all v (cf. Theorem 6.5.12). Green's formula (2.2.5a) provides 1(1.£, v) = fa v.::1udx = 0 for all v with v = 0 on r such that .::1u = 0 follows. Thus, like (2), the variational problem 1(1.£) = min is equivalent to the Dirichlet problem .::1u = 0 in n, 1.£ = II' on r. The so-called Dirichlet principle states that l(u), since it is bounded from below by l(u) 2 0, must take a minimum for some u. According to the above considerations this would ensure the existence of a solution of the Dirichlet problem. In 1870, WeierstraB argued against this line of reasoning, stating that while there may exist an infimum of l(u) over {u E C2(fl) n CO(fl):u = II' on r} it need not necessarily be in this set. For example, the integral J(u):= fo1u 2(x)dx in {u E CO ([0, 1]):1.£(0) = 0,1.£(1) = I} never takes the value inf J(u) = O. Further, the following example due to Hadamard shows that no finite infimum of the Dirichlet integral need exist. Let r and II' be the polar coordinates in the circle fl = Kl(O). The function u(r, 11') = :E~=1 r n1 n- 2sin(n!
W. Hackbusch, Elliptic Differential Equations: Theory and Numerical Treatment, Springer Series in Computational Mathematics 18, DOI 10.1007/978-3-642-11490-8_7, © Springer-Verlag Berlin Heidelberg 2010
144
7.2 Equations with Homogeneous Dirichlet Boundary Conditions
145
The above difficulties disappear if one seeks the solutions in the more suitable Sobolev spaces instead of in C 2 (il) n c<'(il).
7.2 Equations with Homogeneous Dirichlet Boundary Conditions In the following we investigate the elliptic equation (7.2.1a)
Lu=g in il,
L
:L :L (-1)I.8ID.8 aa.8(x)Da
=
(7.2.1b)
lal:=;m 1.8I:=;m
of order 2m (cf. Section 5.3; Exercise 5.3.5d). The principal part of Lis
Lo = (_1)m
:L
D.8aa,8{x)Da .
(7.2.2)
lal=I.8I=m
According to Definition 5.3.4, L is uniformly elliptic in il if there exists such that
:L
aa.8(x)ea+.8 ~
flel 2m
for all x E il, e E
f
>0
(7.2.3)
ffin.
lal=I.8I=m
Attention. In the case that only aa,8 E LOO{il) is assumed, one needs to replace "for all x E il" by "for almost· all x E il" . We assume the homogeneous Dirichlet boundary conditions ou= 0 ( - 0 U=o -
'on
'on
)2
U=O ...
( 0 )m-l -
"on
U=O
onT
'
(7.2.4)
which are only meaningful if r = oil is sufficiently smooth. Note that in the standard case m = 1 (an equation of second order) condition (4) becomes u=o.
Since with u = 0 on r the tangential derivatives also vanish, not only the kth normal derivatives (k ~ m - 1) but also all the derivatives of order ~ m - 1 are equal to zero:
Dau{x)
=0
in x E
r
for
lal ~ m -
1.
(7.2.4')
Condition (4') no longer requires the existence of a normal direction. According to Corollary 6.2.43, (4') can also be formulated as uE HQ{il).
(7.2.4")
Let u E C2m{il)nH(j{il) be a classical solution of (1a) and (4). To derive the variational formulation we take an arbitrary v E Cgo(il) and consider
146
7 Variational Formulation
Since v E C{f(.fl), the integrand vanishes in the proximity of r so that one can integrate by parts:
without boundary terms occurring. Thus we have found the variational formulation
L
1
aa.B(x)(Dau(x))(D.Bv(x)) dx =
lal.I.BI:S;m [J
1
g(x)v(x) dx
(7.2.5)
[J
for all v E
co(n),
since Lu = g. If conversely u E c2m{n) with boundary conditions (4) satisfies condition (5), then the partial integration can be reversed and f[J{g - Lu)vdx = 0 for all v E C{f{n) proves Lu = g. This means that a classical solution of the variational problem (5) with boundary condition (4) is also a solution of the original boundary value problem. Hence the differential equation (la,b) and the variational formulation (5) are equivalent with respect to classical solutions. We introduce the bilinear form a(u, v):=
L
1
aa.B(x){Dau)(D.Bv)dx
(7.2.6)
fa
g(x)v(x)dx.
(7.2.7)
lal.I.BI:S;m [J
and the functional f(v) :=
AB remarked above, the boundary condition (4) for classical solutions u E c 2m(n)nHm(n) means that u E H{{'(.fl). Thus the "variational formulation" or "weak formulation" of the boundary value problem (1), (4) reads as follows: find
u E Hrr(.fl)
with a(u, v)
= f(v) for all v
E Coo{n).
(7.2.8)
A solution of problem (8) which, according to the definition, lies in H{{'(n) but not necessarily in c2m(n), is called a weak solution.
Exercise 7.2.1. (a) Let n be bounded. Show that any classical solution u E c2m(n) n cm(.fl) is also a weak solution. (b) With the aid of Example 2.4.2 show that this statement becomes false for unbounded domains. Theorem 7.2.2.
Let aa.B E LOO{n). The bilinear form defined by (6) is bounded on H{{'(n) x H{{'(n).
7.2 Equations with Homogeneous Dirichlet Boundary Conditions
147
PROOF. Let u,v E Goo(n). The inequality (6.2.5c) yields la(u, v)1 ::;
L Ilaa~IILOO(a)IDauloID~vlo ::; const lulmlvlm. a,~
Since Goo(n) is dense in H{f(n) (cf. Theorem 6.2.10), a{-,.) has an extension to H{f(n) xH{f(n» and is bounded by the same constant (cf. Lemma 6.5.1b) .
•
The function f(v) is also defined and bounded for v E H{f(n) if, for example, 9 E L2(n). According to Exercise 6.5.11, the variational formulation (8) is equivalent to the following one: find u E H{f(n)
with a(u,v) = f(v)
for all v E HO'(n).
(7.2.9)
One can regain the form Lu = f by applying Lemma 6.5.1. Let L E L(H{f(n), H-m(n» and f E H-m(n) = (H{f(n»' be defined by a(u,v) = (Lu, V)H-m(n)XH(J'(a) and f(v) = (f, V)H-m(a)XH(J'(a) for all v E HO'(n). Equation (9) states that Lu = f. (7.2.9') While (la) represents an equation Lu = 9 in CO(n) (Le., for a classical solution), (9') is an equation in H-m(n). Theorem 6.5.9 guarantees unique solvability of Equation (9) if a(·,·) is H{f(n)-elliptic. We first investigate the standard case m = 1 (equations of order 2m = 2). Theorem 7.2.3. Let n be bounded, m = 1, aa~ E LOO(n). Let L satisfy (3) (uniform ellipticity) and be equal to the principal part Lo, i.e., aa~ = 0 for lal + 1.81::; 1. Then the form a(·,·) is HJ(n)-elliptic: a(u,u);::: t'lul~,
f'
> O.
(7.2.10)
PROOF. Since lal = 1.81 = lone can identify a and.8 according to Da = 8/8xi, D~ = 8/8xj with indices i,j E {I,,,,, n}. For fixed x E n use (3) with ~ = Vu(x):
L lal,I~I=l
n
aa~(x)~a+~ =
L
aij(x)~i~j ;::: fl~12 = fIVu(x)12.
i,j=l
Integration over n yields a(u, u) ;::: fIn IVul 2 dx. Since In IVul 2 dx ;::: Gnluli (cf. Lemma 6.2.11), (10) follows with f' = fGn. • Corollary 7.2.4. The condition "n bounded" may be dropped if for a .8 = 0 one assumes aoo(x) ;::: 1J > 0 (instead of aoo = 0). Example 7.2.5. The Helmholtz equation -Llu the bilinear form
+u
=
=
f in n leads to
148
7 Variational Formulation
a(u, v):=
1rf: a
uz.(x)vz.(x) + u(x)v(x)] dx =
i=1
1 a
[(Vu, Vv)
+ uv] dx.
a(u,v) is the scalar product in HJ(D) (and H1(D)). The fact that a(u,u) = lul~ proves the HJ(n)-ellipticity.
Exercise 7.2.6. Let the assumptions of Theorem 3 or Corollary 4 be satisfied, except for the fact that the coefficients aaO and 0.0{3 (Ial = IPI = 1) of the first derivatives are arbitrary constants. Show that inequality (10) holds unchanged. Theorem 3 cannot easily be extended to the case m
> 1.
Theorem 7.2.7. Let the coefficients oj the principal part be constants: aa{3 = const Jor lal = IPI = m. FUrthermore, assume that aa{3 = 0 Jor 0 < lal + IPI :$ 2m - 1, aoo ~ 0 Jor a = P = O. Let L be uniJormly elliptic (cJ. (3)). F'u.rther let either n be bounded or aoo ~
'1 > O.
Then a(·,·) is Hlf(D)-elliptic.
PROOF. We continue u E Hlf(n) through '1.£ = 0 onto JRn • Theorem 6.2.21, Exercise 6.2.22, and inequality (3) show that a(u, '1.£)
-1
aoou2 dx =
a
L
1
L
aa{3
aaf3DauIJf3udx
lal,I{3I=m a lal,I{3I=m =
L
J.
D a uD{3udx
R"
aaf3
L)(ie )au (e)][(i e ){3u(e)]d{
lal,If3I=m
=
J. [ L
aa{3ea+{3l1u(e) 12 d{
R" lal,If3I=m
~ E r leI 2m lu(e)1 2 d{. lIt" Let 0.00 ~ '1 > O. There exists an E' > 0, so that Elel 2m ~ E' E~al::;m lea l2 - '1 for all e E JRn . From this follows EJ leI 2m lu(e)1 2d{ ~ E'lul m - '11'1.£16 and a(u,u) ~ E'lul~ (cf. Lemma 6.2.23). If n is bounded, use Lemma 6.2.11. • Having shown the Hlf(n)-ellipticity of the form a(·,·), we are now able to apply the general proofs after Theorem 6.5.9.
Theorem 7.2.8.
(Existence and uniqueness of weak solutions)!J a(·,·) is H{f(n)-elliptic then there exists a solution u E Hlf(D) oj Problem (9) which satisfies 1
lul m:$ CE IJI-m
(CE from (6.5.5)).
(7.2.11)
7.2 Equations with Homogeneous Dirichlet Boundary Conditions
Since (11) holds for all f E H-m(a) and u is equivalent to
= L-1 f
149
(cf. (9')), inequality (11)
(7.2.11')
The term "variational problem" for (9) goes back to the following statement, inferred from Theorem 6.5.12:
Theorem 7.2.9. Let a(·,·) be an H{f(a)-elliptic and symmetric bilinear form. Then (9) is equivalent to the variational problem find where
u E HQ'(a),
such that
J(u)::; J(v)
for all v E HQ'(a), (7.2.12a)
1 J(v) := 2a(v, v) - f(v).
(7.2.12b)
Attention. If a(·,·) is either not Ho(a)-elliptic or not symmetric, Problem (9) remains meaningful although the solution does not minimize the functional J(u). Example 7.2.10. The Poisson equation -Llu = f in a, u = 0 on leads to a(u,v) = In{Vu(x), Vv (x» dx. For a bounded domain a, a(.,.) is HJ(n)-elliptic (cf. Theorem 3), so that for any f E H-1(a) there exists exactly one (weak) solution u E HJ(a) of the Poisson equation. This is also the solution of the variational problem! In IVul 2 dx - feu) = min.
r,
A weaker condition than Ho(a)-ellipticity is the Ho(a)-coercivity: a(u,u) ~ €Iul~ - Glula.
= 1, and let the coefficients a"'{3 E LOO(a) satisfy condition (3) of uniform ellipticity. Then a(·,·) is HJ(a)-coercive.
Theorem 7.2.11. Let m
PROOF. We write L as L = LI + L Il , with LI satisfying the conditions of Theorem 3, resp. Corollary 4 if a is not bounded, and LII containing only • derivatives of order ::; 1. Then we can apply the following lemma.
Lemma 7.2.12. Let a(·, .) = a' (-, .) + a" (-, .) be decomposed such that a' (', .) is HQ(ll)-elliptic, or perhaps only H{f(ll)-coercive, while a"(u,u)
=
L lal.I.BI$m lal+I.BI<2m
1
a",{3D"'uD{3udx
n
with a",{3 E LOO(.o) contains only derivatives of order::; 2m - 1. Then a(·,·) is also H{f(.o)-coercive.
150
7 Variational Formulation
PROOF. (6.5.12c) follows from (6.2.lOb) with X = Hbal(n), Y = H)fl(n), V = H{f(n), and U = L2(n), so that Lemma 6.5.18c proves the assertion. • The generalisation of Theorem 11 to arbitrary m conditions on the coefficients of the principal part.
~
1 requires stronger
Theorem 1.2.13. (Garding's theorem) Let L be uniformly elliptic (cf. (3)) and assume aa{3 E LOO(n). Purthermore, let the coefficients aa{3 with lal = 1,81 = m be uniformly continuous in n. Then a(·,·) is H[{'(n)-coercive. If conversely aa{3 E C(n) holds for lal = 1,81 = m and aa{3 E LOO(n) otherwise, then from the H{j(n)-coercivity follows uniform ellipticity (3). Details of the proof can be found in Wloka [1, Theorem 19.2]. The proof given there also holds for unbounded n, since the coefficients are uniformly continuous. For the first part of the theorem one uses a partition of unity. The significance of coercivity lies in the following statement. Theorem 1.2.14. Let n be bounded and a(·,·) be H{j(n)-coercive. Then one of the following alternatives holds: (i) Problem (9) has exactly one (weak) solution 11. E H{j(n). (ii) The kernels E = kernel(L) and E* = kernel(L') are k-dimensional for a k E IN, i.e., a(e,v) = 0,
a(v,e*) = 0
for all e E E,e* E E*,v E H{j(n).
Purther, the eigenvalue problem a(e, v) = .>.(e, V)L~(a)
for all v E HO'(n)
(7.2.13)
has countably many eigenvalues which do not accumulate in CIJ.
PROOF. Since for bounded n the embedding V:= H{j(n) c U := L2(n) is compact (cf. Theorem 604.8b), Theorem 6.5.15 is applicable. •
7.3 Inhomogeneous Dirichlet Boundary Conditions Next, we consider the boundary value problem Lu = g in
n,
11.
=
on
r,
(7.3.1)
where L is a differential operator of second order (Le., m = 1). The variational formulation of the boundary value problem reads: find
11.
E
Hl(n) with a(u,v) = f(v)
11.
=
on r
for all v
E
such that HJ(n).
(7.3.2a) (7.3.2b)
7.3 Inhomogeneous Dirichlet Boundary Conditions
151
But according to Section 6.2.5 the restriction ulr of u E Hl(ll) on r is well defined as a function in Hl/2(r). Thus "u = cp on r" must be understood as the equality ulr = cp in Hl/2(r). In contrast to the preceding section one uses a(·,·) in (2b) as a bilinear form on Hl(ll) x HJ(ll). It is easy to see that a(., .) is well-defined and bounded on this product. Remark 7.3.1. For the solvability of Problem (2a,b) it is necessary that: there exists a UO E Hl(ll)
with uolr
= cpo
(7.3.3)
If a function Uo with Property (3) is known, a second characterisation of the
weak solution results: Let Uo
satisfy (3);
find
W
a(w, v) = f'(v) := f(v) - a(uo, v)
E
HJ(ll), such that
(7.3.4a)
for all v E HJ(ll).
(7.3.4b)
Remark 7.3.2. The variational problems (2a,b) and (4a,b) are equivalent. If and ware the solutions of (4a,b), then u = Uo +w is a solution of Problem (2a,b).1f u is a solution of (2a,b), then, for example, UO = u and w = 0 satisfy Problem (4a, b).
Uo
Exercise 7.3.3. Show that
f'
E H-1(ll) for
f' from (4b) and
If'1-1 ~ Ifl-1 + Gsluoh
(7.3.5)
with Gs from la(u,v)\ ~ Gsluhlv\1 (cf. (6.5.1)). Remark 7.3.4. The problem (1) and the variational formulation (2a,b) have the same classical solutions if such exist. PROOF. It suffices to assume v E C~(ll) in (2b). Integration by parts can • be carried out as in Section 7.2 and proves the assertion.
Theorem 7.3.5. (Existence and uniqueness). Let problem (2.9) (with homogeneous boundary values) be uniquely solvable for all f E H-1(ll). Then Condition (3) is sufficient, and necessary, for the unique solvability of Problem (2a,b).
PROOF. If there exists a solution u E H1(ll) of (2a,b) then (3) is satisfied. However, if (3) holds, one obtains via (4a,b) a unique solution since (4b) agrees with (2.9). • Remark 7.3.6. Under the condition
n E CO,l, (3) is equivalent to
cp E Hl/2(r).
(7.3.6)
152
7 Variational Formulation
PROOF. If Uo satisfies Condition (3) then Theorem 6.2.40 shows that cp E H1/2(r). If conversely cp E H1/2(r), then the same theorem guarantees an extension UO E Hl(n) to fl with Uolr = cp and (7.3.7)
•
Let inequality (2.11') hold in the case of homogeneous boundary values. Equation (5) shows lult 5 luoll + Iwlt 5 luoh + (1/1-1 + C'luoh)/f for the solution of Problem (4a,b). The estimate (7) proves Theorem 7.3.7. Let n E CO· 1. Let the bilinear lorm be restricted to Hl(n) x HJ(fl) and let it satisfy (2.11'). Then lor every IE H-1(n) and cp E H 1/2(r) there exists exactly one solution u E Hl(fl) 01 Problem (2a,b) with (7.3.8)
Exercise 7.3.8. Let a(·,·) be symmetric and HJ(n)-elliptic. Show that Problem (2a,b) with I = 0 is equivalent to the variational problem: Find u E H1(fl) with ulr = cp such that a(u, u) becomes minimal (cf. (1.1)).
7.4 Natural Boundary Conditions The bilinear form a(·,·) defined in (2.6) is also well-defined on Hm(fl) x Hm(fl). In analogy to Theorem 7.2.2 there holds Theorem 7.4.1. Assume aal3 E LOO(n). The bilinear lorm defined by (2.6) is bounded on Hm(n) X Hm(.fl): la(u, v)1 5 Ea.l3l1aaI3I1Loo(o)IUlmIVlm lor all u,v E Hm(fl). Now let I be a functional from (Hm(fl))'. Equation (2.7) with U E L2(n), for example, describes such a functional; but (2.7) is only a special case of the functional f subsequently defined in (la), which we want to use as a foundation in the following. Exercise 7.4.2. Let r be sufficiently smooth and let U E L2(n), cp E L2(r) hold. Show that f(v):= iau(x)v(x)dx+
i
cp(x)v(x) dr,
(7.4.1a)
defines a functional in (Hl(fl»' with IIfll(Hl(O»1 5 C(lIullp(S1) + IIcpIlL2(r». This implies 1 E (Hm(fl»' for all m > 1. More precisely, the following also holds:
7.4 Natural Boundary Conditions
IIfll(Hl(O»' :::; C(lIglI(Hl(O»'
+ IIcpIlH-l/2(r»·
153
(7.4.1b)
Frequently, variational problems with a physical background have the form: find u E Hm(n),
a(u,v) = f(v)
such that
for all v E Hm(n).
(7.4.2a) (7.4.2b)
In contrast to the condition u E Hlf(n) from Section 7.2, u E Hm(n) contains no boundary condition. Nevertheless, Problem (2a,b) has a unique solution if a(·,·) is Hm(n)-elliptic. This condition is easy to satisfy. Theorem 7.4.3. Under the conditions of Theorem 7.2.3 or Corollary 7.2.4 a(·,·) is Hl(Q)-elliptic: a(u,u) 2: flul~ for all u E Hl(n). Problem (2a,b) (with m = 1) has exactly one solution which satisfies the estimate (3): (7.4.3)
PROOF. The same as for Theorem 7.2.3 or Corollary 7.2.4, and Theorem 7.2.8. • Corollary 7.4.4. (a) A unique solution which satisfies the estimate (3), also exists if instead of Hl(n)-ellipticity one assumes: a(·,·) is H1(n)-coercive, n E CO,1 is bounded, A = 0 is not an eigenvalue (i.e., a(u, v) = 0 for all v E H1(n) implies u = 0). (b) The combination of inequalities (3) and (1b) results in (7.4.3')
for the solution of (2a,b), if f is defined by (1a) with g E (H1(n))', cp E H-1/2(r). PROOF. (a). According to Theorem 6.4.8b, H1(n) is compactly embedded in £2(Q) so that the statement of Theorem 7.2.14 can be transferred. If A = 0 is not an eigenvalue then £-1 E £«Hl(n»)', H1(n» holds (cf. Theorem 6.4.12). • To find out which classical boundary value problem corresponds to the variational formulation (2a,b), we assume that (2a,b) has a classical solution U E Hm(n) n c2m(Q). Further, v E COO(n) can be assumed also (cf. Lemma 6.5.1b). For reasons of simplicity we limit ourselves to the case m = 1. Under the assumption aa{3 E C1(n) and under suitable conditions on n the following general Green formula is applicable:
154
7 Variational Formulation
a( u, v)
= =
r [t G.tjUill, Villi + t i,j=l
Jn
In [+
aoiUVill,
i=1
+
t
G.toUa;, V + aoouv] dx
,=1
,t1(G.tjUill .)illi - t(lZoiU)ill'
+ t£lioUilli +aooU]Vdx
l [~njG.tjUill' + ~~tzoiu]vdr. s"
(17.4.4)
s
Here, the ~ are the components of the normal direction n define the boundary differential operator
= n(x), x e r. We
nan njG.tj- + L~lZoi, , '-1 ax, s-'-1 ',3-
B:= L
(7.4.5)
when L is described by (2.1b). Equation (4) becomes a(u,v) = InvLudx+ IrvBudr. By the formulation of the problem, a(u, v) agrees with f(v) from (Ia). If we first choose V e HJ(.!J) c Hl(.!J) the boundary integrals drop out and we obtain Lu = 9 as in Section 7.2. By this the identity a(u,v) = f(v) reduces to IrvBudr = Ir«pVdr for all v e H1(.!J). According to Theorem 6.2.40b, vir runs over the set H1/2(r) if v runs over Hl(.!J), so that we have Ir 1/J(Bu - 'P) dr = 0 for all1/J e H1/2(r); thus Bu = 'P. This proves
Theorem 7.4.5. Let r be sufficiently smooth. A classical solution of the problem (2a,b) with f from (1a) is also the classical solution of the boundary value problem (7.4.6) Lu = 9 in n, Bu = 'P on r, and conversely.
The condition Bu = 'P is called the natural boundary condition. This results from the fact that in (2b) (as distinct from (2.9» the function v may assume arbitrary boundary values. Note that the bilinear form determines L as well as B.
Exercise 7.4.6. Show that the bilinear form from Example 7.2.10 for -Llu = 9 has as the natural boundary condition the Neumann condition -au/On = o. Theorem 7.4.7. Let n e CO,l be bounded and a(·,·) be W"'(.!J)-coercive. Then the statements of Theorem 7.2.14 holds with Hm(.!J) instead of H{?(.!J). Example 7.4.8. Let n e CO,l be a bounded domain. The bilinear form a(u, v) = In( (,Vu, Vv) + cuv)dx, associated to the Helmholtz equation -Llu + cu = f
in
n
(c
> 0), au/an = 0 on r,
is H1(.!J)-elliptic since a(u, u) ~ min(1, c)lul~. For c = 0, however, a(·,·) is only H1(.!J)-coercive. As is known from Theorem 3.4.1, the Neumann boundary
7.4 Natural Boundary Conditions
155
value problem for the Poisson equation (i.e., for c = 0) is not uniquely solvable. According to alternative (ii) in Theorem 7, there exists a nontrivial eigenspace E = kernel(L). u E E satisfies a(u, u) = 0, thus Vu = O. Since fl is connected, it follows that u(x) = const and therefore dimE = 1. Since a(·,·) is symmetric, E* := kernel(L') coincides with E. According to Theorem 7, the Neumann boundary value problem a(u, v) = f(v) (v E Hl(fl)) is solvable if and only if f .L E, i.e., f(l) = O. If f(v) := Ing(x)v(x)dx, f(l) = 0 reads as In gdx = O. If, however, f is given by (la), the integrability condition reads f(l) = In 9 dx + Ir cpdr = 0 (this is Equation (3.4.2), in which f should be replaced by -g).
Remark 7.4.9. While the classical formulation of a boundary condition such as au/an = 0 requires conditions on the boundary r, the problem (2a,b) can ba formulated for arbitrary measurable fl. In the following, we proceed in the opposite direction: does there exist, for a classically formulated boundary value problem Lu = g in fl, Bu = cp on r, with given Land B, a bilinear form a(·,·) such that (2a, b) is the corresponding variational formulation? This would mean that the freely prescribed boundary operator B represents the natural boundary condition. For m = 1 the general form of the boundary operator reads
t; bj(x) a + bo(x) n
B
=
aXi
(x E r).
(7.4.7)
With b T = (bl, .. ·, bn) one can also write B = bTV +bo (cf. (5.2.1b,b')). Here bTV is not allowed to be a tangential derivative (cf. Remark 5.2.2):
(b(x), n(x)}
=f 0
for all xE
r.
(7.4.8)
Remark 7.4.10. Let m = 1. Let (8) hold. Let A(x) be the matrix A = (aij) (cf. (5.1.1d)). By passing from Bu = cp to the equivalent scaled equation u Bu = ucp with u(x) = (n(x), A(x)n(x)} / (n(x), b(x)}, one can ensure that (n,ub) = (n, An). Thus in the following it is always assumed that b already satisfies (n, b) = (n, An). PROOF. Because of (8) u is well defined. For Bu = cp and uBu = ucp to be equivalent, u =f 0 is required. This is guaranteed by the uniform ellipticity: (n, An) ~ elnl 2 = e.
Theorem 7.4.11.
(Construction of the bilinear form). Let m = 1.
Let Land B be given by (2.1b) and (7), with b satisfying condition (8). Then there exists a bilinear form a(.,.) on Hl(fl) x Hl(fl) such that to the variational problem (2a,b) co'TTesponds the classical formulation Lu = g in fl, Bu = cp on r.
156
7 Variational Formulation
PROOF. The bilinear form we seek is not uniquely detennined . We will give two possibilities for its construction. First we discuss the absolute term in (7). On the basis of Remark 10 we assume (n, b) = (n, An). (a) Let the vector function P(x) := (P1(X),· . " Pn{x)) E C1(0) be arbitrary. The differential operator
maps every u e C1(n) into zero: L11.1. = O. Thus the operator L can be replaced by L + L1 without changing the boundary value problem. Let a(·,·) be constructed according to (2.6) from the coefficients of L + L 1 . Equation (5) shows that the boundary operator associated with a(·,·) has the absolute term (7.4.9) i=1
If bo = 0, the choice of Pi. = -aoi is successful. Otherwise, two other options are available. (aa) Select Pi such that on r the following holds: (3;(x) = bo(x)n,(x) -aoi(X). Since Inl = 1, the term (9) then agrees with bo(x). The practical difficulty in this method consists in the need to construct a smooth continuation on n of the boundary values Pi(X), x e r . (ab) Set Pi = -aoi and add a suitable boundary integral:
a(u, v) :=
In {~atju:r:.v:r:j+ ~(ato '", +
i
- ao,)u:r:. v +
bo{x)uv dr.
[aoo - ~(ao,):r:, ]uv} dx ,
(7.4.1O)
The integration by parts described above shows that the boundary operator associated to (10) reads (7.4.11) (b) The operator (11) can be written in the form iJ = j)TV +bo with j) = An. Since (n, b) = (n, An) = (n, b) has been assumed already, d := b - b is orthogonal to n. To change j) to b there are again two options. (ba) Define the n x n-matrix A 8 on r by A8 = dnT - ndT , Le., a:j = dtnj - nidj. This A8(x) is skew-symmetric: A8T = _A8. Continue AS(x), which is at first only defined on r, to a skew-symmetric matrix AS E C1(n). [Here we have the same practical difficulty as in step (aa).] The entries of AS define
7.4 Natural Boundary Conditions
Again, L 2 u
157
= 0 holds for all u E G 2 {il), since
Thus L can be replaced by L + L2 without changing the boundary value problem. The coefficients belonging to L + L2 result in a boundary operator B whose derivative terms read Li,j nj{a;,j + aij)a/axi = [(A + A 8 )n]TV. By the construction of AS we have ASn = (dnT - ndT)n = d = b - b , since (n, n) = 1 and (d, n) = O. Since we also have An = b, the derivative term in B gives bTV as desired. The transition L ~ L + L2 does not change the absolute term in 8, so that from (11) follows B = bTV + boo (bb) Let iJ = bTV + bo be given (cf. (11)) such that d = b - b is orthogonal to n. From this the boundary operator T := dTV is the derivative in a tangential direction if n = 2 [resp. in the tangent hyperplane if n ;::: 3]. If r is sufficiently smooth then the restriction vir of v E H1(il) is an element of H1/2(r). By Remark 6.3.14a, one can show that T E L(H1/2(r), H-1/2(r)). Since T(ulr) E H-1/2(r), fr .,pT(ulr) dr) is well-defined for.,p E Hl/2(r), in particular, for .,p = vir with v E H1(il). Thus b(u,v):= £(v1r)T(u1r)dr is a bilinear form bounded on H1(il) x Hl(il). We add b(.,·) to a(·,·) in (10). Integration by parts yields the boundary operator iJ + T = bTV + bo+ (b - b)TV = bTV + bo = B. • Remark 7.4.12. Let the bilinear form (2.6) be Hl(il)-coercive (cf. Theorem 3). Let its coefficients, as well as the boundary r, be sufficiently smooth (E 0 1 ). Then the constructions of the preceding proof again result in an B7jH1 (il)-coercive form.
PROOF. We go through the steps. (a) In step (aa) only terms of lower order are added so that Lemma 7.2.12 is applicable. As for step (ab), see step (bb). (ba) In step (ba) one adds b{u, v) := fa Li)aijuz.vzj + (atj)zjvuz.ldx. Here Lemma 7.2.12 is also applicable, since the skew symmetry of AS results in b(u, u) = fa L(atj)zjuuz,dx. (bb) In the construction in step (bb), Lemma 7.2.12 is applicable analogously. This is easiest to understand in the case n = 2. Let r be described by {(X1(S),X2(S)): 0 :5 s :5 I}. If X1,X2 E G1([O,1]) and x~(O) = x~(1), and if we have d E Gl(r) for the d in T = dTV, then b(.,.) has the representation b(u,v) = f01v(s)r{s)u'(s)ds, where u(s) = U(X1(S),X2(S)), v(s) = V(X1(S),X2(S)), r E G1{[0,1]). Thanks to periodicity, integration by parts yields in b{u, v) = - f~ (rv)'u ds without boundary terms, so that b( u, u) =
! [f01uru' ds - f~ (ru)'u dS] = -! f01 r'u2 ds. This implies
158
7 Variational Formulation Ib(u, u)1 !S
~lIrIlOl([0.1)) lIulrlll~(r) !S 01Irllol([0.1]) lIullt-'(n)
for all 8 E (1/2,1) (cf. Theorem 6.2.4080). Since lul~ !S apply Lemma 6.5.18c.
flul1- Clul~, one can •
The case m ~ 2 has been excluded in this section (except for Theorem 1). Boundary value problems of order 2m require m boundary conditions Bju = CPj on r (j = 1"", m) (cf. Section 5.3). For m ~ 2 the proof of Hm(fl)_ coercivity becomes more complicated. In order to carry over Theorem 7.2.13, one needs in addition the so-called condition of Agmon (cf. Wloka [1, Theorem 19.3], Lions-Magenes [1, p. 210]). The resulting complications can be seen with the aid of the biharmonic equation. Example 7.4.13. (a) To the variational problem: find u E H2(fl) with
a(u, v) := L L1uL1vdx = f(v) := LgvdX+
£
(7.4.1280)
(CP1:: -CP2 v )dr
for all v E JI2(fl), corresponds the classical formulation
.12 u
= 9 in fl,
L1u = CP1
and
a L11£ = CP2 an
(7.4. 12b)
on r.
But the bilinear form a(·,·) is not H 2 (fl)-coercive. (b) To the variational problem: find 1£ E H2(n) n HJ(fl) with
a(1£, v) = f(v):= L gvdx+
£
CP:: dr for all v E H2(fl)
n HJ(fl) (7.4.12c)
(a( " .) as in (1280» corresponds the classical formulation Ll 2u
=9
in fl,
1£ = 0 and .11£ = cP on
r.
(7.4.12d)
The bilinear form is H2(fl) n HJ(fl)-coercive. (c) The boundary conditions in
L1 21£
=9
in fl,
a an
1£ = 0 and -L11£ = cP on
r
(7.4.12e)
are admissible. Nevertheless this boundary value problem cannot be written in the present form as a variational problem. A variational formulation for (12e) reads: Find u E H2(fl) n HJ(fl) such that
a(1£,v)=f(v):= LgvdX- £cpvdr
forallvEH 2(fl) with
::lr=O (7.4.12f)
7.4 Natural Boundary Conditions
159
(a(.,.) as in (12a». But this does not agree with the present concept since and v belong to different spaces.
U
PROOF. The equivalence of the variational and the classical formulation can be shown via integration by parts:
a(u, v)
=
In
vLl 2udx +
1
[LlU:: - 8!u v] dr.
The noncoercivity in part (a) results as follows. Let il c IRn be bounded. For all a E IR, Ua(Xl,··· ,xn ) = sin(axl)exp(ax2) lies in H2(il) and satisfies Llua = 0, hence also a(u a , u a ) = o. If a(·,·) were coercive, there would exist a C with 0 = a(ua,ua ) ~ Elua l2 - Clualo for all a, i.e., lua l2 ::; (C/E)lualo. The contradiction results from lua l2 ~ 182ua/ax~lo = a 21ua lo for sufficiently • large a. Natural and Dirichlet conditions can occur together. In Example 13b, u
=
o is a Dirichlet condition and (aLlu/8n) = cp a natural boundary condition. Even in the case m = 1 both sorts of boundary conditions can occur.
Example 7.4.14. Let "( be a nonempty, proper subset of r. The boundary value problem -Llu = 9
in il,
u = 0 on ,,(,
au/8n
= cp
on r\,,(
(7.4.13)
in the variational formulation reads as follows: find u E H~(il) such that
a(u, v):=
r('\Iu, '\Iv) dx
Jn
= f(v):=
rgvdx + Jr
Jn
cpvdr
r\"r
for all v E H~(il), where H~(il) := {u E Hl(il): u = 0 on "(}. Equation (13) is occasionally termed a Robin problem.
Exercise 7.4.15. Let a(u,v) := In [(V'u, V'v} +cuv] dx with c > 0 on V x V with V:= {u E Hl(il):u constant on r} be defined. Show that (a) a(·, .) is V-elliptic. (b) The weak formulation: u E V, a( u, v) = In gv dx + Ir cpv dr for all v E V corresponds to the problem -Llu+ cu
=9
in .0,
u constant on
r,
l ~: = l dr
cpdr.
(7.4.14)
which is also called an Adler problem. Finally we want to point out the difficulty of classically interpreting a weak solution. In the variational formulation (2a,b) the right-hand sides 9 and cp of the differential equation and the boundary condition are combined in the functional f. In the variational formulation the components 9 and cp are
160
7 Variational Formulation
indistinguishable! u E HI(n) has first derivatives in L2(a), whose restrictions to r do not have to make sense. That is why Bu cannot be defined in general; Bu =
n,
r.
8 The Method of Finite Elements
In Chapter 7 the variational formulation was introduced only for the purpose of proving the existence of a (weak) solution. It will now tum out that the variational formulation is the foundation of a new method of discretisation.
8.1 The Ritz-Galerkin Method Suppose we have a boundary value problem in its variational formulation: Find u E V, so that a(u, v) = f(v)
for all v E V,
(8.1.1)
where we are thinking, in particular, of V = Hlr(D) and V = Hl(D) (cf. Section 7.2, Section 7.4). Of course, it is assumed that a(·,·) is a bounded bilinear form defined on V x V, and that f E V':
la(u,v)1 $ Csllullvllvllv
f
for u,v E V,
E V'.
(8.1.2)
Difference methods arise through discretising the differential operators. Now we wish to leave the differential operator hidden in a(·,·) unchanged. The RitzGalerkin discretisation consists in replacing the infinite-dimensional space V with a finite-dimensional space VN : dimVN
= N < 00.
(8.1.3)
VN equipped with the norm 1I·lIv is still a Banach space. Since VN C V, both a(u, v) and f( v) are defined for u, v E VN. Thus we may pose the problem (4):
Find uN E VN, so that a{uN,v)
= f{v) for all v E VN.
(8.1.4)
The solution of (4), ifit exists, is called the Ritz-Galerkin solution (belonging to VN ) of the boundary value problem (1). To calculate a solution one needs a basis of VN. Let {bl, ... , bN} be such a basis, i.e., (8.1.5) VN = span{b1, ... ,bN}. For each coefficient vector v
= {Vi, ... , VN}T
we define N
Pv:=
L::
Vibi·
(8.1.6)
i=l
W. Hackbusch, Elliptic Differential Equations: Theory and Numerical Treatment, Springer Series in Computational Mathematics 18, DOI 10.1007/978-3-642-11490-8_8, © Springer-Verlag Berlin Heidelberg 2010
161
162
8 The Method of Finite Elements
Remark 8.1.1.
P is an isomorphism between m.N and VN . The inverse
p-l: VN ~ m.N is thus well-defined on VN.
Lemma 8.1.2. Assuming (5) the problem (4) is equivalent to Find UN EVN, so that a(uN,b,)=f(bi )
foralli=I, ... ,N.
(8.1.7)
PROOF. Putting v = b, in (4) gives (7). On the other hand, suppose that v = E~l v,b. E VN is arbitrary. Then (7) and the linearity of a(u N ,.) and f give (4): a(u N , v) - f(v) = a(u N ,
L vib,) - J(L vib,) = L v,(a(uN , b,) - f(b,» = O.
•
We now seek u E m.N so that uN = Pu. The following theorem transforms the problem (4) [resp. (7)1 into a system of linear equations. Theorem 8.1.3. Assume (5). The NxN-matrixL f = (ft, ... , fN)T are defined by L'j := a(bj , b,) fi := f(bi )
(i,j
= (L,j)
= 1, ... , N),
(i= 1, ... ,N),
and the N-vector
(8.1.8a) (8.1.8b)
Then the problems (4) and
Lu=f
(8.1.9)
are equivalent. If u is a solution of (9), then uN := Pu solves the problem (4). In the opposite direction, il uN is a solution 01 (4), then u := p-1u N is a solution 01 (9) (cf. Remark 1).
PROOF. (4) is equivalent to (7). In (7) put uN = Pu = Eujbj ; then a(u N , bi ) = E j uja(bj , bi ) = E j Lijuj = I(bi) = Ii, and thus Lu = f. • In engineering applications where the boundary value problems arise from continuum mechanics (cf. the first paragraph of Section 5.3.1), one calls L the stiffness matrix. The connections between Land a(., .), on the one hand, and between f and f(·), on the other, are clear from
Remark 8.1.4. If (u, v) := Ei U,Vi is the usual scalar product, then a( u, v) (Lu, v), and f(v) = (f, v) with u = Pu, v = Pv.
=
A trivial consequence of Theorem 3 is: Corollary 8.1.5. The Ritz-Galerkin discretisation (4) has a unique solution uN for each f E V' exactly when the matrix Lin (8a) is nonsingular.
8.1 The Ritz-Galerkin Method
163
The Gelfand triple V cUe V' corresponds to the finite-dimensional situation VN C UN c Vk, where the spaces are the same as sets:
However, the three spaces have different norms:
VN has 1I·llv and UN has 1I·lIu as a norm, while the dual norm
IIvllvj. := sup{l(v, u)ul/liullv: 0 =I u E VN}
for v E V'
(in particular for v E Vk = VN ) is defined on V', but is only a norm on Vk, since IIvllvj. = 0 for each v orthogonal to VN. By Lemma 6.5.1 there is an operator associated to a(·, .): VN x VN -+:m.
so that a(u,v)
= (LNU,V)U
for all u,v E VN .
(8. 1. lOa)
Here we are writing (., ')u for (., ·)Vj.XVN (cf. Remark 6.3.12). The map p::m.N -+ VN can also be viewed as p::m.N -+ V since VN C V. The map P* E L(V',:m.N) adjoint to P is defined by (P*u, v)
= (u,Pv)u
for u E V', v E :m.N.
(8.1.10b)
Exercise 8.1.6. (a) Show: The kernel of P* is V.k c U (orthogonal space relative to 1I·llu). The map P*: Vk -+ :m.N is an isomorphism. (b) Since p-l: VN -+:m.N exists, we can define
where 11·11 is the Euclidean norm. Show: The matrix p*p::m.N -+ :m.N has an inverse with spectral norm
(c) Show: QN := P(p*p)-lp*: U -+ U
is the orthogonal projection (relative to 11·lIu) onto UN we have QN E L(V', V).
= VN. In addition
Lemma 8.1.7. Let LN E L(VN, Vk) be the opemtor corresponding to a(·, .): VN x VN -+ :m., and L E L(V, V') that for a(·, .): V x V -+ :m.. The following relationships hold between the opemtors L n , L, and the stiffness matrix L:
164
8 The Method of Finite Elements L = P*LP = P*LNP, LN = P*-lLP- 1 : VN ~ VN, LN is the restriction of QNL and QNLQN to VN ·
PROOF. For all u, v E m,N we have (Lu, v) = a(Pu, Pv)
= (LNPU, Pv)u =
(P* LNPu, v), = (LPu,Pv)u = (P*LPu,v).
•
By Remark 6.3.12 we can write (I,v)u for f(v). If v = Pv, it follows from (lOb) that f(v) = (I, Pv)u = (p* f, v). Using Remark 4 we then have f=P*f
(B.l.lOc)
for the f in the right side of Eq. (9). In the case of a continuous variational problem the V-ellipticity guarantees its unique solvability. The same condition is also sufficient in the discrete case: Theorem 8.1.8. Assume (3). Suppose the bilinear form is V -elliptic: a(u, u) ;::: CEllull~ for all u E V with CE > o. Then the matrix L in (9) is nonsingular and the Ritz-Galerkin solution uN E VN satisfies
IIU
NIl IIv ~ CE IIfllv;, ~ CE IIfllv'.
PROOF. L is nonsingular since for each u 1= 0 we also have Pu thus (Lu, u) = a{Pu, Pu) ;::: CEIiPull~ > 0
(B.l.lla)
f 0, and
and so, in particular, Lu 1= o. By Exercise 6.5.6a a(·,·) is also VN-elliptic with the same constant CEo From Theorem 6.5.B there holds IIL;vll1v;,+-v'N ~ liCE, i.e., IIU N liv = IIL;Vlllv ~ CE1 I1fllv;,. ella) then results from Exercise 8.1.9. Show IIfllVN ~ IIfllv' for any f E V'. Exercise 8.1.10. Show: (a) If a(·,·) is symmetric then so is L. (b) If a(·,·) is symmetric and V-elliptic then L is positive definite. Under the same assumptions the Ritz-Galerkin solution uN solves the following variational problem (cf. Theorem 6.5.12):
J(u N ) ~ J(u) := a(u, u) - 2f(u) for all u E VN . Example 8.1.11. (Dirichlet Problem) The boundary value problem is
8.1 The Ritz-Galerkin Method
-L1u(X, y) = 1 in n = (0,1) x (0,1),
The weak fonnulation is given by (1) with V
In In
a(u,v) := I(v) :=
{'Vu, Vv} dxdy =
In
u=O on
165
r.
= HJ(n), (U3: V3:
+uyVy ) dxdy,
vdxdy.
The functions b1(X,y)
= sin(7rx)sin(7rY),
b2 (x, y)
= sin(37rx) sin(7rY),
bs(x,y)
= sin(7rx)sin(37rY),
b4(X,y)
= sin(37rx)sin(37rY),
fulfil the boundary conditions and so belong to V = HJ(n). They form a basis of V4 := span {bt, ... , b4 }. The matrix elements Lu = a(bi , bi ) can be worked out to be
L11 = 7r 2 /2,
L44
= 911"2/2.
In addition the chosen basis is a(., .)-orthogonal: Lij
=0
for i
=I i,
so that the stiffness matrix L is diagonal. Furthermore one may calculate Ii = I(bi ) = fn bi(x,y) dxdy, getting
Hence u = L -1 f has the components
and the Ritz-Galerkin solution is then uN (x, y) =
:4 (
sin 7rX sin 7ry +
+
;1
:5
(sin 37rx sin 7ry + sin 7rX sin 37rY)
sin 37rX sin 37ryJ.
The Ritz-Galerkin solution and the exact solution for x = y = 1/2 are NIl
u (2' 2)
2848_ 4
= 405 7r
1 1 u(2' 2) =
= 0.07219140 ... ,
L: 7r164 (-It+1' /[(1 + 2v)(1 + 2J.'){(1 + 2v)2 + (1 + 2J.')2)J 00
....1'=0
= 0.0736713 .... Example 8.1.12. value problem be
(Natural boundary conditions) Let the boundary
166
8 The Method of Finite Elements
-Llu(x,y)
= 1I"2COS1l"X in n = (0,1) x (0,1),
The solution is given by u of (1) with V = H1(n),
au/an =
0 on
r.
= cos 1I"X + const. The weak formulation is in terms
a(u,v):= £(ua;va;+u1l v'/J)dxdy, f(v) := 11"2£ v(x,y)cos1l"xdxdy.
The boundary value problem has a unique solution in W:= {v E V:
£
vdxdy
= O}.
The basis functions b1 (x, y)
= x -1/2,
are in W. The stiffness matrix L and the vector f are then
_ [1 1/4 ] L - 1/4 9/80 '
_ [
f -
-2 -(3/2) + 12/11"2
],
so that U
=L
-1
f
=
[ 3 - 60/11"2 ] -20 + 240/11"2 .
The solution is uN (x, y) = (3 - 60/1I"2)(x - 1/2) - (20 - 240/1I"2)(x - 1/2)3. The rutz-Galerkin solution satisfies the boundary condition Bu/on = 0 and the differential equation only approximately: ouN (0, y)/Bn
= 12 -
120/11"2 ~ -0.16.
For x = 1/4 the approximation is u N (I/4, y) = -7/16 + 45/(411"2) 0.70236 ... , whereas u(I/4, y) = COS 11"/4 = 0.7071. .. is the exact value. In the following we shall consider the case in which a(·,·) is no longer V-elliptic, though it is V-coercive. That a(·, .) is V-coercive guarantees that either problem (1) is solvable or A = 0 is an eigenvalue. Even if one assumes V-coercivity and the solvability of the problem (1) one can not deduce the solvability of the discrete problem (4). Example 8.1.13. a(u, v) := J;(u'v' - lOuv) dx is HJ(O, I)-coercive and a(u,v) = f(v) := Jo1 gvdx (v E HJ(O, 1)) has a unique solution. Let V N be spanned by bl(X) = x(l- x) E V = HJ(O, 1) (Le., N = 1). Then the discrete problem (4) is not solvable since L = O. If one replaces the space V in Lemma 6.5.3 with VN , then there follows from Exercise 6.5.4
8.2 Error Estimates
167
Theorem 8.1.14. The problem (4) is solvable for all f E VI and has a unique solution, uN, which satisfies the estimate (8.1.11b) if and only if
inf{sup{la(u,v)l:v E VN, IIvllv
= 1}:u E VN, lIuliv = I} = €N > O.
(8.1.12)
Since (4) is equivalent to the system of equations (9), one has the
Corollary 8.1.15. The matrix L is nonsingular if and only if (12) holds. Exercise 8.1.16. The requirement (12) is equivalent to 1 lIullv S -sup{la(u,v):v E VN, IIvllv €N
= I}
for all u E VN.
(8.1.12' )
and
(LN as in (lOa)).
(8.1.12")
Warning: The condition (12) for VN does not follow from the analogous condition (6.4.5a) for V. See, however, Theorem 8.2.8. The requirement (12/1) guarantees the existence of L-l, but it does not say anything about its condition, cond(L) = IILIIIIL-11I, which is the deciding factor for the sensitivity of the system of equations Lu = f. For example, if one chooses for a( u, v) := J01 u l Vi dx the basis bi = xi (i = 1, ... , N) then one obtains the very badly conditioned matrix lij = ij / (i+ j -1). The conditioning of L is optimal if one chooses the basis to be a(·, .)-orthogonal: a( bi , bj ) = Oij, as is the case in Example 11, up to a scaling factor.
8.2 Error Estimates For difference methods the solution u and the grid function Uh are defined on different sets. The rutz-Galerkin solution, uN, is, on the other hand, directly comparable with u. One can measure the error due to discretisation with lIu - uN IIv or with lIu - uN lIu. Let u be the solution of (1.1): a(u,v) = f(v) for v E V. Suppose - by chance or because of a clever choice of VN - that u also belongs to V N ; then uN := u also satisfies (1.4). That means: The discretisation error is zero if u E VN. We shall now show: The "closer" that u is to VN the smaller is the discretisation error.
168
8 The Method of Finite Elements
Theorem 8.2.1. (Cea) Assume (1.2), (1.3), and (1.12) hold. Let u E V be a solution of the problem (1.1), and let uN E VN be the Ritz-Galerkin solution of (1.4). Then the following estimate holds: (8.2.1) with Cs from (1.2) and eN from (1.12). Note infwEvN lIu-wllv is the distance of the function u from VN; it will be abbreviated in the following to
(8.2.2)
PROOF. If u satisfies a(u,v) = f(v) for all v E V then it does in particular for all v E VN. Since we also have a(uN,v) = f(v) for v E VN , it follows that a(u N - u,v) = 0 for all v E VN.
For arbitrary v, wE VN with IIvllv
(8.2.3)
= 1 we can therefore conclude
a(u N -w,v)=a([u N -u]+[u-w],v)=a(u-w,v)
and la(u N - w, v)1 ::; Csllu - wllvllvllv
= Csllu -
wllv.
From (1.12') we then obtain "uN -wllv ::;
~suP{la(uN -w, v)l:v E VN, IIvllv = 1} ::; (Cs/eN)IIu-wliv. eN
The triangle inequality then gives
IIu -
uNliv ::;
IIu -
wllv + IIw - uNliv ::; (1 + Cs/eN)IIu - wllv.
Since w E VN is arbitrary we deduce the assertion (1).
•
In Theorem 1 the unique solvability of the problem (1.1) was not assumed, but only the existence of at least one solution. If the discretisation error is supposed to converge to zero one makes use of a sequence of subspaces VN. C V that converge to V in the following sense:
Theorem 8.2.2. Let Vi := VN. C V(i E IN) be a sequence of subspaces with .lim d(u, Vi)
$-+00
=0
for all u E V.
(8.2.4a)
Assume that (1.12) holds with eN. ~ f > 0 for all i E IN; in addition assume (1.2) (continuity o/a(·, Then there exists a unique solutionu o/the problem (1.1) and the Ritz-Galerkin solution u i := uN. converges to u:
.».
IIu -
uillv
-+
0
for i
-+ 00.
8.2 Error Estimates
169
Sufficient to ensure (4a) is 00
Vi C V2 C ... C Vi-l C Vi
UVi dense in V.
c ... C V,
(8.2.4b)
i=l
PROOF. (a) Assume first the existence of a solution u. (aa) The estimate (1) implies the convergence: 11'1.£ - uillv ~ (1
+ CslfN.) d(u, Vi) ---+ 0.
(ab) We now wish to show that (4a) follows from (4b). The inclusion Vi-l C Vi implies d(u, Vi) ~ d(u, Vi-i). Now d(u, Vi) will be a null sequence if for each E > 0 there exists an i such that d(u, Vi) ~ E. From the assumption (4b) there is, for each '1.£ E V and E > 0, awE Vi with 11'1.£ - wllv ~ E. Therefore we have w E Vi for an i E IN. That d(u, Vi) ~ 11'1.£ - wllv ~ E then proves (4a). The convergence u i ---+ '1.£ implies the uniqueness of the solution u. (b) The next thing to show is that the image W := {Lv:v E V} C V' of the operator L: V ---+ V' associated to a(·, .) is closed. For each fEW there is a '1.£ E V with Lu = f, so that part (a) of this proof suffices to show the convergence u i ---+ u. Since lIuiliv ~ IIfllv,ll, it follows that lIuliv = lim lIuiliv ~ IIfllv' /€. Let f" E W be a sequence with f" ---+ in V' and f" = Lu". The Cauchy convergence IIf" - f#lIv' ---+ 0 shows 11'1.£" -u#lIv ~ IIf" - f#lIv' I€'---+ 0 so that the limit u· = limu" E V exists. The continuity of L E L(V, V') shows that = lim f" = lim Lu" = Lu*, and thus that E W. Hence W is closed. (c) In order to demonstrate the existence of a solution to the problem (1.1) we have to show the surjectivity of L: V ---+ V'. If L were not surjective (Le., W =j:. V'), there would be an f E W.L with Ilfllv' = 1. Let J v :V ---+ V' be the Riesz isomorphism (cf. Corollary 6.3.7). Set v := -JV1 f E V. It follows that
U
r
r
r
f(v)
= {j,v}v'xv = -(f,J)v' = -1,
a(u,v)
= (Lu,v}v'xv = 0
for all '1.£ E V. Let u i E Vi be the llitz-Galerkin solutions. They must also satisfy a(ui , v) = 0, Le., a(ui , v) - f(v) = 1. We split up v into vi + wi, where vi E Vi, By (4a), with v in place of '1.£, one may guarantee that IIwiliv ---+ O. This shows 1
= a(ui,v) -
f(v)
= a(ui , vi) -
= a(ui,wi ) -
f(v i ) + a(ui,wi ) - f(w i ) f(w i )
and 1 = la(ui,v) - f(v)1 ~ [Cslluillv
+ Ilfllv']
Ilwiliv.
Since lIuiliv ~ IIfllv'/€ is uniformly bounded and Ilwiliv ---+ 0 this amounts to a contradiction. Therefore L must be surjective, so that for each f E V' there exists a solution '1.£ to the equation Lu = f, i.e., to the problem (1.1). •
170
8 The Method of Finite Elements
Corollary 8.2.3. The requirement (1.12) with eN. ;::: l > 0 from Theorem 2 is satisfied with l:= CE i/ a(.,.) is V-elliptic: a(u, u) ;::: CEllull~.
Exercise 8.2.4. Show that (4a) implies that U:l Vi is dense in V. Let QN be the orthogonal projection onto VN (cf. Exercise 8.1.6c). The factors L:V - V', QN:V' - Vlv = VN, L1/:Vlv = VN - VN C V can be composed to give (8.2.5a)
Exercise 8.2.5. Show there is also the representation SN =PL-1p*L.
(8.2.5a')
Lemma 8.2.6. SN is the projection onto VN and is called the Ritz projection. It sends the solutionu o/the problem (1.1) to the Ritz-Galerkin solution uN E VN: uN = SNU. Assuming (1.2) and (1.12) we have IISNllv<-v ~ Cs/eN' A definition
0/ SN
SNU E VN
(8.2.5b)
equivalent to that in (5a) is
and
a(SNu,v)
= a(u,v) for all
v E VN, U E V.
(8.2.5c)
PROOF. (a) Since a(u,v)
= (Lu,v)v'xv = (Lu,v)u = (LU,QNV)U = (QNLu,v)u
for all v E VN, it follows that SNU in (5c) is the Ritz-Galerkin solution for the right-hand side / := QNLu E V', Le., SNU = L·;/QNLu. Conversely one may argue similarly, and so show the equivalence of the definitions (5a) and (5c). (b) (5c) shows that u E VN leads to SNU = u. Thus S~ = SN, i.e., SN is a projection. Inequality (5b) follows from IISNUllv ~ IIQNLullv..:JEN (cf. the proof of (1.11» and IIQNLullvN = IILullvN ~ IILullv ~ Csllullv with Cs from (1.2) (cf. Exercise 8.1.9). • Remark 8.2.1. Let a(·,·) be V-elliptic and symmetric. IIlvlllv := a(v, v)1/2 is a norm equivalent to II· IIv. The Ritz projection SN is, with respect to 11I·llIv, an orthogonal projection onto VN. Thus, in particular, we have (8.2.5d) PROOF. The scalar product associated to 1I1·llIv is a(·, .), so that it is to be shown that a(SNv, w) = a(v, SNW). (5c) implies a(SNv, SNW) = a(v, SNW), since S NW E VN. The symmetry of a(.,·) and exchanging v and W give
8.3 Finite Elements
171
a(SNV'w) = a(W,SNV) = a(SNW,SNV) = a(SNV,SNW), so that a(SNv,w) = a(v, SNW). (5d) results from Remark G.3.8. •
Remark 7 shows once again that the Ritz-Galerkin solution uN = SNU is the best approximation to u in VN in the sense of the norm 111·lllv. This assertion is equivalent to the variational formulation J(u N ) ~ J(v) for all v E VN (cf. Exercise 8.1.lOb). The condition (1.12), with En 2 E > 0, is difficult to prove, except for V-elliptic bilinear forms. However, the following theorem shows that this condition does hold for subspaces approximating well enough. Theorem 8.2.8. Let the bilinear form a(-, .) be V -coercive, where V cUe V'is a continuous, dense, and compact embedding. Let Problem (1.1) be solvable for all f E v'. Assume that (4a) holds for the subspaces V. C V. For larye enough i the stability condition (1.12) is then satisfied with EN. 2 E > o.
The proof of this will be postponed to a supplement to Lemma 11.2.7.
8.3 Finite Elements 8.3.1 Introduction: Linear Elements for
n = (a, b)
A13 soon as the dimension N = dim VN becomes larger the essential disadvantage of the general Ritz-Galerkin method becomes apparent. The matrix L is in general full, i.e., Lij i- 0 for all i,j = 1, ... , N. Therefore one needs N 2 integrations to obtain the values of Lij = a( bj , bi ) = I a ... , whether exactly or approximately. The final solution of the system of equations Lu = f requires O(N3) operations. A13 soon as N is no longer small the general Ritz-Galerkin method therefore turns out to be unusable. A glance at the difference method shows that the matrices Lh which occur there are sparse. Thus it is natural to wonder if it is possible to choose the basis {bi> ... , bN } so that the stiffness matrix Lij = a(bj , bi) is also sparse. The best situation would be that the bi were orthogonal with respect to a(·, .): a(bj,bi ) for i i-j. However, such a basis can be found only for special model problems such as the one in Example 8.1.11. Instead we shall base our further considerations on Remark 8.3.1. Let the bilinear form a(·,·) be given by (7.2.6). Let Bi be the interior of the support supp(b;.) of the basis function bi, i.e., Bi := supp(bi )\8supp(bi ). A sufficient condition that ensures Lij = a(bj , bi) = 0 is BinBj = 0.
PROOF. The integration a(bj,bi )
= Ia ... can be restricted to Bi n B j .
•
In order to be able to apply Remark 1 the basis functions should have as small supports as possible. In constructing them in general one goes about
172
8 The Method of Finite Elements
n
it from the desired goal: one defines partitions of into small pieces, the socalled finite elements, from which the supports of hi are pieced together. Ai; an introduction let us investigate the one-dimensional boundary value problem
-u"(X)
= g(x)
for a < x < b,
u(a)
= u(b) = O.
(8.3.1)
ABsume we have a partition of the interval [a, b] given by a = Xo < Xl < ... < XN+1 = h. Denote the interval pieces by Ii := (Xi-l,Xi) (1 ~ i ~ N + 1). For the subspace VN C HJ(a, b) let us choose the piecewise linear functions:
VN ={u E CO([a, b]): the restriction of u to Ii (1 ~ i ~ N is linear; u(a) = u(b) = O}
+ 1)
(8.3.2)
See the figure below.
X
Xs =b Figure 8.S.1. (a) a piecewise linear function, (b) a basis function
The continuity "u E CO([a, b])" is equivalent to continuity at the nodes Xi (1 ~ i ~ N) : U(Xi + 0) = U(Xi - 0).
Remark 8.3.2. (1 ~ i ~ N): u(x)
U
E VN is uniquely determined by its node values U(Xi)
= [U(Xi)(XHl -
x) + u(xHd(x - Xi)]/[XHl - Xi] for X E I i +1, (8.3.3a)
where u(xo) = U(XN+l) = O. From Theorem 6.2.42 we conclude that VN C HJ(a,b). The (weak) derivative u' E L2(a,b) is piecewise constant: (8.3.3b) The basis functions can be defined (see Figure 1b) by
bi(x)
=
(x - xi-d/(Xi - xi-d { (Xi+1 - X)/(Xi+1 - Xi)
o
for Xi-l < X ~ Xi for Xi < X < Xi+1, otherwise.
1~ i ~ N
(8.3.4)
8.3 Finite Elements
173
Remark 2 then has as a consequence
Remark 8.3.3. One has the representation u = ~!1 u(xi)bi . The supports of the functions bi are 1i - 1 u1i = [Xi-l,Xi+1]' The Bi from Remark 1 is now Bi = (Xi-1,XHt)·
The weak formulation of the boundary value problem (1) is a(u, v) = f(v) with a(u,v):=
lb
f(v):=
u'v' dx,
lb
gvdx
for u,v E HJ(a, b).
By Remark 1 we have, for the matrix elements, Lij 2, since then B. n Bj = 0. For Ii - jl ~ lone obtains L.,.-1 = a(bi _l, bi ) = Lii
= a(bi , bi ) = 1
al•
1al.-lal•
Xi - Xi-1 Xi - Xi-1
(Xi -
Xi_t}-2
-1
1
dx
+
Xi)-2 dx (8.3.5a)
:.;.
= 1/(Xi - Xi-1)
=
dx = -1/(xi - xi-d,
1al.+l (XH1 -
2:._1
L i ,H1
= 0 as soon as Ii - j I ~
+ 1/(XH1 -
Xi),
-1/(XH1 - Xi).
The right-hand side f
1al
= (ft, ... , fN)T is given by (5b):
.+ 1
fi = f(b.) =
gbi dx
%'-1
=
1 Xi - Xi-1
1 alO g(x)(x al.-l
xi-d dx +
1 al.+
1 Xi+l - Xi ""
1
g(X)(XH1 - x) dx.
(8.3.5b)
Remark 8.3.4. The system of equations Lu = f, given in (5a,b), is tridiagonal, thus, in particular, sparse. For an equidistant partitioning Xi := a + ih with h (b - a)/(N + 1) the coefficients are Li,i±1 = -1/h, Li,i = 2/h, fi = h fo [g(Xi + th) + g(Xi - th)](1 - t) dt. If 9 E COCa, b), then by the intermediate value theorem fi = hg(Xi + eih), with leil < 1. Therefore the system of equations Lu = f is, up to a scaling, identical with the difference equation LhUh = !h in Section 4.1, if one defines fh by fh(Xi) := Ii = g(Xi + eih) instead of by fh(Xi) = g(Xi): then L = hLh , f = hfh.
::=
This example shows that it is possible to find basis functions with small supports so that L is relatively easy to calculate. In addition, the similarity is apparent between the resulting discretisation method and difference methods. Finally consider the numerical evaluation of the integral (5b).
Exercise 8.3.5. Show that the Gaussian quadrature formula for with bi(x) as weight function and one support point is:
f:::
1 1
gbi dx
174
8 The Method of Finite Elements
. '= Xi+l f,•. 2
Xi-l
9
(Xi-l
+ X.3 + XH1) .
(8.3.5c)
What is the formula for a partition of equal intervals? If one replaces the Dirichlet boundary conditions in (1) by the Neumann condition then V = Hl(a,b). The subspace VN C Hl(a,b) results from (2) after removal of the condition u(a) = u(b) = O. In order that dim VN = N the numbering has to be changed: The partition of (a,b) is given by a = Xl < X2
< ... <XN = b.
Exercise 8.3.6. Let 8u(a)j8n = go,8u(b)j8n = gb, with a(·,·) as before. Suppose the partition of (a, b) is equidistant: Xi = a + (b - a)(i - 1)/(N - 1). What is the form of the equation Lu = f7 Show that Lu = f has at least one solution if g(x) dx + go + gb = O.
J:
8.3.2 Linear Elements for
n c m?
We shall assume:
n c m,2 is a polygon. AB shown in Fig. 2 we divide n into triangles.
(8.3.6)
Figure 8.8.2. Triangulations of il, t:= number of triangles, N:= number of inner nodes (number of all nodes)
Definition 8.3.7. T:= {Tt, ... , Tt} is called an admissible triangulation of
n if the following conditions are fulfilled:
Ii (1 $ i $ t) are open triangles ("finite elements") Ii are disjoint, i.e., Ii n Tj = 0 for i i j,
U Ti=n, for i
i
j the set Ti
(8.3.7a) (8.3.7b) (8.3.7c)
n Tj
is either
8.3 Finite Elements
i)
empty, or
ii)
a common side of the elements 11 and Tj, or a common edge of the elements 11 and Tj.
iii)
175
(8.3.7d)
Remark 8.3.8. (a) The conditions (7a) and (7c) imply the polygonal shape of il, Le., the assumption (6). (b) Figure 2 shows only admissible triangulations. An example of an inadmissible triangulation would be, e.g., a square partitioned as follows: ~ Let T be an admissible triangulation. The point x is called a node (of T) if x is a vertex of one of the Ti E T. One distinguishes interior and boundary nodes, according to whether x E il or x E ail. Let N be the total number of interior nodes. We define VN as the subspace of the piecewise linear functions: VN := {u E CO(il):u = 0 on ail; on each Ti E
with a linear function, Le., u(x, y)
T
the function u agrees
= ail + ai2x + aiSY on Ti}.
(8.3.8) Remark 2 can be applied to the two-dimensional case under consideration:
Remark 8.3.9. (a) It is true that VN C HJ(il). (b) Each function is uniquely determined by its node values u(Xi) at the interior nodes Xi, 1 ~ i ~ N. PROOF. (a) Example 6.2.5 shows that VN C Hl(il). Since u = 0 on ail, Theorem 6.2.42 proves that VN C HJ(il). (b) Let x, x', x" be the three vertices of 11 E T. The linear function u( x, y) = ail +Ui2 X+Ui3Y is uniquely determined on 11 by the values u(x), U(X/), u(x") .
•
The converse to Remark 9b is as follows:
Remark 8.3.10. Let Xi (1 ~ i ~ N) be the interior nodes of T. For an arbitrary Ui (1 ~ i ~ N) there exists exactly one u E VN with u(Xi) = Ui. It may be written as u = E!l Uibi, where the basis functions bi are characterised by (8.3.9a) If T E T is a triangle with the vertices x, = (X/,y'), x" = (X",y"), then
xi
= (Xi, Yi) [Xi as in (9a)] and
(x - X')(yll - y/) - (y - y')(X" - x') b,(x,y) = ( Xi - X, )( y" - y' ) - ( Yi - yl )( X" - X, )
On all T E
T
that do not have an
Xi
onTo
(8.3.9b)
as a vertex, bi = O.
PROOF. Obviously, using (9b) we get a linear function defined on all T E T, which satisfies (9a). In addition, (9a) forces continuity at all nodes. If the
176
8 The Method of Finite Elements
vertices x j = (Xj, Yj) and x1c = (XIe, YIe) are directly connected by the side of a triangle, then (9b) provides the representation b,(xj + s(x le -xj» = Sbi(XIe) + (1 - s)bi(xj ) with s E [0,1] for both triangles that have this side in common. Thus bi is also continuous on the common edges, so that bi E CO(D). Applying this consideration to two boundary nodes Xl, x" with b,(x/ ) = bi(x") = 0 we derive bi = 0 on an. Thus bi belongs to VN. • Remark 8.3.11. (a) The dimension of the subspace determined by (8) is the number of interior nodes of T. (b) The support of the basis function bi is supp(bi )
= U{T: T
E
T
has Xi as a comer}.
(c) Let Bi be the interior of supp(bi ). There holds Bi n B j the nodes Xi and x j are directly connected by a side.
In
= 0 if and only if
= (Vu, Vv) dx is the bilinear form associated to the Poisson equation. The integmls Lij = a(bj , bi) = Lie (Vbj , Vb,) dx are to be taken over the following Tie: (i) all Tic with Xi as a vertex, if i = j; (ii) all with Xi and xj as vertices, if i #- j. (iii) We have Lij = 0 ifxi and x j are not directly connected by the side of a triangle.
Corollary 8.3.12. a(u, v)
IT"
n
ht"'o-",tJ/-J/O h
Figure 8.S.S. Basis functions for the node (xo, Yo)
We get an especially regular triangulation when we first divide into squares with sides of length h, and then divide these into two triangles (121). The first and second triangulations in Figure 2 are of this sort. We call them "square grid triangulations". The corresponding basis function is depicted in Figure 3. One therefore expects that the matrix L corresponds to a 7-point formula. For the Laplace operator, however, one finds the well-known 5-point formula (4.2.11) from Section 4.2:
8.3 Finite Elements
177
Exercise 8.3.13. Let T be a square grid triangulation. Further let a(u, v) = fa ('V'u, V'v} dx.. The basis functions are described by Figure 3. Show that one has for the entries of the stiffness matrix L Lii = 4,
Lij = -1 if
Xi -
xj
= (0, ±h) or (±h,O), Lij =
° otherwise;
i.e., L agrees with h2Lh from (4.2.8). Although L = h2 Lh holds in the case of the Poisson equation -Llu = g, the finite-element discretisation and the difference method still do not coincide, since h 2 ih has h 2 g(xi ) as components and so differs from Ii = fa gbi dx..
'fJ (0,1)
~~
Y
X'
<[) x2
E (0,0)
x'
x
(1,0)
Figure 8.3.4. Reference triangle T
The integration fT, ... dx over the triangle Ii E T seems at first difficult. However, for each i one can express fT, ... dx as an integral over the reference triangle T in Figure 4. The details are in
Exercise 8.3.14. Let Xi = (Xi,yi) (i = 1,2,3) be the vertices ofT E T, and let T be the unit triangle in Figure 4. Show: (a)
£
v(x,y) dxdy = l(x 2 _x l )(y3 _yl) - (y2 _yl)(x3 _xl)1
£
v(cf>(e,'fJ)) de d'fJ. (8.3.10)
In general one evaluates the integral fT . .. de d'fJ over the unit triangle numerically. Examples of integration formulae can be found in Schwarz [1, §2.4.3] and Ciarlet [1, §4.1]. The necessary ordering of the quadrature formulae is discussed in the same place by Ciarlet [14] and by Witsch [1].
178
8 The Method of Finite Elements
In contrast to the difference methods finite-element discretisation offers one the possibility of changing the size of the triangles locally. The third triangulation in Figure 2 contains triangles that become smaller as they are nearer to the intruding comer. This flexibility of the finite-element method is an essential advantage. On the other hand, one does obtain systems of equations Lu = f with more complex structures, since (a) u can no longer be stored in a two-dimensional array, (b) L cannot be characterised by a star as in (4.2.12). Remark 8.3.15. If one replaces the Dirichlet condition 1.£ = 0 on an by natural boundary conditions, then the following changes take place: Ca) N = dim VN is the number of all nodes (inner and boundary nodes). (b) VN C Hi(n) is given by (S) without the restriction "1.£ = 0 on an". (c) In Remarks 9-11 it should read ''nodes'' instead of "inner nodes" . (d) The matrix elements Lij calculated in Exercise 13 are valid only for inner nodes Xi. For a Neumann boundary value problem in the square n = (0, 1) x (0,1), L coincides with h2 DhLh from Exercise 4.7.Sb. (e) In calculating fi = f(bi) one has to take account of possible boundary integrals over an n supp(b,), if an n supp(b,) =f 0.
8.3.3 Bilinear Elements for
n em?
The difference procedures in a square grid are near to partitioning n into squares of side h (cf. Figure 5a).If, more generally, one replaces the squares by parallelograms one obtains partitions like those in Figure 5a,b. An admissible partition by parallelograms is described by the conditions (7a-d), if in (7a) the expression "triangle" is replaced by ''parallelogram''.
(a)
(b)
Figure 8.3.5. Partition of a into parallelograms
If one were to define the subspace VN by the condition that 1.£ must be a linear function in each parallelogram, then there would be only three of the four vertex values that could be arbitrarily assigned. In the case of the partition in Figure 5b one can see that the only piecewise linear function 1.£ with 1.£ = 0 on an is the null function. Thus in each parallelogram 1.£ must
8.3 Finite Elements
179
be a function that involves four free parameters. We next consider the case of a rectangle parallel to the axes P = (Xl,X2) x (Yl,Y4) and define a bilinear function on P by (8.3.11a) Then u is linear in each direction parallel to the axes - thus, in particular, along the sides of the rectangle. For an arbitrary parallelogram P such as in Figure 6a, the restriction of the function (l1a) to the side of a parallelogram is in general a quadratic function. Therefore one generalises the definition as follows. Let (8.3.11b) be the mapping taking the unit square (0,1) x (0,1) onto the parallelogram P (cf. Figure 6a, b). A bilinear function is defined on P by u(X,Y):= v(4)-l(x,y)),
v(e, "') := (a + f3eKy + 0",).
(8.3.11c)
It is not necessary to calculate v(4)-l(x, y)) explicitly, since all the integrations can be carried out over (0, l)x (0,1) as the reference parallelogram (cf. Exercise 14). Y
x3
'" (1,1)
(0,1)
~ x2~
P
x
(0,0)
(1,0)
Figure 8.S.6. (a) Parallelogram (b) Unit square as reference parallelogram
If 7r = {P!, ... , Pt } is an admissible partition into parallelograms one defines VN C Hl(D) [resp. VN C HJ(D)) by
resp.
VN := {u E C0(1): on all P E 7r, u coincides with a bilinear function} (8.3.12a) VN := {u E CO{n): u = 0 on an; on all P E 7r, u coincides with a bilinear function} (8.3.12b)
Here N = dim VN is the number of nodes [resp. inner nodes).
180
8 The Method of Finite Elements
A bilinear function on P E 7r is linear along each side of P. Continuity in the node points thus already implies continuity in n. Remark 8.3.16. changes.
The Remarks 880, 9, 10, and 11 hold with appropriate
Exercise 8.3.17. Let the bilinear form associated to -..d be a(u,v) = fn{Vu, Vv) dx. Assume n = (0,1) x (0,1) is divided into squares of side h as in Figure 580. What are the basis functions Charac[t':~ed_bi (~a{]? Show that the matrix L coincides with the difference star
k
-1 -1
8 -1
-1 . -1
Figure 8.S.7. A combination of triangles and parallelograms
Remark 8.3.18. Triangle and parallelogram divisions can also be combined. A polygonal domain n can be divided up into both triangles and parallelograms (cf. Figure 7). In this case VN is defined as {u E CO(il):u linear on the triangles, bilinear on the parallelograms}.
8.3.4 Quadratic Elements for
n c m?
Let T be an admissible triangulation of a polygonal domain n. We wish to increase the dimension of the finite-element subspace by allowing, instead of linear functions, quadratics
so that VN
= {u E CO(il):u = 0 on an;
on each Ti E T, u coincides (8.3.14) with a quadratic function}.
8.3 Finite Elements
181
Figure 8.3.8. Nodes for (a) a quadratic ansatz on a triangle, (b) a quadratic ansatz of the serendipity class on a parallelogram
Lemma 8.3.19. (a) Let xl,x2,r be the vert~ces of a triangle T E T, while X 4 ,X5 ,x6 are the midpoints of the sides (cf Figure 8a). Each function quadrntic on t is determined by the values {u(xj):j = 1, ... ,6}. (b) The restriction of the function (13) to a side of T E T gives a onedimensional quadrntic function, which is uniquely determined by three of the nodes lying on this side (e.g., u(x 1),u(x4),u(x2) in Figure 8a). (c) If u is quadrntic on each Ii E T and continuous at all nodes (i. e., vertices of triangles and mid-points of sides), then u is continuous on
n.
PROOF. (a) The quadratic function can be obtained as the uniquely defined interpolating polynomial of the form (13). Part (b) is also elementary. (c) Let T and T be neighbouring triangles in Figure 8a. Since UIT and ul r coincide on xl, x\x2, by part (b) they represent the same quadratic function • on the common side of T and T. We call all [inner] vertices of triangles and mid-points of sides [inner] nodes. By Lemma 18a we can find for each inner node Xi a basis function bi which is quadratic on each T E T and satisfies bi(Xi ) = 1, bi(xj ) = 0 (j -# i). By Lemma 18c bi belongs to V N • This proves Remark 8.3.20. The number of inner nodes is the dimension of VN in 14. Each u E VN admits the expression u = I:!l u(xi)bi , where the basis function belonging to the node is characterised by (ga). In the sequel we wish to assume that, as in Figure 7, both triangles and parallelograms are used in the partition. On the triangles the function u E VN are quadratic. The functions on the parallelograms P must satisfy the following conditions: (a) u(x, y)lp must be uniquely determined by the values at the 4 vertices and 4 mid-points of the sides (cf. Figure 8b). (f3) The restriction to a side of P gives a (one-dimenSional) quadratic function (of the arc length). By condition (a) the ansatz must contain exactly 8 coefficients. The quadratic (13) has only 6 coefficients, while the biquadratic I:osi,j9 atjf:rf
182
8 The Method of Finite Elements
has one parameter too many. If one omits the term ~2'f/2 in the biquadratic ansatz one obtains for the unit square the function
the so-called quadratic ansatz of the serendipity class. The restric= 0,1 [resp. 'f/ = 0,1] give a quadratic function in tions to the sides 'f/ [resp. eJ. The mapping t/J in (llb) (cf. Figure 6a,b) yields the function u(x,y) = v(t/J-l(x,y)) defined on the parallelogram P, which still satisfies the conditions (a) and (fi). If one were to use the full biquadratic ansatz one would one additional node which one could choose as the barycentre of the parallelogram. Cubic ansatzes can be carried out in the same way in triangles and parallelograms (cf. Schwarz [1]).
e
8.3.5 Elements for
n c m.3
In the three-dimensional case assume
n c m.3
is a polyhedron.
(8.3.16)
The triangulation from Section 8.3.2 corresponds now to a division of n into tetrahedra. It is called admissible, if (7a-d) hold in the appropriate sense: (7a) becomes "Ts (1 ~ i ~ t) are open tetrahedra"; in (7d) it now should read "For i -:f j, Ti nTj is either empty, or a common vertex, side, or face of Ts and T/,. Each linear function al + a2x + asy + a4Z is uniquely determined by its values at the 4 vertices of the tetrahedron. As a basis for the space
VN = {u
E
cO(li): u
= 0 on
an;
u is linear on each tetrahedron Ti (1
~ i ~
t)}
one chooses bi with the property (9a): bi(Xi ) = 1, Mxj) = 0 (j -:f i). The support of bi consists in all tetrahedra that share Xi as a vertex. The dimension N = dim VN is again the number of inner nodes (Le., vertices of tetrahedra). As in the two-dimensional case the linear ansatz may be replaced by a quadratic one. Instead of a tetrahedron one can use a parallelipiped or a triangular prism with corresponding ansatzes for the functions (cf. Schwarz
[1)). 8.3.6 Handling of Side Conditions
The space V which lies at the foundation of the whole matter may be a subspace of a simply discretisable space W ::> V. A given function w E W belongs to V if certain side conditions are satisfied. Before we describe this situation in full generality, two examples will be provided as illustrations.
8.3 Finite Elements
183
Example 8.3.21. If one wishes to make the Neumann boundary-value problem -Ll.u = 9 in n and Bu/ an = cp on r uniquely solvable by the addition of the side condition In u(x) dx = 0, one can choose the space V to be
V
= {u E Hl(n):
L
u(x) dx
= o}.
(8.3.17a)
For a bounded domain n the form a(u, v) := In (,V'u, Vv} dx is V-elliptic. The weak formulation (1.1) with I(v) := Ingvdx+ Ir
Example 8.3.22. The Adler problem of Exercise 7.4.15 uses V
= {u E Hl(n): u
constant on r}.
(8.3. 17b)
In both cases W = Hl(n) is a proper superset of V. Let T be a triangulation of n with lin inner nodes and hd boundary nodes. Let W h C W be the space of linear triangular elements (cf. Remark 15): (8.3.18a) The basis functions {bi : 1 :5 i :5 Nh} C Wh will be described as usual by bi(xj ) = oij (cf. (9a». As the finite-element subspace of V we define Vh := W h n V, which is of smaller dimension: (8.3.18b) In the Examples 21 and 22 we have Mh = 1, resp. Mh = lbd difficulty in the numerical solution of the discrete problem (19), Find uhEVh with a(uh,v)=/(v) for all VEVh ,
-
1. The (8.3.19)
begins with the choice of a basis for Vh' It is not possible to use a subset of the functions bb ... , bNh' Thus, for example, in Example 21 no bi belongs to V, and thus also none to Vh, since In bi dx > o. In principle, it is possible in the case of (17a) to construct as new basis functions linear combinations bi := abi1 + f3b i2 E Vh which have again localised (but a bit larger) supports. In the case of (17b) it is still relatively simple to find a practical basis for Vh : The basis functions b, E W h which belong to inner nodes are also in Vh, since bi E HJ(n) c V. As further elements one uses bo(x) := 2::' bj(x), where 2::' denotes the sum over all boundary nodes. In spite of this, even in this case, it would simply be easier if one could work with the standard basis of Who In order to treat the problem (19) with the aid of bi E W h , we reintroduce the notation w h = Pw (w E IRNh a coefficient vector, w h E Wh, cf. (1.6».
184
8 The Method of Finite Elements
Each v E Vh C W h can be written as Pv with v in V := {v E rn.Nh:PV E = P-1Vh . Thus problem (19) is equivalent to:
Vh}
Find u E V with a(Pu, Pv) = f(Pv) for all v E V.
(8.3.19')
From (18a,b) we have dimV = dimVh = Nh- Mh = dimWh - Mh. The spaces Vh and V are described by Mh linear conditions
Nh
2:co;w; = 0 ;=1
(8.3.20)
V= Ker C= {WErn.Nh:Cw=O}, where C
= (co;) is an Mh
(8.3.21)
x Nh matrix. In the case of Example 21 we have Clj
=
fa
bj(x) dx
For Example 22 let xo, ... ,XMh , with Mh = 1bd -1, be the boundary nodes. Then C can be defined as follows:
Mh
= hd-1.
cli
= 1 (1 ~ i
= -1,
CO;
= 0 otherwise.
CMh,O
~
Mh),
The variation of v over V in (19') can be replaced by the variation of w over rn. Nh , if one couples the conditions (20) by using Lagrange multipliers -\ (1 ~ i ~ Mh), which can be put together to form a vector A = (Ai, ... , AMh)' The resulting formulation of the problem is: find u E rn.Nh and A E rn.Mh with Cu
=0
and N a(Pu,Pw) + (A,CW) = f(Pw) for all WE rn. \
where (A, 1')
= I: Ail'i
is the scalar product in
(8.3.22a) (8.3.22b)
rn.Mh .
Theorem 8.3.23. The problems (19) and (22a,b) are equivalent in the following sense. 1fu, A is a solution pair for (22a,b) then u is a solution of (19') and Pu is a solution of (19). Conversely, if u h = Pu is a solution of (19) then there exists precisely one A E rn.Mh such that u and A solve (22a,b). Before we prove this theorem, we give a matrix formulation which is equivalent to (22a,b) Remark 8.3.24. Let Land f be defined as in (1.8a,b). Further, suppose that B := CT with C from (21). Then (22a,b) is equivalent to the system of equations (8.3.22')
8.4 Error Estimates for Finite Element Methods
185
PROOF. Lu + BA = f is equivalent to aWu, Pw) + (A, Cw) = (Lu, w) + (BA, w) = (f, w) = f(Pw) for all wEIR ". Note BTu = 0 is the same as Cu=O. • (of Theorem 23). (a) Assume a solution of (22a,b) is given in terms of u, A. Then Cu = 0 implies u E V and uk := Pu E Vh. Equation (22b) holds in particular for all w E V, so that (19') follows since Cw = O. (b) (19') implies f - Lu E Vol. Since Vol = (KerC)ol = (KerBT)ol = Im(B), there exists a A E IRM " with f - Lu = BA, so that (22'), and thus (22a,b) are satisfied. For the Nh x Mh matrix B with rank Mh we have Ker B = {O}, so that the solution is unique. • PROOF
Note that in the case of Example 21 the system of equations (22') is essentially identical to (4.7.lOa,b).
8.4 Error Estimates for Finite Element Methods 8.4.1 Hl-Estimates for Linear Elements In this section we shall restrict ourselves to the consideration of the linear elements from Section 8.3.2 and therefore assume: is an admissible triangulation of n c IR2, VN is defined by (3.8), if V = HJ(n), VN is as in Remark 8.3.15b, if V = Hl(n).
(8.4.1a) (8.4.1b) (8.4.1c)
T
By Theorem 8.2.1 one has to determine d(u, VN ) = inf{lu-w\1:w E VN}. To do this one begins by looking at the reference triangle from Figure 8.3.4.
Lemma 8.4.1. LetT = {(e, 1]): e,1] ~ O,e +1] ~ I}. For all u E H2(T) there holds
lI u Il12(T)
~ C[ lu(O, 0)1 2 + lu(O, 1)1 2 + lu(I, 0)1 2 +
L
IIDauIl12(T»)' (8.4.2)
lal=2
PROOF. (a) First it has to be shown that the bilinear form a(u, v) := u(O,O)v(O, 0) +u(O, l)v(O, 1) +u(I,O)v(I,O) +
L
(Dau, D aV)L2(T)
lal=2
is continuous on H2(T) x H2(T) and is H2(T)-coercive. The continuity follows from the continuous embedding H2(T) C CO(T), which implies lu(x)1 ~ Clul2 for all x E T (cf. Theorem 6.2.30). Thus we have la(u,v)1 ~ (1 + 3C)luI2IvI2'
186
8 The Method of Finite Elements
Since T E CO. 1, one may apply Theorem 6.4.8b: JP(T) is compactly embedded in H1(T). By Lemma 6.4.13, there exists a constant C 1/ 2 such that
lul~ ~ (4 1'1.£12 + C1/2Iulo) 2 ~ ~Iul~ + 2C~/2Iul~ for all '1.£ E H2(T). Since Elal=2IDaul~ = lul~ -Iul¥, the estimate a(u,u)
~ lul~ -Iul~ ~ ~Iul~ - 2C~/2IuI5,
shows that a(·, .) is H2 (T)-coercive. (b) Since the embedding H2(T) C L2(T) is also compact one may apply Theorem 6.5.15. The operator A E L(H2(T), JP(T)') which is associated with a(·,.) either has an inverse A-1 E L(H2(T)', H2(T» or has A = 0 as an eigenvalue with an eigenfunction 0 =I- e E H2(T). In the latter case e must, in particular, satisfy the equation a(e, e) = O. From this follows that DOle = for all lal = 2, so that e must be linear: e(x, y) = a + f3x + "'(y. Furthermore, from a(e, e) = one may conclude that e(O,O) = e(l,O) = e(O, 1) = 0, so that e = in contradiction to what was just assumed. From Lemma 6.5.3 and Exercise 6.5.6c one may show the H;,2(T)-ellipticity: a(u, '1.£) ~ CElul~ with CE := 1/[ IIA- 1IIk2(T)<-H2(T)' (1 + 30) 1 > and C from part (a). Thus we • have assertion (2) with C:= l/CE •
°
° °
°
Lemma 8.4.2. Let Th := hT := {(x,y): x,y ~ 0, x '1.£ E H2(Th) there holds with 1f31 ~ 2
+y
~
h}. For each
IID13ullh(Th) ~ C{h2-2113I[u2(0, 0) + '1.£2(0, h) + u2(h, 0)1
+ h4-21131
L
(8.4.3)
II DOIU Il12(Th)}'
1011=2 with C independent of '1.£, f3, and h.
PROOF. Let v(~, 11) := u(~h, 11h) E H2(T). The derivatives with respect to (x,y) and to (~,11) = (x,y)/h are related by D~.II = h- I13I D:.". For each 1f31 ~ 2 Lemma 1 gives II D13u Il12(Th) =
JLID~.lIuI2
dxdy =
J£
Ih-I13ID:'"vI2h2
~ d11
= h2-2113II1D13vIl12(T) ~ h2-2113ll1vllk2(T)
~ Ch2-2113I[V2(0, 0) + v 2(1,0)
+ v2(0, 1) +
L
IIDOIvI112(T)j·
1011=2 Since v(O, 0) = '1.£(0,0), v(l, 0) = u(h, 0), v(O, 1) = '1.£(0, h), and IIDl"vIl12(T) = h2I1D~.flUIl12(Th)' one may conclude assertion (3). •
8.4 Error Estimates for Finite Element Methods
187
Lemma 8.4.3. Let T be an arbitrary triangle with Side lengths $ h max ,
Interior angles
~
ao
> O.
(8.4.4)
For each u E H2(T) and 1,81 $ 2, there holds
IID.6 u lli2(T) $ C(ao)
[h~~I.61 L
lu(x)1 2 +
x vel1ex of T
h!t~I.61 lal=2 L IIDaUlli2(T)] , (8.4.5)
where C(ao) depends only on ao and not on u, ,8, or h.
PROOF. Let h $ h max be one of the lengths of the sides of T. In a way similar to that in Exercise 8.3.14a, let cp: Th -+ T be the linear transformation that maps Th onto T. Then V(€,71) := U(CP(€,71)) belongs to H2(Th). Under the conditon (4) we know that the determinant Idetcp'l E [l/K(ao),K(ao)] is bounded both above and below. From .6 2 "" .....8' 2 II D a:,yulI£2(T) $ C1(ao) L.J IIq,f/ v IIL2(Th)' 1.6'1=1.61 (3), and
L
L
II Dl,f/vII12(Th) $ C2 (ao) IID~,yulli2(T)' lal=2 lal=2 there then follows (5). From Lemma 3 follows the important result:
•
Theorem 8.4.4. Assume that conditions (la-c) hold for r, VN, and V. Let ao be the smallest interior angle of all Ti E r, while h is the maximum length of the sides of all 11 E r. Then inf lIu-vIlHlc(O) $ C'(ao)h2- lc lluIlH2(0) for k = 0, 1 and all u E H2(il)n V. lIEVN
(8.4.6) PROOF. For a u E H2(il) [resp. u E H2(il) n HJ(il), if V = HJ(il)] one chooses v := l:i u(xi)bi E VN, i.e., v E VN with v(Xi) = U(Xi) at the [inner] nodes Xi. Since w := u - v vanishes at the vertices of each 11 E r and Da w = nau for lal = 2 (because Dav = 0 for all v E VN), inequality (5) implies the estimate
IID.6 w Il12(0)
=
L
IID.B w Il1 2 (T,)
L [C(ao)h4-21J31 L II Dau Il12(T,)1 lal=2 = h4-21J3IC(ao) L II Dau Il1 $
T,Er
2 (Q)
lal=2 $ h4-21.6IC(ao)lIull~2(0).
188
8 The Method of Finite Elements
Summation over 1.81 :::;; k now proves the inequality (6).
•
Let hT be the longest side of T E T, then h := max{hT:T E T} is the longest side-length which occurs in T. The parameter h is the essential one and will be used as an index in the sequel: T = Th. Denote by PT the radius of the inscribed circle in T E T. The ratio hTI PT tends to infinity exactly when the smallest interior angle tends to zero. If one has the bound max {~I PT: T E T} :::;; const for a family of triangulations, then one calls the family quasi-uniform. In this condition the constant should be the same for all T in the family {Th}. A family {Th} is called uniform if it fulfils the stronger requirement hi min{PT: T E T} :::;; const. A family {Th} which is both unform and for which h -+ 0 is called regular by Ciarlet [1, §3.1]. In the construction of triangulations it should be noted that with increasing refinement (h" -+ 0 ) the interior angle may not also be reduced. Strategies for the systematic construction of quasi-uniform sequences Vh" can be found, for example, in Ha.ckbusch [1, §3.8.2]. The case k = 1 of Theorem 4 may now be written as follows:
Theorem 8.4.5. Assume conditions (1a-c) hold for a sequence of quasiuniform triangulations T". Then there exists a constant C, such that for all h = h" and Vh = Vh" there holds the following estimate: inf 1J€V"
lu - vb : :; Chlul2
for all
u E H2(n) n V.
(8.4.6')
The combination of this theorem with Theorem 8.2.1 gives
Theorem 8.4.6. Assume conditions (1a-c) hold for a sequence of quasiuniform triangulations T". Let the bilinear form fulfil conditions (1.2) and (1.12). Suppose the constants fN := fh" > 0 from (1.12) are bounded below by fh" ~ E > 0 (cf. Corollary 8.2.3). Let Problem (1.1) have the solution u E H2(n) n V. Let u h E Vh be the finite-element solution. Then there exists a constant C, which does not depend on u, h = h" or 11, such that (8.4.7) The combination of the Theorems 5 and 8.2.2 can be written:
Theorem 8.4.7. Assume conditions (la-c) hold for a sequence of quasiuniform triangulations T" with h" -+ O. Let the bilinear form fulfil (1.2) and (1.12) with fh" ~ l > O. Then the problem (1.1) has a unique solution u E V, and the finite-element solution u h " E Vh " converges to u: lu-uh "ll-+0
(11-+00).
(8.4.8)
PROOF. Let u E V and f > 0 be arbitrary. Since H2(n) n V is dense in V, there exists U e E H2(n) n V with lu - ueb :::;; f/2. From (6) and h" -+ 0 we
8.4 Error Estimates for Finite Element Methods
189
see there are v and VE e Vh " with IUE - VEil :$ f./2; thus lu - vEil :$ f.. This proves (2.4a). • Theorem 7 proves convergence without restrictive conditions on u. On the other hand the order of convergence O( h) in the estimate (7) requires the assumption that the solution u e V lies in H2(n). AB will later become clear, this assumption is hardly to be taken as fulfilled in every situation. A weaker assumption is u e V n HB(n) with s e (1,2). The corresponding result is as follows.
Theorem 8.4.8. Assume that in the assumptions of Theorem 6 u e V n H2(Q) is replaced by u e V n HB(Q) with 1 :$ s :$ 2. Let n be sufficiently smooth. Then there holds (8.4.9) The proof uses a generalisation of Theorem 4:
Lemma 8.4.9. Under the assumptions of Theorem 4 and suitable conditions on n, there holds inf lu - vh :$ C"(ao)h B- l lul s for s e [1,2], u e HS(n) n V. (8.4.10) VEVh
PROOF. The proof is based on an interpolation argument that will not be further explained here. Equation (10) holds for s = 2 (cf. (7» and for s = 1, since influ - vI! :$ lu - Olt = lull' From this follows (10) for s e (1,2) with the norm I· Is of the interpolating space [Hl(n) n V, H2(n) n V]s-l (cf. Lions-Magenes [1]), which under suitable assumptions on n coincides with HS(n)nV. • For s = 1 the right-hand side of (10) becomes const ·Iult. In fact the estimate inf{lu-vll:v e Vh } :$ lull is the best possible. To prove this ,choose u .1 Vh (orthogonal with respect to I·h). On the other side s = 2 is the maximal value for which the estimates (9) and (10) can hold. Even u e c=(n) permits no better order of approximation than O(h)! The Ritz projections SN : V ~ VN introduced in (2.5a) will now be written Sh : V ~ Vh. The inequality (9) becomes lu - Shull/luis :$ Ch s - l , and this proves
Corollary 8.4.10. Assume conditions (la-c) forT, Vh and V. Let the bilinear form fulfil (1.2) and (1.12). The Ritz projection Sh satisfies the estimate
III -
Sh IIHI (J2)+-H2(J2)nv :$ Chl.
(8.4.11)
Under the asumptions of Lemma 9 there holds
III -
ShIlHl(.a)+-HS(J2)nV :$ Ch s-
1
for all 1 :$ s :$ 2.
(8.4.11')
190
8 The Method of Finite Elements
8.4.2 L2 and H8 Estimates for Linear Elements
According to Theorem 6 O(h), is the optimal order of convergence. This result seems to contradict the O(h2) convergence of the five-point formula (cf. Section 4.5), since the finite-element method with a particular triangulation is almost identical to the five-point formula (cf. Exercise 8.3.13). The reason for this is that the error u - u h is measured in the I . It norm. The estimate inf{lu -vlo: v E Vh} ::;; Gh21ul2 in Theorem 4 suggests the conjecture that the I ·10 norm of the error is of the order O(h2): lu - uhlo ::;; Gh21u12. However, this statement is false without further assumptions. Up to this point only the existence of a weak solution u E V (e.g., V = HJ(.o) or V = H1(.o» has been guaranteed. But in Theorem 6 we needed the stronger assumption u E H2(.o) n V. A similar regularity condition will also be imposed on the adjoint problem to (1.1) Find u E V with a*(u, v)
= /(v)
for all v E V,
(8.4.12)
which uses the adjoint bilinear form a·(u, v) := a(v, u). For / E L2(il) C V', the value /(v) becomes (f,v)L2(O). The regularity condition is: For each / E L 2 (.o) the problem (12)
(8.4.13)
has a solution u E H2(il) n V with lul2 ::;; GRillo.
In Section 9 we shall see that this statement holds for sufficiently smooth domains .o. Theorem 8.4.11. (Aubin-Nitsche) Assume (13), (1.2), and (1.12) withf'.N = f'.h ~ E > 0, and (8.4.14) Let problem (1.1) have the solution u E V. Let u h E Vh C V be the finiteelement solution. Then, with a constant G1 independent of u and h, (8.4.15a)
If the solution u also belongs to H2(il) independent of u and h, such that
n V,
then there is a constant G2, (8.4.I5b)
Prom Theorem 4, it is sufficent to ensure (14) that Vh be the space of finite elements of an admissible and quasi-uniform triangulation. (cf. Nitsche [1]). For each e := u - u h as the solution of (12) for / = e:
PROOF
a(v,w)
= (e,vk,(o)
E
L2(.o) define w
for all v E V.
E
H2(il) n V (8.4.16a)
8.4 Error Estimates for Finite Element Methods
191
Corresponding to w there is, by (14), a w h E Vh with
Iw - whit ~ Cohlwl2 ~ CoCRhlelo. Equation (2.3) is a(e,v)
(8.4. 16b)
= 0 for all v E Vh; therefore, in particular, we have (8.4.16c)
From (16a-c) we obtain lel~
= (e, e)L2(S1) = a(e, w) = a(e, w - w h ) ~ Csleltlw - whh ~ CslehCoCRhlelo,
and thus (8.4.16d) From (2.1) one deduces
leh = lu -
uhlv ~ (1 + CS/€h) inf
vEV"
lu -
vlv ~ (1 + CS/f"}lulv
so that (15a) follows with C 1 := CsCoCR(l + Cs/'i). If u E H2(il), one may use the inequality (7), leh ~ Chluh, and deduce (15b) with C 2 := CSCOCRC .
•
Corollary 8.4.12. The inequalities (15a) and (15b) are equivalent to the properties (17a) and (17b) of the Ritz projection Sh:
III III -
Shll£2(S1)<-V ~ C1 h, ShIlL2(S1)<-H2(S1)nV ~ C2h2.
(8.4.I7a) (8.4. 17b)
The estimates (17a) and (17b) may also be proved directly. The definition of Eh and the connection between Sh and Sh may be found in Exercise 8.4.13. To a(·,·) let the operator L: V -+ V', the Ritz projection Sh, and the stiffness matrix L be associated. Show: (a) To the adjoint bilinear form a*(-,·) belong the stiffness matrix LT and the Ritz projection (8.4.18a)
(b) There holds (8.4. I4b) PROOF. Second proof of Theorem 11: Let H~(il) := {u E H2(il):u is a solution of (12) for an
f
E
£2(iln
c
V
be the range of £*-1 E £(£2(il), H2(il)). Equip the space H2(il) with the norm 1·12. Since £2(ilY = £2(il) we have that £*-1 E £(£2(il), H;(il)) is
192
8 The Method of Finite Elements
equivalent to L-l E L(H~(!J)',L2(!J)) (cf. Lerruna 6.3.2), so that there is a COl with (S.4.19a) The assumptions (1.2) and (1.12) of Theorem 11 can be brought over without change of the constants to the adjoint problem, so that the statement of Theorem S.2.1 can be written in the form III - Shllv+-H~(O) ::; Cf3h. For the adjoint operator we have the estimate
11(1 - Sh)*IIH~(O)/+-vl ::; Cf3h.
(S.4.19b)
The basic assumption (1.2) may be written (S.4.19c) From Equation (ISb) one may read the expression
1- Sh
= 1- L- 1Si.L = L- 1(1 -
Si.)L = L- 1(1 - Sh)* L.
From (19a-c) one has
III - ShIIL2(0)+-V ::; IIL-111L2(O)+-H~(o)/11(1 - Sh)*IIH~(O)/+-vIIILllvl+-v ::; CaCf3Csh.
(S.4.19d)
Therefore we have proved (17a) with C 1 = CaCf3CS . Since Sh is a projection, so is I - Sh, so that
III - ShIlL2(o)+-H2(o)nv = 11(1 - Sh)211L2(o)+-H2(O)nv ::; III - Sh 1I£2(O)+-vIII - Shllv+-H2(O)nv. From (17a) and (11) we deduce (17b). The regularity condition (13) is weakened in the following theorem:
•
Theorem 8.4.14. Assume (1.2), and (1.12) with fN = fh ~ f > 0, and inequality (10) for all 1 ::; s ::; 2. In the place of(13) we assume H 2- t -regularity for some t E [0,1]. If V = Hl(!J) then assume L*-1 E L(Ht(!J)', H2-t(!J))j if V = HJ(!J), then assume L*-1 E L(H-t(!J), H2-t(.a) n V). Then the finite-element solution satisfies the inequality (0 ::; t ::; 1 ::; 8
::;
2).
(8.4.20)
The Ritz projection satisfies
111- ShIlHt(O)+-H'(o)nv ::; Ct,shs- t .
(S.4.21)
PROOF. (For the case V = Hl(!J)) In (19a-c) replace L2(!J) by Ht(fl), and H~(!J)' by H~-t(fl)' with m-t(ll) := {u E H2-t(ll):u is the solution to f E Ht(ll)'}. Because of (11'), (19b) becomes [with 8 replaced by 2 - t]
8.5 Generalisations
* 11(1 - Sh) IIH~-t(O)/+-VI A
= III -
Shllv+-H~-t(O) ~ C{jh A
1-t
193
.
III -
Ai!, in (19d) one shows that ShIlHt(n)<--v ~ Ch 1- t • Combining this with B 1 -Shllv+-H'(n)nv ~ Ch - (cf. (11')) there follows (21), and thus (20). •
III
For nonlinear boundary value problems, estimates of the errors in terms of other norms (e.g., LOO(n), V(n)) are of interest. For this we refer to Ciarlet [1. §3.3] and Schatz [1].
8.5 Generalisations 8.5.1 Error Estimates for Other Elements The estimates for the errors for linear functions on triangular elements proved in Section 8.4 are also true for bilinear functions on parallelograms and for combinations of both sorts of elements (cf. Section 8.3.7). The proofs are along analogous lines. Even for tetrahedral elements in a three-dimensional region C :m.3 the same results can be carried over. For the proof one notices that u E H2(n) has well-defined nodal values u(Xi) even for n c IR3 , since 2 > n/2 (n = 3, dimension of n; cf. Theorem 6.2.30). For quadratic elements (cf. Section 8.3.4) one expects a correspondingly improved order of convergence. In general one can show the following: If the ansatz function is, in each 1i E T, a polynomial of degree k ~ 1 (Le., u(x) = 2: lvl::;k avxv on Ti ), then
n
d(u, Vh ) = inf{ lu-v!t: v E Vh } ~ Ch k lulk+1 for all u E H k+1(n)nV (8.5.1) (cf. Ciarlet [1, Theorem 3.2.1]). If one uses parallelograms as a basis and uses ansatz functions that are polynomials of degree at least k, then (1) also holds. For example, biquadratic elements and quadratic ansatz functions of the serendipity class fulfil this requirement for k = 2. The result corresponding to Theorem 8.4.6 follows from Theorem 8.2.1:
Theorem 8.5.1. Let Vh fulfil (1), and let the bilinear form satisfy (1.2) and (1.12) with fn =: fh ~ l > O. Assume problem (1.1) has a solution u E V n Hk+1(n). Then the finite-element solution u h E Vh satisfies the inequality (8.5.2) lu - uhh ~ Ch k lulk+l' The Ritz projection satisfies
III -
Shllv+-Hk+l(o)nv ~ Chk.
The estimate (4.9) now holds for s E [1,k] ifu E HB(n)nv. With suitable regularity conditions, there are, as in Theorem 8.4.11, the error estimates
lu - uhlo ~ Ch k+1lulk+l, lIu - u h IIHk-l (0)' ~ Ch2k lulk+1'
(8.5.3a) (8.5.3b)
194
8 The Method of Finite Elements
For (3b) one needs, for instance, Hk+l-regularity: For each f E Hk-1(D) the adjoint problem (4.12) has a solution u E H k +1(D). Remark 8.5.2. Under suitable assumptions the inequality (4.20) holds for 1- k 5 t 5 1 5 s 5 k + 1. For negative t the norm I· It should be understood as the dual norm of H-t(D).
8.5.2 Finite Elements for Equations of Higher Order 8.5.2.1 Introduction: The One-Dimensional Biharmonic Equation All the spaces of finite elements, Vh, constructed so far are useless for equations of order 2m > 2, since Vh ct. Hm(D). According to Example 6.2.5, in order to have Vh C H2(n) it is necessary that not only the function u but also its derivatives Ux, change continuously between elements. The ansatz functions must therefore be piecewise smooth and globally in C 1 (D). As a model problem we introduce the one-dimensional biharmonic equation
ul/I/(x) = g(x) for 0
< x < 1, u(O)
= u'(O)
= u(l) =
u' (I) = 0
which becomes in its weak formulation a(u,v)=f(v) forallvEHg(O,I), where
a(u, v)
=
11
ul/vl/ dx,
f(v)
= fa1 gvdx.
(8.5.4)
Divide the interval n = (0,1) into equal subintervals of length h. Piecewise linear functions (cf. Figure 8.3.1) can be viewed as linear spline functions, so that it is natural to define Vh as the space of cubic splines (with u = u' = 0 for x = 0 and x = 1) (cf. Stoer [I, §2.4) and Stoer-Bulirsch [1, §2.4)). We can take as basis functions B-splines, whose supports in general consist of four subintervals (cf. Figure 1). Since cubic spline functions belong to 02(0,1), they are not just in HJ(O, 1), but even in H3(0, 1) n HJ(O, 1) .
•
.~.
•
Figure 8.5.1. B-spline
Simpler yet than working with spline functions is to use cubic Hermi te interpolation: Vh := {u E C 1(0, l):u cubic on each subinterval,
u(O) = u'(O) = u(l) = u'(l) = O}.
(8.5.5)
8.5 Generalisations
195
To each of the inner nodes Xj = j h there are two basis functions bli and b2i with bH(Xi) = 1, bii(Xi) = 0, b2i(Xi) = 0, b2i (Xi) = 1, bki(Xj) = bki(Xj) = 0 for k = 1,2 (j # i) (cf. Figure 2) .
.6.
.~
(b)
(a)
..
Figure 8.5.2. Hermite basis functions
The support consist of only two subintervals. The expressions are ~ x ~ xi+1. 2 b2i (X) = (h -Ix - xil)2(x - xi)/h for Xi-1 ~ x ~ Xi+1. bji(x) = 0 for x ¢ [Xi-l,Xi+1]'
b1i (x) = (h -Ix - xil)2(2lx - xii
+ h)/h3 for Xi-1
Exercise 8.5.3. Let Uti and U2i (0 < i < l/h) be the coefficients of the expression u(x) = I:i[Utib1i(x) + U2i~i(X)] E Vh. Show that the coefficients of the finite-element solution of the problem (4) are given by the equations h-3[-12u1.i_l + 24uli - 12u1.i+l] + h- 2[-6u2,i_l + 6U2,i+d h-2[6ul,i_l - 6Ul.i+1] +
h- 1 [ 2u2.i_l
= i1£,
+ 8U2i + 2U2,i+l] = hi
(8.5.6)
The system of equations (6) differs completely from the difference equations (cf. Section 5.3.3), since in Equation (6) there appear values Uti of the functions at the nodes together with the values U2i of the derivatives at the nodes.
8.5.2.2 The Two-Dimensional Case The ansatz (5) can be carried over to [1 c if one starts with a partition into rectangular elements. The ansatz function is bicubic:
ne
3
Vh:= {U E 0 1([1): U = Un
= 0 on r,
U=
L
cx."p,x"y'"
",p,=o on each rectangle of the partition.}. At each inner node one may prescribe the four values u, Ua:, u lI ' ua:lI' Consequently there are four unknowns and four basis functions bli(X, y), ... ,b4i(X, y) belonging to each node. The latter are products bji(x)bki(y) (j, k = 1,2) of the one-dimensional basis functions described in Section 8.5.2.1 (cf. MeisMarcowitz [1, Figure 15.22], Schwarz [1, §2.6.1]). If one starts from a triangulation T, a fifth-order ansatz gives what is wanted: u(x) = I:1"1::;5 a"x" on 11 E T. The number of degrees of freedom is
196
8 The Method of Finite Elements
21 (the number of II E Z2 with II ~ 0, 1111:5 5). For the nodes one chooses the points Xl, ... ,x6 in Figure 8.3.Sa. At the vertices xl, x 2,:x3 one prescribes 6 values {Dl'u(xj): IILI :5 2}. The 3 remaining degrees offreedom result from giving the normal derivatives 8u(xi )/8n at the midpoints of the sides x4,x5,x6. Notice that if two neighbouring triangles (T and t in Figure 8.3.8a) have the same vertex values {Dl'u(xj): IILI :5 2, j = 1,2} and the same normal derivative at X4, then both '1.£ and Vu are continuous on the shared side, i.e., Vh C H2(f1). According to whether V = H6(n), V = H2(n) n HJ(n) or V = H2(f1) one has to set '1.£, or the first derivatives, to be zero at the boundary nodes Xi.
Remark 8.5.4. The finite-element space Vh C H2(f1) described here can of course be used for differential equations of the order 2m = 2. 8.5.2.3 Estimating Errors Instead of (5.1) one obtains
where k ~ m depends on the order of the polynomial ansatz (e.g., k = 3 for cubic splines, cubic [resp. bicubic] Hermite interpolation). Here m = 2 for the biharmonic equation. At; in Theorem 8.5.1, there follows from (7) the error estimate (8.5.8a) for the finite-element solutionu h E Vh . Under suitable regularity assumptions one gets (2m - k - 1 :5 t
:5 m :5 s :5 k + 1)
(8.5.8b)
(cf. Remark 8.5.2). The maximal order of convergence 2(k - m) + 2 results for s = k + 1, t = 2m - k - 1 and requires '1.£ E Hk+l(n) n V. In addition each solution of the adjoint problem (12) with f E Hk+1-2m(n) must belong to Hk+l(f1).
8.6 Finite Elements for Non-Polygonal Regions Since the union of triangles and parallelograms only generates polygonal regions a polygonal shape was assumed in (3.6). The finite-element method is, however, in no way restricted to just such regions. On the contrary finite elements can readily be adapted to curved boundaries.
8.6 Finite Elements for Non-Polygonal Regions
197
Figure 8.6.1. Gmvilinear boundaries
Let V = H1(il) and let il be arbitrary. The triangulation T can be so chosen that (3.7a,b,d) hold and the "outer" triangles have two vertices on r = 8il as in Figure 1. In the convex case of Figure 1a one extends the linear functions defined on Tonto T := TUB. In the case that the boundary is concave (see Figure lb) one should replace T by T := T\B. One copes correspondingly in the situation of Figure lc. The nodes and the expressions for the basis functions remain undisturbed by these changes. is the union of the closures of all the inner triangles 11 E T and all modified triangles T which lie on the boundary. All the properties and results of Sections 8.3-8.4 extend to the new situation. The only difficulty is of a practical sort: To calculate L and f one has to work out integrals over the triangle T which involves arcs. Assume now V = HJ (il). The previous construction will not be a success, since the extensions of the linear functions to T will not vanish on the boundary piece r n at. Thus Vh c HJ(il) will not be satisfied. As long as the region n is convex, only the situation shown in Figure 1a occurs, and u h may be extended to BeT by u h = O. In the case of Figure lb, one must set the values at the inner nodes to zero, so that u h = 0 on T = T\B, and in particular u h = 0 on n 8(T\B). All in all, what results is that the support of any u h E Vh is in a polygonal region inscribed in il. One interpretation is the following: In the boundary value problem, il should be replaced by an approximating region ilh C il (cf. Theorem 2.4.6). The finite-element solution described agrees with that which would result from the smaller region. However, the error estimate in Theorem 8.4.4 only holds for the smaller region ilh: inf1JEvh lIu - VIlHl(Oh) ::; ChlluIlH2(O)' Since v = 0 on il\ilh for any v E Vh one should also estimate lIuIlHl(OWh)' Now n\ilh is contained in a strip 8 6 = {x E ilh:dist(x, r) ::; o} of width 0 := max{dist(x, r): x E il\nh}. One can estimate as follows:
n
r
If n\nh consists only of arc segments B as in Figure la (for instance, in the case of a convex region), and if il E C1,l, then 0 = O(h2). From this follows the estimate inf1JEvh lIu - VIlHl(O) S Ch2I1uIlH2(O), as for a polygonal
198
8 The Method of Finite Elements
region. If, however, as in Figure Ib the whole triangle is part of fl\flh , then 8 becomes O(h) and the approximation worsens to inf1JEvh lIu - VIlHl(n) ::; Chl/2I1uIlH2(S1) (cf. Strang-Fix [1, p. 192]). In order to adapt the trianglar or parallelogram elements better to a curved boundary one can use the technique of isoparametric finite elements. From Figure 8.3.4 we see that the reference triangle T may be mapped into an arbitrary triangle f E T by an affine transformation cf> : (e,1]) 1-+ Xl + e(x2 - Xl) + !}(xS - Xl) . The linear [resp. in the case of §8.3.4, quadratic] function u(x) on T can be expressed as the image, vee, 1]) = u(cf>(e, 1]», of a linear [resp. quadratic] function von T. We now replace the affine transformation of triangles cf> by a more general quadratic mapping
cf>(e,1]) := (
al + ~e + as1] + a4e2 + a5e1] + ~1]2 ) : T _ a7 + ase + ag1] + alOe2 + alle1] + al21]2
f
C
JR2 .
The image f is a curvilinear triangle. The coefficients aI, ... , al2 are uniquely determined by the nodes Xl, ... , x 6 of f which are the images of the vertices and midpoints of the sides of the reference triangle T (cf. Figure 2). The triangulation T used so far can be replaced by a "triangulation" by triangles with curved sides, if neighbouring elements have the same arcs as common boundaries and the midpoints also coincide. The ansatz functions on f E f have the form u(x) = v(cf>-l(x», where v(e,1]) is linear [resp. quadratic] in and 1]. The resulting finite-element space is the space of isoparametric linear [resp. quadratic] elements (cf. Strang-Fix [1], Ciarlet [1], Schwarz [1], Zienkiewicz [1]).
e
Figure 8.6.2. Mapping of the reference triangle T to the curvilinear triangle T In general, there is no reason to use curvilinear triangles in the interior of n. AB in Figure 3, one chooses ordinary triangles in the interior (Le., the quadratic transformation cf> is again taken to be linear). A boundary triangle, like f in Figure 3, is, on the other hand, defined as follows: Xl and x 2 are the vertices of f which lie on r. One chooses another boundary point x4 between Xl and x 2 and requires that the boundary af of the triangle cuts the boundary r of the region at Xl, x4, and x 2 .
8.7 Additional Remarks
199
Figure 8.6.S. lsoparametric triangular dissection
Remark 8.6.1. Section 5.2.2 shows that, in the case of other than Dirichlet boundary conditions, the construction of difference schemes is increasingly complicated. Instead one may restrict the usual difference methods to the inner grid points, and near the boundary discretise by using (e.g., isoparametric) finite elements.
8.7 Additional Remarks It is not possible to consider here many of the details about, and modifications of, finite-element methods. In the following, individual cases are just treated summarily. B.7.1 Non-Conformal Elements
Condition (1.3), Vh c V, characterises conformal finite-element methods. V are called non-conformal. An example Discretisations for which Vh of a non-conformal element is the "Wilson rectangle". For this method one starts with a dissection into rectangles R;, = (Xli, X2i) X (YH, Y2i) and assumes the function to be quadratic on R;,:
ct
Vh := {u quadratic on R;" u continuous at vertices of the rectangle R;,}. Notice that u E Vh is required to be continuous only at the nodes, and not in the interior of D. A function which is not continuous along a side of R;, cannot belong to Hi(D): Vh V = Hi(D). A possible basis for Vh is the following. For each node x j , let bj be the basis function belonging to the bilinear elements (cr. Section 8.3.3). The quadratic ansatz chosen here has two
ct
200
8 The Method of Finite Elements
additional degrees of freedom. Because of this one chooses for each rectangle ~ = (Xli, X2i) X (YH, Y2i) two further basis functions
which vanish at the vertices of ~, so that continuity is ensured there if bP) and b~2) are extended by zero outside ~. The first problem which arises for non-conformal elements is the definition of the stiffness matrix L. A definition in terms of Lij = a(bj , bi ), as in (1.8a), is not possible because bj, bi ¢ VI Therefore, in addition to discretising V to Vh, one must replace a(·,·) by a bilinear form ah (-, .): Vh x Vh - IR. For a partition of il into rectangles 14 one defines ah(',') by ah(u,v) :=
I: 1 I: i
aaf3(x)(Dau) (Df3u ) dx,
Ro lal.If3I:5m
if a(·,·) is given by (7.2.6). Notice that a(u,v) = ah(u,v) for u,v E V. Conformal finite-element methods always give consistent discretisations. For nonconformal methods one does not always have consistency. For this problem, in particular for the so-called "patch test" see Strang-Fix [1, p. 174], Ciarlet [1, §4.2], Stummel [2], Gladwell-Wait [1, p. 83-92], and Thomasset [I]. That an ansatz which seems completely reasonable can possibly yield completely false discretisations is shown by Exercise 8.7.1. Let". be an admissible triangulation. Assume that Vh := {u constant on each n E ".}. Show that (a) Vh rt. V := Hl(il). (b) inf{lu - vlo: v E Vh} ~ Chluh for all u E V = Hl(n). (c) Let a(u, v) := fn(''lu, Vv) dx and ah(u, v) := L,T.ET fT. (Vu, Vv) dx. What is the result for Lij = ah(bj , bi )?
Non-conformal elements are, in particular, of importance for equations of higher order, since they are less complicated than conformal elements. 8.7.2 The Trefftz Method
In Theorem 6.5.12 it was shown that for symmetric and V-elliptic bilinear forms the weak formulation (1.1) of the boundary value problem is equivalent to the variational problem Find u E V with J(u)
= min{J(v):v E V}
(8.7.1)
(cf. also Theorem 7.2.9, Exercise 7.3.8). Another variational problem Find W E W with K(w)
= max{K(w):w E W}
(8.7.2)
is called the dual or complementary variational problem to (1), if both have the same solutions u = w E V n W.
8.7 Additional Remarks
201
For example, the Poisson problem, -Llu = f E L2(D) in D, u = 0 on r, leads to J(v) = fa IVvl2dx for v E V := HJ(il). A special dual variational problem is due to Trefftz (1926):
K(w) := -folVwl2 dx for w E W := {w E Hl(D): -Llw = f} (cl. Velte [1, p. 91]). Notice that v E V satisfies the boundary condition v = 0 on r, while for w E W no boundary condition is required though it must satisfy the differential equation Lw = f. 8.7.3 Finite-Element Methods for Singular Solutions The error estimates provided so far, such as for instance (4.7), depend on the assumption u E H2(D) n V, which will be discussed more carefully in Section 9.1. That this assumption does not always hold is shown by Example 8.7.2. (a) The Laplace equation Llu = 0 in the L-shaped region of Example 2.1.4 has the solution u = r2/3 sin«2
8.7.4 Adaptive'Iriangulation The sort of triangulation mentioned in the last paragraph can be most easily obtained by adaptive methods. So far T = Th has been given depending on a given "step size" h. In an adaptive procedure one starts with a triangulation
202
8 The Method of Finite Elements
and uses a-pa>teriori error estimates. Tha>e triangles of TO that lead to the largest errors are then divided further so that a new triangulation T is formed. One repeats this procedure until the errors are satisfactory. For the literature on this we refer to, for example, Rheinboldt's work [1] and the paper of Babuska and Rheinboldt referred to therein, as well as to Eriksson-Johnson
TO
[1). 8.7.5 Hierarchical Bases Let us give the fundamental idea in the one-dimensional case. Let Vh C HJ(O, 1) be, as in (3.2), the space of piecewise linear functions for an equidistant division of the interval n = (0,1) into parts of length h. For h = 1/2 we have dim(VI / 2 ) = 1, and bl as in Figure 1a is the only basis function. Clearly V1/ 4 C V1/ 2 • For V1/ 4 one takes over b1 E V1/ 2 as a basis function, and adds the functions ~, bs as in Figure 1b; {bt.~, b3 } is a basis for Vl / 4 • Similarly Vl / S C V1/ 4. Thus bl,~,bs supplemented by b4, ... ,b7 (cf. Figure 1c) gives a basis - the so-called hierarchical basis - of V1/ S . In contrast to the basis functions that have as yet been used, the supports of the bi are not necessarily of length O(h); e.g., supp(bt} = [0,1). From this it follows that the stiffness matrix has more entries than is otherwise usual. In spite of this, the above choice of basis does have important advantages: (i) A matrix L = L(2h) already calculated for V2h is a submatrix of the stiffness matrix L = L(h) for Vh, so that (a global or local) refinement of the triangulation requires only a minimum of additional effort. (ii) The condition cond(L) = ilL II ilL-III of the stiffness matrix is essentially better than in the standard case (for example, cond(L) = O(log h) instead of O(h- 2 ); cf. Section 8.8). (iii) There are suitable methods available for solving the system of equations Lu = f. Further details on this can be found in Yserentant [1), Bank-DupontYserentant [1), and Hackbusch (9). bs 1\
I
o
1/2 (a) h = 1/2
1
I
o (b) h = 1/4
I
I
I
, I
I \
\
\
\ \
\
\
\
1
o
1
(c) h = 1/8
Figure 8.7.1. A hierarchical basis
8.7.6 Superconvergence In difference methods the error is given by a mesh function Uh - Rhu (Uh: the difference solution, u: the exact solution), while for finite-element methods the
8.8 Properties of the Stiffness Matrix
203
solution u h is defined on the whole region n, and therefore uh - u can be used as the error. A grid function Uh could in principle be extended to a function u h defined on all n, for example, by choosing the values Uh(Xi ) (Xi: a grid point of nh) as the nodal values of the piecewise linear or bilinear finite elements, and thus interpolating Uh with uh := PUh (for P see (1.6». In addition to the errors Uh - Rh u at the grid points, there are now the interpolation errors u - P Rh u as well: u h - u = PUh - U = P(Uh - RhU) - (u - PRhU). As soon as the interpolation errors are of the same order of magnitude as Uh - Rh U, one has for PUh - U the same convergence assertions, otherwise PUh - U displays worse convergence than Uh - RhU. Conversely, for finite-element methods, one can ask if one has the same convergence properties (e.g., IIUh - RhUIl = O(h2), R h: restriction to the nodes) for the values at the nodes Uh = {uh(Xi):xi nodes} or for suitable averaged values. If one has indeed better convergence than for u h -U, one speaks of superconvergence. This notion can also be applied to difference quotients of node values, ifthese are more precise than IDO!uh-DO!ulo ~ Iuh~uh (10:1 = 1). For this see Lesaint-Zlamal [1], Bramble-Schatz [1], Thomee [2], and Louis [1]. Greater precision can be achieved with extrapolation techniques which, under suitable conditions, can be applied to finite-element problems (cf. BlumRannacher [2]).
s.s
Properties of the Stiffness Matrix
The properties of the matrix Lh have been carefully studied for difference procedures since the solvability of the difference equations and the analysis of convergence properties depend on them. In the case of finite-element discretisation we obtain the corresponding assertions in another way. The factors that control the solvability and convergence are the subspace Vh and the operator Lh : Vh -+ V,:, which is associated to the bilinear form a(·,·) : Vh x Vh -+ nt, and which was introduced as LN in (1.lOa). The stiffness matrix L, for a given Vh, depends on the basis {bt, ... } chosen. Let Nh := dim(Vh)'
The isomorphism P : m.Nh -+ Vh defined in (1.6) maps the coefficient vector v to Pv = 2:;:;t Vibi E Vh. For finite elements, in general, the coefficients coincide with the nodal values: Vi = (Pv)(xi ), (1 ::; i ::; Nh). The connection between the matrix L and the operator Lh : Vh -+ V,: is, according to Lemma 8.1.7, (8.8.1) The definition of P* in (1.10b), (P*u, v) = (u, PV)P(O), also depends on the choice of the scalar product (-,.) in m.Nh. Let us choose here the usual
204
8 The Method of Finite Elements Nh
(U, v) =
L
(B.B.2a)
UiVi'
i=l
In Exercise B.1.10 we have already established the following properties of L: If a(·,·) is symmetric [and V-elliptic} then L is symmetric [and positive definite}. Notice that the difference methods do not always have these properties. For the Poisson problem there is a symmetric form, but nonetheless the matrix Lh in Section 4.B.1 is not symmetric (cf. Theorem 4.B.4). The condition cond(L) plays an important role in the solution of the equation Lu = f. We would like to show that, under standard conditions, we have cond(L) = O(h-2m~ (cf. Remark 5.3.10). First we have to decide on the fundamental norm of m. h. Up to a scalar factor
(n: dimension of n c
m.n)
(B.B.2b)
is the Euclidean norm coming from the scalar product (2a). The corresponding matrix norm (B.8.2c) is the usual spectral norm of L. AB an alternative to the norm
II . IIh let us introduce (8.8.3)
In some important cases II· lip and 1I'l!h are equivalent (uniformly in h). AB an example consider the linear elements. Theorem 8.8.1. Let Th, h -+ 0 be a sequence of quasiuniform triangulations. Let Vh be the space of linear elements defined in (3.8) with the usual basis (see (3.9a)). Then there is a constant C p , which does not depend on h, such that (8.8.4)
The basis for the proof of this is Lemma 8.8.2. Let T = {(e,l1): e, 11 ~ O,e +11:5 I} be the unit triangle (see Figure 8.3.4). ffu is linear on T, then
~ [u2 (0, 0) + u 2 (1, 0) + u 2 (0, 1)] :5 :5
1
"6
PROOF. Put U1
[u 2 (0, 0)
Ul
+
u 2 (1, 0)
:= u(O,O),
+ (U2 - ude + (U3 -
U1)17,
U2
+
u 2 (0,
Jh (e, u2
11) ~ dl1
(8.8.5)
I)}.
:= u(1,0),
it follows that
U3
:= u(0,1). From U(e,l1)
=
8.8 Properties of the Stiffness Matrix
flu2(e,1])~d1]=2~(ul
U2
T
205
U3)[~1 ~1 2~] (:~). U3
The eigenvalues of the symmetric 3 x 3 matrix are A1 = A2 = 1 and A3 Therefore the right-hand side is ~ L:~=1 uU24 and ~ 4L:uU24.
= 4. •
To prove Theorem 1 write lIull~ as In (Pu)2 dx = L:T.ETh IT. (Pu)2 dx. Let tPi, : T -. Ti, be the linear maps from T to n as in Exercise 8.3.14a. Let U/,U",U Ill be the values of u = Pu at the vertices of n. The inequality o < C 1h 2 ~ Idet tPH ~ C2h2 and Lemma 2 imply
~l h 2(u /2 +u'12 +u"12 ) S L.IPU I2 dx = IdettP~1 f
L
I(PU)(tPi(e,1]))1 2 ~d1]
s ~2 h2(u/2 + U" 2 + U" /2 ).
(8.8.6)
By definition of a quasiuniform triangulation all the angles of the triangles are ~ ao > O. Let M > 211'/ao. Each node belongs to at least one triangle, and to at most M triangles. Since u /, u", u lll are the components of the vector u, the summation in Equation (6) over all n E Th gives
~lllull~ S lIull~ = L:
1
T.ET T.
(Pu)2 dx
~ M~2I1ull~,
so that inequality (4) holds with Cp := max(MC2/6, 24/C1)1/2.
•
Exercise 8.8.3. Let Vh c HJ(n) be derived from the square grid triangulation (cf. Exercise 8.3.13). Show that V4/311ullp ~ Ilulih v'Sllulip.
s
The matrix (8.8.7) is called the mass matrix when it occurs in engineering applications. Concrete bounds in the inequality (4) have been given by Wathen [1] for various specific elements. Remark 8.8.4. (a) IneqUality (4) is equivalent to IIMII ~ c;'hn,
IIM- 1 11 ~ C;'h- n .
(b) We have lIulip ~ IIh- n MW/ 2I1ullh, lIulih ~ IIhn M- 1 W/ 2I1ullp. PROOF. (a) The calculation lIull~ = In(Pu)2 dx = (Pu, Pu)o = (P*Pu, u) (Mu, u) ~ IIMII(u, u} = IIMllh-nllull~ proves the first inequality in (b). The next follows from Ilull~ = hn(u, u) = hn{Ml/2u,M-lM1/2u} ~ IIhn M- 11I(Ml/2u , Ml/2u} = IIhnM-11I(Mu, u} = IIhnM-lllllull~, since M is positive definite.
=
206
8 The Method of Finite Elements
(b) From h-nllMIi S C~ and hn1lM-11I S ~, using (b), we deduce the inequality (4). Conversely (Mu, u)1/2 = lIulip :5 Cpllullh = Cphn/2(u, U)1/2 implies that h-nllMIi S C~, since M is positive definite (cf. Lemma 4.3.25). Similarly, hn1lM-11I S C~ follows from lIulih Cpllullp. •
s
Since Vh is finite-dimensional Ch := sup{luh/lulo: 0 :f:. u E Vh} is finite. One says that Vh satisfies the inverse estimate if CIl = O(h- 1 ), that is,
(8.8.8) where C1 does not depend on hash - O.
Theorem 8.8.5. Let Th be a sequence of quasiunifonn triangulations. Then the space of finite elements introduced in (3.8) satisfies the inverse estimate. PROOF. As in Theorem 1, the proof follows by transforming the integrals fT.IDO«Pu)1 2 dx for 1i E Th and lal :5 1 into integrals over T. Here we have in addition to transform D = Dx into D~,f/ (cf. the proof of Lemma 8.4.2). •
Theorem 8.8.6. Suppose Cpllullh hold. Then we have
nc
m,n. Assume that (1.2), (8), and lIullp S
IILII S hn-2CsC~CJ.
(8.8.9)
PROOF. Multiply the inequality (u,Lv) = a(Pu,Pv) :5 CslPuhlPv/l S Cs CJh- 2lPulolPvlo
s CsCJh-2C~lIullhllvllh by hn and set u := Lv:
IILvll~
= hn(Lv,Lv) S CscJC~hn-2I1Lvllhllvllh'
•
and thus IILvllh S CsCJC~hn-2I1vllh' that is (9).
Theorem 8.8.7.
II· IIII S Cpll·llp·
Assume n c m,n. Let (1.12) hold with Then there holds
€h ~
13. > 0 and
(8.8.10)
PROOF. We know L- l Section 8.1), so that
= p- 1Lhlp*-l
ilL -lfllh :5 CpliL -lfllp
and IILhlllv,,+-v~ :5
1/€h
s 1/13. (cf.
= CplPL-lflo = CpILh"lp*-lflo
s CpIlLhlp*-lfllv" s (Cp/€)IIP*-lfllv~.
8.8 Properties of the Stiffness Matrix
207
Since IIP*-ifll vh =
sup
I(v, P·-if)ol/lvh =
sup I(P-iv, f} I/lvli O;/'1JE Vh
O;/'1JEVh
= sup I(V, f}I/IPv\t = suph-nllvllhllfllhllPvlo v;/'o
~
v;/'o
Cp sup h-nllvllhllfllhlllvllh v;/'o
= Cph-nllfllh,
we have that IIL-ifllh ~ C~h-nllfllhlf for all fj thus we conclude (10).
•
The combination of Theorems 6 and 7 leads to Theorem 8.8.8. Assume (1.2), (1.12) with fh Then we have
~ f
> 0, (4), m
= 1, and (8).
(8.8.11)
The ideas involved in the proof can, in principle, be carried over to the case 2m > 2, i.e., to boundary value problems of higher order. The inverse estimate becomes (8.8.12)
In order that inequality (4) hold one must be careful in defining the norm 1I·lIh. If all the components Ui of u are nodal values (Pu)(xi ) (as for instance is the case for the spline ansatz for Vh C H2(n)), then 1I·lIh can be defined as in (2b). However as soon as the components Ui of u involve derivatives (Dapu)(xi ), the u1 in (2b) must be replaced by (hlalui)2. For example, when the Hermite functions of (5.5) are used then u has the components Uti and U2i (cf. Exercise 8.5.3), where Uti = (Pu)(xi ) and U2i = .t~ (Pu)(xi ). The appropriate definition of 1I·lIh is lIull~ := hn L:i(U~i + h2u2i) with n = 1, since (0,1) C ]Ri. Exercise 8.8.9. Let V cUe V' be a Gelfand triple and assume Vh C V. Show that the inverse estimate (8.8.13a)
implies
9 Regularity
9.1 Solutions of the Boundary Value Problem in H8(lJ), 8> m 9.1.1 The Regularity Problem The weak formulation of a boundary value problem
Lu = 9 in
n,
Bu =
on
r
(9.1.1)
as u E V,
a(u, v)
= I(v)
for all v E V
(9.1.2)
was, in Section 7, the basis upon which we were able to answer the questions of existence and uniqueness of the solution. Here, by existence of a solution we understand the existence of a weak solution u E V. The error estimates in Section 8.4 made it clear that the statement u E V is not enough. Under this assumption we can only show lIu - uhllv ~ O. The more interesting quantitative estimate lu - uhh = O(h) as in, for example, Theorem 8.4.6 for V = HJ(n) or V = H1(il), requires the assumption u E H2(n) n V. The assertion u E H2(il) or, more generally u E H"(D), is a statement of regularity, i.e., a statement about the smoothness of the solution, which will be examined in greater detail in this section. The regularity proofs in the following sections are very technical. To make the proof ideas clearer, let us sketch the proof of inequality (4) below for the Helmhol tz equation -..:1u + u = I in n c m.2 , u = 0 on r. Step 1: n = m.2 . Since the bilinear form for this situation a( u, v) = (,'Vu, Vv} dx+ uvdx is H1(m.2)-elliptic, (4) holds for 8 = m = 1. We shall prove (4) by induction for 8 = 1,2,3, .. '. To this end we take the derivative of the differential equation with respect to x, -Lluz + U z = Iz. If I E H,,-2(il), then Iz E H B - 3 (il) and the equation -Llv + v = Iz, by the induction assumption, has a unique solution v E H,,-1(il) with Ivl,,-1 $ C~_d/zI8-3 $ C~-11/IB-2. If one sets up this inequality for v = Uz and likewise for v = ttv, the result is lui" $ 1'1,,118-1 + IUzl"-1 + Ittv18-1 $ 3C~_11/18-2' Thus, (4) has been shown for 8.
In
Step 2: il
In
= m.~ = m. x (0, 00). As above, we can obtain the estimate lux 18-1 $
C~_1Ifl"-2' since U x also satisfies
-LlUz+ux
= Ix and the boundary condition
W. Hackbusch, Elliptic Differential Equations: Theory and Numerical Treatment, Springer Series in Computational Mathematics 18, DOI 10.1007/978-3-642-11490-8_9, © Springer-Verlag Berlin Heidelberg 2010
208
9.1 Solutions of the Boundary Value Problem in HB(O),
8
>m
209
r.
= 0 on This is not the case for u.y. But U
n
a
Step 3: Let be arbitrary, but sufficiently smooth. As in Section 6.2.1, is decomposed into (overlapping) pieces ai, which can be mapped into ffi2 or IR~. Correspondingly one splits the solution u into LXiU (Xi is a partition of unity). Then the arguments from steps 1 and 2 prove inequalities for IXiUls which together result in (4). Note that only a sketch of the proof was given. Some of the steps of the proof are incomplete. For example, might not the equation -Llu
Remark 9.1.1. (a) Hm-regularity always holds. (b) Let the variational problem (2) have a unique solution U E H{f(n) with Iulm :s; Coi/l- m for all I E H-m(a). If the boundary value problem is H8_ regular, then the weak solution of (2) with I E Hs-2m satisfies the inequality (9.1.4) (c) Let L be the operator associated with a(·,.). (4) is equivalent to L- 1 E L(HB-2m(n), H8(a» and (4'): (9.1.4') PROOF. (a) Set Cm = 1. (b) In (3) estimate Iulm by Coi/l- m and use the fact that because of 8 ~ m the embedding HB-2m(a) C H-m(a) is continuous such that I/I-m :s; C'l/ls-2m. Thus we obtain (4) with C; = C 8 (1 +CoC'). • The following remark shows that a perturbation of a(·,.) by a smooth term of order < 2m does not change the H 8 -regularity. Remark 9.1.2. Let L be Ht-regular for all t = m, m+ I, ... ,8 E 1N. Let SL be an operator of order :s; 2m -1: SL E L(Hr(n) n Hlt(n), Hr+1-2m(n» for all r = m, m+l,·· ·,s-1. Then L+SL is also Ht-regular for t = m,m+l,···, s. The assumption on SL holds in particular if SL belongs to the bilinear form a"(·,.) in Lemma 7.2.12 and its coefficients aa/3 are sufficiently smooth: for example, aa/3 E cmax{0,s-2m+I/3I} .
210
9 Regularity
PROOF. (By induction) For s = m the statement follows from Remark la. Assume the assertion for s - 1. Let u be the weak solution of (L + oL)u = f E Hs-2m{n), hence also the solution of Lu = J := f - oLu. Under the induction assumption the following already holds:
luls-l ~ C~-tllfI8-2m-l + lulm} such that
IJls-2m ~ Ifls-2m + lIoLlIs-2m+-s-1I uls-1 ~ C~/_dlfls-2m + lulm}. Because of the HS-regularity of L, u belongs to HB(fl) and satisfies luis C~[ IJls-2m + lulm}. Together these inequalities result in statement (4). 9.1.2 Regularity Theorems for
~
•
n = m,n
n
The domain = m,n is distinguished by the fact that it has no boundary, and hence no boundary conditions either.
Theorem 9.1.3. Let m E IN, a(u, v):=
n = m,n.
L
Let the bilinear form
1
aap(x)(Dau)(dv)dx
(9.1.5)
lal.IPI:5m n
be Hm{m,n)-coercive. For some k E IN let the following hold: D'Yaap E LOO{fl)
for all a,p,,,"( with
hi ~ max{O, k + IPI- m).
(9.1.6)
Then every weak solution u E Hm{m,n) of the problem (9.1.7)
(9.1.8)
PROOF. (a) First we must investigate the start of the induction, k
a= ah •i be the difference operator ah.iU{X):=
[u(x+~e;) -u(x-~ei)]/h,ei:
i-th unit vector,l
= 1. Let
~ i ~ n.
Let u be the solution of (7). For v E Hm{m,n) set
d{ u, v) := a( u, ah.iV) + ap{ ah.iU, v), where ap{u, v) := L:lalSm
1.BI=m
JRn aap (Dau )(D!3v ) dx.
(9.1.9a)
9.1 Solutions of the Boundary Value Problem in H 8 (D),
8> m
211
Let u±(x) := u(x± h8j/2). From h 1mn auav dx = 1mn au[v+ - v-] dx = 1]Rn [a-u--a+u+]v dx = -h 1mn av8udx+ 1nn{(a--a)u-+(a-a+)u+}vdx we see that
d(u, v)
L In
=
l"'l~m
mn
aa~(Dau)(D~8v) dx
Ipl~"'-l
-
Linn I"'I~'"
1091='"
~{(aa~ - a-;'~)Dau- + (a~~ - aa~)Dau+}D~vdx.
n
(9.1.9b)
s
Since lavl s Ivl s +l (cf. Lemma 6.2.24), one bounds the first sum in (9b) with Glulmlavlm- 1 ::; Glulmlvl m. One obtains the same bound for the second sum: (9.1.9c) However, if a(·,·) is Hm(JRn)-coercive ap(·,·) is also (cf. Lemma 7.2.12). For CE > 0 and CK from (6.5.10) we thus obtain
GE 18h,iUI~ ::; a p( 8u, au} + C K 18ul5
= -a(u, 8 2u) + d( u, 8u) + CK 18ul5
(9. 1. lOa) The first summand can be transformed according to (7): a(u, 8 2u)o = (f,8 2u)o and bounded by Ifl-m+!182ulm-i. The inequality lavl s Ivl s +! for s = m-1 and v = 8u yields
s
(9. l. lOb) By (9c), the bound for the second summand in (lOa) reads: (9.l.lOc) Finally we have (9.l.1Od)
(lOa-d) yields
Lemma 6.2.24 shows that u E um+ 1 (JRn ) and inequality (8) for k = l. (b) Now let k = 2. In (9a) substitute D'Y8v with bl = 1 for 8v and integrate by parts:
212
9 Regularity
d(u, v) := a(u, D'YBv) - ap(D'Yu, v)
=
L: 10iSm II'ISm-2
1.
aafj(Da u)(JJI1+'Y8v) dx
Rn
- L: 10iSm II'ISm-1
+
L:!.
loiS... 11'1 .....
f.
[D'Y(aafjVOu)}(JJI18v) dx
Rn
[(D'Y8aa,6) (Dau(x -
iSi»
Rn
+ (8a a ,6) (VO+'Yu(x -
ie;»
+ (D'Yaafj(x + tSi)Da 8u)
+ (aafj(x + ie;) - aa,6(X» Da+'YBu] (JJI1v) dx. As in the first part of the proof, from IU,D'YBv)ol ~
Id(u,v)1
~
IfI2-mI D'YBvlm-2
C"lul m+1lvl m and ~
Ifl2-mlvlm
follows
ID'Y8h,i Ulm ~ [lfl-m+2 + (C" + CK)lulm+1}/CE for all hi = I, h > 0, 1 ~ i ~ n. Thus, the differences 18h,iUlm+l are uniformly bounded so that Lemma 6.2.24 proves u E Hm+2(Rft) and lul m+2 :5 C[lfl-m+2 + lulm+1}' If one uses the estimate lulm+1 :5 C[lfl-m+1 + lulm}~ C[lfl-m+2 + lul m}proved in (a), (8) follows for k = 2. (c) An induction argument is carried out analogously for k
~
2.
•
Corollary 9.1.4.
The coercivity in Theorem 3 can be replaced (a) by the sufficient conditions from Theorem 7.2.11; (b) by the assumption that for some ~ the bilinear form ap(u, v) + ~(u, v)o satisfies the assumption (6.5.4a,b).
Corollary 9.1.5. If, in addition, a(.,.) is ~(Rn)-elliptic or if a(-,·) satisfies condition (6.5.4a,b), then in Theorem 3 the estimate (8) should be replaced by (9.1.11)
PROOF. By Remark 2.
•
Corollary 9.1.6. Let a(·,·) be Hm(Rn)-coercive. Let the conditions (6) and f E HAl-m(Rn) be satisfied for k E 1N with k > s+n/2 ~ n/2. Then the weak solution 1£ of (7) belongs to C·(Rn ). Hence for ~ ~ 2m, the weak solution is also a classical solution.
PROOF. The statement results from Sobolev's lemma (Theorem 6.2.30). •
9.1 Solutions of the Boundary Value Problem in HB(O), s
>m
213
Corollary 9.1.7. Let a(.,.) be Hm(ffin)-coercive. Let the conditions (6) and IE H1c-m(llln) be satisfied lor all k E IN. Then the weak solution of problem (7) belongs to coo(ffin). The conditions are satisfied in particular if I belongs to C~(ffin) and the coefficients aa~ are constant. The generalisation of u E H m+1c(ffin) with k E lN to u E Hm+s(ffin) with real s > 0 reads as follows:
Theorem 9.1.8. Let a(.,.) in (5) be Hm(ffin)-coercive. Let s= k
+ e,
0 < e < {) < 1,
k E lN u {O},
For the coefficients (n
= ffin)
aa~ E ct+I~I-m(n)
for k
t:= k
+ {).
let
+ 1,81 ~ m,
aa~ E LOO(n) otherwise. (9.1.12)
Then each weak solution of (7) with IE H-m+s(ffin ) belongs to Hm+s(ffin ), and satisfies the estimate
(9.1.13)
PROOF. The beginning of the induction is given by k = 0, i.e., 0 < s < t < 1. Replace the difference quotient 8 = 8h,j by an approximation to its power 8s :
Ru(x) := Rh,jU(X) := h- s
f:
,,=0
e-"h( -I)" (S)u(X + JLhej) , I-'
where ej is the jth unit vector. Here (~) = 1, and
G) (-1)" = -8(1 - 8)(2 - 8) ...
(I-' -1 - 8)/ JL!
is the binomial coefficient.
Exercise 9.1.9. Let 0 <
8
< 1. Show that (a) The operator adjoint to Rh,j
is
R*u(x)
= Rh,ju(x) = h- s
f:e-"h(
-I)"
,,=0
(b) For any z E V with Izl < 1 we have (1 - z)S (c) For the Fourier transform we have
(s)U(X - JLhej). I-'
= E;;'=o (;H -z)".
!7(Rh,jU)(e) = [(1 - e-h+i~ih)/hjSu(e), !7(Rh,ju)(e) Hint: !7(u(·
+ oej»(e) =
= [(1 -
eieihu(e).
e-h-i{ih)/hjSu(e).
214
9 Regularity
(d) IRh,julT ::; 0ITluIT+S for all T E ffi, h > 0, 1 ::; j ::; n'A u E HT~S(ffin)j similarly IR~,juIT ::; IUIT+S. Hint: Make use of the norms I· IT, and I·IT+S (cf. (6.2.15)), and show 1!1'(Ru)(~)1 ::; (1 + 1~12)s/2Iu(~)I. By analogy with (9) one obtains d(u, v) := a(u, Rh,jV) - ap(Rh,ju, v)
=
:E 101:Sm 1.BI:Sm-l
+
:E
1
aaP(Dau)(Rh,jDPv) dx
R"
f
h-se-"h( -I)"
(8) I'
lol:sm ,,=1 1.BI=m
1
[aap(X - p.hej) - aap(x)][Dau(x - p.hej))(DPv) dx,
R"
since the first summand in (aapDau)(x - p.hej ) = aap(x)Dau(x - p.hej) + [aap(X - p.hej) -aap(x)lDau(x - p.hej) belongs to ap(R*u, v). Since we have laap(x - p.hej) - aap(x)1 ::; O(p.h)t, it follows that Id(u, v)1 ::; 0lulmlvlm
[1 + ht - s ~ e-"h( -I)" (;)p.t]
::; O'lulmlvlm,
for it is true that (~) = 0(p.-s-1) and ~~=l e-"hp.t-s-l same consideration as in Theorem 3 yields IR~,julm ::; O[lRh,jfl-m+lulml ::; O'[lfl-m+s+lulml
=
for all h
O(hs - t ). The
> 0, 1 ::; j
::; n
(cf. Exercise 9d). To obtain this estimate for lul m+ s we write (lul~+s)2 as :E (1 + 1~12)m+slu(~)12~ +
1~19/h
1
1~I~l/h
(1 + 1~12)m+slu(~)12~.
The second integral converges to zero for h -+ O. The first one can be estimated by lul~ plus the sum
o
t
IRh,jul~ ~ 0'
j=l
1
1~I~l/h
(1 +
1~12)m :E 1!1'(Rh,jU)(~)12~ j
= 0" f (1 + 1~12)m:E 1(1 - e-h-i{ih)/hl2slu(~)12~ JI~I~l/h j
= 0" f
JI~19/h
~Cf
JI~19/h
(1 + 1~12)m :Elsin~jhI28IU(~)12~ j h
(1 + 1~12)ml~128Iu(~)12~.
9.1 Solutions of the Boundary Value Problem in HB(lt),
8
>m
215
As in Lemma 6.2.24, '1.1. E Hm+s(JR") and (13) follow. (2) As in the proof of Theorem 3, one carries out induction over k and investigates d(u, v) ;= a(u, RJYYv) - (-1)i'Y 1ap(R* D'Yu, v), hi = k. • Exercise 9.1.10. Generalise Corollary 6 with the aid of Theorem 8.
9.1.3 Regularity Theorems for
{l
= JR~.
The halfspace JR~ in (6.2.18) is characterised by x" > O. As in Section 7, we limit ourselves to the following two cases; either a Dirichlet problem is given for arbitrary m ~ 1, or the natural boundary condition is posed for m = 1.
Theorem 9.1.11. (Homogeneous Dirichlet problem). An analogue to Theorem 3 holds for the Dirichlet problem:
'1.1. E Htr(JR+'),
a(u,v)
= (f,v)o
(9.1.14)
for all v E Htr(JR+').
Theorem 8 can also be carried over if one excludes the values s m-1/2.
= 1/2,3/2, .. "
For the proof we need the following highly technical lemma.
Lemma 9.1.12. Lets >O,s~
H,J, .. ·,m-!}. The norm 1·ls of H8 (JR+.)
is equivalent to
111'1.1.1118 ;= (1'1.1.15 +
L
IDaul~_m)1/2.
(9.1.15)
lal=m
PROOF. The relatively elementary case s
~
m is left to the reader. For
o < s < m too, the proof would be considerably simpler if in (15) one were to
replace the dual norm 1·ls-m of (Ho-S(R+'))' by that of (Hm-s(JRi.))'. Step 1. First we prove the statement for {l = JR" instead of {l = JR+.. According to Theorem 6.2.25a, for {l = JR" the norm III· Ills is equivalent to IlIulll~
;=
(1'1.1.1& +
L
(IDaul~_m)2)1/2.
lal=m
Since (llIulll~)2 =
I lR" [1 + CLlal=m lea I2 )(1 + leI 2 y- mllll(e)1 2 de and
o < Go(l + lel 2 )S ::; 1 + ( L lea I2)(1 + leI 2 )s-m ::; G1(1 + lel 2 y, lal=m
III . 1I1~ and I 'I~ are also equivalent. Step 2. For the transition to (l = JR~ the following extension HS(JR") must be investigated, where x = (x', x,,) E JR";
~; HI! (JR~) -+
216
9 Regularity
(¢U)(x) := u(x)
x E R+,
for L
(4Ju)(x',x n ) :=
L av[u(x', -lIXn) + u(x', -Xn/lI)]
for Xn
< O.
1'=1
We list the properties of 4J as
Exercise 9.1.13. Let the coefficients av of 4J be selected as the solution of the system of equations E~=l av ( lIA: + II-A:) = (_I)A: (0 :::; k :::; L - 1). Show that (a) u E CL-1(R+) yields 4Ju E c L- 1(Rn). (b) 4J E L(HA:(R+), HA:(Rn» for k = 0, 1, ... ,L. (c) The operator adjoint to 4J reads L
(4J*u)(x', xn)
= u(x', xn)+ ~ av [~u(x"
-xn/lI)+lIu(x', -lIXn)]
for Xn
> O.
(d) (8/8xn )A:(4J*u)(x', 0) = 0 for k = 0, 1, ... , L - 2 and u E cA:(Rn). (e) 4J* E L(HA:(Rn), H3(R+» for k = 0,1, ... , L - 1. Hint: cf. Corollary 6.2.43. (f) 4J E L(H8(R+), H8(Rn» for 8 = 1 - L, 2 - L, . .. ,0,1, ... ,L. Step 3. 111·111" :::; OI·IB results from Da E L(H8(R+), H,,-m(R+» (provable via continuation arguments from Remark 6.3.14b) so that 1·1. :::; 11I·llIs remains to be shown. Step 4. luis:::; 14Ju1" :::; 1l1¢U1l1. is true according to Step 1 of the proof. The inequality 1l1¢U11l. :::; Cillulill., which would finish the proof, reduces to
IDa¢UI,,_m :::; OIDaul s _ m for Let
lad = m,
'1.£
E
(9.1.16)
H 8 (lll+).
lal = m. For
(4Jau )(x', xn) := u{x', xn) { E~=l av [( _lI)a nu (x', -lIXn ) + (_ttn u(x', -xn/lI)]
for Xn > 0, otherwise,
one verifies Da4J = 4JaDa. As in Exercise 13f one shows that 4Ja E L(H8(lll+), HB(llln» for 8 = 1 + m - L, 2 + m - L, ... , L - m. This result can be carried 8 E IN (Le., over to real 8 E [1 + m - L, L - m] except for the cases 8 = -1/2, -3/2, ... ) (cf. Lions-Magenes [1, p. 54 ff]). Let L ~ 2m + 1 and v E Hm-"(llln). Since 4J~v E HO"-S(lll+), one infers from (D a4Ju, V)L2(lRn) = (Dau, 4J~V)L2(IR~P the estimate
!-
I{Da¢U, v)ol
= IDaul,,-mll4J~IIH;'-'(R~jJ+-Hm-'(lRn)lvlm-8'
for all v E Hm-8(Rn ), and therefore (16) with C := 114J~IIH;'-'(IR+)+-Hm-'(lRn) = l14>aIlH·-m(lRn)+-H·-m(IR+)'
9.1 Solutions of the Boundary Value Problem in HS(G),
8
>m
217
• PROOF. (of Theorem 11) (a) First let k = 8 = 1. The proof of Theorem 3 can be repeated for the differences ah,jU (j = 1,···, n - 1) and implies the existence of the derivatives au/aXj E Hon(IR+'}, j f n. Thus one has Dau E Hi(IR+.) for all lal = m except for a = (0, ... ,0, m). (b) We set a:= (O, ... ,O,m) E 7l.n, w:= a&aDau,
Fa(v):=
r w(x)Dav(x)dx for lal = m,
1JR'-+-
where a&a is the coefficient from the bilinear form (5). The remainder of the proof runs as follows. In part (c) we will show that (9.1.17)
= (w,Dav)o = (-I)m(Da w,v)o, (17) means that Daw E Hi-m(IR+.) and IDaw\t-m ~ C for lal = m. According to Lemma 12 it follows that w E Hl(IR+.). The coercivity of a(·,·) implies uniform ellipticity of L = Elal=I.BI=m(-I)mD.Baa.BDa, i.e., Eaa.B~a+.B ~ €1~12m (cf. Theorem 7.2.13). For ~ = (0,··· ,0, 1) one obtains a&&(x) ~ €. Hence it follows from wE Hl(IR+.} and D'Yaaa E LOO(nt+'), hi = 1, that D&u E Hl(nt+'). According to part (a) all other derivatives Dau (Ial ~ m, a f a) belong to Hl(nt+.) anyway so that u E H m - i (nt+') has been proved. (c) Proof of (17). For each a f a there exists a "Y with bl = 1, "Yn = 0, ~ "Y ~ a (component-wise inequalities). Integration by parts yields Since Fa(v)
°
for v E CO' (IR+'); and thus
IFa(v)l
~
Calvlm-l with Ca := C[\ulm + ID'Yulml for all v E CO'(IR+').
Here we used the fact that D'Yu E Hm(IR+') according to part (a). Since CO'(IR+') is dense in no-1(IR+'), (17) follows for a f a. There remains to investigate a =
Fa(v)
= a(u, v) -
a. We write
a(u, v) with a(u, v)
=~'
r aa.B (DCtu) (D.Bv ) dx,
1JRn+
where E' represents the summation over all pairs (a, P) (a, P) f (a, a) there exists a "Y with
hl=l,
O~"Y~P,
f
(a, a). For each
a+"Yf(O, ... ,O,m+l).
As above, one integrates each summand with
IPI = m by parts:
218
9 Regularity
for v E CQ'(IR+'), and estimates using C\V\m-l with C = C(u). Altogether one obtains \a(u,v)\ ~ C\V\m-l. Together with \a(u,v)\ = \(f,v)o\ ~ \f\-m+l\v\m-l, (17) also follows for a = a. (d) By induction for (k = 2, ... ) one proves in the same way \Fa(v)\ ~ Clvlm-k, and from this u E Hk+m(IR+'). For real s > 0, s -I- 1/2,···, m-1/2, one proves correspondingly \Fa(v) \ ~ Clvl m- B and hence u E HB+m(IR+'). • The generalisation of Theorem 11 to inhomogeneous boundary values reads as follows.
Theorem 9.1.14. Let the bilinear fonn a(·,·) from (5) be Hir(IR+.)-coercive. For an s > 0, s ¢ {1/2, ... , m - 1/2} either let (6) hold if s = k E 1N, or (12), if s ¢ 1N. Let u E Hm(IR+') be the weak solution of the inhomogeneous Dirichlet problem
= (f,v)o for all v E non(IR+'), alu/ an' = CPI on r = aIR+ for I = 0, 1, ... , m - 1, a(u,v)
(9.l.18a) (9.l.18b)
where f E H-m+B(IR+'),
CPI E Hm+B-I-l/2(r)
(0 ~ I ~ m - 1). (9.1.19)
Then u belongs to Hm+B(IR+') and satisfies the inequality (9.1.20)
PROOF. For m = 1 Theorem 6.2.32 guarantees the existence of Uo E Hm+B(IR+') which satisfies the boundary conditions (18b) (for m > 1 cf. Wloka [1, Theorem 8.8]). w := u - ua is the solution of the homogeneous problem a(w,v) = F(v) := (f,v)o - a(ua,v) (cf. Remark 7.3.2). Theorem 11 may also be carried over to the right-hand side F( v) under discussion here (instead of (f, v)o) and yields w E Hm+B(IR+'). • By similar means one proves
Theorem 9.1.15. (Natural boundary conditions) Let the bilinear form a(·,·) from (5) be Hl(IR+.)-coercive. For s > 0 either let (6) hold if s = k E 1N, or (12) if s ¢ 1N. Let u E Hl(IR+.) be the weak solution of the problem a(u, v)
= f(v):= f g(x)v(x) dx + f cp(x)v(x) dr for all v E Hl(IR+.),
lR+
lr
9.1 Solutions of the Boundary Value Problem in H"(a),
8
>m
219
where gEH
s-1
:=
{HS-l(IR~)
(H1-s(IR+.»)'
for s ~ I} E Hs-1/2(r). for s < 1 •
Then u belongs to H1+B(IR+.) and satisfies the estimate (9.1.21)
If a boundary value problem is in the form (1): Lu = g. Bu = 0 on r = 8IR+.. then according to Theorem 7.4.11 one can find an associated variational formulation and apply Theorem 15.
9.1.4 Regularity Theorems for General Jl c IRn The following theorems show that the above regularity statements also hold for Jl c IRn if Jl is bounded sufficiently smoothly. Theorem 9.1.16. Let Jl E C'+m for some t be Hrf(Jl)-coercive. Let 8 ~ 0 satisfy
~
O. Let the bilinear form (5)
s+I/2¢{1.2.· ..• m}; O:S:;s:S:;t. iftEIN; O:S:;s
e IN,
foralla,{3,'Y withl'Yl:S:;max(O.t+I{3I-m). ift
(9.1.22) aa{J E Ct+I~I-m(Jl) for 1{31 > m - 1. aa{J E LOO(Jl) otherwise. if t ¢ IN.
Then each weak solution u a(u, v)
=
e Hrf(Jl) of the problem
In
f(x)v(x) dx
for all v E H[)({l)
with f E H-m+s (Jl) belongs to Hm+s (Jl) n Hrf (Jl) and satisfies the estimate (9.1.23)
For inhomogeneous boundary conditions 8'u/8nl
=
with
E H m +s- I -
1/ 2 (r)
(I
= 0,···, m -
1)
instead of u E Hrf(Jl), the statement u E Hm(Jl) implies the statement u E Hm+s(Jl) and the estimate (20).
PROOF. (a) Let {Ui; i = O. 1, ... , N} with Ui c Jl be a covering of Jl as in Lemma 6.2.36. Let {Xi}, with Xi = tTl E COO(Jl) and SUPP(Xi) C Ui, be the associated partition of unity from Lemma 6.2.37. There exist derivatives a i E Ct(Ui) which map Ui (i ~ 1) into IR~ such that a i (8Ui n r) c 8IR~. By contrast. UO lies in the interior of Jl so that r n 8Uo = 0. The solution u
9 Regularity
220
can be written as Ei XiU' In part (b) of the proof we will treat XoU, and in part (c) XiU for i 2 1. (b) We set
<4(u,v):= a{Xiu,v) - a{u'Xiv)
(i
= 0, 1, ... , N)
and wish to show the estimate (u E Ho{ll), v E HO-8{ll), 82 1) for
8
= 1. Each summand of do{u,v)
Since Da{xou)
(9.1.24)
has the form
= XoDau + lower derivatives of u, one has
[(D a {Xou)){Dl3v ) - (Dau )(DI3{Xov)))
= L:Cr6 D')'uDli v ,),,6
with hi, 1815 m, hl+1815 2m-I. One integrates Ju o aa{3CyIiD')'uDlivdx with 161 = m by parts, and obtains a bound Clulmlvlm-l. whence (24) follows. The coefficients aal3 can be extended to n in such a way that the corresponding condition (22) is satisfied on n . Denote the resulting bilinear form by Cio{ u, v). Since xo E COO (ll) has a support supp(Xo) C UO, the extension of XOU through Xou(x) = 0 for x E m.n\UO poses no problems. We may formally define do(u, v) for v E Hm-8(m.n ) since only the restriction of v to UO is of any consequence. By (24) do(u,v) can be written, for a fixed u E Ho(ll), in the form
m.
m.
Xou is the weak solution of Cio(xou, v) = a(xou, v) = a(u, xov) + do(u, v) = (I, Xov)o + = (xof + do, V)L2(m.n ) for all v E Hm(m.n).
(do, v)o
Theorem 3 [resp. 8) proves xou E Hm+s(m.n) (thus too xou E Hm+s(ll» and IXoulm+s 5 CUxofl-m+s + ldol-m+s + Ixoulml 5 C'flfl-m+s + Cdlulm + Clulml 5 C"Ufl-m+8 where
8
is still restricted to
8
+ lulml,
(9.1.25a)
5 1.
(c) The same reasoning as for XiU (i = 1, ... , N) shows
a(Xiu,v)
= (xii + <4,v)o
for v E Ho(ll), Idi l-m+ 8 :S Cdlul m •
ill.+.
By assumption the maps a l : UI _ and their inverses (a' )-1 belong to ct+m{Ui) [resp. CHm{ai(Ui))]. Put u{x) := u(x) for x = ai(x), that is = u 0 (a i )-I. In a similar way define o'al3, X.,j, do. In
u
9.1 Solutions of the Bowtdary Value Problem in H·(G),
8
>m
221
one can replace the derivatives D~, D~ with derivatives with respect to the new coordinates, since t ~ 0, and thus obtain the form
which again is H{f(ai(Ui»-coercive and whose new coefficients o'afj satisfy the conditions corresponding to (22). As in (b), o'afj can be continued to 1Rt. :::> ai(Ui) such that the resulting bilinear form "iii(.,.) is H{f(1Rt.)-coercive. One can apply Theorem 11 to
ai(Xfu, v) = a. (XfU, v) = a(Xiu, v) = (Xt.! +~, v)o
= «xd + d.t)/1 det(ai)'I, V)P(R+)
for all
v E Hon( 1Ri-)
which yields XiU E Hm+8(1Rt.) and
IXiUlm+B ~ CsU!I-m+B + Id.tl-m+8 + IXiUlm]. Transforming back (cf. Theorems 6.2.17, 6.2.25g) yields
IXi Ulm+8 ~ Gs[Ifl-m+s + I~ l-m+8 + IXiulmj ~ Gs Ufl-m+8 + Cdlulm + Clulmj, and thus
IXiUlm+8 ~ G~ Ufl-m+s
+ lulm]
for 1 ~ i ~ N,
8
~ 1.
(9. 1. 25b)
(d) (25a,b) hold for 8 ~ 1. Since lulm+s = I:L: Xiulm+8 ~ :L: IXiUlm+8, estimate (23) has been proved for 8 ~ 1. If the conditions of the theorem alIowan 8 E (1,2j, one proves (23) as follows. Since u E Hm+l(.Q) has been proved already, one can estimate the forms d i (u, v) after further integration by parts via 1~(u,v)1 ~ CdlulmHlvlm-s. Accordingly, ~(u,v) = (~,v)o with ~ E H-m+s(.Q) and I~I-m+s ~ Cd lul m+l. If one inserts the above estimate (23) for 8 = 1, one obtains (25a,b) and hence also (23) for 1 < 8 ~ 2. Further induction yields (23) for admissible 8 E (k, k + 1]. (e) The case of inhomogeneous boundary values is treated as in Theorem 14 . Analogously one may prove
•
Theorem 9.1.11. (Natuml boundary conditions) Let.Q E Ct+l with t ~ o. If 0 ~ 8 ~ t E IN or 0 ~ 8 < t ¢. IN, 1Ri- in Theorem 15 can be replaced by n. Exercise 9.1.1S. Transfer Corollaries 5, 6, and 7 to the situation of Theorems 16 and 17. What conditions on .Q must be added?
222
9 Regularity
Corollary 9.1.19. Let n and the coefficients of a(·,·) satisfy the conditions in Theorem 16, resp. 17. An eigenfunction, i.e., a solutionu E V (V = Hlr(fl) in the case of Theorem 16, V = H1(fl) in the case of Theorem 17) of a( u, v)
=0
belongs to Hm+s(fl) for all 0
~
for all v E V, s
~
PROOF. Use Theorem 16 (17) for
t E 1N or 0
f
(9.1.26)
u '" 0 ~
s
< t ¢ 1N.
•
= 0 (and
According to Theorem 6.2.30 (Sobolev's lemma) one obtains sufficient conditions via CA:+~(fl) :::> HA:+~+n/2(fl) for u to be a classical solution from ck+~(n). The minimal conditions for u E CA:+~{n) result from a different theoretical approach, which essentially goes back to Schauder. The following theorem, for example, can be found in Miranda [1, §V]. It shows the Ck+~_ regularity of the operator Lin (5.1.1).
Let k ~ 2, 0 < A < 1. Let fl E Ck+~ be a bounded domain. Let the differential operntor L = E Q.jj{)2/{)x.{)Xj + E Q.j{)/{)Xi +a be uniformly elliptic in n (i.e., (5.1.3a) holds). Let Q.jj, as, a E Ck-2+~(fl) and f E Ck-2+~(n),
Theorem 9.1.20.
The condition n E CHm in Theorem 16 is stronger than necessary. For the Dirichlet problem the Lipschitz continuity of r is already sufficient to obtain the following result.
Theorem 9.1.21. (Netas [1]) Let n E 00. 1 be a bounded domain. Let the bilinear form (5) be Hlr(n)-coercive. Let the following hold: 1/2 ~ t
> s > o.
The coefficients aa{3 E LOO(n) must belong to ct(n) if 1,81 weak solution u E Hlr(n) of the problem a{u, v) =
In
f(x)v(x) dx
=
m. Then the
for all v E H{{'(n)
with f E H-m+s(n) belongs to H{{'+S(n) and satisfies the estimate (23).
The condition of Hlr(n)-coercivity can be replaced by that of uniform ellipticity (7.2.3) (cf. Theorems 7.2.11, 7.2.13). The statement of Theorem 21 cannot be extended to s ~ 1/2 since then u E H{{'+S(n) would contain another boundary condition.
9.1 Solutions of the Boundary Value Problem in H 8 (O), s
>m
223
The proof of Theorem 21 uses an isomorphism R = R* related to Rh,j (cf. proof of Theorem 8) between HO+8(il) and Hlf(il) and also between Hlf(il) and Ho - 8(n) such that the form b(u,v) := a(Ru,Rv) is Ho+S(il)-coercive. It is necessary to prove that btu, v). := a(u, R 2v) is also Ho+S(il)-coercive. We know f e H-m+s(il) implies f := R2f e H-m-s(il). Each solution of a(u, v) = (f, v)o is also a solution of a(u, R 2v) = btu, v) = (i, v)o = (f, R 2v)o so that u e HO+8(il) follows.
9.1.5 Regularity for Convex Domains and Domains with Corners
A domain n is convex if with x',x" e il, x' +t(x" -x') belongs to il for all
o :5 t :5 1 . Convex domains in particular belong to CO,l, but permit stronger regularity statements than Theorem 21.
Theorem 9.1.22. (Kadlec [1]) Let il be bounded and convex. Let the bilinear form (5) be Hlf(n)-coercive. Let the coefficients of the principal part be Lipschitz-continuous: aafj
e CO,l(G)
for allial
= 1.81 = 1;
for the remaining ones let the following hold: D"Yaa{3eLOO(il)
foralla,.8,"'(
with"'(:5I.B1, lal+I.8I<1.
Then every weak solution u
e HJ(il) of the problem
fa
f(x)v(x) dx for all v
a(u, v) with f
=
e HJ(il)
e L2(il) belongs to H2(n) n HJ(il) and satisfies the estimate (9.1.27)
The constant C 1 depends only on the diameter of il. In Section 8.4.2 H2-regularity was required. A generalisation of (27) in the form of Hm+1-regularity for the biharmonic differential equation with m = 2 is known for convex polygons (cf. Blum-Rannacher [1 D. For the Poisson equation the inequality (27) can be reformulated as follows.
Corollary 9.1.23. For the solution of the Poisson equation -L1u = f e L2(il) in a convex domain il with u = 0 on r, the following holds: (9.1.28)
As in Lemma 8.4.1 one shows that the left side of (28) is a norm of H2(n) n HJ(il) which is equivalent to I . 12. Thus, (27) follows. As a model
224
9 Regularity
for the proof of Theorem 22 we carry out the proof for Corollary 23 for the case il em? PROOF. (a) First let us assume that the convex domain is smooth: n E Coo. According to Theorem 16 -Llu = f E L2(il) has a solution u E H2(il) n HJ(il). We want to show
lIIulll~:=
L
IDau l5 ~ ILlul5 for all u E H2(il) n HJ(il).
(9.1.29)
lal=2
It suffices to prove (29) that for all u in the dense subset {u E COO (il): u = 0 on r} = COO(il) n HJ(il). Integration by parts yields
fa (u~2: + U~y +
faiLlu l2 dx dy =
2uz 2: Uyy) dx dy
fa (u~2: + U~y 2~2:yull) + £ = fa (U~1I7 + U~y + 2u~) + £ =
-
dx dy
dxdy
=
IIlulll~ + 2
£
(U2:l17ny -
2
U2:I17 Uyny dr
2
(uzzUyny - uzyUynz ) dxdy
U2:y~)Uy dxdy,
where n = (nll7' ny) is the normal vector. The tangent direction is given by t = (-ny, n ll7 ). Then u l 7l17 ny - ~ynz is the negative tangential derivative -(u",}t. Now U z and uy can be expressed in terms of Ut and Un. Since both u and Ut vanish on r, U II7 = n ll7Un and Uy = nyUn. Hence the boundary integral becomes
2
£
(UZll7 ny - u2:ynll7 )uydr
=-
= -2
£
(nzUn)tnyundr
£
[2u!ny(nll7 )t + nll7ny(u!)e]dr.
Integration by parts of the second summand yields
- t[2u!ny(nzh - (n2:ny)tu!]dr =
£
u![(ny)tnz - (nll7hnyjdr.
The bracketed expression in the last display is the curvature in x E r, which for a convex domain is always ~ O. (28) has thus been proved. (b) Every convex domain il can be approximated monotonically by convex il" E Coo: ill e il2 e ... e il, U" il" = il. We interpret V" := {u E HJ(n) : supp(u) C il,,} as Ritz-Galerkin space V" e HJ(il). Each u E CO>(il) lies in V,. for sufficiently large J.I.. With CO>(il), U" V" is therefore also a dense subset of HJ(il). For every 1/, the Ritz-Galerkin problem provides the solution u" E HJ(il,,) of -Llu" = f in il", u" = 0 on ail". In il\il", u" may be continued by u" = O. Theorem 8.2.2, which is also applicable in the case dim V" = 00, proves lu" - ult -+ 0, where u E HJ(il) is the solution of
9.1 Solutions of the Boundary Value Problem in H·({J),
8> m
225
-Llu = I in a. Theorem 9.1.26 will show that for every p the restriction of u on a,.. belongs to j{J(a,..). For each v E V,. C VI" V ~ p, we have
1 r uzzvdxdYI=1 Jn~
= 1 lim 1'''''''00
r
Jn~
UzVZdxdyl=IJtrrJ,l
~~~ n~
(uv)zvzdxdyl
r (uv)zzvdxdyI5 sup lI(uv)zzIIL2(n~)lIvIlL2(n~).
Jn~
v?,.
From this one infers Llal=2I1Daullh(n~) 5 supv?,,. Llal=2I1Dauvlli2(n~) 5 II/lIi2(n~) 5 II/lIi2(n) (cf. (29)), and obtains (28). • What role the H2-regularity plays for the HJ (a)-projection on a subspace Vh c HJ(a), is shown in Exercise 9.1.24. Let the subspace Vh c HJ(.o) satisfy inf{lu - vh:v E Vh} 5 Cohlul2
for all u E H2(.o) n HJ(.o)
(cf. (8.4.14)). Let the Poisson problem be H2-regular (according to Theorem 21 convexity of .0 is sufficient). Let Qv: HJ (.0) - Vh C HJ (.0) be the orthogonal projection on Vh with respect to I· h. Show that there exists a C 1 such that
lu - Qvulo 5 Cl hlu h for all u E HJ(.o). Hints: (1) With the Poisson problem the boundary value problem -Llu+u = I inn and u = on r are also j{J-regular (cf. Remark 2). The corresponding bilinear form a(·,·) is the scalar product in HJ(.o). (2) Qv agrees with the Ritz projection Sh for a(·, .). (3) Use Corollary 8.4.12. In connection with finite elements one often considers polygonal domains .o. Since polygons belong to CO,l, the Dirichlet problem, according to Theorem 21, is Hm+B-regular with 5 s < 1/2. If the polygon is convex (Le., if the inner angles are 5 11") then as in Theorem 22 one has H2-regularity (m = 1). One obtains results between H3/2 and H2 if the maximal inner angle of the polygon lies between 11" and 211" (cf. Schatz-Wahlbin [1]). If .0 has a reentrant corner the boundary value problem can no longer be H2-regular (cf. Example 2.1.4; with an inner angle a the solution belongs to H1+ S (.o) for S < 11"/0). Stronger regularity properties may be obtained, however, if special compatibility conditions are satisfied in the corners (cf. Kondrat'ev [1]).
°
°
Example 9.1.25. Let u be the solution of the Poisson equation -Llu = I in the rectangle {} = (0,1) x (0,1) with u = on r. Only for s < 3 does IE HB-2(.o) lead to u E HB(.o). Under the additional compatibility condition that I vanish at all corners, 1(0,0) = 1(0, 1) = 1(1,0) = 1(1,1) = 0, however, one can also conclude, for I E HB-2(.o) with 3 < s < 4, that u E HB({}).
°
226
9 Regularity
9.1.6 Regularity in the Interior Up to now all regularity statements have referred to the entire domain il. Instead one can also investigate the regularity of the solution u in ilo CC il. The following theorem shows that the answer depends neither on the smoothness of the boundary nor on the type of boundary condition.
Theorem 9.1.26. Let ilo CC ill C il and s ~ o. Let the bilinear form (5) be HO'(il)-coercive. For the coefficients assume condition (22) with il be replaced by ill and with t ~ 8 e IN or t > 8. Let u eVe Hm(il) be a weak solution of the problem a(u,v) = fa1vdx for all v e V, where the restriction fla1 belongs to Hm+8(ilo) belongs to H-m+8(ild. Then the restriction ofu to and satisfies
no
PROOF. A special covering of il is given by UO = ill. U l = il\ilo. Thus we obtain the assertion from Part b of the proof of Theorem 16. •
9.2 Regularity Properties of Difference Equations The convergence estimates for difference equations in Section 4.4 read, for example, lIu - uhlloo ~ Ch2I1ullc4(o) under the condition that u e C4(D) or u e ca·l(.'O). This regularity assumption is frequently not satisfied (cf. Examples 2.1.3-4). A comparison with the error estimate lu - uhlo ~ Ch21ul2 for the finite-element method suggests that similar estimates also exist for difference solutions. To obtain the latter, one needs to replace the stability estimate IILh"11l2 ~ C (or IILh"llloo ~ C), which corresponds to L-l e L(L2(il), L2(il», by stronger estimates which correspond to L -1 e L(H-l(il), HJ(il» or L-l e L(L2(il), H2(il».
9.2.1 Discrete Hl-Regularity For il C IRn the following grids are defined:
Qh:= {x e IRn:x,
= lIih,
IIi
e Z},
ilh:= ilnQh.
A grid function Vh defined on ilh is extended to Qh by Vh
Vh(X)
=0
for x
e Qh \ilh.
(9.2.1)
= 0: (9.2.2)
The Euclidean norm is now called the L~-norm: (9.2.3a)
9.2 Regularity Properties of Difference Equations
227
It comes from the scalar product
L: Vh(X)Wh(X).
(Vh,Wh)O:= (Vh,Wh)L~ := hn
(9.2.3b)
XEQh
The discrete analogue of HJ(n) is Hl with the norm IVh!t :=
IIVhIIH~ := [IVhl~ + ~ latvhl~] 1/2
where at is the forward difference in the
Xi
t
= 0 on Qh \nh)
(9.2.3c)
direction. The dual norm reads
The associated matrix norms IILhIlHl+-H-1 h h ILhit+-o, etc., are defined by ILhl i +-j := SUP{ILhVhldlvhlj: 0
(Vh
=
ILh!t+--b IILhIlHl+-L2 = h h
Vh satisfies (2)}
for i,j E {-I, 0, I}. (9.2.3d')
Exercise 9.2.1. Show that (a) ILhlo+-o is the spectral norm of Lh (cf. §4.3). (b) The following inverse estimates hold:
(9.2.3e) The matrix Lh yields the bilinear form ah(uh, Vh) := (LhUh, Vh)L2. h
(9.2.4a)
ah(-, .) is said to be Hl-elliptic if aCE> 0 exists such that ah(uh, Uh)
2 CEluhl~ for all Uh and all h > O.
(9.2.4b)
Correspondingly, ah(-,') is said to be Hl-coercive if there exist CE > 0 and C K E lR with ah(Uh, Uh)
2 CEluhl~ - CKluhl~ for all Uh and all h > O.
(9.2.4c)
As defined in Sect. 4.5, Lh (resp. ah(-,'» is said to be L~-stable if
IL;llo+-o ::; Co
for all h > O.
(9.2.4d)
We call Lh Hl-regular if IL;111+-_1 ::; C 1 for all h
> O.
Exercise 9.2.2. (a) Hl-regularity implies L~-stability. (b) IAhl i +-j = IAII-j+--i for i,j E {-I,O, I}. (c) If Lh is Hl-regular, then so is LI.
(9:2.4e)
228
9 Regularity
(d) If Lh and LI are stable with respect to 1·100, Le., IILhlloo :::; Goo, IILIlloo :::; Gl, then L~-stability (4d) follows with Go := (G1Goo )1/2. (e) Hl-ellipticity implies Hl-regularity. The following statement resembles the alternative in Theorem 6.5.15. Theorem 9.2.3. II ah(-,·) is Hl-coercive and il Lh. is L~ -stable, then Lh. is also Hl-regular. PROOF. (a) Let LIuh Coercivity provides
= III. such that ah(Uh.,uh) = (uh,LIuh)O = (uh,lh)O.
IUh.l~ :::; [all. (Uh' Uh) + Gluh.l~lIGE
= [(Uh, A)o + Gluhl~lIGE
:::; G'lIlh.lo + Gluh.lollUh.lo. On the basis of the stability estimate IUh.lo :::; GolAlo we obtain IUhl~ :::; GffiAla. From this one infers ILr- 1h+-o :::; ,fC" := G* and hence IL;;-110+-_1 :::; C* (cf. Exercise 2b). (b) Now let LhUh = III.. According to Part a, one has IUhlo :::; G*llhl-1. By estimating lah(uh,uh)1 = l(fh,Uh)ol :::; IAI-lluhl1 through !GEluhl~ + !GElllhl~lt one obtains the coercivity
•
'Instead of the L~-stability in Theorem 3 one can also assume the solvability of the continuous problem and a consistency condition (cf. Corollary 11.3.5).
By analogy with Lemma 7.2.12 the following may be proved. Exercise 9.2.4. If Lh is Hl-coercive, and if 6Lh := ao + Li bia; + Li a;<;, with laol, Ib,l, 1<;1 :::; const, contains at most first differences, then Lh + 6Lh. is also Hl-coercive. According to Exercise 4 it is sufficient to investigate the principal part of a difference operator as to its coercivity. Example 9.2.5. Let n c m,2 be bounded. On nil. let Lh. be given by the difference method (5.1.18) (with a = 0):
(LhUh)(X, y) = a;; all
(x + ~,y )a;Uh(X,
y)
+ a;; a22
(x, Y + ~) a:Uh(X,
y)
(9.2.5)
for (x, y) E nil., where uh.(x, y) = 0 for (x, y) E Qh. \nh.' Here, let -au. -a22 ~ to > 0 in n. Then ah(·.·) is Hl-elliptic and Lh is Hl-regular.
9.2 Regularity Properties of Difference Equations
229
PROOF. For arbitrary Vh, Wh defined on Qh the following rules of summation by parts hold: (vh,8:'Wh)0 = -(8;'Vh,Wh)0,
(vh,8;'Wh)0 = -(8:'Vh,Wh)0.
(9.2.6)
Hence it follows that ah(uh,uh)
= (Lhuh, Uh)O
=-
(all (-
+~, -) 8:-Uh, 8:-Uh) 0'- (a22 (-.. +~) 8tUh ,8tUh ) 0
~ E[l8:-Uhl~ + 18tuhl~1. As in Lemma 6.2.11 one proves that for bounded fi the norms I· It and IUhh.o := [l8tuhl5 + 8:uhl~F/2 are equivalent (uniformly with respect to h) so that (4b) follows. The H~-regularity results from Exercise 2e. • Exercise 9.2.6. Let fi
c m.2 be bounded. Let the equation
(all uz)z + (a12Uy)z
+ (a12 u z)1/ + (a22Uy)1/ = f
be given on fih by the difference stars (5.1.18/19), where Uh = 0 in Qh \fih . Let the differential equation be uniformly elliptic in n:llii < 0, aUa22 - a12a21 ~ -E(au + a22) > O. Further, let aij E CO(n) hold. Show that for sufficiently small h the associated matrix Lh is H~-regular; for all h > 0, Lh is H~-coercive. Hint: For sufficiently small h the following holds: -au (x
+~, y )d~ - a22 ( x,y + ~)d~ -
[a12(x+~,Y+h) +a12(x+~,Y)]dld2
~ ~(4 + 4), for all dt, d2 E JR. The H~-coercive difference methods constructed so far remain H~-coercive if differences of lower order are added (cf. Exercise 4) or if the principal term (auuz)z + '" is replaced by allu zz + .,. with all E Cl(n). The above difference methods are described by the same difference operator regardless whether the grid points are close to or far from the boundary. The homogeneous Dirichlet boundary condition is discretised by (2): "Uh = 0 on Qh \ fih " . If one wants to approximate the boundary condition more accurately, one needs to select special discretisations in the points near the boundary of fih (cf. Section 4.8.1,4.8.2). One thus obtains an irregularity which makes a proof of H~-regularity, as in Example 5, difficult. We begin with the one-dimensional case. Lemma 9.2.7. Let Lh be the matrix of the one-dimensional Shortley- Weller discretisations of -u" = f on fi, U = 0 on 8fi:
230
9 Regularity
PROOF. (a) First, let n be assumed to be connected. Let the grid points of nh be x, := Xo + lh E n for l = 0, ... ,k > o. Let the boundary points of n be
Xo - s',oh and Xk + Sr,kh with s',o, Sr,k E (0,1). The other factors of Equation (7) in x = Xi are S',i = sr,; = 1. If one takes into account Vh = 0 on ffi\n one obtains the following identity: k
(Vh, LhVh)O := h
L vh(x,)(LhVh)(X,) '=0
= 18+Vhl~ +
+ Vh(Xk)
h-
1{
Vh(XO) [ (~ -
[(~ -
2) Vh(Xk)
+
2) Vh(XO) +
(1 -1~ S,) Vh(Xl)]
(1 -1~ sr) Vh(Xk-l)]} ,
(9.2.8a)
where s, := s',o and Sr := Sr,k. Since Vh(XO) = M+Vh(X-l) and Vh(Xl) = h[8+Vh(X-l) + 8+Vh(XO)) because of Vh(X-t} = 0, the first summand inside the braces can be written as
(9.2.8b)
For a := 8+Vh(X-tl, f3 := 8+Vh(XO) use the inequality -af3 ~ _ja 2 - Af32 with j = (s, + 2)/ s, and note that the function s(1 - S)/[4(1 + S)(2 + s)] is bounded by 0.018 (maximum at S = (y'3 -1)/2). The second summand inside the braces is treated analogously and yields
(Vh' LhVh)O ~ 18+Vhl~
-
0.018h[8+vh(xo)2 + a+Vh(Xk_d 2).
(b) In Part (a) it was assumed that k
(Vh' LhVh)O =
(9.2.8c)
> o. For k = 0 one obtains
h-l~Vh(XO)2 = ~[8+Vh(xo)2 + a+Vh(X_t}2 ~ la+Vhl~, S'Sr
S'Sr
so that (8c) also holds. (c) Let n be arbitrary. Let the components of connection be Ii = (ai, b;) (i E 7l) with bi ::; a;+1. Let Lr) be the discretisation matrix associated with h Each Vh with support in nh = Qh n n can be written as Li vii) where
9.2 Regularity Properties of Difference Equations
Vr) nk
= 0 outside Ii. Let x&i) and
i ) :=
231
X~(i) be the first and last grid points of
nil. n Ii. The statement follows from 18+Vhl~ ~ L{\8+v~)I~ + 1h8+vii)(x~(i)]2} = 2L 18+v~)I~ i
i
• n
c m.2 be bounded. Let Lh be the matrix associated with the Shortley- Weller discretisation of the Poisson equation (cf Section 4.8.1). Then Lh is H~-regular. Theorem 9.2.8. Let
PROOF. Let L~ ILK] be the portion of the x differences [y differences] such that Lh = L~ + LK. The restriction of L~ to a "grid row" {(I/h, y) E n: 1/ E Z} (y is fixed) corresponds to the matrix Lh in Lemma 7. Thus one obtains (VII., L~Vh)O ~ il8tvhl5. With the analogous inequality (VII., LKvh)o ~ !1 8t v h15 one obtains 2(Vh' LhVh)O ~ 18tvhl5 + 18tvhl5 = IVhl~ - IVhl5 ~ Enlvhl~, thus IIL;;-1Ih<-_1 ~ 1/(2En).
n
c m? be bounded. Let the Poisson equation be discretised with the aid of the five-point formula (4.8.14c) at points far from the boundary and by interpolation (4.8.16) at points near the boundary. The associated matrix is H~-regular.
Theorem 9.2.9. Let
PROOF. The corresponding, one-dimensional formulae, except for the scaling factor Sr + Sf ~ 2, agree with (7) so that the proof of Theorem 8 can easily be carried over. Exercise 9.2.10. Show that for a convex domain n c m.2 the matrix Lh of the Shortley-Weller discretisation satisfies the inequality
t vhlfi)·
8 (VII., LhVh)O ~ O.982(18:tVhlfi + 1
Theorems 8 and 9 can be strengthened in the following way: Corollary 9.2.11. Let Lh be as in Theorem 8 or9. Let Dh nil.} be a diagonal matrix with d(x) := min{2sfsr, 2s o su , I}
= diag {d(x):x E
(x E nh ),
where Sl, Sn So, Su come from (4.8.7), respectively (4.8.16). For points far from the boundary, x E nil., evidently d(x) = 1 holds. The matrix L h := DhLh,
232
9 Regularity
which belongs to the re-scaled system of equations Hl-regular: ILh -111+-_1 S C.
LhUh =
f~ :=
Dh/h, is also
PROOF. In the case of the Shortley-Weller method one needs to combine the correction terms (8b) for the x and y directions. The result is that they remain 2= -~[latvhl~ + latvhl~] so that (Vh, Lhvh)O 2= ~[latvhl~ + latvhl~ 12= flvhl~ with f > O. One obtains an analogous estimate for the difference method in Section 4.8.2. • 9.2.2 Consistency
In the following we carry over the estimate lu h - 1.£11 S Chlu\2 S C'hill o, which holds for finite-element solutions, to difference methods. To this end one needs to prove the consistency condition
ILhRh - Rh L I-l+-2 := IILhRh - RhLIIHh"1+-H2(0) S CKh
(9.2.9)
for suitable restrictions
Rh: H2(il)
n HJ(il) --+ H~, R h: L2(il) --+ L~.
To construct the restrictions we first continue 1.£ E H2(il) in ii := EJu E H2(IR?). According to Theorem 6.2.40c one assumes (9.2.lOa) for all 1.£ E with
H2(il) n HJ(il). An analogous continuation Eo: L2(il) --+ L2(IR?)
7 := Eol =
I
on il,
IIEoIIIL2(JR2) S CIIIII£2(I1)
is given, for example, bl, 7 = 0 on IR?\il. Uh' uK: Ctf(JR.2 ) --+ Co(JR. ) be defined by
(UhU)(X, y) = h- 1 f
h/ 2
u(x+€, y) de,
for all IE L2(il) (9.2.lOb) Let the averaging operators
(uKu)(x, y) = h- 1
-~
fh/2 -~
u(x, Y+17) d17· (9.2.11)
The restrictions Rh, Rh are chosen as follows:
Rh := uhu~E2'
Le., (RhU)(X, y)
= h- 2 f
with u = EJu for (x, y) E ilh, Rh := (UhU~)2Eo, Le., (Rhf)(x,y)
h/2 fh/2 fh/2 fh/2
h/2 fh/2 -h/2 -h/2
u(x +~, y + 17) de d17 (9.2.12a)
7(x + €+ ~', y + 17 + 17') de de' d17 d17' -h/2 -h/2 -h/2 -h/2 with I = Eo! for (x, y) E ilh . (9.2.12b)
= h- 4 f
9.2 Regularity Properties of Difference Equations
233
The characteristic properties of the convolutions u~,u~ are the subject of Exercise 9.2.12. Let 8:IJ be the symmetric difference operator (8:IJu) (x, y) := [u(x + hj2, y) - u(x - hj2, y)]jhj 811 is defined analogously. Show that fl:1I_1I:IJ 8_8:IJ_:lJ8 8_811_118 () a UhUh - UhUh' U:IJ - 7fi Uh - Uh7fi' u1I - 8ijUh - Uh8ij' (b)lIu~IIH"(R2)""'H"(1R2) ~ C, lIu~IIH"(R2)+-H"(1R2) ~ C (in particular for k = O,±1,2). (c) lIu~u~vIlL~ := [EXEOh l(u~u~v)(x)12P/2 ~ IIvll£2(R2) for all v E L2(:m?). (d) For a E ao,l(:m?) holds lIauhu~ - u~uKaIlL~+-L2(R2) ~ Chllallco.l(1R2). Here (au~u)(x) = (a(x)uhu)(x) and (uhau)(x) = (uh(au»(x). (e) lIa(u~)V(Uh)1' - (Uh)V(uh)l'aIlL~+-£2(1R2) ~ CVl'hllallco.l(1R2) for v, JL E IN. (f) 111.1. - (uht(Uh)l'uIlH"(R2) ~ CVl'hliuIlH"+1(1R2) for 1.1. E Hk+l(:m?). Consider the differential operator n
L
=
n
L a.j(x){PjaxiaXj + La.(X)ajaxi + a(x). i,j=l i=l
(9.2.13a)
First we assume that n = :m.n and discretise L with the regular difference operator (9.2.13b) Lh = La.j(x)a!a;' + a. (x)a! + a(x) with an arbitrary combination of the ±-signs. Lemma 9.2.13. Let n = :m.n . Let a.j,ai,a E ao· 1(:m.n ). Let Land Lh be given by (13a,b). Then the consistency estimate (9) holds. PROOF. To simplify the notation let us assume that n Rhauu:IJ:IJ
= 2. Let
= (uhu~)2allu:IJ:IJ = all(x)(uhU~)2u:IJ:IJ - 01
with IlolllL~ ~ Chllal1l1co.l(1R2)luI2 (cf. Exercise 12e,b). Let the term corresponding to allUz:IJ in (13b) be, for example, all(Xj h)a,tat. Exercise 12a shows that
with 02(X, y) := -a,tu~uX[u(x - h, y) - u(x, y)] = -UhU~U~[U:IJ(x - h, y) U:IJ(x, y)] and 1102112 ~ lIuK[U:IJ(X - h, y) -U:IJ(X, y)]lIL2(1R2) ~ hluI2 (cf. Exercise 12c,b). Since 8:IJ8:IJ = a:a;;, the error term 02 does not appear if one also approximates a11U:lJfI: by a11(X)ata;;. Finally one obtains au a; a; u~ u~ 1.1. = all at a; Rh 1.1. + all a; a; [U~ u~ - u~ uK)u = a11 a;a; Rhu + alla;UhU~[uK - uh]u:IJ = alla;a; Rhu - all a;03
234
9 Regularity
with 115s1lL~ = II(}"~(}"~[(}"~ - (}"~]U:eIlL~ ~ II[(}"~ - (}"~]uxlIL2(R2) ~ Chlul2 (cf. Exercise 12c,f). Putting this altogether one obtains
82 ] _ [an 8x± 8x± Rh - Rhau 8x2 u
= 51 + au8x± (52 + 5s).
For the first error term the following holds: (9.2.14a) For arbitrary Vh E H~ we have (Vh, a118;(~ + 5s»L2h
= -(8; [auvh], ~ + 5sh2. h
co· 1 (m?)
(9.2.14b)
it follows that 118; [avhlllL~ ~ lIallco'1(R2)lIvhIlL~ lIallco(R2)lIvhIlH~ ~ CllvhIlH~' so that
From a E
lIa118;(~
+ 5s)II H - = 1
h
+
sup I(Vh, au8;(52 + 6s»L21
tlhEH~
h
(9.2.14c)
~ CII 62 + 5s1lL~ ~ C'hluI2'
From (14a,c) we have
± ±
lIa ll8x 8z Rh -
_ 82 Rhall8x2I1Hh"l+-H2(R2) ~ Ch.
(9.2.14d)
Analogously, one shows
± ± lIa;j8x,8xjRh - Rha;j 8 2 /8xi 8xjIlHh"l+-H2(JR2) ~ Ch.
(9.2.14e)
For a;8;'Rh -Rha;8/8xi similar reasoning results in an O(h)-estimate for the norms II . IIHh"l+-Hl (R2) and II ·IIL~+-H2(R2). Both are upper bounds for the larger norm II . IIHh"l +-H2(m,2) such that (9.2.14f) Likewise lI aRh -
RhaII Hh"1+-H2(R2) ~ Ch.
Statement (9) follows from (14e,f,g).
(9.2.14g)
•
When generalising the consistency estimate to more general domains n c m?, the following difficulty arises. The entries of the matrix Lh according to (4.8.7) or (14.8.16) are not bounded by Ch- 2 • Rather, at points near the boundary the inverse of the distance to the boundary point enters, and this distance may be arbitrarily small. One way around this, would be to formulate the discretisation so that the distances between boundary points and points near near the boundary remain, for example, ~ h/2. A second possibility would be a suitable definition of Rh so that the product LhRh appearing in
9.2 Regularity Properties of Difference Equations
235
(9) can be estimated (cf. Hackbusch [2]). Here we choose a third option: Lh is replaced by the re-scaled matrix Lh = DhLh from Corollary 11.
Theorem 9.2.14. Let n E C 2 (or convex) and bounded. For the discretisation of Lu = f for L = -.6 on il with u = 0 on r use the discretisation LhUh according to (4.8.7) or (4.8.16). Let Lh = DhLh be defined as in Corollary 11. Then the consistency estimate (9.2.15)
holds. Here, the matrix Lh from (4.8.7/16) is only taken as an example. The proof will show that the estimate (9), respectively (15), also holds for other Lh if (LhUh)(X), x near the boundary, represents a second difference. First, two lemmas are needed. Let 'Yh C nh be the set of points near the boundary. If Vh is a grid function defined on ilh then we denote by Vh h" the function
(vhh,,)(x) = Vh(X)
for x E 'Yh,
(vhh,,)(x) = 0 for x
E
ilh\'Yh'
Lemma 9.2.15. Let il E CO· l be bounded. Then there exists a C = C(n) independent of h, such that (9.2.16) PROOF. From il E CO· l follows: there exist numbers K E IN and ho > 0 such that for all x E 'Yh, with h ~ ho, not all grid points {x+(vh,JLh): -K ~ v, JL ~ K} lie in n. For each x E ilh select a pair (vo, Po) E 71. 2 with -K ~ Vo, JLo ~ K and x + (voh, JLoh) ¢ n. At this grid point x the functions w~'" (-K ~ v, JL ~ K) are defined by w~O"'O(x) := Vh(X), w~"'(x) := 0 for (v,JL) =1= (VO,JLo). Therefore E~"'=-K w~JA = vhh" is a decomposition with w~"'(x) = Vh(X) or w~JA(x) = o. Without loss of generality let us assume that v > 0 and JL > O. For x E nh we define the chain
xO = X, xl = X + (h, 0), ... ,
XV
=
X + (vh, 0),
x+ (vh,h) , '" , x +JA = x+ (vh,JLh). definition, either w~"'(x) = 0 holds or x v +'" ¢ n,
xv+! =
V
According to the = 0 (cf. (2». In both cases we have
w~"'(xv+"')
\wh"'(x)\ ~ \wh"'(Xo) - whJA(xv+",\
= h 1~[w~"'(xO) ~ h[\a;w~"'(xl)\
w~"'(Xl») + ~[w~"'(xl) - w~"'(x2») + ... 1 + \a';w~"'(x2)\ + ... + \a';Wh(XV )\
+ \a;wh"'(XV +1)\ + ... + \a;WhJJ(Xv+"')\J
i.e.,
236
9 Regularity
and thus Iw~l'lo ~ hUvl18tw~l'lo + IJLII~w~l'lo] ~ 2Khlw~l'h. Summation over v, JL yields estimate (16) for Vh l1'h = 2.., w~1' • • In most cases x E 'Yh already has a direct neighbour in JR2\.!2 of x E 'Yh such that (16) follows with C = ..;2. This holds in particular for convex domains. Lemma 9.2.16. Let Rh be defined by (12a). Let have I(Rh1£)(~,)1 ~ ChIlE21£IlH2(Kh/2(m
where K h / 2 (e):=
for all
~
satisfy (lOa). Then we
eE r,1£ E HJ(.!2) n H2(.!2), (9.2.17)
{e +x E JR2:x E (-~,~) x (-~, ~)}.
PROOF. (a) Let Q = (-!,!) x
Iv(O) -
(-!, i).
~ v(x)dxl ~ C
For v E Jl2(Q) one shows
k(v~:l) + 2v~1I + v~lI)dxdy,
since the leftrhand side vanishes for linear functions 1£(x, y) = a + fix + 'YY. The proof is similar to the one for (8.4.2). Transforming from Q to Qh := h 2h) x ( -2' h 2h) gIves . ( -2' (9.2.18) (b) Let eE
r. Statement (17) follows from (18) with v(x) := u(e + hx) =
(~1£)(e
+ hx),
since 1£(e) =
o.
•
PROOF. (of Theorem 14) (a) Inequality (15) is proved if
I(vh' [L~Rh - DhRhL]1£)ol ~ Ch for all Vh E H~ and 1£ E H2(ll) n HJ(ll) with is split into
IVhh = 11£12 =
(9.2.19a) 1. To this end Vh
In part (b) we show I(v~, [LhRh - DhRhL]1£)ol ~ C1h,
(9.2.19b)
The other steps of the proof, (c) and (d), yield
I(v", [LhRh - DhRhLJ1£)ol ~ C2 h ,
(9.2.19c)
such that (19a) with C = C 1 + C2 follows. (b) Lenuna 15 shows (9.2.19d)
9.2 Regularity Properties of Difference Equations
For v~ one obtains Iv~h ~ IVhh + Iv~h yields Iv~h ~ Ch-llv~lo ~ CCs, thus
237
= 1 + Iv~h. The inverse estimate (3e)
Iv~h ~ C4 •
(9.2.1ge)
Let Lh be the (regular) difference operator on the infinite grid
Qh = {(lIh, p,h): II, p, e 7l}. Since the support of v~ is ilh\'Yh, (V~,L~Wh)O = (v~, LhWh)O holds for all Wh· Furthermore, (v~, DhWh)O = (vK, Wh)O. This proves the first equality in
l(vK, [L~Rh - DhRhL]U)ol
l(vK, [LhRh -
RhL]U)ol ~ C5hlv~hlul2 ~ C5 hC4 C6 := C1h,
=
where u := E2U is the continuation of u to JR2. Further inequalities result from Lemma 13, (1ge), and (9.2.19f)
(cf. (lOa)). Theorem 6.2.40c guarantees the existence of an extension u with (19f) if n e C2. Another sufficient condition for (19f) is the convexity of il. (c) The left side of (19c) splits into (v~, L~RhU)O and (v~, DhRhLu)O. The first part is estimated in part (d). Exercise 12c and (19d) yield the second term I(v~, DhRhLu)ol ~ Iv~lolDhRhLulo ~ C3 hC'ILuio ~
C6hlui2 = C6 h.
(9.2.19g) (d) We set Wh := (L~RhU)hh. Since the support ofv~ is contained in 'Yh, we have (v~,L~RhU)O = (V~,Wh)O. L~ contains differences with respect to the x and y directions. Accordingly we write Wh = w h + w~. In the following we limit ourselves (i) to the Shortley-Weller discretisation, (ti) to the term wh and (ii) to the case that
xe'Yh,
xr=x+(8rh,O)enh (Le.,8r =1),
x':=x-(8,h,O)er.
The other cases should be treated analogously. We set
u := RhU = (7h(7~E2U e H2(JR2). The Shortley-Weller difference in the x direction reads
where dx is the diagonal element of Dh. Since in the equation L~ Uh = fh the variables Uh(e), E r (for example, = x'), have already been eliminated, w~ (x) has the form
e
e
:J)() = wh A:J)() 2dx A( ') X + h2 ( 8, 8, + 8 ) U x .
wh X
r
238
9 Regularity
The factor 2dx /[h 2s,(s, + sr)), due to the definition of D h , remains bounded by 4h- 2. From Lemma 16 and K h/ 2(X') C K 3h / 2(X) one infers IWk(X) - wh(x)1 ~ 4h- 2 Iu(x')I ~ Ch- 1 IluIlH2(K3h/2(X»'
The second divided difference Wh in x
wh(X,y)
= dx
(9.2.19h)
= (x, y) can be represented by
1:,.
g(t)UX 3:(x+t,y)dt
with
for 0 ~ t ~ h, 2(t - h)/(h2 + h 2s,), -2(s,h + t)/(h2s,(1 + s,)), for -s,h ~ t ~ 0, o otherwise. From this one infers g(t)
IWh(X, y)1
={
~ dx [l h
g2(t)
-hili
~ 2h- 1
[1: U~3:(X +
dt] 1/2 [l
h
u!:.,(x + t, y) dt] 1/2
-hili
t,y)
dt f/ ~ Ch- I1ii 2
1
(9.2.19i) IlH2(K,h/2(X»
(cf. (6.2.5a)). From (19h,i) and the corresponding estimate for w~ one obtains IWh(X)1 ~ C7h-1I1iiIlW(Kah/2(X»' such that
IWhl~ = h2
L
L
IWh(X)1 2 ~ C? lliill~2(K'h/2(X» ~ 9C?IIiiIl~2(R2) XE"Yh XE"Yh ~ (3C7 C6 )2 := Cl·
From this follows I(Vh,L~RhU)ol ~ l(vh,Wh)ol ~ IVhlolwhlo ~ CahCs:= Cgh.
(9.2.19j)
•
(19g) and (19j) yield the required inequality (19c).
Remark 9.2.17. The proof steps for Theorem 14 can be carried out in the same manner for more general difference equations (for example, with variable coefficients, as in Lemma 13).
9.2.3 Optimal Error Estimates In the following we compare the discrete solution Uh = LJ;1 fh with the restriction uh := RhU of the exact solution U = L- 1f. From the representation
Uh - Uh
= LJ;1fh - RhU = LJ; 1Uh - Rhf) + LJ; 1R hf = LJ; 1(fh - Rhf) - LJ; 1 (LhRh - RhL)u
one immediately obtains
- Rhu
(9.2.20)
9.2 Regularity Properties of Difference Equations
239
Theorem 9.2.18. Let u E H2(n) hold for the solution of Lu = f. Let the right-hand side fh of the discrete equation L",u", = f", be chosen so that (9.2.21) If, furthermore, L", is H~-regular, and if the consistency condition (9) holds, then u'" satisfies the error estimate
(9.2.22)
Corollary 9.2.19. (a) Inequality (21) is satisfied in particular if one chooses
in:= R",/. (b) The choice fh(x) := I(x) for x E il", (cf (4.2.6b» leads to (21) if I E CO,1(il) or IE H2(n). In these cases the following even holds: If", - R",/lo $; Cllf", - R"'flloo $; C'hllfll c o.l(ii»
resp.
Ifh -
-
Rh/lo $; Ch
2
1/12.
(9.2.23)
(c) In Theorem 18 one can replace the H~-regularity of Lh by that of Lh DhL", (cf Corollary 11), (9) by (15), and (21) by
=
(9.2.21') The proof of (23) is based on the inequality (18).
•
Error estimates of order O(h2) can be derived in the same way if one has consistency conditions of second order. These are, for example, (24a) or (24b): (9.2.24a) (9.2.24b)
Remark 9.2.20. If nil. = Qh (i.e., n = IR.2) or n = (x',x") X (y',y"), the inequalities (24a,b) can be shown in a way similar to Lemma 13. Example 9.2.21. The difference method in Example 4.5.8 shows quadratic convergence. This case can be analysed as follows. Using Remark 20 one shows (24a). In the following section we prove the H~-regularity IL;;112<---0 $; C which for the symmetric matrix under discussion, L h , is equivalent with IL;;110+-_2 $; C (d. (6.3.3». Expression (20) leads to (9.2.25a) The corresponding estimate
240
9 Regularity
(9.2.25b) which is based on (24b), fails due to the fact that the solution in Example 4.5.8 does not belong to H3(n) (cf. Example 9.1.25). The verification of (24b) in the presence of irregular discretisations at the boundary becomes more complicated.
9.2.4 H~-Regularity Under suitable conditions on n, Lu = I E L2(n) has a solution U E H2(n) n HJ(n): luI2 :5 CIllo (cf., for example, Theorem 9.1.22). For the discrete s0lution of LhUh = Ih the corresponding question arises: does IUhl2 :5 CIAlo hold? First one has to define the norm 1·12 of H~. If n = m.2 or if the boundary r coincides with grid lines, one can define
In general, however, one has to use irregular differences at the boundary. At points far from the boundary x E nh \'rh, let DzzUh(X) = ata;Uh(X). At a grid point near the boundary x = (x,y) E 'rh, for example, with x' := (x - 8Ih,y) E r, xr:= (x+h,y) E nh it appears practical to use the divided difference (4.8.5). In general, however, this would violate the inverse estimate IUhl2 :5 Ch-1Iuhh, since 81 may become arbitrarily small. Therefore one defines DzzUh(X) with the aid ofthe grid points (x - 8Ih,y), (x + h,y), (x+ 2h,y): D
2 [Uh(X + 2h, y) - Uh(X + h, y) ( ) ._ zzUh x, Y .- (2 + 81)h h
_ Uh(X + h, y) - Uh(X - 81h, y)] sih '
where Uh(X - slh, y) = o. If, in addition, r passes between (x + h,y) and (x + 2h,y), one needs to replace x + 2h by x + h + srh. If (x + h, y) also no longer belongs to n, set DzzUh(X) := h-2uh(X). The second difference D'II1I is defined accordingly. The mixed difference in the interior reads DZ1l := Near the boundary several choices for DZ1l are possible. The definition of I . 12 reads
a; a;.
Exercise 9.2.22. Prove the inverse estimate
1·12 :5 Ch- 1 1· h.
We call Lh Hl-regular if IL;;-112+-0 :5 C. An equivalent formulation is IUhl2 :5 OIAlo for Uh = L;;-1Ih, Ih E L~.
9.2 Regularity Properties of Difference Equations
241
Exercise 9.2.28. Let Lh be H~-regular, let Lh + 6Lh be Hl-regular and 16Lhl0-1 ~ C for all h. Show that Lh + 6Lh is also H~-regular. Hint: Remark 9.1.2. For the proof of Hl-regularity one can -in analogy to the proof of Corollary 9.1.23- partially sum the scalar product (LhUh, LhUh)O in order to show the equivalence of Ifhl~ = ILhUhl~ and IUhl~. This technique, however, can only be applied to a rectangle fl. Here we use a simpler proof which is also applicable to general domains. Let Ph: L~ - L2(fl) be the following piecewise constant interpolation: (PhUh)(X) := {:hUh(X)
!~ ~ ¢~:
(9.2.26a)
AUh(X,y) := Uh(X',y'), if (x',y') E Qh and
x' - h/2 < x
~ x'
+ h/2,
y' - h/2
(9.2.26b)
< y ~ y' + h/2,
where Uh(X) = 0 for x E Qh\nh (cf. (2».
Lemma 9.2.24. Let fl E CO,I. Then we have IRhPh - 11-1H) ~ Ch,
(9.2.27&)
IPhlo+-o ~ C,
(9.2.27b)
IRh12+-2 ~ C.
(9.2.27c)
PROOF. (a) The estimate (27a) is equivalent to IP':Rh - /10-1 ~ Ch (cf. (6.3.3». Thus we must show IWhio ~ Chluhh for Wh := (P':Rh - l)uh. From the following Exercise 25 we obtain the expression Wh(X) = (q:q~[q:q~ - I]PhUh)(X) for Wh = (P':Rh - l)uh'
x E Qh.
Exercise 12c shows IWhio ~ IwIL2(R2) for w:= (q:q~ -1)Auh. For every
eE (0,1) x (0,1) define the grid function Wh,( E L~ with Wh,(X) := w(x+eh)
for x E Qh. One may check that Wh,(X, y) is a weighted sum of first differences Uh(X, y) - Uh(X', y') where (x', y') E {(x ± h, y), (x, y ± h), (x ± h, y ± h)}. Thus we have IWh,(lo ~ hluhh for all from which one infers
Iwl~ =
1.
R2
e
Iw(x)12 dx = h 2 =
f
J(O, 1)2
f
L
J(O,l)2 xEQ"
Iw(x + eh)1 2 £Ie
IWh,d~£Ie ~ h2luhl~,
thus IWhio ~ Iwlo ~ hluhh. From Wh -Wh
= (Ph -Ph)Rhuh = (Ph -Ph)qhq~PhUh = (Ph -Ph)q:q~Ph(Uhh~)
9 Regularity
242
with i'h
= {x E ilh: K 3h / 2(X) n r =f 0} one infers
IWh - whlo ~ l(Ph -Ph)u~u~lo<--oIPhlo<--oluhhhlo $ CIUhhhlo $ C'hluh\1 (cf. part b of the proof and Lemma 15, in which "th may be replaced by i'h). Together with IWhio $ hluh\1 one obtains IWhio $ (C' + l)hluh\l. (b) (27b) with C = 1 follows from IPhUhlo = IUhlo. (c) Second differences of u = RhU have already been estimated in (19i). (19i) • implies IUhl2 ~ Clul2, i.e. (27c).
Exercise 9.2.25. Show that the adjoint operators for Ph and Rh are Ph: L2(IR2) - L~ and R h: L~ - L2(IR2) with
(P,:u)(x)
= (UhU~U)(X)
for x E Qh,
RhUh = UhUXPhUh.
Furthermore, the following holds: (uhuXAuh)(X) The identity
L;;l
=
= Uh(X) for all x
E
Qh.
R hL- 1Ph - L;;ll(LhRh - RhL)L-1Ph + (RhPh - I)]
yields the estimate IL;;112<--O $ IRh 12<--2 IL- 112<--0 IPh 10<--0 + III2<--lIL;;l\l<---lX -
1
-
[ILhRh - RhLI-l<--2IL- 12<--olPhlo<--o + IRhPh - 11-1<--0]. The inverse estimate from Exercise 22 yields 1112<--1 $ Ch- 1 for the identity I: Hl- H~. Together with the inequalities (27a-c) follows
Theorem 9.2.26. Let Lh be Hl-regular and satisfy the consistency condition (9). Let il be convex or from C2 and let L be H2-regular (i.e., IL- 112<--O $ CJ. Then Lh is also H~-regular.
CO· 1 would be a sufficient condition upon
n
guarantee (27a-c). On the other hand, the condition given agrees with the conditions for Theorem 14 and with the H2-regularity of L. il E
Corollary 9.2.27. In Theorem 26 one can replace (9) and the H~-regularity of Lh by (15) and the H1-regularity of Lh. The result is the H~-regularity of both Lh and L h.
PROOF. For the proof of the H~-regularity of Lh define PhUh(X) := 0 for x E K h/ 2(X), x E "th, and note IDh(RhPh - I)D;;11_1<--O $ Ch. • Exercise 9.2.28. If n E IRn (n $ 3) is bounded and Lh is H~-regular, then Lh is stable with respect to the row-sum norm: IIL;;ll1oo $ C. Hint: Use C- 1lu hlo ~ IIUhlloo ~ CIUhI2.
9.2 Regularity Properties of Difference Equations
243
For general Lipschitz domains n, such as, for example, the L-shaped domain in Example 2.1.4, L [Lh] cannot be H2-regular [H~-regular]. Theorem 9.2.21, however, guarantees HJ+8(n)-regularity for 8 E [0,1/2). For Lh one can define and prove analogously H~+8_regularity (cf. Hackbusch [1]).
10 Special Differential Equations
If the boundary value problems have special properties, one often uses special discretisations for them. We give two examples of this in Sections 10.1 and 10.2. For the first example the variational fonnulation is of decisive importance.
10.1 Differential Equations with Discontinuous Coefficients 10.1.1 Formulation The self-adjoint differential equation
a (£lii(X)aa) + a(x)u(x) = !(x) L a ',i=l Xi Xi n
U
(10. 1.1 a)
in {J
(cf. (5.1.17» occurs often in physics. It can also be written
88
div(A(x) grad '1£) + au =! with A(x) := (£Iii (X»i,j=l, ... ,n, so that for a = 0 and ! = 0 we have the conservation law div q, = 0 for q, := A grad u. In the applications in physics the coefficients £Iii are, in general, constants of the material. The functions tlii can be varying, if the constitution of the material depends on position. As soon as we have several materials in contact with each other, the coefficients tlii(x) may be discontinuous on the boundary of contact 'Y (see Figure 1). The equation (la) can be understood in a classical sense only if aii E G1({J). For discontinuous a.i therefore one uses the variational fonnulation. If one supplements Equation (la) with the Dirichlet condition '1£=0 on
r,
(1O.1.1b)
then the weak formulation is written '1£ E HJ({J) with a(u, v) =
(j,V)L3(O)
for all v E HJ({J),
W. Hackbusch, Elliptic Differential Equations: Theory and Numerical Treatment, Springer Series in Computational Mathematics 18, DOI 10.1007/978-3-642-11490-8_10, © Springer-Verlag Berlin Heidelberg 2010
(10.1.2a)
244
10.1 Differential Equations with Discontinuous Coefficients
245
where a(u,v):=
1
[a(x)u(x)v(x) -
n
t
tlij(X)VX'(X)UX;j(x)]dx.
(10.1.2b)
i,j=1
Note that Equations (2a,b) are defined for arbitrary tlij E LOO(Q).
Figure 10.1.1. An interior boundary 'Y
= 001 n o{h
In the sequel we shall assume the situation to be as in Figure 1: Q is divided by the boundary line "( into two subregions Q 1 and Q2. Let the coefficients tlij be piecewise smooth: aij E C1(Qk) for k = 1,2. Along "( the coefficients may be discontinuous so that the one-sided boundary values
a~;> (x):= ~i.Px tlij(Y),
a~~)(x)
:=
~~ tlij(Y)
(XE"()
YED~
yEDI
may be different. In addition let us assume that the solution U is continuous, U E CO(n), but only that it is piecewise smooth, U E C 1 (Qd and U E C1(n2 ). The one-sided boundary values of the derivatives are denoted u1!)(x) and u~)(x) (x E "(). With these assumptions integration by parts of tlijVx.UXj dx gives the result v(aUX;j)x. dx vaWu1~)ni dr. Since n = (nb n2, ... ) in Figure 1 is the outgoing normal to !lb but the ingotlijV x•UX;j dx ing normal with respect to !l2, in !l2 there results -
- Inl
Inl
I-y
In2
In V( tlijUXj )X. dx + I-y va~~) u1~ ~ dr. Putting this together one has 2
Thus from the variational equation (2a) there follows in addition to the differential equation (la) also the transition equation (10.1.3)
Thus we have shown the following lemma.
246
10 Special Differential Equations
Lemma 10.1.1. Assume aij E C 1(ilk), k = 1,2. If the weak solution '11 of (2a) in il is continuous and piecewise differentiable in ill and il2, then it is the classical solution of the differential equation (la) in ill U il2 = il\'Y. In addition to fulfilling the boundary condition (lb) the solution also satisfies the transition condition (3) on 'Y.
Corollary 10.1.2. If the coefficients are not continuous then the solution of the equation (2a) does not, in general, belong to C2(il), but will have discontinuous derivatives along 'Y . However, the tangential derivative along 'Y may be continuous. Example 10.1.3. Suppose the coefficient of a(u, v) := fol a(x)u'v' dx are given by a(x) = 1 on the interval (O,e) and a(x) = 2 on (e, 1). For this onedimensional example the point plays the role of the curve 'Y. The solution of the equation (2a) with lex) = 1 is
e 1[1l+€x-x + e2 2] u(x)='2 u(x)
2 = '14 [1 +12€ + -ee (1 -
on(O,e),
x) - (1 - x)2 ] on (e,l).
It satisfies -au" = 1 on (0, e) U (e, 1), 1£(0) equation 1· u'(e - 0) = 2 . u'(e + 0).
= '11(1) =
1 and the transition
AB mentioned at the beginning, the differential equation can be written in the form div t/J = / on ill U il2, where t/J := A(x) grad '11 on il and '11 = 0 on r. The transition condition (3) means that (t/J,n) is continuous on 'Y. Since (t/J, t) is also continuous for any tangential direction t, t/J is continuous on 'Y and thus throughout n. The regularity proofs of Section 9.1 can be carried over to the present situation. The smoothness assumptions upon the coefficients of a(.,·) are in each case to be required piecewise on ill and il2; in addition the dividing line (or hypersurface) 'Y must be sufficiently smooth. Then there results the piecewise regularity '11 E Hm+6(ild and '11 E Hm+6(il2 ), insteadofu E Hm+6(il). Bringing in the transition condition (3) one obtains t/J = A(x) gradu E Hm+8-l(il) on the whole region. 10.1.2 Discretisation
When one discretises using finite elements there are the following difficulties: Remark 10.1.4. Linear or bilinear elements give a finite-element solution with the error estimate /'11 - uh/t = O(hl/2). The L2(ll) error bound is in general no better than /'11 - uh/o = O(h).
PROOF. il.., := {t E r: t n 'Y =F 0} consists in all finite elements that are in contact with 'Y. Since '11 has discontinuous derivatives on 'Y, one can have no
10.2 A Singular Perturbation Problem
247
better estimate on il-y than Vu - V v = O( 1) (v E Vh ). The surface area of il-y is of the order O(h), so that there results lu - vI! = O(h1/2). For the bound on lu - uhlo see Example 10.1.5. In Example 3 choose e= (1 + h)/2 and discretise using piecewise linear elements of length h. Then there results an error at x = 1/2 ofu(x) -uh(x) = ah+O(h2) with a ~ 0.00463. Since u-u h behaves linearly in (0,1/2), one obtains the error O(h) not only for the L2 norm lu - uhlo but also for the maximum norm lu - uhl oo and for the L1 norm lIu - uhIlLl(O,l). The usual bounds on the errors lu - uhh = O(h) and lu - uhlo = O(h2) can however be attained. To do this one must adjust the geometry of the triangulation to the curve 'Y. If the dividing line 'Y is piecewise linear, one must choose the triangulation such that 'Y concides with sides of (interior) triangles. If 'Y is curvilinear then it may be approximated by isoparametric elements (cf. §8.6). Similar assertions hold for the difference scheme (5.1.18). Example 10.1.6. Let the discontinuity position ein Example 3 be a grid point, Le., f.jh E IN. Then the difference scheme (5.1.18), which here takes the form h-2 [a((v+!) h) (Uv+l-Uv)
-a((v -!) h) (uv -Uv-l)] = 1
for 1 ::; v ::; h- 1 -1, is suited to the equation from Example 3. In general the error is of the order O(h2). Since the solution of the differential equation is piecewise quadratic here, it is in fact exactly given by the difference solution. On the other hand, if is not a grid point then the error is of the order O( h).
e
In the two-dimensional case one obtains O(h2) difference solutions if 'Y coincides with lines of the grid. Otherwise the error worsens to O(h). Another possible form of difference approximation consists in approximating the differential equations separately in the regions ill and il2 , and then, to handle the unknown values on 'Y to discretise the transition condition (3).
10.2 A Singular Perturbation Problem 10.2.1 The Convection-Diffusion Equation In the following we shall consider the boundary value problem n
-fLlu +
L CiU
z•
=f
in il,
u=o
on
r
(10.2.1)
i=1
for f > O. One calls -fLlu the diffusion term and :E Ciu z• = (c, Vu) the con vection term. For small f the convection term dominates.
248
10 Special Differential Equations
Exercise 10.2.1. Let the coefficients c; be constants. Equation (1) can be transformed by vex) := u(x) exp( - E Cix./2€) into the symmetric HJ(n)elliptic equation -f.Llv + ('2:cV(4€))v
= /exp(-(c,x}/2€)
in
n,
v
= 0 on r. (1O.2.1')
Equation (1) is elliptic for all € > o. That there is a unique solution follows from Exercise 1 (cf. also Theorem 5.1.8). Denote the solution by U£. Remark 10.2.2. u£(x) and oU£(X)/OXi cannot converge uniformly on Ii for € --+ O. PROOF. If U£ and audoxi were continuous in €, one would be able to take the limit € --+ 0 in Equation (1) and one would obtain the first-order differential equation n
- ~ Ci Ott%x.
= / in n
(10.2.2)
i=l
for tto(x) := lim£--to uE(x). This equation is of hyperbolic type, but not compatible with the boundary condition U = 0 on r (cf. Section 1.4), Le., Equation _ (2) has, in general, no solution with tto = 0 on r. Equation (2) is called the "reduced equation". Equations (1) and (2) differ in the "perturbation term" -€Llu. Since the equations, (1) and (2), are of different types one speaks of a "singular" perturbation. The following example will show that tto = li~--to UE exists and satisfies Equation (2), but that the boundary condition tto = 0 is only satisfied on a part ro of the boundary r. Example 10.2.3. (a) The solution of the ordinary boundary value problem -€U"
+ u' = 1
in (0,1),
U(O) = u(l) = 0
(10.2.3a)
is uE(x) = x - (eX / E _l)/(e l / E- 1). On [0,1), UE(X) converges to tto(x) = x. This function satisfies the reduced equation (2), u' = 1, and the left boundary condition uo(O) = 0, but not uo(l) = O. (b) The solution of
-w" - u' = 0 in (0,1),
U(O) = 0, u(l) = 1
(10.2.3b)
is uE(x) = (1 - e- x / E)/(l - e-l/E). In this case, uo(x) := limE-+ouE(x) = 1 satisfies Equation (2), -u' = 0, and uo(l) = 1, but not the boundary condition at x = o. Which boundary condition is fulfilled depends for Equations (3a,b) on the sign of the convection term ±U'. In the many-dimensional case the decisive factor is the direction of the vector c = (ct, ... , en).
10.2 A Singular Perturbation Problem
249
1-+-,~. , __----------------~
1
,,
, :
,,
o~
__________________
Ue
~~
o
1 x
O~-------------------.
o
1
Figure 10.2.1. (a) Solution of Example 30.; (b) Solution of Example 3b
In Figure la,b the solutions of Example 3 are sketched. In the interior U e is close to Uo; only in the neighbourhood of x = 1 (Figure 10.) [resp. x = 0 (Figure Ib)] does U e differ from 1.£0, in order to satisfy the second boundary condition. These neighbourhoods, in which the derivatives of U e attain the order O(I/E), are called boundary layers. For Example 3 the thickness of the boundary layer is of order O(E). Exercise 10.2.4. The interval [1 - e,I], in which the function (ez/e 1)/(e1/e -1) exceeds the value 0 < 'rJ < 1, has the thickness = O(Ellog'rJl).
e
An analysis of singularly perturbed problems is to be found, e.g., in Kevorkian-Cole [1] and Goering et al. [1].
10.2.2 Stable Difference Schemes
= (1,0) is U=cp f in n,
The special case of Equation (1) for c -ELlu + 1.£:11 =
onr.
(10.2.4)
The symmetric difference formula (5.1.10) for Equation (4) is
L. ~ h
-2 [ -E -
M2
~:
-E
+
h/2] .
(10.2.5)
For a fixed E, Lh is consistent of order 2. From Corollary 5.1.16 there follows
Remark 10.2.5. As soon as h < 2E, the difference scheme (5) leads to an M-matrix. The property of being an M-matrix was used in Section 5.1.4 to demonstrate the solvability of the discrete equation. It turns out that the difference
250
10 Special Differential Equations
equations are also solvable for h ~ 210. However, with increasing h/€ there develops an instability which is made clear in Table 1. 'lable 10.2.1
x
0 1/32
2/32 3/32
4/32
...
30/32 31/32
...
0.04 0.22 0.47 0.82
0.02 0.13 0.31 0.73
0 0 0 0
0.87 0.23 0.08 0.06 0.06
1.11 1.89 5.88 49.6 486004
0 0 0 0 0
1
10
0.5 0.1 0.05 0.02
1 1 1 1
0.93 0.97 0.99 0.99
0.87 0.95 0.97 0.99
0.81 0.92 0.96 0.98
0.75 0.89 0.94 0.98
0.01 10-3 10-4 10-5 10-7
1 1 1 1 1
1.00 1.03 4.95 48.6 4859.5
0.99 0.99 0.95 0.94 0.94
0.99 1.04 5.00 48.7 4859.6
0.99 0.99 0.90 0.88 0.88
... ... ...
... ...
... ... ...
Table 1 shows the values uE,h(x,1/2) of the difference solution UE,h of Equation (4) in the unit square n = (0,1) x (0,1) with f = 0 and cp(x, y) = (1- x) Sin(1TY). The grid step is h = 1/32. Fou > h/2 = 1/64 = 0.015625, Lh is an (irreducibly diagonally dominant) M-matrix. Because of the maximum principle (cf. Remark 4.4.4) the values lie between mincp = 0 and maxcp = 1. The solution of the reduced equation (2) is u.o(x,y) = li~-+o uE(x, y) = sin 1Ty. Accordingly the values UE,h(X, 1/2) from the first part of Table 1 approach the limiting value sin 1T /2 = 1. In the second part of the table we have 10 < h/2. Thus Lh is not an M-matrix. Notice that uE,h(31/32, 1/2) > 1, i.e., even the maximum principle does not hold. In addition one can see that UE,h(X, 1/2) converges to 1 - x for the even multiples x = lIh, and thus not to u.o(x, y) = sin 1Ty. For the intervening odd multiples x = lIh an oscillation of amplitude O(h 2 /2€) develops. The resulting difficulty is similar to that for initial value problems for stiff (ordinary) differential equations: If one keeps 10 constant and lets h go to zero, the convergence assertions of Section 5.1.4 hold. However, if 10 is very small, the condition h < 210, without which one does not obtain a reasonable solution, cannot in practice be satisfied. One way out of this difficulty has already been described in Section 5.104. One must approximate the convection term by suitable one-sided differences (5.1.14)
(10.2.6)
where
ct := max{O, £:i}, ci := min{O, £:i}.
10.2 A Singular Perturbation Problem
251
Remark 10.2.6. If one discretises Equation (1) using (6) then one obtains, for all € > 0 and h > 0, an M-matrix L h . For fixed €, the scheme has consistency order 1. Exercise 10.2.7. The discretisation of Equation (3b) corresponding to (6) is Lh = h- 2 [-€ 2€ + h -€ - h j and this gives the discrete solution Uh(X) = [1 - (1 + h/€)-re/hj/[1_ (1 + h/€)-l/hj. If one applies the difference operator (6) to a smooth function one obtains the Taylor series (10.2.7)
The O(h) term hlcllurere + hlc21UvY is called the numerical viscosity (or the numerical ellipticity) since it amplifies the principal part. A second remedy consists in replacing the parameter € by a discretisation using €h ~ €. If one chooses €h := max{€, hlctl/2, hlc21/2} or €h
h
= € + 2' max lcd,
(10.2.8)
then the symmetric difference method (10.2.9)
leads to an M-matrix. It is true that the convection term has been discretised to a second order of consistency, but the error of the diffusion term is O(€h -€), which, for the case h > 2€/lIcll oo which is relevant in practice, amounts to O(h). Instead of (7) one has LhU = Lu - (€h - €)L\u + O(h2).
(10.2.10)
The difference -(€h - €)L\u is called the artificial viscosity. In the one-dimensional case the methods using numerical and artificial viscosities do not differ: Remark 10.2.S. If, in the one-dimensional case, on chooses €h according to the second alternative in (8), then the difference formulae (6) and (9) coincide.
10.2.3 Finite Elements The difficulties described in the previous section are not restricted to difference methods. Exercise 10.2.9. Show that
252
10 Special Differential Equations
(a) Linear finite elements on a square grid triangulation applied in the case of Equation (4) give a discretisation that is identical with the difference method Lh
= E [ -1
-1 4 -1
-1
]
+ h
[0 -1/3 -1/6
-1/6 0 1/6
1/6] 1/3 0
(10.2.1180)
(cf. Exercise 8.3.13). (b) For bilinear elements (cf. Exercise 8.3.17) one obtains
Lh
=
i [ -1-1-1
-1 8 -1
-1] -1 + -1
1~
(c) One-dimensional linear elements for difference formula
[-1 0 +1] -4 0 +4 . -1 0 +1 -Ett."
+
tt.'
(10.2. lIb)
= f lead to the central (10.2.11c)
The Exercises 9b,c show that finite-element methods correspond to central difference formulas, and thus can equally well lead to instability. The method of artificial viscosity corresponds to the finite-element solution of the equation -Eh Lltt. + (c, grad tt.) = f for appropriate Eh. As in Exercise 9b one may show Remark 10.2.10. If one sets Eh := maxiE, ICllh, Ic2lh} and uses bilinear elements then the discretisation of Equation (1) leads to an M-matrix. On the other hand the matrix (1180) has different signs in the sub-diagonal and in the super-diagonal, so that it is not possible to have an M -matrix for any value of Eh. The analogues of one-sided differences are more difficult to construct. One approach is to combine So finite-element method for the diffusion term with a (one-sided) difference method for the convection term (cf. Thomasset [1, §2.4]). A second possibility is the generalisation of the Galerkin method to the Petrov-Galerkin method, in which the discrete solution of the general equation (8.1.1) is defined by problem (12): Find
tt. E
Vh , so that a(tt., v)
= f(v)
for all v E
Wh
(cf. Fletcher [1, §7.2J, Thomasset[l, §2.2]). Here we have
dim Vh
= dim Wh
(but in general Vh =f Wh).
(10.2.12)
11 Eigenvalue Problems
11.1 Formulation of Eigenvalue Problems The classical formulation of an eigenvalue problem reads Le=>.e
inD,
Bje=O
onr (j=I,···,m).
(11.1.1)
Here L is an elliptic differential operator of order 2m, and Bj are boundary operators. A solution e of (1) is called an eigenfunction if e =1= o. In this case, >. is the eigenvalue associated with e. As in Section 7, one can replace the classical representation (1) by a variational formulation, with a suitable bilinear form a(., .): V x V ~ IR taking the place of {L,B j }: Find e E V
with a(e,v)
= >.(e,v)o
for all v E V.
(11.1.2a)
(u, v)o = I.f2 uti dx is the L2( D)-scalar product. Strictly speaking one ought to replace (., ·)0 by (-, ·)u where U is the Hilbert space of the Gelfand triple V cUe V' (cf. Section 6.3.3). But here we limit ourselves to the standard case U = L2(D). The adjoint eigenvalue problem is formulated as
Find e E V
with a(v,e)
= X(v,e)o
for all v E V.
(1l.1.2b)
Definition 11.1.1. Let>. E CU. By E(>') one denotes the subspace of all e E V which satisfy Equation (1)[ resp. (2a)]. E(>.) is called the eigenspace for >.. With E*(>') one denotes the corresponding eigenspace of Equation (2b). >. is called an eigenvalue if dimE(>') ~ 1. Theorems 6.5.15 and 7.2.14 already contain the following statements:
c L2(Q) be continuously, densely and compactly embedded (for example, let V = H{f'(D) with bounded D). Let a(-, .) be Vcoercive. Then the problems (2a,b) have countably many eigenvalues>. E CU which may only have an accumulation point at 00. For all >. E
Theorem 11.1.2. Let V
dimE(>.)
= dimE*(>') < 00.
W. Hackbusch, Elliptic Differential Equations: Theory and Numerical Treatment, Springer Series in Computational Mathematics 18, DOI 10.1007/978-3-642-11490-8_11, © Springer-Verlag Berlin Heidelberg 2010
(11.1.3)
253
254
11 Eigenvalue Problems
Exercise 11.1.S. In addition to the assumptions of Theorem 2 let a(·,·) be symmetric. Show that all eigenvalues are real and problems (2a) and (2b) are identical so that E()") = E*()"). For ordinary differential equations of second order (Le., n c IR 1 , m = 1) it is known that all eigenvalues are simple: dim E()..) = 1. This statement is incorrect for partial differential equations as the following example shows.
Example 11.1.4. The Poisson equation -Lle = )..e in the rectangle (0, a) X (O,b), with Dirichlet boundary values e = 0 on r, has the eigenvalues ).. = (v7r/a)2 +(J.L7r /b)2 with V, JL E IN. The associated eigenfunction reads e(x, y) = eV,~(x, y) := sin(v7rx/a) sin(J.L7ry/b). In the case of the square a = b one obtains for 1/ =1= J.L eigenvalues ).. = )..v,I-' = )..1-',11, which have multiplicity at least 2 since eV'I-' and el-',v are linearly independent eigenfunctions for the same eigenvalue. A triple eigenvalue, for example, exists for a = b, ).. = 507r2 / a2 : E()") = span { e i ,7, e7 ,i, e5,5}.
The eigenfunctions e E E()"), by definition, belong to V. The regularity investigations of Section 9.1 immediately result in a stronger regularity.
Theorem 11.1.5. Let V = H{f(n) with m ;::: 1, or V = Hm(n) with m = 1. Under the assumptions oj Theorems 9.1.16 [resp. 9.1.17] we have E()")
c
Hm+lI(n).
PROOF. Along with a(·, .), a),(u, v) := a(u, v) - )..(u, v)o also satisfies the assumptions. Since a),(e, v) = 0 for e E E()..), v E V, the statement follows from Corollary 9.1.19. • Besides the standard form (2a) there are generalised eigenvalue problems. An example is the Steklov problem
-Lle = 0 in
n, Be/an =)..e
on
r,
whose variational formulation reads e E Hi(D), fo(Ve, Vv}dx = >-'frevdr (v E Hi(n». One can show that all eigenvalues are real and that the statements of Theorem 2 hold.
11.2 Finite Element Discretisation 11.2.1 Discretisation Let Vh C V be a (finite-element) subspace. The Ritz-Galerkin (resp. finiteelement) discretisations of the eigenvalue problems (1.2a,b) read
eh E Vh,
a(en,v) = )..h(en,v)o for all v E Vh,
(I1.2.1a)
11.2 Finite Element Discretisation
255
(11.2.1b) The discrete eigenspaces Eh(Ah), Ei.(Ah) are spanned by the solutions of the problems (la) [resp. (lb)]. As in Theorem 11.1.2, dimEh(Ah) = dimE;;(Ah) holds. If a(.,·) is symmetric, then Eh(Ah) = Ei.(Ah). As in Section 8.1, the formulation (la,b) can be transcribed into matrix notation. Remark 11.2.1. Let e and e* be the coefficient vectors for eh = Pe and e*h = Pe* (cf. (8.1.6)). The eigenvalue problems (la,b) are equivalent to (11.2.1') where the stiffness matrix L is defined as in Theorem 8.1.3 and M by (8.8.7). Since in general M #- I, (I') involves generalised eigenvalue problems. Exercise 11.2.2. Show that (a) M is positive definite and possesses a decomposition M = AT A (for example, A = Ml/2 or the Cholesky factor). (b) The first problem in (I') is equivalent to the ordinary eigenvalue problem La = Aha with L := (AT)-lLA- 1 , a = Ae. The second problem in (I') corresponds to LTa* = Aha* with a* = Ae*. When investigating convergence, one must watch out for the following difficulties: (i) A uniform approximation of all the eigenvalues and eigenfunctions by discrete eigenvalues and eigenfunctions is impossible since the infinitely many eigenvalues of (1.2a) are set against the only finitely many of (la). It is only possible to characterise a fixed eigenvalue A of (1.2a) as an accumulation point of discrete eigenvalues {Ah: h < O}, and to set up estimates for IA - Ahl. (ii) If A and Ah are the continuous, resp. discrete, eigenvalues then dimE(A) = dimEh(Ah) need not hold. It is preferable to limit oneself to the case of simple eigenvalues where dimE(A) = dim Eh(Ah) = 1. If dimE(A) = k > 1, it may well be that the multiple eigenvalue A is approximated by several discrete eigenvalues A~), i = 1,···, k: dimE(A) = E~=l dimEh(A~». The error estimates of IA - A~) 1 are then generally worse than for simple eigenvalues. Only for the mean value .x := E~=l A~) /k does one obtain the usual estimates (cf. Babuaka-Aziz [1, p. 338]). Exercise 11.2.S. Let the eigenvalue problem be as in Example 11.1.4 with a = b and A = 5071"2/ a2 • Let Vh consist of linear elements over a square grid triangulation. Show that to the given triple eigenvalue A corresponds a double eigenvalue A~l) and a single eigenvalue distinct from it, A~2), with limh......O A~) = A (i = 1,2). Hint: the nodal values of the discrete eigenfunctions agree with the continuous eigenfunctions e 1,7, e7,l, and e5 ,5.
256
11 Eigenvalue Problems
11.2.2 Qualitative Convergence Results This section concerns the question as to whether Ah -+ A and eh -+ e for h -+ O. The rate of convergence will be discussed in Section 11.2.3. The basic assumptions are the following: Let a(-.·): V x V
-+
lR be V-coercive,
(1l.2.2a)
c
L 2 (n) be continuously, densely, and compactly embedded, (1l.2.2b) so that the Riesz-Schauder theory is applicable (Theorem 11.1.2). Furthermore, let a sequence of subspaces Vh, (14 -+ 0) be given which increasingly approximate V (cf. (8.2.4a)): Let V
(l1.2.2c) We define a~(·,·): V x
V
-+
cr:,
W(A):=
a~(u, v) :=
a(u, v) - A(U, v)o,
(l1.2.3a)
la~(u, v)l,
(1l.2.3b)
inf
sup
uEV
tiE V
lIullv=l .I1t1l1v=l
Wh(A):=
inf
sup
UEVh
tlEVh
la~(u,v)l.
(l1.2.3c)
lIullv=l IItlllv=l
Exercise 11.2.4. Let L and Lh be the operators associated with a(·, .): V x V -+ lR and a(., .): Vh x Vh -+ lR (cf. (8.1.lOa)). Show that (i) If A is not an eigenvalue we have (11.2.4) (cf. Lemma 6.5.3 and Exercise 8.1.16). (ii) W(A) and Wh(A) are continuous in A E cr:. (iii) If in (3b,c) one replaces a~(u, v) by a~(v, 11.) one obtains the variables W* (A) and wh (A) which correspond to the adjoint problem. The following holds: W*(A) = W(A) and wh(A) = Wh(A). Hint: Use Lemma 6.5.17. With the aid of (4) and Theorem 6.5.15 one proves the following connection between W(A), Wh(A), and the eigenvalue problems.
Remark 11.2.5. Let (2a,b) hold. A is an eigenvalue of (1.2a) if and only if W(A) = 0, and Ah is eigenValue of (la) if and only if Wh(Ah) = o. Lemma 11.2.6. Let a(·,·) be V-coercive (cf. (2a)). Then there exists a J.L E lR such that aJ.'(-'·) is V-elliptic. In addition, then w(J.L) ~ l/CE and Wh(J.L) ~ l/CE hold with CE > 0 from Definition 6.5.13.
11.2 Finite Element Discretisation
257
Let A c CD be compact. Let (2a--c) hold. Then there exist numbers C > 0 and .,,(h) > 0, independent of A E A with limh-to .,,(h) = 0 such that
Lemma 11.2.7.
WheAl
~
weAl
CW(A) - .,,(h),
~
CWh(A) - .,,(h)
for all A E A. (11.2.5)
PROOF. Let the operators Z = ZeAl: V - V and Zh = Zh(A): V - Vh be defined as follows:
z := Z(A)U is the solution of al'(z, v) = (A - JL)(u, v)o for all v E V, (11.2.6a)
zh:= Zh(,x)U is the solution of al'(zh,v)
= (A -
JL)(u,v)o for all v E Vh , (1l.2.6b)
where JL is chosen according to Lemma 6. It shows
IIZlIv+-v' From a>.(u, v)
s Cz ,
IIZhllv+-vl S Cz
= al'(u - z, v)
w(A)lluliv
s
(11.2.6c)
and the definition of weAl one infers
la>.(u, v)1 =
sup
for all A E A.
sup
vEV
vEV
IIvllv=l
IIvllv=l
lal'(u - z, v)1
s Csllu - zllv
(11.2.6d) with Cs := IIL-JLlllvl+-v. For an arbitrary U E Vh one infers from a>.(u, v) = a,.(u - zh, v), Lemma 6, and (6d) that sup
la>.(u, v)1 =
vEV"
IIvllv=l
~
C;/lIlu -
sup
lal'(u - z\ v)1 ~ C;/llu - zhll v
vEV"
IIvllv=l
zllv
-liz -
zhllvl ~ C;/[CS1W(A)
-liZ -
Zhllv+-vl lIullvi
thus WheAl ~ (CECS)-lw(A) -liZ - Zhllv+-v JCE. From this follows the first part of (5) with C = (CECS)-l > 0 and .,,(h) = liZ - Zhllv+-vjCE, if lim sup II ZeAl - Zh(,x)lIv+-v h-tO>'EA
= O.
{11.2.6e)
The proof of (6e) is carried out indirectly. Its negation reads: there exist € > 0, ~ E A, h. - 0 with II Z(Ai) - Zh.(~)lIv+-v ~ € > O. Then there exist Ui E V with (11.2.6f) Due to (2b) and the compactness of A there exists a subsequence Aj E A, E V with limAj = A*, limuj = u* E V' in V'. (2c) and Theorem 8.2.2 show IIIZ(A*) - Zhi(A*)u*lIv - O. Together with (6c) we obtain
Uj
258
11 Eigenvalue Problems
II[Z(,xj) - Zhj(,xj)1ujllv ~ IHZ(,xj) - Z(,x*)]ujllv + II [Zhj (,xj) - Zhj(,x*)]ujllv + IIZ(,x*)[Uj - u*111v + IIZhj(,x*)[Uj - u*]/lv + II[Z(,x*) - Zhj(,x*)]u*IIv ~ 2GI,xj - ,x*1 + 2Gzlluj - u*IIv + II[Z(,x*) - Zhj(,x*)]U*/lv -+ 0 1
in contradiction to (6f). For the proof of the second part of (5) one replaces (6d) and the following estimate by Wh(,x)lluh/lv ~
lap.(uh - zh, v)1 ~ Gsllu h - zhll v
sup
for all uh E Vh,
1JEV" 1I1Jllv=l
laA(uh,v)1
sup 1JEV 1I1Jllv=l
=
sup 1JEV 1I1Jllv=l
lap.(uh - z,v)l;::: GE/llu h - zllv
for all u h E Vh.
;::: Gj;/[GS1Wh(,x) - IIZ(,x) - Zh(,x)llv+-vllluhllv
Let u E V with /lu/lv sup 1JEV 1I1Jllv=l
= 1 be selected such that
laA(u,v)1
=
inf
sup
laA(u,v)1
= w(,x).
uEV 1JEV lIullv=l 111Jllv=l
Since sup laA(u - uh, v)1 ~ Gsllu -uhllv, it follows for arbitrary u h E Vh that w(,x) ;::: GWh(,x) - IIZ(,x) - Zh(,x)IIv+-v - Gsllu - uhllv.
•
From (2c) and (6e) follows the second part of (5).
A corollary to Lemma 7 is Theorem B.2.B.1f Problem (B.1.1) for all f E V' is solvable, then ,x = 0 cannot be an eigenvalue, i.e., w(O) > O. Thus it follows that loN, ;::: WN,(O) ;::: lo := !Gw(,x) > 0 for sufficiently large i.
A second corollary concerns the convergence of the eigenvalues.
Theorem 11.2.8. Let (2a-c) hold. If ,xh, (i -+ 00, hi -+ 0) are discrete eigenvalues of (la) with ,xh, -+ Ao then Ao is an eigenvalue of (1.2a). PROOF. If Ao were not an eigenvalue then w(,x) ;::: 'f/O > 0 would be in the lo-neighbourhood of K£(Ao), since w(,x) is continuous (cf. Exercise 4(ii». There would exist ho > 0 such that "1(h) ~ G1}o/2 for all h ~ ho (G and "1(h) from (5». For all ,xh, E KE(,xo) with hi ~ ho the contradiction follows from (5): 0= Wh,(,xh.) ;::: GW(,xh. - "1(h i )
;:::
1 G'f/O - "2 G1}o
1
= "2G"10 > O.
•
Lemma 11.2.9. (Minimum Principle) Let (2a,b) hold. The functions w(,x) and Wh(,x) in the interior of A C () have no proper positive minimum.
11.2 Finite Element Discretisation
259
PROOF. Let L be the operator associated with a(·, .). Let >'*, with w(>.*) > 0, be an arbitrary point in the interior of A. For sufficiently small € > 0 we have Ke(>'*) C A and w(>.) > 0 in Ke(>'*). Thus (L - >.I)-1 is defined in Ke (>. *) and holomorphic. Cauchy's integral formula says (L - >'1)-1 =
~1 21n
( -
>.)-l(L - (I)-ld(
for all >. E Ke(>'*).
J8K.
From this one infers
i.e., w(>.) ~ min{w«():( E 8Ke(>'*)} (cf. Exercise 4(i». Thus, w(>.) cannot assume a proper minimum in Ke(>'*). For Wh(>') the conclusion is the same .
•
The converse of Theorem 8 is contained in
Theorem 11.2.10. Let (2a-{!) hold. Let >.0 be the eigenvalue of (1.2a). Then there exist discrete eigenvalues >'h of (1 a) (for all h) such that limh-to >'h = >'0.
PROOF. Let € > 0 be arbitrary. According to Theorem 11.1.2, >'0 is an isolated eigenvalue: w(>.) > 0 for 0 < I>' - >.01 ~ € (€ sufficiently small). Since w(>.) is continuous and 8KE (>.0) is compact, we have that PE := min{w(>'): I>' - >.01 = €} is positive. Because of (5) and w(>'o) = 0 one obtains for sufficiently small h
for all >. E 8KE(>'0)' Thus Wh(>') must have a proper minimum in KE(>'O)' By Lemma 9 the minimal value is zero. Thus there exists a >'h E Ke(>'o) which is a discrete eigenvalue, Wh(>'h) = o. • The convergence of the eigenfunctions is obtained from Theorem 11.2.11. Let (2a-{!) hold. Let eh E Eh(>'h) be discrete eigenfunctions with lIeh llv = 1 and limh-to >'h = >'0. Then there exists a subsequence eh;. which converges in V to an eigenfunction e E E(>.o):
eE E(>.o), lIeh;. - ellv -
0
(i - (0),
lIellv =
1.
PROOF. The functions eh are uniformly bounded in V. Since V C L2(il) is compactly embedded (cf. (2b», there exists a subsequence eh ;. which converges in L2(il) to an e E L2(il): (11.2.7a)
260
11 Eigenvalue Problems
We define z = Z(Ao)e, zh. = Zh.(Ao)e according to (6a,b). According to Theorem 8.2.2 there exists an hl(E) > 0 such that
liz -
(11.2.7b)
zh'lIv ::5 E/2
for h. ::5 ht(f). The function e h • is a solution of
a,. (eh• , v) = (Ah. - JL) (eh• , t' )0
for all v E Vh•.
(11.2.7c)
A combination of eh• = Zh.(Ao)e (Le., (6b) for A = Ao) and (7c) yields
a,.(zh. - eh., v)
= F.(v) := (Ao - JL)(e _e h., v)o - (Ah. -
Ao)(eho, v)o (11.2.7d)
for all v E Vh.. Since IIF.II{" -+ 0 because Ah. -+ Ao and (7a), there exists an h2(f) > 0 such that IIFillv 1 ~ f/[2CE ] (CE from Lemma 6) and (11.2.7e) for hi ~ h2(f). (7b) and (7e) show that liz - eh'lIv ~ f for hi ~ min{ht(f), h2(f)}; thus liffii-+oo eh • = z in V. Therefore lime h• = z in L2(Q) c V also holds. (7a) proves z = e E V such that e = z = Z(Ao)e becomes aCe, v) = .\o(e, v)o. Therefore e = lime h• is an eigenfunction of (1.2a). In particular, lIellv = limllehollv = 1. • Exercise 11.2.12. Let (2a~) hold. Let Ah,eh,Ao, and e be as in Theorem 11. Show that (a) IfdimE(.\o) = 1 then also limh-+Odim Eh(Ah) = 1. (b) Let dimE(Ao) = 1. Then we have lime h = e in V for eh := eh/(e\e)v if I(eh,e)vl ~ 1/2 and eh := eh otherwise.
11.2.3 Quantitative Convergence Results The geometric and algebraic multiplicities of an eigenvalue Ao of (1.2a) agree if (11.2.8) dim Ker(L - .\01) = dim Ker(L - .\01)2. Lemma 11.2.13. Let (2a,b) and dimE(.\o) = 1 hold. Then (8) is equivalent to (e, e*)o t- 0 fOT 0 t- e E E(Ao), 0 t- e* E E*(Ao). PROOF. We have Ker(L -.\01) = E(.\o) = span {e}. dim Ker(L -AoI)2 > 1 holds if and only if there exists a solution v E V for (L - .\o1)v = e. According to Theorem 6.5.15c this equation has a solution if and only if (e, e*)o = O. • Let E(Ao) = span {e}, E*(Ao) and e* can be normalised so that
= span {e*}.
(e,e*)o
= 1.
Under the assumption (8) e (11.2.8')
11.2 Finite Element Discretisation
We define V:= {v e V: (v,e*)o = O}, V' = {v' e V': (v/,e*)o be the dual norm for 11·11" = 1I·lIv. For Problem (9): For
I e V',
find u
e
Vwith a),,(u,v)
= (f,v)o
for all v
261
= O}. Let 1I·lI v, e V,
(11.2.9)
one defines the variable corresponding to (3b)
w(>.):=
inf
la),,(u,v)l.
sup
(11.2.10)
uE" vE" lIullv=1 Ilvllv=l
Lemma 11.2.14. Let (2a,b), (8), and dimE(>.o) = 1 hold. Then there exists an € > 0 such that w(>.) ~ C > 0 lor alii>' - >.01 S €. Problem (9) has exactly one solution u e V with lIullv
s II/lIv'/w(>'),
il w(>.) > O.
t:
ce.
PROOF. Let V -+ V' be the operator associated with a(·, .): V x V -+ For 0 < I>' - >.01 € (€ sufficiently small) (L - >.1)u = I has a unique solution u e V. From I e V' follows
s
0= (f,e*)o = ([L - >.I]u,e*)o = (u, [L - >.I]*e*)o
= (Xo -
X)(u, e*)o,
i.e., u e V. Thus there exists (£ - >'1)-1: V' -+ V as the restriction of (L - >.1)-1 to V' C V'. For>. = >'0 Problem (9) has a unique solution according to Theorem 6.5.15c. Then there exists (£ - >'1)-1 for all >. e KE(>.o). According to Remark 5 (with V instead of V) , w(>.) must be positive in KE(>'O). The continuity of w(>.) proves w(>.) ~ C > O. In analogy to (4) one has lIullv = lIull" II/II",/w(>'). The bound by II/lIv'/w(>') results from
s
s
Exercise 11.2.15. Show that
II/lIv'
~ II/lIv' for all I
e V'.
Lemma 11.2.16. Let (2a-c), dimE(>'o) = 1, and condition (8) hold. Let >'h be discrete eigenvalues with limh--+O >'h = >'0. According to Exercise 12b there exists an eh e Eh(>'h) and e*h e Eh(>'h) with eh -+ e e E(>.o), e*h -+ e* e E* (>.0), (e, e*)o = 1. This enables one to construct the space Vh := {v h e Vh: (v h, eh)o = O} and the variable Wh(>'):=
inf
sup
uE "" Ilullv=1
vE "" II vllv=1
la),,(u, v)l.
Then there exists a C > 0 independent of hand>' e (() such that Wh(>') CWh(>')' For sufficiently small € > 0 and h, Wh(>') ~ 1J > 0 for all I>' ->'01 S PROOF. (a) First statement: there exists min{lIv + ae*hllv: a E
ce} ~ IIvllv/C
ho > 0 and C such that
or all v
e Vh
and all 0
< h ~ ho.
~ €.
262
11 Eigenvalue Problems
The proof is carried out indirectly. The negation reads: there exists a sequence Cti E 4J, Itt. -+ 0, Vi E Vh, with IIViliv = 1 and IIVi + aie*h'IIv -+ O. Thus there exist subsequences with Cti -+ a*, Vi -+ v* in L2(il). Evidently, Wi := Vi + aie*h, must have the limit w* = lim Wi = v* + a*e* in L2(il). Since lim IIWill£2(n) :::; CHm IIwiliv = 0, it follows that w* = 0, and thus v* = -a*e*. From 0 = lim (Vi, e*h,)o = (v*, e*)o = -a* IIe* 115 one infers a* = O. Thus the contradiction follows from 1 = lim IIviliv :::; limllwillv + limIiCtie*h'IIv = lim IIWiliv = O. (b) Second statement: Wh(A) ~ CWh(A) with C from (a). Because a>.(u,v) = a>. (1.£, V + ae*h) for 1.£ E Vh we have Wh(A)
= inf
sup
Ia>. (1.£, v) I/(IIuliv IIvllv)
O#uEV" O#tJEV"
~C
inf
sup maxla>.(u, v +ae*h)II[IIullvllv + ae*hllvl
O#UEV" O#tJEV" aEt:
= C inf
sup la>.(u, w)I/[IIullvllwllvl
~ C
sup
O#uE V" O#wE V"
inf
Ia>. (1.£, w)lIlIIullvllwllvl = CWh(A).
O#uE V" O#wE V"
(c) Let f > 0 be chosen such that Ao is the only eigenvalue in KE(Ao). For sufficiently small h, Ah is the only discrete eigenvalue in Ke(Ao). In the proof of Theorem 10 we have already used Wh(A) ~ rl' > 0 for A E 8K(AO), h:::; hO(f). From Part b follows that Wh(A) ~ 'TJ := 'TJ'C > 0 for A E 8KE(AO). According to Exercise 12a, a>. (u, v) = (I, v)o (v E Vh) is solvable for each f E Vh and all A E KE(AO) such that Wh(A) = 0 is excluded. Lemma 9 shows that Wh(A) ~ 'TJ > 0 in Ke(Ao). •
Exercise 11.2.17. Let (2a,b) hold. The functions 1.£, V E V satisfy (1.£, v)o =fi O. Let d(u, Vh) be defined as in (8.2.2). Show that if d(u, Vh) is sufficiently small then there exists a u h E Vh with
Lemma 11.2.18. Let (2a-c) hold. Let Ao be an eigenvalue with (8) and dimE(Ao) = 1. For sufficiently small h there exists eh E Eh(Ah) with
lie -
ehllv :::; C[lAo - Ahl + d(e, Vh)l.
PROOF. Let zh := Zh(Ao)e be the solution of (6b). Since e = Z(Ao)e, one has (11.2.11a) For all v E Vh we have
11.2 Finite Element Discretisation
263
al'(zh-eh, v) = (Ao - /l)(e,v)o - (Ah - /l)(eh,v)o = (Ao - Ah)(e, v)o + (Ah - /l)(zh - eh, v)o + (Ah - /l)(e - zh, v)o, SO that a>." (zh_e h, v) = (Ao - Ah)(e, v)o + (Ah - /l)(e - zh, v)o for all v E Vh. (11.2.11b) According to Lemma 16 we have I(e*,e*h)ol ~." > 0 for sufficiently small h. eh can be scaled so that (zh - eh , e*h)o = o. (llb) corresponds to Problem (9). Lemma 14 proves
Ilzh - ehllv ~ Wh(Ah)-lC[lAo - Ahl + lie - zhllv ) ~ C'[ lAO - Ahl + d(e, Vh)). Together with (lla) one obtains the statement.
•
Theorem 11.2.19. Let (2a---c) hold. Let AO be an eigenvalue with (8) and dimE(Ao) = 1. Let e E E(AO), e* E E*(Ao), llellv = 1, (e, e*)o = 1. Then there exist discrete eigenvalues Ah with (11.2.12) PROOF. Choose u h according to Exercise 17 such that lIe* - uhllv ~ 2d(e*, Vh), (e* - uh,e)o = o. Discrete eigenvalues Ah -+ AO exist by Theorem 10. From
o = a>.o (eh, e*) = a>." (e h , e*) -
(AO - Ah)(eh, e*)o =a>.,,(eh,e* _uh) -(Ao-Ah)(e\e*)o = a>." (e h - e, e* - uh) - (AO - Ah)( eh, e*)o = a>." (e h - e, e* - uh) - (AO - Ah)[(e, e*)o + (e h - e, e*)oJ
follows lAo - Ahl ~ C'lIe h - ellvllle* - uhllv + lAO - Ahll. By Lemma 18 there exists an eh E Eh (Ah) such that
lAO - Ahl ~ C'C/[lAo - Ahl + d(e, Vh)](IAo - Ahl + 2d(e*, Vh)]. From this one obtains (12) with C = 3C'C" for sufficiently small h, since lAo - Ahl-+ 0, d(e, Vh ) -+ 0, d(e*, Vh) -+ o. • Theorem 11.2.20. Under the assumptions of Theorem 19 there exist for e E E(AO), e* E E*(Ao) discrete eigenfunctions eh E Eh(Ah), e*h E Ei.(Ah) with (11.2.13)
PROOF. Insert (12) with d(e*, Vh ) estimate in (13) follows analogously.
~
const into Lemma 18. The second •
264
11 Eigenvalue Problems
In the following, let V C Hl(il). Theorem 11.1.5 proves E(AO) C H1+ S(il),
E*(>.o) C H1+ S(il).
(l1.2.14a)
for all u E E(>.o) U E*(AO).
(1l.2.14b)
Also, let (14b) hold (cf. (8.4.10)): d(u, V,,) ~ Ch8I1uIlHl+"(G)
Corollary 11.2.21. Let (2a), (2c), (14a,b) hold. Let Ao be the eigenvalue with (8) anddimE(Ao) = 1. Then there exists A",e" E E,,(A,,), e*" E Eh(A,,) such that
Occasionally eigenfunctions may have better regularity than is proven for ordinary boundary value problems. For example, let -Lle = Ae be in the rectangle il = (0,1) x (0,1) with e = 0 on First, Theorem 11.1.5 implies e E H2(il) n HJ(il), thus e E CO(il) (cf. Theorem 6.2.30). Thus e = 0 holds in the corners of il. According to Example 9.1.25 it follows that e E HS(il) for s < 4. As in Section 8.4.2 one obtains better error estimates for e - e" in the L2-norm. The proof is postponed until after Corollary 29.
r.
Theorem 11.2.22. Let (2a-c), (8), dimE(Ao) = 1, and (14a,b) with s = 1 hold. Let a(·,·) and a*C·) be H2- regular, i.e., for f E L2(il) , a~(u,v) = (I,v)o and a~(v,u*) = (v,J)o (v E V", JL from Lemma 6) have solutions u, u* E H2(il). Let e E E(Ao) and e* E E*(Ao). Then there exist A", e" E E,,(A,,), and e*" E Eia(A,,) with
lie - e"IIL2(G) ~ C'h2, If(14a,b) also hold with some s
> 1,
lIe* - e*"IIL2(G) ~ C'h2. one must replace C'h2 by C'h1+B.
11.2.4 Complementary Problems In Problem (9) we have already encountered a singular equation which nevertheless was solvable. In the following let >.0 be the only eigenvalue in the disc Kr (Ao). The equation a). (u, v) = (I, v)o
for all v E V
(11.2.16a)
is singular for A = Ao. For A ~ >.0 Equation (16a) is ill-conditioned. In the following we are going to show that Equation (16a) is well-defined and wellconditioned if the right-hand side f lies in the orthogonal complement of E*(Ao):
11.2 Finite Element Discretisation
f 1- E*(AO)
,i.e., (f,e*)o = 0 for all e* E E*(AO».
265
(l1.2.16b)
In the case of A = AO , with u, u + e (e E E(AO)) is also the solution. The uniqueness of the solution is obtained under the conditions (8) and (16c): (l1.2.16c)
u 1. E*(>.o).
Remark 11.2.23. Let (2a,b) and (8) hold. Let >.0 be the only eigenvalue in Kr(AO). Then (16a,b) has exactly one solution u for all IA - Aol ~ r which satisfies (16c). There exists a C independent of f and A such that lIuliv ~
Cllfllv
1 •
PROOF. This follows from Lemma 14 in which the assumption E(AO) = 1 is not necessary. • The finite element discretisation of Equation (16a) reads: Find u h E Vh with a>. (u h , v) = (f' v)o
for all v E Vh.
(11.2.17)
In general, Equation (17) need not be well-defined, even assuming (16b). For the sake of simplicity we limit ourselves in the following to simple eigenValues: dimE(>.o) = 1. Equation (17) is replaced by (18a): Find u h E Vh
with a>.(uh,v)
= (f(h),v)O for all v E Vh
(11.2.18a)
with f(h) 1. Ei.(Ah),
(1l.2.18b)
uh 1. Ei,,(Ah).
(l1.2.1Sc)
Exercise 11.2.24. Show that pSa-c) ~s equivalent to: Find u h E Vh with a>.(u\ v) = (f(h) , v)o for all v E Vh with Vh = Vh n Ei,,(Ah).i as in Lemma 16. Lemma 16 proves the Remark 11.2.25. Let (2a-c), (S), dim E(AO) = 1 hold. Let AO be the only eigenvalue in Kr(AO). Then there exists an ho > 0 such that, for all h ~ ho and all A E Kr(AO), the Problem (lSa,b) has a unique solution u h = Uh(A) which satisfies the additional conditions (lSc). Further there exists a C independent of h, A, and f(h) such that lIuhllv ~ Cllf(h)lIv. If Ei.(Ah) =1= E*(AO), f from (16b) need not satisfy condition (lSb). If eh E Eh(Ah) and e*h E Ei.(Ah) with (e h, e*h)o = 1 are known, one can define
f(h) := Qhf := f - (f' e*h)oe h . f(h) satisfies (lSb) since Qh represents the projection on E;;(Ah).i.
(11.2.19)
266
11 Eigenvalue Problems
Exercise 11.2.26. Let u.l E(Ao), dimE(>.o) (eh,e*h)o = 1. Show that
= dim Ell. (>'11.) = 1, Ilehl/v = 1,
d(u, Eh(>'h)l. n VII.) := inf{IIu - vhl/v, vII.
E
VI., vII. .l Ei,.(>'h)}
S C[d(u, VII.) + l/ulivinf{l/e* - e*hllvl:e*
E
E*(Ao)}].
Theorem 11.2.27. Let (2a-e), (8), dimE(>.o) = dim Eh(>'h) = 1 hold. Let Ao be the only eigenvalue in Kr(Ao}. Let h be sufficiently small such that Uollowing Remark 25) the Problem (18a-e) is solvable. For the solutions u and u h 01 (16a-e) and (18a-e) the error estimate
I/u-uhllv S C[d(u, Vh)+I//IIvdnf{l/e*-e*hl/vl:e*
E
E*(Ao)}+II/(h)-/l/v/] (11.2.20)
holds, with C independent 01 I, 1(11.), and h. PROOF. Repeat the proof of Theorem 8.2.1 for a>.(-,·) instead of a(·, .). Here one must choose w E VII. with w.l Eh(>'h). Furthermore, (8.2.3) becomes
a>.(uh - u, v) = (f(h) - I, v)o for all v EVil.. EN agrees with Wh(>') ~ "1 > 0 (>. E Kr(Ao))) (cf. Lemma 16). I/u - wllv is estimated with the aid of Exercise 26, with IIuliv S II/IIvl being added. •
Corollary 11.2.28. 111(11.) is defined by (19), then inequality (20) becomes IIu - uhllv S C'[d(u, VII.) + I//livl inf{lle* - e*hl/v: e* E E*(Ao)}]. (12.2.21a)
PROOF.
li h) - Ilivl S CI(f, e*h)ol = CI(f, e* - e*h)ol S Cl//livll/e* - e*h IIv.
•
Corollary 11.2.29. II additionally the assumptions u E H1+ a(n), (14a), and d(u, VII.) S Ch 8 1UI1+a hold lor u E H1+ S (n) then (21a) yields the estimate (1l.2.21b) It remains to add the PROOF. (Theorem 22) For e E E(Ao) there exists ell. E Eh(>'h) with I := e - eh .l E(>.o) and 1/11 = I//I/v S Cha = Ch. According to Remark 25 the problem a>.o (v, w) = (v, 1)0 has a solution w .l E* (>'0) for all v E V. The assumption of regularity yields w E H2(n), Iwl2 S Cillo such that wh E VII. exists with w h .l Ei,.(>'h), Iw - whit S Chlwl2 S C'hi/io. The value
11.3 Discretisation by Difference Methods
267
a>"o(f,w h ) = a>"o(e,w h ) -a>"o(eh,wh ) = 0 -a>"o(eh,wh ) = (-'0 - Ah)(eh, wh)o - a>"h (eh, w h) = (-'0 - Ah)(eh, wh)o can be bounded by Ch21whlolehlo ~ C'h 2 1flo (cf. (15)). From Ifl~ = (f,f)o = a>"o(f,w) = a>"o(f,w - w h ) + (AO - Ah)(e\wh)o ~ C[C'hlfltlflo + h21flol
and Ifb ~ Ch one infers Iflo ~ C'h2 • The same method is then applied to le* - e*hlo. • Exercise 11.2.30. Formulate the conditions for and the proof of the error estimate lIu-UhllL2(s) ~ Ch21ul2 (u, u h from Corollary 28). Hint: Decompose u - u h in 11 + 12 with 11 with 11 .1 E*(-'o) and 12 E E(AO).
11.3 Discretisation by Difference Methods In the following we limit ourselves to the case of a difference operator of the order 2m = 2. The differential equation Lu = f with homogeneous Dirichlet boundary condition is replaced, as in Sections 4 and 5, by the difference equation LhUh = fh. The eigenvalue equations Le = Ae, L*e* = >':e* are discretised by (11.3.1)
Lh is the transposed and complex-conjugate matrix to Lh . The general assumptions of the following analysis are:
v = HJ(n), n E co· 1 bounded,
(11.3.2a)
a( u, v) = (Lu, v)o is HJ (n)-coercive,
(1l.3.2b)
Consistency condition ILhRh -
RhLI-h-2 ~ Ch.
(11.3.2c)
Condition (2c) has been discussed in Section 9.2.2. Assume furthermore that Lh is Hk-coercive. For suitable J-t E m
L",.h := Lh - J-tI (I: identity matrix) is thus Hk-regular: (1l.3.2d) Furthermore, let
L", := L - J-tI H 2(n)-regular,
(11.3.2e)
Le., IL;112+-0 ~ C. The boundedness of Land Lh reads ILI-1+-1 ~ C,
IL h l-1<-l ~ C.
(11.3.2f)
268
11 Eigenvalue Problems In (9.2.26a,b) we have introduced prolongations Ph: L~ -+ L2(J2). Now we need a mapping Ph:H~ -+ HJ(J2):
-+
L2(JR.2) and
Ph:L~
u(x) = {AUh(X)
if K h/2(X) C fl, otherwise, PhUh(X) := (O"hO"~U)(X) (x E J2),
o
where K h/ 2(X) = {y E JR2 : IIx - Ylloo < h/2}, Ph are defined according to (9.2.26b) and O"hO"~ according to (9.2.11). Check that PhUh E HJ(fl) and !Phh+-1 S C.
(11.3.2g)
Let Rh and Rh be defined as in (9.2.12a,b). They satisfy (11.3.2h)
Exercise 11.3.1. Show that: IRh - Rhlo+-1 S Ch,
II - RhPhlo+-1 S Ch,
II - PhPh10+-1 S Ch,
!Ph - Rh 10+-1 S Ch, (1l.3.2i)
II - RhPh10+-1 S Ch,
II - R;;Rhlo+-l S Ch. (11.3.2j)
The first inequality in (2j) is equivalent to I(PhUh,PhVh)O - (Uh,Vh)ol
s Chluhlolvh/l.
(11.3.2j*)
Hint: Exercise 9.2.12 and Lemma 9.2.15. Lemma 11.3.2. Let (2a,c,g,h,i) hold. (a) It is true that
lim Iu - PhRhUh h-+O
= h-+O lim Iu -
R;;Rhult = 0 for all U E HJ(fl), (1l.3.2k) (11.3.21)
lim I(Rh - Rh)ulo = 0 for all U E L 2(.o), h-+O 1 l~ I [Lh •.xRh - RhL.x]ul-l = 0 for all U E Ho (fl), A E GJ. (11.3.2m) (b) If U E HJ(.o) and limh-+o IRhulo
=0
then U =
o.
PROOF. (a) The proof of (2k,l,m) follows the same pattern, which will be demonstrated for the case of (21). For f > 0 one has to show I(Rh - Rh)ul Sf for h S h(f). Since HJ(J2) is dense in L2(J2) there exists au E HJ(J2) with Iu-ulo S f/[2IRh -Rhlo+-o) such that I(Rh -Rh)(U-u)lo S f/2. By (2i) there follows I(Rh - Rh)ulo Chlu!t f/2 for h S h(f) := f/[2C!u!t). Altogether this gives I(Rh - Rh}:.£lo f. (b) From (2k) one infers 0 = limh-+O(Rhu, Ph'u)o = limh-+o(PhRhu, u)o = (u,u)o thus u = o. •
s
s
s
Exercise 11.3.3. Let Ah := I - 8;;8; - 8:8; and A:= 1- .1. Show that
11.3 Discretisation by Difference Methods
lul~
269
IUhl~ = (AhUh, Uh)O, IL~ul~l = (v,L~u)o for v = A-IL~u E HJ(n),
= (Au, u)o,
IL~,hUhl~l = (Vh, L~,hUh)O for Vh = Ah"l L~,hUh' lim I(AhRh - RhA)ul-l = 0 for all U E HJ(n).
h->O
The following analysis is tailored to the properties of the difference methods under discussion. We have tried to avoid a more abstract theory that is applicable to finite elements as well as difference methods. This kind of approach can be found in Stwnmel [I] and Chatelin [I]. The variable Wh(A) is now defined by
Wh(A):=
sup I(L~,hUh' vh)ol =
inf
IUh!1=llvhll=l
inf
IL~.hUhl-l.
(11.3.3)
IUh!1=l
As in Exercise 11.2.4 we have Wh(A) = 0, if A is an eigenvalue of L h; Wh(A) = 1/ILi".ilt+--l otherwise.
(11.3.3')
The analogue of Lemma 11.2.7 reads Lemma 11.3.4. Let A C C[) be compact. Let (2a-m) hold. Then there exist variables C > 0 and '11(h) - 0 (h - 0), independent of A E A, such that
Wh(A)
~
CW(A) - '11(h),
W(A) ~ CWh(A) - '11(h)
for all A E A, h > o. (11.3.4)
PROOF. (a) Since A is compact, W(A) is continuous, and Wh(A) is equicontinuous in A, it is sufficient to show that limh->owh(A) ~ CW(A), and W(A) ~ Climh->owh(A) for all A E A with C > o. (b) Define for A E A and Uh with IUhh = 1 and IL~,hUhl-l = Wh«A)
u:= PhUh,
Zh:= (A - JL)L;,iuh,
Z = (A - JL)L;lu
with JL from (2e). Without loss of generality it can be assumed that J.t We have
¢ A.
zit = 11\(Uh - Zh) + PhZh - zit ~ IPhlt+-lluh - Zh!t + IPhZh - zit, IPhZh - zit = IPh[Zh - RhZ) - [I - PhRh]zlt
Iu -
~ IPhh+-llA - JLIIL;.i(.UhL,. - L,.,h R h]L;lu + L;,i[1 - '&"Ph]Uhh
+ 1(1 -
PhRh)zlt - 0 (h - 0)
(cf. (2gj,m» so that (11.3.5a)
270
11 Eigenvalue Problems
As in (2.6d) one obtains Iu - zit ~ C 2W(A)lUIt,
C 2 > o.
(11.3.5b)
From
Wh(A) = IL,x,hUhl-l = ILJ',hUh + (JL - A)Uhl-l ~ (LJ',hUh + (JL - A)Uh' Uh)O ~ -IJL - Alluhl~ + (LI-',hUh, Uh)O ~ -CAluhl~ + CEluhl~ = -CAluhl~ + CE with CA := max{IJL - AI: A E A} > 0 follows
IUhl~ ~ CAl[CE -Wh(A»). Either Wh(A) ~ CE/2 from which the statement results directly, or Wh(A) $ C E /2 which yields (11.3.5c) We want to show that there exists ho > 0 and Cp = Cp(Co) such that for all Uh with IUhlo ~ COIUhh and h $ ho.
IUhh $ CplPhuhh
(11.3.5d)
The negation of (5d) reads: there exists Uh with IUhh = 1, IUhlo ~ Co and IPhUhh -- 0 (h -- 0). From IRhPhUhlo $ IRhPhUhh $ CIPhUhh -- 0 and IUh -RhPhUhlo ~ II -RhPhlo~lluhh ~ Ch -- 0 follows IUhlo -- 0 in contrast to IUhlo ~ Co. Thus we have (5d). Together with LJ',h(Uh - Zh) = L,x,hUh, (2d) and (5a,b,d) yield the first of the inequalities (4):
Wh(A) = IL,x,hUhl-l = ILJ',h(Uh - zh)l-l ~ CEluh - Zhlt ~ CECllu - zit - 0(1) ~ C EC l C2W(A)lult - 0(1) ~ CW(A) - 0(1) with C:= CEClC2/Cp > O. (c) Let
W(A)
~
> 0 be arbitrary; U E HJ(.f1) with Iuh = 1 can be selected such that IL,xul-l - f. Set Uh := Rhu. According to Exercise 3 it holds that
f
IL,xul~l
= (v, L,xu)o for v = A- l L,xu E HJ(.f1),
IL,x,hUhl~l = (Vh' L,x,hUh)O
A = 1-.1,
for Vh:= Ah l L,x,hUh.
From
Rhv - Vh
= Ahl[AhRh -
RhL,xU - L,x,hUh
RhAjA- l L,xu + Ah1[RhL,x - L,x,hRhju -- 0
= [RhL,x - L,x,hRhju -- 0,
(v, L,xu)o - (RhV, RhL,xU)O = ([I - RhRhjv, L,xu)o -- 0, for h -- 0 (cf. (2m), (2k», one infers IL,x,hUhl-l -- IL,xul-l and
W(A) ~ IL,xul-l -
f
~ IL,x,hUhl-l -
f -
0(1) ~ Wh(A) -
f -
0(1)
(h -- 0)
11.3 Discretisation by Difference Methods
for each
€
> O. Thus w{>.)
~
limh ......owh{>.) has been proved.
271
•
Corollary 11.3.5. Under the assumptions of Lemma 4 the following holds: If there exists L-;:1 for all >. E A then there exists an ho > 0 such that L>..h for all >. E A and h :s he is H~-regular: sup{IL-;:~hi--1: >. E A, h ::; he} ::; c .
.
PROOF. We have assumed w{>.) > 0 in A so max{w{>.): >. E A} =: '" > O. Choose he according to Lemma 4 such that Wh(>') ~ Cw{>.) - !C'" ~ !C'" for h::; ho. Then IL.\~h-+-l ::; 2/(C",) for all >. E A, h ::; he. •
.
The proof of Theorem 11.2.8 can be carried over without change and results in
Theorem 11.3.6. Assume (2a-m). If >'h. (hi -+ 0) are discrete eigenvalues of Problem (1) with >'h, -+ >.0 then >'0 is an eigenvalue of (1.2a). Lemma 11.2.9 and Theorem 11.2.10 can also be carried over without changes.
Theorem 11.3.7. Let (2a-m) hold. Let >.0 be an eigenvalue of (1.2a). Then there exist discrete eigenvalues >'h of (1) Vor all hI such that limh-+o >'h = >'0. Theorem 11.3.8. Let (2a-m) hold. Let eh be discrete eigenfunctions with lehh = 1 for >'h, where >'h -+ >.0 (h -+ 0). Then there exists a subsequence eh. such that Phoeh. converges in HJ{n) to an eigenfunction 0 ~ e E E(>.o). Purther, we have leh, - Rh,elt -+ o.
PROOF. Because IPhehh :s C (cf. (2g) the functions eh := Pheh are unifomIly bounded. HJ(n) is compactly embedded in L2(n) (cf. (2a) and Theorem 6.4.8a) such that a subsequence e h, converges in L2(n) to an e E L2{n): leh • - elo -+ O. We have in particular IRhe - ehlo ::; IRh(e - eh ) - (RhPh - I)ehlo
-+
0 (h = hi
-+
0
-+
0).
Estimate (2c) yields IRh Z - zhlo ::; IRhZ - zhh ::; Chlelo
for Z = {>'o - J.L)L;;.le,
Zh:=
(>.0 - J.L)L;,~Rhe.
From L J4 •h(Zh - eh) = (>'0 - J.L)(Rhe - eh) + (>'h - >.o)eh -+ 0 in H;;l follows IZh - ehl! -+ 0 such that IRh(e - z}lo -+ 0 (h = ht -+ 0) results. Lemma 2b shows that e = Z E HJ(n), and 0 = e E E(>.o) is excluded because of Ilim Rh,eh = llimehh = 1. •
Theorem 11.3.9.
Let (2a-m) hold. Let e h be the solution of the discrete eigenvalue problem Lheh = 'Xhe h with lehlt = 1, limh......O >'h = >.0. Then there
11 Eigenvalue Problems
272
exists a subsequence eh. such that Ph.et in HJ(.!1) converges to an eigenfunction 0 # e* E E*(AO)' Furthermore, leh. - Rh.e*h -+ O. Proof Sketch. The proof is not analogous to that of Theorem 8 since the consistency condition (2a,m) does not necessarily imply the corresponding statements for the adjoint operators. One has to carry out the following steps: (a) e*h := Pheh -+ e* converges in L2(.!1) for a subsequence h = kt -+ O. (b) For Z = (AO - p.)L~-le* and Zh := (Ao - P.)L~:hleh the following holds:
Z - RhZh
= (Ao -
p.)L~-l[(e* - R;;'e;;') + (R;;'L~.h - L,.Rh)]L~:hle;;'.
For each v E L2(.!1) we obtain
(v, Z - R;;'Zh)O
= (Ao -1'){(L;lv, e* - e*h)o + ([P'; - Rh]L;lv, eh)o
+(L;.k[L,..hRh - RhL"lL;lv, eh)o} -+ 0 for h = kt -+ 0 (cf. (2i), (2c». (c) IZh - ehh -+ 0 may be inferred from L~.h(Zh - eh) = (AO - Ah)eh -+ O. In particular, (v, RhZh - Rhe;;')o -+ 0 for each v E £2(n). (d) (v,R;;'e;;' - Rheh)O = ([Rh - Rhlv,eh)o -+ 0 for each v E L2(.!1) (cf. (21». (e) (v, Rheh - e*)o -+ 0 for each h = hi -+ O. (f) From (b) and (e) follows (v, Z - e*)o = 0, thus Z = e* E E*(Ao), where e* # o. Exercise 11.3.10. Carry over Exercise 11.2.12 to the present situation. In Lemma 11.2.16 we defined Wh(A). Now set
Vh := {Uh: (uh,eh)h = O} = {eUl., L;;'eh = Aheh' Ah W(A) := inf sup I(L~.hUh' vh)ol/(Iuhh IVhh).
-+
Ao, (11.3.6)
O#U"EV" O#v"EV"
A basic condition for the following is dim Eh(Ah) = 1,
Eh(Ah) = span {eh},
E;;'(Ah) = span {eU.
(11.3.7)
Here
E(Ah) := {Uh E H~: LhUh = AhUh},
E;;'(Ah):= {Uh E H~: LhUh = XhUh}.
Exercise 10 shows that (7) is valid for h
~
ho if dimE(Ao)
= 1.
Lemma 11.3.11. Let (2a-m), dimE(Ao) = 1, and (2.8) hold. Then there exist ho > 0 and a C > 0 that is independent of h ~ ho and A E 4J such that Wh(A) ~ CWh(A) for h ~ ho. For sufficiently small f > 0 and h, Wh(A) ~ 'TJ > 0 for alllA - Aol ~ f. PROOF. (a) There exists a C > 0 such that
11.3 Discretisation by Difference Methods
273
Indirect Proof: Assume that there exists a sequence Vh. with hi --+ 0, IVh.h = 1, ai e CU, Who := Vh. + aie'h., IWh.ll --+ O. For a subsequence of {hd
ai
--+
a*,
Ph.Vh.
--+
v*
and Ph.et
--+
e* -lOin L2(fJ),
converge. From 0 = (vh.,e'h.)o = (vh.[I - Ph'.Ph.]e;;')o + (Ph.Vh.,Ph.e'h.)o--+ (v*,e*)o one infers (v*,e*)o' = 0, a* = (w*,~*)o/(~*,e*)o = 0, V* = 0'. The contradiction results from 1 = lim IVh. It = lim IWh. It = o. (b) The rest of the proof runs as in Lemma 11.2.16. • Lemma 11.3.12. Let (2a-m), dimE(Ao) = 1, and (2.8) hold. One may choose 0 -I e e E(-\o) and eh e Eh(Ah) so that IRhe - ehlt $ C[h+ 1-\0 - Ahll· PROOF. We have that e = (-\o-IJ)L;le e H2(n)nHJ(fJ). Assume also that Zh := (-\0 - JL)L;,lRhe. The inequality (9.2.22) implies IRhe - zhlt $ Chlel2. For sufficiently small h we have (eh, e'h) -10 so that one can scale eh such that (eh - zh,e'h)o = O. From
L:>.",h(eh - Zh) = (Ah - Ao)Rhe + (Ah - JL)(Zh - Rhe) = (Ah - -\o)Rhe + (Ah - JL)[(Zh - Rhe) + (Rh - Rh)ej one infers that leh - zhh $ CIIAh - -\01 + h] (cf. Lemma 11) since I(Rh Rh)el-l = O(h). The statement follows from this. • Lemma 11.3.13. Under the assumptions of Lemma 12 we have the inequality IAh - -\01 $ Ch. PROOF. Choose eh, eh such that (Rhe - eh, eh)o = (eh, Rhe* - eh)o = O. For the Rayleigh quotient Xh := (LhRhe, Rhe*)o/(Rhe, Rhe*)O we then have
I~h - Ahl $ C1Rhe - ehltlRhe* - ehll $ €h[h + lAO - Ahll with €h := CIRhe* - ehlt· From (2c, j) one infers that
(LhRhe, Rhe*)O - (Le, e*)o = ([LhRh - RhLje, Rhe*)O + (Le, [R'hRh - Ije*)o = O(h)
and I~h - Aol = O(h) such that IAh - Aol $ Ch + €h\Ah - -\0\. For sufficiently small h we have €h $ (cf. Theorem 9, Exercise 10) thus \Ah -Aol $ 2Ch. •
!
Lemmata 12 and 13 give
274
11 Eigenvalue Problems
Theorem 11.3.14. Let (2a-m), E(AO) = span {e}, and (2.8) hold. There exists eh E Eh(Ah) with IRhe - ehh S Ch. Since IRhe* - ehh = 0(1) or even IRhe* - ehh = O(h), according to Theorem 11.2.19 one expects that lAo - Ahl = o(h) [resp. O(h2)). In general, this estimate is false as the following counterexample shows. Example 11.3.15. The eigenvalue problem -u" + u' = AU in (0,1) with u(O) = u(l) = 0 has the solution u(x) = exp(Ax)sin(1rx) with A = 1/2. The eigenvalue is Ao = 1r2 + 1/4. One calculates that the discretisation a+ u = AU has the eigenva.lue Ah = h- 2[2 - cos(1rh)(eA'h
+ e-A'h) -
-a-a+u+
isin(1rh)(e A'h - e-A'h))
+ h-l[cos(1rh)eA'h -1 + isin(1rh)e A'h] = 1r2 + ~ + (~ _ 3;2) h + O(h2) with A' := 10g(1 such tha.t
lAo - Ahl turns out no better than O(h).
h)/(2h)
12 Stokes Equations
12.1 Systems of Elliptic Differential Equations In Example 1.1.11 we have already stated the Stokes equations for - LlUl + 8p/8x l = ft, - LlU2 + 8p/8x2 = 12, - 8ul/8xl - 8u2/8x2 = o.
n ern?: (12.1.1ad (12.1.1a2) (12.1.1b)
In the case of n c rn.s another equation -Llus + 8p/8xs = is needs to be added, and the left-hand side of (1b) must be supplemented by -8us/8xs. A representation independent of the dimension can be obtained if one takes together (U1' U2) [resp. (Ul, U2, us)] as a vector u satisfying, in n, the equations -Llu+Vp=f -divu= O.
(12.1.2a) (12.1.2b)
Here, div is the divergence operator n
divu = 'L8ud8x., i=l
and n is both the dimension of n c mn and the number of components of u(x) E rn.n . Only n S 3 is of physical interest here. In fluid mechanics the Stokes equation describes the flow of an incompressible medium (neglecting the inertial terms) and describes the velocity field. With x E rn.n , Ui(X) is the velocity of the medium in the Xi direction; the function p denotes the pressure. Up to now we have not formulated any boundary conditions. In the following we limit ourselves to Dirichlet boundary values: u=O
onr.
(12.1.3)
This implies that the flow vanishes at the boundary. No boundary condition is given for p. Since both the pairs (u,p) and (u,p+const) satisfy the Stokes equations (2a,b), (3), one obtains: Remark 12.1.1. p is determined only up to a constant by the Stokes equation (2a,b) and the boundary conditions (3).
W. Hackbusch, Elliptic Differential Equations: Theory and Numerical Treatment, Springer Series in Computational Mathematics 18, DOI 10.1007/978-3-642-11490-8_12, © Springer-Verlag Berlin Heidelberg 2010
275
276
12 Stokes Equations
The Stokes equations have been chosen as an example of a system of differential equations. It remains to investigate whether Equation (2a,b) is elliptic in a sense yet to be defined. Even though the functions Ui, for given p, are solutions of the elliptic Poisson equations -LlUi = Ii - Op/OXi, in determining p, (2a,b) do not provide an elliptic equation, in any current sense of elliptic. A general system of q differential equations for q functions U l , " ' , u q can be written in the form q
LLijuj
= Ii
in
n c IR7l
(1 ~ i ~ q)
(12.1.4a)
j=1
with the differential operators
Lij =
L
caDa
(1 ~ i,j ~ q).
(12.1.4b)
lal:5k.; The equations (4a) are summarised as Lu = I where L is the matrix (Lij) of differential operators and U = (Ul,'" ,uq)T, I = (ft,···, Iq)T. The order of the operator Lij is at most ~j. Let the numbers ml. ... , mq, mL ... , m~ be chosen so that (12.1.5) ~j ~ ffii + mj (1 ~ i,j ~ q). As the principal part of Lij one defines
L~:=
L
caDa.
lal=m.+mi
< mi + mj, Ca = 0 holds for lal = mi + mj, and thus L~ characteristic polynomial associated with L~ reads:
If ~j
= O. The
and forms the matrix function
Definition 12.1.2. (Agmon-Douglis-Nirenberg [1]) Let (5) hold for mj. The differential operator L = (Lij) is said to be elliptic in x En if
ffii,
(12.1.6a) L is said to be uniformly elliptic in
n if there exists an
f
> 0 such that (12.1.6b)
with 2m := 2:;=1 (mi +mi). To be more precise one should call L elliptic with indices mi, mj, since the definition does depend on mi, mj. In connection with
12.1 Systems of Elliptic Differential Equations
277
this problem, as well as for an additional condition for x E r, q = 2, see the original paper of Agmon-Douglas-Nirenberg [1]. Further information on this subject can be found in Cosner [1].
Exercise 12.1.3. (a) Show that the numbers mi, mj are not unique. If mi and mj satisfy the inequality (5), then so do ffii -k and mj+k. The definition of L~ is independent of k. (b) Show that for q = 1, Le., for the case of a single equation one recovers from (6a) the Definitions 1.2.3a [resp. (5.1.3a)]; (6b) corresponds to (5.1.3a'). (c) For a first-order system (Le., kij = 1, mi = 1, mj = 0), (6a) coincides with Definition 1.3.2. In order to describe the Stokes equations in the form (4a, b) we set q:=n+l,
Lii = -..:1,
u=(Ul, ... ,un,p)T,
Liq = -Lqi = O/OXi for 1
The orders are numbers
~i =
2,
~q
= k qi = 1 (i
~
f=(ft"",fn,O?, ~
i ~ n,
Lij = 0 otherwise.
n), and k ij = 0 otherwise. The
mi = m~ = 1 for 1 ~ i ~ n = q - 1,
mq = m~ = 0
satisfy (5). L~ coincides with Lij and is independent of x: L{;(~) = _1~12,
L:;(~) = -L~(~)
= ~i
for i ~ n,
Lf;(~)
=0
otherwise.
From this we see
IdetLP(~)1 = /det
[
o
_1~12
0
-6 so that (6b) is satisfied, with
co
= 1.
Exercise 12.1.4. In elasticity theory the so-called displacement function n c rn.3 _ rn.3 is described by the Lame system:
u:
J.L..:1u + (,\ + J.L)Vdivu = f
in il,
U
= cp
on
r
(ft, >. > 0). Show that this system of three equations is uniformly elliptic: Idet L P (~)I = J.L2 (2J.L + '\)1~16.
For the treatment of Stokes equations we will use a variational formulation in the next section. For reasons of completeness we point out the following transformation.
Remark 12.1.5. Let n = 2 and thus U = (UbU2). Because divu = 0 there exists a so-called stream function i.P with Ul = oi.P/OX2, U2 = -0iP/OX1.
12 Stokes Equations
278
Insertion in Equation (1al,2) results in the biharmonic equation Ll2cp = a/2/aXl - afdaX2. The boundary condition (3) means VCP = 0 on r. This is equivalent to acp/an = 0 and acp/at = 0 on r where a/at is the tangential derivative. acp/at = 0 implies cp = const on r. Since the constant may be chosen arbitrarily, one sets cp = acp/ an = 0 on r.
12.2 Variational Formulation 12.2.1 Weak Formulation of the Stokes Equations Since U = (Ub···, un) is a vector-valued function, we introduce HA(n) := HJ(,G) x HJ(n) x ... x HJ(J])
(n-fold product).
A corresponding definition holds for H-l(n), H2(n), etc. The norm associated with H6(n) will again be denoted by I· It in the following. According to Remark 12.1.1 the pressure component p of the Stokes problem is not uniquely determined. In order to determine uniquely the constant in p = V+const, we standardise p by the requirement p dx = o. That is the reason why in the following p will always belong to the subspace L~(n) C L2(n):
In
L~(n) := {p E L2(n):
L
p(x) dx = O}.
To derive the weak formulation we proceed as in Section 7.1 and assume that and p are classical solutions of the Stokes problem (1.2a,b). Multiplication of the ith equation in (1.2a) with Vi E c8"(n) and subsequent integration implies that U
In =
l
fi (X)Vi (x) d(x) = LI-Llui(x) + ap/aXi]Vi(X) dx
(12.2.1a)
{(VUi(X), VVi(X)} - p(X)aVi(X)/aXi} dx for Vi E coc(n), 1:CS; i
:cs;
n.
Summation over i now gives
l
{(Vu(x), Vv(x)) - p(x) div, vex)} dx =
l
(f(x), vex)) dx,
(12.2.1a')
where the abbreviation
is used. Equation (1.2b) is then multiplied with some q E L~(n) and integrated, giving
12.2 Variational Formulation
-In
q(X) divu(x) dx = 0 for all q E
L~(il).
279
(12.2.1b)
With the bilinear fonus a( u, v) :=
In
(Vu(x), Vv(x)} dx for u, v E H6( il),
b(P,v) := - iP(X)diVV(X)dx for P E
L~(il),v E HA(il),
(12.2.2a) (12.2.2b)
we obtain the IVweak formulation of the Stokes problem as (3a-c): Find
U
a(u,v)
E HA(il)
and P E L~(il)
+ b(P,v) = f(v):=
b(q, u)
=0
In
such that
(12.2.3a)
(f,v)dx
for all v E H6(fl), (12.2.3b)
for all q E L~(il).
(12.2.3c)
In (3b) we first replace "v E H6(il)" by "v E G~(il)". Since both sides of (3b) depend continuously on v E H6(il) and since G~(il) is dense in H6(il), (3b) follows for all v E HJ(il). Remark 12.2.1. A classical solution u E G2(il)nHfi(il), P E Gl(il)nL~(il) of the Stokes problem (1.2a,b), (1.3) is also a weak solution, i.e., a solution of (3a-c). If conversely (3a-c) has a solution with u E G 2 (!}), P E Gl(!}), then it is also the classical solution of the boundary value problem (1.2a,b), (1.3). PROOF. (a) The above considerations prove the first part. (b) Equation (3c) implies divu = O. Let i E {I,··., n}. In Equation (3b) one can choose v with Vi E G8"(fl), Vj = 0 for j ¥= i. Integration by parts recovers (la) and hence the ith equation in (1.2a). •
12.2.2 Saddlepoint Problems The situation in (3a-c) is a special case of the following problem. We replace the spaces H6(il) and L~(il) in (3a-c) by two Hilbert spaces V and W. Let a(·, .): V x V b(., .): W x V -
m. m.
a continuous bilinear form on V x V,
(12.2.4a)
a continuous bilinear form on W x V,
(12.2.4b)
11 E V',
hE W'.
(12.2.4c)
In generalisation of (6.5.1) we call b(., .): W x V if there exists a Gb E m. such that
m. continuous (or bounded),
Ib(w,v)1 ~ Gbllwllwllvllv for all w E W;v E V. The objective of this chapter is to solve the problem (5a-c):
280
12 Stokes Equations Find v E V
and w E W
with
(12.2.5a) (12.2.5b) (12.2.5c)
a(v,x) + b(w,x) = h(x) for all x E V; b(y,v) =h(Y) forallyEW. Formally, (5a-c) can be transformed to the form Find u E X
with c(u, z)
= f(z)
for all z EX
(12.2.6a)
if one sets: X
=V
x W;
c(u, z) := a(v, x) + b(w, x) + bey, v)
f(z)=/t(x)+h(Y)
and
for u= (:).z= (:).
(12.2.6b)
Exercise 12.2.2. Show that (a) c(·, .): X x X -+ lR is a continuous bilinear form. (b) Problems (5a-c) and (6a,b) are equivalent. That the variational problems (5) and (6) must be handled differently than in Section 7 is made clear by the following remark. Remark 12.2.S. The bilinear form c(·,·) in (6b) cannot be X-elliptic. PROOF. We have c(u, u)
= 0 for all u = (~).
•
In analogy to (6.5.9) we set
J(v, w) := a(v, v) + 2b(w, v) - 2h(v) - 2h(w), and therefore J(v, w) = c(u, u) - 2f(u) for u = ( : ) . For v E V, w E W we know J( v, w) is neither bounded below nor above. Therefore the solution v*,w* of (5a-c) does not give a minimum of J; however, under suitable conditions, v·,w* may be a saddle point.
Theorem 12.2.4. Let (4a-c) hold. Let a(-,·) be symmetric and V-elliptic. The pair v* E V, w* E W is a solution of the problem (5a-c) if and only if J(v*,w):5 J(v*,w*):5 J(v,w*)
for all v E V;w E W.
(12.2.7)
Another equivalent characterisation is J(v*,w*)
= minJ(v,w*) = wEW max min J(v, w). tJEV tJEV
PROOF. (aa) Let v*,w* solve (5). The expression in brackets in
(12.2.8)
12.2 Variational Formulation
281
J(v, w*) - J(v*, w*) = a(v* - v, v* - v) -2[a(v*, v* - v)
+ b(w*, v* -
v) - ft(v* - v)]
vanishes because of (5b). Since a(v* - v, v* - v) > 0 for all v* =I- v E V, the second inequality in (7) follows. One also proves the converse as for Theorem 6.5.12: If J(v, w*) is minimal for v = v* then (5b) holds. (ab) If v* is a solution of (5c), then J(v*, w*) - J(v*, w) = 2[b(w* - w, v*) - h(w* - w)]
vanishes for all w, which proves the first part of (7) in the stronger form J(v*,w) = J(v*,w*).If, however, v* is not a solution of (5c), there exists a wE W such that 6 := J(v*, w*) - J(v*, w) =I- O. Since J(v*, w*) - J(v*, w) = -6 for w := 2w* - w, the first inequality cannot be valid. For the converse define w± = w* 1= w. The first part of (7) implies 0:5 J(v*, w*) - J(v*, w±)
= ±2[b(w, v*) - h(w)]
for both signs, and so b(w,v*) = h(w). Since wE W is arbitrary, one obtains (5c). (ba) We set j(w) := minvEvJ(v,w). According to Theorem 6.5.12, j(w) = J(vw, w), where Vw E V is the solution of (5b). If Vw and Vw' are the solutions for wand WI, it follows that a(vw - vw',x) = F(x) := b(w - wl,x)
Since
IIFllv' :5 Cbllw -
for all x E V.
wlllv and IIvw - vw'llv :5 C/IIFllv' one obtains
IIvw - vw,lIv :5 Cllw - w/llv
for all w, WI E W.
(12.2.9a)
(bb) By using the definition of Vw in (5b) we can write: J(v*, w*) - J(vw, w) = a(v*, v*) + 2b(w*, v*) - 2ft(v*) - 2h(w*) - [a(vw, vw) + 2b(w, vw) - 2ft (vw) - 2h(w)]
= 2[ft(vw - v*) - b(w, Vw - v*) - a(vw, Vw - v*)] + a(vw - v*, Vw - v*) + 2[b(w* - w, v*) - h(w* - w)] = a(vw - v* ,Vw - v*) + 2[b(w* - w, v*) - h(w· - w)]. (12.2.9b) (bc) Let v*, w* be a solution of (5a-c). Because of (5c) the expression in brackets in (9b) vanishes and we have J(v*, w*)
(5b) gives
V w•
= J(vw, w) + a(vw - v*, Vw - v*)
= v*,
~
J(vw, w)
= j(w).
and so J(v*,w*) =j(w*)
= maxj(w)j wEW
(12.2.9c)
282
12 Stokes Equations
i.e., (8) holds. (bd) Now let v*, w* be a solution of (8). If in (9b) one sets w = w*, Vw = Vw' one obtains from J(v*,w*) = j(w*) that v* = Vw" Hence a(vw-v*,vw-v*) = a( Vw - V w' , Vw - vw') depends quadratically on IIw - w* Ilv (cf. (9a». The variation over w := w* - AY (A E 1R, yEW arbitrary) gives 0=
d~j(w* - Ay)/.~=O = 2[b(y, v*) -
f(y»),
and so (5c) is proved. (5b) has already been established with v* =
Vw "
•
12.2.3 Existence and Uniqueness of the Solution of a Saddlepoint Problem To make the saddlepoint problem (5a-c) somewhat more transparent we introduce the operators associated with the bilinear forms: with a(v,x) = (Av,x)v1xv for v,x E V, (12.2.lOa) B E L(W, V'), B* E L(V, W')
A E L(V, V')
= (Bw, x)v'xv = (w, B*X)WXW" (12.2.10b) = (Cu,z)x'xx for u,Z E X. (12.2.lOc) now has the form Cu = f, while (5a-c) can be written
with b(w, x)
C E L(X,X') with c(u,z) Thus problem (6a,b) as
Av+Bw=
11
B*v =/2.
(12.2.11a) (12.2.11b)
If one assumes the existence of A-l E L(V', V), one can solve (lla) for v: (12.2.12a) and substitute in (llb): B* A-1Bw
= B*A- l 11- h·
(12.2.12b)
The invertibility of A is in no way necessary for the solvability of the saddlepoint problem. However, it does simplify the analysis, and does hold true in the case of the Stokes problem. Remark 12.2.5. (a) Under the assumptions
A- l E L(V', V),
(B*A-1B)-1 E L(W', W)
(12.2.13)
the saddlepoint problem (5a-c) [resp. Equations (1Ia, b») are uniquely solvable. (b) A necessary condition for the existence of (B* A -1 B)-l is
B E L(W, V') is injective.
(12.2.14)
12.2 Variational Formulation
283
PROOF. (a) Under the assumption of (B* A-1B)-1 E L(W', W), (12b) is uniquely solvable for w, and then (12a) yields v. (b) The injectivityof B* A-1 B implies (14). • Attention. In general, B: W -+ V' is not bijective so that the representation of (B* A-1 B)-l as B-1 AB*-l is not possible. The example of the 3x 3 matrix C =
(1-1 -1) ( A B). 0 WIth A 1 and BT
=
B =
1 shows that a system (1)
of the fonn (lla,b) can be solvable even with a singular matrix A. Therefore the assumption A -1 E L(V', V) is not necessary. A closer look reveals the subspace V: bey, v) = 0 for all yEW} c V, (12.2.15) which, as we noted before, in general is not trivial. The kernel of a continuous mapping is closed so that V, according to Lemma 6.1.17, can be represented as the sum of orthogonal spaces:
Vo
:= ker B* = {v E V: B*v =
V
O} = {v
E
= Vo ED VJ.. with VJ..:= (Vo)J...
(12.2.16a)
Exercise 12.2.6. Let (16a) hold. Show that (a) V' can be represented as V' = Vti ED Vi, where Vti := {v' E V': v' (v) = 0 for all v E VJ..},
(12.2.16b) (12.2.16c)
Vi := {v' E V':v'(v) = 0 for all v E Vo}.
As a nonn on Vo and Vi one uses II . IIv'. (b) The Riesz isomorphism Jv: V -+ V' maps Vo onto Vo and VJ.. onto Vi. (c) Vo and Vi are orthogonal spaces with respect to 1I·lIv'. (d) The following holds:
IIv'II~, = Ilvtill~'o + Ilvl.lI~'.J..
for v'
= vti +Vl.,vti E Vti,Vl.
E Vi. (12.2.16d)
The decompositions (16a,b) of V and V' define a block decomposition of the operator A: A=
Aoo E L(Vo, Vti),
[1:
AoJ.. E L(VJ.., Vti),
1:~], AJ..o E L(Vo, Vi),
AJ..J.. E L(VJ.., Vi).
Here, for example, Aoo is defined as follows: Aoovo
= vti for
Vo E
Vo if Avo = vti + Vl. with vti E Vti, Vl.
E Vi·
The corresponding decomposition of B* into (Bo, Bl) is written as (0, B*), since = 0 according to the definition of Yo. Conversely, we have range (B) C Vi, so that B = (~). System (lla,b) thus becomes
Bo
284
12 Stokes Equations
Aoovo + Aol. v l. = flO Al.ovo + Al.l. Vl. + Bw = fa B*vl. = h where v
= Vo +Vl., Vo E Yo,
Vl. E Vl., it
= flO + fa,
(12.2.17a) (12.2.17b) (12.2.17c) ito E VcJ, fa E V.f.
Theorem 12.2.7. Let (4a-e) hold. Let Va be defined by (15). A necessary and sufficient condition for the unique solvability of the saddlepoint problem (5a-e) for all f1 E V' is the existence of the inverses AiXl E L(V~, Yo),
B- 1 E L(vI, W).
(12.2.18)
PROOF. (a) (17a-e) represents a staggered system of equations. (18) implies B*-1 = (B-1)* E L(W', Vl.) so that one can solve (17c) for Vl. = B*-1 h. Vo E Va is obtained from (17a): Vo = AO01 UlO - Aol.vl.). Finally, w results from (17b). (b) In order to show that (18) is necessary, we take ito E VcJ arbitrary, fa = 0 and h = O. By hypothesis here we have a solution (vo, Vl., w) E Va X Vl. X W. B*vl. = 0 implies Vl. E Va, so that Vl. = 0 because Vo n Vl. = {o}. Thus Aoovo = flO has a unique solution Vo E Va for each flO E V~. Since Aoo: Va VcJ is bijective and bounded, Theorem 6.1.13 shows that E L(Vo, Va). If one takes fa E Vi arbitrary and flO = 0, h = 0, one infers Vl. = 0 and Vo = 0, so that Bw = fa has a unique solution w E W . .Ai?, we did for A oo , one also infers that B-1 E L(Vi, W). •
AOl
The formulation of conditions (18) in terms of the bilinear forms results in the Babuska-Brezzi conditions: inf{sup{la(vo,xo)l:xo E Va, IIxoliv
= 1}:vo E Va, IIvoliv = I}
~
a > 0, (12.2.19a) sup{la(xo, vo)l: Xo E Va, IIxoliv = I} > 0 for all 0 =I Vo EVa, (12.2.19b) inf{sup{lb(w,x)l:xE V,lIxliv = l}:WE W,lIwllw = 1}~,8> O. (12.2.19c) Exercise 12.2.8. Show that (19a) [resp. (19c») are equivalent to (19a') [resp. (19c'»): for all Vo E Yo, (12.2.I9a') for all wE W. (12.2.I9c')
sup{la(vo,xo)l:xo E Va,lIxoliv = I} ~ allvollv sup{lb(w,x)l:x E V, IIxliv
= I} ~ ,8l1wllw
Lemma 12.2.9. Let (4a,b) hold. Let Vo be defined by (15). Then conditions (18) and (19a-e) are equivalent. Here we have IIAiXlllvo<-v~ $ I/a, IIB- 1l1w<--vl $ 1/,8.
12.2 Variational Formulation
PROOF. Because of (16d) and b(w, x) the form inf{sup{lb(w,x)l:x e V..L, IIxliv
285
= 0 for x e Vo one can write (19c) in
= 1}:w e W, IIwllw = I} 2 (3 > O. (12.2.19d)
For 0 =I x e V..L one has that x ¢ Yo, therefore according to (15): sup{lb(w,x)l:we W,
IIwllw = I} > 0
for all 0 =I x e V..L.
(12.2.1ge)
As in the proof of Lemma 6.5.3, we obtain the equivalence of (19a,b) with •
Aol e L(V&, Yo) and of (19d,e) with B-1 e L(Vl., W).
Corollary 12.2.10. (a) Condition (19b) becomes superfluous if a(·, .) is symmetric on Vo x Yo or if Lemma 6.5.17 applies. (b) Each of the following conditions is sufficient for (19a,b) and hence also e L(V&, Yo): for
AoJ
a(·, .): Yo x
Yo ---+ ill. is Yo-elliptic:
a(vo,vo) 2 allvoll~ for all Vo e Yo, a(·, .): V x V ---+ ill. is V -elliptic.
(12.2.20a) (12.2.20b)
PROOF. (a) As in Lemma 6.5.17. (b) (20b) implies (20a); (20a) yields (19a,b). • Exercise 12.2.11. Show that under the assumptions (4a--c) , (18) is also equivalent to the existence of C- l e L(X',X) (cf. (lOc)). Find a bound for IIC-lllx+--x' in terms of IIAolllvo+--v~, IIAllvl+--VI, and IIB-lllw+--vl'
12.2.4 Solvability and Regularity of the Stokes Problem Conditions (19a,b) (Le., problem:
AOol
e L(V&, Yo)) are easy to satisfy for the Stokes
Lemma 12.2.12. Let D be bounded. Then the forms (2a,b) describing the Stokes problem satisfy the conditions (4a,b) and (19a,b). PROOF. (4a,b) is self-evident. According to Example 7.2.10, In('V'u, \7v} dx is HJ(D)-elliptic. From this follows the HA(D)-ellipticity of a(·, .). Corollary lOb proves (19a,b). • It remains to prove condition (19c), which for the Stokes problem assumes the form
SllP{lk w(x)divu(x)dxl:u e HA(D), lull = I} 2 (3lwlo for all w e
L~(D),
(12.2.21)
12 Stokes Equations
286
or
IIVwIIH-l(.a) ~
.8l1wIlL2(0)
for all
w E L~(a).
(12.2.21')
Lemma 12.2.13. Sufficient and necessary for (21) is that for each w E L~(il) there exists 1.1. E H6(il) such that
(12.2.21") PROOF. (a) For w E L~(a) select 1.1. such that (21") holds and set u := u/luh. The left-hand side in (21) is ~ fa w(x) divu(x)dx = Iwl6!luh ~ .8lwlo. (b) If (21) holds one infers as in Section 12.2.3 the bijectivity of B*: V.L ~ W with IIB*-lllv.l.+-w ~ 1/.8. Therefore 1.1.:= B*-lw satisfies condition (21") .
•
Necas [2) proves
Theorem 12.2.14. Condition (21) is satisfied if a E CO,l is bounded. Under this assumption, the Stokes problem
-Llu+Vp=f,
-divu=gina,
1.1.=0 onr
(12.2.22)
has a unique solution (u,p) E H6(n) X L~(il) for f E H-l(a) and 9 E L~(a), with (12.2.23) 11.1.11 + 1P10 $ Cn[ Ifl-1 + Iglo). Remark 12.2.15. Under the conditions that n = 2 and that a E (jl is a bounded domain...the existence proof can be carried out as follows.
PROOF. We need to prove (21"). For w E L~(a) solve -Ll
1.1.2:= -
Clearly, 1.1.1,1.1.2 E Hl(il). Let the normal direction at x E r be n(x) = (n1(x), n2(x»T. The tangent direction is thus t(x) := (n2(x), -nl(x»T. For 1.1. = (1.1.1. u2f one obtains
(1.1., n)
=
-<pa;n1-t/Jyn1 -
= -8
(1.1., t) = -<pa;n2 - t/JlJn2 +
= 0 on r
12.2 Variational Formulation
287
•
H 2-regularity
The above proof uses the of the Poisson problem and requires corresponding assumptions on ll. Theorem 14 also assumes II E CO,l. Since the Poisson equation -L1u = f is solvable for any domain II which is contained in the disk KR(O), or at least in a strip {x E IRn : Ixl! < R}, resulting in the inequality luh ~ CRlfl-l, one might conjecture that a similar result holds for the Stokes problem. Counterexample 12.2.16. For € E (0,1) let lle = {(x,y):-1 < x < 1, y < € + (1 - €)Ixl} (cf. Figure 1). All lle E CO,l are located in the rectangle (-1,1) x (0,1). Nevertheless, there exists no f3 > 0 such that (21/1) holds for all € > 0, wE L2(lle).
o<
o
-1
1
Figure 12.2.1.
n.
PROOF. We select w E L~(lle) such that w(x, y) = 1 for x> 0, w(x, y) = -1 for x ~ o. Let the inequality (21/1) hold for lle with f3e > o. Let u E Hli(lle) with luh ~ Iwlo/ f3e be chosen according to Lemma 13. We continue u by u = 0 onto IR2. For the restriction on x = 0 we have according to Theorem 6.2.28
Let X(y) = 1 for 0 < y < € and x(y) U1(O,y)X(Y) and Ixlo =.,fi, we have
=
0 otherwise. Since U1(0,y)
=
lfo u (0,y) dy l = 11m. U (0,y)x(y) dy l ~ IU1(0,·)lolxlo ~ 2C/€If3e. e
1
1
Let at = {(x,y) E lle:x > O} and'Y := {(x,y):x = 0, 0 < y < €} allt\alle• Because w = 1 in llt, and because divu = w we have
=
288
12 Stokes Equations
1
1/25 [ Iw(x)1 2 dx = [ w divudx = [ divudx = (u, n} dr Jnt Jnt Jnt ~
=-
i
U1 dr =
-1€
u1(0,y)dy.
The last two inequalities result in 1/25 2C,ff./f3E, from which we infer that f3E > 0 cannot be independent of €. • Exercise 12.2.17. Construct a domain il located on the strip m. x (0,1) in which the Stokes equations are not solvable. Hint: Join the domains ill/v (11 E 1N) from Figure 1. As for the case of scalar differential equations one obtains stronger regularity of the Stokes solution u,p if one assumes more than f E H-1(il). Theorem 12.2.18. Let il be bounded and sufficiently smooth. Let u and p be the (weak) solution of the Stokes problem (22) with f E Hk(il), g E Hk+1(n) n L~(n) for k E 1N U {O}. Then we have u E Hk+2(il) n H6(il), p E Hk+1(il) n L~(il) and there exists a C depending only on il such that (I2.2.23b)
•
PROOF. Cf. LadyZenskaja [1, Chap.III, §5]
In analogy to Theorem 9.1.22 it is sufficient to assume the convexity of n in order to obtain u E H2(il) and p E H1(il) from f E L2(il). Theorem 12.2.19. (Cf Kellogg-Osborn (1)) Let n c m.2 be bounded and convex. If f E L2(il), then the Stokes equation (1.2a,b) has a unique solution u E H2(n) n HA(n), p E H1(n) n L~(il), which satisfy the estimate (I2.2.23c) For the more general problem (22) with g the solution satisfies
¥= 0 in a convex polygonal domain (I2.2.23d)
if f E L2(n) and g E L~(n) n Hgcn). Here, HJ(il) is the subspace of H1(n) with the following (stronger) norm: 2 IIgIlHl(n):= 1 "'" LJ liD a gIlL2(n)
2 11/2 + 116-1 gIIL2(n)
with
lal=l
6(x) := min {Ix - el: e E r
comers of the polygon n}.
12.2 Variational Formulation
289
12.2.5 A Vo-elliptic Variational Formulation of the Stokes Problem Vo C H6(.G) has been defined in (15) by Vo := {u E H6(.f1):divu = o}. As kernel for the mapping B* = -div E L(HA(.f1), L~(.f1», Vo is a closed subspace of Hfi(.f1), Le., again a Hilbert space for the same norm I· \1. In the following we investigate the problem Find u E
Yo with a(u, v) = f(v) for all v E Yo,
(12.2.24)
where a(u, v) := fo('\7u(x), Vv(x») dx. Problem (24) has the same form as the weak formulation of the Poisson equations -Ll'Ui = fi (i = 1,···, n), only here H6(.G) has been replaced by Yo. Lemma 12.2.20. Let.G be a bounded domain (or bounded in one direction; cf. Exercise 6.2.12b). The form a(·,·) is Yo-elliptic. The constant CE > 0 in a(u, u) ~ CElul~ depends only on the diameter of .f1. In particular, problem (24) has a unique solution u E Yo with lull CE11flv~ for f E
vo·
:s
PROOF. The HA(.f1)-ellipticity of a(·,·) (cf. Lemma 12) carries over to Yo c H6(.f1) (cf. Exercise 6.5.6a). This implies the other statements (cf. Theorem .. 6.5.9) Theorem 12.2.21. Let.f1 E CO,l be a bounded domain. Assume f E H-1(.f1). Then the solution u E Yo of problem (24) coincides with the solution component u of the mixed formulation (3a-c). PROOF. According to Exercise 6 one can split f E Hfi(il) in such a way that
o, hEV', fo(v) =0 for v EV1., h(v) =0 for
f=fo+h, foEV
VEYo.
In (24) one can replace f(v) by fo(v). u E Vo c H6(.f1) results in -Llu E H-1(.f1). The part of -Llu that belongs to Vi is g1. E Vi with 91.(V1.) := a(u, V1.)
for all V1. E V1.
= (Yo)1..
Theorem 14 proves the condition (21'), which results in the bijectivity of B: L~(.a) -+ Vi for B = V (cf. Lemma 13). p := B-1(h - 91.), by definition, satisfies
= (Bp,V1.)H-l(O) XH W"l) = h(v1.) - 91. (V1.) = h(v1.) - a(u,v1.) all v 1. E V1.. Furthermore, since div Vo = 0 for Vo E Vo, it follows that
b(p,v1.)
for
= 0 for all Vo E Vo. be split into v = Vo + V1.
b(p, vo)
For arbitrary v E HA(.G), to v 1. E V1., one obtains b(p, v)
= b(p, vo) + b(p, V1.) = h(v1.) -
with Vo E
Yo and
a(u, V1.) = f(v1.) - a(u, v) + a{u, vo)
290
12 Stokes Equations = I(vi.) - a(u, v)
+ I(vo) =
I(v) - a(u,v),
by (24). Since also b(w,u) = 0 for all w E L~(n) because u E Yo, u and p satisfy the variational formulation (3a-c) of the Stokes problem. • Note that Problem (24) is solvable for all bounded domains although Problem (3a-c) depends more sensitively on n (cf. Counterexample 16). Theorem 21 shows that only the component p has a domain-dependent bound Iplo ~ 0 0 1/1-1 while luh ~ 01/01-1 ~ 01/1-1 holds for all n c KR(O). Strictly speaking, the variational problem (24) is not equivalent to the Stokes problem since, for example, for n from Exercise 17 the Stokes equations are not solvable, whereas Problem (24) definitely has a solution. The original formulation (3a-c) may be interpreted as Equation (24), into which one has introduced the "side condition" div u = 0 using the Lagrange function p (cf. Section 8.3.6).
12.3 Mixed Finite-Element Method for the Stokes Problem 12.3.1 Finite-Element Discretisation of a Saddlepoint Problem One would have an ordinary Ritz-Galerkin discretisation if in the variational formulation (2.24) one were to replace the space Vo by a finite-dimensional subspace Vh C Vo. But this is not as easy as it sounds. Exercise 12.3.1. Let the square n = (0,1) x (0, 1) be triangulated regularly as in Figure 8.3.2, 5a or decomposed into grid squares. Let V~1) c HJ(n) be the space of the finite triangular elements (cr. (8.3.8)) [resp. of bilinear elements (cf. (8.3.12b))]. Define the corresponding subspace for the Stokes problem as Vh := {u = (ut, U2): U1, U2 E V~1) and divu = O} c Va. Show that Vh contains only the null function. The remaining procedure is thus oriented toward the weak formulation (2.3a-c). The space X = V x W is replaced with Xh = Vh X Who The discrete problem Find v h E Vh and w h E Wh with a(vh,x) + b(w\x) = hex) for all x E Vh, bey, v h ) = hey) for all y E W h,
(12.3.1a) (12.3.1b) (12.3.1c)
is called a mixed Ritz-Galerkin problem [resp. a mixed finite-element problem if Vh and W h are formed from finite elements]. To the formulation (2.6a,b) corresponds the equivalent way of writing (1a-c) Find xh E Xh with c(x\ z)
= I(z)
for all z E Xh.
(12.3.1')
12.3 Mixed Finite-Element Method for the Stokes Problem
291
In the case of the Stokes equations, (2.3a-c) provides the desired solution which satisfies the side condition divu = O. However, the finite-element solution of (la-c) generally does not satisfy the condition divuh = O. One can view v h from (la-c) as the solution of a nonconformal finite-element discretisation of (2.24), as is shown in the following exercise. Exercise 12.3.2. Let h = 0 in (lc)j set VO,h := {x E Vh: b(y,x) = 0 for all y E W h }. Show that each solution v h in (la-c) is also a solution of Find v h E Vo,h with a(v\x) = !1(x) for all x E VO,h.
(12.3.2)
Since in general Vo,h ct. Vo (cf. (2.15», (2) is a nonconformal discretisation of (2.24). Let {by, .. " b~V,h} and {br', . . " b~W,h} be suitable bases of Vh and Wh, where NV,h = dim Vh, NW,h:= dim Who (12.3.3a) Let the coefficients of v E Vh and W E Wh be v and w: NV,h V = Pvv:=
L: Vibr,
W
NW,h -- P Ww..- "" L..J Wi biW .
i=l
(I2.3.3b)
i=l
As in Section 8.1, we prove
Theorem 12.3.3. The variational problem (Ia-c) is equivalent to the system of equations
[~r B;] [:] = U~]
(I2.3.4a)
where the matrices and vectors are given by
Ah,ij := a(bj,bf), fi,i := !1(bf),
(1::; i,j::; NV,hj 1::; k::; NW,h) (I2.3.4b) (1::; i ::; NV,h; 1 ::; k ::; NW,h)' (I2.3.4c)
Bh,ik:= b(br',bf)
h,k:= h(br')
The connection between (Ia-c) and (4a-c) is given by v h = Pvv, For u := f := (~~) one obtains the system of equations
C:'),
Chu=fj
h
C :=
[~r ~h],
wh = Pww. (12.3.4a/)
which corresponds to the formulation (1').
12.3.2 Stability Conditions When selecting the subspaces Vh and Wh one has to be careful, for even seemingly reasonable spaces lead to singular or unstable systems of equations (Ia-c). The former occurs in the following example.
292
12 Stokes Equations
Example 12.3.4. Let the Stokes problem be posed for the L-shaped domain in Figure 8.3.2. For all three components Ut. U2, and p we use piecewise linear triangular elements over the triangulation T which is given by the second or third picture of Figure 8.3.2. Here, let Ul = U2 = 0 on r and let p E Wh satisfy the side condition p E L~(n), i.e., foPdx = O. Then the discrete variational problem (Ia--e) is not solvable. PROOF. The second [third] triangulation in Figure 8.3.2 has 5 (10) inner nodes Xi, with each of them carrying values U1(xi) and U2(X i ). Thus dim Vh = 2· 5 = 10 [dim Vh = 20]. Because of the side condition foPdx = 0 the dimension of Wh is smaller by one than the number of inner and boundary nodes: dim Wh = 21 - 1 = 20 [dim Wh = 21]. In both cases the statement results from Lemma 12.3.5. It is necessary for the solvability of (Ia-c) that NV,h i.e., dimVh ~ dimWh.
~
NW,h,
PROOF. According to Theorem 3 the solvability of (la-c) is equivalent to the nonsingularity of the matrix eh in (4a'). Elementary considerations show that rank (eh) ~ NV,h + rank (Bh)' Since Bh is an NV,h x NW,h-matrix it follows from NV,h < NW,h that rank (eh) < NV,h + NW,h and hence e h is singular. • Thus an increase in the dimension of Wh does not always lead to an better approximation of w E W. The choices of Vh and Wh must be mutually adjusted. The inequality dim Vh ~ dim Wh corresponds to the requirement that B in (2.lOb) needs to be injective, but not necessarily to be surjective. In order to formulate the necessary stability conditions, we define (12.3.5)
(cf. Exercise 2). VO,h is the discrete analogue of the space Vo in (2.15). The conditions, which can be traced back to Brezzi [I], read: inf{sup{la(vo,xo)l: Xo E VO,h, Ilxollv
= 1}:vo E VO,h, II voII V = I} ~ ah > 0, (12.3.6a)
inf{sup{lb(w,x)l: x E Vh, IIxll v
= 1}:w E Wh, IIwllw =
I} ~ Ph> O.
(12.3.6b) Theorem 12.3.6. Let (2.4a--e) hold and dim Vh < 00. The Brezzi conditions (6a-b) are necessary and sufficient for the solvability of the discrete problem (la-c). The solution u h = (vn, w h ) E Vh X Wh satisfies lIu h llx := [lIvhll~
+ IIwhll~ ]1/2 ~ Ghllfllx'
:= Gh[lIftllv'
+ IIhll~/]1/2,
(12.3.7) where G h depends on ah, Ph and on the bounds C a and C b in la( v, x) I ~ Callvllvllxllv and Ib(w,x)1 ~ Cbllwllwllxllv. If
12.3 Mixed Finite-Element Method for the Stokes Problem
293
(12.3.6c) holds for all parameters h of a sequence of discretisations, then the discretisation is said to be stable and Ch remains bounded: Ch ~ C for all h.
PROOF. Theorem 12.2.7 and Lemma 12.2.9 are applicable with VO,h, Vh, Wh instead ofVo, V, W. (2.19a,c) then are the same as (6a,b), while (2.19b) follows from (2.19a) because dim VO,h < 00 (cf. Exercise 6.5.4). • Condition (6a) is trivial for the Stokes problem: Exercise 12.3.7. Show that Condition (6a) is always satisfied with a constant independent of h, if a(·, .): V x V -+ m. is V-elliptic.
O!h
12.3.3 Stable Finite-Element Spaces for the Stokes Problem 12.3.3.1 Stability Criterion For the Stokes problem Vh C H6(D), W h C L~(D) must hold. In a bounded domain a(·,·) is H6(D)-elliptic so that (6a) is satisfied with O!h ;::: CE > 0 . It is somewhat more difficult to prove the conditions (6b,c), which for the Stokes problem assume the form
sup{lb(P,u)l:u E Vh, lull = I};::: ~Iplo for all p E Wh
(12.3.6d)
wherein f3 > 0 must be independent of p and h. It is simpler to prove the modifiedoondition
sup{lb(p, u)l: u E Vh , lulo = I} ;::: ,8lPh for all p E W h
(12.3.8)
(,8 > 0 independent of p and h). Since in (8) the norm 1P11 occurs, the latter requires Wh C H1(D). This excludes, for example, piecewise constant finite elements. Together with the usual assumptions, Condition (8) is sufficient for stability. Theorem 12.3.S. Let n E CO,l be bounded. Let the Poisson problem be H2_ regular (it is sufficient that n be convex; cf. Theorem 9.1.22). Let Vh satisfy the approximation property
and the inverse estimate
(12.3.9b) Then Condition (8) is sufficient for the Brezzi condition (6b,c).
294
12 Stokes Equations
PROOF. (a) For given p E Wh there exists u E HA(J?) with luh = 1 and b(p, u) 2:: .8lplo (cf. Theorem 12.2.14 and (2.21». According to Exercise 9.1.24, the orthogonal H6(n)-projection u h of u on Vh satisfies the conditions u
= u h + e,
luhh ~ luh = 1,
lelo ~ Chlull = Ch.
From b(P,uh ) = b(p,u) - b(p,e) 2:: .8lplo - b(p,e) 2:: .8lplo -lPltlelo 2:: .8lploChlPh and luhh ~ lone infers (12.3.10) (b) Because of (8) and (9b) there exists u* E Vh with lu*lo = 1 and
From this follows sup{lb(P, vh)l: vh E Vh, Iv hll = 1} 2:: ,8hlph
with,8:= /3/Cr.
(12.3.11)
Hone multiplies (10) with ,8/(C+,8) and (11) with C/(C+,8) the sum reads: (12.3.12) Since!!. is independent of p and h, (6d) [Le., (6b,c)] has been proved.
•
12.3.3.2 Finite-Element Discretisations with the Bubble Function In the following let J? be a polygonal domain and Th an admissible triangulation. Example 4 shows that linear triangular elements make no sense for u and p. We increase the dimension of Vh by adding to it the so-called "bubble functions" or "bulb functions" . On the reference triangle T = {(~,1I):~,1I > 0,~+1I < I} the bubble function is defined by u(~, 11)
:= ~1I(1 - ~ -11)
in T,
u = 0 otherwise.
(12.3.13a)
The name derives from the fact that u is positive only in T and vanishes on aT and outside. The map ¢: T ~ T to a general T E Th (cf. Exercise 8.3.14) results in the expression (12.3.13b) for the bubble function on T.
Exercise 12.3.10. Let Th be a quasiuniform triangulation. Show that there exists a C > 0 independent of h such that Ii'ui'(x,y)dxdy 2:: C Ii' dxdy.
12.3 Mixed Finite-Element Method for the Stokes Problem
295
We set vl:linear combinations of the linear elements E HJ(n) and the bubble functions for T E Th, (12.3.14) Vh := vl x vl, Wh : linear elements E L~(n). For the side condition Wh C L~(n) see Section 8.3.6. Since uT E HJ(n), we have Vh C H6(n). Theorem 12.3.11. Let Th be the quasiuniform triangulation on a bounded polygonal domain n. Let Vh and W h be given by (14). Then the stability condition (8) is satisfied. Under the further conditions that n E co,1 and that the Poisson problem be H2(n)-regular the Brezzi condition (6b,c) also holds.
PROOF. (a) Choose an arbitrary P E Vp = (PzIT,PvIT)' We set
v:=
L
_
TETh
(pzIIT~T) Pv TUT
so that liilo
E
Vh ,
Who
On every T E
Th,
Vp is constant:
v:= v/lvlo (uT bubble function (13b) on T),
= 1. Exercise 10 yields
Since I'\lplo and !Pit are equivalent norms on the subspace H1(n) n L;(n) it follows that Ib(P,v)1 ~ G'lphlVplo and Ib(P,ii)1 ~ G'lpl1lVplo/Jvlo. In a similar way one shows that Ivlo $ G"IVplo and obtains Ib(p, ii)1 ~ .8!P1t with j3 := G'IG" independent of h. The left-hand side in (8) is ~ Ib(p, ii)l, so that (8) follows. (b) (9a) is satisfied for Vh (cf. Theorem 8.4.5). The same holds for the inverse estimate (9b) (cf. Theorem 8.8.5). Therefore (6b,c) follows from Theorem 8 .
•
A general result on the stabilisation by bubble functions can be found in Brezzi-Pitkiiranta [1].
296
12 Stokes Equations
12.3.3.3 Stable Discretisations with Linear Elements in Vh If one wants to avoid bubble functions, one must increase the dimension of Vh in some other way. In this section we shall consider for Vh and W h two different triangulations Th/2 and Th. By decomposing each T E Th as in Figure 1, through halving the sides, into four similar triangles, one obtains Th/2' We define: Vh C H5(!~) : linear elements for triangulation
Th/2'
(12.3.15)
Hi(!l) n L~(!]) : linear elements for triangulation
Wh C
Th,
or
Vh
c
HM!l) : quadratic elements for triangulation
Th.
Wh C Hi(!]) n L~(!]) : linear elements for triangulation
Figure 12.3.1. Triangulations
Theorem 12.3.12. (16).
Th
and
(12.3.16) Th.
Th/'J
Theorem 11 holds analogously for Vh
X
Wh in (15) or
PROOF. (a) Let Vh x W h be given by (15). For each inner triangular side 1 of the triangulation Th, there exist two triangles Ti1" T21' E Th with 1 = Ti/' nT2/' (cf. Figure 2). Let a/at be the derivative in the direction of 1, let a/8n be the directional derivative perpendicular to it. There exists a" and b/, with a~ + b~
= 1,
8j8x
= a/,8jan + b/,8j8t,
8j8y
= b/,8j8n -
a/,8jat.
In contrast to 8pj8n, /:)Pjat is constant on Ti/' U T2/' U 1. We denote its value by Pel/,' The mid-point x/' of 1 is a node of Th/2. We define the piecewise linear function u/' over Th/2 by its values at the nodes u/,(x1') and set
= Pth,
u,,(xj
)
=0
at the remaining nodes
v := I: (~~)PthU/' E Vh, v:= v/lvlo.
(12.3.17a)
(12.3.17b)
l'
The sum
2::/' extends over all interior sides of Th. In Ti/' U T2/' we have
12.3 Mixed Finite-Element Method for the Stokes Problem
297
so that b(p, v)
=
f
I: 'Y
(Vp,
T17UT27
~ Gh 2
I: IPth1 'Y
U'Y
dx
2•
If 1'1> 1'2, 1'3 are the sides of T E
G'h2l:~=1 Ipth.1
f
(!:JPtl'YU'Y) dx = I: IPthl 2 'Y T17UT27
Th (Th
is quasiuniform!), then frlVpI2 dx :::;
2.
From this one infers that b(p, v) ~ Gil IplllVplo, as in the proof of Theorem 11, and finishes the proof analogously. (b) In the case of quadratic elements given by (16) one has the same nodes as in (a) (cf. Figure 8.3.8a and Figure 1). Use (17a,b) to define the quadratic • function v E Vh and carry out the proof as in (a).
a/at -"""/ a/an ''''''''''----/
Figure 12.3.2. ,,(, T1'Y'
T2'Y
12.3.3.4 Error Estimates In the following, the condition (8.1.12) should be replaced by the stability condition (6a-c). In the place of the approximation property (8.4.6') we now have the inequalities inf{lv - vhh: v h E Vh} :::; Ghlvl2 for all v E V
= H5(n),
(12.3.18a)
inf{jp - phlo:ph E W h } :::; Ghlph for all pEW = L~(n).
(12.3.18b)
Condition (18a) is equivalent to condition (9a) in Theorem 8. The following theorem applies to general saddlepoint problems, and can be reduced to Theorem 8.2.1.
Theorem 12.3.13. Let U = (:) EX = V x W be the solution of (2.5a-c) [resp. (6a)]. Let the discrete problem (1a-c) with Xh = Vh X W h C X satisfy
298
12 Stokes Equations
the Brezzi condition (6a-c) and have the solution u h exists a variable C independent 01 h such that
= (v\ w h ). Then there (12.3.19)
PROOF. The Brezzi condition (cf. Theorem 6) yields lIuhllx ::5 Cll/llx' for all right-hand sides f E X' in (2.5a-c), in particular for all I E Xh = X~. The above inequality means IILh"lllx" ....x:. ::5 C for the operator L h: X h --.. X~ which belongs to c(·, ,):Xh x Xh --.. nt. According to Exercise 8.1.16 IILh"lllx" ....x:. ::5 C is equivalent to Condition (8.1.12) for c(-,·):X x X --.. nt [instead of a(·, .): V X V --.. nt] with fN = l/C. Theorem 8.2.1 yields the statement (19). •
<;), <;:)
For the Stokes problem with instead of 11. = inequality (19) is now rewritten as follows: 111. - uhl~ + IP - phl~
(::'),u h
::5 C 2 inf{lu - uhl~ + IP - phl~: u h E Vh,ph
= (::,:) ,
E Wh}
or 111. -uhh
+ IP -
ph 10 ::5 v'2Cinf{lu -uhh
+ IP -
phlo:uh E Vh,ph E Wh }. (12.3.20)
Theorem 12.3.14. Let the Stokes equation (1.2a,b) have a solution 11. E H2(il)nHa(.!1), p E Hl(n)nL~(il) (cf. Theorem 12.2.19). Forthe subspaces Vh C Haeil) and Wh C L~(n) let the Brezzi condition (6a-c) and the approximation conditions (18a,b) be satisfied. Then the discrete solution uh,ph satisfies the estimate (12.3.21)
PROOF. Combine inequalities (20) and (18a,b).
•
Using the same reasoning as in the second proof for Theorem 8.4.11 with c(·, .) instead of a(-, .) one proves
Theorem 12.3.15. For each f E L2(n), 9 E L~(n) n Hl(il) let the Stokes problem -..du + Vp = I, -divu = 9 have a solution 11. E H2(il) n HMn), p E HI(il) n L~(n) with 111./2 + Iph ::5 C[l/lo + /gh]. Under the assumptions 01 Theorem 14 we then have the estimates
+ IP - phl_l ::5 c' h[luh + Iplo], 111. -uhlo + /p - phl_l ::5 C" h 2!1ul2 + /ph] 111. - uhlo
lorthefinite element solutions. Here
(12.3.22a) (12.3.22b)
1P1-1 is the dual norm lor Hl(n)nL~(n).
Corollary 12.3.16. Combining (22b) and (21) one obtains
12.3 Mixed Finite-Element Method for the Stokes Problem
299
(12.3.22c)
A new exposition of the mixed finite-element discretisation can be found in Brezzi-Fortin [1].
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Index
Note: The page numbers given in bold-face refer to occurrences of the term printed emphatically on the page indicated. Adjoint problem, 141, 190, 196 Adler problem, 159, 183 Agmon, condition of, 158 Analytic, 20 BabuSka condition, 139 Babuaka-Brezzi condition, 284 Banach space, 110, 111, 112, 113, 131, 135, 137, 161 Basis, 161, 165, 166, 171, 182, 183, 204, 291 hierarchical, 202 Basis condition, 171,172,173,175, 176, 177, 180, 181, 183, 200 Beltrami operator, 14 Biharmonic equation. See Differential equation Bilinear form, 137, 154, 155, 227, 279 adjoint, 138, 190, 191 continuous or bounded, 138, 146, 157, 161, 185, 279, 280 corresponding to a pde of order 2m, 146, 210 corresponding to the Helmholtz equation, 147, 154, 208 corresponding to the Poisson equation, 149, 154, 165, 176, 180, 183 extension of a, 138 operator associated to a. See Operator, 138, 140, 142, 163, 209
symmetric, 138, 141, 149, 152, 164, 170, 200, 204, 208, 280, 285 V-elliptic (Hm-elliptic, etc.), 140, 141, 147, 148, 149, 152, 153, 154, 159, 164, 170, 183, 200, 204, 212, 228, 280 V-coercive (Hm -coercive, etc.), 141, 142, 143, 149, 150, 153, 154, 157, 158, 166, 171, 185, 210, 211, 212, 213, 218, 219, 222, 223, 226, 227, 228,229, 256,267 Block, block matrix, block partitioning, 283 Boundary condition, 158, 226, 275 Dirichlet, 27, 35, 38, 65, 85, 86, 96, 145, 150, 159, 164, 174, 178, 215, 222, 229 discretisation of the, 100 first, 35, 96 mixed, 96 natural, 152, 154, 155, 159, 165, 178, 218, 221 Neumann, 35,65,96, 154, 174, 183 periodic, 98 second, 96 third,96 Boundary element method (BEM), 37 Boundary layer, 249
308
Index
Boundary value, 7,9, 10, 12, 26 Boundary value problem, 13, 19,27, 28,85 Brezzi condition, 292, 293, 295, 298 Bubble function, 294, 295,296 Cauchy-lliemann equations, 4, 7 Chequer-board. See Ordering Coefficient vector, 161, 255 Compact (cf. embedding, nine-point formula, operator), 185, 201, 257, 269 Complementary problem, 264 Complete, 24, 112 Completion, 112, 117, 122, 125 Condition, 108, 204 Cone condition, 136 Conservation form, 244 Convex. See Domain Consistent, consistency, 59, 64, 69, 93, 94, 200, 228, 232, 233, 235, 239, 242, 267 Convection-diffusion equation, 247 Convergence, 168 of the difference method, 59, 60, 61, 69, 83, 84, 95, 102, 108, 239 of the eigenfunction, 255,259, 260, 271, 272 of the eigenvalues, 255, 258, 260 super, 202, 203 Coordinate transformation. See Domain Corner, re-entrant (cf. L-shaped domain), 34, 223, 225 Covering, 127 Dense, 10, 112, 113, 114, 115, 117, 119, 123, 141, 169, 170, 224 Derivative, 104 conormal, 96, 201 normal, 15, 65, 95, 96, 145 tangential, 96, 99, 103, 145, 224, 246,278 weak, 115, 116, 172 Diagonalisable, real, 6, 7 Diagonally dominant, 47, 48, 49, 52 irreducibly, 47, 48, 49, 52, 53, 92, 93,250 Difference (divided), 38, 49, 62, 79, 240 backward or left-sided, 38, 56, 65, 72,91,260 forward or right-sided, 38, 91, 227,250 second, 38, 80, 238, 240
symmetric, 88, 70, 71, 102,233, 249, 251, 252 Difference method, 38, 59, 65, 90, 91,93,94,102, 108, 173, 177, 178,202,204,247,249,267, 269 13-point, 105, 109 of higher order, 62 Difference operator, 10, 44, 121, 210, 213, 228, 233 Difference star (stencil), 44, 62, 91 Differential equation(s) biharmonic, 74, 95, 103, 104, 105, 106, 108, 158, 194, 218, 223,278 of first order, 1, 3, 4, 6, 277 of order 2m, 104, 108, 145, 158 of second order, 2, 5, 8, 12, 85, 90, 150 ordinary, 1,2, 3, 38, 248, 254 partial, 1, 2, 7 system of, 3, 4, 6, 7, 105, 275, 276 Differential operator, 5, 59, 60, 75, 85,89,104,253,276 formally adjoint, 92, 95 boundary, 95, 154, 155, 156 Dirichlet integral, 144 Dirichlet principle, 144 Divergence (operator), 275 Domain, 12 convex, 79, 197, 223, 224, 225, 231, 235, 236, 242, 288, 293 exterior, 86 general, 78, 100, 109, 196, 234, 241,243 L-shaped, 13, 34, 127, 201, 243 normal, 15, 16, 19, 28, 33 transformation of the, 6, 89 unbounded, 23, 146, 147 Double-layer potential. See Potential Dual form, 131, 134 Dual space, 130 Eigenfunction, 222, 258, 254,259, 263,271 Eigenvalue, 5, 6, 7, 10, 47, 51, 52, 53, 137, 142, 150, 258, 254, 255, 256, 258, 259, 260, 261, 262, 264, 265, 271, 274 Eigenvalue problem, 7, 11, 137, 150, 253 adjoint, 253, 256 elliptic, 253, 254, 255, 274 Elliptic, 4, 5, 6, 7, 9, to, 11, 12, 85, 104,276
Index uniformly, 86, 87, 88, 89, 104, 145, 147, 148, 149, 150, 155, 222, 229, 276, 277 V -. See Bilinear form, 248, 256, 280, 285, 289, 293 Embedding compact, 185, 136, 142, 143, 150, 153, 171, 186, 253, 256 continuous, 118, 131, 133, 134, 136, 137, 141, 143, 171, 185, 253,256 dense, 131, 133, 134, 137, 141, 171, 253, 256 Error estimates for difference methods, 190, 238, 239 for eigenvalue problems, 260, 263, 264 for finite element methods, 185, 190, 193, 196, 246, 266, 297, 298 for Ritz-Galerkin methods, 167 Estimate. See Error estimate, inverse estimate Existence. See Solution Extension. See Bilinear form, operator of a function, 123, 129, 215, 237 Extrapolation (method), 61, 84 Finite elements, 171, 172, 174,251, 254 constant (bi)cubic, 182, 194, 195, 196, 296 (bi)linear, 172, 174, 175, 178, 179, 182, 183, 185, 190, 194, 199, 204,246,247,252,295,296 isoparametric, 198,247 mixed, 290, 299 nonconformal, 199, 200, 291 of the serendipity class, 181, 182, 193 quadratic, 181, 182, 193 Finite element method, 24 Five-point formula, 40, 41, 53, 60, 62, 92, 176, 190, 231 Form. See Bilinear form, dual form, sesquilinear form Fourier expansion, 9 Fourier transform, 110, 119, 120, 124, 125, 126, 135, 213 Functional, 132, 140, 146, 152 Fundamental solution, 16, 17, 28 Galerkin method. See Ritz-Galerkin, Petrov-Galerkin Garding, Theorem of, 150
309
Gel/fand triple, 41, 188, 134, 136, 141, 163, 207, 253 Gerschgorin, Theorem of, 46, 48 Gradient, 15 Green's formula, 15, 29, 144, 153 Green's function of the first kind, 28, 29, 30, 31, 33, 34, 95, 105, 144 discrete, 55, 72, 95 Green's function of the second kind, 85 Grid, 89, 40, 226 Grid function, 89, 44 Grid points close to the boundary, 63, 79, 83 far from the boundary, 42, 63, 79 neighbouring, 41, 42, 45, 80 Grid size (width, step size), 89 Harmonic, 18, 15, 16, 19, 20, 22, 23, 24, 25, 144 Harnack, Theorem of, 20 Heat equation, 3,4, 5, 7, 8, 10 Helmholtz equation, 147, 154, 208 Hilbert space, 110, 114, 115, 117, 122, 132, 134, 279, 289 Holder continuous, 29, 80, 110, 123, 125 Hyperbolic, 4, 5, 6, 7, 8, 9, 10, 11, 248 Improperly posed. See Well-posed Inclusion, 135, 136, 142 Initial-boundary value problem, 8, 9,207 Initial value problem, 2, 3, 7, 8, 9, 38 Instationary problem, 10, 11 Integral equation method, 36, 87 Interpolation, 84, 124 Hermite, 194, 196, 207 spline, 194, 196, 207 Inverse estimate, 206, 227, 240, 293 Irreducible, 45, 54 Irreducibly diagonally dominant. See Diagonally dominant Lagrangian (factor), 184, 290 Lame differential equations, 277 Laplace equation (of potential equations), 2, 12 Laplace operator, 12, 14, 131, 176 Lexicographical ordering. See Ordering Lipschitz continuity, 23, SO Map, mapping. See Operator Mass matrix, 205
310
Index
Maximum (-minimum) principle, 17, 18, 19, 20, 27, 86, 103, 250, 258 Mean-value property, 11, 18, 19, 20, 27,54 Mehrstellen method, 64, 83 M-matrix, 45,49,50,51,52,53, 64, 68, 81, 83, 84, 91, 92, 93, 100,106,249,250,251,252 Multi-index, 30, 104 Neighbour. See Grid points Neumann condition. See Boundary condition Nine-point formula, 90, 180 compact, 63, 63 Nodal points, 112, 115, 178, 181, 203 Nodal values, 112, 195 Norm, 29, 30, 50 dual, 131, 132, 134, 163, 194, 215, 227, 298 Euclidean, 14, 51, 57, 110, 163, 204,226 equivalent, 111, 117, 123, 204, 295 matrix,50 associated, 50, 204, 227 maximum, 46, 110 operator, 111 row sum, 50, 59, 170, 242 Sobolev, 117, 122 Sobolev-Slobodeckii, 122, 247 spectral, 51, 59, 204 supremum, 23,110,112 vector, 50 Normal (direction), 15, 116, 145, 224 Normal system, 105 Normed space, 110 Operator, 111 adjoint, 132, 163, 213, 216, 242 associated. See Bilinear form, 209, 256,282 compact, 135, 136, 137 difference. See Difference operator differential. See Differential operator dual,131 selfadjoint, 132 Ordering (of the grid points), 42, 43,45 lexicographical, 42, 67, 70 chequer-board,43 Orthogonal (cf. projection), 114 Orthogonal space, 114, 163, 283
Parabolic, 4, 5, 5, 7, 8, 9, 10, 11 Partition of unity, 128, 130, 150, 209,219 Petrov-Galerkin method, 252 Plate equation (cf. biharmonic), 85, 103 Poisson equation, 21, 34, 35, 38, 41, 57, 103, 149, 155, 177, 201, 225, 231, 254, 276, 287, 289, 293,295 Poisson's integral formula, 19, 20, 22 Polar coordinates, 13, 14, 15, 97, 98, 144 Positive definite, 5, 52, 53, 93, 106, 164, 204, 255 Positive semidefinite, 52, 81 Potential, double-layer (dipole), 36 single-layer, 36 volume, 36 Potential equation (cf. Laplace equation), 2, 4, 5, 7, 9, 12, 13, 14, 16, 28, 35 Precompact, 135 Principal part, 6, 12, 95, 145, 147, 148, 150, 223, 228, 251, 216 Projection, 132, 170 orthogonal, 132, 133, 163, 170, 225,294 Ritz, 110, 189, 191, 192, 193, 225 Rayleigh-Ritz. See Ritz Reduced equation, 248, 250 Reference triangle (reference element), 10, 177, 179, 185, 198, 204,294 Regularity, 190, 192, 196, 208, 209, 215, 219, 222, 223, 225, 232, 246, 254, 264, 267, 285, 287, 288, 293, 295 discrete, 227, 228, 229, 230, 231, 239, 240, 241, 242, 243, 271 Riesz representation theorem, 132 Riesz isomorphism, 132, 283 Riesz-Schauder theory, 131, 142 Ritz-Galerkin method (cf. solution), 161, 171, 290 Robin problem, 159 Saddle-point (problem), 279, 280, 282, 284, 290, 297 Scalar product, 52, 114,115,117, 121,132,148,162,170,227 Separation (of variables), 3, 11 Serendipity class. See Finite elements Sesquilinear form, 138
Index Seven-point formula, 91, 92, 176 Shortley-Weller discretisation, 78, 80, 81, 83, 229, 231, 237 Side condition, 182, 183, 290, 291 Single-layer potential. See Potential Singular perturbation, 247, 248 Singularity function, 14, 16, 56 discrete Sobolev, lemma of, 125, 212, 222 Sobolev space (cf. norm), 114, 117, 122, 128, 133, 134, 136, 145 Solution classical, 27, 30, 32, 145, 146, 147, 154, 212, 222, 246, 279 existence of a, 9, 13, 23, 29, 32, 35, 37, 67, 137, 142, 144, 151, 155, 161, 286, 292 existence and uniqueness of a, 137, 141, 142, 147, 148, 149, 150, 151, 152, 153, 162, 164, 166, 167, 168, 171, 183, 188, 222,248,282,284,285,286, 288, 289, 292 Ritz-Galerkin, 164, 165, 166, 168, 170 uniqueness of the, 18, 19, 26, 27, 35, 86, 87, 88, 148, 149 weak, 27, 146, 150,159, 190, 208, 209, 210, 212, 213, 218, 219, 222, 223, 226, 246, 279 Sparse matrices, 44, 171, 173 Spectral radius, 47 Spectrum, 137, 142 Stable, stability, 59, 64, 70, 90, 94, 95, 102, 103,227,228,242, 249,252,291,293 Star (stencil), 178, 180 Stationary problem, 10 Steklov problem, 254 Step size. See Grid size Stiffness matrix, 162, 163, 165, 171, 177,191,200,202,203,255 Stokes equations, 4, 103,275,278, 279,282, 285, 286, 288, 290, 291,292, 293, 298 Support, 115, 171, 173, 176, 182, 183, 194, 195, 202 Time, 8, 9, 10, 11 Trace (of a function), 3, 123, 129 Trace (of a matrix), 87 Transformation. See Domain, Fourier Transformation theorem, 119, 129 Transition condition, 245, 246, 247 Trefftz'method, 200
311
Triangulation, 174 adaptive, 201 admissible, 174, 175, 179, 185, 190 quasi-uniform, 188, 190, 204, 206, 294,295 regular, 188 uniform, 188 Type (of a pde), 1,4,5,6,7,10 Type invariance, 6 Uniqueness. See Solution Variational formulation (weak formulation), 144, 146, 149, 150, 158, 161, 165, 166, 173, 183, 194, 208, 244, 253, 254, 278, 288 Variational problem, 141, 152, 159, 164,171 dual/complementary, 200 Viscosity, artificial, 251, 252 numerical, 251 Wave equation, 2, 3, 5, 8, 10 Weak formulation. See Variational formulation Well-posed (problems), 23 Wilson's rectangle, 199