Linear and Semilinear Partial Differential Equations: An Introduction 9783110269055, 9783110269048

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De Gruyter Textbook Precup · Linear and Semilinear Partial Differential Equations

Radu Precup

Linear and Semilinear Partial Differential Equations An Introduction

De Gruyter

Mathematics Subject Classification 2010: 35-01, 35J, 35K, 35L, 35B, 35C, 35D, 47H, 47J.

ISBN 978-3-11-026904-8 e-ISBN 978-3-11-026905-5 Library of Congress Cataloging-in-Publication Data A CIP catalog record for this book has been applied for at the Library of Congress. Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available in the Internet at http://dnb.dnb.de. ” 2013 Walter de Gruyter GmbH, Berlin/Boston Printing and binding: Hubert & Co. GmbH & Co. KG, Göttingen ⬁ Printed on acid-free paper Printed in Germany www.degruyter.com

Non multa, sed multum

Preface

- What would be the best textbook for a brief, rapid, first and core introduction into partial differential equations (PDEs)? - How to select and organize materials for a new textbook in PDEs, to offer not just first elements in this complex field, but also some opening toward further study, theoretical research, mathematical modeling, and applications? The first question is explicitly or just implicitly put by oneself, by any young person interested or just constrained to take contact with PDEs. The second question should be addressed to oneself by any author interested to produce a new introductory course in PDEs. Answering to any of the two above questions is a very difficult task. There are several excellent texts in PDEs, each of them with its own balance of classicmodern, elementary-advanced, and theoretic-applicability. Here are some of them: Barbu [2], Bers–John–Schechter [4], Brezis [5], Courant–Hilbert [8], DiBenedetto [10], Egorov–Shubin [11], Evans [12], Folland [13], Friedman [14], Garabedian [15], John [19], Jost [20], Logan [25], Mihlin [26], Mikhailov [27], Mizohata [28], Nirenberg [32], Petrovsky [35], Rauch [41], Schwartz [45], Shimakura [46], Sobolev [49], Tikhonov–Samarskii [53], and Vladimirov [54]. Writing this book, I had in mind the above questions. The result is a book in three parts which is intended to conform to the Latin phrase “Non multa, sed multum” (“not many, but much”, “not quantity, but quality”). In Part I, the reader finds an accessible, elementary introduction to linear PDEs, in the framework of classical analysis, without using the notion of distribution. However, I have considered useful in introducing, in this first part, the notion of generalized or weak solution of a boundary value problem. Weak solutions are sought in larger spaces than the common spaces of continuously differentiable functions, which are here introduced more naturally, by the completion with respect to the corresponding energetic norms. Part I can be used for a first standard one-semester course in PDEs for mathematics students. Part II addresses to students who follow a second partial differential equations course. Here, distributions and many more results on Sobolev spaces (some of them with laborious proofs, optional for a first reading) are presented and used for

viii

Preface

treating boundary value problems for elliptic, parabolic, and hyperbolic equations. The goal is to prepare readers for studying more advanced treatises and research papers. Finally, Part III is for an upper master course and introduces the reader to semilinear equations. Here the theory is mainly based on abstract results of nonlinear functional analysis. The author’s favorite approach is the operator method which allows us to treat unitary, following the same program, nonlinear problems for both evolution (heat, wave, time-dependent Schr´’odinger) and stationary equations. The linear theory of PDEs is involved in the construction of operator equations and in the identification of the properties of the solution operators associated to boundary value problems for nonhomogeneous equations. These properties are then combined with those of the nonlinear perturbing terms of the semilinear equations to make possible the application of fundamental principles from the theory of nonlinear operators. In Part III our intention was: first to give a strong motivation for the linear theory, and second to introduce the readers into the modern theory of nonlinear PDSs and to sensitize them to recent theoretical research. An exiting panorama over the main themes and major contributions to the theory of partial differential equations can be found in the remarkable work by H. Brezis and F. Browder, Partial differential equations in the 20th century, Publications du Laboratoire d’Analyse Numérique 17 (1998), fasc. 4, 1–85; Adv. Math. 135 (1998), 76–144. I hope that by splitting the book into three parts I have made a useful work, accessible to several reader categories: graduate, master and PhD students and young researchers interested in PDEs, nonlinear functional analysis, applied mathematics, and mathematical modeling. Cluj, November 2011

Radu Precup

Notation

Rn

Euclidian space of dimension nI inner product: .x; y/ D x  y D

n P j D1

xj yj I norm: jxj D

n P j D1

!1=2 xj2

r

Open ball of Rn of center x and radius r Open ball of the normed space X of center x and radius r Area of unit sphere of Rn ; !n D 2 n=2 = .n=2/ n P j˛j n; Operator ˛1 @˛2 ; ˛ 2 N D ˛j j˛j ˛ @x1 @x2 :::@xn n j D1   @u @u @u Gradient: ru D @x ; @x ; :::; @x n 1 2

@u @

D .ru; / ;



Laplace’s operator (Laplacian): u D

Br .x/ Br .xI X / !n D˛

Ck

.˝/

C k .˝/   C01 ˝ H01 .˝/ H 1 .˝/ Lp .˝/ L1 .˝/ S X X0 C k .Œ0; T  I X /

 2 Rn ; jj D 1

n P j D1

@2 u @xj2

Space of k times continuously differentiable functions on the open set ˝  Rn Space of functions u 2 C k .˝/ whose partial derivatives until order k extend continuously on ˝  D ¹u 2 C 1 ˝ W u D 0 on @˝º (˝  Rn bounded open) R 1=2 R .u; v/0;1 D ˝ ru  rv dxI juj0;1 D ˝ jruj2 dx     Completion of C01 ˝ ; j:j0;1 @u 2 L2 .˝/º Completion of ¹u 2 C 1 .˝/ W u; @x j  i1=2 hR  with respect to norm a .u; u/1=2 D ˝ jruj2 C u2 dx

D ¹u W ˝ ! R W u measurable and jujp integrableº R 1=p .1  p < 1/ jujLp D ˝ jujp dx Space of essentially bounded measurable functions jujL1 D inf ¹M W ju .x/j  M a.e. x 2 ˝º Schwartz space Banach space with norm j:jX Dual of X Space of functions u W Œ0; T  ! X which are k times continuously differentiable on Œ0; T 

x

Lp .0; T I X / D .˝/ E .˝/ D 0 .˝/ E 0 .˝/ S0 L1loc .˝/ H m .˝/ H0m .˝/ H m .˝/

Notation

Space of measurable functions u W Œ0; T  ! X with 1=p R T p 0 such that jg .y/  g .x0 /j  " for every y 2 †ı WD ¹y 2 @B W jy  x0 j  ıº : Then, for jx  x0 j  ı=2 and

34

Chapter 3 Elliptic Boundary Value Problems

y 2 @B n †ı ; we have jx  yj  jx0  yj  jx  x0 j  ı  ı=2 D ı=2 and also using (3.19), we obtain Z R2  jxj2 ju .x/  g .x0 /j  jg .y/  g .x0 /j jx  yjn dy !n R @B Z R2  jxj2 D jg .y/  g .x0 /j jx  yjn dy !n R †ı  Z n C jg .y/  g .x0 /j jx  yj dy @Bn†ı

 n 2 R  jxj2 ı !n Rn1 ;  " C 2M 2 !n R where M D jgjC.@B/ : This shows that if jx  x0 j is sufficiently small, and so jxj is close enough to R; then we have ju .x/  g .x0 /j  2"; which is the desired result. Now we can conclude that the Dirichlet problem for Laplacian and ball is wellposed, that is, its solution exists, is unique and depends continuously on the data. Poisson’s formula is the base of Perron’s method for proving the existence of the solution of the Dirichlet problem for Laplace’s equation on general domains (see section “Complements”).

3.8

Dirichlet’s Principle

  Let ˝  Rn be a bounded open set and let f 2 C ˝ : Consider the Dirichlet problem for Poisson’s equation with homogeneous boundary condition ² u D f in ˝ (3.20) uD0 on @˝: By a classical solution of problem (3.20), we mean a function u 2 C 2 .˝/ which pointwise satisfies equalities (3.20). As we have seen, problem (3.20) has at most one classical solution. Let us consider the following space of functions: ¯   ®   C01 ˝ D u 2 C 1 ˝ W u .x/ D 0 for all x 2 @˝ and define  energy functional associated to the Dirichlet problem,  the 1 E W C0 ˝ ! R; by  Z  1 2 E .u/ D jruj  f u dx: ˝ 2

Section 3.8 Dirichlet’s Principle

35

We have the following theorem of variational characterization of the classical solution of the Dirichlet problem: n Theorem 3.12 (Dirichlet’s principle). Let     ˝  R be a bounded open set of 1 1 2 class C and let u 2 C ˝ \ C0 ˝ : Then the following statements are equivalent:

(i) u is the classical solution of problem .3:20/: (ii) u satisfies the variational identity Z   .ru  rv  f v/ dx D 0 for all v 2 C01 ˝ :

(3.21)

˝

(iii) u is the strict absolute minimum point of the energy functional, i.e.   E .u/ < E .w/ for every w 2 C01 ˝ with w ¤ u: Proof. (i) ) (ii): Assume that u is the classical solution of problem (3.20). Then, multiply Poisson’s equation in (3.20) by any v 2 C01 ˝ ; integrate over ˝ and apply the first Green’s formula, to derive (3.21). (ii) ) (i): From (3.21), again by using Green’s first formula, one obtains the equality Z ˝

.u C f / v dx D 0

  which holds for all v 2 C01 ˝ : Since u C f is continuous on ˝; we deduce that u C f D 0 in ˝: Thus u solves (3.20). Before proving the equivalence (ii) , (iii) we shall derive the expression E .u C v/ of the energy functional for a sum of two arbitrary functions u; v 2 C01 ˝ W  Z  1 E .u C v/ D jr .u C v/j2  f .u C v/ dx 2  Z˝  1 1 2 2 D jruj C jrvj C ru  rv  f u  f v dx 2 ˝ 2 Z Z 1 .ru  rv  f v/ dx C D E .u/ C jrvj2 dx: (3.22) 2 ˝ ˝ (ii) ) (iii): Assume that u satisfies the variational identity (3.21). Let w 2 C01 ˝ : Then using (3.22) we deduce that Z 1 E .w/ D E .u C .w  u// D E .u/ C jr .w  u/j2 dx 2 ˝  E .u/ :

36

Chapter 3 Elliptic Boundary Value Problems

Hence u minimizes E: WeRclaim that the last inequality is strict in case that w ¤ u: Indeed, the integral ˝ jr .w  u/j2 dx is equal to zero if and only if r .w  u/ is identically zero in ˝; hence if w  u is a constant. But since both functions u; w are zero on @˝; this constant can not be other than zero. Then w D u; which was excluded from the beginning.   (iii) ) (ii): Assume that u minimizes E: Let v 2 C01 ˝ be any function. Then E .u/  E .u C t v/ for every t 2 given by

R:

Thus t D 0 is a minimum point of the function g W

R ! R;

g .t / D E .u C t v/ : Let us now check differentiability of g at t D 0: Indeed, using (3.22) we have g .t /  g .0/ t

D D

E .u C tv/  E .u/ t Z

t .ru  rv  f v/ dx C 2 ˝

Z ˝

jrvj2 dx

and letting t ! 0 we obtain 0

g .0/ D

Z ˝

.ru  rv  f v/ dx:

Thus the classical Fermat’s theorem applies and gives g 0 .0/ D 0; that is (3.21). Remark 3.2. If we define for E the directional derivative at u in the direction v; by E .u C t v/  E .u/ ; E 0 .uI v/ WD lim t !0C t then Z .ru  rv  f v/ dx E 0 .uI v/ D ˝

and (ii) in Theorem 3.12 expresses the property of u of being a critical point of E; in the sense that   E 0 .uI v/ D 0 for every v 2 C01 ˝ : According to Theorem 3.12, two possibilities exist to prove the existence of solutions to the Dirichlet problem: either

Section 3.9 The Generalized Solution of the Dirichlet Problem

37

(a) to show the existence of a function u satisfying the variational identity (3.21), or (b) to prove the existence of a function u which minimizes the energy functional. Unfortunately  is not possible in general. The reason is that the space    this 1 ˝ \ C0 ˝ is too poor for that. We show in Section 3.9, the way this space can be enriched (completed) in order that the existence problem can be solved. C2

3.9

The Generalized Solution of the Dirichlet Problem

  According to (3.21), we shall endow the linear space C01 ˝ (where ˝  Rn is a bounded open set) with the inner product .:; :/0;1 defined by Z    .u; v/0;1 D ru  rv dx u; v 2 C01 ˝ (3.23) ˝

and the corresponding norm j:j0;1 Z

1 2

juj0;1 D .u; u/0;1 D

jruj dx 2

˝

 12

  u 2 C01 ˝ :



(3.24)

With these notations, the variational identity (3.21) can be written as    .u; v/0;1  .f; v/L2 D 0 v 2 C01 ˝ and the energy functional is E .u/ D

1 2 juj  .f; u/L2 2 0;1

  u 2 C01 ˝ :



In both formulas we may consider more generally that f 2 L2 .˝/ : The presence of the term juj20;1 =2 in the expression of the energy functional confers to norm j:j0;1 the name of energetic norm. Using the same arguments like in the proof of Theorem 3.12, we obtain the next result: n 2 Proposition   3.2. Let ˝  R be a bounded open set, f 2 L .˝/ and 1 u 2 C0 ˝ : Then the following two statements are equivalent:

(i) u satisfies the variational identity   .u; v/0;1  .f; v/L2 D 0 for all v 2 C01 ˝ :

(3.25)

38

Chapter 3 Elliptic Boundary Value Problems

(ii) u is the strict absolute minimum point of the energy functional, i.e.   E .u/ < E .w/ for every w 2 C01 ˝ ; w ¤ u:   Compared with Theorem 3.12, this result suggests that a function u 2 C01 ˝ which satisfies identity (3.25), or, equivalently, minimizes the energy functional, be considered generalized solution   of the Dirichlet problem (3.20). Notice that 1 even in the larger space C0 ˝ ; the problem of the existence of a solution, that is of an element u with (3.25),    can not be solved. The reason is that the  property prehilbertian space  C01 ˝ ; .:; :/0;1  is not complete. Let us denote by H01 .˝/ ; .:; :/0;1 the completion of the prehilbertian space  1   C0 ˝ ; .:; :/0;1 : Recall that any metric space .X; d / admits a completion; completion of a space is an expansion that includes the limits of all Cauchy sequences, including the ones that do not converge. It is the “smallest” complete metric space which isometrically includes X: The key of completion is the equivalence relation on the set of all Cauchy sequences. One says that two Cauchy sequences .xk / and .yk / of elements from X are equivalent if d .xk ; yk / ! 0 as k ! 1:

b

For any Cauchy sequence .xk / we let .xk / be the class of all Cauchy sequences e be the set of all classes of equivalent which are equivalent to .xk / : Let X e if we identified any Cauchy sequences. We may speak of the embedding X  X c element x 2 X to .x/; the class of all Cauchy sequences which are equivalent to the constant sequence .x/ : Furthermore, metric d can be extended from eX e as follows: X  X; to X   d .xk /; .yk / WD lim d .xk ; yk / :

b b

k!1

It is not difficult to check that this definition does not depend on the choice of   e e representations of the equivalence classes, that d is a metric on X ; X; d e : For details we refer to is complete and X is isometrically embedded in X Kantorovich–Akilov [21]. The next result allows us to identify the elements of H01 .˝/ to functions from 2 L .˝/ ; and thus to regard H01 .˝/ as a subspace of L2 .˝/ : Theorem 3.13 (Poincaré’s inequality). Let ˝  Rn be a bounded open set. Then there exists a positive constant C only depending on ˝ such that Z Z u2 d x  C 2 jruj2 dx for all u 2

C01

  ˝ :

˝

˝

Section 3.9 The Generalized Solution of the Dirichlet Problem

39

Proof. Let C > 0 be chosen such that ³ ²   C ˝  x 2 Rn W x D x1 ; x 0 ; jx1 j  ; x 0 2 Rn1 : 2   Let u 2 C01 ˝ and extend u and ru to vanish outside ˝: Notice that function u2 is of class C 1 in ¹x 2 Rn W x D .x1 ; x 0 / ; jx1 j  C2 ; x 0 2 Rn1 º: For any x 0 2 Rn1 and  C2  x1  0; we have Z x1   @u2  0  u2 x1 ; x 0 D s; x ds @x1 C 2 Z x1   @u  0  D 2 u s; x 0 s; x ds; @x1 C 2 whence, by using Hölder’s inequality, we obtain !1=2 Z Z 0    0 2 0 2 u s; x ds u x1 ; x  2 C 2



0

C 2

@u  0  s; x @x1

2

!1=2

ds

:

Since the right-hand side does not depend on x1 ; integration with respect to x1 from  C2 to 0 yields Z

0

C 2

  u2 x1 ; x 0 dx1

Z  C

0

C 2

u

2



0



s; x ds

!1=2 Z



0 C 2

Now square and simplify to obtain Z Z 0   2 0 2 u x1 ; x dx1  C C 2

0

C 2



2

@u  0  s; x @x1

2

@u  0  s; x @x1

!1=2

ds

ds:

Rn1

with respect to x 0 yields   Z @u 2 u2 dx  C 2 dx: . C2 ;0/Rn1 . C2 ;0/Rn1 @x1

Integration over Z

Adding the corresponding inequality for 0  x1  Z

u dx  C 2

˝

Z  2 ˝

@u @x1

2

dx  C

C 2

we obtain

Z 2 ˝

jruj2 dx:

:

40

Chapter 3 Elliptic Boundary Value Problems

Poincaré’s inequality can be put under the form    jujL2  C juj0;1 u 2 C01 ˝ :

(3.26)

  This inequality implies that any sequence of elements from C01 ˝ which is Cauchy with respect to norm j:j0;1 ; is also Cauchy in L2 .˝/ ; and thus convergent in L2 .˝/ : Moreover, this shows that if .uk / ; .vk / are two   inequality sequences of elements from C01 ˝ which are Cauchy with respect to norm j:j0;1 and equivalent in the sense that juk  vk j0;1 ! 0 as k ! 1; then their limits in L2 .˝/ coincide. Thus there is an one-to-one correspondence between the elements of the completion H01 .˝/ (classes of equivalent Cauchy sequences) and some elements of L2 .˝/ : In this sense, H01 .˝/ can be viewed as a linear subspace of L2 .˝/ : In addition, the Poincaré inequality (3.26) extends by density to the whole space H01 .˝/ : Thus   (3.27) jujL2  C juj0;1 u 2 H01 .˝/ : It follows from this inequality that any convergent sequence from H01 .˝/ also converges in L2 .˝/ to the same limit. One says that the embedding H01 .˝/  L2 .˝/ is continuous. In fact a stronger result holds, namely that any bounded sequence in H01 .˝/ has a subsequence that converges in L2 .˝/ ; i.e. the embedding H01 .˝/  L2 .˝/ is compact. Indeed, the Arzelà-Ascoli theorem (see e.g.    DiBenedetto [10, p.17] and Precup [38]) guarantees that the embedding C01 ˝  C ˝   is compact. On the other hand, the embedding C ˝  L2 .˝/ is obviously   continuous. As a result the embedding C01 ˝  L2 .˝/ is compact too. This   property remains true by completion of C01 ˝ : The space H01 .˝/ is called a Sobolev space and represents the energetic space of the Dirichlet problem (3.20). It is a Hilbert space and the following inclusions hold and are dense:   C01 ˝  H01 .˝/  L2 .˝/ : In what follows we shall denote inner product .:; :/0;1 and norm j:j0;1 ; by .:; :/H 1 and j:jH 1 ; respectively. 0 0 It is now obvious that the energy functional can be extended to the energetic space H01 .˝/ ; as follows: E W H01 .˝/ ! R; E .u/ D As above one can prove the next result:

1 2 juj 1  .f; u/L2 : 2 H0

Section 3.9 The Generalized Solution of the Dirichlet Problem

41

Proposition 3.3. Let ˝  Rn be a bounded open set, f 2 L2 .˝/ and u 2 H01 .˝/ : Then the following statements are equivalent: (i) u satisfies the variational identity .u; v/H 1  .f; v/L2 D 0 for all v 2 H01 .˝/ : 0

(3.28)

(ii) u is the strict absolute minimum point of the energy functional, i.e. E .u/ < E .w/ for every w 2 H01 .˝/ ; w ¤ u: By analogy with Theorem 3.12, we can now define the notion of a generalized solution of the Dirichlet problem. Definition 3.3. Let ˝  Rn be a bounded open set and let f 2 L2 .˝/ : By a weak solution (or generalized solution) of Dirichlet problem (3.20), we mean a function u 2 H01 .˝/ which satisfies identity (3.28). Theorem 3.14 (existence and uniqueness of the weak solution). Let ˝  Rn be a bounded open set. Then for each f 2 L2 .˝/ ; problem .3:20/ has a unique weak solution. Proof. We have to prove the existence of a unique function u 2 H01 .˝/ satisfying identity (3.28). To this end, we apply the Riesz representation theorem (see,  1 .˝/ .:; ; :/H 1 and the e.g. Kantorovitch–Akilov [21]) to the Hilbert space H0 0 linear functional F W H01 .˝/ ! R; F .v/ D .f; v/L2 : The continuity of F follows from Poincaré’s inequality as follows: jF .v/j  jf jL2 jvjL2  C jf jL2 jvjH 1 : 0

Thus there exists a unique u 2 H01 .˝/ with F .v/ D .u; v/H 1 0

for all v 2 H01 .˝/ :

Hence .u; v/H 1 D .f; v/L2 for all v 2 H01 .˝/ ; that is u is the (unique) weak 0 solution of problem (3.20).

42

Chapter 3 Elliptic Boundary Value Problems

We finish this section by some considerations about the inner product .:; :/H 1 0   and norm j:jH 1 on H01 .˝/ : If u; v are two functions from C01 ˝ ; then 0 the numbers .u; v/H 1 and jujH 1 are expressed according to (3.23) and (3.24), 0 0 by two Riemann integrals involving the gradients of u; v; that is the first-order partial derivatives of the two functions. A natural question that arises is whether such integral representations of the inner product and norm are possible in the general case when u; v 2 H01 .˝/ : The answer is affirmative, once we manage to give a meaning to the first-order partial derivatives of any function from the Sobolev space H01 .˝/ : Towards this end, let us fix our attention on an arbitrary function u 2 H01 .˝/ : From what was shown above, it follows that  u is the 1 2 limit in L .˝/ of a sequence .uk /k1 of elements from C0 ˝ ; which is Cauchy with respect to the energetic norm. Hence Z jr .uk  um /j2 dx ! 0 as k; m ! 1: ˝

Obviously, for each j 2 ¹1; 2; :::; nº ; one has ˇ Z ˇ Z ˇ @uk @um ˇ2 ˇ ˇ  dx  jr .uk  um /j2 dx: ˇ @xj ˇ ˝ @xj ˝   k Thus the sequence @u is Cauchy in L2 .˝/ : Let uxj be its limit in @x j

k1

L2 .˝/ ; and define @u WD uxj @xj

.j D 1; 2; :::; n/ :

It is not hard to check that the definition is correct since it does not depend on the choice of the representation .uk / of u: Thus, to any function from H01 .˝/ one can associate “first-order partial derivatives”. These are functions from L2 .˝/ and make possible the representation formulas Z .u; v/H 1 D 0

˝

ru  rv dx; jujH 1 D 0

Z

jruj dx 2

˝

 12

for any u; v 2 H01 .˝/ : We note that this time, the integrals are in the sense of Lebesgue.

3.10

Abstract Fourier Series

The aim of this section is to present the basic properties of the Fourier series in an abstract Hilbert space. The theory is then applied in the next section, where

Section 3.10 Abstract Fourier Series

43

an approximation procedure for the weak solution of the Dirichlet problem is presented. This procedure is based on the expansion of any function in terms of eigenfunctions of the Laplace operator. Let H be a real Hilbert space endowed with inner product .:; :/ and norm j:j : Let .k /k1 be an orthonormal system of elements from H; that is ²   0 if k ¤ j k ; j D 1 if k D j: Definition 3.4. For any element u 2 H; the series 1 X

.u; k / k

kD1

is called the Fourier series of the element u in terms of the orthonormal system .k /k1 : Its coefficients, the real numbers .u; k / ; are called the Fourier coefficients of u with respect to .k /k1 : Proposition 3.4. (a) For each u 2 H; the sum of the square of its Fourier coefficients with respect to .k /k1 is finite and not greater than juj2 ; i.e. 1 X

.u; k /2  juj2 ;

u 2 H .Bessel’s inequality/:

(3.29)

kD1

(b) The Fourier series of any element is convergent. Proof. (a) For each m; since .k /k1 is orthonormal, we have ˇ2 ˇ m ˇ ˇ X ˇ ˇ .u; k / k ˇ 0  ˇu  ˇ ˇ kD1 ! m m X X .u; k / k ; u  .u; k / k D u kD1 m X

D juj2 

kD1

.u; k /2 :

kD1

Hence

m X kD1

.u; k /2  juj2 :

(3.30)

44

Chapter 3 Elliptic Boundary Value Problems

P1 2 This shows that the positive numerical series kD1 .u; k / is convergent and also proves (3.29). (b) From ˇ ˇ2 1 0 ˇ mCp ˇ mCp mCp X X ˇ X ˇ ˇ .u; k / k ˇˇ D @ .u; k / k ; .u; k / k A ˇ ˇkDmC1 ˇ kDmC1 kDmC1 D

mCp X

.u; k /2

kDmC1

one infers that the sequence of partial sums of the Fourier series P is Cauchy in H; 1 2 as is in R; the sequence of partial sums of the numerical series kD1 .u; k / : Since H is complete, the sequence of partial sums of the Fourier series is convergent, that is the Fourier series is convergent. Proposition 3.5 (minimum property of the Fourier coefficients). Let u 2 H: Then for every m 2 Nn ¹0º and ak 2 R; k D 1; 2; :::; m; the following inequality holds: ˇ ˇ ˇ ˇ m m ˇ ˇ ˇ ˇ X X ˇ ˇ ˇ ˇ .u; k / k ˇ : ak k ˇ  ˇu  ˇu  ˇ ˇ ˇ ˇ kD1

kD1

Proof. Square to find that the inequality to be proved becomes juj2  2

m X

ak .u; k / C

kD1

m X

ak2  juj2 

kD1

m X

.u; k /2 :

kD1

But this is equivalent to the obvious inequality m X

.ak  .u; k //2  0:

kD1

Theorem 3.15. Let .k /k1 be an orthonormal system of elements from the Hilbert space H: Then the following statements are equivalent: (i) The Fourier series with respect to .k /k1 of any element u 2 H converges to u itself. (ii) If for some u 2 H; all Fourier coefficients are zero, i.e. k D 1; 2; :::; then u D 0:

.u; k / D 0;

Section 3.11 The Eigenvalues and Eigenfunctions of the Dirichlet Problem

(iii) Parseval’s equality juj2 D

1 P

45

.u; k /2 holds for all u 2 H:

kD1

(iv) Any element u 2 H can be approximated as well we wish by finite linear combinations of elements of the system .k /k1 : Proof. (i) ) (ii): According to (i) one has 1 X

.u; k / k D u;

u 2 H:

(3.31)

kD1

Then, if for some u; all Fourier coefficients .u; k / are zero, it follows from (3.31) that u D 0: (ii) ) (i): For any u we have ! 1 X .u; k / k ; j D 0; j D 1; 2; ::: : u kD1

P Then (ii) guarantees u  1 kD1 .u; k / k D 0: (i) , (iii): Use (3.30) to find ˇ2 ˇ 1 1 ˇ ˇ X X ˇ ˇ .u; 'k / 'k ˇ D juj2  .u; 'k /2 : ˇu  ˇ ˇ kD1

kD1

(i) ) (iv): Trivial, since the partial sums of the Fourier series are finite linear combinations of elements of the system .k /k1 : (iv) ) (i): Follows from the minimum property of the Fourier coefficients. An orthonormal system .k /k1 satisfying any one of the four equivalent conditions from Theorem 3.15 is said to be complete. Our goal in Section 3.11 is to show that a complete orthonormal system can be associated to the Dirichlet problem in L2 .˝/ ; as well as in H01 .˝/ :

3.11

The Eigenvalues and Eigenfunctions of the Dirichlet Problem

In this section the weak solution of the Dirichlet problem (3.20) will be constructed using the Fourier series method and the complete orthonormal system in H01 .˝/ of the eigenfunctions of the Dirichlet problem.

46

Chapter 3 Elliptic Boundary Value Problems

Definition 3.5. A real number is an eigenvalue of the Dirichlet problem for ; if the problem ² u D u in ˝ (3.32) uD0 on @˝ has a nonzero weak solution. Such a solution is called an eigenfunction. Thus, 2 with

R

is an eigenvalue if and only if there exists u 2 H01 .˝/ n ¹0º .u; v/H 1 D .u; v/L2 0

for all v 2 H01 .˝/ :

(3.33)

Notice that the null function is a solution of (3.32) for every : Hence the interest here is to find those values of for which (3.32) also has nonzero solutions. Proposition 3.6. (a) The eigenvalues of the Dirichlet problem are positive numbers. (b) The eigenfunctions corresponding to different eigenvalues are orthogonal both in L2 .˝/ and H01 .˝/ : Proof. (a) Let be an eigenvalue and let u be an eigenfunction corresponding 2 2 to : If in (3.33), we let v D u; we obtain jujH 1 D jujL2 : Since, as an 0

eigenfunction, u ¤ 0; we deduce that > 0: (b) Let 1 ; 2 be two different eigenvalues and let u1 ; u2 be two eigenfunctions corresponding to 1 and 2 ; respectively. Then .u1 ; v/H 1 D 1 .u1 ; v/L2 ; .u2 ; v/H 1 D 2 .u2 ; v/L2 0

0

for all v 2 H01 .˝/ : Choose v D u2 in the first equality and v D u1 in the second one, to find

1 .u1 ; u2 /L2 D 2 .u1 ; u2 /L2 D .u1 ; u2 /H 1 : 0

(3.34)

Since 1 ¤ 2 ; we infer that .u1 ; u2 /L2 D 0; i.e. u1 ; u2 are orthogonal in L2 .˝/ : Finally (3.34) implies .u1 ; u2 /H 1 D 0; i.e. the orthogonality of u1 ; u2 in H01 .˝/ :

0

Theorem 3.16. The Dirichlet problem has a sequence . k /k1 of eigenvalues and correspondingly, a sequence .k /k1 of eigenfunctions, normal in L2 .˝/ ; i.e. jk jL2 D 1 for all k; and the following properties hold:

Section 3.11 The Eigenvalues and Eigenfunctions of the Dirichlet Problem

47

(a) 0 < 1  2  :::  k  kC1  ::: and k ! 1 as k ! 1I (b) .k /k1 is orthonormal and complete in L2 .˝/ I  (c)



p1 k k k1

Proof. Let

is orthonormal and complete in H01 .˝/ :

° ± 2 1

1 D inf jujH 1 W u 2 H0 .˝/ ; jujL2 D 1

(3.35)

0

and let .uk / be a minimizing sequence, i.e. 2 uk 2 H01 .˝/ ; juk jL2 D 1; juk jH 1 ! 1 as k ! 1: 0

Since the embedding H01 .˝/  L2 .˝/ is compact, we may assume, eventually passing to some subsequence, that uk ! 1 in L2 .˝/ ; where 1 is a certain function from L2 .˝/ : It is clear that j1 jL2 D 1: Furthermore, the identity   2 2 2 2 juk  um jH 1 C juk C um j 1 D 2 juk j 1 C jum j 1 H H H 0

0

0

0

implies   2 2 2 2  2 C ju ju j j juk  um jH 1 m H 1  1 juk C um jL2 ! 0 k H1 0

0

0

as k; m ! 1: Thus the sequence .uk / is Cauchy in H01 .˝/ : Consequently, 2 uk ! 1 in H01 .˝/ ; whence 1 D j1 jH 1 ; that is the infimum in (3.35) is 0 reached Observe now that 1 is an eigenvalue and 1 is a corresponding eigenfunction. Indeed, for each v 2 H01 .˝/ ; the function g .t / WD

2 j1 C t vjH 1 0

2 j1 C t vjL 2

defined in a neighborhood of the origin where j1 C t vjL2 ¤ 0; attains its minimum at t D 0: Hence its derivative at t D 0 is zero, i.e. h i g 0 .0/ D 2 .1 ; v/H 1  1 .1 ; v/L2 D 0: 0

This shows that 1 D 1 1 in the weak sense. Next, we show that 1 is the smallest eigenvalue. Indeed, if is any eigenvalue and  a corresponding eigenfunction with jjL2 D 1; and if in the identity   .; v/H 1 D .; v/L2 v 2 H01 .˝/ 0

48

Chapter 3 Elliptic Boundary Value Problems

2 we set v D ; then we obtain D jjH 1 : Taking into account the definition of 0

1 ; we may infer that 1  : The second eigenvalue can be obtained as ° ± 2 1

2 D inf jujH 1 W u 2 H0 .˝/ ; jujL2 D 1; .u; 1 /L2 D 0 : 0

Similarly, one can show that the infimum is reached by some 2 and that 2 is an eigenvalue and 2 a corresponding eigenfunction. At step k; having already selected the eigenfunctions 1 ; 2 ; :::; k1 ; define ° ±   2 1 .˝/ ; W u 2 H D 1; u;  D 0; j D 1; k  1

k D inf jujH juj 2 1 j L2 L 0 0

and choose a function k for which the infimum is reached. Clearly, . k / is a nondecreasing sequence. Assume it would be bounded. Then 2 1 from k D jk jH 1 we would have that sequence .k / is bounded in H0 .˝/ : 0

Since the embedding H01 .˝/  L2 .˝/ is compact, we may conclude that .k / has a subsequence converging in L2 .˝/ : This, however, is not possible, since 2 2 2 jk  m jL 2 D jk jL2 C jm jL2  2 .k ; m /L2 D 2:

Thus k ! 1 as k ! 1:   The fact that the system p1 k is orthonormal in H01 .˝/ follows from a k consequence of (3.33), namely s ! 1

k 1 .k ; m /L2 : D p k ; p m

m

m

k 1 H0

To prove completeness, fix any function v 2 H01 .˝/ : Then, for every integer n  2; we let wn WD v 

n1 X

.v; k /L2 k D v 

kD1

n1 X kD1

1 .v; k /H 1 k : 0

k

Since .k / is orthonormal in L2 .˝/ ;   wn ; j L2 D 0 for j D 1; 2; :::; n  1; whence, due to the expression of n ; 2 2 jwn jH 1  n jwn jL2 : 0

(3.36)

Section 3.11 The Eigenvalues and Eigenfunctions of the Dirichlet Problem

49

On the other hand, ˇ2 ˇ n1 ˇ ˇ X 1 ˇ ˇ 2 .v; k /H 1 k ˇ jvjH 1 D ˇwn C 0 ˇ ˇ 0

k

H01

kD1

D

wn C

n1 X kD1

D

2 jwn jH 1 0

C

n1 X 1 1 .v; k /H 1 k ; wn C .v; k /H 1 k 0 0

k

k kD1

n1 X kD1

! H01

1 2 2 .v; k /H 1  jwn j 1 : H0 0

k

Hence 2 jwn jL 2 

1 1 2 jwn jH jvj2 1 : 1  0

n

n H0

Now let n ! 1 and use n ! 1; to deduce that wn ! 0 in L2 .˝/ : Then (3.36) implies ! 1 1 X X 1 1 .v; k /L2 k D v; p k (3.37) vD p k



1 k k H kD1 kD1 0

with convergence guaranteed in L2 .˝/ : But since all Fourier series are convergent, it follows that this series (look atits second  expression) converges in 1 1 .˝/ too. This proves that the system p k is complete in H01 .˝/ : H0 k

Finally, the completeness of .k / in L2 .˝/ follows from (3.37) and the density of H01 .˝/ in L2 .˝/ : p Remark 3.3. From (3.35) it follows that 1= 1 is the smallest constant for which Poincaré’s inequality (3.27) holds for all u 2 H01 .˝/ : Thus 1 jujL2  p jujH 1 0

1



 u 2 H01 .˝/ :

(3.38)

The next result gives a series representation of the weak solution of the Dirichlet problem in terms of its eigenvalues and eigenfunctions. Theorem 3.17. Let ˝  Rn be a bounded open set and . k / ; .k / be the eigenvalues and eigenfunctions of the Dirichlet problem, like in Theorem 3:16: Then, for every f 2 L2 .˝/ ; the unique weak solution u of problem (3:20/ can be expanded in terms of eigenfunctions 1 X .f; k /L2 k ; uD

k kD1

(3.39)

50

Chapter 3 Elliptic Boundary Value Problems

where the series convergence holds in H01 .˝/ ; and consequently in L2 .˝/ too. Proof. Let u 2 H01 .˝/ be the weak solution of problem (3.20). Then ! 1 1 .u;  / 1 X X k H 1 1 0 u; p k k u D p k D



k 1 k k H kD1 kD1 0

1 X .f; k /L2 D k :

k kD1

3.12

The Case of Elliptic Equations in Divergence Form

In this section the variational method for the Laplacian is extended to the larger class of elliptic operators in divergence form, i.e. to operators of the form   n X @ @u aj k .x/ C a0 .x/ u Lu D  @xk @xj j;kD1

(3.40) D  div .A .x/ ru/ C a0 .x/ u; where A .x/ is the nth-order square matrix of elements aj k .x/ and the following conditions are satisfied: aj k ; a0 2 L1 .˝/ ; aj k D akj ; a0  0 and n X

aj k .x/ j k   j j2 for all 2 Rn and a.e. x 2 ˝:

(3.41)

j;kD1

The constant  > 0 is called an ellipticity constant. The key of this extension lies in the fact that for a bounded ˝; the bilinear form 0 1 Z n X @u @v @ aj k a .u; v/ D C a0 uv A dx (3.42) @x j @xk ˝ j;kD1   defines an inner product on C01 ˝ such that the corresponding norm 8 0 1 9 12 n 0 in ˝: By a classical solution of problem (3.46), when ˝ is of class C 1 and f 2 C.˝/; we mean a function u 2 C 2 .˝/ for which equalities (3.46) are pointwise satisfied. Let us attach to the Neumann problem, the energy functional  Z    1 1 2 2 a C u  f u dx: E W C 1 ˝ ! R; E .u/ D jruj 0 2 ˝ 2 In analogy with the Dirichlet problem, the following variational characterization of the classical solution holds for the Neumann problem. Theorem Let ˝  Rn be a bounded open set of class C 1 and let  3.18.  2 u 2 C ˝ : Then the following statements are equivalent: (i) u is the classical solution of problem .3:46/: (ii) u satisfies the variational identity Z   .ru  rv C a0 uv  f v/ dx D 0 for all v 2 C 1 ˝ :

(3.47)

˝

(iii) u is the strict absolute minimum point of the energy functional, i.e.   E .u/ < E .w/ for all w 2 C 1 ˝ ; w ¤ u: Proof. (i) ) (ii): Assume that u 2 C 2 .˝/ is a classical solution of problem (3.46). Then multiply by v 2 C 1 .˝/; integrate over ˝ and use the Green’s first formula, to obtain Z Z Z @u .ru  rv C a0 uv/ dx  v d D f v dx: (3.48) ˝ @˝ @ ˝ Since @u=@ D 0 on @˝; one obtains Z Z .ru  rv C a0 uv/ dx D f v dx;

(3.49)

that is (3.47). (ii) ) (i): From (3.49), one has Z Z .u C a0 u  f / v dx 

(3.50)

˝

˝

˝

@˝

@u v d D 0 @

Section 3.13 The Generalized Solution of the Neumann Problem

53

    for all v 2 C 1 ˝ : In particular, this holds for every v 2 C01 ˝ and gives Z   .u C a0 u  f / v dx D 0; v 2 C01 ˝ : ˝

Then, by a standard argument, u C a0 u  f D 0 on ˝: Turning back to (3.50), we find that Z   @u v d D 0; v 2 C 1 ˝ ; @˝ @ whence again by the standard argument, @u=@ D 0 on @˝: The equivalence (ii) , (iii) can be proved similarly as for Theorem 3.12.   This theorem suggests to endow the space C 1 ˝ with the inner product molded after the differential operator u C a0 u; Z    .ru  rv C a0 uv/ dx u; v 2 C 1 ˝ a .u; v/ D ˝

and the corresponding energetic norm 1 2

a .u; u/ D

²Z  ˝

2

jruj C a0 u

2



dx

³ 12 :

(3.51)

Let H 1 .˝/ be the completion of the space ³ ² @u 1 2 2 2 L .˝/ ; j D 1; 2; :::; n u 2 C .˝/ W u 2 L .˝/ ; @xj with respect to norm (3.51).  It is clear  that this  space contains as a subspace the completion of the space C 1 ˝ ; a .:; :/ : One can, however, prove (see 1 H 1 .˝/ coincides with the Adams [1, p. 53–56])  if ˝ is  1  that  of class C ; then completion of C ˝ ; a .:; :/ : The space H 1 .˝/ is the energetic space of the Neumann problem (3.46). It is a Hilbert space and the following embeddings hold:   C 1 ˝  H 1 .˝/  L2 .˝/ : Also, the energy functional can be extended to H 1 .˝/ ; as 1 E W H 1 .˝/ ! R; E .u/ D a .u; u/  .f; u/2 2 and, as above, one can establish the following result:

54

Chapter 3 Elliptic Boundary Value Problems

Proposition 3.7. Let ˝  Rn be a bounded open set, f 2 L2 .˝/ and u 2 H 1 .˝/ : Then the following statements are equivalent: (i) u satisfies the variational identity a .u; v/  .f; v/L2 D 0 for all v 2 H 1 .˝/ :

(3.52)

(ii) u is the strict absolute minimum point of the energy functional, i.e. E .u/ < E .w/ for all w 2 H 1 .˝/ ; w ¤ u: In analogy with Theorem 3.18, we define the notion of a generalized solution of the Neumann problem. Definition 3.6. Let ˝  Rn be a bounded open set and let f 2 L2 .˝/ : By a weak (or generalized) solution of the Neumann problem .3:46/; we mean a function u 2 H 1 .˝/ which satisfies identity .3:52/: Theorem 3.19 (existence and uniqueness). Let ˝  Rn be a bounded open set. Then for each f 2 L2 .˝/ ; problem .3:46/ has a unique weak solution. Proof. Apply the Riesz representation theorem to the Hilbert space .H 1 .˝/ ; a .:; :// and to the linear functional F W H 1 .˝/ ! R;

F .v/ D .f; v/L2

whose continuity can be established as follows: First we have jF .v/j  jf jL2 jvjL2 : On the other hand, condition a0  m > 0 yields Z Z  12  12 1 2 2 v dx p a0 v dx jvjL2 D m ˝ ˝ Z    12 1 2 2  p jrvj C a0 v dx m ˝ 1 1 D p a .v; v/ 2 : m Then

1 1 jF .v/j  p jf jL2 a .v; v/ 2 ; m which proves the continuity of F: Now, the Riesz representation theorem guarantees the existence of a unique u 2 H 1 .˝/ with F .v/ D a .u; v/ for all v 2 H 1 .˝/ : Thus a .u; v/ D .f; v/L2 for every v 2 H 1 .˝/ ; i.e. u is the (unique) weak solution of problem (3.46).

Section 3.14 Complements

55

A similar eigenvalues and eigenfunctions theory for the Neumann problem is also possible having, as a consequence, the eigenfunctions expansion formula of type (3.39) for the weak solution of the Neumann problem. Remark 3.4. It should be stressed that, compared to the Dirichlet boundary condition "u D 0 on @˝ " which is imposed directly on the elements of the D 0 on @˝ " is hidden energetic space, the Neumann boundary condition " @u @ in the equation of variation (3.52) (recall the proof of implication (ii) ) (i) in Theorem 3.18). This is the reason that the Neumann boundary condition is said to be a natural boundary condition. A more detailed presentation of the Sobolev spaces, in general, and of spaces and H 1 .˝/ ; in particular, will be done in Part II.

H01 .˝/

3.14

Complements

3.14.1

Harnack’s Inequality

Harnack’s inequality states that for any nonnegative harmonic function, the minimum and the maximum values on any compact of its domain of definition are comparable. Theorem 3.20. Let u be a nonnegative harmonic function on ˝: Then for any two concentric balls Br D Br .x0 / and BR D BR .x0 / with 0 < r < R and B R  ˝; we have   n2 n2 R R Rr RCr (3.53) u .x0 /  u .x/  u .x0 / RCr RCr Rr Rr for all x 2 Br : Proof. Applying eventually a translation we may assume x0 D 0: Based on Poisson’s formula and on the mean value theorem of harmonic functions, for each x 2 BR ; we have R2  jxj2 u .x/ D !n R

Z

R2  jxj2 u .y/ d   y n !n R @BR jx  yj Z 2 2 R  jxj 1 n2 D u .y/ d nR !n Rn1 @BR .R  jxj/ n2  R R C jxj u .0/ ; D R  jxj R  jxj

Z @BR

u .y/ dy .jyj  jxj/n

56

Chapter 3 Elliptic Boundary Value Problems

whence the second inequality in (3.53) follows. For the first inequality in (3.53), one starts from Z Z R2  jxj2 R2  jxj2 u .y/ u .y/ u .x/ D n dy  n dy !n R !n R @BR jx  yj @BR .jxj C jyj/ and one continues in a similar way. Using (3.53) we deduce that for any two points x1 ; x2 2 Br ; we have   RCr n u .x2 / : (3.54) u .x1 /  Rr This allows us to prove the following result. Corollary 3.7 (Harnack’s inequality). For every connected open set ˝  Rn and any compact K  ˝; there is a constant C only depending on ˝ and K such that uC u (3.55)

max K

min K

for every nonnegative harmonic function u 2 C 2 .˝/ : Proof. Let 0 < R < dist .K; @˝/ and r D R=2: Since ˝ is connected and K is compact, we can find a finite sequence of closed balls B R .x1 / ; B R .x2 / ; :::; B R .xm / included in ˝ such that the balls Br .x1 / ; Br .x2 / ; :::; Br .xm / cover K and are connected in the sense that Br .xk / \ Br .xkC1 / ¤ ; for k D 1; 2; :::; m  1: We claim that (3.55) holds with C D 3nm : For this, it is sufficient to show that u .x/  3nm u .y/ for every points x; y 2 K: We arrive at this inequality after successively applying (3.54) at most m times. We note that Harnack’s inequality implies the strong maximum principle of u; x0 2 ˝; then the function harmonic functions. Indeed, if u .x0 / D

sup ˝

v WD u .x0 /  u is harmonic and nonnegative on ˝ and

vD0 min K

for every

compact K  ˝ containing x0 : Then, from (3.55) it follows that v  0 on K: Thus v D 0 on K; that is u is constant on K: Since K is arbitrary we may infer that u is constant on ˝: Corollary 3.8. Any nonnegative harmonic function on a connected open set is either null, or positive on that set. Proof. Assuming the contrary we may choose a compact K such that 0 and

u D 0; min K

which is in contradiction with (3.55).

u> max K

Section 3.14 Complements

3.14.2

57

Hopf’s Maximum Principle

The generalization of the maximum principle to the case of general elliptic operators is due to E. Hopf. We shall consider the operator Lu D

n X j;kD1

X @2 u @u aj k .x/ C bj .x/ C c .x/ u @xj @xk @xj n

j D1

D M u C c .x/ u; where we assume that aj k ; bj ; c 2 C.˝/; aj k D akj ; c .x/  0 on ˝; and

n P

aj k .x/ j k   j j2

j;kD1

for all x 2 ˝ and 2 Rn

(condition of uniform ellipticity) for some  > 0: Theorem 3.21. Let ˝  Rn be a connected open set and u 2 C 2 .˝/ \ C 1 .˝/: Assume that M u  0 on ˝ and there exists x0 2 @˝ such that u .x0 /  u .x/ for all x 2 ˝: In addition assume that ˝ satisfies the interior ball condition at x0 ; that is, there exists an open ball B  ˝ with x0 2 @B: Then either u is constant on ˝; or @u .x0 / > 0: @ Proof. First we shall prove a weak version of the theorem, under the additional assumption u .x/ < u .x0 / for all x 2 ˝: We chose a concentric ball B0 with B 0  B: Assume without loss of generality that the origin is their center (otherwise make a translation). Denote by R the radius of B: We shall write for simplicity r D jxj : Consider the function 2

2

v .x/ D e˛r  e˛R ; where ˛ > 0: We have v D 0 on @B; v > 0

in B and

@v < 0 on @B: @

58

Chapter 3 Elliptic Boundary Value Problems

To find M v we compute the first- and second-order derivatives @v 2 D 2˛xj e˛r ; @xj

@2 v 2 2 D 4˛ 2 xj xk e˛r  2ıj k ˛ e˛r : @xj @xk

Here ıj k is the symbol of Kronecker. Then 1 0 n X   2 2 aj k .x/ xj xk  C ˛ A e˛r  4˛ 2 r 2  C ˛ e˛r : M v  @4˛ 2 j;kD1

It follows that for a sufficiently large ˛; M v > 0 on the annular region B n B0 : Let us fix such a number ˛: Then, for every " > 0; one has M ."v C u/ > 0 on B nB0 : This implies (see Evans [12, p. 328] for details) that the function "v C u attains its maximum over the compact B n B0 ; only on boundary @.B n B0 /: Since u < u .x0 / on @B0 ; we may chose " > 0 such that "v C u < u .x0 / on @B0 : Hence "v C u attains its maximum on @B: Since v D 0 on @B; it follows that the maximum of "v C u on the compact B n B0 is reached at x0 : Then @ ."v C u/ .x0 /  0: @ .x0 / < 0; we deduce the desired result Since @v @ the proof after the next theorem.

@u @

.x0 / > 0: We shall complete

The generalization of Theorem 3.4 is the following result. Theorem 3.22 (Hopf’s strong maximum principle). Let ˝  Rn be a connected open set and u 2 C 2 .˝/ : Assume that M u .x/  0 for all x 2 ˝: If there is a point x0 2 ˝ with u .x0 / D sup u; then u is constant on ˝: ˝

Proof. Let m D

sup u ˝

and M D ¹x 2 ˝ W u .x/ D mº : It is clear that M is

nonempty (x0 2 M) and closed in ˝: Hence it suffices to show that M is open. Let x  2 M and consider a ball B R .x  /  ˝: It suffices to prove that BR=2 .x  /  M: For this, let x 2 BR=2 .x  / be arbitrary. Then   R ı D dist .x; M/  dist x; x  < : 2 We shall prove that ı D 0: Assume the contrary, i.e. ı > 0 and look at the restriction of u to B ı .x/ : We have u 2 C 2 .B ı .x// and from the definition of

Section 3.14 Complements

59

ı; there is x1 2 @Bı .x/ \ M such that u .x1 / D m and u < m on Bı .x/ : Then, from the weak version of Theorem 3.21 already proved, .@u=@/ .x1 / > 0: On the other hand x1 belongs to ˝ and is a maximum point of u; hence ru .x1 / D 0: The contradiction we arrived proves that ı D 0: Complete proof of Theorem 3.21. Under the assumption of Theorem 3.21, if u is not constant, then according to Theorem 3.22, u .x/ < sup u D u .x0 / ˝

for all x 2 ˝: Thus the additional assumption of the weak version of Theorem 3.21 is satisfied and the proof is complete. Corollary 3.9 (Hopf’s weak maximum principle). Let ˝  Rn be a bounded open set and u 2 C 2 .˝/ \ C.˝/: If Lu  0 on ˝ and u  0 on @˝; then u  0 on ˝: Proof. Assume the contrary. Then the open set ˝ 0 WD ¹x 2 ˝ W u .x/ > 0º is nonempty. Also, since c .x/  0; we have M u D Lu  c .x/ u  0 on ˝ 0 : In addition u D 0 on @˝ 0 : Thus on each connected component of ˝ 0 ; function u is nonconstant and attains its supremum at interior points, which contradicts Theorem 3.22.

3.14.3

The Newtonian Potential

A special function appears in (3.3) and (3.7) being defined in terms of the fundamental solution of Laplace’s equation: Z N .x  y/ f .y/ dy: V .x/ D ˝

It is called Newtonian potential of density f: The basic property of the Newtonian potential is of being a solution of Poisson’s equation V D f: Theorem 3.23. Let ˝  Rn be a bounded open set. (a) If f is a bounded measurable function on ˝; then V 2 C 1 .Rn / and for every x 2 Rn ; we have Z @N @V .x/ D .x  y/ f .y/ dy; j D 1; 2; :::; n: (3.56) @xj @x j ˝

60

Chapter 3 Elliptic Boundary Value Problems

(b) If f 2 C 1 .˝/; then V 2 C 2 .˝/ and V D f

in ˝:

(3.57)

Proof. (a) From Remark 3.1, it follows that the function: Z @N .x  y/ f .y/ dy Vj .x/ D ˝ @xj is well defined on Rn : To prove that Vj D @V =@xj we consider a function  2 C 1 .R/ with the following properties: 0    1; 0  0  2;  .t/ D 0 for t  1 and  .t / D 1 for t  2: For each " > 0; define Z V" .x/ WD

˝

N .x  y/ " f .y/ dy;

where " D  .jx  yj ="/ : One can immediately see that V" 2 C 1 .Rn / and Z @ @V" .x/ D Vj .x/  ¹.1  " / N .x  y/º f .y/ dy: @xj jxyj2" @xj It follows that

ˇ ˇ ˇ ˇ ˇVj .x/  @V" .x/ˇ ˇ ˇ @xj ˇ ˇ  Z ˇ @N ˇ 2 ˇ ˇ  sup jf j ˇ @x .x  y/ˇ C " jN .x  y/j dy j jxyj2" ² 2n" for n  3  sup jf j n2 4" .1 C jln 2"j/ for n D 2:

Consequently, as " ! 0; V" and @V" =@xj converge to V and Vj ; respectively, uniformly on Rn : Thus V 2 C 1 .Rn / and Vj D @V =@xj : (b) Let x0 2 ˝ be an arbitrary point and B D Br .x0 / with B  ˝: We have Z V .x/ D Vr .x/ C N .x  y/ f .y/ dy; ˝r

where ˝r D ˝ n B and Vr .x/ D

Z B

N .x  y/ f .y/ dy:

Section 3.14 Complements

61

Clearly, function V .x/  Vr .x/ is harmonic in B: We wish to show that Vr 2 C 2 .B/ and Vr .x0 / ! f .x0 / as r ! 0: To this end, observe that from the first part of this theorem, Vr 2 C 1 .Rn / and Z Z rx N .x  y/ f .y/ dy D  ry N .x  y/ f .y/ dy: rVr .x/ D B

B

Integration by parts yields Z Z 1 rVr .x/ D  N .x  y/ f .y/ .y  x0 / dy C N .x  y/ rf .y/ dy; r @B B where we used  .y/ D .y  x0 / =r: The first integral clearly defines a function from C 1 .BI Rn /: Based on the first part, the second integral also belongs to C 1 .BI Rn /: It follows that Vr 2 C 2 .B/: Furthermore, again using (a) we find for x 2 B W Z 1 divx .N .x  y/ f .y/ .y  x0 // dy Vr .x/ D  r @B Z C divx .N .x  y/ rf .y// dy B Z 1 .rx N .x  y/ ; y  x0 / f .y/ dy D  r @B Z .rx N .x  y/ ; rf .y// dy: C B

Since 1 r

Z

.rx N .x0  y/ ; y  x0 / f .y/ dy Z 1 D  f .y/ d ! f .x0 / !n r n1 @B and

@B

ˇ ˇZ Z ˇ ˇ ˇ .rx N .x  y/ ; rf .y// dy ˇ  C jx0  yj1n dy  C r ! 0 ˇ ˇ B

B

as r ! 0; we have Vr .x0 / ! f .x0 / as r ! 0: Hence (3.57) holds. The name of potential comes from physics, where one says that a vector (force) field F is derived from a potential if there exists a scalar function V with rV D F:

62

Chapter 3 Elliptic Boundary Value Problems

For example, according to Coulomb’s law, an electric charge q placed at a point y 2 R3 acts on the unit charge placed at point x with the force q .x  y/=jx  yj3 : If now we consider a distribution of electric charges in the region ˝; of density f .y/ ; then by summation, the total force acting on the unit charge placed at x will be Z xy f .y/ dy: FD 3 ˝ jx  yj We observe that the potential of this vector field is the function Z 1 V .x/ D f .y/ dy: ˝ jx  yj Theorem 3.23 makes possible to reduce the study of boundary value problems for Poisson’s equation, to that of the corresponding problems for Laplace’s equation. Indeed, if u is a solution of Poisson’s equation u D f

in ˝;

where f 2 C 1 .˝/ \ C.˝/; and V is the Newtonian potential of density f; then the function v D u  V solves Laplace’s equation v D 0

in ˝:

In addition, it is important for the Dirichlet, Neumann and Robin boundary value problems, that V 2 C 1 .Rn / and so we may speak about the values of V and @V =@ on the boundary of ˝: In Section 3.14.4 we shall describe the method of subharmonic functions, or Perron’s method, for proving the existence of the solution of the Dirichlet problem for Laplace’s equation.

3.14.4

Perron’s Method

We noted that the harmonic functions on an interval .a; b/ are the linear functions on .a; b/ : We shall call a function subharmonic in .a; b/ ; if it is continuous and convex on .a; b/ : Recall that a continuous function u on .a; b/ is convex if Jensen’s inequality u .x  r/ C u .x C r/ u .x/  2 is satisfied for every subinterval Œx  r; x C r  .a; b/ : A function u is superharmonic in .a; b/ if u is subharmonic in .a; b/ ; or equivalently, if it is continuous and concave on .a; b/ : These notions can be extended for functions of several variables as follows.

Section 3.14 Complements

63

Definition 3.7. A function u 2 C.˝/ is said to be subharmonic in ˝ if Z 1 u .y/ d for every B r .x/  ˝: u .x/  !n r n1 @Br .x/ The function u 2 C .˝/ is superharmonic in ˝ if u is subharmonic in ˝: From the mean value theorem of harmonic functions, it follows that the sum of a subharmonic function with a harmonic one yields a subharmonic function. We now present some of the basic properties of the subharmonic functions. Theorem 3.24. Let u 2 C 2 .˝/ : The function u is sub(super)harmonic in ˝ if and only if u  ./ 0 on ˝: Proof. If u  0 on ˝; then applying (3.11) to the ball B D Br .x/ and taking into account the expression of Green’s function for sphere and its positivity, we obtain Z Z @G .x; y/ dy u .x/ D  u .y/ G .x; y/ dy  u .y/ @y B @B Z @G .x; y/ dy   u .y/ @y @B Z @G .0; y  x/ dy D  u .y/ @y @B Z 1 D u .y/ d; (3.58) !n r n1 @B which shows that u is subharmonic in ˝: Conversely, inequality (3.58) implies Z Z u .y/ G .x; y/ dy D u .y/ G .0; y  x/ dy  0: B

Then

B

Z u .x/ B

G .0; y  x/ dy 

Z B

G .0; y  x/ Œu .x/  u .y/ dy;

whence the conclusion u .x/  0 follows if we apply the second mean value formula for the integral from the right-hand side, we divide by the integral from the left and we pass to the limit r ! 0: Theorem 3.25 (maximum principle of subharmonic functions). Let ˝  Rn be a connected open set and u 2 C .˝/ be a subharmonic function in ˝: If there exists x0 2 ˝ such that u.x0 / D sup u; then u is constant on ˝: ˝

64

Chapter 3 Elliptic Boundary Value Problems

Proof. Use the definition of subharmonic functions and the reasoning from the proof of Theorem 3.4. An interesting application of the maximum principle is the following theorem of characterization of the harmonic functions. Theorem 3.26. A function u 2 C .˝/ is harmonic if and only if it is both subharmonic and superharmonic. Proof. The necessity follows from the mean value theorem of harmonic functions. For sufficiency, let B D Br .x/ a ball with B  ˝: Let v 2 C 2 .B/ \ C.B/ be the unique solution guaranteed by Theorem 3.11, of the Dirichlet problem ² v D 0 in B vDu on @B: Then the functions u  v and v  u are subharmonic in B and zero on @B: From the maximum principle of subharmonic functions we have u  v on B: Hence u is harmonic in B and, since the ball is arbitrary, u is harmonic in ˝: Corollary 3.10. Let .uk / be a sequence of harmonic functions in ˝ which converge to a function u uniformly on each compact included in ˝: Then u is harmonic in ˝: Proof. Clearly u 2 C .˝/ : Furthermore, let B D Br .x/ be any ball with B  ˝: The functions uk being harmonic satisfy Z 1 uk .y/ d ; uk .x/ D !n r n1 @B whence, using the uniform convergence on B of uk to u; we find Z 1 u .y/ d: u .x/ D !n r n1 @B This shows that u is both subharmonic and superharmonic in ˝; thus harmonic in view of Theorem 3.26. Theorem 3.27. Let u be a subharmonic function in ˝ and B an open ball with B  ˝: Let u be the harmonic function in B satisfying u D u on @B: Then the function ² u .x/ ; x 2 B (3.59) U .x/ D u .x/ ; x 2 ˝ n B; called the harmonic modification of u on B; is also subharmonic in ˝ and u  U:

Section 3.14 Complements

65

Proof. Let B 0 be an arbitrary ball with B 0  ˝ and let h be an harmonic function in B 0 with U  h on @B 0 : First we remark that u  u is subharmonic in B and null on @B: Consequently, from the maximum principle of subharmonic functions, u  u  0 on B; whence u  U on the whole ˝: It follows that the subharmonic function in B 0 ; u  h satisfies u  h  0 on @B 0 : The maximum principle implies u  h  0 on B 0 : Since u D U on ˝ n B; we then have U  h on B 0 n B: On the other hand, U  h is harmonic in B \ B 0 and U  h  0 on @ .B \ B 0 / ; whence U  h on B \ B 0 : Thus U  h on B 0 and, due to the arbitrary choice of B 0 ; U is subharmonic in ˝: Also observe that if two functions u1 ; u2 are subharmonic in ˝; then the function u WD ¹u1 ; u2 º is also subharmonic in ˝: In what follows ˝  Rn is a connected bounded open set and g 2 C .@˝/ : A subharmonic function v 2 C.˝/ is said to be a subfunction of g; if v  g on @˝: Similarly, a superharmonic function v is a superfunction of g; if v  g on @˝: From the maximum principle it follows that any subfunction of g is less or equal than any superfunction of g: In particular, a constant function k is a g .k  g/: Let subfunction (superfunction) of g; if and only if k 

max

max @˝

min @˝

Sg be the set of all subfunctions of g: Clearly if u 2 C 2 .˝/ \ C.˝/ is the solution of the problem ² u D 0 in ˝ uDg on @˝;

(3.60)

then v  u for every subfunction v of g: This suggests to look for the solution u under the form v .x/ ; x 2 ˝: (3.61) u .x/ D

sup

v2Sg

This is the basic idea of Perron’s method. Lemma 3.2. The function u given by .3:61/ is harmonic in ˝: Proof. For each v 2 Sg ; the function v 

max g @˝

is subharmonic in ˝ and

nonpositive on @˝: From the maximum principle, we then have v 

g max @˝

on

˝; which shows that function (3.61) is well defined. Let x0 2 ˝ be an arbitrary point and .vk / be a sequence of functions from Sg such that²vk .x0 / !³u .x0 / as k ! 1: Replacing eventually vk by the function

max

vk ;

min g @˝

belonging to Sg ; we may assume that .vk / is bounded in C.˝/:

; also

66

Chapter 3 Elliptic Boundary Value Problems

Let B D BR .x0 / be a ball with B  ˝ and let Vk be the harmonic modification of vk on B; conforming to (3.59). From Poisson’s formula, Z vk .y/ R2  jx  x0 j2 x 2 B: Vk .x/ D n dy ; !n R @B jx  x0  yj It follows that on each ball B r .x0 / included in B; the sequence of continuous functions .Vk / is bounded and equicontinuous. Then, according to the Arzelà– Ascoli theorem, it contains an uniformly convergent subsequence. Choosing r D R  1=k and using a diagonal procedure we can find a subsequence of .Vk / ; for simplicity also denoted by .Vk / ; which converges uniformly on every compact subset of B to some function v 2 C .B/ : From Corollary 3.10, v is harmonic in B: In addition, since Vk 2 Sg ; one has v  u on B: Also, from vk  Vk we deduce u .x0 /  v .x0 / : Hence u .x0 / D v .x0 / : We claim more, namely that v D u on B: To this end, assume that v .x1 / < u .x1 / for some point v .x1 / : Denoting v 2 Sg with v .x1 / < b x1 2 B: Then there exists a function b wk WD v ; Vk º and considering its harmonic modification Wk (on B), by ¹b the above procedure we can find a subsequence of .Wk / which converges to a harmonic function w satisfying v  w  u on B and v .x0 / D w .x0 / D u .x0 / : Thus, the function w  v is harmonic in B and attains its maximum at the interior point x0 2 B: Now the maximum principle gives w D v on B; and from Wk .x1 /  wk .x1 /  b v .x1 / ; we deduce the contradiction w .x1 / D v .x1 / > v .x1 / : Therefore, u is harmonic in B and since B was v .x1 /  b arbitrary chosen in ˝; u is harmonic in ˝:

max

The next step is to prove that the function u given by (3.61) is continuous on ˝ and satisfies u D g on @˝; provided that @˝ has suitable geometric properties. These are expressed by means of the notion of barrier. Let x0 2 @˝: We say that a function w 2 C.˝/ is a barrier at x0 (with respect to ˝/; if w is superharmonic in ˝; w > 0 in ˝ n ¹x0 º and w .x0 / D 0: Notice that the property of a point x0 of admitting a barrier is a local property. Indeed, if x0 admits the barrier w 2 C.˝ \ B r .x0 // with respect to ˝ \ Br .x0 /; then the function ² ¹w .x/ ; mº for x 2 ˝ \ Br=2 .x0 / w e .x/ D m for x 2 ˝ n Br=2 .x0 / ; ¯ ® where m D w .x/ W x 2 ˝; r=2  jx  x0 j  r ; is a barrier at x0 with respect to ˝: A point x0 2 @˝ is said to be regular if there exists a barrier at x0 :

min

min

Section 3.14 Complements

67

Lemma 3.3. Let u be the harmonic function given by .3:61/: If the point x0 2 @˝ is regular, then u .x/ ! g .x0 / as x ! x0 : Proof. Let " > 0 and M D

max jgj : @˝

Let w be a barrier at x0 : Using the

continuity at x0 of g and w and the barrier’s properties, we deduce that there are positive numbers ı and k with jg .x/  g .x0 /j < " if x 2 @˝; jx  x0 j < ıI kw .x/  2M if x 2 ˝; jx  x0 j  ı: The function g .x0 / C " C kw is a superfunction of g; while g .x0 /  "  kw is a subfunction of g: Using the definition of u and the fact that each subfunction is dominated by any superfunction, we deduce g .x0 /  "  kw .x/  u .x/  g .x0 / C " C kw .x/ ;

x 2 ˝:

Then ju .x/  g .x0 /j  " C kw .x/ ;

x 2 ˝:

Now the conclusion follows since w .x/ ! 0 as x ! x0 : An immediate consequence of Lemmas 3.2 and 3.3 is the following theorem. Theorem 3.28. The necessary and sufficient condition for problem .3:60/ to have a solution in C 2 .˝/ \ C.˝/ for every g 2 C .@˝/ ; is that all the points of the boundary @˝ be regular. Proof. The sufficiency is immediate from Lemmas 3.2 and 3.3, the solution of problem (3.60) being the function u given by (3.61). For necessity, let x0 2 @˝ be any point of the boundary. Then the solution of problem (3.60) which corresponds to the function g .x/ D jx  x0 j is a barrier at x0 ; as it can be immediately seen. Hence x0 is regular. Next we present some sufficient conditions for that all the points of the boundary be regular. (a) The exterior segment condition (for the case n D 2/: One says that ˝  Rn satisfies the exterior segment condition at x0 2 @˝; if there is a point x1 2 Rn n ¹x0 º with

x1 C .1  / x0 … ˝

for every 2 Œ0; 1 :

68

Chapter 3 Elliptic Boundary Value Problems

When n D 2; this condition is sufficient for the point x0 to be regular. Indeed, using the polar coordinate system . ; '/ ; with pole x0 and polar axis of direction x0  x1 ; one can verify that the function ´  2 ln  2 for x ¤ x0 ln  C ' w .x/ D 0 for x D x0 is a barrier at x0 with respect to ˝ \ Br0 .x0 / ; where r0 < 1:

(b) The exterior cone condition. One says that ˝  Rn satisfies the exterior cone condition at x0 2 @˝; if there exits an open convex cone C with vertex at the origin and a number r0 > 0 such that ˝ \ .x0 C C / \ Br0 .x0 / D ;: This condition is equivalent to the existence of a point x1 2 two numbers 0 < 0 <  and r0 > 0 such that

Rn n ¹x0 º

and

˝ \ Br0 .x0 /  ¹x W angle .x0  x1 ; x  x0 / < 0 º : One can prove (see Dautray–Lions [9, vol. 2]) for n  3; that if ˝ satisfies the exterior cone condition at x0 ; then x0 is regular and admits a barrier of the form w .x/ D r  . / ; where r D jx  x0 j and  D angle .x0  x1 ; x  x0 / : Obviously, the exterior cone condition implies the exterior segment condition. For example, if ˝ is convex, then it satisfies the exterior cone condition. (3) The C 1 smoothness condition. If ˝ is of class C 1 ; then the exterior cone condition is satisfied at every point of the boundary. In conclusion, we have the following existence result for the Dirichlet problem, which finishes the discussion about the well-posedness of this boundary value problem. Theorem 3.29. Let ˝  Rn be a connected bounded open set with boundary consisting only of regular points. Then for every f 2 C 1 .˝/ \ C.˝/ and g 2 C.@˝/; there exists a unique solution u 2 C 2 .˝/ \ C.˝/ of the Dirichlet problem ² u D f in ˝ uDg on @˝:

3.14.5

Layer Potentials

In (3.3) and (3.7) two types of functions appear being given in terms of the fundamental solution of Laplace’s equation and of the surface integral. They are:

Section 3.14 Complements

69

- Single-layer potential of density ˛; Z W0 .x/ D N .x  y/ ˛ .y/ dy I @˝

- Double-layer potential of density ˇ; Z @N .x  y/ ˇ .y/ dy : W .x/ D  @˝ @y Layer potentials have an obvious physical significance: the single-layer potential is the potential of the electric field induced by a charges distribution on the surface @˝ with density ˛I the double-layer potential is the potential of dipoles distributed over the surface @˝ with density ˇ and oriented in the direction of the exterior normal : In what follows, we assume that ˝  Rn is a bounded open set of class C 2 and ˛; ˇ 2 C.@˝/: We mention without proof the most important properties of the layer potentials. For details we refer the reader to Barbu [2], DiBenedetto [10], Folland [13], Mihlin [26] and Vladimirov [54]. Theorem 3.30. (a) The function W is harmonic in continuous.

Rn n @˝

and its restriction to @˝ is

(b) For each point x 2 @˝; one has W .z/ ! W .x/  12 ˇ.x/

as z ! x; z 2 ˝I

W .z/ ! W .x/ C 12 ˇ.x/ as z ! x; z 2 Rn n ˝:

(3.62)

Formula (3.62) shows that the double-layer potential is discontinuous at any point x of the boundary for which ˇ.x/ ¤ 0: In this case, we have the jump formula W C .x/  W  .x/ D ˇ .x/ : Here W C ; W  are the limits of W from the exterior and interior of ˝; respectively. Before stating the main result on the single-layer potential, we introduce the following notations for a function u 2 C .Rn / \ C 1 .Rn n @˝/ and a point x 2 @˝ W @u .x/ W D lim  .x/  ru.x C t  .x//; t !0 @  @u .x/ W D lim  .x/  ru.x C t  .x//: @ C t !0C

70

Chapter 3 Elliptic Boundary Value Problems

Theorem 3.31. (a) The function W0 is harmonic in

Rn n @˝

and is continuous on

Rn :

(b) For each point x 2 @˝; one has Z 1 @N @W0 .x/ D .x  y/ ˛ .y/ dy  ˛ .x/ @  @ 2 Z@˝ x @W0 1 @N .x/ D .x  y/ ˛ .y/ dy C ˛ .x/ : C @ 2 @˝ @x Thus, for the single-layer potential, one has the jump formula @W0 @W0 .x/   .x/ D ˛ .x/ C @ @ for every x 2 @˝: Together with the Newtonian potential, the layer potentials can be used for the proof of the existence of solutions to elliptic boundary value problems, as sketched in the next section.

3.14.6

Fredholm’s Method of Integral Equations

Assume that ˝  Rn is a bounded open set of class C 2 ; so that all the above properties of the layer potentials hold. Theorem 3.30 suggests to look for the solutions of the interior and exterior Dirichlet problems (Di ) and (De ) under the form of a double-layer potential, while Theorem 3.31 suggests for the solutions of the interior and exterior Neumann problems (Ni ) and (Ne ), the form of a single-layer potential (we point out that all these problems are considered with respect to Laplace’s equation). More exactly, we seek solutions of (Di ) and (De ) in the form Z @N .x  y/ ˇ .y/ dy u .x/ D  @˝ @y and solutions of (Ni ) and (Ne ) in the form Z N .x  y/ ˛ .y/ dy : u .x/ D @˝

In view of Theorems 3.30 and 3.31, the four boundary value problems are equivalent to the following linear integral equations: Z 1 @N .x  y/ ˇ .y/ dy D g .x/ ; x 2 @˝I (Di )  ˇ .x/  2 @˝ @y

Section 3.15 Problems

71

(De )

1 ˇ .x/  2

(Ni )

1  ˛ .x/ C 2

(Ne )

1 ˛ .x/ C 2

Z @˝

@N .x  y/ ˇ .y/ dy D g .x/ ; @y

x 2 @˝I

@˝

@N .x  y/ ˛ .y/ dy D g .x/ ; @x

x 2 @˝I

@˝

@N .x  y/ ˛ .y/ dy D g .x/ ; @x

x 2 @˝:

Z Z

Thus the problem of the existence of the solutions of the four boundary value problems is reduced to the existence of solutions ˇ and ˛; respectively, of the above Fredholm integral equations of the second kind. The study of these equations is based on the Fredholm alternative for completely continuous linear operators in Banach spaces. Details can be found in Barbu [2], Dautray–Lions [9], Folland [13], Mihlin [26], and Vladimirov [54].

3.15

Problems

Differential operators 1.

Equation (1.6) is said to be in the canonical form if aj k 2 ¹1; 0; 1º for every j and k: If akk D 1 (or 1) for all k 2 ¹1; 2; :::; nº and aj k D 0 for j ¤ k; the equation is elliptic; if akk are all equal to 1 (or all equal to 1) except at least one equal to 0 and aj k D 0 for j ¤ k; then the equation is parabolic; if there are at least two opposite coefficients, the equation is hyperbolic. Details can be found in Vladimirov [54]: Reduce the following equations to canonical form and precise their type: (a) uxx 2uxy 3uyy Cux D 0 (b) 10uxx 6uxy Cuyy Cux Cuy D 0 (c) uxx  4uxy C 4uyy  2ux D 0 (d) uxx  xuyy D 0:

2.

Find all the solutions of the equations: (a) a2 uxx  uyy D 0 (b) 3uxx C 2uxy  uyy D 0: Hint. First reduce to canonical form.

3.

Find the expression of Laplacian in cylindrical and spherical coordinates in R3 ; and in polar coordinates in R2 : Hint. Changing of coordinates yields   @ @u C u D 1 @ @

1 @2 u 2 @' 2

C

@2 u @z 2

(in cylindrical coordinates ; ' and z)

72

Chapter 3 Elliptic Boundary Value Problems

u D u D

1 @ r 2 @r

1 @  @



 @u C @

1 @2 u 2 @' 2

(in polar coordinates and ')     1 @u @ r 2 @u C sin  C 2 @r @ r sin  @

@2 u 1 r 2 sin2  @' 2

(in spherical coordinates r;  and ').

4.

Find all radial solutions, i.e. of the form u .x/ D v .jxj/ ; of Laplace’s equation in Rn n ¹0º : Hint. Solve the ordinary differential equation v 00 C

n1 0 v D0 r

for r > 0;

where v D v .r/ ; r D jxj : Obtain the solutions u .x/ D c1 jxj2n C c2 for n  3; u .x/ D c1 ln jxj C c2 for n D 2:

Boundary value problems 5.

Solve the Dirichlet problem on a rectangular domain 8 0 < x < a; 0 < y < b < u .x; y/ D 0; 3 x u .x; 0/ D sin a ; u .x; b/ D 0; 0  x  a : u .0; y/ D u .a; y/ D 0; 0 < y < b: Hint. Use the method of separation of variables. Look for nonidentically zero harmonic solutions with separable variables, i.e. in the form u .x; y/ D X .x/ Y .y/ with X .0/ D X .a/ D Y .b/ D 0: Substituting into the equations yields the conditions X 00 C X 00

Y  Y

D 0; 0  x  aI X .0/ D X .a/ D 0I D 0; 0  y  bI Y .b/ D 0:

Find that nontrivial solutions exist only for

D k WD .k=a/2 ; namely Xk .x/ D sin .kx=a/ ; Yk .y/ D sinh .k .b  y/ =a/ ; k 2 N n ¹0º :

Section 3.15 Problems

73

Then we look for the solution of the problem in the form of a linear combination of functions Xk .x/ Yk .y/ : Since sin

3

x 3 x 1 3x D sin  sin ; a 4 a 4 a

we obtain u .x; y/ D

6.

.by/ 3.by/ 3 1 x sinh a 3x sinh a  : sin sin 3b 4 a 4 a sinh b sinh a a

Solve 8 0 < x < ; 0 < y <  < u D 0; u .x; 0/ D A sin x; u .x; / D 0; 0  x   : u .0; y/ D B sin 3y; u .; y/ D 0; 0 < y < : Hint. The solution is sought in the form u D u1 C u2 ; where u1 and u2 are harmonic in .0; /  .0; / and each of them satisfies only one nonhomogeneous boundary condition.

7.

Find the solutions (in polar coordinates) of the following problems: (a) u D 0 for < R; u D cos2 ' for D RI (b) u D 0 for < R; @u=@ D A cos ' for D RI (c) u D 0 for > R; u D cos 3' for D R; juj  C I (d) u D 0 D R2 :

for R1 < < R2 ; u D A for D R1 ; u D

cos '

for

Hint. Look for nontrivial solutions of Laplace’s equation in the form u . ; '/ D Z . / ˚ .'/ : Using the Laplacian in polar coordinates yields ˚ 00 .'/ C ˚ .'/ D 0; 2 Z 00 . / C Z 0 . /  Z . / D 0: Since ˚ .' C 2/ D ˚ .'/ ; the first equation has nontrivial solutions only for D k WD k 2 ; namely ˚k .'/ D ak cos k' C bk sin k'; k 2 N: Furthermore, find Zk . / D ck k C dk k for k  1 and Z0 . / D c0 ln C d0 for k D 0:

74

Chapter 3 Elliptic Boundary Value Problems

The solution of the boundary value problems inside the sphere ˝ D ¹x 2 R2 W D jxj < Rº is sought in the form 1  k X a0 .ak cos k' C bk sin k'/ uD ; C 2 R kD1

¯ ® outside the sphere, i.e. on ˝ D x 2 R2 W > R ; in the form  k 1 X R a0 .ak cos k' C bk sin k'/ C ; uD 2 kD1

¯ ® and inside the spherical layer ˝ D x 2 R2 W R1 < < R2 ; in the form 1  1    X X u D a ln CbC ak k C bk k cos k'C ck k C dk k sin k' kD1

kD1

(see Problem 1 at the end of Chapter 8 for justifying the above formalism). 8.

Find the steady-state temperature distribution u . ; '; z/ inside an infinite cylinder of base radius R; if the lateral surface is kept at the temperature u .R; '; z/ D A cos ': Hint. It is expected that function u does not depend on z: Hence it suffices to solve the problem in a transverse section of the cylinder. Thus, the problem becomes: u D 0 for < R; u D A cos ' for D R:

9.

Show that there is no solution u 2 C 2 .B/; B D BR .0/ ; to the Neumann problem ² u D 1 in B @u D 0 on @B: @ Hint. Use Green’s first formula.

Harmonic functions 10.

Prove that if u is harmonic in the open set ˝  Rn ; then for any closed ball B r .x/  ˝; the value u .x/ at the center of the ball is equal to the average value of u in the ball Br .x/ ; that is Z n u .x/ D u .y/ dy !n r n Br .x/

Section 3.15 Problems

75

(“solid” version of the mean value formula). Hint. Integrate from 0 to r with respect to ; the formula Z n1 !n  u .x/ D u .y/ dy : @B .x/

11.

A harmonic function in (Liouville’s theorem).

Rn

which is bounded below or above, is constant

Hint. Assume without loss of generality u  0: Let x1 ; x2 2 Rn be two arbitrary points and r1 D r2 C jx1  x2 j  r2 > 0: Then B2 D Br2 .x2 /  Br1 .x1 / D B1 and using the solid mean value formula, one finds  n Z Z n r1 n u .y/ dy  u .y/ dy D u .x1 / : u .x2 / D n n !n r2 B2 !n r2 B1 r2 Letting r2 ! 1 yields u .x2 /  u .x1 / : Similarly, u .x1 /  u .x2 / : Another proof of Liouville’s theorem can be done by using Theorem 3.20. 12 . Let u be a harmonic function in a connected open set ˝  Rn : If u vanishes in an open nonempty subset of ˝; then u is identically zero in ˝: Hint. Let BR .x0 /  ˝ be a ball where u vanishes. Using Harnack’s inequality show that u vanishes along any continuous path starting from x0 : 13 . For each ˛ 2 Nn n ¹0º ; there is a constant C D C .n; ˛/ such that if u is a bounded harmonic function on ˝  Rn ; then jD ˛ u .x/j 

C .dist .x; @˝//j˛j

sup juj ; ˝

x 2 ˝:

Hint. If @˝ D ;; that is, ˝ D Rn ; the inequality to be proved reduces to the property of D ˛ u of being identically zero, which is true from Liouville’s theorem. Assume @˝ ¤ ;: Based on Corollary 3.3, it suffices to estimate the first order derivatives. For r < dist.x; @˝/ ; one has Z Z @u n n @u .x/ D .y/ dy D u .y/ j .y/ d @xj !n r n Br .x/ @xj !n r n @Br .x/ ˇ ˇ and the inequality follows since ˇj ˇ  1 and r mes.@Br / D n mes.Br / :

76

Chapter 3 Elliptic Boundary Value Problems

14 . Prove that the result from Problem 13 holds for the following value of constant C W j˛jŠ : C .n; ˛/ D .ne/j˛j e Solution. Assume that the formula is valid for ˛ and prove that it remains true for ˇ with jˇj D j˛j C 1: For such a ˇ; we have D ˇ u D @D ˛ u=@xj for some j: We fix a number  2 .0; 1/ and use the "solid" mean value formula for function D ˇ u and ball B r .x/ ; where, as above, r < dist.x; @˝/ : We obtain Z n @D ˛ u ˇ dy D u .x/ D n !n . r/ B r .x/ @xj Z yj  xj n dy : D ˛ u .y/ D n !n . r/ @B r .x/ jy  xj By our assumption,  ˛

jD u .y/j 

ne .1  / r

j˛j

j˛jŠ sup juj ; e ˝

y 2 @B r .x/ :

It follows that ˇ ˇ  ne j˛jC1 1 j˛jŠ ˇ ˇ ˇ sup juj : ˇD u .x/ˇ  j˛j r .1   /  e 2 ˝ Finally, choose  D 1= jˇj and use the inequality     1 j˛j 1 jˇ j 1  1  e: jˇj jˇj 15 . Any harmonic function u in ˝ is locally analytic, i.e. for each x 2 ˝; there is a neighborhood V of x such that the Taylor series X D ˛ u .x/ ˛

˛Š

.x  x/˛ ;

 ˛ Q where .x  x/˛ D j D1 xj  xj k and ˛Š D jnD1 ˛j Š; converges to u .x/ uniformly on V: Qn

Hint. From Taylor’s theorem one has u .x/ D

X D ˛ u .x/ .x  x/˛ C ˛Š

j˛jm

X jˇ jDmC1

D ˇ u . / .  x/ˇ ; ˇŠ

Section 3.15 Problems

77

for some 2 Br .x/ : Since ˇ  ˇ ˇ ˇ ˇ ˇD u . /ˇ 

jˇ j jˇjŠ sup juj ne2nC1 r e ˝ ne

and jˇjŠ  enjˇ j ˇŠ; we deduce ˇ ˇ ˇ ˇ ˇ ˇ ˇD u . /ˇ ˇˇ ˇ ˇ.  x/ˇ ˇ  enjˇ j sup juj : ˇŠ ˝ Thus, the remainder in Taylor’s formula can be estimated as follows: ˇ ˇ 1 0 ˇ ˇ X ˇ X ˇ ˇ . / D u n.mC1/ A ˇ .  x/ˇ ˇˇ  @ e sup juj ˇ ˇŠ ˝ ˇ ˇjˇ jDmC1 jˇ jDmC1  .m C 2/n en.mC1/ sup juj ˝

and tends to zero as m ! 1 ; uniformly in x 2 Br .x/ : 16 . Let .uk / be a sequence of harmonic functions in the connected open set ˝: If there is an harmonic function u0 with u0  uk on ˝ for all k; and a point x0 2 ˝ such that the sequence of real numbers .uk .x0 // is bounded above, then, for each compact subset K of ˝; .uk / has a subsequence which converge uniformly on K: Solution. We may assume that uk  0 on ˝; for every k (otherwise take uk  u0 instead of uk /: Let K  ˝ be compact with x0 2 K: Harnack’s inequality shows that max uk  C min uk  C uk .x0 /  C 0 : K

K

Hence the sequence .uk / is bounded in C .K/ : The conclusion will follow from the Arzelà–Ascoli theorem once we prove that the sequence is equicontinuous on K: Let x 2 ˝ and B D B r .x/  ˝: Denote mk .r/ D min uk ; B

Mk .r/ D max uk : B

Applying (3.54) to the functions uk  mk .r/ and Mk .r/  uk we find Mk .r=2/  mk .r/  3n .mk .r=2/  mk .r// ; Mk .r/  mk .r=2/  3n .Mk .r/  Mk .r=2// : Let $k .r/ WD Mk .r/  mk .r/ ; the oscillation of uk on B: Then $k .r/ C $k .r=2/  3n .$k .r/  $k .r=2// ;

78

Chapter 3 Elliptic Boundary Value Problems

whence $k .r=2/   $k .r/ ; It follows that

where  D

3n  1 < 1: 3n C 1

  $k r=2j   j $k .r/   j Mk .r/ :

  Consequently, $k r=2j ! 0 as j ! 1; uniformly in k: This implies the equicontinuity of .uk / on any compact subset K of ˝:

The maximum principle 17.

Using the maximum principle for Laplacian, prove the following maximum principle for the operator u C c .x/ u; with c .x/  0 W Let ˝  Rn be a connected open set, c 2 C .˝/ a function satisfying c .x/  0 on ˝: If u 2 C 2 .˝/ is such that u C c .x/ u  0 in ˝; sup u > 0 and there is a point x0 2 ˝ with u .x0 / D sup u; then u is ˝

˝

constant on ˝:

² ³ Hint. Show that the set M WD x 2 ˝ W u .x/ D sup u is both open ˝

and closed in ˝: To prove that it is open, consider an arbitrary point x 2 M: Since u .x/ D sup u > 0; we may chose a connected open ˝

neighborhood ˝ 0  ˝ of x such that u .y/  0 for every y 2 ˝ 0 : Hence c .y/ u .y/  0 on ˝ 0 ; whence u  0 in ˝ 0 : Now Theorem 3.4 implies that u is constant on ˝ 0 : 18.

Let ˝  Rn be a bounded open set and c 2 C.˝/: Prove that if c .x/  0 on ˝; then for the operator u C c .x/ u the weak maximum principle holds, namely: If u 2 C 2 .˝/ \ C.˝/; u C c .x/ u  0 on ˝ and u  0 on @˝; then u  0 on ˝: Hint. Assume the contrary. Then

sup u ˝

D max u > 0 and we may apply ˝

the result from the previous problem. 19.

If for the operator u C c .x/ u the maximum principle holds, then inf c < for every real number for which a function  2 C 2 .˝/ \ ˝

C.˝/ exists such that  C  D 0;  > 0 in ˝ and  D 0 on @˝:

Section 3.15 Problems

79

Hint. Otherwise we would have c .x/  on ˝: Then  C c .x/    C  D 0 in ˝ and since   0 on @˝; we should have   0 on ˝; a contradiction. 20.

State and prove a uniqueness theorem for the Dirichlet problem ² u C c .x/ u D f .x/ in ˝ uDg on @˝;

(3.63)

where c .x/  0 for every x 2 ˝: 21.

Let ˝  Rn be a bounded open set, c 2 C .˝/ satisfying c .x/  0 on ˝; f 2 C.˝/ and g 2 C.@˝/: Prove that there existsa constant  C > 0 only depending on ˝; such that if u 2 C 2 .˝/ \ C ˝ is the solution of (3.63), then jujC.˝/  jgjC.@˝/ C C jf jC.˝/ : Hint. Follow the proof of Theorem 3.7, replacing the weak maximum principle for Laplacian by the analog result for the operator u C c .x/ u:

22.

23.

Let ˝  Rn be an open set such that Rn n ˝ is a bounded set. Let c 2 C .˝/ satisfy c .x/  0 on ˝; f 2 C .˝/ and g 2 C .@˝/ : Prove that there exists at most one solution u 2 C 2 .˝/ \ C.˝/ of the exterior Dirichlet problem 8 < u C c u D f in ˝ uDg on @˝ : u .x/ ! 0 for jxj ! 1: ¯ ® Let ˝ D x 2 R2 W jxj < 1 and assume that u 2 C 2 .˝/ \ C.˝/ is the solution of the problem ² u D 0; jxj < 1 u D 2 ln .2 C sin '/  sin '; jxj D 1: Find the extreme values and the extremum points of u: Hint. According to Corollary 3.5, the extreme values of u are reached only on @˝: Thus the problem reduces to the problem of finding the extreme values and the extremum points of the function 2 ln .2 C t /  t; t 2 Œ1; 1 :

80

24.

Chapter 3 Elliptic Boundary Value Problems

Consider the nonlinear Dirichlet problem ² u .x/ D f .u .x// ; x 2 ˝ u .x/ D 0; x 2 @˝

(3.64)

in a bounded open set ˝  Rn : Prove that if f 2 C .R/ and jf .t /j  M for every t 2 R; then there exists a constant C > 0 only depending on ˝ and M; such that jujC.˝/  C for every solution u 2 C 2 .˝/ \ C.˝/ (generally not unique) of problem (3.64). Solution. For a solution u denote v .x/ WD f .u .x// : Clearly u solves the linear Dirichlet problem u D v in ˝; u D 0 on @˝: Then, from Theorem 3.7, jujC.˝/  c jvjC.˝/  cM D C: 25.

Assume that in (3.64), f 2 C 1 .R/ and is nondecreasing on R: Prove that if u; v 2 C 2 .˝/ \ C.˝/ are two solutions of problem (3.64) with u  v on @˝; then u  v on ˝: Solution. We have  .u  v/ .x/ D f .u .x//  f .v .x// D  .u  v/ .x/ c .x/ ; ´

where c .x/ D

f .u.x//f .v.x// v.x/u.x/ f 0 .u .x// ;

; if u .x/ ¤ v .x/ if u .x/ D v .x/ :

One has c 2 C .˝/ and c .x/  0 for all x 2 ˝: Thus  .u  v/ C c .x/ .u  v/ D 0 in ˝ and u  v  0 on @˝: Now the weak maximum principle for the operator w C c .x/ w guarantees u  v  0 on ˝:

Green’s function 26.

Prove that Green’s function of the Dirichlet problem for Laplacian satisfies the following inequalities: 0 < G .x; y/ < N .x  y/

for all x; y 2 ˝; x ¤ y:

Hint. G .x; :/ is harmonic in ˝n¹xº ; zero on @˝ and G .x; y/ ! C1 as y ! x: Using the maximum principle, it follows that G .x; y/ > 0: For the second inequality apply the maximum principle to the function ˚ .x; :/ D G .x; :/ C N .x  :/ :

Section 3.15 Problems

27.

81

Prove that function ˚ is continuous on ˝  ˝: Hint. Using the continuity of function ˚ .x0 ; :/ and the maximum principle for the harmonic function ˚ .x0 ; :/  ˚ .x; :/ gives  

j˚ .x; y/  ˚ .x0 ; y0 /j j˚ .x0 ; y0 /  ˚ .x0 ; y/j C j˚ .x0 ; y/  ˚ .x; y/j ˇ    ˇ ˇN x0  y 0  N x  y 0 ˇ j˚ .x0 ; y0 /  ˚ .x0 ; y/j C

max

y 0 2@˝

! 0 as .x; y/ ! .x0 ; y0 / : 28 . Prove that Green’s function is symmetric in its two variables, that is, G .x; y/ D G .y; x/ for all x; y 2 ˝; x ¤ y: Hint. Let x1 ; x2 2 ˝; x1 ¤ x2 and r > 0 so small that the closed balls of radius r with centers x1 and x2 are disjoint and included in ˝: Green’s second formula applied

to the harmonic functions  G .x1 ; :/ ; G .x2 ; :/ and to the open set ˝ n B r .x1 / [ B r .x2 / ; gives ³ ² Z @G @G .x2 ; y/  G .x2 ; y/ .x1 ; y/  G .x1 ; y/  @y @y @Br .x1 / ³ ² Z @G @G .x2 ; y/  G .x2 ; y/ .x1 ; y/ : D G .x1 ; y/ @y @y @Br .x2 /

d

d

Next observe that the limit as r ! 0 of the first and the fourth integrals is equal to zero. Also, as r ! 0; one has Z @˚ .x1 ; y/  ! 0; G .x2 ; y/ @y @Br .x1 / Z @˚ .x2 ; y/  ! 0: G .x1 ; y/ @y @Br .x2 /

d d

Hence

lim r!0

Z

d

@N .x1  y/  @y @Br .x1 / Z @N .x2  y/ ; D G .x1 ; y/ r!0 @Br .x2 / @y G .x2 ; y/

lim

which gives G .x2 ; x1 / D G .x1 ; x2 / :

d

82

29.

Chapter 3 Elliptic Boundary Value Problems

  For each x 2 ˝; find the function ˚ .x; :/ 2 C 2 ˝ which satisfies y ˚ .x; y/ D 0 for y 2 ˝; ˚ .x; y/ D N .x  y/ for y 2 @˝; in the following cases: (a) ˝ D ¹x 2 Rn W x1 > 0º (half-plane) (b) ˝ D ¹x 2 Rn W x1 > 0 and jxj < Rº (half-ball).

Weak solutions of the Dirichlet problem 30.

Prove that under the assumptions (3.41) and if a0  0; the norma .u; u/1=2  given by (3.43) is equivalent to norm juj0;1 ; on the space C01 ˝ :

31.

Consider the one-dimensional Dirichlet problem ² u00 C u D f in .0; 1/ u .0/ D u .1/ D 0: Write down the corresponding energy functional E; compute E 0 .uI v/ D

lim

t !0C

t 1 .E .u C t v/  E .u// ;

and give the definition of the weak solution and some of its characterizations. 32.

Write down the energy functional and give characterizations of the weak solution for each of the following problems:   ´ u C 1 C jxj2 u D 1 in ˝ (a) uD0 on @˝ ² u C u D jxj in ˝ (b) uD0 on @˝:

33.

Write down the Dirichlet problem whose energy functional is  Z  1 2 2 (a) E .u/ D jruj C u C jxj u dxI ˝ 2  Z  1 1 2 2 (b) E .u/ D jruj C u C jxj u dx: 2 ˝ 2

Section 3.15 Problems

83

x1 x2 .ln jxj/˛ ; with 0 < ˛ < 1; 34 . Verify that the function   2  u .x1 ; x2 / D satisfies u 2 C R and u … C 2 R2 : Therefore, the “gain of two” regularity result for the solution Poisson’s  of the Dirichlet problem for m 2 equation (take ˝ D B1 0I R ) does not hold in the case of C spaces. However, it holds in Sobolev spaces (see Part II, Corollary 8.2) and also in Hölder spaces (see Gilbarg–Trudinger [17, p. 52]).

Fourier series: eigenvalues and eigenfunctions 35.

Prove that the following systems of functions from the classical theory of Fourier series, are orthonormal and complete: (a)

p1 ; p1  2

sin

(b)

p1 ; 

cos

q (c)

q

2 

2 

sin

x;

p1 

q

x; 

cos

2 

x;

cos

p1 

sin

2x;

p1 

cos

2x; ::: in L2 .; / I

2x; ::: in L2 .0; / I

in L2 .0; / :

kx k1

Hint. (a) The completeness follows from the Weierstrass type theorem of approximation of continuous functions on Œ;  with equal values at the ends of the interval, by trigonometric polynomials (see e.g. A. F. Bermant and I. G. Aramanovich [3], p.731); also use the density of this space of functions in L2 .; / ; as well as proposition (iv) of Theorem 3.15. (b) For completeness it suffices to show that from .u; cos kx/L2 .0;/ D 0; k D 0; 1; :::; where u 2 L2 .0; / ; it follows u D 0: If e u is the even extension of u to Œ;  ; then .e u;

cos

kx/L2 .;/ D 0; k D 0; 1; :::

.e u;

sin

kx/L2 .;/ D 0; k D 1; 2; ::: :

Applying the result from (a) gives e u D 0 on Œ;  ; hence u D 0 on Œ0;  : 36.

Expand into Fourier series with respect to the systems in Problem 35, the following functions: 1I xI sin2 x:

84

37.

Chapter 3 Elliptic Boundary Value Problems

Solve the eigenvalues and eigenfunctions problem ² u D u in ˝ uD0 on @˝; for: (a) interval® ˝ D .0; a/ (b) rectangular domain ˝ D .0; a/  .0; b/ ¯ (c) disk ˝ D x 2 R2 W jxj < R : q  2 2 kx Hint. (a) k D k ;  D k a a sin a .k D 1; 2; :::/ : (b) Use the separation of variables method looking for eigenfunctions of the form u .x1 ; x2 / D A .x1 / B .x2 / : Substitution leads to    2 jx2 k 2 j 2 kx1 sin

kj D C ; kj D p sin a b a b ab .k; j D 1; 2; :::/ :  q  q jx2 kx1 2 2 ; are orsin b Since the systems of functions a sin a b 2 2 thonormal and complete one can  in L .0; a/ and L .0; b/ ; respectively, 2 prove that the system kj is orthonormal and complete in L .˝/ : This implies that numbers kj represent all the eigenvalues of the problem. (c) In polar coordinates, the problem reads as follows: ´ 2 @ u @2 u C 1 @u C 12 @' for < R 2 C u D 0 @ @2 uD0 for D R:

One looks for nonzero solutions in the form u . ; '/ D A . / B .'/ ; with the natural condition B .' C 2/ D B .'/ : One finds the problems B 00 C B D 0 in .0; 2/ ; B .' C 2/ D B .'/ I   1 0  00 A C A C  2 A D 0; A .R/ D 0; jA .0/j < 1: The first problem admits nontrivial solutions only for  D k 2 ; namely Bk.'/ D ˛k cos k' C ˇk sin k': Furthermore, substitution y . / WD A p yields 

p   

R D 0; jy .0/j < 1: 2 y 00 C y 0 C 2  k 2 y D 0; y This is Bessel’s equation. One of its solutions, finite at the origin,is the p

R D 0 so-called Bessel’s function of order k; Jk . / : Equality Jk

Section 3.15 Problems

85

p gives R D jk ; where jk .j D 1; 2; :::/ are the positive zeros of function Jk : Thus we have found the eigenvalues and eigenfunctions: jk

kj D

u1kj . ; '/ D Jk

jk R

!2

R

! cos k'; u2kj . ; '/ D Jk

jk R

! sin k':

For details on Bessel functions we refer the reader to Vladimirov [54]. 38.

Using eigenvalues and eigenfunctions solve the Dirichlet problem u D f in ˝; u D 0 on @˝; if: (a) ˝ D .0; / and f .x/ D sin x C 2 sin 2x (b) ˝ D .0; / and f .x/ D 1 (c) ˝ D .0; /  .0; / and f .x1 ; x2 / D sin x1 sin 3x2 (d) ˝ D .0; /  .0; / and f .x1 ; x2 / D x1 x2 : Hint. Apply (3.39).

39.

For each f 2 L2 .˝/ denote by ./1 f; the weak solution of the Dirichlet problem ² u D f in ˝ uD0 on @˝: Prove the inequalities: ˇ ˇ ˇ 1 ˇ ˇ./ f ˇ

H01

ˇ ˇ ˇ ˇ ˇ./1 f ˇ

1  p jf jL2 ;

1

L2



1 jf jL2

1

(3.65) (3.66)

p and show that 1= 1 and 1= 1 are the best constants for inequalities (3.65) and (3.66), respectively. Solution. In the definition of the weak solution .u; v/H 1 D .f; v/L2 ; v 2 H01 .˝/ 0

chose v D u and take into account the expression of 1 given by (3.35). It follows that 1 2 jujH jf jL2 jujH 1 ; 1 D .f; u/L2  jf jL2 jujL2  p 0 0

1

86

Chapter 3 Elliptic Boundary Value Problems

whence, dividing by jujH 1 one obtains (3.65). For the second inequality, 0 use Parseval’s equality 2 jujL 2 D

1 X

2 .u; k /L 2 :

kD1

Furthermore, use .u; k /H 1 D k .u; k /L2 ; 1  k .k  1/ and 0 again Parseval’s equality, to deduce 2 jujL 2

D

1 1 X X 1 1 2 2 .u; / .f; k /L D  2 k H1 2 0

k

2k

kD1



kD1

1 1 X 1 2 2 .f; k /L jf jL 2 : 2 D 2 2

1

1 kD1

This gives (3.66). Finally observe that for f D 1 ; the equality holds in both (3.65) and (3.66). For other problems, the reader is referred to Pikulin–Pohozaev [36] and Vladimirov [55].

Chapter 4

Mixed Problems for Evolution Equations

The next two chapters are dedicated to the heat equation @u  u D f @t and the wave equation @2 u  u D f: @t 2 Here u and f are functions of variables x and t; and u is computed only P 2 with respect to the spatial variables, that is, u D jnD1 @ u2 : @xj

Recall that if f and u depend only on x; and not on t; then the two equations reduce to Poisson’s equation, u D f: For this reason, the elliptic equations are said to describe stationary processes and equilibrium states.

4.1

The Maximum Principle for the Heat Equation

Consider the boundary value problem 8  u D f < Lu WD @u @t u .x; 0/ D g0 .x/ : uD0

in Q in ˝ on †;

(4.1)

where ˝  Rn is a bounded open set, 0 < T  1; Q WD ˝  .0; T /;

† WD @˝  .0; T /

and functions f; g0 are given. Denote by B the parabolic boundary of cylinder Q; i.e. B WD .@˝  Œ0; T / [ .˝  ¹0º/ : Problem (4.1) is a physical model for the temperature distribution u in an homogeneous body ˝I The Cauchy initial condition gives information about temperature distribution in the body at initial time t D 0; while the Dirichlet condition expresses the fact that zero temperature is maintained on its boundary @˝:

88

Chapter 4 Mixed Problems for Evolution Equations

By a classical solution of problem (4.1) we mean a function u D u .x; t/ 2 which pointwise satisfies the three equalities, where ² ³ @u @u @2 u 2;1 ; 2 C.Q/ : C .Q/ D u W Q ! R W u; ; @t @xj @xj @xk

C 2;1 .Q/

The uniqueness of the classical solution as well as its continuous dependence on the data f and g0 will follow from the next theorem. Theorem 4.1 (maximum principle for the heat equation). Let ˝  bounded open set, 0 < T  1 and u 2 C.Q/ \ C 2;1 .Q/: If

Rn

be a

Lu  0 in Q and u  0 on B; then u  0 in Q: Proof. First we assume that T < 1: For an arbitrary " > 0; we consider the function v .x; t/ WD u .x; t/  "t: Then Lv D Lu  " < 0 in Q: Let .x0 ; t0 / 2 Q be a maximum point of v; i.e.

v .x0 ; t0 / D

max v:

If

Q

.x0 ; t0 / 2 Q; then the first-order partial derivatives of v at .x0 ; t0 / are zero, while the second-order ones are nonpositive. Then @v .x0 ; t0 / D 0 and v .x0 ; t0 /  0: @t Hence Lv .x0 ; t0 /  0; which is a contradiction. Thus .x0 ; t0 / … Q: If t0 D T and x0 2 ˝ (hence x0 is an interior point), then we should have @v .x0 ; t0 /  0 and v .x0 ; t0 /  0: @t So, in this case again, we have Lv .x0 ; t0 /  0; a contradiction. Therefore .x0 ; t0 / 2 B and

max u Q

 

Letting " ! 0 yields

max v C "T D max v C "T B Q u C "T  "T: max B max u  0: Q

Hence u  0 in Q: When T D 1; as above, we obtain that for each 0 < T 0 < 1; u  0 in QT 0 ; where QT 0 D ˝  .0; T 0 / : Consequently, u  0 in Q:

Section 4.1 The Maximum Principle for the Heat Equation

89

Remark 4.1. For T < 1; Theorem 4.1 can be equivalently restated in the following form: If u 2 C.Q/ \ C 2;1 .Q/ and Lu  0 in Q; then u: max u D max B

(4.2)

Q

Indeed, it suffices to apply Theorem 4.1 to function u  M; where M D

u: max B

As a consequence we have the following result about uniqueness and continuous dependence on the data of the classical solution of the Cauchy–Dirichlet problem (4.1), of the same type as Theorem 3.7. Corollary 4.1 (uniqueness and estimation). Problem .4:1/ has at most one classical solution u: In addition, if f 2 L1 .Q/; then ju .x; t/j  jg0 jC.˝/ C c jf jL1 .Q/

for every .x; t/ 2 Q;

(4.3)

where c is a constant only depending on ˝: Proof. (1) If u1 ; u2 are two classical solutions, then u WD u1  u2 satisfies Lu D 0 in Q and u D 0 on B: Theorem 4.1 implies u  0 and u  0 in Q: Hence u D 0 in Q; so u1 D u2 : (2) Apply Theorem 4.1 to the functions   v .x; t/ WD ˙ u .x; t/  jg0 jC.˝/  2ı  x1 Cı jf jL1 .Q/ ;

e

e

where ı > 0 is chosen such that jx1 j  ı for every x  D .x1 ; x2 ; :::; xn / 2ı  x1 Cı : 2 ˝: Finally the desired estimation holds with c D

max e ˝

e

Theorem 4.1 represents the weak maximum principle for the heat equation, the analogue of the weak maximum principle for the Laplacian (Theorem 3.5). Equality (4.2) shows that function u attains its maximum on boundary B; however, it is not excluded to attain it in Q as well. For instance, u can be a constant function on Q: A strong maximum principle guarantees that this is the unique possibility. The statement and proof of the strong maximum principle for the heat equation, and for more general parabolic type equations, can be found, for example, in Barbu [2], Dautray–Lions [9, cap. V] and Protter–Weinberger [39]. As in the case of elliptic problems, instead of treating directly the problem of the existence of classical solutions, it is more convenient to proof first the existence of generalized solutions and then to come back to classical solutions via regularity theorems. We will proceed this way in what follows.

90

4.2

Chapter 4 Mixed Problems for Evolution Equations

Vector-Valued Functions

The presence of variable t having the meaning of time justifies the name of evolution equations given to such kind of equations. In fact, when analyzing such equations it is often useful to confer a privileged role to variable t: Thus, to a scalar function u .x; t/ one associates the mapping t 7! u .t/ WD u .:; t/ ;

(4.4)

whose value at each point t is the function u .:; t/ of variable x: Hence u .t / .x/ D u .x; t/ : In what follows, the meaning of symbol u; either of a real function of variables x and t; or of a function of variable t with values functions of variable x; will be clear from the context. Obviously, when working with such kind of mappings it is important to precise the space of functions where the mappings take values in, and also some of their properties, such as measurability, continuity, differentiability, integrability, etc. In this section, we shall deal more generally with mappings with values in a vector space, i.e. with vector-valued functions. More exactly, we shall consider mappings with values in a Banach space. Let X be a Banach space with norm j:jX and I an interval of real numbers. We denote by C.I I X / the space of all continuous functions u W I ! X: If I is a compact interval Œa; b ; then C .Œa; b I X / is a Banach space with respect to the uniform norm jujC .Œa;b IX / D

max ¹ju .t /jX W

t 2 Œa; bº :

Let u W I ! X be an arbitrary function. One says that u is differentiable at a point t0 2 I; if the limit

du .t / D u0 .t / D lim 0 t!t dt 0

0

1 Œu .t /  u .t0 / t

exists in X: Denote by C 1 .I I X/ the space of all functions u from I to X which are differentiable on I and with continuous derivative u0 W I ! X: More general, C k .I I X / is the space of all functions u from I to X which are k times differentiable on I and with continuous k-order derivative. If X is a Hilbert space with inner product .:; :/X and u 2 C 1 .I I X / ; then 2 (Exercise) the scalar function ju .t /jX belongs to C 1 .I / and

d ju .t /j2 D 2 u0 .t / ; u .t / X X dt

for all t 2 I:

(4.5)

Other basic notions for real functions can be extended in a natural way for functions with values in X; by replacing, whenever necessary, the absolute value

Section 4.3 The Cauchy–Dirichlet Problem for the Heat Equation

91

ju .t /j with norm ju .t /jX : Such are the notions of a measurable function and of an (Riemann or Lebesgue) absolutely integrable function. For 1  p < 1 and 1  a < b  1; we shall consider the space p L .a; bI X / of all measurable functions u W .a; b/ ! X for which ! p1 Z b p < 1: jujLp .a;bIX / WD ju .t /jX dt a

For the rest of this chapter, X will be one of the spaces   C m ˝ ; L2 .˝/ ; H01 .˝/ ; where ˝  Rn : We conclude this section by two results concerning uniformly convergent series of functions. Compared to the usual framework of classical analysis, here the series are composed moreP generally of vector-valued functions. 1 A series of functions kD1 uk .t / ; with uk 2 C .Œa; b I X/ ; is uniformly convergent to a function u W Œa; b ! X if for ˇPeach " > 0; thereˇis N D N ."/ ˇ .which does not depend on t/ such that ˇ m kD1 uk .t /  u .t / X < " for all m > N and t 2 Œa; b : The necessary and sufficient condition that the series be uniformly convergent ˇ is that for each " > 0 there exists N D N ."/ such ˇP ˇ ˇ mCp that ˇ kDmC1 uk .t /ˇ < " for all m > N; p  1 and t 2 Œa; b (Cauchy’s X criterion). The following results hold: P1 10 If the series of functions kD1 uk .t / ; uk 2 C .Œa; b I X / ; is uniformly convergent to u; then u 2 C .Œa; b I X / : P1 1 .Œa; b I X / ; is convergent at a point 20 If the series (1) kD1 uk .t / ; uk 2 C P 0 t0 2 Œa; b and the series of derivatives (2) 1 kD1 uk .t / is uniformly convergent on Œa; b ; then series (1) is uniformly convergent on Œa; b ; its sum u .t / belongs to C 1 .Œa; b I X / and u0 .t / is the sum of series (2). For proofs we refer the reader to S.M. Nikolsky, [31], Vol. 1, p. 435.

4.3

The Cauchy–Dirichlet Problem for the Heat Equation

  Let us observe that if u 2 C 2;1 Q is a classical solution of problem (4.1),     f 2 C Q and g0 2 C ˝ ; then the mappings t 7! u .t/ WD u .:; t/ ;

t 7! f .t / WD f .:; t/

92

Chapter 4 Mixed Problems for Evolution Equations

      also denoted by u and f; belong to C 1 Œ0; T  I C 2 ˝ and C Œ0; T  I C ˝ ; respectively (note that when T D 1; by Œ0; T  we shall understand the interval Œ0; 1/). In addition we have 8 0 < u .t /  u .t/ D f .t / for t 2 Œ0; T  u .0/ D g0 (4.6) : u .t / .x/ D 0 for x 2 @˝; t 2 Œ0; T  : Thus, (4.6) can be viewed as another form ofwriting problem    (4.1). From (4.6) we have that for each t 2 Œ0; T  ; u .t / 2 C 2 ˝ \ C01 ˝ and ² u .t/ D f .t/  u0 .t / in ˝ u .t / D 0 on @˝: This, according to Dirichlet’s principle (in case that we assume additionally that ˝ is of class C 1 ) yields Z     ru .t/  rv  f .t/ v C u0 .t/ v dx D 0 for every v 2 C01 ˝ : ˝

Hence ´ .u0 .t / ; v/L2 .˝/ C .u .t / ; v/H 1 .˝/ D .f .t / ; v/L2 .˝/ ; t 2 Œ0; T  0 u .0/ D g0 :

(4.7)

This problem still has a sense if more generally f .t / ; u0 .t / 2 L2 .˝/ and g0 ; u .t / ; v 2 H01 .˝/ ; without any smoothness condition on ˝: We shall use Fourier’s method looking for the solution of (4.7) in the form u .t / D

1 X

uk .t / k ;

(4.8)

kD1

where k are the eigenfunctions of the Dirichlet problem considered in Section 3.11, and uk are real functions which are going to be determined. Formal substitution in (4.7) yields 8 P ° ± 1 0 .t / . ˆ .t / . ; v/ C u ; v/ u 1 2 < k k k L .˝/ kD1 H0 .˝/ D .f .t / ; v/L2 .˝/ k ˆ : P1

kD1 uk

.0/ k D g0 : (4.9)

Since .k ; v/H 1 .˝/ D k .k ; v/L2 .˝/ ; the first equality becomes 0

1 X kD1



 u0k .t / C k uk .t/ .k ; v/L2 .˝/ D .f .t / ; v/L2 .˝/ :

Section 4.3 The Cauchy–Dirichlet Problem for the Heat Equation

This gives us

1 X 

93

 u0k .t / C k uk .t / k D f .t /

kD1

and furthermore

u0k .t / C k uk .t / D fk .t / ;

where fk .t / is the Fourier coefficient of f .t / with respect to k ; that is, fk .t / D .f .t / ; k /L2 .˝/ : On the other hand, the second equality in (4.9) yields uk .0/ D g0k ; where g0k is the Fourier coefficient .g0 ; k /L2 .˝/ : Therefore, uk is the solution of the Cauchy problem ² 0 uk .t / C k uk .t / D fk .t / ; t 2 Œ0; T  uk .0/ D g0k : We can easily solve this problem and obtain Z t k t k g0 C ek .ts/ fk .s/ ds; t 2 Œ0; T  : uk .t / D e

(4.10)

0

Thus we may say that the solution of the Cauchy–Dirichlet problem (4.1) is completely determined by (4.8) and (4.10). However, the result obtained by the above formalism needs a more careful investigation which has to give answers at the following questions: (a) Does series (4.8) converge for every fixed t ‹ What is the functional space where the convergence takes place? (b) What are the properties of function u defined by (4.8)? (c) In what sense is u defined by (4.8), a solution of the Cauchy–Dirichlet problem?   and f 2 C Œ0; T  I L2 .˝/ ; .4:8/ and Theorem 4.2. For g0 2 H01 .˝/  .4:10/ define a function u 2 C Œ0; T  I H01 .˝/ : Moreover, if T < 1; then 2 C T jf j2C .Œ0;T IL2 .˝// : juj2C .Œ0;T IH 1 .˝//  2 jg0 jH 1 .˝/ 0

0

(4.11)

Proof. Assume that T < 1: When T D 1; the same reasoning applies on each finite subinterval Œ0; T 0  of Œ0; 1/:

94

Chapter 4 Mixed Problems for Evolution Equations

1 (a)  First we prove that u 2 C.Œ0; T I H0 .˝//: Remember that the system p1 k is orthonormal and complete in H01 .˝/ : According to (4.8) we k have 1 1 p X X k uk .t / k D

k uk .t / p : u .t/ D

k kD1 kD1

Denote vk .t / WD

p

k uk .t / and

Hence

k

1 X

u .t / D

1 WD p k 2 H01 .˝/ :

k

vk .t /

k:

(4.12)

kD1

It suffices to prove that series (4.12) is uniformly convergent. We have ˇ ˇ2 ˇ mCp ˇ mCp X ˇ X ˇ ˇ ˇ .t / v D vk2 .t / : k kˇ ˇ ˇkDmC1 ˇ 1 kDmC1 H0

Thus the problem reduces to the uniform convergence on Œ0; T  of  2the series  of P1 2 2 .t / 2 we .a v  2 a C b real functions : Using the inequality C b/ kD1 k deduce vk2 .t / D k u2k .t / "  2 k e 

2k t

2k t

 2 k e  2 k



g0k

2



g0k



g0k

Z C

2 2

T 0

Z C

k .t s/

e 0

Z C

t

t 0

2k .t s/

e

fk .s/ ds

ds

Z

t 0

2 #

.fk .s// ds



2

.fk .s//2 ds:

(4.13)

P1 2 Hence the uniform convergence of the functional series kD1 vk .t / is reduced to the convergence of the numerical series 1 1 Z T  2 X X .fk .s//2 ds: and

k g0k kD1

Since g0k D .g0 ; k /L2 D

kD1

0

1 1 .g0 ; k /H 1 D p .g0 ; 0

k

k

k /H 1 0

;

Section 4.3 The Cauchy–Dirichlet Problem for the Heat Equation

95

2 from Parseval’s equality, the sum of the first series is jg0 jH 1 : Also 0

1 X

2 .fk .s//2 D jf .s/jL 2 ;

0  s  T;

(4.14)

kD1

  where the convergence is uniform since f 2 C Œ0; T  I L2 .˝/ ; as follows from Dini’s theorem (see, e.g. W. R. Rudin [43] and V. Smirnov [48]). Hence series (4.14) can be integrated term by term and so 1 Z X kD1

T 0

.fk .s// ds D

Z

1 T X

2

0

.fk .s// ds D

Z

T

2

0

kD1

2 jf .s/jL 2 ds:

Furthermore, inequality (4.11) follows from the estimations 2 ju .t/jH 1 D 0

1 X

vk2 .t / 

kD1

Z

2 D 2 jg0 jH 1 C 0

1 X



2 k g0k

2

Z C

kD1 T 0

0

T

! .fk .s//2 ds

2 jf .s/jL 2 ds

2 2  2 jg0 jH : 1 C T jf j C .Œ0;T IL2 .˝// 0

The next theorem is concerning with the requirement that u 2 C 1 .Œ0; T  I L2 .˝//; which is a natural condition in view of (4.7).  2  1 .˝/ ./ 2 L Theorem 4.3. Let g : If either f 2 C.Œ0; T  I H01 .˝//; 0   1 2 or f  2 C Œ0; T I L .˝/  ; then .4:8/  and .4:10/ define a function u 2 C Œ0; T  I H01 .˝/ \ C 1 Œ0; T  I L2 .˝/ ; solution of problem .4:7/: Proof. Assume that T < 1: Here again, when T D 1; the same reasoning applies on each finite subinterval Œ0; T 0  of Œ0; 1/: We have to prove that  u 2 C 1 Œ0; T  I L2 .˝/ : To this end, it suffices that the differentiated series P 1 0 kD1 uk .t / k is uniformly convergent on Œ0; T  ; which in its turn reduces to 2 P1  0 the uniform convergence of the series of real functions uk .t / : kD1   We begin by observing that if g0 2 ./1 L2 .˝/ ; then a function h 2 L2 .˝/ exists with .g0 ; w/H 1 D .h; w/L2 0

for all w 2 H01 .˝/ :

This together with the definition of eigenvalues and eigenfunctions gives

k g0k D k .g0 ; k /L2 D .g0 ; k /H 1 D .h; k /L2 : 0

96

Chapter 4 Mixed Problems for Evolution Equations

Then function uk .t / can be represented as follows: Z t k t k uk .t / D e g0 C ek .t s/ fk .s/ ds 0 Z t   k ek .ts/ fk .s/  k g0k ds D g0 C 0 Z t D g0k C ek .ts/ fkh .s/ ds; 0

where

  f h .t / D f .t /  h and fkh .t / D f h .t / ; k

L2

Then u0k

Z .t / D

fkh .t /

 k

0

t

:

ek .t s/ fkh .s/ ds:

We discuss the cases:     (a) Assume f 2 C Œ0; T  I H01 .˝/ and g0 2 ./1 H01 .˝/ : Then   f h 2 C Œ0; T  I H01 .˝/ : One has 2  Z t 2  0 D fkh .t /  k ek .ts/ fkh .s/ ds uk .t / 0  Z t Z t 2 2  h 2 2k .t s/ h  2 fk .t / C k e ds fk .s/ ds 

 2 fkh .t /

2

Z C k

0

0 T 

2 fkh .s/ ds

0

(4.15)

 0 2 P and thus the uniform convergence in L2 .˝/ on Œ0; T  of the series 1 kD1 uk .t / P1  h 2 reduces to the uniform convergence of the series and to the kD1 fk .t /   2 P1 R T h ds: The sum of convergence of the numerical series kD1 0 k fk .s/ ˇ ˇ2 ˇ ˇ the first series is ˇf h .t /ˇ 2 ; and the convergence is uniform on Œ0; T  since  L f h 2 C Œ0; T  I L2 .˝/ : The convergence of the second series follows from the convergence !2 1 1 1 2   2 X X X 1 h h h f .s/ ; p k

k fk .s/ D

k f .s/ ; k 2 D L

k H01 kD1 kD1 kD1 ˇ ˇ2 ˇ ˇ D ˇf h .s/ˇ 1 ; (4.16) H0

  which is uniform on Œ0; T  since f h 2 C Œ0; T  I H01 .˝/ :

Section 4.3 The Cauchy–Dirichlet Problem for the Heat Equation

97

    (b) Assume f 2 C 1 Œ0; T  I L2 .˝/ and g0 2 ./1 L2 .˝/ : Making the change of variable t  s D ; we obtain Z t k ek fkh .t   / d; uk .t/ D g0 C 0

whence, since f h 2 C u0k

 1

 Œ0; T  I L2 .˝/ implies fkh 2 C 1 Œ0; T  ; we have

.t / D e

k t

Z fkh .0/

C

D ek t fkh .0/ C

t

0

Z

t

0



0

ek fkh .t   / d 

0

ek .ts/ fkh .s/ ds:

Then u0k .t /2 Z t  0 2 ! 2 Z t   2 fkh .0/ C e2k .ts/ ds fkh .s/ ds 0

0

Z t  0 2  2 1 h  2 fk .0/ C fkh .s/ ds

k 0 Z T  0 2  2 1 h  2 fk .0/ C fkh .s/ ds:

1 0 Now the uniform convergence on Œ0; T  of series the convergence of the numerical series 2

1  X

fkh .0/

kD1

and

1 Z X kD1

T 0

P1  kD1

(4.17) 2

u0k .t /

follows from

2   0 fkh .s/ ds:

ˇ ˇ ˇ h ˇ2 The sum of the first series is ˇf .0/ˇ 2 and the convergence of the second series L is a consequence of the convergence 1  0 2 ˇˇ 0 ˇˇ2 X h fk .s/ D ˇˇ f h .s/ˇˇ ; kD1

L2

  which is uniform on Œ0; T  since f h 2 C 1 Œ0; T  I L2 .˝/ :     (c) The case f 2 C Œ0; T  I H01 .˝/ and g0 2 ./1 L2 .˝/ can be reduced to cases (a), (b) if one represents u in the form v C w; with v and w the solutions of problem (4.7) for the data Œf; 0 and Œ0; g0  ; respectively.

98

Chapter 4 Mixed Problems for Evolution Equations

The above results justify the following definition. Definition 4.1. By a weak problem .4:1/; we   Cauchy–Dirichlet   solution of the mean a function u 2 C Œ0; T  I H01 .˝/ \ C 1 Œ0; T  I L2 .˝/ which satisfies .4:7/ for all v 2 H01 .˝/ :  2  1 Theorem 4.4. Let g0 2 ./ L .˝/ : If either f 2 C.Œ0; T  I H01 .˝//  or f 2 C 1 Œ0; T  I L2 .˝/ ; then the Cauchy–Dirichlet problem .4:1/ has a unique weak solution. Proof. Existence: already established by Theorem 4.3. solutions of (4.1).Then the funcUniqueness: Assume that  u1 ; u2 are weak   tion u WD u1  u2 2 C Œ0; T  I H01 .˝/ \ C 1 Œ0; T  I L2 .˝/ and for every v 2 H01 .˝/ satisfies ´ .u0 .t / ; v/L2 C .u .t / ; v/H 1 D 0; t 2 Œ0; T  0 u .0/ D 0: Choose v D u .t/ and use (4.5) to obtain   1 d 2 2 2 0 D u0 .t / ; u .t / L2 C ju .t/jH ju .t /jL 1 D 2 C ju .t /j 1 ; t 2 Œ0; T : H 0 0 2 dt 2 It follows that ddt ju .t/jL 2  0 for all t 2 Œ0; T : Hence the function t 7! 2 ju .t/jL2 is nonincreasing on Œ0; T : This function is continuous on Œ0; T  since u 2 C.Œ0; T I L2 .˝//; and vanishes at t D 0 since u .0/ D 0: Hence it is identically zero on Œ0; T : Thus u .t / D 0 for all t 2 Œ0; T  ; that is, u1 D u2 :

Additional regularity propertiesof the weak solution will be presented in Part II.  1 .˝/ 2 .˝/ ; we say that the function u 2 Œ0;  2 H For g and f 2 C T I L 0 0   C Œ0; T  I H01 .˝/ defined by (4.8) and (4.10) is a presolution of the Cauchy– Dirichlet problem. Notice that in (4.1), the Dirichlet boundary condition can be replaced by other conditions. For example, we may consider the Neumann boundary condition “@u=@ D 0 on †”, and the corresponding Cauchy–Neumann problem for the heat equation, for which an analogue theory can be worked out.

Section 4.4 The Cauchy–Dirichlet Problem for the Wave Equation

4.4

99

The Cauchy–Dirichlet Problem for the Wave Equation

The ideas from the previous section can also be used for solving in a generalized sense the Cauchy–Dirichlet problem for the wave equation: 8 2 ˆ in Q < Lu WD @@t u2  u D f @u (4.18) u .x; 0/ D g0 .x/ ; @t .x; 0/ D g1 .x/ in ˝ ˆ : uD0 on ˙: From a physical point of view, problem (4.18) is a model for small vibrations of an elastic membrane (string, in one space dimension). The Dirichlet boundary condition “u D 0 on †” expresses the fact that the membrane is fixed along its contour (the ends of the string are fixed, in one space dimension), while the Cauchy initial value conditions “ uj t D0 D g0 ; u t j t D0 D g1 ” express the initial positions and initial velocities of the points of the membrane (string). By a classical solution of (4.18), we shall mean a function u 2 C 2 .Q/ which satisfies the four conditions in a pointwise manner.   Definition 4.2. Let f 2 L2 0; T I L2 .˝/ ; g0 2 H01 .˝/ and g1 2 L2 .˝/ : By a weak solution of problem (4.18) we mean a function u 2 C Œ0; T  I H01 .˝/  1 \ C Œ0; T  I L2 .˝/ such that for each v 2 H01 .˝/ ; the mapping t 2 Œ0; T  7! .u0 .t/ ; v/L2 is absolutely continuous and satisfies ´ d 0 dt .u .t / ; v/L2 C .u .t/ ; v/H01 D .f .t / ; v/L2 for a.e. t 2 Œ0; T  ; u .0/ D g0 ; u0 .0/ D g1 : (4.19) It is immediately seen that if ˝ is of class C 1 ; then classical solutions are also weak solutions. As for the heat equation, an existence, uniqueness, and representation result holds for the weak solution of the Cauchy–Dirichlet problem related to the wave equation. Theorem 4.5 (existence, uniqueness, representation). Let ˝  Rn be a bounded open set and 0 < T  1: Assume that f 2 L2 0; T I L2 .˝/ D L2 .Q/ ; g0 2 H01 .˝/ and g1 2 L2 .˝/ : Then the Cauchy–Dirichlet problem .4:18/ for the wave equation has a unique weak solution. In addition, the solution admits the representation 1 X uk .t / k ; (4.20) u .t / D kD1

100

Chapter 4 Mixed Problems for Evolution Equations

where uk .t / D g0k cos

p 1

k t C p g1k sin k t

k Z t p fk .s/ sin k .t  s/ ds;

p

1 Cp

k

(4.21)

0

. k /k1 ; .k /k1 are the eigenvalues and eigenfunctions of the Dirichlet problem for the operator ; with jk jL2 .˝/ D 1; and g0k D .g0 ; k /L2 .˝/ ; g1k D .g1 ; k /L2 .˝/ ; fk .t / D .f .t / ; k /L2 .˝/ : Proof. Existence: We use the Fourier method looking for the solution in the form (4.20). Formal substitution in (4.19) yields the following conditions on the coefficients uk .t / W ² 00 uk .t / C k uk .t / D fk .t / ; t 2 Œ0; T  (4.22) uk .0/ D g0k ; u0k .0/ D g1k : The solution of this Cauchy problem isgiven by (4.21). Next we series  show that  1 1 2 (4.20) really defines a function u 2 C Œ0; T  I H0 .˝/ \ C Œ0; T  I L .˝/ ; which is the weak solution ofthe problem.  (a) First prove that u 2 C Œ0; T  I H01 .˝/ : For this it suffices to prove that series (4.20) is uniformly convergent on Œ0; T  ; if T < 1 and on each finite subinterval Œ0; T 0  of Œ0; 1/; when T D 1: Assume T < 1: Since 1 X

1 p X k uk .t / k D

k uk .t / p

k kD1 kD1   and the system p1 k is orthonormal and complete in H01 .˝/ ; we only k P1 2 have to prove the uniform convergence of the series kD1 k uk .t / : From (4.21) we have 2 ! Z t  2 p 1  k 2 1 2 k uk .t /  3 g0 C g C fk .s/ sin k .t  s/ ds

k 1

k 0    Z t 2 1  k 2 t k 2 (4.23) g  3 g0 C C f .s/ ds :

k 1

k 0 k

Thus the problem reduces to the convergence of the numerical series 1 1  1 Z T  2 X 2 X X k k g1

k g0 ; and fk2 .s/ ds: kD1

kD1

kD1

0

Section 4.4 The Cauchy–Dirichlet Problem for the Wave Equation

101

2 The first series converges to jg0 jH 1 (see the proof of Theorem 4.2), the second 0 RT 2 2 one to jg1 jL 2 and the third one to 0 jf .s/jL2 ds:   .˝/ : We have to prove the uniform conver(b) Prove that u 2 C 1 Œ0; T  I L2P 1 u0 .t / k ; or equivalently the uniform gence of the differentiated series P1  0 kD12 k convergence of the series kD1 uk .t / : We have

u0k

Z t p p p p k k .t / D  k g0 sin k t C g1 cos k t C fk .s/ cos k .t  s/ ds; 0

whence     Z t  2 2 2  0 k k 2 fk .s/ ds : uk .t /  3 k g0 C g1 C t

(4.24)

0

The conclusion now follows by using the already presented arguments. (c) Prove that for each v 2 H01 .˝/ ; the function .u0 .t / ; v/L2 is absolutely continuous on Œ0; T  and satisfies (4.19). This follows from the similar property of function u0k .t/ : Indeed, from u00k C k uk D fk ; one has u0k

.t / D

u0k

Z .0/ C

t 0

Œfk ./  k uk . / d:

Then  u0k .t/ k ; v L2 Z t  0  Œ.fk . / k ; v/L2  k .uk ./ k ; v/L2  d D uk .0/ k ; v L2 C 0 Z th i   .fk . / k ; v/L2  .uk . / k ; v/H 1 d: D u0k .0/ k ; v L2 C 

0

0

Denote by sm .t / and P P1Sm .t / the partial sums of m order of the series 1 .t/ f and  k kD1 k kD1 uk .t / k : Then by summation we obtain 



0 .t / ; v L2 Sm

D





0 .0/ ; v L2 Sm

C

Z th 0

.sm ./ ; v/L2  .Sm . / ; v/H 1 0

i d:

(4.25) uniformly with respect to  2 Œ0; T  ; it Since Sm ./ ! u ./ in follows that Z t Z t .Sm . / ; v/H 1 d ! .u . / ; v/H 1 d: H01 .˝/ ;

0

0

0

0

102

Chapter 4 Mixed Problems for Evolution Equations

 0  0 .t / ! u0 .t / in L2 .˝/ ; we deduce that Also, from Sm Sm .t / ; v L2 ! .u0 .t / ; v/L2 for every t 2 Œ0; T  : Finally, from sm ./  f . / ! 0 in L2 .˝/ ; for every  2 Œ0; T  ; we have that .sm . /  f . / ; v/L2 ! 0 pointwise. On the other hand, Bessel’s inequality gives j.sm ./  f . / ; v/L2 j  jsm . /  f . /jL2 jvjL2  2 jf . /jL2 jvjL2 ; where function  2 Œ0; T  7! jf ./jL2  belongs to L1 .0; T / as follows from our hypothesis f 2 L2 0; T I L2 .˝/ : Now Lebesgue’s dominated convergence theorem guarantees that .sm . /  f ./ ; v/L2 ! 0 in L1 .0; T / : Consequently, Z t Z t .sm . / ; v/L2 d ! .f . / ; v/L2 d: 0

0

Passing to the limit in (4.25), we obtain the formula Z th i   0  0  .f . / ; v/L2  .u ./ ; v/H 1 d; u .t / ; v L2 D u .0/ ; v L2 C 0

0

which shows that the function .u0 .t / ; v/L2 is absolutely continuous on Œ0; T  ; and whose differentiation leads to relation (4.19). Uniqueness: From (4.23) and (4.24) we deduce inequalities   Z t 2 2 2 2 .s/j  3 C C t d s ; (4.26) jg jg j j jf ju .t/jH 1 0 H1 1 L2 L2 0 0 0   Z t ˇ 0 ˇ2 2 ˇu .t /ˇ 2  3 jg0 j2 1 C jg1 j2 2 C t jf .s/jL2 ds : L L H 0

0

If u1 ; u2 are any two solutions, then u WD u1 u2 is the solution corresponding to the data g0 D g1 D 0; f D 0: The first inequality in (4.26) implies that u .t / D 0 for every t 2 Œ0; T  ; and so u1 D u2 :

4.5

Problems

The maximum principle for the heat equation 1.

Let ˝  Rn be a bounded open set and 0 < T < 1: Let u 2 C.Q/ \ C 2;1 .Q/ be a solution of the homogeneous (in absence of any heat sources) heat equation Lu D 0: If m D inf u and M D sup u; B

B

then m  u .x; t/  M for every .x; t/ 2 Q: Hint. Apply Theorem 4.1 to functions v WD m  u and w WD u  M:

Section 4.5 Problems

2.

The lateral surface of a homogeneous ball of radius R is kept at zero temperature. The initial temperature at any point x of the ball is jxj˛ .R  jxj/ ; where ˛  1: Find an upper bound of the temperature in the ball, for any moment of time. Hint. M D

3.

103

max

0rR

r ˛ .R  r/ :

If u; b u are the classical solutions of the mixed problem (4.1) correspondb 2 L1 .Q/; g 0 2 C.˝/; f ing to the data g0 2 C.˝/; f 2 L1 .Q/ and b respectively, then ˇ ˇ ˇ bˇˇ g 0 jC.˝/ C c ˇf  f : ujL1 .Q/  jg0  b ju  b 1 L

.Q/

Hint. Apply Theorem 4.2 to the function v WD u  b u: 4.

  If u 2 C 2;1 Q is the solution of problem (4.1), where T D 1; then for each T 0 < 1; we have ju .x; t/j  jg0 jC.˝/ C t jf jC.Q

T0/

;

.x; t/ 2 QT 0 :

Hint. Apply Theorem 4.2 to cylinder QT 0 and functions ˙ ujg0 jC.˝/  t jf jC.Q 0 / : Compare this estimation to global estimation (4.3) estabT lished only for functions f which are bounded on ˝  Œ0; 1/: As in the elliptic case, the maximum principle is a strong tool for the qualitative analysis of the solutions of linear and nonlinear boundary value problems. The next examples come to illustrate this assertion. 5 .

Let ˝ D ¹x 2 Rn W jxj < 1º and h 2 C Œ0; 1/: Consider the Cauchy– Dirichlet problem   ´ jxj2 @u .t/  u D h u  2n  1 in ˝  .0; 1/ @t u .x; t/ D

1 2n

on .@˝  Œ0; 1// [ .˝  ¹0º/:

Prove that the problem has at most one classical solution u and 1 R t h.s/ds jxj2 0 : C 2n 2n   Hint. Consider the function v WD u  jxj2 = .2n/ H .t / ; where H .t / D Rt v D 0 and 0 h .s/ s: Show that Lv D 0 in Q D ˝  .0; 1/ ; 0  u .x; t/ 

d

v D 1= .2n/ : max B

e

e

min B

104

6 .

Chapter 4 Mixed Problems for Evolution Equations

Let ˝  Rn be a bounded open set, Q D ˝  .0; 1/ ; h 2 C.Q/ \ L1 .Q/ and ˛ 2 .0; 1/ : Prove that if u 2 C.Q/ \ C 2;1 .Q/ is a nonnegative solution, bounded on B of the nonlinear equation (with a sublinear nonlinearity) @u  u D h .x; t/ u˛ ; @t then jujL1 .Q/ 

1 1˛

sup u C B



c jhjL1 .Q/

1  1˛

:

Here c is a constant depending only on ˝: .Boundedness on B implies boundedness on Q/ Solution. Let ı > 0 be such that jx1 j  ı for every x 2 ˝: For any finite T > 0; consider the function   v WD u  sup u  e2ı  ex1 Cı jhu˛ jC.QT / ; B

where QT D ˝  .0; T / : Apply Theorem 4.1 to obtain v  0 in QT : It follows that u .x; t/  sup u C c jhu˛ jC.QT / B

for all .x; t/ 2 QT : 1

From Young’s inequality we deduce jchj u˛  ˛u C .1  ˛/ jchj 1˛ ; and so  1  u .x; t/  sup u C ˛ jujC.QT / C .1  ˛/ c jhjL1 .Q/ 1˛ ; .x; t/ 2 QT : B

Hence u .x; t/ 

1 1˛

sup u C B



c jhjL1 .Q/

1  1˛

; .x; t/ 2 QT :

Since T < 1 has been chosen arbitrary, the inequality extends to whole Q: 7.

Prove that if u 2 C.Q/ \ C 2;1 .Q/ is bounded on B and solves the equation @u Lu WD  u D au in Q D ˝  .0; 1/ ; @t

Section 4.5 Problems

105

where a > 0; then ju .x; t/j  eat sup juj for every .x; t/ 2 Q: B

An upper bound independent on t does not exist, as shows the example: u t  uxx D 2u; 0 < x < ; for which u D et sin x is a solution. Hint. Let v WD Problem 1.

eat u:

Check that Lv D 0 and apply the result in

Mixed problems for the heat equation 8.

Find the solution of the problem 8 0 < x < ; t > 0 < u t  uxx D f .x; t/ ; u .x; 0/ D g0 .x/ ; 0 0 0 0: Application: g0 D A sin xl 1 sin 3x : l Hint. Apply Theorem 4.3 and use Problem 3.37. Find 1 X

u .x; t/ D

2

j;kD1

where aj k D

11.

4 l2

 jx1 2 2 aj k e. l / .j Ck /t sin l

Z lZ

l

g0 .x1 ; x2 / sin 0

0

jx1 l

sin

sin

kx2 ; l

kx2 dx1 dx2 : l

Let u be the weak solution of problem (4.1) for T D 1 and f D 0: Prove the following asymptotic behavior: u .t / ! 0 in H01 .˝/ as t ! 1: Hint. With the notations from the proof of Theorem 4.2, in view of (4.13), we have  2 vk2 .t /  2 k e2k t g0k : Since t et  Then

e1

for every t  0; 2 k e2k t  .et /1 for t > 0:

2 ju .t /jH 1 0

D

1 X kD1

vk2 .t / 

1 2 t > 0: jg0 jL 2 ; et

Section 4.5 Problems

12.

107

Let u be the weak solution of the mixed problem (4.1) for T D 1 and f .t/  f .0/ D f0 2 L2 .˝/ : Prove that u .t / ! v in H01 .˝/ as t ! 1; where v is the weak solution of the Dirichlet problem v D f0 in ˝; v D 0 on @˝ (If the heat density of the internal sources is constant in time, then, for large t; the effect of the initial temperature distribution is negligible and the thermal regime can be considered as stationary). Hint. The function w WD u  v solves the mixed problem 8 @w in Q < @t  w D 0 w .:; 0/ D g0  v in ˝ : wD0 on †: According to the previous problem, w .t / D u .t /  v ! 0 in H01 .˝/ as t ! 1:

13.

Prove that the weak solution u of problem (4.1) for f D 0; satisfies the conservation law

1 2 ju .t /jL 2 .˝/ C 2

Z

t 0

2 ds D ju .s/jH 1 .˝/ 0

1 2 jg0 jL 2 .˝/ ; 2

t 2 Œ0; T  :

2 Hint. Integrate from 0 to t the identity .u0 .s/ ; u .s//L2 Cju .s/jH 1 D 0: 0

Mixed problems for the wave equation 14.

Solve the problem of free vibrations of a homogeneous string whose ends are fixed 8 < u t t  a2 uxx D 0; 0 < x < l; t > 0 u .x; 0/ D g0 .x/ ; u t .x; 0/ D g1 .x/ ; 0 < x < l : u .0; t/ D u .l; t/ D 0; t > 0 in the following cases: (a) g0 D A sin .x= l/ (initial deflections), g1 D 0 (initial velocities) (b) g0 D x .l  x/ ; g1 D 0 (c) g0 D 0; g1 D A: Hint. Use Fourier’s method presented in the proof of Theorem 4.5. Then  1  X kx kat kat C bk sin sin ; ak cos u .x; t/ D l l l kD1

108

Chapter 4 Mixed Problems for Evolution Equations

where 2 ak D l

15.

Z

l 0

kx 2 dx; bk D g0 .x/ sin l ka

Z

l

g1 .x/ sin 0

kx dx: l

Solve the problem of forced vibrations of a homogeneous string whose ends are fixed 8 < u t t  a2 uxx D f .x; t/ ; 0 < x < l; t > 0 u .x; 0/ D g0 .x/ ; u t .x; 0/ D g1 .x/ ; 0 < x < l : u .0; t/ D u .l; t/ D 0; t > 0 if: (a) f D A sin .x= l/ ; g0 .x/ D B sin .2x= l/ ; g1 D 0 (b) f D A; g0 D g1 D 0:

16.

Solve the problem of free vibrations of a rectangular membrane .0; a/  .0; b/ fixed at its edge. Application: u j t D0 D A sin .x1 =a/ sin .2x2 =b/ ; u t j t D0 D 0: Hint. Solve

17.

8 < u t t  u D 0; 0 < x1 < a; 0 < x2 < b u j t D0 D g0 .x/ ; u t j t D0 D g1 .x/ : u jx1 D0 D u jx1 Da D u jx2 D0 D u jx2 Db D 0:

Prove that the weak solution of problem (4.18) for f D 0; satisfies the conservation law ˇ 0 ˇ2 2 2 2 ˇu .t/ˇ 2 C ju .t/jH D jg1 jL ; t 2 Œ0; T  : 1 2 .˝/ C jg0 j 1 L .˝/ .˝/ H .˝/ 0

0

 2 Hint. From u00k .t /C k uk .t / D 0; it follows that the function u0k .t / C

k u2k .t/ is constant on Œ0; T  : Hence 

u0k .t/

2

 2  2 C k u2k .t / D g1k C k g0k ;

whence the desired identity follows by summation.

Chapter 5

The Cauchy Problem for Evolution Equations

In this chapter the heat equation and the wave equation are considered in the entire space Rn requiring only initial conditions at time t D 0: The solving of the corresponding Cauchy problems as well as the analysis of their solutions will be based on the Fourier transform. The essential of the method is that by Fourier transform, linear partial differential equations with constant coefficients become linear ordinary differential equations with constant coefficients, thus explicitly solvable equations. Then, applying the inverse Fourier transform, the explicit solution of the partial differential equation is obtained. In fact, the same idea of reducing PDEs to ODEs was the base of Fourier’s series method in Chapter 4.

5.1

The Fourier Transform

In this chapter, we shall consider more generally complex-valued functions and consequently, all functional spaces will be assumed to be complex linear spaces.

5.1.1

The Fourier Transform on L1 .Rn /

Definition 5.1. For f 2 L1 .Rn / ; the Fourier transform of f denoted by T Œf  is the function defined by Z 1 f .x/ eixy dx; y 2 Rn : T Œf  .y/ D n=2 n R .2/ Pn Here x  y D j D1 xj yj :

Notice that the function T Œf  is continuous and bounded on Rn : Indeed, if k ! y 0 in Rn ; then the sequence of functions f .x/ eixy converges to 0 f .x/ eixy at each point x 2 Rn and is dominated by the integrable function jf .x/j : Now the continuity of T Œf  at y 0 follows from Lebesgue’s dominated convergence theorem. On the other hand, Z 1 1 .x/j d x D jf jf jL1 ; jT Œf  .y/j  .2/n=2 Rn .2/n=2 yk

110

Chapter 5 The Cauchy Problem

which proves that T Œf  is bounded on Rn ; and also that the Fourier transform defines a continuous linear operator T W L1 .Rn / ! L1 .Rn / ; with jT Œf jL1 

5.1.2

1 .2/n=2

jf jL1 :

Fourier Transform and Convolution

Definition 5.2. The convolution f g of two functions f; g W Rn ! C is defined (when it exists) by Z Z   .f g/ .x/ D f .x  y/ g .y/ dy D g .x  y/ f .y/ dy x 2 Rn :

Rn

Rn

The basic result concerning the existence of the convolution of two functions is the following theorem. Theorem 5.1. If f 2 L1 .Rn / and g 2 Lp .Rn / .1  p  1/ ; then f g 2 Lp .Rn / and jf gjLp  jf jL1 jgjLp : Proof. The case p D 1 is trivial. Let p < 1 and q be the conjugate exponent of p; that is p1 C q1 D 1: From Hölder’s inequality we obtain ˇ ˇZ ˇ ˇ ˇ ˇ .x .y/ f  y/ g d y ˇ ˇ n R  q1 Z Z  jf .x  y/j dy

Rn

jf .x  y/j jg .y/j dy p

Rn

 p1

The conclusion follows if we raise to the power p and we integrate over

:

Rn :

The next theorem is an efficient tool for the use of convolution. R Proposition 5.1. Let g 2 L1 .Rn / ; a D Rn g .x/ dx and for each k 2 ¹0º ; gk .x/ D k n g .kx/ : Let f 2 Lp .Rn / :

Nn

(a) If 1  p < 1; then f gk ! af in Lp .Rn / as k ! 1: (b) If p D 1 and f is uniformly continuous on uniformly on Rn ; as k ! 1: The proof needs the following lemma.

Rn ;

then f gk ! af

Section 5.1 The Fourier Transform

111

Lemma 5.1. If f 2 Lp .Rn / ; where 1  p < 1; then jh f  f jLp ! 0 as h ! 0: Here for h 2 Rn ; .h f / .x/ D f .x  h/ : Proof. For f 2 C01 .Rn / ; h f ! f even uniformly. If f 2 Lp .Rn / and " > 0; we choose a g 2 C01 .Rn / with jf  gjLp < "=3 : It follows that jh f  h gjLp < "=3 and jh f  f jLp  jh f  h gjLp C jh g  gjLp C jg  f jLp < " whenever jhj < ı" : Proof of Proposition 5.1. The change of variable kx D y yields to Z Œf .x  y/  f .x/ gk .y/ dy .f gk / .x/  af .x/ D n ZR  

 f x  k 1 y  f .x/ g .y/ dy: D

Rn

(a) Then we have p q L1

p

Z

j.f gk / .x/  af .x/j  jgj As a result jf gk  af

p jLp

p q L1

 jgj

Rn Z

Rn

ˇ ˇ   ˇf x  k 1 y  f .x/ˇp jg .y/j dy: ˇ ˇ ˇk 1 y f  f ˇp p jg .y/j dy: L

ˇp ˇ The sequence of functions y 7! ˇk 1 y f  f ˇLp jg .y/j .k  1/ is dominated by .2 jf jLp /p jg .y/j 2 L1 .Rn / and, according to Lemma 5.1, tends pointwise to zero as k ! 1: Then Lebesgue’s dominated convergence theorem guarantees that it tends to zero in L1 .Rn / : (b) For each ı > 0; there is a compact W with Z jg .y/j dy < ı:

Rn nW

Then sup

Rn

x2



j.f gk / .x/  af .x/j

sup

Rn ; y2W

x2

ˇ  ˇ  ˇf x  k 1 y  f .x/ˇ

whence the conclusion.

Z

Rn

jg .y/j dy C 2 jf jL1 ı;

112

Chapter 5 The Cauchy Problem

A basic property of the Fourier transform is that of transforming the convolution of two functions into an usual product of functions. More exactly, elementary computation yields the following result. Theorem 5.2. If f; g 2 L1 .Rn / ; then n

T Œf g D .2/ 2 T Œf  T Œg :

5.1.3

The Fourier Transform on the Schwartz Space S .Rn /

Interesting properties of the Fourier transform hold on a particular subspace of L1 .Rn / ; namely on the Schwartz space S D S .Rn / of all smooth functions which together with all their derivatives decay rapidly to zero as jxj ! 1: More exactly, ³ ² ˇ ˇ   ˇ ˇ S D f 2 C 1 Rn W sup ˇx ˛ D ˇ f .x/ˇ < 1 for all ˛; ˇ 2 Nn :

Rn

Here by x ˛ we have denoted the monom x1˛1 x2˛2 ::: xn˛n : For example, for every a > 0; the function f W Rn ! R given by 2

f .x/ D eajxj

belongs to S: Note that if a function f is in S; then all its derivatives D ˛ f also belong to S: S is a complex linear space. The functions in S and all their derivatives tend to zero faster than any power x ˛ as jxj ! 1: A convergence can be introduced on S as follows: fk ! f in S

if

x ˛ D ˇ fk .x/ ! x ˛ D ˇ f .x/ uniformly on for every ˛; ˇ 2 Nn :

Rn;

This convergence corresponds to a countable set ¹pm W m 2 Nº of seminorms on S; where ˇ ˇ X ˇ ˛ ˇ ˇ .f 2 S/ : sup ˇx D f .x/ˇ pm .f / D j˛j;jˇ jm

Rn

Notice that for 1  p  1; one has S  Lp .Rn / both algebraically and topologically. Indeed, if f 2 S; then jf .x/jp  C

n Y

1 ; 2 1 C x j j D1

Section 5.1 The Fourier Transform

whence p

jf jLp .Rn /  C

113

n Z Y j D1

1 dxj D C  n < 1: 2 1 C x R j

In addition, if fk ! f in S; then fk ! f uniformly on follows that fk ! f in Lp .Rn / :

Rn ;

whence it

Proposition 5.2. If f 2 S; then T Œf  2 C 1 .Rn / and for every ˇ 2 Nn ; the following formulas hold: h i   (5.1) D ˇ T Œf  .y/ D T .ix/ˇ f .x/ .y/ y 2 Rn ; i h   (5.2) T D ˇ f .y/ D .iy/ˇ T Œf  .y/ y 2 Rn : h i Proof. Compute D ˇ T Œf  and T D ˇ f by differentiation under the integral sign and integration by parts jˇj D ˇ1 C ˇ2 C ::: C ˇn times, respectively. Proposition 5.3. (a) If f; g 2 S; then Z

Rn and

Z f .x/ T Œg .x/ dx D Z

Rn

g .x/ T Œf  .x/ dx

(5.3)

Z

Rn

f g dx D

Rn

T Œf  T Œg dx:

(b) If 1

2

h .x/ D e 2 jxj ;

(5.4)

(5.5)

then T Œh D h; that is, h is a fixed point of the operator T: Proof. (a) Fubini’s theorem (see [5], Theorem 4.5) is used. Formula (5.4) can be deduced from (5.3) by replacing g with T Œg and taking into account that T 1 Œg D T Œg: (b) We have Z Z P 2 1 1  12 jxj2 ixy  1 .xj Ciyj /  12 jyj2 T Œh .y/ D e dx D e 2 dx .2/n=2 Rn .2/n=2 Rn n Z Y 2 1  1 .xj Ciyj / e 2 dxj D h .y/ .2/n=2 j D1 R n Z 1 Y z 2 D h .y/ n=2 dz: p e  j D1 Im zDyj = 2

114

Chapter 5 The Cauchy Problem

R p 2 Since R et dt D ; the conclusion will follow as soon as we prove that Z Z 2 z 2 e dz D ez dz: (5.6) Im zDa

Im zD0

To this end, let R > 0 and the rectangular contour CR with vertices A .R; 0/, B .R; 0/ ; C .R; a/ and D .R; a/ : According to Cauchy’s theorem, Z 2 ez dz D 0: CR

Hence

Z DC

z 2

e

dz D

Z

2

AB

ez dz:

Now (5.6) is obtained if we pass to the limit as R ! 1: Theorem 5.3. If f 2 S; then T Œf  2 S and the operator T W S ! S is a continuous linear bijection with inverse Z 1 1 f .x/ eixy dx D T Œf  .y/ D T Œf .x/ .y/ : T Œf  .y/ D .2/n=2 n

R

.The Fourier transform is an automorphism of S:/ Proof. (a) First we show that T Œf  2 S: For this we use (5.1) and (5.2) to deduce h i y ˛ D ˇ T Œf  .y/ D y ˛ T .ix/ˇ f .x/ .y/ i h D .i/jˇ j y ˛ T x ˇ f .x/ .y/ h i D .i/j˛jCjˇ j .iy/˛ T x ˇ f .x/ .y/ h i  D .i/j˛jCjˇ j T D ˛ x ˇ f .x/ .y/ : h i  Then we take into account T D ˛ x ˇ f .x/ 2 L1 .Rn / : (b) Next we prove that

 T 1 ŒT Œf  D T T 1 Œf  D f for every f 2 S: For each k 2 Nn ¹0º ; let g . / D eixk

2 jj2

;

2 Rn :

Section 5.1 The Fourier Transform

115

Then, based on Proposition 5.3(b), T Œg .y/ D

Z

1

ek

2 jj2

ei.yx/ d

R .2/ D k n h .k .x  y// D hk .x  y/ ; n=2

n

where h is the function given by (5.5). Using (5.3) we obtain Z Z k 2 jj2 ix e e T Œf  . / d D g . / T Œf  . / d (5.7) n Rn R Z f .y/ T Œg .y/ dy D .f hk / .x/ : D

Rn Since

Z

R

n

h .x/ dx D .2/n=2 T Œh .0/ D .2/n=2 h .0/ D .2/n=2 ;

Proposition 5.1 implies f hk ! .2/n=2 f as k ! 1; uniformly on Rn : On the other hand, it is clear that for each x; the first integral in (5.7) tends to .2/n=2 T 1 ŒT Œf  .x/ : Therefore T 1 ŒT Œf  D f: The reader is advised to use simple changes of variables to prove the following calculation formulas for the Fourier transforms of translation and dilation: T Œf .x  x0 / .y/ D eiyx0 T Œf .x/ .y/ ;   T Œf . x/ .y/ D j jn T Œf .x/ 1 y

(5.8) (5.9)

.x0 2 Rn ; 2 R n ¹0º/ : An immediate consequence of Theorem 5.2 is the next result. Theorem 5.4. If f; g 2 S; then f g 2 S and D ˛ .f g/ D f D ˛ g D D ˛ f g



 ˛ 2 Nn :

Proof. The fact that f g belongs to S follows from T 1 ŒS D S; n

f g D .2/ 2 T 1 ŒT Œf  T Œg and from the quite obvious property that the product of two functions in S also belongs to S: Furthermore, we have

 n T D ˛ .f g/ .y/ D .iy/˛ T Œf g .y/ D .2/ 2 .iy/˛ T Œf  .y/ T Œg .y/

116

Chapter 5 The Cauchy Problem

and





n T f D ˛ g .y/ D .2/ 2 T Œf  .y/ T D ˛ g .y/ n

D .2/ 2 .iy/˛ T Œf  .y/ T Œg .y/ : Hence

 



T D ˛ .f g/ D T f D ˛ g ;

whence D ˛ .f g/ D f D ˛ g: To justify the assertion made at the beginning of this chapter, namely that linear PDEs with constant coefficients reduce to ODEs after applying the Fourier transform, let us consider the equation X ˇ a˛ˇ Dx˛ D t u .x; t/ D f .x; t/ ; x 2 Rn ; t 2 I  R; j˛jCˇ m

where Dx˛ D

@j˛j @ˇ ˇ and D D : t @x1˛1 @x2˛2 ::: @xn˛n @t ˇ

If we apply the Fourier transform only with respect to the spatial variable x (considering the variable t as fixed) and use (5.2), then we find for each y 2 Rn ; the equation 0 1 m X X ˇ @ a˛ˇ .iy/˛ A D t .T Œu .t / .y// D T Œf .t / .y/ : ˇ D0

j˛jmˇ

For a fixed y; this is a linear ordinary differential equation with constant coefficients, in the independent variable t and unknown function v .t / WD T Œu .t / .y/ :

5.2

The Cauchy Problem for the Heat Equation

This section deals with the Cauchy problem for the nonhomogeneous heat equation ² @u  u D 0 in Q D Rn  .0; 1/ @t (5.10) u .x; 0/ D g0 .x/ in Rn :

Physically, the problem models the heat conduction in the whole space Rn when is known the initial temperature distribution g0 : By a classical solution of problem (5.10) we mean a function u 2 C 2;1 .Q/ which satisfies (5.10). The next lemma is useful for the proof of the existence of the classical solution.

Section 5.2 The Cauchy Problem for the Heat Equation

117

Lemma 5.2. Let I be an interval of real numbers. Assume u 2 C .I I S/ : .x; t/ exists at any point Then u 2 C 1 .I I S/ if and only if the derivative @u @t @u n .:; t/ is from R  I; @t .:; t/ 2 S for every t 2 I and the mapping t 7! @u @t continuous from I to S: In this case u0 .t / D

@u .:; t/ ; @t

t 2 I:

Proof. Necessity: If u 2 C 1 .I I S/ ; then 1 Œu .t C /  u .t /  u0 .t / ! 0 in S as  ! 0: 

(5.11)

In particular, for every x 2 Rn ; 1 Œu .t C  / .x/  u .t / .x/  u0 .t / .x/ ! 0 as  ! 0:  .x; t/ exists and is equal to u0 .t / .x/ : Hence derivative @u @t Sufficiency: We have to prove (5.11). One has @u    1 Œu .t C / .x/  u .t/ .x/ D x; t ;  @t where jt   t j  j j : Now (5.11) follows by using the continuity of a function from I to S:

@u @t

.:; t/ as

Theorem 5.5 (existence, uniqueness, representation in S). Let g0 2 S: Problem .5:10/ has a unique solution u in C 1 .Œ0; 1/I S/: In addition, u 2 C 1 .Œ0; 1/I S/ and for t > 0; one has the representation Z (5.12) u .x; t/ D N .x  y; t/ g0 .y/ dy;

Rn

where N .t / .x/ D N .x; t/ D

1 .4 t/n=2

e

jxj2 4t

;

x 2 Rn ; t > 0:

(5.13)

Proof. Uniqueness and representation: Assume that u 2 C 1 .Œ0; 1/I S/ is a solution of the Cauchy problem. Then, for each t  0; ² 0 u .t /  u .t / D 0 in Rn (5.14) u .0/ D g0 ;

118

Chapter 5 The Cauchy Problem

where all functions are in S: Since from u 2 C 1 .Œ0; 1/I S/ it follows u 2 C 1 .Œ0; 1/I S/; we infer that u0 2 C 1 .Œ0; 1/I S/; that is, u 2 C 2 .Œ0; 1/I S/: Step by step, we finally obtain u 2 C 1 .Œ0; 1/I S/: Via Fourier transform, equalities (5.14) are equivalent to ´ d 2 n dt T Œu .t/ .y/ C jyj T Œu .t / .y/ D 0; y 2 R (5.15) y 2 Rn : T Œu .t/ .y/j t D0 D T Œg0  .y/ ; Here we used the following properties whose proofs are exclusively based on the linearity and continuity of the operator T W S ! S W w 2 C 1 .Œ0; 1/I S/ ” T Œw 2 C 1 .Œ0; 1/I S/;

d  d dt T Œw .t / D T dt w .t / : Now we solve the Cauchy problem (5.15) for a linear ODE with constant coefficients, where y is considered to be fixed, as a parameter, and we find 2

T Œu .t/ .y/ D T Œg0  .y/ ejyj t :

(5.16)

Applying the inverse Fourier transform we obtain the representation h i 2 u .t / .x/ D T 1 T Œg0  .y/ ejyj t .x/ ; x 2 Rn ; t  0:

(5.17)

Existence: It is sufficient to show that under the hypothesis g0 2 S; the function w given by 2

w .t / .x/ D T Œg0  .x/ ejxj

t

belongs to C 1 .Œ0; 1/I S/ and satisfies (5.15). First we remark that for every t > 0; w .t / 2 S as a product of two functions in S (note that w .t / … S for t < 0). For the continuity of w at a point t0 2 Œ0; 1/; we need to show that x ˛ D ˇ w .t / .x/ ! x ˛ D ˇ w .t0 / .x/ uniformly on

Rn

as t ! t0 :

Clearly x ˛ D ˇ w .t/ .x/ has the form X 2 p .x; t/ ejxj t D T Œg0  .x/ ; j jjˇ j

where p .x; t/ are polynomials in variables x and t: From Lagrange’s theorem of finite increments applied with respect to t; we obtain ˇ ˇ ˇ ˇ ˛ ˇ ˇx D w .t / .x/  x ˛ D ˇ w .t0 / .x/ˇ ˇ ˇ  ˇX @ ˇ 2 2 ˇ p .x; /  jxj p .x; / D T Œg0  .x/ˇˇ ejxj ; D jt  t0 j ˇ @t

Section 5.3 The Cauchy Problem for the Wave Equation

119

where  lies between t and t0 : If we assume that jt  t0 j  1; then the a polynomial q .x/ only absolute value of @p =@t  jxj2 p is dominated by P q .x/ D T Œg0  .x/ is depending on t0 : Since T Œg0  2 S; the function bounded on Rn : Thus ˇ ˇ ˇ ˇ ˛ ˇ ˇx D w .t / .x/  x ˇ D ˇ w .t0 / .x/ˇ  C jt  t0 j in Rn : Similarly one can prove that the function 2 @w D  jxj2 T Œg0  .x/ ejxj t @t

belongs to C.Œ0; 1/I S/: The conclusion w 2 C 1 .Œ0; 1/I S/ now follows from Lemma 5.2. Formula (5.12): Direct calculation in (5.17), for t > 0; gives Z Z 2 1 g0 .z/ eiyz ejyj t eiyx dz dy u D n .2/ Rn Rn Z Z 2 1 D g0 .z/ ejyj t Ciy.xz/ dy dz: n .2/ Rn Rn Furthermore, the change of variables y D .2t /1=2 yields Z Z p 1 1  12 jj2 i.xz/= 2t .z/ g e e d dz u D 0 n n .4 t/ 2 Rn .2/ 2 Rn Z jxzj2 1 D g0 .z/ e 4t dz: n .4 t/ 2 Rn We note that (5.12) defines the classical solution of the Cauchy problem even under more general conditions on g0 : Function N .x; t/ defined by (5.13) for t > 0 and extended by zero for t  0; is called the fundamental solution of the heat equation. Note that N 2 C 1 .Rn  .0; 1// and jN .t/jL1 .Rn / D 1 for every t > 0: Also, observe that (5.12) can be written as a convolution of two functions from S; more exactly, u .x; t/ D .g0 N .t// .x/ ; x 2 Rn ; t > 0:

5.3

The Cauchy Problem for the Wave Equation

In this section we apply the Fourier transform to solve the Cauchy problem for the wave equation ´ 2 @ u  u D 0 in RnC1 @t 2 (5.18) .x; 0/ D g1 .x/ in Rn : u .x; 0/ D g0 .x/ ; @u @t

120

Chapter 5 The Cauchy Problem

From a physical point of view, the problem is a model for wave propagation (stress waves in solids, electromagnetic waves, water waves, sound waves, etc.) when the initial regime at time t D 0 is known.   A classical solution is a function u 2 C 2 RnC1 which satisfies the two equations on RnC1 : Theorem 5.6 (existence, uniqueness, representation in S). If g0 ; g1 2 S; then problem .5:18/ has a unique solution in C 2 .RI S/: In addition u 2 C 1 .RI S/ and the following representation holds:  sin .jyj t / 1 .x/ : (5.19) T Œg0  .y/ cos .jyj t / C T Œg1  .y/ u .t/ .x/ D T jyj In particular, for x 2 Rn and t > 0; we have (i) for n D 1 .D’Alembert’s formula/W u .x; t/ D

1 1 Œg0 .x C t/ C g0 .x  t / C 2 2

Z

xCt

g1 .y/ dy:

(5.20)

xt

(ii) for n D 2 .Poisson’s formula/W Z Z 1 g0 .y/ dy g1 .y/ dy 1 @ C : u .x; t/ D q q 2 @t B t .x/ 2 B t .x/ 2 2 2 2 t  jx  yj t  jx  yj (5.21) (iii) for n D 3 .Kirchhoff’s formula/W   Z Z 1 1 @ 1 u .x; t/ D g0 .y/ d C g1 .y/ d: (5.22) 4 @t t @B t .x/ 4 t @B t .x/ Proof. Assume that u 2 C 2 .RI S/ is the solution of the problem. Then, for every t 2 R; we have ² 00 u .t /  u .t / D 0 in Rn ; u .0/ D g0 ; u0 .0/ D g1 ; all functions which appear being in S: These equalities, via the Fourier transform, are equivalent to 8 2 d T Œu .t / .y/ C jyj2 T Œu .t / .y/ D 0; ˆ ˆ < dt 2 (5.23) T Œu .t / .y/j t D0 D T Œg0  .y/ ; ˆ ˆ ˇ : d ˇ dt T Œu .t / .y/ tD0 D T Œg1  .y/

Section 5.3 The Cauchy Problem for the Wave Equation

121

for all y 2 Rn : Solving we obtain w .t/ .y/ WD T Œu .t / .y/ D T Œg0  .y/ cos .jyj t / C T Œg1  .y/

sin .jyj t /

jyj

;

(5.24)

whence representation (5.19). Notice that for all t 2 R and not just for t  0; as was the case for the heat equation, functions cos .jyj t/ and sin .jyj t / = jyj as well as all their partial derivatives with respect to y are smooth and bounded. Also, their derivatives with respect to t grow at most polynomially in y: We note that at y D 0 no singularity appears since sin .jyj t / = jyj ! t as y ! 0:

By this we have also proved the existence of the solution once we show that function w given by (5.24) belongs to C 2 .RI S/ and satisfies (5.23). The argument is similar to that for the heat equation and for this reason we omit to reproduce it. To obtain representation (5.20) we use (5.8) and we deduce  1 Œg0 .x C t / C g0 .x  t / .y/ T 2 1 D .T Œg0 .x C t / .y/ C T Œg0 .x  t / .y// 2  1  iyt D e C eiyt T Œg0  .y/ 2 D T Œg0  .y/ cos .yt/ : (5.25) R xCt On the other hand, if we denote v .x/ WD 12 xt g1 .y/ dy; then v 0 .x/ D 1 2 .g1 .x C t /  g1 .x  t// and as above we have 

 1  iyt T v 0 .y/ D e  eiyt T Œg1  .y/ D iT Œg1  .y/ sin .yt/ : 2 Since T Œv 0  .y/ D iyT Œv .y/ ; it follows T Œv .y/ D T Œg1  .y/ Now (5.19) and (5.26) give (5.20).

sin .yt/

y

:

(5.26)

122

Chapter 5 The Cauchy Problem

To obtain (5.22) we perform calculation Z Z Z 1 ixy g1 .z/ dz .y/ D e g1 .z/ dz dx T 3 @B t .x/ @B t .x/ .2/ 2 R3 Z Z t2 ixy D e g1 .x C t z/ dz dx 3 @B1 .0/ .2/ 2 R3 Z Z t2 D eixy g1 .x C t z/ dx dz 3 3 .2/ 2 @B1 .0/ R Z D t2 T Œg1 .: C t z/ .y/ dz @B1 .0/ Z 2 D t eit zy T Œg1  .y/ dz @B1 .0/ Z D t 2 T Œg1  .y/ eit zy dz @B1 .0/

and using spherical coordinates we find Z 2 Z Z it zy e d z D d' @B1 .0/

0

eit jyj cos  sin  d

0

ˇ D 1 ˇ eit jyj cos  ˇ D 2  D0 it jyj  4 sin .jyj t / 2  itjyj e : D  eit jyj D it jyj t jyj

Z

Hence



T @B t .x/

g1 .z/ dz .y/ D 4 t T Œg1  .y/

This implies   Z  @ 1 g0 .y/ d .y/ D T @t t @B t .x/

@ @t



1 T t

sin .jyj t /

jyj

Z

:

(5.27)



g0 .z/ dz .y/ @B t .x/

D 4T Œg0  .y/ cos .jyj t / :

(5.28)

Clearly (5.27) and (5.28) give (5.22). Finally, Poisson’s formula (5.21) can be obtained from Kirchhoff’s formula by Hadamard’s method of descent, which consists in regarding the solution of the Cauchy problem for n D 2 as solution independent of x3 of the Cauchy problem for n D 3; and in carrying out the elementary transformations to reduce the integral over the sphere to an integral over the disk (for details, see, for example, Vladimirov [54] and DiBenedetto [10, p. 308]).

Section 5.4 Nonhomogeneous Equations: Duhamel’s Principle

123

Remark 5.1. The solution of the Cauchy problem for the homogeneous wave equation can be represented in terms of convolution, in the form   u .t / .x/ D g0 N 0 .t / .x/ C .g1 N .t // .x/ ; (5.29) where N .t/ D

1

T 1 n=2 .2/



sin .jyj t /



jyj

:

A simple exercise based only on the definition of the Fourier transform shows that: (i) for n D 1; x 2 R and t > 0; we have 1 N .t / .x/ D H .t  jxj/ : 2 Here H is the Heaviside function ² 0 for x  0 H .x/ D 1 for x > 0I (ii) for n D 2; x 2 R2 and t > 0; we have ´ p1 2 t 2 jxj2 N .t / .x/ D 0

for jxj < t for jxj  t:

Notice that N .t / 2 L1 .Rn / for every t > 0; in both dimensions n D 1 and n D 2: In the case n D 3;  such a representation of the solution in the form (5.29) is not 1 3 possible in L R : However it becomes possible if we extend the convolution and Fourier transform to the larger space of the tempered distributions (see Part II). We note that (5.19) gives the classical solution of the Cauchy problem even under more general conditions on data g0 ; g1 :

5.4

Nonhomogeneous Equations: Duhamel’s Principle

Recall that nonhomogeneous equations of the type Lu D f arise from mathematical modeling of real processes subjected to external sources or stimuli. Thus, in the case of the nonhomogeneous heat equation u t  u D

124

Chapter 5 The Cauchy Problem

f .x; t/ ; the term f has the physical interpretation as the density of the external heat sources, that is, the amount of heat produced by the sources in the unit volume per unit time. In the case of the nonhomogeneous wave equation u t t  u D f .x; t/ modeling the small vibrations of an elastic membrane, the term f .x; t/ stands for the density (force per unit mass) of the external force. In this section we shortly discuss the Cauchy problem for the nonhomogeneous heat and wave equations. Let us first consider the case of the heat equation: ² @u  u D f; x 2 Rn ; t > 0 @t (5.30) u .x; 0/ D g0 .x/ ; x 2 Rn : We look for the solution in the form u D u1 C u2 ; where u1 satisfies the nonhomogeneous equation and the null initial condition u1 .x; 0/ D 0; while u2 solves the homogeneous equation under the given initial condition, i.e. u2 is the solution of problem (5.10). Since u2 was already determined, it remains to solve (5.30) for g0 D 0; that is, ² @u  u D f; x 2 Rn ; t > 0 @t (5.31) u .x; 0/ D 0; x 2 Rn : The idea of Duhamel’s principle consists in looking for the solution of (5.31) under the form Z t v .x; t  s; s/ ds; (5.32) u .x; t/ D 0

where the function v D v .x; t; s/ depending on an additional variable s 2 will be determined. To this end, we formally substitute into (5.31) and obtain Z t .v t .x; t  s; s/  v .x; t  s; s// ds D f .x; t/ : v .x; 0; t / C

R;

0

This equality clearly holds if v satisfies the following conditions ² v t .x; t; s/ ds  v .x; t; s/ D 0 .x 2 Rn I t; s  0/ .x 2 Rn ; s  0/ : v .x; 0; s/ D f .x; s/ Note that by this method, the source term f is turned into an initial state. Furthermore, according to the result in Section 5.2, Z N .x  y; t/ f .y; s/ dy: v .x; t; s/ D

Rn

Here N is the fundamental solution of the heat equation. Thus, the solution of problem (5.31) is Z tZ N .x  y; t  s/ f .y; s/ dy ds: (5.33) u .x; t/ D 0

Rn

Section 5.5 Problems

In conclusion, the solution of (5.30) is Z jxyj2 1 .y/ g u .x; t/ D e 4t dy 0 .4 t/n=2 Rn Z tZ 2 f .y; s/  jxyj 4.t s/ dy ds: e C n=2 0 Rn .4 .t  s//

125

(5.34)

It can be written in a convolution form as e N / .x; t/ .t > 0/ ; u .x; t/ D .g0 N .t// .x/ C .f e is where the first convolution acts in Rn ; the second one in RnC1 and f the extension by zero of f for t  0: One can observe that if g0 2 S and f 2 C 1 .Œ0; 1/I S/ ; then this formula defines a function u 2 C 1 .Œ0; 1/I S/ which is the classical solution of the Cauchy problem. We note that in this case again, (5.34) gives us the classical solution even under more general conditions on data g0 ; f: In a similar way, using Duhamel’s principle, we solve the Cauchy problem for the nonhomogeneous wave equation ´ 2 @ u  u D f in RnC1 @t 2 .x; 0/ D g1 .x/ in Rn : u .x; 0/ D g0 .x/ ; @u @t The solution is sought in the form u D u1 C u2 ; where u1 is given by (5.19), while u2 solves the problem ´ 2 @ u  u D f in RnC1 @t 2 (5.35) .x; 0/ D 0 in Rn u .x; 0/ D @u @t and is represented as (5.32). Replacing formally in (5.35), the following conditions on v .x; t;  / are required: ² v t t .x; t; s/  v .x; t; s/ D 0 v .x; 0; s/ D 0; v t .x; 0; s/ D f .x; s/ : This problem, parameterized by s; is solvable as shown by Theorem 5.6.

5.5 1.

Problems Show that: 2

(a) eajxj 2 S for every a > 0 m  (b) 1 C jxj2 … S for m 2 N:

126

2.

Chapter 5 The Cauchy Problem

Compute T ŒH .R  jxj/ ; where H R > 0: Hint. T ŒH .R  jxj/ .y/ D D

3.

Let u .x/ D H .x/ ex ; x 2 tion: (a) u

1 p 2

Z

is the Heaviside function and

R R

ei x y dx

r ˇR 1 ei x y ˇˇ 2 sin Ry : D p ˇ  y 2 i y R

R:

Find the Fourier transform of the func-

(b) u .x/ (c) u .x  x0 / (d) ei x x0 u .x/ (e) u .x/ sin x: 4 .

Prove that if u 2 C k .Œ0; T  I S/ ; then u 2 C 1;k .Rn  Œ0; T / : .:; t/ D Hint. Let k D 1: From u 2 C 1 .Œ0; T  I S/ it follows that @u @t u0 .t/ 2 S: Hence the partial derivatives Dx˛ D t1 u exist and are continuous for every ˛ 2 Nn : Thus the problem reduces to proving that Dx˛ u 2 C 1 .Œ0; T  I S/ : We have    @ @u @ u .x; t C  /  u .t / .x; t/ D lim !0 @xj @t @xj        @u @u @ @  .x; t/ ; D lim D x; t !0 @xj @t @xj @t function from Œ0; T  to S: Hence the funcsince u0 is a continuous  @ @u tion @t @x .:; t/ exists, belongs to S and depends continuously on j

t: Lemma 5.2 implies

@u @xj

2 C 1 .Œ0; T  I S/ : In general, Dx˛ u 2

C 1 .Œ0; T  I S/ : 5.

Solve the Cauchy problem (5.10) for n D 1 and: (a) g0 D eax

2

(b) g0 D H .1  jxj/ : Hint. Apply (5.12).

Section 5.5 Problems

6.

127

Solve the Cauchy problem (5.30) for n D 1 and: 2

(a) g0 D eax ; f D et C t (b) g0 D sin x; f D et sin x: Hint. (a) The function v D u C et C t 2 =2 satisfies the homogeneous heat equation. 7.

Write down the representation formula for the solution of the Cauchy problem for the equation   u t  a2 u D f .x; t/ x 2 Rn ; t > 0 : Hint. The function v .x; t/ D u .ax; t/ satisfies the equation v t  v D f .ax; t/ and v .x; 0/ D g0 .ax/ : We obtain Z 2 1  jxyj 2 t dy 4a .y/ g e u .x; t/ D  n=2 n 0 R 4a2  t Z tZ 2 f .y; s/  jxyj 4a2 .ts/ dy ds; e C n=2  0 Rn 4a2  .t  s/ or in the convolution form e N / .x; t/ .t > 0/ ; u .x; t/ D .g0 N .t // .x/ C .f where the fundamental solution of the equation u t  a2 u D 0 is N .x; t/ D 

8.

9.

Solve the problem ²

1 4a2  t

n=2



e

jxj2 4a2 t

.t > 0/:

u t  a2 uxx D sin x; x 2 R; t > 0 u .x; 0/ D 0; x 2 R:

Solve the Cauchy problem (5.18) for n D 1 and:   (a) g0 D 1= 1 C x 2 ; g1 D 0   (b) g0 D 0; g1 D 1= 1 C x 2   (c) g0 D g1 D 1= 1 C x 2 (d) g0 D jxj2 ; g1 D 0: Hint. Use D’Alembert’s formula.

128

10.

Chapter 5 The Cauchy Problem

Solve the Cauchy problem for the nonhomogeneous wave equation if n D 1 and: (a) g0 D g1 D 0; f D H .t / sin t (b) g0 D 0; g1 D ex ; f D H .t/ .x C t / :

11.

Solve the initial value problem ² x 2 R; t > 0 u t t  a2 uxx D sin x; u .x; 0/ D u t .x; 0/ D 0; x 2 R:

Part II

Modern Theory

Chapter 6

Distributions

In Part I several notions of weak or generalized solutions have been introduced for a number of boundary value problems for partial differential equations. These allowed us to speak about solution even when data were less smooth, as is the case most frequently in mathematical physics. They also allowed us to apply abstract results of functional analysis for proving the existence of solutions. Two questions remain: what is the sense that the weak solution satisfies the partial differential equation and in what case the weak solution is a classical one. The answer to the first question is that the weak solution satisfies the equation in the sense of the theory of distributions. To give an answer to the second question, we shall need regularity results for weak solutions, which are mainly based on embedding theorems for Sobolev spaces. As we shall see in this Part II, the theory of distributions and, in particular, the theory of Sobolev spaces, offers the most natural framework for the study of partial differential equations.

6.1

The Fundamental Spaces of the Theory of Distributions

Let ˝  Rn be an open set. For a function ' W ˝ ! R; one defines the support as the closure in ˝ of the set of points where ' does not vanish, i.e. supp ' D ¹x 2 ˝ W ' .x/ ¤ 0º (closure in ˝). Thus, by definition, supp ' is a subset of ˝; closed in ˝; but not necessarily closed in Rn : Using the consecrated notations of the theory of distributions, we shall design by D .˝/ and E .˝/ the linear spaces C01 .˝/ and C 1 .˝/ ; respectively, where C01 .˝/ D ¹' 2 C 1 .˝/ W supp ' is a compact subset of

Rn º :

Hence, a function ' 2 C 1 .˝/ belongs to C01 .˝/ if and only if a compact subset K of Rn exists with K  ˝ and ' D 0 on ˝ n K: The space E .˝/ is endowed with a convergence as follows: 'k !' in E .˝/ if and only if, for each multi-index ˛ and any compact K  ˝;

132

Chapter 6 Distributions

the sequence D ˛ 'k converges to D ˛ ' uniformly on K: Notice that this convergence comes from the endowment of E .˝/ with the family of seminorms ® pm;K W K is compact, K  ˝ , m 2 Nº ; where pm;K .'/ D

X j˛jm

sup x2K

jD ˛ ' .x/j ;

' 2 E .˝/ :

In D .˝/ we define the convergence as follows: 'k ! ' in D .˝/ if and only if there exists a compact K  ˝ such that supp 'k  K for all k; supp '  K and D ˛ 'k converges to D ˛ ' uniformly on K for every ˛ 2 Nn : Clearly the following inclusion: D .˝/  E .˝/ holds both algebraically, as linear spaces, and topologically, in the sense that 'k ! ' in D .˝/ implies 'k ! ' in E .˝/ : An important role in the theory of distributions, as well as in the theory of Lp spaces is played by the function ´ 1  n jxj2 1 for jxj < 1 e 2 D R ; .x/ D 0 for jxj  1: Obviously supp D B 1 .0/ : We shall denote by 1 ; the function c ; where the normalizing constant c is taken such that Z .x/ dx D 1: c jxj1

Also, they are important those functions obtained from 1 by a change of variables which makes the support become the closed ball of radius 1=k ; namely k .x/ D k n 1 .kx/ : One has k 2 D



R

n



Z ; supp k D B 1=k .0/ and

jxj1=k

k .x/ dx D 1:

The sequence of functions k ; k 2 N n ¹0º ; is called, from reasons that we are going to see immediately, a regularizing sequence. The spaces D .˝/ ; E .˝/ and S .Rn / are called the fundamental spaces of the theory of distributions. Their elements are also named test functions.

Section 6.2 Distributions: Examples; Operations with Distributions

6.2

133

Distributions: Examples; Operations with Distributions

Definition 6.1. A distribution on an open set ˝  Rn is a linear functional u W D .˝/ ! R; which is continuous in the sense that 'k ! ' in D .˝/ implies u .'k / ! u .'/ in

R:

(6.1)

The set of all distributions on ˝ is denoted by D 0 .˝/ and is naturally organized as a linear space. Often, the value (or the action) u .'/ of a distribution u 2 D 0 .˝/ on a test function ' 2 D .˝/ will be denoted by .u; '/ : The distribution space D 0 .˝/ is also endowed with a convergence; thus, we say that uk ! u in D 0 .˝/ if .uk ; '/ ! .u; '/ for all ' 2 D .˝/ : 0 This convergence ® comes from the¯ endowment of space D .˝/ with the family of seminorms p' W ' 2 D .˝/ ; where

p' .u/ D j.u; '/j ;

6.2.1

u 2 D 0 .˝/ :

Regular Distributions

Let f 2 L1loc .˝/ : One may associate to function f a distribution uf defined by Z   f .x/ ' .x/ dx; ' 2 D .˝/ : uf ; ' D ˝

Since, whenever for two functions f; g 2 L1loc .˝/ with uf D ug ; one has f D g a.e. on ˝ (see Brezis [5, Corollary 4.24]), then we may identify distribution uf to function f: Thus, L1loc .˝/ can be viewed as a linear subspace of D 0 .˝/ : A distribution u is said to be regular or of function type if there exists a locally integrable function on ˝ .i.e. f 2 L1loc .˝//; such that u D uf : Similarly, a distribution u is of class C m if u D uf for some function f 2 C m .˝/ : Thus we have the following inclusions of linear spaces: D .˝/  C m .˝/  L1loc .˝/  D 0 .˝/ ; Also

D .˝/  Lp .˝/  L1loc .˝/  D 0 .˝/ ;

m 2 N: 1  p  1:

134

Chapter 6 Distributions

For example, the Heaviside function H W R ! R; H .x/ D

²

0; x  0 1; x > 0

is a regular distribution on R; called the Heaviside distribution. Its action on a test function ' 2 D .R/ is given by Z Z 1 .H; '/ D ' .x/ dx D ' .x/ dx: Œ0;1/ \ supp '

0

Those distributions which are not regular are called singular distributions.

6.2.2

The Dirac Distribution

To each point x0 2 ˝; one may associate a distribution ıx0 2 D 0 .˝/ defined by .ıx0 ; '/ D ' .x0 / ; ' 2 D .˝/ ; called the Dirac distribution at x0 : For x0 D 0; it is simply denoted by ı: One can easily prove that functional ıx0 is really a distribution on ˝; and that (see Problems) it is singular. Notice that the Dirac distribution on Rn is the limit in D 0 .Rn / of a sequence of test functions. More exactly, k ! ı in D 0 .Rn / as k ! 1: Indeed, if ' 2 D .Rn / ; then Z . k ; '/ D k .x/ ' .x/ dx jxj k1

D ' . k /

Z jxj k1

k .x/ dx

D ' . k / ! ' .0/ : We shall see later that any distribution on ˝ is the limit in D 0 .˝/ of a sequence of test functions in D .˝/ ; that is, D .˝/ is sequentially dense in D 0 .˝/ :

6.2.3

Differentiation

Standard operators and operations in D .˝/ can be extended to D 0 .˝/ : Such are the translation and the partial derivative operators. The recipe of these extensions is simple and consists in passing the operator from distribution onto the test function.

Section 6.2 Distributions: Examples; Operations with Distributions

135

For instance, if u 2 D 0 .Rn / and y 2 Rn ; then the translate of u by y; denoted y u; is defined as follows: if u is a regular distribution, i.e. u 2 L1loc .Rn / ; then Z Z u .x  y/ ' .x/ dx D u .x/ ' .x C y/ dx D .u; y '/; .y u; '/ D

Rn

where

Rn

  y ' .x/ D ' .x C y/ :

This suggests that for any distribution u 2 D 0 .Rn / ; the translate by y be defined by   (6.2) .y u; '/ D .u; y '/; ' 2 D Rn :

To define the derivative @u=@xj of a distribution u 2 D 0 .Rn / ; we start from the equality       1

1

hej u  u ; ' D u; hej '  ' ; h h

where by ej we have denoted the vector in Rn with all components zero except the j th component which is equal to 1: Since as h ! 0; the test function in the right-hand side tends to @'=@xj ; then it is natural to define the derivative @u=@xj by       @u @' ; ' 2 D Rn : ; ' D  u; @xj @xj This motivates the following general definition. Definition 6.2. For any distribution u 2 D 0 .˝/ and any multi-index ˛ 2 Nn ; the derivative D ˛ u is a distribution on ˝ given by     ˛ D u; ' D .1/j˛j u; D ˛ ' ; ' 2 D .˝/ : The next proposition shows that the distribution derivative is a generalization of the classical derivative. Proposition 6.1. If f 2 C m .˝/ ; then for every ˛ 2 have uD ˛ f D D ˛ uf :

Nn

with j˛j  m; we

Proof. It is sufficient to prove the formula for D ˛ D @=@xj : This is left to the reader as a good exercise.

136

Chapter 6 Distributions

For example, the derivative of the Heaviside distribution H is the Dirac distribution, i.e. H 0 D ı: Indeed, for any test function ' 2 D .R/ ; we have Z 1  0    H ; ' D  H; ' 0 D  ' 0 .x/ dx 0

D ' .0/  ' .1/ D ' .0/ D .ı; '/ : Therefore, even if function H is not differentiable on R in the classical sense, it is, however, differentiable in the sense of distributions. As a conclusion, any distribution is differentiable as many times as we wish.

6.2.4

Multiplication by a Smooth Function

Let u 2 D 0 .˝/ be a distribution and let a W ˝ ! R be an infinitely differentiable function. Then the product au is a distribution on ˝; given by .au; '/ D .u; a'/ ;

' 2 D .˝/ :

It is an easy exercise to check the product rule for the derivative of au; namely @ .au/ @a @u D uCa : @xj @xj @xj

(6.3)

The fundamental property of distributions of being infinitely many times differentiable, together with the operation of multiplication by smooth functions make possible the extension of the classical differential operators to the larger space of distributions. Thus we can define the operator L D L .D/ W D 0 .˝/ ! D 0 .˝/ ; X a˛ .x/ D ˛ u; Lu D j˛jm

where a˛ 2 C 1 .˝/ : The action of distribution Lu over the test function ' is the same as the action of u over the test function L .D/ '; where L .D/ is the transpose operator given by X .1/j˛j a˛ .x/ D ˛ : L .D/ D j˛jm

Thus, for every u 2 D 0 .˝/ and ' 2 D .˝/ ; we have .L .D/ u; '/ D .u; L .D/ '/ : In particular, the following formula holds for the Laplacian: .u; '/ D .u; '/ ;

' 2 D .˝/ :

Section 6.2 Distributions: Examples; Operations with Distributions

6.2.5

137

Composition with a Smooth Function

If  W ˝ ! ˝1 is a diffeomorphism of class C 1 and u 2 D .˝1 / ; then u ı  2 D .˝/  D 0 .˝/ and for each ' 2 D .˝/ ; we have Z .u ı ; '/ D u . .y// ' .y/ dy Z˝ ˇ  ˇ u .x/ ' 1 .x/ ˇdet D1 .x/ˇ dx; D ˝1

where D1 .x/ is the Jacobian matrix of the transformation y D 1 .x/ : This motivates to define for a general distribution u 2 D 0 .˝1 / ; the distribution u ı  2 D 0 .˝/ by ˇ ˇ .u ı ; '/ D .u; ˇdet D1 ˇ ' ı 1 /; ' 2 D .˝/ : (6.4) For example, if ˝ D ˝1 D Rn and  .x/ D x  y; where y 2 Rn is a given point, then u ı  is the translation y u and (6.4) becomes (6.2). If  .x/ D x; where ¤ 0; then u ı  is the dilatation of u by ; and for every ' 2 D .Rn / ;    .u . x/ ; ' .x// D j jn u .x/ ; ' 1 x : In this formula, as well as in others which follow, the notation u .x/ is used to show that the distribution u acts over test functions of variable x: In what follows, we shall denote by Ru the dilatation of u by D 1; and we shall call it the reflection of u: Hence   .Ru; '/ D .u; ' .x//; ' 2 D Rn :

6.2.6

Convolution

In Chapter 5, the convolution of two functions on Z Z .f g/ .x/ D f .x  y/ g .y/ dy D

Rn

Rn

Rn

have been defined by

f .y/ g .x  y/ dy

(6.5)

and a basic result, Theorem 5.1, about its existence has been proved. From this theorem, in particular, we have that the convolution of any two test functions f; g 2 D .Rn / exists and f g 2 D .Rn / I viewed as a distribution on Rn ; the function f g acts over a test function ' as follows: .f g; '/ D .g; .Rf / '/ : This suggests to extend the definition of the convolution to the case where g is a general distribution on Rn : Thus, the convolution of a function

138

Chapter 6 Distributions

f 2 D .Rn / with a distribution u 2 D 0 .Rn / is the distribution denoted f u; defined by   .f u; '/ D .u; .Rf / '/ ; ' 2 D Rn : For an example, let us compute f ı: One has .f ı; '/ D .ı; .Rf / '/ D ..Rf / '/ .0/ Z f .x/ ' .x/ dx D .f; '/ : D

Rn

Hence f ı D f: An easy exercise yields to the differentiation rule of the convolution:   D ˛ .f u/ D f D ˛ u D D ˛ f u: The next result shows that the convolution of a function in D .Rn / with a distribution is always a regular distribution of class C 1 : Proposition 6.2. If f 2 D .Rn / and u 2 D 0 .Rn / ; then f u 2 C 1 .Rn / and .f u/ .x/ D .u; x .Rf // ; x 2 Rn : Proof. Let g .x/ D .u; x .Rf // : If xk ! x; then xk .Rf / ! x .Rf / in For a given number h > 0; deD .Rn / : As a result, g is a continuous function.   note by jh the divided difference operator hej  I = h which approximates the differential operator @=@xj : Then   jh g D u; x jh .Rf / and x jh .Rf / ! x

@ .Rf / @xj

as h ! 0:

It follows that g 2 C 1 .Rn / and

  @g @ .Rf / : D u; x @xj @xj

Step by step, one can prove that g 2 C 1 .Rn / and   D ˛ g D u; x D ˛ .Rf / ; ˛ 2 Nn :

Section 6.2 Distributions: Examples; Operations with Distributions

139

Next we show that the distribution associated to function g coincides with f u: For this, we represent .g; '/ as limit of a sequence of Riemann sums, as follows: X ˛    .g; '/ D lim ' u; ˛=k .Rf / k n k k!1 ˛2Zn  ˛  X  D lim u; ' ˛=k .Rf / k n : k k!1 ˛2Zn Then one shows that X ˛    ' k n ˛=k .Rf / ! .Rf / ' in D Rn ; k ˛2Zn whence, finally, one finds the equality .g; '/ D .u; .Rf / '/ D .f u; '/ : Proposition 6.3. Let  2 D .Rn / be such that 0    1;  .x/ D 1 for jxj  1 and  D 0 for jxj  2: For each k0 2 nN n ¹0º ; consider  1 .x/ k .x/ WD  k x : Then, for every distribution u 2 D .R / ; we have   k u ! u and k . k u/ ! u in D 0 Rn as k ! 1: Consequently, each distribution on Rn is limit in D 0 .Rn / of a sequence of test functions in D .Rn / ; i.e. D .Rn / is sequentially dense in D 0 .Rn / : Proof. For every ' 2 D .Rn / ; we have

. k u; '/ D .u; .R k / '/ : Furthermore, direct calculation of classical analysis yields .R k / ' ! ' in D .Rn / : The second convergence can be proved similarly. Finally the last assertion is based on k . k u/ 2 D .Rn / :

6.2.7

Distributions of Compact Support

Let u 2 D 0 .˝/ and ! be an open subset of ˝: We say that u is equal to zero on !; if .u; '/ D 0 for all ' 2 D .!/ : By the support of the distribution u 2 D 0 .˝/ ; we mean the set supp u WD ˝ n ¹x 2 ˝ W u is equal to zero on a neighborhood of xº : Clearly, supp u is closed in ˝:

140

Chapter 6 Distributions

It is immediately seen that supp ıx0 D ¹x0 º ; supp .D ˛ u/  supp u; and if u 2 L1loc .˝/ ; then supp u D ¹x 2 ˝ W u .x/ ¤ 0º

.closure in ˝/:

Proposition 6.4. If u 2 D 0 .˝/ and ' 2 D .˝/ have disjoint supports, then .u; '/ D 0: Proof. For each point y 2 supp ' we choose an open !y  ˝ such that y 2 !y and u is equal to zero on !y : Then we choose a nonnegative function hy 2 D .˝/ with supp hy  !y and hy .y/ > 0: The collection of sets  ®  ¯ hy > 0 WD x 2 ˝ W hy .x/ > 0 .y 2 supp '/ is an open cover of the compact set supp ': Let h1 ; h2 ; :::; hm be the corresponding functions of a finite subcover. For each j 2 ¹1; 2; :::; mº ; define the function [  'j D 'hj = .h1 C h2 C ::: C hm / on hj > 0 'j

D 0 otherwise.

  We have 'jP2 D .˝/ and supp 'j  supp hj  !j : Hence u; 'j D 0; and since ' D 'j ; we infer that .u; '/ D 0: The next proposition allows us to identify the set of all distributions on ˝ of compact support, with E 0 .˝/ ; the set of all continuous linear functionals on E .˝/ : Proposition 6.5. (a) The restriction to D .˝/ of a continuous linear functional on E .˝/ ; is a distribution on ˝ of compact support. (b) Each distribution on ˝ of compact support extends uniquely to a continuous linear functional on E .˝/ : Proof. (a) The fact that the restriction to D .˝/ of any continuous linear functional on E .˝/ is a distribution on ˝ is a consequence of the algebraictopological inclusion D .˝/  E .˝/ : To show that the support of such a distribution is compact, it is sufficient to remark that a linear functional u W E .˝/ ! R is continuous if and only if there are m D m .u/ 2 N and c D c .u/  0 such that, for every ' 2 E .˝/ ; X jD ˛ 'j ; j.u; '/j  c j˛jm

max K m

Section 6.2 Distributions: Examples; Operations with Distributions

141

S where K1  K2  ::: are compact subsets of ˝ with ˝ D Kj .then supp u  Km /: The sufficiency of the condition is immediate. For necessity, assume the contrary. Then, for each m D c 2 N n ¹0º ; there would exist a 'm 2 E .˝/ with X j.u; 'm /j > m jD ˛ 'm j : j˛jm

max K m

We may assume that the right-hand side of the above inequality is equal to 1: Then one can see that 'm ! 0 in E .˝/ as m ! 1; while j.u; 'm /j > 1; which contradicts the continuity of u: (b) Choose a function 2 D .˝/ with D 1 in a neighborhood of supp u: Then, according to Proposition 6.4, for every ' 2 D .˝/ ; we have .u; '/ D .u; '/ : It is clear that the functional e u W E .˝/ ! R; .e u; '/ D .u; '/ ; is an extension of u to E .˝/ : Finally the uniqueness of the extension is a consequence of the density of D .˝/ in E .˝/ : The next result shows that distributions of compact support extend to distributions on Rn : In addition, the extension by zero outside ˝ is unique. Proposition 6.6. Let u 2 E 0 .˝/ : Then there is a unique distribution e u 2 0 n u on D .˝/ and supp e u  ˝: D .R / with u D e Proof. Let e u be such an extension. Consider a function 2 D .˝/ with supp  ˝ and D 1 in a neighborhood of the compact supp u: Then, for every ' 2 D .Rn / ; .e u; '/ D .e u; '/ D .u; '/ ; whence the uniqueness of the extension and its expression. Proposition 6.7. The space D .˝/ is sequentially dense in D 0 .˝/ : Proof.SChoose a sequence of open bounded sets ˝k with ˝ k  ˝kC1 and ˝ D ˝k ; and a sequence of functions k 2 D .˝kC1 / with k D 1 on k

˝k : Let u 2 D 0 .˝/ : One can prove that k u ! u in D 0 .˝/ : Denote by uk 2 E 0 .Rn / ; the unique extension by zero of k u: Choose a sequence jk 2 N n ¹0º with 1=jk < dist .˝k ; ˝kC1 / : Then jk uk 2 D .˝kC1 / and jk uk ! u in D 0 .Rn / and therefore in D 0 .˝/ : Note that for u 2 D 0 .Rn / and v 2 E 0 .Rn / ; the convolution u v D v u can be defined by   .u v; '/ D .u; ' .Rv// ; ' 2 D Rn :

142

6.2.8

Chapter 6 Distributions

Weyl’s Lemma

We have the following regularity result for those functions which satisfy Laplace’s equation in the sense of distributions. Proposition 6.8. If u 2 L1loc .˝/ satisfies the equation u D 0 in D 0 .˝/ ; then u 2 C 2 .˝/ : ® ¯ Proof. For each k 2 N n ¹0º ; consider the set ˝k WD x W B 1=k .x/  ˝ : Then, for any ' 2 C01 .˝k / ; k ' 2 D .˝/ and k u 2 C 1 .˝k / : We have . . k u/ ; '/ D . k u; '/ D .u; .R k / '/ D .u;  ..R k / '// D .u; .R k / '/ D 0: As a result,  . k u/ D 0 on ˝k : Then, from the mean theorem of harmonic functions, we have Z n . k u/ .x C y/ dy . k u/ .x/ D !n r n Br .0/ for every x 2 ˝k and small enough r: From Proposition 5.1 we know that k u ! u in L1 .˝ 0 / as k ! 1; for every ˝ 0 with ˝ 0  ˝: Then we may infer that k u ! u uniformly on any compact set included in ˝: Now Corollary 3.10 guarantees that u is an harmonic function on ˝:

6.3

The Fourier Transform of Tempered Distributions

In Chapter 5, we have defined the Fourier transform on the space L1 .Rn / of complex-valued functions by Z 1 T Œf  .y/ D f .x/ eixy dx; y 2 Rn ; .2/n=2 Rn P where, recall, x  y D jnD1 xj yj : Also, we proved that T is an automorphism of the space S: Recall the formulas: T Œf g D .2/n=2 T Œf  T Œg ; h i D ˇ T Œf  D T .ix/ˇ f .x/ ; i h T D ˇ f .y/ D .iy/ˇ T Œf  .y/ ;

(6.6) (6.7) (6.8)

Section 6.3 The Fourier Transform of Tempered Distributions

143

T Œf .x  x0 / .y/ D eiyx0 T Œf .x/ .y/ ;   T Œf . x/ .y/ D j jn T Œf .x/ 1 y ; T

1

(6.9) (6.10)

Œf  .y/ D T Œf .x/ .y/ D T Œf  .y/ :

(6.11)

In what follows we shall extend the Fourier transform to the larger space S 0 .Rn / of tempered distributions. In particular, we shall analyze the Fourier transform on L2 .Rn / :

6.3.1

The Fourier Transform on S 0 .Rn /

Definition 6.3. A tempered distribution on u W S .Rn / ! C:

Rn

is a continuous linear functional

We denote by S 0 D S 0 .Rn / the space of all tempered distributions on Rn and we say that uk ! u in S 0 as k ! 1; provided that .uk ; '/ ! .u; '/ for every ' 2 S: Since D .Rn /  S .Rn /  E .Rn / algebraically and topologically, then the following inclusions hold       E 0 Rn  S 0 Rn  D 0 Rn both algebraically and topologically. Also, from S  Lq .Rn / algebraically and topologically, we deduce that     Lp Rn  S 0 Rn algebraically and topologically, for 1  p  1: Notice that any constant function c belongs to S 0 ; but c … S if c ¤ 0: The Fourier transform operator T extends to S 0 as a continuous linear invertible operator (automorphism of S 0 /; in the following way: T W S 0 ! S 0; .T Œu ; '/ D .u; T Œ'/ ;

' 2 S:

Similarly, the inverse Fourier transform operator on S 0 is defined by 

T 1 Œu ; '

1 0 ! S 0; T W S 1  D u; T Œ' ;

' 2 S:

For an example, let us compute the Fourier transform of the Dirac distribution ı: For every ' 2 S; we have .T Œı ; '/ D .ı; T Œ'/ D T Œ' .0/ Z 1 ' .x/ eix0 dx D D n=2 n R .2/

1 .2/n=2

! ;' :

144

Hence

Chapter 6 Distributions

T Œı D .2/n=2 and T 1 Œ1 D .2/n=2 ı:

We advise the reader to prove that (6.7)–(6.11) remain true in S 0 ; and that (6.6) also holds for f 2 D .Rn / and g 2 S 0 .Rn / :

6.3.2

The Fourier Transform on L2 .Rn /

We already know that S  L2 .Rn /  S 0 and that the Fourier transform operator T is an automorphism of S as well as of S 0 : A natural question arises: what can be said about the Fourier transform on L2 .Rn /‹ The answer is given by the next proposition. Proposition 6.9 (Plancherel). The Fourier transform is an automorphism of L2 .Rn / : In addition   jujL2 D jT ŒujL2 ; u 2 L2 Rn ; i.e. T is an isometry of L2 .Rn / : Proof. Let u 2 L2 .Rn / : We have T Œu 2 S 0 and using (5.4), j.T Œu ; '/j D j.u; T Œ'/j D j.u; T Œ'/L2 j  jujL2 jT Œ'jL2 D jujL2 j'jL2 ;

' 2 S:

This inequality extends by density to all functions ' 2 L2 .Rn / : It follows that T Œu is a continuous linear functional on the Hilbert space L2 .Rn / : Consequently,   T Œu 2 L2 Rn and jT ŒujL2  jujL2 : The converse inequality is obtained by replacing u with T 1 Œu :

6.3.3

Convolution in S 0

In Section 5.1 we have seen that if f; g 2 S; then f g 2 S and i h f g D T 1 .2/n=2 T Œf  T Œg :

(6.12)

Also, in the previous section we talked about the convolution of a test function with a distribution on Rn : Using the same formula we can define the convolution of a function from S with a tempered distribution as follows:

Section 6.4 Problems

145

If f 2 S and g 2 S 0 ; then by the convolution f g we mean the tempered distribution given by .f g; '/ D .g; .Rf / '/ ;

' 2 S:

Formula (6.12) remains true in this more general case (Exercise). The reader who is interested in broadening his or her knowledge in distributions and Fourier transform may consult Gelfand–Shilov [16], Schwartz [44] and Vladimirov [54].

6.4

Problems

1.

A number of properties of the distributions have been given without explanation; their proofs are simple and in general require only the manipulation of definitions. They represent a first class of exercises in the theory of distributions. For instance, prove that the following functionals are distributions (i.e. are linear and continuous): ( a) D ˛ u W D .˝/ ! R; .D ˛ u; '/ D .1/j˛j .u; D ˛ '/ ( b) ı W D .Rn / ! R; .ı; '/ D ' .0/ :

2.

Prove that the Dirac distribution ı is not regular.

Hint. Assume it is regular. Then there is a function f 2 L1loc .Rn / with Z   .ı; '/ D f .x/ ' .x/ dx D ' .0/ for all ' 2 D Rn :

Rn

1

Choose ' .x/ WD 'k .x/ D .kx/ ; where .x/ D e jxj2 1 for jxj < 1 and .x/ D 0 for jxj  1; and obtain Z 1 .ı; 'k / D f .x/ e k2 jxj2 1 dx D 'k .0/ D e1 : jxj 0; x2 > 0º: Prove: ( a) aı D a .0/ ı for every a 2 C 1 .Rn / ( b) x k ı .m/ D 0 for m D 0; 1; :::; k  1: Let u; v 2 D 0 .Rn / : If ' .Rv/ 2 D .Rn / for every ' 2 D .Rn / ; then the functional   ' 2 D Rn 7! .u; ' .Rv// is a distribution on denoted u v:

Rn ;

called the convolution of distributions u; v and

Prove that if the convolution u v exists, then the convolutions D ˛ u v; u D ˛ v also exist and D ˛ .u v/ D D ˛ u v D u D ˛ v:

8.

P ˛ be a differential operator with constant Let L .D/ D j˛jm a˛ D coefficients. By a fundamental solution of the equation L .D/ u D 0; we mean a distribution N 2 D 0 .Rn / such that L .D/ N D ı:

Section 6.4 Problems

147

Prove that if N is a fundamental solution of the equation L .D/ D 0; f 2 D 0 .Rn / and the convolution N f exists, then L .D/ .N f / D f; i.e. u D N f is a solution of the nonhomogeneous equation L .D/ u D f: 9.

Show that the ordinary differential equation with constant coefficients Lu WD u.m/ C

m1 X

ak u.k/ D 0

(6.13)

kD0

admits the fundamental solution N .x/ D z .x/ H .x/ ; where H is the Heaviside function and z is the solution of the Cauchy problem ² Lz D 0 z .0/ D z 0 .0/ D ::: D z .m2/ D 0; z .m1/ D 1: Hint. One has .zH /0 D z 0 H CzH 0 D z 0 H Czı D z 0 H Cz .0/ ı D z 0 H: Similarly, .zH /00 D z 00 H; :::; .zH /.m1/ D z .m1/ H and .zH /.m/ D z .m/ H C ı: It follows that L .zH / D L .z/ H C ı D ı: 10.

Prove that the solutions in D 0 .a; b/ ; 1  a < b  1; of the homogeneous equation (6.13) coincide with the classical solutions. Hint. It suffices to discuss the case Lu WD u0 I in the general case rewrite 0 the equation 0 as a differential system of the form v C Av D 0; and then xA D 0: as v e Equality u0 D 0 is equivalent to .u; ' 0 / D 0 for all ' 2 D .a; b/ : Rb Hence .u; / D 0 for every 2 D .a; b/ satisfying a dx D 0: Rb Choose  2 D .a; b/ with a dx D 1: Then ! Z Z Z b

.u; '/ D u; 

a

' dx C '  

b

a

' dx

b

D .u; /

a

' dx D .c; '/ :

Hence u D c: 11.

Solve in D 0 .R/ the equation u00  a2 u D ıx0 : Hint. Use Problems 8, 9 and 10. It follows that u D C1 sinh ax CC2 cosh ax CN ıx0 D C1 sinh ax CC2 cosh ax CN .x  x0 / ; where N .x/ D a1 H .x/ sinh ax:

148

Chapter 6 Distributions

12 . Prove that function N given by (3.1) is a fundamental solution of Laplace’s equation. 13 . Prove that the function N .x; t/ D

H .t / .4 t /n=2

e

jxj2 4t

is a fundamentalsolution  of the heat equation Lu WD u t  u D 0; i.e. LN D ı in D 0 RnC1 : 14 . Show that a fundamental solution of the wave equation Lu WD u t t u D 0 is  H .t/ 1 sin jyj t .x/ : T N .x; t/ D jyj .2/n=2 For n D 1; N .x; t/ D 12 H .t  jxj/ I for n D 2; N .x; t/ D

H .t jxj/ 2

p

t 2 jxj2

I

.t / for n D 3; N .x; t/ D H 4 t ı† t .x/ ; where †R is the sphere of radius 3  at  distribution on R ; defined by R centered R the origin and ı†R is 3the ı†R ; ' D †R ' .x/ dx; ' 2 D R :

Hint. For solving Problems 12-14 we refer the reader to Vladimirov [54]. 15.

Compute the Fourier transform of distribution u 2 S 0 for: (a) u D ıx0 , (b) u D x ˛ , (c) u D D ˛ ı:

16.

Prove that for any f 2 S 0 ; there exists a unique solution u 2 S 0 of the equation u  u D f; namely  1 T Œf  : u D T 1 1 C jxj2 In particular, if f 2 S; then u 2 S:

Chapter 7

Sobolev Spaces

From Part I, we already know the Sobolev spaces H01 .˝/ and H 1 .˝/ ; which were attached to the energetic norms corresponding to the Dirichlet and Neumann problems, under homogeneous boundary conditions. More exactly, for a bounded open subset ˝ of Rn ; the space H01 .˝/ has been defined as the completion   of C01 ˝ with respect to the corresponding energetic norm of the Dirichlet problem; also H 1 .˝/ has been defined as the completion of the space ¹C 1 .˝/ W u; @u=@xj 2 L2 .˝/ ; j D 1; 2; :::; nº with respect to the energetic norm of the Neumann problem. Hence, in both cases, the Sobolev space has been defined, by completion, starting from a subspace of differentiable functions in the classical sense. In this chapter, we shall present an other way of defining Sobolev spaces, starting this time from a superspace and using derivatives in the sense of distributions. Thus, for example, H 1 .˝/ will be the space of all functions u in L2 .˝/ whose derivatives @u=@xj ; j D 1; 2; :::; n; in the sense of distributions, also belong to L2 .˝/ ; that is, are regular distributions represented by functions from L2 .˝/ :

7.1

The Sobolev Spaces H m .˝/

Definition 7.1. Let ˝  Rn be an open set and m 2 N: The Sobolev space H m .˝/ is defined as follows ® ¯ H m .˝/ D u 2 L2 .˝/ W D ˛ u 2 L2 .˝/ for j˛j  m : We note that in the above definition, the derivative D ˛ u is in the sense of distributions. Therefore, the elements of H m .˝/ are those functions from L2 .˝/ whose derivatives in the sense of distributions, until order m; also belong to L2 .˝/ : If u 2 H m .˝/ and ˛ is a multi-index with j˛j  m; then for every ' 2 D .˝/ ; we have Z Z     ˛   ˛ j˛j u .x/ D ˛ ' .x/ dx: (7.1) D u .x/ ' .x/ dx D .1/ D u; ' D ˝

˝

150

Chapter 7 Sobolev Spaces

The space H m .˝/ is a linear subspace of L2 .˝/ : It is endowed with the inner product X   .u; v/H m D (7.2) D ˛ u; D ˛ v L2 j˛jm

and the corresponding norm 0 jujH m D @

X

1 12 2 A : jD ˛ ujL 2

j˛jm

An important particular case is m D 1; when ² ³ @u 1 2 2 2 L .˝/ ; j D 1; 2; :::; n H .˝/ D u 2 L .˝/ W @xj and inner product and norm are: .u; v/H 1 D .u; v/L2

 Z n  X @u @v .uv C ru  rv/ dx; C ; D @xk @xk L2 ˝ kD1

jujH 1 D

Z  ˝

2

u C jruj

2



dx

 12 :

Proposition 7.1. The space H m .˝/ endowed with inner product .7:2/ is a Hillbert space. Proof. The result is a direct consequence of the completeness of L2 .˝/ : Indeed, if .uk / is a Cauchy sequence in H m .˝/ ; then in view of the expression of the norm of H m .˝/ ; we have that sequences .uk / ; .D ˛ uk / .j˛j  m/ are Cauchy in L2 .˝/ : Then there are u; u˛ 2 L2 .˝/ .j˛j  m/ such that uk ! u and D ˛ uk ! u˛ in L2 .˝/ .j˛j  m/: It remains to show that u˛ D D ˛ u .j˛j  m/ : From (7.1), we have     ˛ D uk ; ' L2 D .1/j˛j uk ; D ˛ ' L2 ; whence we deduce   .u˛ ; '/L2 D .1/j˛j u; D ˛ ' L2 ; that is,

  .u˛ ; '/ D .1/j˛j u; D ˛ ' :

Thus u˛ D D ˛ u and so uk ! u in H m .˝/ ; as we wished.

Section 7.1 The Sobolev Spaces H m .˝/

151

Notice that the following inclusions hold: C01 .˝/  H m .˝/  L2 .˝/  L1loc .˝/  D 0 .˝/ : The next result gives a characterization of the elements of H m .˝/ : Theorem 7.1. Let u 2 L2 .˝/ : The necessary and sufficient condition for that u belongs to H m .˝/ is to exist a constant C such that ˇ ˇZ ˇ ˇ ˇ u D ˛ ' dx ˇ  C j'j 2 for all ' 2 C 1 .˝/ ; ˛ 2 Nn ; j˛j  m: (7.3) L 0 ˇ ˇ ˝

Proof. Necessity: If u 2 H m .˝/ ; then according to Hölder’s inequality, one has ˇ ˇ ˇ ˇZ Z ˇ ˇ ˇ ˇ ˛ ˇ u D ˛ ' dx ˇ D ˇ.1/j˛j ' D u dx ˇˇ  jujH m j'jL2 : ˇ ˇ ˇ ˝

˝

Sufficiency: Let ˛ 2 Nn : Inequality (7.3) shows that the linear functional R ' 7! ˝ uD ˛ ' dx is continuous on the dense subspace C01 .˝/ of L2 .˝/ : Based on the Hahn–Banach theorem, it can be extended to a continuous linear functional F on L2 .˝/ : Now, from the Riesz representation theorem, there exists a v˛ 2 L2 .˝/ such that .F; '/ D .v˛ ; '/L2 for every ' 2 L2 .˝/ : In particular,

Z ˝

u D ˛ ' dx D

Z ˝

v˛ ' d x

for all ' 2 D .˝/ ;

whence D ˛ u D .1/j˛j v˛ 2 L2 .˝/ : Hence u 2 H m .˝/ : Remark 7.1. According to Theorem 7.1, a function u from L2 .˝/ belongs to H 1 .˝/ if and only if there is a constant C > 0 with ˇ ˇZ ˇ ˇ @' ˇ u dx ˇˇ  C j'jL2 for all ' 2 C01 .˝/ ; j D 1; 2; :::; n: ˇ ˝ @xj We note that, more generally, for 1  p  1; one defines the Sobolev space W m;p .˝/ by W m;p .˝/ D ¹u 2 Lp .˝/ W D ˛ u 2 Lp .˝/ for j˛j  mº : This space endowed with norm jujW m;p D

X j˛jm

jD ˛ ujLp ;

152

Chapter 7 Sobolev Spaces

or with the equivalent norm

P

p

˛ j˛jm jD ujLp

1=p ; is a Banach space. Clearly

W m;2 .˝/ D H m .˝/ : Remark 7.2. According to a result by Meyers and Serrin (see Adams [1, p. 52]), for 1  p < 1; the space W m;p .˝/ coincides with the completion of ¹u 2 C m .˝/ W D ˛ u 2 Lp .˝/ for j˛j  mº with respect the norm jujm;p D P ˛ j˛jm jD ujLp : Thus, for m D 1; Definition 7.1 is equivalent to the definition of H 1 .˝/ given in Section 3.13.

7.2

The Extension Operator

It is often useful that a certain property of the elements of H m .˝/ be first established for H m .Rn / : Thus will be important to define an extension operator u 2 H m .Rn / : which to each function u 2 H m .˝/ will associate a function e Such an extension operator can be constructed if ˝ is smooth enough. In what follows we shall use the notations: ® ¯ RnC D x 2 Rn W x D .x 0 ; xn / ; x 0 2 Rn1 ; xn > 0 ; ¯ ® Q D x 2 Rn W x D .x 0 ; xn / ; x 0 2 Rn1 ; jx 0 j < 1; jxn j < 1 ; QC D Q \ RnC ; ® ¯ Q0 D x 2 Rn W x D .x 0 ; 0/ ; x 0 2 Rn1 ; jx 0 j < 1 :

Theorem 7.2. Let ˝  Rn be an open set of class C 1 with bounded boundary, or ˝ D RnC : Then there exists a linear operator P W H 1 .˝/ ! H 1 .Rn / and a constant C > 0 only depending on ˝; such that the following conditions are satisfied for every u 2 H 1 .˝/ W (i) .P u/ .x/ D u .x/ ; x 2 ˝I (ii) jP ujL2 .Rn /  C jujL2 .˝/ I (iii) jP ujH 1 .Rn /  C jujH 1 .˝/ : Remark 7.3. Inequality (iii) expresses the continuity of P: Also, (ii) expresses the continuity of P with respect to the L2 norm. For the proof we need the following lemma.

Section 7.2 The Extension Operator

153

Lemma 7.1. If u 2 H 1 .QC / ; then its extension by reflection ²   u .x 0 ; xn / if xn > 0 u x 0 ; xn D u .x 0 ; xn / if xn < 0 belongs to H 1 .Q/ and p ˇ ˇ ˇu ˇ 2 D 2 jujL2 .QC / ; L .Q/

ˇ ˇ ˇu ˇ

H 1 .Q/

D

p

2 jujH 1 .QC / :

(7.4)

Proof. It is clear that u 2 L2 .Q/ and that the first equality from (7.4) holds. The proof will be complete if we shall establish the following formulas:   @u  @u D for j D 1; 2; :::; n  1; (7.5) @xj @xj @u D @xn



@u @xn

˘ ;

(7.6)

  where @u=@xj is the extension by reflection of @u=@xj ; and for any function f defined on QC ; ²   f .x 0 ; xn / if xn > 0 f ˘ x 0 ; xn D f .x 0 ; xn / if xn < 0: For their proof let us consider a function  2 C 1 .R/ such that  .t / D 0 for t < 1=2 and  .t/ D 1 for t > 1: Also consider the sequence of functions .k / from C 1 .R/ ; where k .t / D  .k t / : Let ' 2 D .Q/ : One can immediately see that for 1  j  n  1; Z Z @  @' u dx D u dx; @xj Q QC @xj .x 0 ; xn / D ' .x 0 ; xn / C ' .x 0 ; xn / : In general … D .QC / ; howwhere ever, k WD k .xn / .x 0 ; xn / 2 D .QC / : Hence Z Z @ .k / @u u dx D  k dx: @xj QC @xj QC Since @ .k / =@xj D k @ =@xj ; we have Z Z @ @u uk dx D  k @xj QC QC @xj

dx:

154

Chapter 7 Sobolev Spaces

Letting k ! 1 and using Lebesgue’s dominated convergence theorem, we find Z Z @ @u u dx D  dx: QC @xj QC @xj Thus

Z

@' u dx D  @xj Q 

Z QC

@u @xj

dx D 

Z  Q

@u @xj



' dx;

whence it follows (7.5). To prove (7.6), let ' 2 D .Q/ : We have Z Z @  @' u dx D u dx; @xn Q QC @xn where  .x 0 ; xn / D ' .x 0 ; xn /  ' .x 0 ; xn / : As above, k  2 D .QC / and Z Z @ .k / @u u dx D  k  dx: @xn QC QC @xn One has @ .k / =@xn D k @=@xn C k0 .kxn / : We claim that Z uk0 .kxn /  dx ! 0 as k ! 1: QC

Indeed, since  .x 0 ; 0/ D 0; there is a constant M such that j .x 0 ; xn /j  M jxn j on Q: Furthermore, if we denote C D sup j0 .t /j ; we obtain t 2Œ0;1

ˇ ˇZ Z ˇ ˇ ˇ ˇ uk0 .kxn /  dx ˇ  kM C juj xn dx ˇ ˇ ˇ QC xn 2.0; k1 / Z  MC juj dx ! 0 .as k ! 1/ : xn 2.0; k1 / Therefore, for k ! 1; we find Z Z @ @u u dx D   dx: QC @xn QC @xn Finally observe that

Z QC

@u  dx D @xn

Z  Q

@u @xn

˘ ' dx:

Section 7.2 The Extension Operator

155

Proof of Theorem 7.2. The conclusion of Lemma 7.1 remains true, with the same proof, if instead of Q; QC we take Rn and RnC ; respectively, which proves Theorem 7.2 for ˝ D RnC : If ˝ is of class C 1 with bounded, equivalently, compact boundary, then there is a finite open cover .Uk /1km of @˝ and invertible mappings k W 2 C 1 .U k /; k .QC / D Uk \ ˝ Q ! Uk such that k 2 C 1 .Q/; 1 k and k .Q0 / D Uk \ @˝: Consider the partition of unity subordinated to the m S cover ¹U0 ; U1 ; :::; Um º ; where U0 D ˝ n U k ; that is the functions kD1

0 ; 1 ; :::; m with the following properties: k 2

C01 .Uk / ;

0  k  1;

m X kD0

k D 1 on ˝ [

m [

Uk :

kD1

Pm Pm Let u 2 H 1 .˝/ : We may represent u D kD0 k u D kD0 uk ; with uk WD k u: We shall now define the extension to Rn of each function uk for k D 0; 1; :::; m: (a) The extension of u0 is defined by ² u0 .x/ ; x 2 ˝ e u0 .x/ D 0; x 2 Rn n ˝: According to (6.3), @e u0 =@xj D 0 @u=@xj C u@0 =@xj in D 0 .Rn / ; whence it 1 follows that e u0 2 H .Rn / and u0 jL2 .Rn / D jujL2 .˝/ ; je

u0 jH 1 .Rn /  C jujH 1 .˝/ ; je

where constant C only depends on the norms of 0 and r0 : (b) Extension of uk ; 1  k  m: Let vk .y/ WD uk .k .y// ; y 2 QC : One has vk 2 H 1 .QC / and according to Lemma 7.1, its extension by reflection vk belongs to H 1 .Q/ : Now we bring back vk on Uk by defining wk .x/ WD   .x/ ; x 2 Uk : We have wk 2 H 1 .Uk / ; wk D uk on Uk \ ˝ and vk 1 k jwk jL2 .Uk /  C juk jL2 .Uk \˝/ ;

jwk jH 1 .Uk /  C juk jH 1 .Uk \˝/ ;

where C does not depend on u: Finally the extension to Rn of function uk having all the desired properties, is ² k .x/ wk .x/ ; x 2 Uk e uk .x/ D 0; x 2 Rn n U k : 1 1 n In conclusion, Pm the extension operator we looked for is P W H .˝/ ! H .R / ; P u D kD0 e uk :

156

Chapter 7 Sobolev Spaces

For example, we can use the extension operator to prove the following density result. Proposition 7.2. Let ˝  Rn be an open set of class C 1 : Then the set of the restrictions to ˝ of all functions from C01 .Rn / is dense in H 1 .˝/ : Proof. Let u 2 H 1 .˝/ : ( a) Case ˝ D Rn W We have k . k u/ ! u in H 1 .Rn / as k ! 1: Here k is like in Proposition 6.3. (b) Case @˝ bounded: k . k P u/ converges to P u in H 1 .Rn / and so to u in H 1 .˝/ : (c) Case @˝ unbounded: Let " > 0: Then there is a k0 with ˇ ˇ ˇk u  uˇ 1 < "=2: 0 H .˝/ Consider now an extension v 2 H 1 .Rn / of k0 u: Then there is a w 2 C01 .Rn / with jw  vjH 1 .Rn / < "=2: It is clear that jw  ujH 1 .˝/ < ": Remark 7.4. It follows from above that C01 .Rn / is dense in H 1 .Rn / : Also, 1 if ˝ is bounded of class C 1 ; then C 1 .˝/ is dense  n  in H .˝/ :   n 1 One can prove also for ˝ D RC that C0 RC is dense in H 1 RnC : However, in general, C01 .˝/ is not dense in H 1 .˝/ :

7.3

The Sobolev Spaces H0m .˝/

Definition 7.2. The Sobolev space H0m .˝/ is the closure of C01 .˝/ in H m .˝/ : The space H0m .˝/ endowed with the inner product of H m .˝/ ; is a Hilbert space. Remark 7.5. Remark 7.4 guarantees the equality H01 .Rn / D H 1 .Rn / : However, in general, H01 .˝/ is a proper subspace of H 1 .˝/ : In case that ˝ is bounded (or at least bounded in one direction), the space can be endowed with inner product Z .u; v/H 1 D ru  rv dx

H01 .˝/

0

˝

and corresponding norm Z jujH 1 D 0

jruj dx 2

˝

 12 ;

Section 7.3 The Sobolev Spaces H0m .˝/

157

equivalent to j:jH 1 : To prove the equivalence of norms j:jH 1 and j:jH 1 on H01 .˝/ ; first observe that jujH 1  jujH 1 I then use Poincaré’s inequality.

0

0

The theorem which follows offers sufficient conditions for a function in H 1 .˝/ to belong to the subspace H01 .˝/ : Theorem 7.3. Let ˝  Rn be an open set and u 2 H 1 .˝/ : Each one of the following conditions is sufficient for that u 2 H01 .˝/ W (i) supp u is a compact included in ˝I (ii) u 2 C.˝/ and u D 0 on @˝: Proof. Assume (i). Consider a bounded open set ˝0 of class C 1 such that supp u  ˝0 ; ˝ 0  ˝; and a function ˛ 2 C01 .˝0 / with ˛ D 1 on supp u .˝0 and ˛ can be obtained using a C 1 partition of unity). Clearly ˛u D u: According to Proposition 7.2, there is a sequence .uk /  C01 .Rn / with uk ! u in H 1 .˝0 / : It follows that ˛uk ! ˛u in H 1 .˝/ ; where ˛uk 2 C01 .˝/ : Hence, ˛u D u 2 H01 .˝/ : Assume (ii). First we shall analyze the case when supp u is bounded. Choose a function G 2 C 1 .R/ such that jG .t /j  jt j for every t 2 R; G .t / D 0 for jt j  1 and G .t/ D t for jtj  2: It is easily seen that uk D G .ku/ =k 2 H 1 .˝/ and supp uk  ¹x 2 ˝ W ju .x/j  1=kº ; that is, supp uk is a compact included in ˝: It follows that uk 2 H01 .˝/ : On the other hand, using Lebesgue’s dominated convergence theorem, one can see that uk ! u in H 1 .˝/ (Exercise). Consequently, u 2 H01 .˝/ : In case that supp u is unbounded, then we consider the sequence of the truncations k u of u; where functions k are those in Proposition 6.3. As above, k u 2 H01 .˝/ ; and, on the other hand, k u ! u in H 1 .˝/ : Hence u 2 H01 .˝/ : The next theorem shows that in the case when ˝ is of class C 1 ; the condition u D 0 on @˝ is also necessary for that u 2 H01 .˝/ \ C.˝/: Theorem 7.4. Let ˝  Rn be an open set of class C 1 : If u 2 H01 .˝/\C.˝/; then u D 0 on @˝: Proof. By local charts our problem can be reduced to the following one: if u 2 H01 .QC / \ C.QC /; then u D 0 on Q0 : Toward this end, consider a sequence .uk /  C01 .QC / with uk ! u in H 1 .QC / : For .x 0 ; xn / 2 QC ; we have ˇZ ˇ Z xn ˇ ˇ ˇ  0 ˇ @uk  0 ˇ ˇ ˇ xn @uk  0  ˇ ˇuk x ; xn ˇ D ˇ ˇ ˇ dt: ˇ ; t d t ; t x x ˇ ˇ ˇ ˇ @x @xn n 0 0

158

Chapter 7 Sobolev Spaces

Hence, if 0 < "  1; then ˇ Z Z " Z "ˇ Z ˇ @uk  0 ˇ ˇ  0 ˇ 1 ˇ ˇuk x ; xn ˇ dxn dx 0  x ; t ˇˇ dt dx 0 : ˇ " jx 0 j 0 is such that ju"  ujLq .˝/ < "=3: We have jh u  ujLq

 jh .u  u" /jLq C jh u"  u" jLq C ju"  ujLq "  jh u"  u" jLq C 2 : 3 To estimate jh u"  u" jLq ; denote ˝h D ¹x 2 ˝ W x  h 2 ˝º : Then Z Z q ju" .x  h/  u" .x/jq dx C ju" .x/jq dx: jh u"  u" jLq D ˝h

˝n˝h

For the first integral, if we write 1=q D ˛=1 C .1  ˛/ =2 ; where 0 < ˛  1 .recall that q < 2 / and we use the interpolation inequality, we obtain ˛ 1˛ jh u"  u" jLq .˝h /  jh u"  u" jL 1 .˝ / jh u"  u" j 2 L .˝h / h  1˛ ˛  C jhj˛ ju" jH 1 .˝/ 2 ju" jL2 .˝/

 C0 jhj˛ ; where C0 depends only on " being independent of u: For the second integral, we apply Hölder’s inequality and find q

q

ju" jLq .˝n˝h /  j1jLr .˝n˝h / ju" jL2 .˝/ ; where 1=r C q=2 D 1: It follows that there is a ı" > 0 independent of u; such that jh u"  u" jLq < "=3 if jhj < ı" : Therefore jh u  ujLq < " for jhj < ı" and u 2 B: Theorems 7.7 and 7.8 remain true if in place of H 1 .˝/ one takes H01 .˝/ ; with an open, respectively a bounded open ˝; without any smoothness assumption on ˝: The explanation relies on the existence of an extension operator P W H01 .˝/ ! H 1 .Rn / for any open set ˝: More precisely, we have the following result. Proposition 7.3. Let ˝  the extension by zero

Rn ²

u .x/ 0

for x 2 ˝ for x 2 Rn n ˝

 B @u ;

j D 1; 2; :::; n:

e u .x/ D belongs to H 1 .Rn / and @e u D @xj

be an open set. Then, for every u 2 H01 .˝/ ;



@xj

(7.10)

  Section 7.6 The Embedding of H m .˝/ into C ˝

165

Proof. Since u 2 H 1 .˝/ ; then the extensions by zero e u; @u ; and @x j 1 n H . /:

R

respectively, obviously belong to

R

L2 .

n/ :

A



@u @xj

of u

We now prove that e u2

Since u 2 then there exists a sequence .uk /  C01 .˝/ with uk ! u in H 1 .˝/ : Then, for every ' 2 C01 .Rn / ; we have ˇ ˇ Z ˇ ˇZ ˇ ˇ ˇ ˇ @uk ˇ  juk j 1 ˇ uk @' dx ˇ D ˇ ' d x H .˝/ j'jL2 .˝/  C j'jL2 .˝/ : ˇ ˇ @xj ˇ ˇ ˝ @xj ˝ H01 .˝/ ;

Passing to the limit k ! 1 yields ˇZ ˇ ˇZ ˇ ˇ ˇ ˇ @' ˇˇ ˇ u @' dx ˇ D ˇ u dx  C j'jL2 .˝/  C j'jL2 .Rn / : ˇ ˇ ˇ ne ˝ @xj R @xj ˇ This, in view of Remark 7.1, guarantees e u 2 H 1 .Rn / :

We note that the converse of Proposition 7.3 is also true, namely: If ˝ is of u 2 H 1 .Rn / ; then u 2 H01 .˝/ (see Brezis [5, class C 1 ; u 2 L2 .˝/ and e Proposition 9.18]). Thus, for the space H01 .˝/ we have the following results. Theorem 7.9. Let ˝  Rn be an open set. Then (a) H01 .˝/ is continuously embedded into Lq .˝/ for n  3 and q 2 Œ2; 2 ; or n D 2 and q 2 Œ2; C1/: (b) If in addition ˝ is bounded, then H01 .˝/ is compactly embedded into Lq .˝/ for n  3 and q 2 Œ1; 2 /; or n D 2 and q 2 Œ1; C1/: Notice that in the literature the number 2 1 D .n C 2/ = .n  2/ ; for n  3; is said to be a critical exponent.

7.6

  The Embedding of H .˝/ into C ˝ m

We shall present a result about the continuous embedding of the Sobolev space H m .˝/ into C.˝/: In this respect we use a characterization of the elements of H m .Rn / in terms of the Fourier transform. Proposition 7.4. For each m 2 N; ² ³ m  n  n   n 2 2 m 2 2 H R D u 2 L R W 1 C jxj T Œu 2 L R : ˇ ˇ m ˇ ˇ 2 2 ˇ In addition ˇ 1 C jxj T Œuˇˇ

L2

is an equivalent norm with jujH m :

166

Chapter 7 Sobolev Spaces

Proof. Let u 2 L2 .Rn / : Then, according to (6.8),     u 2 H m Rn ” D ˛ u 2 L2 Rn ;

   ” T D ˛ u 2 L2 Rn ; j˛j  m   ” x ˛ T Œu 2 L2 Rn ; j˛j  m   ” w0 .x/ .T Œu/2 2 L1 Rn 1   ” w02 T Œu 2 L2 Rn ;

j˛j  m

P ˛ 2 where w0 .x/ D j˛jm jx j : Furthermore, we remark that instead of the weight w0 ; one may consider any other smooth function w for which two positive constants c1 ; c2 exist  with c1 w0  w  c2 w0 : In particular, we may choose w .x/ D 1 C jxj2

m

:

The following regularity result holds for H m .Rn / provided that m is large enough. Theorem 7.10. If m > n=2; then     H m Rn  C R n and

sup juj

R

n

 C jujH m for every u 2 H m .Rn / :

Proof. The key of the proof is the relation 

1 C jxj2

 m2

2 L2



Rn



which is true if and only if m > n=2: This immediately follows after passing to spherical coordinates: Z 1 Z  m  m n1 d x D !n r dr: 1 C r2 1 C jxj2

Rn

0

Assume that m > n=2 and u 2 H m .Rn / : Then, from 

1 C jxj2

 m2

 T Œu ;

1 C jxj2

 m2

2 L2



Rn



;

Section 7.7 The Sobolev Space H m .˝/

167

it follows that T Œu 2 L1 .Rn / : Furthermore, we have ˇ ˇ ju .x/j D ˇT 1 ŒT Œu .x/ˇ D jT ŒT Œu .x/j Z n n 2 2  .2/ jT ŒujL1 D .2/ jT Œu .x/j dx Rn Z   m2   m2 n 1 C jxj2 1 C jxj2 D .2/ 2 jT Œuj dx Rn ˇ ˇ  m2 ˇ ˇ T Œuˇˇ  C jujH m : (7.11)  C0 ˇˇ 1 C jxj2 L2

From this, also using Lemma 5.1, we obtain ju .x C x0 /  u .x0 /j  C jx u  ujH m ! 0 whence u 2 C .Rn / : In addition (7.11) shows that

as x ! 0;

sup juj

Rn

 C jujH m :

Clearly, the previous result implies the inclusions Hm



Rn



\

 Ck Hm



Rn



Rn





for m > k C

 C1



Rn



n ; 2

:

m0

Theorem 7.11. Let ˝  Rn be an open set of class C m ; with bounded boundary, or ˝ D RnC : If m > n=2; then the following continuous embedding holds: H m .˝/  C.˝/: Proof. Use the extension operator. More complete results on continuous and compact embeddings for the Sobolev spaces W m;p .˝/ can be found in Adams [1] and Brezis [5].

7.7

The Sobolev Space H m .˝/

Definition 7.4. Let m 2 N: The space H m .˝/ consists of all distributions f on ˝ for which there is a constant C with j.f; '/j  C j'jH m

for all ' 2 D .˝/ :

(7.12)

Proposition 7.5. The space H m .˝/ can be identified with the dual of H0m .˝/ :

168

Chapter 7 Sobolev Spaces

Proof. If f 2 H m .˝/ ; then as a distribution, f is a linear functional on C01 .˝/ : Formula (7.12) shows that this functional is continuous with respect to norm j:jH m on C01 .˝/ : Since C01 .˝/ is dense in H0m .˝/ ; then f admits a unique extension to a continuous linear functional on H0m .˝/ : Hence f is an element of the dual of H0m .˝/ : Conversely, if g is a continuous linear functional on H0m .˝/ ; then there is a constant C with jg .'/j  C j'jH m for every ' 2 H0m .˝/ : It follows that, for each compact K  ˝; there is a constant C 0 > 0 such that X ˛ sup jD ' .x/j (7.13) jg .'/j  C j'jH m  C 0 j˛jm

x2K

for every ' 2 D .˝/ satisfying supp '  K: Here we have taken into account the inequality p jD ˛ 'jL2 .˝/   .K/ jD ˛ 'j1 ; where  .K/ is the Lebesgue measure of K: Inequality (7.13) guarantees that the restriction of g to D .˝/ is continuous, i.e. 'k ! 0 in D .˝/ implies g .'k / ! 0: Hence the restriction of g to D .˝/ is a distribution f on ˝: We have j.f; '/j D jg .'/j  C j'jH m for every ' 2 D .˝/ ; which proves f 2 H m .˝/ : We shall identify L2 .˝/ to its dual, but not H0m .˝/ to H m .˝/ : Since is continuously embedded into L2 .˝/ ; then, passing to dual spaces, we have that L2 .˝/ is continuously embedded in H m .˝/ : Therefore, the following continuous (and dense) embeddings hold: H0m .˝/

H0m .˝/  L2 .˝/  H m .˝/ : Note that, seen as an element of H m .˝/ ; a function f 2 L2 .˝/ is identified with the continuous linear functional on H0m .˝/ given by u 2 H0m .˝/ 7! .f; u/L2 : The elements of H m .˝/ can be represented by means of functions from as shows the next theorem.

L2 .˝/

Theorem 7.12. For every functions g˛ 2 L2 .˝/ ; j˛j  m; the distribution X .1/j˛j D ˛ g˛ f D (7.14) jajm

is an element of H m .˝/ :

Section 7.7 The Sobolev Space H m .˝/

169

Conversely, for each f 2 H m .˝/ ; there are functions g˛ 2 L2 .˝/ ; j˛j  m such that .7:14/ holds and 0 jf jH m D @

X

1 12 2 A : jg˛ jL 2

j˛jm

Here jf jH m stands for the norm of f in H m .˝/ ; the dual of .H0m .˝/ ; j:jH m /: Proof. If g 2 L2 .˝/ and j˛j  m; then D ˛ g is a distribution on ˝ and for every ' 2 D .˝/ ; ˇ ˛ ˇ ˇ ˇ ˇ D g; ' ˇ D ˇ g; D ˛ ' ˇ  jgj 2 jD ˛ 'j 2  jgj 2 j'j m : L L L H Hence D ˛ g 2 H m .˝/ : It follows that a sum of derivatives of order  m of functions from L2 .˝/ ; also belongs to H m .˝/ : To prove the converse assertion, denote A WD ¹˛ 2 Nn W j˛j  mº and  A consider the Hilbert space E WD L2 .˝/ equipped with inner product .g; h/E WD

X

.g˛ ; h˛ /L2 ;

˛2A

where g .˛/ D g˛ and h .˛/ D h˛ : Also consider the linear mapping T W H0m .˝/ ! E; given by   T u D D ˛ u ˛2A : Since jT ujE D juj is an isometry between H0m .˝/ and the m ; then T  Hm  subspace E0 WD T H0 .˝/ of E: Now let f be any element of H m .˝/ : Then the linear functional   h 2 E0 7! f; T 1 h .here .f; v/ stands for the value of functional f at v/ is continuous on E0 and its norm is equal to jf jH m : From the Hahn–Banach theorem, it can be extended to a continuous linear functional G on E; with jGjE 0 D jf jH m : Furthermore, Riesz’s representation theorem guarantees the existence of a g D .g˛ /˛2A 2 E such that .G; h/ D .g; h/E for every h 2 E; and 0 jGjE 0 D jgjE D @

X

j˛jm

1 12 2 A : jg˛ jL 2

170

Chapter 7 Sobolev Spaces

For any ' 2 D .˝/ ; we have   .f; '/ D f; T 1 T ' D .G; T '/ D .g; T '/E X  X    D g˛ ; D ˛ ' L2 D g˛ ; D ˛ ' j˛jm

D

X

.1/

j˛j

j˛jm

 D g˛ ; ' :



˛

j˛jm

Hence f D

P

j˛jm .1/

j˛j

D ˛ g˛ :

According to the Riesz representation theorem, for every Hilbert space .H; .:; :/H /; the mapping u 2 H 7! .u; :/H 2 H 0 is an isomorphism between H and its dual H 0 ; which allows the identification of H with H 0 : In the case of spaces H0m .˝/ and H m .˝/ ; this isomorphism can be expressed by means of a differential operator. Theorem 7.13. The operator L W H0m .˝/ ! H m .˝/ ; LD

X

.1/j˛j D 2˛

j˛jm

is an isometric isomorphism between H0m .˝/ and H m .˝/ : Proof. From Theorem 7.12, operator L is well defined. Also, for all u 2 H0m .˝/ and ' 2 D .˝/ ; one has .Lu; '/ D .u; '/H m :

(7.15)

Thus, the action of the distribution Lu over a function from D .˝/ .dense in H0m .˝// is given by the value at that function of the continuous linear functional .u; :/H m defined on H0m .˝/ : In this sense, one may say that Lu D u in H m .˝/ ; where H0m .˝/ is equipped with inner product .:; :/H m : Attention! the exact sense of assertion “Lu D u in H m .˝/” is given by (7.15); actually, it is clear that in general Lu ¤ u in D 0 .˝/ : L is an isometry since jLujH m D D

j.Lu; v/j j.Lu; v/j D sup v2C01 .˝/n¹0º jvjH m v2H0m .˝/n¹0º jvjH m sup

sup

v2C01 .˝/n¹0º

j.u; v/H m j D jujH m : jvjH m

Section 7.7 The Sobolev Space H m .˝/

171

Remark 7.6. Theorem 7.13 guarantees that for each f 2 H m .˝/ ; there is a unique function u 2 H0m .˝/ with Lu D f; that is, the problem ² P j˛j D 2˛ u D f in D 0 .˝/ ; j˛jm .1/ m .˝/ u 2 H0 has a unique solution. This function can be considered to be the generalized (or weak) solution of the problem ² P j˛j D 2˛ u D f in ˝ j˛jm .1/ uD0 on @˝: In addition jujH m D jf jH m : It is useful to analyze the special case m D 1; that is the case of spaces and H 1 .˝/ ; when ˝ is bounded and H01 .˝/ is endowed with the R 1=2 : equivalent norm jujH 1 D ˝ jruj2 dx

H01 .˝/

0

Theorem 7.14. Let ˝  Rn be a bounded open set. For every functions gj 2 L2 .˝/ ; j D 1; 2; :::; n; the distribution f D

n X @gj @xj

(7.16)

j D1

is an element of H 1 .˝/ : Conversely, for each f 2 H 1 .˝/ ; there are functions gj 2 L2 .˝/ ; j D 1; 2; :::; n such that .7:16/ holds and 1 12 n X ˇ ˇ2 ˇgj ˇ 2 A : D@ L 0

jf jH 1

j D1

  Here jf jH 1 stands for the norm of f in H 1 .˝/ ; the dual of H01 .˝/ ; j:jH 1 : 0

Remark 7.7. Assume that ˝ is bounded. Then for every vector-valued function g 2 L2 .˝I Rn / ; the distribution f D  div g belongs to H1 .˝/ : Conversely, for each f 2 H 1 .˝/ ; there is a vector-valued function g 2 L2 .˝I Rn / with f D  div g and jf jH 1 D jgjL2 .˝IRn / : Theorem 7.15. Let ˝  Rn be a bounded open set. Then the  operator  W 1 .˝/ 1 H0 ! H .˝/ is an isometric isomorphism between H01 .˝/ ; j:jH 1 and its dual H 1 .˝/ :

0

172

Chapter 7 Sobolev Spaces

Remark 7.8. More general, every elliptic operator L of type (3.40), is an isometric isomorphism between .H01 .˝/ ; a .:; :// and its dual H 1 .˝/ ; where the inner product a .:; :/ is defined by (3.42). Also we may say that Lu D u in H 1 .˝/ ; in the sense that .Lu; '/ D a .u; '/ for all ' 2 D .˝/ :

7.8

Fourier Series in H 1 .˝/

This section prepares the following Chapters 10–12 and makes possible the extension of the reasonings from Chapter 4, to the case of functions f with values in H 1 .˝/ : Parseval’s equality and the completeness property of eigenfunctions k are here generalized to H 1 .˝/ : Also, since in Chapter 12 all functions will be considered with complex values, the above mentioned generalization will be presented, more generally, for complex-valued functions, although the same notations as for the real case are used to denote the corresponding spaces of complexvalued functions. Thus, in this section, L2 .˝/ is the space of all complex-valued measurable R functions u with ˝ ju .x/j2 dx < 1 endowed with inner product and norm Z .u; v/L2 D

˝

u .x/ v .x/dx;

Z jujL2 D

ju .x/j dx 2

˝

 12 :

Also the Sobolev space of complex-valued functions H01 .˝/ is endowed with inner product and norm ! Z n X 1 @u @v .u; v/H 1 D dx; jujH 1 D .u; u/H2 1 : 0 0 @xk @xk 0 ˝ kD1

As usual by H 1 .˝/ we denote the dual of H01 .˝/ ; that is the space of all continuous linear complex-valued functionals on H01 .˝/ : The duality between H01 .˝/ and H 1 .˝/ is defined as follows: for f 2 H 1 .˝/ and u 2 H01 .˝/ ; .f; u/Rstands for the valued of f at uI in particular, if f 2 L1loc .˝/ ; then .f; u/ D ˝ f udx; and if f 2 L2 .˝/ ; then .f; u/ D .f; u/L2 : For the complex-valued case,  is still an isometry between spaces H01 .˝/ and H 1 .˝/ : Recall that by k and k .k D 1; 2; :::/ we have meant the eigenvalues and eigenfunctions of : Thus ² k D k k in ˝ k D 0 on @˝:

Section 7.8 Fourier Series in H 1 .˝/

173

We shall assume that jk jL2 D 1: Then the systems .k /k1 ;





p1 k k k1

are

H01 .˝/ ;

L2 .˝/

orthonormal and complete in and respectively. In addition, recall Poincaré’s inequality, which is also true for complex-valued functions: 1 jujL2  p jujH 1 ; 0

1

u 2 H01 .˝/ :

(7.17)

The basic result is the generalization to the space H 1 .˝/ over C; of Parseval’s equality and of the completeness property of eigenfunctions k : More exactly we have the following result. Theorem 7.16. (a) For any u 2 H 1 .˝/; one has 1 X

uD

.u; k /k

 in H 1 .˝/



(7.18)

kD1

and

1 X 1 2 j.u; k /j2 D jujH 1 :

k

(7.19)

kD1

(b) If u 2 L2 .0; T I H 1 .˝//; then uD

1 X

.u; k / k

 in L2 .0; T I H 1 .˝// :



kD1

Proof. (a) We use the fact that  is an isometry between spaces H01 .˝/ and H 1 .˝/ : Thus, if u 2 H 1 .˝/ ; then ./1 u 2 H01 .˝/ and equality (7.18) is equivalent to ./1 u D

1 X

.u; k / ./1 k



 in H01 .˝/ :

kD1

Since ./1 k D

1  k k

  and .u; k / D ./1 u; k

can be rewritten as ./1 u D

1 X kD1

1 ./1 u; p k

k

H01

! H01

; the last equality

  1 p k in H01 .˝/ :

k

174

Chapter 7 Sobolev Spaces

 But this equality is true since the system in

H01 .˝/ :

p1 k k

 is orthonormal and complete

Equality (7.19) is equivalent to Parseval’s equality in H01 .˝/ for the function ./1 u: (b) According to (7.19) one has 2 ju .t /jH 1

1 X 1 D j.u .t / ; k /j2

k

for a.e. t 2 Œ0; T  :

kD1

Thus the problem reduces to the convergence in L1 .0; T / of the sequence of partial sums. This happens by the Lebesgue dominated convergence theorem since 2 the partial sums are dominated by the function t 7! ju .:/jH belonging to 1 1 L .0; T / : Remark 7.9. In summary: if the eigenfunctions k are normalized in L2 .˝/ ; then: (i) one has

p

1

k ; jk jH 1 D p I

k  p  k ;

k k are orthonormal and complete in

jk jL2 D 1; jk jH 1 D 0

 (ii) the systems .k / ;

p1 k

L2 .˝/ ; H01 .˝/ and H 1 .˝/ ; respectively; (iii) due to the embeddings H01 .˝/  L2 .˝/  H 1 .˝/ and to the completeness of the system .k / in all the three spaces, the Fourier series of a function u 2 H01 .˝/ with respect to the three systems from (ii), in L2 .˝/ ; H01 .˝/ and H 1 .˝/ ; respectively, that is, X

.u; k /L2 k ; ! X 1 1 u; p k p k ;

k

k H01   X p p u; k k 1 k k ; H

are identical and can be written as X .u; k / k ;

Section 7.9 Generalized Solutions of the Cauchy Problems

175

where by .u; k / we mean the action of u as an element of H 1 .˝/ over k Note that the scalar product in H 1 .˝/ is given by   .u; v/H 1 WD ./1 u; ./1 v 1 : H0

7.9

Generalized Solutions of the Cauchy Problems

In Chapter 5 we studied the Cauchy problem for the heat and wave equations. However the space S where we worked is too restrictive for applications. Remembering that the Fourier transform realizes an isometry also in the case of the space L2 .Rn / ; it is natural to put the problem of extending the results in Chapter 5, to problems with initial data in L2 .Rn / : Clearly, in this situation we cannot expect classical solutions and we shall only speak about weak solutions. We start with an existence, uniqueness, and representation result for the weak solution of the Cauchy problem for the heat equation. Theorem 7.17 (existence, uniqueness, and L2 representation). If g0 2 L2 .Rn / ; then there exists a unique function     u 2 C.Œ0; 1/I L2 Rn / \ C 1 ..0; 1/I L2 Rn / such that

²

u0 .t /  u .t / D 0 in D 0 .Rn / for t > 0; u .0/ D g0 :

(7.20)

In addition u 2 C 1 ..0; 1/I L2 .Rn // and the representation formula (5.17) holds. Proof. Uniqueness and representation. Let u be any function satisfying all the conditions in the statement. Since u .t/ 2 S 0 ; (7.20) is equivalent via Fourier transform in S 0 ; to (5.15). As in the proof of Theorem 5.5 one can obtain the representation (5.17), with the mention of the fact that, this time, the Fourier transform applies to functions in L2 .Rn / : Existence. First observe that the function w from the proof of Theorem 5.5 satk / 2 L2 .Rn / for all t > 0 isfies: w .t / 2 L2 .Rn / for every t  0; and @ w.x;t @t k and k  1: Also, similar arguments based on Lagrange’s theorem of finite increments, like those in the proof of Theorem 5.5, yields w 2 C.Œ0; 1/I L2 .Rn // k and @ w.x;t/ 2 C..0; 1/I L2 .Rn //: The details are left to the reader. @t k One can prove that the property u0 .t /  u .t / D 0 in D 0 .Rn / for t > 0; is equivalent to the fact that u .x; t/ satisfies the heat equation in D 0 .Rn  .0; 1// ;

176

Chapter 7 Sobolev Spaces

and also to the fact that for every v 2 H 1 .Rn / ; .u .t / ; v/L2 .Rn / 2 C 1 Œ0; 1/ and

d .u .t/ ; v/L2 .Rn / C .ru .t/ ; rv/L2 .Rn IRn / D 0 dt

for all t  0:

An increase in regularity for the weak solution u; at t D 0; is obtained provided that g0 satisfies some smoothness conditions. Thus, one can prove that if g0 2 H m .Rn / ; then u.k/ 2 C.Œ0; 1/I H m2k .Rn // for 0  k  m=2: As for the heat equation, we may speak of weak (generalized, or distributional) solutions of the Cauchy problem for the wave equation, under more general assumptions on the data. Theorem 7.18 (existence, uniqueness, and L2 representation). If g0 2 H 1 .Rn/ and g1 2 L2 .Rn/ ; then there exists a unique function u 2 C 1 RI L2 .Rn / \ C RI H 1 .Rn / such that u .0/ D g0 ; u0 .0/ D g1 and for every v 2 H 1 .Rn / ; .u .t/ ; v/L2 .Rn / 2 C 2 .R/ and

d2 .u .t/ ; v/L2 .Rn / C .ru .t / ; rv/L2 .Rn IRn / D 0 for all t 2 R: dt 2

(7.21)

In addition, the representation formula (5.19) and the following energy conservation law ˇ 0 ˇ2 ˇu .t /ˇ 2 n C jru .t /j2 2 n n D jg1 j2 2 n C jrg0 j2 2 n n (7.22) L .R IR / L .R IR / L .R / L .R / hold. Proof. Existence and representation. We have to show that formula  sin .jyj t / 1 .x/ T Œg0  .y/ cos .jyj t / C T Œg1  .y/ u .t / .x/ D T jyj     defines a function u 2 C 1 RI L2 .Rn / \C RI H 1 .Rn / such  that1 u .0/ D g0 ; 0 n u .0/ D g1 and (7.21) holds. First we prove that u 2 C RI H .R / and @u=@t 2 C RI L2 .Rn / ; where @u .x; t/ D T 1 ŒT Œg0  .y/ jyj sin .jyj t / C T Œg1  .y/ cos .jyj t / .x/ : @t

Section 7.9 Generalized Solutions of the Cauchy Problems

177

The fact that for a fixed t; u .t/ belongs to H 1 .Rn / can be proved as follows: u .t /

    @ u .t / 2 L2 Rn .j D 1; 2; :::; n/ H 1 Rn ” @xj    @ ” T u .t / 2 L2 Rn @xj   ” yj T Œu .t/ .y/ 2 L2 Rn   yj sin .jyj t / 2 L2 Rn : ” yj T Œg0  .y/ cos .jyj t / C T Œg1  .y/ jyj 2

Since g0 2 H 1 .Rn / ; then iyj T Œg0  .y/ D T



  @ g0 .y/ 2 L2 Rn ; @xj

and due to the boundedness of cos .jyj t / ; yj T Œg0  .y/ cos .jyj t / 2 L2 .Rn / : yj yj sin .jyj t/ is bounded, we have T Œg1  .y/ jyj sin .jyj t / 2 L2 .Rn / : Also, since jyj Thus u .t / 2 H 1 .Rn / : .:; t/ 2 L2 .Rn / : The check of Similarly, it can be proved the relation @u @t this inclusion, the proof of the continuity with respect to t of functions u .t / ; @u .:; t/ ; as well as the proof of (7.21) are left to the reader. @t To show the energy conservation law, we start from the equalities: 

T u0 .t / .y/ D T Œg0  .y/ jyj sin .jyj t / C T Œg1  .y/ cos .jyj t / ; jyj T Œu .t / .y/ D T Œg0  .y/ jyj cos .jyj t / C T Œg1  .y/ sin .jyj t / : These, for the real parts, yield 

2

Re T u0 .t / C jyj2 .Re T Œu .t //2 D jyj2 .Re T Œg0 /2 C .Re T Œg1 /2 :

Use a similar equality for the imaginary parts and add the two relations side by side to obtain ˇ ˇ 0  ˇT u .t / .y/ˇ2 C jyj2 jT Œu .t / .y/j2 D jT Œg1  .y/j2 C jyj2 jT Œg0  .y/j2 ; which is the energy conservation law at each point y: Finally integration with respect to y gives the desired global law (7.22). Uniqueness immediately follows from the energy conservation law. Notice that property  (7.21) is equivalent to the fact that u .x; t/ satisfies the wave equation in D 0 RnC1 (Exercise).

178

Chapter 7 Sobolev Spaces

We also note that a gain in regularity for the weak solution is guaranteed provided that g0 2 H mC1 .Rn / and g1 2 H m .Rn / ; where m  1: As regards the Cauchy problem for the nonhomogeneous heat equation, we can generalize the result in Section 5.4 to the case of functions from L2 .Rn / ; in the following way: We denote S .t / g0 ; the solution of (5.10) for g0 2 L2 .Rn / : Hence S .t/ g0 D g0 N .t / for t > 0; S .0/ g0 D g0 : Theorem 7.19. If f 2 L1loc .Œ0; 1/I L2 .Rn //; then there exists a unique function v 2 C.Œ0; 1/I L2 .Rn // such that v .0/ D 0 and v t  x v D f in D 0 .Rn  .0; 1// : Moreover, the following representation holds: Z t ŒS ./ f .t  / .x/ d v .t / .x/ D 0 Z tZ   e D f N .x; t/ D f .y; / N .x  y; t  / dy d: 0

Rn

The family .S .t// t 0 of linear operators from L2 .Rn / into itself, has the following properties: S .t C s/ D S .t/ S .s/

for all t; s  0;

S .0/ D I (identity operator), lim

t !0

S .t / g0 D g0

for every g0 2 L2 .Rn / :

In addition, from (5.17), we have that operators S .t / are nonexpansive, i.e.   (7.23) jS .t / g0 jL2  jg0 jL2 for every g0 2 L2 Rn : One says that .S .t // t0 is a continuous semigroup of linear contractions on the Hilbert space L2 .Rn / : For the abstract theory of semigroups of continuous linear operators and its applications to evolution equations, we refer the reader to Brezis [5], Cazenave–Haraux [7], Dautray–Lions [9, chapter 17] and Pazy [34]. For an other approach to the Cauchy problem, see, for example, Barbu [2] and Vladimirov [54]. We conclude this section by an application to the semilinear problem ² @u .x; t/  u .x; t/ D f .x; u .x; t// for x 2 Rn ; t > 0 @t (7.24) u .x; 0/ D 0 for x 2 Rn :

Section 7.9 Generalized Solutions of the Cauchy Problems

179

Theorem 7.20. Assume that f W Rn  R ! R satisfies the following conditions (i) f .:; u/ is measurable for each u 2 R; and f .:; 0/ 2 L2 .Rn / I (ii) there is a  0 such that jf .x; u/  f .x; v/j  a ju  vj

for all u; v 2 R and a.e. x 2 Rn :

Then the Cauchy semilinear problem (7.24) has a unique weak solution u 2 C.Œ0; 1/I L2 .Rn //: Proof. We have to prove that the problem has a unique solution on any finite interval Œ0; T  : The Cauchy problem on Œ0; T  is equivalent to the fixed point equation u D A .u/ in the Banach space C Œ0; T  I L2 .Rn / ; for the operator Z A .u/ .t / D

t 0

S .t   / f .:; u . // d; t 2 Œ0; T  :

Using (7.23), we have for every number  > 0; Z t jf .:; u .//  f .:; v . //jL2 d jA .u/ .t /  A .v/ .t /jL2  0 Z t  a ju . /  v . /jL2 d 0 Z t e e ju . /  v ./jL2 d D a 0   a t  max e ju ./  v . /jL2 : e  2Œ0;T

It follows that max e t jA .u/ .t /  A .v/ .t /jL2 

t 2Œ0;T

a max e t ju .t /  v .t /jL2 :  t 2Œ0;T

Now choose  > a; so that the operator A become  a contraction  on the space   2 n  t C Œ0; T  I L .R / equipped with norm max ju .t /jL2 e ; and the cont2Œ0;T

clusion follows from Banach’s contraction principle.

Chapter 8

The Variational Theory of Elliptic Boundary Value Problems 8.1

The Variational Method for the Dirichlet Problem

Theorem 7.15 guarantees that if ˝  Rn is a bounded open set, then for each f 2 H 1 .˝/ there exists a unique function u 2 H01 .˝/ with u D f; i.e. the problem ² u D f in D 0 .˝/ u 2 H01 .˝/ has a unique solution. This function u can be considered as generalized, or weak solution of the Dirichlet problem ² u D f in ˝ (8.1) uD0 on @˝: Since L2 .˝/  H 1 .˝/ ; a natural question arises: in case that f 2 L2 .˝/ ; does this notion of a weak solution coincide to the analog notion introduced in Part I? The answer is affirmative as shown the following theorem of variational characterization of the weak solution. Let us attach to the Dirichlet problem (8.1) with f 2 H 1 .˝/ ; the energy functional 1 2 E W H01 .˝/ ! R; E .u/ D jujH 1  .f; u/ : 0 2 Here .f; u/ stands for the value of the continuous linear functional f on function u: Recall that when f 2 L2 .˝/ ; .f; u/ coincides with .f; u/L2 : Proposition 8.1. Let u; v 2 H01 .˝/ : (a) The real function of a real variable t 2 R 7! E .u C t v/ 2 R is differentiable at t D 0 and 0

E .uI v/ D

lim

t !0C

E .u C tv/  E .u/ D t

Z ru  rv dx  .f; v/ ˝

D .u; v/H 1  .f; v/ : 0

Section 8.1 The Variational Method for the Dirichlet Problem

181

(b) If v D ' 2 D .˝/ ; then the number E 0 .uI '/ is equal to the action of the distribution  .u C f / on the test function '; i.e. E 0 .uI '/ D  .u C f; '/ : Proof. (a) Simple calculation based on the properties of inner product yields i t2 h 2 (8.2) E .u C tv/ D E .u/ C t .u; v/H 1  .f; v/ C jvjH 1 : 0 0 2 Now the expression of the derivative in (a) follows immediately. (b) If ' 2 D .˝/ ; then, using derivatives in the sense of distributions, we have Z n Z X @u @' .u; '/H 1 D ru  r' dx D dx 0 @xj @xj D

˝ n  X j D1

@u @' ; @xj @xj



j D1 ˝

D  .u; '/ :

Therefore, E 0 .uI '/ D  .u C f; '/ : Theorem 8.1 (Dirichlet’s principle). Let f 2 H 1 .˝/ and u 2 H01 .˝/ : Then the following statements are equivalent: (i) u D f in D 0 .˝/ : (ii) E 0 .uI v/ D 0 for all v 2 H01 .˝/ : (iii) E .u/ < E .w/ for every w 2 H01 .˝/ ; w ¤ u: Proof. (i))(ii): If u D f in D 0 .˝/ ; then for every ' 2 D .˝/ ; one has .u C f; '/ D 0; whence, taking into account Proposition 8.1(b), we find that E 0 .uI '/ D 0: Then, by density, E 0 .uI v/ D 0 for all v 2 H01 .˝/ : (ii))(i): From (ii), in particular, we have E 0 .uI '/ D 0 for all ' 2 D .˝/ : Then by Proposition 8.1, it follows that .u C f; '/ D 0 for every ' 2 D .˝/ ; that is u C f D 0 in D 0 .˝/ : (ii))(iii): Let w 2 H01 .˝/ ; w ¤ u: Using (8.2) for t D 1 and v D w  u; we obtain 1 2 E .w/ D E .u C .w  u// D E .u/ C E 0 .uI w  u/ C jw  ujH 1 0 2 1 2 D E .u/ C jw  ujH 1 > E .u/ : 0 2 (iii))(ii): If u is the absolute minimum point of E; then for each fixed v 2 H01 .˝/ ; one has E .u/  E .u C tv/ for all t 2 R: Now Proposition 8.1 (a) and Fermat’s theorem guarantee E 0 .uI v/ D 0:

182

Chapter 8 Variational Theory of Elliptic Problems

A function u 2 H01 .˝/ which satisfies any one of the three equivalent conditions from Theorem 8.1, is called a weak, or generalized solution of the Dirichlet problem (8.1). From Theorem 7.15, the weak solution exists and is unique. This conclusion can be also obtained if we apply the Riesz representation theorem to the continuous linear functional F W H01 .˝/ ! R; F .v/ D .f; v/ ; like in the proof of Theorem 3.14. Moreover, even in this more general setting, the eigenfunctions expansion of the weak solution holds, namely 1 X .f; k / k uD

k kD1

where the series convergence is in H01 .˝/ : Indeed, if f 2 H 1 .˝/ ; then u D ./1 f 2 H01 .˝/ and ! 1 X 1 1 1 ./ f; p k uD p k

k

k H1 kD1 0

  (with series convergence in H01 .˝/), while ./1 f; k

H01

D .f; k / :

Addition regularity properties of the weak solution can be established provided that ˝ and f are smooth. Theorem 8.2. Let u 2 H01 .˝/ be the weak solution of problem .8:1/: (a) If ˝ is of class C mC2 and f 2 H m .˝/ .m 2 N/; then u 2 H mC2 .˝/ and there exists a constant C > 0 only depending on ˝ and m such that jujH mC2  C jf jH m : In particular, if m > n=2; then u 2 H mC2 .˝/  C 2 .˝/: (b) If ˝ is of class C 1 and f 2 C 1 .˝/; then u 2 C 1 .˝/: The proof of this result is laborious and will be presented in Section 8.4. The next result answers the question: under which conditions the weak solution is a classical one? Theorem 8.3. If ˝ is of class C 1 ; f 2 C.˝/ and u 2 H01 .˝/ \ C 2 .˝/ is the weak solution of problem .8:1/; then u is a classical solution.

Section 8.1 The Variational Method for the Dirichlet Problem

183

Proof. Since ˝ is of class C 1 and u 2 H01 .˝/ \ C 2 .˝/; from Theorem 7.4, we have that u D 0 on @˝: On the other hand, based on Dirichlet’s principle, u satisfies the equation u D f in D 0 .˝/ : But since distributions u and f are regular and correspond to continuous functions, and the embedding C.˝/  D 0 .˝/ is one-to-one, it follows that u satisfies Poisson’s equation in the classical sense. We have illustrated on the case of the Dirichlet problem (8.1), the stages to be accomplished when the variational method is applied to general boundary value problems. These stages are as follows: Stage 1. Define the notion of a weak (generalized) solution. This first requires the definition of the functional space where solutions are looking for. Second, the definition of the weak solution should be done in a such way that classical solutions be covered. Stage 2. Use the variational method to establish the existence and uniqueness of the weak solution. By this approach, the solution appears as an extremum point, or more generally, as a critical point of a functional associated to the problem. Stage 3. Investigate the regularity of the obtained solution in terms of data smoothness. Stage 4. Return to classical solutions. Find sufficient smoothness conditions on data, to guarantee that the weak solutions are classical. For the rest of this section, we deal with the Dirichlet problem with a nonhomogeneous boundary condition. Theorem 8.4. Let ˝ be a bounded open set of class C 1 and let  D @˝: Then for each f 2 H 1 .˝/ and  2 H 1=2 . / ; there exists a unique function u 2 H 1 .˝/ which solves ² u D f in D 0 .˝/ uj D : Here notation uj is used to designate the trace of u on : Proof. From the definition of the space H 1=2 . / ; there is a function u1 2 H 1 .˝/ with u1 j D : Remark 7.7 guarantees that u1 2 H 1 .˝/ : Then the problem ² u D f C u1 in ˝ uD0 on @˝ has a unique weak solution u0 2 H01 .˝/ : Clearly u D u0 C u1 is the function we were looking for.

184

Chapter 8 Variational Theory of Elliptic Problems

The function guaranteed by the previous theorem can be considered the weak solution of the nonhomogeneous Dirichlet problem ² u D f in ˝ uD on @˝ for f 2 H 1 .˝/ and  2 H 1=2 . / : The proof of Theorem 8.4 shows that the nonhomogeneous Dirichlet problem can be reduced, via the trace operator, to the homogenous Dirichlet problem. To conclude this section, let us note that all the above results, as well as the variational method which was used, can be adapted to the case of elliptic operators in the divergence form: Lu D 

n X j;kD1

@ @xk

  @u aj k .x/ C a0 .x/ u @xj

D  div .A .x/ ru/ C a0 .x/ u:

8.2

The Variational Method for the Neumann Problem

In this section we follow for the Neumann problem, the same steps as in the case of the Dirichlet problem. In Part I we have considered the Neumann problem ² u C a0 .x/ u D f in ˝ (8.3) @u D0 on @˝ @ where ˝  Rn is a bounded open set, a0 2 C.˝/; a0 .x/  m > 0 in ˝: By a classical solution, when ˝ is of class C 1 and f 2 C.˝/; we have understood a function u 2 C 2 .˝/ which satisfied pointwise the equalities (8.3), and by a weak, or generalized solution we have meant a function u 2 H 1 .˝/ satisfying the variational identity a .u; v/  .f; v/L2 D 0; Z

Recall that a .u; v/ D

˝

v 2 H 1 .˝/ :

(8.4)

.ru  rv C a0 uv/ dx:

We have seen that for each f 2 L2 .˝/ ; the weak solution of the Neumann problem exists, is unique and minimizes the energy functional 1 E W H 1 .˝/ ! R; E .u/ D a .u; u/  .f; u/2 : 2

Section 8.2 The Variational Method for the Neumann Problem

185

As for the Dirichlet problem, for a test function ' 2 D .˝/ ; we have E 0 .uI '/ D a .u; '/  .f; '/L2 D  .u  a0 u C f; '/ where .u  a0 u C f; '/ stands for the action of the distribution ua0 uCf on the test function ': It follows that the weak solution of the Neumann problem satisfies the equation u C a0 u D f in the sense of distributions, i.e. u C a0 u D f

in D 0 .˝/ :

(8.5)

The converse assertion does not hold, that is, not any function u 2 H 1 .˝/ satisfying (8.5) is a generalized solution of the Neumann problem. The reason is that C01 .˝/ ¤ H 1 .˝/ : This allows us to assert that property (8.4) implicitly gives a certain meaning to the boundary condition @u=@ D 0; enough to guarantee the uniqueness of the solution. Additional regularity properties for the weak solution of the Neumann problem can be established if one takes into account the embedding theorems for Sobolev spaces. Thus, we may state the following regularity theorem. Theorem 8.5. Let u 2 H 1 .˝/ be the weak solution of the Neumann problem .8:3/: (a) If ˝ is of class C mC2 ; a0 2 C m .˝/ and f 2 H m .˝/ ; then u 2 H mC2 .˝/ and there is a constant C > 0 only depending on ˝ and m such that jujH mC2  C jf jH m : In particular, if m > n=2; then u 2 C 2 .˝/: (b) If ˝ is of class C 1 ; a0 2 C 1 .˝/ and f 2 C 1 .˝/; then u 2 C 1 .˝/: The next result gives sufficient conditions for that the weak solution of the Neumann problem be a classical solution. Theorem 8.6. Let ˝ be of class C 1 and f 2 C.˝/: If u 2 H 1 .˝/\C 2 .˝/ is the weak solution of problem .8:3/; then u is the classical solution of the Neumann problem. Proof. From u C a0 u  f D 0 in D 0 .˝/ ; and u C a0 u  f 2 C.˝/; it follows that u satisfies the equation u C a0 u  f D 0 in the classical sense. We now prove that @u=@ D 0 on @˝: From the definition of a weak solution, one has Z .ru  r' C a0 u'  f '/ dx D 0; ' 2 C 1 .˝/  H 1 .˝/ : ˝

186

Chapter 8 Variational Theory of Elliptic Problems

This together with the first Green’s formula and the already proved equality uC a0 u  f D 0; yield Z Z Z @u .' u C ru  r'/ dx D .u  a0 u C f / ' dx D 0: ' d D @˝ @ ˝ ˝ Z

Thus

@˝

@u ' d D 0 for all ' 2 C 1 .˝/: @

(8.6)

This implies @u=@ D 0 on @˝: Indeed, otherwise, it would exist a point x0 2 @˝; with @u=@ .x0 / ¤ 0: To fix ideas, assume that @u=@ .x0 / > 0: Then there is " > 0 with @u .x/ > 0 for all x 2 B " .x0 / \ @˝: @ If now in (8.6) we choose ' .x/ D " .x  x0 / ; we obtain Z Z @u @u .x/ " .x  x0 / d D .x/ " .x  x0 / d > 0; 0D @˝ @ B " .x0 /\@˝ @ a contradiction. Hence @u=@ D 0 on @˝ and thus u is a classical solution.

8.3

Maximum Principles for Weak Solutions

The classical weak maximum principle, Theorem 3.5, can be generalized for functions in the Sobolev space H 1 .˝/ : To this end, a sense should be conferred to the inequalities u  0 in ˝ and u  0 on @˝; when u 2 H 1 .˝/ : Definition 8.1. A distribution u 2 D 0 .˝/ is said to be nonnegative and we designate it by u  0 in D 0 .˝/ ; if its action on every nonnegative test function is a nonnegative number, i.e. .u; '/  0 for every ' 2 D .˝/ with '  0 in ˝: This notion allows us to define a partial-order relation in D 0 .˝/ ; as follows: Let u; v 2 D 0 .˝/ : We say that u  v; if distribution v  u is nonnegative, that is v  u  0 in D 0 .˝/ : Obviously, this relation in D 0 .˝/ is reflexive and transitive; to prove that it is antisymmetric we have to show that inequalities u  0 and u  0 imply u D 0: To this end, let us consider any test  C  C function ' 2 D .˝/ : We have ' D '  ' and 'k D ' k  .'  /k ;

Section 8.3 Maximum Principles for Weak Solutions

max

where ' C D ¹0; 'º ; '  D regularized k f: Now from

187

max ¹0; 'º ; and for any function f; fk

is its

j.D ˛ 'k / .x/  .D ˛ '/ .x/j ˇ ˇ  D ˇ k D ˛ ' .x/  .D ˛ '/ .x/ˇ Z  jD ˛ ' .y/  D ˛ ' .x/j k .x  y/ y

d

jxyj< k1

we immediately find that 'k ! ' in D .˝/ ; as k ! 1: since   Then, .u; / D 0 whenever 2 D .˝/ ;  0; and functions ' C k ; .'  /k belong to D .˝/ and are nonnegative, we have

  C    .u; 'k / D .u; '/ D u; ' k  .u; .'  /k / D 0:

lim

lim

k!1

k!1

Hence .u; '/ D 0 for every ' 2 D .˝/ ; i.e. u D 0; as wished. Notice that if a distribution u is regular, i.e. u 2 L1loc .˝/ ; then its positivity in the sense of Definition 8.1 reduces to the positivity of u as a function, i.e. to u .x/  0 for a.e. x 2 ˝: Definition 8.2. Let u 2 H 1 .˝/ : We say that u  0 on @˝ if u 2 H01 .˝/ : Similarly, u  0 on @˝ if uC 2 H01 .˝/ ; or equivalently if u  0 on @˝: A preorder (reflexive and transitive) relation can now be introduced in H 1 .˝/ if for any u; v 2 H 1 .˝/ we let u  v on @˝ whenever v  u  0 on @˝: It is clear that if u  0 on @˝ and u  0 on @˝; then u 2 H01 .˝/ : We shall prove that the converse of this assertion is also true having as a consequence the fact that the relation  on @˝ induces an order relation in the quotient space H 1 .˝/ =H01 .˝/ :



"

Proposition 8.2. Let u 2 H 1 .˝/ : Then uC ; u 2 H 1 .˝/ and ´ @u @uC .x/ if u .x/ > 0 @xj .x/ D @xj 0 if u .x/  0 ´ 0 if u .x/  0 @u .x/ D @u .x/ if u .x/ < 0: @xj @xj In particular, if u 2 H01 .˝/ ; then uC ; u 2 H01 .˝/ : For the proof we need the following lemma.

(8.7)

188

Chapter 8 Variational Theory of Elliptic Problems

Lemma 8.1. Let f 2 C 1 .R/ be a function with bounded derivative and let u 2 H 1 .˝/ : Then f ı u 2 H 1 .˝/ and   @u @ .f ı u/ D f0ıu ; @xj @xj

j D 1; 2; :::; n:

(8.8)

If in addition f .0/ D 0 and u 2 H01 .˝/ ; then f ı u 2 H01 .˝/ : Proof. We have f 0 ıu 2 L1 .˝/ and @u=@xj 2 L2 .˝/ ; whence .f 0 ı u/ @u=@xj 2 L2 .˝/ : It remains to prove relation (8.8), i.e. Z Z  @u  0 @' .f ı u/ dx D  ' dx; ' 2 D .˝/ : (8.9) f ıu @xj @xj ˝ ˝ To this end, fix any test function ' 2 D .˝/ : We shall prove (8.9) assuming without loss of generality, that ˝ is bounded of class C 1 (otherwise we replace ˝ by a bounded set ˝ 0 of class C 1 ; with supp '  ˝ 0  ˝/: Then, from Proposition 7.2, it follows that there is a sequence .uk /  C01 .Rn / with uk ! u in H 1 .˝/ ; as k ! 1: It is clear that for each k; one has Z Z  0  @uk @' .f ı uk / dx D  ' dx : f ı uk @xj @xj ˝ ˝ Now (8.9) is obtained if we pass to the limit as k ! 1 and we prove that f ı uk ! f ı u

and

 0   @u  @uk f ı uk ! f0ıu @xj @xj

in L2 .˝/ :

The first convergence follows from the estimation Z Z ˇ 0 ˇ2 2 2 ˇ ˇ jf ı uk  f ı uj dx  sup f juk  uj2 dx  C juk  ujL 2 .˝/ : ˝

˝

For the second convergence we use the inequality ! 12 ˇ Z ˇ ˇ 0  @u ˇ2  @uk  0 ˇ f ı uk ˇ dx  f ıu ˇ @xj @xj ˇ ˝ ˇ2 ! 12 Z ˇ ˇ 0ˇ ˇ @uk ˇ @u ˇ ˇ  sup ˇf ˇ ˇ @x  @x ˇ dx j j ˝ ! 12 ˇ ˇ Z ˇ 0 ˇ2 ˇ @u ˇ2   0 ˇ ˇ f ı uk  f ı u ˇ ˇ C : ˇ @x ˇ dx j ˝

Section 8.3 Maximum Principles for Weak Solutions

189

The first integral goes to zero as k ! 1 since uk ! u in H 1 .˝/ : The second integral also tends to zero, as a consequence of Lebesgue’s dominated convergence theorem, if we remark that f 0 ı uk ! f 0 ı u a.e. in ˝; eventually only for a subsequence (recall that from uk ! u in L2 .˝/ we have the pointwise convergence a.e. to u of a subsequence of .uk //: Assume in addition that f .0/ D 0 and u 2 H01 .˝/ : Then there exists a sequence .uk /  C01 .˝/ with uk ! u in H 1 .˝/ : From the first part of the lemma, f ı uk 2 H 1 .˝/ ; and since supp f ı uk  supp uk  ˝; Theorem 7.3 guarantees f ı uk 2 H01 .˝/ : Also, as above, f ı uk ! f ı u in H 1 .˝/ : Consequently, f ı u 2 H01 .˝/ : Proof of Proposition 8.2. Clearly uC ; u ; as well as the functions from the right-hand side of equalities (8.7), belong to L2 .˝/ : Let us now compute the distributional derivative @uC =@xj : For " > 0; consider the function f" 2 C 1 .R/ with bounded derivative, ´   2 C "2 1=2  "; t > 0 t f" .t / D 0; t  0: Apply Lemma 8.1 to obtain for any ' 2 D .˝/ ; Z Z u @u @' .f" ı u/ dx D  ' dx :   1=2 @xj @xj .u>0/ u2 C "2 ˝ Pass to the limit as " ! 0; to find Z Z @' @u uC dx D  ' dx @x @x j j .u>0/ ˝ which proves (8.7) for uC : Assume now that u 2 H01 .˝/ : Since f" .0/ D 0; one has f" ıu 2 H01 .˝/ : In addition, one immediately can check that f" ıu ! uC in H 1 .˝/ as " ! 0: Thus, uC 2 H01 .˝/ : The corresponding results for u are deduced using the equality u D .u/C : We are now ready to state and prove the weak maximum principle for functions from H 1 .˝/ : Theorem 8.7 (weak maximum principle for weak solutions). Let ˝  Rn be a bounded open set and u 2 H 1 .˝/ : If u  0 in D 0 .˝/ and u  0 on @˝; then u  0 a.e. in ˝:

190

Chapter 8 Variational Theory of Elliptic Problems

Proof. Since u  0 in D 0 .˝/ ; for every ' 2 D .˝/ with '  0; we have  n  X @u @' .u; '/ D  D  .u; '/H 1  0: ; 0 @xj @xj j D1

By density, we deduce .u; v/H 1  0 0

every v 2 H01 .˝/ with v  0: Now choose v D uC (recall  that @˝ means uC 2 H01 .˝/) and take into account that uC ; u H 1 D 0

for on (8.7)), to find

ˇ ˇ2     0  u; uC H 1 D uC  u ; uC H 1 D ˇuC ˇH 1 : 0

0

Hence

uC

u0 0 (see

 0 and so u D

u

0

 0 a.e. in ˝:

The next result about the positivity of the weak solution of the Dirichlet problem is a direct consequence of Theorem 8.7. Corollary 8.1. Let ˝  Rn be an open set and let f 2 H 1 .˝/ with f  0 in D 0 .˝/ : If u 2 H01 .˝/ is the weak solution of the Dirichlet problem ² u D f in ˝ uD0 on @˝; then u  0 a.e. in ˝: Remark 8.1. Corollary 8.1 tells us that the linear operator ./1 W H 1 .˝/ ! H01 .˝/ is positive (or isotone). Remark 8.2. Theorem 8.7 and Corollary 8.1 remain true for operator u C cu in the place of u; if and only if c < 1 ; where 1 is the first eigenvalue of the Dirichlet problem for the operator  (see Problems 8.2 and 8.3). The reader is advised to state and prove the maximum principle for a general elliptic operator in the divergence form. A generalized version of the strong maximum principle, Theorem 3.4, is also known (see Gilbarg–Trudinger [17, Theorem 8.19] for a proof): Theorem 8.8 (strong maximum principle for weak solutions). Let ˝  Rn be a connected open set and let u 2 H 1 .˝/ be such that u  0 in D 0 .˝/ : If for a closed ball B  ˝ one has ess

sup u B

then function u is constant in ˝:

D ess

sup u; ˝

Section 8.4 Regularity of Weak Solutions

8.4

191

Regularity of Weak Solutions

A regularity property of the weak solution of the Dirichlet problem is given by the next theorem. Theorem 8.9. Let ˝  Rn be an open set of class C 2 and with bounded boundary, or ˝ D RnC : Let f 2 L2 .˝/ and u 2 H01 .˝/ be a function with the property Z Z .ru  rv C uv/ dx D f v dx for all v 2 H01 .˝/ : (8.10) ˝

˝

Then u 2 H 2 .˝/ and there exists a constant C only depending on ˝ such that jujH 2  C jf jL2 : Proof. We shall discuss several cases: (a) the case ˝ D Rn I (b) the case ˝ D RnC I (c) the general case, with subcases: (i) interior regularity, i.e. in a bounded open subset ˝ 0 with ˝ 0  ˝ (when we shall be inspired by the case ˝ D Rn /I ( ii) regularity near the boundary (when we shall be inspired by the case ˝ D RnC /:

(a) The case ˝ D Rn : For each h 2 Rn n ¹0º ; denote .Dh u/ .x/ D

u .x C h/  u .x/ : jhj

If in (8.10) we choose v D Dh .Dh u/ ; we find 2 jDh ujH 1  jf jL2 jDh .Dh u/jL2  jf jL2 jDh ujH 1 :

Hence jDh ujH 1  jf jL2 ; whence ˇ ˇ ˇ ˇ ˇDh @u ˇ  jf j 2 ; L ˇ @x ˇ 2 k L

k D 1; 2; :::; n:

For h D t ej ; where ej is the unit vector of axis Oxj ; we deduce that ˇZ ˇ ˇ ˇ Z ˇ ˇ ˇ @u @' ˇˇ @u ˇ ˇ dx ˇ D ˇ lim D t ej ' dx ˇˇ ˇ t!0 ˝ @xk ˝ @xk @xj ˇ Z ˇ ˇ @u ˇˇ ˇ  lim ˇ' Dt ej @x ˇ dx t !0 ˝

 jf jL2 j'jL2

k

192

Chapter 8 Variational Theory of Elliptic Problems

for all ' 2 C01 .Rn / and j D 1; 2; :::; n: Now, thanks to Remark 7.1, we have that @u=@xk 2 H 1 .Rn / for every k; and so u 2 H 2 .Rn / :

(b) The case ˝ D RnC : The reasoning from Case 1 remains valid if direction h is parallel to the boundary of semispace ˝I Indeed, this happens since for h k @˝; we immediately can see that u 2 H01 .˝/ implies Dh u 2 H01 .˝/ : Thus the choice v D Dh .Dh u/ in (8.10) is still possible. Therefore, in this case again, for every ' 2 C01 .˝/ ; we have ˇ ˇZ ˇ @u @' ˇˇ ˇ dx ˇ  jf jL2 j'jL2 (8.11) ˇ ˝ @xk @xj but only for 1  j  n  1 and 1  k  n: Since Z Z @u @' @u @' dx D dx @x @x @x j j @xk ˝ ˝ k it follows that inequalities (8.11) hold for all j; k with j  n  1 or k  n  1: It remains to convince ourselves that an inequality of type (8.11) also holds for j D k D n: Indeed, from (8.10), we can deduce that there is a constant C > 0 with ˇ ˇ ˇ ˇ n1 Z ˇ ˇZ Z ˇ ˇ X ˇ ˇ @u @' @u @' ˇ .f '  u'/ dx ˇˇ dx ˇˇ D ˇˇ dx C ˇ ˝ @xn @xn ˝ ˇ j D1 ˝ @xj @xj ˇ  C jf jL2 j'jL2 : (c) The general case. (i) Interior regularity. Let ˝ 0 be a bounded open set with ˝ 0  ˝ and consider a function 0 2 C01 .˝/ such that 0 D 1 in a neighborhood of the compact ˝ 0 : Then the function v WD 0 u extended by zero outside ˝ belongs to H 1 .Rn / and solves in the distributions sense the equation v C v D g; where   g D 0 f  2r0  ru  u0 2 L2 Rn : According to Case 1, v 2 H 2 .Rn / and jvjH 2 .Rn /  C1 jgjL2 .Rn / : One can immediately see that jgjL2 .Rn /  C2 jf jL2 .˝/ : Thus, u 2 H 2 .˝ 0 / and jujH 2 .˝ 0 /  C jf jL2 .˝/ :

(ii) Regularity near the boundary. We shall consider, for simplicity, the case when ˝ is bounded. As in the proof of Theorem 7.2, we take a C 1 partition

Section 8.4 Regularity of Weak Solutions

193

P .k / of the unity and we represent u D m kD0 k u: We shall prove that k u 2 H 2 .˝/ for 1  k  m .0 u 2 H 2 .˝/ as already shown at case (i)). Recall that k 2 C01 .Uk / and there is a bijection  W Q ! Uk such that  2 C 2 .Q/;  D 1 2 C 2 .U k /;  .QC / D Uk \ ˝ and  .Q0 / D Uk \ @˝: Note that (see Theorem 7.3) v WD k u 2 H01 .˝ \ Uk / and v is the weak solution of the problem ² v D k f  k u  2rk  ru  uk  g in ˝ \ Uk vD0 on @ .˝ \ Uk / ; where g 2 L2 .˝ \ Uk / and jgjL2 .˝\Uk /  C jf jL2 .˝/ : The key idea of the proof is to make the change of variable y D  .x/ ; aimed to transport the problem from ˝ \ Uk to QC : So let w .y/ D v . .y// for y 2 QC : Then v .x/ D w . .x// for x 2 ˝ \ Uk : Let us show that w is the weak solution of an elliptic problem of the form 8   n <  P @ a .y/ @w D e g .y/ in QC ij @xj @xi (8.12) i;j D1 : wD0 on @QC ; where e g D .g ı / jdet Dj ; D is the Jacobi matrix of function  .y/ and functions aij 2 C 1 .QC / satisfy for a certain ˛ > 0; the ellipticity condition n X

for y 2 QC ; 2 Rn :

aij .y/ i j  ˛ j j2

i;j D1

To this end, let 2 H01 .QC / : Denote ' .x/ D 1 .˝ Clearly ' 2 H0 \ Uk / : We have

. .x// for x 2 ˝ \ Uk :

n n X X @' @w @i @ @j @v D and D : @xl @yi @xl @xl @yj @xl i D1

j D1

Thus Z

Z ˝\Uk

rv  r' dx D

˝\Uk

Z D

QC

Z D

n X @i @j @w @ dx @xl @xl @yi @yj

i;j;lD1 n X

i;j;lD1 n X

@i @j @w @ jdet Dj dy @xl @xl @yi @yj

aij .y/

QC i;j D1

@w @ dy; @yi @yj

(8.13)

194

Chapter 8 Variational Theory of Elliptic Problems

where aij .y/ D jdet Dj

n X @i @j : @xl @xl

lD1

Notice that aij 2 C 1 .QC / and since the Jacobi matrices D and D are nonsingular, there exists ˛ > 0 with n X i;j D1

n n X X @i aij .y/ i j D jdet D .y/j i @xl lD1

 ˛ j j2

i D1

for y 2 QC and 2 Rn : On the other hand, one has Z Z Z .g ı / jdet Dj dy D g' dx D ˝ \ Uk

!2

QC

QC

e g dy:

(8.14)

From (8.13) and (8.14) it follows that w is the weak solution of problem (8.12). Next we shall prove that g jL2 .QC / ; w 2 H 2 .QC / and jwjH 2 .QC /  C je whence coming back to the variable x; it will follow that v D k u 2 H 2 .˝/ and jvjH 2 .˝/  C jf jL2 .˝/ : For this, let D Dh .Dh w/ with h k Q0 and jhj small enough that 2 H01 .QC / I we note that supp w  ¹.y 0 ; yn / W jy 0 j < 1ı; 0 < yn < 1ıº for some ı > 0: Now using the fact that w is the weak solution of problem (8.12), we deduce that n Z X i;j D1 QC

 Dh

@w aij @yi



@Dh w dy D @yj

Z QC

e g Dh .Dh w/ dy:

We have Z QC

e g Dh .Dh w/ dy  je g jL2 jDh .Dh w/jL2  je g jL2 jDh wjH 1 : 0

On the other hand   @w @Dh w @w .y/ C Dh aij .y/ .y/ ; .y/ D aij .y C h/ Dh aij @yi @yi @yi

(8.15)

Section 8.4 Regularity of Weak Solutions

whence n Z X

 Dh

i;j D1 QC

@w aij @yi



195

@Dh w 2 dy  ˛ jDh wjH 1 C jwjH 1 jDh wjH 1 : (8.16) 0 0 @yj

From (8.15) and (8.16) by using Poincaré’s inequality, we find g jL2 .QC / /  C je g jL2 .QC / : jDh wjH 1 .QC /  C0 .jwjH 1 .QC / C je 0

(8.17)

From this, as in Case 2, we deduce that for every  2 C01 .QC / ; and all i; j with i  n  1 or j  n  1; we have ˇ ˇ ˇ ˇZ Z ˇ ˇ ˇ @w @ ˇˇ @w ˇ ˇ ˇ dy ˇ D ˇ lim D t ej dy ˇ ˇ ˇ t !0 QC @yi ˇ ˇ QC @yi @yj ˇ ˇ ˇ Z ˇ @w ˇˇ ˇ Dt ej dy ˇ D ˇ lim ˇ t !0 QC @yi ˇ  C je g jL2 .QC / jjL2 .QC / : Consequently, from Riesz representation theorem, ˇ 2 ˇ ˇ @ w ˇ @2 w 2 ˇ  C je g jL2 .QC / 2 L .QC / and ˇˇ @yi @yj @yi @yj ˇL2 .QC /

(8.18)

if i  n  1 or j  n  1: It remains to investigate the distribution @2 w=@yn2 : For this, we use again the fact that w is a weak solution. Since for  2 1 .Q / 1 C0 C ; we have =ann 2 H0 .QC / ; it follows n Z X i;j D1

whence

@w @ aij @yi @yj QC



 ann



Z d

yD

QC

e g

 dy ; ann

   @w @ dy ann @yi @yj ann QC   Z X Z   @w @ D dy: e g dy  aij @yi @yj ann QC ann QC Z

i Cj n=2; then u 2 C 2 .˝/: If ˝ is of class C 1 and f 2 C 1 .˝/; then u 2 C 1 .˝/: Proof. The proof is achieved by induction over m 2 Theorem 8.9. Assume now that m D 1:

N:

For m D 0 one has

The case ˝ D Rn : For each k 2 ¹1; 2; :::; nº ; the following relation holds:   Z Z     @u @f @u r  rv C v dx D v dx; v 2 C01 Rn : (8.19) @xk @xk ˝ ˝ @xk

To see this it suffices to replace v by @v=@xk in (8.10). Then, from Theorem 8.9, it follows that ˇ ˇ ˇ 2 ˇ ˇ @f ˇ ˇ @ u ˇ  n @2 u 1 ˇ Cˇ ˇ 2H R and ˇˇ ˇ @x ˇ 2 @xk @xj @xk @xj ˇH 1 k L for all 1  j; k  n: Hence u 2 H 3 .Rn / and jujH 3  C jf jH 1 :

The case ˝ D RnC : We also use equality (8.19) and the addition remark that if f 2 H 1 .˝/ ; u 2 H 2 .˝/ \ H01 .˝/ and u satisfies (8.10), then @u=@xk 2 H01 .˝/ for 1  k  n  1: Indeed, for h k @˝; we have jDh ujH 1  jujH 2 :

If we choose h D ˛ek ; where 1  k  n  1; there exists a sequence ˛l ! 0 with D˛l ek u ! g weakly in H01 .˝/ as l ! 1: Consequently, if we take the limit in the equality .Dh u; '/L2 D .u; Dh '/L2 ;

' 2 C01 .˝/

Section 8.4 Regularity of Weak Solutions

we obtain .g; '/L2

  @' D u; ; @xk L2

197

' 2 C01 .˝/ :

Thus @u=@xk D g 2 H01 .˝/ for 1  k  n  1: From Theorem 8.9 we deduce that ˇ 2 ˇ ˇ @ u ˇ @2 u 1 ˇ  C jf j 1 for j C k < 2n: 2 H .˝/ and ˇˇ H @xk @xj @xk @xj ˇH 1 It remains to show that @2 u 2 H 1 .˝/ 2 @xn

ˇ 2 ˇ ˇ@ uˇ and ˇˇ 2 ˇˇ  C jf jH 1 : @xn H 1

To this end, observe that from (8.10) one has X @2 u @2 u D f  u C @xn2 @xj2 j D1 n1



in D 0 .˝/ ;

where the distribution in the right-hand side is, based on the above considerations, a function from H 1 .˝/ ; whose norm in H 1 .˝/ is dominated by C jf jH 1 : The general case. With the notations from Theorem 8.9, we have aij 2 g 2 H 1 .QC / : Next we proceed like in the previous case. One C 2 .QC / and e concludes that for 1  k  n  1; the function w e D @w=@yk is the weak solution of the problem 8     n n P @aij @w @e g <  P @ a @e @ w C D in QC ij @yi @yj @yk @yj @yk @yi i;j D1 i;j D1 : w eD0 on @QC : g jL2 .QC / ; we find Using Theorem 8.9 and inequality jwjH 2 .QC /  C je ˇ 2 ˇ @ w ˇ ˇ @y @y k

j

ˇ ˇ ˇ @e ˇ w ˇ D ˇˇ  C je g jH 1 .QC / @yj ˇH 1 .QC / H 1 .QC /

ˇ ˇ ˇ ˇ

for 1  k  n  1 and 1  j  n: Finally, from (8.12), we have   X @ann @w @2 w @ @w ann 2 D e gC C aij in D 0 .QC / @yn @yn @yn @yj @yi i Cj 0 in ˝: Proof. (a) According to Theorem 8.9, for each smooth bounded open subset ˝ 0 with ˝ 0  ˝; we have k 2 H 2 .˝ 0 /: Furthermore, Corollary 8.2 implies successively that k 2 H 3 .˝ 0 / ; k 2 H 4 .˝ 0 / and so on. Thus k 2 T 1 m 0 1 .˝ 0 / : It follows that  2 C 1 .˝/ : k mD1 H .˝ / : Consequently, k 2 C (b) We now prove that 1 is simple, i.e. 1 < 2 ; and that 1 admits a positive eigenfunction 1 : To this end, first note that if the minimum in (3.35) is reached for some function 1 ; then it is also reached for its absolute value j1 j :

Section 8.5 Regularity of Eigenfunctions

199

Then from the proof of Theorem 3.16, one deduces that both 1 and j1 j are eigenfunctions. Hence  j1 j D 1 j1 j in ˝: It follows that the function p

.x; t/ D j1 .x/j e

1 t

.x 2 ˝; t 2 R/

is a nonnegative harmonic function in ˝  R: Now Harnack’s inequality, more exactly Corollary 3.8, implies > 0 in ˝  R; whence j1 j > 0 in ˝: Then 1 D j1 j (or 1 D  j1 j). Thus each eigenfunction corresponding to 1 is either positive, or negative in ˝: As a result, any two eigenfunctions corresponding to 1 cannot be orthogonal in L2 .˝/ : This shows that the linear space generated by the eigenfunctions corresponding to 1 is one-dimensional, i.e. 1 is simple. (c) Finally, the fact that k 2 L1 .˝/ follows from the next lemma, if we take into account that both functions kC ; k belong to H01 .˝/ (see Proposition 8.2). Lemma 8.2. Let > 0 and u 2 H 1 .˝/ : (a) If u C u  0 in D 0 .˝/ and u  0 on @˝; then ˇ ˇ u .x/  C ˇuC ˇ 2 a.e. in ˝: L

(8.21)

(b) If u C u  0 in D 0 .˝/ and u  0 on @˝; then u .x/  C ju jL2

a.e. in ˝:

Proof. We shall prove assertion (a); assertion (b) is then obtained if we replace u by u in (a). For ˛  1 and N > 0; consider the function h 2 C 1 Œ0; 1/; ² ˛ t ; 0t N h .t / D ˛N ˛1 t C .1  ˛/ N ˛ ; t > N and denote

Z g .t / D uC

t

h0 .s/2 ds:

0

2 H01 .˝/ : Then, in view of Lemma 8.1, the Since u  0 on @˝; one has functions h ı uC ; g ı uC also belong to H01 .˝/ : Now from u C u  0 in D 0 .˝/ ; we deduce that Z .ru  rv  uv/ dx  0 for every v 2 H01 .˝/ ; v  0: ˝

200

Chapter 8 Variational Theory of Elliptic Problems

Choose v D g ı uC and use (8.7) together with inequality g .t /  tg 0 .t / .t  0/; to deduce Z Z ˇ2    ˇ g 0 uC ˇruC ˇ dx 

ug uC dx ˝ Z˝    uC g uC dx ˝

Z



   2 g 0 uC uC dx;

˝

Z

that is,

˝

ˇ  C ˇ2 ˇrh u ˇ dx 

Hence

ˇ  C ˇ ˇh u ˇ

H01



Z ˝

ˇ 0  C  C ˇ2 ˇh u u ˇ dx:

p ˇ   ˇ

ˇh0 uC uC ˇL2 :

(8.22)

Consider for simplicity only the case n  3: By Sobolev’s embedding theorem, Theorem 7.9, we obtain ˇ  C ˇ ˇ   ˇ ˇh u ˇ 2  c ˇh0 uC uC ˇ 2p=.p2/ ; L L where 2 D 2n= .n  2/ ; p > n is fixed and c D c .n; ˝/ : Letting N ! 1; we find ˇ Cˇ ˇ ˇ ˇu ˇ ˛2  .c ˛/ ˛1 ˇuC ˇ 2˛p : L L p2

Denote q D 2p= .p  2/ and r D n .p  2/ = Œp .n  2/ : We have 2 < q < 2 ; r > 1 and ˇ ˇ ˇ Cˇ ˇu ˇ ˛qr  .c ˛/ ˛1 ˇuC ˇ ˛q : L

uC

L

.˝/ yields the stronger relation uC 2 2 This shows that relation ˛qr .˝/ : Consider now successively ˛ D 1; r; r 2 ; r 3 ; :::; r N 1 ; where N is L any positive integer, to deduce ˇ Cˇ ˇu ˇ

N Lr q



N 1 Y



cr m

L˛q

r m ˇ C ˇ ˇu ˇ

Lq

ˇ ˇ ˇ ˇ  c a r b ˇuC ˇLq  C ˇuC ˇLq :

mD0

Here aD

N 1 X mD0

r m ; b D

N 1 X mD0

mr m and C D C .n; p; ˝/ :

(8.23)

Section 8.6 Problems

201

Since for each m  1; there is N with m  r N q; it follows that uC 2 T m 1m C ˇuC ˇLq in ˝0 : Then, for every N; we would have ˇ ˇ ˇ ˇ ˇ ˇ C ˇuC ˇLq  ˇuC ˇLr N q .˝/  ˇuC ˇLr N q .˝

1

0/

 M  rN q ;

ˇ ˇ whence letting N ! 1; we would derive the contradiction C ˇuC ˇLq  M: It is clear that (8.24) gives uC 2 L1 .˝/ ; and together with (8.22) yields (8.21).

8.6 1.

Problems The method of separation of variables leads us in case of the Dirichlet problem ² u D 0 for < R uDg for D R on the disc ˝ D ¹x 2 R2 W jxj < Rº; to the expression uD

1  k X a0 .ak cos k' C bk sin k'/ C : 2 R

(8.25)

kD1

Formally, for D R; we require 1 X a0 .ak cos k' C bk sin k'/ D g: C 2 kD0

The system of functions cos k'; sin k' being orthogonal and complete in L2 .0; 2/ ; we find for g 2 L2 .0; 2/ W 1 ak D 

Z

2

g .'/ cos k' d'; 0

1 bk D 

Z

2

g .'/ sin k' d': 0

202

Chapter 8 Variational Theory of Elliptic Problems

Prove that if g 2 C 1Œ0; 2 and g .0/ D g .2/ ; then series (8.25)  is convergent in C ˝ and defines a function u 2 C 1 .˝/ \ C ˝ ; harmonic in ˝; whose restriction to @˝ coincides with g: Hint. We use the classical result on Fourier series saying that if g 2 C 1 Œ0; 2 and g .0/ D g .2/ ; then the Fourier series of g converges to g uniformly on Œ0; 2 (see, for example, S.M. Nikolsky, [31], Vol. 2, p. 205, Theorem 15.5.2). Denote by Sm the partial sum of series (8.25) and by sm the partial sum of the Fourier series of g: Then ² Sm D 0 in ˝ Sm D sm on @˝: Also, using (3.9), we deduce that ˇ ˇ ˇ ˇ ˇSmCp  Sm ˇ  ˇsmCp  sm ˇ

C .@˝/

C.˝/

! 0 as m ! 1:

2 D .˝/ ; Hence Sm ! u in C.˝/ for some function u 2 C.˝/: If then 0 D .Sm ; / D .Sm ;  / ! .u;  / D .u; / ; which shows that u satisfies Laplace’s equation in the sense of distributions. Weyl’s lemma, Proposition 6.8, guarantees that u is harmonic in ˝: 2.

Let ˝  Rn be a bounded open set, u 2 H 1 .˝/ and c < 1 : Prove that if u C cu  0 in D 0 .˝/ and u  0 on @˝; then u  0 a.e. in ˝ .the maximum principle holds for the operator u C cu if c < 1 /: Hint. Assume the contrary. Then uC ¤ 0 and from ˇ ˇ2 ˇ ˇ2  ˇuC ˇH 1 C c ˇuC ˇL2  0 0

we deduce that

ˇ C ˇ2 ˇu ˇ 1 H0 ˇ ˇ2  c < 1 ; ˇuC ˇ 2 L

which contradicts (3.35). 3.

Conversely, if the maximum principle holds for the operator u C cu; then c < 1 : Hint. Assume that c  1 : Then, since '1 > 0 in ˝; we should have '1 C c'1  '1 C 1 '1 D 0 in ˝; '1  0 on @˝; whence '1  0 in ˝; a contradiction.

Section 8.6 Problems

4.

5.

203

Prove that if c < 1 ; then the linear operator .  cI /1 W H 1 .˝/ ! H01 .˝/ is positive, i.e. f 2 H 1 .˝/ and f  0 imply .  cI /1 f  0 (equivalently, isotone). Let ˝  Rn be a bounded open set, c < 1 ; c0 > 0 and u 2 H01 .˝/ such that u C cu C c0  0 in D 0 .˝/ and u  0 on @˝: Then there is an " > 0 depending only on c; c0 ; such that u  "1 a.e. in ˝: Hint. Let v WD "1 : Since 1 D  1 1 and 1 2 L1 .˝/ ; we can choose " > 0 small enough that v C cv C c0  0 in ˝: Then  .v  u/ C c .v  u/  0 in D 0 .˝/ and v  u  0 on @˝: The maximum principle now gives v  u  0 a.e. in ˝:

6.

If f 2 L2 .˝/ and u 2 H 1 .˝/ is the weak solution of the Neumann problem ² u C u D f in ˝ @u D0 on @˝; @ then

ess inf f  u .x/  ess sup f ˝

˝

for a.e. x 2 ˝ (Maximum principle for the weak solution of the Neumann problem). Hint. Let M D ess sup f: According to Proposition 8.2, .u  M /C 2 ˝

H 1 .˝/ : Then Z Z ˇ ˇ2 ˇ Cˇ .f  u/ .u  M /C dx ˇr .u  M / ˇ dx D ˝ Z˝ .M  u/ .u  M /C dx  ˝ Z ˇ ˇ2 ˇ Cˇ .u /  M D  ˇ ˇ dx: ˝

It follows that .u  M /C D 0; that is, u  M:

Part III

Semilinear Equations

Part III Semilinear Equations

207

The goal in this Part III is to suggest how the theory of linear partial differential equations can be used to treat boundary value problems for semilinear equations. Nonlinear boundary value problems represent a vast, expanding and extremely captivating domain of the theory of partial differential equations. They arise from mathematical modeling of lots of real processes from physics, biology, chemistry, engineering, economics, etc. Their investigation requires results of the theory of linear PDEs, functional analysis, operator theory, topology, measure theory, etc. Our approach to nonlinear problems will be mainly based on the operator method. This consists in rewriting the boundary value problem, say of the type ² Lu D F .u/ u 2 D .L/ .L W D .L/  X ! Y; F W X ! Y / ; as a fixed point problem, namely u D L1 F .u/ ; u 2 X; and then in applying abstract results of nonlinear functional analysis, such as Banach’s contraction principle, Schauder’s and the Leray–Schauder fixed point theorems, critical point theorems, monotone iterative methods etc. Clearly, the key of the method is the inverse of the operator L; that we call the solution operator. It associates to each element f 2 Y; the unique solution u of the problem for the corresponding nonhomogeneous equation, i.e. ² Lu D f u 2 D .L/ : Also, the properties of the solution operator are essential for the applicability of the abstract results of nonlinear analysis. In this book, L is a linear partial differential operator of elliptic, parabolic, or hyperbolic type and its domain D .L/ is defined by taking into account boundary and initial conditions. In Chapter 3 we have defined the solution operator for the Dirichlet problem and the elliptic operator L D ; simply denoted by ./1 ; as a mapping from L2 .˝/ to H01 .˝/ ; while in Chapter 8 we have extended it from H 1 .˝/ to H01 .˝/ : This extension will allow us in the next Chapter 9, to treat semilinear equations with nonlinearities having a superlinear growth. Similarly, for evolution equations, it will be useful to extend the notion of solution by considering the space H 1 .˝/ instead of L2 .˝/ : We shall give this extension, and thus define the solution operator, for the heat equation (Chapter 10), wave equation (Chapter 11) and Schrödinger equation (Chapter 12). Then the operator method specifically applies to semilinear problems for these types of equations, following the same programme.

Chapter 9

Semilinear Elliptic Problems

In this chapter, we shall restrict ourselves to discuss the solvability of the semilinear Dirichlet problem ² u D g .x; u; ru/ C f in ˝ (9.1) uD0 on @˝; where ˝  Rn is a bounded open set, f 2 H 1 .˝/ and g W ˝  RnC1 ! R is a given function. Here the term g.x; u; ru/ perturbs f converting the linear equation u D f into a semilinear equation. We seek weak solutions, i.e. functions u 2 H01 .˝/ with g .:; u; ru/ 2 H 1 .˝/ and .u; v/H 1 D .g .x; u; ru/ C f; v/ 0

for all v 2 H01 .˝/ :

We shall present some existence results for problem (9.1), obtained by using the Banach, Schauder and Leray–Schauder fixed point theorems, the monotone iterative method and the critical point technique. Let us first note that problem (9.1) is equivalent to the fixed point problem u D A .u/ ; where and

u 2 H01 .˝/ ;

(9.2)

A D ./1 .B C f / B W H01 .˝/ ! H 1 .˝/ ; B .u/ D g .:; u; ru/ :

Obviously, the first problem which has to be clarified is the correct definition of the operator B: The following section is devoted to this problem.

9.1

The Nemytskii Superposition Operator

Definition 9.1. Let ˝  RN be a bounded open set and g W ˝  Rn ! Rm be any function. By the superposition operator or Nemytskii operator associated to g; we mean the mapping Ng which assigns to each function w W ˝ ! Rn ; the function Ng .w/ W ˝ ! Rm ; given by Ng .w/ .x/ D g .x; w .x//

.x 2 ˝/ :

Section 9.1 The Nemytskii Superposition Operator

209

The properties of the operator Ng depend upon the properties of the function g: Here we are interested that operator Ng apply the space Lp .˝I Rn / ; or more generally, the space Lp1 .˝I Rn1 /  Lp2 .˝I Rn2 / with n1 C n2 D n; into the space Lq .˝I Rm / : Definition 9.2. We say that a function g W ˝  Rn ! Rm satisfies the Carathéodory conditions if: (i) function g .:; y/ W ˝ ! Rm is measurable for every y 2 Rn I (ii) function g .x; :/ W Rn ! Rm is continuous for a.e. x 2 ˝: Proposition 9.1. If g satisfies the Carathéodory conditions, then Ng maps measurable functions into measurable functions. Proof. Let w W ˝ ! Rn be a measurable function. Then there is a sequence .wk / of functions with a finite number of values such that wk .x/ ! w .x/ as k ! 1; for a.e. x 2 ˝: In view of condition (ii), we then have Ng .wk / .x/ D g .x; wk .x// ! g .x; w .x// D Ng .w/ .x/

(9.3)

as k ! 1; for a.e. x 2 ˝: On the other hand, function wk having a finite number of values, can be represented as a finite sum X wk .x/ D j .x/ yj ; j

where yj 2 Rn ; j is the characteristic function of a subset ˝j  ˝; [ ˝D ˝j and ˝i \ ˝j D ; for i ¤ j: j

It follows that

ˇ   Ng .wk /ˇ˝ D g :; yj : j

Then, according to condition (i), we have that the restriction of Ng .wk / to each subset ˝j is measurable. Hence Ng .wk / is measurable on ˝: Finally (9.3) implies that Ng .w/ is measurable, as a limit of measurable functions. Theorem 9.1. Let ˝  RN be a bounded open set, g W ˝  Rn ! Rm and 1  p; q < 1: If g satisfies the Carathéodory conditions and there exists c 2 RC and h 2 Lq .˝I RC / such that p

jg .x; y/j  c jyj q C h .x/

210

Chapter 9 Semilinear Elliptic Problems

for every y 2 Rn and a.e. x 2 ˝; then the operator     Ng W Lp ˝I Rn ! Lq ˝I Rm ; Ng .w/ D g .:; w/ is well defined, continuous, and bounded. Moreover p ˇ ˇ    q ˇNg .w/ˇ q  c C jhjLq .˝/ w 2 Lp ˝I Rn : jwj m Lp .˝IRn / L .˝IR /

(9.4)

Proof. (1) First we show that Ng is a bounded well-defined operator. Indeed, Proposition 9.1 guarantees that Ng maps measurable functions into measurable functions. Furthermore, if w 2 Lp .˝I Rn / ; then using the triangle norm inequality we obtain  q1 ²Z  Z q ³ q1 ˇ ˇ p ˇNg .w/ˇq dx q  c jw .x/j C h .x/ dx ˝

˝

p

 c jwjLq p C jhjLq ;

which proves inequality (9.4), shows that Ng .w/ 2 Lq .˝I Rm / and that Ng is bounded, i.e. maps bounded sets into bounded sets. (2) To prove that Ng is continuous we need Vitali’s lemma: Lemma 9.1 (Vitali). Let ˝  RN be a bounded open set and let .uk / be a sequence of functions from Lp .˝I Rn / .1  p < 1/ with uk .x/ ! u .x/ as k ! 1; for a.e. x 2 ˝: Then u 2 Lp .˝I Rn / and uk ! u in Lp .˝I Rn / as k ! 1; if and only if for each " > 0 there exists a ı" > 0 such that Z (9.5) juk jp dx < " D

for all k and every D  ˝ with  .D/ < ı" : Let wk 2 Lp .˝I Rn / be such that wk ! w in Lp .˝I Rn / as k ! 1: Then there exists a subsequence .wk 0 / of .wk / with wk 0 .x/ ! w .x/ for a.e. p x 2 ˝; and the necessity part of Vitali’s lemma implies that jwk 0 jLp .DIRn / < " whenever  .D/ < ı" : Then, on one hand Ng .wk 0 / .x/ ! Ng .w/ .x/ for a.e. x 2 ˝; and on the other hand, if  .D/ < ı" ; we have p ˇ ˇ ˇNg .wk 0 /ˇ q  c jwk 0 jLq p .DIRn / C jhjLq .D/ L .DIRm / 1

 c " q C jhjLq .D/ : Hence the conditions of Vitali’s lemma are also satisfied for the sequence of functions Ng .wk 0 / : Then the sufficiency part in Vitali’s lemma guarantees that q m that the whole sequence Ng .wk 0 /  ! Ng .w/ in L .˝I R q/ : It follows Ng .wk / converges to Ng .w/ in L .˝I Rm / : Thus Ng is continuous.

Section 9.2 Application of Banach’s Fixed Point Theorem

211

Remark 9.1. In a similar way, one can prove the following result: If g W ˝  Rn ! Rm satisfies the Carathéodory conditions, n D n1 C n2 and there are c1 ; c2 2 RC and h 2 Lq .˝I RC / such that ˇ ˇ p1 ˇ ˇ p2 ˇ  ˇ ˇg x; y 1 ; y 2 ˇ  c1 ˇy 1 ˇ q C c2 ˇy 2 ˇ q C h .x/ for all y 1 2 Rn1 ; y 2 2 Rn2 and a.e. x 2 ˝; then the operator       Ng W Lp1 ˝I Rn1  Lp2 ˝I Rn2 ! Lq ˝I Rm ;     Ng w 1 ; w 2 .x/ D g x; w 1 .x/ ; w 2 .x/ is well defined, continuous, and bounded. Moreover, ˇ  1 2 ˇ ˇ 1 ˇ pq1 ˇ 2 ˇ pq2 ˇNg w ; w ˇ q ˇ ˇ ˇ ˇ C c  c w 1 2 w Lp2 .˝IRn2 / C jhjLq .˝/ : Lp1 .˝IRn1 / L .˝IRm /

9.2

Application of Banach’s Fixed Point Theorem

Before stating the existence results, it is useful to recall some basic results from the theory of Sobolev spaces. Lemma 9.2. Let ˝  Rn be a bounded open set. Then ˇ ˇ ˇ ˇ (a) ˇ./1 f ˇ 1 D jf jH 1 for every f 2 H 1 .˝/ I H0

(b) jujL2 

p1 1

(c) jf jH 1 

jujH 1 for every u 2 H01 .˝/ I

p1 1

0

jf jL2 for every f 2 L2 .˝/ :

Proof. (a) The equality expresses the property of the operator ./1 of being an isometry between H 1 .˝/ and H01 .˝/ (see Theorem 7.15). (b) The relation is Poincaré’s inequality. (c) Let f 2 L2 .˝/ and u D ./1 f: Then u D f and so .u; v/H 1 D .f; v/ D .f; v/L2 for all v 2 H01 .˝/ : 0

For v D u; we obtain 2 jujH 1 D .f; u/L2  jf jL2 jujL2 ; 0

212

Chapter 9 Semilinear Elliptic Problems

whence using (b) we deduce 1 2 jujH jf jL2 jujH 1 : 1  p 0 0

1 Thus

1 jujH 1  p jf jL2 0

1 ˇ ˇ ˇ ˇ and (c) is proved since jf jH 1 D ˇ./1 f ˇ 1 D jujH 1 : H0

0

Now we are ready to state and prove an existence and uniqueness result for the semilinear problem (9.1). Theorem 9.2. Let ˝  Rn be a bounded open set, f 2 H 1 .˝/ and g W ˝  RnC1 ! R a function with the following properties: (i) g satisfies the Carathéodory conditions, i.e. g .:; y/ W ˝ ! R is measurable for each y 2 RnC1 and g .x; :/ W RnC1 ! R is continuous for a.e. x 2 ˝I (ii) g .:; 0; 0/ D 0 and there are constants a; b 2 RC with jg .x; u; v/  g .x; u; v/j  a ju  uj C b jv  vj for every u; u 2pRI v; v 2 Rn and a.e. x 2 ˝I (iii) a= 1 C b= 1 < 1; where 1 is the smallest eigenvalue of the Dirichlet problem for the operator : Then problem .9:1/ has a unique solution u 2 H01 .˝/ : In addition u is the limit in H01 .˝/ of the sequence of successive approximations uk WD Ak .u0 / ; k  1; for every u0 2 H01 .˝/ : Proof. The Carathéodory conditions guarantee that for every measurable function w W ˝ ! RnC1 ; the function g .:; w .:// W ˝ ! R is also measurable.  If 1 .˝/ 2 nC1 and u 2 H0 ; then the couple w WD Œu; ru belongs to L ˝I R so is a measurable function. It follows that the function B .u/ is measurable. In addition jB .u/ .x/j D jg .x; u .x/ ; ru .x//j D jg .x; u .x/ ; ru .x//  g .x; 0; 0/j  a ju .x/j C b jru .x/j : Since a juj C b jruj 2 L2 .˝/ ; we deduce that B .u/ 2 L2 .˝/ : Then, in view of L2 .˝/  H 1 .˝/ ; we have B .u/ 2 H 1 .˝/ : This shows that the operator B is well defined from H01 .˝/ to H 1 .˝/ :

Section 9.3 Application of Schauder’s Fixed Point Theorem

213

Next we prove that conditions (ii), (iii) imply that A is a contraction on Indeed, for every u1 ; u2 2 H01 .˝/ ; using Lemma 9.2, we obtain

H01 .˝/ :

ˇ ˇ ˇ ˇ jA .u1 /  A .u2 /jH 1 D ˇ./1 ŒB .u1 /  B .u2 /ˇ

H01

0

1 D jB .u1 /  B .u2 /jH 1  p jB .u1 /  B .u2 /jL2

1  1  p a ju1  u2 jL2 C b jr .u1  u2 /jL2 .˝IRn / :

1 Since 1 jr .u1  u2 /jL2 .˝IRn / D ju1  u2 jH 1 and ju1  u2 jL2  p ju1  u2 jH 1 ; 0 0

1 we deduce that



jA .u1 /  A .u2 /jH 1  0

 a b Cp ju1  u2 jH 1 : 0

1

1

The conclusion now follows from the well-known Banach fixed point theorem.

9.3

Application of Schauder’s Fixed Point Theorem

For the next result we shall assume instead of the Lipschitz condition that function g .x; u; v/ has a growth at most linear in u and v: Under this less restrictive condition, the conclusion derived from Schauder’s fixed point theorem will be weaker, namely, we shall obtain only existence and neither uniqueness and nor approximation of the solution. First we recall the notion of a completely continuous operator required by Schauder’s fixed point theorem. Definition 9.3. An operator T W D  X ! Y; where X and Y are Banach spaces, is said to be completely continuous, if it is continuous and maps any bounded subset of D into a relatively compact set of Y: It follows from the definition that every completely continuous operator is a bounded operator (i.e. maps bounded sets into bounded sets). Also, we immediately can see that the composition of two or more bounded and continuous operators, of which at least one is completely continuous, is completely continuous too. From what follows will result the extremely important role of compact embedding theorems in guaranteeing the complete continuity of the nonlinear operators associated to boundary value problems.

214

Chapter 9 Semilinear Elliptic Problems

Theorem 9.3 (Schauder). Let X be a Banach space, D a nonempty closed convex bounded subset of X and let A W D ! D be a completely continuous operator. Then A has at least one fixed point. For a proof and additional information about completely continuous operators, we refer the reader to Precup [38]. Theorem 9.4. Let ˝  Rn be a bounded open set, f 2 H 1 .˝/ and g W ˝  RnC1 ! R a function with the following properties: (i) g satisfies the Carathéodory conditions; (ii) there exist a; b 2 RC and h 2 L2 .˝I RC / such that jg .x; u; v/j  a juj C b jvj C h .x/

(9.6)

for every u 2 R; v 2 Rn and a.e. x 2 ˝I (iii) a1 C pb < 1: 1

Then problem .9:1/ has at least one solution u 2 H01 .˝/ : Proof. Clearly ./1 f 2 H01 .˝/ : Next we shall deal with the operator ./1 B: We have B D J ı Ng ı P; where

  P W H01 .˝/ ! L2 ˝I RnC1 ; P .u/ D Œu; ru ;

Ng is the Nemytskii operator associated to g;   Ng W L2 ˝I RnC1 ! L2 .˝/ ; Ng .w/ .x/ D g .x; w .x// ; and J is the embedding operator J W L2 .˝/ ! H 1 .˝/ ; J .u/ D .u; :/L2 : It is clear that P is a continuous linear operator, hence bounded. Conditions (i) and (ii) guarantee that Ng is well defined, continuous, and bounded. Furthermore, the embedding of L2 .˝/ in H 1 .˝/ being compact, we have that J is a completely continuous linear operator. Consequently, the composed operator B D J ı Ng ı P is completely continuous from H01 .˝/ to H 1 .˝/ : Thus the composition of B with the continuous linear operator ./1 W H 1 .˝/ ! H01 .˝/ is a completely continuous operator from H01 .˝/ to itself. Therefore, the operator A W H01 .˝/ ! H01 .˝/ ; A .u/ D ./1 .B .u/ C f /

Section 9.4 Application of the Leray–Schauder Fixed Point Theorem

215

is completely continuous. Next, for each u 2 H01 .˝/ ; we have ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ jA .u/jH 1  ˇ./1 B .u/ˇ 1 C ˇ./1 f ˇ 1 0

H0

H0

D jB .u/jH 1 C jf jH 1 1  p jB .u/jL2 C jf jH 1

1  1   p a jujL2 C b jr .u/jL2 .˝IRn / C jhjL2 C jf jH 1

  1 a b  Cp jujH 1 C c; 0

1

1 where

p c D jf jH 1 C jhjL2 = 1 :

This together with (iii) implies the existence of a sufficiently large R > 0 such that jujH 1  R implies jA .u/jH 1  R: 0

0

H01 .˝/

Thus A is a self-map of the closed ball of with center at origin and radius R: The conclusion now follows from Schauder’s fixed point theorem.

9.4

Application of the Leray–Schauder Fixed Point Theorem

A justified question is if the existence of solutions to problem (9.1) can be established when function g .x; u; v/ satisfies (9.6) without the restriction that constants a; b are sufficiently small. Such kind of results can be deduced from the Leray–Schauder fixed point theorem if we assume a sign condition. Moreover, a superlinear growth of g in u and v is possible. Theorem 9.5 (Leray–Schauder). Let .X; j:jX / be a Banach space, R > 0 and A W B R .0I X / ! X a completely continuous operator. Assume that jujX < R .strictly/ for every solution u of the equation u D A .u/ and every 2 .0; 1/ : Then A has at least one fixed point. The proof and several applications can be found in monographs O’Regan– Precup [33] and Precup [38].

216

Chapter 9 Semilinear Elliptic Problems

Theorem 9.6. Let ˝  Rn .n  3/ be a bounded open set, f 2 H 1 .˝/ and g W ˝  RnC1 ! R a function with the following properties: (i) g satisfies the Carathéodory conditions; (ii) there exist a; b; ˛; ˇ 2 RC with 1  ˛ < 2  1 D .n C 2/ = .n  2/ ; 1  ˇ < 2= .2 /0 D 1 C 2=n and h 2 L2 .˝I RC / such that jg .x; u; v/j  a juj˛ C b jvjˇ C h .x/ for all u 2 R; v 2 Rn and a.e. x 2 ˝I (iii) the sign condition u g .x; u; v/  0

holds for all u 2 R; v 2 Rn and a.e. x 2 ˝: Then problem .9:1/ has at least one solution u 2 H01 .˝/ :

min

Proof. Let p1 D 2 =˛; p2 D 2=ˇ and p D ¹p1 ; p2 º : The restrictions on exponents ˛; ˇ guarantee that .2 /0 < p1  2 and .2 /0 < p2  2: Hence .2 /0 < p  2; which implies that the embedding Lp .˝/  H 1 .˝/ is compact. Consequently, the embedding operator p

J W L .˝/ ! H

1

Z

.˝/ ; .J .u/ ; v/ D

d

uv x ˝

.u 2 Lp .˝/ ; v 2 H01 .˝// is completely continuous. Furthermore, from  ˛p  ˛p1 D 2 ; we have L2 .˝/  L˛p .˝/ ; while from ˇp  ˇp2 D 2; that L2 .˝/  Lˇp .˝/ : Thus the linear operator   P W H01 .˝/ ! L˛p .˝/  Lˇp ˝I Rn ; P .u/ D Œu; ru is well defined and continuous. Also, the Nemytskii operator   Ng W L˛p .˝/  Lˇp ˝I Rn ! Lp .˝/ ; Ng .u; v/ D g .:; u; v/ is well defined, continuous, and bounded, and satisfies the inequality ˇ ˇ ˇ ˇ ˇ ˇ ˇNg .u; v/ˇ p  a ˇjuj˛ ˇ p C b ˇˇjvjˇ ˇˇ C jhj p L L L p D

˛ a jujL ˛p

C

L ˇ b jvjLˇp .˝I

Rn / C jhjLp :

Thus the operator B D J ı Ng ı P is well defined and completely continuous from H01 .˝/ to H 1 .˝/ : Therefore, the operator A D ./1 .B C f /

Section 9.4 Application of the Leray–Schauder Fixed Point Theorem

217

from H01 .˝/ to H01 .˝/ is completely continuous. Next, we show that there exists a number R > 0 such that jujH 1 < R

(9.7)

0

for every solution u 2 H01 .˝/ of the equation u D A .u/ and every 2 .0; 1/ : To this end, assume that u 2 H01 .˝/ n ¹0º solves u D A .u/ ; for some

2 .0; 1/: Then .u; v/H 1 D .B .u/ C f; v/ for every v 2 H01 .˝/ : 0

Choosing v D u and taking into account B .u/ 2 Lp .˝/ ; the sign condition and < 1; we obtain 2 jujH 1 D Œ.B .u/ ; u/ C .f; u/  .f; u/ < jf jH 1 jujH 1 : 0

0

Obviously, here we have assumed that f ¤ 0; the case f D 0 being trivial. Hence jujH 1 < jf jH 1 and (9.7) holds with R D jf jH 1 : Now the conclusion 0 follows from the Leray–Schauder fixed point theorem. Remark 9.2. Theorem 9.6 remains true even if, more generally, instead of the sign condition (iii) we assume that a constant c < 1 exists such that u g .x; u; v/  c juj2 for all u 2 R; v 2 Rn and a.e. x 2 ˝: Indeed, in this case, we have 2 jujH 1 0

Z D Œ.B .u/ ; u/ C .f; u/ D

u g .x; u; ru/ dx C .f; u/ ˝   c 2  c jujL < juj2 1 C jf jH 1 jujH 1 : 2 C jf jH 1 jujH 1 0 0

1 H0

Since c < 1 ; we deduce that jujH 1 < jf jH 1 = .1  c= 1 / : 0

Remark 9.3. Theorem 9.6 is true for n D 2 and n D 1; without any restrictions on the exponents ˛; ˇ  1:

218

9.5

Chapter 9 Semilinear Elliptic Problems

The Monotone Iterative Method

The monotone iterative method, or the method of lower and upper solutions, reduces the problem of the existence of solutions to the existence of a lower solution u and an upper solution u: The basic tool for this method is the maximum principle. We shall present this method for problem (9.1), where the right-hand side of the equation does not depend on ru; that is, for the problem ² u D g .x; u/ C f in ˝ (9.8) uD0 on @˝; where f 2 H 1 .˝/ and g W ˝  R ! R: We shall consider the case n  3: Definition 9.4. By a lower solution of problem (9.8), we mean a function u 2  0 H 1 .˝/ such that Ng .u/ 2 L.2 / .˝/ and ² u  g .x; u/ C f in D 0 .˝/ u0 on @˝: By an upper solution of problem (9.8), we mean a function u 2 H 1 .˝/ such  0 that Ng .u/ 2 L.2 / .˝/ and ² u  g .x; u/ C f in D 0 .˝/ u0 on @˝: Theorem 9.7. Let ˝  Rn be a bounded open set, f 2 H 1 .˝/ ; and let u; u be respectively a lower and an upper solution of problem .9:8/ with u .x/  u .x/ for a.e. x 2 ˝: If the function g W ˝  R ! R satisfies the Carathéodory conditions and is increasing in its second variable, i.e. g .x; u1 /  g .x; u2 /

(9.9)

for u .x/  u1  u2  u .x/ and a.e. x 2 ˝; then problem .9:8/ has at least one weak solution u 2 H01 .˝/ satisfying u .x/  u .x/  u .x/ for a.e. x 2 ˝: Proof. Let u 2 H 1 .˝/ be any function with u .x/  u .x/  u .x/ for a.e. x 2 ˝: The Carathéodory conditions guarantee that the function Ng .u/ D g .:; u .:// is measurable. Furthermore, (9.9) implies Ng .u/  Ng .u/  Ng .u/ in ˝; whence 0  Ng .u/  Ng .u/  Ng .u/  Ng .u/ :

Section 9.5 The Monotone Iterative Method

219  0

This together with Ng .u/Ng .u/ 2 L.2 / .˝/ imply that the measurable func 0  0 tion Ng .u/  Ng .u/ belongs to L.2 / .˝/ : As a result Ng .u/ 2 L.2 / .˝/ :  0 Since L.2 / .˝/  H 1 .˝/ ; it makes sense   A .u/ D ./1 Ng .u/ C f and v WD A .u/ 2 H01 .˝/ : We have

²

and

²

u  g .x; u/ C f u0

in D 0 .˝/ on @˝

v D g .x; u/ C f vD0

in D 0 .˝/ on @˝:

Consequently ²

 .v  u/  g .x; u/  g .x; u/  0 in D 0 .˝/ vu0 on @˝:

Then the maximum principle guarantees that v  u  0 in ˝: Similarly, u  v  0 in ˝: Thus u  u  u implies u  A .u/  u: k In particular, u  A .u/  u; whence, step-by-step, one finds  that u  A .u/  u for all k: In addition, it is easy to see that the sequence Ak .u/ is increasing. Then, from Beppo Levi’s theorem of monotone convergence, we may infer that   there exists a function u 2 L2 .˝/ with Ak .u/ ! u in L2 .˝/ as k ! 1: It remains to show that A .u/ D u: In addition we shall prove that the  convergence  1 .˝/ also holds in H0 : Indeed, the monotonicity of the sequence Ak .u/ and 

its convergence to u in L2 .˝/ guarantee that Ak .u/ .x/ !  u .x/ fora.e. x 2 ˝: Then the Carathéodory continuity condition yields Ng Ak .u/ .x/ !    Ng .u/ .x/ for a.e. x 2 ˝: But the sequence Ng Ak .u/ is itself monotone.   Hence, again by virtue of Beppo Levi’s theorem, Ng Ak .u/ ! Ng .u/ in  0

of the operator L.2 / .˝/ ; and so in H 1 .˝/ : Next using the continuity  1 1 1 k ./ from H .˝/ to H0 .˝/ ; we infer that A A .u/ ! A .u/ 

in H01 .˝/ ; so in L2 .˝/ too. Now the uniqueness of the limit implies that A .u/ D u: Remark 9.4. Theorem 9.7 remains true if operator u is replaced by u  cu; with c < 1 :

220

Chapter 9 Semilinear Elliptic Problems

Lower and upper solutions can be obtained for different classes of functions g (see the section of Problems). More details about the method of lower and upper solutions, including an alternative proof without monotone iterations, can be found in R˘adulescu [42, Chapter 1]. An abstract version for operator equations is given in Precup [38].

9.6

The Critical Point Method

Definition 9.5. Let X be a normed linear space and let E W X ! R be any functional. The derivative of E in direction v 2 X at u 2 X; denoted by E 0 .uI v/ ; is defined as E 0 .uI v/ D

E .u C t v/  E .u/ ; t !0C t lim

if it exists. In case that E 0 .uI v/ exists for every v 2 X and the functional E 0 .uI :/ is linear and continuous on X; we say that E is differentiable at u and we denote E 0 .u/ D E 0 .uI :/ : Thus E 0 .u/ 2 X 0 and .E 0 .u/ ; v/ D E 0 .uI v/ for all v 2 X: If E is differentiable at any u 2 X; then E is said to be differentiable. For example, consider the energy functional associated to Dirichlet problem (8.1) with f 2 H 1 .˝/ ; E W H01 .˝/ ! R; E .u/ D

1 2 juj 1  .f; u/ : 2 H0

We already know that E 0 .uI v/ D .u; v/H 1  .f; v/ 0

for all u; v 2 H01 .˝/ : Hence E is differentiable and E 0 .u/ D .u; :/H 1  .f; :/ : 0

Recall that by a weak solution of Dirichlet problem (8.1) we have understood a function u 2 H01 .˝/ with E 0 .uI v/ D 0 for every v 2 H01 .˝/ : This means that E 0 .u/ D 0; that is, u is a critical point of the energy functional. The aim of this section is just to give an idea about the use of the critical point method for the treatment of the semilinear problem (9.8) with f 2 H 1 .˝/ : A weak solution to this problem is a function u 2 H01 .˝/ such that g .:; u .:// 2 H 1 .˝/ and .u; v/H 1  .g .:; u/ C f; v/ D 0 for all v 2 H01 .˝/ : 0

Section 9.6 The Critical Point Method

221

A first question is if, as in the linear case, a differentiable functional on H01 .˝/ exists in a such way that its critical points coincide with the weak solutions of the semilinear problem (9.8). The answer is positive and given by the following theorem. Theorem 9.8. Let ˝  Rn be a bounded open set, n  3; f 2 H 1 .˝/ and let g W ˝  R ! R satisfy the Carathéodory conditions and the growth inequality  (9.10) jg .x; u/j  c juj2 1 C h .x/ for every u 2 R and a.e. x 2 ˝; where c 2 RC and h 2 L.2 / .˝/ : Let G W ˝  R ! R be the primitive of g with respect to the second variable, which vanishes at the origin, i.e. Z u g .x; / d .x 2 ˝; u 2 R/ : G .x; u/ D  0

0

Then the functional E W H01 .˝/ ! R given by  Z  1 2 (9.11) E .u/ D jru .x/j  G .x; u .x// dx  .f; u/ ˝ 2 ! Z u.x/ Z 1 2 g .x; / d dx  .f; u/ D jru .x/j  ˝ 2 0 is differentiable and 

 E 0 .u/ ; v D .u; v/H 1  .g .:; u/ C f; v/ 0

(9.12)

for all u; v 2 H01 .˝/ : Remark 9.5. For v D ' 2 D .˝/ ; (9.12) becomes   0 E .u/ ; ' D .u  g .:; u/  f; '/ : Thus, as a distribution on ˝; E 0 .u/ D u  g .:; u/  f: Hence E 0 D   Ng  f: Proof of Theorem 9.8. We have to prove that  Z  G .x; u C tv/  G .x; u/  g .x; u/ v dx ! 0 as t ! 0C : t ˝

(9.13)

222

Chapter 9 Semilinear Elliptic Problems

Let

G .x; u C tv/  G .x; u/  g .x; u/ v: t Using Lagrange’s mean value theorem, we may write .x; t/ D

.x; t/ D Œg .x; u C t .x/ v/  g .x; u/ v where  .x/ 2 Œ0; 1 : The Carathéodory condition of continuity implies .x; t/ ! 0 as t ! 0C for a.e. x 2 ˝: On the other hand, using (9.10), for t < 1; we obtain     j .x; t/j  jvj c ju C t vj2 1 C c juj2 1 C 2 jhj h i    jvj c .juj C jvj/2 1 C c juj2 1 C 2 jhj : Notice that the right-hand side in this inequality does not depend on t and is a function from L1 .˝/ : Indeed, this follows from Hölder’s inequality since  u; v 2 H01 .˝/  L2 .˝/ and c .juj C jvj/2

 1

C c juj2

 1

 /0

C 2 jhj 2 L.2

.˝/ :

Now Lebesgue’s dominated convergence theorem guarantees that in L1 .˝/ as t ! 0C ; that is, (9.13).

.:; t/ ! 0

Formula (9.11) defines the energy functional of the semilinear problem (9.8), while (9.12) shows that its critical points are the weak solutions of that problem. In the linear case, the energy functional has a unique critical point which is its unique minimizer. In the nonlinear case, the Dirichlet problem can have several solutions, and correspondingly, the energy functional, several critical points. An other question is now, by what kind of methods can we identify critical points of the energy functional (9.11). The simplest way is to imitate the linear case assuming conditions on g guaranteeing the existence of a minimizer of the energy functional, and to rely on the following extension of Fermat’s theorem. Proposition 9.2. Let X be a normed linear space and E W X ! R be any functional. If u 2 X is a relative minimizer of E and E is differentiable at u; then E 0 .u/ D 0: Proof. Assume u is a relative minimizer of E: Then there is a ball Br .u/ of center u and radius r > 0 such that E .w/  E .u/ for all w 2 Br .u/ : Let v 2 X be arbitrary with jvj D 1: For every t 2 Œ0; r/; we have E .u C tv/  E .u/ and E .u  t v/  E .u/ :

Section 9.6 The Critical Point Method

223

The first inequality yields E 0 .uI v/  0; while the second one, E 0 .uI v/  0: Then, since E 0 .uI v/ D .E 0 .u/ ; v/ and E 0 .uI v/ D .E 0 .u/ ; v/ D  .E 0 .u/ ; v/ ; we deduce that .E 0 .u/ ; v/ D 0: Since v was arbitrary, we can infer thatE 0 .u/ D 0: The next proposition gives sufficient conditions for a functional to have a global minimizer. Proposition 9.3. Let X be a reflexive Banach space and E W X ! R a functional. Assume that for each 2 R; the level set .E  / WD ¹u 2 X W E .u/  º is closed convex. In addition, assume that E is coercive, i.e. E .u/ ! 1 as juj ! 1: Then E is bounded from bellow on X and attends its infimum. Proof. Let m D infX E .u/ : Clearly 1  m < 1: Consider a minimizing sequence .uk / in X; i.e. E .uk / ! m as k ! 1: The coerciveness of E implies that .uk / is bounded. Since X is reflexive, there exists a weakly convergent subsequence of .uk / : Without a distinct notation for this subsequence, we may assume that uk ! u0 weakly, for some u0 in X: Let a be any real number with m < a: Then, there is a ka such that E .uk /  a; that is, uk 2 .E  a/ ; for all k  ka : Recall that for each weakly convergent sequence, there is a sequence of convex combinations of terms from the initial sequence, which is strongly convergent to the same limit. Let .e uk / be that seuk ! u0 strongly. quence of convex combinations of terms of .uk /kka ; with e The convexity of the set .E  a/ guarantees that e uk 2 .E  a/ ; while the fact that .E  a/ is closed, implies u0 2 .E  a/ : Hence E .u0 /  a: Letting a # m; we find E .u0 /  m; which first shows that m > 1 and then that E .u0 / D m: Now we are ready to state and prove an existence result for the Dirichlet problem (9.8). Theorem 9.9. Let ˝  Rn be a bounded open set and n  3: Let f 2 H 1 .˝/ and g W ˝  R ! R: Assume that g satisfies the Carathéodory conditions and the growth condition jg .x; u/j  c juj2

 1

C h .x/

(9.14)

for every u 2 R and a.e. x 2 ˝; where c 2 RC and h 2 L.2 / .˝/ : If in addition g .x; 0/ D 0 and g .x; :/ is decreasing on R for a.e. x 2 ˝; then problem .9:8/ has a unique weak solution u 2 H01 .˝/ : Moreover, u minimizes the energy functional.  0

224

Chapter 9 Semilinear Elliptic Problems

2 jujH 1 .f; u/ is continuous and convex on 0 R H01 .˝/ : We shall prove that the same is true for the functional  ˝ G .x; u/ dx: From (9.14) we find that there exists a constant c1 and a function h1 2 L1 .˝/ ; with  jG .x; u/j  c1 juj2 C h1 .x/

Proof. It is clear that the functional

1 2

for all u 2 R and a.e. x 2 ˝: Then Theorem 9.1 on Nemytskii’s superposition  operator implies the continuity of NG D G .:; u .:// from L2 .˝/ to L1 .˝/ :  Let uk ! u in H01 .˝/ : Then uk ! u in L2 .˝/ ; whence G .:; uk .:// ! G .:; u .:// in L1 .˝/ : Hence ˇ Z ˇZ ˇ ˇ ˇ .G .x; uk /  G .x; u// dx ˇ  jG .x; uk /  G .x; u/j dx ! 0; ˇ ˇ ˝

˝

R

i.e. the functional  ˝ G .x; u/ dx is continuous. It follows that E is also continuous and thus, the level sets .E  / are closed. On the other hand, since g .x; :/ is decreasing on R R ; we have that G .x; :/ is concave on R: This implies that the functional  ˝ G .x; u/ dx is convex on H01 .˝/ : Then, as a sum of convex functionals, E is convex as well. Consequently, its level sets .E  / are convex. Finally we show that E is coercive. Indeed, since g .x; :/ is decreasing and g .x; 0/ D 0; we have G .x; u/  0 for every u 2 R: Then E .u/ 

1 2 juj 1  jf jH 1 jujH 1 ; 0 2 H0

whence E .u/ ! 1 as jujH 1 ! 1: 0 Now Propositions 9.3 and 9.2 imply the existence of a solution minimizing the energy functional. The uniqueness of the solution is a consequence of the strict convexity of the functional E; guaranteed by the strict convexity of the term 2 jujH 1 (Exercise). 0

In the last four decades, critical point theory has had a remarkable development. Several methods were introduced to find critical points which are not extremum points. The reader wishing to initiate himself or herself into these methods can see the monographs Motreanu–R˘adulescu [29], Precup [38] and R˘adulescu [42]. Critical point techniques applied to partial differential equations are presented in Rabinowitz [41] and Struwe [50]. Other methods for the treatment of nonlinear boundary value problems are based on the theory of monotone operators in the sense of Minty and Browder (see Barbu [2] and Lions [23]). A good introduction to nonlinear boundary value problems is the work of Taylor [51, vol. 3].

Section 9.7 Problems

9.7

225

Problems

1.

Give examples of functions g which satisfy all the conditions of each of Theorems 9.2, 9.4, and 9.6.

2.

Check that the function g.x; u; v/ D a jujp1 u satisfies the conditions of Theorem 9.6 if a 2 RC and 1  p < 2  1: Prove that, for each f 2 H 1 .˝/ ; there exists at least one weak solution to the problem ² u D a jujp1 u C f in ˝ uD0 on @˝:

3.

Prove that in Theorem 9.7, if in addition u; u 2 L1 .˝/ ; i.e. there exist constants m; M with m  u .x/  u .x/  M for a.e. x 2 ˝; then the assumption that g .x; u/ is increasing in u 2 Œu .x/ ; u .x/ ; for a.e. x 2 ˝; can be replaced by the requirement that a constant K  0 exist such that the function g .x; u/ C Ku is increasing in u 2 Œm; M  ; for a.e. x 2 ˝: Hint. Write the equation under the form uCKu D g .x; u/CKuCf and take into account Remark 9.4.

4.

Prove that if f 2 L1 .˝/ and g .x; u/ D cu  a jujp1 u; where c < 1 and a 2 RC ; then the function u D .  cI /1 c0 is an upper solution, while u D u is a lower solution of problem (9.8), where c0 D jf jL1 : Hint. The maximum principle gives u  0: Then u D cu C c0  cu C f  a jujp1 u in D 0 .˝/ : Hence u is an upper solution.

5.

Consider the problem 8 < u D g .u/ in ˝ u>0 in ˝ : uD0 on @˝:

(9.15)

Assume that g 2 C 1 .RC / ; g .0/ D 0; g 0 .0/ > 1 and limu!1 g .u/ =u < 1 : Prove that there exists an upper solution u and a lower solution u D "1 with 0 < "1  u: Then prove the existence of a solution with u > 0 in ˝:

226

Chapter 9 Semilinear Elliptic Problems

Hint. Let c be any number satisfying limu!1 g .u/ =u < c < 1 : Then there is R > 0 such that g .u/  cu for every u > R: Function g being continuous on Œ0; R ; there is c0 > 0 with g .u/  c0 for every u 2 Œ0; R : Thus g .u/  cu C c0 for all u 2 RC : Then u D .  cI /1 c0 is an upper solution. On the other hand, from g 0 .0/ > 1 we have g .u/  1 u for every u 2 Œ0; ı : Since 1 2 L1 .˝/ and 1 > 0 in ˝ (see Theorem 8.10), there exists " > 0 with 0 < "1  ı in ˝: Then  ."1 / D "1 D " 1 1  g ."1 / ; i.e. u D "1 is a lower solution. To prove that "1  u use the relations u D cu C c0 and  ."1 /  g ."1 / ; which give  .u  "1 /  c .u  "1 /  c0 C c "1  g ."1 /  0: Now the maximum principle guarantees that u  "1  0: 6.

Prove that, under the assumptions of Problem 9.5, u  u for every solution u of problem (9.15).

7.

Show that function g from Problem 9.2 satisfies all the assumptions of Theorem 9.9 for 1  p  2  1:

8.

Give existence results of the type of Theorems 9.2, 9.4, 9.6 for the case where operator B from H01 .˝/ to H 1 .˝/ is a general one, i.e. for the boundary value problem ² u D B .u/ C f in ˝ uD0 on @˝; where f 2 H 1 .˝/ and B W H01 .˝/ ! H 1 .˝/ :

Chapter 10

The Semilinear Heat Equation

10.1

The Nonhomogeneous Heat Equation in H 1 .˝/

In this section, we present an existence and uniqueness result for the weak solution of the Cauchy–Dirichlet problem for the nonhomogeneous heat equation.   Theorem 10.1 (J. L. Lions). Assume that f 2 L2 0; T I H 1 .˝/ and g0 2 L2 .˝/ : Then there exists a unique function     (10.1) u 2 L2 0; T I H01 .˝/ \ C Œ0; T  I L2 .˝/ such that for each v 2 H01 .˝/ ; the function t 7! .u .t / ; v/L2 is absolutely continuous on Œ0; T  and ´ d dt .u .t / ; v/L2 C .u .t/ ; v/H01 D .f .t / ; v/ for a.e. t 2 Œ0; T  (10.2) u .0/ D g0 : Moreover, one has 1 1 2 2 ju .t /jL jg0 jL 2  2 C 2 2

Z

t 0

2 ju ./jH 1 0

d D

Z 0

t

.f . / ; u . // d

(10.3)

for all t 2 Œ0; T  : Proof. We proceed like in Section 4.3 looking for u in the form (4.8), i.e. u .t/ D

1 X

uk .t / k :

(10.4)

kD1

Replacing formally in (10.2) we obtain expression (4.10) of coefficients uk .t / ; that is Z uk .t / D ek t g0k C

t

ek .ts/ fk .s/ ds;

0

where, this time, fk .t / D .f .t/ ; k / : Recall that g0k D .g0 ; k /L2 :

228

Chapter 10 The Semilinear Heat Equation

(a) First we show that series (10.4) defines a function u 2 C.Œ0; T  I L2 .˝//: To this end it suffices to prove that the series converges in L2 .˝/ ; uniformly with respect to t 2 Œ0; T  : Let us consider the partial sums of series (10.4), m X

sm .t / D

uk .t / k :

kD1

We have

ˇ ˇ2 ˇ mCp ˇ mCp X ˇ X ˇ ˇ2 ˇ ˇsmCp .t /  sm .t /ˇ 2 D ˇ ˇ .t / u D u2k .t / :  k k ˇ ˇ L ˇkDmC1 ˇ 2 kDmC1 L

Thus  2of the2 series P1 the2 problem has been reduced to the uniform convergence 2 .t / Œ0;  .a u  2 a Cb and on T : Using the inequality C b/ kD1 k Hölder’s inequality, we obtain " 2 #  2 Z t 2 k k .t s/ e fk .s/ ds uk .t /  2 g0 C 0



 2 g0k

2

C

1

k

Z

t 0

fk2 .s/ ds:

(10.5)

Hence it remains to prove the convergence of the numerical series 1  1 Z T 2 X X 1 2 k and fk .s/ ds: g0 0 k kD1

kD1

2 Parseval’s equality guarantees that the sum of the first series is jg0 jL 2 ; while (7.19) implies 1 X 1 2 2 f .s/ D jf .s/jH 1 :

k k kD1   On the other hand, from f 2 L2 0; T I H 1 .˝/ ; it follows that the function 2 1 s 7! jf .s/jH 1 belongs to L .0; T / : Now Lebesgue’s dominated convergence theorem guarantees that Z T 1 Z T X 1 2 2 fk .s/ ds D jf .s/jH 1 ds:

0 0 k kD1

Also note that (10.5) yields Z 2 ju .t /jL 2



2 2 jg0 jL 2

C

t 0

2 jf .s/jH 1 ds;

t 2 Œ0; T  :

(10.6)

Section 10.1 The Nonhomogeneous Heat Equation in H 1 .˝/

229

  (b) Next we prove that u 2 L2 0; T I H01 .˝/ : We have ´       0 .t / ;  sm j L2 C sm .t / ; j H 1 D f .t / ; j ; j D 1; 2; :::; m 0 sm .0/ D g0m

(10.7)

Pm

where g0m D

k kD1 g0 k :

It follows that   0 2 sm .t/ ; sm .t / L2 C jsm .t /jH 1 D .f .t / ; sm .t // ; 0

or equivalently

1 d 2 2 jsm jL 2 C jsm j 1 D .f; sm / : H 0 2 dt

Integration gives Z t 1 1 2 2 2 jsm . /jH jsm .t /jL2  jg0m jL2 C 1 d 0 2 2 0 Z t Z t .f ./ ; sm .// d  D jf . /jH 1 jsm . /jH 1 d: 0

0

(10.8)

0

Then, using Hölder’s inequality and jg0m jL2  jg0 jL2 ; we infer that 2 jsm jL 2 .0;t IH 1 .˝//  0

1 2 (10.9) jg0 jL 2  jf jL2 .0;t IH 1 .˝// jsm jL2 .0;tIH 1 .˝// ; 0 2

whence we find the boundedness of the sequence .sm / in L2 .0; T I H01 .˝//: u weaklyin Passing eventually to a subsequence, we may assume that sm ! e  1 .˝/ 2 L 0; T I H0 : We already know that sm ! u in C Œ0; T  I L2 .˝/ : Then, for each " > 0; there is m" such that jsm .t /  u .t /jL2  " for all t 2 Œ0; T  and m  m" : Recall that for each weakly convergent sequence, there exists a sequence of convex combinations of its elements, which is strongly convergent  c to the same limit. Let sm be such a sequence of convex combinations of   elements of .sm /mm" which strongly converges in L2 0; T I H01 .˝/ to e u: c .t / ! e u .t / Again passing if necessary to a subsequence, we may assume that sm 1 .˝/ 2 in H0 (so in L .˝/ too) for a.e. t 2 Œ0; T  : On the other hand, we have c .t /  u .t /j u .t /  u .t /jL2  jsm L2  " for all t: Passing to the limit we obtain je " for a.e. t: Since " was arbitrary,it follows that u .t / D e u .t / for a.e. t 2 Œ0; T  : Thus u 2 L2 0; T I H01 .˝/ : (c) From (10.7) we find for m  j; Z t       sm ./ ; j H 1 d sm .t / ; j L2  g0m ; j L2 C 0 0 Z t   f . / ; j d: D 0

230

Chapter 10 The Semilinear Heat Equation

Letting m ! 1 we obtain     u .t/ ; j L2  g0 ; j L2 C Z t   f ./ ; j d: D

Z

t



u . / ; j

0

 H01

d

(10.10)

0

Since any v 2 H01 .˝/ is the limit in H01 .˝/ of a sequence of linear combinations of the elements k ; using (10.10), we find Z t .u .t/ ; v/L2  .g0 ; v/L2 C .u . / ; v/H 1 d 0 0 Z t .f ./ ; v/ d: D 0

This shows that the function t 7! .u .t/ ; v/L2 is absolutely continuous and

d .u .t / ; v/L2 C .u .t / ; v/H 1 D .f .t / ; v/ for a.e. t 2 Œ0; T  : 0 dt (d) The equality u .0/ D g0 is obtained from sm .0/ D g0m by letting m ! 1: (e) For uniqueness, assume that u and v are two functions enjoying all the properties from the statement. Then for w D u  v and any j; we have Z t     w .s/ ; j H 1 ds D 0; t 2 Œ0; T  : w .t / ; j L2 C 

But w .s/ ; j 

 H01



0

0

D j w .s/ ; j



w .t / ; j

Z

 L2

C j

t



L2

: Hence

w .s/ ; j

0

 L2

ds D 0; t 2 Œ0; T  :

  It follows that w .t / ; j L2 D C ej t : Since w .0/ D 0; we find that C D 0;   that is w .t/ ; j L2 D 0 for every t 2 Œ0; T  : Thus w .t / D 0 for all t 2 Œ0; T  ; whence u D v: P (f) Proof of relation (10.3). Let f m .t / D m kD1 fk .t / k : Then, for every m; j; one has       0 (10.11) sm .t / ; j L2 C sm .t / ; j H 1 D f m .t / ; j : 0   Let " 2 .0; 1 be any number. From sm ! u in C Œ0; T  I L2 .˝/ and  f m ! f in L2 0; T I H 1 .˝/ ; it follows that there exists M" such that jsm  ujC .Œ0;T IL2 .˝//  "; jf m  f jL2 .0;T IH 1 .˝//  "

Section 10.1 The Nonhomogeneous Heat Equation in H 1 .˝/

231

  1 2 for all m  c M" : Since sm ! u weakly in L 0; T I H0 .˝/ ; there is a sequence sk k1 of convex combinations of elements of the sequence .sm /mM"   with skc ! u as k ! 1 strongly in L2 0; T I H01 .˝/ : Consequently, skc .t / ! u .t / in H01 .˝/ and also in L2 .˝/ for a.e. t 2 Œ0; T  : Let c   be the sequence of convex combinations of f m .m  M" / ; with fk k1   the same coefficients as for skc k1 : Clearly, for every k  1; the following estimations hold: ˇ c ˇ ˇs  uˇ  "; k C Œ0;T IL2 .˝/ ˇ c ˇ ˇ k ˇ  ": fˇ 2 ˇ f L .0;T IH 1 .˝//  c Passing eventually to a subsequence, we may assume that f k ! f" weakly ˇ ˇ   in L2 0; T I H 1 .˝/ and ˇs c .0/ˇ 2 ! " : We have k

L

j"  jg0 jL2 j  "; jf"  f jL2 .0;T IH 1 .˝//  ": From (10.11) we first obtain    c     0 skc .t / ; j 2 C skc .t/ ; j H 1 D f k .t / ; j ; 0

L

and then    0 skc .t/ ; skc .t /

L2

  c ˇ2 ˇ C ˇskc .t /ˇH 1 D f k .t / ; skc .t / ; 0

whence, by integration, ˇ2 1 ˇˇ c ˇˇ2 1ˇ sk .t / L2  ˇskc .0/ˇL2 C 2 2

Z 0

t

ˇ ˇ c ˇs . /ˇ2 1 d D k H 0

Z t  c  f k . / ; skc . / d: 0

Letting k ! 1; we deduce that Z t Z t 1 2 1 2 2 .f" ./ ; u . // d ju ./jH 1 d D ju .t /jL2  " C 0 2 2 0 0 for almost every t: We have ˇ ˇ ˇ 2 2 ˇ ˇ"  jg0 jL 2 ˇ  " ." C jg0 jL2 /  " .2 jg0 jL2 C 1/ ;

232

Chapter 10 The Semilinear Heat Equation

ˇZ ˇ Z T ˇ T ˇ ˇ ˇ .f" ./  f ./ ; u . // d ˇ  jf" . /  f . /jH 1 ju . /jH 1 d ˇ 0 ˇ 0 ˇ 0  jf"  f jL2 .0;T IH 1 .˝// jujL2 .0;T IH 1 .˝// 0

 " jujL2 .0;T IH 1 .˝// : 0 Then ˇ ˇ Z t Z t ˇ ˇ1 2 ˇ ˇ ju .t/j2 2  1 jg0 j2 2 C .f ./ . // ./j d   ; u d  ju 1 ˇ ˇ2 L L H0 2 0 0 ˇZ ˇ ˇ ˇ T ˇ 1 ˇˇ 2 ˇ ˇ 2 ˇ .f . / . / .//  C  f ; u d  ˇ ˇ ˇc" ˇ"  jg0 jL 2 " ˇ 0 ˇ 2 for a.e. t and some constant c independent of t and ": From this we may infer that the continuous function from the left-hand side is zero on Œ0; T  : Thus (10.3) is proved. Another proof can be found in Temam [52, p. 68]. A third proof based on an isomorphism theorem of the same type as Theorem 7.15, is given in Lions– Magenes [24, p. 257]. Remark 10.1. Equality (10.3) implies 1 d 2 2 for a.e. t 2 Œ0; T  : ju .t/jL 2 C ju .t /j 1 D .f .t / ; u .t // H0 2 dt Definition 10.1. By a weak or generalized problem 8 @u < @t  u D f u .x; 0/ D g0 .x/ : uD0   where f 2 L2 0; T I H 1 .˝/ and g0 2 satisfying (10.1) and (10.2).

(10.12)

solution of the Cauchy–Dirichlet in Q in ˝ on †;

(10.13)

L2 .˝/ ; we mean a function u

Remark 10.2. If u is a weak solution of problem (10.13), then u satisfies the equation u0  u D f in the sense of the theory of distributions with values in H 1 .˝/ (see Temam [52, p. 67]).

Section 10.2 Regularity Results

10.2

233

Regularity Results

In this section which could be skipped on first reading, we analyze the regularity of the weak solution of the Cauchy–Dirichlet problem for the heat equation. Just a simple look to the representation (10.4) for the homogeneous case f D 0; suggests that, due to the exponential factors ej t ; the solution should have remarkable properties of regularity. We shall prove that this is indeed the case and that heat equation has a strong regularity effect over the initial datum g0 : We shall see that the solution u .x; t/ is of class C 1 with respect to x; for every t > 0; even if g0 is discontinuous. To discuss the regularity of the solution it is convenient to consider the following subspaces of X0 D L2 .˝/ W ® ¯ Xm D u 2 H01 .˝/ W u 2 Xm1 ; m D 1; 2; :::. Clearly ® ¯ Xm D u 2 H01 .˝/ W u; 2 u; :::; m1 u 2 H01 .˝/ ; m u 2 L2 .˝/ and ::: XmC1  Xm  :::  X1  H01 .˝/ : Lemma 10.1. (a) The space Xm endowed with the inner product   .u; v/Xm D m u; m v L2 is a Hilbert space. (b) For any u 2 Xm it is true the following generalization of Parseval’s equality 2 D jujX m

1 X

 2

j2m u; j L2 :

(10.14)

j D1

Proof. (a) First, one can check for the mapping .u; v/Xm that all the axioms of 2 D .u; u/Xm D 0; then the inner product are satisfied. For instance, if jujX m  m1  1 .˝/ m m1 0D uD  u : Since  u 2 H0 ; in view of the existence and uniqueness theorem for the weak solution of the Dirichlet problem, we have  0 D m1 u D  m2 u : Repeating the above argument we finally obtain u D 0: For the completeness, let .vk /k1 be a Cauchy sequence in Xm : Then the sequence .m vk /k1 is Cauchy in L2 .˝/ and so convergent to some v m 2

234

Chapter 10 The Semilinear Heat Equation

L2 .˝/ : Based on Theorem 3.14, there exist functions v m1 ; v m2 ; :::; v 0 with v j 1 D v j for j D m; m  1; :::; 1: Obviously v 0 2 Xm and vk ! v 0 in Xm ; as k ! 1: Thus, Xm is complete. (b) The result will follow from Parseval’s equality once we have shown that     m  u; j L2 D .1/m jm u; j L2 : Indeed,       m  u; j L2 D  m1 u; j H 1 D  j m1 u; j L2   0   D j m2 u; j H 1 D ::: D .1/m jm u; j L2 : 0

 2



Lemma 10.2. If f 2 L 0; T I L2 .˝/ and g0 2 H01 .˝/ ; then the solution u of problem (10.13) has the additional property u 2 L2 .0; T I X1 / :

Proof. We come back to the proof of Theorem 10.1. From  0    0 ; j H 1 ; sm ; j j L2 D sm 0         sm ; j j H 1 D sm ; j j L2 D sm ; j L2 D sm ; j X1 ;    0  f; j j L2 D f; j L2 ; we obtain 

0 .t / ; sm .t / sm

 H01

2 C jsm .t /jX D .f .t / ; sm .t //L2 1

whence 1 d 2 2 jsm jH 1 C jsm jX1 0 2 dt

It follows that

D .f; sm /L2  jf jL2 jsm jL2  1 2 2  : jf jL2 C jsm jX 1 2

d 2 2 2 jsm jH 1 C jsm jX1  jf jL2 : 0 dt

Section 10.2 Regularity Results

235

Integration gives Z 2 jsm .t /jH 1 0

C

t

0

Z 2 jsm .s/jX 1

2 jsm .0/jH 1 0

ds 

C

Z

2  jg0 jH 1 C 0

T 0

t 0

2 jf .s/jL 2 ds

2 jf .s/jL 2 :

This implies that the sequence .sm / is bounded in L2 .0; T I X1 / : Furthermore, as at step (b) in the proof of Theorem 10.1, we obtain that u 2 L2 .0; T I X1 / : The following regularity result both in t and x holds. Theorem 10.2. Let p 2 N and g0 2 H01 .˝/ : If g0 2 Xp and f 2  Tp k .0; T I X H ; then pk kD0 u2

p \

    C k Œ0; T  I Xpk \ L2 0; T I XpC1 :

kD0

Proof. We can show this by mathematical induction with respect to p: For p D 0; the statement is that g0 2 H01 .˝/ and f 2 L2 0; T I L2 .˝/ ; imply  u 2 C Œ0; T  I L2 .˝/ \ L2 .0; T I X1 / ; which is true in view of Theorem 10.1 and Lemma 10.2. We assume the resultT holds for p and we provethat it is also true for pC1: Let pC1 g 0 D g0 C f .0/ 2 g0 2 XpC1 and f 2 kD0 H k 0; T I XpC1k : Then b   Tp 0 k b u of Xp and f D f 2 kD0 H 0; T I Xpk : Consequently, the solution b b the corresponding problem to data b g 0 and f has the property b u2

p \

    C k Œ0; T  I Xpk \ L2 0; T I XpC1 :

kD0

Let

Z u .t / D g0 C

From b u2

Tp kD0

u2

0

t

b u .s/ ds:

  C k Œ0; T  I Xpk we deduce that

p \ kD0

\   pC1   C kC1 Œ0; T  I Xpk D C j Œ0; T  I XpC1j ; j D1

(10.15)

236

Chapter 10 The Semilinear Heat Equation

    while from b u 2 L2 0; T I XpC1 we have u 2 C Œ0; T  I XpC1 : Hence pC1 \

u2

  C k Œ0; T  I XpC1k :

kD0

Finally one can see that u is the solution which corresponds to data g0 and f; whence   u  f 2 L2 0; T I XpC1 : u D u0  f D b   Therefore u 2 L2 0; T I XpC2 : We shall conclude with regularity results for the homogeneous heat equation, i.e. for f D 0: In this case, the solution is u .t/ D

1 X

ej t g0 j ; j

t 2 Œ0; T  :

j D1

  We have already seen that if g0 2 L2 .˝/ ; then u 2 C Œ0; T  I L2 .˝/ : Moreover, we have the following result. Theorem 10.3. (a) If g0 2 L2 .˝/ ; then u 2 C k ..0; T I Xm / for all k; m 2 N:   (b) If g0 2 Xp ; then u 2 C k Œ0; T  I Xpk for 0  k  p:   T Xp ; then u 2 C 1 Œ0; T  I Xp for every p 2 N: (c) If g0 2 p2

N

Assume in addition that ˝ is of class C 1 : Then: (d) If g0 2 L2 .˝/ ; then u 2 C 1 .˝  Œt0 ; T / for 0 < t0 < T: (e) If g0 2 C 1 .˝/ and the following compatibility conditions hold: g0 D 0; m g0 D 0 on @˝ for every m 2 N;

(10.16)

then u 2 C 1 .˝  Œ0; T / .so u is a classical solution/: Proof. (a) We have to prove that for every l; the lth-order derivative series 1 X j D1

.1/l jl ej t g0 j j

(10.17)

Section 10.2 Regularity Results

237

is uniformly convergent in Xm on each subinterval Œt0 ; T   .0; T : Notice that the eigenfunctions j belong to all spaces Xm since m 2 N:

m j D .1/m jm j ;

Based on this relation and taking into consideration the expression (10.14) of the norm on Xm ; the uniform convergence of the series (10.17) of functions with values in Xm ; reduces to the uniform convergence of the series of real functions 1 X

2.lCm/ 2j t

e

j



j

2

g0

:

(10.18)

j D1 2.lCm/ 2j t e

Note that for t  t0 ; there exists a C D C .l; m; t0 / > 0 with j

 j e2j t0  C; for all j 2 Nn ¹0º : Thus the functional series (10.18) is dominated by the convergent numerical series 2.lCm/

C

1  X

j

2

g0

2 D C jg0 jL 2 :

j D1

Hence series (10.18) is uniformly convergent on Œt0 ; T  : (b) For m D p  k and l  k  p; series (10.18) can be written as 1 X

2.lk/ 2j t

j e



2p

j



j g0

2 :

j D1

 1 and e2j t  1 for j 2 Nn ¹0º and t  0; this Since j functional series is dominated by by convergent numerical series 2.lk/

2.lk/

2.lk/

1

1 X

2p

j



j

g0

2

2.lk/

D 1

2 : jg0 jX p

j D1

(c) An immediate consequence of (b). (d) Apply (a) and take into account that if ˝ is of class C 1 ; then, in view of Theorem 8.2, (10.19) Xm  H 2m .˝/  C k .˝/ both algebraically and topologically, for every m; k with 2m > k C n=2: 1 (e) T From g0 2 C .˝/ and (10.16), based on Theorem 7.3, we find that g0 2 Xp : Now we apply (c) and we use the embeddings in (10.19). p2

N

238

Chapter 10 The Semilinear Heat Equation

The above existence and regularity results can be extended to more general problems of the form 8   n P @ @u @u ˆ ˆ .x/  a D f .x; t/ in Q jk < @t @xj @xk j;kD1

ˆ u .x; 0/ D g0 .x/ ˆ : uD0

in ˝ on †:

Obviously in this case j ; j will be the eigenvalues and eigenfunctions of the Dirichlet problem for the elliptic operator replacing the Laplacian. Also, in the proofs, spaces Xm will be redefined correspondingly.

10.3

Application of Banach’s Fixed Point Theorem

According to Theorem 10.1 one may associate to the Cauchy–Dirichlet problem 8 @u < @t  u D f in Q (10.20) u .x; 0/ D 0 in ˝ : uD0 on † the solution operator of the heat equation,       S W L2 0; T I H 1 .˝/ ! L2 0; T I H01 .˝/ \ C Œ0; T  I L2 .˝/ ; given by Sf D u; where u is the weak solution of problem (10.20). The estimation theorem which follows implies, on the one side, the continuous dependence of f and g0 of the solution of problem (10.13), and,  on the other side, the nonexpansivity of the solution operator S; from L2 0; T I H 1 .˝/ to  1 .˝/ 2 2 1 2 and from L 0; T I H .˝/ to C Œ0; T  I L .˝/ : L 0; T I H0   Theorem 10.4. Let f 2 L2 0; T I H 1 .˝/ and g0 2 L2 .˝/ : If u is the solution of problem (10.13), then for every t 2 Œ0; T  ; one has q jujC .Œ0;t IL2 .˝//  2 jg0 j2 C jf j2   q 1 2 2 jf j C jf j C 2 jg0 j ; jujL2 .0;t IH 1 .˝//  0 2 where jf j D jf jL2 .0;t IH 1 .˝// and jg0 j D jg0 jL2 .˝/ : In particular, for g0 D 0; the following inequalities hold for every t 2 Œ0; T  W

and

jujC .Œ0;t IL2 .˝//  jf jL2 .0;t IH 1 .˝// ; jujL2 .0;t IH 1 .˝//  jf jL2 .0;t IH 1 .˝// : 0

(10.21)

Section 10.3 Application of Banach’s Fixed Point Theorem

239

Proof. The first inequality is relation (10.6). For the second one, let us remark that from (10.9) it follows that:   q 1 2 2 jf j C jf j C 2 jg0 j ; jsm jL2 .0;t IH 1 .˝//  0 2 whence c jsm jL2 .0;t IH 1 .˝// 0

  q 1 2 2  jf j C jf j C 2 jg0 j : 2

Letting m ! 1 we obtain the desired inequality. Consider now the semilinear problem 8 @u < @t  u D ˚ .u/ in Q u .x; 0/ D g0 .x/ in ˝ : uD0 on †:

(10.22)

From Banach’s contraction principle, we derive the following existence and uniqueness result. Theorem 10.5. Let g0 2 L2 .˝/ and let     ˚ W C Œ0; T  I L2 .˝/ !L2 0; T I H 1 .˝/ be any mapping for which there is a constant a 2 RC such that (10.23) j˚ .u/ .t /  ˚ .v/ .t/jH 1  a ju .t/  v .t /jL2 a.e. on Œ0; T  ;   for every u; v 2 C Œ0; T  I L2 .˝/ : Then problem (10.22) has a unique solution u, i.e.     u 2 L2 0; T I H01 .˝/ \ C Œ0; T  I L2 .˝/ and for each v 2 H01 .˝/ ; the function .u .t / ; v/L2 is absolutely continuous on Œ0; T  and ´ d dt .u .t / ; v/L2 C .u .t/ ; v/H01 D .˚ .u/ .t / ; v/ a.e. on Œ0; T  u .0/ D g0 : Proof. Let u0 be the solution of problem (10.22) which corresponds to ˚ D 0: We have to solve the fixed point problem   u D u0 C .S ı ˚ / .u/ ; u 2 C Œ0; T  I L2 .˝/ :

240

Chapter 10 The Semilinear Heat Equation

The conclusion will follow from Banach’s fixed point theorem once we have proved that the operator     A W C Œ0; T  I L2 .˝/ ! C Œ0; T  I L2 .˝/ ; A .u/ D u0 C .S ı ˚ / .u/   2 .˝/ : Let u; v 2 Œ0;  is a contraction with respect to a suitable norm on C T I L  C Œ0; T  I L2 .˝/ : Using Theorem 10.4 we deduce that Z

jA .u/ .t/ 

t

2 j˚ .u/ .s/  ˚ .v/ .s/jH 1 ds 0 Z t 2  a2 ju .s/  v .s/jL 2 ds:

2 A .v/ .t/jL 2



0

  We fix a number  > a2 =2 and we consider the norm on C Œ0; T  I L2 .˝/ given by   ju .t/jL2 .˝/  t : kuk D

max

t 2Œ0;T

e

Then Z 2 2 2 jA .u/ .t /  A .v/ .t /jL 2  a ku  vk

 Divide by

e2 t

a2 ku  vk2 2

t 0

e2s ds

e2 t :

and take the maximum for t 2 Œ0; T  to obtain a kA .u/  A .v/k  p ku  vk : 2

p Since a= 2 < 1; the operator A is a contraction with respect to norm k:k : Example 10.1. Let ‰ W L2 .˝/ ! H 1 .˝/ be an operator such that j‰ .u/  ‰ .v/jH 1  a ju  vjL2 for all u; v 2 L2 .˝/ and some a 2 RC : Then the mapping     ˚ W C Œ0; T  I L2 .˝/ ! L2 0; T I H 1 .˝/ defined by ˚ .u/ .t / D ‰ .u .t//

   u 2 C Œ0; T  I L2 .˝/ ; t 2 Œ0; T 



satisfies the assumption of Theorem 10.5.

(10.24)

Section 10.4 Application of Schauder’s Fixed Point Theorem

Example 10.2. Let W ˝  R ! R be a function with for every  2 R; .:; 0/ 2 H 1 .˝/ and j .x; 1 / 

241

.:; / measurable

.x; 2 /j  a0 j1  2 j

for all 1 ; 2 2 R; a.e. x 2 ˝ and some constant a0 2 RC : Then the operator ‰ W L2 .˝/ ! H 1 .˝/ given by ‰ .u/ D

.:; u .://

satisfies the condition from the previous example. Indeed, one can represent ‰ .u/ D with  .x; / D

.x; / 

.:; 0/ C  .:; u .://

.x; 0/ : Notice that

j .x; /j  a0 j j . 2 R; a.e x 2 ˝/ ; whence according to Theorem 9.1, the superposition operator  .:; u .:// maps L2 .˝/ into L2 .˝/ ; and in addition j .:; u .://   .:; v .://jL2  a0 ju  vjL2 : The embedding L2 .˝/  H 1 .˝/ being continuous, there will exist a constant a 2 RC such that (10.24) holds.

10.4

Application of Schauder’s Fixed Point Theorem

The following lemmas will be used in the proof of the complete continuity of the solution operator S; making it possible for us to apply Schauder’s fixed point theorem. Lemma 10.3 (Arzelà–Ascoli). Let .B; j:jB / be a Banach space. A subset F of C .Œ0; T  I B/ is relatively compact if and only if F .t / WD ¹f .t / W f 2 F º is relatively compact in B for every t 2 Œ0; T  ; and F is equicontinuous, i.e. for each " > 0; there is a ı > 0 such that jf .t1 /  f .t2 /jB  " for every f 2 F and t1 ; t2 2 Œ0; T  with jt1  t2 j  ı: For a proof see, e.g. O’Regan–Precup [33, p. 72].

242

Chapter 10 The Semilinear Heat Equation

Lemma 10.4 (Lions). Let X; B and Y be Banach spaces and the following embeddings hold: X  B compactly; B  Y continuously. Then, for each  > 0 there exists N  0 such that jujB   jujX C N jujY

for all u 2 X:

(10.25)

Proof. For each n 2 N; consider the set Un D ¹u 2 B W jujB <  C n jujY º: S The sets Un are open in B; Un  UnC1 and B D n2N Un : The unit sphere † of X being relatively compact in B; there is a number N with †  UN : Hence jvjB <  C N jvjY for every v 2 X with jvjX D 1: Finally, for an arbitrary u 2 X; u ¤ 0; take 1 u to obtain (10.25). v WD jujX Lemma 10.5. Let X; B and Y be as in Lemma 10.4 and let 1  p  1: If a set F is bounded in Lp .0; T I X / and relatively compact in Lp .0; T I Y / ; then F is relatively compact in Lp .0; T I B/ : ® ¯ Proof. For an arbitrary given " > 0; there exists a finite subset fj of F such that for each f 2 F; there is an fj with ˇ ˇ ˇf  fj ˇ p  ": L .0;T IY / Inequality (10.25) implies ˇ ˇ ˇ ˇ ˇ ˇ ˇf  fj ˇ p ˇf  fj ˇ p ˇf  fj ˇ p   C N L .0;T IB/ L .0;T IX / L .0;T IY /   c C N ";

where c is the diameter of F in Lp .0; T I X/ : For any "0 > 0; if we choose "0 "0 and " WD 2N ; we find that  WD 2c ˇ ˇ ˇf  fj ˇ p  "0 : L .0;T IB/ This shows that F is relatively compact in Lp .0; T I B/ :

Section 10.4 Application of Schauder’s Fixed Point Theorem

243

Theorem 10.6. The solution operator S of the heat equation is completely continuous from L2 .0; T I H 1 .˝// to L2 .0; T I Lp .˝// for 1  p < 2 when n  3; and for every p  1 if n D 1 or n D 2: Proof. Assume first that .2 /0  p < 2 : Then H01 .˝/  Lp .˝/  H 1 .˝/ ; where the first embedding is compact and the second one is continuous. According to Lemma 10.5it is sufficient to prove that for each bounded subset  M of 1 2 1 2 L 0; T I H .˝/ ; the set S .M / is bounded in L 0; T I H0 .˝/ (which is true by Theorem 10.4) and relatively compact in L2 0; T I H 1 .˝/ : We shall prove even more, namely that S .M / is relatively compact in C.Œ0; T  I H 1 .˝//: Indeed, from (10.21) one has that S .M / is bounded in C.Œ0; T  I L2 .˝//: Then, for each t 2 Œ0; T  ; the set S .M / .t / is bounded in L2 .˝/ and thus relatively compact in H 1 .˝/   : It remains to show that S.M / is 1 equicontinuous in C Œ0; T  I H .˝/ : Let f 2 M and u D S .f / : With the notations from the proof of Theorem 10.1, we have 0 .t /  sm .t / D f m .t / : sm

Then ˇ ˇ 0 ˇs .t /ˇ 1  jsm .t /j 1 C jf m .t /j 1 D jsm .t /j 1 C jf m .t /j 1 : H H H m H H 0

Consequently ˇ 0 ˇ m ˇs ˇ 2 m L .0;T IH 1 .˝//  jsm jL2 .0;T IH01 .˝// C jf jL2 .0;T IH 1 .˝// :     Since sm ! u in L2 0; T I H01 .˝/ ; f m ! f in L2 0; T I H 1 .˝/ as  2 0; T I H 1 .˝/ and .M / m! 1; and since M and S are bounded in L   0 L2 0; T I H01 .˝/ ; respectively, we deduce that jsm jL2 .0;T IH 1 .˝//  C for every m; where constant C does not depend on f: On the other hand, Z t1 0 .t / dt; sm .t1 /  sm .t2 / D sm t2

whence jsm .t1 /  sm .t2 /jH 1

ˇZ ˇ  ˇˇ

t1 t2

ˇ ˇ ˇ ˇ 0 ˇs .t /ˇ 1 dt ˇ : m ˇ H

Furthermore, Hölder’s inequality implies that p p ˇ 0 ˇ ˇ 2  C jsm .t1 /  sm .t2 /jH 1  jt1  t2 j ˇsm jt1  t2 j: 1 L .0;T IH .˝// Letting m ! 1 yields ju .t1 /  u .t2 /jH 1  C

p

jt1  t2 j;

whence we can conclude that S .M / is equicontinuous in C.Œ0; T  I H 1 .˝//:

244

Chapter 10 The Semilinear Heat Equation

Finally, the case 1  p < .2 /0 follows from the previous case since  0 L.2 / .˝/  Lp .˝/ (recall that ˝ was assumed to be a bounded set). The reader interested in compactness criteria in Lp .0; T I B/ is referred to the excellent paper by J. Simon, [47]. The following existence result is a consequence of Schauder’s fixed point theorem. Here the Lipschitz condition (10.23) on the nonlinear term ˚ is weakened to a condition of at most linear growth. Theorem 10.7. Let g0 2 L2 .˝/ and let     ˚ W L2 0; T I L2 .˝/ !L2 0; T I H 1 .˝/ be a continuous mapping for which there exists a constant a 2 RC such that the following inequality holds: j˚ .u/ .t /  ˚ .0/ .t/jH 1  a ju .t /jL2 for a.e. t 2 Œ0; T    and all u 2 L2 0; T I L2 .˝/ : Then problem (10.22) has at least one solution. Proof. We look for a fixed point of the operator     A W L2 0; T I L2 .˝/ ! L2 0; T I L2 .˝/ ; A .u/ D u0 C .S ı ˚ / .u/ : Theorem 10.6 and the fact that ˚ is a bounded operator, guarantee the complete continuity of A: It remains toprove the existence of a bounded closed convex subset D of L2 0; T I L2 .˝/  ; with A .D/  D: 2 Let u 2 C Œ0; T  I L .˝/ : As in the proof of Theorem 10.5, for  > 0 we obtain a kA .u/  A .0/k  p kuk : 2 It follows that: a kA .u/k  kA .0/k C p kuk : 2 Choosing  > a2 =2 and a positive number  p  R  kA .0/k = 1  a= 2 ;   ¹u 2 C Œ0; T  I L2 .˝/ W we deduce that A is a self-map of the set D0 D  kuk  Rº: Let D be the closure of D0 in L2 0; T I L2 .˝/ : Clearly, D is a bounded closed convex nonempty subset of L2 .0; T I L2 .˝//: In addition,  the continuity of A from L2 0; T I L2 .˝/ to itself implies that A .D/  D: Now Schauder’s fixed point theorem can be applied.

Section 10.5 Application of the Leray–Schauder Fixed Point Theorem

245

Example 10.3. Let ‰ W L2 .˝/ ! H 1 .˝/ be a continuous operator such that (10.26) j‰ .u/  ‰ .0/jH 1  a jujL2 for all u 2 L2 .˝/ and some a 2 RC : Then the mapping     ˚ W L2 0; T I L2 .˝/ ! L2 0; T I H 1 .˝/ given by ˚ .u/ .t / D ‰ .u .t //



   u 2 L2 0; T I L2 .˝/ ; t 2 Œ0; T 

satisfies all the assumptions of Theorem 10.7. .:; / is Example 10.4. Let W ˝  R ! R be a function such that .x; :/ is continuous for a.e. x 2 ˝; .:; 0/ 2 measurable for every  2 R; H 1 .˝/ and there is a0 2 RC with j .x; / 

.x; 0/j  a0 jj

for all  2 R and a.e. x 2 ˝: Then the operator ‰ W L2 .˝/ ! H 1 .˝/ given by ‰ .u/ D

.:; u .://

satisfies all the conditions from the previous example.

10.5

Application of the Leray–Schauder Fixed Point Theorem

We conclude this chapter by an existence result for a superlinear problem. Theorem 10.8. Let n  3; g0 2 L2 .˝/ ; f 2 H 1 .˝/ and let W˝ R ! R be a function such that .:; u/ is measurable for each u 2 R; .x; :/ is continuous for a.e. x 2 ˝; and there exist a; ˛ 2 RC ; 1= .2 /0  ˛ < 2  1 with (10.27) j .x; u/j  a juja for all u 2 R and a.e. x 2 ˝: In addition assume that u

.x; u/  0

(10.28)

for all u 2 R and a.e. x 2 ˝: Then problem (10.22) with ˚ .u/ .t / D .:; u .t// C f has at least one solution.

246

Chapter 10 The Semilinear Heat Equation

Proof. Let p D ˛ .2 /0 : From 1= .2 /0  ˛ < 2  1 D 2 = .2 /0 ; one guarantees that the solution operator S has 1  p < 2 : Then Theorem 10.6  2 is completely continuous from L 0; T I H 1 .˝/ to L2 .0; T I Lp .˝// : We look for a fixed point of the operator     A W L2 0; T I Lp .˝/ ! L2 0; T I Lp .˝/ ; A .u/ D u0 C .S ı ˚ / .u/ : Notice that A .u/ D u0 C S .f / C S .˚0 .u// ; where ˚0 .u/ .t/ D N .u .t // and N is the superposition operator associated to : Assumption (10.27) guarantees, via Theorem 9.1, that the operator N is well-defined continuous  0 and bounded from Lp .˝/ into L.2 / .˝/ and ˇ ˇ ˇN .v/ˇ 2 0  a jvj˛ p : L . / L

Thus, we infer that the operator ˚0 is well-defined continuous and bounded from  0 L2 .0; T I Lp .˝// to L2 .0; T I L.2 / .˝// and consequently, from L2 .0; T I  Lp .˝// to L2 0; T I H 1 .˝/ : Therefore A is completely continuous. Next we show that there exists R > 0 with jujL2 .0;T ILp .˝// < R

(10.29)

for every solution u of the equation u D A .u/ and every 2 .0; 1/ : Indeed, if u D A .u/ for some 2 .0; 1/ ; then from (10.12) and (10.28) we deduce

d 2 2 D . .:; u .t // C f; u .t // ju .t /jL 2 C ju .t /j 1 H0 dt  .f; u .t //

 jf jH 1 ju .t /jH 1 : 0

By integration we obtain Z

T 0

Z 2 ju .t/jH 1 0

dt  c

T 0

! 12 2 ju .t /jH 1 0

dt

C c0;

where the constants c and c 0 do not depend on u and : Hence jujL2 .0;T IH 1 .˝//  C; 0

.˝/ ; one immediately obtains an estimation of  whence, since type (10.29). The conclusion now follows from the Leray–Schauder principle. H01 .˝/

Lp

Section 10.5 Application of the Leray–Schauder Fixed Point Theorem

Example 10.5. The function .x; u/ D  juj˛1 u .u 2 R/ where 1= .2 /0  ˛ < 2  1 satisfies all the assumptions of Theorem 10.8.

247

Chapter 11

The Semilinear Wave Equation

11.1

The Nonhomogeneous Wave Equation in H 1 .˝/

  In Section 4.4, we have proved that for f 2 L2 0; T I L2 .˝/ ; g0 2 H01 .˝/ 1 and g1 2 L2 .˝/ ; there  exists a unique weak solution u 2 C Œ0; T  I H0 .˝/ 1 2 \ C Œ0; T  I L .˝/ of the Cauchy–Dirichlet problem for the wave equation 8 2 ˆ in Q < @@t u2  u D f @u (11.1) u .x; 0/ D g0 .x/ ; @t .x; 0/ D g1 .x/ in ˝ ˆ : uD0 on † and in addition the following estimations hold:   Z t 2 2 2 2 jf .s/jL2 ds ; ju .t/jH 1  3 jg0 jH 1 C jg1 jL2 C t 0 0 0   Z t ˇ 0 ˇ2 2 ˇu .t /ˇ 2  3 jg0 j2 1 C jg1 j2 2 C t jf .s/jL2 ds : L L H 0

(11.2)

0

  In applications, the space L2 0; T I L2 .˝/ might be too small, therefore an extension of the definition of weak solution, as well as of the existence and uniqueness result to larger spaces of functions with values in H 1 .˝/ appears to be of a notable interest.   Definition 11.1. Let f 2 L2 0; T I H 1 .˝/ ; g0 2 L2 .˝/ and g1 2 H 1 of problem   (11.1) we mean a function u 2  .˝/ : 2By a weak 1solution 1 C Œ0; T  I L .˝/ \ C Œ0; T  I H .˝/ such that Z T Z T   .u .t / ; h .t //L2 dt D .f .t / ; v .t // dt C .g1 ; v .0//  g0 ; v 0 .0/ L2 0

0

(11.3)    for every h 2 L 0; T I L2 .˝/ ; where v 2 C Œ0; T  I H01 .˝/ \ C 1 .Œ0; T  I L2 .˝// is the solution of the problem 8 2 ˆ in Q < @@t 2v  v D h @v v .x; T / D @t .x; T / D 0 in ˝ ˆ : vD0 on †  2

whose existence and uniqueness is guaranteed by Theorem 4.5.

Section 11.1 The Nonhomogeneous Wave Equation in H 1 .˝/

249

Notice that the “strange” condition (11.3) can be better understood by a formal argument: the equation u00  u D f is multiplied in H 1 .˝/ by v and then integrated over Œ0; T  ; and similarly, the equation v 00  v D h is multiplied by u and then integrated. These yield Z T  0  0   0   u .t / ; v 0 .t / dt u .T / ; v .T /  u .0/ ; v .0/  Z D

.u .t / ; v .t // dt C

0

0

Z

T

T 0



.f .t / ; v .t // dt;

   v 0 .T / ; u .T /  v 0 .0/ ; u .0/ 

Z D

0

Z

T

.v .t/ ; u .t// dt C

T 0

Z

T

0



 u0 .t / ; v 0 .t / dt

.h .t / ; u .t // dt:

Substraction, together with u .0/ D g0 ; u0 .0/ D g1 ; v .T / D v 0 .T / D 0 gives (11.3). The generalization of Theorem 4.5 to functions f with values in H 1 .˝/ is the following result contained in the monograph Lions–Magenes [24].   Theorem 11.1 (Lions–Magenes). Assume that f 2 L2 0; T I H 1 .˝/ ; g0 2 L2 .˝/ and g1 2 H 1 .˝/ : Then problem (11.1) has a unique weak solution in the sense of Definition 11.1. Moreover, the following inequalities hold:   Z t 2 2 2 2 jf .s/jH 1 ds ; ju .t /jL2  3 jg0 jL2 C jg1 jH 1 C t 0   Z t ˇ 0 ˇ2 2 ˇu .t /ˇ 1  3 jg0 j2 2 C jg1 j2 1 C t jf .s/jH 1 ds : (11.4) L H H 0

Proof. As in Section 4.4, we look for the solution in the form u .t/ D

1 X

uk .t / k ;

(11.5)

kD1

where p p 1

k t C p g1k sin k t

k Z t p 1 fk .s/ sin k .t  s/ ds Cp

k 0

uk .t / D g0k cos

(11.6)

and, this time, fk .t / D .f .t / ; k / ; g0k D .g0 ; k /L2 and g1k D .g1 ; k / :

250

Chapter 11 The Semilinear Wave Equation

  2 (a) We first show that series (11.5) defines a function u P2 2C Œ0; T  I L .˝/ : This reduces to the uniform convergence of the series uk .t / ; which follows from the estimation    Z t 2 1  k 2 1 2 2 k Ct fk .s/ ds : g uk .t /  3 g0 C

k 1 0 k This estimation the  also implies  first inequality in (11.4). 1 1 (b) u 2 C Œ0; T  I H .˝/ : This reduces to the uniform convergence of the 2 P 1  0 .t / : From u series  k k

u0k

Z t p p p p k k .t / D  k g0 sin k t C g1 cos k t C fk .s/ cos k .t  s/ ds 0

we obtain the estimation

   Z t 2 2 1  0 1  k 2 1 2 k Ct fk .s/ ds u .t /  3 g0 C g

k k

k 1 0 k

which gives the conclusion and also the second inequality in (11.4). (c) To derive (11.3) we use the reasoning after Definition 11.1 which is correct, not just formal, when applied to partial sums sm .t / D g0m D

m X

uk .t / k ;

kD1 m X

f

m

.t / D

0

fk .t / k ;

kD1

g0k k ;

m X

g1m D

kD1

Hence Z T Z .sm .t / ; h .t//L2 dt D

m X

g1k k :

kD1

T





f m .t / ; v .t /

0

dt C.g1m ; v .0//

  g0m ; v 0 .0/ L2 :

Letting m ! 1 yields the result. (d) Uniqueness. Let u1 ; u2 have all properties from Theorem 11.1. Then for h D u D u1  u2 ; (11.3) gives Z T 2 ju .t/jL 2 dt D 0; 0

whence u D 0: Hence u1 D u2 : A better regularity of the solution in x is possible if f is smooth in t and the initial data g0 ; g1 are correspondingly regular.

Section 11.1 The Nonhomogeneous Wave Equation in H 1 .˝/

251

  Theorem 11.2. Assume that f 2 W 1;2 0; T I H 1 .˝/ ; i.e. f is absolutely continuous from Œ0; T  to H 1 .˝/ and its derivative f 0 belongs to  L2 0; T I H 1 .˝/ : Also assume that g0 2 H01 .˝/ and  g1 2 L2 .˝/ : 1 Then the  weak solution  u of problem (11.1) belongs to C Œ0; T  I H0 .˝/ 1 2 \ C Œ0; T  I L .˝/ and for some constant c > 0 W   Z t ˇ 0 ˇ2 2 2 2 2 ˇ ˇ f .s/ H 1 ds ; ju .t /jH 1  c jg0 jH 1 C jg1 jL2 C jf .0/jH 1 C 0 0 0   Z t ˇ ˇ ˇ 0 ˇ2 ˇu .t/ˇ 2  c jg0 j2 1 C jg1 j2 2 C jf .0/j2 1 C ˇf 0 .s/ˇ2 1 ds : (11.7) L H L H H 0





0

Proof. (a) To show that u 2 C Œ0; T  I H01 .˝/ ; as at step (a) inP the proof of Theorem 4.5, we have to prove the uniform convergence of the series

k u2k .t / : First, the last term in (11.6) gives after integration by parts Z t Z t p p 1 d 1 cos k .t  s/ ds fk .s/ sin k .t  s/ ds D fk .s/ p

k 0 ds

k 0 p 1 1 fk .0/ cos k t D fk .t / 

k

k Z t p 1 fk0 .s/ cos k .t  s/ ds:  Rt

k

0

Next, using fk .t / D fk .0/ C 0 fk0 .s/ ds; we obtain     Z t  2 2  0 2 1 1 2 2 k k .fk .0// C fk .s/ ds ;

k uk .t /  c k g0 C g1 C

k

k 0 whence the desired conclusion and the first inequality in (11.7).  (b) u 2 C 1 Œ0; T  I L2 .˝/ : This follows from the estimation of u0k .t /2 : Indeed, we have Z t p p p p u0k .t / D  k g0k sin k t C g1k cos k t C fk .s/ cos k .t  s/ ds 0 p p p k k D  k g0 sin k t C g1 cos k t Z t p d 1 sin k .t  s/ ds fk .s/ p ds

k 0 p p p p 1 D  k g0k sin k t C g1k cos k t C p fk .0/ sin k t

k Z t p 1 fk0 .s/ sin k .t  s/ ds: Cp

k 0

252

Chapter 11 The Semilinear Wave Equation

It follows that u0k

    Z t  2 2  0 2 1 1 2 k k .t /  c k g0 C g1 C .fk .0// C fk .s/ ds ;

k

k 0 2

whence the desired result and the second inequality in (11.7).

11.2

Application of Banach’s Fixed Point Theorem

According to Theorem 11.1, we may associate to the Cauchy–Dirichlet problem 8 2 ˆ in Q < @@t u2  u D f @u u .x; 0/ D 0; @t .x; 0/ D 0 in ˝ ˆ : uD0 on † the following solution operators:   S0 W L2 0; T I H 1 .˝/ ! C   S W L2 0; T I H 1 .˝/ ! C

 S0 f D u; Sf D u; u0 ;

   Œ0; T  I L2 .˝/ \ C 1 Œ0; T  I H 1 .˝/ ;   Œ0; T  I L2 .˝/  H 1 .˝/ ;



where u .t / D

1 X kD1

uk .t / k

1 and uk .t / D p

k

Z

t

fk .s/ sin 0

p

k .t  s/ ds:

The next theorem expresses, on the one hand, the continuous dependence on data of the solution of the Cauchy–Dirichlet problem for the wave equation, and on the other hand, implies the continuity of the linear operators S0 and S: Theorem 11.3. Let u be the solution of problem .11:1/:   (a) If f 2 L2 0; T I H 1 .˝/ ; g0 2 L2 .˝/ and g1 2 H 1 .˝/ ; then for every t 2 Œ0; T  ; ˇ ˇ2 juj2C .Œ0;t IL2 .˝// C ˇu0 ˇC .Œ0;t IH 1 .˝//    2 2 2 c 2 jg0 jL : 2 C jg1 jH 1 C jf j 2 L .0;t IH 1 .˝//   (b) If f 2 L2 0; T I L2 .˝/ ; g0 2 H01 .˝/ and g1 2 L2 .˝/ ; then for every t 2 Œ0; T  ;

Section 11.2 Application of Banach’s Fixed Point Theorem

253

  ˇ ˇ2 2 2 2 : juj2C .Œ0;t IH 1 .˝// Cˇu0 ˇC .Œ0;t IL2 .˝//  c 2 jg0 jH 1 C jg1 jL2 C jf j 2 2 L .0;tIL .˝// 0

 1;2

0



(c) If f 2 W 0; T I H 1 .˝/ ; g0 2 H01 .˝/ and g1 2 L2 .˝/ ; then for every t 2 Œ0; T  ; ˇ ˇ2 juj2C .Œ0;t IH 1 .˝// C ˇu0 ˇC .Œ0;t IL2 .˝// 0   ˇ 0 ˇ2 2 2 2 2 ˇ ˇ .0/j : C C  c jg0 jH 1 C jg1 jL f jf 2 H 1 L2 .0;tIH 1 .˝// 0

Here c stands for a constant depending only on T: Proof. The result is a direct consequence of inequalities (11.4), (11.2) and (11.7) applied to each subinterval Œ0; t of Œ0; T  : In what follows we shall denote by F0 ; F the spaces H01 .˝/  L2 .˝/ and L2 .˝/  H 1 .˝/ endowed with norms: jwjF0 D



2 jujH 1 0 .˝/

C

2 jvjL 2 .˝/

 12

; jwjF

  12 2 2 D jujL2 .˝/ C jvjH 1 .˝/

for w D Œu; v : The next result on the continuity of the solution operator S is a direct consequence of Theorem 11.3. Theorem 11.4. The solution operator S is continuous:   (a) from L2 0; T I H 1 .˝/ to C .Œ0; T  I F / and j.Sf / .t/jF  c jf jL2 .0;t IH 1 .˝// ; t 2 Œ0; T  I   (b) from L2 0; T I L2 .˝/ to C .Œ0; T  I F0 / and j.Sf / .t /jF0  c jf jL2 .0;tIL2 .˝// ; t 2 Œ0; T  I   (c) from W 1;2 0; T I H 1 .˝/ to C .Œ0; T  I F0 / and j.Sf / .t/jF0  c jf jW 1;2 .0;tIH 1 .˝// ; t 2 Œ0; T  :  1=2 2 0 2 Here jf jW 1;2 .0;T IH 1 .˝// D jf .0/jH is 1 C jf jL2 0;T IH 1 .˝/ . /   1;2 1 0; T I H .˝/ : the norm of the space W

254

Chapter 11 The Semilinear Wave Equation

We now consider the semilinear problem   8 2 @ u @u ˆ < @t 2  u D ˚ u; @t

in Q

(11.8) .x; 0/ D g1 .x/ in ˝ on †:   Here, in general, we assume that ˚ W C .Œ0; T  I F / ! L2 0; T I H 1 .˝/ ; g0 2 L2 .˝/ ; g1 2 H 1 .˝/ ; and we are interested in weak solutions. u .x; 0/ D g0 .x/ ; ˆ : uD0

@u @t

The following existence and uniqueness result is derived from Banach’s fixed point theorem and is the analogue of Theorem 10.5. Theorem 11.5. (a) Assume that g0 2 L2 .˝/ ; g1 2 H 1 .˝/ ;   ˚ W C .Œ0; T  I F / ! L2 0; T I H 1 .˝/ and j˚ .w1 / .t/  ˚ .w2 / .t /jH 1  a jw1 .t /  w2 .t /jF for a.e. t 2 Œ0; T  ; all w1 ; w2 2 C .Œ0; T  I F / and some constant a 2 RC : Then there exists a unique weak solution u of problem .11:8/: (b) Assume that g0 2 H01 .˝/ ; g1 2 L2 .˝/ ; ˚ W C .Œ0; T  I F0 / ! L2 .0; T I L2 .˝// and j˚ .w1 / .t /  ˚ .w2 / .t /jL2  a jw1 .t /  w2 .t /jF0 for a.e. t 2 Œ0; T  ; all w1 ; w2 2 C .Œ0; T  I F0 / and some constant a 2 RC : Then there exists a unique weak solution u of problem .11:8/: (c) Assume that g0 2 H01 .˝/ ; g1 2 L2 .˝/ ; ˚ W C .Œ0; T  I F0 / ! W 1;2 .0; T I H 1 .˝// and ˇ ˇ j˚ .w1 / .0/  ˚ .w2 / .0/jH 1 C ˇ˚ .w1 /0 .t /  ˚ .w2 /0 .t /ˇH 1  a jw1 .t /  w2 .t /jF0 (11.9) for a.e. t 2 Œ0; T  ; all w1 ; w2 2 C .Œ0; T  I F0 / and some constant a 2 RC : Then there exists a unique weak solution u of problem .11:8/:

Section 11.2 Application of Banach’s Fixed Point Theorem

255

Proof. Let u0 be the solution of problem (11.8) corresponding to ˚ D 0 and let w0 D u0 ; u00 : We have to solve the fixed point problem w D w0 C .S ı ˚ / .w/ : (a) The conclusion will follow from Banach’s fixed point theorem once we have proved that the operator A W C .Œ0; T  I F / ! C .Œ0; T  I F / ; A .w/ D w0 C .S ı ˚ / .w/ is a contraction with respect to a suitable norm on C .Œ0; T  I F / : Let w1 ; w2 2 C .Œ0; T  I F / : From Theorem 11.4(a) we have jA .w1 / .t /  A .w2 / .t /jF

D jS .˚ .w1 /  ˚ .w2 // .t /jF Z t 1=2 2  c d s j˚ .w1 / .s/  ˚ .w2 / .s/jH 1 0

Z  ac

t 0

jw1 .s/ 

w2 .s/j2F

ds

1=2 :

Take any number  > a2 c 2 =2 and consider on C .Œ0; T  I F / the norm   kwk D jw .t /jF  t :

max

t 2Œ0;T

e

Then Z jA .w1 / .t /  A .w2 / .t /jF

 ac kw1  w2 k 

ac p kw1  w2 k 2

t

0

e ds

1=2

2s

e t ;

whence we deduce that ac kA .w1 /  A .w2 /k  p kw1  w2 k : 2

(11.10)

Since pac < 1; the operator A is a contraction with respect to norm k:k : 2 (b) Use similar arguments for the operator A from C .Œ0; T  I F0 / to itself. (c) First observe that if we let D WD ¹w 2 C .Œ0; T  I F0 / W w .0/ D w0 .0/ D Œg0 ; g1 º ;

256

Chapter 11 The Semilinear Wave Equation

then A .D/  D since by the definition of S; .Sf / .0/ D 0 for every f: Next we use the above arguments. According to Theorem 11.4(c), for every w1 ; w2 2 D; we have jA .w1 / .t /  A .w2 / .t/jF0 D jS .˚ .w1 /  ˚ .w2 // .t /jF0   12 Z t ˇ2 ˇ 2 0 0 ˇ ˇ ˚ .w1 / .s/  ˚ .w2 / .s/ H 1 ds  c j˚ .w1 / .0/  ˚ .w2 / .0/jH 1 C : 0

Furthermore, (11.9) implies j˚ .w1 / .0/  ˚ .w2 / .0/jH 1  a jw1 .0/  w2 .0/jF0 D 0 and also

ˇ ˇ ˇ˚ .w1 /0 .s/  ˚ .w2 /0 .s/ˇ 1  a jw1 .s/  w2 .s/j : F0 H

Hence Z jA .w1 / .t /  A .w2 / .t /jF0  ac

0

t

jw1 .s/  w2 .s/j2F0 ds

1=2 :

Finally, if we consider as at part (a) a suitable norm k:k on C .Œ0; T  I F0 / ; we obtain a contraction condition on D; of type (11.10). Clearly, endowed with the metric induced by norm k:k ; the set D is a complete metric space on which Banach’s fixed point theorem applies to A: Example 11.1. Let ‰ W F ! H 1 .˝/ be any mapping for which there exists a constant a 2 RC with (11.11) j‰ .w1 /  ‰ .w2 /jH 1  a jw1  w2 jF for all w1 ; w2 2 F:   Then the mapping ˚ W C .Œ0; T  I F / ! L2 0; T I H 1 .˝/ given by ˚ .w/ .t/ D ‰ .w .t //

.w 2 C .Œ0; T  I F / ; t 2 Œ0; T /

satisfies the assumption of Theorem 11.5(a). .:; / is Example 11.2. Let W ˝  R ! R be a function such that 1 .:; 0/ 2 H .˝/ and there exists a0 2 RC measurable for each  2 R; with j .x; 1 /  .x; 2 /j  a0 j1  2 j for every 1 ; 2 2 defined by

R

and a.e. x 2 ˝: Then the mapping ‰ W F ! H 1 .˝/

‰ .w/ D

.:; u .://

.w D Œu; v 2 F /

has the property from the previous example.

Section 11.3 Application of the Leray-Schauder Fixed Point Theorem

257

Indeed, we may write ‰ .w/ D

.:; 0/ C  .:; u .://

where  .x; / D .x; /  .x; 0/ : Notice that  satisfies j .x; /j  a0 jj . 2 R; a.e. x 2 ˝/ ; whence, from Theorem 9.1, the superposition operator  .:; u .:// maps L2 .˝/ into itself, and in addition j .:; u1 .://   .:; u2 .://jL2  a0 ju1  u2 jL2  a0 jw1  w2 jF : The embedding L2 .˝/  H 1 .˝/ being continuous there will exist a constant a 2 RC such that (11.11) holds. The reader is advised to give similar examples based on Theorem 11.5(b) and (c).

11.3

Application of the Leray-Schauder Fixed Point Theorem

Theorem 11.6. Let 1  p < 2 if n  3 and p  1 if n D 1 or n D 2:   2 0; T I L2 .˝/ from L to (a) The operator S0 is completely continuous   C .Œ0; T  I Lp .˝// and from W 1;2 0; T I H 1 .˝/ to C .Œ0; T  I Lp .˝// :   (b) The operator S is completely continuous from L2 0; T I L2 .˝/ to C .Œ0; T  ; Lp .˝/  H 1 .˝/ and from W 1;2 0; T I H 1 .˝/ to C.Œ0; T  I Lp .˝/  H 1 .˝//: Proof. We first remark that function u D S0 f is the solution of problem (11.1) with g0 D g1 D 0: Hence we are in the case g0 2 H01 .˝/ and g1 2 L2 .˝/ : Also, when n  3; it suffices to prove the result for .2 /0  p < 2 ; the case  0 1  p < .2 /0 being a consequence, since L.2 / .˝/  Lp .˝/ (recall ˝ is bounded). Hence, assume that .2 /0  p < 2 :   2 0; T I L2 .˝/ ; re(a) We shall prove that for each bounded subset M of L   spectively of W 1;2 0; T I H 1 .˝/ ; the set S0 .M / is  bounded in C.Œ0; T  I H01 .˝// and relatively compact in C Œ0; T  I H 1 .˝/ ; and after that we shall apply Lemma 10.5. Indeed, the estimations inTheorem 11.3(b)  and, respectively 1 .˝/ .M / Œ0;  in C T I H and the equicon(c), imply the boundedness of S 0 0   tinuity of S0 .M / in C Œ0; T  I L2 .˝/ ; whence the desired conclusion.   (b) Let M be bounded in L2 0; T I L2 .˝/ ; respectively in W 1;2 .0; T I H 1 .˝//: We already know that S0 .M / is relatively compact in C.Œ0; T  I

258

Chapter 11 The Semilinear Wave Equation

Lp .˝//: We have to show additionally that S0 .M /0 is relatively compact in C Œ0; T  I H 1 (c), S0 .M /0 is  .˝/ : 2From Theorem 11.3(b) and respectively bounded in C Œ0; T I L .˝/ : Then, the compact embedding L2 .˝/  H 1 .˝/ implies that S0 .M /0 .t/ is relatively compact in H 1 .˝/ : Finally, for the  0 equicontinuity of S .M / in C Œ0; T  I H 1 .˝/ ; take any f 2 M and u D S .f / : With the notations from the proof of Theorem 11.1, we have 00 .t /  sm .t / D f m .t / : sm

Then ˇ ˇ 00 ˇs .t /ˇ 1  jsm .t /j 1 C jf m .t /j 1 D jsm .t /j 1 C jf m .t /j 1 : H H H m H H 0

Consequently ˇ 00 ˇ m ˇs ˇ 2 m L .0;T IH 1 .˝//  jsm jL2 .0;T IH01 .˝// C jf jL2 .0;T IH 1 .˝// :     Since sm ! u in L2 0; T I H01 .˝/ ; f m ! f in L2 0; T I H 1 .˝/  as 2 1 m!  1; and since  M and S .M / are bounded in00 L 0; T I H .˝/ and L2 0; T I H01 .˝/ ; respectively, we deduce that jsm jL2 .0;T IH 1 .˝//  C for every m; where constant C does not depend on f: On the other hand, Z t1 0 0 00 .t / dt; sm .t1 /  sm .t2 / D sm t2

whence

ˇZ ˇ 0 ˇ ˇ ˇs .t1 /  s 0 .t2 /ˇ 1  ˇ m m ˇ H

t1 t2

ˇ ˇ ˇ 00 ˇ ˇs .t /ˇ 1 dt ˇ : m ˇ H

Furthermore, Hölder’s inequality implies that p p ˇ ˇ ˇ 0 ˇ ˇs .t1 /  s 0 .t2 /ˇ 1  jt1  t2 j ˇs 00 ˇ 2 m m m L .0;T IH 1 .˝//  C jt1  t2 j: H Letting m ! 1 yields p ˇ ˇ 0 ˇu .t1 /  u0 .t2 /ˇ 1  C jt1  t2 j; H   which implies the equicontinuity of S .M /0 in C Œ0; T  I H 1 .˝/ : The next theorem is a general existence principle for the perturbed problem (11.8) which immediately follows from the Leray–Schauder fixed point theorem.

Section 11.3 Application of the Leray-Schauder Fixed Point Theorem

259

Theorem 11.7. Let g0 2 H01 .˝/ and g1 2 L2 .˝/ : Assume that 1  p < 2 if n  3 and p  1 if n D 1 or n D 2: In addition assume that ˚ .u; v/ D ˚0 .u/ C ˚1 .u; v/ ; where the mappings     ˚0 W C Œ0; T  I Lp .˝/ ! L2 0; T I L2 .˝/   ˚1 W C .Œ0; T  I F / ! L2 0; T I L2 .˝/ are such that the following conditions are satisfied: (i) ˚0 and ˚1 are continuous and bounded; (ii) there exists R > 0 such that jujC .Œ0;T ILr .˝//  R .r D

max ¹2; pº/

ˇ ˇ and ˇu0 ˇC .Œ0;T IH 1 .˝//  R

for all solutions u of the problem 8 2   ˆ in Q < @@t u2  u D ˚ u; @u @t u .x; 0/ D g0 .x/ ; u t .x; 0/ D g1 .x/ in ˝ ˆ : uD0 on †

(11.12)

and every 2 .0; 1/ :   Then problem (11.8) has at least one solution u 2 C Œ0; T  I H01 .˝/ \ C 1 .Œ0; T  I L2 .˝//: Proof. Problem (11.8) is equivalent to the fixed point problem in L2 .0; T I L2 .˝//;   (11.13) f D ˚0 .u0 C S0 f / C ˚1 u0 C S0 f; .u0 C S0 f /0 ; where u0 is the solution of (11.1) corresponding to f D 0: Indeed, f solves (11.13) if and only if u D u0 C S0 f is a solution of (11.8).   The operator S0 being completely continuous from L2 0; T I L2 .˝/ to C .Œ0; T  I Lp .˝// ; and ˚0 being continuous and bounded from C.Œ0; T  I Lp .˝// to L2 0; T I L2 .˝/ ; we have that  mapping ˚0 .u0 C S0 f / is 2 0; T I L2 .˝/ to itself. Furthermore, we have completely continuous from L

   u0 ; u00 : From ˚1 u0 C S0 f; .u0 C S0 f /0 D ˚1 .w0 C Sf / ; where w0 D   2 0; T I L2 .˝/ Theorem 11.6 (b), S is completely continuous from L to    / is continuous and T I F : Also ˚ C Œ0; T  I L2 .˝/  H 1 .˝/ D C .Œ0; 1   bounded from C .Œ0; T  I F / to L2 0; T I L2 .˝/ : Hence the   composition 2 2 Con˚1 .w0 C Sf / is completely continuous from L 0; T I L .˝/ to itself.  0 / .u / C S f C˚ C S f; C S f u is comsequently, the mapping ˚0 .u 0 0 1 0 0 0 0   pletely continuous from L2 0; T I L2 .˝/ to itself.

260

Chapter 11 The Semilinear Wave Equation

The conclusion now follows from the Leray–Schauder principle if we observe   2 2 that (ii) guarantees the boundedness in L 0; T I L .˝/ of the set of all the solutions of the equations f D Œ˚0 .u0 C S0 f / C ˚1 .w0 C Sf / ; 2 .0; 1/ :

(11.14)

Indeed, f is a solution of (11.14) if and only if u D u0 C S0 f is a solution of (11.12). Then (ii) together with the boundedness of the operators ˚0 and ˚1 implies, via the relation

  f D ˚0 .u/ C ˚1 u; u0 ; the existence of a bound for f: We conclude by an existence result for a class of superlinear problems. Theorem 11.8. Let g0 2 H01 .˝/ ; g1 2 L2 .˝/ ; ˚0 .u/ .t / D  ju .t /jq2 u .t / 2 if n  3; q  32 if n D 1 or n D 2; and  2 RC : where 32  q < 2 C n2 Assume that ˚1 is like in Theorem 11.7. In addition assume that

j˚1 .w/ .t /jL2 .˝/  h .t / C a jw .t /jF0 for every w 2 C .Œ0; T  I F0 / ; where h 2 L2 .0; T I RC / and a 2 RC : Then problem ˚ .u; v/ D ˚0 .u/ C ˚1 .u; v/ has at least one solution  .11:8/ with 1 u 2 C Œ0; T  I H0 .˝/ \ C 1 Œ0; T  I L2 .˝/ : continuous Proof. Let p D 2 .q  1/ : Then the operator  ˚0 is2well-defined,  p and bounded from C .0; T I L .˝// to C Œ0; T  I L .˝/ : Also, from 32  2 .n  3/ ; one has 1  p < 2 : q < 2 C n2 Let now u be a solution of problem (11.12), let w D Œu; u0  and let  

f WD ˚0 .u/ C ˚1 u; u0 :   Clearly f 2 L2 0; T I L2 .˝/ : Under the notations from the proof of Theorem 11.1, we have 00 .t /  sm .t / D f m .t / sm 0 .t / in the sense of L2 .˝/ and integration, whence, after multiplication by sm we obtain Z t ˇ 0 ˇ  m  0 ˇs .t/ˇ2 2 C jsm j2 1  0 C 2 .s/ L2 ds; f .s/ ; sm m L H 0

0

Section 11.3 Application of the Leray-Schauder Fixed Point Theorem

261

2 2 where 0 D jg0 jH 1 C jg1 jL2 : Letting m ! 1 we find 0

ˇ 0 ˇ2 ˇu .t /ˇ 2 C ju .t /j2 1  0 C 2 L H 0

Since





˚0 .u/ .s/ ; u0 .s/

L2

Z

t 0

D



 f .s/ ; u0 .s/ L2 ds:

 d q ju .s/jLq ; q ds

we have Z

t





˚0 .u/ .s/ ; u0 .s/

0

L2

ds D  

  q q ju .t /jLq C jg0 jLq q q

 q jg0 jLq : q

Consequently jw .t/j2F0

ˇ2 ˇ 2 D ˇu0 .t /ˇL2 C ju .t/jH 1 0 Z t       C2 ˚1 u; u0 .s/ ; u0 .s/ L2 ds 0 Z t ˇ  ˇ h .s/ C a jw .s/jF0 ˇu0 .s/ˇL2 ds;   C2 0

q

where  D 0 C 2 q jg0 jLq : Furthermore jw .t/j2F0

 C

Z th 0

Z

 cCb

t 0

i h2 .s/ C .1 C 2a/ jw .s/j2F0 ds jw .s/j2F0 ds:

2 Here c D  C jhjL 2 .0;T / and b D 1 C 2a: Now Gronwall’s inequality implies the existence of a constant R > 0 with jw .t /jF0  R for all t 2 Œ0; T  : Since F0  Lr .˝/  H 1 .˝/ ; condition (ii) in Theorem 11.7 is satisfied. The conclusion now follows from Theorem 11.7.

The reader is left to think at possible applications of Schauder’s fixed point theorem.

Chapter 12

Semilinear Schrödinger Equations

This chapter deals with weak solvability of the Cauchy-Dirichlet problem for the perturbed Schrödinger equation: 8 < u t  iu D ˚.u/ in Q u.x; 0/ D g .x/ in ˝ (12.1) : uD0 on †: Here ˝  Rn is a bounded open set and ˚ is a general nonlinear operator which, in particular, can be a superposition operator, a delay operator, or an integral operator. Specific Schrödinger equations arise as models from several areas of physics. The problem is a classical one and our goal here is to make more precise the operator approach based on abstract results from nonlinear functional analysis. More exactly, we shall precise basic properties, such as norm estimation and compactness, for the (linear) solution operator associated to the nonhomogeneous linear Schrödinger equation and we shall use them in order to apply the Banach and Schauder theorems to the fixed point problem equivalent to problem (12.1). A similar programme has been applied in the previous chapters to discuss nonlinear perturbations of the heat and wave equations. Compared to Chapters 10 and 11, here all the spaces consist of complex-valued functions.

12.1

The Nonhomogeneous Schrödinger Equation

Consider the Cauchy–Dirichlet problem equation 8 < u t  iu D f u.x; 0/ D g .x/ : uD0

for the nonhomogeneous Schrödinger in ˝  .0; T / in ˝ on @˝  .0; T /:

(12.2)

As for the heat and the wave equations, we shall apply Fourier’s method. For the first result, we shall work unitary in one of the spaces H01 .˝/ ; L2 .˝/ ; 1 H .˝/ ; which is denoted by V; endowed with the corresponding inner product .:; :/V and norm j:jV and we shall assume that jk jV D 1 for every k:

Section 12.1 The Nonhomogeneous Schrödinger Equation

263

We already know that the system .k / of eigenfunctions of the Dirichlet problem for ; is complete in each of the three spaces (remember Remark 7.9). In what follows the notation C 1 .Œ0; T  I V / is used to denote the space of all functions u with ./1 u 2 C 1 .Œ0; T  I V / : Theorem 12.1. Assume that g 2 V and f 2 C .Œ0; T  I V / : Then there exists a unique function u 2 C .Œ0; T  I V / \ C 1 .Œ0; T  I V / with u .0/ D g and u0 .t/  iu .t/ D f .t / in V In addition ju .t/j2V

 2

Z jgj2V

Ct

t 0

jf

.s/j2V

ds

.t 2 Œ0; T / :

(12.3)

for all t 2 Œ0; T  :

(12.4)



Proof. (a) We look for a solution in the form u .t/ D

1 X

uk .t / k :

(12.5)

kD1

If we formally replace into (12.2) we obtain Z t ik t uk .t / D e gk C eik .t s/ fk .s/ ds;

(12.6)

0

where fk .t / D .f .t / ; k /V and gk D .g; k /V : (b) Series (12.5) defines a function u 2 C .Œ0; T  I V / : Indeed its uniform convergence reduces to the uniform convergence of the series of real functions P 2 juk .t /j : We have Z t 2 ! 2 2 (12.7) juk .t /j  2 jgk j C jfk .s/j ds 0

  Z t 2 2  2 jgk j C t jfk .s/j ds : 0

Thus the uniform convergence of the series gence of the numerical series X

2

jgk j

and

P

XZ

T 0

juk .t /j2 is reduced to the converjfk .s/j2 ds:

264

Chapter 12 Semilinear Schrödinger Equations

According to Parseval’s equality, the sum of the first series is jgj2V ; while of the RT second one 0 jf .s/j2V ds since f 2 C .Œ0; T  I V / : (c) u 2 C 1 .Œ0; T  I V / : We have to prove that v WD ./1 u 2 C 1 .Œ0; T  I V / : One has v .t/ D

1 X

1

uk .t / ./

kD1

1 X 1 k D uk .t / k :

k kD1

Thus the problem is to show the uniform convergence in V of the series of derivatives 1 X 1 0 u .t / k :

k k kD1

P

This reduces to the uniform convergence of the series u0k

.t / D i k e

ik t

Z gk C fk .t /  i k

t e

ˇ ˇ 0 ˇu .t /ˇ2 : We have k

1 2k

ik .t s/

fk .s/ ds:

0

Then ˇ2 1 ˇˇ 0 uk .t/ˇ  2

k

3

2k



Z

2k

2

2

jgk j C jfk .t /j C

1  3 jgk j2 C 2 jfk .t /j2 C t

1

t 2k Z

t 0



t 0

2

jfk .s/j

d

s

!

jfk .s/j2 ds :

So we are finished since X X jfk .t /j2 D jf .t /j2V jgk j2 D jgj2V and uniformly on Œ0; T  : (d) Identity (12.3) follows if we pass to the limit in the corresponding identity for partial sums. (e) Uniqueness. Assume u1 ; u2 are two such functions. Then u WD u1  u2 satisfies u .0/ D 0 and u0 .t /iu .t/ D 0 in V; i.e. ./1 u0 .t /Ciu .t / D 0 in V: Denote v WD ./1 u: Then v .0/ D 0 and v 0 .t / C i u .t / D 0 in V: In particular, this gives  0  v .t / ; v .t/ V C i .u .t / ; v .t //V D 0:

Section 12.1 The Nonhomogeneous Schrödinger Equation

265

Observe that .u .t/ ; v .t//V 2 RC (verify for each space L2 .˝/ ; H01 .˝/ and H 1 .˝//; and that   1 d Re v 0 .t / ; v .t/ V D jv .t /j2V : 2 dt Hence

d jv .t /j2V D 0 on Œ0; T  : dt

Thus jv .t /j2V is constant and since v .0/ D 0; we have v .t /  0; hence u D 0; that is u1 D u2 : (f) Relation (12.4) is an immediate consequence of (12.7). If f is less regular in x and more regular in t; then we have   Theorem 12.2. Assume that g 2 H01 .˝/ and f 2 C 1 Œ0; T  I H 1 .˝/ : Then there exists a unique function     u 2 C Œ0; T  I H01 .˝/ \ C 1 Œ0; T  I H 1 .˝/ with u .0/ D g and u0 .t /  i u .t/ D f .t / in H 1 .˝/ .t 2 Œ0; T / : In addition 2 ju .t /jH 1 0

ˇ 0 ˇ2 ˇu .t /ˇ

H 1

  Z t ˇ 0 ˇ2 2 2 2 ˇ ˇ f .s/ H 1 ds ;  4 jgjH 1 C jf .t/jH 1 C jf .0/jH 1 C t 0 0   Z t ˇ 0 ˇ2 2 2 ˇ ˇ  3 jgjH 1 C jf .0/jH 1 C t (12.8) f .s/ H 1 ds : 0

0

Proof. As above we look for a solution in the form (12.5), where this time we assume that jk jL2 D 1: Hence in (12.6), fk .t / D .f .t / ; k / and gk D .g; k / : First note that using derivative fk0 ; uk .t / can be written as Z t ik t uk .t / D e gk C eik .t s/ fk .s/ ds 0 Z t 0 i eik .t s/ fk .s/ ds D eik t gk 

k 0 i

i

fk .t / C eik t fk .0/ D eik t gk 

k

k Z t i C eik .t s/ fk0 .s/ ds:

k 0

266

Chapter 12 Semilinear Schrödinger Equations

  (a) u 2 C Œ0; T  I H01 .˝/ : This reduces to the uniform convergence of series P

k juk .t /j2 which follows from the estimation

k juk .t /j2

! Z t ˇ 0 ˇ2 1 1 t 2 2 ˇf .s/ˇ ds  4 k jgk j C 2 jfk .t /j C 2 jfk .0/j C 2

k

k

k 0 k   Z t 1 1 1 ˇˇ 0 ˇˇ2 2 2 2 D 4 k jgk j C fk .s/ ds : jfk .t /j C jfk .0/j C t

k

k 0 k   (b) The relation u 2 C 1 Œ0; T  I H 1 .˝/ follows from the estimation 2

ˇ ˇ2 Z t ˇ ˇ2 ˇ 1 1 ˇˇ 0 .t s/  i  t  i  k ˇi k e k gk C fk .t /  i k e fk .s/ ds ˇˇ uk .t /ˇ D ˇ

k

k 0 ˇ ˇ2 Z t 0 ˇ ˇ 1 ˇ ik t ik .t s/ .t / .s/ D g C f f  ds ˇ e  i k e k k k ˇ

k ˇ 0 ˇ2 ˇ Z t ˇ 1 ˇˇ i k t ik t ik .t s/ 0 D gk C e fk .0/ C e fk .s/ ds ˇˇ i k e ˇ

k 0   Z t 1 1 ˇˇ 0 ˇˇ2 2 2  3 k jgk j C fk .s/ ds : jfk .0/j C t

k 0 k We shall associate to the Cauchy–Dirichlet problem with g D 0; for the Schrödinger equation, the following solution operator: S

W

C.Œ0; T  I H 1 .˝// ! C.Œ0; T  I H 1 .˝//;

C.Œ0; T  I H 1 .˝// 7! Sf 2 C.Œ0; T  I H 1 .˝//; 1 X .Sf / .t / D uk .t / k ; jk jL2 D 1; f

2

kD1 Z t

uk .t/ D

e

0

ik .t s/

fk .s/ ds; fk .t / D .f .t / ; k / :

Hence Sf is the unique function u satisfying the conditions of Theorem 12.1, for g D 0:

12.2

Properties of the Schrödinger Solution Operator

It follows from (12.4) that the solution operator S is a continuous linear self-map of each of the spaces:       C Œ0; T  I H01 .˝/ ; C Œ0; T  I L2 .˝/ and C Œ0; T  I H 1 .˝/ :

Section 12.2 Properties of the Schrödinger Solution Operator

267

Also, from Theorem 12.2,        S C 1 Œ0; T  I H 1 .˝/  C Œ0; T  I H01 .˝/ \ C 1 Œ0; T  I H 1 .˝/ : Theorem 12.3. (a) Let f 2 C .Œ0; T  I V / : Then for every t 2 Œ0; T ; p j.Sf / .t /jV  2t jf jL2 .0;tIV / :

(12.9)

Here, as above, V is any of the spaces L2 .˝/ ; H01 .˝/ and H 1 .˝/ :   (b) Let f 2 C 1 Œ0; T  I H 1 .˝/ : Then for every t 2 Œ0; T ;   ˇ 0 ˇ2 2 2 ˇ ˇ  12 jf .0/jH j.Sf / .t/jH 1 1 C t f L2 0;t IH 1 .˝/ . / ; 0   ˇ ˇ ˇ ˇ ˇ.Sf /0 .t /ˇ2 1  3 jf .0/j2 1 C t ˇf 0 ˇ2 2 : H H L .0;t IH 1 .˝// Proof. (a) Simple consequence of (12.4). (b) Use (12.8) and Z t f 0 .s/ ds; f .t / D f .0/ C 0



whence jf

2 .t /jH 1

 2 jf

Z 2 .0/jH 1

Ct

t

ˇ 0 ˇ2 ˇf .s/ˇ

0

H 1



ds :

As regards the complete continuity of the Schrödinger solution operator, we have Theorem 12.4. Let 1  p < 2 if n  3 and p  1 if n D 1 or n D 2: The solution operator is completely continuous: (a) from C.Œ0; T  I H01 .˝// to C.Œ0; T  I Lp .˝//I (b) from C 1 .Œ0; T  I H 1 .˝// to C.Œ0; T  I Lp .˝//: Proof. (a) Obviously, if M is a bounded subset of C.Œ0; T  I H01 .˝//; then S .M / is bounded in C.Œ0; T  I H01 .˝//; as shows Theorem12.3 (a). Consequently, for each t; S .M / .t/ is a bounded subset of H01 .˝/ ; thus relatively compact in Lp .˝/ : It remains to prove that S .M / is equicontinuous in C .Œ0; T  I Lp .˝// : To this end, let f 2 M and u D Sf: We have ˇZ ˇ juk .t1 /  uk .t2 /j D ˇˇ 2

t1

ik .t s/

e t2

ˇZ ˇ2 ˇ ˇ ˇ fk .s/ ds ˇ  jt1  t2 j ˇˇ

t1 t2

ˇ ˇ jfk .s/j ds ˇˇ : 2

268

Chapter 12 Semilinear Schrödinger Equations

Then, via Parseval’s equality, we obtain Z 2 ju .t1 /  u .t2 /jH 1  jt1  t2 j 0

T

0

2 jf .s/jH 1 : 0

This together with ju .t1 /  u .t2 /jLp  c ju .t1 /  u .t2 /jH 1 0

proves the equicontinuity of S .M / in C .Œ0; T  I Lp .˝// : (b) If M is bounded in C 1 .Œ0; T  I H 1 .˝//; then from Theorem 12.3 (b), S .M / is bounded in C.Œ0; T  I H01 .˝// and in C 1 .Œ0; T  I H 1 .˝//: This implies that S .M / is relatively compact in C Œ0; T  I H 1 .˝/ : Then, for .2 /0  p < 2 .n  3/ ; from Lemma 10.5, we deduce that S .M / is relatively compact in C.Œ0; T  I Lp .˝//: The case 1  p < .2 /0 is now a simple  0 consequence in view of the inclusion L.2 / .˝/  Lp .˝/ .˝ being bounded/ :

12.3

Applications of Banach’s Fixed Point Theorem

Our first existence and uniqueness result for the semilinear problem (12.1) is established by means of Banach’s fixed point theorem. Theorem 12.5. Let g 2 L2 .˝/ and     ˚ W C Œ0; T I L2 .˝/ ! C Œ0; T I L2 .˝/ be a mapping for which there exists a 2  a constant 2 inequality holds for all u; v 2 C Œ0; T I L .˝/ W

RC

such that the following

j˚.u/.t/  ˚.v/.t /jL2  a ju.t /  v.t /jL2 for every t 2 Œ0; T :

(12.10)

Then there exists a unique function     u 2 C Œ0; T  I L2 .˝/ \ C 1 Œ0; T I L2 .˝/ such that ² 0 u .t /  i u .t/ D ˚ .u/ .t / in L2 .˝/ for every t 2 Œ0; T  ; u .0/ D g:

Section 12.3 Applications of Banach’s Fixed Point Theorem

269

Proof. Let u0 be the solution of problem (12.1) corresponding to ˚ D 0: We have to solve the fixed point problem   u D u0 C .S ı ˚ / .u/ ; u 2 C Œ0; T  I L2 .˝/ : The conclusion will follow from Banach’s fixed point theorem once we have shown that the operator     A W C Œ0; T  I L2 .˝/ ! C Œ0; T  I L2 .˝/ ; A .u/ D u0 C .S ı ˚ / .u/   is a contraction with respect to a suitable norm on C Œ0; T  I L2 .˝/ : Let u; v 2  C Œ0; T  I L2 .˝/ : Using (12.9) we obtain 2 jA .u/ .t/  A .v/ .t/jL 2

2 D jS .˚ .u/  ˚ .v// .t /jL 2 2  2t j˚ .u/  ˚ .v/jL 2 .0;t IL2 .˝// :

Let  > T a2 be a fixed number and consider the norm on C.Œ0; T  I L2 .˝//;   kuk D ju .t /jL2  t :

max

e

t 2Œ0;T

Furthermore, Z j˚ .u/  ˚

2 .v/jL 2 .0;t IL2 .˝//

t

d

2 j˚ .u/ .s/  ˚ .v/ .s/jL 2 s 0 Z t 2 2 2s 2s  a ju .s/  v .s/jL 2 s 0 Z t 2s s  a2 ku  vk2

D

e e

d

0



a2 ku  vk2 2

Hence 2 jA .u/ .t/  A .v/ .t/jL 2 

Dividing by

e2 t

e d

e2 t :

t a2 ku  vk2 

e2 t :

and taking the maximum over Œ0; T  yields p Ta kA .u/  A .v/k  p ku  vk : 

p p   Since T a=  < 1; the operator A is a contraction on C Œ0; T  I L2 .˝/ with respect to the norm k:k :

270

Chapter 12 Semilinear Schrödinger Equations

Example 12.1. Let ‰ W L2 .˝/ ! L2 .˝/ be any mapping for which there exists a constant a 2 RC with (12.11) j‰ .u/  ‰ .v/jL2  a ju  vjL2 for all u; v 2 L2 .˝/ :     Then the mapping ˚ W C Œ0; T  I L2 .˝/ ! C Œ0; T  I L2 .˝/ given by     ˚ .u/ .t / D ‰ .u .t // u 2 C Œ0; T I L2 .˝/ ; t 2 Œ0; T  satisfies the assumption of Theorem 12.5. .:; / is Example 12.2. Let W ˝  C ! C be a function such that 2 measurable for each  2 C; .:; 0/ 2 L .˝/ ; and there is a constant a 2 RC with j .x; 1 /  .x; 2 /j  a j1  2 j for all 1 ; 2 2 defined by

C

and a.e. x 2 ˝: Then the mapping ‰ W L2 .˝/ ! L2 .˝/ ‰ .u/ D

.:; u .://

satisfies the condition from the previous example. Indeed, we may write ‰ .u/ D

.:; 0/ C  .:; u .://

where  .x; / D .x; /  .x; 0/ : Notice that j .x; /j  a j j (for all  2 C and a.e. x 2 ˝/; so the superposition operator  .:; u .:// maps L2 .˝/ into L2 .˝/ : In addition j .:; u .://   .:; v .://jL2  a ju  vjL2 ; whence (12.11). For the next result, denote    ®  ¯ C01 Œ0; T I H 1 .˝/ D u 2 C 1 Œ0; T I H 1 .˝/ W u .0/ D 0 : Theorem 12.6. Let g 2 H01 .˝/ and     ˚ W C Œ0; T I H01 .˝/ ! C01 Œ0; T I H 1 .˝/ be a mapping for which there exists  a constant a 2 RC such that the following inequality holds for all u; v 2 C Œ0; T I H01 .˝/ W ˇ ˇ ˇ˚.u/0 .t/  ˚.v/0 .t /ˇ 1  a ju.t /  v.t /j 1 for every t 2 Œ0; T : (12.12) H H 0

Section 12.3 Applications of Banach’s Fixed Point Theorem

271

Then there exists a unique function     u 2 C Œ0; T  I H01 .˝/ \ C 1 Œ0; T I H 1 .˝/ satisfying ² 0 u .t /  iu .t / D ˚ .u/ .t / in H 1 .˝/ for every t 2 Œ0; T  ; u .0/ D g:

Proof. We have to solve the fixed point problem for the operator     A W C Œ0; T I H01 .˝/ ! C Œ0; T I H01 .˝/ ; A .u/ D u0 C .S ı ˚ / .u/ : From Theorem 12.3(b), we have 2 jA .u/ .t /  A .v/ .t/jH 1 0

2 D jS .˚ .u/  ˚ .v// .t /jH 1 0 Z t ˇ ˇ ˇ˚ .u/0 .s/  ˚ .v/0 .s/ˇ2 1 ds  12t H 0 Z t 2  12t a2 ju .s/  v .s/jH 1 ds: 0

0

As above, if we consider  > 6T a2 and norm   kuk D max ju .t /jH 1 e t 0

t2Œ0;T

  on C Œ0; T I H01 .˝/ ; we obtain r

6T ku  vk :      Example 12.3. Let ˚0 W C Œ0; T I H01 .˝/ ! C Œ0; T I H 1 .˝/ be any mapping for which there is a constant a 2 RC such that kA .u/  A .v/k  a

j˚0 .u/ .t/  ˚0 .v/ .t/jH 1  a ju .t /  v .t /jH 1 0





for all u; v 2 C Œ0; T I H01 .˝/ and t 2 Œ0; T  : Then the mapping ˚ given by Z t ˚0 .u/ .s/ ds ˚ .u/ .t/ D 0

satisfies all the assumptions of Theorem 12.6.

272

12.4

Chapter 12 Semilinear Schrödinger Equations

Applications of Schauder’s Fixed Point Theorem

The next existence results are based on Schauder’s fixed point theorem. The Lipschitz condition on the nonlinear term ˚ in Theorem 12.5 is weakened to a condition of at most linear growth. Theorem 12.7. Let g 2 H01 .˝/ ; p 2 Œ1; 2 / and     ˚ W C Œ0; T I Lp .˝/ ! C Œ0; T I H01 .˝/

RC

such that the

j˚.u/.t /  ˚.0/.t /jH 1  a ju.t /jLp for every t 2 Œ0; T :

(12.13)

a continuous mapping for which there exists a constant a 2 following inequality holds for every u 2 C .Œ0; T I Lp .˝// W 0

Then there exists at least one function     u 2 C Œ0; T  I H01 .˝/ \ C 1 Œ0; T I H 1 .˝/ such that ² 0 u .t /  iu .t/ D ˚ .u/ .t/ in H 1 .˝/ for every t 2 Œ0; T  ; u .0/ D g:

Proof. We look for a fixed point of the operator     A W C Œ0; T I Lp .˝/ ! C Œ0; T I Lp .˝/ ; A.u/ D u0 C .S ı ˚ /.u/: Theorem 12.4 and the continuity and boundedness of the operator ˚ guarantee the complete continuity of A: It remains to find a nonempty, bounded, closed and convex subset D of C .Œ0; T I Lp .˝// with A.D/  D: Let u 2 C .Œ0; T I Lp .˝// : As in the proof of Theorem 12.5, one obtains jA .u/ .t/  A .0/ .t /jLp

 c jA .u/ .t /  A .0/ .t /jH 1 0 p  c 2t j˚ .u/  ˚ .0/jL2 .0;t IH 1 .˝// : 0

Since Z

0

t

2 j˚ .u/ .s/  ˚ .0/ .s/jH 1 ds 0 0 Z t 2 2  a ju .s/jL p ds;

2 D j˚ .u/  ˚ .0/jL 2 .0;t IH 1 .˝//

0

Section 12.4 Applications of Schauder’s Fixed Point Theorem

we have

Z jA .u/ .t/ 

2 A .0/ .t/jL p

2 2

 2a c t

If in C .Œ0; T  I Lp .˝// we consider the norm  kuk D ju .t/jLp

max t 2Œ0;T

r

then we deduce kA .u/  A .0/k  ac

kA .u/k  kA .0/k C ac

0

e t

2 ju .s/jL p ds:

 ;

T kuk : 

r

Hence

t

273

T kuk : 

If we choose  > a2 c 2 T; then we can find a large enough R > 0 such that kuk  R implies kA .u/k  R: Thus, Schauder’s fixed point theorem applies. Similarly, we have Theorem 12.8. Let g 2 H01 .˝/ ; p 2 Œ1; 2 / and     ˚ W C Œ0; T I Lp .˝/ ! C01 Œ0; T I H 1 .˝/ be a continuous mapping for which there exists a constant a 2 RC such that the following inequality holds for every u 2 C .Œ0; T I Lp .˝// W ˇ ˇ ˇ˚.u/0 .t/  ˚.0/0 .t /ˇ 1  a ju.t /j p for every t 2 Œ0; T : (12.14) L H Then there exists at least one function     u 2 C Œ0; T  I H01 .˝/ \ C 1 Œ0; T I H 1 .˝/ satisfying ² 0 u .t /  u .t/ D ˚ .u/ .t / in H 1 .˝/ ; for every t 2 Œ0; T  ; u .0/ D g:

i

274

Chapter 12 Semilinear Schrödinger Equations

Example 12.4. Let     ˚0 W C Œ0; T I Lp .˝/ ! C Œ0; T I H 1 .˝/ be a continuous mapping for which there is a constant a 2 RC such that j˚0 .u/ .t /  ˚0 .0/ .t /jH 1  a ju .t /jLp for all u 2 C .Œ0; T I Lp .˝// and t 2 Œ0; T  : Then the mapping ˚ given by Z t ˚0 .u/ .s/ ds ˚ .u/ .t/ D 0

satisfies all the assumptions of Theorem 12.8. For other methods and results in semilinear evolution equations we refer the reader to the books Barbu [2], Cazenave [6], Cazenave–Haraux [7], Lions [23], Temam [52] and Vrabie [56].

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Index

Arzelà–Ascoli theorem, 241 barrier, 66 Bessel (equation), 84 (function), 84 (inequality), 43 bounded operator, 210 Boussinesq equation, 20 canonical form, 71 Carathéodory conditions, 209 Cauchy problem, 11, 12 Cauchy-Dirichlet problem, 11, 12 characteristic polynomial, 6 classical solution, 34 coercive functional, 223 compact embedding, 160 complete system, 45 completely continuous operator, 213 completion, 38 conservation law, 13 continuity equation, 14 continuous (embedding), 159 (semigroup) , 178 convection equation, 15 convection-diffusion equation, 16 convolution, 110, 137 critical (exponent), 165 (point), 36, 220 D’Alembert formula, 120 diffusion (constant), 15 (equation), 15 dilatation, 137

Dirac distribution, 134 directional derivative, 4 Dirichlet (principle), 35, 181 (problem), 10 distribution , 133 (derivative), 135 (space D 0 .˝/), 133 (space E 0 .˝/), 140 divergence , 3 (form), 50 (theorem), 5 double-layer potential, 69 Duhamel principle, 124 eigenfunction, 46 eigenvalue, 46 elliptic equation, 6 energetic (norm), 37 (space), 40 energy functional, 34 equation of order m, 6 extension by reflection, 153 exterior boundary value problem, 11 exterior cone condition, 68 exterior segment condition, 67 Fisher equation, 16 Fourier (coefficient), 43 (method), 92 (series), 43 (transform), 109 Fredholm method, 70 fundamental solution , 146 (of Laplace’s equation), 22 (of heat equation), 119

Index fundamental spaces, 132 Gagliardo lemma, 160 Gauss theorem, 21 Gauss–Ostrogradski theorem, 5 Ginzburg-Landau equation, 19 gradient, 3 Green (formulas), 21 (function), 30 Hadamard method of descent, 122 harmonic (function), 4 (modification), 64 Harnack inequality, 56 heat equation, 7 Heaviside function, 134 homogeneous equation, 6 Hopf (strong maximum principle), 58 (weak maximum principle), 59 hyperbolic equation, 7 integration by parts formula, 5 interpolation inequality, 161 isotone operator, 190 Kirchhoff formula, 120 Klein-Gordon equation, 19 Laplace equation, 4 Laplacian, 4 layer potential, 68 Leray-Schauder theorem, 215 linear partial differential equation, 5 Lions lemma, 242 Liouville theorem, 75 logistic equation, 16 lower solution, 218 maximum principle, 189 maximum principle for heat equation, 88 mean value theorem, 25 minimum property, 44

279 monotone iterative method, 218 multi-index, 3 natural boundary condition, 55 Navier-Stokes system, 19 Nemytskii superposition operator, 208 Neumann problem, 10 Newton theorem, 26 Newtonian potential, 59 nonnegative distribution, 186 normal vector, 5 operator (D ˛ ), 3 (method), 207 (of order m), 6 orthonormal system, 43 parabolic equation, 6 Parseval equality, 45 partial differential equation of order two, 6 Perron method, 62 Plancherel theorem, 144 Poincaré inequality, 38 Poisson (equation), 4 (formula), 33, 120 (kernel), 33 principal symbol, 6 quasilinear equation, 6 radial solution, 72 reaction-diffusion system, 17 reflection, 137 regular (distribution), 133 (point), 66 regularizing sequence, 132 Rellich–Kondrachov theorem, 163 Riemann-Green theorem, 23 Robin problem, 10 Schauder fixed point theorem, 214 Schrödinger equation, 18

280 Schwartz space, 112 separation of variables, 72 sine-Gordon equation, 19 single-layer potential, 69 singular distribution, 134 smooth set, 4 Sobolev (embedding theorem), 160, 163 (space H m .˝/), 167 (space H 1 .˝/), 53 (space H m .˝/), 149 (space H01 .˝/), 38 (space H0m .˝/), 156 (space W m;p .˝/), 151 solution operator, 207 space   (C01 ˝ ), 34 (H 1=2 . /), 159 (D .˝/), 131 (E .˝/), 131 (S 0 ), 143 specific heat, 15 strong maximum principle, 26, 190 subfunction, 65

Index subharmonic function, 63 superfunction, 65 superharmonic function, 63 support of a distribution, 139 support of a function, 131 tempered distribution, 143 test function, 132 thermal conductivity, 15 trace, 159 translation, 137 Tricomi equation, 7 uniform ellipticity, 57 upper solution, 218 Vitali lemma, 210 wave equation, 7 weak (maximum principle), 28 (solution), 41 well-posed boundary value problem, 12 Weyl lemma, 142