Strongly Elliptic Systems and Boundary Integral Equations Partial differential equations provide mathematical models of many important problems in the physical sciences and engineering. This book treats one class of such equations, concentrating on methods involving the use of surface potentials. It provides the first detailed exposition of the mathematical theory of boundary integral equations of the first kind on non-smooth domains. Included are chapters on three specific examples: the Laplace equation, the Helmholtz equation and the equations of linear elasticity. The book is designed to provide an ideal preparation for studying the modern research literature on boundary element methods. Dr. McLean received his PhD from the Australian National University, and is currently a Senior Lecturer in Applied Mathematics at the University of New South Wales.
Strongly Elliptic Systems and Boundary Integral Equations WILLIAM McLEAN University of New South Wales
AMBRIDGE
UNIVERSITY PRESS
PUBLISHED BY THE PRESS SYNDICATE OF THE UNIVERSITY OF CAMBRIDGE The Pitt Building, Trumpington Street, Cambridge, United Kingdom CAMBRIDGE UNIVERSITY PRESS The Edinburgh Building, Cambridge CB2 2RU, UK http://www.cup.cam.ac.uk 40 West 20th Street, New York, NY 10011-4211, USA http://www.cup.org 10 Stamford Road, Oakleigh, Melbourne 3166, Australia Ruiz de Alarc6n 13, 28014 Madrid, Spain © Cambridge University Press 2000 This book is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 2000 Printed in the United States of America Typeface Times Roman 10/13 pt.
System l TEX2e
[TB]
A catalog record for this book is available from the British Library.
Library of Congress Cataloging in Publication Data McLean, William Charles Hector, 1960Strongly elliptic systems and boundary integral equations / William McLean. cm.
p.
Includes index.
ISBN 0-521-66332-6 (hc.). - ISBN 0-521-66375-X (pbk.) 1. Differential equations, Elliptic.
2. Boundary element methods.
1. Title.
QA377.M3227
2000
515'.353 - dc2l
99-30938
CIP
ISBN 0 521 66332 6 hardback ISBN 0 521 66375 X paperback
To Meg
Contents
page xi
Preface 1.
Introduction
1
Exercises
15
2. Abstract Linear Equations
17
The Kernel and Image Duality Compactness Fredholm Operators Hilbert Spaces Coercivity Elementary Spectral Theory Exercises
18
20 27 32 38
42 45
52
3. Sobolev Spaces Convolution Differentiation Schwartz Distributions Fourier Transforms Sobolev Spaces - First Definition Sobolev Spaces - Second Definition Equivalence of the Norms Localisation and Changes of Coordinates Density and Imbedding Theorems Lipschitz Domains Sobolev Spaces on the Boundary
57-
58 61
64 69 73
75 79 83 85
89
96
The Trace Operator
100
Vector-Valued Functions
106
Exercises
107
vii
Contents
viii 4. Strongly Elliptic Systems
The First and Second Green Identities Strongly Elliptic Operators Boundary Value Problems Regularity of Solutions The Transmission Property Estimates for the Steklov-Poincare Operator Exercises 5. Homogeneous Distributions Finite-Part Integrals Extension from R" \ {01 to ll8" Fourier Transforms Change of Variables Finite-Part Integrals on Surfaces Exercises 6. Surface Potentials Parametrices Fundamental Solutions
The Third Green Identity Jump Relations and Mapping Properties Duality Relations Exercises 7. Boundary Integral Equations
Operators on the Boundary Integral Representations The Dirichlet Problem The Neumann Problem Mixed Boundary Conditions Exterior Problems Regularity Theory Exercises 8.
The Laplace Equation Fundamental Solutions Spherical Harmonics Behaviour at Infinity Solvability for the Dirichlet Problem Solvability for the Neumann Problem Exercises
Contents 9. The Helmholtz Equation Separation of Variables The Sommerfeld Radiation Condition Uniqueness and Existence of Solutions A Boundary Integral Identity 10.
ix
276 277
280 286 289
Exercises
293
Linear Elasticity
296
Korn's Inequality Fundamental Solutions Uniqueness Results Exercises
297
Appendix A. Extension Operators for Sobolev Spaces Exercises
299 301
305
309 315
Appendix B. Interpolation Spaces The K-Method The J-Method Interpolation of Sobolev Spaces Exercises
317
Appendix C. Further Properties of Spherical Harmonics
334
Exercises
318 321
329 333
338
References
341
Index
347
Index of Notation
353
Preface
The study of integral equations in connection with elliptic boundary value problems has a long history, during which a variety of approaches has emerged. Rather than attempt a broad survey, I have chosen to pursue in detail just one
approach, in which both the differential and integral formulations of a given boundary value problem are viewed abstractly as linear equations involving a bounded operator from a Hilbert space into its dual. The decisive property of this operator is that its associated sesquilinear form is positive and bounded below, apart perhaps from a compact perturbation.
In the classical Fredholm method, the solvability of the Dirichlet and Neumann problems is proved by reformulating them as integral equations of the second kind. Here, we effectively reverse this strategy, deriving key properties of the boundary integral equations from previously established results for the associated partial differential equations. Moreover, our approach leads to Fredholm integral equations of the first kind. The theory of such first-kind integral equations can be traced back to Gauss (see Chapter 1), and developed into the form presented here during the 1970s, in the work of Nedelec and Planchard [74], [76]; Le Roux [56], [57], [58]; and Hsiao and Wendland [42]. Those authors were all studying Galerkin boundary element methods, and al though this book does not deal at all with numerical techniques, it is written very much from the perspective of a numerical analyst. A major difficulty in a work such as this is the large amount of background material needed to present the main topics. Aware that readers differ in their prior knowledge, I have tried to adopt a middle path between, on the one hand, writing a textbook on functional analysis, distributions and function spaces, and on the other hand just stating, without proof or exposition, a litany of definitions and theorems. The result is that more than one-third of the text is made up of what might be considered technical preliminaries. My hope is that the book will be suitable for someone interested in finite or boundary element methods who xi
xii
Preface
wants a deeper understanding of the relevant non-numerical theory. I have aimed to keep the exposition as simple, concise and self-contained as possible, while at the same time avoiding assumptions that would be unrealistic for applications. Thus, I felt it essential to allow non-smooth domains, to consider systems and not just scalar equations, and to treat mixed boundary conditions. Here is an outline of the contents. Chapter 1 has two purposes. Firstly, it attempts to sketch the early history of the ideas from which the theory of this book developed. Secondly, it serves to introduce those ideas in an informal way, and to acquaint the reader with some of the notation used later.
The second chapter presents topics from linear functional analysis that are immediately relevant to what follows. I assume that the reader is already familiar with elementary facts about the topology of normed spaces, and of a few fundamental, deeper results such as the open mapping theorem and the Hahn-Banach theorem. Chapter 3 develops the theory of Sobolev spaces on Lipschitz domains. After a quick treatment of distributions and Fourier transforms, we study in detail fractional- and negative-order spaces based on L2. These spaces play an essential role in nearly all of the subsequent theory. In Chapter 4, we begin our investigations of elliptic systems. A key tool is the first Green identity, used to arrive at the abstract (weak) formulation of a boundary value problem mentioned above. The centrepiece of the chapter is the Fredholm alternative for the mixed Dirichlet and Neumann problem on a bounded Lipschitz domain. We go on to prove some standard results on regularity of solutions, including the transmission property. The final section of the chapter proves some difficult estimates of Ne6as [72] that relate the H I -norm of the trace of a solution to the L2-norm of its conormal derivative. These estimates are used later when showing that, even for general Lipschitz domains, the basic mapping properties of the surface potentials and boundary integral operators hold in a range of Sobolev spaces. Chapter 5 is something of a technical digression on homogeneous distributions. As well as dealing with standard material such as the calculation of Fourier transforms, we include results from the thesis of Kieser [48], including the change-of-variables formula for finite-part integrals. Chapters 6 and 7 form the heart of the book. Here, we study potentials and boundary integral operators associated with a strongly elliptic system of partial differential equations. Our overall approach is essentially that of Costabel [14], allowing us to handle Lipschitz domains. The first part of Chapter 6 deals with parametrices and fundamental solutions, and uses the results of Chapter 5. We then prove the third Green identity, and establish the main properties of the
Preface
xiii
single- and double-layer potentials, including the familiar jump relations. Chapter 7 derives the boundary integral equations for the Dirichlet, Neumann and mixed problems, treating interior as well as exterior problems. The Fredholm alternative for the various boundary integral equations is established by showing positive-definiteness up to a compact perturbation, a property that is intimately related to the strong ellipticity of the associated partial differential operator.
Chapters 8-10 treat three of the simplest and most important examples of elliptic operators. For these specific cases, we can refine the general theory in certain respects. Chapter 8 deals with the Laplace equation, and includes a few classical topics such as spherical harmonics and capacity. Chapter 9 deals with the Helmholtz (or reduced wave) equation, and Chapter 10 gives a brief treatment of the linearised equilibrium equations for a homogeneous and isotropic elastic medium.
The book concludes with three appendices. The first of these proves Calder6n's extension theorem for Sobolev spaces on Lipschitz domains, including the fractional-order case. The second gives a rapid but self-contained treatment of interpolation spaces and establishes the interpolation properties of Sobolev spaces on Lipschitz domains. The third proves a few facts about spherical harmonics. At the end of each chapter and appendix is a set of exercises. These are of various types. Some are simple technical lemmas or routine calculations used at one or more points in the main text. Others present explicit solutions or examples, intended to help give a better feeling for the general theory. A few extend results in the text, or introduce related topics. Some mention of what I have not covered also seems in order. Many books treat Fredholm integral equations of the second kind. Wellknown older texts include Kellogg [45] and Giinter [35], and we also mention Smirnov [95] and Mikhlin [65, Chapter 18]. Problems on non-smooth domains are treated by Kral [49] and Burago and Maz'ya [6], using methods from ge-
ometric measure theory, and by Verchota [ 102] and Kenig [46], [47] using harmonic analysis techniques. Works oriented towards numerical analysis include Kress [50], Hackbusch [36] and Atkinson [3]. Boundary value problems can also be reformulated as Cauchy singular integral equations, as in the pioneering work by Muskhelishvili [71 ]; for a modem approach, see Gohberg and Krupnik [28] or Mikhlin and Pr6l3dorf [66].
Even for boundary integral equations of the first kind, the material presented in this book is by no means exhaustive. For instance, Costabel and Wendland [ 15] have generalised the approach used here to higher-order strongly
elliptic equations. One can also study boundary integral equations as special cases of pseudodifferential equations; see, e.g., Chazarain and Piriou [10]. We
xiv
Preface
make contact with the theory of pseudodifferential operators on several occasions, but do not attempt a systematic account of this topic. Other significant matters not treated include the LP theory for p # 2, various alternative boundary conditions, especially non-linear ones, and a detailed study of the dominant singularities in a solution at corner points or edges of the domain.
During the period I have worked on this book, the Australian Research Council has provided support for a number of related research projects. I thank David Elliott for reading an early draft of the complete manuscript and making a number of helpful suggestions. I also thank Werner Ricker and Jari Brandts for the care with which they read through later versions of some of the chapters. Alan McIntosh and Marius Mitrea helped me negotiate relevant parts of the harmonic analysis literature. Visits to Mark Ainsworth at Leicester University, U.K., to Youngmok Jeon at Ajou University, Korea, and to the Mittag-Leffler Institute, Stockholm, provided valuable opportunities to work without the usual distractions, and made it possible for me to complete the book sooner than would otherwise have been the case. Needless to say, I am also indebted to many other people, who helped by suggesting references, discussing technical questions, and passing on their knowledge through seminars. Sydney,
December 1998
1
Introduction
The theory of elliptic partial differential equations has its origins in the eighteenth century, and the present chapter outlines a few of the most important historical developments up to the beginning of the twentieth century. We concentrate on those topics that will play an important role in the main part of the book, and change the notation of the original authors, wherever necessary, to achieve consistency with what comes later. Such a brief account cannot pretend to be a balanced historical survey, but this chapter should at least serve to introduce the main ideas of the book in a readable manner. To limit subsequent interruptions, we fix some notational conventions at the outset. Let n denote a bounded, open subset of W (where n = 2 or 3 in this chapter), and assume that the boundary r = 80 is sufficiently regular for the outward unit normal v and the element of surface area dcr to make sense. Given a function u defined on S2, we denote the normal derivative by a,,u or au/8v. Sometimes we shall work with both the interior and the exterior domains (see Figure 1)
SZ-=S2 and
S2+=R'\(S2-U11),
in which case, if the function u is defined on S2}, we write
y±u(x) = av u (x) =
lim
u(y) and
lim
v (x) grad u (y)
y-+z,yES2t
for x E F,
whenever these limits exist. The Euclidean norm of x E 1R" is denoted by Ix I. The prototype of an elliptic partial differential equation is Au = 0, where A denotes the Laplace operator (or Laplacian), defined, in n dimensions, by 11
a2 tc
Au(x) _ L /-t 8x 1
(1.1)
Introduction
2
Figure 1. Interior and exterior domains n- and S2+ with boundary r.
When Au = 0 on 0, we say that the function u is harmonic on Q. In two dimensions, there is a close connection between the Laplace equation and complexanalytic functions. Indeed, u + iv is differentiable as a function of the complex variable xl + ix2 if and only if u and v satisfy the Cauchy-Riemann equations,
-
au
av
and
ax2
ax,
au
av
ax2
ax,
(1.2)
in which case Du = 0 = Av and we say that u and v are conjugate harmonic functions. The pair of equations (1.2) appeared in Jean-le-Rond d'Alembert's Essai d'une Nouvelle Theorie de la Resistance des Fluides, published in 1752. At around the same time, Leonhard Euler derived the equations of motion for an irrotational fluid in three dimensions. He showed that the fluid velocity has the form grad u, and that for a steady flow the velocity potential satisfies Au = 0. This work of d'Alembert and Euler is discussed by Truesdell [100]; see also Dauben [ 18, p. 3111.
In 1774, Joseph-Louis Lagrange won the Prix de 1'Academie Royale des Sciences for a paper [51] on the motion of the moon; see also [30, pp. 478-479, 1049]. This paper drew attention to two functions that later came to be known as the fundamental solution,
G(x, y) =
1
4n 1x
-yl
for x, y E R3 and x # y,
(1.3)
and the Newtonian potential,
u(x) = I G(x, y) f(y) dy.
(1.4)
J
Up to an appropriate constant of proportionality, G(x, y) is the gravitational
Introduction
3
potential at x due to a unit point mass at y, and thus u is the gravitational potential due to a continuous mass distribution with density f. The Coulomb force law in electrostatics has the same inverse-square form as Newton's law of gravitational attraction. Thus, u also describes the electrostatic potential due to a charge distribution with density f ; mathematically, the only change is that f may be negative.
In a paper of 1782 entitled Thiorie des attractions des sphiroides et de la figure des planetes, Pierre Simon de Laplace observed that the Newtonian potential (1.4) satisfies Au = 0 outside the support of f, writing Du in spherical polar coordinates. Later, in a paper of 1787 on the rings of Saturn, he gave the same result in Cartesian and cylindrical coordinates. Birkhoff and Merzbach [7, pp. 335-338] give English translations of relevant excerpts from these two works. By transforming to polar coordinates centred at x, i.e., by using the substitution y = x + pw where p = Iy -x 1, it is easy to see that the Newtonian potential (1.4) makes sense even if x lies within the support of f , because dy = p2 d p dw.
However, the second partial derivatives of G are 0(p-3), and this singularity is too strong to allow a direct calculation of Au by simply differentiating under the integral sign. In fact, it turns out that
-Au =f everywhere on R3, an equation derived by Simeon-Denis Poisson [7, pp. 342346) in 1813; see Exercise 1.1 for the special case when f is radially symmetric. Poisson made other important contributions to potential theory. A paper [18, p. 360] of 1812 dealt with the distribution of electric charge on a conductor Q. In equilibrium, mutual repulsion causes all of the charge to reside on the surface r of the conducting body, and r is an equipotential surface. The electrical potential at x E R3 due to a charge distribution with surface density * on r is given by the integral
SL*(x) =
J'
G(x, y)f(y)day,
(1.5)
so SL / is constant on r if Mfr is the equilibrium distribution. The function SL * is known as the single-layer potential with density *, and satisfies the Laplace equation on the complement of 1, i.e., on S2+ U S2-. Although SL * is continuous everywhere, Poisson found that its normal derivative has a jump discontinuity:
8v SL*-8: SL r=-,/r onr. Exercise 1.2 proves an easy special case of this result.
(1.6)
4
Introduction
A further stimulus to the study of the Laplace equation was Jean-BaptisteJoseph Fourier's theory of heat diffusion. In 1807, he published a short note containing the heat equation,
au --aAu=0, at
where u = u(x, t) is the temperature at position x and, time t, and a > 0 is the thermal conductivity (here assumed constant). For a body S2 in thermal equilibrium, au/at = 0, so if one knows the temperature distribution g on the bounding surface r, then one can determine the temperature distribution u in the interior by solving the boundary value problem
Au=0 onQ, u=g on 1.
(1.7)
This problem later became known as the Dirichlet problem, and for particular, simple choices of 0, Fourier constructed solutions using.separation of variables; see [7, pp. 132-138]. His book, Theorie analytique de la chaleur, was published in 1822. In 1828, George Green published An Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism [31], [33, pp. 1-115];
an extract appears in [7, pp. 347-358]. In his introduction, Green discusses previous work by other authors including Poisson, and writes that although many of the artifices employed in the works before mentioned are remarkable for their elegance, it is easy to see they are adapted only to particular objects, and that some general method, capable of being employed in every case, is still wanting.
Green's "general method" was based on his two integral identities:
1 grad
udx=J
wavdcr
- Jew/udx
(1.8)
and
In
(wAu - uAw) dx =
J
(w au - u av
dQ,
(1.9)
where u and w are arbitrary, sufficiently regular functions. Using (1.9) with
Introduction
5
w(y) = G(x, y), he obtained a third identity,
1 G(x, y)Au(y) dy - fr u(y) a-G(x, y) day
u(x)
av,,
Jsz
+
G(x, y)
Jr
au (y) day ev
for x E Q.
(1.10)
Actually, Green derived a more general result, showing that (1.10) is valid when G (x, y) is replaced by a function of the form
Gr(x, y) = G(x, y) + V (x, y), where V is any smooth function satisfying Ay V (x, y) = 0 for x, y E Q. In other words, Gr(x, y) has the same singular behaviour as G(x, y) when y = x,
and satisfies AyGr(x, y) = 0 for y # x. Green gave a heuristic argument for the existence of a unique such Gr satisfying Gr(x, y) = 0 for all y E I': physically, Gr(x, y) represents the electrostatic potential at y due to a point charge at x when I' is an earthed conductor. This particular Gr became known as the Green's function for the domain 0, and yields an integral representation formula for the solution of the Dirichlet problem (1.7),
u(x) = -
J
g(y) aav Gr(x, y) day for x E 0.
(1.11)
In practice, finding an explicit formula for Or is possible only for very simple
domains. For instance, if St is the open ball with radius r > 0 centred at the origin, then
Gr(x, y) =
1
4irIx - yl
1
r
47r IxIIx= - yl
where xO = (r/Ix1)2x is the image of x under a reflection in the sphere 1'. In this case, the integral (1.1 l) is given by
u(x)=
1
47rr flyl=r
forIxI
a formula obtained by Poisson [18, p. 360] in 1813 by a different method. Green also used (1.10) to derive a kind of converse to the jump relation (1.6), by showing that if a function u satifies the Laplace equation on S2+ U Q , is continuous everywhere and decays appropriately at infinity, then u = SL i,
where l'=-(avu-avu).
Introduction
6
Green's Essay did not begin to become widely known until 1845, when William Thomson (Lord Kelvin) introduced it to Joseph Liouville in Paris [99, pp. 113-121]. Eventually, Thomson had the work published in three parts during 1850-1854 in Crelle's Journal fair die reine and angewandte Mathematik [31].
Meanwhile, Poisson and others continued to apply the method of separation of variables to a variety of physical problems. A key step in many such calculations is to solve a two-point boundary value problem with a parameter
),>0, (adu I +(b-Xw)u=0 forO<x < 1, -d dx dx du
dx du
dx
- mou =0 at x =0,
(1.12)
-m1u=0 atx=1,
where a, b and w are known real-valued functions of x such that a > 0 and w > 0, and where mo and m 1 are known constants (possibly oo, in which case the boundary condition is to be interpreted as u = 0). The main features of the
problem (1.12) can be seen in the simplest example: a = w = 1 and b = 0. The general solution of the differential equation is then a linear combination of
sin(/x) and cos('Jx), and the boundary conditions imply that the solution is identically zero unless the parameter X satisfies a certain transcendental equation
having a sequence of positive solutions X1 < X2 < A3 < ... with Xj -+ oo. In the general case, the number Aj was subsequently called an eigenvalue for the problem, and any corresponding, non-trivial solution u = Oj of the differential equation was called an eigenfunction. For the special case a = w = 1, Poisson showed in 1826 that eigenfunctions corresponding to distinct eigenvalues are orthogonal, i.e., I
I OJ(x)4k(x)w(x) dx = 0
if X O Ak,
and that all eigenvalues are real; see [62, p. 433]. A much deeper analysis was given by Charles Francois Sturm in 1836, who established many important properties of the eigenfunctions, as well as proving the existence of infinitely many eigenvalues. Building on Sturm's work, Liouville showed in two papers from 1 836 and 1837 that an arbitrary function f could be expanded in a generalised
Introduction
7
Fourier series, 00
where c; _
f (x) = E cj¢j (x),
fo Oj (x) f (x)w(x) dx
i_i
f0 oj(x)2w(x) dx
thereby justifying many applications of the method of separation of variables. Excerpts from the papers of Sturm and Liouville are reproduced in [7, pp. 258281]; see also [62, Chapter X]. Carl Friedrich Gauss wrote a long paper [26] on potential theory in 1839; see also [7, pp. 358-361] and Liitzen [62, pp. 583-586]. He re-derived many of Poisson's results, including (1.6), using more rigorous arguments, and was apparently unaware of Green's work. Gauss sought to find, for an arbitrary conductor 0, the equilibrium charge distribution with total charge M, i.e., in mathematical terms, he sought to find a function * whose single-layer potential SL * is constant on IF, subject to the constraint that fr * do = M. Introducing an arbitrary function g, he considered the quadratic functional Jg (0) = f(Sq5 - 2g)¢ do,,
where So = y+ SL 0 = y- SL 0 denotes the boundary values of the singlelayer potential, or, explicitly, SO (x) =
f
r
G(x, y)cp(y)dar
for x E F.
In the case g = 0, the quantity Jg(4)) has a physical meaning: it is proportional to the self-energy of the charge distribution 0; see Kellogg [45, pp. 79-80]. One easily sees that Jg (¢) is bounded below for all 0 in the class VM of
functions satisfying fr 4) dv = M and 0 ? 0 on T. Also, Jg(0 + BO) _ Jg (0) + S Jg + O (S4)02), where the first variation of Jg is given by BJg = BJg(0, Scp) = 2
fr
(S¢ - g) 30 do,.
Suppose that the minimum value of Jg over the class VM is achieved when 0 = 4/r. It follows that SJg (>/r, SO) = 0 for all Ski satisfying fr BO dQ = 0 and * + S4) > 0 on I', and therefore Si/r - g is constant on any component
of r where * > 0. Gauss showed that if g = 0, then Mfr > 0 everywhere on I', and thus deduced the existence of an equilibrium potential from the existence of a minimiser for Jo. He also showed that this minimiser is unique,
Introduction
8
and gave an argument for the existence of a solution 'i/i to the boundary integral equation
S>/i = g on I'.
(1.13)
The single-layer potential of this +/i is the solution of the Dirichlet problem for the Laplace equation, i.e., u = SL i/r satisfies (1.7). In a series of papers from 1845 to 1846, Liouville studied the single-layer potential when I' is an ellipsoid, solving the integral equation (1.13) by adapting his earlier work on the eigenvalue problem (1.12). Let w be the equilibrium
density for T, normalised so that Sw = 1. Liouville showed that if I' is an ellipsoid, then
S(wi/rj)=ji i/rj
forj=1,2,3,...,
where the i/rj are Lame functions, and the ,
AI /L2?µ3?...>0
are certain constants satisfying
withpj -40 as j -;oo.
He established the orthogonality property
f 1fj (x)*trk(x)w(x) doix = 0 for j # k, and concluded that the solution of (1.13) is i/r(x) = w(x)
cj *j W, i=1
(x)g(x)w(x) dax where c j - fr'lr' fr *j (x)2w(x) dvx
In his unpublished notebooks (described in [62, Chapter XV]) Liouville went a considerable distance towards generalising these results to the case of an arbitrary surface r, inventing in the process the Rayleigh-Ritz procedure for finding the eigenvalues and eigenfunctions, 20 years before Rayleigh [97] and 60 years before Ritz [98]. During the 1840s, Thomson and Peter Gustav Lejeune Dirichlet separately advanced another type of existence argument [7, pp. 379-387] that became
widely known on account of its use by Riemann in his theory of complex analytic functions. Riemann introduced the term Dirichlet's principle for this method of establishing the existence of a solution to the Dirichlet problem, although a related variational argument had earlier been used by Green [32]. If
Introduction
9
one considers the functional
J(v) = I 1gradv12dx
Js
for v in a class of sufficiently regular functions Vg satisfying v = g on T, then it seems obvious, because J (v) > 0 for all v E Vg, that there exists a u E Vg satisfying
J(u) < J(v) for all v E Vg.
(1.14)
Given any w such that w = 0 on r, and any constant h, the function v = u + hw belongs to Vg, and, assuming the validity of the first Green identity (1.8), simple manipulations yield
J(v) = J(u) - 2h
J0
wLu dx + h2J(w).
Here, the constant h is arbitrary, so the minimum condition (1.14) implies that
1 wLu dx = 0
whenever w = 0 on T.
By choosing w to take the same sign as Au throughout 0, we conclude that u is a solution of the Dirichlet problem for the Laplace equation. Conversely, each solution of the Dirichlet problem minimises the integral. Dirichlet also established the uniqueness of the minimiser u. In fact, if both u 1 and u2 minimise
J in the class of functions Vg, then the difference w = u I - u2 vanishes on t, and, arguing as above with h = 1, we find that J (u 1) = J (u2) + J (w). Thus, J (w) = 0, so w is constant, and hence identically zero, implying that ui = U2
ono. In 1869, H. Weber [1041 employed the quadratic functional J (v) in a Rayleigh-Ritz procedure to show the existence of eigenfunctions and eigenvalues for the Laplacian on a general bounded domain. He minimised J(v) subject to two constraints: v = 0 on r, and f n v(x)2 dx = 1. If we suppose that a minimum is achieved when v = u , then by arguing as above we see that
fwLuidx=0
whenever w = 0 on I' and
jw(x)ui(x)dx=0.
Here, the extra restriction on w arises from the second of the constraints in the minimisation problem. Weber showed that -Dug _ ,k1u, on Q, where
X, = J(ul). In fact, for an arbitrary v satisfying v = 0 on r, if we put
Introduction
10
a= fnvu,dxandw=v-au,,thenw=O on r, and fnwu,dx=0, so by the first Green identity,
v(-Du,) dx = Js
i
Jc
(w + au,)Au, dx
au, =a J(u,)-ui frdv/ remembering that u, = 0 on F. Next, Weber minimised J (v) subject to three constraints: the two previous ones and in addition fn vu, dx = 0. The minimiser u2 is the next eigenfunction, satisfying -Du2 = A2u2 on S2, where 1A2 = J(u2) > A,. Continuing in this fashion, he obtained sequences of (orthonormal) eigenfunctions uj and corresponding eigenvalues Aj, with 0 < A, < A2 a.3 <
--
Although simple and beautiful, Dirichlet's principle (in its naive form) is based on a false assumption, namely, that a minimiser u E V. must exist because J (v) > 0 for all v E V8. This error was pointed out by Karl Theodore Wilhelm WeierstraB [7, pp. 390-391] in 1870, and the same objection applies to the variational arguments of Gauss, Liouville and Weber. During the period from 1870 to 1890, alternative existence proofs for the Dirichlet problem were devised by Hermann Amandus Schwarz, Carl Gottfried Neumann and Jules Henri Poincare; see Girding [25] and Kellogg [45, pp. 277-286]. We shall briefly describe the first of these proofs, Neumann's Methode des arithmetischen Mittels, after first introducing some important properties of the double-layer potential,
DL *(x) = f *(y) aay G(x, y) day r
for x
t'.
A surface potential of this type appears in the third Green identity (1.10), with = uIr; note the similarity with the general Poisson integral formula (1.11). The double layer potential has a very simple form when the density is constant on F. In fact
DL 1(x) =
1
0
for x E Q-, for x E Q+,
(1.15)
as one sees by taking u = 1 in (I.10) if x E 0-, and by applying the divergence theorem if x E 52+. Obviously, DL * is harmonic on Sgt, but the example /r = 1 shows that the double-layer potential can have a jump discontinuity, and it turns
Introduction
11
out that in general
y+DL*-y-DLi/r=* on r; cf. (1.6). Thus, if we let
Tt/r=y' DL*+y-DLi/r,
(1.16)
then
onI'.
(1.17)
The operator T may be written explicitly as
Ti/r(x) = -i/i'(x)+2J [*(y) -ilr(x)]avyG(x, y)dcy forx E F, and we see in particular that Ti = -1, in agreement with (1.15). Neumann's existence proof built on earlier work by A. Beer [4], who, in 1856, sought a solution to the Dirichlet problem (1.7) in the form of a doublelayer potential u = DL +/r. Beer worked in two dimensions, and so used the fundamental solution
G(x, y) = I log 27r
for x, y E R2 and x
-A 1
Ix
y.
In view of (1.17), the boundary condition y-u = g on F leads to the integral equation
-i/r + Ti/r = 2g on I'.
(1.18)
The form of this equation suggests application of the method of successive approximations, a technique introduced by Liouville in 1830 to construct the solution to a two-point boundary value problem; see [62, p. 447]. Beer defined
a sequence *o, *l, i/r2, ... by
*o = -2g and i/ri = T'Jri-i - 2g
for j ? 1,
which, if it converged uniformly, would yield the desired solution * = limi,oo ilri . However, Beer did not attempt to prove convergence; see Hellinger and Toeplitz [38, pp. 1345-1349]. The kernel appearing in the double layer potential has the form 8
I BvyG(x'y)=T"
Ix - yI"
Introduction
12
where T2 = 2rr is the length of the unit circle, and T3 = 4ir is the area of the unit sphere. For his proof, Neumann [77] assumed that Q- is convex. In this case, vy (x - y) < 0 for all x, y E F, so
min* < -(T *)(x) < max */r
for X E I',
and it can be shown that (provided the convex domain 0- is not the intersection of two cones) for every continuous g there exists a constant a8 such that
maxI(Tmg)(x) XEr
- (-1)ma.I < Cr"`,
with 0 < r < 1,
where the constants C and r depend only on F. We define a density function 00
/r = E(T2jg + T 2j+1 g), j=0
noting that the series converges uniformly on I' because
IT2jg+T2j+lgl
IT2jg
_ (_1)2jagl + IT2j+lg
- (-1)2j+la81
C(r2j + r2j+l). Also, the identity m
m
g+ T >(T2jg + T2j+lg) = T2m+2g + J:(T 2jg +
T2j+lg)
j=0
j=0
implies that g + T /i = ag + i/r, so by (1.17) we have y- DL it = (a8 - g). 2 function Therefore, the desired solution of the Dirichlet problem (1.7) is the u = a8 - 2 DL i/r. In a paper of 1888 dealing with the Laplace equation, P. du Bois-Reymond [20] expressed the view that a general theory of integral equations would be of great value, but confessed his inability to see even the outline of such a theory. (This paper, incidentally, contains the first use of the term "integral equation", or rather Integralgleichung.) The various results known at that time all seemed to rely on special properties of the particular equation under investigation. Only during the final decade of the nineteenth century did a way forward begin to emerge. In 1894, Le Roux [55] successfully analysed an integral equation of the form x
fJ n
K(x, y)u(y) dy = f (x) for a < x < b,
Introduction
13
with a sufficiently smooth but otherwise quite general kernel K, and a righthand side satisfying f (a) = 0.'He constructed a solution by first differentiating with respect to x, and then applying the method of successive approximations. Two years later, Volterra [103, Volume 2, pp. 216-262] independently considered the same problem, using the same approach, and remarked in passing that the integral equation could be looked upon as the continuous limit of an n x n linear algebraic system as n -* oo. Volterra's remark was taken up by Ivar Fredholm [23] in a short paper of 1900,
which was subsequently expanded into a longer work [24] in 1903. Fredholm considered an integral equation of the form U(X) + A.
J0
K(x, y)u(y) dy = f (x) for 0 < x < 1,
(1.19)
with a general continuous kernel K and a complex parameter),. As motivation,
he mentions a problem discussed a few years earlier in an influential paper of Poincare [82], namely, for a given function f on F to find a double-layer potential u = DL * satisfying
y-u - y+u = X(y-u + y+u) + 2f on r. In view of (1.16) and (1.17), this problem amounts to finding a density function
* satisfying
-*-).T* = 2f
on F.
(1.20)
The special case X = -1 and f = g is just Beer's equation (1.18) arising from the interior Dirichlet problem, and similarly A = +1 and f = -g gives the analogous equation arising from the exterior Dirichlet problem. Poincare had shown that both equations are solvable for a wide class of smooth but not necessarily convex domains. Fredholm began his analysis of (1.19) by introducing a function D (A) defined by the series
D (X) = 1+ A J K (y, y) dy 0
A2
K(yi, yi)
+ 2! Jo Jo K(y2, yi)
K(yi, y2) K(y2, y2)
(1.21)
which he called the determinant of the integral equation. In fact, if we put
xj = j/n for 1 < j < n, and replace the integral in (1.19) by the obvious
Introduction
14
Riemann sum, then we obtain the discrete system ),
u(x j) +n> K(xj, xk)u(xk) = f (xi) for 1 < j < n k=1
whose determinant can be written as X-"
It
n k=1
K (xki , xk i)
K (xk, , xk, )
2!n2 k,=1 k,=1 K(xk,, xk,)
K(xkz, xk,)
An
+ nln
n
n
K(xki,xki)
... K(xki+xk,)
E ... E
ki=1
K (xk,
xk ) I
Formally at least, in the limit as n -+ oo the determinant of the discrete system tends to D(A). (This heuristic derivation does not appear in Fredholm's papers, but see [38, p. 1356] and [19, p. 99].) Fredholm proved that the series (1.21) converges uniformly for X in any compact subset of the complex plane, and so defines an entire function. By generalising Cramer's rule for finite linear systems, Fredholm showed that if D(A) 0 0, then (1.19) has a unique continuous solution u for each continuous f. He applied this result to the boundary integral equation (1.20), and so proved the existence of a solution to the Dirichlet problem on any bounded C3 domain in the plane.
Fredholm also gave a complete account of the case when D(k) = 0, by considering the transposed integral equation
v(x)+A
I
I
K(y,x)v(y)dy=g(x) for0 <x < 1,
(1.22)
which has the same determinant as the original equation (1.19). He proved that if D(A) has a zero of multiplicity m at A = Ao, then for this value of the parameter the two homogeneous equations, i.e., (1.19) and (1.22) with f and g identically zero, each have m linearly independent solutions. In this case, the inhomogeneous equation (1.19) has a (non-unique) solution u if and only if fo1 f (x)v(x) dx = 0 for every solution v of the transposed homogenenous equation. The above dichotomy in the behaviour of the two integral equations,
corresponding to the cases D(A) # 0 and D(A) = 0, is today known as the Fredholm alternative.
The simplicity and generality of Fredholm's theory made an immediate and lasting impression, not least on David Hilbert, who, during the period 1904-1906, made important contributions that later appeared in his influential book [39] on integral equations. Hilbert was especially interested in the
Exercises
15
case when the kernel is symmetric, i.e., when K is real-valued and satisfies K (y, x) = K (x, y) for all x and y. The zeros of the determinant are then purely real, and f o r m a nondecreasing sequence a.i, X 2 ,--- , counting multiplicities. For each j there is a non-trivial solution i/rj of the homogeneous equation with A = Al, and the sequence i/ri, *2, ... can be chosen in such a way that the functions are orthonormal:
J
f
,/r j
(x)1//k(x)dx = 8jk
=
1
if j = k,
10
if J ' 54 k.
Of course, i/rj is an eigenfunction of the integral operator with kernel K, and the corresponding eigenvalue is 1/a,1. Hilbert proved the identity i
t
1
-1
f f K(x, y)u(x)v(y) dx dy =
t
* (x)u(x) dx
j>I x
f
f(y)v(y) dy,
0
which is the continuous analogue of the reduction to principal axes of the quadratic form associated with a real symmetric matrix. He also studied the convergence of eigenfunction expansions. Our story has now arrived at a natural stopping point. The period of classical analysis is about to be overtaken by the geometric spirit of functional analysis.
By 1917, F. Reisz [87] had effectively subsumed Fredholm's results in the general theory of compact linear operators, a topic we shall take up in the next chapter.
Exercises
1.1 Show that if f is a radially symmetric function, say f (y) = F(r) where r = Iyj, then the Newtonian potential (1.4) is radially symmetric, and is given by u(x) = U(p), where p = IxI and
U(p) =
1f P
a
F(r)r2 dr +
f
00
F(r)r dr.
p
o
Hence verify Poisson's equation:
-Au(x)=---
--
(P2UF(p)=f(x)
16
Introduction
1.2 Let r = (y E R3 : IYI = a} denote the sphere of radius a > 0 centred at the origin. Show that if the density i is constant on F, then the single-layer potential (1.5) is radially symmetric, i.e., a function of p = Ix I. Show in particular that
SL 1(x) =
ifp < a, a2/p ifp > a, a
and verify that the jump relation (1.6) holds in this case.
1.3 Fix x, y E 0 with x # y, and for any sufficiently small c > 0 let 9E denote the region obtained from S2 by excising the balls with radius E centred at x and y. By applying the second Green identity (1.9) to the functions Gr(x, .) and Gr(y, ) over E. and then sending e 0, show that Gr(x, y) = Gr(y, x).
2
Abstract Linear Equations
Later in this book, we shall reduce elliptic boundary value problems, and also their equivalent boundary integral formulations, to operator equations of the form Au = f, with A a bounded linear operator from a Hilbert space into its dual. The ellipticity of the partial differential equation will imply that A is the sum of a positive-definite operator and a compact operator (the latter possibly zero). Our aim now is to study such operators abstractly, using techniques from functional analysis. We begin by considering some topics that can be understood more clearly in a less restricted setting. In fact, we shall develop the concept of a Fredholm operator acting between two Banach spaces, even though it would suffice for our later applications to consider only operators with index zero acting between Hilbert spaces. At the end of the chapter is a short treatment of spectral theory, covering just the simplest cases, namely, self-adjoint operators that are compact or have a compact inverse. We shall use c and C to denote small and large generic constants, whose values may change even within a single chain of estimates, but with c always bounded away from zero, and C always bounded away from infinity. If II II and II' are norms on a vector space X, then we write 11
IIu1I
IIu1I'
forallu E X,
to indicate equivalence of the norms, i.e., c II u II x < II u II' < C II u II x for all u E X.
The reader should also note that our sesquilinear forms, and in particular our inner products, are conjugate-linear in the first argument, and linear in the second.
A familiarity with basic concepts and results from general topology and linear functional analysis is assumed, but some effort will be made to refresh the reader's memory. The theorems that we cite without proof can be found in virtually any textbook on functional analysis; Yosida [106] and Simmons [94] will serve as our standard references. 17
Abstract Linear Equations
18
The Kernel and Image Suppose that X and Y are complex vector spaces, and let A : X -+ Y be a linear map. The kernel (or null space) of A is the subspace of X defined by
kerA={u EX: Au=O}, and the image of A is the subspace of Y defined by
im A = (f E Y : there exists u E X such that f = Au}. Given f E Y, we can seek a solution u E X to the linear equation
Au = f. It follows at once from the definitions above that a solution exists if and only if f E im A, in which case u is unique modulo ker A, i.e., any two solutions differ by an element of the kernel. Thus, the inverse A-1 exists if and only if kerA = {0} and im A = Y. Recall that if W is a subspace of X, then the elements of the quotient space X1 W are the cosets u + W = {u + w : w E W), and the vector space operations in X/W are given by
A(u+W)_(Au)+W and (ul+W)+(u2+W)_(uI+u2)+W, for A E C and u, u1, u2 E X. The dimension of X/W is called the codimension of W in X. Every linear map A : X -). Y induces an isomorphism
A/: X/ ker A --- imA defined by
A/ (u + ker A) = Au
for u E X.
Assume now that X and Y are normed spaces. A linear map A : X -+ Y is continuous if and only if it is bounded, i.e., if and only if Il Au ll r < C II u II x for all u E X. The vector space G(X, Y) consisting of all such bounded linear maps is itself a normed space, with
IlAullr IIAIIc(x.Y) = sup OuEX Ilullx
Recall that if Y is a Banach space (i.e., if Y is complete), then so is ,C(X, Y).
The Kernel and Image
19
If W is a closed subspace of X, then we can make X/ W into a normed space by defining
Ilu+Wllx/w = WEW inf
Ilu+wllx
Furthermore, X/ W is aBanach space when X is aBanach space. If A E £(X, Y),
then ker A is closed, and the induced isomorphism is bounded, i.e., A/ E £(X/ ker A, im A). When is the inverse (A/)-' bounded? We can answer this question with the help of the open mapping theorem.
Theorem 2.1 Suppose that X and Y are Banach spaces, and let A E £(X, Y). If im A = Y, then A is an open mapping, i.e., A maps each open subset of X to an open subset of Y.
The proof of this result uses a Baire category argument, and can be found in [106, p. 75] or [94, p. 236]. Since a function Y -+ X is continuous if and only if the pre-image of every open set in X is open in Y, and since a closed subspace of a Banach space is again a Banach space, the next corollary follows at once.
Corollary 2.2 Suppose that X and Y are Banach spaces. If A E £(X, Y), then the following conditions are equivalent: (i) The subspace im A is closed in Y. (ii) The induced map A/ : X/ ker A -* im A has a bounded inverse. (iii) There is a constant C such that
Ilu +kerAllx/kerA < CIIAulir foru E X. In particular, there exists a bounded inverse A-' E £(Y, X) if and only if im A = Y and kerA = {0). When A is as in Corollary 2.2, our problem Au = f is essentially well posed: for each f E im A, the solutions form a coset u + ker A that depends continu-
ously on f. Of course, in applications, one typically starts with a concrete integral or differential operator A, and then seeks complete spaces X and Y such that A : X -+ Y is bounded and im A is closed. Satisfying both conditions simultaneously requires a tight fit between operator and spaces. If II II r is too strong (i.e., too big) relative to II II x, then the space Y will be too small to serve as the codomain, and typically one obtains an unbounded operator by restricting
Abstract Linear Equations
20
A to some dense subspace of X. However, if II . II r is too weak (i.e., too small), then the space Y will be too large, and im A will fail to be closed. If V and W are subspaces of X, and if each u E X can be written as u = v.+ w for a unique v E V and a unique w E W, then X is said to be the (internal) direct sum of V and W, and we indicate this fact by writing X = V e W. A projection
is a bounded linear operator P : X -+ X having the property that P2 = P. If p
is a projection, then (I - P)2 = I - P and P(I - P) = (1- P)P : 0, where I is the identity operator, i.e., l u = u. Thus, I - P is also a projection, and by writing u = Pu + (I - P)u, we obtain adirect sum decomposition X = V ® W, where V and W are the closed subspaces
V=imP=ker(1-P) and W=im(I- P) =kerP. This state of affairs is described by saying that P is the projection of X onto V, parallel to W. Exercise 2.2 shows that there exists a projection onto every finitedimensional subspace, and onto every closed subspace with finite codimension.
Duality
If X is a normed space, then we denote its dual space by X*. Thus, X* £(X, C) is the space of bounded linear functionals g : X -). C. We shall write (g, u) = g(u) for the value of the functional g E X* at the vector u E X. By the definition of the norm in £(X, C), I (g, u)I
Ilgllx = sup I(g' u)I OOUEX
(2.1)
IlullX
The dual space X* is a Banach space even if X is not complete. A key tool in the study of duality is the Hahn-Banach theorem, one version of which is as follows. Theorem 2.3 If W is a subspace of a normed space X, then everyfunctional in W * can be extended to a functional in X* having the same norm.
For a proof, see [106, p. 1061 or [94, p. 228]. Zorn's lemma or another equivalent of the axiom of choice is needed unless one introduces some extra assumption(s), such as that X is separable. We now establish a few simple consequences of Theorem 2.3 that will be used later. If u E X and W e X, then the distance between u and W is defined
Duality
21
by
dist(u, W) = inf 11 u - wllx; WEW notice that if W is a closed subspace, then dist(u, W) = 11 u + W 11 X/ w.
Theorem 2.4 Let W be a subspace of a normed linear space X, and let u E X. If dist(u, W) > 0, then there exists a functional g E X* such that
(g, u) = dist(u, W),
11811x = 1 and (g, w) = 0
for w E W.
Proof Put d = dist(u, W), and assume that d > 0. It follows that u 0 W, and we may form the direct sum Wl = W ® span{u}. Define g E W* by
(g, w + Au) = Ad
for w E W and A E C,
and observe that
I(g,w+),u)I
forty E W and ),
0,
so II g 11 w; < 1. Moreover, given c > 0, there is,a w E W such that d < II u - w II x <
d+E,andthusd= I(-1)d1= I(g, w - u)I -11g11w1. 11w - ullx < IIg11w; (d+E), implying that IIg II w; > d/(d + E). Hence, 11g II w; = 1, and we can complete the
proof by applying Theorem 2.3.
Corollary 2.5 Let W be a closed subspace of X, and let u E X. If is 0 W, then there exists a functional g E X* such that
(g, u) > 0,
Ilgllx = 1 and (g, w) = 0
for w E W.
Proof If W is closed and u f W, then dist(u, W) > 0. Corollary 2.6 If 0
0
u E X, then there exists a functional g E X* such that
(g,u)=Ilullx and Proof. Take W = (0), so that dist(u, W) = IIu11x.
0
Corollary 2.7 The dual space X* separates the points in X, i.e., for all u, u E X, if u # v, then there exists a functional g E X* such that (g, u) # (g, v).
Corollary 2.8 Let is E X. If (g, u) = 0 for all g E X*, then u = 0.
Abstract Linear Equations
22
Consider X** _ (X*)*, the second dual of X, and define c : X --> X** by
(cu, g) _ (g, u)
for U EX and g E X*.
From (2.1) and Corollary 2.6, we see that II Lu II X.- = II u II x, sot is an isometric isomorphism from X onto a subspace t (X) in X**. This fact allows us to identify
X with i (X), and write
X C X**. Obviously, X is closed in the complete space X** if and only if X is itself complete. If X = X**, then X is said to be reflexive; thus, every reflexive space is complete. For any linear map A : X Y, the transpose At : Y* -+ X* is the linear map defined by
(Atv, u) = (v, Au)
for all v E Y* and u c- X.
With the help of Corollary 2.6, we can show the following.
Lemma 2.9 The transpose At is bounded if and only if A is bounded. In fact, IIA`II,c(Y*,x*) = IlAllc(x.Y)
Proof If A is bounded, then the definition of At gives I(Aty, u) I = I(v, Au) I _< IIvIIY*IiAllc(x,Y)Ilullx
for u E X and v E Y*,
so IIA`vllx < IiAllc(x,Y)IIvIIY* and hence IIA`Ilc(Y*,x*) < IiAllc(x.Y) Conversely, suppose that At is bounded, and let u E X. If Au 0 0, then by Corollary 2.6 there is a v E Y* such that
(v, Au) = IlAullx
and
IIUIly* = 1.
Using the definition of At once again, we have
IlAullx = I(Atv,u)I < IiAtvllx*Ilullx IlAtllc(Y*.x*)IIvIIY*Ilullx = IIA`Ilc(r*,x*)Ilullx,
and since the inequality II Au II x < II At II c(Y*,x*) II u II x is trivial if Au = 0, we deduce that IIAIIr(x,Y) < I1Atllc(Y*,x*) 0
Duality
23
In studying solutions of the equation Au = f , it is helpful to consider at the same time the transposed equation A`v
=g
for a given g E X* and an unknown v E Y*. To describe the relationship between
the two equations, we use the following terminology. For any subset W c X, the annihilator W a is the closed subspace of X* defined by
Wa=(gEX*:(g,u)=0foralluEW). Dually, for V C X* the annihilator IV is the closed subspace of X defined by
aV={uEX:(g,u)=0forallgE V}. Lemma 2.10 The kernels and images of A and At satisfy
ker A = a(im A`)
and
ker A` _ (im A) a.
Proof. Applying the various definitions gives
a(im A`) = (uEX : (g, u) = 0 for all g E im A`)
= (uEX (A`v,u) =0forallvEY*) = [U EX : (v, Au) = 0 for all v E Y*)
={uEX:Au=0}=kerA, and a similar argument shows that (im A) a = ker At.
0
One sees directly from the definition of the annihilator that
W c a(W a)
for any subset W C X,
V c (aV) a
for any subset V C X*,
and likewise
so by Lemma 2.10,
im A c a(ker A`)
and
im A` c (ker A) a.
The question now arises as to when the reverse inclusions also hold. The next two lemmas will help to provide us with the answer.
Abstract Linear Equations
24
Lemma 2.11 Let X be a normed space. A subset W c X satisfies W = (W a) if and only if W is a closed subspace of X. Proof. The condition is obviously necessary. To prove sufficiency, assume that W is a closed subspace of X, and let u E X. If U V W, then we can find g E X* as in Corollary 2.5. Since g E W' and (g, u) # 0, we see that u a (W a). Thus, a(W a) C W, and the result follows.
Dual to Lemma 2.11 is the result that a subset V C X* satisfies V = (a V) a if and only if V is a weak* closed subspace of X*; see Schechter [91, p. 192]. However, we shall not use this fact.
Lemma 2.12 Suppose that W is a subspace of a normed space X. (i) There exists an isometric isomorphism J1 : X*/W a -+ W* given by
for g E X* and w E W.
(Ji (g + W a), w) = (g, w)
(ii) If W is closed, then there exists an isometric isomorphism J2
: W'
(XI W)* given by
(J2$,u+W) = (g, u)
forgEWa and uEX.
Proof. LetgEX*. For any hEWaand wE W, we have (g, w) = (g+h,w), so J1 is well defined and I (Ji (g + W a), w) I < fig + h II x II w II x, implying that
I(J,(g+ We), w)I ::S i w, II
IIg+ WalIx*/w,Ilwllw
Hence, IIJ1(g + Wa)IIw* <_ fig + W all x*/w
Next, if J, (g + W a) = 0, then g E W a, and so ker Jl = {0 + W a), i.e., Ji is one-one. To see that J1 is onto, let g E W*. By Theorem 2.3, there is a functional $1 E X* such that $1 = g on W, and 11 g, 11 x. = IISllw Hence, 4 g = J1(gI + W a) E im J1, and moreover JjJ1 1gjjx*/w.' =1Ig1 +
I19111x* = Ilgllw*,
completing the proof of (i). Turning to part (ii), we assume that W is closed and let g E W'. For any u E X and W E W, (g, u) = (g, u + w), so J2 is well defined, and I (J2g, u + W) II g II x* II u + w II x, implying that
(J2$, u + W) 1 <_ Ilgllx* inw 11 u + wllx = Ilgllx* 11 u + W 11 x1 w.
Duality
25
II g 11x Obviously, ker J2 = {0}, so J2 is one-one. To see that J2 is onto, let h e (X/ W)*. We define a linear functional g : X -k C
Hence, lI J2g II (x/ w)
by (g, u) = (h, u + W), and observe that
I(g,u)I <
IIh11(x/w)*I1 uIIx
Hence, g is bounded with 11 g II X. < 11 h II (x/ w)-. Also, if u E W, then u + W =
0 + W so (g, u) = 0, showing that g E W a and h = J2$ E im J2. Finally, 11J2'h11Wa =11gIIx <
0
so J2 is an isometric isomorphism.
We are now ready to prove the main result for this section. Here, the key point is that if the image of A is closed, then a necessary and sufficient condition for the equation Au = f to be solvable is that the given right-hand side f be annihilated by every solution v of the homogeneous transposed equation Atv = 0.
Theorem 2.13 Suppose that X and Y are Banach spaces. For A E £(X, Y), the following conditions are equivalent: (i) im A is closed in Y. (ii) im At is closed in X*. (iii) im A = a(ker At). (iv) im At _ (ker A) a. When these conditions hold, there are isometric isomorphisms (Y/ im A)*
ker At
and X*/ im At -- (ker A)*.
Proof. We have seen that A gives rise to a bounded linear operator,
A/:X/kerA-*imA. Likewise, At gives rise to a bounded linear operator (At)/ : Y*/ ker At --* im At,
related to the transpose of A/,
(A/)t : (im A)* -+ (X/ ker A)*,
Abstract Linear Equations
26
in the following way. Since ker At = (im A) a, and since ker A is a closed subspace of X, Lemma 2.12 yields isometric isomorphisms
J, : Y*/ker At -* (imA)*
and
J2 : (kerA)a -+ (X/kerA)*,
so ifuEXandvEY*, then ((Al)tJ, (v + ker A'), u + ker A) = (J, (v + ker A'), At(u + kerA))
= (J, (v +kerA`), Au) = (v, Au) = (A'v, u) = ((A`)/ (v + kerA`), u)
= (J2 (A)/ (v + kerA), u + kerA), where, in the final step, we used the fact that im(At)/ = im At C (ker A)'. Thus,
(AI)tJ, = J2(At)l : Y*/kerA` -+ (X/kerA)*,
(2.2)
and we have the chain of equivalences im A is closed
A/ has a bounded inverse (A/)t has a bounded inverse
(Corollary 2.2) (Exercise 2.3)
= (At)/ has a bounded inverse (2.2)
= im At is closed
(Corollary 2.2),
showing that (i) is equivalent to (ii). Next, Lemmas 2.10 and 2.11 show that (i) implies (iii). Since every annihilator is closed, (iii) implies (i), and (iv) implies (ii). Hence, the equivalence of the four conditions will follow if we prove that (i) implies (iv). Thus, assume that im A is a closed subspace of Y. To prove (iv), it suffices to show the inclusion (ker A) a C im At, so let g 'E (ker A) a. We define a bounded linear functional v on im A by
(v, f) = (JZg, A7' f) for f E im A, and then use the Hahn-Banach theorem to extend v to a bounded linear functional u on all of Y. In this way,
(g, u) = (JZg, u +kerA) = (JZg, A/ 'Au) = (v, Au) = (u, Au) = (Atv, u) for all u E X, so g = Atv E im At, and thus (ker A)' C im At, as required.
Compactness
27
Finally, if (i)-(iv) hold, then Lemma 2.12 yields isometric isomorphisms
X*/imA` = X*/(kerA)a -+ (kerA)*
and
ker A'= (im A)' -> (Y/imA)*.
0
In many applications, it is natural to work with a space Z that is only isomorphic to X* in the following way. Suppose there is a bilinear form Z x X -+ C that is bounded in the obvious sense, i.e., I((g,u))I
CIIgIIzIluIx
forgEZ and uEX,
allowing us to define a bounded linear operator t : Z -+ X* by
(tg, u) = ((g, u)) If t has a bounded inverse, i.e., if t is an isomorphism of Banach spaces, then we call Z a realisation of the dual space X*. In this case, II tg II x - II g Il z
sup
O0uEX
I ((g' u)) I
IIulix
and we routinely identify Z with X* by suppressing the distinction between g and tg, and between and
Compactness Let us recall some facts about compact subsets of metric spaces. Suppose that a set X is equipped with a metric Ix. An open cover of a subset W C X is a family of open subsets of X whose union contains W. We say that W is compact if every open cover of W has a finite subcover. Every compact set is closed. If W, the closure of W, is compact, then W is said to be relatively compact. For e > 0, an E -net for W is a finite subset (w 1, ... , c W with the property that for each w E W there exists an index i = i(w) E (1, ... , n} such that Iw, w, Ix < e. If W has an E-net for every e > 0, then W is said to be totally bounded. Every totally bounded set is bounded. Theorem 2.14 In any metric space X, the following three statements are equivalent:
(i) The subset W is relatively compact. (ii) Every sequence in W has a subsequence that converges in X. (iii) The subset W is totally bounded. In particular, any relatively compact set is bounded.
28
Abstract Linear Equations
For a proof, see [106, p. 13] and [94, pp. 120-125]. Suppose now that the whole metric space X is compact. The set C(X) of all continuous functions f : X --+ C is a Banach space with norm
11fIIc(x) =maxIf(x)1. A subset F C_ C(X) is said to be equicontinuous if for every E > 0 there exists a S > 0 such that, for all x, y E X and for every f E F, Ix, YI x < 8
implies
If (x) - f (y) l < E.
The importance of this property stems from the following theorem of Arzela and Ascoli. Theorem 2.15 Let X be a compact metric space. A subset of C (X) is relatively compact if and only if it is bounded and equicontinuous.
For a proof, see [106, p. 85] or [94, p. 126]. There is an analogous characterisation of the relatively compact subsets of Lp(R"); see [106, p. 275].
Theorem 2.16 For 1 < p < oo, a subset W is relatively compact in L p (R") if and only if the following three conditions are satisfied: (i) W is bounded, i.e., IIf IILn(a"") < C for f E W.
0 as h - 0, (ii) W is p-mean equicontinuous, i.e., IIf ( + h) - f IIL,,(tt"") uniformly for f E W. (iii) III II L,,(R"\B,,) _-* 0 as p --+ oo, uniformly for f E W, where Bp = {x E R : IxI < p]. Suppose now that X and Y are normed spaces. A linear operator from X into Y is said to be compact (or completely continuous) if it maps every bounded subset
of X to a relatively compact subset of Y. Every compact operator is bounded. Also, any linear operator with a finite-dimensional image is compact, because in a finite-dimensional normed space every bounded set is totally bounded. It follows from Theorem 2.14 that a linear map K : X -+ Y is compact if and only if every bounded sequence ui in X has a subsequence up such that Ku J. converges in Y. In the light of the Arzela-Ascoli theorem, if K : C [0, 1 ] -+ C [0, 1 ] is compact, then we expect Ku to be smoother than u, and so it is not surprising that many integral operators are compact. Similarly, we shall see later that, in the case of a partial differential operator acting between appropriate Sobolev spaces, the lower-order terms give rise to only a compact perturbation of the principal part.
Compactness
29
Given our earlier study of duality, it is natural to ask about the compactness of the transpose.
Theorem 2.17 Consider a linear map K : X Y*--4X*.
Y and its transpose Kt
(i) If K is compact, then Kt is compact. (ii) If Kt is compact, and if Y is complete, then K is compact. Proof Assume that K is compact, and take a bounded sequence wJ in Y*. To prove (i), it suffices to show that a subsequence of KtwJ converges in X*. We denote the closed unit ball in X by U = {u E X : Ilu llx -< 11, and let vJ denote the restriction of the functional wJ to the compact set K(U). For f = Ku and U E U, we have
lvJ(f)I = 1(w J,Ku)l <
S C,
because the w j are bounded in Y *, and II u II x < 1. Thus, l vJ (f) 1 < C for f E
K(U), and
Ivi(fi) - vi(f2)I = 1(wi, .fi - f2)I < Cllf - f2llY
for fi, f2 E K(U),
so the v j are bounded and equicontinuous. Applying the Arzela-Ascoli theorem,
we deduce that a subsequence vJ, converges uniformly on K(U). Given any nonzero u E X, put u = II U II x 1 u E U, and observe that
(Kiwi, -Ktwk,, u)l = IlullxI(wj' -Wk,, Ku)I = IlullxlvJ,(Ku) -Vk,(KU)I, so
IIKtw3, - Ktwk-11x* < max Ivi'(f) - Vk,(f)I. fEK(U)
Therefore, the subsequence Ktwj, is Cauchy in the complete space X". X** and To prove (ii), we assume now that K* is compact, and let tX : X L y: Y - Y** be the natural imbeddings. For all u E X and g E Y*,
(tyKu, g) = (g, Ku) = (K*g, u) = (txu,
K*g) = (K**txu,
g),
which shows that tyK = K**LX. Suppose that uJ is a bounded sequence in X. The sequence tx u J is bounded in X**, and K** : X** --s Y** is compact by part (i), so there is a subsequence u j, such that K**txu J, converges in Y**. Thus, Ku J. = t- y' K**txu J, is Cauchy, and hence convergent, provided Y is complete.
0
Abstract Linear Equations
30
We now turn our attention to operators of the form I + K : X - X, where I is the identity operator and K is compact. The results obtained for this special case will be used in the next section to deduce important properties of a much wider class of operators. As we saw in Chapter 1, Fredholm developed the
first general theory of equations of the form u + Ku = f, albeit with K a concrete integral operator. His method of Fredholm determinants used only the techniques of classical analysis, but is briefly described by Riesz and Sz.-Nagy
in their well-known textbook on functional analysis [88, pp. 172-176]. This book also sets out an abstract theory due to its first author [87], and which we now follow. Until the end of this section, we write for brevity II II = II . lix; no other norms occur.
In the next lemma, one thinks of u as being nearly orthogonal to W, even though the norm might not arise from an inner product.
Lemma 2.18 If W is a closed subspace of a normed space X, and if W 0 X, then for each E E (0, 1) there exists u E X such that 11u II = 1 and dist(u, W) >
1 - E. Proof Choose any v E X \ W, put d = dist(v, W), and note that d > 0 because W is closed. Given E E (0, 1), choose WE E W such that d < II v - wE II < d/(1 - E), and put u = Ilv - wE II- (v - wE). Obviously (lull = 1, and for all W E W,
Ilv-WEll(u-w)=v-wE-Ilv-wEllw=v- (an element of W), so II v - wE II II u - w II > dist(v, W) = d, and hence II u - w II ? dl ll v - wE II >
1 - E. For emphasis, the symbol C is used below to denote strict inclusion, i.e., for any sets V and W, we write V CW if and only if V c W and V W.
Theorem 2.19 Let X be a normed space, assume that K : X -+ X is compact, and define A : X X by
A=I+K. (i) For each n > 0, the subspace V, = ker A" is finite-dimensional. (ii) For each n > 0, the subspace W = im A" is closed. (iii) There is a finite number r such that
(0)=VocVic...CV,.=Vr+1 = ..
Compactness
31
and
W'- =W'-+1=
.
(iv) X =Vr®Wr. Proof. Suppose for a contradiction that V, is not finite-dimensional. Using Lemma 2.18, we can recursively construct a sequence u j in V, such that
for j#k.
IIujII=1 and Iluj-ukll? i
Since K is compact, there is a subsequence up and an element 4) E X such that
Ku p -+ 0. But u j + Ku j = Au j = 0 so up = -Kuj,
-0, which is
impossible because u j, is not a Cauchy sequence. Hence, V1 must be finitedimensional. Part (i) follows at once, because
A"=I+L
where
L(n)Km, m
(2.3)
M=1
and L is compact by Exercise 2.6. (The case n = 0 is trivial because A° = I.)
To prove that W1 is closed, suppose that f j = Au j -+ f. Let d j = dist(u j, ker A), and choose V j E ker A such that dj < 11uj
- vjll < (1 + j-1)dj.
(2.4)
If the d j are bounded, then so is the sequence w j = u j - v j , and there is a subsequence w j, such that Kw j, -+ 0. Since Aw j = Au j = f j, it follows that W j? = fj, - Kw j, --> f - 4), and thus f = lim Awj- = A(f - 0) E W1. If the d j are not bounded, then by passing to a subsequence we can assume that dj -+ 00
anddj>0.Define wj=lluj-vjll-1(uj-vj)sothat llwjll=landKwj,-*0. Since IIAwjII=IIuj -vjII-1IIA(uj-vj)II
ituj - vjIIfj = uj - vj+IIuj - vjII0=uj - (an element of kerA), so the definition of d j gives II u j - v j 11 11 *j II ? d1, and hence II *j II ? d j /
II u j - vj I I ? 1/(.l + j-1) by (2.4). But *j, -+ 0, and we conclude from this contradiction that the d j must be bounded. Hence, f E W1, so W1 is closed, and part (ii) follows by (2.3).
One easily verifies that V g and that if Vr = V,+,, then V,z for all n > r. Suppose for a contradiction that no such r exists, i.e., assume
Abstract Linear Equations
32 V" C
for all n. By Lemma 2.18, we can choose u,, E V ,,+l such that II u,, II _
1 and dist(u,,, V") > 2. If n > m, then
Ku,,, - Ku = u" - (Au" + u,,, - Au,,,) = u,, - (an element of so II Ku,,, - Ku,, II > dist(u,,, V,,) > ; and hence no subsequence of Ku,, converges. This contradiction implies that V = V,,+1 for some n, and we define r = min{n : V,, = Next, one easily verifies that W,, 2 W,,+1, and that if Wr, = Wr,+1 for some r', for all n > r'. Suppose for a contradiction that no such r' exists, then W,, = i.e., assume W,, W,,+i for all n. By Lemma 2.18, we can choose u E W such that 11 u, II = 1 and dist(u,,, W,,+1) > 2. If n > m, then
Ku" - Ku,,, = u,,, - (Au,,, + u,, - Au,,) = u,,, - (an element of W,,,+1), so II Ku - Ku,,, II > dist(um, W.+1) > !,giving the desired contradiction. We may therefore define r' = min{n : W,, = W,,+1 }. At this point, a simple lemma is needed: for each k > 0 and f E W,.,, the equation Aku = f has a unique solution u in W,.,. Indeed, u exists because Wr'+k = W,.,; to prove uniqueness, suppose for a contradiction that the homogeneous equation Aku = 0 has a non-trivial solution u = u 1 E Wr'. For n > 1, we choose recursively u,, E W,.' such that Au,, = un_1. Since A"u,, = Aul = 0
but An-1u" = u1 # 0, we see that u E V but u,, 0 V, 1, a contradiction if
n > r. To complete the proof of part (iii), we now show that r = r'. If f E V,.'+1, then the homogeneous equation Au = 0 has a solution u = Ar' f E Wr,, and by the argument above, u = 0, so f E V,.-. Therefore, Vr,+1 c Vr,, and hence r' > r. In particular, if r' = 0 then r = r' = 0. Suppose now that r' > 1, and choose f = Ar'-1 V E W,,_1 such that f 0 Wr,. The equation Ar'u = Ar'v has a solution u
in W,.,, and we have A' (v - u) = 0, but Ar -1(v - u) = f - Ar'-1 u
0 because Are-1 U E W2r-_1 = Wr'. Thus, v - u E Vr, but v - u 0 V,.-_1, showing Vr-_1 and hence r' < r. that Vr,
Finally, we turn to part (iv). The homogeneous equation Aru = 0 has only the trivial solution in Wr, so Vr n Wr = (0). Given f E X, let u E Wr be the solution of the equation A2'u = Ar f , andput v = A''u E W,.. Since Ar (f - v) _
Arf -A2ru=0, we have f =(f -v)+vEVr+Wr. Fredholm Operators Throughout this section, we shall assume that X and Y are Banach spaces. A bounded linear operator A : X --± Y is said to be Fredholm if
Fredholm Operators
33
1. the subspace im A is closed in Y; 2. the subspaces ker A and Yl im A are finite-dimensional. In this case, the index of A is the integer defined by
index A = dim ker A - dim (Y1 im A). As a consequence of Theorem 2.13, A is Fredholm if and only if At is Fredholm, in which case
index A = dim ker A - dim ker At = - index At. F. Noether [80] introduced the term "index" in the above sense for a concrete class of singular integral operators. In finite dimensions, the index depends only on the spaces, and not on the operator.
Theorem 2.20 If A : C" -* C'" is a linear map, then A is Fredholm, and index A = n - m. Proof. By performing elementary row and column operations, we can find bases for C" and C' relative to which the action of A is given by an m x n diagonal matrix [ajk] with entries
ajk =
I ifl<j=k
otherwise,
where r = dim im A = dim im At is the rank of A. The number of zero columns
in the matrix [ajk] is dim ker A = n - r, so
index A = dim ker A - dim(C" / im A) = (n - r) - (m - r) = n - m. 0 The preceding result might lead one to expect the following.
Theorem 2.21 If A : X -* Y and B : Y -* Z are Fredholm, then so is BA
X -- Z, and index(BA) = index A + index B.
Abstract Linear Equations
34
Proof The operator BA factors naturally into a product of Fredholm operators, each of which is either one-one or onto, as follows:
uHu+kerAHAuHAu+kerBH BAu, X -* X/ ker A - Y -+ Y/ ker B -k Z. Therefore, it suffices to consider two special cases:
(i) im A = Y and ker B = {0) (onto followed by one-one). (ii) ker A = {0) and im B = Z (one-one followed by onto).
In case (i), ker(BA) = ker A and im(BA) = im B, so the result is obvious. Thus, assume case (ii). Since Y/ im A and ker B are finite-dimensional, there is a finite-dimensional subspace YA c Y and a closed subspace YB c Y such that
Y=YA0imA=YBED ker B. We define
Y2=imAflkerB,
Y1=imAnYB,
Y3=YAfYB,
Y4=YAnkerB,
so that Yl is closed; Y2, Y3 and Y4 are finite-dimensional; and
imA=Y1ED Y2i
kerB=Y2®Y4,
Y=Y1®Y2ED Y3ED Y4.
Next, we define
X1 = A-1(Y1),
X2 = A-1(Y2),
Z1 = B(YI),
Z2 = B(Y3).
By restriction, the operators A and B define Banach space isomorphisms 2--
X2 ^'Y2,
Y3-_Z2,
so X1 and Z1 are closed; X2 and Z2 are finite-dimensional; and
X = X1 ®X2,
ker(BA) = X2,
Z = Z1 ®Z2,
im(BA) = Z1.
We see at once that BA is Fredholm, with
index(BA) = dim ker(BA) - dim[Z/ im(BA)] = dim X2 - dim Z2 = dim Y2 - dim Y3 = dim(Y2 ® Y4) - dim(Y3 ® Y4)
= dim ker B - dim(Y/ im A),
Fredholm Operators
35
and, remembering that dim(Z/ im B) = 0 = dim ker A, the formula for the index of BA follows. The next result encapsulates, in the present abstract setting, the main conclusions of Fredholm's original theory of second-kind integral equations.
Theorem 2.22 If A = I + K, where K : X -> X is compact, then A : X -* X is Fredholm and index A = 0. Proof. Theorem 2.19 shows that ker A is finite-dimensional and that im A is closed. Moreover, since Kt X * -+ X* is compact by Theorem 2.17, we may apply Theorem 2.19 to At = I + Kt, and deduce that ker AI (and hence also X/ im A) is finite-dimensional. Thus, A is Fredholm. To compute the index,
let r be as in Theorem 2.19. If r = 0, then ker A = {0} and im A = X, so index A = 0. If r > 1, then since X is the direct sum of ker A' and im A', we have ker A'' a- X/ im A' and hence index A' = 0. Theorem 2.21 gives index A' = r index A, so again index A = 0.
Suppose A E £(X, Y) and B E ,C(Y, X). If BA =I+ Kx where Kx : X -* X is compact, then B is called a left regulariser for A. Likewise, if AB = I + Ky
where Ky : Y -+ Y is compact, then B is called a right regulariser for A. If B is both a left and a right regulariser for A, then we say that B is a two-sided regulariser for A. Notice that by Theorem 2.17, B is a left (respectively, right) regulariser for A if and only if Bt is a right (respectively, left) regulariser for At. The next lemma leads to a characterization of Fredholm operators.
Lemma 2.23 Let A E £(X, Y).
(i) If A has a left regularises, then ker A is finite-dimensional, and im A is closed.
(ii) If A has a right regularises, then im A is closed, and Y/ im A is finitedimensional.
Proof. Suppose that B is a left regulariser for A. Since ker A c ker(BA), and since BA = I + Kx has a finite-dimensional kernel by Theorem 2.19, we see that ker A is finite-dimensional. Now suppose for a contradiction that im A is not closed. By Corollary 2.2, there is a sequence uj E X such that 11U; +kerAllx/kerA = 1
and
IIAujlly -+ 0.
By Exercise 2.4, we can assume that the uj are bounded, and hence obtain a subsequence uj, such that Kxuj, converges. Since Auk' -± 0, it follows that
Abstract Linear Equations
36
BAu y --s 0 and so u!- = BAu j. - Kxu , converges; say uj' u. On the one hand, Au = lim Au/' = 0, so u E kerA, but on the other hand u/' + kerA -* u + ker A, so II u + ker A II x/ ker A = 1, a contradiction. Thus, (i) holds, and then
(ii) follows by duality using Theorem 2.13. Theorem 2.24 For A E 12(X, Y), the following are equivalent:
(i) A is Fredholm. (ii) A has a left regulariser and a right regularises. (iii) A has a two-sided regulariser.
Proof. Since (iii) trivially implies (ii), and since Lemma 2.23 shows that (ii) implies (i), it suffices to prove that (i) implies (iii). Thus, assume A is Fredholm. Since ker A and Y/ im A are finite-dimensional, there is a closed subspace XA c
X and a finite-dimensional subspace YA g Y such that X = XA ® ker A and Y = YA ® im A. Let P denote the projection of X onto ker A parallel to XA, and let Q denote the projection of Y onto YA parallel to im A. The operators P and Q are compact because their images are finite-dimensional. We define a Banach space isomorphism A I : XA -+ im A by A l u = Au for U E XA, and
then define B = A7'(I - Q) E £(Y, X). Since
BAu = Ai'(I-Q)Au = Ai'Au = Ai'Ai(I-P)u = (I-P)u foruEX, and
ABf = AAi' (1 - Q) f = A1AI '(1 - Q) f = (I - Q) f for f E Y, the operator B is a two-sided regulariser for A.
Corollary 2.25 If B is a two-sided regulariserfor A, then index B = - index A.
Proof Since A is a two-sided regulariser for B, it follows that B is Fredholm, and then Theorems 2.21 and 2.22 imply that index B + index A = 0. As an important application of regularisation, we now prove that the index is stable under compact perturbations, and thereby generalise Theorem 2.22.
Theorem 2.26 If A : X --+ Y is Fredholm, and if K : X -* Y is compact, then their sum A + K : X Y is Fredholm, and index(A + K) = index A.
Proof. By Theorem 2.24, A has a two-sided regulariser B E £(Y, X), i.e., BA = I + Lx and AB = I + Ly with Lx : X -3 X and Ly : Y Y compact.
Fredholm Operators
37
Since
B(A+K) = I +(Lx+BK) and (A+K)B = I +(Ly,+KB), and since Lx + BK : X --> X and Ly + KB : Y
Y are compact by Exercise 2.6, we see that B is also a two-sided regulariser for A + K. Hence, by Theorem 2.24 and Corollary 2.25, the operator A + K is Fredholm and has the same index as A. 0 Suppose that X is equipped with a conjugation, i.e., with a continuous, unary
operation u H it satisfying
u+u=it +v, .u=Iii
and
(u)=u,
for all u, v E X and A E C. The third condition implies that the map u H u has a continuous inverse, and we have IIuIIx
Ilullx.
In practice, when X is a function space, u is just the usual pointwise complex conjugate, i.e., u(x) = u(x) for all x. In general, a conjugation on X induces a conjugation on X*, defined by (g, u) = (g, u), and this expression defines a bounded sesquilinear form on X* x X, which we denote by
(g, u) _ (g, u). If X is reflexive, and if we identify X with X** in the usual way, then naturally defined on X x X*, and satisfies
is
(u, g) = (g, u). Now consider a linear map A : X -+ Y. If X and Y are each equipped with a conjugation, then we define the adjoint A* : Y* -+ X * by
(A*u, u) _ (v, Au)
for v E Y* and U EX.
Throughout this and the preceding two sections, we could have worked with and A* in place of and At. For instance, we can state the celebrated Fredholm alternative as follows. Theorem 2.27 Assume that A : X Y is Fredholm with index A = 0. There are two, mutually exclusive possibilities:
Abstract Linear Equations
38
(i) The homogeneous equation Au = 0 has only the trivial solution u = 0. In this case,
(a) for each f E Y, the inhomogeneous equation Au = f has a unique solution u E X;
(b) for each g E X*, the adjoint equation A*v = g has a unique solution v E Y*.
(ii) The homogeneous equation Au = 0 has exactly p linearly independent solutions u1, . . . , u p for some finite p > 1. In this case, (a) the homogeneous adjoint equation A*v = 0 has exactly p linearly independent solutions v1, ... , vp; (b) the inhomogeneous equation Au = f is solvable if and only if the right-hand side f satisfies (vj, f) = O for j = 1, ... , p; (c) the inhomogeneous adjoint equation A*v = g is solvable if and only if the right-hand side g satisfies (g, uj) = 0 for j = 1, ... , p. Proof The result follows at once from Theorem 2.13 and the definition of the index.
Hilbert Spaces Let H be a vector space equipped with an inner product induced norm by
IIullH=
H, and denote the
(u,u)H foruEH;
see Exercise 2.8. Recall that H is said to be a Hilbert space if it is complete with respect to II II H. Remember also our convention that inner products are conjugate-linear in the first argument.
For a general normed space X, given u E X and W C X, we call w E W a best approximation to u from W if dist(u, W) = II u - w II x. Also, recall that the set W is said to be convex if Au + µv E W whenever u, v E W and a., p. E [0, 1 ]
with I + p. = 1. The following property of Hilbert spaces turns out to have important consequences. Lemma 2.28 Suppose that W is a non-empty, closed, convex subset of a Hilbert space H. For each u E H, there exists a unique best approximation to u from W.
Proof. Let d = dist(u, W), and choose a sequence v E W such that 11u vn II H -+ d. By the parallelogram law (Exercise 2.8),
II(um -U)+(U-vn)IIH+II(v,, 2IIvm -ull,+211u-Un1IH,
-U)-(U-un)IIH
Hilbert Spaces
39
and this equation simplifies to give H
IIv,n
=2
uIIH + 2IIu - v,, IIX -4
Since W is convex, we have z (v,,, +
11 12
U II
(v 111
v
+
u II'H.
E W, so H
> dist(u, W) = d,
and hence 11 V.,
- u,IIH < 2(Ilv,,, -up H -d2)+2(lly -ullH -d2).
Therefore, v,1 is a Cauchy sequence, and since W is closed, we deduce that
v -+ w for some w E W, with Ilu - wIIH =
Ilu - v IIH = d, as
required. To show uniqueness, suppose that wI, W2 E W satisfy Ilu - W1 11H = d = IIu - W21111-
Using the parallelogram law as above, with v,,, and v replaced by wl and W2, we find that
I1w1-w2IIH<2(I1w,-u11H-d2)+2(11w2-uIIH-d2)=0, So W1=w2.
O
We now focus on the special case when W is a subspace of H. Write u 1 v if u and v are orthogonal, i.e., if (u, V) H = 0. More generally, for a subset W c H,
write ulWif (u,v)H=0for all vEW. Lemma 2.29 Let W be a subspace of an inner product space H, and let u E H. A vector w E W is a best approximation to u from W if and only if u - w 1 W.
Proof. Suppose that dist(u, W) = ll u - w II H, and let v E W. In showing that (u-w, v) = 0, we can assume II v II H = 1. Since W is a subspace, w + v E W, and so, with d = dist(u, W),
d2 < flu-(w+v)IIH = IIu-wllH-2Re(u-w,v)H+1
=d2-[Re(u-w,v)H]22+[1-Re(u-w,v)H]2
Abstract Linear Equations
40
giving Re(u - w, V)H = 0. Replacing v with iv, we have
0 = Re(u - w, iv)H = Re[i(u - w, v)H] = - Im(u - w, v)H, showing that (u - w, V) H = 0, as required. To prove the converse, suppose that u - w 1 W. For every v E W, we have
w-VEW so(u-W,w-v)H =0, giving Ilu-vl12
= II(u-w)+(w-v)IIH = Ilu-w112H+2Re(u-W,w-V)H+11w-v1lH = flu - wllH + 11w -
V112
> 11U - wpH.
Hence, d2 > flu - wllH, implying that dist(u, W) = IIu - wIIH If W is a closed subspace of a Hilbert space H, then we can define P : H -+ H by Pu = w, where w E W is the best approximation to u from W. With the help of Lemma 2.29, we see that P is linear, and that
P2=P,
IIPIIG(H,H)=1,
imP=W, kerP=W1,
where the closed subspace W1 = {u E H : u complement of W in H. Thus,
I
W} is called the orthogonal
H=W® W. The operator P is called the orthogonal projection of H onto W. Given U E H, the inner product determines a bounded linear functional i 1u E H*, given by
(tiu,v) =(u,V)H forvEH. In fact, from the Cauchy-Schwarz inequality (Exercise 2.8),
I(u,v)HI
IIuIIHIIvIIH,
(2.5)
: H -+ H* is a conjugate-linear isometry. The next result, known as the Riesz representation theorem, shows that t] is onto, and hence invertible. we see that II c i u II H = II u II H, so L l
Theorem 2.30 Let H be a Hilbert space. For each f E H* there exists a unique u E H such that
(f, v) _ (u, v)H for all v E H. Furthermore, 11 .f 11 H = 11 U 11 H
Hilbert Spaces
41
Proof Consider the closed subspace W = ker f . If W = H, then f = 0 and so u = 0. Thus, we suppose that W # H, and choose uo E W1 with uo # 0. Since uo 0 W, we have (f, uo) # 0. Put u l = (f, uo)-'uo so that (f, u l) = 1, and observe that for every v E H,
(f,v-(f,Out) =(f,v)-(f, v)(f,u1}=0. Therefore, U 1 E W -L and v - (f, v) u 1 E W, implying
0=(u1,v-(f,v)u1)H=(u1,v)-(f,v)IIu111H, so u =
11U111 H2
u 1 has the required property. Uniqueness is immediate from
Exercise 2.9.
The Riesz representation theorem shows that we can make H* into a Hilbert space by defining (f, g)H' = (t1 1g, 11 1f)H,
and then obtain a conjugate-linear isometry 12 : H*
H** given by
(12f, g) = (f, g)H The composite map 1211 : H -+ H** is linear and invertible, with
(12t1u,f)=(t1u,f)H==(LI'f,u)H=(f,u). Thus, 1211 coincides with the natural imbedding of H in H**, and we see that every Hilbert space is reflexive. When H is equipped with a conjugation, the norm in the dual space can be written as I (f, u)I 00-EH IIuIIH
11P H' = sup
for f E H* and u E H.
It follows that we can then define a linear isometry 1 : H --* H* by
(tu, v) _ (u, v)H
for u, v E H,
or equivalently, (tu, v) _ (u, V)H. Using t, we can identify H with H*, obcoincides with the sesquilinear serving that in this way the inner product form We conclude this section with one other important fact about Hilbert spaces. (See [106, p. 126] or [94, p. 233] for generalisations to Banach spaces.) Given
Abstract Linear Equations
42
any sequence ul in H, if
(ui,v)H-+ (U, V)H
foreachvEH,
then we say that the uj converge weakly to u in H, and write uj Exercise 2.9 shows a weak limit is unique, when it exists.
u in H.
Theorem 2.31 Let H be a Hilbert space. Every bounded sequence in H has a weakly convergent subsequence.
Proof. Let uj be a bounded sequence in H. To begin with, we assume that H is separable, i.e., we assume that there exists a countable dense subset S = {v], v2, ...}. The Cauchy-Schwarz inequality shows that the sequence
of scalars (up vl)H is bounded, so there is a subsequence ui of uj such that
vI)H converges. Likewise, there is a subsequence u of u such that
(u , V2) H converges. Proceeding in this fashion, we obtain successive subsequences u!, such that v!)H has a finite limit as j -+ oo, for each fixed k and l satisfying 1 < I < k. Hence, the diagonal subsequence u has the property that lim j,c. V)H exists for each v E S. Since S is dense in H, and since the u are bounded, Exercise 2.11 shows that u -- u. Now drop the assumption that H is separable. We let Ho denote the closure of span{u3 }, and let P denote the orthogonal projection of H onto Ho. Since Ho is separable, there exists u E Ho such that (up v)H -+ (u, V)H for each v E Ho.
Hence, if v E H, then (ui, V)H = (ui, PV)H -+ (u, Pv)H = (u, V)H, or in other words uj - u in H.
Coercivity Consider a Hilbert space V and a bounded, sesquilinear form (D : V x V
C.
Thus, '(u, v) is conjugate-linear in u, is linear in v, and satisfies Ic(u, v)I < CIIuIIvIIVIIv
foru, V E V.
We can therefore define a bounded, linear operator A : V -* V* by (Au, v) = (D (u, v)
for u, v E V.
Conversely, each bounded linear operator from V into V* determines a unique
bounded sesquilinear form on V. In this context, we do not identify V with its dual V*, but we do identify V with its second dual V**, so the adjoint
Coercivity
43
A* : V -+ V * is given by
(A*u, v) = (u, Av) = (Av, u)
for u E V and v E V.
We also define V : V x V - C, the adjoint of (D, by 4)*(u, v) = 4)(v, u),
so that c* corresponds to A*, i.e., (A*u, v) = c*(u, v). Given f E V*, one can seek a solution u E V to the equation Au = f, or equivalently, to the variational problem
(h (u, v) = (f, v) for ally E V.
(2.6)
We say that (D and A are positive and bounded below on V if
Re 4(u,u)>CIIully foru E V. Notice that Re *(u, u) = Re b(u, u), so 4 and A are positive and bounded below on V if and only if the same is true of V and A*. In this case, we can apply the celebrated lemma of Lax and Milgram [54]. Lemma 2.32 If the bounded linear operator A : V ed below, then it has a bounded inverse A-1 : V*
V * is positive and boundV.
Proof Since cIIuIIv
Ilully ; CIIAu1Iv,
foru c- V.
Hence, ker A = (0), and Corollary 2.2 shows that im A is closed. Likewise, ker A* = (0), so in fact im A = V* by Theorem 2.13. Combining Theorem 2.26 and Lemma 2.32, we immediately obtain the main result for this section.
Theorem 2.33 If A = AO + K, where AO : V V * is positive and bounded below, and K : V V * is compact, then A : V -+ V * is Fredholm with zero index, and hence the Fredholm alternative holds for the equation Au = f.
If A* = A, or equivalently, if 1 * = 4), then we say that A is self-adjaim, and that 4 is Hermitian. In this case, (Au, u) is real for all u E V, and if
(Au, u) > 0 for all u E V \ (0),
Abstract Linear Equations
44
then we say that A is strictly positive-definite. Such an operator A determines a new inner product and norm on V,
(u, v)A = c(u, v)
and
IIUIIA =
(u, u)A
In many applications, (u, U)A can be interpreted as a measure of the ener of the system whose state is represented by u, so is often called the energy inner product associated with the operator A. The boundedness of A immediately implies that the energy norm is weaker than the norm in V, i.e., II U II A < C II u II v, and it is easy to see that the two norms are equivalent if and
only if A is positive and bounded below on V. Thus, one can view the Riesz representation theorem (Theorem 2.30) as a special case of Lemma 2.32. Now consider two Hilbert spaces, V and H, such that V is a dense subspace
of Hwith IIullH <_ Cllully
forallu E V.
Assume that H is equipped with a conjugation inducing, by restriction, a conjugation on V. In particular, II u II v - II u II v as well as II u II H Il u ll H We identify H with its dual H*, but do not identify V with V*. It will help to keep in mind
the standard example where H = L2(Q), and V is a closed subspace of the Sobolev space H I (2). With the assumptions above, the inclusion map V -+ H is bounded and oneone with dense image. Consequently, the same is true for the transposed map
H = H* -* V*, as one sees from Lemma 2.10 and Exercise 2.5. Using the transposed map to identify H with a dense subspace of V*, we write
VcHcV*, and say that H acts as a pivot space for V. Note that the original meaning of (u, v), as the inner product of the vectors u and v in H, is consistent with its second meaning, as (u, v) for a functional u E V* and a vector v E V, i.e., the two interpretations agree if u E V. We say that 4) and A are coercive on V (with respect to the pivot space H) if
Re 4)(u,u)>cIIuII,-CIIUIIH foruEV.
(2.7)
The next result then follows at once, as a special case of Theorem 2.33.
Theorem 2.34 Let H act as a pivot space for V. If 4) is coercive on V, and if the inclusion map V -+ H is compact, then A : V -+ V * is a Fredholm operator with zero index.
Elementary Spectral Theory
45
Elementary Spectral Theory Suppose that X is a dense subspace of a normed space Y, and that A : X -+ Y is a (possibly unbounded) linear operator. The resolvent set of A consists of those ). E C for which the operator Al - A has a bounded inverse on Y, or, more precisely, for which
1. Al - A : X -a im(,lI - A) is one-one and onto; 2. im(AI - A) is dense in Y;
3. II(AI-A)-'.fIlr
ker(AI-A)={o EX : Aq=X¢} is precisely the eigenspace corresponding to the eigenvalue A. In general, it can happen that some elements of the spectrum are not eigenvalues of A.
Now suppose that H is a pivot space for V, take X = V and Y = V*, and assume that A : V -+ V* is self-adjoint. The eigenvalues of A must be purely real, and any two of its eigenvectors with distinct eigenvalues must be orthogonal, as the following elementary arguments show. If Ao = A0 and
¢ # 0, then 'XIIo112
so A
= (0, ?4) = (0, AO) = (AO, 0) =
()'O,
0) =1110112,
i.e., A is real. Also, if A41 = A 14' and A402 = A24'2, with A l # A2,
then
(A1 -12)(01,02) = (1101, fit) - (4'i, X24'2) = (A4',, 02) - (0i, A4'2) = 0, so (4'1, 02) = 0.
Next, we further restrict our attention to the case when V = H = V *, and consider a compact self-adjoint operator K : H ->. H. The spectral theory for such operators follows in a remarkably simple manner from the next lemma. Here, and in the proof of the next theorem, II II always denotes either the norm
in H or the norm in £(H, H); no other norms occur.
Abstract Linear Equations
46
Lemma 2.35 If K : H -+ H is compact and self-adjoint, then (i) at least one of the numbers II K II and - II K II is an eigenvalue of K, (ii) every eigenvalue of K lies in the closed interval [- IIKII, 11 K 11 ].
Proof We can assume that II K II > 0, because otherwise the result is trivial. Choose a sequence uj in H such that
Ilujll = 1.
and
IIKuill - IIKII
On the one hand, the self-adjointness of K implies that
(K2uj, uj) = (Kuj, Ku,i)
= II Kuj 112,
so
II K2uj
- IIK112u; 112 = IIK2u1
112 -211K11211Ku;112+ 11K114
11KIl2(IFKII`
- IIKu;ll2),
and therefore
K2u;-IIKIl2u;-+
0.
On the other hand, the compactness of K implies that, after passing to a subsequence uj,, there exists u E H such that
K2u;- -+ IIKIl2u. Hence, II K II2uj' -+ II K II2u, so uj, --* u, and we have
K2u = IIKIl2u
and
(lull = 1,
which means that u is an eigenvector of K2 with eigenvalue 11K112. Finally, since
(K + IIKII)v = 0,
where v = (K - IIKII)u,
we see that either v = 0, in which case IIKII is an eigenvalue of K with eigenvector u, or else v rA 0, in which case -11 K II is an eigenvalue of K with eigenvector v. Thus, (i) holds, and (ii) follows at once from Exercise 2.15.
Recall that W denotes the closure of a subset W.
Elementary Spectral Theory
47
Theorem 2.36 If K : H -+ H is a compact, self-adjoint linear operator on a Hilbert space H, then there exist (possibly finite) sequences of vectors 11x1, f2, *3, ... in H, and of real numbers A1, µ2, µ3, ... , having the following properties: (i) Each irk is an eigenvector of K with eigenvalue g j. (ii) The eigenvectors ii, *2, 'Y3, . .. are orthonormal. (iii) The eigenvalues satisfy I A I I > Il.L21 > 1/131 > ... > 0. (iv) If the sequences are infinite, then µj -+ 0 as j -+ oo. (v) For every u E H,
µj (*j, u)*j,
Ku = i>1
with convergence in H when the sum is infinite. (vi) We have the orthogonal direct sum decomposition
H=WE) kerK,
where W = span[*,, f2, ...}.
In particular, if ker K = {0} then W is dense in H.
Proof Assume IIKII > 0, because otherwise the result is trivial. Note also that if A is any non-zero eigenvalue of K, then the corresponding eigenspace ker(AI - K) = ker(I - .X-1 K) is finite-dimensional by Theorem 2.19. Thus, we have orthonormal bases *1, ... , *1i1 and *1+1, ... , *,,,, for the eigenspaces associated with the eigenvalues II K II and -11 K 11, respectively. (It may happen
thatl=Oorm1,butm1 > 1.) Put µ1=...=p.,=IIKII
and
11t+1=...=µ,n, =-IIKII,
so that
K*jµj*j, (*i,*k)=Sjk and Iµ 1=IIKII
forj,kE{1,...,in1}.
Let VI = span[*1, ... , *,,,, }, and observe that if u E Vi then Ku E V1 because
(Ku,*1)=(u,K*i)=Aj(u,*i)=0 forl < j <m1. Hence, the restriction K1 = K) v, is a compact, self-adjoint linear operator on VI L, and we claim that
IIKIII < IIKII.
Abstract Linear Equations
48
Indeed, it is clear that II K1 II < II K II; suppose for a contradiction that 11 K, II = 11 K 11. By Lemma 2.35, K1 has an eigenvalue µ with I A I = II K II, so there exists
a non-zero vector * E Vi with K, i/r = it*. However, Ki/r = K, * = p4,, so E V1, which is impossible because v, fl Vi = {0}. If II K1 II > 0, then we repeat the construction above to obtain *j and p j form, + 1 < j < m2 satisfying
Ki/rj = jt ij
f o r j, k E (1,
and (*j, *k) = sjk
... , m2},
with I µj I = II K, II form, + 1 < j < m2. Putting V2 = span(*,, ... , *"12 11 the restriction K2 = K I v_ is a compact, self-adjoint operator on V2 , and IIK211 < IIKiII.
There are two possibilities: either we can continue indefinitely, constructing an infinite sequence of operators K1, K2, K3, ..., or else the process halts after (say) r steps because II K, II = 0. In either case, it is clear that (i), (ii) and (iii) hold. To prove (iv), suppose that the sequences are infinite, and observe that
IIK'j/u'j - K1k/pk11 = 11 *j - *kll = 'A2- if j
k,
has no convergent subsequence. Since K is so the sequence j= compact, it follows that the sequence (*j/µ
ll*j/µjII - 00. Next, for any u E H, if K1 exists then m,
u = E(i/rj, u)i/rj + yr
with v1 E V1-L,
j=1
where V, = span(*,,...,
as above, and so rn,
Ku = EAj(*j, u)*j + K1v1. j=1
If II Kr II = 0 for some r, then we are done with the proof of (v). Otherwise, IIKlv1II
< IIKIIIIIv,II < IIKIIIIIull = Iµm,Illull,
so K1vt -+ 0 as l -+ oo, and again (v) holds. Finally, consider. (vi). If U E ker K, then A j (*j, u) = (K*j, u) _ (*j, Ku) 0 and so (i/rj, u) = 0 for all j, implying (w, u) = 0 for all w E W, and hence
Elementary Spectral Theory
49
also for all w E W, i.e., u E W1. Conversely, if u E W1, then (ifj, u) =0 for all j, so Ku = 0 by (v), i.e., u E ker K. Thus, ker K = W1.
Theorem 2.36 has its historical origins in the Hilbert-Schmidt theory of integral operators with symmetric kernels. Our next result applies to self-adjoint partial differential operators; see Theorem 4.12. In part (ii), the term "complete
orthonormal system" means that the Oj are orthonormal and span a dense subspace of H; consequently, 00
u=E(Oj,u)¢j foruEH, j=1
with the sum converging in H.
Theorem 2.37 Let H act as a pivot space for V, and assume that H is infinitedimensional and that the inclusion map V -+ H is compact. If the bounded linearoperator A : V -+ V* is self-adjoins and coerci ve, then there exist a sequence of vectors 4 , 02,.03, ... in V and a sequence of real numbers ,tt, A2, X3.... , having the following properties:
(i) For each j > 1, O j is an eigenvector of A with eigenvalue A. (ii) The eigenvectors Oi, 02,403, ...forma complete orthonormal system in H. (iii) The eigenvalues satisfy X I < A2 < A3 < ... with A j -+ oo as j -+ oo. (iv) For each u E V, 00 Au = T,),j(0j, u)oj (convergence in V*).
j=t
Proof Let the constant C be as in the coercivity bound (2.7), so that the bounded, self-adjoint linear operator A + C : V -* V* is positive and bounded below. By Lemma 2.32, the inverse (A + C)-' : V* -+ V exists and is bounded, and the restriction K = (A + C)-' I v is compact from V into V, because the inclusion V C V* is compact. The operator K is also self-adjoint with respect to the energy inner product for A + C, because
(Ku, v)A+c = ((A + C)Ku, v) _ (u, v) _ (u, (A + C)Kv) _ (u, Kv)A+c
for u, V E V.
We apply Theorem 2.36, noting that *j = (A + C) K *j = A j (A + C) * and, since the *j are orthonormal with respect to the energy inner product for A + C,
(trj, *k) _ ((A+C)Kifj, yak) = (AjO'j,*k)A+C =µj6jk.
Abstract Linear Equations
50
In particular, A j = II *j II2 > 0, and we define
4j =
1
µj
*j
and
k=1 - C, Aj
so that (0j, 4k) = Sjk and Aq5j = (A + C)1 5j - Ccj = Xjoj. The eigenfunctions ilrj, and hence also the Oj, span a dense subspace of V because ker K = {0}. Parts (i)-(iii) now follow because V is dense in H, and A j .. 0 as j -* oo. Moreover, if u E V, then 00
00
,/, u = j(1/rj, u)A+Cifj = E(4j, u)Oj,
j=1
(2.8)
j=1
with the sum converging in V, and'since A : V -+ V* is bounded, we obtain the expansion of Au in part (iv). Corollary 2.38 If, in addition to the assumptions of Theorem 2.37, the operator A is strictly positive-definite, i.e.,
(Au, u) > 0 for all u E V \ {0}, then
(i) the eigenvalues are all strictly positive: 0 < a.1 < A2 < A3 < ...; (ii) the operator A is positive and bounded below on V, and the energy norm for A is given by 00
IIullA =
(Au, u) =(AJJU)I2 11
fore E V;
j=1
(iii) the inverse operator A-1 : V* -f V exists and is bounded, self-adjoint and positive and bounded below, with energy norm
(A-'f, f) = 1: X-
1/2
00
11f IIA
=
1I(.pj, f)12
forf EV*.
j=1
Proof. The positive-definiteness of A implies that ,l j = (A¢ j, O j) > 0 for all j. Since zero is not an eigenvalue of A, we have ker A = {0}, and hence by Theorem 2.34 the inverse A-1 : V* -+ V exists and is bounded. By Exercise 2.3, A-1 is also self-adjoint, and it is easy to see that A-1 is strictly positivedefinite. Thus, the two energy norms make sense and satisfy II U 11 A C II U II V
Elementary Spectral Theory and II f II A
51
< C II f II v.. Furthermore, since the sum (2.8) converges in V,
for f E V' and u E V.
(f, u) = 1:00 (f,-0j)(0j, u) j=1
Taking f = Au gives the formulae for (1 U II A and II f 11 A-1, because (f, 0j) _
(Au, g5j) _ (u, A0j) = Aj(u, Oj) and hence (0j, u) = Aj 1(0j, f). Finally, by the Cauchy-Schwarz inequality, oc
I(f,u)1 <
a 1/2(f, 0 1 ) A 2 ( 0 , u)
II f II
C II
CIIfIIA-1Ilully, _<
C II
V
II
II
II
A and
and bounded below. Our final result can be viewed as a special case of the Fredholm alternative.
Corollary 2.39 Let k E C and f E V*, and consider the equation
(A -),)u = f, where the assumptions of Theorem 2.37 hold.
(i) If l 0 (X 1, 12, A3, ... }, then the operator A - A : V --+ V* has a bounded inverse, and the unique solution u E V is given by u=(A-A)-,
f=E(0j,.f4j 00
j=1
Xj - JL
(ii) If .l E (A1, ,X2, A3, ...}, then a necessary and sufficient condition for the existence of a solution u E V is that
(0j, f) = 0 for all j with X j = X. In this case, the general solution is
u- 1: ajoj +E (0j,f)0j, xi=x
Al 741
'Xj -'X
where the a j are arbitrary constants. The infinite sums in parts (i) and (ii) converge in V.
Proof It is instructive to avoid making direct use of Theorem 2.27. Without loss of generality, we can assume that A j > 0 for all j, as in Corollary 2.38.
Abstract Linear Equations
52
Define a linear operator Bx by
Bx f =
(0j) f) x #z
Aj -
and observe that B, : V* -+ V is bounded because, by Corollary 2.38,
),;I (0j, IVBxf ll
Bxf)12 = v - 0,I(O;, A x;#x
J=1
Aj -'X
2
A'
( j -'X ) x;5ex sup
f)12
f) 12 Aj 1I(0j, E x,x
<- C11f112
Since the sums w
(A - A)Bx f = B, (A - A)f =
(0j, f)Oj and f = 1: (0j, f)'Oj x,Ox
,i=1
converge in V*, we see that if A is not an eigenvalue of A, then Bx = (A - )')-1.
Assume now that A is an eigenvalue. If a solution u E V exists, then the right-hand side f satisfies
(cbj, f) = (0;, (A - A)u) _ ((A - A)Oj, u) = (A! - A)(0J,u) = 0
whenever A3 = A.
Conversely, if (0j, f) = 0 whenever Aj = A, then (A - X) B, f = f so Bx f is a solution, and to complete the proof it suffices to show that
ker(A - A) = span( 1 : Aj = A).
(2.9)
In fact, (A - A)u = 0 if and only if (0j, (A - A)u) = 0 for all j, and since (0j, (A - A)u) = ((A - A)41, u) = (Ai - A)(4i,u),
we conclude that u E ker(A - A) if and only if u 1 4, whenever A! # A. Hence, (2.9) holds because of (2.8).
Exercises 2.1
Show that for a normed space X to be complete it is necessary and sufficient that, for every sequence uj in X, if the numerical sum EJ__1 IIuj I1x
Exercises
53
converges in R, then the vector sum E 'I u j converges in the norm of X, i.e., there exists a u E X such that II u - . I u j Ilx --> 0 as m -k oo. 2.2 Let X be a normed space. (i) Show that the vectors u I, ... , un E X are linearly independent if and only if there exist functionals gi, ... , gn E X* such that
(gj, Ilk) = Sjk =
1
if j = k,
0
ifi 0k.
(ii) Show that with uj and g j as in (i), we can define a projection P : X X by If
Pu = >(gj, u)uj
for u E X.
j=1
(iii) Let V = span{u t, ... u } and w = n;=1 ker g j. Show that P is the projection of X onto V, parallel to W. (iv) Suppose that W is a closed subspace of X, and that the cosets u I+ W, ... , un + W form a basis for the quotient space X1 W. We may therefore define functionals g1, . . . , g,, in (XI W)* by (9j, uk+W) = Sj k . Show that any set of representatives {u 1, ... , un } must be linearly independent, and deduce that n
Pu=E(gj,u+W)uj j=1
2.3
is the projection of X onto V = span {u1, ... , parallel to W. Let A E £(X, Y), where X and Y are normed spaces. Show that A has a bounded inverse if and only if Al has a bounded inverse, in which case
(A-I)t = (A')-
2.4
Let W be a closed subspace of a normed space X, and let E E (0, 1). Show
that each coset u+ W E X/W has a representative v E u+ W satisfying (1- E)IIVIIx S H U + W ll x/w :5 llvllx
2.5
Let X be a normed space. (i) Show that a subset W c X is dense if and only if W a = {0}. (ii) Suppose now that V is a subset of the dual space X*. (a) Show that if V is dense, then aV = {0}. (b) Show that if aV = (0), and if X is reflexive, then V is dense.
54 2.6
Abstract Linear Equations Let X, Y and Z be Banach spaces.
(i) Show that if K : X -* Y and L : X -+ Y are compact, then so are K+L:X Y and µK : X -a Y, for any scalar A. (ii) Show that if A : X -* Y is bounded, and if K : Y -- Z is compact, then KA : X -> Z is compact. (iii) Show that if K : X -- Y is compact, and if A : Y -+ Z is bounded, 2.7
then A K : X --), Z is compact. Assume that X and Y are Banach spaces. (i) Show that if (I All c(x,x) < 1, then 1 - A : X inverse, and
II(1- A)-' Ilc(x vi <
X has a bounded
I
1- IIAIIc(x,x)
(ii) Show that if A E .C(X, Y) has a bounded inverse A-' E C(Y, X), and if E E C(X, Y) satisfies II EII c(x,Y) < 1/IIA-' IIc(Y,x), then A + E has a bounded inverse, and II (A + E)-' II c(Y,x) <-
2.8
IIA-' II c(Y,x)
1 - IIA-' Ilc(Y,x)IIEIIc(x,Y)
(iii) Suppose that B E £(Y, X) is a two-sided regulariser for a Fredholm operator A E £(X, Y). Show that if IIEIIC(x.Y) < 1/11 BJJc(yx), then A + E : X -+ Y is Fredholm, and index(A + E) = index A. (iv) Let t H A, be a continuous function from [0, 1] into £(X, Y). Show that if A, is Fredholm for each t E [0, 1], then index A, is constant for t E [0, 1], and in particular index Al = index A0. Let H be an inner product space, and let u, v E H. (i) Prove the Cauchy-Schwarz inequality (2.5). [Hint: first reduce to the case when IIuIIH = IIVIIH = (ii) Deduce the triangle inequality II U + V II H
2.9
II U II H + II V II H
(iii) Prove the parallelogram law: II u + v II H + II u - v II y = 211 u II H + 2IIvIIH Let H be an inner product space and let u E H. Show that if (u, v)H = 0
forallvEH,then u=0. 2.10 Let H be a Hilbert space. (i) Show that if W is a closed subspace of H, and if P is the orthogonal
projection of H onto W, then P is self-adjoint, i.e., (Pu, V)H = (u, Pv)H for all u, V E H. (ii) Show that if P : H -+ H is a self-adjoint projection, and if W = im P, then P is the orthogonal projection onto W.
Exercises
55
2.11 Let u j be a bounded sequence in a Hilbert space H. (i) Prove that if the sequence of scalars (u1, v)H converges for each v in a dense subset of H, then there exists a unique u E H such that u j -k u in H, and moreover II u II H < lim sup Jam. II u J II H. (ii) Prove that u in H if and only if u j - u in H and II U j II H -->
j-u
II u ll
2.12 Let H1 and H2 be Hilbert spaces. (i) Show that if A : H1 H2 is bounded, and if uJ - u in H1, then
Au -k Au in H2. (ii) Show that if K : H1 - H2 is compact, and if uJ - u in H1, then AuJ
(iii) Prove (conversely) that if a bounded linear operator K : H1 -* H2 has the property that uJ -k 0 in H1 implies KuJ - 0 in H2, then K is compact. 2.13 Suppose that V is a real Hilbert space, and that A : V -f V* is a bounded, self-adjoint linear operator. Assume further that A is strictly positivedefinite. Let f E V*, and define the quadratic functional Jf : V -+ IR by
Jf(v) = !(A v, v) - (f, v).
i
(i) Show that if Au = f, then Jf(v) = Jf(u) +BV) 211v - u11A. (ii) Show that Jf(v + Sv) = Jf(v) + (Av - f, + 1113V112 for all v, SVEV. (iii) Deduce that u E V satisfies Au = f if and only if Jf (u) = min
Ev
Jf(v). 2.14 Suppose that b is a bounded sesquilinear form on a Banach space X. Show that if 4) is positive and bounded below, then X is in fact a Hilbert
space. [Hint: consider (u, v) = 1(u, v) + (D(v, u).] 2.15 Let A : X -+ X be a bounded linear operator on a Banach space X. Show that if It E C satisfies lµ l > II A Ilc(x.x), then the operator AI - A : X X has a bounded inverse.
2.16 Let K : H -- H be a compact, self-adjoint linear operator on a Hilbert space H, and let it,, /12.... and *1, *2, ... be the eigenvalues and eigenvectors of K, as given by Theorem 2.36. Suppose also that K is strictly positive-definite:
(Ku,u)>0 foruEH\(0), and define V to be the Hilbert space obtained by completing H in the energy norm /T
Ilully=
(Ku,u)= (/.LJl(*J. u)12 J=1
1 /2
56
Abstract Linear Equations
(i) Show that the inclusion map H -> V is compact with dense image. Deduce that V* C H C V, i.e., H is a pivot space for V*.
(ii) Show that K has a unique extension to a bounded linear operator K : V -* V*, and that this extension has a bounded inverse.
(iii) Obviously, i/,j E V for all j, and µ, > µ2 > ... > 0. Show that
Ku = EAj(*r, u)'/ij
for U E V,
j=1
with the sum converging in V *.
2.17 Let H act as a pivot space for V, and let A : V -+ V* be a self-adjoint, Fredholm operator with index zero. Define Vo = V fl (ker A)1, where I means the orthogonal complement in H. (i) Show that V = Vo ® ker A and V * = im A ® ker A, where the direct sums are orthogonal with respect to (ii) Let A0 = AI V., and show that A0 : Vo -+ Vo im A is invertible.
(iii) Show that if, in addition, A is coercive and (Au, u) > 0 for all
uEVo\(0),then (Au, u) > c II u 11 v
for all u E Vo.
and thus A0 is positive and bounded below on V0.
3
Sobolev Spaces
In the context of variational methods, one naturally seeks a solution to a linear second-order elliptic boundary value problem in a space of functions that are square-integrable and have square-integrable first partial derivatives. Physically,
the functions in such a Sobolev space typically represent the system states for which the total energy is finite. Before commencing our study of elliptic problems, we shall therefore treat Sobolev spaces as a separate topic. Liitzen [61]
discusses the historical developments that led to the modem ideas of weak or distributional derivatives, on which the theory of Sobolev spaces is built; see also Dieudonne [19, pp. 248-252]. The first four sections of this chapter cover relevant parts of the theory of distributions and Fourier transforms. The reader will need to understand this material at a practical level before proceeding any further. Next, we define
the Sobolev space Wp(2) based on Lp(Q), for 1 < p < oo, but soon focus almost exclusively on the case p = 2. The space HS(R"), which coincides with WI (RI) = W2 (R"), is then defined via the Fourier transform in the usual way, and after that we introduce HS (0) and HS (0) for a general open set cZ C ". We go on to develop the standard density, imbedding, duality and trace theorems, mostly assuming that c2 is a Lipschitz domain. The classical reference
for this theory is Lions and Magenes [59]; however, our approach is more along the lines of Chazarain and Piriou [14, Chapter 2]. For results on Sobolev
spaces WP (0) with p # 2, we refer to Adams [2], Bergh and Lofstrom [5] and Grisvard [34]. Only one result in this chapter is not completely standard: a theorem of Costabel [ 14] stating that, for a Lipschitz domain 0 with boundary r, the trace operator is bounded from HS(SZ) to H-1/2(F) for 2 <s < (not 2 just for 2 < s < 1). This fact will be used when we consider the mapping
properties of boundary integral operators.
57
Sobolev Spaces
58
Convolution In this section only, SZ may be any (Lebesgue) measurable subset of R" (n > 1)
with strictly positive measure; throughout the rest of the book, cZ is always assumed to be open. The Banach space L p (St) is defined in the usual way, with
f
Iu(x)Ipdxl
IIIIL,,(n) = ( \ sz
I/p
)
/
for l < p < oo,
and with II u II La, (ca) equal to the essential supremum of u over Q. We define
(u, v)n = I u(x)v(x) dx
J
whenever the integrand belongs to LI (S2), and usually just write (u, v) if 0 W. Let p* denote the conjugate exponent to p, i.e., 1
=
1
P+P**=1,
(3.1)
and recall Holder's inequality: I(u, v)SZI < IIUIIL,.M)IIVIIL,,(sz)
for 1 < P:5 00.
By Exercise 3.1, IIUIIL'.(c) =
sup
I(u, V) S21
for U E Lp.(2) and 1
O#vEL,,(1) IIVIIL,,(fa) (3.2)
so L . ( 7 ) is isometrically imbedded in the dual of L p (0). In fact, Lp.(S2) = [L p(c2)]*
for 1 < p < oo (but not p = oo);
(3.3)
see Yosida [106, p. 115]. If the (measurable) functions u and v are defined on the whole of R", then we define their convolution u * v by
(u * v)(x) =
r
J
u(x - y)v(y) dy
whenever the integrand belongs to L I (W). Convolution turns out to be a very effective tool for approximating non-smooth functions by smooth ones, a tech-
nique that we shall use repeatedly. The substitution y = x - z shows that (u * v)(x) exists if and only if (v * u)(x) exists, in which case they are the same. Thus, convolution is commutative:
v*u=u*V.
Convolution
59
The following theorem gives simple criteria for the existence of the convolution.
Theorem 3.1 Let 1 < p < oo, 1 < q < oo and 1 < r < oo, and suppose that
-+-=1+-. p q r 1
1
1
(3.4)
If u E LP (lR") and v E Lq (R"), then u * v exists almost everywhere in ]R" and belongs to Lr(]R" ), with
RR" IIu * VIIL,.(R") < II1IIL,(R")
(3.5)
Proof. First note that if p = q = r = 1, then J
J
lu(x-Y)v(y)Idydx= f v(Y)I(fR Iu(x-y)Idx)dY
=f
Iv(Y)IIIuIIL,(R")dy = IIuIIL,(R'1)IIVIIL,(R"),
"
so u * v exists almost everywhere and belongs to L i (]R"), and (3.5) holds. Now let u E La (R"), v E Lq (1R") and 0 E L,.. (W), for p, q and r satisfying (3.4), with r > 1. Assume first that u, v and 0 have compact supports, and so belong to L 1 (R). Since 1 l p + l /q + 1/r ; = 2, if we put ilr (x) (-x), then by Exercises 3.2 and 3.3, I(u * v,4)I = I(u * v * c)(O)I <- IIuIILp(R")IIUIILq(R")II1kIIL,*(R") = IIuIILp(R")I IVIILg(R")IIOIIL,*(R")
Now drop the assumption that u, v and 0 have compact support, define
if Ix I < j, 0 if Ix I > j, 1
Xr (x)
and put uj = X1 u, vj = Xjv and ¢j = Xj 4,. Since (I1J I * It; 1, I4) I) -< IIuj IIL,,(R") III; IILq(R")II4); IIL,.(R")
< IIuIILp(R")IIVIILq(R")II4IIL,.(R")
for all j, the monotone convergence theorem implies that I(u * v, 4))I _< (lul * Ivl, I4I) <_ IIuIIL,,(R")IIvIILq(R")II4IILrs(R"), giving the desired bound for 11U * V II L,. (R")
D
Sobolev Spaces
60
Taking r = oo in (3.4), we see that u * v is bounded if u E L p (IR") and v r= L. (IR"). To prove a stronger result, we introduce the p-mean modulus of continuity, 1/p
p
cvp(t,u)=sup(J lu(x+h)-u(x)Ipdxl Ihl
fort>0 and
/
Ilk"
1 < p < oo. It can be shown that, for 1 < p < 00, if u E L p (IR" ), then cop (t , u) as t J, 0.
3.6)
0
Theorem 3.2 Let I < p < 00. If U E L p (IR") and V E L p. (IR"), then u * v is uniformly continuous on IR". If, in addition, p 1, then (u * v)(x) -* 0 as I x l -+ oo.
Proof By Holder's inequality,
[u(x - z) - u(y - z)lv(z) dz
I(u*v)(x)-(u*v)(Y)I = IRIS
< wp(Ix - YI, u)IIvIILP.(R'), so u * v is uniformly continuous on W. Assume now that p Since p and p* are both finite, there exists R > 0 such that
f
u(Y)Pdy < Ep
and
1, and let c > 0.
v(Y)I'dy < E',
J yl>R
yl>R
and by Holder's inequality,
u(x - y)v(y) dy
I(u * v)(x)I
u(x - Y)v(Y)dy
flyl>R
<
1/P
(
Iu(x - y)IpdY)
IIVIIL,,.(tt.-)
Iyl
1/p'
+ IIUIIL,(R") (LI>R v(y)P' d) 1/p
< (fly-xl
IIUIILN(R")E.
Iflxl>2Randly-xl
2R-R= R. Thus, KU * v)(x)I < EIIvllLp.(R") + IIUIILn(R")E
andso(u*v)(x)-*OaslxI
oo.
for Ixl > 2R,
Differentiation
61
Differentiation Let S2 be a non-empty open subset of R". If a function u : Q -* C is sufficiently smooth for them to exist, then we denote the partial derivatives of u by 'a
aau(x)
(ax,
I
an
...
(ax")u(x),
..., a,,) is a multi-index, i.e., an n-tuple of non-negative integers. The order of the partial derivative aau is the number where a = (al,
lad = al + ... + a", and we put u(k)(x;
Y) lal=k
. a,,! and ya = y;'
where a! = a1 ! be written as
(3.7)
a 8,U(X)Y"
yin. In this way, Taylor's formula may
k
1 u1i)(x;
u(x + y) =
I!
y) + 1 k! J0 (1 - t)kulk+i)(x + ty; Y) dt; 1
(3.8)
see Exercise 3.5. For any integer r > 0 we let
Cr(S2) = (u : 8'u exists and is continuous on S2 for jaI < r), and put
C°°(Q) = n cr(ci). r>0
The support of u, denoted by supp u, is the closure in S2 of the set (x E 0 u (x) 0 0}. If K C= SZ, i.e., if K is a compact subset of 0, then we put
CK (0) _ {u E Cr (0) : Supp U C K}
and CK (Q) =
I
CK (S2). I
r>O
Next, we define
Ccomp(S2) _ {u : u E Cr (S2) for some K C 0} and
C mp(0) _ {u : u E CK (0) for some K C= c }. Exercise 3.6 gives some examples of compactly supported C°O functions.
62
Sobolev Spaces
We now show that differentiation commutes with convolution; see also Exercise 3.10.
Theorem 3.3 Let r > 0 and 1 < p < oo. If U E Ccomp (R") and V E Lp (R), then u * v E Cr (R") and
aa(u*v)=(a1u)*v for lot I < r.
Proof The case r = 0 follows from Theorem 3.2. Suppose r = 1, and consider the difference quotient with respect to the Ith variable,
A, hu(x) =
u(x + he,) - u(x)
for l < 1 < n and h
'h
where e,, ... , e" are the standard basis vectors for R". It is easy to see that 01,h (u * v) = (A1,,, u) * v, so for every x E
IR
11
I[o1,h(u * VA(x) - [(81u) * ul(x)I = 1[(A1,hu - a1u) * vl(x)I III/Au - a,uIIL,,.(R")II
IIL,,(R,').
Since u is C1 and has compact support, we have Dj,i,u -> aju in Lp.(R"), and thus 81 (u * v) = (a1u) * v. A simple inductive argument proves the result for the general case r > 1.
The connection between convolution and approximation by smooth functions, mentioned earlier, comes about via the following construction. Let * E Cop (R") satisfy
*>0onIlk",
*(x)=0forIx1> 1,
Jf *(x)dx=1;
(3.9)
such a function exists by Exercise 3.6. We define
kE(x)_e' r(E-lx)
forxER" and E>0,
(3.10)
so that, as one sees via the substitution x = Ez, `YE> 0 on R",
*E(x)=0 for Ixl>E,
r
J
*,(x) dx = 1.
These properties of 'VE mean that (`YE * u)(x) is a kind of local average of u around x, and for this reason 'YE * u ti u when c is small.
Differentiation
63
Theorem 3.4 Let 1 < p < oo. If `/. E is as above, and if u E L p (R" ), then
and II'/'E * U - uII L,(R") < co,,(, Eu),
11 *c * uII Lp(1R") < IIUIIL,,(R")
u in Lp(R") as e 4. 0.
so VE * u
Proof. Since 11 *E
IIL,(R") =
1, we see from Theorem 3.1 that
IIi *UIILn(R") < II1IreIIL,(R")IIUIILp(R") = IIUIIL,,(R")
Also,
'- * u(x) - u(x) = u * 'VE(x) - u(x)
0,., 0,., (Y) dY R"
and so, for any 0 E L p. (W'), Holder's inequality implies that
I(*E*u-u,0)I
=I fYISe
R*E(Y)
f [u(x - y) - u(x)]O(x) dx dy
I
J
'Y E(Y)Wp(E,
u)IIOIILr,+(R^) dy = Wp(E,
Y1 <E
and the desired estimate for ikE * u - u follows.
Corollary 3.5 If I < p < oo, then C,,,,,p(S2) is dense in Lp(SZ).
Proof. We choose any K j C S2 such that KI S; K2 e ... and 0 = U1 K); for example,
Kj = {x E S2 : dist(x, R" \ SZ) > 1/j and Ixj < j).
Let U E L(), define U j E L(R) by u (x) =
u(x) if x E Kj,
ifx E1R' \Kj,
0
and consider the function uj,E =
YkE * u j.
Theorem 3.3 shows that U j., E
C a ,p(R"), Exercise 3.4 shows that supp u j,E C Q
if E < dist(K1, R" \ 0),
and by Theorem 3.4 Iluj.E - UIILE,(c2) S IIUE.j - uj IIL,(R1') + IIuJ - uII L,,(s )
< Wp(e, U j) + IIuIIL,,(Q\KJ)
64
Sobolev Spaces
Since II U II L,,(s,\K) --). 0 as j -+ oo, and since cop (E, uj) -* 0 as c -* 0 for each fixed j, the result follows. We can also use convolution to smooth out the characteristic function of a set.
Theorem 3.6 Let F be a closed subset of R". For each E > 0, there exists XE E C°°(R') satisfying
0 < X, ,(x) < 1 and I aaXE (x) I < CE-IaI
if x E F, if 0 < dist(x, F) < E,
XE (x) = 0
if dist(x, F) > E.
XE (x) = 1
Proof Define vE E L°O(]R") by J1
VE(x)
0
if dist(x, F) < E, if dist(x, F) > E,
so that vE = I on a neighbourhood of F, and choose 1/r E C mp(R) satisfying (3.9). With the help of Exercise 3.4, one sees that the function XE _ *E/a * VE/2 has the required properties.
Schwartz Distributions
A (measurable) function u : 0 -* C is said to be locally integrable if u is absolutely integrable on every compact subset of Q. We denote the set of all such functions by L1,1oc(52). The following observation is the starting point for the theory of distributions.
Theorem 3.7 If u, v E L1,10 (0) satisfy
1 u(x)q(x) dx = J v(x)o(x) dx for all O E C mp(S2), then u = v almost everywhere on Q. Proof. Let K C= 52, and choose an open set 521 such that K C= S21 c S21 Cc Q.
We define f E L1(R") by
f(x)=
J
u(x) - v(x) if x E 521, 0
ifxER"\521,
'YE be as in Theorem 3.4. There is an co > 0 such that if x E K and 0 < E < co then (x - -) E C mp (521) and therefore (Vi, * f) (x) = (f, *E(x - .))tt,, = (u - v, 'YE(x - .))a = 0. Since '/FE * f -* f in L1(R"),
and let
Yk E
Schwartz Distributions
65
it follows that f = 0 almost everywhere on K, i.e., u = v almost everywhere on K. 0 Theorem 3.7 shows that a locally integrable function u is uniquely determined by its associated linear functional 4) H (u, 0) a. We wish to introduce a larger
class of linear functionals on C,'O.P (0). Following the notation introduced by Schwartz [92], put
E(Q) =
C°O(S2),
DK(S2) = CK (S2),
V(S2) = C mP(S2),
for any K C= Q. Since we want our functionals to be continuous in an appropriate sense, we now define convergence of sequences in each of these function spaces. No deeper properties of the underlying locally convex topologies will be used; cf. [106]. Let (4j) J _ , be a sequence in E (S2). We write
cj -+ 0 in S(Q) if, for each compact set K and for each multi-index a,
8'Oj -* 0 uniformly on K. When, for a fixed K, this condition holds for all a, and in addition supp c! c K for all j, we write
cj - 0 in DK (S2). 0 in D(7) means that 4j -+ 0 in DK(S2) for some K C= 0. Convergence to a non-zero function is then defined in the obvious way. For instance, cj --* ¢ in DK (0) means ¢ E DK (0), 4j E DK (0) for all j, and ¢j - 4) -3 0
Finally, 4j
in DK (S2).
Consider an abstract linear functional f : D(52) - C. For the moment, we write e(0) to denote the value off at ¢ E D(S2). If f is sequentially continuous, i.e., if for every sequence ¢j in D(S2), c1 -* 0 in D(S2)
implies t(4 f) -+ 0,
then t is called a (Schwartz) distribution on Q. In this context, the elements of D(Q) are referred to as test functions on 0. The set of all distributions on S2 is denoted by D*(0). Associated with each u E L 1,1 (0) is the linear functional to defined by (Lu)(4)) = (u, 4))c
for 0 E D(S2).
(3.11)
66
Sobolev Spaces
It is clear that to : D(7) -f C is sequentially continuous, and hence a distribution on 92. Furthermore, Theorem 3.7 shows that the linear map i : L 1,1"C (0) --+
D*(S2) is one-one. Hence, we may identify u with tu, and thereby make L1,1oc(92) into a subspace of D*(92). Those distributions that are not locally integrable can then be viewed as generalised functions.
As an example, fix x E R", and let E E D*(92) be the associated point evaluation functional defined by
f(0) = 0(x) for 0 E D(R"). It is not possible to represent this e by a locally integrable function. To see why,
suppose for a contradiction that £ = to for some u E L1.1oc(R"). It follows from Theorem 3.7 that u = 0 almost everywhere on R" \ (x}, and since the set (x) has measure zero, this means that u = 0 almost everywhere on R". Hence, ¢(x) = (u, 4)) = 0 for all 0 E D(R"), a contradiction. Following the convention introduced by Dirac, we denote the point evaluation functional for x by S., or in the case x = 0, just by S. In keeping with the philosophy that distributions are generalised functions, we henceforth write (u, 0) n for the value of u E 'D* (0) at 0 E D(92), whether or not u is locally integrable on 0. If 92 = R", then we usually omit the subscript, and just write (u, 0); for instance,
(S. ,0) =0(x) for0 E D(R"). Suppose that 921 is an open subset of 0, and for any ¢ E D(921) let E D(92) denote the extension of ¢ by zero. For any distribution u E D*(92) the restriction u I n, E D* (921) is defined by
(uIn,, 0)si, = (u, )st for 0 E D(9Z1). We say that u = 0 on S21 if u I a, = 0, and define supp u to be the largest relatively closed subset of Q such that u = 0 on n \ supp U. If U E L1,1,, (92), then the distributional support of u is the same as its essential support as a function, i.e., supp u is the largest closed set such that u = 0 almost everywhere on S2 \ supp u. We define E*(92) in the obvious way, i.e., a linear functional u : E(Q) -+ C
belongs to E* (Q) if it is sequentially continuous: (u, 4 )n -+ 0 whenever ¢J -} 0 in E(92). Exercise 3.8 shows that the inclusion D(92) g E(S2) is continuous and dense, so E*(92) S; D*(92). In fact, we can characterise E*(92) as follows.
Schwartz Distributions
67
Theorem 3.8 The space E* (S2) coincides with the space of distributions having compact support, i.e., E* (Q) = [U E D* (0) : supp u C= S2).
Proof Suppose that u E D*(S2) and supp u C= S2. By Theorem 3.6, there is
a X E D(l) with X = 1 on a neighbourhood of supp u. We define a linear functional ii on E(52) by putting for ¢ E E(SZ),
and claim that
(i) O; -+ 0 in E(S2) implies (u, c;)Q --). (u, O)g; (ii) (il, ¢)o = (u, O)n for ¢ E D(S2). Together, these facts mean that u = u E E* (S2) when E* (S2) is viewed as a subspace of D*(S2). We remark that u does not depend on the choice of X. To prove (i), assume Oj --+ 0 in E(0). It follows at once that XOj XO
in D(cz), so (u, oi)n = (u, XOi) - (u, xO)si = (u, 0)cz. To prove (ii), assume ¢ E D(S2). Since (1 - X)0 = 0 on a neighbourhood of supp u, we have (u, (I - X)0)a = 0 and therefore (u, 0) _ (u, 0) 52 (u, (1 - x)O)si _ (u, A2. Conversely, let u E E*(S2), and suppose for a contradiction that supp u is not compact. Choose an increasing sequence of compact sets KI c K2 c .. . with S2 = U,° , K j, then u I a\KJ # 0 for all j. Thus, we can find 4 E V (S2) such that (u, cj)n = 1 and supp4J Cc n \ K3. It follows that t -+ 0 in E(S2), 0, a contradiction. so (u, /j)n
The restriction map u r-+ u In, is just one of many linear operations that can be extended from functions to distributions. For instance, to define partial differentiation of distributions one formally integrates by parts:
(ac'u, cn = (-1)I0" (u, aacc
for u E D*(S2) and 0 E D(0).
Here, the sequential continuity of 8au follows at once from the fact that if 0 in D(S2), then 8a0j ->- 0 in D(S2). Also, we define the complex ¢j conjugate u E D*(S2) of U E D*(S2) by
(u, 95) = (u, 0)
for 0 E D(S2),
Sobolev Spaces
68
and we generalise the meaning of the inner product in L2(0),
(u, On =
Jsi
u(x)v(x)dx,
by putting
(u, 4)n = (u, ¢)n
for u E D*(0) and 0 E D(S2).
When S2 = l(8", we just write (u, 0). The convolution of a distribution with a test function is defined in the obvious way,
(u * 0)(x) = (u, i0(x - ))
forx E R1, U E
D*(R")
and 0 E D(R"). (3.12)
Thus, in particular, S * 0 _ 0, or formally fyt. 8(x - y)o (y) dy = 0(x), so we can think of 8(x - y) as the continuous analogue of Kronecker's Stk. Finally, multiplication of a distribution u E D*(S2) by a smooth function * E COO(Q) is defined by
(ifu, q )n = (u, *0)j2
for0 E D(S2).
In each of the above examples, Theorem 3.7 guarantees that the generalised concept is consistent with the classical one. For instance, if the classical partial derivative aj u exists and is locally integrable on 0, then 8i (cu) = c (a; u), where i is the imbedding defined in (3.11). Distributions are a powerful conceptual tool for the study of partial differential equations, and provide, in particular, a very effective system of notation. Fortunately, we shall require few technical results from the theory of distributions. However, the following fact will be used.
Theorem 3.9 Suppose that u E D* (0) and X E Q. If supp u C (x }, then for some m > 0,
u = E ac a"3,, on 0, 1" I
where the coefficients are given by a" = (-1)I"1 (u, (. - x)') /a!. Proof Denote the open ball with centre x and radius c > 0 by
BE(x)=(YER" : IY - xI < E}, and choose co > 0 such that the closed ball K = B,,,(x) is a subset of 0. By
Fourier Transforms
69
Exercise 3.9, there exists an integer m > 0 such that I (u, ¢) I < C 1] sup 18"4) I
for 0 E DK (S2).
(3.13)
laI
Given an arbitrary 4) E E(S2), we write the Taylor expansion of 0 about x as 0 _ 01 + 02, where m
41(Y) _ J=l
li 0(j) (X; Y - x) _ j!
la15,n
a
a"4)x)(Y - x)a
and
f(l - t)m(x + t(y - x); Y - x) dt.
02(Y) = m!
Since (u, 0) = (u, 01) + (u, 02) and (u, 01) _ I"I 5,n
e"4)(x)( -x)a) 1(u, e d
aa(8*3
(-1)10'Iaa8"46(x) la l <m
,
laI<m
it suffices to show that (u, 02) = 0. By Theorem 3.6, we can choose a function X E C°°(R") such that x (y) = 1
for IYI < 2, and x(y) = O for iyl > I. Define XE(Y) = X(E-,(y -X)) so that XE = 1 on B,/2 (x), and xE = 0 outside BE(x). Hence, (1 0 on a neighbourhood of suppu = {x}, so (u, (1 - XE)42) = 0 and therefore Since laaXEI < CE-I"l and (u,102) = (u, [x + (1 - XE)14)2) = (u, 18014611 <
CE"'+1-lal
on B, (x), the estimate (3.13) implies that
I (u, 02)1 < C E sup Ia"(xE4)2)I < C E lal5nn K
lal<m
Elal-m
sup
18"4)2(Y)i < CE
lY-1:5's
for 0 < e < Eo. Thus, (u, 02) = 0, as required.
0
Fourier Transforms To motivate the definition of the Fourier transform, we begin with some heuristic
remarks on multiple Fourier series. Given L > 0, we say that a function u 1[8" -> C is L -periodic if
u(x + kL) = u(x)
for x E R" and k E V,
or in other words, if u is L-periodic in each of its n variables. We can think of
70
Sobolev Spaces
such a function as being defined on the additive quotient group TL = R" / (L7G" ),
and introduce the L2 inner product
(u, v)T =
u(x)v(x) dx,
J
where integration over TL just means integration over any translate of the cube (0, L)". Using separation of variables, we can easily find the normalised L-periodic eigenfunctions of the Laplacian (1.1): for k E Z", the function 'bk(x) = satisfies
-Aok = (27rIkI)2k
on TL,
and II0kIIL2(rL) = 1. Since -0 is self-adjoint, we expect from Theorem 2.37 that the Ok will form a complete orthonormal system in L2(TL). Thus, for a general L-periodic function u, we should have
u(x) = 1: (Ok, u)r bk(x) =
1L"
uL(kIL)e'22r(k/c).X,
(3.14)
kEZ
kEZ"
where
UL( ) =
JaI
e-ibrg-xu(x)dx.
The analogous expansion for a non-periodic function can be viewed as arising in the limit as L -+ 00. If U E L 1(1R' ), then we define its Fourier transform u =
2u by
u(t;) _ x {u(x)) =
fp
dx fort E
!lam"
and expect from (3.14) that, under appropriate conditions, u = .F*u, where.F* is the adjoint of the integral operator T, i.e.,
u(x) _
-,x{u(t)} = f
(3.15)
for x E W. In fact, we can readily prove the following.
Theorem 3.10 If both u and u = 'Fu belong to Li(R"), then the Fourier inversion formula (3.15) is valid at every point x where u is continuous.
Fourier Transforms
71
Proof. Let *(x) = e-"1"12 and `YE (x) = E-"* (e -'x). Exercise 3.11 shows that 1/r is invariant under the Fourier transform: F* = 1/r = -7-1*. Therefore, by and .F*1i E _ 'YE In other words, the inversion Exercise 3.12, formula is valid for *E, implying that
j
J
d
d =
fR. Cfa
R11
l.
d dY
= fit" u(Y) J
f,u(Y)*E(x - Y)dy 1 as c y 0 for each 1f Since 7' dominated convergence theorem that
fRna(')ei2'
E IR", we deduce from the
d = lim(1/*E * u)(x).
(3.16)
E10
Assuming u is continuous at x, let co > 0 and choose So > 0 such that Iu(x - y) - u(x)I < co for IYI < So. Observe that fR YE(Y) dy = E(0) = 1, so for 0 < E < co,
f
[u (x - y) - u(x)]*E(Y)dy
Iu(x - y) - u(x)IfE(Y)dy lyl<&o
+
f
Iu(x - y) - u(x)I*E(y) dY
yl>-So
< Eo f *e (Y) dY
+
(f Iu(x - y) - u(x)I dy) \ at^
sup *E(Y) lyl?ao
"(60/E)'
Eo +2IIuIIL,(R^)E-"e
and the result follows at once from (3.16).
0
Corollary 3.11 If both u and u are continuous everywhere and belong to L 1(IR!1), then.F*.Fu = u = .F.F* u.
72
Sobolev Spaces
We now consider the actions of .7 and 1* on the Schwartz space of rapidly decreasing, C°° functions, 8(W1)
E C°°(R") : sup Ix"afl¢(x)I < oo XER"
for all multi-indices a and
}.
Sequential convergence in this space is defined by interpreting the statement
¢j -+ 0 in S(R't) to mean that, for all multi-indices a and 0,
x"0O(x) -+0 uniformly forx E R". Elementary calculations show that if 0 E S(R"), then and
F:,,g[(-i27rx)"O(x)} (3.17)
so the Fourier transform defines a (sequentially) continuous linear operator
.F : S(R") -* S(R"). Moreover, by Corollary 3.11, this operator has a continuous inverse, namely
the adjoint.1'' : S(R") -* S(R"). By Exercise 3.8, the inclusions D(R") C S(RI) C E(R") are continuous with dense image, so we have
E*(1(g") c S*(R") c D*(
),
and the elements of S*(R"), i.e., the continuous linear functionals on S(IR'1), are called temperate distributions. A sufficient condition for a function u E LI,,,,(R") to be a temperate distribution is that it is slowly growing: u(x) _ O (Ix I'') as Ix I -+ oo, for some r. The formulae
(J u, 0) = (u, TO) and
(.F*u, ') = (u, F 4 )
are obviously valid if both u and 0 belong to S(R"), and serve to define extensions
S* (R") -). S*(R")
and
J'" : S* (R") -> S*(k8")
We also have the following result, known as Plancherel's theorem.
Sobolev Spaces - First Definition
73
Theorem 3.12 The Fourier transform and its adjoint determine bounded linear operators
.F: L2(R") -a L2(R") and F* : L2(R") -+ L2(
n
with .F-' = .F". Furthermore, these operators are unitary: (.Fu, .Fv) _ (u, v) = (.F*u, .F*v) for u, v E L2(Il8").
Proof. If u, v E S(R"), then (.Fu,.Fv) = (.F*.Fu, v), and .F*.Fu = u by Corollary 3.11, because u E S(W) c LI(W), so (.Fu, Fv) = (u, v). In particular, taking v = u, we see that II.FuIIL2(Rn) = IIuIIL2(RI) for U E S(Rn). Corollary 3.5 implies that S(R") is dense in L2 (R"), so F has a unique extension from S(R") to a unitary operator on L2(Illi"). Furthermore, this extension satisfies (.Fu, 0) _ (u,.Fo) for all ¢ E S(lR"), consistent with the definition of Fu as a temperate distribution. In other words, the extension from S(R") agrees with the restriction from S*(If8"). Similar arguments yield the same reO
sults for .F*.
Corollary 3.13 The Fourier transform preserves the L2-norm: III fIL,(R'") _ IIuIIL2(1R")
a
Another important fact about the Fourier transform is its effect on convolutions: if u, v E L (Ilk") then
V) W) =
J
J
u(x - y)v(Y)dydx
f f e-1br(x-y)-4u(x - y)
aw
J
dy
= h()v() Sobolev Spaces - First Definition Suppose I < p < oo, and let 0 be a non-empty open subset of R". The Sobolev space WP' (n) of order r based on Ln(S2) is defined by
Wp(E2) = {u E Lp(Q) : 8Yu E L,(c2) for Iaj < r}. Here, of course, 8au is viewed as a distribution on n, so the condition 8"u E L p (Q) means that there exists a function g,, E L p (n) such that (u, 8"O) s- = (-1)I"I (ga, O)n for all 0 E D(S2), or equivalently 8'u = tga where t : L1j,(S2)
Sobolev Spaces
74
-+ D*(0) is the imbedding defined by (3.11). Such a function ga is often described as a weak partial derivative of u. The completeness of L p (S2) implies that WP (0) becomes a Banach space on putting
Ilullw;,(12) _
(tr1PdX)
1/p
To define Sobolev spaces of fractional order, we denote the Slobodeckil seminorm by
lu(x) - u(y)Ip lulu,p>sa =
CJsi J
oo
I x- y I"+P"L
dx dy
)1/p
for 0 < µ < 1.
(3.18)
Notice that the integrand is the pth power of lu(x) - u(y)I/lx - yliL+"/p, so for p = oo we get the usual Holder seminorm. For s = r + µ, we define
Wp(l) = {u E W,(2): Iaaulµ.p.n < oo for lal = r}, and equip this space with the norm
Ilullw;(n) =
E Iaaulµ.P", Ia1=r
For any integer r > 1, the negative-order space WT (S2) is defined to be the space of distributions u E D* (Q) that admit a representation
u = E as fa
with f, E L p(S2).
(3.19)
lal
This space is equipped with the norm \ i/p
Ilullw,,r(n) = inf G IlfallLP(sz)) IaI
where the infimum is taken over all representations of the form (3.19). Using Holder's inequality, it is easy to verify that I(u, u)QI < Ilullw;.(cz)llVIIw"(cI)
where p* is given by (3.1).
foru E WW-.r(0) and V E D(0),
Sobolev Spaces - Second Definition
75
In this book, we will rarely use Sobolev spaces with p 2, and so adopt the abbreviation W1 (Q) = W2 (Q). For any integer r > 0, the norm in W"(n) arises from the inner product (u, V) W, (Q) _
J
a" u (x)a' v(x) dx,
and likewise if s = r + p then the norm in W' (0) arises from the inner product (u, v) W, (a)
f fn
_ (u, v)W,
IaI=r si
[aau(x) - aau(y)l[aav(x) - a,v(y)] dx dy. yI11+2u
Ix -
Thus, W' (0) is a Hilbert space for all real s > 0.
Sobolev Spaces - Second Definition In this section, we introduce a second family of Sobolev spaces, which later will turn out to be equivalent to the one given in the preceding section. For S E R, we define a continuous linear operator ,75 : S(RI) S(lR"), called the Bessel potential of order s, by
d for x E R.
J5u(x) = fR11 (1 +
In this way,
x-. (,75u(x)} _ (1 + I I2)s"2u( ),
(3.20)
so under Fourier transformation the action of ?S is to multiply u by a function that is D(I IS) for large . We can therefore think of ?SEas a kind of differential
operator of orders; cf. (3.17). Notice that for all s, t
Js+' = JS Jt ,
R,
= J--, Jo = identity operator.
It follows from (3.20) that
(Yu, v) = (u, ,75v)
and
(.75u, v) _ (u, .7sv)
for all u, v E S (R"), giving a natural extension of the Bessel potential to a linear operator ?5 : S* (W) -+ S* (WI) on the space of temperate distributions.
Sobolev Spaces
76
For any s E R, we define HS(R"), the Sobolev space of order s on R", by Hs(R??) = {u E S*(JRf) : JSu E L2(R")),
and equip this space with the inner product (u, v)H=(R") = (JSu, 3Sv)
and the induced norm (3.21)
(u, u)H3(R") =
Notice that the Bessel potential
,7S : HS(R") - L2(R") is a unitary isomorphism, and in particular, since J°u = u,
H°(R") =L2(R"). Several facts about HS(R") follow immediately from standard properties of L2(R ). For instance, HS(R") is a separable Hilbert space, and D(R") is
dense in HS(R") because ,75[S(R')] = S(R") is dense in L2(R"), and the inclusion D(R") c S(W) is continuous with dense image. Also, one sees from (3.2) and (3.3), with p = 2, that H-S(R") is an isometric realisation of the dual space of HS (RII ), i.e.,
H-S(R") = [HS(R")]*
fors E R,
(3.22)
and sup
I(u, v)1
OOVEH3(R") IIuIIH=(R")
=
sup
1(u, v)I
o#vEH=(R") IIVIIHI(R")
for U E H-'(RI). Plancherel's theorem (Theorem 3.12) and (3.20) imply that IIUIIH=(R,t) =
j(1 + 12u2d, "
so if s < t then IIuIIH=(R") < IIuIlH,(R") and hence H'(R") c HS(R"). This inclusion is continuous with dense image. For any closed set F C R", we define the associated Sobolev space of order s by
HF.=(uEHS(R"):suppucF),
Sobolev Spaces - Second Definition
77
whereas for any non-empty open set 7 c Rn we define
HS(S2) _ {u E D*(S2) : u = UIn for some U E HI (R")). We see at once that H. is a closed subspace of HS (RI), and is therefore a Hilbert
space when equipped with the restriction of the inner product of H''(R"). A Hilbert structure for HS (0) is defined with the help of the orthogonal projection
P=P,,n: HS(R")-f which satisfies
PUI n = 0 and (I - P)UI n = UI n
for all U E HS(R").
Noting that if U I n = 0 then P U = U because U E HI \n, we see that a well-defined inner product on H' (S2) arises by putting
(u, u)HI(n) = ((1 - P)U, (I - P)V)HX(R,,)
if u = U112 and v = VI
for U, V E HS (R" ). Notice that the induced norm satisfies IIuIIH=(n) =
(u, u)H°(n)
= vin
uEHt(R»)
IIUIIHS(R,I),
(3.23)
because if U In = u then IIUIIHs(R") = IIPUIIHa(R") + II(1- P)UIIH
(R
II(I - P)UIIHt(R") =
The map U r-* (1 - P)Ul n is a unitary isomorphism from the orthogonal complement of onto HS(c2). Therefore, H4(S2) is a separable Hilbert space. Also, (3.23) shows that the restriction operator U H UIn is continuous from HS (R") to HS (0), and thus the space
D(7)=(u:u=Ulnforsome UED
11)
I
is dense in HI (Q) because D(R") is dense in H' (R'). We also define two other Sobolev spaces on 0, HS(S2) = closure of V(S2) in HS(R"), Ho (S2) = closure of D(S2) in HS(S2),
which we make into Hilbert spaces in the obvious way, by restriction of the
78
Sobolev Spaces
inner products in HS (R") and in HS (0), respectively. It is clear that HS (S2) c
H
and
H" (S2) c Ho (S2),
and later we shall establish the reverse inclusions subject to conditions on 92 and s. Note that an element of H is a distribution on R", but, provided the n-dimensional Lebesgue measure of the boundary of n is zero, the restriction operator u H uIn defines an imbedding
H&cL2(0) fors?0.
(3.24)
(If U E H and uIa = 0, then suppu c a2 = SZ \ S2, implying u = 0 on R".) In general, if s < -Z, then Hazy {0} no matter how smooth the boundary of 0, so we cannot imbed H. in a space of distributions on S2; see Lemma 3.39. The necessity of introducing more than one kind of Sobolev space on S2 can be seen already from the next theorem, which extends our earlier observation
that H-'(R") is an isometric realisation of the dual space of H`(R"); see also Theorem 3.30. Theorem 3.14 Let S2 be a non-empty open subset of R", and let s E R.
(i) If fl` (Q) = HI'r, and if we define (u, v) a = (u, V) for U E H-1(S2) and v = V IQ with V E HI (R"), then H''(0) is an isometric realisation of HS (0)*. (ii) If HS (S2) = Ham, and if we define
(u, v)n = (U, v) foru = UIn with U E H-'(R"), and V E H5(S2), then H-5(n) is an isometric realisation of H5(S2)*.
Proof First note that
(u, V) = 0 if u E D(Q), V E HS(R") and VIsi = O, so (u, v)c is well defined for u E H-s(S2) and V E H'(S2). We claim that (lu)(v) _ (u, v)Q defines an isometric isomorphism t : H-S(S2) -* HS(S2)*. Indeed,
I(lu)(v)I = I(u, V)I
IIUIIH-S(R,1)IIVIIH-(R")
whenever v = VIQ,
so I(lu)(v)I < IIUIIH-r(n)IIVIIH'(a) and hence IItuIIH-(Q)» < IIuIIH-=(O). Fur-
thermore, given a functional f E H3(S2)*, the map V H f(V In) is bounded
Equivalence of the Norms
79
on HS (R") because the restriction map V i-3 V I n is bounded from H3 (RI)
onto H'(Q). We know already that H-3(R!) is an isometric realisation of [HS (R")]*, so there exists u E
H-S (R") satisfying
f(Vlc) = (u, V) for all V E HS(R"), and
It(Vlra)I o#VEH'(R")
II V II Hs(a^)
If V E D(R \ n), then V I = 0, so (u, V) = 0, showing that suppu c S2, i.e., u E His. Assume now that H-S(S2) = H 3. In this case, u E and we see from
f(v) = t(Vlg) = (u, V) = (tu)(v)
for v = Vlo and V E HS(R'l)
that f = tu, sot is onto. Finally,
It(Vln)I < implying that IIuIIH-,(n) = IIulIH-'(a°) <_ IItuIIHs(c) Thus, t is an isometry. Part (ii) then follows because every Hilbert space is reflexive.
Equivalence of the Norms
When 0 = R" and s > 0, a simple argument based on Plancherel's theorem shows the equivalence of the two kinds of Sobolev norms; see also Exercise 3.13. We use the abbreviation
(LI I Ixx)
Ylu(Y)I2
n+2µ
dxdy /!1/2
for the Slobodeckii seminorm (3.18) with p = 2, and write I u 1, , = the usual way.
Lemma 3.15 If 0 < .t < 1, -then Iulµ
=
a".,f
where
apt, = Ijwl=1 lei2t - 1 >0
l2
dco dt.
I u I1,,.
in
80
Sobolev Spaces
Proof. Define the forward difference operator Sh by
Shu(x) = u(x + h) - u(x), and consider the Fourier transform
I)i ( ).
.FShu( ) _
(3.25)
By making the substitution h = y - x, applying Plancherel's theorem and then reversing the order of integration, one finds that
f
2=
IIShUIIL_(Rn)
l
Ihl'-u+n
FC"
f
f
dh = R
IhI2µ+n
We transform the inner integral to polar coordinates, letting h = pw, where p = I hl and co = h/IhI. Since dh = pi-1 dp dw,
f
112
'H
Ihl2u+n
A=f
p-2µ-1 >o
lei2rrpl;."
- 112 dcodp = au l
I2µ,
0
where we used the substitution p = I
that f
f
1-1 t
and exploited radial symmetry. Note
lei2'
(,,j_1 ' - 112 dw is O(t2) as t J. 0, and is 0(1) as t Too, so that aµ is a finite, positive real number for 0 < µ < 1.
Theorem 3.16 Ifs ? 0, then W'(R') = H' (RI) with equivalent norms. Proof. Let r be a non-negative integer, and let 0 < p < 1. In view of (3.17), Plancherel's theorem gives IIu11w'.(R")
_
j br( )lu( )led
1:
lal
where
(I +
br( ) _
1512
)r,
proving the result if s = r. By Lemma 3.15, ifs = r + µ, then IIUIIW(Rn) + E laauIA
=
f
lal=r
')I2d +
,
fRn
(1 +
la l=r
fR'
)12d
Equivalence of the Norms
81
Corollary 3.17 For any non-empty open subset 0 C R", there is a continuous inclusion Hs (Q) C Ws (S2)
for s > 0.
Proof. Given U E Hs (Q), we can find U E Hs (R l) such that u = U I si and IIu11HIM) = IIU11H=(R") By the theorem, U E
Ws(1Rl), so is E W1 (Q) and
Ilullw=(n) _< IIUIIws(w°) - IIUIIHs(a") = IIul1H1(sz)
The next theorem shows that the reverse inclusion holds if there exists an extension operator for 0. The existence of such operators is proved in Appendix A, under appropriate assumptions on S2; see also Theorem 3.30.
Theorem 3.18 For any non-empty open set Q C R" and any real s > 0, if there exists a continuous linear operator E : Ws (0) -> Ws (R") such that Eu I n = u
for all u E Ws (0), then
H'(Q) = W3(c) with equivalent norms.
Proof. If U E Ws(S2), then is = Uln for U = Eu E Ws(R") = Hs(R"), so u E Hs(S2) and IIuIIH,'(n) _< IIUIIH=(R") - IlEullwS(R,,) < Cllullws(j2),
giving a continuous inclusion W1 (S2) c H3(0). No extension operator is needed to establish the corresponding result when s is a negative integer.
Theorem 3.19 For any non-empty open set 7 C R", and for any integer r > 0,
H-r(2) = W-r(2) with equivalent norms.
Proof First consider the case 0 =1R' , and recall that [H'(R"))* = H-r (Rn) We define a Banach space isomorphism J : Wr(W") --+ H -1 (WI) by (Ju, v) = (u, v)Wr(R,.)
foru, v E Wr(R") = Hr(Rn),
Sobolev Spaces
82
and introduce another inner product and norm for H-r (R),
((u, v))-r = (J-'u, J-'v)wr Obviously, Illu III-r
and
((u, u))-r =
IIIuIII-r =
IIJ-iullwr(tt").
flu II H-' (>a")
If U E H-r(R") and V E H" (R), then
(u, v) _
(J-lu,
where fa = (-1)Ialaa J-1u E W-r(R") with IIuIIW-r(R^)
(aaJ-lu, aav) _
v)wr(R") =
(aafa, v),
IaI
IaI
Wr-I«I(R")
c L2(R"). Thus, u = F-Ial
-
< > IIfaIIL2(R") IaI
IIa(rJ-lull2 laI
(J-lu, J-`u)Wr(Rn)
= IIIUIII?r
Conversely, if u E W-r(R") has a representation u = F-lal
(aafa, v)I = 11: (-1)'aI(fa, 8av)
I(u, v)I = I
IaISr
Ial:Sr
1/2 112L,
C
(R")
Ilyllwr(R^)
IaI
for ally E Wr(1R")=Hr(Rn),sou E H-r(R") andIIUIIH-r(R")
Now consider the case when S2 # R", and let u E H-r(S2). We choose U E H-r(R") = W-r(RL8") such that u = U112 and Ilullx -r(Q) = IIU11x-1 (R^), and then choose Fa E L2(R") such that 80T.,
U=
and
IIUIIv-r(tt") < E IaI
IaI
Put fa = Fa I a E L2(Q), so that u =
IaI
W-r(S2), and IIfalIL2(n) <
IIuIIW-r(n) < IaI-
<
U112
E IaI
IIFaIIL2 z(R")
H II UII-r(an) =
IIUI12
H-r(n)'
IaI as f", on 0 with fa E Conversely, suppose that u E W -r (0) satisfies u = L2 (S2). Extend fa by zero outside Q, and denote this extension by Fa E L2(IR").
Localisation and Changes of Coordinates
83
Put U = EIaI
IIFotIIL2(R")
IaIsr
-
IIJ
L2(n)'
IaI
Thus, U E H-r(7) and IIuIIH-rM) < CIIuIIw-r(n)
Localisation and Changes of Coordinates Our next result shows that multiplication by a smooth cutoff function defines a bounded linear operator on HI (Q) or on F11 (0).
Theorem 3.20 Suppose that 0 E Ccomp(R") for some integer r > 1, and let IsI < r. If U E HS(S2), then 4u E H5(S2) and II-0UIIH5(n) <- CrliIOiiW,(R")IIuIIH-(n).
The same holds with HS (S2) replaced by Hs (S2).
Proof. Suppose first that Q = R". In view of Theorem 3.16, the result is clear for s = r, and hence by the duality relation (3.22), for s = -r. The case -r < s < r then follows by interpolation, using Theorem B.7 (from Appendix B). An alternative proof, leading to a different norm for 0, is given in Exercise 3.16. Now let 7 be any open subset of IE,i;", and let it = U I st for some U E Hs (R").
Since Ou = (0U)In or, strictly speaking, q5nu = (¢U)In, we see that II0uIIH5(n) < IIOUIIH=(R") <- CrII4llwr (R")IIUIIH1(R"),
and the desired estimate follows after taking the infimum over U. With HS (S2) replaced by F11 (Q), the proof is even easier because IIouIIH$(s2) = II0u1IHI (R") and IIuIIH=(n) = IIuIIHs(R").
0
In conjunction with Theorem 3.20, the following notion is often useful; cf. Exercise 3.19. A partition of unity for an open set S C R" is a (finite or infinite) sequence of functions 01, 02, ... in C°O(R") such that
1. Oi > 0 on R" for each j; 2. each point of S has a neighbourhood that intersects supp 4j for only finitely many j ;
3. J:j >t Oj(x) = 1 for each x E S.
84
Sobolev Spaces
Notice that condition 2 implies that the sum in condition 3 is finite for each x E S. If S is not open, then we say that the cj form a partition of unity for S if they form a partition of unity for some open neighbourhood of S. Suppose now that W is an open cover for S, i.e., W is a family of open sets for which S C U W. We say that a partition of unity (4)> 1 is subordinate
to W if for each j there exists W E W such that supp cj C W. Theorem 3.21 Given any open cover W of a set S C R", there exists a partition of unity (q5j) j>1 for S subordinate to W. Moreover, the 4j can be chosen in such a way that supp .ij is compact for each j.
Proof Put 0 = U W. We begin by constructing open n-cubes Q 1, Q2.... with the following properties: (i) the family { Q j l j> 1 is an open cover of n;
(ii) for each j there exists a set W E W such that Q j C W; (iii) each point of 9 has a neighbourhood that intersects only finitely many of the Q p Let 01 C 02 C SZ3 C . be a strictly increasing sequence of bounded open sets whose union is 0 and that satisfy SZj C= S2 j+1 for each j > 1. For convenience,
we put 0-1 = S20 = 0, and then define the compact sets K j = S2 j \ S2 j-1 for j > 1. Since K j does not intersect S2 j_2, given x E K j we can find an open cube G centred at x with G j,s C= W \ S2 j-2 for some W E W. The family {G j.X : x E K j } is an open cover for K j, so by compactness we can extract a finite subcover Q p After relabelling the cubes, we obtain a countable family Uj> 1 Qj = (Q1, Q2, ... } with the required properties (i)-(iii). (The set Q j is disjoint from each cube in Q i+2 U Q j+3 U .... and so intersects only cubes in the finite family Q1 U U Qj+1.) For each j > 1, we can use Exercise 3.6 to construct a function 1/rj E C ,p (]Il;") satisfying i/ij > 0 on Q j, and ij = 0 on R" \ Q p. Property (iii) of the Q j implies that the sum 11(x) = > j> 1 *j (x) defines a function E CO° (S2), and property (i) implies that > 0 on Q. Hence, we obtain the desired partition of unity by defining q 5j (x) = *j (x) / ' (x) for x E 0, and O j (x) = 0 otherwise.
Corollary 3.22 Given any countable open cover {W1, W2, ...} of a set S C R", there exists a partition of unity (P1, 02, ... for S having the property that supp q5j C Wj for each j > 1.
Density and Imbedding Theorems
85
Proof. Let 01, 02, ... be a partition of unity for S subordinate to the given open cover, define the index sets 11 = {k > 1 : supp 0k C= WI) and
Ij = {k > 1 : supp Ok C= W3 and k f I, U ... U Ij _ 1) for j > 2, and then put O! (x) = >kEI; Ok(x) for j > 1. We will now show that the Sobolev spaces on ]R" are invariant under sufficiently regular changes of coordinates; u o K denotes the composite function defined by (u o K)(x) = u[K(x)].
Theorem 3.23 Suppose that K : IR" -* ]R" is a bijective mapping and r is a positive integer, such that a"K and FK-1 exist and are (uniformly) Lipschitz on R" for J a I < r - 1. For 1 - r < s < r, we have u E H'(1R") if and only if u o K E HS(IR"), in which case 11U
0 K11 H-
(RI-)
11U11
S(R")
Proof. It suffices to show that for 1 - r < s < r, 11U 0KIIii
(R1-) < Cr11UIIH'(R-')
If s = r, then the estimate follows directly from the chain rule, because H" (IR") = W r (]R" ). The same estimate holds for s =1- r because H 1-r (]R") [Hr-1 (R'l)]* and
(u o K, v)
= (u, (v o K-1)I det(K
The case 1 - r < s < r then follows by interpolation, using Theorem B.7.
Density and Imbedding Theorems We saw earlier that D(SZ) is dense in H8(S2), but it is easy to find examples where D(S2) is not dense in Ws (0); see Exercise 3.18. However, we have the following theorem of Meyers and Serrin [64]. The proof relies on a technical lemma. Lemma 3.24 Lets E ]R and c > 0. For each u E HS (I8") there exists v E D(W' ) satisfying 11U
- v1IH-I(tt'-) < E
and
supp u c {x E ]R" : dist(x, suppu) < E).
Proof. See Exercises 3.14 and 3.17.
Sobolev Spaces
86
Theorem 3.25 For any open set 0 and any real s > 0, the set W5(S2) fl g (o) is dense in Ws(9).
Proof. Define a strictly increasing sequence of bounded open sets W1 C W2C ... by Wj = {x E S2 : Ix1 < j and dist(x,R" \ 0) > 1/j), and choose a partition of unity 01, 02, ... as in Corollary 3.22. Let U E WS (S2) and c > 0. For each j, the function 4j belongs to D(S2), so Oju E WI (R-1) = HI (W) and we can apply Lemma 3.24 to obtain a sequence (v)1 of functions in D(S2) satisfying
IItju - vj11wT(n) :
2jj
and
suppvj c Wj+i.
(Here, we use the fact that 110j u - vj II w=(n) = II /. ju - vj 11 w'(ttn).) Define v(x) _ F_1 v j (x), and note that this sum is finite for x in any compact subset
ofS2,sovEE(c2)and 00
Ilu - v11w(n) = T(Oju
00
j=1
E
1
- vj) W (SZ)
j=1
0 Next, we prove the Sobolev imbedding theorem, which shows that if s is large enough then the elements of HS (1[8") can be thought of as continuous functions, and the elements of Ho (S2) as functions that vanish on the boundary of n.
Theorem 3.26 Suppose 0 < tt < 1. If U E H"t 2+u (R` ), then u is (almost everywhere equal to) a Holder-continuous function. In fact, IUWI <- C II u lI H-1n+,t (R,1)
and
-YI'
lu(x) - u(y)I < for x, y E R".
Proof By the Fourier inversion formula (Theorem 3.10) and the CauchySchwarz inequality, if u E S(R") and X E R", then Iu(x)I <
f
dl;
where
C2 = f (1 +
112)-n/2-u
d < oo.
Density and Imbedding Theorems
87
Now let u E H"/2+1, (1[8"). We choose uj E S(R") such that uj -+ u in H"/2+4 (R"), and observe that, by the estimate above,
lu/(x) - uk(X)I < cllu/
-
Therefore, we can define a uniformly continuous function U : R" C by U (x) = lim1.._, 00 u/ (x) for x E R". Let l E D(R" ). On the one hand, (uj, 0) -+ (u, q5) because u/ -+ u in H"/2+u (R"), and on the other hand (uj, q5) -+ (U, q0)
because uj -a U uniformly on IR". We conclude from Theorem 3.7 that, for almost all x E I[8", u(x) = U(x) and
Iu(X)I = 1U(X)1 =j-,0o lira luJ(x)I -< Cj-+oo lira Similarly, we see from (3.25) that Shu(x) = u(x + h) - u(x) is bounded by
f
dl; < MA(h)11u11 H,t/2+u(R,,)
where Mm (h)2 =
f
(1 +
1412)-n/2-u
-1 I2 d
It is clear that M,u (h) < C for all h E R", and if 0 < l h 1 < 1 then M1, (h)2 < C
f
(I +
h12d
ICI<1/Ihl
+4f
(1 +
112)-n/2-u
d
t1>1/Ihl
< C1h12 (i + f l
1/IhI
l
p1-2u dp J + C Jr00 p-1-2u dp 1/IhI /l
< Clh12(l + 1h12u-2) + Clhl2u < CIh121' , SO lShu(X)I <
for allx E R".
We shall also prove an important compactness result that originated in a paper of Rellich [84].
Theorem 3.27 Assume -oo < s < t < oo. (i) If K is a compact subset of I(8", then the inclusion HK c HK is compact. (ii) If S2 is a bounded, open subset of R", then the inclusion Hr (0) C HS (S2) is compact.
Sobolev Spaces
88
Proof To prove part (i), let (u)1 be a bounded sequence in HK for some compact set K C= R". We want to show that a subsequence converges in HK. Choose a cutoff function X E D(R) satisfying X = 1 on K, so that
uj(
)=Xuj(t)=f
n
Using the Cauchy-Schwarz inequality and Peetre's inequality (see Exercise 3.16), we find that
(I + ItI2)`Ii{{j(f)I2 < 2111
(1 + It
\J
-
r1I2)1nhIX(t
-17)I2dr1)
X (f
Likewise, since 81u j = (a"j() * u j = j( * u j where X,, (x) = (-i27rx)"X (x), we have I4I2)hIa°`uj(t)I2
(1 +
< 21t"IIX.,112
.
It follows, in particular, that the sequence of Fourier transforms (j) ° 1 is uniformly bounded and equicontinuous on any compact subset of JR". Let K1, K2, K3,... be an increasing sequence of compact sets with U ° 1 K, = JR" . By the Arzela-Ascoli theorem (Theorem 2.15), there is a subsequence i that converges uniformly on K1. From this subsequence, we can extract a subsequence u that converges uniformly on K2. Continuing in this way, we obtain successive subsequences such that u'. converges uniformly on K,. For brevity, we now denote the diagonal subsequence ui by u j. Thus, the Fourier transforms u j converge uniformly on any compact subset of Ill;", and it suffices to show that the functions u j themselves are Cauchy in HK . Given e > 0, we first choose R large enough so that
f
(I + ItI2)SIij(t) - uk(t)I2 d4 1>R
(1 +
(I + ItI2)`Iuj(t)
R2)S-t
fn
IIuj - ukllH,(R-.)
(I + R2)(-3
- 2(IIuj
2
(1 +
- uk(t)12d Il ukll
E
2
Lipschitz Domains
89
for all j, k > 1, and then choose N large enough so that
f
(1 + 11;12)5 lug
uk (i; ) 12 dt < 2
for all j, k > N.
I
) < E for j, k > N, completing the proof of part (i). i is a bounded sequence in H'(S2). By hypothesis, Suppose now that 0 is a bounded subset of R", so we can find a compact set K C=1[8" together Hence, I l u 1 - Uk i i
with a cutoff function x E DK (R") such that cZ C K and x = 1 on 0. For each j, choose U1 E H'(]R) With II UjIIH'(R") = IIUJ IIH'(n3- By Theorem 3.20, is bounded in HK, so by part (i) there is a subsequence, the sequence (x x Uj E HK again denoted by (x UU )_1, that converges in H. Put U = limb.
and u = UIn E H5(S2), and observe that Ilu1 - UIIHs(n) = II(xU.i - U)IQIIHs(Q) <- IIXU; - UII H' -
Thus, uj -* u in HS (S2), proving part (ii).
O
Lipschitz Domains Denote the boundary of the open set S2 by
r=au =s2n(w\Sl). Thus far, no use has been made of any regularity assumption on IF, but henceforth we shall require that, roughly speaking, the boundary of S2 can be represented locally as the graph of a Lipschitz function (using different systems of Cartesian
coordinates for different parts of r, as necessary). The simplest case occurs when there is a function : R"-1 -* R such that
S2={x
ER"-I}.
all
(3.26)
If 4 is Lipschitz, i.e., if there is a constant M such that
MIx' - y'I
for all x', y' E
(3.27)
then we say that 0 is a Lipschitz hypograph.
Definition 3.28 The open set cZ is a Lipschitz domain if its boundary r is compact and if there exist finite families (WJ} and (0j) having the following properties:
Sobolev Spaces
90
(i) The family (Wj) is a finite open cover of I', L e., each W J is an open subset
of w, and I' C U; Wj. (ii) Each Qj can be transformed to a Lipschitz hypograph by a rigid motion, i.e., by a rotation plus a translation. (iii) The set Q satisfies W3 n s2 = w j fl S2j for each j. Notice that if 0 is a Lipschitz hypograph as in (3.26), then
I'={x ERn-1 :x
E
nI
We also remark that although, in our definition, the boundary of a Lipschitz domain must be compact, the domain itself may be unbounded. In particular, if SZ is a bounded Lipschitz domain, then R" \ SZ is an unbounded Lipschitz domain. Sometimes, a different smoothness condition will be needed, so we broaden
the above terminology as follows. For any integer k > 0, we say that the set (3.26) is a Ck hypograph if the function : lR' -+ ]f8 is Ck, and if 81 is bounded for lad < k. In the obvious way, we then define a Ck domain by substituting "Ck" for "Lipschitz" throughout Definition 3.28. Likewise, for 0 < It < 1, we define a domain by adding the requirement that the kth-order partial derivatives of be Holder-continuous with exponent µ, i.e.,
18' (x') - d' (Y') i < Mix' - A,
f o r all x', y' E Rn-1 and IaI = k.
Hence, a Lipschitz domain is the same thing as a CO, 1 domain. Notice that in the definition of a Ck or domain, we can assume if we want that has compact support, because r is always assumed to be compact. The class of Lipschitz domains is broad enough to cover most cases that arise in applications of partial differential equations. For instance, if k > I and r, is a compact, (n -1)-dimensional Ck submanifold of R", then 0 is a Ck domain and hence also a Lipschitz domain. Furthermore, any polygon in RI or polyhedron in RI is a Lipschitz domain. One can construct many other examples using the fact that if K : R" - * R" is a C 1 diffeomorphism and if 0 is a Lipschitz domain, then the set K(S2) is again a Lipschitz domain. Figure 2 shows some examples of open sets that fail to be Lipschitz domains: (i) is disqualified because of the cusp at the point A; (ii) because of the crack B C (a Lipschitz domain cannot be on both sides of its boundary); and (iii) because in any neighbourhood of the point D it is impossible to represent r as the graph of a function. For a Lipschitz domain, in fact even for a CO domain, a much stronger density result than Theorem 3.25 holds.
Lipschitz Domains
91
(i)
0
Figure 2. Examples of regions that fail to be Lipschitz domains.
Theorem 3.29 If 0 is a CO domain, then (i) D(S2) is dense in WS (Q) for s > 0; NO D(S2) is dense in H or in other words HS (S2) = Hn for S E R. Proof. Suppose to begin with that n is of the form (3.26) for some continuous function : ]R"_' - ]R having compact support.
Lets > 0, u E WI (0) and e > 0. For S > 0, we define
ua(x)=u(x',x"-S)
and
528=(x ER" :x,,
so that ua E WS(Qs). Since a"us = (a' u)8, we can choose 3 small enough so that Ilu-uslsalIW$(si) < 2
and then choose a cutoff function X E E(]R") satisfying X = 1 on 0 and X = 0 on ]R" \ 52(8/2), so that Xus E WS (]R"). Hence, by Theorem 3.16, there exists V E D(]R") such that E
Ilxua-VIIWs(atn)<2.
Sobolev Spaces
92
implying that the restriction v = V In satisfies
IIu - vllwa(n) = Ilu - usln + (Xus - V)Inllwl(n) IIu - U lnllw$(n) + IIXus - Vllw=(RlI) < E.
Thus, D(S2) is dense in WS(Q). To complete the proof of part (i), we suppose now that S2 is a Co domain, and let be a finite open cover of r as in Definition 3.28. Define one
additional open set WO = {x E Q : dist(x, R" \ S2) > S}, choosing a small enough 8 > 0 so that S2 c_ U'=o Wj. We may assume that Wj is bounded for I < j < J, but WO will be unbounded if 0 is unbounded. Let 0 be a partition of unity for S2 such that supp j c W j for all j. By the case of a Co hypograph considered above, if 1 < j < J then there exists vj E D(l) such that Iloju - vj II w.,(n) < E/(J + 1). In fact, by Lemma 3.24, the same is true also when j = 0, because 4ou E W1 (R"). Put v = F_J=o V j E D(S2); then J
11U - v1Iw.(n) = E(Oju j=o
IIOjU-vj11Wx(n) <E.
- vj)
W (n)
j=o
To prove (ii), assume once again that 0 is of the form (3.26), and let s E R, u E HS (S2) and r > 0. This time we put us (x) = u (x', x,, + 8), and observe that us E HI (R") and suppus c {x E R" : x" (x') - 8}. Choose S small enough to ensure IIu -US IIH.(ttl') <
2
E,
and then apply Lemma 3.24 to find v E D(Q) satisfying 11us - vIIH=(R,I) <
2
Since 11u -vIIH=(n)= 11(U - us) + (us - v)IIHs(R) < E, we see that 1)(0) is a dense subspace of HS (S2). As with part (i), the result carries over to Co domains with the help of a partition of unity. Theorems 3.29 and A.4 allow us to apply Theorems 3.14 and 3.18, and hence deduce the following important result.
Theorem 3.30 If Q is a Lipschitz domain, then
(i) HS (Q)* = H-S (S2) and HS (S2)* = H-S (0) for all s E R; (ii) WS(S2) = HS(Q) for all s > 0.
Lipschitz Domains
93
We remarked earlier that HS (S2) C Ho (S2). The next two technical lemmas will enable us to establish the reverse inclusion apart from certain exceptional values of s.
Lemma 3.31 Suppose U E D(R). If 0 < s < Z, then f f °Cx-z.`Iu(x)12dx < CS J 0
0
,
, (u(x) -u(Y)12 dxdy. Jo Ix r
Y11+2s
If, in addition, u (0) = 0, then the inequality holds also for 1 < s < 1.
Proof. Observe that the double integral converges for 0 < s < 1. For x > 0, define
v(x)
1
x
u(y)dy f'[u(x) -u(y)]dy=u(x)-J x X
o
and
w(x) = r°° X
V
dY Y
Since v'(x) = u'(x) - x-1v(x) and w'(x) = -x-1v(x), we see that
u'(x) = v'(x) - w'(x) forx > 0. Furthermore, u has compact support, and both v(x) and w(x) tend to zero as x tends to infinity, so
u(x)=v(x)-w(x) forx>0. 'By the Cauchy-Schwarz inequality, Iv(x)12 <
x
f
Iu(Y) - u(x)12 dY,
0
implying co x-l-2s
00
x-2SIv(x)12dx < fo fo
=
f
f X Iu(Y)
u(x)12dydx
0
x-1-2slu(Y)u(x)12dxdy y
°° Iu(x) - u(Y)12
°° Jo
Jo
Ix - y11+2s
dxdy
94
Sobolev Spaces
for 0 < s < 1. By Exercise 3.20, °O
1
x-2slw(x)I2dx < /
00
1
(2
- s)
fors <
x-2slv(x)I2dx
I':
z'
and the first part of the lemma follows. One easily verifies by Taylor expansion that v(x)
0 as x -* 0+, so w(x) _ v (x) - u (x) -+ -u (0) as x -* 0+. Thus, if u (0) = 0 then w (0) = 0, implying that
w(x)=w(x)-w(0)=w(x)-V(Y)dy=- f x!(Y)dY J0° Hence, by Exercise 3.20,
I
00
x-2slw(x)l2dx <
f
I (s-
y
Y
Jo
00
2
x-2slv(x)I2dx
fors >
2'
2)
giving the second part of the lemma. Lemma 3.32 If S2 is a Lipschitz domain and u E E) C2), then
fn
dist(x, I')-2slu(x)I2dx < CIIuIIH,( for0 < s < Z < s < 1.
If, in addition, u = 0 on T, then this inequality holds also for 2
Proof. We prove the result for S2 a Lipschitz hypograph given by x, < (x'). If y E I' and M is a Lipschitz constant for , as in (3.27), then yn = f (y') and so
VX') - Xnl = IYn - X. + (x') - (Y')I < Ix,1 - Yni + Mix' - Y'I <
1+M2Ix-yl,
implying that Xn dist(x, r) > (x') 1+M2
for X E 0.
5 <;(x') f
Using the substitution x _ (x') - t, followed by Lemma 3.31, we see that
J
dist(x, r)-2slu(x)l2 dx < (1 + M2)s f
=C u
[
(x') -
jt'Iu(x'(x') - t)12 d t dx'
Lipschitz Domains
I
C
R"-t
95
r
r
Y<{(x')
z<M')
Iu(x' ,
y)-u(x'
,
z)12
Iy - z11+2s
x dydzdx', so if u = UIn where U E D(R"), then, arguing as in the proofs of Lemma 3.15 and Theorem 3.16,
1
dist(x, I')-2jlu(x)12 dx < C J
R
I U(x', x"I i h
U(x) 12
dx dh
JJ
f
00
_ C J R" I U(h)f'`
Ihll+zs
00
=cJR-
dh d
and the result follows because D(RI) is dense in H' (R ?)
Notice that the value s = 1 is excluded in the two lemmas above; cf. Exercise 3.22. Also, recall our earlier discussion of the imbedding (3.24). Theorem 3.33 Lets > 0. If S2 is a Lipschitz domain, then
P(Q) = fu E L2(52) : u E H'(R")) C Hp(S2), where u denotes the extension of u by zero:
u(x) =
u(x)
if x E 92,
0
ifxER"\S2.
In fact,
H's(Q) = Ho(S2)
provideds 0 {2, 2, 2,
}
Proof. For the moment, we think of the elements of H' (S2) as distributions
on R. If U E Hs(S2), then the restriction v = ulst belongs to L2(S2), and u = v as a distribution on R", so D E H'(R"). Conversely, if u E L2(S2) and u E H' (R"), then supp u c S2, so u E H = HS (S2). We have already seen that HS (S2) c _Ho (S2).
Now view HI(Q) as a subspace of L2(c2), and let u E D(S2). For any integer r > 0, Theorems 3.16 and 3.30 give IIuIIHr(p) = IIuIIHr(R")
IlaauIILz(R") Ial5r
=
IIUI1Hr(si),
-
Ila"uIIL,(n) IaI
96
Sobolev Spaces
so Hr(0)=Hp(0).For s=r+µwith 0
I
+
Ix -
xR
IaI=r
a
_
aau(x) - aau(y)l2
dx d y
l aau(X) - aau(y)12
2
Ila uIILZ(sz) Ial=r
+
y12µ+n
Jf2 xf2
Ix - y12µ+n
dxdy
1 aau(X)12 dx dy Jf2x(R11\s2) IX - yl2IL+n IaI-r
+
laau(y)12 Ian=r
(R"\12)xS2 Ix - yl2u+n
dxdy
= Hull W$(Q)+2 E J laau(X)I2wµlX)dX, IaI=r
2
where the weight w12 is defined by
dy
wµ(x) = fR,l\f2 lx - y12µ+n
for x E Q.
Introducing polar coordinates about x, we see that
wµ (x) < C dist(x,
I')-24
for X E S2,
so by Lemma 3.32, "(S2) H < C1,11U112..(n) 1: f w(x)laau(x)12dx < CN, E IlacU112 IaI=r
IaI=r
provided s
integer + z . Hence, in this case, Ho (2) c Hs (Q).
Sobolev Spaces on the Boundary Any Lipschitz domain SZ has a surface measure a, and an outward unit normal v
that exists a-almost everywhere on F. In fact, by Rademacher's theorem [83], [ 105, Theorem 11 A, p. 272], if lR' ' -->. R is Lipschitz, then is Frechetdifferentiable almost everywhere with Ilgrad0lLo.(w"-I) < M,
Sobolev Spaces on the Boundary
97
where M is any Lipschitz constant for , as in (3.27). If 0 is the Lipschitz hypograph (3.26), then
do:, =
a (x'), 1) 1 + grad (x')12 dx' and v (x) = (- grad
for x E F. (3.28) The divergence theorem can then be proved in a straightforward manner.
Theorem 3.34 If 0 is a Lipschitz domain, and if F : R" -+ R" is a C' vector field with compact support, then
I divFdx=J
r
Proof Assume to begin with that 0 is a Lipschitz hypograph (3.26). For 1 < k < n - 1, we define Uk : R11-1 -+ R by Ox')
U k (x') =
J
Fk (x', x,,) d x , , . 00
f
akuk(X') = Fk(X',
J
S(-x')
8kFk(x)dxn, 00
and fR-,-, ak uk (x') dx' = 0 because Uk has compact support. Thus,
I
<('(x')
akFk(x) dx = -
f "- Fk(x',
--
dx' for 1 < k < n -
with
f
a,, F.(x)dx= f
Fn(X',y(X'))dX,
so that
jdivF(x)dx z
= fat-- F(x',
x')) (-grad (x'), 1) dx' =
JF. v da.
Now let 0 be a Lipschitz domain, and take a partition of unity (.0j) J_o for S2 as in the proof of Theorem 3.29. Since q5o E D(22), it is obvious that
f
ak(0OFk) dXk = O, 00,
sr Sobolev Spaces
98
so fn div(00F) dx = 0. Hence, from the case of a Lipschitz hypograph treated above, J
f
J
r
r
divFdx=>2J div(OjF)dx=>2J v- (b1F)dcr= j v Fdc,
a
If c2 is a Lipschitz hypograph, then we can construct Sobolev spaces on its boundary I' in terms of Sobolev spaces on R"-1, as follows. For U E L2(F) _ L2(F, Q), we define
forx' EI[8", put
for0 < s < 1,
HS(T) _ {u E L2(I') : u E HS(R"-')) and equip this space with the inner product (u, v)Hs(r) _ (u;, VC) H' (R"- 1)
Recalling that dci is given by (3.28), we put IIu1IH-q(r) _ (Iut
for0 < s < 1,
1 + Igrad
and then define H-S(l') to be the completion of L2(F) in this norm. It follows that H-1(F) is a realisation of the dual space of HS(F), with sup
IIuIIH-5(r)
I(u, v)rI
OovEH'(r) IIVIIH'(r)
=
I(u, v)rI
sup o#veH'(r) IIUIIHtr)
for. IsI < 1,
where
(u, Or =
Jr
u(x)v(x)da(x) and (u, v)r =
Jr
u (x) v (x) da (x).
If K (Sl) is a Lipschitz hypograph for some rigid motion K : R" --+ ]R", then we define HS(F) in the same way except that u[K-1(x', Ox'))}.
Suppose now that SZ is a Lipschitz domain. Using the notation of Definition 3.28, we choose a partition of unity 10j) subordinate to the open cover ( Wj )
of F, i.e., we choose O j E D(W j) satisfying > j 4 (x) = 1 for all x E F. The
Sobolev Spaces on the Boundary
99
inner product in Hs (r) is then defined by (u, V)Hs(r) _
(Oiu,fj v)H-(r1),
(3.29)
where r, = 8 S2,. Theorems 3.20 and 3.23 imply that a different choice of { W, },
{S2, } and {0, ) would yield the same space Hs (r) with an equivalent norm, for Is l < 1. If 0 is C't-" fork > 0, then HS (r) is well defined for Is l < k. We shall also require Sobolev spaces defined over only a part of the boundary
of n. Consider a disjoint union
r=r1UIIU1-2,
(3.30)
where 1`1 and 172 are disjoint, non-empty, relatively open subsets of r, having 11 as their common boundary in r. When S2 is a Lipschitz hypograph, we call (3.30) a Lipschitz dissection of r if there is a Lipschitz function Q : Ri-2 -+ R such that
{x E r : xi_1 < Q(x")},
TI= 172= {x Er:xi_1 >Q(X")),
where x" = (x1, ... , xn_2); for n = 2, the function q reduces to a constant. In the obvious way, we extend the notion of a Lipschitz dissection to the case when S2 is the image under a rigid motion of a Lipschitz hypograph.
Next, suppose that n is a Lipschitz domain. We say that (3.30) is a Lipschitz dissection of r if, in the notation of Definition 3.28, there are Lipschitz dissections 8 S2, = r,, u l1, U r2, such that
W,nr1 = W,nr1,,
W,nfI= W,nII,, W,nr2= W,nr2,,
for all j. We remark that, in this case, the subsets r 1 and 1`2 are not necessarily
connected. LetD(r1) = 1-0 E D(r) : supp0 c r1}; by defining
Hs(r1) = {U1r, : U E Hs(r)), Hs(171) = closure of D(171) in HI (r),
Ho (r) = closure of D(rI) in Hs(ri), the properties of Sobolev spaces on Lipschitz domains in R carry over to Sobolev spaces on r1, subject to the condition that Is I < 1, or Is I < k if (3.30) is Ck-1.1 in the obvious sense.
Sobolev Spaces
100
The Trace Operator In studying boundary value problems, we shall need to make sense of the restriction u I r as an element of a Sobolev space on r when u belongs to a Sobolev space on 7. The main idea is contained in the following lemma.
Lemma 3.35 Define the trace operator y : D(IR") -* D(IR"-') by
yu(x) = u(x', 0) forx' E
IR
n-
Ifs > 2, then y has a unique extension to a bounded linear operator
y : H$(1R") -4 H" '/2(1R"-'). Proof. For u E D(1R"), the Fourier inversion formula (3.15) gives
f
Yu(x') = f,'
u(',n)
(f
dt = J
f
"
oo
and so 00
Yu(')=J 0000 u(', n)dSn=J
(1+1 12)-S/2(1+1 12)s'%u 00
Applying the Cauchy-Schwarz inequality, we obtain the bound
f
W
(1 +
where, using the substitution t;,, = (1 + oo
Ms( ') =
1W1I2)'/2t,
f
00
dtn
00 (1+I 112+I n12)S
(1 +I
112)S-1/2
The integral with respect to t converges because s >
dt (1+t2)S.
oo
so if we write MS =
MS (0) then
(1 +
MS
f
(I +
do
00
Integrating over ' E 1R" gives MSIIUIIHs(Rn),
and since D(1R") is dense in H5 (R"), we obtain a unique continuous extension
for y.
The Trace Operator
101
The lemma above is sharp in the sense that
Hs-l/2(R"-') = {yu : u E HS(R")) for s > 1, because y has a continuous right inverse rio, as we now show.
Lemma 3.36 For each integer j > 0, there exists a linear operator
77j:S(1R"-')-* S(R") satisfying
a" (rij u) (x', 0) =
8",u(x')
if a = j,
0
if
for x' E Ri-', U E S(R"-)) and any multi-index a = (a', a"). Moreover, rij has a unique extension to a bounded linear operator i7 j :
Hs-j-1/2(Rn-i) -+ H' (W') for S E R.
Proof Choose a function O j E D (R) satisfying 9 j (y) = yj /j! for I y j < 1, and define z1
u(x) =
dl;' forx ER".
(1 +
JR1
Since 9j(k) (0) = S jk, we see that
aa(77ju)(x', 0) =
J
j
gives
as required. The substitution x = (1 +
11 u() _
(1
+ ,i )j/2
a"'u(x')s j"
ei2nt' X'
f4 (
L(1 + I
I)
x] dxn
j[(1 +
_
(1
gives
and the substitution 4n = (1 +
4-1 X
J
I
,12)1+l
[(1+1 12)-i/2 n]I2d ndS 00
Sobolev Spaces
102
where 00
CS = J 0 (1 +t2)SIdj(t)I2dt < oo,
for all s E R. For Sobolev spaces on domains, we can now prove the following.
Theorem 3.37 Define the trace operator y : D(S2) -+ D(F) by
Yu = uIr If SZ is a
Ck-1,1 domain,
and if 2 < s < k, then y has a unique extension to a
bounded linear operator
y : H$(Q) -+
(3.31)
HS-112(r),
and this extension has a continuous right inverse.
Proof Since H5(]R") is invariant under a Ck-1.1 change of coordinates if 1 - k < s < k, one sees, via a partition of unity and a local flattening of the boundary, that if z < s < k then IIYUIIHs-1/2(r) <_
CIIUIIHs(R'
if u = Uisi for U E D(R").
Hence, IIYuIIHs-112(r) < CIIuIIHscQ> for all u E D(O), and we obtain a unique continuous extension because D(S2) is dense in Hs (S2). A right inverse for y can be pieced together using the same partition of unity, by means of the operator '1o from Lemma 3.36.
The preceding theorem applies, in particular, for 1 < s < 1 if 0 is a Lipschitz domain. For technical reasons, we shall require the following, stronger result of Costabel [14].
Theorem 3.38 If S2 is a Lipschitz domain, then the trace operator (3.31) is bounded for 1 < s < z Proof. It suffices to consider a Lipschitz hypograph (3.26). Let U E D(1R"), and put
u(x) = u (x', (x') + x") for x = (x', x")
E lR".
The Trace Operator
103
By definition, IIYUIIH'-112(1) =
and to estimate the right-hand side, we introduce the notation 00
u(x', n) = f
x,) dx, 00
for the partial Fourier transform of u (x) with respect to x12.Observe that uC(x', 412) = ei2- ,,C(x')u(x', n), so
n)IIL2(R"-') = Ilk', n)IIL,(R"-') and
n)IIH< Cliu(,
C ,ZIIu(,
n)I12
00
IIuIIEs = f
n)I1Hl(R°-1)]d
,t
00
f then
where as(h) = I nI25 + I nI2s-2(1 + RUCHES <_ CIIu11E'
fors E R.
It is easy to see that
fors > 1,
11uI1E5 < CIIu11Hs (R.")
and we claim
IIu(,
CIIuflE,
In fact, the substitution n = (1 + d
f7a, (
" = C5(1 + )
1
,12)-(s-1/2),
for z < s < Z.
yields where CS =
f
00 00
t -2(l+ t t2)'
104
Sobolev Spaces
with C, < oo for 2 < s < 2, so, applying the Cauchy-Schwarz inequality,
Ilu(,
0)II2
Hs-1i2(Ra-1)
f
J R°-r
(1
J - -l
(1 +
I
hI2)s-I/2I f
2
00
u( '> n) dSn 00
f (J-00 as ()
)
00 a,()II ( )12dn)
d'
00
=Csf as()Iu()12d =C'11U112Es 1'
Thus, we see that Ilut(, 0)IIHf-II2(RI-r) -< CIIul1El < C11U11El < CIIUIIHS(R")
for 1 < s < 2
which, combined with Theorem 3.37, shows that the trace operator (3.31) is
bounded for 1 < s < 2 . The next lemma is a version of a standard fact [41, p. 47] about distributions supported by a hyperplane, and will allow us to characterise Ho (S2) using the trace operator. The symbol ® means the tensor product of distributions, so, formally, (vj ® SWjW)(x) = vj(x')SWD(x"). In the proof, we use the notation R+ = {x E R'r : x" > 0) for the upper half space.
Lemma 3.39 Consider the hyperplane F = {x E R" : x = 0).
(i) Ifs > - 2, then HF = {0). (ii) If s < - 2, then HF is the set of distributions on R" having the form
u = E uj ®S(j)
with vj E
H''+j+l/2(R'r-t).
(3.32)
0:5j<-S-1/2
Proof. First we show that any u of the form (3.32) belongs to Hi.. Obviously, supp u c F, and since {v j (x')S(j) (xn)) = D j (') (i2ir n)j, we have I1vj ®3(j,112
R) =
f(l + n
The substitutions = (1 + 1'12)I/2t gives vj
®scj>II 2 Ht(R")
where Ci.,s =
f
=
00
(1 +t2y12rrt12' dt,
J 00
and here Cj,, < 00 for s + j < - Z , so u E Hi..
Cj,sIIVjIIHs+i+h/2(R"-1),
(3.33)
The Trace Operator
105
Next, we show that if u E HF, and if 0 E D(R") satisfies
8;¢(x',0)=0 for0< j < -s - Z, then (u, 0) = 0. Indeed, by Exercise 3.22, if we define
0(x) ifx E R't, 0
otherwise,
then 0± E H(R). It follows that there is a sequence (0)n° 1 in D(R \ F) converging to in H_s (R") as m -+ oo. Hence, (u, 0) = Urn,. (u, ¢,") = 0. In particular, u = 0 if u c- HF. for s > - ? . Suppose now that u E H F . with -k - 2 < s < -k - f o r an integer k > 0, z and assume 0 < j < k. Let ri; be as in Lemma 3.36, and define v; E D* (][8"- ) by
(v1,4) _ (-1)j(u, rl;0) for d E D(R't-'). Observe that I(v;, 0)1 <_ IIu11Hs(R )II?1;OIIH-J(ut't) <_ so IIVi IIH=+;+U=(R
Il2(RI-I),
') < CIIuIIHI (tt"), and that k
u-
(u, p)
vi
for 0 E D(R")
;-o where k
p = 0 - E'1;>/il with *;(x') = dO(x', 0). ;=o
Since 8; p (x', 0) = 0 for 0 < l < k, and since -s - z < k + 1, we have (u, p) = 0 and so (3.32) holds.
It only remains to deal with the case s = -k - 2. The argument above HFk-"2 shows that if u E then u = I _O v; ®S(J) with v; E Hj-k(IEP"-') for 0 < j < k - 1, and with Vk E H-E(IRH_L) for any E > 0. Thus, it suffices to show that vk = 0. In fact, (3.33) implies v; ®S(j) E for 0 < j < k - 1, giving Vk ®S(k) E H-' '/2 (R"). However, 2 -k-J/2-. IIv k ®S (k' lIH
and
, = k,_k_1 2_E IIv kII HC'
oo as c --> 0, giving the desired result.
Theorem 3.40 Assume that cZ is a Ck-1 1 domain.
i
H-k-1/z(Wi)
106
Sobolev Spaces
(i) If 0 < s < 2, then Ho (1) = HS (S). (ii) If 1 < s < k, then Ho (S2) = {u E H'(S2) : y (8au) = 0 for J a < s - ? Proof It suffices to deal with a half space Q = R+. We will use the characterisation of the dense subsets of a Banach space given by Exercise 2.5. First suppose 0 < s < z . If W E HS (7)* = H-S (S2) satisfies (w, -0) = 0 for all q5 E D(S2), then supp w is a subset of the hyperplane F = {x E R" : x" = 01, and we conclude from part (i) of Lemma 3.39 that w = 0. Hence, D(E2) is dense in HS (S2), i.e., Ho (S2) = HS (S2).
Next suppose that s > Z, and let E = {u E HS(7) : y(8au) = 0 for (aI < }, noting that this definition makes sense because y o 8' H' (Q) -z HS-111(F) for kal < s - 2. Let f E E* satisfy £(o) = 0 for all ¢ E D(Q). By the Hahn-Banach theorem, there is a w E HS (S2)* = H-S (S2) such that f(o) = (w, gyp) for all 0 E E. Since W E HFS, part (ii) of Lemma 3.39
s-
shows that w = Eo<j<s-1/2 Vi ® 3(j) with vj E
H-s+j+1/2(R"-1)
Hence, for
every q5 E E,
> (-l)J(vj, Y(3
)) = 0,
o<j<s-1/2
and so D(S2) is dense in E, proving part (ii).
0
Vector-Valued Functions Thus far, the present chapter has dealt only with spaces of scalar-valued (generalised) functions. The results obtained extend in a straightforward manner to spaces of vector-valued functions
u: S2-+ c", and this final section does no more than establish some notational conventions. We denote the space of compactly supported, C"'-valued, C°O test functions by
D(Q)," = D(S2; C').
The (sequentially) continuous linear functionals on D(O)"' are then the ("valued distributions on 0, and we view these objects as generalised C"'-valued functions, by writing
(u, v)n = 1 u(x) v(x) dx,
Jn
(3.34)
where the dot denotes the bilinear form on cm whose restriction to R" coincides
Exercises
107
with the standard Euclidean inner product, i.e., "1
U.V=EuJVJ. j=1
The set of all Cm -valued distributions on SZ is denoted by D* (0)"' = D* (S2; C'").
We think of u as a column vector or m x I matrix, and let u* denote the row vector or 1 x m matrix obtained by transposing the complex conjugate u. Using matrix multiplication, we may then write the standard unitary inner product in CC' as m
u*V=U.V=Eujvp I=1
In this way, the sesquilinear form associated with the bilinear pairing (3.34) is given by
(u, On =
u(x)*v(x) dx. fn
Of course, if u and v are square-integrable functions from S2 to CC', then (u, v)n is their inner product in L2 (S2)m = L2 (S2; C"'). The definitions of the vector Sobolev spaces on S2, W'(S2)' = Wp (S2; (C'" ),
H3 (S2)" = HS (S2; (C'"),
HS (S2)m = HS (S2; (Cm),
should now be obvious. Likewise for the vector Sobolev spaces on r. Occasionally, we shall encounter normed spaces of matrix-valued functions, such as L,, (S2)mx,n = L,,(12; C">°"), whose meaning should also be obvious.
Exercises 3.1 Suppose that u E Lp(S2) satisfies I(u, v)nl < MIIvIILp.(n)
for all V E Lp.(S2).
(i) Show that if 1 < p < oo, then II u 11 L,,(sa) < M. [Hint: take v = ] sign(u) lulp-1
(ii) Show that if p = oo, then for every measurable set E C 0 with I E l > 0, the mean value of l u I over E is bounded by M, i.e., E1
I11JElu(x)Idx<M.
Deduce that 11U IIL (n) < M. [Hint: take v = sign(u) XE, where XE is
the characteristic function of E.]
Sobolev Spaces
108
3.2 Prove that convolution is associative:
(u*v)*w=u*(v*w) foru,v,wEL1(R") 3.3 For 1 < j < k, let f j E L 1(R") be a compactly supported, non-negative 1, and let 0 < X j < 1. Fix X E 1I8", and
function satisfying II fj II L,
define
g(A)=(fi' *...* f,R)(x) forA=(At,...,Ak). (Take f to be identically 1 outside as well as inside supp f3.) ° that g (e j) = 1 where ee j E Rk is the vector with compo(i) Show nents (ej )1 = 1 - S jl. (ii) Use the fact that, for any positive a j and any non-negative A j and µj, k
k
k
j=1
j=I
j=1
H ajt-r)xj+ruj = exp (1 - t) L log a^' + t L log aµ' to show that the function g : [0, I ]k --> [0, oo) is convex. (iii) Deduce that g(A) < 1 if At +...+Ak = k -1. [Hint: A Aj)ej.] (iv) Hence show that k
(1-
1
if E-=k-1. j=1;pj 3.4 Show that supp(u * v) c supp u+ supp v = {x + y : x E supp u and y E supp U).
3.5 Recall the notation (3.7). (i) Show that
(d)k dt
u(x + ty) = u(k)(x + ty; y).
[Hint: k!/a! equals the number of permutations of k = IaI objects when there are a j objects of type j, for I < j < n, and it is assumed that objects of different types are distinguishable, but objects of the same type are indistinguishable.] (ii) Use integration by parts to verify Taylor's formula for a function of one variable: k
f (s) =
f(j)(0)
E j j=o
1
Sk+t
Si +
k1
1
J
(1 - t)k f(k+I)(ts) dt.
Exercises
109
(iii) By taking f (s) = u(x + sy), derive Taylor's formula (3.8) for a function of n variables.
3.6 Define f : R -+ R by
f(t) =
e-' 1'
10
if t > 0,
ift<0.
(i) Show that f (i) (t) -* 0 as t y, 0, for each j > 0. (ii) Deduce that f E C°°(R).
(iii) Construct a CO0 function g : IR - R with g > 0 on (-1, 1) and with supp g = [-1, 1]. (iv) Construct a COG function * E C mP(R") satisfying (3.9). 3.7 Let S2 = (0, 1), choose any 0 E D(S2) not identically zero, and define
¢j E D(S2) by 0, (x) = 0 (j -' x). Show that -Oj -+ 0 in E(Q), -but not in D(S2).
3.8 Establish that the inclusions D(S2) C E(S2) and D(R'1) C S(R") E(1E8") are continuous with dense image, by proving each of the following statements:
(i) If ¢j - 0 in D(S2), then qj -+ 0 in E(S2).
(ii) If 4i -+ 0 in D(R), then ¢f - 0 in S(R"). (iii) If 4j -* 0 in S(R"), then Oj - 0 in E(R"). (iv) Let Ki C K2 C be an increasing sequence of compact sets whose union is S2, and let Xi E D(S2) satisfy Xi = 1 on Kj. If 0 E E(S2), then Xicb -+ 4) in E(S2).
(v) Let X E D(W) satisfy X (x) = 1 for IxI < 1, and define Xf E D(R") for each positive integer j by X,i (x) = X (j _'X). If 0 E S (W), then
XjO -* 0 in S(W). 3.9 Consider a linear functional f : D(S2) -+ C. Show that a is sequentially continuous (and hence a distribution on S2) if and only if for each compact set K C S2 there exists an integer m > 0 such that
It(4))I -S CK,," E sup laaol for all 4) E DK(S2). Ia1
[Hint: to prove the necessity of the condition, suppose for a contradiction
that there is a K for which no such m and CK,,,, exist, and deduce the existence of a sequence c 0 in DK(S2) such that £(4j) = 1.] 3.10 Prove from the definition (3.12) that if u E D*(W) and 0 E D(R' ), then
u*0isC°O,with 8a (u * 0) = (a' u) * 46 = u * (a"4)).
Sobolev Spaces
110
3.11 Show that
.Fx,
'1 {e-n1x1' 1 =
l
J
J
1
.%_)
00
e-i2n jxj-nx dx = e-nl J
l''
00
3.12 For * E L) (RI), show that (E-lx)l)
_
and
3.13 Let k be a positive integer. We denote the kth-order forward difference operator by 8 , and then define the kth-order L2 modulus of continuity by k (t, a) u) = Sup II SI U II Lz (RI I)
for t > 0.
IhI
(i) Adapt the proof of Theorem 3.16 to show that if 0 < s < k, then 2
2 IIuIIHA(R")
ti IIuIIL2(R") +
f
2
Ilsh uiiLZ(R,,) " IhI2T+n
dh.
(ii) Show that [cvk(t, u)]2 < Ck J
(iii) Deduce that for 0 < s < k, kU I
u L,(R. 11
fRIhI'-s+n
) dh
f
00
[COk(t' u)]2 t2s+)
dt
00
r l221s[cok(2-j, u)]2. j==-oo
3.14 Let Xj E D(]R") be as in part (v) of Exercise 3.8, and lets E R. Show that if u E Hs(]R"), then Xju -3 u in Hs(IR") 3.15 Consider a distribution u E V* (0) with supp u c K C= Q. (i) Show that there is a unique u E D*(]R") satisfying It = u on 0, and u = 0 on 1R" \ K. [Hint: use a cutoff function X E D(Q) with X = 1 on K.] (ii) Show that if s E ]R and u E Hs (S2), then II u II Ha (Ruu) < Cs. K II u II Hs (n)
3.16 Give an alternative proof of Theorem 3.20 for 0 = R", as follows.
Exercises
11 l
(i) Prove Peetre's inequality:
fore, n E R' ands E R.
(1+1e12)s < 21s'(1+11; _1 I2)1sl(1+InI2)s
(ii) Use the relation 4u =
* u to deduce that if u E Hs(R'), then
Ou E Hs(R") with II0UIIH.'(Rn) < CsIIUIIH.(Rf), where Cs = 21sJ12
f
(1 + I
de.
3.17 Let * and *' be as in (3.9) and (3.10). (i) Show that if u E D* (RI), then the convolution uE = Vrr * u belongs to £ (Rn) and
supp uE c {x E R" : dist(x, supp u) < E}. (ii) Lets E R, and show that if u E Hs (R71) then and
IIUEIIH-(R'I) <_ IIuIIH-I(R'I)
l o IIuE - UII HS(R.,) = 0.
3.18 Show that if S2 is the crack domain shown in Figure 2(ii), then D(S2) is not dense in HI (Q) for s > n/2. 3.19 Let U E V* (Q) and S E (i) Suppose W is an open set and X E E) (W). Show that if u E HI (w n 0), then XU E HI (0) and 11Xu1IH,(92) < Cx,s1Iu1IHs(wns ).
(ii) Suppose (¢i )
is a partition of unity of the type used in the proof of Theorem 3.29. Show that u E H-'(9) if and only if 95i u E Hs (Wi n o)
for 0 < j < J, in which case I IIUIIH=(n)
^' L .i=o
3.20 Prove the following inequalities, due to Hardy: for a > 0 and 1 < p < oo,
dY\ dx LJ'(x-" 0
Jo
X
If(Y)I Y
I/P
<
x
a
f
IY-"f(Y)IP
dy1'lP
o
JJ
and
00( CJo
Cx",1
d
(oo
If(Y)I
PdxlhhP
y) x
J
1(foo < a
d
1/P
.
ly"f(Y)IP y
l
[Hint: make the substitution y = xt in the inner integral, and then apply Minkowski's inequality, i.e., think of t H f (- t) as a map from (0, 1) or (1, oo) into a weighted LP space on (0, oo).]
Sobolev Spaces
112
3.21 Let1
fb`
p
p x-a f a If(t)I dtl/ dx < ( p- 1
b
p
If(t)Ipdt,
I
Ja
1
f (b -x , If(t)I dt)pdx < \
u
p
/paIbIf(t)Ipdt
.p-
[Hint: use Exercise 3.20.] (ii) Show that, for u E D(R2), fffb a
fblu(x,x)-u(Y,Y) p('Jdx)(-v/-2 dy)
x-Y
a
(p2p 1
)'f'f
Y)I P+Iazu(x,Y)IP]dxdy.
u(x, x) - u(y, y) = fx a2u(x, t) dt + f: aIu(t, y) dt y <x.] [Hint:
for
(iii) Let yu(x') = u(x', 0) for x' E R"-1 and u E D(IR"). Show that
ff , -,
IYu(x') - Yu(Y')I p dx'dy' < C,,,, Ix' - y' I p
J
i=
R
18fu(x)Ip dx.
(iv) Show that, for u E D(IR"), IIYUIIL,,(W'-1) < CIIUIIW;(R")
8 (X u) (x) dx for a suitable function
[Hint: write y u (x') X(X") .1
(v) Deduce that y : WW(R")
for each integer k >"0.I.
3.22 Consider the half space Q = {x E R,1 : x" < 01. Let U E D(S2), and define U(X)
U (X) =
if x" < 0,
0
(i) Show that Ck."(1 + It'I)-k(1 + 14"1) - I for every k > 0. (ii) Hence show that U E Hi = HS (S2) for s < 1
(iii) Show that if a,u(x', 0) = 0 for 0 < k < j, then U E Hr (S2) for s < j + 3 .
4
Strongly Elliptic Systems
We are now ready to begin our study of boundary value problems for linear elliptic systems of second-order partial differential equations. The first task is to explain how, via the first Green identity, such problems fit into the abstract scheme treated in Chapter 2. We then define the class of strongly elliptic operators, and investigate when such operators are coercive. After that, an existence and uniqueness theorem for weak solutions in H' (S2)" is given, expressed in the form of the Fredholm alternative. Next, we prove regularity of the solution on the interior and up to the boundary, under appropriate assumptions on the data and the domain. We also prove the transmission property, which will be used later to show regularity at the boundary of surface potentials for smooth domains. The final section of the chapter presents some rather technical estimates relating the H1-norm of the trace and the L2-norm of the conormal of
derivative. These estimates will allow us to prove some limited regularity of surface potentials for Lipschitz domains.
The First and Second Green Identities Suppose that Q is a non-empty open, possibly unbounded subset of ll8", and consider a linear second-order partial differential operator P of the form n
it
It
Pu=->>8J(Ajkaku)+j:AjBJu+Au onc2,
(4.1)
j=1
j=1 k=1
where the coefficients
Ajk = [apgJ,
AJ
= `a',q],
A = [apgl
are functions from S2 into C""", the space of complex m x m matrices. Thus,
I < p < m and 1 < q < m, and P acts on a (column) vector-valued function 113
Strongly Elliptic Systems
114
U : Q -- C'" to give a vector-valued function Pu : cZ -,, Cm, whose compo-
nents are M
M
(POP
- L+ L Lr nn
nn
aJ (ap4aku4) +
j=1 k=1 R=1
nn
j=1 q=1
M
ap9u9
ap9aju4 + 9=1
fort
r
12
D(u, V) =
J
(
n
n
n
E E(Ajkaku)*ajv + J:(Ajaju)*v + (Au)*v dz.
(4.2)
j=1
j=1 k=1
(Recall that * denotes the conjugate transpose of a matrix or vector.) It will always be assumed that the coefficients A jk, A j and A belong to L,"' (S )"'X'n, so that 0 is bounded on HI (Q)m: 14) (u, v)I
_
CIIu1IH1(92)'" IIvIIH- (a)
foru, V E HI(S2)m.
If, in addition, the leading coefficients A jk are Lipschitz, then P : H2(Q)"' -± L2 (2)' is a bounded linear operator. When the lower-order terms are dropped from P, we are left with the principal part Po, which can be written in divergence form as n
Pour
where 13ju = EAjkaku. j=1
(4.3)
k=1
If 0 is a Lipschitz domain, then the conormal derivative
>2 vjy(Sju) on I',
is defined by (4.4)
j=1
where, as usual, v is the outward unit normal to 0, y is the trace operator for 0, and r = 8S2 is the boundary of Q. The conormal derivative arises naturally via the following lemma, known as the first Green identity. Lemma 4.1 If S2 is a Lipschitz domain, and if the coefficients Ajk are Lipschitz, then
1'(u, v) = (Pu, v)o +
yv)r
foru E H2 (0)"' and V E HI(S2)m
The First and Second Green Identities
115
Proof. By the divergence theorem (Theorem 3.34), if w E C.IpmP(S2) then
I 8jwdx = Jr
vj wda,
and one sees with the help of the density and trace results in Theorems 3.29 and 3.38 that this formula holds in fact for any w E H 1(S2). Taking w = (,t3 j u)*v, we obtain n
a [(B;u)*v]dx = fr vjy[(Bju)*v]dc
J
for U E H2(0)"' and V E H (SZ)n',
and the result follows after summing over j, because by (4.3), n
n
aj[(Bju)*v] = -(Pou)*v + 1:(Bju)*ajv. j=1
J=1
0
In order to state a dual version of Lemma 4.1., we define It
Bju = EAkjaku+A*u, k=1
and put n
n
n
aku) - Eaj(A*u)+A*u
P*u = j=1 k=1
j=1
n
_-E8jBju+A*u on S2, j=1 and
_ B vu =
n
E vjy(Bju) on P.
(4.5)
j=1
In fact, by arguing as before, but now with w = u*B j v, one easily verifies the following identity. Lemma 4.2 If c2 is a Lipschitz domain, and if the coefficients A jk and A j are Lipschitz, then
4 (u, v) = (u, P*v)n + (Yu, Bvv)r
foru E H1(S2)m and V E H2(c2)n'.
Strongly Elliptic Systems
116
Thus,
(Pu, v)n =' (u, v) = (u, P*v)Q because yv =
for U E Hz(S2)'" and V E D(S2)"',
0 on T. Hence, if u E H1(S2)"', then we can define Pu
as a distribution on S2 by
(Pu, v)n = t (u, v) for v E D(c2)'", even if the coefficients Ajk are not Lipschitz, but only belong to Loo(S2)''"Likewise,
we can define the distribution P*u for any u E H 1 (S2)' by
(P*u, v)u = (D*(u, v)
for u E D(S2)'".
The operator P* is called the formal adjoint of P. Its principal part is given by n
n
!t
(P*)ou=-E1: aj(A*
j=1
j=1 k=1
which coincides with the formal adjoint of Po, allowing us to write
Po = (P*)o = (Po)*. However, the conormal derivative of u relative to P* is n
n
vjy(Axjaku) = 9,u -
vjy(A!u), j=1
j=1
which coincides with I3,u if and only if A j = 0 for all j. One says that P is formally self-adjoint if P* = P, i.e., if the coefficients of P satisfy
A* = Ajk, In this case, l3
Aj = O,
A* = A.
(4.6)
13, and the sesquilinear form (4.2) is Hermitian, i.e.,
(D(v, u) _ 4(u, v)
for u, v E H'(SZ)"'.
The next lemma will allow us to extend the definition of the conormal derivative.
Lemma 4.3 Suppose that S2 is a Lipschitz domain. If U E H1(S2)m and f E H-I (Q)m satisfy
Pu = f
on 0,
The First and Second Green Identities
117
then there exists g E H-1/2(r)'" such that
0(u, v) = (f, v)n + (g, Yv)r for v E
HL(S2)m.
Furthermore, g is uniquely determined by u and f, and we have IIgIIH-"2(r)" < C
IIUIIH'(2)1° + C11f 11x-I(ny
Proof By Theorem 3.37, there exists a bounded linear operator n : H 1 /2 (r)^' HI(E2)1 satisfying v for all v E H'/2(r)"'. Since [Hl(S2)"']* _ H-' (S2)"' and [Hh/2(r)"']* = H-1/2(11)"', we can define g E H-'/2(11)'" by
(g, w)r = 0 (u, 77 w) - (f, r)w)a for w E Ht/2(11)'". Given V E H1(c2)"', consider the function vo = v - 77y v. Since yvo = 0, we have vo E Ho (S2)'" by Theorem 3.40, so there is a sequence o in E)(0)"' that converges to vo in H 1 (0)1. Hence, using the fact that Pu = f on Q,
c(u, vo) = h
(D(u, 0;) =1lim (f, 0,i)s2 = (f, vo)Q,
and by the definition of g,
1(u, v) = (D(u, vo + )7Yv) = (f, vo)n + (g, Yv)r + (f, r1Yv)o
= (f, On + (g, Yv)r, as required. Finally, if g, and $2 both satisfy the conclusions of the lemma, then
for the difference g, - $2 E H-1/2(11)"' we have (g, - 92, yv)r = 0 for all V E Hl (2)'", and therefore (g, - $2, w)r = O for all w E H1/2(11)'", implying
91 = g2 Note that g is not uniquely determined by u alone, but depends on the choice off . The problem is that we could have Pu = f, = f2 on S2, with the difference f, - f2 a non-zero distribution on R" having support in F. However, provided f is clear from the context, we shall write g and call this distribution
the conormal derivative of u. In particular, if Pu E L2(0)"', then we always define B ,u by making the natural choice Pat
f - {0
on S2,
on R" \S2,
thereby ensuring consistency with the original definition of the conormal deriva-
tive. We extend the definition of 13,u in the same fashion, with the help of Lemma 4.2. The next theorem follows at once; cf. (1.8) and (1.9).
Strongly Elliptic Systems
118
Theorem 4.4 Let 0 be a Lipschitz domain, and suppose that u, v E H I (Q)n'.
(i) If Pu E L2 MY', then the first Green identity holds:
c(u, v) _ (Pu, v)n + (13u, Yv)r (ii) If P*v E L2(S2)"', then
c(u, v) _ (u, P*v)Q + (Yu, B
v)r
(iii) If both Pu and P*v belong to L2(S2), then the second Green identity holds:
(Pu, On - (u, P*v)st = (Yu,13,,v)r - (13,u, Yv)r Strongly Elliptic Operators Let V be a closed subspace of H1(Q)"', such that V is dense in L2(0)"'. Following the terminology established in Chapter 2, we say that (D and P are coercive on V if Re 4) (u, u) > cIIUIIHI(n)s.
- CIIuIIi,(0),0I
for U E V.
Obviously, in this context L2 (S2) acts as the pivot space for V. When seeking to determine whether or not a given differential operator is coercive, we can ignore the lower-order terms, and consider just the sesquilinear form corresponding to the principal part, n
4>0(U, v) =
E(Ajkaku)*8jvdx.
Jj=.1
k=1
Lemma 4.5 The differential operator P is coercive on V if and only if its principal part Po is coercive on V.
Proof. For any 6 > 0, we have I'D (u, v) -'Do(u, v)I -< C11U11HI (n),,, IIv11L,(n)" < C(EIIU1121
so if Po is coercive, i.e., if Re co (u, u) > c II u 11 H
+E-'IIVI1L2(s2)
Al" - C II u II L2(2)"' , then
Re4D(u,u) > and by choosing E sufficiently small, we see that P is coercive. The converse is proved in the same way. 0
Strongly Elliptic Operators
119
The differential operator P is said to be strongly elliptic on S2 if "
It
Re>2>2 [Ajk(x)
for all x E 0, l; E R" and >) E C"'.
k 1 *Sj77 >
j=1 k=1
(4.7)
Depending on the subspace V and the regularity of S2, this purely algebraic condition on the leading coefficients is often necessary and sufficient for P to be coercive.
Theorem 4.6 Assume that the coefficients Ajk are (bounded and) uniformly continuous on Q. The differential operator P is strongly elliptic if and only if it is coercive on Ho (S2)1'1.
Proof. Suppose that P is strongly elliptic. First we consider the special case when the leading coefficients A jk are constant. Let U E Ho (S2)'" = H' (S2)"',
i.e., let u E H1(R")"' with supp C S2. Since .F,,k[aju(x)} Plancherel's theorem implies that
f (Ajkaku)*aju dx = (27r )2 R11
)J* ju(S) dS,
JR"
so, taking ri = u(l) in (4.7), Re(Do(u, u)
(27r
)2
f
c>2
g
f. Iajul2dx
j=1
CIIUI12'(2)u' -CIIUIIL2(s2)"
By Lemma 4.5, we see that P is coercive on Ho (S2)"' To handle the general case, let c > 0 and choose 8 > 0 such that
max IA jk(x) - Ajk(Y)I < E j,k
for Ix - yI < S.
(4.8)
Cover SZ with a locally finite family of open balls B1, B2, B3, ... , each of radius S. (If 0 is bounded, then the family of balls will be finite.) Since the diameters of the balls are bounded away from zero, we can assume that for each d > 0, there is a number Nd such that any given set of diameter less than d intersects at most Nd balls. By Corollary 3.22 and Exercise 4.6, we can find real-valued functions 01, 02, 03, ... in C,1,,mp(W) with 01 > 0 and supp,0) c B1, such that 01(x)2 = 1, 1>1
>2cbi(x) < C, and >2I8j0,(x)I < C, 1?1
!?1
forx E S2.
Strongly Elliptic Systems
120
Note that the number of non-zero terms in each sum is finite, and is bounded independently of x. Since [Ajkak(4lu))*aj(Oly) = 01 (Ajkaku)*ajll + (akOl)0!(Ajku)*ajv + O1(aj0l)(Ajkaku)*v + (akOl)(ajO,)(Ajku)*v, we have 0o(0lu, 01u) < CIIuIfHI(s2)'n IIu1IHI(2)n, 1> 1
and also Re (Do(u, u)
>
Re4 o(01u, 01u) - CIIu1IHI(n)n' IIuIIL,(9),1 . 1>1
Let fio denote the sesquilinear form obtained from (Do by freezing the coefficients Ajk (x) at x = x1, the centre of the ball BI, and observe that
4'o(0tu, 01u) - Co(01u, 0tu) = f {[Ajk(x)
- Ajk(x!)Jak(01u)}*aj(olu) dx.
From the special case considered earlier, we know that (Do is coercive on Ho (S2),
with constants independent of 1, and by (4.8),
IAjk(x) - Ajk(x!)I < E forx E BI, so
Re to(Otu, 01u)
Re (Do(01u, 01u)
-
6110,UI12
> (c -E)IIfiluIIHa(n),n -CII01uIIL,(n)"1, and
Re Do (u,u)> 1>1
-C
1101U112
),n -
CIIuIIHo(S2)n, IIu11L2(szyn
1>1
Since the 0,2 form a partition of unity,
II0IUIIL2(n) = f E0l(x)2lu(x)I2dx = IIUIIL2(n),n !>i
l>1
Strongly Elliptic Operators
121
and
Il018;u + (a;o,)ull2L2(),H IIajul1L2(n)N, - C11
uIIHo(Q),,,11u11L2(n)"'-
Using the inequality ab < (E'a2 + b2/E'), we see that 2
/
Re(Do(u, u) > (c - E - E')IIUIIHo(n)m - Cf l +
1
E,
)IIU112,(si)u'
so P is coercive on Ho (S2).
To prove the converse, take a real-valued cutoff function >/r E C mp(R") satisfying
*>0onR", Jr=0forlxl>1,
*(x)2dx=1. fRII
Let xo E 0, and put *E (x) _ E-"j21/r(E-l (x small,
- xo)), so that for E sufficiently
E C mp(S2) with
r 'VE >
0 for Ix - xol > e,
0 on n,
J
/rE(x)2 dx = 1.
Thus, '1/!E (x)2 converges to 8 (x - x0) as c 4. 0. Consider the function uE(x) _ 1E(x)e`t.XI?. Since
IIUEIlL2(n)"' = 1171,
and ajtfE = E-'(aj*)E, we have
and since ajuE =
Now,
(AjkakuE)*ajuE =
'YE (Arjk
+i1IE(ak /E)(Ajk17)* J
-',,/E(aj'YE)(Ajk k?I)*17 + (ak
so if we define
A'k
=
f1/Ie(x)2Ajk(x)dx,
t
E)(aj*E)(AjkY1 !)*TI,
Strongly Elliptic Systems
122 then
ReE
C(E-2
j11 > ReIo(uE, uE) -
j=1 k=1
If we now assume that P is coercive on Ho (0)', then Re'Do(uE, uE) _> c11 uEII HacWN - CIIuEllc2(si),,, >-
c(E-2 +
CIrjJ2,
implying that n
Re
C(l + E-2 +
kr1)* j17 ?
it
j=1 k=1
Now replace by tl; where t > 0, divide through by t2, and send t -+ 00 to obtain (4.7) with A)k in place of A,k. Since Ask -+ Ajk(xo) as c y 0, we conclude that P is strongly elliptic. 0
For scalar problems, i.e., when m = 1, the strong ellipticity condition (4.7) simplifies to n
Re
1: 1: Ajk(x)44k4'j > it
for allx E S2 and
E R",
j=1 k=1
and the next result is usually sufficient for establishing that the differential operator is coercive on the whole of H 1(S2), not just on Ho (S2); cf. Exercise 4.1.
Theorem 4.7 Assume that P has scalar coefficients (i.e., m = 1), and that P is strongly elliptic on 0. If the leading coefficients satisfy Ak j = A jk
for all j and k,
on S2,
then P is coercive on H 1(S2).
Proof Define F : 0 x C" --* C by !Y
n
F(x,
Ajk(x) k `;j. j=1 k=1
The symmetry condition on the leading coefficients implies that
F(x, + ir1) = F(x, ) + F(x, >)) for x,17 E R",
Strongly Elliptic Operators
123
and so by strong ellipticity, Re F(x, l;) > cI:;12 for all l; E C" (not just for L"). Hence, for all u E H' (S2),
E
cgradull?.().
Re (DO (u, u) = j Re F(x, gradu) dx z
El
In a similar fashion, when m > lit is easy to see that P is coercive on H' (S2)" if Akj = Ask and n
n
n
E[A,k(x)k]*;
Re
j=1 k=I
>c
lei l2
j=1
for all x E cZ and 1, ... , l; E R'",
(4.9)
but this assumption excludes some strongly elliptic operators with important applications; see Exercise 10.3. Using an approach due to Ne6as [72, pp. 187-195], we shall prove a sufficient condition for coercivity on H' (S2)"` in the case when the leading coefficients can be split into sums of Hermitian rank-1 matrices, i.e.,
t Ajk =
brj r=1
where the blj are (column) vectors in C". It follows that Po must be formally self-adjoint, and that L
(Do(u, v) =
J
2
It
uNvdx
where Nu = E b* 8j u.
(4.10)
J=I
1=1
Note that the first-order differential operator N acts on a vector-valued function u to produce a scalar-valued function Nu, and that L
b0(u, u) =
QNulli2cn) >_0
for u E H' (p)'".
r=1
An important example of a strongly elliptic operator of this type is described in Chapter 10; see in particular Theorem 10.2. The proof of coercivity is based on the following technical lemma, whose proof turns out to be surprisingly difficult.
Strongly Elliptic Systems
124
Lemma 4.8 If l is a Lipschitz domain, then for any integers p > 0 and q > 1, and for any u E D(S2), IIuIIH-P(s2) <- CIIUIIH-1-q(92) + C T, 11 VU 11 H-1-11 (9) IctIsq
Proof. Since D(S2) is dense in HS (S2) if s < 0, it suffices to consider u E D(S2). Plancherel's theorem gives
C
f
m
(i \ + lai=q Ir
(1 +
2 <- C I I u I I H
l2)Iu(t)l2d
C E II a°` u II H-n-a (R") , IaI=q
so by Exercise 3.15, IIuIIH-1(n) <- CK (lullH_P (n) + 57, II a" u II H
(lz)
ifsuppu c KC=S2.
Ial=q
In doing away with the dependence on K, we can assume that 7 is a Lipschitz hypograph, given by x < (x'). Our strategy is to make a change of variable x = x(y), with x E 92 and y in the negative half space R.
For E > 0, let
'NE be as in (3.9) and (3.10), introduce the C°° function f (y', c) = (`YE * ) (y'), and define
KE (Y) = (Y', f (Y', -Ey) + y,1) Since grad
E L,,. (R"- 1), we find that a
(
for y < 0.
(ayn-1)«,,-, (aE
I ) «I ...
ay,
C )a,r f (Y', E) < EIaI-1
for laI > 1.
(4.11)
Thus, a,(KE)n(y) = 1 - Ea. f (y', -Ey11) = 1 + O(E), and we now fix c small enough so that c < 8,1(KE),1(Y) < C
for y < 0,
and write K = KE. In this way, K (y) is a strictly increasing function of y, E (-oo, 0), with K. (y) f ' (y') as y71 T 0, and so K : lR" - 0 is a C°O diffeomorphism, and it can be shown using (4.11) that Ia"K(Y)I <
CI-1 IYn l
for lal > 1.
Strongly Elliptic Operators
125
In the substitution x = K(y), we have x' = y', so the Jacobian is simply
detDK(y) =
K. (y).
Also, by differentiating the equation xn = f (x', one sees that ay,
-ajf(x', -Eyn)
axj
1 - Eanf (x', -Eyn)
yn with respect to x,
and by differentiating with respect to x,,, one sees that ay" ax,,
1
11 -Eanf (x', -Eyn)I
The higher-order partial derivatives of K giving
<_ C.
can be estimated in a similar fashion,
afor lal > 1.
aaK-1(x)
lyn l
Let us now show that CIIuIIH-P-,(f2),
11U 0
(4.12) Ilaj(u OK) IIH-,, (J^) <
CllajuHHH-P-,(n) +
Cllanu!IH-P-I(-2).
For 0 E D(R" ), (u o K, 0)w,h I =
I
fn u(x)(cp o K-1)(x) det
DK-1(x)
dx
(g)II(0oK-1)detDK-1IIH,+i(n)+
aju(x) +
for 1 < j
and since
aj(u O K)(y) = t anU(X)a,, K. (Y)
for
f
< n-
1,
- n,
we have
(aj(u o K), O),,
CllajullH-n-'(s2)II(0 o K-1) detDK-1lIHP+i(O) +Clla,,UIlH-r-1(S2)II(4ajKn) oK-1 detDK-1IIH,,+l(s2),
where only the second term on the right is present if j = n. With the help of Theorem 3.33 and Exercise 4.3, we find that both II(10o K-1) detDK-11IH+,+t(n)
and
II (0ajK,,) o K-1 detDK-1IIHp+i(S2)
Strongly Elliptic Systems
126
are bounded by IYnl2(jal-p-1)Iact 5 (Y)12dY
C lal
implying that (4.12) holds. Also, if q5 E V(S2) then I (u o K, (0 o K) det DK)R
(u,
< C II u o K II H-p(R"-) II (0 0 K) det DK 11 ft,(R") and
II(,OOK)detDKpHP(w) < C
IaIp f
x')
I
2
Ho
CII0II2 SO 11 u 11 H-1(1z) -< C II u o K II H-p (RID.
(S2),
Hence, using the Seeley extension operator E
from Exercise A.3, we have IIUIIH-p(n) < CIIE(u o K)IIH-p(R,n) n
< C II E (u 0 K) II
C:II aj E(u 0 K) 11
H-p-I
H-p-I (Ra )
j=1 n
C E 1181(u 0 K) II H-p (Z)
< C II u 0
j=1 n
<- CIIullH-p(Q)+C11aju11H-p m), j=1
which proves the result when q = 1. The general case now follows by induction D on q.
Theorem 4.9 Assume that 0 is a Lipschitz domain and that (Do has the form (4.10), and put n
forl
The operator P is coercive on H 1(0)'n if, for 1 < r < m, there is an integer q,- > 1 such that, for every multi-index a with J a I = q,. + 1, there exist polynomials Q1, ... , QL, each homogeneous of degree qr (with scalar coefficients), satisfying L
L 1=1
taer for all E Rn,
Strongly Elliptic Operators
127
where er is the rth standard basis vector in C'. (Note that Ql depends on r and a.) Proof. Let U E D(SZ)"' and 0 E D(S2), and let Q1 be the partial differential operator corresponding to Q1(4) in the same way that N corresponds to N1( ); thus, under Fourier transformation,
and T
fix- 4{Nu(x)} = Nl(i2n1)u()
Suppose P and 8 are multi-indices with IPI = I and 181 = qr. Applying Plancherel's theorem, and taking a = p + 8 and ur = e,**, u, we have
(81ur, aa4')n = (d ur, 810) = f lq.-1(2n L
fN
t(i2rr )u( )Q,(i2
"
L
dr;
)d
L
_ >(Nu, Q10) _ 1=1
>(N1 u, Q10)a, 1=1
so L
I (anal ur, 4'Tl <_ C L IINruIIL?(n)IIQ14IILZ(sa) 1=1
L
< C:
II L2 (sa)11011,9q'. (a)"
1=1
implying that L
IlNullf,(la) = Ccpo(u, u).
C 1=1
Thus, by applying Lemma 4.8 with p = 0, IlalurllLA(sa)
C11aflU,112
E 11aaaflU,112 I«I :sq,
Ila"a#urll H v,(n) +c II ur II L,(Q)
+ C4)o(u, u),
and we have only to sum over 10 1 = I and 1 < r < m .
128
Strongly Elliptic Systems
Boundary Value Problems In this section, we shall see how the Fredholm alternative allows us to answer fundamental questions of existence and uniqueness for the solutions of elliptic boundary value problems. Suppose that the boundary of the domain S2 has a Lipschitz dissection
r= ID UIIUFN, and that boundary conditions of Dirichlet and Neumann type are specified on I'D and FN, respectively. Thus, our task is to find u E H' (S2)" satisfying
Pu = f on S2, (4.13)
YU = gD on FD,
l3vu = gN on rN,
for given f, gD and gN. Using our definitions of Pu and B,u as distributions, we see that (4.13) is equivalent to
yu = gD on rD and c(u, u) = (f, v)D + (gN, yv)rN
for v E HD(S2)"', (4.14)
where
HD(S2)"' _ {v E H' (Q)' : yv = 0 on rD}. Recalling Theorem 2.27, we expect that if there exist non-trivial solutions to the homogeneous problem
Pu = 0
on S2,
yu=0
onrD,
0
on FN,
(4.15)
then we ought to consider also the homogeneous adjoint problem,
P*v=0
on S2,
yv = 0
on rD,
O
on rN.
(4.16)
Theorem 4.10 Assume that 0 is a bounded Lipschitz domain, and that P is coer-
cive on HAM)"'. Let f E H-' ( 7)m, gD E HI/2(rD)m and gN E H- 1/2 (r,4 yn, and let W denote the set of solutions in H' (S2)' to the homogeneous problem (4.15). There are two mutually exclusive possibilities:
Boundary Value Problems
129
(i) The homogeneous problem has only the trivial solution, i.e., W = (0). In this case, the homogeneous adjoint problem (4.16) also has only the trivial solution in H 1(S2)'", and for the inhomogeneous problem (4.13) we get a unique solution u E H 1(S2)". Moreover, CII f
CH
DIIH'1'tro)- + CIIgNIIH-""2(rN)^'
(ii) The homogeneous problem has exactly p linearly independent solutions, i.e., dim W = p, for some finite p > 1. In this case, the homogeneous adjoint problem (4.16) also has exactly p linearly independent solutions, say v1, ... , VP E H 1(Sl)"', and the inhomogeneous problem (4.13) is solvable in H 1(0)'" if and only if
for I < j < p.
(v1, f )Q + (Yv1, 9N)rN = (Evv1, 9D)rp
When the data satisfy these p conditions, the solution set is u + W, where u is any particular solution. Moreover
IIu+WIIH'(i)"lw
Proof. Put V = HD(S2)"' and H = L2(S2)"', and consider the operator A : V -+ V * determined by (D in the usual way. We note that since S2 is bound-
ed, Theorem 3.27 applies, so the inclusion V C H is compact. Hence, by Theorem 2.34, A is Fredholm with index 0. Furthermore, each distribution fo E H-1 (0) gives rise to a unique functional Fo E V*, defined by (Fo, v) _ (fo, v)n for v E V (although different fo's can give the same F0 if r'D 0 0). Thus, for uo E V and fo E H-1(Q), the equation Auo = F0 is equivalent to
forvEV.
yuo=OonFD and 4)(uo,v)=(fo,v)n
(4.17)
To handle the inhomogeneous Dirichlet condition in (4.13), we write u = H1/2(r')n, and : H1/2(r),n _+ H1(S2)m uo + iEgD, where E : H1/2(rD)"' --+ are extension operators constructed with the help of Theorems A.4 and 3.37. 1
In this way, yu = gD on rD if and only if yuo= 0 on r'D. Choose any extension gN E H-1/2(r')"' of gN, and define fo ER -1 (0)"' by
(fo, v)c = (f, v)n - ID(r7EgD, v) + (gN, Yv)r
for v E H1(S2)'n,
so that (4.14) is equivalent to (4.17), or in other words, so that (4.13) is equivalent
Strongly Elliptic Systems
130
to the equation Auo = FO. We remark that uo can be thought of as a solution of
Puo = fi
on Z,
yuo = 0
on I'D,
13vuo = 81
on FN,
where f, = f - PiEgD and gi = gN -13vri EgD. Notice that for v E V, I(Fo, v) I = I(fo, v)r21 < C11f
11N-I(Q)
IIvjjHi(n)'" + CIIrjEBDIIH'(52)"" IIviiW(s2)",
+ CII8NIIH-'r(rN)- IIYvIIH1/?(rN),"
< C(11f IIH-x(52)" +
II9DIIH"/2(rD)" +
II8N11H-./2(rN)"')IIviIH'(n)'n
so
IIFoIIv- <
C11f11X-'(s2)"'
+CIIBDIIH'/'-(FD)"" +CIIBNIIH-1/'-(rN)"'
The homogeneous problem (4.15) is equivalent to Au = 0, so W = ker A. Likewise, the homogeneous adjoint problem (4.16) is equivalent to
yv = 0 on FD and 4)(u, v) = 0
for U E V,
which in turn is equivalent to A*v = 0. Thus, in case (i) there exists a unique uo = A-' FO, and hence also a unique u = uo + r?EgD, with IIuiIH'(s2)"H < IIuoIIv + IInEgD11H'(s2)"' -< CIIF0IIv + CI18DIIH'/'-(r0)-,.
In case (ii), the equation Auo = FO is solvable if and only if (Fo, vj) = 0 for 1 < j < p. Since the first Green identity gives 4)(?7EgD, vj) = (rlEgD, P*vj)c + (Yr1EgD, Evvj)r = (8D, 13vvj)rD,
and since (RN, yvl)r = (gN, yvj)rN, it follows that (4.13) is solvable if and only if
(f, vj)n - (8D, kvj)ro + (8N, Yvj)rN = 0
for 1 < j < p.
When these p conditions are satisfied, Il uo + W ll v/ w < C II Fo II v , and by choosing w E W such that Il uo + w II v = II uo + W II v/ w, we have
IIu+whlH'(s2)-" < Iluo+wlly+IIlEBDIIHI(r2)-' :S CIIFolfv*+CIlgDIIHI/2(rD)",,
giving the required estimate for IIu + W II H 1(Q)m/ w
0
Boundary Value Problems
131
Note that for a pure Dirichlet problem, i.e., when I, = I'D, any strongly elliptic operator with leading coefficients in W,,,.(Q)mxm is coercive on V = HD02)"' = Ho (S2)"' by Theorem 4.6. Also, we may take f E V* = H- I (Q)1n, because no use is made of the conormal derivative. For a pure Neumann problem, i.e., when r = FN, the proof of the theorem above is greatly simplified because
u=uo E V=H'(U)"' andFo= fo EV*=H-'(S2)"` Next in importance after the Dirichlet and Neumann problems is the third boundary value problem:
Pu = f
on S2,
=g
on F.
(4.18)
Here, the coefficient B is a known function (matrix-valued, if m > 1). The third boundary value problem reduces to a pure Neumann problem if B is identically zero. In fact, for non-zero B, (4.18) is only a compact perturbation of the Neumann problem, and we can state the Fredholm alternative in terms of the homogeneous problem
Pu=O B,u+Byu =0
on S2,
(4.19) on IF,
and the homogeneous adjoint problem
P*v=0
on S2,
(4.20)
on F, as follows.
Theorem 4.11 Assume that 0 is a bounded Lipschitz domain, that the difLet ferential operator P is coercive on H' (0)m, and that B E
f E H-'(S2)"' and g E H-'l/2(F)m, and let W denote the set of solutions in H' (c2)"' to the homogeneous problem (4.19). There are two mutually exclusive possibilities.
(i) The homogeneous problem has only the trivial solution, i.e., W = (0). In this case, the homogeneous adjoint problem (4.20) also has only the trivial solution in H' (0)"', -and for the inhomogeneous problem (4.18) we get a unique solution u E H' (S2)". Moreover, IIUIIa (2)" < CII.f
CIIgIIH-w(r)-.
Strongly Elliptic Systems
132
(ii) The homogeneous problem has exactly p linearly independent solutions, i. e., dim W = p, for some finite p > 1. In this case, the homogeneous adjoint problem (4.20) also has exactly p linearly independent solutions, say U1, ... , up E H' (S2)'", and the inhomogeneous problem (4.18) is solvable in H' (0)"' if and only if
(vi, f)n + (Y vi, g)r = 0 for 1 < j < p. When the data satisfy these p conditions, the solution set is u + W, where u is any particular solution. Moreover IIu + WII H'(S2)"'/w < Cll f 111,-i(ny" + CIIgIIH-"(r)
Proof Put (Au, v)S2 = fi(u, v) and (Ku, v)Q = (Byu, yv)r for u, v E H' (0)"'. The operator A : H' (S2)"' - . H-' (S2)' is bounded, and for any 0 <
E <'
I (Ku, v)!al <_
so K : H 1/2+E (S2)'"
C II uII H1r_+f(sa)'" IIv11H'(a),",
H-' (0)"' is bounded and hence K: H' (S2)'"
H-' (S2)'" is compact. Since A is coercive on H' (S2)'", it follows that A + K : H' (S2)"' -* H-' (S2)'" is a Fredholm operator with index 0. By the first Green identity, a function u E H' (0)"' is a solution of (4.18) if and only if
((A + K)u, v)n = (.f, On +
Yv)r + (Byu, Yv)r,
so (A + K)u = F where (F, v)o = (f, v)n + (g, yv)r. The estimate I(g, Yv)rI < CIIg11H-'/'--(r)I" IIY-IIH""2(r)_< C11811H-"12(r)IlullH'(a)"
shows that F E F1 -I (S2)'", so the results follow directly from Theorem 2.27 after noting that, by the dual version of the first Green identity (Lemma 4.2), the adjoint of A + K is given by
(u, (A + K)*v)Q = (D(u, v) + (Byu, Yv)r
= (u, P*v)n + (yu,
B*yv)r,
assuming in the usual way that P*v is a specified distribution in H-' (0)m. We conclude this section with a result on the spectral properties of elliptic operators.
Theorem 4.12 Assume that 0 is a bounded Lipschitz domain, that P is coercive on HD(0)"1, and that BE Lc.(I')""". If P is formally self-adjoint, and if
Regularity of Solutions
133
B* = B, then there exist sequences of functions 01, 02, -03, ... in H'(0)'", and o f real numbers A1, A2, A3, ... , having the following properties:
(i) Each 4i is an eigenfilnction of P with eigenvalue Aj:
PC = A;o; on 0, YO/ = 0
on rD,
13,'q5;+Byq;=0
on rN.
(ii) The eigenfunctions 01, 02, 03, ... form a complete orthonormal system in L2(0)"'
(iii) The eigenvalues satisfy -C < Al < A2 < A3 < ... with AJ -+ 00 asj -+ oo. (iv) For u, v E HD(SZ)"', 00 ' (u, v) _ EA!(u, 0j)n(0i, v)n
and
j=1 00
E('; +
C)1(.0;, u)n12.
j=1
Proof Put H = L2(2)and V = HD(2)n', and consider the operator A V -+ V * defined by
(Au, v)n = c(u, v) + C(u, v)cy + (Byu, yv)rN
for u, v E V.
Since A is self-adjoint, as well as positive and bounded below on V for C sufficiently large, the desired results follow at once from Theorem 2.37 and Corollary 2.38.
Regularity of Solutions A key feature of the elliptic equation Pu = f is that, away from the boundary, the solution u is smoother than the right-hand side f. We shall prove this fact, and also a result on regularity up to the boundary, using a method introduced by Nirenberg [78] based on estimates of the lth partial difference quotient,
AI.hu(x) =
u(x+he1)-u(x) h
for l <1
(4.21)
(Here, el denotes that lth standard basis vector for R".) The method of proof relies on the fact that 8; commutes with 't.h for any j and 1, and also on the
Strongly Elliptic Systems
134
next lemma, which allows one to deduce L2-estimates for a,u from uniform L2-estimates for O,,hu. Lemma 4.13 Let 1 < l < n, and for brevity write 1j u II = II U II L,.(Ru)m
(i) Ifa,uEL2(R")'",then II A1,011 < Ilatullforallh E ]R, and I1At,hu-atoll -
0 ash
0.
(ii) If there is a constant M such that II LX t,h u II < M for all h sufficiently small,
then a,u E L2(R")' and Ilatull < M. Proof. To prove part (i), suppose that atu E L2(1R")'". We have i
falu(x + the,) dt,
I!,hu(x) =
so the Cauchy-Schwarz inequality implies that IAI.hu(x)12
-
18,u(x+thet)I2dt. 0
By integrating with respect to x, reversing the order of integration, and making
the substitution y = x + the,, we obtain the estimate IIDt,hull2 < Ilatull'. It atu in L2(IR")' if u is smooth with compact support, so is clear that D,,hu for each c > 0 we choose uE E D(R")"' such that 11a,uE - atull < E. From the estimate
Ilot.hu - atoll < Il ot.h(u - uE)II + IIAI,huE - atuEll + IlalUE - atoll (l/t.huE - atuEll +211atu - a,ull < II Ll.hUE - atuEll +2E, we have lim
II A,,h u - at a ll < 2E, showing that A,,,, u -+ al u in L2 (]R" )'".
To prove part (ii), assume that IIAI,hull < M. By Theorem 2.31, there is
a sequence hj - 0 and a function v E L2(1R")"' such that O1,h,U - v in L2(R")",, i.e.,
lim (A,,hru, 0) = (v, 4)
for each ¢ E L2(1R")'Ja00
If ¢ E D(]R")'", then
-(u,
lira (u, Al,-h;th) = lira (A1,h;u, 46) _ (v, 0),
1>00
J-),00
SO a,u = v E L2(R)'". Part (i) now implies that A,,hu -* a,u in L2(IR")'", and thus ll atu ll = limb->o II AI.hu ll < M.
0
Regularity of Solutions
135
In stating the next lemma, we use the summation convention, i.e., we sum any repeated indices from 1 to n, except where indicated otherwise. The proof involves only routine calculations, and is left to the reader.
Lemma 4.14 Let [P, Q] = PQ - QP denote the commutator of P with Q. (i)
The commutator of P with any pointwise multiplication operator x : u t-+ x u is a differential operator of order 1 (not 2):
[P, x]u = (ajx)(Aju - Ajkaku) - aj[(akx)Ajku]. (ii) The commutator of P with a; is a differential operator of order r + 1 (not
r + 2): r f(a,-PAjk)apaku P
P
-
r=1 P-1
Cr) P
[(ai-PAi)aPaju + (a; -PA)apu].
(iii) The commutator of 1t.h with any pointwise multiplication operator u
x u is given by
[Al.h, X]u = (Al,hX)u( +het). We require two other properties of At,h; again, the proof is left as an exercise for the reader. Lemma 4.15 Assume that supp u c s2 fl (92 - he,) and supp v c 92 fl (92 + het ).
(i) If u, v E L2(92)m, then (Al,hu, v)SZ = -(u, At,-hv)SZ (ii) If u, v E H' (9Z)"' and if the coefficients of P satisfy IAl,hAjkI < C and I L I,h A j I < Con 0, then
I4'(At,hu, v) + 1(u, Lt,-hv)I < We now prove regularity in the interior.
Theorem 4.16 Let 921 and 02 be bounded, open subsets of R", such that 921 C 922, and assume that P is strongly elliptic on 922. For any integer r > 0, if u E H1(922)' and f E Hr (922)"' satisfy
Pu = f on 02,
Strongly Elliptic Systems
136
and if the coefficients of P belong to Cr,l (S22)mx"1, then u E Hr+I(Q I)," and
UUIIHI-+2<_
+Ciif1IHIcn,r"
Proof. Choose a real.-valued cutoff function x E D(522) with x = I on 521. Part (i) of Lemma 4.14 gives
P(XU)=fl
with
IIfi IIL,(sz2)" _< IIXf IIL2(sz2) + CIIulIH'(nl)",
(4.22)
so
(Dsz,(Xu, v) = (ff, v)n2
for V E Ha (S22)"
Thus, by part (ii) of Lemma 4.15 followed by part (i) of Lemma 4.13,
I'sz2(Ai,h(Xu), v)j
I'a2(Xu, Ar,-hv)I +
IIVIIH,(sz,)°-
= I(fl, At.-hv)Q,I + CIIXUIIH'(sz,)" IIVIIH'(O,)°-
(IIfi II+ IIxUIIH'((z2Y')IIvIIH'(p2)" if h is sufficiently small and v has compact support in 522. The operator P is coercive on Ho (02)"' by Theorem 4.6, so by taking v = Ai.h (x u) we obtain the estimate 2
IIAi,h(xu)IIH'(n2)" < CIIAr,h(Xu)IIL,(02) + (IIfi IIL,(sa,) + IIxuIIH'(sz2)-")IIAI,h(Xu)IIH'(sz,)-
Applying the inequality ab <
(Ea2 + E-i b2) with c sufficiently small, we 2
conclude that IIAI.h(xu)IIH(S 2)"
CIIA1,h(xu)IIL2(sz2) + Cll f IIL2+ CIIxuIIH(n,) (4.23)
CIIuIIHl(22) + CII!IiL2(s22)'
Part (ii) of Lemma 4.13 now gives the result in the case r = 0.
Proceeding by induction on r, we let r > I and choose an open set 0' satisfying S21 C= 0' C S21 C= 522. By the induction hypothesis, II u II H'+-(sz; )H < C II u II H' (02)" + C II f II H'- t (sz,) N
,
and by part (ii) of Lemma 4.14,
P(81 u) = f2 on SZi,
(4.24)
Regularity of Solutions
137
with
IIf2IIL2(nI)M ` IIar f
ciiuIIH'+'(2 )m < IIf
IIHr(Q2)nr + C11ullH'(s22)n,.
Since f2 E L2 (0'0"', we can apply the result for r = 0, and deduce that IIa uII
CI1a1 ul1HI(2,),,, + CIIf211L2(2,)n1
CIIUIIH'(f22)n, +CIIfIIHr(2,).n,
so the induction goes through.
Next, we consider regularity up to the boundary. As well as assuming con-
ditions on f and the coefficients of P, we require extra smoothness for the trace or conormal derivative of u (i.e., for the boundary data), and also for the boundary itself. In the proof, we estimate the tangential derivatives of u using difference quotients as above, but another trick is needed to deal with the normal derivatives. Hence the next lemma. Lemma 4.17 If P is strongly elliptic, then each diagonal leading coefficient has a uniformly bounded inverse, or equivalently, I nI < CIAj;ril (no sum over j)
for ij E C. Proof Take l;j = Sir in (4.7), and get Re(Arrri)*>) > cIi]I2. Refer to Figure 3 for an illustration of the sets used in the next theorem.
Theorem 4.18 Let G 1 and G2 be open subsets of IE?, such that G 1 C= G2 and G i intersects the boundary F of a Lipschitz domain 0. Put
SZ1 =GlnP, 02=G2n0, I72=G2nF,
Figure 3. The sets used in Theorem 4.18.
Strongly Elliptic Systems
138
and suppose, for an integer r > 0, that F2 is C'+'- 1. Assume further that P is strongly elliptic on 522, that u E H1(S22)"' and f E Hr (Q2)"' satisfy
Pu = f
on 522,
and that the coefficients of P belong to Cr" 1(S 2),n xm
(i) If yu E Hr+3/2(r2),", then u E Hr+2(S21)'" and IIUIIHr+2(nI)m < CIIUIIH'(si,)n +CIIYUIIH"+a1'-(r2)"
(ii) If P is coercive on H1(S2)m, and if
E H''+1/2(1'2)"', then u E
Hr+2(521)"' and IIu1IHi-}2(si,)nt -< CIIu1IHIM,)N1
Proof After a
+CIIB,,UIIHI+I/2(r2)" +CIIfIIHl-
Cr+1.1 change of coordinates, we can assume that S2 is the half
space xn < 0 (see Exercise 4.2). To prove part (i) in the case r = 0, weassume first that y u = 0 on F2. Choose a cutoff function X E V (G2) satisfying X= 1 on G 1; then X U E Ho (S22), and by arguing as in the first part of the proof of Theorem 4.16 we see that Ila/(xu)IIHI(n2).I, < CIIuIIHI(n,)"' + CII f 42(n2)°'
for 1 < I < n - 1.
We cannot estimate an (X u) in this way, because Al,h (X u) might not belong to Ho (02) if l = n. However, the partial differential equation can be rewritten as
-Ann an u = f3,
(4.25)
where n
f3 = f + (anA"")8,1u + E aj(Ajk3ku) - > Ajaju - Au, j=1
(J.k)0(n.n)
so by Lemma 4.17, 82u11L2(a1)" 11
C n-I Il81ullH-(92,)'»,
C11 f IILZ(sll)N + C11 uII H'(n,)u +
(4.26)
!=1
giving the desired estimate for II U11 H2(S2, )^,
For non-zero yu E H3/2(r2)m, we use Lemma 3.36 to find w E H2 (02)' satisfying y w = y u on F2, and II w II H2($i2)m < C II Y u II H3/2(1'2) n . The differ-
ence u - w E H'(02) "' satisfies P(u - w) = f - Pw on 522, and y (u - w) = 0
Regularity of Solutions
139
on 1`2, so by the argument above,
flu - w1IH2(n,)1,, < CllU - wllHl(n,),,, +C11f -PW11L,(n,)u', and therefore
IIUIIH2(n,)<- CIIUIIH'(n2)"' + CIIf'IIL2(n2)"' + CII1IIH2(n3)1",
completing the proof of part (i) for r = 0.
Proceeding by induction on r, we let r > 1 and choose an open set G' satisfying G1 C- G C G C= G2. By the induction hypothesis, IlullH'''(n',),,, < CIIUIIH'(n2)' -f-
CIIYUIIHr+I/2(17,),,, +CII f
where 0', = G,, fl a If 1 < I < n - 1, then y (81 u) = 81(y u), so it follows from (4.24) and the result for r = 0 proved above that
C11 al UIIH'+C11 al }'U ll
H3/2(r2)+C11f211L2(nj).,,.
Thus, Cllf IIH'(si,)n,
11 al U11H2(n1)n, -< CIIUIIH'(n2)^' +
for1<1
= af3 + p
(rtai+2u P
)
Hence, n-t
Ilan+2U IIL2(12,)n, < CIIfIIHr(n,),,, +Cflu11Hr+1(n,)^' +C1:
II8/UIIHr+1(n,p,11
1=1
(4.27)
and the induction goes through. Turning to part (ii), the first Green identity applied to the equation P(Xu) _ fl gives
'(Xu, v) = (f1, v)n2 + (l,(Xu), Yv)r, for all v E H'(SZ2)"'.
(4.28)
Here, fl is the same as in (4.22) from the proof of interior regularity, but now x does not vanish on the boundary, so we need coercivity on H' (U)"', not just on Ho (S2)"'. If 1 < 1 < n - 1, then we can repeat the argument leading to (4.23). The only change is the presence of an extra term involving
Strongly Elliptic Systems
140
the conormal derivative of xu. In this way, we arrive at the following estimate for the tangential difference quotients: IIAI,h(Xu)IIHI(122) < CIIUIIH+ C11 f l1i,(12),,, + CI(13v(Xu),
YAl.-hv)r2I'
where v = El,h (X u). Let Fi = G1 fl I'2. We can assume that supp X C G1, and
then, since [B,, X]u = vjy[(akX)Ajku], IIx8vulli1/2(r;)N, + IIvjY[(akX)AjkUIIlHin(r,,),n < CIIBvuIIH-/2(r2),- + CIIYuIIHI/2(r2)^,.
By Exercise 4.4, IIYAl,-hVIIH-'/2(r;)" = IIAl.-hYVIIH-1/2(r,,)
II81YVIIH- /2(r;)n,
CIIYVIIH'/2(r,,)< CIIvIIH'(12),,,, so
(13v(Xu), YA1,-hv)r,l = I(13v(xu), YL1.-hv)r, (4.29)
ll8v(xu)IIHm/2(r;)IIYAl.-hVIIH-'/2(r,)'"
C(IIBvuIIH"/2(r2),- + IIuIIH-(122))IIEl,h(Xu)IIH'(922)" '
and thus, IID1,h(xu)IIH'(s22)" :5
+CIIBvuIIH'/2(r2y, +CIIfIIc2(Q2)
CIIuIIH'(122)"
forl <1
for r = 0. For r > 1, we make the induction hypothesis that CIIUIIH'(n2)"
IIullH'
+CIIB,uIIH'-"/2(r2),,, +CIIfIIH'-'(n2)°,.
Since
r-I
(B"u)
al
13 (al u)
-YP
( )(a'Aflk)(afaku)
\Pl
-,
we see th at if 1 < 1 < n - 1, then II
IIBvuIIHr+i/'_(r,),,, + CIIUIIHr+t(12',),"
The Transmission Property
141
and so, with f2 as in (4.24),
IlaruIIHZ(nl),,,
fort
The Transmission Property In our later study of surface potentials and boundary integral equations, we will often consider two equations simultaneously, both involving the same op-
erator P, but with one over the interior domain S2- = Q and the other over the complementary exterior domain S2+ = R" \ (Q U F). We then have the disjoint union
=S2+UrUQ-, with I, = 8 S2+ = 8 Q- the common boundary of the two domains, as depicted in Figure 1 of Chapter 1. The vector v will always denote the outward unit normal to c2-, and therefore the inward unit normal to 52+. (If one thinks of S2+ as locally the half space x > 0, then v = e,,.) We denote the one-sided trace
operator for n' by y±, and the sesquilinear form and conormal derivative and L3v , respectively. On account of our convention for SZ} by '1 regarding the meaning of v, the first Green identity (Lemma 4.1) takes the form
(D+(u, v)
= (Pu, On- T (B}u, y±v)r
for u E H2( 1)"' and V E Hl(S2t)"',
(4.30)
and dually (Lemma 4.2)
+(u, v) = (u, P"v)12t : (y±u, BV Or
for u E H) (S2±)"' and VE
H2(i)m.
(4.31)
As explained in the discussion following Lemma 4.3, we extend the definitions of B} and B to make these identities valid for u, v E H 1(S2#)"' whenever in the former case Pu, or in the latter case P*v, belongs to k-1 (01)'In this setting, it is convenient to think of u as a function defined on the whole
of R", but possibly having jumps in its trace and conormal derivatives across
Strongly Elliptic Systems
142
the surface 1,. We denote these jumps by
[ulr = y+u - y-u,
[Bvul r = 13,,'u - Bvu,
[Bul r =13 u -
provided both of the corresponding one-sided quantities make sense. When the jump vanishes, the + or - superscript is redundant and will often be dropped, i.e., we shall write
yu = y+u = y-u if [u]r' = 0, and similarly for the conormal derivatives. For brevity, we sometimes write
u = uln= to emphasise that we are considering the pieces of u separately.
Lemma 4.19 Let f } E H-1(SZ±)"', define f = f + + f - E H_I (RU)"', and suppose that u E L2(R")"' with u± E H' (SZ±)'". If
Pu} = f ± on Sgt,
(4.32)
then
c+(u+, v) + 4-(u-, v) _ (f, v) - ([B,u]r, yv)r for v E H'
(]E8")"',
(4.33) and
(Pu, 0) = (f, 0) + ([ulr, BvO)r - ([Bvulr, YO)r for 0 E D(R")"'. (4.34)
Thus, Pit = f on R" if and only if [ulr = [Bvulr = 0. Proof. Equation (4.33) follows at once from (4.30). The definition of Pu as a distribution on R!' gives (Pu, 0) = (u, P*O) = (u+, P*O)Qa- + (u-, P*O)n-, and by (4.31),
(u}, P*4)n- = 4±(ut, 4) + (Y:-u, BvO)r, so (4.34) follows from (4.33).
0
We can now prove the transmission property for (4.32): if the jumps in u and its conormal derivative are smooth, and if the right hand sides f + and f - are
The Transmission Property
143
smooth up to the boundary, then the restrictions u+ and u- must be smooth up to the boundary. Jumps are also allowed in the coefficients of P, although we shall not use this fact in later chapters. The proof again uses Nirenberg's method of difference quotients. Results like the following theorem, but involving more general types of transmission conditions on F, appear in work by Schechter [90] and Roitberg and Seftel [89]. Theorem 4.20 Let G 1 and G2 be bounded open subsets of R" such that G 1 C= G2 and G 1 intersects 17, and put
S2 =G;ns2l and
for j = 1, 2.
Suppose, for an integer r > 0, that x2 is
Pu± = f±
Cr+1. t, and consider the two equations on S22 ,
r+1
where P is strongly elliptic on G2 with coefficients in C'' 1(SZ )! L2(G2)"' satisfies
u± E H1[ulr E H'+312 (x2"
If U E
[B,,u]r E H(x2)"'
,
and if f ± E Hr(S22)"', then U-1- E H''+2(Q+)'" and IIu+II Hr+2(n ),,, + II U
GIIu+IIHI(n
),,,
II
+CIIU IIHI(1
+ C II [u]r2 II
)8;
C II [Bvulr, II Hl+',Z(r,)-
+CIIf+IIH,(Q2),,, +CIIf Proof As in the proof of Theorem 4.16, it suffices to consider the case when SZ+ is the half space x" > 0. Suppose to begin with that [u]r = 0 on r2i so that u E H1(G2)r by Exercise 4.5. We fix a cutoff function X E D(G2) such that X = .l on G 1, then, as with (4.22), (x u:-)
on Q:'
where the functions fl and f satisfy II f1 IIL,(Q: (Q:)- < C11 Xf IIL,(si})r, +C1lu}IIH1(0) .
By Lemma 4.19,
4) G,(Xu, v) = (fl, V) G, -
Yu)r,
for V E Ho (G2)"',
Strongly Elliptic Systems
144
an equation eerily like (4.28). Repeating the argument from the proof of interior regularity, but with an extra term involving [B (Xu)]r, we see that < CHUIIH(GZ)" + CIIf1IIL2(G2)"
11Al.h(XU)112
+ CI ([B,, (X 01r, y A,, -h v) r2
where v = O,./,(Xu); cf. (4.23). If 1 < 1 < n - 1, then the argument leading to (4.29) shows that I([Bv(xu)]r, Yol,-hv)r, I < (II[Evu)rlIHI/2(r2)"' + IIUIIH'(G2) X Ilol.h(Xu)Ilk'(G2)1'-,
and therefore Il a,(xu)IIH'(G2)" -< CIIUIIH'(G2)" + CII[E,1u]rHIH"12(r2)"' + Cll f1IIL2(G2)
Of course, 8"(Xu) is generally not in HI(G2)'", because [8u]r # 0, but as in (4.25),
f3
on n2" ,
with
CIIftIIL2(7
)1" + CIIuIIH'(nt),,, + C> IIa,UIIH'(s2'),,,. 1=I
Hence, WE E H2(7 )'" and the desired estimate holds for r = 0.
For non-zero [ulr E H312(F2)'", we use the extension operator 770 of Lemma 3.36 to construct w E H2(G2)"' satisfying
Yw = [u]r on I'2
and
IIB,wIIH'12(r2)'" + IIwIIH2(G2)1,, < CII[u]rl!H3P(r2)'".
Consider the function
I u+
UI =
onQ
u- +w on Q2.
Since
Put = )
f
on Q21
f-+Pw oncz
,
with [u1]r = [u]r - w = 0 on I'2, and [B ullr = [Buu]r -
E
HI/2(I'2)m, the preceding argument applies, showing that ui E H2(S2 )"'.
Estimates for the Steklov-Poincare Operator
145
Therefore, u± E H2(0)"' and Ilu
Ilui IIH2cn;
IIH2co, .
Ilu1 Ilx'csa >>° + Ilwllx'csi .^
< CIIu1IIHt(G2)1" + C11[8,,ui]r1iH"r(r2)" + Cflf+IIL2c2;>^'
+Cllf +PzIIL2(n)" +Ilwllx2(n;),,,, which, because CIIwitH2(n2)ur
< CII[u]rllx3/2cr2yn,,
shows that the desired estimate for the case r = 0 holds also when [u]r 0 0. An inductive argument like the one used for part (ii) of Theorem 4.18 takes care
ofr> 1. Estimates for the Steklov-Poincare Operator Consider the semi-homogeneous Dirichlet problem,
Pu = 0
on Q,
(4.35)
yu=g onl' and the adjoint problem
P*v = 0
on cl,
yv=ci
onr.
4.36)
If the fully homogeneous problem has only the trivial solution in H' (S2)"', i.e., if g = 0 implies u = 0, then under the usual assumptions we are able to define solution operators
U:gHu and
V:OF-).v,
with
U: H'/2(r)m -)-H'(S2)m
and V: H'/2(r )ra -+ H'(P)"'.
We can also form the Steklov-Poincare operators 13,U : g H 0H that satisfy
H'12(r)n, -+ H-1/2(1')'' and
13,V :
Hi/2(r)"'
--
(4.37) and 13,V :
H-'/2(1')"'. (4.38)
Strongly Elliptic Systems
146
The purpose of this section is to prove that, under certain conditions, 13,U
are also bounded from H' (r)m to L2a fact that will be used
and
later in our study of surface potentials and boundary integral operators; see Theorem 6.12. Notice that
(Bvug, O)r = t'(Ug, V-0) = (g, B,VO)r,
(4.39)
so (B,U)* =13 V. If SZ is at least C", then the regularity estimates of Theorem 4.18 apply, and we can extend (4.37) and (4.38) as follows. C'+'.1 domain, for some integer Theorem 4.21 Assume that SZ is a bounded, If r > 0, and that P is strongly elliptic on S2, with coefficients in Cr.1 (0) the Dirichlet problem (4.35) has only the trivial solution in H 1(SZ)'" when g = 0, then the solution operator has the mapping property
U : Hs+1/2(r)m
Hs+l (S2)m
for 0 < s < r + 1,
and the Steklov-Poincare operator has the mapping property
B,U : Hs+1/2(r)m -+ Hs-1/2(r )m for -r - 1 < s < r + 1.
Proof. The case s = 0 is covered already in (4.37) and (4.38). Part (i) of Theorem 4.18 shows that U : H'+3/2(r)m -+ H'+2(Q)m, and thus 8,U: Hr+3/2(r)m -+ H'+1/2(r)m, which means that the result holds for s = r + 1. Boundedness for the range 0 < s < r + 1 now follows by interpolation, i.e., by Theorems B.8 and B. 11. The same arguments apply to V and B, V, so, in view of (4.39), we can extend X3VU in a unique way to a linear operator that is
bounded for -r - 1 < s < 0. Our task is much harder when c2 is permitted to be Lipschitz. In this case, with the help of the following integral identity. we will estimate Here, for the sake of brevity, we use the summation convention, i.e., we sum any repeated indices from 1 to n. Lemma 4.22 Assume that cZ is Lipschitz, and that the leading coefficients Ajk belong to W111 (c2)m"". For any real-valuedfunctions h 1, h2,
... , h" E W1 m),
and for' any u, v E H2(Q)m,
v,y{[(hlAJk - hJAlk - hkAjl)aku]*aiv} dx Jr
_ f {(Djkaku)*ajv +
(hk8ku)*(Pov)} dx,
Estimates for the Steklov-Poincare Operator
147
where
Djk = at(hIAjk) - (alhj)Alk - (alhk)Aji. Proof By the divergence theorem, it suffices to show that 8({[(hlAjk - h jAlk
=
(DjkakU)*ajv
- hkAjr)aku]*ajv} + (Pou)*(hjajv) + (hkaku)*(Pov).
In fact, the left-hand side expands to a sum of five terms,
[8r(hlAjk - h jAlk
- hkAjl)(aku)]*ajv
+ [(hlAjk - hkA j,)8,aku]*ajv + [ - h jAlkalaku]*ajv
+ [(h,Ajk - h jAik)aku]*alajv + [ - hkAjIaku]*a,ajv. The second term vanishes because its factor (...) is skew-symmetric in 1 and k, and likewise the fourth term vanishes because its factor (...) is skew-symmetric in I and j. The third term equals
h j [Pou + (alA1k)aku]*ajv = (Pou)*(hj81v) + [hj(atArk)aku]*ajv, and the fifth term equals
-(hkaku)*Aj*iatajv = (hkaku)*[Pov + (a,A;t)ajv] _ (hkaku)*(Pov) + [hk(a,Ajl)aku]*ajv, so we get the desired right hand side with
Djk = aI(hlAjk -hjArk -hkAji)+hj8lAlk+hkalAjl.
0
Rellich [85] used a special case of the above identity to obtain an integral representation for the Dirichlet eigenvalues of the Laplacian. Subsequently, Payne and Weinberger [81] generalised the Rellich identity to handle secondorder elliptic systems with variable coefficients, and used it to bound the errors in certain approximations to (D(u, u) and pointwise values of u, when u is the solution to a Dirichlet or Neumann problem. In what follows, we use the arguments of Necas [72, Chapitre 5]. We will use certain first-order partial differential operators of the form
Qu = Qky(aku) with Qk E L,,(P)mxm If vk Qk = 0 on I', then such a Q is said to be tangential to r.
Strongly Elliptic Systems
148
Lemma 4.23 Assume that 0 is a Lipschitz domain, and let u E CCIomP(S2)m.
(i) If Q is afirst-order, tangential differential operator, then IIQiIIL,(r)-n < CIIYullH-(r),,,.
(ii) The normal and conormal derivatives of u satisfy IIt3vuIIL2(r) n < Cllau/avlIL2cryn +
CIIYuIIHi(ryn,
and, when P is strongly elliptic on 2, I1au/a))1L,(r),n <- CIIBvu1IL,cr)" +CIIYUIIH1tr>n,.
(iii) For1<j
n-l
Q,(x', ox')) _
Ql(x', (x'))81 (x'), l=1
and so n-1
Q1(x', (x'))[alu(x', (x')) + anu(x',
Qu(x', (X )) t=1
n-1
_
t-t
Q1(x', 0x'))aiut(x'),
where ut(x') = u(x', C(x')). Part (i) follows at once. Next, consider the identity (vj Ajkvk)
au av
=
Qu,
where Qu = vj(Ajivivk - Ajk)y(aku).
The first half of part (ii) is immediate, because the differential operator Q is tangential to r. The second part follows because the condition (4.7) for strong ellipticity implies that the m x m matrix v1 AJk Vk is uniformly positive-definite,
so Il(vjAjkvk)au/avllL,(r)n
cllau/avIlL,.(n- Finally, to prove part (iii), we
Estimates for the Steklov-Poincare Operator
149
observe that
Y(a;u) = vj
au Qju, av +
Where Qju = (S.ik - VJVk)y(aku),
and the differential operator Ql is tangential to T.
Concerning the next theorem, we remark that only part (i) is actually used later.
Theorem 4.24 Assume that 0 is Lipschitz, that P is strongly elliptic on n, and that the leading coefficients AJk are Lipschitz and satisfy Aki = Ask.
(Thus, the principal part of P is formally self-aajoint.) Let U E H 1 (S2)' and f E L2(S2)'" satisfy
Pu= f on Q. (i) If yu E H' (1')"', then
E L2(S2)'", and
IllvullL2(r)" < CIIYuII H'(r)-' + CIIuIIW(Q)"" + CIIf IIL,(sa)--
(ii) If P is a scalar operator (i.e, if m = 1), if AkJ = Alk (so the leading coefficients are purely real), and if fi,u E L2(r), then yu E H1(17), and IlYulluI(r) <- CIIEUuIIL,(r)+CIIuIIHI(sz)+CIIfIIL,(sz)
Proof Suppose in the first instance that u E H2(S2)m. Taking v = u in Lemma 4.22, and noting that Po = Po, we obtain
vIy{[(h,A;k - hJA,k - hkAjl)aku]*aju} dx Jr
= 12 { (Djkaku)*8 j u + 2 Re [(Pou)* (h; a ju)] } dx.
(4.40)
We split the integrand on the left-hand side into a sum of three terms,
(Qju)*(a;u) + (aku)*Qku - vl[(h,Ajk)aku]*aju, where Qj and Qk are tangential differential operators, defined by
Qju = v((hlAjk - hkAJl)aku and
Qku = vr(h,A*k -
hjAk)3ju.
Strongly Elliptic Systems
150
The matrix A = v j A jk vk E L,,. (I')'n xm is Hermitian and uniformly positiveexists definite on r, because P is strongly elliptic. Thus, A.' E and is uniformly positive-definite. A short calculation shows that
(Ajkaku)*aju =
u + (Qju)*aju,
where Q'' u = (A jk - vP A jp Av' Aqk vy) ak u. The operator
is also tangential,
so if we choose the h1 such that
hjv1>c,
(4.41)
then
da
I113" ull2Z(rro < C J
r
< CIIYajUIIL,(r)"(IIQjuIIL,(r)- + IIQ'UIIL,(r)- + IIQ';uHIL,(r) +CIIuIIH(n),., +CII7'ouliL2(s)''IIUIIHI(n)
By Lemma 4.23, IIYajuIIL,(r)"N
cllau/avjIL,tr)- +CIIYuIIH1(ryST CIIL.u1IL2(r)" + CIIYulIHI(r)-,
and so CII5vuIIL,(n-11YullH-(r)- +CIIYuIIHI(ry, +CIIuIIHI(n)w + CIIPouIIL2(n),"l111IIHI(n)Since II Pou II L, (n)., < II f 11L,(92)' + C 11U 11 HI (U)"',
the estimate in part (i) follows.
Now drop the requirement that u E H2(SZ)'", i.e., allow u E H' (c2)1, but assume y u E H 1(I')". It suffices to consider S2 of the form x" < (x'), where is Lipschitz with compact support. By Theorem 4.6, the operator P is coercive on Ho (0)'", so for A sufficiently large,
4)(u. u)
clIuIIH1(nr.
for u E Ho (l)'".
Choose a sequence Sr E CI (R"-1) such that 1.
S, -*
2. r <
in L,,,, (]Ri-1), and grad r -- grad in L p (IR"-') for I < p < oo; on 1R"-', and II grad r C, for all r;
3. r(x') = (x') = 0 for Ix'I sufficiently large.
Estimates for the Steklov-Poincare Operator
151
We let Or denote the set of points x E IR " satisfying x < r (x'), and put rr = a Or . Obviously, 12,. c 0. Define g : 0 -> C'" by
g(x) = Yu(x', C(x')), so that g(x) is independent of x". We easily verify that II9IIH-(n)-
CIIYUII H'(r)"1
and
IIYrgHIH,(rr)< CIIYUIIH'(r)m,
where y,. is the trace operator for 1'r, and where, in the second estimate, the constant C is independent of r. The operator P + A is positive and bounded below on Ho (S2)"', and hence also on Ho (0r)'n, so there is a unique solution ur E H1(52r)"' to the Dirichlet problem
(P+A)ur = f +Au on Or, Yrur =Yrg
on 1`r.
Moreover,
Yrur = Yrg, II urII H'(S2)'' < IIurIIH'(52,.)"' + II9IIH'(S2\52,)'"
< CIIYuIIH'(r)r' +CIIf +AuIIL,(Q)nI.
Since rr is smooth, we have ur E H2(S2r)"' by Theorem 4.18, and so the argument in the first part of the proof shows that IIBvurIIL,(r,)"' < CIIYrurIIH'(r,.)," +CIIurIIH'(sZ)" +CIIf +AuIIL2(92r)"'
< CIIYUIIH'(r)'" + CIIuIIH'(n)"" 4 CIIf IIL,(n)111,
(4.42)
with the constants again independent of r. We claim that ur converges to u in Hi (S2)"'. Let Or denote the sesquilinear form on Or, and apply the first Green identity to obtain
(Dr(ur, v) + A(ur, v)sz,. _ (f + Au, v) Q,. + (B"ur, Yrv)r,
for V E
H1(SZr)"',
(4.43) and
(D (u, v) + ), (u, v) s2 = (f + ), u, v)n + (B u, yv)r
for v E
H1(St)"'
152
Strongly Elliptic Systems
Hence, if V E Ho (Q)"', then
(D(ur - u, v) + A,(ur - U, On = [(Dr(ur, v) + A(ur, V)str]
- [c(u, v) + JA(u, v)Q] + [('(ur, v) - 4>r(ur, V) + ,X(ur, v)n\SZ,] = (Ever, Y,.v)r, - (f +,Xu, v)S \s2, + [4)(ur, v) - (Dr(ur, v) + ),(u,-, v)12\S2r],
so by taking v = ur - u and remembering that u, = g on Q \ S2r, IIU,.
-U112
(Q)",
< CII8A-IILZ(rr)"IIYr(ur -u)IILZ(rr),n +C(IIf +AuIIL,(s2\sz,.>>n + IIgIIH'(S2\nr)'n)IIur - UIIH0(i2),n.
Define
wr (x') =
1 + I gra4-(x')12
and w (x') =
1 + I grad (x') I2,
and note that 1 < wr (x') < C, 1 < w (x') < C, and wr -+ w in LP (I
"-l)
for 1 < p < oo. We have II Yr(ur - u)IILz(r,)"' = J n- I u(x',
f
< CII
(x,))
-
U (X,,
S' (X,))
12
(x') "-'
(x') - r(x')1
4
I8"u(x', x")12 dx" dx' (XI)
11 L,. (R"-I)IIUIIH(4.44)
which, in combination with (4.42), implies that II ur - u 11 Ho (Q). -+ 0, as claimed. The sequence of functions *r (x') =13,, ur (x', r (x')) is bounded in L2 (]R"-1)'"
because IIY'rIIL,(RH-i),n <_ J
I-,
I13vur(x',
r(x,))I2wrx')dx'
= II13,u,112 r,.)n
so by Theorem 2.31, after passing to a subsequence we can assume that iJrr converges weakly to a function * in L2(Ri-1)"', i.e., (t/rr, v) -+ (+/r, v) for each v E L2(IRa-1)"'. Define (x', (x')) = ,lr(x'), and let v E D(SZ)"', say. We have
(t3vur,Yrv)rr =
J fR
(x!)*v(x', (x'))w(x') dx' _ (, Yv)r,
Estimates for the Steklov-Poincare Operator
153
and therefore, sending r -* oo in (4.43) gives (D(u, v)
v)sz +
for V E D(S2)'n
Yv)r
Hence, Bvu = j E L2(I')m, and by Exercise 2.11, the estimate of part (i) follows from the uniform estimate (4.42) for B,ur. To prove (ii), assume m = 1, and suppose once again that u E H2(SZ) and that the h1 satisfy (4.41). This time, we write the integrand on the left-hand side of (4.40) as
uzh,A jkdku8ju - Bvuh j8ju - hkakuBvu, and hence obtain, using strong ellipticity,
E IIYajUIIL,(r) < C j=1
fr
hrv1Y{Ajk8ku 8ju}dv n
<- CIIBvuIIL,(r)
IIYajUIIL2(r) j=1
+
CIIPoUIIL2(17)IIUIIHl(s2)
Thus, n CIIUIIH,
II YajUII L,(r) < CIIBvuIIL2(r) +
(-) + CII f IIL2(n),
j=I
and the estimate of part (ii) follows. Next, we allow u E H1(c2) and assume Bvu E L2(F). Define gr E L2(Fr) by
gr(x', r(X')) = Bvu(x', (x'))w(x')/wr(x'), and let Ur E H1(c r) be the solution of the Neumann problem (with A. sufficiently large, as before)
(P + A.)ur = f +),U On Or, Bvur = gr
on Fr.
We easily verify that IIgrIIL2(rr) < CIIBvUIIL2(r), and by Theorem 4.7, 2
(Pr(U, v)+A(v, On, ? CIIVIIH t(sar)
with the constant independent of r in both cases. Thus, we have the uniform
Strongly Elliptic Systems
154 bounds
IIUrIIH'(22,) < CII f + AUIIH--(SZr) +
CIIgrIIH-112(p,)
< C1IBvUIIL,(r) + CIIuIIH'(0) + CIIf IIL,(o),
and, since Ur E H2(2r) by Theorem 4.18, our earlier argument gives n
:IIYajur11L,(rr) -< CIIg1IIL2(rr)+CIIUrIIHI(S2r)+C11f1IL2(nr) j=1
CIIBvuIIL,(r)+CIIurIIH'(s2r)+CIIfIIL,(sa). (4.45) Using the method in the proof of Theorem A.4, we can extend u,. to a function in H' (R") in such a way that II U, II H' (52\52,.) -* 0. To show that ur converges to u in H' (Q), we observe that for v E Hl (SZ),
(D(Ur - U, V) + ),(Ur - U, v)S2 = [4)r(Ur, v) + a.(Ur, V)nr]
- [c(u, v) + A(U, v)cz]
+ [
(gr, Yrv)rr, - (E.,u, Yv)r - (f + A.u, v)c\i2r + cc2\a'(Ur, v) +A(Ur, v)12\nr. Since
I (gr, Yrv)rr - (13,, u, Yv)r12
5vu(x', (x'))[v(x'' Sr(x')) C11L3"U112 L2(r)
-III
L.(1"-')IIVII
- v(x', (x'))]w(x') dx'
2
2 H'(n\Qr)'
we see by taking v = Ur - u and again applying7I eorem 4.7, this time over S2 instead of SZr, that
IIU.-uIIH'(S2) C II
and so ur -+ u in H' (0). Define *r (x')
C II
11 H'
(yrur)(x', r(x')) and *(x') =
(yu)(x', (x')).Bywriting*,-(x')-1(x') _ [ur(x', r(x'))-Ur(x', 0x'))]+ (Ur - u) (x', (x')), and estimating the first term in a manner similar to (4.44), we obtain
I( r- r,v)I
[IIr-Ili( -)IlurllH'(s\n,)+IIY(ur-u)IIL2(r)]IIvIIL,(JR'I-I),
Estimates for the Steklov-Poincare Operator
155
and therefore *r + t// in L2(R"-' ). Finally, by (4.45), II*rlIH'(R"-') -< CIIurIIHt(r,) <- CIIB,,uIIL2(r) + CIIuIIHI(S2) +CIIfflL2(n),
so Exercise 2.11 shows that 1 E H' (R"-') with II * II H ' (]"-') < lira sup II ,y r II H ' M-' ), implying that u E H' (1'), and that the estimate of part (ii) holds.
It is now a simple matter to obtain the desired mapping properties for the Steklov-Poincare operators. We can also introduce a meaningful concept of a solution u = Ug E L2 (&2)' for g E L2(F)"` (but see also the sharper mapping property for U proved in Theorem 6.12). Theorem 4.25 Assume that S2 is a Lipschitz domain, that P is strongly elliptic on S2, and that the coefficients AJk and AJ are Lipschitz (not just Lam). If
Akj = AJk,
and if the solution operators (4.37) exist, then the Steklov-Poincare operato satisfy BvU :
Hs+1/2(r,)"'
+ Hs-1/2(r,)n'
and 9,V: H.s+112(r,)'n -* Hs-1/z(r,)+n
for - 2 < s < Z. Furthermore, the solution operators have bounded extensions U : L2 (1,),n
L2(2)n'
and V : L2(F)m -* L2(S2)m.
Proof. For s = 0, the first part of the theorem was established in (4.38). We obtain the result for s = 1 by applying part (i) of Theorem 4.24 to (4.35) and (4.36). The case s = -? then follows by duality, using (4.39). Finally, interpolation gives the complete range -1 < s < 1.
In the second part of the theorem, it suffices to consider U. Let g E H 1/2 (r),
and f E L2(Q)m, and put u = Ug E H' (S2)m. Our assumptions imply the existence of a unique w E H' (S2)m such that
P*w = f
on S2,
yw=0 on1. The first Green identity gives, on the one hand,
(D(u, w) = (Pu, w)n + (13vu, Yw)r = 0,
Strongly Elliptic Systems
156
and on the other hand, (D(u, w) = (u, P*w)n + (Yu,
(u, f)n + (g,
so
I(u, f)nI = I(g, Bvw)r
IIglIL2(r)'"IIBpwIIL2(r)'"
By part (i) of Theorem 4.24, IIBvwIIL2(r)'" _< CIIYWII H'(r)'" + CIIwIIH'(n)"' + Cll f IIL2(n)" < CIIfllL2(n)",,
and hence 1(u, f) n l < C II g II
L2 (r)m
II f II L, (n)m , implying that IlUg it L2 (n)m =
0
IIuIIL2(n)'" <- C118IIL2(r)m
Exercises
4.1 ShowthatPu = -8j (Ajk0ku)+Aj8ju+Au,where Ajk = 2(Ajk+Akj) _
Akiand Aj =Aj+28k(AJk-Akj). S2 be a C' diffeomorphism. We denote the Jacobian of the coordinate transformation x = K (y) by
4.2 Let K : S2K
J(Y) = I detK (Y)I =
18(x1,
... ,
and write UK = uoK and vK = voK, so that u (x) = UK (y) and v(x) = vK (y). (i) Let 0 be the sesquilinear form (4.2). Show that c (u, v) = cK (UK, vK ), where 4>K is the sesquilinear form with coefficients
Ajk(Y) _ (ay Ars (x) 8xk) J(Y), \
r
s
, Aj(Y) _ (Ar(X)2)J(Y) ax,
AK(Y) = A (x)J(y).
(ii) Let PI denote the differential operator with the coefficients in part (i).
Show that Pu = f on 0 if and only if PKuK = fK on UK, where
fK(Y) = f(x)J(Y) (iii) Show that P is strongly elliptic (respectively, coercive) on Q if and only if PI is strongly elliptic (respectively, coercive) on 7K. 4.3 Show that if f (0) = 0 and a > 2, then
t-af
00
(Jo
I
1/2
(t) 12
dt)
[Hint: use Exercise 3.20.]
1/2
1
<
t'-« f,(t)12 dt) a -
(J0
I
.
Exercises
157
4.4 Show for all s E R that if al u E HS (R' )" `, then the difference quotient (4.21)
[Hint: l e" - 11 s 101 for O E R.] satisfies II A1,hu 11 Il a,U I! 4.5 Suppose that u E L2 (W) and u± = u I r} E H I (II8'f). Show that
(a, u, 0)
= (a;u+, 0)i + (a;u-, 0)A +
J0 ([u]r, Yt)r
if l<j
where I' = R"-1 x {0}. Deduce that u E HI(R") if and only if [u]r = 0. 4.6 Show that in Corollary 3.22 we can choose the functions Oj for the partition of unity in such a way that 0, (x) = i/rj (x)2, where 1/r; E C mp t [Hint: start by considering g in Exercise 3.6.] 4.7 Consider the classical cooling-off problem [19, p. 56]: au
at - aLu = 0 av+byu=0
u = uo
on S2,
fort > 0,
on r, fort>0, on S2,
when t = 0,
where a and b are positive constants. Let 401, 02, ... and. 11, A2, ... be the eigenfunctions and eigenvalues of the stationary problem, as in Theorem 4.12, i.e.,
-a 0Oj = ,Xj¢j on 0, a0;
-}-by¢j =0
av
on!,
and derive the series representation 00
u(x, t) = > e-"(0 , uo)ccpj(x) In what sense does this sum converge?
for x E Q and t > 0.
5
Homogeneous Distributions
For a E C, a function u : R" \ (0}
u(tx) = t°u(x)
C is said to be homogeneous of degree a if
for all t > 0 and x E 1R" \ t0}.
(5.1)
To extend this concept to distributions, we introduce the linear operator Mr, defined by
MMu(x) = u(tx)
for 0 o t E'1R and x E IR",
and observe that for every u E L 1,10 (1R" ),
(Miu, 0) = Its-"(u, M1ltfi)
fort # 0 and 45 E D(1R").
(5.2)
If U E D*(IR") then (5.2) serves to define Mtu E D*(1R"), because M11, : D(1R") -* D(RI) is continuous and linear. We then say that u ED*(R) is homogeneous of degree a on IR" if M,u = t°u on 1R" in the sense of distributions, for all t > 0. This chapter develops the theory of homogeneous distributions, using Hadamard's notion of a finite-part integral to extend homogeneous functions on 1R" \ {0} to distributions on R. We consider in some detail the Fourier trans-
form of, and the change of variables formula for, such finite-part extensions. A technique used several times is to reduce the general n-dimensional case to a one-dimensional problem by transforming to polar coordinates. Most of the material that follows can be found in standard texts such as Schwartz [92], Gel'fand and Shilov [27], and Hormander [41], but the final two sections dealing with finite-part integrals on surfaces - include some less well-known results from the thesis of Kieser [48]. The results of this chapter will be applied later in our study of fundamental solutions of elliptic partial differential operators, and of boundary integral operators with non-integrable kernels. 158
Finite-Part Integrals
159
Finite-Part Integrals We begin our study of homogeneous distributions by focusing on the simplest example, namely, the one-dimensional, homogeneous function xa
10
if x > 0,
ifx <0.
If Re a > -1, then x+ is locally integrable on R, and is obviously homogeneous of degree a as a distribution on R, not just as a function on R \ {0}. To deal with
the interesting case Re a < -1, we use the following concept, introduced by Hadamard [37] in the context of Cauchy's problem for hyperbolic equations.
..., bN+2 be complex numbers, with Re a1>0andaf#0 for I<j
Definition 5.1 Let al, ... , aN and b1,
and
aj 54 ak
whenever j # k.
(5.4)
If a function g satisfies N bl g(E) _ L Eai
+ bN+1 log E + bN+2 + 0(1) as c 4. 0,
1=1
then the term bN+2 is called the finite part of g(e) as c .{ 0, and we write
f.p. g(E) = bN+2 E.J,O
When no singular terms are present, i.e., when b1 = ... = bN+t = 0, the finite part is just the limit of g(E) as c it exists, the finite part is unique.
0. The next lemma shows that, when
Lemma 5.2 Let a 1, ... , aN and b 1, ... , bN+2 be complex numbers, and assume that the aj satisfy (5.3) and (5.4). If N
lim E b' + bN+l log E + bN+2 = 0, Eaj CIO
(j=1
thenbj =Ofor1 < j
Homogeneous Distributions
160
with X j E R \ {0}, so JE-"j I =I exp(-iA. j log E) I= 1 for I< j < N, and we see at once that bN+l = 0. Moreover, there exist a sequence o and real numbers 01, ..., ON such that Em 4. 0 and limmi 6. j = eloj for 1 < j < N. Thus, putting E = Em ex and sending m -+ oo, we see that N
Lb
je-i(.Ljx+oj) + bN+2
= 0 for x E R.
(5.5)
j=1
By (5.4), we may assume without loss of generality that X 1 < k2 < . . . < AN, and by analytic continuation we may replace x with ix in (5.5) to obtain N
bj ezjx-iej + bN+2 = 0
for x E R.
j=1
oo, to conclude that bN = 0. Otherwise, X1 must be negative, so we divide through by e^tx and send x -+ -oo to conclude that b1 = 0. Repeating this argument, one sees that b j = 0 for 1 < j < N, and then (5.5) reduces to bN+2 = 0. Now suppose µ > 0. Since If AN > 0, then we divide through by eA"x and send x
N
1im E" ELO
N (--j + bN+1109 E + bN+2 j=1 E"j
= lim 4"0
biE"j-µ Rea j=µ
it follows as above that b j = 0 if Re a j = it, and repetition of this argument shows that b j = 0 if Re a j > 0, which gets us back to the case when µ = 0.
0 ForE> 0and0ES( H ), we define the integral 00
dx fs
and its finite part,
Ha (cj) = f.p. Ha.E (O) Ejo
The following lemma establishes the existence of Ha (0), and shows that we can define a temperate distribution f.p. X'- E S*(R), called the finite part extension of x+, by putting
(f.p. x+, fi(x)) = Ha(d)
for 0 E S(R).
Obviously, if x is restricted to R \ {0}, then f.p. x+ coincides with x+.
Finite-Part Integrals
161
Lemma 5.3 For any integer k > 1 and any 0 E S(R),
Ha(0) =
if Re a > -k - 1 and
(-1)kH°+k(0(k)) (a + 1)(a + 2) ... (a + k)
-1,-2,...,-k,
a
but 00
H-1(q) = -J
0'(x)logxdx 0
and
1
H-k-1 (0) =
1 O(k)(0) +
j
k! j=1
k!
1H-1
(O(k))
Proof. Integration by parts gives
_E) a
-
H + 1O,)
if a# -1.
Suppose first that Re a > -k - 1 and a ; -1, -2, ... , -k. By Taylor expansion, k-1 O(j) Ea+1O E)
=
i0)Ej+a+1
+0(6 Rea+k+l ),
j=0
so if Ha+1(q') exists, then Ha+1(0')
a+l and the first part of the lemma follows by induction. Next, integration by parts and Taylor expansion give 00
-0(E) logE -
_ -0(0) logE -
f0'(x)logxdx
J0
00 O'(x)
logx dx + O(E log,-),
implying the formula for H-1(0). Finally, when a H-k-1,, (
_-
E-ko
= -k - 1,
(E) + H-k W)
Homogeneous Distributions
162
and by Taylor expansion, k-1
(1)
1=1
J'
(k)
k
so if H_k (4') exists then (k)
+ Xk ( k
The formula for H_k_ 1(0) follows by induction.
'O
The next lemma shows that the distribution f.p. x+ is homogeneous of degree a on R, except when a is a negative integer.
Lemma 5.4 If c E S(R) and t > 0, then t-1Ha(Ml/tO) = taHa(O) fora # -1, -2, -3,
...,
t-1 H k-1(M'/to) = t-k-1 H k-1(46) + (t-k-1 log t)
(k)
but
ki0)
for any integer k > 0.
Proof If Re a > -1, then it suffices to make the substitution x = ty to obtain
t-'Ha(Mi/tcb) = t-' f OOxa¢(x/t)dx = to
f00
Ya,O (y) dy = taH.
-1, -2, ... , -k. For cf. (5.2). Now suppose that Re a > .-k - 1 and a brevity, write bk = (-1)k/[(a + 1) (a + k)]; then by Lemma 5.3, t-'Ha(Mi/t4) = t-1bkHa+k((M11 )(k)) = t-'bkHa+k (t-kMl/t4(k))
= bkt-k-lHa+k(M1/tO(k)) = bktaHO+k(O(k)) = taHa(t) However, when a = -1, we have
t-1H-l (Ml/A = -t-' J
(Ml/to)'(x) logx dx
0
_ -t-2
00
J0
0'(t-1x) l o g
dx,
Finite-Part Integrals
163
and the substitution x = ty gives 00
t-1 H-1 (MI/to) = -t-1 J
O'(Y) log(tY) dy = t-1 H_I (0) + 0 (O)t-' log t. 0
1/j, then
In general, if we let ck = (1/k!)
t-'H-k-1 (M11,O) =
1
t-'Ck(M1/iO)(k)(0) + k! k
= Ckt-k-1o(k)(O) + t kl
H_i ((M1/t0)(k))
1
H_I(M1/,0(k))
= t-k-1 (CkO(k) (0) + I H-1(O(k)))
+
0(ki 0) t-k-1
log t,
giving the second formula in the lemma. We shall also have use for the homogeneous function 0
x° = (-x)+ _
1xIa
if x > 0, if x < 0,
and its finite-part extension, E
(f.p. X!, 0 (x)) = f.P
fO -00
= f.p.
Ix JaO (x) dx
f xa0(-x)dx = Ha(M-t4b) E
It is easy to verify that
f.p.(-x)+ = f.p. x,,
(5.6)
and that f.p. xa is homogeneous of degree a on R, except when a is a negative integer. Indeed, Lemma 5.4 shows that if t > 0, then
f.p. (tx)f = to f.p. xt for a
-1, -2, -3, ... ,
but
f.p.(tx) fk-i = t-k-1 f.p.
x±k-1
+ (F1)k
t-k
log t 8(k) k
(x)
for any integer k > 0.
(5.7)
This loss of homogeneity does not occur in the case of the function x-k-1
Homogeneous Distributions
164
Indeed, the finite-part extension
x-k-10(x) dx
(f.p. x-k-1, 4b(x)) = f.p. E(O J XI>E
= H-k-1(0) + (-1)k+1H-k-I(M-10), satisfies x+k-1 + (_ l)k+1
f.p. x-k-1 = f.p.
f.p.
x_k-1
(5.8)
and so, by (5.7) and (5.6), f.p.(tx)-k-1 = t-k-1 f.p.x-k-1
if 0
(5.9)
t E IR.
One can formally integrate by parts k + 1 times to express (f.p. x-k-1, 0(x)) as a convergent integral.
Lemma 5.5 For q5 E S(R) and for any integer k > 0, 00 (f.p.x-k-1,,0(x))
r 0(k+1)(x)logIxI dx.
= k J o0
Proof. By (5.8) and Lemma 5.3,
(f.p.x-k-1, 0(x)) = H_k-1(0) + (-1)k+1H-k-I (M-1l) kk
E1
[O(k)(0) + (-1)k+I (M-10)(k)(0)]
k!J=1
(_l)k+1(M-1,)(k)),
+ k1 H-1 (.(k) + and thus, because (M_10)(k) = (-1)kM_1q5(k),
(f.p.x-k-1, q5(x)) = kI H_1 (0(k)
=
1
k1
f
°O
{
- M-I0(k)) (k+l)(x)
+.O(k+l)(-x)] logx dx,
giving the desired formula.
0
Later, when studying the Fourier transforms of f.p. xt and f.p. x-k-1, we shall encounter the distribution (x ± i0)', defined by
((x f MY', 0(x)) = li m foo (x ± iy)°4(x) dx, with the branch of z° = exp(a log z) chosen so that -ir < arg z < it.
(5.10)
Finite-Part Integrals
165
Lemma 5.6 The formula (5.10) defines a temperate distribution (x ± i0)' E S*(R), given by
(x ± i0)' = f.p. x+ + e}""` f.p. x° f o r a # -1, -2, -3, ... , and
(_ I k+l (x ± i0)-k-I = f.p. x-k-1 ± i7r
(k)
S
k
(x) for any integer k > 0.
Proof If Re a > -1, then we may apply the dominated convergence theorem to obtain the first formula. If -k - 1 < Re a < -k for some integer k > 0, so that Re(a + k) > -1, then integration by parts gives 00
(-1)k (x ± iy)a+k
foo
(a + ])(a+ 2)...(a + k)
f 00 (x + iy)aQb(x) dx =
01k) (x)
dx,
so in the limit as y 4. 0, we have (-1)k (xa+k + efin(a+k)xa+k)
((x f i0)a, fi(x)) _
(a + l)(a + 2) ... (a + k)
'
(k)
dk x+ k + (-1)kefinaxa+k dxk
and we have only to apply Exercise 5.3. However, when a = -k - 1,
L:x ± iy)14 (x) dx =
f log(x ± iy)
1(x) dx,
and since -n < arg(x ± iy) < it, yy im
log(x ± iy)
logx
ifx > 0,
log Ix I ± i7r
if x < 0.
Thus, by Lemma 5.5,
((x ±
i0)-k-1,
0(x)) =
00 flogIxI1)(x)dx
kI 1
k
_
° f(±iir)(x)dx
(f.p.x-k1, 0(x)) -
which yields the desired expression for (x ±
i0)-k-1.
ki (fiir) 0(k)(0),
Homogeneous Distributions
166
Extension from W'\{0} to 1[t" If U E L1,1oc(R" \ {0)), then we can try to define the finite-part extension f.p. u as a distribution on R' by writing
(f.p. U, 0) = f.p. f
u(x)¢(x) dx for O E D(R").
XI>E
In the special case when the finite part of the integral is just a limit, we speak of the principal value p.v. u, i.e.,
(p.v. U, ) = limJ
40 IXI>E
u(x)O (x) dx.
Suppose now that u E C°°(R' \ (0)) is homogeneous of degree a. By in-
troducing polar coordinates p = lxi and w = x/p, so that x = poi and dx = pn-1 dp dco, we find that
f
u(x)cb(x) dx = J
pat"-1-0 (pce)) dp dcv.
u(co) p>E
X I>E
This formula prompts us to define the linear operator Ra by Racb(x)
= f.p.
40 JE
po+n-1cb(px) 00
dp = (f P
p+++n-1,
(px))
for X E R" \ {0),
(5.11)
so that
(f.p. U, 0) =
JI of=1
u(c))Ra¢(co) dcv
for ¢ E S(R").
(5.12)
Here, to justify taking the finite part inside the integral with respect to co, it suffices to check that the o(1) term in the expansion of fE00 pa-n+l0(pco) dp tends to zero uniformly for lwJ = 1. As a consequence of Lemma 5.4, we are able to prove the following.
Theorem 5.7 Suppose that u E C°°(R" \ 0) is a homogeneous function of degree a.
(i) If a -n, -n - 1, -n - 2, ... , then f.p. u is the unique extension of u to a homogeneous distribution on W.
(ii) If a = -n - k for some integer k > 0, then a homogeneous extension
Extension from R" \ (0) to R"
167
exists if and only if u satisfies the orthogonality condition
whenever l a l= k.
wa u (co) d w= 0
(5.13)
In this case, the homogeneous extensions of u consist of all distributions of the form
f.p. u + Y' ca 8.8, Ial=k
with arbitrary coefficients ca E C.
Proof Consider
f
(Mr f.p. U, 0) = t-r" (f.p. U, MI/r4>) =
u(w)t-"RaMI/r4>(w) dcv.
IwI=I
Since RaMI/ra(w) = Ra4>(t-ico), we see from Exercise 5.4 that for a as in part (i), U(w)t-n(t-1)-a-lzRa4)(w)
(Mr f.p. U, 0) = f
dw = (ta f.p. u, 0).
wl=1
However, if a = -n - k, then (Mr f.p. U, 0) U(w)t-"
((t_t)kR_il_k4>(cv)
-
(t-1)klog(t-1)
Iwl=1
E aao(0)wa a
dw
lal=k
f.p. U, 0) +
t-"-k
as of
log t
a.
Ial=k
w"u(w) dw,
Iwl=1
so f.p. u is homogeneous on R" if and only if (5.13) holds. To settle the question of uniqueness, and to show the necessity of the orthogonality condition for existence if a = - n - k, let u E V* (R") be any extension of u. Since u - f.p. u = 0 on R" \ 0, Theorem 3.9 implies that
ii -f.p.u=Ecaa,3, a
where the sum is finite. The result follows because 8a8 is homogeneous of degree -n - lad; see Exercises 5.1 and 5.2.
Homogeneous Distributions
168
The next theorem complements the one above, and introduces a particularly important class of homogeneous functions satisfying the orthogonality condition (5.13).
Theorem 5.8 Suppose that u E C°O(R" \ {0)) is a homogeneous function of degree -n - k, for some integer k > 0. If u has parity opposite to k, i.e., if U (_X) = (_1)k+1 U (X) for x E R" \ {0},
(5.14)
then
(f.p. U, 0) =
p-k-1, O(pw)) dw for ¢ E S(R"),
12 f
Iwl=1
and f.p. u is the unique extension of u to a homogeneous distribution with parity opposite to k.
Proof. The parity condition (5.14) implies that (f.p. U, ci} =
f
u(-m)R-"-k.(-w) dw
Iwl=1
f
(-1)k+1u(w)R-n-k41(-w) dw,
wl=1
and by (5.6),
R-n-ko(-w) _ (f.p.
p+k-1,
(f.p.
p_k-1, O(pw)),
so we have
f =2 f
(f.p. U, ) = 1
2
u(w)[R-n-k O(w) + (-1)k+1R-n-k(-w)] dw
w l=t
u(w)(f.p. p+k-1 + (_1)k+1 f p
p- k-1, 0) dw,
wl=1
giving the desired formula; recall (5.8). The homogeneity of f.p. u follows from Theorem 5.7 because the parity condition (5.14) implies the orthogonality condition (5.13). Alternatively, one sees from (5.9) that
M, f.p. u = Itl-"t-k f.p. U
on R" fort E lR \ {0),
and in particular, f.p. u has parity opposite to k. Finally, if Ia I = k then F 8 has the same parity as k, so f.p. u is the only homogeneous extension of u having the opposite parity to k. 0
Fourier Transforms
169
The uniqueness results in Theorems 5.7 and 5.8 yield a simple proof of the following fact.
Theorem 5.9 Suppose that u E C°O(R" \ {0}) is a homogeneous function of degree a, and assume, if a = -n - k for some integer k > 0, that u has parity opposite to k. Then for any multi-index a, 8 " f.p. u = f.p.(a"u)
on I[8".
P r o o f By Exercise 5.2, if a # -1, -2, -3, ... , then a" f.p. u and f.p.(a"u) are homogeneous extensions of a"u with degree a - I a I , and must therefore coincide. If a = -n - k but u has parity opposite to k, then
a"u(-x) _(-1)k+I"I+1a"u(x) for 0 0 x E W, so 8"u is homogeneous of degree -n - (k + lal) and has parity opposite to k + la l. Thus, a" f.p. u and f.p.(a"u) are again homogeneous extensions
0
of a"u, and both have parity opposite to k, so they must coincide.
Fourier Transforms Our aim in this section is to compute the Fourier transform of the finite-part extension of a homogeneous function. Following the pattern of previous sections, the one-dimensional distributions f.p. xf and f.p. x-k-1 will be treated first. In order to state the next lemma, we require the gamma function,
'(a)
=f
00
forRea>0.
In the usual way, r is extended by means of the identity
r(a + l) = ar(a) to a meromorphic function on C having simple poles at 0, -1, -2, ... , and satisfying r (k + 1) = k ! for any integer k > 0. T:r_.4 {f.p. x' 1. If a # -1, -2, -3, ... , then
Lemma 5.10 Let IIa
}
r1Q ()
_
r(a + 1) (±i2ir)"+1
(
i0)
,
but for any integer k > 0,
IZ}k-1( ) _
(:Fi2 r
k
(1o27rll ± 12 sign() - r'(1) - E
I
j=1 I
(When k = 0, the empty sum over j is interpreted as zero.)
Homogeneous Distributions
170
Proof To begin with, suppose that Re a > -1. For any 17 > 0, the function e-2nglxlx11 belongs to LI(R), and 00
Fx,
{e-2nglxlXa l = fJ J
t
dx.
Making the substitution z = 27r (ri f ii )x, and then applying Cauchy's theorem to shift the contour of integration back to the positive real axis, we find that o
[27r(ii±i
_
(±i27r)a+l
e-zzadz
Jo
J
1
00
a1??)--1
1'(a + 1) (
Sending ri J, 0, we obtain the first part of the lemma for Re a > -1, and the
case Re a < -1 then follows by analytic continuation; see Exercise 5.6. For the second part of the lemma, consider 00
H_k-1,E
)=f
(a-i2Trl
e-ibrtxx-k-1 dx,
E
where f,.00 is interpreted as limn. fE if k = 0. Suppose l; E R}, and make ±i27rIr; Ix to obtain the substitution z = H-k-I,E(e-i2n1;,)
_
(i27rf)k
f fi00 Jti2ir I IE
e-:.z-k-1 dz.
Applying Cauchy's theorem (and Jordan's lemma, if k = 0), we see that fi0c
2,rI
i2n1? IE
f.
o00
e-zz-k-1 dz =
e-zz-k-1
dz = f 1
4
=±
e-zZ-k-I dz
- fc.,
e-zz-k-1 dz,
IE
f
e-zz-k-1
resse-zz-k-1
=
dz
(_I)_ k
o l2
(i27rj)k (H_k_I.2rIIE() T
k
i7r (-1)k 2
k!
)
Fourier Transforms
171
By Exercise 5.7 together with Lemma 5.3, we have f-
k' 0) log 27r l
P (0)
(P
k!
- (-1)k k
k
- - tog27r ICI + k! H-t 1
1
i=t J
E i=t k
1
J
- l0 g27r ICI+e-xlogxdx
I
(-i27rt)k
i7r
2 sign(t)-1
kt
(f.p. xk-', fi(x)) = f.p. f
x-k-t f
J
E
oo
,
k (1)-E _1
i=t
00
(n-k-t'
I
I
e-i2n1XO(t) dt dx
00
00 Elo
foo
where the final step is justified because the o(1) term in the expansion of can be bounded by f (E)g(t), with f (E) = o(1) as c 4. 0, and g(') having only polynomial growth as It I oo. The formula for n±k-i (t) H_k-t.E(e-i2"t')
is now established, and the one for II-k-t (t) then follows by (5.6) because the 0 Fourier transform commutes with M_ t; see Exercise 5.8.
As an immediate consequence of Lemmas 5.6 and 5.10, we obtain the formulae below; see also Exercise 5.10. Corollary 5.11 F o r a n y integer k, let 1-lk() =
.
_
1 nk(t,) = (-i2zr s'kl(t) and l1-k_, (t) _ ( )k
{f-p. xk}. If k > 0, then 1227r 7r
tk sign(t).
Turning to the general, n-dimensional setting, we require the following technical lemma.
Homogeneous Distributions
172
Lemma 5.12 Let a E C and U E D* (R" ). If u is homogeneous of degree a on R" \ {O}, i.e., if Mtu = tau on Rn \ {0} for all t > 0, then u E S*(R"). If, in addition, u is CO° on R" \ (0), then u is C°° on R' \ (0). Proof Let* E C mP (R" \f0}) be as in Exercise 5.12, and define X E C mP (R" ) by X(x)=1-J1
*(tx)
dt
forxER".
0
Put uO = Xu and ul = (1 - X)u; then uo E S* and uo E C°°(R") because uo has compact support, so it suffices to consider u I. We have
(ui, ) = (j' ir(tx)
a tu(x),
0(x)=
f '(u,
OM1
fr)
dt
for q E D(Rn),
(u, OM *) = (u, Mt (*Ml/t-O)) = (t-"Ml/tu, rMl/to)
= t-n-a(u, *M,1, 0). Let K = supp *; then there is an integer k such that
1(u, iMi,t4)I < C E max I aa(*Mltt4) I< C E t-I°`I max I a"O(t-lx)I lal
xEK
Ial
Choose an integer r > n + Re a; then because 0 gi~ K, t-IaI
max I
a"0(t-lx)I
xEK
< Ct-IaIXEK max E IxI6aa0(t-1x)1 lfll=r+Ial
< Ctr
E max IfiI=r+Ial
l(t-lx)I
aa4(t-lx)I,
XEK
so
I(ul, 0 :5 C
f
E sup 1: I#I=r+lal YER"
1
tr-n-Rea
dt
J0
showing that ul E S* Now make the additional assumption that u is CO° on R \ (0). Since X = 1 on a neighbourhood of 0, we see that u 1 is C°° on R", and (RI).
laaul(x)I < C(1 +
IxI)Rea-lal
forx E R".
Fourier Transforms
173
Hence, if L81 > n + ja I + Re a, then the function 8. [(-i2zrx)a u 1(x)] belongs to L 1(1 ' ), and we deduce from (3.17) that (i2iri; )O 8"u i
is continuous
onR R.
Lemma 5.12 shows that the Fourier transform of a homogeneous distribution always exists as a temperate distribution. Furthermore, the following holds. Theorem S.13 Let a E C. If u is a homogeneous distribution of degree a on I[2",
then its Fourier transform u is a homogeneous distribution of degree -a - n
on R. Proof. For t > 0 and 0 E D(R ),
(Mru, 0) = (Mr.Tu, ) = t-"(u, .FM1/ro), and by Exercise 5.8, t-"(u, .FM11r4) = t-"(u, t"M,.Fo) = t-"(M111u, F4) _ (t-n-au, ,F4) = (t-n-a Fu, 95),
so Mrir. = t-n-au.
If U E C°°(lf8" \ (0)) is homogeneous of degree a, then by Lemma 5.12 the finite-part extension f.p. u is a temperate distribution on 1l8", and, recalling (5.12), the Fourier transform of f.p. u is given by
(.F f.p. U, 0) = (f.p. U, ) =
dco
JIwI=1
for g5E S( (5.15)
We can express Raq in terms of the one-dimensional Fourier transform in Lemma 5.10.
Lemma 5.14 If X E 1[8" \ {0}, then
Rac(x) = (n +n-1(S
x), ( ))
for 4 E S(R").
40+11-1, (px)). To express the Proof By (5.11), we have (f.p. function p H c (px) as a one-dimensional Fourier transform, we make the substitution = i;1 + tx/Ix12, where t = i; x and thus '_r is the orthogonal
projection of onto the hyperplane normal to x. In this way,
(px) = J 0 f tl=o
t l dal dt
_
(p),
Homogeneous Distributions
174
where
0,'(0 =
1+
Ixl JX =o
IX12x
dal,
so
Rac(Px) = (f.p p++n 1, O^x (P)) = (na+n-! OS) Now see Exercise 5.11.
Together, (5.15) and Lemma 5.14 show that the Fourier transform of f.p. u is given by
f.p.u,q5)= f
))dw for
u(w)(n
Iwl=
ES(R'), (5.16)
and similarly, the inverse Fourier transform of f.p. u is given by
u(w)(II +,:-1(-x w), q5(x)) dcv for O E S(R").
(.F* f.p. U, 0) Iwl=1
(5.17)
If Re a < I - n so that
E L 1,10c (R), then we may write
(f.p. u(x)) = and
-,C{f.p. u(i )} =
L
IIa+n-1(
co)u(a)) do)
I=1
Iwsee
f
]Ia}n-1(-x w)u(w) dco;
(5.18)
Exercise 5.12 for alternative formulae that do not require any restriction on a.
Change of Variables
We wish to investigate the effect of a change of variable x = K(y), where K : S2" -+ S2 is a Cx diffeomorphism satisfying K (O) = 0,
and S2" and S2 are open neighbourhoods of 0 in R". For any E > 0, let BE={yER,':lyl<E}
Change of Variables
175
denote the open ball in 1R" with radius c and centre 0, so that if u E L 1.10c (Q \ (0} )
and 0 E D(Q), then
J
u(x)O (x) dx = J
(5.19)
Theorem 5.15 below implies the existence of the finite part of the right-hand side of (5.19) when u is homogeneous, or, more generally, when u is the sum of finitely many homogeneous functions and a remainder term that is integrable on Q.
Theorem 5.15 If U E C°°(1R" \ (0}) is homogeneous of degree a, then there exist E > 0 and functions wo, w1, w2, ... and Ro, R1, R2, ... with the following properties: (i) For every m > 0, the composite function u o K admits the expansion m
u(K(Y)) _ > wi(y) + R,. (y) for0 < IYI < E. i=o
(ii) For every j > 0, the function wj is C°O on IR" \ (0) and homogeneous of degree a + j :
wj (ty) = t'
wi (y)
fort >0 and y ER' ' \ {0}.
(iii) For every m > 0, the function R, is CO° on BE \ (0) and, for every multi-index a, IaaRm(y)1 <
C".alylO+m+l-IaI
for0 < IYI <
c.
If, in addition, a = -n - k for some integer k > 0, and u satisfies the parity condition (5.14), then for all j > 0 the parity of the function wj is opposite to that of k - j, i.e.,
wj(-x) = (-1)k-'+1 w! (x) for x E IR" \ {0). Proof Since K is a diffeomorphism, the derivative K'(0) : morphism, and since K (O) = 0, K(Y) = K'(O)y + C(IYI2)
R" is an iso-
as y -- 0.
Letting h(y) = K(y) - K'(0)y, we see that there exists E > 0 such that cIYl < IK'(0)Y + th(y)l < CI yI
for 0 < Iyl < E and 0t1.
Homogeneous Distributions
176
Thus, by Taylor expansion of u about K'(0)y,
lI u(i) (K'(0)Y; h (Y)) + Rm.I (Y),
U (K (Y)) _
j!
j=0
where
1 JrI (1 - t)mu(m+I)(K'(0)y + th(Y); h(Y)) dt.
R,,,,1(y) =
M!
0
In turn, Taylor expansion of h about 0 allows us to write m
h(Y) _ >2 hr(Y) + R,,,,2(Y), r=2
where
hr(Y) = 1 K(r)(0;
r!
and
Y)
R..2 (Y) =
1 m!
ft J 0
(1 - t),,,K(,,,+I)(ty; Y) dt.
Hence,
_
m
u(P) (K'(0)Y; h(Y)) _ > ... >2 u(P) (K'(0)Y; hr, (Y), ... , hrp (Y)) + Rm.p(Y), r,=2
r,,=2
and we see that
w3(y) = > l u(P) (K'(0) Y; hr, (Y), ... , hr,, (Y)), I
P where the sum is over all p > 0, rl > 2, ... , rp > 2 satisfying rl + ...+ rp - p = j
(Notice that j ? p > 0.) Since a"u is homogeneous of degree a - Ial, and hr is homogeneous of degree r, it follows that w j is homogeneous of degree
a - p + rI +
+ rp = a + j. The estimates for Rm(y) follow from the
bounds CIYIa+m+I-laI
IaaRm,t(Y)I <
and
IaaR,,,,2(Y)I < Cmin (1,
IYIm+I-lal)
for 0 < lyl < E. Finally, if a = -n - k and u(-x) = (-1)1+lu(x), then
8"u(-x) =
(-1)1+l-I"I8"u(x)
and so the term wj is homogeneous of de-
gree -n - (k - j), with W j (-Y) =
=
pi (-K'(0)Y; hr, (-Y), ..., hrp (-y)) > u(P) (-1)k+l-P+r,+...+rPwj(Y)
because hr(-y) = (-l)rhr(y).
0
Change of Variables
177
Now consider the left-hand side of (5.19). Since K(QK \ BE) = 2 \ K(BE), the question arises as to whether f.p. Eyo
JS2\K(BE)
u(x),o (x) dx = f.p. Eyo
r
JS2\BE
u(x)o(x) dx.
In fact, Exercise 5.7 shows that these two finite-part integrals can differ, even if K is linear, when a is an integer less than or equal to -n. Once again, we seek first to understand the one-dimensional case.
Lemma 5.16 Suppose n = 1. If a # -1, -2, -3, ..., then 00
f.p.
f
xacb(x) dx = (f.p. x+, 0(x))
(5.20)
(f)
and for any integer k > 0, f.p.
JR\K(BE)
x-k-10(x) dx = (f.p. x-k-1, 0(x))
(5.21)
Proof The case Re a > -I is trivial, so suppose that Re a > -k - 1 and a # -1, -2, ... , -k for some integer k > 1. Integration by parts gives [K (E)]a+10 (K (E))
Ha+1,K(E) ('Y')
a+1
a+1
'
and by Taylor expansion, [K(E)]a+10 (K(E)) =
k-1 0(f)(0)
E
[K(E)] a+j+1 + O(Ea+k+1 )
j=o and
k-l Ea+l+l +
[K (E)]a+j+l =
Cjl l=j
for some cjl E R. Thus, f.p.[K(E)]a+l0(K(E)) = 0, and we have -1
f' P' Ha,K(E)(0) = OE O
f.p. Ha+1K(E)(h'), a+140
Homogeneous Distributions
178
provided the right-hand side exists. Induction on k yields f.p. Ha.K(E)(W) = f.p. Elo
CIO
(-1)kHa+k,K(E)(T (k))
(a+ 1)(a+2)...(a+k) (-1)k Ha+k (O(k) )
(a + 1)(a +
k)'
which, by Lemma 5.3, shows that (5.20) holds. To prove (5.21), let where
f.p. Jk,E(b),
x-k-10 (x) dx.
Jk,E(0) = f \K(BE)
E4.O
If we can show that
Jk(0) =
ki f
(k+1)
(x) log IxI dx,
(5.22)
then (5.21) will follow by Lemma 5.5. Integration by parts gives
JO,E(4) = 0(K(-E)) log IK(-E)I - O(K(E)) log IK(E)I
'(x) log IxI dx, R\K(B,)
and by Taylor expansion, 4 (K (±E)) log IK (±E) I = ¢ (0) log IK'(0)E I + O (clog
1
f,
so (5.22) holds for k = 0. If k > 0, then E(K(E)) k[K(E)]k
=
Jk-1,CW) k
(K (-6)) k[K(-E)]k
+
k
k"¢(i)(0) /
+ f=p
J!
1
[K(E)]k-i
1
1
[K(-E)]k-iJ
+0(6).
Given any integer m > 1, we can define a C°° function f : SZ -+ IR by
(K(y))a=
1
[K'(0)]m
ifyES2\0,
ify=0,
Change of Variables
179
and so f. p. E JO
1
{K(±E)j'n
L p. EO
f (±E) =
f
(±E)m
cnn)
(0) mI
Thus,
Jk(0) =
Jk-l(0') k
and (5.22) follows by induction on k.
For c sufficiently small, the set K(BE) is approximately ellipsoidal and can be described using the function g in the next lemma. Recall that Sn-' = {w E R" : I w I = 1 } denotes the unit sphere in R". Lemma 5.17 There is an co > 0 and a C°O function g : (-EO, EO) X
n S-1
-o- IIB,
such that
K(BE)fl(pw:0
(5.23)
Moreover, g satisfies
g(E, w) = -g(-E, -w) and
IaEg(E, w)I
Cj,
(5.24)
foriEi 0. Proof. Choose po > 0 such that tw E cZ for It1 < po and IwI = 1, and define f (t, w) = sign(t) IK-' (tw)1. Since K-' (0) = 0, Taylor's theorem gives
K-' (tw) = tLw + t2M(t, w), where
L = (K-')'(0)
and
M(t, w) =
f
(1 - s)(K')2(stw; w) ds.
Thus, f (t, co) = t I Lw + tM(t, w) I,
Homogeneous Distributions
180
and, because La) 0 0 for Iwl = 1, there is a pi E (0, po) such that f (t, co) is Cx for Itl < pi and Iwl = 1. The uniform bound 8t f (0, (o) = I L(vI ? c > 0 means that we can apply the implicit function theorem and define g by
f(t,w)=E # t=g(E,w), for I E I < Eo, with co > 0 independent of w. The relation (5.23) follows because, for Ix I < co,
K(x)=pw
p=g(Ixl,w),
and because g(E, c)) is a monotonically increasing function of E. Finally, (5.24)
holds because f (t, w) = -f(-t, -w) and because I a/ f (t, w)1 < Cj for Itl < pi and lwl = 1. We are at last in a position to prove the main result for this section; see Kieser [48, Satz 2.2.12] for the original proof using the calculus of pseudodifferential operators.
Theorem 5.18 Let K : S2" -+ 0 be a diffeomorphism with K(0) = 0, as above,
and suppose that u E C°°(]R" \ 0) is homogeneous of degree a. If a # -n,
-n - 1, -n - 2, ... , or if a = -n - k (k > 0) but u satisfies the parity condition (5.14), then
f.p. J "\K(B,) u(x)c(x) dx = (f.p. u, EO
)
for
E
and hence
f f.p. J \B, u(x)4(x)dx = f.p. Elo E.(.O
u (K (y)) 0 (K (y)) I detK'(y)I dy \B,
for 0 E D(Q). Proof Using the notation in Lemma 5.17 and making the substitution x = pw, we have 00
f
u(x)cb(x) dx = "\K(B,)
JIml=1
u(w)
Jg(E.W)
pa+n-10(pw) dp dw
for0<E <Eo. If a # -n - k, then by Lemma 5.16, f.p.
oo
pa+n-io(pw) dp = Et.o Jg(E,o)
(f.p.
4(pw)) = Ra (w),
Finite-Part Integrals on Surfaces
181
and the result follows at once from (5.12). If a = -n - k and u satisfies (5.14), then
f f
f
u(w)
00
p-k-'O
(pcv)dpdo)
Jg(E,w)
=J
u(-w) J
=
f
u(w) I
g(-E.w)
p-k-)¢(pw)dpdw
-oo
Iw1=)
=
dpdcv
g(E.-w)
L.
21 fl-1=1 u(w)fR\K.(B,)
where Kw is the one-dimensional diffeomorphism defined by K (E) = g (E, w). Hence, it suffices to apply Theorem 5.8 and Lemma 5.16.
Finite-Part Integrals on Surfaces Although we shall avoid making explicit use of the results in this section they nevertheless give some insight into the nature of the surface potentials and boundary integral operators encountered in the later chapters. Consider a C°° surface of the form
F=
(5.25)
and assume that the origin has been chosen so that (O) = 0. By (3.28), if and 0 E D(r), then uE
fr u(x)fi(x)
do. =
f
wi-I
dx'.
u(x', (x'))O(x', (x')) 1 +
For finite-part integrals on 1, we have the following result of Kieser [48, Satz 4.3.9].
Theorem 5.19 Let I' be a C°O surface passing though the origin, as above, and -n + 1, suppose that u E C°°(]R" \ (01) is homogeneous of degree a. If a
-n, -n - 1, ... , or if a = -n + 1 - k (k > 0) but u satisfies u(-x) _ (-1)k+)u(x), then f.p. Elo
f
u(x)O(x) dox \B.
= f.p. E.IA
for 0 E D(r).
f IX'1>E
u(x', (x'))O(x', (x')) 1 +
dx'
Homogeneous Distributions
182
Proof We introduce polar coordinates in R"-', writing
x' = rw,
r = Ix'I,
co = x'/r E S!,-2.
For X E F, we have IX 12 = r2 + (r(0)2, and so
x E F \ BE f r l +
E.
Since (0) = 0, there exists EO > 0 and a C°O function g : (-EO, EO) X S"-2 R such that 1' \ BE = { (x', (x')) : x' = rw, r > g (E, co) and w E Sn-2 }
for 0 < E < co,
with g(E, w) _ -g(-E, -co); cf. Lemma 5.17. The result follows by arguing as in the proof of Theorem 5.18, remembering that now the integral is over RI-I instead of R". Finite-part integrals on surfaces arise as boundary values of functions of the form
f (x) _ f u(x - y),/i(y) day for x E R" \ IF, r
(5.26)
where the integral is divergent if x E F. We shall consider u of the following type.
Assumption 5.20 The distribution u is of the form
u(x) = Jct%x[v(4))
with
P( )
where p and q are homogeneous polynomials satisfying
(degree of p) - (degree of q) = j - 1 for some j > -n + 2, and in addition
q(); 0 forall ER' \{0). Thus, V E C°O(R" \ {0)) is homogeneous of degree j - 1, and so is locally integrable on lid" because j - 1 > -n + 1. By Theorem 5.13, we see that u is a homogeneous distribution on ][l;" of degree -n + 1 - j, and by Lemma 5.12, u is C°° on R" \ {0}. Also, by Exercise 5.8,
v(-l;) = (-1)'-'v(l;)
and
u(-x) = (-1)j-lu(x).
(5.27)
Finite-Part Integrals on Surfaces
183
Figure 4. Integration contours in the definition (5.28) of f }.
To begin with, we investigate the simplest case, when I' = R' ' and so
f(x) =
f
u(x' - Y', x")ir(Y) dy'
for x,,
0.
The next lemma gives an alternative representation for f in terms of the Fourier transform of *. We denote the upper and lower complex half planes by
C+={zEC:Imz>O} and C-={zEC:Imz<0}, put
and let C± and C,- be the closed, semicircular contours shown in Figure 4. If w(z) is continuous for z in the closed half plane Ct U R, and analytic for Iz I > ro and z E C}, then we denote the integral of w over C} by
f w(z) dz =
f
cr
w(z) dz
for r >ro.
By Cauchy's theorem, this integral is independent of r.
Lemma 5.21 If u satisfies Assumption 5.20, and if i,r E S(Ri-1), then
f (x) =
forx E R±,
where
m±(', x") =
f
f
Furthermore,
t-ix,,) = tim±(',
fort >0 and i'540,
(5.28)
Homogeneous Distributions
184 and
m+(-C, -xn) = (_1)r-'m-(', xn). Proof The function f has a natural extension to a distribution on R", namely the convolution u * (i/r 0 8), where (Vr (9 8)(y) = +/r(y')S(y"). Hence, for all 0 E S(II8"),
(f, 0) = (u * (tk 0 8), .F.F*O) = (F[u * ( ® S)], f*O)
= J '*O. Suppose now that supp o c R". In this case, for each i;' the function ¢( ', ) is continuous on C± U R, is analytic on C±, and satisfies where
bounds of the form
4(01:5 CM,N(1 + WD-MO + It U-" fore' E R` I and
E C} U
Hence,
H t"
f
d'n oo
and to shift the contour of integration in the "-plane, we consider the poles of putting Z(i') = ( " E C : q('', l") = 0 ). Since q is homogeneous, we have Z(ti;') = and since the coefficients of the polynomial q(i', ) are continuous functions of ',
for
f
f
00
d ,= =
v( 00
=
f v()() d = f f f
v(
)ei2'
x0 (x) dx d "
x,:) dx.
yR 'RI,
the substitution z = ti;, gives
Finally, since
m±(t ', t-ixn) =
f fv(t)ei2ht x t
dz
trm f(i;, x),
Finite-Part Integrals on Surfaces
185
the substitution z = - gives
and since
M+(-C, -xn) =
dz
Jfc;
f
C,
by (5.27).
which equals
Since m±(k', is a C°C function of x,,, we see that the restrictions f I w. and f IR,. can be extended to C°O functions on R. We now consider the one-sided boundary values of f on the hyperplane x, = 0, given by
f.+(x') =
lim
Z--s(X'.o),ZEWI
f (z) = fR11
-I
mf(',
')e'2rtX d ' for X' E R!'- 1.
Theorem 5.22 Suppose that u satisfies Assumption 5 .20, and that i E S(Il8i-1)
(i) If j < -1, then f+ W) = .t--(x') =
f
u(x' - y', 0)f(y') dy'.
(5.29)
Rn-1
(ii) If j > 0, then
f+ (x') + f_(x') = 2 f.p. E4O
f
IX'-Y'l>E
u(x' - y', 0)i(y') dy',
and the jump in f across the hyperplane x = 0 has the form
f+(x') - f-(x') = E c«a°*(x'), IaI=i
for some coefficients Cc, E C.
Proof Using the sum and difference of m+ and m_, we define us(x')
where
ms( ') =
0) + m-( ', 0),
and
ud(x') =
where md(') = m+(,', 0) - m-(', 0),
Homogeneous Distributions
186
so that
f++f_=u5*ifr
f+-f_=ud*1/r.
and
By Lemma 5.21, m5 and and are homogeneous of degree j, and satisfy
(-1)'-lm5(4') and md(- ') = (-1)'md('), so us and Ud are homogeneous distributions on R" of degree -j - (n - 1), and satisfy
u5(-x') = (-1)1-1u5(x')
and
ud(-x') = (-1)!ud(x').
Since u(x - y) is C°° for x # y, it is easy to see that if x'
supp *, then (5.29) holds (even if j > 0), and in particular f+ (x') - f- (x') = 0. Therefore
suppud(x' - ) c {x'}. If j < -1, then Ud E L111,,(Rn-1) so Ud = 0. If j > 0, then, with the help of Theorem 3.9 and Exercise 5.1, we deduce from the homogeneity of Ud that
Ud*t/t= Ecaaa*, lal=i
for some ca E C. Furthermore, u5(x' - y') = 2u(x' - y', 0) for x'
y', because (5.29) holds when x' f supp *, so the homogeneous distributions u5 and 0) are equal on R"-1 \ {0}. Since both have degree -(n - 1) - j 2f.p. and parity j - 1, we see by Theorem 5.8 (applied in R11- 1, not R") that in fact D 0) as distributions on R' 1. us = 2 f.p. Suppose now that r is the graph of a C°° function We denote the epigraph and hypograph of by
Q' _ {x E 1R" : x" > C(x')}
: IRi-1 -+ R, as in (5.25).
S2- = {x E 1R" : x < (x')},
and
and denote the boundary values oof the function f defined in (5.26) by
ft(x) =
lim
J u(z - y)*(y) day for x E F. r
It is possible to generalise Theorem 5.22 as follows.
Theorem 5.23 As before, let r be the CO° surface (5.25), with (0) = 0. If u satisfies Assumption 5.20, and if r E D(r), then the restrictions f Iu+ and f In-
can be extended to C°O functions on 1R1. Moreover, for x = (x', (x')) E F we have
(i) if j < -1, then /'
f+(x) = f-(x) = J u(x - y)f(y) day; r
Exercises
187
(ii) if j > 0, then
u(x - y)f(y)dav f+(x)+f-(x) =2f.p.J 40 r\e, (x) and, for some coefficients ba E C°O (Rn-1),
f+(x) - f-(x) = E ba(x')8 I (x', (x')). Ia1
We shall not prove this result, because techniques from the theory of pseudodifferential operators would be required; cf. Kieser [48, Satz 4.3.6] or Chazarain and Piriou [ 10, p. 280]. In all subsequent proofs, we avoid using Theorem 5.23. It does, however, help to account for some of our results in Chapter 7.
Exercises
5.1 Show that the Dirac distribution 8 E D* (R") is homogeneous of de-
gree -n. 5.2 Show that if u E CcO(R" \ {0}) is homogeneous of degree a E C as a function on R" \ (0), then for any multi-index a the partial derivative 8"u is homogeneous of degree a - lad on R" \ {0). Show further that if a E D*(R") is homogeneous of degree a as a distribution on R", then 8"u is homogeneous of degree a - IaI on R". 5.3 Show that
d -
f.p. x} = ±a f.p. x f
1
for a
-1, -2, -3, ... ,
-
but
d f.p. x fk =- Fk-f .p. xtk-1 ± (k i)k
6(k) (x)
for any integer k > 1.
Deduce that
dx
f.p. x-k = -k f.p.
x-k-1
5.4 Recall the definition (5.11) of Ra0. Show that if 0 E D(R"), t > 0 and
xEIR"\{0},then R0¢(tx) = t-a-"RaO(x)
for a O -n, -n - 1, -n - 2, ...,
Exercises
188 but
R-n-k 0(tx) = tkR-n-k4 (x) - tk logt E
8 0i )x
I"I=k
a!
for any integer k > 0.
5.5 Show that if u is a homogeneous function in C°°(]RI \ {0}), and if * E
C' (R"), then* f.p. u = f.p.(*u). 5.6 Show that for each 0 E S(R), the function a H H,,(0) is analytic for a ¢ -1, -2, -3, ... , with all simple poles, and residues ck)(0)
res
a=-k-l Ha(4) _
kI
5.7 Show that if A > 0 and 0 E S(R), then Otki0)
f.p. H_k-l,AE(O) = H-k-1(0) -
logA
Ej0
for any integer k > 0. 5.8 Show that for t 0,
M,T= ItI-"FM111,
M,.7 r'*= ItI-"-T*MI1t,
,'Mr = ItI-"M111.P, F*Mt = ItI-"Mllt.'F*. 5.9 Show that f.p. El0
J
x°-1 a-z dx = F(a)
for a E C \ (0, -1, -2, ...}.
E
5.10 Prove Corollary 5.11 directly, i.e., without using Lemma 5.10. For the first part, use (3.17), and for the second part, show that H_A_I (e- i2a
)=
(-12Jt)k+1 2kl
k
for any integer k > 0.
sign (s)
5.11 Show that if u E L1,10c (R) and 0 E D(R"), then
u(t)oa(t)dt,
.ifR." u(a x)q5 (x) dx = FOO
where
f
0. (t) = l
al I
.,.L=O O
(x' + talla) dxl, I
Exercises
189
and dx 1 is the surface element on the (n - 1)-dimensional hyperplane
normal to a. Show further that 0a E D(R), and that if u is a distribution on R, then u(a x) makes sense as a distribution on I[8". 5.12 Derive alternative formulae to (5.12), (5.16) and (5.17), as follows. (i) Show that there exists f E C mP (0, oo) satisfying
r
00
f (t)
dt t
00
=1
and fo
f (t) log t dt = 0.
[Hint: look for f in the form f (t) = Cg (At) for appropriate constants
C>0and),>0.] (ii) Deduce that the cutoff function 1i (x) = f (Ix 1) belongs to Cmp (R" \ {0}), and satisfies
I
W
(tx) t = 1 and
f0*
(tx) logt
dlog
IxI
for x E R" \ {0}.
(iii) Show that if u E C0(R" \ {0}) is homogeneous of degree a, then (f.p. U, ¢) =
J
u(x)*(x)RQ¢(x) dx
for O E S(W").
(iv) Deduce that
TX,g{f.p.u(x)} = (n
*(x)u(x))
and
)7":(f-p-U(0) = (nt+n-1(-S . x), / (f)u( )) 5.13 Show that if u E C°°(Il8" \ (0)) is homogeneous of degree -n - k for some integer k > 0, and if u satisfies the parity condition (5.14), then
",(f.p.u( )} _
fl-k-1 (-x co)u(w)dco
-1
2 1.1=i (i2ir)k+1
4k!
J WI=i
sign(x CO) (X w)ku(w) dw.
5.14 Suppose that K E C°°(R" \ {0}) is homogeneous of degree -n, and that
K(w)dw=0. Iw1=1
Exercises
190
Define
K(x-y)u(y)dy forx ER",
KEu(x)=J ly-xl>E
whenever this integral exists and is finite, and put
Ku(x) = limKEu(x), whenever this limit exists. (i) Show that if u is Holder-continuous and has compact support, then
Ku (x) = I
K(x - y)[u(y) - u(x)] dy
Ix-yl
where px > 0 is any number such that u (y) = 0 for Ix - y I > Px (ii) Show, with the assumptions of (i), that KEu -* Ku uniformly on compact subsets of R. (iii) Show that p.v. K exists and is a homogeneous distribution of degree -n on R. [Hint: see Theorem 5.7.] (iv) Show that Ku = (p.v. K) * u. C II u II Hs (R") for -oo < s < oo, and (v) Deduce that II Ku II H (w') that I Ku IK < C I u I ,,, for 0 < .t < 1. [Hint: use Theorem 5.13 and Lemma 3.15.]
6
Surface Potentials
Following the notation of Chapter 4, we consider a second-order partial differential operator n
n
n
Pu = -TL: aj(Ajkaku) +>Ajaju + Au. J=1 k=1
J=1
From this point onwards, we shall always assume that P has C°O coefficients and is strongly elliptic on W. Thus, Alk, Aj and A are (bounded) C°O functions from R" into CmXI, with the leading coefficients satisfying n
n
Re >2>2[Ajk(x)4k71]* jrl ? CIA121,712
for x,
E R" and rl E Cm .
j=1 k=1 (6.1)
In this and subsequent chapters, we shall develop integral equation methods for solving boundary value problems involving P. Such methods require a twosided inverse for P, or, more precisely, they require a linear operator !9 with the property that
PGu = u = GPu for u E £*(R")"'.
(6.2)
Since P is a partial differential operator, it is natural to seek g in the form of an integral operator:
1
CJu(x) = f G(x, y)u(y) dy for x E W.
(6.3)
R's
The kernel G is said to be a fundamental solution for P, and the same term is also applied to the operator g, although we shall sometimes refer to the latter as a volume potential. We shall also work with a kind of approximate fundamental solution, known as a parametrix, that is generally easier to construct. 191
Surface Potentials
192
The plan of the chapter is as follows. The first two sections set out the main properties of parametrices and fundamental solutions, emphasising the simplest case when P has constant coefficients. Next, we prove the third Green identity, in which the single- and double-layer potentials arise. Following the approach of Costabel [14], we then prove the jump relations and mapping properties of these surface potentials for the case of a Lipschitz domain. The final section of the chapter establishes some relations between the surface potentials associated with P and those associated with P*.
Parametrices A smoothing operator on R" is an integral operator
)Cu(x) _ ! K(x, y)u(y)dy forx E R", whose kernel K is C°O from R" such K satisfies
into C" xm a it is easy to see that any
E(
K : E* (R")'"
Conversely, it can be shown that every continuous linear operator from E* (R") into E(IR")m has a C°° kernel; see [10, p. 28].
A linear operator G : E*(Rn)"' -+ D* (W)"' is called a parametrix for P if there exist smoothing operators K 1 and K2 such that
PGu = u - Klu and GPu = u - /C2u
for U E E* (W')'.
(6.4)
Roughly speaking, a parametrix allows us to invert P modulo smooth functions. Later, we shall write G as an integral operator as in (6.3), and refer also to its kernel G (x, y) as a parametrix for P. When P has constant coefficients, we can easily construct a parametrix via the Fourier transform. Indeed, let PO and P (t;) denote the polynomials corresponding to Po and P, respectively, i.e., n
Po(e) =
(27r)2
j Ajk k j=1 k=1
and
P( j=1
Parametrices
193
For any u E S* (R")'",
and
.7=x_{Pou(x)} = Po( )u( )
{Pu(x)} =
..F
and the strong ellipticity condition (6.1) can be written as for
Re77*Po(')'1 ?
E R" and 17 E C"'.
then Thus, if b = Re r7*b < 17711b1, giving follows that the £2 matrix norm of Po(e)-1 satisfies the bound IPo( )-11 <
cII2
for 0
(6.5) Ib1. It
E R",
(6.6)
so we can find po > 0 such that n 1
Po(t)-l(i27r)>2AJ
J
4 and
1
1
AI < 4
for ICI > po,
and hence C
IP(A)-1I <
ICI'-
(6.7)
for ICI > po.
Fix a cutoff function X E C mP (R") satisfying
and define
Gu(x) =
for I I < 2po,
1
X
,
{[1
- X()lP( )-lu(g)}.
(6.8)
We observe that g is an integral operator as in (6.3) with kernel
G(x, y) = G(x - y),
where G(z) =
{[1
-
)-1 }.
(6.9)
Theorem 6.1 If the strongly elliptic operator P has constant coefficients, then the formula (6.8) defines a parametrix for P, and moreover G : Hs-1(Rn)m -+ Hs+1(R")"'
for -oo < s < oo.
Proof It is easy to see that G : S(R")' -+ S(R")"'
and
9 : S*(R")m -+ S*(R")m,
Surface Potentials
194
and since.F{PGu} = u - xu = .F{GPu} the condition (6.4) is satisfied with K1 u = 1C2u = .F* {x u}, or equivalently, with
Ki(x, Y) = K2(x, Y) = ( 0" X)(x - Y) This kernel is C°° because x has compact support. Also, the estimate (6.7) implies that
((1 + I42)S+,I [1-
IIcuIIH,+.(R")m = fR"
CJR'l(1+I I2>SF,I[1-x( )]I I-2 ()I2d C II u I I H- I (W,),,,
proving the desired mapping property.
In the general case when P is permitted to have non-constant coefficients, a parametrix g can be constructed using the symbol calculus from the theory of pseudodifferential operators; see Chazarain and Piriou [10, pp. 221-224]. The mapping property of Theorem 6.1 remains valid locally, i.e., given any fixed cutoff functions X1, X2 E Cm p(R"), X1 CJX2 :
H'-'(R")
HS+1(1[8")"`
for -oo < s < oo.
(6.10)
The next lemma will help us to describe the behaviour of the kernel G (x, y).
Lemma 6.2 Suppose that v E C°O (ll8" \ {0}) is homogeneous of degree -j for
some integer j > 1. If
u(x) = T4*'.'(f.p. v()}, then the distribution u is locally integrable on 1[8", and is CO° on 118" \ {0}. Moreover,
(i) if 1 < j < n - 1, then u is homogeneous of degree j - n; (ii) if j > n, then u(x) = u1(x) +u2(x)log1xI, where u 1 and u2 are homogeneous of degree j - n and C°O on R" \ {0}, with u2 a polynomial.
Parametrices
195
Proof Part (i) follows at once from Lemma 5.12 and Theorem 5.13, because v is locally integrable on R". If j > n, then by (5.18),
II±-x w)v(w) dw,
u(x) _ ICI=1
and part (ii) follows from Lemma 5.10.
0
We state the next theorem for the general case, but give a proof only for P having constant coefficients.
Theorem 6.3 Assume that P is strongly elliptic with C°° coefficients on R". There exists a parametrix 9 for P whose kernel admits an expansion of the form N
G(x,y)Gj(x,x-y)+RN(x,y),
(6.11)
j=U
for each N > 0, where the functions Go, GI, G2, ... and Ro, RI, R2, ... have the following properties:
(i) For each j > 0, the function G j is C°D on Il8" x (Rn \ (0)), and has the same parity as j in its second argument, i.e., G j (x, -z) = (-1)j G j (x, z). (ii) If 0 < j < n - 3, then G j (x, z) is homogeneous in z of degree 2 - n + j. (iii) If j > n - 2, then G j has the form G j(x, z) = Gj1(x, z) + G j2(x, z) log Izl, where G j I (x, z) and G j2(x, z) are homogeneous in z of degree 2 - n + j, with G j2(x, z) a polynomial in z.
(iv) If0 0. (v) If N + 1 > n - 2, then RN is C2-"+N on R", and
aaRN(x, y) = O(Ix - y12-n+(N+1)--IaI log Ix _ y;) as Ix - yI
0,forIal >2-n+(N+1).
Proof As mentioned above, we assume P has constant coefficients, and consider G given by (6.9). The choice of po ensures that there exists an expansion
Surface Potentials
196
of the form
P( )-1 =
for ICI > Po, j=o
with Vj E CO0(R" \ (0))"'11 rational and homogeneous of degree -2 - j. We define
Gj(z) _ and apply Lemma 6.2 to obtain (ii) and (iii). The expansion (6.11) serves
to define RN, and we see that G j (-z) = (-1)jG j (z) because Vj (-t) _ (-1)jV1(4) Write RN(x, y) = RN,1(x - y) + RN,2(x - y) + RN,3(x - y), where N+I
RN 1 = -X E f.p. V3,
00
RN.2 = f.p. VN+I,
RN,3 = (1-X) E V3.
j=0
j=N+2
We see that RN, I is C°O on R" because RN, I has compact support. Parts (ii) and (iii) imply that RN,2(X, Y) = RN,2(x - y) = GN+1 (x - y) satisfies conditions (iv)-(v). To deal with the remaining term RN,3 (X, Y) = RN,3(x - y), we use the bounds
ITx-, {(-127rz)fla"RN,3(z)}I = la
)I
C(1 +
if -2 - (N + 2) + I al < -n - 1, and thus 8" RN,3 is continuous on I[8" if la l < 3 - n + N.
Indeed, taking fi = 0, we see that .'{8"RN,3} E
Furthermore, by summing over I Ig I = r > 0, we see that I z I' 8" RN, 3 (z) is
bounded for z E IE8" if -2 - (N + 2) + lal - r = -n - 1, i.e., if -r = 2 - n + (N + 1) - Jul < 0, and thus laaRN,3(z)l < CIzl2-n+(N+l)-Ial if
lal>2-n+(N+1). Notice in particular that the parametrix G(x, y) in Theorem 6.3 is CO0 for x # y. We can use this fact, together with the mapping property (6.10), to extend the interior regularity result of Theorem 4.16. Theorem 6.4 Let 521 and 02 be open subsets of R", such that SZ I C= 02, and
assume that P is strongly elliptic with C°° coefficients. For s, t E R, if u E H' (522)"' and f E H-'(02)' satisfy
Pu = f on 522,
Fundamental Solutions
197
then u E HS+2 (cZ 1)' and IIuIIH=+'(sz1),,, < CIIa1IHI(n2)' + C11 f
Proof. Choose an open set 2 c 02 such that SZ1 C= 0 and SZ C Q2, and then choose a cutoff function X1 E C mp(SZ2) such that x1 = 1 on Q. We have
X1U - LC2(Xiu) = GP(xiu) = G(xif) +G[Pxiu - XIPul, and thus IIuIIH+2(&2,)1r' _ IIK2(Xiu) +G(Xif) +G[PXIU - X1Pu]IIH=+2(c1)m
Since K2 is a smoothing operator, II1C2(X1u)IIHs+2(Q,)"' < CIIuIIH,(n2)"', and
it follows from (6.10) that
II(XIGXI)f IIHs+2(n,)-n <
CIIfIIHs(S22)°; Finally, since Pxlu - x1Pu = 0 on S2, and since G(x, y) is C°O for x ¢ y, we have II G[Pxiu - x1Pu]IIHx+2(sn,)^ < CIIUIIH'(02)" 11
One interesting consequence of the above result is that the parametrix is unique modulo smoothing operators. Corollary 6.5 If G1 and g2 are parametrices for P, then !91 - 92 is a smoothing operator, and hence G1 - G2 is C°C on R" x ]R". In particular, it follows that the mapping properties (6.10), and the expansion in Theorem 6.3, are valid for every parametrix of P. Two further consequences are now obvious.
Lemma 6.6 If G is a parametrix for P, then g : D(R")"'
£(R11)"'.
Theorem 6.7 The operator G is a parametrix for P i f and only if G* is a parametrix for P*.
Fundamental Solutions A parametrix g (or its kernel G) is said to be a fundamental solution for P if (6.4) holds with 1C1 = 0 = LC2, i.e., if
PGu = u = GPu
for u E E*(R")'".
Surface Potentials
198
When P has constant coefficients, it is natural to seek a convolution kernel
G(x, y) = G(x - y) with G E S* (LR" )"""" . Indeed, taking Fourier transforms we see that such a G
is a fundamental solution if and only if
I, or equivalently,
PG = 8 on R".
(6.12)
When the polynomial P(i) is homogeneous, i.e., when P( ) = P0(e), then we can easily construct a fundamental solution as follows. Theorem 6.8 Assume that? has constant coefficients and no lower-order terms
(A1=OandA=0). (i) If n = 2, then the formula
f
G(z) ='P*->z{f.p.
[r'(1) - log2rrlw zI]P(w)-' dw
wl=
defines a fundamental solution for P. (ii) I fn > 3, then G (z) = P (l;)-' } defines a fundamental solution for P, and G is homogeneous of degree 2 - n. (iii) If n = 3, then G in (ii) has the integral representation G(z)
=
2z1
f l P(w1)-1 dwl,
where Sz = {w1 E S2 : wl z = 01 is the unit circle in the plane normal to z, and dwj is the element of arc length on Sz . Proof. We see from (6.6) that P (l;) is invertible for C°O and homogeneous of degree -2 for E R" \ {0}.
0. Thus, P (t;)-' is
Suppose n = 2. We know from Lemma 5.12 that f.p. P(l;)-' is a temperate distribution, so G(z) is well defined as the inverse Fourier transform of f.p. P(l;')-', and by Exercise 5.5, f.p.
I = .Fz,. (3(z)},
Fundamental Solutions
199
implying that G is a fundamental solution for P. Since P (-w) = P (w), we see from (5.18) that
G(z) =
12 f.1=1 [II±,(-w z) + II±1(w z)]P(co)-' dc),
II±I() = F'(1) -
sign(se),
2
giving the integral representation for G (z), and completing the proof of part (i). Part (ii) is clear from Theorem 5.13, because for n > 3 the function P ()-1 is locally integrable on R", and thus homogeneous of degree -2 as a distribution on lid".
To prove (iii), we note that by Exercise 5.12, if n = 3 then
G(z) = (IIa (-
()P( )-1)
z),
Moreover, since P(-i) =
for z E R3 \ {0}.
and since we can choose * so that
(- . z) + IIo ( z), ()P( Z(rI Observe that I1 (- z) = IIo (' z) and x+ + x° = 1, so fl + 11 G(z) =
.F{ I) = S, and therefore, if we define cp (') = 1/r () P (t) proof of Lemma 5.14, then, by Exercise 5.11,
G(z) =
_
z),()P()-1) =
Z(8( 1
21zl
t t=0
2(8, 0z) =
=
and Oz (t) as in the
2c6z(0)
P 1)-1 d l
( Y'
Introducing polar coordinates in the plane normal to z, i.e., putting l = pwl with p = and wl = l/p E Si'-, we find that
f
l Z=0
(l)P(Sl)-'d
1 = J J'1E
(Awl)P(Awl)
>O
Sl
(f o
which yields the desired formula for G(z).
Adpdwl
(Awl) dp) P(wl)-' dw1, A
Surface Potentials
200
Of course, in part (i) of Theorem 6.8 we can simply take (6.13)
zl)P(co)-1 dcv,
G(z) = Js (log IW1
because P = Po annihilates constants. We shall not prove any other general existence result for fundamental solutions, although Chapter 9 treats a particular example with A 0 0. Dieudonne [19,
pp. 253-256] discusses the history of existence proofs for fundamental solutions of general classes of partial differential operators. Gel' fand and Shilov [27, p. 122] give a reasonably simple proof for scalar elliptic operators of arbitrary order with constant coefficients, and Hormander [41, Theorem 7.3.10] considers arbitrary (not necessarily elliptic) partial differential operators with constant coefficients. Miranda [67, Theorem 19, VIII] treats second-order elliptic equations with variable coefficients.
The Third Green Identity Let us recall the notation used in our discussion of the transmission property (Theorem 4.20). The set S2- is a bounded Lipschitz domain in JE81, SZ+ is the complementary, unbounded Lipschitz domain, r = a S2+ = 8 S2-, and we have
the sesquilinear forms ct = Ogf, defined by n
n
n
cI(u, v) = f E E(A jkaku)*ajv + (A jaju)*v + (Au)*v dx; t j=1 k=1 j=1 cf. (4.2). The one-sided trace operators for SZ+ and S2- are denoted by y+ and y-, respectively, so that
y±u = (U±) I r
when u = U}1c2± for some Ut E D(1[8")"'
In the usual way, we extend y+ to a bounded linear operator
y} : Hs(c2±)m -*
Hs-1/2(r)m
for 2 < s < 2
The one-sided conormal derivatives of a function u E H2(S2±)m are defined by BV
n j=1
n
n
Vjy} EAjkaku
u
(k=1
and 13
vjy j=1
EA* k=1
with the usual generalisation via the first Green identity, as in Lemma 4.3. Remember also our convention that the unit normal v points out of S2- and into 52+.
The Third Green Identity
201
When u is defined on the whole of R, we sometimes write ut = uIn± for its restriction to Q±. To avoid redundant + and - signs, we write the onesided traces as y+u and y-u instead of y+u+ and y-u-, and similarly for the one-sided conormal derivatives. The jumps in these quantities are denoted by
[u)r=Y+u-y u, Au)r=Bvu-B-u, and we often indicate that a jump vanishes by dropping the + or - superscript; for instance, we write
yu = y+u = y-u if [u]r = 0. The first new symbols are y*, the adjoint of the two-sided trace operator, and B*, defined by
(Y*, 0) = (Vr, YO)r and
for 0 E E(R")m (6.14)
0)
By Theorem 3.38, y0 E HI-Emm for 0 < c < 2, so y*4r makes sense as a distribution on R" for any * E HE-' (F)'". Similarly, since 9.-0 E L.(1')"', makes sense as a distribution on R" for any * E Li (I')"'. we see that Obviously,
supp y*Vr c supp /r c r and
supp B* fr c supp * c r.
Using y* and B*, we can restate Lemma 4.19 as follows.
Lemma 6.9 Let f } E H-I (S2±)'" and put f = f+ + f - E
H_1 (1i$
n )n' and
suppose that u E L2(Rn)'" with u± E H1(SZt)"' If
Pu± = f } on Q±, then
Pu = f + B*[u]r -
on lR".
(6.15)
Now let C be a parametrix for P. Thus, there are smoothing operators 1C1 and 1C2 such that
P9u = u - ICiu and cPu = u - IC2u
for U E E*(Rn)"',
and of course ICI = 0 = IC2 if G is a fundamental solution. Motivated by Lemma 6.9, we define the single-layer potential SL and the double-layer
Surface Potentials
202
potential DL by
SL=Gy* and DL =GBv. Applying G to both sides of (6.15), the third Green identity follows immediately.
Theorem 6.10 If, in addition to the hypotheses of Lemma 6.9, the function u
has compact support in 1R" (and thus also f has compact support in R"), then
u = g f + DL[u]r -
K2u
on R".
The definitions above mean that
(SL V,, 0) = O1i, Yc*0)r and (DL i/r, 0)
1Z\ ifl BAG*0)r' for 0 E E( H
so by considering test functions with supp 0 C= R" \ F, and recalling from Theorem 6.3 and Corollary 6.5 that G(x, y) is C°° for x # y, one obtains the integral representations
SL*(x) =
f
G(x, y)i/i(y) dc,,
(6.16)
DL*r(x) = f [Bv.,G(x, y)*]*1/r(y) day,
(6.17)
for x E 1R" \ r. Notice also that
PSL Jr=y*Vr-ICI y*Vr and PDLKlon R, (6.18)
and hence
P SL Vr = -Ki y** and P DL
on SZ}.
(6.19)
In particular, if G is a fundamental solution, then P SL Vr = 0 = P DL Vr on 52±.
Jump Relations and Mapping Properties The surface potentials SLib and DL* are C°O on S2± because G(x, y) is C°° for x # y. We shall now investigate their behaviour at the boundary F. The results in the next two theorems, for general Lipschitz domains, are from the paper of Costabel [14].
Jump Relations and Mapping Properties
203
Theorem 6.11 Fix a cutoff function x E C o ,(R). The single-layer potential SL and the double-layer potential DL give rise to bounded linear operators
XSL: H-1/2(r )m
H1(l[$n)m,
X DL : Hl/2(r)m
HI(S2:)m,
ySL : H-t/2(r),n
H1/2(r)m,
yt DL : HI/2(r)m
HI/2(11)m,
H-1/2(r)",
B DL : H1/2(r)m
H-1/2(r),n,
B} SL : H-1/2(r)m
and satisfy the jump relations
[SL f]r = 0 and [B SL 1/rl r = -1/r
for* E H-1/2(r)m,
[DL Tf]r = Vr and [B DL *lr = 0
for* E HI/2(r)m.
and
Proof Choose a second cutoff function xt E D(W), satisfying Xt = 1 on a neighbourhood of 0- U r. For * E D(r)m and 0 E D(li8")'n, (xgxtY**,
(xSL 1/r, 4)) =
4)) = (*, Y(xiG*X)4))r,
and by Theorem 3.38 and (6.10),
y : HI (R n)m -* H1/2(r)m
and
X19*X : H-'(R)' -+
HI (Rn)nt,
(6.20) so IIY(xtg*x)4)IIH112(Rn)>-< <
CII4IIH-'(Rn)m. Hence,
I(XSLi/r, 4)I <- CIILrIIH-"2(r)
implying that
I I X SL *I I H I ta pn
II)IIH-'(W)"'
$ C I I r I I H-"i2 (r),,, .
This inequality proves the
mapping property for X SL. The mapping property for ySL and the first jump re-
lation [SL*]r = 0 follow at once because y : HI (Rf)m -+H1 /2(r)m; cf. Exercise 4.5. The mapping property for By SL is proved using the first Green identity. In fact, since P SL 1/r = -Kt y*ilr on 52-, we see by applying Lemma 4.3 that
IIB- SL IIH-itr< C11 SL1/rIIH(n-)` + CIlk] Y*1II If 95 E HI
then
OCIY*+lr, 4))a-I = I(*, YICI4))rl <_ CII'P
II4II
Hcn-n,,
Surface Potentials
204
CIIlfIIH-1/2(rp.,. Hence, using the mapping property of SL just proved, IIBy SLiIIH_112(V)"' < CIIiIIH-,r-(r),,,. Essentially the same argument, applied over SZ+, proves the mapping property for B+ SL. The only complication is that the cutoff function Xi is needed to ensure that X, SL * E H' (S2+)"; the details are left to the reader. SO II)Cly*
The jump [B SL *]r is found by applying the formula (6.15) to the function u = Xi SL 1/r. Indeed, since [u]r = 0 and B: u = B SL 1/', if S2' is an open neighbourhood of SZ- U r on which X, = 1, then
Pu = -1CJy*'tif - y*[B, SL'+/r]r on S2', whereas by (6.18),
Pu = y*c/r - JC, y** Thus,
on S2'.
[B, SL *1r) = 0, or in other words,
(ilr + [Bv SL if]r, y0)r = 0 for all O E D(R")m, and so * + [13 SL*]r = 0. All properties of the single-layer potential have now been established.
To handle the double-layer potential, we choose k > 0 large enough so that P +X is positive and bounded below on H, '(Q-)'. (Such a ), exists by Theorem 4.6.) Thus, the Dirichlet problem
(P +))u = 0
on S2-, (6.21)
y-u=g onr
has a unique solution u E H' (12_)m for each g E H112(I')m, and we can define the solution operator U : g i-+ u. Recall that this operator was discussed in the final section of Chapter 4. Let * E D(r)'" and define u e L2(R")"' by
u=
U* on S2-,
to
on 52+.
Since Pu = -1Au on Sgt, the third Green identity (Theorem 6.10) gives
u = -AGu + DL[u]r -
1C2u
and since y+u = BV +u = 0,
[u]r = -yU,/r = -z/r
and
[B,u]rr
on R",
Jump Relations and Mapping Properties
205
so
DL,/r=SLBVU r-u-AGu+K2u onR".
(6.22)
The mapping property of XSL and the mapping property (4.38) of B. U imply that IIxSLl5 U*IIH'(R")-' -<
IIH-"(r')n' <
and by (6.10), CIIU
IIx9uIIH2(R'"),,, = II (xcXI)uII
Thus, the mapping property (4.37) of U gives II X DL
II H (sz.+ )m < C II * II H ,n (r)""
The mapping property of y' DL follows at once from (6.20), and we obtain the mapping property of B DL by again using (6.22): IIB
DL,fIIH-,n(r),,, < JIB' SLB-UiI,IIH-'/2(r),,, + IIB:V'-uIIH
(r),,,
+x.IIBv GuII H--n(r),,, + IIBdK2uIIH--n(r)
< CHBu trill
H--r_(r),,, + CII(xlQxl)uIIHI(xu),A
+ IIu1IL2(R")m
-< CII*IIH"12(r)," +CII1!IIH'(sz-)< CIIiIIHii2(1-)m.
Finally, [Gu]r =
0 because Gu E H2(R), and
[SL By if *lr = 0 and
[13 SL 13V U*lr = -Bv U*,
so by (6.22),
[DL,/rlr = -[ulr = * and [13 DL *]r = -Bv U,/r -
r
_ -13v u = 0, proving the jump relations for DL.
O
Next, we show that the mapping properties of the surface potentials can be extended to a range of Sobolev spaces. Note, however, that the results below are not quite the best possible; see the discussion following the proof.
Theorem 6.12 Fix a cutoff function X E C mp (R'), and assume that -1 <
s<2,
Surface Potentials
206
(i) For the single-layer potential, we have XSL : Hs-1/2(r)m -+ Hs+l(Rrt)m, ySL : Hs-1/2(r)m --+ Hs+1/2(r)m
(ii) If P satisfies the hypotheses of Theorem 4.25, then the solution operator for the Dirichlet problem (6.21) satisfies U : Hs+1/2(r)m -+ Hs+I (Q-) M,
and for the single- and double-layer potentials we have BV SL :
Hs-I/2(r)nn
Hs-1/2(r)m,
XDL : Hs+1/2(r)m -+ Hs+l (Qf)M, Hs+1/2(r)m, y} DL : Hs+1/2(r)m
B DL : Hs+1/2(r)m
Hs-1/2(r)m
Proof We prove (i) by generalising the corresponding part in the proof of Theorem 6.11. Using, instead of (6.20),
y : H-s+l (R1)m _ H-s+ 1/2(r)'n X1JC*X :
H-s-I (Rf)m
and
H-5+1 (r')"',
(6.23)
we have IIY(XIGX)OIIH-+L/z(r)-, < CII0IIH-f-t(R,,), , and hence
I(XSLXi*,O)I = I(*1 Y(Xig*X)O)rl To prove (ii), let V : to (6.21):
CII
l
H v be the solution operator for the dual problem
(P* + A)v = 0
on Q-,
y-v=0 on1. By Theorem 4.25, B0U :
H3+1/2(r)m
-+ Hs-1/2(r)m and B0V : Hs+1/2(r)m _+ Hs-1/2(r)m, (6.24)
and also
U : L2(r)m -* L2(0-)m.
(6.25)
Jump Relations and Mapping Properties
207
We choose a number p large enough so that Ix I < p/2 for all x E S2-, and put
Q ={xES2+:IxI
up : H'/2(I')m -a H'
(0p)"'
defined by Up +g = u, where u is the unique solution of the Dirichlet problem
(P+A)u = 0 on Qp,
y+u=g onl', yp u = 0
on l'p,
and yp is the trace map from H' (Q +)R' onto H'/2(l'p)' . Now let g E D(I')"', and define w E L2 (W)"' by on S2-,
Llg
w= U,g onSZp, 0
on S2+ \ S2p+.
Define u on FP to be the inward unit normal to Q+, and let SLp denote the singlelayer potential on the (disconnected) surface 8S2p+ = F U T. Since [w]rur,, = 0 and
-.kw
on c2
,
Pw = -Aw on S2p , 0
on S2+\S2p,
the third Green identity (Theorem 6.10) implies that
w = c(-)Aw) - SLp[Bvw]rur,, + Kew on R". Using (6.25), we see that 119(-Aw) + K2W
CIIgwIIH2(n-)., +
CIIwIIL2(n-)
<- C II w L2(C2-UQ ), III +CIIU gIIL2(n+)"< CIIUOIIL2(0-)-
_< CI1811L2(ry <- CIIgI1Hf+-12(r)-
Surface Potentials
208
and by part (i) and (6.24), IISLp[13vw]rur,IlH=+-(n-)- < CII[Bvw]rur,11Hs-1i2(rur,)n'
< C11Bvug11H=-v2(r)" +C11Bvup
g1IH.,-!ntrp,1
+C11BvUp giIH=-112(r,) s7 C11911 H=+,n(r),,, .
Hence, IIldgIIH,+i(n-). = 11wIIH.,+-c92-gym < C IIgllH-+-n(F)
.
Next, consider the operator B. SL, and let *, ¢ E D(I')m. By the second Green identity (Theorem 4.4),
((P + )) SL *, Vc)n_
- (SL
(P* + A)VO)
_ (Y SL *,13-v V-0) r - (13v SL *, Y VO) r,
and since P SL i/r = -)C1y*i/r and (P* +.l)VO = 0 on S2-, and y-VO _ 0 on r', we see that (13- SL ti, O)r = (y SL,/r,13- Vcb)r + (r, YKi Vcb)r - X(SLi/r, VO)n-. Therefore, I(Bv SLR, o)rl < CIIY SLitIIH.,+w(r) ,s Ilav vOIIH-,--n(r)+ CII / II Hf-112trr,l II YKi V-PII H-=+W(r)-
+ CII SL i/rIIL2(n-)N, CII*IIHs-u2(r)(IIOIIH-,+1i2(t)'' +
and since C II O II H-s+112(r)Hi , the mapping property of 5- SL follows. We can handle 13+ SL in essentially the same way. The results for the double-layer potential now follow from (6.22). Indeed, C11XDL i11HI+I(Wk)"'
CIIxSL13v
4'(Q )" + CIIxullH:+1 (nf)N,
+C11x(KZU - XgU)11H2(R )C11Bv U*11H=-112(r)µ +C11U'IIH-+' (n-)-, + C11U11L2(W) CII
II Hs+"n(rr +
CIIu*IIH.,+1(n-)- < CIIillH.-u2(ryi,
and
II B: DL C1113
II
SLXivUSGIIHS-1/2(F) + 11B' uIIHs_t/2(r),
Jump Relations and Mapping Properties
209
+x1IBy9u11L,(r),,, +
CIlB,u*1IH-n(r),,, CIIfIIH.,+i,z(r),,, +
C11* IIHs+-n(r),,,
proving the mapping property for B DL = BI DL.
O
The proof of Theorem 6.12 breaks down if s = ±2. In particular, the trace operator no longer satisfies the mapping property in (6.23); cf. Theorem 3.38. Nevertheless, it turns out that all of the conclusions of Theorem 6.12 are valid for - 2 < s < Z , even when the principal part of P is not formally self-adjoint,
as required in Theorem 4.25. However, the proofs for the cases s = ±z are difficult, and rely on techniques from harmonic analysis. Of course, the estimates for s = f2, in combination with the interpolation property of the Sobolev spaces, yield a proof of the case - z < s < i independent of the one given above. A detailed discussion of harmonic analysis techniques for elliptic equations on nonsmooth domains is beyond the scope of this book. Nevertheless, in view of the importance of the mapping properties for s = ± , a few pointers to the i the survey by Jerison literature may be appropriate. Two useful early papers are and Kenig [43] on the Dirichlet and Neumann problems for the Laplacian on
a Lipschitz domain, and the study by Fabes, Jodeit and Riviere [21] of the classical method of surface potentials for the Laplacian on a C' domain. A key ingredient in [21] is the fact, proved by Calder6n [9], that if F is C' (or even Lipschitz but with a sufficiently small Lipschitz constant), then the Cauchy integral defines a bounded linear operator on L2(F). Subsequently, Coifman,
McIntosh and Meyer [11] extended Calderon's result to the case when r is Lipschitz (without restriction on the size of the Lipschitz constant), after which Verchota [102] was able to extend the results of [21] to Lipschitz domains. Later, other elliptic equations were treated, including the Stokes system [221 and the
equations of linear elasticity [17]; see also the survey paper of Kenig [47]. Mitrea, Mitrea and Taylor [68], [69] have only recently treated general strongly elliptic systems. Further historical and bibliographical details up to 1991 appear in a monograph by Kenig [47]. In most of these papers, not only the boundedness of the various boundary operators is of concern, but also their invertibility, a question we shall address in Chapter 7. Also, estimates of the surface potentials in Sobolev norms on the domain do not appear explicitly in most cases. Instead, bounds are proved for the nontangential maximal functions of the potentials and their derivatives, from which the Sobolev estimates follow; see Jerison and Kenig [43, pp. 62-63], [46, p. 145] and [44, Theorem 4.1].
Surface Potentials
210
When r is smooth locally, we can use the transmission property to show that the mapping properties in Theorem 6.12 hold also for s > 1. Recall Figure 3 of Chapter 4.
Theorem 6.13 Let G 1 and G2 be bounded open subsets o G 1 C= G2 and G 1 intersects r, and put
92: =G;nQ} and F =G1fl l
)I
such that
forj=1,2.
Suppose, for some integer r > 0, that 172 is Cr+1.1, and let
E H-r-2(r)"' and (i) If*Ir2 E
- 2 <s < r + 1.
HS-1/2(r2)"', then SLi/ E HS+1(cf)m and II SL `r II W+I (S2t)n-
C II 'Y II H'r-2(r)'n + C II 'Y II H'-1j2(r2)m .
(ii) If 1// Ir2 E HS+112(1'2)', then DL i/r E Hs+' (S2})"' and II DL
11 H.,+'(o
CII'Y
IIH-r-2(r).u + CII*IIH'+112(r2),,,.
Proof. Since the fundamental solution G(x, y) is C°O for x
y, we can as-
sume without loss of generality that supp ,/r C= ['3/2i where 1'3/2 = G3/2 n r and G1 C= G3/2 C G3/2 C= G2. For -Z < s < 2, we can repeat the proof of Theorem 6.12, noting that the mapping properties (6.24) now hold by Theorem 4.21, so no extra assumptions about P are necessary.
For s = r + 1, part (i) follows from Theorem 4.20 because the single-layer
potential u = SL * satisfies Pu = -K1 y*i/r on 01 with uIS2 E (SZZ [ulr = 0, [l3vulr, - -VIE Hr+1/2(1')"' and 1Cly*tfi E Hr(S2z )m. We then obtain (i) for the full range of values of s by interpolation, viewing SL as a linear operator from HS-1/2(1'3/2)n' into HS +' (S21)" Similarly, part (ii) follows because if s = r + 1 , then the double-layer po-
tential u = DL1fr satifies Pu = -1C1on S2:L with ui E Hr+3/2(['2)m, (Evulr = 0 and )C113'1/r E Hr(c )m. Mr,
E H'(S22 )m
Corollary 6.14 Fix a cutoifunction X E C mp(R"). If the whole of IF is
Cr+1.1
for some integer r > 0, and if -? < s < r + 1, then X SL : Hs-1/2(r)m
Hs+1(cf)m and X DL : Hs+1/2(r)m -+ H5+1(Qt)m.
Duality Relations
211
Duality Relations Recall from (4.30) that the first Green identity for S2} takes the form
(D±(u, v) = (Pu, v)n* + (BV 'u, Y}v)r, and from (4.31) that we have also the dual version
I}(u, v) = (u, P*v)nt + (Y}u,13v v)r.
(6.26)
Let f t E H-i (S2})"', V E L2(R")m and v} = vjc E HI (SZ})"', and suppose that
P* v} = f t on Sgt. Putting f = f + + f - E H-` (R")"', and arguing as in Lemma 4.19, we find that
([ulr,
(P*v, ) _ (f, ') +
for O E D(R")"',
or in other words,
P*v = f + 13*[v]r - Y*[8pvlr
on R",
(6.27)
which is the dual version of (6.15). Recall from Theorem 6.7 that G* is a parametrix for P*. Indeed, if ICI and 1C2 are as in (6.4), then
P*9*u = u -1C2u and 9*P*u = u -1Ciu.
(6.28)
Accordingly, we define SL and DL, the single- and double-layer potentials associated with P*, by SL = G*y*
and
DL = G*13*.
Assuming that v has compact support, we may apply g* to both sides of (6.27), and obtain another version of the third Green identity,
v = 9-f + DL[vlr -
1Ci v
on R";
(6.29)
cf. Theorem 6.10. The definitions of SL and DL mean that (SL
Yg0)r and (DL i/r, 0)
Bu
b)r
for q5 E
D(R")"',
Surface Potentials
212
and we see that for x E 1W' \ r, SL VI (X) =
jG(y,x)*lfr(y)dcry
(6.30)
and
DLVr(x) = f [Bv,yG(y, x)]*lIr(y) day;
(6.31)
cf. (6.16) and (6.17). Theorems 6.11-6.13 on the jump relations and mapping properties of SL and DL carry over in the obvious way to SL and DL, with 13, taking the place of B,,; thus,
for* E H-'/2(r)n,
[SL,*i]r = 0 and [Br, SL,f ]r and
[DL ik]r = * and
for* E HI/2(r)m
0
We remark that if P is formally self-adjoint, then 1(9 + G*) is a self-adjoint parametrix for P, which means that g can be chosen to satisfy g* _ g. Obviously, in this case SL = SL and DL = DL. The traces of the single-layer potentials SL and SL satisfy the following duality relation.
Theorem 6.15 If
E H-1/2(r')"1, then
(Y SL*, Or
Y §L
Or.
Proof. Fix a cutoff function X E D(R") with X = 1 on a neighbourhood of r. The operator XC*X : H-1(R")m -> H1(R")m is the adjoint of XcX : H-I(R")"' -+ H1(111;")m, so, noting that
y : H1(]Rn)m -4 H1/2(r)m
and
y* : H-I/2(r')"' -+ H-1(1[8")m,
we have
(y SL*, O)r = (XGXY*4,, Y*c) = (Y*
.
Xc*XY*,) = (9/r, Y SLO)r.
0 The next lemma will help us to relate the trace of the double-layer potential and the conormal derivative of the single-layer potential. The functions KI and K2 are the kernels of the smoothing operators 1C1 and K2 in (6.4).
Duality Relations Lemma 6.16 Suppose U E D(R")"', and put i/r = yu. If X E c2±, then
±DL i/r(x) = b ( G
)*, u) + fn K 2
=f
y)u(y) dy
[G(x, y)Pu(y) + K2(x, y)u(y)] dy
and
±DL* (x) =
u)+f
K,(y,x)*u(y)dy
j [G(y, x)*P*u(y) + Ki(y, x)*u(y)] dy
=
Proof. Taking u = Sx in (6.28), we see that P* G (x, )* = Sx - K2 (x, )* on R". Thus, for X E cZ±, the first Green identity gives
-(K2(x,
u)n ±
(x,
r=
VF(G(x,')*,
u)
= (G(x, )*, Pu),, ± (G(x, )*, We see from (6.16) and (6.17) that
(G(x, )*,13vu)r =
and
DL *(x),
proving the formulae for ± DL * (x). The formulae for ± DL * (x) follow in x) = Sx - KI (., x) on W. 0 the same way, because Theorem 6.17 Suppose U E H' (R")'", and put * = yu E Hh/2(1')"'. (i) For -0 E H-1/2(10"',
ytDLi/r)r =
u) + (Kzy**, u)n = f(l3 SLR, +/r)r
and
±(O, y} DL1r)r =
u) +
u)nc = ±(l3 SL 4r,
)r
(ii) For 0 E H1/2 (ryn' ±(-0, By DL*) r = (P =F (D L 0, u) + (1C2Bv0,
u)c _ ±(Bv DL 0, r)r
and
Ci
u) +
u)si$ _ +(B DL¢, )r
Surface Potentials
214
Proof Theorem 6.3 implies that ayk G(x, y) is locally integrable on R" for 1 < k < n. Hence, the function x H c1 (G (x, )*, u) is continuous on R", and we can show that ¢*(x)(D:F (G (x, )*, u) dcx
=
(SLq5, u).
Jr
For instance, the first of the three terms arising from the definition of '-F is
f
r
O(x)* f (AJk(y)aykG(x, y)*)*a,u(y)dyd6x
=
f
7T
ayk
(fr G(x, y)*O(x) ddx)*A;k(y)8;u(y) dy
(AJkak SLO)*aJu dy. QT-
Thus, by Lemma 6.16,
u) +
Y}DL )r =
fr
O(x)*
f
K2 (x, y)u(y)dydrx
rt$
u) + zf (fr
K2(x,
y)
(x) dax)*u(y) dy
_ (D'(SL0, u) + (1C2* Yu) and the first Green identity (6.26) gives I
yFu)r -(K*Y*4, u)nr ± (I3 SLR, )r,
(SL 0, u) _ (P* SL 0, u) ± (Bv SL
proving the first half of part (i); the second half holds by a similar argument.
To prove part (ii), we use the second formula for DL f in Lemma 6.16, followed by part (i) and the first Green identity. Indeed,
-±(O, B± DL Or = ±(O, By
fr
O(x)*B..X
f
{
[G(x, y)7'u(y)
+K2(x, y)u(y)] dydo,
_ ±(Y DL
(DLO, Pu)nT +
u)
u) + (K2B*0, u)cc, and since P* DLO = -K2B*o on Q:F, another application of the first Green
Exercises
215
identity gives
(DI(DL0, u) + (1C*B*O, u)s = ±(13v DL 0, 1G')r The second half of part (ii) is proved in the same way.
Exercises 6.1 With the notation of Theorem 6.3, show that if P* = P, then G1 = 0. 6.2 By thinking of G (y, x)* as a parametrix for P*, we can apply Theorem 6.3 to obtain an expansion N
G(y, x)* = E G;(x, x - y) + RN(x, y), i=o
where the Gj have the obvious properties. Show that Go (x, z) = Go (x, -z)*.
6.3
Here is another way of deriving the jump relations for the single- and double-layer potentials, assuming that the basic mapping properties of Theorem 6.11 are known.
(i) Show that if * E H-1/2(I')"' and u = SL *, then
([ulr, By ')r = ([Bvul r + i/r, Y0)r
for 4b E D(R")
[Hint: use Lemma 6.9 in combination with (6.18) and (6.19).]
(ii) Show that if 1/r E H1/2(r)"' and u = DL i/r, then
([ulr
- ", B4)r = ([Bvulr, Y')r
for O E D(W)"(iii)
Assume that S2- is C2. Show that if f c- H1/2(F)"' andg E H-1/2(F)"' satisfy
(f, By ')r = (g, YO)r for 0 E D(R")"`, then f = 0 and g = 0. (Hint: Since D(][8")"' is dense in H 2 (W)"', we can use a C2 coordinate transformation to reduce to the case 0:1- _
Rt. For r7l as in Lemma 3.36, if 0 E D(R"-1)"' and 0 = rill/r, then y¢ = 0 and 9,0 = Costabel [14, Lemma 3.5] gives a proof for Lipschitz domains.] 6.4 Fix a cutoff function X E C mp(1R"), and show that
X DL : Ha-1/2(F)' -* Hs(]R')"' [Hint: adapt the proof of Theorem 6.12 (i).]
for-! < s < 2.
216
Surface Potentials
6.5 Let the hypotheses of Theorem 6.13 be satisfied, and suppose that f I E
H- (S2±)"' with f+ having compact support. As in Lemma 6.9, we put f = f + + f - E H-' (R")"', and note that 9f E H101c (R")"' by (6.10), and [9f ]r = 0 by Exercise 4.5. (i) Show that f ]r = 0 if f E L2(IR") (ii) Prove regularity of the volume potential up to the boundary: if f 11 st2 E HT (S22 )'", then 9f" E Hr+2(E2 :)"' and
II9fIIIHr+2(n')m < ClIfIII+ GIIfIIH,«z,,,,. [Hint: use Theorem 4.20.]
Boundary Integral Equations
Using the properties of the surface potentials established in Chapter 6, we can reformulate boundary value problems over the domain n- or S2+ as integral equations over the boundary F. To describe these reformulations, we begin by defining four boundary operators (three if P is formally self-adjoint) in terms of traces and conormal derivatives of surface potentials, and by showing how to write them as integral operators, in some cases with non-integrable kernels. Next, the pure Dirichlet and Neumann problems for the interior domain Stare shown to be equivalent to boundary integral equations of the first kind, for which the Fredholm alternative is valid. The case of mixed boundary conditions is more complicated, because one obtains a 2 x 2 system of integral equations. We establish the Fredholm alternative for this system only when P is formally self-adjoint. The next section treats exterior problems, i.e., boundary value problems for the unbounded domain Q+. In such cases, a suitable radiation condition must be specified, to force appropriate behaviour of the solution at infinity. Finally, we study regularity of the solution to the integral equation when the surface and the data are suitably smooth (at least locally).
Throughout this chapter, G is always a fundamental solution (not just a parametrix) for P, and we implicitly assume whenever P is formally selfadjoint that G(y, x)* = G(x, y).
Operators on the Boundary Recall from (6.16) and (6.17) that the single- and double-layer potentials associated with P are given by
SL*(x) = DL *(x)
Jrr
(x, y) f(y) da,,,
= f[&).G(x, y)*]*fr(y) do, 217
Boundary Integral Equations
218
and recall from (6.30) and (6.31) that the ones associated with P* are given by SL 1/r (x) = DL 1/r (x)
=
jG(y,x)*1fr(y)day, (7.2)
f[13L,yG(y,x)]*1/i(y)day,
for x E W \ r. We will see that all traces and conormal derivatives of these potentials can be expressed in terms of four boundary operators, namely
R = -B, DL : H1/2(r )m - H-1/2(r)m S = ySL : H-1/2(r)m -* H1/2(r)m, H1/2(r)m, T = y+DL +Y-DL: H1/2(r)m T = y+ DL + y- DL : H112(r)m - H 1/2(r)'".
(7.3)
These mapping properties were proved in Theorem 6.11. (The reader may now
wish to turn to the first section of Chapthr 8 and look at the explicit forms for R, S and T in the simplest and most familiar case, i.e., when P = -A.) The duality relations in Theorems 6.15 and 6.17 show that the adjoints of the operators in (7.3) are given by DL : H1/2(r)m
R*
-.. H-1 /2(r)m,
S* = ySL : H-112(r)m -+ H1/2(r)m, (7.4)
T* = B+ SL +BV SL : H-1/2(F)m
H112(r)m,
T* = BV SL +B SL:
H-112(r)n.
H-112(I-)m
From the definitions above, and the jump relations in Theorem 6.11, we obtain the following expressions for the traces and conormal derivatives of the single- and double-layer potentials:
ySL* = S1/r,
ySL1/r = S*1/r,
SL 1r = 1(+1/r+T
B: SL 1/r = 1(+1/r+T*1/i),
(7.5)
y} DL 1/r =
(f1* r + T1/r),
y± DL 1(r = '-2 (f1/r + T>/r),
Z
B DL 1/r = -R1i,
B DL 1/r = -R*1/r.
If the partial differential operator P is formally self-adjoint, then
SL = SL,
DL = DL,
B, = B,,,
T = T,
S* = S,
R* = R,
Integral Representations
219
and so the eight relations in (7.5) reduce to four:
Y-1 DL ,/r = 1(f,' + T*),
ySL i/r = Si/r.
13, DL 1/r = -R*.
13:' SL l/r = Z (::F* + T**),
Theorem 6.12 implies at once that the mapping properties in (7.3) and (7.4)
extend to a range of Sobolev spaces as stated in the theorem below. Note, however,'our discussion of the end-point cases s = ±1 following the proof of Theorem 6.12.
Theorem 7.1 For -1 < s < Z, S:
Hs-1/2(r)en - Hs+1/2(r)m
and
S* :
Hs-1/2(r)m
-+ H$+1/2(r)m , (7.6)
and if P satisfies the assumptions of Theorem 4.25, then
R Hs+1/2(r)m
Hs-1/2(r)m,
T Hs+1/2(r)m
Hs+1/2(r)m,
R* : Hs+1/2(r)m -+ Hs-1/2(r)m, T* Hs-1/2(r)rn -* Hs-1/2(r)n', (7.7)
T : Hs+1/2(r)m
Hs+1/2(r)m,
T* : Hs-112 (nn,
Hs- 1/2(r)ln.
For smooth domains, a larger range of values of s is allowed; cf. Exercise 7.8. Cr+1.1 for Theorem 7.2 If r is same integer r > 0, then the mapping properties in (7.6) and (7.7) hold for -r - 1 < s < r + 1.
Proof The mapping properties for 0 < s < r + 1 follow from Theorems 6.13 and 3.37. We then get the estimates for -r - 1 < s < 0 by duality.
Integral Representations We shall now derive integral representations for each of the eight boundary operators in (7.3) and (7.4).
For p > 0 and x E R", let B,, (x) denote the open ball with centre x and radius p. If n > 3, then, by Theorem 6.3, the leading term Go in the homogeneous expansion of G has degree 2 - n. If n = 2, then Go contains a logarithm. Consequently,
IG(z, y) I dvy < rnB, (X)
CE
for Z E IR" and n > 3,
CE (1 + I log E I)
fort E B, (x) and n = 2,
Boundary Integral Equations
220
and it is easy to see that if, say, * E L,, (r)n`, then
Sr(x) = J G(x, y)if(y) day and S*Vr(x) =
J
G(y, x)*l/r(y) da,, (7.8)
for x E F. Hence, S and S* are integral operators on F with weakly singular kernels. To handle the other six boundary operators, we define
,,*(x) = 2 fr T, **(x) =
\B,(x)
[ B&.yG(x, Y)*]*f(Y) day,
2f \B,(x) B,.xG(Y,x)*y+(Y)day,
TE(x) = 2 /
r\B,(x)
[Bv,yG(Y, x)]**(Y) day,
TEr(x) = 2 J
Bv.., G(x, Y)l(Y) day,
r\Be (x)
REik (x) = - f
R:f(x) = - f
\B,(x)
Bv,x[Bu,yG(x, Y)*]**(Y) day,
Bv,x[Bv,yG(Y,x)]* (Y)day; \B, (x)
cf. the integral formulae for the single- and double-layer potentials given in (7.1) and (7.2). Recalling the definition of By from (4.3) and (4.4), and the definition of By from (4.5), we see that the kernels of the last six integral operators above are given explicitly as follows:
2[B,,,G(x, y)*]* = 2[an+kG(x, Y)Akj(Y) + G(x, y)Aj(Y)]vj(y), 2Bv,xG(y, x)* = 2[an+kG(Y, x)Akj (x) + G(y, x)Aj (x)]*v j (x),
2[Bv,yG(y, x)]* = 2vj (Y)[Ajk(Y)akG(Y, x)]*,
2Bv,xG(x, y) = 2vj(x)Ajk(x)8kG(x, y), (7.9)
-Bv,x[Bv,yG(x, y)*]* _ -vj(x)[Ajk(x)akan+nnG(x, y)Anr1(Y)]u1(Y) - vj (x)[Ajk(x)akG(x, y) Al (y)] vi (y), -Bv.x[Bv,yG(y, x)]* =
x)A,,,1(x)]*vl(x)
- vj(Y)[Ajk(Y)akG(Y,
x)Al(x)]*vl(x).
Here, we have used the summation convention, and note that an+kG(x, y) = a),,. G(x, y).
Integral Representations
221
In general, the six kernels in (7.9) are all strongly singular on the (n - 1)dimensional surface F, because the leading term in the homogeneous expansion
of VG is of degree 2 - n - Ia I. To investigate what happens as e .0, suppose that SZ- is given locally
by x < (x'), and define the directional derivative (x,
d (x', h') = lim
+ t h') - (x')
tlo
t
For X E F, i.e., for x _ (x'), we shall say that r is uniformly directionally differentiable at x if
(x' + h') _ (x') +
h') +o(Ih'I)
as Ih'I - 0.
(7.10)
h') is homogeneous of degree 1, but not necessarily linear, Note that in h'. In order to state our next theorem, we define two subsets of the unit sphere §n-1 c lf8n
T+(x) = {ro E S"-' : w >
' co))
T -(x) = [co E S"-' : w <
d )j.
In the simplest case, when is differentiable at x', we have grad (x'), and so by (3.28),
T±(x)={wES"-1:±v.v(x)>01.
h') = h' (7.11)
Theorem 7.3 Letx E F, suppose that r is uniformly directionally differentiable at x, and define
af(x) = f a}(x) =
f
an+kGo(x, w)AkJ(x)wJ du , a(x)
8,,+kGo(x, -w)*A jk(x)wj dw,
TT- (.x)
where Go(x, x - y) is the leading term in the homogeneous expansion of G(x, y). For* E D(IP)', y" DL * (x) = ±a t (x) i/i (x) + E0 2 TE r (x),
y} 5L *(X) = fa}(x)1(x) + E O 2T
(x),
Boundary Integral Equations
222
and so
Ti/r(x) = [a+(x) - a-(x)l ii(x) +
Elm
TEf(x),
T*(x) = [a+(x) - a-(x)l W (X) + li TE'(x) o Proof Suppose 11r = yu where u E
y+DL*(x) =
f
We know from Lemma 6.16 that r\B.(x)(G(x, .)*3 u),
and since P*G(x, )* = 0 on UT \ BE (x), the first Green identity gives 1012R\B,(x)(G(x, ), u) = 1TE1/f(x) + f
day, 2
naBE(x)
where v is the outward unit normal to S2- \ BE (x) and the inward unit normal
toSZ+\BE(x).Suppose y E c flaBE(x)andputy = x + Ew, where co E Sn-1 Observe that v(y) _ +W, so by (7.9) andTheorem 6.3, [8v,yG(x, y)*]* _ [an+kG(x, x + EW)Akj(X + E(0)
+ G(x, x +EW)Aj(x + EW)](+Wj) +an+kGO(x, -EW)Akj (X)&)j +
O(1
+ I)logEI) ifn=2,
10(6 Z-" )
if n > 3.
Now put
T,±(X) = (W E S" :X+EWES2}}, so that, noting an+kGO(x, -z) = -an+kGO(X, z),
fna
[Bv,yG(x, y)*]*u(y) day B, (x)
=+
f
Taxan+kGO(X, EW)Akj(x)WjU(X + Eco)e'
+O(E(1+IlogEI))
1 O(E)
ifn =2, ifn > 3.
' dW (7.12)
Since an+k Go (x, z) is homogeneous in z of degree 1-n, and since (7.10) implies that
ma ([ }(x) \ l E (x)l U[ 1 E (x) \ T (x)l J= 0, A
E
Integral Representations
223
we see that
lim]
[13,,,yG(x, y)*]*u(y)day =+a+(x)Vr(x),
40 S2T-naBEcx)
giving the formula for y t DL * (x). The expression for T * (x) then follows immediately from the definition of Tin (7.3). The formulae for y± DL *(X) and T >/r (x) follow by a similar argument, with the help of Exercise 6.2.
When I' is sufficiently smooth, the preceding results for T and T simplify, and we can deal with the other four boundary operators; cf. Theorem 5.23. Theorem 7.4 Let X E I' and Mfr E D(P)"'.
(i) If I' has a tangent plane at x, then
and T*(x) = limTEl/r(x).
Tif(x) = lim TE* (x) CIO
CIO
(ii) If r is C1,'` (with 0 < µ < 1) on a neighbourhood of x, then T*l/r(x) = 1imTE Vr(x)
and
T**(x) = 1imTEi/r(x).
E40
CIO
(iii) If r is C2 on a neighbourhood of x, then Ri/r(x) = f.p. RE1/r(x)
and
R*1/r(x) = f.p. RE*(x). CIO
w), and since TI(x) is given Proof. Since a"+kGo(x, -cv) = by (7.11), we see that a-(x) = a+ (x) and so TEi/r(x) -* T i/r(x). In the same way, a+(x) = ii-(x), so T,1/r(x) -+ Ti/r(x). Part (ii) follows from part (i) because, cf. (7.12), the combination
B,,,xG(y, x)* + 13,,,yG(x, y)* = [vi (x)Akj(x)* - vj (y)Aki
(y)*]
x a"+kGo(x, x - y)*
+
=
J 0(1+lloglx-yil) ifn=2, 1 O(Ix - y12-")
O(Ix
-
if n > 3,
you+1-n)
is only a weakly singular kernel on F, and [13,,,yG(x, y)*]* is the kernel of T.
224
Boundary Integral Equations
We now deal with the hypersingular operators. By Lemma 6.16,
R,f (x) = -BL DL * (x) = -B' SLBvu(x)
Em([X3v,.,G(x, )]*, Pu),f ,
where S2E = Sgt \ BE (x). Since P* [Bv x G (x, )]* = 0 on S2E , the second Green identity (Theorem 4.4) gives
+([13v,xG(x, .)]*, Pu)n
= -(Bv[Bv,xG(x, )]*, Yu)a,E *
+ ([B.,,xG(x, .)]
Bvu)a52, .
From (7.9), we see that ')]*,
-(Bv[Bv,xG(x,
yu)anz = RE1f(x)
± SEtnaBf(x)
Bv,x [Bv,yG(x,
Y)*]*u(Y)
day
and
([Bv.xG(x, )]*,Bvu)asa{
-
B,,,xG(x,Y)Bvu(Y)dory, (x)
where v(y) is the outward unit normal to BE(x). Thus,
Rilr(x) =
2
[::FBvu(x) + T*Bvu(x)] + im(REf(x) + CIO
fJ
2T
*Bvu(x)
{Bv,x [Bv.yG(x, Y)*]*u(Y) B, (x)
- A xG(x, y)Bvu(Y)} derv I, and by arguing as in the proof of Theorem 7.3 and noting that co), we find that
(7.13)
Go (x, -(0) _
B,,,xG(x, y)B,u(Y) dcy S2 naBE(x)
f
vj(x)Ajk(x)8kG(x, x + Ew)wl8lu(x + Ew)En-1 dw. (x)
Differentiating the expansion in Theorem 6.3 with respect to x, one obtains
as the leading term
x - y), and since
z) is odd and
Integral Representations
225
homogeneous of degree 1 - n as a function of z, we have Bv,xG(x, y)13vu(y) day, S2+naBf(x)
=
I
-2 1
Vj (x)Ajk(X)a.+kGo(X, w)a,u(x)wl do) + 0(E).
wI-1
Hence, taking the average of the + and - expressions for Ri/r(x) in (7.13), we are left with R,/i(x) = lim RE*(x) +
1
L-naBEx )
13V,x[ VyG(x, y)`]*it(y) day *
1
2 L+fl8B2 (x)
Bv,x[9v.YG(X,y)*]u(y)da,,
1
In view of Theorem 6.3 and (7.9), if we let u,,,l(y) = A,,,,(y)u(y) and ul(y) _ A,(y)u(y), then -,Bv,x[Bv.YG(x,
y)*]*u(y)
= vj(x)Ajk(x)[an+kan+,nGO(x, x - y)un:l(x) + akan+1GO(x, x - y)umt(X) + an+kan+mG1 (x, x - y)unsl (X)
an+ka,,+mGo(x, x - y)8pum!(X)(yp - xp)
- an+kGO(X, x - y)u!(x)]vl(y)
+ 0(jx - y12-',). For the leading term, we apply Exercise 7.2 with f (w) = Aml(x)wl to obtain
f
i naBf(x)
=E
Go (x, -w)
an+kan+mGo(x, x - y)v!(y) day
fT-(x)
8n+kam+kG0 (x, w)wl do) + 0(e).
Each of the remaining strongly singular terms in the integrand has the form
f (x, x - y), where f (x, z) is even and homogeneous of degree 1 - n as a function of z. Since
lim J e10
S2{naBE(.r)
f (x, x - y) day: =
JT'(x)
f (x, co) dw,
the contributions from T+(x) and T- (x) cancel, and part (iii) follows.
Boundary Integral Equations
226
The Dirichlet Problem We now show how the single- and double-layer potentials allow a pure Dirichlet problem to be reformulated as a boundary integral equation of the first kind with a weakly singular kernel.
Theorem 7.5 Let f E H-1(S2-)m and g E H1/2(f')"'. (i) If U E H '(Q-)' is a solution of the interior Dirichlet problem
Pu = f y-u = g
on St-,
(7.14)
on 1',
then the conormal derivative * = BV _U E H-1/2(F)m is a solution of the boundary integral equation
(g + Tg) - y9 f on r,
Sl/r =
(7.15)
2
and u has the integral representation
u=Gf -DLg+SLi/r on Q-.
(7.16)
(ii) Conversely, if ilr E H-112(1F)m is a solution of the boundary integral equation (7.15), then the formula (7.16) defines a solution u E Hi (l-)"' of the interior Dirichlet problem (7.14).
Proof As in Theorem 6.10, we view f as a distribution in H-'(R)' with supp f C 52--; cf. Theorem 3.29 (ii). Suppose that u E H1(Q-) satisfies (7.14), and define u = 0 on the exterior domain Q+. Applying Theorem 6.10, we obtain the representation formula,
u=Gf -DLy-u +SLB.u on Q-,
(7.17)
and then by (7.5),
y-u = y9 f - (-y-u + Ty-u) + SB- u i
on F.
Part (i) now follows from the boundary condition y-u = g. To prove (ii), suppose that i/i E H-112(I')m satisfies (7.15), and define u by (7.16). The mapping property (6.10) of the volume potential, together with those of the surface potentials given in Theorem 6.11, imply that G f , DL g and
SL i/r all belong to H1(Q-), so u E H1(SZ-)m. By (6.2), we have PG f = f on lR", and by (6.19), we have P DL g = P SL 0 on SZ-, so Pu = f on Q-. Finally, y-u = g by (7.5). The next theorem shows that the boundary integral equation (7.15) satisfies
the Fredholm alternative; cf. Theorem 2.33. The method of proof was first
The Dirichlet Problem
227
used by Nedelec and Planchard [76], [74], Le Roux [56], [57], and Hsiao and
Wendland [42], for the case when P is the Laplacian. These authors were all concerned with error estimates for Galerkin boundary element methods, in which context positivity up to a compact perturbation is of fundamental importance for establishing stability.
Theorem 7.6 The boundary operator S = y SL admits a decomposition
S=So+L, in which So : H-1/2 (r)m - H1/2(r')m is positive and bounded below, i.e.,
Re(So*, Or ? CII IIH-1n(r)for* E H-1/2(r)m, and in which L : H-1/2 (r')m -+ H 1/2 (r)m is a compact linear operator. Hence,
S : H-1/2(r)m -) H1/2(r')m is a Fredholm operator with index zero.
Proof. Put u = X SL i/r and v = X SL 4), where 0 E H-1/2(r)" and X E C mp(R") is a cutoff function satisfying X= 1 on a neighbourhood of S2-. Since Si/r = yu and 4 _ -[Bp,v]r, and since Pv = 0 on Q-, the first Green identity implies that
(SQL, fi)r = (yu, By v - By v)r = 'Do- (u, v) + Dn+(u, v) + (L1', 4))r, where
(L1l, 4))r = -(u, Pv)o+.
(7.18)
We have [u]r = 0 so u E H 1(R")m, and hence the strong ellipticity of P implies that Re (Do- (u, u) + Re (Dsy+ (u, u) = Re OR,. (u, u) ? C II U II h l (R,,),,, + (L2i, Or,
(7.19)
where
(L2l,10)r = -C(u, v)R,,. Furthermore, II
IIH-,ncr)y = IIBV u - By uIIH-"n(r)'" < CIIuIIHI(R),-,
(7.20)
Boundary Integral Equations
228
so if L = L 1 + L2 and So = S - L, then So is positive and bounded below, as required. By Theorem 3.27, to show that L : H-1/2(r')"' -+ H112(r')'" is compact, it suffices to show that L:
HE-1(r')"'
is bounded for 0 < E < 1. In fact,
-* H1-E (F)"'
1 has the form
L
L1ifr(x) = f K1(x, y)*(y)dory, r where K1 is C°° on a neighbourhood of r x F, because G(x, y) is C°O forx # y, and Pv has compact support in SZ+. To deal with L2, we apply the CauchySchwarz inequality and obtain I(L2f,' )rl < CIIuIILZ(R"")"IIVIILZ(R")"'
By the mapping property of the single-layer potential in Theorem 6.12, Ilu1ILZ(R")nr < CllullH'+112(Rn)", < CIIfIIH'-I(r)", so
I(L2i,')rl <- CII1IIHI-1(ry"
0
and hence IIL21IIH1-'(r)'" < C11*11HE-1(r)"'
Applying the Fredholm alternative to the boundary integral equation (7.15), we see that a solution exists if and only if
(i(g + Tg) - Y9 f, Or = 0
(7.21)
for every solution 0 E H-1/2(F)of the homogeneous adjoint equation S*q =0. Every such 0 has the form 0 = 9,v where v E HI (Q-)' satisfies
P*v=0 on Q-,
y-v=0 onF, and since (g + Tg) = g + y- DL g, the condition (7.21) is equivalent to 2
(g, B,,v)r = (y [Gf - DLg],gv v)r = O (G f - DLg, v)
by (6.26), since P*v = 0,
= (P[c f - DLg), v)Q_
since y-v = 0,
= (f, On-,
The Neumann Problem
229
which is the same as the condition obtained in Theorem 4.10 for solvability of the (pure) Dirichlet problem.
The Neumann Problem The pure Neumann problem can also be reformulated as a boundary integral equation of the first kind, but this time the kernel is hypersingular.
Theorem 7.7 Let f E H-1 ($Z-)"' and g E H-1 /2(F)'". (i) If U E H 1(Q-)'" is a solution of the interior Neumann problem
Pu = f on Q-,
(7.22)
B.u =g on r,
then the trace i/r = y-u E H 1/2 (F)"' is a solution of the boundary integral equation R,/r
2(g-T*g)-B-9f on l',
(7.23)
and u has the integral representation
u = G f - DL i/r +- SL g
on St-.
(7.24)
(ii) Conversely, if* E H1/2(F)' is a solution of the boundary integral equation (7.23), then the formula (7.24) defines a solution u E HI (Q-)"' of the interior Neumann problem (7.22). Proof. Essentially, one repeats the proof of Theorem 7.5, interchanging g and*, and taking the conormal derivative of the Green representation formula (7.17), instead of its trace. In fact, by (7.5),
B-u =B-9f +Ry-u+2(B-u+T*B-u) on F, 0
giving (7.23).
Next we show that the Fredholm alternative is valid for the boundary integral equation (7.23). As in the proof of Theorem 6.12, we use the notation SZP = {x E Q+ : lxi < p} = n+ n BP,
(7.25)
where the number p is large enough so that IxI < p/2 for all x E 0-. The argument below follows a similar pattern to the one for the Dirichlet problem (Theorem 7.6).
Boundary Integral Equations
230
Theorem 7.8 If P is coercive on H 1(S2-)' and on H' (Q+)"', then the boundary
operator R = -B DL is coercive on H1/2(I')"', i.e., for 1/1 E H1/2(r)m.
Re(R*, O r ? cII*IIHI/2(r)Hence,
R : H1/2(r)'" --* H-1/2(r)m
is a Fredholm operator with index zero.
Proof Let * and 0 belong to H 1/2 (r)' and put u = X DL Vr and v = X DLO, where X E C o ,P(R") is a cutoff function satisfying X= I on a neighbourhood of S2-, with supp X Cc B,,. Since R1/r = and 4) _ [yv]r, and since Pu = 0 on n-, the first Green identity implies that
(Ri,r, 4))r = (-Bu, y+v - Y_Or = (Du-(u, v) + c0P (u, v) + (L,*, 4)r, where (L1*, 0) r =
-(Pu,
By hypothesis,
Re On-(u, u) + Re Z +(u, u) ? c(IIu1IHI(Q-),,, +
(L2'cr, Or,
where (L2*, Or = -C[(u, v)Q- + (u, v)QP ], and we have II
IIY+u - Y-uIIH112(r)< C(IIuIIHl(n-),,, + IIuIIH(9 ),,,).
Hence, if we put L = L, + L2 and Ro = R - L, then R = Ro + L and Re(Ro'1G', Or ? cII IrII
for ir E H1/2(1')"'.
To complete the proof, we will show that
L : H-1/2(r )m --* H1/2(r)m C11
H-l /2 (r),n < C II * II L2 (r),,, In fact, since 1G II L, is an integral operator on F whose kernel is C°° on a neighbourhood of r x r, it suffices to consider L2. By the Cauchy-Schwarz inequality,
is bounded, and so I (L iG', fir') r I S
I (L2'cr, fi)rI <
C(IIuttL2(Q-)H,IIvIIL,(sz-)'N + IIuIIL2(c1P)m IIVIIL2(oP)-n),
and by the mapping property of the double-layer potential in Exercise 6.4, II L2(Q+)m < CII *11 H--n(r)m,
Mixed Boundary Conditions
231
so
I(L2J, Orl < CIIiIIH-I12(r)m1IOIIH-w(rr,, and hence lIL2*IIHI/2(r)", < CII
Mixed Boundary Conditions Elliptic problems with mixed Dirichlet and Neumann boundary conditions can be reformulated as 2 x 2 sytems of boundary integral equations. As in Chapter 4, let
r=rDunurN be a Lipschitz dissection of the boundary. We write
when supp 1 c rD U n,
SDDI _ (S!)Iro and TN*D'k and
RNN* _ (Rf)IrN and TDNr _ (T')Iro
when supp'k
rN U II.
It follows from (7.3) that, at least for s = 0, SDD :
Hs-1/2(rD)m
Hs+112(rD)m,
RNN : Hs+1/2(FN)m
Hs-1/2(FN)"
,
TND
:
Hs-1/2(rD)n1
Hs-1/2(rN),n,
TDN Hs+1/2(rN)m -+ Hs+1/2(rD)m;
cf. Theorems 7.1 and 7.2.
Theorem 7.9 Let f E H-1(0-)'n, gD E H1/2 (I'D)' and gN E H-1/2(1 N)", and consider the mixed boundary value problem
Pu= f
on Q-'
Y _U = gD
on rD, on rN.
BV -U = gN
(7.26)
Extend the Dirichlet and Neumann data to the whole of r, in such a way that gD E H1/2(r)' and gN E H-112 (l'), and define hD E H 1/2 WD)... and hN E H-1/2(FN)m by
on rD, hD = (gD + TgD) - SgN - Y9f 2 hN = Z(gN-T*9N)-RgD-,Q;9f onrN.
Boundary Integral Equations
232
(i) If U E H' (S2-)"' is a solution of (7.26), then the differences Y/D =By U - gN E H-1/2(rD)m
y'N = y U - gD E H1/2(rN)m
and
satisfy SDD
NND ['-i T
-'TDN
*D
hD
RNN
*N
hN
(7.27)
and u has the integral representation
-DL(+/rN'+'gD)+SL(>/rD+gN) on Q_-
(7.28)
E H-'/2(1'D)" and N E H1/2(FN)m satisfy the system (ii) Conversely, if of boundary integral equations (7.27), then the formula (7.28) defines a solution u E H' (S2-)"' of the mixed boundary value problem (7.26).
Proof. Let U E H' (r)"` be a solution of (7.26). Since B. u = 1'D + gN and
y-u =>/rN+gD,weseefrom (7.17)that u = 9f - DL( *N + gD) + SL( *D + gN)
on Q-,
=0
and since u satisfies the boundary conditions, *D = 0 on 1'N, and on 1'D. Hence, by (7.5),
gD = y u = yGf - 2 (-gr + TDN*N + TgD) + SDD7GD + SgN
on FD,
and
gN = BV u = B-9f + RNN*N + RgD + 2
(gN
+ TND 1rD + T*gN)
Of IN,
giving
SDDY'D - 2TDNYN = (gD +TgD) - SgN - yCf 2
on rD,
TND*D + RNN1N = 2 (gN - T*gN) - RgD - 5; g f on I'N,
which is just the 2 x 2 system (7.27). This argument proves part (i).
Conversely, suppose that V'D E H1/2(rD)"' and *N E H-'/2(I'N)m satisfy (7.27). By (6.10) and Theorem 6.11, the equation (7.28) defines a function u E H' 02-)'", and obviously Pu = f on S2-. Finally, by working backwards through the calculations above, we see that y-u = gD on I'D, and ;t3; u = gN on 1'N, proving part (ii).
Mixed Boundary Conditions
233
By putting SDD
A=
IT* 2
ND
1
-2TDN
nb
[*D]
Y
*N
h=
hD hN
RNN
we can write the system (7.27) as
AO =h, and by putting (0, O)rDxrN = (1kD, OD)r0 + (*N, ON)rN,
we have (A/, 0)roxrN = (SDD*D, OD) rD - I(TDNfN, IOD)rD + 2 (TNDtD, ON)rN + (RNN1N, ON)rD.
When P* = P, a simple argument shows that the Fredholm alternative is valid for (7.27). Theorem 7.10 Let H = H-1hI2(F D)"' X H1/2(rN)"'. If 2 isformallyself-adjoint, and if P is coercive on H' (S2-)"' and on H' (S2+)', then
A=Ao+L, where AO : H -* H* is positive and bounded below, i.e.,
Re(Ao,o,'tb)rOxrN ? cIIijIH for' E H, and where L : H -+ H* is a compact linear operator. Hence,
A : H -+ H* is a Fredholm operator with index zero. Proof Let So be as in Theorem 7.6, and let Robe as in the proof of Theorem 7.8.
Thus, S=So+LsandR=Ro+LR,where Ls: H-1/2(r)"
H1/2(r)m
and LR : H1/2(r),n _, H-112(F),n
Boundary Integral Equations
234
are compact linear operators. Noting that T = T because P is formally selfadjoint, we define
A0 _ [(S01rD)Ir.D - ITDN*N W
Lzb =
and
(Ls1D)Iro (LRl N)IrN
2TNDWD -r (R0 N)I rN
In this way, A = Ao + L, the operator L : H --> H* is compact, and
(Aoi, Y')rDxrN = ((So D)Irp,
D)ro
- I(TDNZG'N, *D)rp
fN)rN + ((RokN)IrN, *N)rN.
+ Since
(TN)*o, *N)rN = (*D, TDN*N)rN = (TDN*N,'D)rp, and since supp 1D c I'D U 11 and supp 1/'N c rN U 11, it follows that
Re(Aoi, tP)roxrN = (So1D, y'D)r +
V'N)r
+cDI1NIIH,/2(r),,, = cII
IIH,
as required.
Exterior Problems Integral equation methods are particularly suited to boundary value problems posed on the exterior domain 52+. It turns out that, in general, the solution will
not belong to H1(S2+)"', but only to HI (cZ)'" for each finite p, where 0v is defined as in (7.25). Furthermore, to make use of the third Green identity on S2+ we require a somewhat stronger result than Theorem 6.10, that incorporates a suitable radiation condition. In other words, some assumption about the behaviour of the solution at infinity is needed, and here we shall adopt the approach of Costabel and Dauge [16].
Lemma 7.11 Let u E D*(S2+)"`. If Pu has compact support in Q+, then there exists a unique function .Mu E C°°(R")"' such that
Mu(x) = f G(x,
dory
- j [Ev.yG(x, y)*]*u(y) dory
(7.29)
,
for x in any bounded Lipschitz domain S2 such that S2 U supp Pu C= S2 and where r 1 = 8 SZ I .
Proof. First note that, by Theorem 6.4, the distribution u is C°O on S2+ \ supp Pu. Given X E R11, we define .M u (x) to equal the right-hand side of (7.29)
Exterior Problems
235
with r, the boundary of any ball Q- = B. centred at the origin with radius p large enough to ensure that S2- U supp Pu C- Bp and x c Bp. By applying the second Green identity over an annular region of the form Bp, \ B,o, , one sees that the definition of Mu(x) is independent of the choice of p, because Pu = 0 = P*G(x, )* on Bp, \ Bp, . Similarly, to see that (7.29) holds for x in any bounded Lipschitz domain 01 with Q- U supp Pu C 0, , we apply the second Green identity over Bp \ S2- for any p such that Q7 C Bp. Notice that PMu = 0 on R". We now give the desired version of the third Green identity for Q+. Theorem 7.12 Suppose that f E H-i (9+)m has compact support, and choose po large enough so that
T UI'C=Bpo
and
supp f C-52+x.
If U E D* (52+)m satisfies
Pu = f
on SZ+,
and if the restriction of u to 52+A belongs to H l (S2+)"', then
u = 9f + DL y+u - SL B' u + Mu on 52+.
(7.30)
Proof By Theorem 6.10 with u- identically zero, the representation formula (7.30) holds on the bounded Lipschitz domain S2+ fl &2I, with Mu(x) given by (7.29) for x E SZ
.
(Keep in mind that v is the inward unit normal to S2+ fl
0, on r, but the outward unit normal on r I.) Before considering boundary value problems on 52+, we need to know how M acts on volume and surface potentials.
Lemma 7.13 Fix Z E R". If u(y) = G(y, z), then Mu = 0 on R". Proof Let z, x E R' with x # z. We choose a bounded Lipschitz domain S2such that z E S2- and X E 52+, and define
G(x, y)*
v(y)-{0
for y E Q-
foryE52+
Since P*v = 0 on SZ±, the third Green identity (6.29) gives
v(z) = SL B,, v(z) - DL y-v(z),
Boundary Integral Equations
236 or equivalently,
G(x, z)* =
fr
G(y, z)*B,,,yG(x, y)* day -
Jr
[B,,,,G(y, z)]*G(x, y)* dQy.
Thus,
G(x, z) = J [t3.,,yG(x, y)*]*G(y, z) day -
f
G(x, y)BV.yG(y, z) day,
or equivalently,
u(x) = DL yu(x) and therefore Mu(x) = 0 by Theorem 7.12, because Pu = 0 on SZ+.
Lemma 7.14 Let f E E* (R")'°. If U = 9f, then Mu = 0 on R". Proof. Obviously, Pu = 0 on R'1 \ supp f , and since G(x, y) is C°O for x # y, one sees from Lemma 7.13 that
Mu(x) _ (M.G(x, y), f (y)) = 0 for x E R" \ supp f . Since PMu = 0 on 1R", it follows from the third Green identity that Mu = 0
onR". To state the main result for this section, it is convenient to introduce the notation H11
(Q+),,,
= {u E D* (S2+)m U I Q+ E H 1(SZp)"` for each finite
p > 0 such that Q- C BPI.
We point out that Exercise 7.4 gives some simple sufficient conditions on u for
ensuring that .Mu = 0, in the case when P = Po. Theorem 7.15 Suppose that f E H-1 (SZ+)m has compact support. (i) Let g E H 1/2(I')"'. If U E H11 (Q+)'" is a solution of the exterior Dirichlet problem
Pu = f on St+,
y+u=g onr, Mu = 0
on lR",
(7.31)
Exterior Problems
237
then the conormal derivative Mfr = B+ U E H-112(r)'" is a solution of the boundary integral equation
Si/i = y9 f - (g - Tg) on r,
(7.32)
;
and u has the integral representation
u=Gf+DLg-SL,/i one
.
(7.33)
Conversely, if ili E H-112(I')ry` is a solution of the boundary integral equation (7.32), then the formula (7.33) defines a solution u E H11 (S2+)m of the exterior Dirichlet problem (7.31).
(ii) Let g E H-112(x)"'. If U E Hi, (S2+)'" is a solution of the exterior Neumann problem
Pu = f on 52+,
B+u=g on r, Mu = 0
(7.34)
on R",
then the trace* = y+u E H 112 (r ),n is a solution of the boundary integral equation
Ri/i = By G f - i (g + T*g) on r,
(7.35)
and u has the integral representation
u = G f + DL *- SL g on Q+.
(7.36)
Conversely, if ili E H1/2(I')'" is a solution of the boundary integral equation (7.35), then the formula (7.36) defines a solution u E HIOc, (Q+)... of the exterior Neumann problem (7.34). (iii) Let gD E H1/2(r)and gN E H-1/2(I')", and define hD E H1i2(rD)"' and hN E H-1/2(rN)m by
hD = Y9 f - SgN - (gD - TgD)
on FD,
2
hN = BV +9f - RgD - (gN + T *gN) 2
on FN.
Boundary Integral Equations
238
If u E H11 (S2+)"' is a solution of the exterior mixed problem
Pu = f
on SZ+,
Y+u=gD on I'D, l3+ u = gN
on FN,
Mu=0
on R",
(7.37)
then the differences
*D = l3+ u - gN E H-1/2(rD)n'
and *N = Y+u - gD E
H112(pN)'n
satisfy
-
SDD
RNN
TND
hD
1/JD
Z TDN
j[Nj=[hN]'
(7.38)
and u has the integral representation
u = 9f + DL(i/rN + 9D) - SL(*D +,N)
on Q+.
(7.39)
Conversely, if ilrD E H-1/2 (rD)"' and *N E H1/2(rN)' satisfy the system of boundary integral equations (7.38), then the formula (7.39) defines a solution u E H11 (S2+)"' of the exterior mixed problem (7.37).
Proof. Suppose that u E Hil (S2+) is a solution to the exterior Dirichlet problem (7.31). By Theorem 7.12,
u = 9f + DL y+u - SL l3+ u on 52+,
(7.40)
and then by (7.5),
Y+u = yG f + (y+u + T y+u)
- S13+ u
on F.
2
Using the boundary condition y+u = g, and putting /r = B+ u, we arrive at the boundary integral equation (7.32) and the integral representation (7.33). To complete the proof of (i), suppose conversely that * E H-1/2(F)m satisfies (7.32), and define u by (7.33). Together, (6.10) and Theorem 6.11 im-
ply that u E H11 (S2+)"', and it follows from (6.2) and (6.19) that Pu = f on 52+. Also, y+u = g by (7.5). Finally, Lemma 7.14 shows that M9 f = 0,
M DL g = MG9vg = 0 and M SL Vr = MGy"i/r = 0, so Mu = 0, and hence u is a solution of (7.31).
Regularity Theory
239
The proof of part (ii) proceeds in the same way, except that one takes the conormal derivative of both sides of (7.40), instead of the trace, to obtain
B'u=8'Gf -Ry+u-z(-l3vu+T*B-,) on F. The proof of part (iii) is similar to that of Theorem 7.9.
Regularity Theory In Theorem 4.18, we proved local regularity up to the boundary for solutions to elliptic partial differential equations. A simple argument based on this result yields the following local regularity estimates for the boundary integral equations. Recall the definition (7.25) of the set Q+ Theorem 7.16 Let G 1 and G2 be bounded open subsets of R" such that G 1 C= G2
and G 1 intersects r. Put
Sgt-= G; n c2} and r = Gl n S2+ and suppose, for some integer r > 0, that 1`2 is
for j = 1, 2, Cr+1,1
(i) If */r E H-1/2(I')'" and f E Hr+3/2(r2)"` satisfy
Si/r= f onF2, then t/r E Hr+l/2(rl)"' and IIkIIH1+'/2(r,)< GIIiIIH-"2(r)"
+CIIfIIHr+3/2(r2)m
(ii) If P is coercive on H1(Q-)"' and on H1(c2)"', and if i// E H1/2(I')' and f E Hr+1/2(x2)"' satisfy
R* = f
on I'2,
then i/r E Hr+3/2(F1)"' and II*IIHI+3/2(r,)
< CII
IIH'/2(r), + Cll f IIH1+i/2(r2)"-
Proof. Let G3/2 be a bounded open subset of R" such that G1 C= G3/2 C G3/2 C= G2,. and put 523/2 = G3/2 n Sgt.
Boundary Integral Equations
240
In case (i), the single-layer potential u = SL * E H 1(S22 )"' satisfies
Pu=O one , yu= f on r2, so using the jump relation for I31 SL * (Theorem 6.11), together with the trace estimates (Theorem 3.37) and elliptic regularity (Theorem 4.18), we find that JIB, u
- By uIIH'+1r-(r,)
CIIuIIHr'+2(n3/,)1" + CIIuIIH'+2(niZ)m
CIIuIIH'(sz2)", +CIIu1IH'(n )", +CIIf1IHr+3r_(r2),,,.
The estimate follows because II u II H' (s2±)m < C II *II H-'/2(r)m
In case (ii), the double-layer potential u = DL i E H' (S22 )"' satisfies
Pu=0 onc22, l3,u = f on r2, so
IIfilH'+sn(r1)", = IIY+u - y-UII
CIIu and finally IIuIIH'(12
)m
< C11
IIuII
H'(Q )", + CII f 1I
IIH'/2(r)m
D
We saw in Theorem 7.2 that the mapping properties of the boundary integral operators hold for an extended range of Sobolev spaces when F is smoother than just Lipschitz. The regularity result just proved allows us also to extend the Fredholm property for R and S.
Theorem 7.17 If I' is
C''+1.1 for
some integer r > 0, then
H5+1/2(r) is Fredholm with index zerofor -r < s (i) S : Hs-1/2 (r)"' r, and ker S does not depend on s in this range; (r)m HS-1/2 (r)"' is Fredholm with index zero for -r -1 < (ii) R : s < r + 1, and ker R does not depend on s in this range.
-
Proof We know from Theorem 7.6 that S is Fredholm with index zero when
s = 0. Thus, let 01, ..., ¢p be a basis for kerS in H-1/2(r)'". In fact, by Theorem 7.16 these basis functions belong to
Hr+1/2(r)",, and we can choose
Exercises
241
them to be orthonormal in L2(r')m. The same reasoning applied to S* yields an orthonormal basis 01, ... , BP for ker S*, with each 9 j E H''+1 j2 (r)"` . We can therefore define a compact linear operator
for -r < s < r,
K : HS-1/2(r)m by P
Ki/r = E(Oj, '1` j=1
and a bounded linear operator
A = S + K : HS-112(r)r -+ H:+1/2(f
)m
for -r < s < r.
The operator A is certainly Fredholm with index zero when s = 0, and since
(0j, Silr)r = 0 and (0j, K,lr) = (0j, Or
for all * E H-1/2(r)"',
it is easy to see that the homogeneous equation A* = 0 has only the trivial H-t/2(r)m. Thus, the inhomogeneous equation A*/r = f has a solution in unique solution * E H-112(1,)m for every f E H1j2(F)"`. Furthermore, if f E Hr+112(r)m, then Si/r E Hr+1/2(x)"1 because Kt/r E Hr+112(ryn, and so * E Hi-112(r)"' by Theorem 7.16. It follows that A has a bounded inverse for s = r, and the same is true of A*. Hence, by interpolation and duality, A is invertible for -r < s < r. Therefore, since K is compact, the operator S must be Fredholm with index zero for -r < s < r. Also, ker S does not depend on s, H-r-1/2(1,)"1 satisfies Si/r = 0, then A* = Kiln E Hr+i'2(r)m because if * E A-1 Kulr E Hr-1/2(x)m. The proof of (i) is now complete. and thus i/r = Part (ii) may be proved using the same approach. The allowed range of s is larger because the basis functions for ker R belong to Hr+3/2(r)m
0
Exercises 7.1 Suppose that S2 is a C°O hypograph x < (x'), and let K (x, y) be any one of the six kernels in (7.9). Show that
K(x,y)u(y)day
f.p. J E,o r\B,(f x)
= f.p. ElO
X,-y'I>E
K(x, y', (y'))i(y', (y')) 1 + I grad (y')I2 dy'
242
Boundary Integral Equations
for x = (x') and'* E D(1')'. [Hint: apply Theorems 5.15, 5.19 and 6.3.]
7.2 Suppose that S2- is a C2 hypograph given by yn < C(y'), and assume (without loss of generality) that
(0) = 0 and
grad (0) = 0.
Put 'T'E = {w E Si-1 : Ew E S2+},
and show the following. (i) There exists a = a(E, r7) such that rE+
= IN E Sn-1 co = (r7 cos 9, sin 9), 17 E S"-2, a (E, r7) < 0 < 7r/2).
(If n = 2, then 17ES0={+1,-1}.) (ii) The function a satisfies lima (6, r7) = 0
a=
and
E
o
as ac
1
2
` n-1 n-1
(E r7 cos a).]
(iii) With S+'-' = [Co E S"-1 : (0n > 0}, n/2
f
T,+
f(w)dw =
sin0)dOdr7 10E.0 f (i?(E,nf(r7cos0, Cos 0,
5'-2
=f X
n-1 n-1
f (w) dw - 46
a, a, (0) p=1 q=1
+
f (17, 0)r7p r7q d 77 + 0 (E2). J1-2
(iv) In part (iii), the term in E vanishes if f (-w) = - f (w) for all co E Sn -1.
7.3 Show that the right-hand side of the boundary integral equation (7.23) satisfies
z (8 - T"8) - Ci- Gf, dr)r = 0
for every solution 0 E H1/2(1')'" of the homogeneous adjoint problem
Exercises
243
R*4 = 0 if and only if f and g satisfy
(g, y v)r + (f, v),- - = 0 for every solution v E H 1(Q-)"' of
P*v=0 onQ
,
onl'. [Hint: see the discussion following Theorem 7.6.] 7.4 Assume that P has no lower-order terms (i.e., P = PO), and that G (x, y) _ G(x - y) is as in Theorem 6.8. Show that if u satisfies the hypotheses of Lemma 7.11 and if, as Ix I -+ oo,
u(x) = o(l) and 13,u(x)
o([Ixl log IxI]-1)
when n = 2,
o(IxI-1)
when n > 3,
then Mu = 0 on R". Here, v is the outward unit normal to the ball Bp.
7.5 Show that if Pu = 0 on R, then Mu = u on R. [Hint: apply the third Green identity over the ball BP.] 7.6 The Calderon projection is the linear operator PC defined by
PCO =
y-(SL*2 -DL*I)
,
where
13v (SL *2 - DL *1)
_
*1 *2
Theorem 6.11 implies the mapping property
Pc : H1/2(r)n' X H-1/2(r)m -+ H1/2(r)m
X
H-I/2(r)m
and if 0 = Pct/i, then the function u = SL *2 - DL *1 satisfies
Pu = 0
on Q-,
y-u=01 on I',
B u=42 onr, so u = SL 4)2 - DL 01 by Theorem 6.10. Hence, Pc4) = 45, or in other words PC20 = Pct', demonstrating that Pc really is a projection. (i) Show that
PC=
2(I - T)
S
ER
2
(I + T'*)
Boundary Integral Equations
244
(ii) Deduce that
SR = 4(I - T2), ST* = TS, RT = T*R, RS = 4[I - (T*)2].. 7.7 Recall the definition of the Steklov-Poincare operators 13 V from (4.38). Show that if S : H-'t2(I')'" -+ H1/2(P)"' is invertible, then we have the representations
B,U = R+ 4(I +T*)S-'(1 +T) = 2(I +T*)S-1 and
B,,V = R* + (1 + T *)(S*)-1(1 + T) = (I + T *)(S*)-' 2
4
7.8 Let S2± = R1, and think of r as R"-'. Also, assume P = Pa with constant coefficients, so that is a homogeneous quadratic polynomial, and let G(x, y) = G(x - y) be the fundamental solution given by Theorem 6.8. (i) By applying Lemma 5.21, show that if n > 3, then
S*(x') = J
dal
ms(
for x' E R"-',
where, with the notation (5.28),
The function ms is called the symbol of S. (ii) In the same fashion, show that for n > 2,
R*(x') = J
mR( ')
(')e12ngxd",
where n
MR( O =
J
it
E1: j=1 k=1 n
n
_ -(2X)2 E E Anj (J
ff
SjSkP(S', n)-'
j=1 k=1
(iii) From the homogeneity properties
ms(t') =
and mR(t:;') = tms(1;')
fort > 0,
Exercises
245
deduce that S:
Hs-1/2(RR-1)m
Hs+1/2(R"-I)m
Hs+1/2(wn-1)m
Hs-1/2(Rn-1)m
and
R:
for all s E R. (iv) Let X E C mP(lR") satisfy X = I on a neighbourhood of zero, and consider
u(x) = Ti-1s1l1 - X{OJv( )). where v is rational and homogeneous of degree j - 1, as in Assump-
tion 5.20, but with j < -1 + n. (Thus, v is not locally integrable on RI.) Modify Lemma 5.21 accordingly, and hence show that the mapping property of S in part (iii) holds also when n = 2.
The Laplace Equation
Our development of the general theory of elliptic systems and boundary integral equations is now complete. In this and the remaining two chapters, we concentrate on three specific examples of elliptic operators that are important in applications. This chapter deals with the Laplace operator in R", denoted by If
A
a2 J=1
The Laplacian constitutes the simplest example of an elliptic partial differential operator, and its historical role was discussed already in Chapter 1. After deriving the fundamental solution for the Laplacian, we shall introduce a classical tool from potential theory: spherical harmonics. These functions turn out to be eigenfunctions of the boundary integral operators associated with the Laplace
equation on the unit ball. They are also useful for studying the behaviour of harmonic functions at infinity, leading to simple radiation conditions. The final section of the chapter investigates ker S and ker R, and the sense in which S and R are positive-definite. The operator P = -A has the form (4.1) with constant, scalar coefficients
Ajk = Sjk,
Aj = 0,
A = 0.
Associated with -A is the Dirichlet form (Pn (u, v) =
f grad u grad v dx,
and the conormal derivative (4.4) is simply the normal derivative,
=
au a
. V
Obviously, -A is formally self-adjoint and strongly elliptic; see (4.6) and (4.7).
246
Fundamental Solutions
247
Fundamental Solutions The Fourier transform of -Du is P(4)u(4) where P(l;) = (27r )2 1 1; 12, so by Theorem 6.8 a fundamental solution for -A is
G(x,y)=G(x-y),
where G(x)=4
1
whenn=3.
Exercise 8.1 is the corresponding calculation for n = 2. To give a fundamental
solution for a general n, we denote the surface area of S"-', the unit sphere in W, by
r
r
cia-
7rn/2
dcv=2
Iwl=1
r(n/2)
;
(8.1)
see Exercise 8.2.
Theorem 8.1 A fundamental solution for the operator -A is given by
G(x) =
when n > 3,
1
1
(n - 2)T
IxIn-2
and, for any constant r > 0, by
G(x) = Zn log Ixl
when n = 2.
Proof The Laplacian is radially symmetric (see Exercise 8.3), so it is natural to seek G in the form G(x) = w(p) where p = Ix 1. Since AG = 0 on R" \ (0), Exercise 8.4 shows that w must satisfy
Ld
PIT-t dp p
-l dw
dp = 0
for p > 0,
so
w(p) =
an
1
+ bn
when n > 3,
n - 2 pn-2
or 1
w(p) = a2 109 - + b2 p
when n = 2,
for some constants an and bn. The choice of b is arbitrary, but an is fixed by the requirement that G satisfy (6.12), i.e., by the requirement that -AG = 8
248
The Laplace Equation
on R", or in other words
-(G, AO) _ O(0) for O E D(1R").
(8.2)
Any test function 0 E D(R") has compact support, so we can apply the second Green identity (1.9) over the unbounded domain {x :IxI > c), and arrive at the formula
-
f
G(x)Oq(x) dx =
f
0(x)a,G(x) da,r - J
(x) do-,
X I=E
X I>E
(8.3)
where yr = -x/E. Since grad G(x) = dp IXI = - IXI"
for n > 2,
(8.4)
we have 8,G(x) = -(x/E) grad G(x) = a,, -0` for IxI = E. Thus, by the mean-value theorem for integrals,
f
Ej J
fi(x) dax =
(x)8vG(x) dox = a"
IxI=E
xI=E
for some xE satisfying IxE I = E, whereas
O(E)
JIxI=E
G(x)8vO(x) da,, = 10(cllog,-I)
if n > 3, if n = 2.
Thus, if a" = 1/Ta, then (8.2) follows from (8.3) after sending f 0. An alternative method of determining a" is to apply the third Green identity
to the constant function u = I over the unit ball. One obtains the formula DL 1(0) = -1, from which it again follows that a" = 1 / T . 0 Throughout the remainder of this chapter, G(x, y) = G(x - y) will always denote the fundamental solution from Theorem 8.1. Recalling the definitions of the boundary integral operators S, R and T given in (7.3), we see first that by (7.8), Si/r(x) =
*(Y)
1
(n - 2)Tn Jr Ix
-
yI"-2
d6y
when n > 3,
and
S* (x) =
1* 2,
r
(y) log Ix
day YI
when n = 2,
249
Fundamental Solutions for x E I". By (7.9) and (8.4), the kernel of T is
2a,,,yc(x, y) -
2 v,, (x - y)
T.
IX - Y111
and the kernel of R is 1
.yG(x,y)=
Ix-yln+2
'
then
for n > 2. Notice that if r is
for x, y E r,
so T has an integrable kernel, allowing us to write
T*(x)
=
forxEr;
Y)
2
- An
see Theorem 7.4, and cf. Exercise 8.6. If r is C2, then
R1/r(x) = vl f.p. 6g0
n I
J
f
r
vx . UY'n
\B,(.,,) IX - Y1 yx
V/(y) day
(y - x) vy (x - y) Ix - y1
daY
for x E r; see (7.9). For later use, we conclude this section by considering the eigenvalues of -A.
Theorem 8.2 Let S2 be a bounded Lipschitz domain and let r = rD U 1Z U rN be a Lipschitz dissection of r = 8 Sl. If U E H 1(S2) is a non-trivial solution of
-Au = Au on 7, yu = 0 on rD, 8u=0 onrN, then A > 0. If, in addition, rD intersects every component of r, then A > 0.
Proof Applying the first Green identity, we have 4> (u, u) = (Au, On + (8 u, Yu)r = AIIulI .2(n) I
and obviously 4 (u, u) = II grad u 112,(Q) > 0 and IIU II L2(n) > 0, so A > 0.
Moreover, if A = 0 then OQ(u, u) = 0, implying that gradu = 0 on n, and hence u is constant on each component of 0.
0
250
The Laplace Equation
The same argument yields a uniqueness theorem. Corollary 8.3 Let 0 be a bounded Lipschitz domain. If U E H I (Q) is a solution of the homogeneous mixed problem
-Au=0 one, y u = 0 on I'D,
a,u = 0 on FN, then u is constant on each component of 92. If, in addition, rD intersects every component of r, then u is identically zero on 0.
We remark that, for the first part of each of the preceding two results, the homogeneous boundary conditions could be replaced by the weaker assumption yu)r < 0. that
Spherical Harmonics For each integer m > 0, let P,,, (]R") denote the set of homogeneous polynomials of degree m in n variables, i.e., the set of functions u of the form
u(x) = E aaxa for x E R,
(8.5)
IaI=m
with coefficients as E C. A solid spherical harmonic of degree m is an element of the subspace
xm(R")= (uEPm(1R"):Au =0on W). Apart from the results involving the boundary integral operators, our general approach to the study of spherical harmonics is essentially that of Miiller [70]. Let
M(n, m) = dim Pm(R") and N(n, m) = dimf,"(IR") for n > I and m > 0. (8.6) By a standard combinatorial argument, the number of non-negative integer + a" = m is solutions al, . . . , a" to the equation al +
M(n, m) _
1
1
(m+n- ), n
(8.7)
Spherical Harmonics
251
and since each u E P," (R") has a unique representation m
u(x) _
>2Vk(xI)Xn -k
E Pk(R"-'),
(8.8)
forn > 2 and m > 0.
(8.9)
with Uk
k=0
we see at once that
M(n, m) _ >2 M(n - 1, k) k=0
Also, Po = 71o is just the space of constant functions, and P1 = WI is just the space of homogeneous linear functions, so
M(n, 0) = N(n, 0) = 1 and M(n,1)=N(n,l)=n
forn> 1.
Taking the Laplacian of (8.8), we find after some simple manipulations that m
Au (x) =
[A'vk(x') + (m - k + 2) (m - k + 1)uk_2(x')]Xnl-k k=2
forn>2 and m>2, where A' is the Laplacian on IRn-1. Thus, U E 7-(R") if and only if vk-2(X')
-A'vk (x')
(m-k+2) (m-k+1) fort
with the choice Of V,,,-l EP,,-I(
n
) and vn, E Pm (R
(8.10)
1) being arbitrary.
Hence,
N(n, m) = M(n - 1, m - 1) + M(n - 1, m)
forn>2 and m> 1, ( 8.
11 )
and so it follows from (8.9) that m
N(n,m)=>2N(n-1,k)
forn> 1 and m> 1.
(8.12)
k=0
Furthermore, since (
ifm=Oorl,
1
N(1, m) = t 0 ifm > 2, we have
N(2, m) _
1
2
ifm = 0, if m> 1,
(8.13)
The Laplace Equation
252
and in view of (8.7) and (8.11),
N(n , m) _
2m + n2 2
n-
forn > 3 and m > 0.
(m n+ n 3 3l
(8.14)
In particular, N(3, m) = 2m + 1. Put
7-l," (&'-') = {
: * = u lso-t for some u E H. (R")I.
Corollary 8.3 implies that the restriction map u H u Is- i is one-one, and hence is an isomorphism from 7I,,, (R") onto 7l", (S"- ), so
dimfl,,,(S"-') = N(n, m) forn > 2 and m > O. An element of 7-1,,, (S' ') is called a surface spherical harmonic of degree m.
We now show that 7-l": (S'-') is an eigenspace of each of the four boundary integral operators on S". Later, in Theorem 8.17, we shall see that the spherical harmonics account for all of the eigenfunctions of these operators.
Theorem 8.4 If T = S"-i forn > 3, then T = T * _ - (n - 2) S, Ri/r =
m(m+n-2) /r and SiJr =
2m+n-2
1
2m+n-2
fori/r E 7L (8.15)
Proof. Observe that
forx,yEF,
vy.=y and
(8.16)
so the kernel of T is
2 vy (x-y) =-(n-2)G(x,y), r" Ix - yl" and therefore T = T * = - (n - 2) S. Now suppose that Jr = y u E 71,,, ; with u E 7-l, (Rn). Euler's relation for homogeneous functions, u(y) = mu(y), implies that au
av
=mi/r on IF,
-1) y; a;
(8.17)
and therefore, applying Theorems 7.5 and 7.7,
S(mi/r) =
Ti/r) 2
and
R1* r = 2
T*(mi/r)}.
(8.18)
Spherical Harmonics
253
Since T = T* = -(n - 2) S, it follows that mSi/r = [i/r - (n - 2)Si/r]
Rir = 2m[i/r + (n - 2)S*],
and
Z
0
giving (8.15).
The same method of proof yields the following result in two dimensions.
Theorem 8.5 If F = S', then
T* = T** =
27r
fr k(y) dcr for*
E
L2(F),
(8.19)
and when'i/r E R,, (S'), Ri/r = 0,
Ri/r= 2ir,
Si/r = (log r)i t,
Si/r= 2ir, 1
T ilr = T *ifr = -,lr for m = 0;
Ti/r=T*1/r=0 form> 1.
Proof The relations (8.16) remain valid for x, y E S" kernel of T is now I
1
it Ix-y12
27r
(8.20)
(8.21)
when n = 2, but the
forx,yEF,
implying (8.19). To prove (8.20), suppose that m = 0. We see at once from (8.18) that R +' L= 0. Furthermore, i/r is constant, so by symmetry Si/r is also constant,
and then uniqueness for the solution of the interior Dirichlet problem shows that SL * is constant on the disc Q-. At the origin, we have
SL*(0) =
27r
Ivl=i
log
IYI*(Y)day = (logy)
.
so SL i/r = (log r)ilr on SZ-, and hence Sii = y- SL a/r = (log r)i/ on r. If m > 1, then using (8.17) and the divergence theorem,
T*
T*v/r
27rm, r
8vdc
2rrmfst-Dudx=O.
Thus, (8.21) follows at once from (8.18). We now consider spherical harmonics that are invariant under rotation about the nth coordinate axis.
254
The Laplace Equation
Lemma 8.6 Given m > 0, there exists a unique function u satisfying (i) U E H. (R"); (ii) if A E R"" is an orthogonal matrix satisfying Ae = e, then
u(Ax) = u(x) forx E R"; (iii) U(en) = 1. In fact,
u(x) =
1
ri,-I
(x + ix'
j7)... drl.
(8.22)
fs"-2
Proof. One easily verifies that (8.22) defines a function u satisfying (i), (ii) and (iii). To show uniqueness, suppose that A E IR""" satisfies the assumptions of (ii). It follows that A has the block structure
A= where A' E
1[l;("-l) "t"-ti is orthogonal. Thus, with uk as in (8.8),
u(Ax) _ E vk(A'x')x, -k, k=0
and so the conclusion of (ii) means that vk (A'x') = vk (x') for all x' E IEI;n-1 which in turn means that vk (x') depends only on Ix'j. Hence, every it E P (R")
satisfying (ii) has the form (8.8) with vk(x') = dklxIk, where dk = 0 if k is odd. Exercise 8.4 shows that
O'Ix'Ir = r(n - r
- 3)Ix'jr-2,
so the condition (8.10) for u E T(m (R") holds if and only if the coefficients satisfy
-k(n + k - 3) dk-2 = (m - k + 2)(m, - k + 1) dk
for2
(8.23)
If m is even, then dm is arbitrary and d,,,_ i = 0, whereas if m is odd, then dm_I
is arbitrary and d", = 0. Since u(e") = do, the normalisation condition (iii) fixes u with the requirement that do = 1.
0
Spherical Harmonics
255
With the notation of Lemma 8.6, we define Pm (t) = Pn, (n, t), the Legendre polynomial of degree m for the dimension n > 2, by
1 - t2 + ten, for -1 < t < 1 and r?E Sii-2,
where to =
Pm(t) = u(w),
1 and u (-x) _
noting that u (w) is independent of q. Observe that since u
(-1)nlu (x), the Legendre polynomials satisfy
form>0 and n>2.
Pm(1)=1 and P,n(-t)=(-1)mPm(t) Also, we have the explicit representation nl
Pm(n, t) = >dk(n, m)(1 _
t2)k/2tm-k,
k=O
where the coefficients dk = dk(n, m) are determined by the recurrence relation (8.23) with the starting values do = 1 and d1 = 0.
When n = 2, the integral (8.22) becomes just the sum over n E S° _ (-1, +1), with To = 2, so we find that
1 - t2)m + (t - i 1 - t2)m].
P. (2, t) = 2 [(t + i
(8.24)
If t = cos 0, then t ± i-,,/1 - t2 = e±'O, so P. (2, cos 0) = cos m46,
and therefore Pm (2, t) is the mth Chebyshev polynomial of the first kind. Exercise 8.8 shows that Pm(3, t) is the usual Legendre polynomial of degree m. The fundamental solution for the Laplacian can be expanded in terms of these polynomials, as follows. Theorem 8.7 Let bm = bm (n) be the coefficients in the Taylor expansion 00
(1 _ 1z)n-2
=
bnzm for I z I < 1.
(8.25)
m=O
If 0 < IxI < IYI and x y = Ix IIyI cos0, then 1
IX
00
> bm Pm (n, cos 0)
- yln-2 = n1=0
Ix lIn lyln-2+n1
for n > 3,
(8.26)
The Laplace Equation
256
and m
00
log Ix
= log
1
A
IYI
E
+ in=1
M
Pm(2, cos 0)
.
IYI-
Proof. By Taylor expansion about x = 0, we have 00 1
Ix - y In-2
_
Fm(x, y)
(8.27)
for IxI < IYI,
M=O
where
F.(x,Y) = > a, (y)x"
and
a. (y) =
1
IaI
1
lal=m
Note that F. (x, y) is homogeneous of degree m in x, and of degree - (n -2) -m in y. Taking the Laplacian of (8.27) with respect to x, we see that 00
EOxFm(x,Y)=0 for IxI < IYI, ,n=0
so by uniqueness of the coefficients in a Taylor expansion, Fm
y) E 7-1m (ill;")
for each integer in > 0. Moreover, if A E R" ,n is an orthogonal matrix, then
jAx - Ayl=Ix - Yl,and so Fm (Ax, Ay) = F,,, (x, y).
In particular, Fn, (Ax, y) = Fm (x, y) if Ay = y, and therefore by Lemma 8.6,
Fm (w, C) = b,n P,,, (co ) for co,
E S" -1,
and for some constant b,,,. Thus, Fm (x, Y) = Ixln
IYI-(,r-2)-m
Fm (ti' IyI l
brPm(COS0)
In-2+m '
(8.28)
and by choosing x and y so that I y I = 1 and cos 0 = 1, we have Ix - yI = (Ix12 - 2IxI + 1)1h/2 = 1 - IxI, so the b,n are as in (8.25). The proof of (8.26) is now complete. When n = 2, we proceed in the same way, except that now Ial
a,(Y)_ (ai
8y logIYI'
Spherical Harmonics
257
so if m > 1, then F,n (x, y) is homogeneous of degree m in x, and of degree -m in y. It follows that for some constants bn m
1
Fo(x, y) = log
and F. (x, y) = b,n Pm (2, Cos 0) Ix
lYl
l
form > 1,
iyitm
and by choosing lyl = 1 and cos0 = 1, we see that b,n = 1/m because log
I
lxl = -log(l - lxl)
m 00 m=1
m
for lxl < 1.
The next theorem gives some expansions of general harmonic functions in terms of spherical harmonics. Recall the definition of Mu given in Lemma 7.11.
Theorem 8.8 Write x = pco, where p = Ixl and w = x/p. (i) If /u(x) = O for p < po, then there exist
u(x) = EP "'
(w)
E fn,(Sa-1) such that
for p < P0.
m=0
(ii) If Du (x) = 0 for p > P0, and if M u = 0 on R", then there exist
E
such that
U(X) = EP 00 2
(w)
for p > Po,
when n > 3,
,n=0
and 00
U (X) = (log P)1o(w) +
p-I"Yin. (w)
for p > Po,
when n = 2.
m=1
Proof Let S2- = Bp,,, the open ball of radius po and centre 0, and consider the special case when u = SL 0 for some 0 E L 1(I'). By (8.27), we get the desired expansion for p < po, with ,n (w) _
(n - 2)Tn
Fm(we Y)O(Y) day.
If n > 3, then by interchanging x and y in (8.27) we see that 00
Ix - yin-2
= L F,n(y, x) m=0
for Ixi > IYI,
The Laplace Equation
258
and by (8.28),
F
X)
2+2,,
yl
(XI)
so u has the desired expansion for p > po, with
f
IYIn-2+2mFm(w, y).0 (Y)day. 1 ..(w) = (n - 2)Tn r When n = 2, the only essential change is to the terms with m = 0, which are
Jr
Fo(x, y)o (Y) day = - J (log IYI)O(Y) dcry
if p < po,
Fo(y, x)o (Y) day = -(log p) J 0 (Y) day r
if p > po.
r
and
Jr
Likewise, any double-layer potential u = DL 0 has expansions of the desired form, with *o = 0 in part (ii), because if n > 3, then a
=0 E ava F(x,y)
1
avy IX
M=0
forlxl
Y
and 8
1
°°
1
avy Ix - An' =
a n 2+2,n F. (x, Y)] avy IYI -
for IXI > IYI
In the general case, if Au(x) = 0 for lxl < po, then by applying the third Green identity over SZ- = Ba, , where p, < po, we see that an expansion of the desired form holds for p < pl. In fact, this expansion is valid for p < po, because the ikm cannot depend on the choice of p I. This completes the proof of part (i), and part (ii) follows in a similar fashion with the help of Theorem 7.12.
Behaviour at Infinity Fix po > 0, and define z
xa
_ \Izll
x
for x 0 0.
(8.29)
259
Behaviour at Infinity We call xu the inverse point of x with respect to the sphere aB,,,,. Notice (xa)O = x
and
> po if and only if lx I < po, xO = x if and only if lx I = po, and IxO I < po if and only if Ix I > po. For any subset E C_ III" \ {0}, we write EO _ {xa : x E E}, I
and for any function u defined on E, we define uO, the Kelvin transform of u, by uO(x4)
(?i)i_2u(x) .
(8.30)
Exercise 8.9 shows that
Lu3(xo) _
n+2 xllAu(x),
IPoll
a fact that yields a simple characterisation for the radiation condition Mu = 0; see (7.29) for the definition of Mu.
Theorem 8.9 Suppose that Au(x) = O for Ix I > Po. (i)
When n > 3, the function u satisfies Mu = O on R" if and only if
u(x) = O(Ix12-")
as Ixl
oo.
(ii) When n = 2, the function u satisfies Mu = 0 on R'1 if and only if there is
a constant b such that
u(x) = blog lxl + O(Ixl-') as lxl -* oo. (iii) When n = 2, the function u satisfies u (x) = O (1) as Ix I -a oo if and only if there is a constant b such that
u(x) = b + O(Ixl-') as 1x1 - oo. In this case, Mu = b. Proof The Kelvin transform ul is harmonic on the punctured ball Bp, \ (0), so by applying the third Green identity over an annular domain Bp, \ Bp,, with 0 < pi < p2 < po, we see that uO = v° + v°°, where vO(xa) is harmonic for Ixui < po, and v°O(xa) is harmonic for IxV I > 0, with Mv°° =0 on R".
The Laplace Equation
260
To prove part (i), let n > 3 and write w = x/Ixl = xd/I xd I. By Theorem 8.8, there are surface spherical harmonics *,0, and of degree m, such that 00
v°(xd)
= Y Ixd Im *° ((O)
for Ixd I < P0,
,n=0
and 00
v°O(xd) _
(w)
Ix0
for Ixal > 0.
,n=0
Suppose that u(x) = as I x I -+ oo. This assumption means that ud is bounded at zero, and thus r10 must be identically zero for all m > 0. Hence, O(Ixi2-n)
ud=v° and u(x)
(A) n-2 ud(xa) _ E Po-2+2,n 00
x
(w)
for IxI > p0.
,n=0
We conclude that 8"u(x) = O(Ixl2-"-1,I) for all a, and so Mu = 0 on R" by Exercise 7.4. Conversely, if Mu = 0 on R" then u(x) = O(Ixi2-") by Theorem 7.12.
Now suppose that n = 2. By arguing as above, it is easy to see that if u(x) =bi log lxi +b2 + O(Ixl-1) as lxl -+ oo, then the function 5(x)= u (x) - b1 log Ix I - b2 satisfies Mu = 0, and therefore Mu = b2M 1= b2 by Lemma 7.13 and Exercise 7.5. The converse again follows by Theorem 7.12.
0 Solvability for the Dirichlet Problem
We know from Theorems 7.6 and 7.8 that, for any bounded Lipschitz domain Q-, the boundary integral operators
S : H-1/2(F) - H1/2(1') and R : H'/2(f') -+
H-1/2(1.)
are Fredholm with index zero. The following uniqueness theorem for the exterior Dirichlet problem will help us to investigate ker S. We shall see that complications arise when n = 2. Theorem 8.10 A function u E H11 (Q+) satisfies
Au = 0
on Q+,
y+u = 0 u(x) = if and only if u = 0 on Q+.
on r, O(Ix12-")
as IxI -* 00
(8.31)
Solvability for the Diriehlet Problem
261
Proof. Suppose that (8.31) holds. Applying the first Green identity over S2p = 7+ fl Bp for p sufficiently large, we have (DS2P (u, u) _ -(av u, Y+u)r
=
- fa
da B,,
(-f'_-u(Pw))u(Pw)P"_' dw. dp
Theorems 8.8 and 8.9 imply that
u(pw) = O(p2-n) and
-d u(pw) _ dp
0(p")
if n > 3,
O(p-2)
if n = 2,
so
fzP0(p-')
Igrad ul2dx=fiS2+(u,u)
O(p2-n)
if n > 3,
ifn = 2. (
Sending p -+ oo, we deduce that grad u = 0 on Q+, and thus u is constant on each component of 52+. Since y+u = 0, it follows that u = 0 on 52+. The converse is obvious. 0
Corollary 8.11 Let* E H-1"2(F) satisfy S* = 0 on r. (i) Ifn > 3, then i/i = 0.
(ii) Ifn=2and(l,+/i)r=0,then*/r=0. Proof. The single-layer potential u = SL * satisfies
Au=0 on Q-'-, y±u = 0 on l', and as IxI -+ oo, we have u(x) = O(Ixl2-") when n > 3, but
it(x)=-- (1,OrlogIxl+O(Ixl ') whenn=2. Thus, provided we assume that (1, Or = 0 when n = 2, it follows from Theorem 8.10 that it is identically zero, and hence . =
0.
For the Laplacian, we can prove a stronger version of Theorem 7.6.
Theorem 8.12 Let
V = H-'/2(F)
ifn > 3,
262
The Laplace Equation
and
V =[ * E H-112(1') : (1, lr')r = 0 } if n = 2. The boundary operator S satisfies
(Si,r, Or = J grad SL 1k grad SL 4, dx for r1i E V and O E H-1/2(r), (8.32)'
and is strictly positive-definite on V, i.e.,
(S*, *) r > 0 for all * E V \ (0). Proof. Let 1lr E V and 0 E H-112(r). If p is sufficiently large, and if we put
u = SL * and v = SL 0, then (as in the proof of Theorem 7.6) the jump relations and the first Green identity imply that
(SL, 4,)r = (yu, av v - av v)r = n-(u, v) + ('n+ (u, v) + I Bp BP
The integral over aBp is O(p2-") if n > 3, and is O(p-1) if n = 2. Thus, in either case, we obtain (8.32) after sending p -+ oo.
It is obvious from (8.32) that (S*, Or > 0. Moreover, if (S*, Or = 0,
0
then grad
Corollary 8.13 If n > 3, then S is positive and bounded below on H-1/2(r), i.e.,
(Si, Or ? CII
IIH-iIa(r)
for all',/r E H-112(x).
(8.33)
Proof. By Theorem7.6,S:H-1/2(I') -* H"2 (F) is Fredhoim with zero index, and since S is strictly positive-definite, ker S = (0}. Hence, S has a bounded inverse. Since S-1 is self-adjoint, and since the inclusion H1/2(r) c_ L2(r) is compact by Theorem 3.27, the result follows from Corollary 2.38 with A = S-1.
0 By modifying the boundary integral equation Silr = f and adding a side condition, we obtain a system that is always uniquely solvable, even when n = 2.
Lemma 8.14 Given any f E H1/2(r) and b E C, the system of equations
S*+a= f and (1,*)r=b, has a unique solution ,/r E H-1i2(r) and a E C.
Solvability for the Dirichlet Problem
263
Proof Introduce the Hilbert space H = H-1/2(1') x C, identify the dual space H* with H1"2(I') x C by writing ((i/r, a), (0, b))
0)r + ab,
and define a bounded linear operator A : H --* H* by
A(*, a) = (S* + a, (1, if) r). In this way, A is self-adjoint, and we now show that A has a bounded inverse.
Let So and L be as in Theorem 7.6, so that S = So + L with So invertible and L compact as operators from H-112(f) to H1/2(I'). We define Ao(,b, a) = (So,/r, a)
and
K(,f, a) = (a + L+/r, (l, i/r)r
- a),
H* comso that A = AO + K, with AO : H H* invertible, and K : H pact. By Theorem 2.26, A is invertible if the homogeneous system A (*, a) _ (0, 0) has only the trivial solution. In fact, if
S* + a =0 and
(1,
')r = 0,
then (Si/r, Or = (-a, *)r = -a(1, *)r = 0, so (r = 0 by Theorem 8.12,
and inturna= -S* = 0. Theorem 8.15 There exists a unique distribution,/req E
H-1/2
(F) such that S*eq
is constant on r, and (1, t/req)r = 1. If n > 3, then S*eq > 0.
Proof Let
9'eq be the solution of the system in Lemma 8.14 when f = 0 and b = 1. Thus, Slfeq = -a is constant on r, and by Theorem 8.12, if n > 3, then
-a = -a(1, ifeq)r = (S*eq, lkeq)r > 0. The distribution *eq is real-valued, and is called the equilibrium density for F. If n > 3, then the reciprocal of the positive constant S*eq is called the capacity of r, a quantity we denote by Capr, so that 1
Capr
= Sz/req
when n > 3.
This terminology has its origins in electrostatics: if an isolated conductor carries
a charge Q in equilibrium, so that the potential V is constant throughout the conductor, then the ratio Q/ V does not depend on Q, and is called the capacitance. Mutual repulsion causes all of the charge to lie on the boundary of the conductor, so (with appropriately normalised units) the electrostatic potential is
The Laplace Equation
264
SL 1/r, where i/r is the surface charge density. Thus, Q = ,fr l/r da and V = Si,
and in the case of a unit charge Q = 1, we have /r = *eq, so the capacitance is the reciprocal of S*eq. Now consider the case n = 2, and write S = S, to indicate the dependence on the choice of the parameter r in the fundamental solution from Theorem 8.1. The equilibrium density 1/req is the same for all r, but not so the constant Sr1/req. Since Srl/eq is not always positive, one introduces the logarithmic capacity,
Capr = e2'', so that 1
2n
1
log Capr
_
S1 *,q
when n = 2.
Notice that Sri//
Si* +
(127r, Or
log.1,r,
and hence S,.1/req = 2I log
r CaPr
In particular, Sr1leq = 0 if and only if r = Capr.
Theorem 8.16 Consider S,.: H-1/2(I') -3 H1/2(F) when n = 2. (i) The operator S,. is positive and bounded below on the whole of H-1/2(F)
if and only if r > Capr. (ii) The operator S,. has a bounded inverse if and only if r 0 Capr. Proof. For brevity, put ar = Sr1/req = (27r)-l log(r/Capr). Let Vr E H-1/2(F),
define *0 = * - (1, *)r*eq, and observe that
i/I = *o + (1,1)rieq,
(1, *o)r = 0
and Sr.
/ = Sr'Yo + a,.(1, ifr)r
Also, since (Sri/ro, *eq) _ (*o, Srl, eq)r = 0, we have
(Sri, *)r = (Sr*o, *o) r +a,(1/r. 1)r(1, if)r.
(8.34)
If r < Capr, then (Sr 1/req, +feq) r = ar < 0. To complete the proof of (i), suppose
that r > Capr, or equivalently, ar > 0. By Theorem 8.12, both terms on the right-hand side of (8.34) are non-negative, and the first is zero if and only if *o = 0. Thus, (Sr*, Or 0, with equality if and only if *o = 0 and (1, Or = 0, i.e., if and only if /r = 0. Hence, Sr is strictly positive-definite on
Solvability for the Dirichlet Problem
265
the whole of H-112(F). Arguing as in the proof of Corollary 8.13, we conclude that Sr is positive and bounded below on H-1/2(r).
Turning to part (ii), we note that if r = Capr, then Sr cannot be invertible because Sr lyeq = 0. Thus, suppose that r
Capr and S,. ly = 0. We have Sr ly0 = -ar(l, *)r, hence (S,.1/ro, fo)r = 0, and therefore ly0 = 0 by Theorem 8.12.
In turn, (1,1/r)r = 0 because ar 0 0, giving 1/r = 0. Thus, the homogeneous equation has only the trivial solution, and Sr is invertible.
In the case of the unit sphere r = Sn-1, it is clear from symmetry that 1/req takes the constant value I/ T,,, and in view of Theorems 8.4 and 8.5,
ifn = 2,
1
{(n-2)T,, ifn>3. Further properties of Capr are given in Exercises 8.10 and 8.11, and in the books of Hille [40, pp. 280-289] and Landkof [52]. We conclude this section with an interesting application of Theorem 2.36.
Theorem 8.17 If m 0 1, then 11,,, (Si -1) and f-l1 (Sn-1) are orthogonal to L2(Sn-1). Furthermore, the orthogonal direct sum each other as subspaces of ®O0 (Si-1) is dense in 7-f m=0 m L2(S"-1)
Proof. Recall from Theorems 8.4 and 8.5 that l,n (Sn-1) is an eigenspace of S. Hence, the orthogonality of 7-l n, (Sn-1) and Iii (Sn-1) follows at once from the
fact that S is self-adjoint. We may assume, by choosing r > 1 if n = 2, that S is strictly positive-definite on
H-112(Sn-1).
Since the inclusion
L2(Sn-1) c
L2(Sn-1) is compact H-1/2(Sn-1) is compact, the operator S: L2(Si-1) -+ with ker S = (0). Hence, the eigenfunctions of S span a dense subspace of L2(Si_1),
and to complete the proof it suffices to show that every eigenfunction of S is a spherical harmonic. Suppose for a contradiction that ly E L2(r) is a non-trivial solution of Sly =
µ1y on Si-1 for some (necessarily positive) µ, and that * 1 7-l,n(Si-1) for every m > 0. It follows from Theorem 7.2 that * E C°°(Si-1), so by Theorem 6.13 the single-layer potential u = SL * is C°O up to the boundary of the unit ball. If ly,,, E 7-1,,, (S-1) is as in part (i) of Theorem 8.8, then for 0 < p < 1,
f
- 1=1
00
SL*(pw)1/r(w)dw = 11 p,n m=1
and so, sending p f 1, we see that (S*, l tradiction.
f
,n(W)ly(co)do) = 0,
fJwI=1
0, implying, ly = 0, a con-
The Laplace Equation
266
Solvability for the Neumann Problem Solutions of the Neumann problem for the Laplacian are unique only modulo constants; more precisely, the following holds.
Theorem 8.18 A function u E H' (Q-) satisfies
Au = 0 on cZ-,
(8.35)
a-u=0 onr
if and only if u is constant on each component of 12-. Likewise, u e Hloc(S2+) satisfies
Au=0 avu=0 u(x) = if and only if u is constant on each component of S2+ and, when n > 3, is zero on the unbounded component of 52+.
Proof The first part of the theorem is a special case of Corollary 8.3. The exterior problem is handled in a similar fashion, by applying the first Green identity over S2p , and arguing as in the proof of Theorem 8.10.
Let Q, , ... , Op be the components (i.e., the maximal connected subsets) of S2-, and define
of =
1
on S2 ,
0
on S2- \ S2i ,
for 1 < j < p. Thus, the functions v1, ..., vi,, form a basis for the solution space of the homogeneous interior Neumann problem (8.35).
Theorem 8.19 Let f E H(S2-) and g E H-1/2(p). The interior Neumann problem
-Au = f
a,u=g
on Q-,
(8.36)
onF,
has a solution u E H1(S2-) if and only if the data f and g satisfy
Jsz,
f dx +
f
asp,
gda=0 for t<j
-
in which case u is unique modulo the subspace span {ul, ... , vP}.
(8.37)
Solvability for the Neumann Problem
267
Proof In the present case, the conditions in part (ii) of Theorem 4.10 reduce to
(vi, f)n- + (y vi, 8)r = 0 for 1 < j < p. Let
r denote the components of r, and define the function Xi on r
by
on I'i,
onF\ri, for 1 < j < q. The lack of uniqueness for the Neumann problem gives rise to a lack of uniqueness for the corresponding boundary integral equation; cf. Exercise 8.17.
Theorem 8.20 A function i E H112(r) satisfies R,lr = 0 on r if and only if 1/r is constant on each component of r. Thus, the functions X I, ... , Xy form a basis for ker R.
Proof. If Rill = 0, then the double-layer potential u = DL i/r satisfies
Au = 0
eau=0 u(x) = O(Ix12-") so u is constant on each component of R" \ F, and thus f = [u]r is constant on each component of r. To prove the converse, it suffices to observe that, by the third Green identity, 1
DL Xi
0
inside I'i, outside I 'j,
(8.38)
so RXi = 8 DL Xi = 0 on F. Corresponding to Theorem 8.12 for the operator S, we have the following result on the positivity of R; see also Exercise 8.17.
Theorem 8.21 The operator R : H112(r) -- H-1"2(I') satisfies,
(R*,O)r =
f `r grad DLi grad DL0dx for ,/r, 0 E H1/2(I')
(8.39)
The Laplace Equation
268 and
(Ri, Or ? cll*IIH'r(r) for* E H1/2(P) such that (Xi, Or = 0 for 1 < j < q. Proof. Let i/i, 0 E H'12(1'), and put u = DL i/i and v = DL ¢. As in the proof of Theorem 7.8, we find that for p sufficiently large,
(Ri, Or =
y+v
- y v)r
= (Dn+ (u, v) + fin- (u, v) +
f
da.
aB,,
Since gradu(x) = O(Ixl-") and v(x) = O(Ixi1-11) as IxI - oo, the integral over 8BP is 0(p-"), and we obtain (8.39) by sending p -+ oo. It follows at once from (8.39) that (R 4r, 0r > 0. Moreover, if(Ri/i, r =0, then grad u = 0 on R' \ T, so u is constant on each component of R" \ r, implying that * = [ulr is constant on each component of F. Hence, by Theorem 8.20, (Rill, ill) r = 0 implies Ri/i = 0. Therefore, the coercive, self-adjoint operator R is strictly positive-definite on the orthogonal complement of ker R, and we may appeal to part (iii) of Exercise 2.17.
Exercises 8.1
Let G (x, y) = G (x -y) be the fundamental solution given by Theorem 6.8
when n = 2 and P(4) = (2ir)21'I2. (i) Show that
G(x) =
1
2ir
1
log lxI -F
I"(1) -log27r 27r
- (2,r)2 J 1
2ir
log I sin BI dB.
(ii) Use the substitution 0 = 20 to show that 2n
I 8.2
Deduce (8.1) from
(L: 8.3
log I sin 0 I d9 = -27r log 2.
= Tf
re`dt) ePep"dp = 0
Show that if A E Rnxn is an orthogonal matrix, and if v(x) = u(Ax), then Lv(x) = (Du)(Ax). In particular, when u is harmonic, so is v.
Exercises 8.4
269
Show that if u(x) = w(p), where p = IxI, then
Au(x)dew+n-ldwp
dpe 8.5 8.6
dp \
dp /
T. Jr
(x yl1y) [.
(y)
- *(x)] da,.
Deduce from (8.12) that the numbers N(n, m) have the generating function 00
E N(n,
1+z
m)z,,,
Z)11-1
,,,=o
8.8
p'
dp
d
Show that T1 = -1 and T*+//eq = -*eq. Assume the hypotheses of Theorem 7.3, take P =-A, and let X E r. (i) Show that a:: (x) = -a [T+(x)]/T,,. (ii) Show that limEl.o TE 1(x) _ -1 - [a+ (x) - a-(x)] = -2a+ (x). (iii) Deduce that
T,l/(x) = -*(x) + 8.7
1
for IzI < 1.
Let b,,, = b,,, (n) be as in Theorem 8.7. Show that when n > 3, the Legendre polynomials have the generating function 00
Ebm(n)P.(n,t)zm = m =0
1,
1
(1 - 2tz + Z2)(n-2)/2
and by differentiating both sides with respect to z, derive the three-term recurrence
Po(t) = 1,
b1 P1(t) = (n - 2)t,
(m + 1)b,,,+1 P,n+1(t) - (2m + n - 2)t b,,, P,,, (t)
+(m + n - 3)b,,,-1'n-1 (t)=0 form > 1. Show likewise that when n = 2, 00
,,,=1
1m P»,(2, t)zm = log
1
1 - 2tz + ze
for IzI < 1
and
Po(2, t) = 1 ,
P1(2, t) = t,
Pm+1(2, t) - 2t P,,, (2, t) + Pm-1(2, t) = 0
form > 1.
The Laplace Equation
270
8.9
Recall the definition (8.29) of the inverse point xd with respect to the sphere a Ba, .
(i) Show that
_
axk
po) l2
axk
x;xk (s;k - 2IX11
CIXI
and
" axi ax;
E
4
=
PO
aXJ axk
1=1
Ixl
8jk'
(ii) Consider two curves x =x(t) and y = y(t) that intersect at a point a E II8n when t = 0. Show that
4dx dy
dxO
dy=
PO
dt
dt
(Ia)
dt
dt
when t = 0,
and deduce that the mapping x i--> xI is conformal, i.e., anglepreserving. (iii) Show that if E is a plane or sphere, then so is Ed. [Hint: consider the
equation a Ix 12 + b x + c = 0, where the coefficients a and c are scalars, but b is a vector.] (iv) Think of = xx as a system of orthogonal curvilinear coordinates for x = , and deduce that -2,q
Au =
ICI)
[(PO)2j,-4 au
n
j=j
Next, establish the identity
P, 2-n a
2n-4 au
a; (ICI)
(ICI)
(po) n 2
au a
-2a; a;
a2
+ a; (ICI)
ICI n-2
a2
ICI) PO
n-2
uA
n-2
a2
u
(Iii) PO
and finally conclude that
aupo)
n+2 n
ICI
1
n-2
a2
n+2
n
po
a; (ICI) u =(IXI)
where u° is the Kelvin transform (8.30).
1
a2ud
a;'
Exercises
271
8.10 Let n > 3, and consider the exterior Dirichlet problem
Au = 0
on T2+,
y+u=1
on I', u(x)=O(Ix12-n)
aslxl -+ oo.
Show that the unique solution is u = Capr SL *eq, and deduce that
Capr = -
f
8+ u dcr.
This result can sometimes be used to compute Capr; see Landkof [52, p. 165]. 8.11 Suppose that I' is a simple, closed curve in the complex z-plane, and let w = f (z) define a conformal mapping of S2+ U r onto I w I > 1. Since f is one-one on cZ+, it must have a simple pole at oo; see Markushevich [63, pp. 90-91]. Thus, there is a constant pr such that
f(z)= Z +O(1) asz -* oo. Pr
Moreover, we can assume that pr is real, because the domain I w I ? 1 is invariant under rotation, i.e., under multiplication by e`9 for any angle 0. The constant pr, is then known as the external conformal radius. (i) Show that the real-valued function u defined by f = eh'+", satisfies
Au =0
on Q+,
y+u = 0 u(z) = log
on I', IZI
Pr
+0(l) as z - oo.
(ii) Hence show that
u = log
r
Capr
- 2n SL *eq,
and deduce that pr = Capr. (iii) Suppose that a > b > 0, and let r be the ellipse
Iz-cl+lz+cl=2a, where c= a2-b2. Thus, a and b are the semimajor and semiminor axes, respectively,
The Laplace Equation
272
and c/a is the eccentricity. Verify that the formula
z=
a+b 2
w+
a-b 2w
defines a conformal mapping w l-4 z of the region I w > 1 onto 52+, and deduce that Capr = 1(a + b). For further examples, see Landkof [52, p. 172]. 8.12 Prove the (crude) bound
Capr
ifn = 2,
diam(1') (n - 2)T,,
_-
diam(I')"-2
if n > 3.
8.13 For any a > 0, we write ar = {ax : x E 17). By expressing the equilibrium density for al' in terms of the equilibrium density for r, show that
ifn = 2,
a Capr Caper = t&3_2Capr
if n > 3.
8.14 Derive the following variational characterisation of the capacity: min (Si, ,/,EH-'12(r). (l.*)r=l
Or = (Sieq, *eq)r
12n
log
Cap,
r
Capr
ifn = 2,
ifn>3.
LI
[Hint: if (1, *)r = 1, then fr = *o + *eq with (1, fo)r = 0.] 8.15 Let Q+ = C \ [-1, 1]. The formula
- 1Iw+w Z-2 1
defines a conformal mapping w H z of the region I w I> 1 onto Q+. [This mapping is a degenerate case of the one in Exercise 8.11 (iii) above.] Define
fo(z) = log
wr
and f,n(Z) =
2m w'"
form > 1,
and let un, (x, y) = Re fm (z),
where z = x + iy.
Exercises
(i) Show that w = z + ru
273
z2 - 1, and that
z2 - 1 = ::+-i
fl - x2
for-1 <x < 1.
(ii) Hence show that if -1 < x < 1, then 8:':U",
(x, 0) =
vlimp ifmz) T1
=
1 -x2
x
1
when in = 0,
2(x±i 1 -x2 -'n
when m > 1.
(iii) Let r = (-1, 1), and show that u,,, = SL 1/r,,,, where Vr,,, = -[8vu,,, Jr is given by
fo()
2
and d *,,,(x) =
1 - xz
P. (2-, x) 1
form > 1.
xz
Here, P(2, x) is the Chebyshev polynomial (8.24). (iv) Deduce that for x E r,
r
1
2rrlog\ Ix-YI/ (v) Assume r
if m = 0,
2 log 2r
Pm (2, y)
1-y2 dy
1
l)m(2,X)
if m > 1.
Z, and express the solution of Si/ = f as a series
involving the Fourier-Chebyshev coefficients of f . 8.16 Let c2+, r and w H z be as in Exercise 8.15, but now put
um(x, y) = Re(2iwm+i
)
form > 1.
(i) Show that um = DLi/r,,,, where 1/r,,, = [u,,,Jr is given by *.(x) = Q,,, (x) VI-1- -
x2 and Qm (cos 8) =
sin(m + 1)0' sin 8
(The function Q,,, is the mth Chebyshev polynomial of the second kind.) (ii) Deduce that f.p.
27r EyO
Jc-,,l)\B,(x) (x - y)
2
Qm(Y)
1 -y2dy =
m-
Q,n(x)
2
for-1<x<1. (iii) Hence obtain an explicit series solution for the equation RI/r = f .
The Laplace Equation
274
8.17 Let f E
and g E H-1/2(1'), and put
h=Z(g-T"g)-a1gf, so that Ri/r = h ig the boundary integral equation corresponding to the interior Neumann problem (8.36) for the Laplacian; see Theorem 7.7. Define an operator
R, : H1/2(F) - H-1'2(r) by q
R1i/r = Ri/i +>2(X,, *)rXj, i=1
are as in Theorem 8.20. (i) Show that R, is self-adjoint, and that
where X1,
, Xq
(R1r, i)r ? cIII/FI12hIp()
for all *E H1/2(1').
(ii) Show that TXJ = -Xj, and deduce that
(Xi,h)r=±
f
fdx+J gdQ fort < j
si; nsz-
where S2j is the bounded, simply connected domain enclosed by l',, with the + (-) case occurring when v points out of (into) S2j. (iii) Explain why 0+ has q - p bounded components (and one unbounded component).
(iv) Suppose q = p = 1. Show that the unique solution * of R I* = h satisfies Ri/r = h if and only if the data f and g satisfy
fn -
fdx+
r
gdQ=0,
the necessary and sufficient condition for the existence of a solution to (8.36).
(v) Now suppose q = 2 and p = 1, with r1 and f'2 labelled so that they are the boundaries of the unbounded and bounded componenents of Q+, respectively. Show that the result in (iv) is still valid provided either f = 0 and g I r, = 0, or g 1 r2 = 0. How can we proceed for general f and g? 8.18 Let S2- be the lower half space {x E R" : x" < 0) so that I' = R't-1, and let t/r, -0 E
D(Rn-1)
Exercises
275
(i) Show that the double-layer potential for the Laplacian maybe written as DL X/r
= -(a,, G) * (* 0 8).
(ii) Deduce that
a,DL1/r=1/r®S-}-G*(,L'Vr®S), where A' _ E"-, 8? is the Laplacian in
IR
-1
(iii) Hence derive the identities n-1
R* = -SD'*
and
(Ri, Or = >(Saj*, ajo)r. i=1
8.19 Show that in the case of the Laplacian, the functions ms and mR of Exercise 7.8 are given by
ms(')=
1
4n 1t'I
and mR(')=irI'I
8.20 Let r = r0 U 11 U r1 be a Lipschitz dissection of r, so that F0 is an (oriented) open surface, and define So ' = (S*)Iro and R0i/r = (R*)Iro if supp c/r c r0. (i) Show So : H-1J2(F0) - H1/2(ro) and R0 : H112(ro) -+ H-1/2(ro) are Fredholm operators with index zero. H-1/2(ro) (ii) Show that there exists a unique distribution /r q E such is constant on ro, and (1, *q) ro = 1. We use *,,0 to define that Capro, the capacity of r0, in the same way as for a closed surface. (iii) Prove for n > 3 that So is positive and bounded below on H1/2 (2r0). 1/ (F) (iv) Let n = 2. Prove that S0 is positive and bounded below on H if and only if r > Capro, and that So : H-1/2(r0) --* H1/2(Fo) is invertible if and only if r 54 Capro. -1/2 (v) Prove that R0 is positive and bounded below on H(F0). 8.21 Show that the logarithmic capacity of a line segment equals one-quarter of its length. [Hint: use Exercises 8.13 and 8.15.] 8.22 As in Theorem 7.10, the interior, mixed problem for the Laplacian leads to a 2 x 2 system of boundary integral equations of the form Aib = h, where _ ( SDD -2TDN A
-
L 2 TND
RNN
Show that if S2- is connected, and if r > Caprp in case n = 2, then A : H - H* is positive and bounded below on the space
H = H-1/2(FD) x H1/2(rN)
9
The Helmholtz Equation
Consider the scalar wave equation, 82U 8t2
-c2LU=0.
(9.1)
We obtain a time-harmonic solution U(x, t) = Re[e-'aru(x)] if the spacedependent part u satisfies the Helmholtz equation,
-Du - k2u = 0,
(9.2)
with k = A/c. In this setting, the wave number k is real, but in the theory that follows we shall allow the coefficient k2 to be any non-zero complex number. It is convenient to assume, without loss of generality, that
0 < arg k «.
(9.3)
This chapter begins by showing how separation of variables leads to Bessel's equation, and by deriving a fundamental solution G(x, y) = G(x - y). (The presence of the lower-order term in the Helmholtz equation means that we cannot apply Theorem 6.8 to obtain G.) Next, we discuss the well-known Sommerfeld radiation condition, and proceed to establish an existence and uniqueness theorem for the exterior problem. The final section of the chapter derives an integral identity that connects the sesquilinear forms associated with the boundary operators S and R. The books of Colton and Kress [12], [13] give more detailed treatments of the Helmholtz equation, emphasising integral equations of the second kind. The sesquilinear form associated with the Helmholtz operator P = -A - k2 is
(Dn(u,v)=J
J
n 276
Separation of Variables
277
and since P* = -,L - k2,
au.
13"u = B"u = a"u =
av
Obviously, P is strongly elliptic. Also, P is self-adjoint if and only if k is real.
Separation of Variables We begin our investigation of the Helmholtz equation (9.2) by seeking solutions of the form
u(x) = f (kp)i/r(co) where x = pw and p = Ix1.
(9.4)
It will be convenient to introduce the Beltrami operator, As -i, a differential operator on the unit sphere defined by
where j (x) = i/r (co) for x 0 0.
As,,-,* = (ai)IS-
Thus, i%r is the extension of i/r to a homogeneous function of degree 0.
Lemma 9.1 If u has the form (9.4), and if z = kp, then
Du(x) = k2( f"(z)i(w) +
1
f
f,(z)i(w) + 1 (z)As-(w)). Z2
Z
Proof Since ap/axe = x3/p, we find that XJ2
p)2f"(z)+kp2
a,u(x) =
p
+ 2kpr
f'(z) I(x)
f(z)a; (x).
Being homogeneous of degree 0, the function j satisfies Ej=1 xj aj >%r (x) = 0, so
Du(x) = (k2f"(z) + k Finally, because A
n
p
1 f'(z))i(x) + f (z)Oi(x)
is homogeneous of degree -2,
fi(x) = p-2A(w) = giving the desired formula for Au(x).
k2 z2
278
The Helmholtz Equation
We now restrict our attention to the case when * is a surface spherical harmonic, because * is then an eigenfunction of the Beltrami operator. Lemma 9.2 If 1/r E Nm (Sn-1), then - As,,-i * = m (m + n - 2) *. Proof Let v/r = u is., for a solid spherical harmonic u E 7-lm (RI). Since u (x) = p'"* (co), we see that u has the form (9.4) with k = 1 and f (z) = z"'. Applying Lemma 9.1, it follows that Au (x) = pm-2 [m (m + n - 2) * (to) + As,-, * (co)J, which is identically zero.
Lemmas 9.1 and 9.2 show that when * E 7-1m(Sn-1), the function (9.4) is a solution of the Helmholtz equation (9.2) if and only if f is a solution of
f"(z)+n
lf'(z)+I1-m(m Z2 -2)) f(z)=0.
z
(9.5)
This ordinary differential equation can be transformed, by putting
g(z) =
zQhi2,-1 f(z),
into Bessel's equation of order µ,
g"(z) + z g'(z) +
C1 -
)(z) = 0,
(9.6)
with n µ=m+2-1.
Let J, denote the usual Bessel function of the first kind of order µ, which has the series representation (-1)P(z/2)µ+2P
J (z) _ P=O
P ! F (µ + p + 1)
for I arg z l «
(9.7)
The Bessel function of the second kind, Yµ, is defined by YN, (z) =
J,, (z) cos rr µ - J-4 (z) sin rrµ
for I arg i l < 7r,
if µ 0 7L,
and by Y,,, (z) = 4lim Y. (z)
for I arg z I < 7r,
if m E Z.
The functions J. and YN, form a basis for the solution space of Bessel's
Separation of Variables
279
equation (9.6), so the functions
j. (z) =j m (n, Z) =
rr Jm+(n/2)-1 (z) 2
Z(n/2)-1
and
Ym (Z) = Ym (n+ z)
= Vf7r2 Y,n+(n/2)-1(z) Z(n/2)-1
form a basis for the solution space of the original differential equation (9.5). If n = 3, then j,,, and y,,, are known as spherical Bessel functions.
Lemma 9.3 Let u have the form (9.4), where 1 E Hn,
(S"-1)
(i) If f (z) = j," (n, z), then -Au - k2u = 0 on R". (ii) If f (z) = y,,, (n, z), then -Au - k2u = 0 on Ilk" \ 10). Proof The preceding argument shows that in each case, u is a solution of the Helmholtz equation on R" \ 101. The series (9.7) shows that J.(z)/z"` is an entire function of z2, and therefore the same is true for jm(n, z)/z"'. Since pm*(w) is a homogeneous polynomial of degree m in x, it follows that if f = j,,,, then u is CO° on W, proving (i).
0 As z -* 0, the singular behaviour of the non-negative integer- and halfinteger-order Bessel functions of the second kind is given by
?n [-r'(l) + log(z/2)][1
Yo(z) _
O(z2)] + O(z2),
and, for any integer m > 1, Yµ (z) =
-r(µ) 7r(z/2)
[1 + O(Z2)
11
-2 ( m
/2)'7
z
1
1
og(z/2) [ + O (z2)l 1
i f /.c = m,
if.t=m-2,
to
where each O(z2) term is an entire function of z2; see [1, p. 360]. We are therefore able to show the following.
Theorem 9.4 For any constant a E C, a fundamental solution G(x, y) _ G(x - y) for the Helmholtz operator -A - k2 on R" is given by G(x) =
2(2.,7r)("-1)/2
[-Yo(n, klxI) + ajo(n, klxl)].
(9.8)
The Helmholtz Equation
280
If n = 3, then this expression reduces to
cos(klxl) +a sin(klxl)
G(x)
(9.9)
47rIx1
Proof. By (8.1), if n > 3 and µ = (n/2) - 1, then
I'(µ) _ F(1 + p)//1 _ 7r2-u
7r2-9
1'(n/2) (n - 2)7r2-n/2
27rn12/Tn
2(27r)(n-I)/2
(n - 2)Tn
(n - 2)7r2-n/2
2 7r
Thus, if Go(x - y) denotes the fundamental solution for the Laplacian given in Theorem 8.1, then for n > 2, kn-2 2(2n)(n-l)/2Yo(n,
kl xl) = Go(x)[1 + f (x)] + g(x) as Ixl -+ 0,
where f (x) = O (Ix 12) and
const + O(Ixl2)
g(x) =
0
const x log(klxl/2) [1 + O(Ix12)]
for n = 2, for odd n > 2, for even n > 2;
again, the O (IX 12) terms are analytic functions of Ix 12. To prove that G is a fundamental solution, we let w = S + (A + k2)G and show that w = 0 on Rn. In fact, since Ho(S"-I) consists of the constant functions, Theorem 9.3 implies that w is independent of the coefficient a, and that w = 0 on Rn \ {0}. Thus,
it suffices to show that w is locally integrable on R. We know already that -OGA = S, so w = k2GA + (A + k2)(GA f + g), therefore, as lxI -* 0, O(log Ixl)
w(x) =
{
O(Ixl2-n)
if n = 2, if n > 3,
and the result for a general n follows. In the particular case n = 3, we can use Exercise 9.1.
The Sommerfeld Radiation Condition When dealing with exterior problems for the Helmholtz equation, it is convenient to introduce the Hankel functions of the first and second kind, Hu1)(z) = JN,(z) +iYN,(z)
and
Hµ)(z) = Jµ(z) - iY,.(z),
The Sommerfeld Radiation Condition
281
which form an alternative basis for the solution space of Bessel's equation. We also put h
(z) = Jm (z) + 1Ym (z)
and
(Z) = Jm (Z) - lYm (Z),
writing h(1) (n, z) and h(2) (n, z) when it is desirable to indicate the dimension n explicitly. If n = 3, then and h,(nt) and h,(n2) are known as spherical Hankel
functions.
From the standard asymptotic expansions for the Bessel functions - see Abramowitz and Stegun [1, p. 364] or Gradshteyn and Ryzhik [29, pp. 961962] - we find that [e'tz-(2'+"- 1).,/4)] +
1
h,(11) (Z) =
z
OC
(n-1)/2
1)],
{\
Z1
(9.10) Z
(n-1)/2 Ce-i[z-(2m+n-1)n/4)] + O Z
hm2) (z) =
L
1\\
J,
J
as z - oo with -rr < arg z < 7r. By Lemma 9.3, the function u(x) = hI')(kp)i/r(w)
forx = pw, *
E'H.(Sn-1)
(9.11)
is a solution of the Helmholtz equation, and the corresponding time-harmonic solution of the wave equation (9.1) satisfies Re
[e-tar h(;) (kp)
* (w)}
= pcos[kp - At - (2m + n - 1)nr/4] + O(p-(n+1)/2) as p - * oo. Physically, this solution is an outgoing or radiating wave; if h ;2j is used in place of h;,;), then we obtain an incoming wave.
Definition 9.5 Write x = pcw with p = I x I and co E Sn', and let u(x) be a solution of the Helmholtz equation for p sufficiently large. We call u a radiating solution if it satisfies the Sommerfeld radiation condition
um
au
p(n-1)/2
C
uniformly in co.
p
- iku) = 0,
The Helmholtz Equation
282
The derivatives of h;,;) and h(2) have the asymptotic behaviour
d l) dhn,z
=
_
1
z (n-1)/2
-i z(n-l)/2
dz
2nn-1),r/4)1
[e'
+
o\(
i, (9.12)
[e_2m+n1)a/4)1 +
p(n-l)/2 (au ap
_ iku) = o 1 ),
\p
/
(9.13)
and is therefore a radiating solution. Also, by taking a = i in (9.8), we obtain a radiating fundamental solution, kn-2
klxl),
G(x) =
(9.14)
and, as a special case, eikI.t
G(x) =
47rIxI
when n = 3.
(9.15)
Recalling our assumption (9.3), we note that if Im k > 0 then G (x) has exponential decay at infinity. The following theorem reveals the connection between Definition 9.5 and our earlier treatment of radiation conditions for general elliptic equations; recall the definition of the operator M given in Lemma 7.11. Theorem 9.6 Let u be a solution of the exterior Helmholtz equation
-Au-k2u=0 onQ+, and suppose that M is defined using the radiating fundamental solution (9.14).
(i) If Mu = 0, then u satisfies (9.13). (ii) If u satisfies lim
P-'a% aBr,
then Mu = 0.
au
-ap- iku
2
dcr = 0,
(9.16)
The Sommerfeld Radiation Condition
283
Hence, the requirement Mu = 0 is equivalent to the Sommerfeld radiation condition.
Proof. By enlarging S2- if necessary, we can assume that u is C°O on SZ+. By
Theorem 7.12, if Mu = 0, then u = DL y+u - SL 8v u on Q+, so part (i) follows from Exercise 9.4. Assume now that u satisfies (9.16). We claim that
Iul2dcr=O(1) asp-+ oo.
(9.17)
LBp
In fact,
-au ap
2
2
8u
iku
I
ap
+ IkI2IU12 +2Im1
au
kavul,
and by applying the first Green identity over the bounded domain SZp = S2+ n Bp, we see that
f(gradu12-k21uI2)dx=fBpavuda- f a-vuda.
(9.18)
(In the first integral on the right, v is the outward unit normal to S2p+, but in the second integral v is the inward unit normal.) Multiplying (9.18) by k, and taking the imaginary part, we obtain
- f Im(kaa u) da = Im(k) -
Ik12Iu12) dx
JBPI
m(kavu) da
and so
'
2
Im(k) f12 (Igradu12 + Ik12Iu12) dx + j 2
- f Iml k a 4 uJ da
as p
\I aU I
+
Ik121u12 I dQ
(9.19)
oo.
The claim (9.17) now follows from our assumptions that k # 0 and Im k > 0. To complete the proof of (ii), we simply write
Mu(x) =
G(x,
Ja BPp
-JB
A
iku(y)] day
[a,,. G(x, y) - ikG(x, y)]u(y) day
284
The Helmholtz Equation
for p sufficiently large, and then apply the Cauchy-Schwarz inequality, noting that the radiating fundamental solution G (x, y) = G (x - y) satisfies
f IG(x, Y) 12 day < C sP
JaBP
Ix -
yl-(n-u day .5 C
and
f Ia,i,yG(x, y) - ikG(x, Y)I2 day < C J aP
Ix -
YI-(n+1) dory
< Cp-'-
aBP
0 We are now in a position to give an expansion of the fundamental solution in spherical harmonics or Legendre polynomials.
Theorem 9.7 For m > 0, let {*n,p : 1 < p < N(n, m)} be an orthonormal basis for The radiating fundamental solution G(x, y) = G(x - y) given by (9.14) has the expansion oo N(n,m)
G(x, y)
=ikn-2
hm)(klxl)srnep(xllxl)jm(k!Y!)/mp(Y/IYI) M=O p=1
ikn-2
T.
00
N (n, m) h(,) (k l x l)jm (kl y l) P, (n, cos 0) m=0
forlxI>IYI>0, where
cos 8 = IxIIYI
Proof Let
E 1-1,,, (Si-1), and put
u(y) = j,n(kp)i/r(co) and v(y) = h;;)(kp)*(w)
for y = pa.
By Lemma 9.3, we have -Au - k2u = 0 on W, and -Av - k2v = 0 on R' \ {0). Hence, Mu = u by Exercise 7.5, and therefore by Theorem 7.12,
f
[a",yG(x, y)u(y) - G(x,
day = 0 for IxI > p,
(9.20)
aP
whereas My = 0 by Theorem 9.6, so
f [a,,,yG(x y)v(y) - G(x, aP
day = v(x)
for lxl > p.
(9.21)
The Sommerfeld Radiation Condition
285
It follows from Exercise 9.3 that cu
m) = dzt m z
j,n(z)ddz
x (9.20) - j,,,(kp) x (9.21) gives the equation
so the combination
f
_
for lxl > p, P
or equivalently,
ik,=-2j».(kp)h;)(klxl)*(x/Ixl)
G(x, pW)/(w) da) =
for lxl > p.
By Theorem 8.17, we have 00 N(n.m)
G(x, y) = G(x, pw) = E E f m=o p=1
.
, G(x, prl)*,np(n) dri *,np(o)),
implying the first expansion, and the second then follows by Corollary C.2. However, so far we have proved only convergence in the sense of It is easy to see from the definition of the Legendre polynomial P," (n, t) immediately following Lemma 8.6 that
L2(S"-I).
I P. (n, t) I < 1
for -1 < t < 1,
so the mth term in the second expansion is bounded by N(n, m)Ih(;)(klxl)j,,, (k I y 1)1. Using the standard large-argument approximations [ 1, p. 365),
Jµ(z) = Yp (z)
(?-)[1+o1] µ \ ez
and
} [I + 0(1)] asµ -+ 00
(with fixed z), we find after some calculation that
iy,n(klxl)jm(klyl) [1 +o(1)]
-i
(k1
In 1)
2m+n-2(klxl)
[I + 0(1)]
asm -±oo.
Since (8.14) shows that N(n, m) = O(m"-2) as in -+ oo with n fixed, the expansions converge pointwise for Ix I > I y 1.
286
The Helmholtz Equation
Uniqueness and Existence of Solutions Theorems 4.12 and 8.2 imply that for each Lipschitz dissection r = I'D U 11 U rN there exist interior eigenvalues 0 < Al < A2 < , and corresponding interior eigenfunctions 01, -02, ... in H1 (S2-), satisfying
-L\/b1 =,,joj on S2-,
Y-0j = 0
on 1'D,
aOj=0
On PN,
with O j not identically zero. Therefore, by Theorem 4.10, the interior problem
-Du -k2u = f
on n-,
Y -U = gD
on FD,
au u = gN
on rN,
(9.22)
has a unique solution u E H1(S2-) for each f E H-1(S2-), gD E H1/2(rD) and gN E H-1/2(FN) if and only if k2 is not an interior eigenvalue. Otherwise, a solution exists if and only if the data satisfy
(4j, Do- + (Y-0j, gN)rN = (a, 4'j, gD)rp for all j such that Aj = k2. Notice that since X j > 0, if Im k > 0 then k2 cannot be an interior eigenvalue, and so (9.22) is uniquely solvable. The following result of Rellich [86] will help us to prove uniqueness for exterior problems.
Lemma 9.8 For any real wave number k > 0, if
-Au-k2u=0 on R"\BPo and if lim
P-+oo
f
I u (x) I2 do. = 0,
(9.23)
xI=P
then u = 0 on R" \ BPo. Proof. Let (1//mp : 1 < p < N(n, m)) be an orthonormal basis for Theorem 6.4 shows that u (x) is C°° for Ix I > po, so
7-lm(S"-1)
oo N(n,m)
u(x) = E E fmp(kP)hI/mp(co) M=O p=1
for x = pw, p > po and w E S)'-
Uniqueness and Existence of Solutions
287
where
fmp(t) = J u(k-1 tw)rmp(w) dw for t > kpp. n-I
The sum converges in L2(Sn-1), and 00 N(n,m)
Iu(x)I2 dax Ixl=p
=j
Iu(Aw)I2An-1
n -i
dw = E
An-1
Ifmp(kP)I2.
M=O P=1
Since u satisfies the Helmholtz equation, the function fmp is a solution of (9.5), and hence fmp(t) = amlphin)(t) + amphm)(t) for some constants amp and am2p. By (9.10),
A"-1I fmp(kP)I2 =
lan pe2i(kp-(2m+n-1)a/4] +amPl2
+ O(A-1),
so the assumption (9.23) implies that a npeie +am2p = 0 for all real 0. Therefore, amp = a, ;,p = 0, which means that fn p is identically zero. Lemma 9.9 Suppose that u E H1oc (S2+) is a radiating solution of the Helmholtz equation, i.e., suppose
-Au-k2u=0 on Q+ and
lim A(n-1)l2 au
- iku) = 0.
(9.24)
A
If
Im(k f (8v u)u de) > 0, r
then u = 0 on 0+. Proof If Im k > 0, then we see from (9.19) that fsi+ I u (x) 12 dx 0 as p oo, and thus u must be identically zero on 52+. If Im k = 0, then we can apply Lemma 9.8 because (9.19) shows that f xl=p I u (x) I2 d cx -* 0 as p oo, and k > 0 by our assumption (9.3). The desired uniqueness theorem follows at once.
288
The Helmholtz Equation
Theorem 9.10 I fu E H11 (S2+) is a solution of the homogeneous exterior mixed problem
-Au - k2u = 0 y+u = 0
on 52+,
on I'D,
a+u=0 onrN, and if u satisfies the Sommerfeld radiation condition (9.24), then u = 0 on
W.
It is now possible to deduce existence results for exterior problems using boundary integral equations. For brevity, we treat only the pure Dirichlet problem; but see also Exercise 9.5. Theorem 9.11 If f E H-' (S2+) has compact support, and if g E H 1/2 (I'), then the exterior Dirichlet problem for the Helmholtz equation,
-Au - k2u = f on 52+, y+u = g
on r,
has a unique radiating solution u E HioC(SZ+).
Proof. We have already proved uniqueness. By Theorems 7.15 and 9.6, a solution exists if and only if there exists i/r E H-1/2(r) satisfying
Si/r = y9 f - 2(g - Tg) on F.
(9.25)
(In the usual way, to define 9f we view f as a distribution on Rwith supp f c 52+.) Theorem 7.6 shows that the Fredholm alternative is valid for this boundary
integral equation, and by Theorem 7.5 the set ker S* consists of all functions of the form a. v where v E H' (S2-) is a solution of the interior homogeneous adjoint problem
-Lv-k2v=0 on Q-, y-v=0 on T. Using (7.5) and the second Green identity, we find that for all such v,
YGf - 2(g - Tg))r =
Y (Gf +DLg))r
= (y v, a. (9f +DLg))r _((_A_ k2)v, gf + DLg)o_ + (v, (-A - k2) (g f + DL g)) n_ .
A Boundary Integral Identity
289
Each of the three terms on the right vanishes, because y-v = 0 on F, (-A k2)v = O on S2-, and (-t - k2)(G f +DL g) = f = 0 on S2-. Thus, * exists, as required.
We remark that the solution fr of the boundary integral equation (9.25) is unique if and only if k2 is not an interior Dirichlet eigenvalue for -A.
A Boundary Integral Identity In this section, we derive a remarkable identity connecting the hypersingular boundary integral operator R with the weakly singular operator S, associated with the Helmholtz equation in 1[83. This identity, together with analogous ones for other elliptic equations, was introduced by Nedelec [75] as a way of avoiding the evaluation of hypersingular integrals in Galerkin boundary element methods
involving R. It will be convenient in what follows to work with the bilinear instead of the sesquilinear form form For any scalar test function w E D(I83) and any vector-valued test function W E D(I83)3, we have the identities
div(uW) = (grad u) W + u div W, div(wF) = F F. grad w + (div F)w,
div(F x W) _ (curl F) W - F curl W, if, say, u : JR3 -* C and F : R3 -± C3 are C'. Consequently, the divergence theorem implies that
(grad u, W) = -
J
u div W dx,
arty
(div F, w) = -
Ja3
F grad w dx,
(curl F, W) = f F curl W dx,
Jrt;
so for distributions u E D*(R3) and F E D* (R3)3,
(gradu, W) = -(u, divW), (div F, w) = - (F, grad w), (curl F, W) = (F, curl W). Given a scalar test function
E D(I83), we shall write Or = Ojr, and define
290
The Helmholtz Equation
yt and at by
(Or, fr)r and (8v0r,
(YtOr,
for * E D(R3);
(Or, 8v5v )r
cf. (6.14). The following identities hold in the sense of distributions.
Lemma 9.12 If 0, * E D(R3), then div y`(Orv)
-a'. Or and curl yt(Orv)
-y`(v x y grad 0)
on R3.
Proof We have (div YL(Orv), w) = -(Yt(Orv), grad w) = -(Orv, y grad w) r
_ -(Or, avw)r = (-8.'Or, w), which proves the first part of the lemma. Next,
(curl y`(4rv), W) = (ryt(Orv), curl W) = (Or v, y curl W)r
=
Jr
v y (0 curl W) da,
so by the divergence theorem,
(curl y`(Orv), W) = F
div(cb curl W) dx. JL
From the identities
div (0 curl W) = grad 0 curl W + 0 div curl W = grad
curl W + 0
and
div (grad 0 x W) = (curl grad ) W - grad
curl W = 0 - grad ¢ curl W,
we have div(O curl W) = -div(grad 0 x W). Thus,
(curl y`(Orv), W) = f
Jsrzt
_- J
div(grad 0 x W) dx
r
which proves the second part of the lemma.
fr
v
(grad o x W) da
A Boundary Integral Identity
291
Now fix a 0 E D(R3), and define
u = DL Or on R3
and Ft = grad u} on
using, in the double-layer potential, any fundamental solution of the form (9.9). We also construct a locally integrable vector field F : R3 _+ C3 by putting
F = I F+
F-
on 52+,
on S2-.
In this way, the support of the distribution F - grad u is a subset of 1'. Two further technical lemmas are required. Lemma 9.13 As distributions on R3,
u = -div SL(Or v),
div F = -k2u,
grad u = F + y`(Or v),
curl F = yt(v x y grad 0).
Proof. Using Lemma 9.12, we find that since 8j commutes with the convolution operator G,
u = DL Or = 9(8vOr) _ c(-div Y`(Or v))
= -divc(y`(Orv)) _ -divSL(Orv). Next, since
(F-grad u,W) =J3(F.W+udivW)dx and
f(F. W+udivW)dx =f } f div(u±W)dx = F zt
r
v y}(uW)da,
it follows by the jump relation for the double-layer potential, [u] r = [DL 'r ] r = Or, that
(F-gradu, W) =- fr v -(cbrYW)do = -(Orv,YW)r = (-Y`(Orv),W), so grad u = F + y`(tr v). Another application of Lemma 9.12 gives
div F = div[grad u - y`(Orv)] = Au + 8tbr,
292
The Helmholtz Equation
and since -Au - k2u = (-A - k2)GB,t,¢r = a,`,Or, we see that div F = -k2u. Finally, since curl grad u = 0, the second part of Lemma 9.12 implies that
curl F = curl[gradu - yt(Orv)] = yt(v x y grad 0). Lemma 9.14 The vector potential
A = g(curl F) = SL(v x y grad 0) satisfies
div A = 0 and curl A = F - k2 SL(Orv)
on R3.
Proof Since div curl F=O, we have div A = div g (curl F) = 9 (div curl F) 0, whereas since
curl curl F = grad div F - A F = grad(-k2u) -AF
= -k2[F + Yt(cbrv)] -AF = (-OF - k2F) - k2yt(Orv), we have
(curl A, W) = (curl g(curl F), W) = (9 (curl curl F), W)
= ((-A - k2) F - k2yt(Or v), G W) = (F, (-Lx - k2)gW) - (k2c(YtOrv), W)
= (F - k2 SL(4,rv), W), as claimed.
We can now prove the main result for this section; see also Exercises 8.18 and 9.6.
Theorem 9.15 Let G (x, y) = G (x - y) where G is given by (9.9). If 0, i/r E D(R3), then
(ROr, +/rr)r = (S(v x Y grad-0), v x y grad *) r - k2(S(Orv), frv)r Proof. Using the definition of R and the first Green identity, we see that
(Ror, fr)r = (-av u, fr)r = +
Jet
(grad u . grad * - k2ui/i) dx,
Exercises
293
and on 01 we have
grad u grad aJr - k2ui/r = [curl A + k2 SL(Orv)] grad * + k2 div SL(4rv)* = curl A grad * + k2 { SL(0r v) grad + [div SL(Or v)]i/r }
= div[A x grad 1/r + k2 SL(¢rv)*]. Hence, the divergence theorem gives
(Ror, *r)r = - J v y[A x grad
dQ
r _ (yA, v x y grad *)r - k2(S(Orv),1/r'rv)r,
0
and finally y A = S(v x y grad 0).
Exercises 9.1 Show from the series definitions of JI12 (z) and Y112 (z) that the zero-order spherical Bessel functions may be written as
Jo(3, z) _
sin z z
and
yo (3, z) _
-cos z z
9.2 Prove Theorem 9.4 by showing that as Ix I -+ 0,
a;G(x) = a;Go(x) + +
I O(Ixl log lxi) O(Ix13-n)
if n = 2, if n > 3,
and then arguing as in the proof of Theorem 8.1. 9.3 Show that if f and f2 are solutions of the differential equation (9.5), then their Wronskian
W = W (fl, f2) _
fi
f2
fl'
f2
is a solution of
dW +n-1W=0. dz
z
Deduce that W = const/z"-', and in particular W
(h(,'), (2)) n, h n?
= -2i
Z"-I
[Hint: use (9.10) and (9.12).]
and W Um, ym) _
294
The Helmholtz Equation
9.4 Let G(x, y) = G(x - y) be the radiating fundamental solution given by (9.14), and write x = pco with p = Ix I and w E
Sn-1
(i) Show that Ix - yI = IxI - w y + O(IxI-1) as IxI
oo, uniformly
foryEF. (ii) Use (9.10) and (9.12) to show that k(n-3)/2e-i(n-3)Yr/4
SL 1Jr(x) =
2(27r)(n-1)/2
x
eikp
p(n-1)/2
(f(f
e-ikmy*(y)day+O(p-I)
and
k(n-3)/2e-i(n-3)n/4 DL 11f (x) =
2(2.7r)(':-1)/2
xC
eikp
p(n-1)/2
f (a,,ve-ikw,y)i(y)
r
day + O(p-1)l
as Ix I -* oo, uniformly in w. (iii) Show that SL f and DL f satisfy the Sommerfeld radiation condition. (iv) Deduce that if u E Him (Q+) is a radiating solution of the -Au -k2 u = 0 on 52+, then there exists a unique function uc,. E CO°(Si-I) such that eikp
p(n-1)/2 [uoo(w) + O(p-1)]
as p -+ oo,
uniformly in w. The function u,,. is called the far field pattern of u. [Hint: use Theorem 7.12.] (v) Show that if u... = 0 on Sn-1, then u = 0 on Sa+. [Hint: use Lemma 9.8.]
9.5 Show that if f E H-I (S2+) has compact support, and if g E H-1/2(I'), then the exterior Neumann problem for the Helmholtz equation,
-Au - k2u = f on 52+, 8v u = g
on l',
has a unique radiating solution u E HIa, (S2+). [Hint: reformulate the
Exercises
295
problem as a boundary integral equation involving the hypersingular operator R, and apply the Fredholm alternative as in the proof of Theorem 9.11.] 9.6 Show that, in two dimensions, the identity of Theorem 9.15 takes the form
(ROr, *r)r = (Sai0, atf)r - k2(S(Orv), irrv)r for 0,' E D(R2) where r = (-V2, vi) is the tangent vector to I' satisfying v x r = e3, and
a,0 = r
grad 0 denotes the tangential derivative of 0.
10
Linear Elasticity
In the preceding two chapters, we considered the simplest and most important examples of scalar elliptic equations. Now we turn to the best-known example of an elliptic system, namely, the equilibrium equations of linear elasticity. For the history of these equations, we refer to the article by Cross in [30, pp. 10231033], the introduction of the textbook by Love [60], and the collection of essays by Truesdell [101]. Our aim in what follows is simply to show how the elasticity equations fit into the general theory developed in earlier chapters. Necas and Hlavd6ek [73] give a much more extensive but still accessible treatment of these equations, without, however, discussing boundary integral formulations. Let u denote the displacement field of an elastic medium. Mathematically, u : 7 -+ C', so m = n in our usual notation, and physically u is R"-valued and the dimension n equals 3. In Cartesian coordinates, the components of the (infinitesimal) strain tensor are given by
Ejk(u)=Z(ajuk+akuj) forj,kE(l,2,...,n), and we denote the components of the stress tensor by E jk. Thus, using the summation convention, the kth component of the traction over r is vi E jk. (The
traction is the force per unit area acting on S2 through the surface T.) If f is the body force density, then in equilibrium we have aJ Eik + fk = 0.
(10.1)
For a linear homogeneous and isotropic elastic medium, the stress-strain relation is E jk (u) = 2µE jk (u) + A(diV u)s jk,
where the Lame coefficients µ and X are real constants. We can write the 296
Korn's Inequality
297
equilibrium equations (10.1) in our standard form Pu = f by putting
Pu = -a3Bju
and
(13ju)k = Ejk(u).
Notice that the conormal derivative has a direct physical meaning:
B, u = traction on r, and since
aj Ejk =,1 aj(ajuk + akU j) + a.ak(div u) = µajajuk + (µ + ).)ak(diV U), the second-order partial differential operator P can be written in the form
Pu = -µ/u - (µ + A) grad(div u).
(10.2)
This chapter begins with a proof of Korn's inequality, thereby establishing that P is coercive on H 1(S2)" . After that, we derive the standard two- and three-dimensional fundamental solutions. The third and final section discusses uniqueness theorems and the positivity of the boundary integral operators, but only for the three-dimensional case.
Korn's Inequality We see from (10.2) that the Fourier transform of Pu is P u ( ), where P ( is the homogeneous, R"'-valued quadratic polynomial with jk-entry
)
Pjk( ) = (27r or, letting I, denote the n x n identity matrix,
(202[µI I21" + (µ + A)
(10.3)
Thus,
fori ElR' andnEV, and therefore P is strongly elliptic if and only if
µ > 0 and 2µ +.l > 0;
(10.4)
cf. (6.5). Landau and Lifshitz [54, p. 11] explain the physical significance of (10.4). Since
(BjU)*ajV = Ejk(u)aJVk =2pEjk(it)ajVk+X(dlvu)akVk = 21.LE jk (u)Ejk (v) + A(div u) (div v),
Linear Elasticity
298
the sesquilinear form associated with P is
(pc(u, v) = f [21LEjk(u)Ejk(v)+A(divu)(divv)]dx. For a physical displacement field u : 0 -+ 1R3, the quadratic form
2Osz(u, u) = 2 f Ejk(u)Ejk(u)dx z
is the free energy of the elastic medium within S2; see Landau and Lifshitz [53, p. 12]. It will be convenient to let grad u denote then x n matrix whose jk-entry entry is ajuk, and to write
IIE(u)Ili2(sz)"xn = f Ejk(u)Ejk(u)dx and z
J
8juk8jukdx.
Notice that
cl z(u, u) =
21L11E(u)IIi2(U)"xn
+ I1divu11t2(12).
(10.5)
If the Lame coefficients satisfy (10.4), so that P is strongly elliptic, then we know from Theorem 4.6 that (D is coercive on Ho (S2)". In fact, a stronger result known as Korn's first inequality holds. Recall that, by Theorem 3.33, Hi (Q)" = Hp (S2)" if S2 is Lipschitz.
Theorem 10.1 If 0 is an open subset of R", then IIE(u)112 (n)1xn
grad u11L2foru E
>_
(Q) n.
all
Proof. It suffices to consider u E D(O)". As in the first part of the proof of Theorem 4.6, we apply Plancherel's theorem to obtain
JR" Ejk(u)Ejk(u)dx = fR#l
=j 2
kuj)(17T)(5juk+} kuj)d
I . ul2) d "
f
2
fR- Iaju12dx. j=1
The result follows, because we can replace R" by 0 in both of the integrals with respect to x.
0
Fundamental Solutions
299
A much deeper result is Korn's second inequality (or just Korn's inequality). Theorem 10.2 If S2 is a Lipschitz domain, then IIE(u)IlL,()nxn ? cllgradUHL,(f)nxn - CIIuIIL2ISZ),7
foru E H'(SZ)n.
Proof. The left-hand side has the form (4.10) with L = n2, and for convenience we change the index set, writing N1ku = EJk (u) instead of N u. Since
.AIku = 2(a;uk + aku;) = 2(ekaj +ei ak)u, we have
(Nfku (x) } = Njk (i2rc )u (l; ), where NJk (
) = 2 (ek1 + e Sk).
The desired inequality holds because the hypotheses of Theorem 4.9 are sat-
isfied, with q,. = 1 for 1 < r < m = n. For instance, when n = 3 and
r=1, =2el,
1N11(e)T =tiel,
-6 N22
2NI1(S)T = 12e1,
-1N23(e)T +3N12()T +2N,3( )T = 2e3e1,
t3 N, ]( )T = 1e3e1,
-1N33( )T +3N13()T + 3N31( )T = t3e1,
)T
and the other cases can be handled in the same way.
In view of (10.5), Korn's second inequality implies the following result, which allows us to apply the general theory of elliptic systems. Necas and Hlavacek [73, pp. 46-49] give simple physical arguments showing that both Lames coefficients should be positive for typical elastic materials.
Theorem 10.3 For any Lipschitz domain S2, the elasticity operator (10.2) is coercive on H I (S2)" if µ > 0 and X > 0.
Fundamental Solutions Since the polynomial (10.3) is homogeneous, we may obtain a fundamental solution for the elasticity operator using Theorem 6.8; see Landau and Lifshitz [54, p. 30] for an alternative approach.
Linear Elasticity
300
Theorem 10.4 If µ # 0 and 2µ + A # 0, then a fundamental solution G (x, y) G(x - y) for the elasticity operator (10.2) is given by T
G(x)
8nµ(2µ + .X)
C(3µ + X)
1 13 + (µ +,X) IxI3) I
when n = 3, (10.6)
and by G(x)
=
1
4gµ.(2µ +X)
((3p' + .) log
1 I2 + (µ + A) xxTI2) 1
I
when n = 2.
IX
Proof By (10.3), if CO E S", then
P(w) = (2n)2[µ1 + (µ +,l)WWT ] and so by Exercise 10.1, z
p(w)-I =
(27r)
µ(2µ + ),)
[(2µ + ),)I - (µ +))WWT ].
Assume that n = 3, and choose an orthonormal basis ?71, 712 E JR3 for the plane normal to x, so that the unit circle in this plane has the parametric representation SY : w1 = (cos 0)x1i + (sin 0) 712
for -7r < B < n.
By Theorem 6.8(iii), z
G(x)
- I µ(2µ + X) J [(2µ
+ ),)I3 - (µ + )A)wlwl] dw1,
and since
fs
xx T
+ (sine 0)r1271i] dO = 7r (7JI 17T + 712711) = 7r (13
- IxI2 )'
the formula for G(x) follows. Suppose now that n = 2; by (6.13), the formula
G(x) =
-(2n)
z
µ(2µ +,X)
J((log Iw sI
xI)[(2µ + ),)12
- (µ +)-)wwT ] dw
Uniqueness Results
301
defines a fundamental solution. The vectors
ni =
x
x,
1
=
Ixl - Ixl
X,21
and
X2
J7,
Ixl L
form an orthonormal basis for R`, and by putting co = (cos 9)r71 + (sin 0) 112 we see that
Js
log Iw xI dco =
T
_n
log(Ix cos01) dO = 2.7r log IxI + const
1
and
(log 1w x l )WWT dco
Js'
=J log(Ixcos01)[(cos20)17,riT +(co59sin9)(ri,riz +rl2rli) + (sin2 9)7721i2 ] d9
J
"
1 +cos 20 2
1og(IxCOSel)
T
,+
1 - cos 20
T1 dB.
2
2 J/
n
An integration by parts gives ;r
f
log (Ix cos 91) cos 20 d0
=
-,
rr
,
IT
sin O sin 20 dO = Cos 9 2
sine 0 dO = rr, n
and one easily verifies that
xxT
and
'?Irli +rl2riz =12
ntrli -7722 =2Ixle -12,
so
f
1
(log lc) xl)wwTdco= 1
r
log(IxcosOl)dO12+
/ 2xxT
I 7rI
1x12
12)
= rr log Ix 112 + it x 12 + const,
leading to the stated formula for G(x).
U
Uniqueness Results Throughout this section, we assume that the components of u are real, and treat only three-dimensional problems.
Linear Elasticity
302
To apply the Fredholm alternative (Theorem 4.10) to the mixed boundary value problem in linear elasticity, we must determine all solutions of the homogeneous problem. As a first step, we show that the only strain-free displacement fields are the infinitesimal rigid motions, and that such displacement fields are also stress-free.
Lemma 10.5 Let S2 be a connected open subset of R3. A distribution u E D*(52;R3) satisfies E(u) = 0 on S2 if and only if there exist constant vectors a, b E R3 such that
u(x)=a+bxx forxES2.
(10.7)
Moreover, in this case E(u) = 0 on S2.
Proof. Let B = [b,3] E 83x3 denote the skew-symmetric matrix defined by Bx = b x x, i.e.,
B=
0
-b3
b2
b3
0
-b1
-b2
b1
0
Ifu(x)=a+b xx,then ajuk =bkj,so Ejk(u) = 1 (bkj + bjk) = 0. To prove the converse, assume that E(u) = 0 on S2. Since, with the notation of (3.9) and (3.10),
E(*,*u)='YE*E(u) on(xES2:dist(x,I')>e}, we can assume that u E C`O(S2)3. The diagonal entries of the strain tensor are
just Ejj(u) = ajitj (no sum over j), so a1u1 = a2u2 = 83113 = 0
on 0.
Since the off-diagonal entries of the strain tensor also vanish, we can show that
a2u1=83u1=0,
82
a1u3=a2u3=0
on Q;
for instance, a uk = al (2Elk (u) - aku 1) = -ak (al u 1) = 0. Therefore, Fu = 0 on 0 if (a I > 3, implying that u is a quadratic polynomial in x. In fact, the vanishing of the partial derivatives listed above shows that u must have the form ul(x) = al + b12X2 + b13x3 + C1X2X3,
u2(X) = a2 + b21X1+ b23X3 + C2X1X3. u3 (X) = a3 + b31x1 + b32x2 + c3X1X2,
Uniqueness Results
303
for some constants aj, bik and cj. Since E(u) equals 0
b12+b2,
b13+b31
b,2 + b2,
0
b23 + b32
b, 3 + b3,
b23 + b32
0
1
1
+ 2
0
(C) + C2)X3
(Cl + C3)X2
(Cl + C2)x3
0
(C2 + C3)x1
(Cl + C3)X2
(C2 + C3)xl
0
we conclude that bfk = -bk! and c! = 0 for all j, k. Finally, if u has the form (10.7), then E(u) = 2AE(u) + A(divu)13 = 0 because E(u) = 0 and div u = E!j (u) = 0. O Theorem 10.6 Assume that 0 is a bounded, connected Lipschitz domain in and that the Lame coefficients satisfy
A>0 and A>0.
(10.8)
Let W denote the set of solutions in H' (S2; R3) to the homogeneous, mixed boundary value problem
-p Au - (µ + A) grad(div u) = 0 on 0,
yu =0
on I'D,
0
on F.
(10.9)
(i) If I'D # 0, then W = {0}, i.e., (10.9) has only the trivial solution. (ii) If I'D = 0, so that (10.9) is a pure Neumann problem, then W consists of all functions of the form (10.7) for a, b E 1R3.
Proof. If U E H' (Q)3 is a solution of (10.9), then c(u, v) = 0 for all v E H1(Q)3 such that yv = 0 on I'D. In particular, by taking v = u and recalling (10.5), we see that 2 tjIE(u)IIL,(Q)3x3 +AIIdIvuIIL,(g2) = 0.
Our assumptions on a and A then imply that E(u) = 0 on Q, and so u has the form (10.7). If FD # 0, then, because rD is relatively open in r, we can find x, y, z E rD such that x - y and z - y are linearly independent. From the three equations
a+bxx=0, a+bxy=0, a+bxz=0,
Linear Elasticity
304
it follows that b x (x - y) = 0 and b x (z - y) = 0, and we conclude that b = 0. (Otherwise, x - y and z - y would both be scalar multiples of b.) In turn, a = 0, and part (i) is proved. If I'D = 0 and u has the form (10.7), then E (u) = 0 on S2, by Lemma 10.5, and so u is a solution of the homogeneous Neumann problem, i.e., u E W. 0 To conclude this section, we consider the boundary integral operators S : H-'t2(F; 1183)
H112(F; 1183)
and
R : H't2(r; R3) -+
H-1/2(1,; R 3)
associated with the three-dimensional elasticity operator (10.2), and defined using the standard fundamental solution (10.6). Theorem 10.7 Assume that the Lame coefficients satisfy (10.8).
(i) The weakly singular boundary integral operator is positive and bounded below on the whole of its domain, i.e.,
(Si, Or ? cIIrf1H-,r-(r)3 for all Ir E H-1/2(r; R3). In particular, ker S = (0). (ii) The hypersingular boundary integral operator is positive and bounded below on the orthogonal complement of its null space. Indeed, if SZ- is connected, then the six functions Xi : I' -+ 1183 (1 < j < 6) defined by
forxEI'and1
X1(x)=el and X,+3(x)= e1xx form a basis for ker R, and we have (R1G, Or >- cII*II
1r_(r)3
for* E H1t2(r; R3) such that (Xj, Or = 0 for 1 < j < 6. Proof. By Theorem 10.3, the assumptions on µ and X ensure that the elasticity Since, in three dimensions, the operator (10.2) is coercive on H1(Q'; elastic single-layer potential satisfies R3).
SL*(x) = O(Ixl-1) and gradSLt/r(x) = O(Ixl-2)
as Ixl -+ oo,
we can argue as in the case of the Laplacian (Theorem 8.12) to show that
(S*, O)r = (Df[2tEJk(SL +(SL 1/i, SL O) + - (SL i*r, SL 0)
=llr)Eik(SLgi) +A(divSL*r)(divSL0)]dx
Exercises
305
for all i/r, q5 E H-112(x)3. In particular,
(S*, 1G)r = 2p lE(SL )IILz(R3)3x3 +.klldivSL,GllLz(R3) > 0. Moreover, if (S*, i)r = 0 then E(SL ir) = 0 on 1R3, so SL*(x) = a +b x x for some a, b E R3. In fact, a = b = 0 because SL * (x) -+ 0 as Ix I - + oo, and hence i/r = -[By SL *]r = 0. Part (i) now follows by arguing as in the case of the Laplacian (Corollary 8.13). Turning to the hypersingular operator, we can argue as in Theorem 8.21 that (Ri/r, cb)r = (D+(DLt/r, DL.O) + (D-(DL,/r, DL-0)
for all i/r, 95E H1/2(I')3,
andinparticular (R1/r, Or ? 0. Moreover, if (Rir,1/r)r = 0, then E(DL i/r) _ 0 on St}, implying that E (DL /r) = 0 on SZ} and hence Ri/r = -B, DL t/r = 0 on r. Exercise 2.17 then shows that R is positive and bounded below on the orthogonal complement of ker R. To complete the proof of part (ii), we assume that Sl- is connected. It suffices to show that t/r E ker R if and only if ti = u I r for some u of the form (10.7).
Let Ri/r = 0 on F. Since (Ri/r, Or = 0 we have E(DL ifr) = 0 on Was above, and so by Theorem 10.6, there are vectors a±, b± E 1R3 such that DL i/r (x) = a± + b± x x for x E SZ±. In fact, a+ = b+ = 0 because DL i/r (x) -* oo, and hence /r = [DL i/r] r = u I r where u is given by (10.7) with 0 as Ix I
a= -a_ andb=-b_. Conversely, suppose that ilr = u I r where u (x) = a + b x x for x E R3. 0 on I', the third Green identity for u reduces Since Pu = 0 on Sl, and
to u = - DL i/r on SZ-, implying that 0 = -Br, DL i/r = Ri/r on I', i.e., i/r E ker R.
Exercises
10.1 Show that ifa00,a+b#0andwES"-',then (aI + bwwT)-1
= a(a + b) [(a + b)I
- bwwT ].
10.2 Nitsche [79] gives the following elementary proof of Kom's second inequality for C1 domains. (The same paper also extends the method of proof to cover Lipschitz domains, with the help of a regularised distance.) Suppose first that S2- is the hypograph x" < (x'), and let u E H 1(S2-)". Given s > 0, we define us E H 1(S2+)" by reflection in
Linear Elasticity
306
the surface x _ (x'), i.e.,
us(x) = u(x', (x') - s[x" - (x')]) for x > (x'), and then consider v E H I (S2+)" of the form
I au''+bu''. vj =
pu,+qu,',
ifl < j
ifj=n,
for unspecified constants a, b, p, q, s and t such that s > 0 and t > 0. (i) Show that y+v = y-u if and only if
a+b=1 and p+q=1. (ii) Show that if x E Q+, then
(aju)r(x)
(ajus)(x) _
for 1 < j < n - 1, for j = n.
+(1 -s(a"u)s(x)
(iii) Deduce that for I < j < n - I and I < k < n - 1,
Ejk(v) = aEjk(u)s +bEjk(u)' +
all
(a"Uj)Sak0 + 2(1 + t)[(anukYaj + (anuj)'akf], whereas
E,,,,(v) = -spE, (u)S - tgEnn(u)'(iv) Show that if
p = -sa and q = -tb, then for 1 < j < n - 1, Ep,(v) = pEj,,(u)s +qEj"(u)' + + 2(1 + t)E,,,,(u)'8j
(1 +s)E..(u)sajg 2
(v) Show that IlusllL,cn+) = s-I IIuIIL2c01.,,
Exercises
307
and deduce that, with M = max l < j <,1_ 1
Il aj
11 cx (!rt°-' )'
CI IIE(u)IIL2(9-)"" +C2M
where Cl and C2 depend only on a, b, p, q, s, t and n.
(vi) Choose (say) a = -2, b = 3, p = 4, q = -3, s = 2 and t = 1, so that these constants satisfy the conditions in parts (i) and (iv), and define U E H 1 (R")' by
U= I u L
u
on 2
,
on 52+.
Show, by applying Korn's first inequality (Theorem 10.1) to U, that
2(1 +C1)IIE(u)Ili + 2C2 M2 Ilgrad U112
and deduce that if M < 1 / 2C2 then
II grad u II L, (sI-)"" x <
(vii) Finally, use a partition of unity to prove Kom's second inequality (Theorem 10.2) for a C 1 domain (with compact boundary). 10.3 For n = 2, the elasticity operator (10.2) has the form 2
2
Pu = ->
aj(Ajkaku)
j=I k=1
where 21L+ ,X
All
o],
01, µ
L
A22 = Thus, f o r Z ; I = [ 2
t1
121T,
µ 0
0
2µ+.1]'
2 = R2I ,221T E R2,
2
j=1 k=1
I
= 2[fitl
22
12
2I1
4µ+2), µ+X µ+A 4µ+2), 0
0
0
0
0
0
0
0
2µ µ+A µ+A 2µ
Linear Elasticity
308
Assuming P is strongly elliptic, so that the Lame coefficients satisfy (10.4), show that the condition (4.9) fails to hold if h > µ or if -2µ <
A < -5µ/3. 10.4 Show for n = 3 and for u e C I (7)3 that the surface traction can be written as
-µ(v x curl Or. [Hint: first establish the identity (v x curl u)k = P J 8k U f - V J 8 J uk .]
10.5 Consider the pure Neumann (i.e., pure traction) problem in linear elasticity:
-µEu - (µ + X) grad(div u) = f on 0, on1. Show that a solution exists if and only if the resultant force and the resultant torque both vanish, i.e.,
r
J f(x)dx+J g(x)dcx = 0 r
and
xx f(x)dx+ f xxg(x)dar=0. Jn
r
Appendix A Extension Operators for Sobolev Spaces
We say that E : WP(S2) -* W, (1R") is an extension operator if E is bounded and satisfies
Euln = u for u E Wp(S2). For our purposes, the significance of such operators stems from Theorem 3.18.
If Q is Lipschitz, then a construction due to Calderbn [8] yields, for each integer k > 1, an extension operator Ek : WP (S2) -3 WP (R) that is bounded for 1 < p < oo. Using a different method, Stein [96, p. 181] obtained an extension operator E : W PI (S2) -+ WP (118"), not depending on k > 0, and bounded
for 1 < p < oo. In the case when 0 is smooth, there is a simpler construction, due to Seeley [93]; see Exercise A.3. In the main result of this appendix, Theorem A.4, we shall use a modified version of Calderon's extension. Suppose that S2 is a hypograph, S2= (x
x _ 1), and the function
: 1R11- I -- R is Lipschitz:
Mix' - y'i for x', Y' E R"-'.
(A.2)
A crude extension operator is obtained simply by reflection in the boundary of S2.
Theorem A.1 If 0 is the Lipschitz hypograph (A.1), and if
Eou(x) =
for x E S2,
U (X)
1u(x',2 (x')-x") forx c- W' \S2,
then Eo : W p (S2) --* Wp (IR") is bounded for 0 < s < 1 and I < p < oo. 309
Extension Operators for Sobolev Spaces
310
Proof Puti = (x',
xn). One easily verifies that x E 0 if and only if x E R" \ 0, and vice versa, with x = x. Moreover, the Jacobian determinant of the transformation x H x is identically equal to -1. Thus, II Eou II L,,(W'\n) _ IIUIILP(n), and IIEouIIL,,(R11) = 211uIILn(12) Also, if x E R" \ 0, then
for 1 < j < n - 1, for j = n,
8;u(2)
ajEou(x)
-anu(2)
implying that II8;UIILP(s2) +2MIIanUIIL,,(52) { I1
fort < j < n - 1, f j = n.
u11 z,,(9)or
8
Hence, Eo : W, (0) --+ WP (R") is bounded if s = 0 or 1. Assume now that 0 < s < 1 and 1 < p < oo. Recalling the definition (3.18) of the Slobodeckii seminorm, we write IEouIf pR,,
=lulsps2+I(+12+I3,
where
I,
lu(x) - u(Y)IP
Ix -
fL >x').
Yln+ps
dxdy ,
n
Iu(2)-u(y)Ipdxd
I213 = Since I2, that
Y,
n+ps
Iu(x) fL,<(X),
>t(Y')
v,
Ix - Y 112+p$
dxdY.
/f 4M2 IX - y1, we see
- 5 I < 21(x') - (y')I + Ixn -
Iz-j'I<2 1+M21x-yI, and so
lu(2)-u(Y)Ipdxd 11
Ix - yIn+Ps
< (2 1 +
y
p
M2)n+Ps
IL
(X7, ?5,<;(Y')
I u(x) - u(y)1 dz f
IX-
d"Y
=C
uIs P.
I
If x,, > (x') and y < (y'), then
-Y'I,
Extension Operators for Sobolev Spaces
311
so
Ix -
+21x,, - (x')1 < CIx -y I,
Iy +x,, - 2C(x')I < ly -
and hence
I2-
Iu(x)-u(Y)IPdxd y, Ix - YI,1+Ps
ff
Iu(x) -u(Y)Ip
Ix - y1n=+PS
X...
dz dY= cIuIP P.sz,
Similarly, 1 3 < C I u I ° P n, giving the desired bound for Eou.
Finally, consider the remaining case 0 < s < 1 and p = oo. If x > (x') and y" > (y'), then I Eou(x) - Eou(y)I = Iu(z) - u(Y)I < Iuls.oo.nlx - YI'`
< [2V-1 + M2lsluls,oo,szlx - yls.
If x > (x') and y < (y'), then Eou(x) - Eou(y)I = Iu(i) - u(y)l < Iuls,..nIi - yls < CIuls,.,S21x - yls, and similarly if x < (x') and y > (y'), then I Eou (x) - Eou (Y) I < C l u (s,oo,5
Ix - yIs. To obtain an extension operator for s > 1, we shall use the Sobolev representation formula; recall the notation of (3.7). Lemma A. 2 For each integer k > 1, if i/r E C°°(S"-') satisfies
f
f(co) dco =
-1=1
(-1)k (k - 1)!'
(A.3)
and if u E C o ,(R" ), then
u(x)=f t^
* \ I_
u(k) (x
I'
+ y; Y) dy for x E R".
Proof By Exercises A.1 and 3.5,
(k_I()( 00
u(x)
(k -11)!
dp
°o
k
(k- 1)!
x + pco) dp
pk-lu(k)(x+pco;co)dp.
10
312
Extension Operators for Sobolev Spaces
Multiplying both sides by * (w) and integrating with respect to co, we see that 00
u(x) = f
*(c)) f
uI=1
p-flu(x + pco; pw) p"-1 dp dm.
0
We use the abbreviation I u 1 µ for the Slobodeckil seminorm on R" with p = 2.
Lemma A.3 If K E C u E D(R"), then
\ (0}) is homogeneous of degree 1 - n, and if
(
for-oo<s
IIa,(K * u)IIHs(II) < CIIUIIH=(RH) and
laj(K*u)lu
81(K * u)(x) = (K * aju)(x) = lim) E1,0
Iy-XI>E
K(x - y)aj u(y) dy,
and since
a;.y[K(x - y)u(y)] = -a;K(x - y)u(y) + K(x - y)a;u(y), the divergence theorem gives
xj fly_Xj=45
y' K(x - y)u(y) day = - j
y -XI>E
E
+f
ai K(x - y)u(y) dy
K(x -y)a;u(y)dy.
ly-xl>E
Making the substitution y = x - Eco and using the homogeneity of K, we see that the left-hand side equals
f
cojK(Eco)u(x
Eco)&'
dco = f
wl=t
coK(w)u(x - Eco)
WI=t
so, defining aj = f.I=) cod K (co) dw, we have
af(K*u)(x)=aju(x)+lim E,,0
f
fly-xI>E
a;K(x-y)u(y)dy.
Extension Operators for Sobolev Spaces
313
Hence, by Exercise 5.14, the desired estimate for aj (K * u) will follow at once if we show that
ajK(co)dw=0.
(A.4)
Choose a cutoff function X E C 1P(0, oo) satisfying f0 X (p) dp/p = 1. Integration by parts gives
f ajK(x)X(Ix1) dx = ,.
fR"
K(x)X'(Ix1) X dx, 1X I
and by transforming to polar coordinates, we have aj K(Pw)X (P)Pi-1 dp dw fp>O jwl=rI
f
"'1=1
p>o
K(Pw)X'(P)wjpn-1
dpdw,
which, using the homogeneity of aj K and K, simplifies to
f
0
00
X(P)dp f P
ajK(co)dco=-f' X'(p)dp
f
K(w)wjdo.
I"I_1
IWI=1
Since ff ° X (p) dp/p = 1 and fo x'(p) dp = -X (0) = 0, the function Kj satisfies (A.4), as required.
We are now in a position to obtain the desired extension operator.
Theorem A.4 Assume that 0 is a Lipschitz domain. For each integer k > 0, there exists an extension operator Ek : WZ (Q) -* W2 (R") that is bounded
forks
V = {yER":y"<-Mly'IJ, and observe that by (A.2),
x+VCQ forxEQUT. Fix an integer k > 1 and a function * E CO0 (S"-1) such that (A.3) holds, with
supp * c Sn-1 n V.
Extension Operators for Sobolev Spaces
314
Let X E C ,-..p [0, oo) be a cutoff function satisfying X = 1 on a neighbourhood of 0, and replace u (x + y) by X (I Y I )u (x + y) in the proof of Lemma A.2 to obtain the identity k
u(x) _ E 1=0
Jv
*,(y)u('(x + y; y) dy
for x E S,
where *1 (Y) =
(k)jlk_l_fl(k...l)(1l),(_)
forO
Let Eo be the extension operator from Theorem A. 1, assume u E V(SZ), and put
ua = Eo(a"u) We define
f 'Eku(x) 1=0
Jv
t.I ua(x
r(Y)
+ y) y' dY = >2 (pa * ua)(x)
Ial=1
lal
for'x E R", where
"a (Y) = I' *,(-y) (-y)' for la l =1, so that EkuIk = u. If Ial
forlal
so by Theorem A.1, IIEkull
(R-) _
IIaflEkullL2(R") < C lal
IRI
CE lal
118aU112
- CIIUIIW,(n).
Exercises
315
Also, if 0 < j.c < 1 then
iflal
< k, if lal = k and I1I = k,
CIIfIIL2(R1)
_ 10R. * f)Im,>R < {
Clf lµ,p
implying that
Ia1Ekal
C
>Rn <_
I" I=k
I
CIIUI12
)+C
la"ul'2`.,
=CIIUI1wk+
).
El
I"I=k
Exercises A.1 Show that, for any integer k > 1, if f : [0, oo) -+ C is a Ck function with compact support, then
_ .f (0) =
(k
k
-11) Jo
o0
tk-1 fck) (t)
dt.
o
A.2 In Exercise A.3, we shall need a sequence (,lk)k° satisfying 00
E 2'klk = (-1)' for all j > 0.
(A.5)
k--O
(i) Let ao, ..., aN-1 be distinct complex numbers, let b be any complex number, and consider the N x N linear system N-1
E(ak)'xk=b' for0< j
Show that the unique solution is given by N-1 xk =
aj - b for 0 < k < N - 1.
1=0
aj - ak
h 4k.
[Hint: by Cramer's rule, xk is the ratio of two Vandermonde determinants.]
(ii) Show that 00
00
i=o
1=1
1
Extension Operators for Sobolev Spaces
316
(iii) Deduce that the sequence ,kk =
00
1 + 2-j
1 1 - 2k-j i=0 jek
satisfies k
Ixkl
1
1
C 2k(k-1)/2
and is a solution of the infinite linear system (A.5). A.3 Seeley [93] has given a simple construction of an extension operator for the half space c2 = R. Let the sequence (,kk) be as in Exercise A.2, and define 1
u(x)
Eu(x) _
00
if X. < 0,
Xku(x',
if x > 0.
I k=° (i) Show that E : D(Q) -+ D(R"). (ii) Show that E : Wn (S2) -+ WP (R") is bounded for s > 0 and I< p
A.4 Let 0 < µ < 1 and E > 0, and choose a cutoff function X E C mp[0, co) satisfying x (y) = 1 for 0 < y < 1. Define the C°'µ epigraph n = {(x, Y) E R2 : y > IxI'`}, and the function
u(x, Y) =
y'-Ex
(y)
for (x, y) E Q.
2(µ-' - 1). (i) Show that u E W2 (S2) if E < (ii) Show that u l C°'- (Q) if I - E < X < 1. (iii) Deduce that no extension operator from W2 (S2) to W2(llt2) exists. [Hint: use Theorem 3.26.]
Appendix B Interpolation Spaces
Suppose that Xo and X 1 are normed spaces, and that both are subspaces of some
larger (not necessarily normed) vector space. In this case, X0 and X1 are said to form a compatible pair X = (Xo, X1), and we equip the subspaces Xo n X1 and Xo + X 1 with the norms IIuIIX,)1/z
IIulIxonx, = (IIuIIXa + and
IIuIlxa+x, = inf {(IIuoIIX0 +
11U1 11X21)1/2
u = uo + ul where uo E Xo and u1 E X1 }.
Notice that for j = 0 and 1,
X0 nX1 cXj cX0 +X1, and these inclusions are continuous because IIuIlxa+x, < II U II x1 < II ulIxonx,
If XI c Xc,then XonX, =X1 andXc+X1 =X0. In this appendix, we present a general method for constructing, from any given compatible pair X, a family of normed spaces Xe,q = (Xo, X 1)e.q
for 0 < 9 < 1 and I < q < oo,
each of which is intermediate with respect to Xo and X1, in the sense that Xo n X1 c Xe,q c Xo + X1. 317
(B.1)
Interpolation Spaces
318
Moreover, we shall see that XB,q has the following interpolation property. Take
a second compatible pair of normed spaces Y = (Yo, Y1), and two bounded linear operators Ao : Xo -+ Yo
and
A, : X 1 -+ Y1.
If Aou = Alu
for U E Xo n X1,
then AO and A 1 are said to be compatible, and there is a unique bounded linear operator AB : XB,q -* YY,q
such that
ABU =Aou = Alu foruEXonXl.
(B.2)
We will also show that if X0 and X1 are Sobolev spaces based on L2, then so is X8.2
For technical reasons, it is convenient to construct Xe,q in two different ways. Thus, we shall define two spaces, Ko,q (X) and Jo,q (X), and show X B,q = KB,q (X) = Je.q (X ),
with equivalent norms. The K-method will be used to prove the interpolation properties of HS (c2), after which the interpolation properties of HS (Q) follow by a duality argument that relies on the J-method. We conclude by considering the interpolation properties of HI (F). For more on the theory of interpolation spaces, see Bergh and Lofstrom [5].
The K-Method The K -functional is defined for t > 0 and u E Xo + X 1 by K(t, u) = inf {(IIuoIIX0 + t2IIu111X,)112
u = uo + u l where uo E X0 and U1 E X1 }. When necessary, we write K (t, u; X) to show explicitly the choice of the compatible pair X = (Xo, X1). For fixed t > 0, the K-functional is an equivalent
The K-Method
319
norm onX0+X1:
K(t, u + v) < K(t, u) + K(t, v)
K(t,.Xu) = IXIK(t, u), and
min(l, t)Ilullxo+x, <- K(t, u) < max(l, t)Ilullxo+x, If we fix u, then the K-functional is a non-decreasing function of t, and in fact, for all positive s and t,
min(1, t/s)K(s, u) < K(t, u) < max(l, t/s)K(s, u). Next, we define a weighted Lq-norm,
dt lq
00
llflle,q
=
` It-e.f(t)Iq r
for 0 < 9 < 1 and 1 < q < oo,
0
with the obvious modification when q = oo, Il f Ile.00 = ess sup It-If (t) I. t>0
This weighted norm has an important dilatation property, namely,
Ilt H f(at)Ile,q =aellflle,q fora > 0.
(B.3)
Now define
Ko,q(X) = (u E Xo + X1 :
II
u)IIe,q < oo},
and put II u11 K., (x) =
u) III,?,
where the constant Ne,q > 0 may be any desired normalisation factor. As the default value, we take
ifl q
I
Nq
Ilmin(l, )Ile,q
I[qO(1-9)Vk if q = oo, 1
and thereby simplify the statement of the next lemma.
(B.4)
Interpolation Spaces
320
Lemma B.1 Assume the normalisation (B.4).
(i) If U E Xa fl Xj, then u E K°,q(X) and II'IIxe.,(X) -< IIu11X-°IIuIIX, < Ilullxonx,
(ii) If U E K°,q(X), then u E Xo + X,, with
K(t,u)
Ilullxo+x, < IIu11xe.,(x)
Proof We can assume u 54 0. Put a = II u II x, / II a II x-, so that
K(t, u) < min (Ilullxo, tllullx,) = Ilullxo min(l, at),
(B.5)
and hence u)Ilo,q < Ilullxoa°II min(l, )Ile,q = Ilullxo°IIuIIX,/N°,q, implying the inequality II u II xe,, (x) < IIuIIX-°IIuIIX,. Next, taking p = (1 - 0)-', we have (Ilu!Ix-
Hull X-°IIuIIX, <
< [(1
p
s)p
+
(II
u
IP*,)r
= (1 - 9)Ilullxo +alluIIx,
- 9)2 +02]'/211u11xonx,,
completing the proof of (i). In a similar fashion, the inequality min(1, s/t) K(t, u) < K(s, u) implies that t-" 11 min(l, )II°,7K(t, u) <-
u)Ilo,q,
and so K(t, u) < t9 11 u11X,.q Finally, because min(l, t) 11 u11 X" +X, < K(t, u) we 0 have II u II xo+x, /No,q < II K u)11 °.q, which completes the proof of (ii).
Lemma B.l shows that X°,, = K°,q(X) satisfies (B.1), with continuous inclusions. The next theorem establishes the interpolation property. Here, the choice of normalisation is irrelevant (provided it is the same for X and Y). Theorem B.2 If the bounded linear operators AO : Xo Yo and A, : X 1 -+ Yi are compatible, then there exists a unique bounded linear operator A° : K°,q (X) -+ Ko,q (Y) satisfying (B.2). In this case, if Ma and M1 are positive constants such that
IlAjullrj < Mjllullxj
fore E Xj and j =O, 1,
then IIAoullxo.g(r) < Mo-9M°IIuIIxo.q(X)
foru E Ko,q(X).
The J-Method
321
Proof If AB exists, then it must satisfy Aeu = Aouo + A1u1 whenever u = uo +u1 with u1 E Xj, so uniqueness is clear, and Exercise B.1 shows that A9u is in fact well defined in Yo + Y1. Put a = M1 /Mo, so that K(t, Aeu; Y) < (IlAouo11Y0 +t21IA1u111Y,)1,2 < (Mo11 uo112 +t2Mi 11u111x,)1/2
< Mo(Iluollzo + (at)211 u111x1/2.
Hence, K(t, Au : Y) < MOK(at, u; X), and so by (B.3), M0 _B
II As u II K0.q (Y) < Moae 11 u ll K, (X) =
MB II U KII g cx> 11
O
The J-Method Our second construction of Xe,q uses the Bochner integral for functions taking
values in a normed space, and we digress briefly to review a few pertinent definitions and facts; see Yosida [106, pp. 130-136] for more details. Let (S, A) be a measure space and let Z be a normed space. We say that a function f : S -+ Z is finite-valued if there exist finitely many vectors Uk E Z and mutually disjoint p-measurable sets Ek c S with A(Ek) < oo such that f takes the constant value Uk on Ek, and is identically zero on S \ Uk Ek. In this case, the integral of f is a well-defined vector in Z, given by
Is
f (t) dµt =
uk/) (Ek) k
A function f : S -* Z is said to be strongly Z-measurable if there exists a sequence f,,, : S --> Z of finite-valued functions such that ,,,(t) converges to f (t) in the norm of Z for A-almost all t E S. In this case, the real-valued function t t-+ II fn (t) - f (t) II z is measurable, and if the condition
lim fSI1 m-.oc
fn(t)-f(t)Ilzdlkt=0
is satisfied, then f is said to be Z-integrable. The integral of such an f is well defined, as a vector in the completion of Z, by taking the limit of the integrals of the f,,,. A strongly Z-measurable function f : S -+ Z is Z-integrable if and only if the real-valued function t i-+ II .f (t) II z is integrable, in which case
f s
.f(t)dA,
< z
f
s
11f(t)Ilzd,ir
Interpolation Spaces
322
Resuming our consideration of Xo,q, we define the J functional,
J(t, u) = (IIuIIXo
fort>O and uEXoflXi.
+t211ul1z.)1/2
For each fixed t > 0, the J-functional is equivalent to the usual norm on Xo fl X1,
and for each fixed u E Xo fl X1, the function t -+ J(t, u) is non-decreasing. Furthermore,
min(1, t/s)J(s, u) < J(t, u) < max(1, t/s)J(s, u) and
K(t, u) < min(1, t/s)J(s, u).
(B.6)
We define Je,q (X) to be the subspace of Xo + X 1 consisting of those vectors u that possess a representation
u=
f
00 f(t)t
(B.7)
0
for some function f satisfying 00
11 f (t) 11 xo+x,
d t<
oo and
fb
II f (t) 11 xonx,
d t<
00
for0
= inf 11t H J(t, f (t)) 11 o.q,
where the infimum is taken over all possible representations (B.7).
Theorem B.3 If 0 < 0 < I and I < q < oo, then Je,q (X) = Ke,q (X) with equivalent norms.
Proof Suppose that u E Je,q(X) has a representation (B.7). Using (B.6), we see that
K(t, u) <
f f
00
K(t, f (s))
o
=
o
d < fo s
00
min(1, 1/s)J(ts,
00
min(l, t/s)J(s, f(s))
f(ts))d s
ds s
The J-Method
323
so °O
f min(1, l/s)s011 J(.,.f())II8,q 0
ds s
and hence Jo00
min(1, 1/s)se
II UII J9.q(X)
-
ds
l11 IIJ9.,(x
Conversely, suppose that u E Ko.q (X), and let c > 0. For each m E Z there is a decomposition u = uon, + U I., with urn, E Xj and where t,,, =2n'
(1+E)K(tm,u)2,
Iluon,IlXo+t
By Lemma B.1, K (t,,, , u) ::S C t e I I u I I Ka.4 (x), so I l uom l l xn
= 0(t,',,) and II u,n, ll x, _
O(te-'). In particular, IIuo,nI1xo -* 0 as m -+ -oo, and Ilu1,,,11x, m -,- +oo. Define f : (0, oo) -+ Xo fl X, by f(t) _ UOm - Uo,m-1 - Ul.m-1 - Uln, log 2 log 2
0 as
for t,,,-1 < t < tm,
so that
u- J r-M
= U-
f(t)dt
(UOn,
-M
Xo+X,
-
UO.m-1) Xo+X,
= Ilu - UO.M' + u0,-M II xo+x, = II u0.-M + U 1.M' II xo+x, IIu1,M'IIX,)1/2,
(Iluo,_MIIX0 +
implying the representation formula (B.7). Moreover, if tm_1 < t < tn then
(log2)2J(t,1(t))2 = Iluon, - uo.rn-IIIXo +t211u,,n,-1- u,mUx, < 2[IIu0,nII2X0
Iluo,m
112
2[(l + E)K(t,n, u)2 + 4(1 + E)K(t,,,_,, u)2]
< 2(1 + E)[(tm/t)2 +4]K(t, u)2 < 16(1 + -)K(t, u)2, and so (log 2)11J(.,f('))IIe,q <4 IIu11Ja.4(x) <
E Je,q(X)and
u)IIo,q.
We now show that the K- and J-functionals are dual to each other.
Interpolation Spaces
324
Lemma B.4 If X0 fl X 1 is dense in X0 and in X1, then Xo fl X 1 is dense in Xo + X1, and X * = (Xo, X i) is a compatible pair of Banach spaces. Moreover,
Xo + X- = (Xo fl X1)*
and Xo fl x * = (Xo + X1)*,
with equal norms. In fact, for each t > 0,
K(t, g; X*) =
J(t, g; X*) =
I (8, u)I
sup
0 uExonx, J(t- , u; X) sup '05kuEXo+X,
and
I (g, u) I
K(t-1, u; X)
Proof. Given u = uo + u 1 E Xo + X1, the density assumption means that we can find sequences uo,,, and u 1 in Xof1X1 such that uj,» -+ uj in Xj as m --* oo. Define u», = uom + U lm E Xo fl X1, and observe that um - u = (uo,,, - uo) + (ulm-u1),So llu»,-uIlX0+x, Um U in Xo + X1, showing that Xo fl X 1 is dense in Xo + X 1. Also, the continuous dense inclusion Xo fl X2 c Xi gives an imbedding X! c (Xo n X1)*, so Xo and X i can be viewed as subspaces of (Xo fl X 1)*, and hence form a compatible pair.
Let g E Xo + X i , and take any decomposition g = go + gl with gj E X . For U E Xo fl x 1, the Cauchy-Schwarz inequality implies that
I(8 u)I
(so, u)+(81, u)I
so I(g, u)I < K(t, g; X*)J(t-1, u; X). In particular, by putting t = 1 we see that g e (Xo fl X1)* and llgH(xonx,)- < llgllxo+x; Conversely, let g E (Xo fl X 1) * and put
M, =
sup o#uExonx,
I(8,u)I
J(t-1, U'; X)
We equip the product space Xo x X 1 with the norm II(uo, ul)Ilxoxx, = (IIuollX0
+t-211U1112)'11,
and denote the diagonal subspace by W = {(u, u) : u E Xo fl X1 }. Now define a linear functional g w on W by
(gw, (u, u)) = (g, u)
for it E Xo fl X1,
The J-Method
325
and observe that
I(gw, (u, u))I < M,J(t-', U; X) = M, II (u, u)Ilx0Xx
so the Hahn-Banach theorem gives (go, g1) E Xo x X = (Xo x X 1)* such that
(gw, (u, u)) = ((go, gi), (u, u)) = (go, u) + (gi, u) for u E Xo fl X1, and II(go, g1)Ilxoxx; _ (IIgoIIXo +t211g11IX1/2 = M,.
Hence, g = go + gi E Xo + Xi, and K(t, g; X*) < M. In particular, putting t = 1 gives Ilgllxo+x; < M1 = IIgll(xonx,)»
Next, let g E Xo fl X. If u = uo + u1 E Xo + X1, then 2»
2
2 112(IIu ollXo 2 I(g, u)I = I(g, uo)+(g, u1)1 _ (Ilgllxo+t IISIIX,)
+t-211u111X')1I2,
so I (g, u) I < J (t, g; X*)K(t-', u; X), and by putting t = 1 we see that g E (Xo + X1)* with Ilgllxo+x,)» <_ Ilgllxonx;
Conversely, let g E (Xo + X1)*, and put
M, =
Ifu X, then l(g,u)I
sup
I(g,u)I
O#UEXO+X, K (t-1
, u; V X)
<M,K(t-1-,u; X)<Mtmin(IIullxo,t-1IIulIx,),show-
ing that g E Xo fl X j . Given E > 0, we can find u1 E Xj such that
(g, uj) < Ilgllxj < (1 +E)(g, uj)
and
IIuj IIx, = 1.
(In particular, (g, uj) isreal.)PutAj = 1181Ix;,sothat 11811X. < (1+E)(g. AJu) and J(t, g; X*)2 < (1 + E)(g, ),ouo + t2)1u1)
_< (1 +E)M,K(t-1, Aouo+t2A1u1; X)
< (1
+E)Mt(IR,Xouol12
(1 + E)Mt(IIgIIXo +
+t-2IIt2A1u11IXYi2
(1 + E)M,J(t, g; X*).
Hence, J(t, g; X*) < Mt, and in particular IISIIxanx; < IIg11(x,,+x,)
0
Interpolation Spaces
326
Theorem B.5 Assume that Xo fl X1 is dense in Xo and in X1. If 0 < 0 < I and 1 < q < coo, then X0 fl X 1 is dense in Xe,q and (Xo, X i
where
.q = (Xo, X')e.q-,
*+ 1
1
q*
q
1
= 1,
with equivalent norms.
Proof Let u E Je,q (X) have a representation (B.7), and define u,,, E Xo fl X
1
by f(t)GIt
un, =
JI/m
t
Since
e
u 11 J".' (X) < (JO.1,,I1)U(,fl.OO) I t
J (t, f (t))
-) t
jdt
11q
we see that u,,, -+ u in Je,q (X), and hence Xo fl x 1 is dense in Xe,q. Thus, the inclusions Xo fl x 1 c Xe,q c Xo + X 1 are continuous and have dense images, so by Lemma B.4,
Xo fl Xi c (X9,q)* c Xo + x
.
It suffices to show that there are continuous inclusions c KK,q(X)*
and
Je,q(X)* 9 Ke.q*(X*)
Let g E Je,q=(X*), and take any 0 : (0, oo) -+ Xo fl x* such that
8=
f
000(t) dt, t
with the integral converging in Xo + X* = (Xo fl X1)*. For any u E Ke,q(X), I(8, u)I _5
f
00
1(0(t), u)I dt <
f
x J(t, 0(t); X*)K(t-1, u; X)
dt
< 11tH J (t, 0 (t); X*) 11 e,q 11 t H K (t-', u; X) II _e.q' so 1(g, u)1:5 1181IJA.,,.(x*) Ii t H K(r', u; X)11-e,q showing that g E Ke,q (X )* and II g II K .q (x)
II
u; X) 110,q,
NO, q' II B II JJ.q«
Now let g E Jo.q(X)*. Given c > 0, we can use Lemma B.4 to find a piecewise-constant function >1t : (0, oo) -+ Xo fl X1 such that (g, sli(t)) is real,
The J-Method
327
and
(g, fi(t))
K(t-', g; X*) < (1 +E)
J(t, fi(t); X) For any measurable function f : (0, oo) -f (0, oo) such that II f Ile,q < oo, the integral
f (t) * (t)
00
of = J
dt
J(t, *(t); x) t
defines a vector U f E Je,q (X) satisfying II u f II Jd.q cx) < II f Il o,q We choose f
so that
I
00
K(t-',
dt
g; X*)f(t) t = lit H K(t-' , g; x*)II _e ,q =jjK(-.g;X*)jjo,7»
and II f 11 0,7 = 1. Thus,
11
g; X*)11O.q» <
f
00
(1 + E)
o
(g, fi(t))
f(t) dt t
J(t, *(t); x)
(1
= (1 + E)(g, Uf)
(X) _< (1
showing that g E Ko,q»(X*) and
D
The next result is known as the reiteration theorem.
Theorem B.6 Let 1 < q < oo, and for j = 0 and 1 put (Xej,q
Y'=jl X3
if0
If 00 # 91, then (Yo, Yi),1.q = (Xo, Xi)e.q
for 9 = (1 - ri)9o + n91 and 0 < i < 1.
Proof. Let U E K,1,q(Y). For any uj E Yj such that u = uo + u1, we have K(t, at; X) < K(t, uo; X) + K(t, u1; X) < Ct°'Iluo11yo + Cto' IIu111r,
<
Cto°[IIuoIIY°+t2(0,-O,)IIu111r,]112,
where, in the second step, we use Lemma B.1(ii) if 0 < 9j < 1, or (B.5) if 9j E {0, 11. Hence,
K(t, u; X) < Cto0K(to'-o0, u; Y),
Interpolation Spaces
328
and therefore the substitution t = t°' -°o gives
< C f 0" It-°t°0K(t°,-°°,
q
f0
f
=C
u; Y) Iq
dt t
It-nK(t,u;Y)Ig101-9o1
dr
(B.8)
t
= C191 -00,11Ul1K,q(Y)
for 1 < q < oo, with II u II
K9..,,(x)
< C sup t-ete° K (te' -e0, u; Y) = C sup r-" K (r, u; Y) = C 11U II K,,..(Y) t>o
r>o
in the limiting case q = oo. Thus, Y,1,q c Xe,q.
Conversely, let u E X°,q, and take any representation of the type (B.7). By (B.6), 00
t°OK(t°'-e°, u; Y) < f
t°OK(te,-e0, f (s); Y) ds S
o
<
r
t°° min (1, (t/s)el -eo) J(se,-eo, f(s); Y)
Jo
d s
Using Exercise B.2 if 0 < 9j < 1, or just the definition of the J-functional if 9j = j, we see that 11f (s) II Y; < Cs-0j J(s, f (s); X) and so j (SO, -°D, f (S); Y) =[11f(S)11 y°+ S2(e, -e0)
II f (S) II Y1
]1/2 < CS-26° j (S' .f (s); X)
Thus, with the help of the substitution s = at,
x teOK(te'-e0, u; Y) < C
Jo
= C f "0
min ((t/s)°°, (tls)°')J(s, f (s); X) mina-°°,
a-'91)J (at, f (at); X)
ds s
da ,
and hence by (B.8), IIUIIK,,.a(Y) =
1
let-eol
11t i-+ t °o K(t°' -°°, u; Y)II°,q
too
< CJ min(a-OD,a-°')IIt )-+ J(at, f(at); X)II °,q da Q
0
p
=CJ
0
a ° min(a-°0, a-A')
It It H J(t, f(t); X) II °,q
Interpolation of Sobolev Spaces
329
Finally, taking the infimum over all f satisfying (B.7) gives
II u II K,,.,, (Y)
C11 u11 j,,, (x), proving that Xo,q c Y,?,q.
Interpolation of Sobolev Spaces If Xo and X1 are Sobolev spaces on R", then the K-functional can be computed explicitly via the Fourier transform, allowing us to prove the following.
Theorem B.7 If so, s! E R, then (HS0(I8"), H`' (Il8"))8,2 = H'(R")
for s = (1 - 9)so + 9s1 and 0 < 9 < 1.
Moreover, the Sobolev norm (3.21) equals the Ko.2-norm if, instead of the default
normalisation (B.4), we take NB,q
=
(2sinir9\'2 J
(B.9)
n
Proof. Let u = uo + u1 with ui E H''i (1R") for j = 0 and 1, and observe that
f [(1 + I
+t2I1U1112H
I2)S°luo( )I2
+ t2(1 + I 12)s' Iu 1(4)12]
Since u () = uo
u
we see from Exercise B.4 that for each
integrand is minimized when uo and u 1 are such that
t2(1 +
t2(1 +
(1 +
t2(1 +
(1 +
It follows that t2(1 +
K(t, u; HS0(18"), HS' (R"))2
J =
(1 +
t2(1 + 112)s
j(l + ,I
where
a(') = (1 + 1
+t2
lu()I2d
the
Interpolation Spaces
330
Thus,
IIK(, u)IIa2 =
f
(1 + 1
11f 110.2]21u( )12 d = Il.f
,1
and we have only to apply Exercise B.S.
We also have an interpolation theorem for Sobolev spaces on domains. Theorem B.8 Let S2 be a non-empty open subset of R. If so, s 1 E R, then
(HS0 (0), HS' (52))9,2 = H'(Q)
for s = (1 - 9)so + 9s1 and 0 <0 < 1.
Moreover, the Sobolev norm (3.23) equals the KB,q-norm if the latter is normalised by (B.9). Proof Put X j = HS' (S2) and Yj = Hs; (IIR") Let U E HS (a), and choose U E HS (R") such that u = U I n and II U 11 H= (2) _ II U II H'(R^) Take any decomposition U = Uo + U1 with Uj E Yj, and observe that since u = uo + u1, where u j = Uj Isi E X j, we have
K(t, u; x)2 <
IIuoIIX° +t211u111X,
IIUO112 +t211U111y,,
and thus K (t, u; X) < K (t, U; Y). It follows by Theorem B.7 that u E K9,2 (X) with IIuIIKB.2(x) <- II UII K0.2(Y) =
IIUIIHI(Q)
Conversely, let u E K9,2(X), and take a decomposition u = uo + ul with U j E X j. Choose U j E Y j such that u j= U j I sz and II u j II x1 = II Uj II r; , and put
U = Uo+U1 E Yo+Y1. We have K(t, U; Y)2 < IIUoIIy° 2 +t21IU1IIY, = IIuoIIX° +t21Iu11IX1 ,
so K(t, U; Y) < K(t, u; X) and hence 11U11 K,., (y) < I1 u11KB.2(x) Since u = U
Theorem B.7 implies that IIuffH'(S) < IIUIIHsr(R") = IIUIIK0.2(Y) < IIUIIK9.2(x)
0 An interpolation result for HS (S2) follows at once, by duality; see Corollary 3.30 and Theorem B.S. Theorem B.9 Assume that 0 is a Lipschitz domain. If so, s1 E 1R, then
(R' m), HS' (S2))e 2 = HS(Q) with equivalent norms.
for s = (1 - 9)so + 9s1 and 0<0 < 1,
Interpolation of Sobolev Spaces
331
Our final result deals with Sobolev spaces on the boundary r = 8 Q. The proof makes use of the reiteration theorem (Theorem B.6) and a simple interpolation property of pivot spaces.
Lemma B.10 If H is a pivot space for V, then H = (V, V*) 1/2,2. Proof Let Y = (V, V *)1/2. z. By Theorem B.5 and Exercise B.7, Y* = (V*, V)1/2.2 = (V, V*)1-1/2.2 = Y, so if u E Y, then II u II H = (u, u) < II u II Y II u II Y < C II u II .. Hence, we have a
continuous, dense inclusion Y C H, and dually, H = H* c Y* = Y. Theorem B.11 Assume that 0 is a Cr-1,1 domain for some integer r ? 1. If so, S1 E R satisfy Iso I < r and is, I < r, then
(Hs0 (F), Hs' (r))0,2 = HS (r)
for s = (1 - 9)so + 9s1 and 0 <0 < 1, (B.10)
with equivalent norms.
Proof Suppose in the first instance that r is a Cr-1,1 hypograph, and recall the definition of Hs (F) for the two cases 0 < s < r and -r < s < 0, given in the discussion after Theorem 3.34. The interpolation property follows at once from Theorem B.7 if both so and sl belong to [0, r], or if both belong to [-r, 01. Furthermore, we claim that
(H-r(r), Hr(r))9 2
=
H2e-1(F)
for 0 < 0 < 1.
Indeed, the case 0 = z follows from Lemma B.10, after which the cases 0 < 0 < 1 and 1 < 0 < 1 follow with the help of Theorem B.6. Another application of Theorem B.6 then shows that (B.10) holds when one of the sj belongs to [0, r] and the other to [-r, 0). Suppose now that r is the boundary of Lipschitz domain in the sense of Definition 3.28. Thus, recalling (3.29), IIuIIHJ(r)
_
II01uII2W(r,).
If we put
X = (HS°(r),
HS' (r))
and
X1 = (HsG(rt), HS' (rt)),
Interpolation Spaces
332
then K9,2(XI).= Hs(rl) for each 1, and the problem is to show that Ke,2(X) _
Hs (r). Let U E HS(r). We choose cutoff functions Xi E C o p(R'1) satisfying XI = 1 on a neighbourhood of supp lpl, and supp X/ c W1, so that 4ru = X101u.
Take any decomposition 01 u = uro + u11 with Ulj E Hsi (I,1), and observe that
U = 2biU = EXIU1o+EX1U11, l
I
I
so 2
2
+ t2
H i (r)
H o (r)
C Y(IIXIUiOI Jpo(r) + t211Xru11 IIHr, (r)} I
C E(IlurollHxo(r,) + t211Un IIH,,(rv) I
Thus,
K(t, u; X)2 < C >2 K(t, 01u; X1)2, and therefore u E Ke,q (X) with IIUIIK9.Q(x)
Ilb1UIiK,(x,)
C
=IIuIIH=(r)
Conversely, let u E Ke,q (X) C Hmin(so,s,) (r). Taking any decomposition u = uo + ul with uj E Hs1(r), we have Olu = Oluo + (plug with O1u1 E Hsi (F,), so E[1101uoIIHw(r,)
K(t, 01 u; X1)2
+ tell
01UI
IIHs'(r,)]
l
1
=
IIUO112
+ t211u111H., (r)
and hence >1 K(t, 01 u; X1)2 < K(t, u; X)2. It follows that ¢IU E Ke,2(XI) _ HI (rl) for each 1, with IIuIIH= (r)
IIp1uIIKe.2(x,) -<
IIUI1K2e.
2(X)
El
Exercises
333
Exercises B.1 Suppose that AO : Xo -)- Yo and A : X 1 --* Y1 are compatible, and let u E Xo + X 1. Show that Aouo + A 1 u does not depend on the choice of 1
uo E X0 and U 1 E X 1 satisfying u = uo + u 1. B.2 Show that J(t, u) > to IIuIIK.,4(x). [Hint: use (B.6).] B.3 Give a direct proof of the interpolation property for Je,q (X), i.e., show that
Theorem B.2 holds if Ke,q is replaced by Jo,q. [Hint: J(t, Af (t); Y) < MoJ(at, f (t); X), where a = M1 /Mo.] B.4 Fix real numbers Ao > 0 and A 1 > 0, and a complex number z. Show that min (AoIzo12 + A1Iz112) =
z=zo+zi
AOA1
Ao + Al
Iz12,
and that the minimum is achieved whenAozo = A1z1-= AoA1z/(Ao+A1). B.5 Use contour integration to show that cc t 1-26
,r
1 + t2 dt
for 0 < 9 < 1.
2 sin r9
0
B.6 Show that if Xo and X 1 are complete, then so are Xo fl X 1, X0 + X 1 and
Xe,q for 0 < 9 < 1 and 1 < q < oo. [Hint: use Exercise 2.1.] B.7 Show that (X1, Xo)B,q = (Xo, X1)I-o,q for0 < 9 < 1 and I < q < oo. B.8 Assume that H, V and A satisfy the hypotheses of Corollary 2.38, and equip V with the energy norm II II A. By arguing as in the proof of Theorem B.7, show
K(t, u; H, V )2
= E 1+?
t2I(0;,u)12,
and deduce that for the normalisation (B.9), CO
IIUIIK9.z(H.v) =
E'Xjl(4),,
j=1
u)12 = IIAB"2uIl2.
Appendix C Further Properties of Spherical Harmonics
We shall prove a result (Corollary C.2) used in Chapter 9, and also construct the classical spherical harmonics, which form an orthogonal basis for xm Recall the definition of the Legendre polynomial Pn, (n, t) in the discussion following Lemma 8.6. (S2).
Theorem C.1 The orthogonal projection Q,n : L2 (5"-1) -> xm (S' -1) is given by the formula Q,n
,/ (CO) = N(n, m)
Pm (n, w 1) (r1) d?l
T. 4-1 form > 0, w E S' ' and it E
L2(S"-1).
Proof. Let (rmp : 1 < p < N(n, m)) be an orthonormal basis for
xm(S'
)
By part (i) of Exercise C.1, Qm 1F' (w) =
f
_
K(w, n)i(n) dn,
where
N(n,m)
'I, 'I' K(w, n) = E Y'mp(w) lmp(n) p=1
If A E
nxn
is an orthogonal matrix, then
J
*(Aw) dco =
J
*(w) dco for r E L1(S"-1),
and, by Exercise 8.3, the function co 1-* u(Aco) belongs to Rm(S") whenever u E xm (Sn-1). It follows by part (iii) of Exercise C. 1 that
K(Aco, An) = K(w, n) for co, n E Sr-1, 334
Further Properties of Spherical Harmonics
335
and in particular, when Aen = e we have K(Aw, en) = K(cv,
Since
w H K (co, e,,) belongs to ln, (S"-' ), if u is as in Lemma 8.6 then
K(w, en) = au(w) = aP.. (n, w en)
for w E S"_ 1,
where a = 1/K(en, en). Given r! E Sn-1, choose an orthogonal matrix A E Rnxn such that Ail = en; then K(w, rl) = K(Aw, en) = aP.,(n, Acv en)
Finally, since K(w, w) = aP,n(n, 1) = a, we see that N(n,m)
aTn=
f ,-
K(w,co)dw= E Il*mpllL,
,
p=1
giving a = N(n, m)/Tn.
O
The proof above also establishes the addition theorem.
-
Corollary C.2 If (if,np : 1 < p < N(n, m)) is an orthonormal basis for xm (Sn-1), then N(n,m)
'
for w, )l E Sn-'
(CO) Y',np (11)
P=I
The next result is known as the Funk-Hecke formula.
Theorem C.3 Let f : [-1, 1] -+ C be a continuous function. If m > 0, then
f (w rl)(n) drf = A(w) for* E 71 fS.-I
where
= Tn-1
f
f (t)Pm(n, t)(1 -
t2)(n-3)/2
dt.
(C.1)
Proof We begin by showing that f (l; w) Pm (n, rl co) d w = A Pm (n, l; rl)
fore, rl E Sn-' .
(C.2)
rl) to be the left-hand side of (C.2), and observe that for a fixed l;, Define Also, if A E RI"" is an the function >) r-# F(l;, r7) belongs to
Further Properties of Spherical Harmonics
336
orthogonal matrix, then F(A4, Arl) = F(4, rl), so by arguing as in the proof of Theorem C.1 we see that ,XP,,, (n, rl) for some constant .l. Since
Pm(n,1)=1, I = XPm(n, 1) = F(en, en) =
f
f(en w)P,n(n, en W) do),
S`n-I
and the formula (C.1) follows by Exercise C.2. Thus, (C.2) holds, and upon multiplying both sides of this equation by *(q), where 1/r E xm (Sn- t ), and integrating with respect to q, we have
f
S.-I
f(
. w)
\
Sn
-
Pm(n, n
w)*(rl) drl dw
I
fs.-I
P. (n,
rl)*(rl) dil.
Applying Theorem C.l, the result follows at once. We now begin our construction of an explicit orthogonal basis for H. (S! I- ),
starting with the case n = 2. Remember that 1-1o(Sn-1) consists of just the constant functions on S", so it suffices to deal with m > 1. Lemma C.4 Let 9 be the polar angle in the usual parametric representation of the unit circle S1, co = (cos9, sin 9).
If m > 1, then N(2, m) = 2 and the functions cosm9
1rm1(w) _
and
*,n2 (W)
1/7-r
r7r
sin m9
form an orthonormal basis for Hm (S1)
Proof. We have seen already in (8.13) that N(2, m) = 2 form > 1. Define two real-valued solid spherical harmonics u 1, u2 E Nm (R2) by u1(x) + iu2(x) = I (x1 + ix2)m,
and observe that 1lrn,1 and i/r,,i2 are the corresponding surface spherical harmonics in 11m (S1), i.e., 1/ m p = u p is, for p = 1 and 2. One readily verifies that *,,l and
1/rm2 are orthonormal in L2(S1).
Any non-trivial function Amj (n, t) satisfying the conclusion of the next theorem is called an associated Legendre function of degree m and order j for the dimension n.
Further Properties of Spherical Harmonics
337
Theorem C.5 Assume n > 3, let 0 < j < m, and define
A,nj(n, t) = (1 - t2)jj2Pm-j(n +2j, t).
(C.3)
If i/r E ?-l j (Si-2) and
where co =
W(co) = A,nj (n, t)i/r(r1),
then
1 - t2 n + ten and rl E gi-2,
E fm(Sn-1)
Proof For any 1(r E L i (Sn-2), the formula
u(x) = f-2(xn + ix' )'n
(
) dt
defines a solid spherical harmonic of degree m, whose restriction to S"-' may be written as
u(w)=f (t+i 1-t2 r! If we now assume that and Exercise C.5,
E
?j j (S, -2), then by Theorem C.3, Exercise C.4(ii)
t
u(w) = Tn-2*(r1) f (t + i l - t2 s)mPj(n - 1, s)(1 = cl (17) (1 - t2)'/2
J
(t + i l - t2 1
s)m-I (1
-
s2)(n-4)/2
ds
s2)r+(n-4)/2
ds
= c2i(11)(1 - t2)j/2Pm- j(n + 2j, t) = c2`1(w), where the constants cl and c2 depend on n, m and j. Thus, 41 E ?-lm claimed.
(S"), as
Combining Theorem C.5 and Exercise C.6, and recalling (8.12), we see how to construct an orthogonal basis for W n (S"-1) by recursion on the dimension n. Corollary C.6 If {7(rjp
x j (Sn-2), and if
:
I < p < N(n - 1, j)) is an orthogonal basis for
q'mjp(w) = Amj(n, t)'Yjp(q),
where w =
1 - t2 q + ten and r1 E Sn-2,
then {Wmjp:O < j < m and 1 < p < N(n - 1, j)) is an orthogonal basis
Further Properties of Spherical Harmonics
338
for H,n (Sn-1), and
N(n + 2j, in - j) T +2l! 1 II*jP ,(S,,-2) In particular, taking n = 3 and using a basis for fl3(S1) of the type in Lemma C.4, we arrive at the classical spherical harmonics. Theorem C.7 Use the standard parametric representation for the unit sphere S2,
w = (sin 0 cos 0, sin 0 sin 0, cos 0) for 0 < 0 < 2ir and 0 < 0 < rr.
If in > 1, then N(3, m) = 2m + 1 and the functions *mo (w) = Pn, (3, cos 0),
*mj (w) = (sin Ql )' Pm_j (3 + 2j, cos 0) cos j O for 1 < j < m, YIm.m+i (w) = (sin 1)i
Pm_j (3 + 2j, cos ¢) sin jO for 1 < j < m,
form an orthogonal basis for lm (52), With II1f,no II L2(s2) = 4n' and T3+2i
It
'I, II
n=; II L2cs2) = IIY'm.m+J HL,(S2) =
N(3 + 2j, m - j) T2+2j
for 1 < j < m.
Exercises C.1 Suppose that (S, p) is a measure space, and let Q be the orthogonal projection from L2(S, µ) onto a finite-dimensional subspace V. (i) Show that if {¢p)n 1 is an orthonormal basis for V, and if we define N
K(x, y)
LOp(x)Op(y), p=1
then
Qu(x) =
Js
K(x, y)u(y) dµy
for x E S and u E L2(S, µ).
(ii) Show that the kernel K does not depend on the choice of the orthonormal basis for V.
(iii) Suppose that a group G acts on S, and that µ and V are invariant under G, i.e., if g E G, then
u(gx) d,i = J f u(x) d/.L s
s
for u E L1 (S, A),
Exercises
339
and
x H u (gx) belongs to V whenever u E V.
Show that the kernel K is invariant under G, i.e., if g E G then K(gx, gy) = K(x, y) for x, y E S. C.2 Show that if f is, say, continuous on Si-1, then
f Jg f
(co)
f
dw =
I I
Jg,-2
f
1 - t2 >7 + te,i) d>) (1 - t2)(n-3)/2 dt.
C.3 Prove the orthogonality property of the Legendre polynomials:
f
T.
1
1
P., (n, t)PI(n, t)(1 - t2)(n-3)l2dt =
N(n, m) Tn_1
1
smI
form >Oandl >0. [Hint: use Theorems C.1 and C.3.]
C.4 For n > 2 and m > 0, let m
P n
(t) = (1 -
t2)-(n-3)/2-d
dtm (1
-
t2)m+(n-3)/2
(i) Show that pm is a polynomial of degree m. (ii) Use integration by parts to show that if, say, f E C'n [-1, 1], then
f
f(t)Pm(t)(1 - t2)(n-3)/2dt 1
j f ('n) (t) (1 -
t2)m+(n-3)/2
dt.
(iii) Deduce the orthogonality property fJ
Pm(t)PI(t)(1 -
t2)3/2 dt = 0
if m
(iv) Find pm(1), and conclude from Exercise C.3 that
(-),n
P(n, t) _ (n+2m-3)(n+2m-5)...(nresult known as the Rodrigues formula.
1)
C5 Show that
P.(n,t)=
1
Tn-t
g,-z
(t+i
foralla)
ES"-2,
Further Properties of Spherical Harmonics
340
and then derive the Laplace representation,
P. (n, t) =
Tn-1
j (t + i
1 - t2 s)m (1 -
s2)(n-4)12 ds
for n > 3.
[Hint: use Theorem C.3 with f (s) = (t + i 1 - t2 s)m and 1/r = 1 E ? jO(Sn-2).)
C.6 Suppose n > 3, and let Am j (n, t) be the associated Legendre function (C.3).
Show that for *1, *2 E C(Sn-2), if and
IPi(W) = Amj(n, where W =
42(W) = Amj(n,
1 - t2 + ten, then
('P1, W2)L20°-i) =
1
N(n + 2j, m - j)
(*I,
2)
Tn+2j-1L2(5-z).
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Index
Adjoint (formal) of a differential operator, 116 of an abstract linear operator, 37, 43 of conormal derivative operator, 201 of trace operator, 201 Annihilator of a subspace, 23 Arzela-Ascoli theorem, 28 Associated Legendre function, 336, 340 Beer, A., 11 Beltrami operator, 277 Bessel function, 278 spherical, 279. 293 Bessel potential, 75 Bochner integral, 321 Boundary integral equation and logarithmic capacity, 264, 275 for Dirichlet problem, 226 for exterior problem, 236 for problem with mixed boundary conditions, 231 for Neumann problem, 229,242 side condition for, 262. 274 Boundary integral operator (Schwartz) kernel of, 220, 223 adjoint of, 218 arising from a self-adjoint differential operator, 218 arising from the Laplacian, 248, 249 definition of, 218, 233 mapping properties of, 219, 245 on an open surface, 275 symbol of, 244, 275
Calderdn, A. P., 309 Calderbn projection, 243 Capacity, 263 and exterior Dirichlet problem, 271 behaviour under dilatation, 272
logarithmic, 264 and conformal mapping, 271 of a line segment, 275 of an ellipse, 272 of an open surface, 275 variational characterisation of, 272 Cauchy-Riemann equations, 2 Cauchy-Schwarz inequality, 54 Chebyshev polynomials, 255 explicit solutions in terms of, 272, 273 Codimension, 18 Coercivity abstract definition of, 44 for differential operators, 118 change of coordinates, 156 on H 1(12), 122, 126 on Ha (0), 119 for elasticity operator, 298, 299 for hypersingular integral operator, 230 Compact linear operator, 28 properties of, 54 transpose of, 29 Compact subsets of LN, 28 Compatible pair of normed spaces, 317 Completely continuous linear operator, see compact linear operator Conjugate exponent, 58 Conjugation, 37 Conormal derivative, 114 L2 estimates, 149 generalised, 117 relative to the formal adjoint, 116 Convergence in D(S2), 65 in E(S2), 65
in E(S2) but not in D(P), 109 in S(]R' ), 72 Convex set, 38
347
348
Index
Convolution, 58 and approximation, 63, 111 and differentiation, 62. 109 and Fourier transform, 73 associativity of, 108 support of, 108 Cooling-off problem, 157 Costabel. M., 102, 202 Cutoff function, 83
d'Alembert, J., 2 Density of Cm p(S2) in L p (c ). 63 of D(S2) in E(S2), 109
of D(Q) in Hs(9), 77 of D(R") in S(R"), 109 of D(S2) in H'(Q), 111 of S(R11) in E(W), 109 of D(S2) in W" (Q), 91 of Ws (0) fl E(92) in Ws(S2), 86 Dirac delta function(al), 66 convolution with, 68 homogeneity of, 187 Dirichlet form, 246 Dirichlet problem, 4 solution operator for, 145 Dirichlet's principle, 8 abstract form, 55 Dirichlet, P., 8 Distribution, 65, 109 multiplication with a smooth function, 68
of form u(a x), 189 partial derivative of, 67 temperate, 72 with compact support, 67 Divergence theorem, 97 Domain Lipschitz, see Lipschitz domain 89 of class Ck, 90 of class 90 Double-layer potential, see surface potential du Bois-Reymond, P., 12 Dual space, 20 of Lo, 58, 107 of a Sobolev space, see Sobolev spaces, duality relations realisation of, 27
Eigenfunction expansions, see spectral theory Eigenvalue, 45 Elasticity operator, 297 Energy inner product, 44 example from linear elasticity, 298 Epigraph, 186, 316 Equicontinuous set of functions in C(X), 28 in LP, 28
Equilibrium density, 263 Euler, L., 2 Extension operator, 81, 309, 313 non-existence of, 316 Seeley, 316
External conformal radius, 271 Far-field pattern, 294 Finite part, 159 Finite-part extension differentiation of, 169 homogeneity of, 162, 166, 168 in n dimensions, 166
ofx'k-1, 164 of x+, 160 Finite part integral change of variables formula, 177, 180 hypersingular integral operator. 223 on a surface, 181 Fourier transform, 70 inversion theorem, 70 of a homogeneous distribution, 172, 173, 189
of a temperate distribution, 72 of partial derivatives, 72 of f.p. u, 174 Fourier, J.-BA., 4 Fredholm alternative, 14, 37, 43 for boundary integral equations, 226, 228, 229, 240 for coercive operators, 44 for the mixed boundary value problem, 128 for the third boundary value problem, 131 relation to eigensystem, 51 Fredholm equation of the second kind, 13 abstract theory, 30, 35 Fredholm operator, 32 Fredholm, 1., 13 Fundamental solution, 191, 197 for elasticity operator, 300 for Laplacian, 2, 11, 247, 268 for the Helmholtz operator, 279 radiating, 282 integral formula for, 198 series expansion of, 255, 284 Funk-Hecke formula, 335
Gamma function, 169, 188 Gauss, C. F., 7 Generalised function, 66 Green identity first, 4, 114, 116, 1 1 8, 141
first, dual version of, 115, 118, 141 second, 4, 118 third, 5, 202
dual version of, 211 with radiation condition, 235
Index Green's function, 5 symmetry of, 16 Green, G., 4, 8
Hadamard, J., 159 Hahn-Banach theorem, 20 Hankel function, 280 spherical, 281, 293 Hardy's inequalities, l 11, 112 Harmonic analysis techniques, 209 Heat equation, 4 Helmholtz equation, 276 radiating solution of, 281 Hermitian sesquilinear form, 43, 116 Hilbert space, 38 best-approximation properties, 38, 39 dual of, 41, 42 weak sequential compactness in, 42 Hilbert, D., 14 Holder's inequality, 58 Homogeneous distribution, 158 derivatives of, 187 Homogeneous function, 158 and change of variables, 175 derivatives of, 187 orthogonality condition for, 167 parity condition for, 168, 169, 175 Hypograph, 186 of class C't, 90 Lipschitz, see Lipschitz hypograph Image, 18 Imbedding of S* (W) in D* (W), 72
of L11(2) in D`(S2), 66 Index of a Fredholm operator, 33 compact perturbation, 36 homotopy, 54 small perturbation, 54 Inner product, 38 Interpolation of normed spaces, 318 duality properties, 324, 326
J-method, 322 K-method, 319 reiteration theorem, 327 Inverse point with respect to a sphere, 259, 270 J-functional, 322 Jump relations, see surface potential K-functional, 318 determined by a positive-definite, self-adjoint operator, 333 for Sobolev spaces, 329 Kelvin transform, 259, 270 Kelvin, Lord, see Thomson, W.
349
Kernel (null space), 18 of boundary integral operator, 240 Helmholtz equation, 288 Laplacian, 267 linear elasticity, 304 Kom's inequality first, 298 second, 299, 305 Lagrange, J.-L., 2 Lame coefficients, 296 Laplace equation, 1 Laplace operator, 246 eigenvalues of, 249 rotational invariance of, 268 Laplace, P. S., 3 Lax-Milgram lemma, 43 Le Roux, J., 12 Legendre polynomials, 255 generating function for, 269 Laplace representation, 340 orthogonality property, 339 recurrence relation for, 269 Rodrigues formula for, 339 Liouville, J., 6-8 Lipschitz dissection, 99 Lipschitz domain definition, 89 non-examples of, 90 outward unit normal to, 96 surface element for, 96 Lipschitz hypograph, 89 Locally integrable functions. 64
Meyers-Serrin theorem, 85 Modulus of continuity, 60, 110 Mollifier, see convolution Multi-index, see partial derivative Ne&as, J., 123, 126, 147 Nedelec, J. C., 289 Neumann, C., 10, 12 Newtonian potential, see volume potential Nirenberg, L., 133 Nitsche, J. A., 305 Noether, F., 33
Open mapping theorem, 19 Orthogonal complement, 40 Orthogonal projection, see projection, orthogonal
Parametrix, 192 adjoint of, 197, 211 behaviour of kernel, 195 mapping property for, 193, 197
350
Index
Partial derivative, 61 weak, 74 Partition of unity, 83 and Sobolev norm, 111, 331 Peetre's inequality, 88, 110 Pivot space, 44 interpolation property of, 331 use of L2, 118 Plancherel's theorem, 73 Poincard, H., 10, 13, 145 Poisson integral formula, 5 Poisson's equation, 15
Poisson, S: D., 3-6 Positive and bounded below, 43 boundary integral operator, 262, 264, 267, 275 Potential electrostatic, 3, 5, 263 gravitational, 3 surface, see surface potential vector, 292 volume, see volume potential Principal part, 114 and coercivity, 118 Principal value, 166, 190 Projection, 20 orthogonal, 40, 54
Quotient norm, 19 Quotient space, 18 Radiation condition, 234, 243 for the Laplacian, 259 Sommerfeld, 281, 283, 294 Reflexive Banach space, 22, 37 Regulariser, 35 Regularity theory for boundary integral equations, 239 interior, 135,196 up to the boundary, 137 Relatively compact set, 27 Rellich-Payne-Weinberger identity, 146
Rellich, F., 87, 147, 286 Riesz representation theorem, 40 Riesz, F., 15 Rigid motion (infinitesimal), 302 Scalar wave equation, 276 Self-adjoint operator, see adjoin Separation of variables, 4, 277 Sequential compactness, 27 Sesquilinear form, 42 arising from a boundary integral operator,
261,275 arising from an elliptic differential operator, 114
arising from the Helmholtz operator, 276 arising from the Laplacian, 246 Single-layer potential, see surface potential Singular integral operator, 190, 312 Slobodeckii seminorm, 74, 79 and Fourier transform, 79 Smoothing operator, 192 Sobolev imbedding theorem, 86 Sobolev representation formula, 311 Sobolev spaces compact imbeddings, 87 definition via Bessel potentials, 76 definition via weak derivatives, 74 density theorems, see density duality relations, 76, 78, 92, 98 equivalent norms for, 96, 110
Hs(1R") = W(W), 80 Hs(Q) = Ws (0), 81, 92 Hs (S2) = Ho (S2), 95,112
Hs(S2)=H' 91 interpolation properties of, 329-331 invariance under change of coordinates, 85 of negative order, 74 of vector-valued (generalised) functions, 106
on the boundary of a domain, 98, 99 Sommerfeld radiation condition, see radiation condition, Sommerfeld Spectral radius, 55 Spectral theory for coercive self-adjoin[ operators, 49 for compact self-adjoint operators, 47, 55 for self-adjoint elliptic differential operators, 132 Spectrum of a linear operator, 45 Spherical harmonics, 250, 252 addition theorem for, 335 and boundary integral operators, 252 and the Helmholtz equation, 279 classical, 338 eigenfunctions of the Beltrami operator, 278 for the circle, 336 orthogonal basis for, 337 orthogonal projection onto, 334 L2(S"-1), orthogonality in 265 series expansion in, 257 Stein, E. M., 309 Steklov-Poincare operators, 145 estimates for a C"+1 , 1 domain, 146 estimates for a Lipschitz domain, 155 representations in terms of boundary integral operators, 244 Strain tensor, 296 Stress tensor, 296 Stress-strain relation, 296 Strictly positive-definite operator, 44
Index Strongly elliptic differential operator, 119 change of coordinates, 156 linear elasticity, 297 with constant coefficients, 193 Sturm-Liouville problem, 6 Sturm, C. F., 6 Successive approximations, method of, 11 Support essential, 66 of a distribution, 66 of a function, 61 Surface area of unit sphere, 247, 268 Surface potential double layer, 10, 202 traces of, 221 duality relations, 212, 213 for self-adjoint differential operator, 212, 218 jump relations for, 3, 11, 186, 203, 215 mapping properties of, 203, 205, 210 single layer, 3, 201 traces and conormal derivatives of, 218 Tangential differential operator, 147 Taylor expansion, 61 Test function, 65 Thomson, W., 6, 8 Totally bounded set, 27 Trace operator, 100 and Ho (f ), 105
351
and surface potentials, 209 for Ct- 1, 1 domains, 102 for Lipschitz domains, 102 112 from Wk (R") to W; one-sided, 141 right inverse for, 101 Traction, 296 alternative formula for, 308 as conormal derivative, 297 Transmission property, 142, 143 for surface potentials, 183, 186, 210 Transpose of a linear operator, 22 inverse of, 53 Uniformly directionally differentiable surface, 221
Uniqueness theorem for the Helmholtz equation, 288 for the Laplace equation, 260 mixed boundary conditions, 250 Neumann problem, 266 Volterra, V., 13 Volume potential, 2, 191 behaviour at the boundary, 216
Wave number, 276 Weak convergence, 42, 55 Weber, H., 9 WeierstraB, K., 10
Index of Notation
Functional Analysis A*
A/ At
dist(u, W) im A
(g, u) )A II
11A
ker A
£(X, Y) ®
(g, u)
u1v
uIW ti spec(A) Wa
av
uj - u (X0, X1)e,q X*
adjoint of A, 37 induced map on cosets modulo ker A, 18 transpose of A, 22 distance from point u to set W, 21 image (range) of linear operator A, 18 same as (g, u), 37 energy inner product for A, 44 energy norm for A, 44 kernel (null space) of linear operator A, 18 space of bounded linear operators from X to Y, 18 direct sum, 20 value of functional g E X* at u E X, 20 u is orthogonal to v, 39 u is orthogonal to the set W, 39 equivalence of norms, 17 spectrum of A, 45 subspace of X* that annihilates W c X, 23 subspace of X that annihilates V C X*, 23 uj converges weakly to u, 42 interpolation space, 318 dual space of X, 20
Theory of Distributions Ccomp°O
(Q)
Comp(c2)
space of C°O functions with compact support in 0, 61 space of Cr functions with compact support in 0, 61 353
Index of Notation
354
D(S2) D(S2)
space of infinitely differentiable functions on 0, 61 space of functions in CO°(S2) having support in K, 61 space of r times continuously differentiable functions on 0, 61 space of functions in C` (S2) having support in K, 61 C mp(Q) with sequential convergence defined, 65 space of restrictions to 7 of functions in D(IR"), 77
S
same as So, 66
CO0 (Q)
CK (S2)
C' (n) CK (S2)
S(R")
Dirac delta function(al) at x, 66 CK (S2) with sequential convergence defined, 65 space of Schwartz distributions on S2, 65 space of distributions with compact support in 92, 66 C°°(0) with sequential convergence defined, 65 finite-part extension of u, 166 principal value of u, 166 finite-part integral of xa i (x) over the half line x > 0, 160 abbreviation for (u, v)g when S2 = R", 68 same as (u, v)s2, 68, 107 space of locally integrable functions on 0, 64 dilatation operator, 158 tensor product of functions or distributions, 104 abbreviation for (u, v)n when 0 =1R", 66 integral (generalised, if necessary) of u v over S2, 58, 66, 106 Fourier transform of f.p. xt, 169 pa+'t-(x) over p > 0, 166 finite-part integral of Schwartz class of rapidly decreasing Coo functions on 1R", 72
x+
xa if x > 0, but 0 if x < 0, 159
f.p. X"
f.p. x-k-1
finite-part extension of x+, 160 (x l" if x < 0, but 0 if x > 0, 163 finite-part extension of x° , 163 finite-part extension of x-k_1 for an integer k > 0, 164
(x ± i0),
164
SX
DK (S2)
-D. (0)
E. (0) E(S2) f.p. U P.V. U H a ((a)
(u, v)
(u, On L1.1°c(0)
A ® (u, v)
(u, v)c
n () Ra0
X0
f.p. X'.
Sobolev Spaces HF HS (R") HS (I-) HS (S2)
HS (S2)'"
space of distributions in HI (RI) with support in F, 76 Sobolev space on 1R" (definition via Bessel potential), 76 Sobolev space on 17, 98 space of restrictions to S2 of distributions in HS (R" ), 77 space of HS functions on Q with values in Cm, 107
Sobolev Spaces
Ho (0) H` (Q)
closure of D(S2) in HS (S2), 77
HS (52)"'
space of HS functions on S2 with values in Cm, 107 Bessel potential of order s, 75 Slobodeckil seminorm, 74
is
HA.P,0
355
closure of D(2) in HS (I8"), 77
I
Wp(S2) Wp(S2)m
abbreviation for W2 (S2), 75 Sobolev space of orders > 0 based on LP(0), 73, 74 space of WP functions on S2 with values in C"', 107
Differential and Integral Operators AS"-1
B; B,,
BV 13*
B; B,,
DL DL Eik(u) G (x, y) G
G(x - y) Gj(x, x - y)
[u]r
P PO
(P = On
4)t R
M S
Elk
Beltrami operator on the unit sphere, 277 ft h component of generalised flux or traction, 114 conormal derivative, 114 conormal derivative from S21, 141 adjoint of B,,, 201 dual version of B;, 115 dual version of 8,,, 115 double-layer potential, 10, 202 dual version of DL, 211 strain tensor, 296 fundamental solution or parametrix, 2, 191 volume potential, 191 G(x, y) in translation-invariant case, 193 jth term in homogeneous expansion of G(x, y), 195 jump in across r, 142
jump in BA across P, 142 jump in u across I', 142 Laplace operator, 1 second-order differential operator, usually strongly elliptic, 113 principal part of P, 114 sesquilinear form arising from P for the domain £2, 114 abbreviation for (Dnt, 141 boundary integral operator, conormal derivative of DL, 218 operator arising in radiation conditions, 234 boundary integral operator, trace of SL, 7, 218 stress tensor, 296
Index of Notation
356 S§L_ L
T T
U V
single-layer potential, 3, 202 dual version of SL, 211 boundary integral operator, sum of one-sided traces of DL, 11, 218 dual version of T, 218 solution operator for the Dirichlet problem, 145 solution operator for the adjoint Dirichlet problem, 145
Other Symbols a!
IaI
8"u ya C+
CCapr u*v A1,h
dQ
u = ,'Fu u = .P*u
t y
factorial of the multi-index a, 61 order of the partial derivative determined by a, 61 partial derivative of u determined by the multi-index a, 61 monomial determined by the multi-index a, 61 complex upper half plane Im z > 0, 183 complex lower half plane Im z < 0, 183 capacity of IF, 263 convolution of u and v, 58 difference quotient in lth variable with step size h, 62 element of surface area on 1, 1, 97 Fourier transform of u, 70 inverse Fourier transform of u, 70 (common) boundary of S2 = St- and of 52+, 1, 89, 141 trace operator for S2, 100, 102
Yadjoint of y, 201 Y}
rD FN hml), h(2)
Ju j," LP (S2)
M(n, m) N(n, m)
trace operator for Sgt, 1, 141 portion of r with Dirichlet boundary condition, 128 portion of r with Neumann boundary condition, 128 spherical Hankel functions, 281 Bessel function of the first kind, 278 spherical Bessel function of the first kind, 279 Lebesgue space of pth-power-integrable functions on 0, 58 dimension of P,, (R"), 250
v
dimension of R. (R"), 250 Euclidean norm in R" or unitary norm in C'", 1 outward unit normal to 9 = S2-, 1, 97, 141
0
domain in 1R", 1
Sgt p*
interior (-) and exterior (+) domains, 1, 141 conjugate exponent to p, 58
1X I
Other Symbols
generalised Legendre polynomial of degree m for the dimension n, 255 space of homogeneous polynomials of degree m, 250
P. (n, t)
P. (R")
upper half space x > 0, 183 lower half space x < 0, 183
]R+
]R"
xm (W)
f. (S'
I)
u*
uv utkI(x; y)
T. T}(x) xa Yu ym
F(a) HM H(2)
f
space of solid spherical harmonics of degree m, 250 space of surface spherical harmonics of degree m, 252 transpose of the complex conjugate of the vector u, 107 dot product of vectors u and v, 107 kth Fr6chet derivative of u, 61 Kelvin transform of u, 259 surface area of S"- 1, 247 221
inverse point of x with respect to a sphere, 258 Bessel function of the second kind, 278 spherical Bessel function of the second kind, 279 gamma function, 169 Hankel functions, 280 special contour integral, 183
357
