Problems on Partial Differential Equations [1 ed.] 978-3-030-14733-4

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Problem Books in Mathematics

Maciej Borodzik Paweł Goldstein Piotr Rybka Anna Zatorska-Goldstein

Problems on Partial Differential Equations

Problem Books in Mathematics Series Editor: Peter Winkler Department of Mathematics Dartmouth College Hanover, NH 03755 USA

More information about this series at http://www.springer.com/series/714

Maciej Borodzik • Paweł Goldstein • Piotr Rybka • Anna Zatorska-Goldstein

Problems on Partial Differential Equations

123

Maciej Borodzik The University of Warsaw Warsaw, Poland

Paweł Goldstein The University of Warsaw Warsaw, Poland

Piotr Rybka The University of Warsaw Warsaw, Poland

Anna Zatorska-Goldstein The University of Warsaw Warsaw, Poland

ISSN 0941-3502 ISSN 2197-8506 (electronic) Problem Books in Mathematics ISBN 978-3-030-14733-4 ISBN 978-3-030-14734-1 (eBook) https://doi.org/10.1007/978-3-030-14734-1 Library of Congress Control Number: 2019935542 Mathematics Subject Classification (2010): 35-01, 35A, 35J, 35K, 35L, 46-01, 46E, 46F © Springer Nature Switzerland AG 2019 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG. The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

In our teaching practice of presenting modern introductory courses on partial differential equations (PDEs for short), we have been facing a shortage of books with problems for independent study, which could be also used by lecturers. This book aims at bridging this gap. We use the term “modern introduction to PDEs” on purpose. This means that the emphasis is on the theory based on the Sobolev spaces and weak derivatives. We cover mostly the theory of linear equations of the elliptic, parabolic, and hyperbolic types. A few examples of nonlinear equations are included. We also consider elements of the classical theory of solutions to PDEs, e.g., harmonic functions, the method of characteristics, or closed formulas for solutions of low-dimensional wave equations. In particular, we present the separation of variables method, which is a motivation for the Galerkin method. This book presents the necessary theoretical material in a nutshell. This background is always highlighted at the beginning of a section devoted to particular topics. We avoid presenting the proofs of theorems we use; however, we give references to the relevant sources. For this reason, this book cannot be used as a textbook on PDEs, but it is a companion text, because we present worked-out problems and problems for independent studies. Our experience suggests that it can cater to many introductory or mid-level courses.

v

Introduction

Partial differential equations (PDEs for short) form a vast section of mathematical analysis. It suffices to say that the entire theory of analytic functions of one complex variable is a study of properties of solutions of the Cauchy–Riemann system of two linear first-order PDEs. We must also keep in mind that examining the properties of solutions of any particular equation may be a goal of long-term research programs conducted by large teams of mathematicians. Such programs may sometimes have a theoretical perspective; sometimes they may be oriented toward physics or engineering and focus on certain applications. We hope that the reader, who gets through this text and who solves a considerable number of problems from all chapters of this book, shall appreciate and understand our perspective. Considering the vastness of the field, a lecturer conducting an introductory course on PDEs for students of mathematics faces a challenging task. We do not insist that differential equations are a particularly difficult field of mathematics, but we claim that in order to understand various concepts and grasp the meaning of the variety of theorems, it is crucial to master a large apparatus of analytical tools. Students who wish to befriend PDEs should try to observe this topic from a certain distance, which allows to see the association between pure and applied mathematics. One of the possible ways of conducting lectures on PDEs is building upon achievements of the twentieth-century elementary functional analysis, distribution theory, and Sobolev spaces as well as the concept of weak solutions. This toolbox and the resulting language open an opportunity for a broader approach, and at the same time, they are indispensable for everyone who would care to get acquainted with PDEs to an extent that goes beyond an introductory one-semester course. This way of organizing material was accepted by the authors of this text. As a result, abstract but actually relatively simple functional analysis tools form the basis of this work. Particular equations (augmented with appropriate boundary or initial conditions) and a description of properties of their solutions take, in this approach, the role of examples. Sometimes, they are just an illustration of more or less typical applications of the exposed theory.

vii

viii

Introduction

This volume grew out of authors’ classroom practice. While teaching PDE courses at different levels, we were facing the lack of a problem book, and we needed a supply of exercises at different levels of difficulty. As a result, we shared among us the problems we found in the literature and the ones we designed. In 2009, Prof. Jacek Mi˛ekisz, who was the leader of a project cofinanced by the EU European Social Fund and the Polish Ministry of Science and Higher Education, inspired us to gather these problems in a book freely available online to help younger generations of lecturers and their students. This is how the first Polish edition of this book emerged. Later, we decided to revise the book thoroughly and to translate it into English to make it widely accessible. We expanded existing chapters by adding a substantial amount of new material. Several sections, e.g., Sect. 5.3 on parabolic equations, were written from scratch. As a result, this book can serve as a problem book for PDE courses at basic and intermediate levels. Instructors and students of advanced calculus and functional analysis can also benefit from Chaps. 1 and 2 as well as Sect. 3.1 of this book. Although our emphasis has been on weak solutions, in the book we also cover classical methods, so that the book may prove helpful for different approaches at a PDE course. L. Evans’ Partial Differential Equations [8] is a textbook which is closest in spirit to our problem book. The structure of the book as follows. Each chapter is divided into four sections: a theoretical introduction, worked-out problems, exercises for self-study, and a small section on bibliographic comments. By design, there is no answer section. To some extent, we follow the classification of second-order equations into elliptic, hyperbolic, and parabolic, even though this division is not exhaustive, in general. We begin the book with a recollection of integration by parts formulas, including the theory of differential forms. It is virtually impossible to state the weak form of a linear problem in the divergence form without the Gauss formula. We have noted in our teaching practice that students appreciate recalling this part of material before starting the real work. Since our intention was to study weak solutions, we had to devote an entire chapter to the Fourier transform, the distribution theory, and the Sobolev spaces. In fact, from the point of view of needed techniques, this is the key chapter of this book. A student should get acquainted with this material before tackling the weak formulation of PDEs. A separate chapter is devoted to methods common to different types of equations. We begin this chapter with the necessary introduction to the theory of weak convergence in Hilbert spaces. This is sufficient when we aim at understanding linear equations. There are, however, parts of the text devoted to nonlinear problems, e.g., in Sect. 3.3.3. Thus, we need some information about weak convergence in Banach spaces, as well. The separation of variables is covered in the majority of textbooks on partial differential equations. Nonetheless, it is necessary to make the book complete. The separation of variables is a classical technique. At the same time, its generalization is the Galerkin method, which is a modern tool of analysis and computations.

Introduction

ix

The latter method justifies the numerical computations based on the finite element method. We present it as a recipe how to deal with different kinds of possible problems. We also give a number of worked-out examples, including also nonlinear equations. The last two chapters reflect to some extent the classification of second-order equations as elliptic, hyperbolic, and parabolic ones, which is complete in the case of two variables. Chapter 4 is devoted to a broad range of linear second-order elliptic problems, with a prominent place occupied by harmonic functions. The last chapter deals with evolution equations: we study hyperbolic and parabolic problems as well as the first-order equations. The list of included topics is long, but at the same time we omitted a number of areas, e.g., we do not approach in any deeper sense the calculus of variations and we do not introduce viscosity solutions. Our practice shows that these topics, although important, are not a part of a standard PDE course even at an intermediate level.

The Content of the Chapters Chapter 1 presents the necessary material from mathematical analysis. The central topic is integration by parts. The subsection on convolution prepares the reader to deal with various kinds of approximations, which are developed in this course. Chapter 2 introduces the fundamental objects of the modern theory of PDEs, which are distributions, the Fourier transform, and the Sobolev spaces, in which we look for weak solutions of various problems. The remaining part of this book is built on Chap. 2, which must be mastered before tackling the content of Chaps. 3–5. Methods of finding approximated solutions are developed in Chap. 3. Here, we present the theory of weak convergence and the Galerkin method in various settings. We also solve simple PDEs by writing explicit, to some extent, formulas for solutions. In Chap. 4, we study solvability of elliptic problems whose prominent examples are the Poisson and Laplace equations. We study boundary value problems. We present the classical approach based on the Green’s function which is useful today from the theoretical point of view. We also devote our attention to harmonic functions, also from the point of view of complex analysis. However, most important for the book is the development of weak solutions based on the Lax–Milgram lemma, which shows the power of functional analytic methods. The last Chap. 5 is heterogeneous, because it is devoted to evolutionary equations, whose solutions behave quite variously. We study here the problems, where the initial conditions as well as the boundary data may be imposed. The hyperbolic equations propagate signals with finite speed. Moreover, the solutions are not smoother than the initial data. To some extent, the first-order problems treated here may be regarded as hyperbolic problems. Finally, we deal with parabolic problems, which share many features with elliptic equations. The most important one is that both types of equations smooth out the data. In addition, we see that the speed of

x

Introduction

signal propagation is infinite in case of parabolic problems, which is in a striking contrast with hyperbolic equations. Finally, we should warn the reader that problems about particular equations are scattered throughout the book. For example, problems on hyperbolic or parabolic equations are to be found not only in Chap. 5 but also in Sect. 1.1 devoted to integration by parts, in Sect. 2.1 on Fourier transform, or in Sect. 3.2 dealing with the separation of variables method.

About the Authors Two teams have worked on this book. The Polish original version was written by: Maciej Borodzik, the University of Warsaw Tomasz Cie´slak, Mathematical Institute, PAS Piotr B. Mucha, the University of Warsaw Piotr Rybka, the University of Warsaw Witold Sadowski, the University of Warsaw Paweł Strzelecki, the University of Warsaw Agnieszka Tarasi´nska, the University of Warsaw Anna Zatorska-Goldstein, the University of Warsaw However, after a lapse of years since the inception of the project, the above team has changed significantly—some people changed the employer or got an important administrative position. The new team acquired a new member and now it consists of the following persons: Maciej Borodzik, the University of Warsaw Paweł Goldstein, the University of Warsaw Piotr Rybka, the University of Warsaw Anna Zatorska-Goldstein, the University of Warsaw. This new team reworked and revised the Polish original and added completely new sections. Finally, we would like to thank our families for support and those who read the preliminary version of the text and shared with us their remarks, which helped to improve the book. They are Karol Bołbotowski, Iwona Chlebicka, Łukasz Chomienia, Szymon Górka, Wojciech Górny, Michał Mi´skiewicz, Tomasz Piasecki, and Magdalena Rybka. Warsaw, Poland Warsaw, Poland Warsaw, Poland Warsaw, Poland The Fall/Winter of 2018

Maciej Borodzik Paweł Goldstein Piotr Rybka Anna Zatorska-Goldstein

Notation

Let us quickly review the notation and conventions used throughout the book. (A) Calculus • N is the set of natural numbers, including zero. • R+ is the set of positive real numbers. • For a sequence {an }∞ n=1 of real numbers, we denote by lim inf an (respectively, lim sup an ) the limit inferior (respectively, the limit superior), of the sequence an . That is, lim inf an = lim n→∞

n→∞



 inf am

m>n

(respectively, lim sup an = lim n→∞

n→∞



 sup am ). m>n

• AB = (A \ B) ∪ (B \ A) is the symmetric difference of sets A and B. (B) Multiindices • α ∈ Nn is called an n-dimensional multiindex. • |α| = α1 + · · · + αn for α ∈ Nn . • For a smooth function f , we write Dα f =

∂ |α| f . . . . ∂xnαn

∂x1α1 ∂x2α2

• For (x1 , . . . , xn ) ∈ Rn , the monomial x α is defined by x α = x1α1 . . . xnαn . (C) Geometry and topology in Rn : • A region is an open connected subset of Rn . • B(x, r) is an open ball in Rn with center x and radius r. • n is the unit normal vector pointing outward at the boundary of a region. xi

xii

Notation

• Rn+ = {x ∈ Rn : xn > 0}. • The orientation of the boundary of a region  in Rn with smooth boundary is induced using the rule “outward first.” This means that if e1 , . . . , en−1 form an oriented basis of the space Tx ∂ (for x ∈ ∂), then n , e1 , . . . , en form an oriented basis of Tx  = Rn . This orientation is used in Stokes’ theorem. (D) Functions • Unless specified otherwise, a smooth function is a function that is infinitely differentiable. • C0r () is the vector space of C r -smooth functions vanishing in the complement of an open set ; r may be equal ∞. (E) Measure and integration • Ln denotes the n-dimensional Lebesgue measure in Rn . • σS is the induced measure on a submanifold S ⊂ Rn . Often, when S is clear from context, we simply write σ . • δa denotes the Dirac delta distribution at a. • The abbreviation a.e. stands for almost everywhere (with respect to Lebesgue measure, unless stated otherwise). (F) Linear spaces, inner products, norms • If V is a vector space, then the dual vector space is denoted by V ∗ . In particular, we write (W01,2 ())∗ for the dual space to the Sobolev space W01,2 (). However, following the customary practice, we denote the space of distributions by D and the space of tempered distributions by S . • |x| denotes the Euclidean norm of x in Rn . • f X is the norm of an element f of a normed space X. • x · y = nk=1 xk yk is the scalar product of x, y ∈ Rn . • f, g H denotes the inner product of elements f, g of a Hilbert space H , in particular in L2 () (see below). If no ambiguity arises, then the subscript H is suppressed. • (φ, v) is the duality pairing, where v ∈ V and φ ∈ V ∗ and V ∗ is the dual vector of space of V . However, there is one exception to the above rules on duality pairing and inner product notation: following customary practice, we denote the value of a distribution T on a test function φ (i.e., a particular case of duality pairing) by angle brackets: T , φ • 1/p

 f Lp ≡ f p =

|f |p 

,

Notation

xiii

whenever 1 ≤ p < ∞, f :  → Rn is a measurable function and  is a measurable subset of RN . • Lp (, Rn ) = {f :  → Rn : f is measurable and f p < ∞}. Actually, Lp () is the space of equivalence classes of measurable functions, i.e., if f p , gp < ∞ and LN ({x ∈  : f (x) = g(x)}) = 0, then we identify f and g. || · ||p is a norm on Lp (, Rn ) making it a Banach space; see [20, Ch. 3]. • ||f ||∞ = ess sup |f | ≡ inf{t > 0 : LN ({x ∈  : |f (x)| > t}) = 0}, whenever f :  → Rn is a measurable function and  is a measurable subset of RN . • L∞ (, Rn ) = {f :  → Rn : f is measurable and f ∞ < ∞}. L∞ () is the space of equivalence classes of measurable functions, which are equal a.e. Again || · ||∞ is a norm on L∞ (, Rn ), making it a Banach space with this norm; see [20, Ch. 3]. • Lp () = Lp (, R), p ∈ [1, ∞].

Contents

1

Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.1 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.1.1 Theoretical Background . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.1.2 Worked-Out Problems .. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.1.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2 Convolutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.1 Theoretical Background . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.2 Worked-Out Problems .. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.3 Bibliographical Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

1 2 2 7 11 15 15 16 22 26

2 Distributions, Sobolev Spaces and the Fourier Transform . . . . . . . . . . . . . . 2.1 The Fourier Transform .. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.1.1 Theoretical Background . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.1.2 Worked-Out Problems .. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.1.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.2 The Theory of Distributions . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.2.1 Theoretical Background . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.2.2 Worked-Out Problems .. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.2.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.3 Sobolev Spaces .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.3.1 Theoretical Background . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.3.2 Worked-Out Problems .. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.3.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 2.4 Bibliographical Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

27 27 28 32 39 42 42 47 50 52 52 64 71 76

3 Common Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.1 Weak Convergence .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.1.1 Theoretical Background . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.1.2 Worked-Out Problems .. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.1.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

79 80 80 93 95

xv

xvi

Contents

3.2 The Separation of Variables Method . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.2.1 Theoretical Background . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.2.2 Worked-Out Problems .. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.2.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.3 Galerkin Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.3.1 Theoretical Background . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.3.2 Worked-Out Problems .. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.3.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.4 Bibliographical Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

101 101 108 114 118 118 119 132 134

4 Elliptic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.1 Classical Theory of Harmonic Functions . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.1.1 Theoretical Background . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.1.2 Worked-Out Problems .. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.1.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.2 Weak Solutions .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.2.1 Theoretical Background . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.2.2 Worked-Out Problems .. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.2.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 4.3 Bibliographical Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

135 135 135 143 149 156 156 163 170 177

5 Evolution Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.1 First-Order Equations and the Method of Characteristics .. . . . . . . . . . . . 5.1.1 Theoretical Background . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.1.2 Worked-Out Problems .. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.1.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.2 Hyperbolic Problems.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.2.1 Theoretical Background . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.2.2 Worked-Out Problems .. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.2.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.3 Parabolic Equations .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.3.1 Theoretical Background . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.3.2 Worked-Out Problems .. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.3.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.4 Bibliographical Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

179 179 179 189 198 202 203 207 215 227 227 233 238 244

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 247

Chapter 1

Preliminaries

This may be surprising, but the theory of weak solutions to PDEs is based on the integration by parts formula, whose simplest form is below, 

b a

du ϕ dx = − dx



b

u a

dϕ dx + u(b)ϕ(b) − u(a)ϕ(a). dx

This formula helps us to move the derivative from u, a candidate for a weak solution, to ϕ, which is a smooth function. In this chapter we recall the classical multidimensional statements of the integration by parts formulas, including the Stokes formula, which unifies all partial results. For this reason we pay attention to the exposition of the differential forms. This material is the content of Sect. 1.1. Section 1.2 is devoted to a topic in analysis, which is indispensable for the PDEs, when we use methods based on regularization. Namely, we present the convolution and its properties. The most important one is: the convolution of any locally integrable function and even of a distribution, with a smooth function of compact support results in a smooth function.

© Springer Nature Switzerland AG 2019 M. Borodzik et al., Problems on Partial Differential Equations, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-14734-1_1

1

2

1 Preliminaries

1.1 Integration by Parts 1.1.1 Theoretical Background Basic Formulas In the one-dimensional case the following fundamental result is valid for any continuously differentiable function f : [a, b] → R: 

b

f (x) dx = f (b) − f (a).

(1.1)

a

The formula reads that the increment of a smooth function on the interval [a, b] is the integral of its derivative over [a, b]. Applying (1.1) to a product of functions we obtain the formula for integration by parts:  a

b

f g dx +



b

fg dx = f (b)g(b) − f (a)g(a).

(1.2)

a

In order to state (1.2) in higher dimensions we need to generalize the notion of the full derivative (that is, generalize the f in (1.1)) as well as the notion of the increment of a function (that is, the meaning of f (b) − f (a)). The first way to define higher dimensional integration by parts is to start with an open subset  ⊂ Rn and a function F :  → Rn , which we might interpret as a vector field on  whose coordinates are (F1 , . . . , Fn ), where Fk is the k-th coordinate of the function F . Definition 1.1.1 The divergence of the vector field F is given by div F =

n  ∂Fk k=1

∂xk

.

(1.3)

Intuitively, the divergence tells us whether the flow of the vector field F is expanding (volume increasing) or contracting (volume decreasing). This follows from the formula LF dvol = div F dvol, where LF is the Lie derivative and dvol is the volume form. The Lie derivatives are not covered in the present book, we refer an interested reader, e.g., to [3]. The first way to generalize (1.1) is the following. Suppose  is a bounded region in Rn , whose boundary is of class C 1 . For x ∈ ∂ define n(x) to be the unit normal vector to ∂, pointing outwards. For example, if  = {x : g(x) > 0} for a ∇g(x) . continuously differentiable function g and ∇g(x) = 0, then n (x) = − ||∇g(x)||

1.1 Integration by Parts

3

Theorem 1.1.2 (Gauss’ Law) Under the above assumptions on , if F :  → Rn is of class C 1 , then 

 div F dLn = 

F · n dσ∂ , ∂

where Ln denotes the n-dimensional Lebesgue measure and σ∂ is the hypersurface measure. Remark 1.1.3 For a smooth hypersurface S (in Theorem 1.1.2 S = ∂), the integral F · n  dσS can be interpreted as the flux (flow rate) of the vector field passing S through S.

Surface Integrals In Theorem 1.1.2 we saw the integral dσ (∂). Let us spend a while recalling the construction and showing how to compute integrals over submanifolds of Rn . We will restrict ourselves to submanifolds that are parameterized. Let M ⊂ Rn be a smooth k-dimensional submanifold, U ⊂ Rk an open subset, and G : U → M a parameterization, that is, a smooth bijection whose differential has rank k at each point of U . Let g : M → R be a measurable function, that is, we require that the composition g ◦ G : U → R is measurable. We define 



g(G(x)) det DG(x)T · DG(x) dx,

g dσM = M

(1.4)

U

where DG is the differential, ·T means the matrix transposition. Note that DGT ·DG is the Gram matrix of DG. As a special case, if k = n − 1 and M is a graph of a function f : M = {(x1 , . . . , xn−1 , f (x1 , . . . , xn−1 )) : (x1 , . . . , xn−1 ) ∈ U }, then 

 M

g dσM =

U

 g(x1 , . . . , xn−1 , f (x1 , . . . , xn−1 )) 1 + fx21 + · · · + fx2n−1 dx1 . . . dxn ,

∂f where fxk is the shorthand for ∂x . k Another special case is when k = {(x1 (t), . . . , xn (t)) ∈ Rn : t ∈ (a, b)} and



 g dσM = a

M

Here x˙i denotes

dxi dt .

b

1. Then, we can write M

 g(x1 (t), . . . , xn (t)) x˙12 + · · · + x˙n2 dt.

=

(1.5)

4

1 Preliminaries

Remark 1.1.4 In the literature there are many types of notation for the integrals over submanifolds, like dvol(M), dM, dσ , dlk (M), dHk , and many others. In this book we use dσM . If the choice of the manifold is clear from the context, we often write simply dσ .

A Quick Review of Differential Forms For the reader’s convenience we quickly recall the calculus of differential forms. Formally speaking, a differential form is a section of an exterior power of the cotangent bundle. This way of thinking can be quite confusing at the beginning, hence we give a very down-to-earth definition. Definition 1.1.5 Let M be a manifold. A differential form of degree k, or shortly, a k-form is an object that in local coordinates x1 , . . . , xn of M is given as ω=



fi1 i2 ...ik (x)dxi1 ∧ · · · ∧ dxik ,

1≤i1 0 we write 

∞ −∞

 g(x − y)fn (y) dy =

−ε

g(x − y)fn (y) dy

−∞  ∞

+

 g(x − y)fn (y) dy +

ε

ε −ε

g(x − y)fn (y) dy.

In the first two integrals we estimate |g(x − y)| by M. We obtain   ∞  ε  ∞  −ε     g(x − y)f (y)dy − g(x − y)f (y)dy f (y)dy + M fn (y)dy. ≤ M n n n   −∞

−ε

−∞

ε

1.2 Convolutions

19

For any ε > 0 we choose n sufficiently large so that ∞ δ ε fn (y) < 4M . Then    

−ε

−∞ fn (y)


0 such that |g(x − y) − g(x)| < 4δ if |y| < ε. For n adjusted to ε as above we obtain    

ε −ε

   δ ε δ g(x)fn (y)dy  ≤ fn (y)dy < 4 4 −ε −ε

 g(x − y)fn (y)dy −

ε

and    

  3 g(x − y)fn dy − g(x) fn (y)dy  ≤ δ. 4 −∞ −ε 

∞

∞

Furthermore, since

−∞ fn (y)dy

 1≥

ε

= 1, we have ε

−ε

fn (y)dy > 1 −

δ . 2M

Suppose x is such that g(x) > 0 (the cases g(x) = 0 and g(x) < 0 are analogous and left to the reader). Then    ∞ δ 3δ 3δ g(x) 1 − − < g(x − y)fn (y)dy ≤ g(x) + . 2M 4 4 −∞ As g(x) < M we have g(x)(1 − g(x) − But this means that 

∞

−∞

5δ < 4



δ 2M )

∞ −∞

> g(x) − 2δ , hence

g(x − y)fn (y)dy ≤ g(x) +

3δ . 4

g(x − y)fn (y)dy → g(x) uniformly in x.

♦ We are still using the definition of fn from (1.11). We would like to prove that, at least if g is compactly supported, fn ∗ g converges to g in the L1 -norm. To this end we need a preparatory result.

20

1 Preliminaries

Problem 1.2.7 Let g be a bounded measurable function on R with compact support S. Suppose M is such that √ |g(x)| < M for almost all x ∈ R. Prove that, if dist(x, S) > R > 0 and R n > 1, we have: √ 2M |fn ∗ g(x)| ≤ √ e−R n . π

Solution Let x satisfy dist(x, S) > R. We have   |fn ∗ g(x)| = 

∞

−∞

   fn (y)g(x − y)dy  ≤

∞ −∞

fn (y)|g(x − y)|dy.

We now estimate g(x − y) by M if |y| > R. Note that by assumption g(x − y) = 0 for |y| ≤ R. 

∞ −∞

 fn (y)|g(x − y)|dy ≤ M

−R −∞

 +

∞

 fn (y)dy = 2M

R

∞

fn (y)dy. R

Plug in the definition of fn to get 

∞

2M R

2M fn (y) = √ π



∞

e−ny

R

2

√ 2M ndy = √ π



∞ √

e−u du. 2

R n

If u √ > 1, we have e−u < e−u , hence we can bound the last integral from above by −R n , completing the proof. ♦ e 2

Problem 1.2.8 Let fn be as in (1.11) and g ∈ L1 (R). Prove that the sequence fn ∗g converges to g in the L1 norm. Solution Suppose first that g is a continuous function supported in [−R, R]. By Problem 1.2.6 we know that the sequence fn ∗ g converges to g uniformly. The problem is that the uniform convergence alone does not imply the L1 convergence if the domain is unbounded. We act as follows. Fix ε > 0. Take n large enough so ε that |fn ∗ g(x) − g(x)| < 8R for all x ∈ R. Increase n if necessary to ensure that √ √M e −2R n < ε , where M = sup |g(x)|. By Problem 1.2.7 we have for x > 2R: 8 πn √ 2M |fn ∗ g(x)| ≤ √ e−x n . π

Therefore 

∞ 2R

2M |fn ∗ g(x)|dx ≤ √ π



∞ 2R

e−x

√ n

√ 2M ε dx = √ e−2R n ≤ . 4 πn

1.2 Convolutions

21

Analogously, we show that 

−2R

ε . 4

|fn ∗ g(x)|dx ≤

−∞

We write 

∞ −∞

 |fn ∗ g(x) − g(x)| dx =  +

2R −2R

−2R −∞

|fn ∗ g(x)|dx 

∞

|fn ∗ g(x) − g(x)|dx +

|fn ∗ g(x)|dx

2R

ε ≤ + 4



2R −2R

ε ε ε ε ≤ + + = ε. 8R 4 2 4

Now assume that g is an arbitrary function in L1 (R). Fix ε > 0 and find a function h which is continuous, compactly supported and such that ||g − h||L1 < 4ε , see Problem 1.2.14. Choose n large enough so that ||fn ∗ h − h||L1 < ε2 . Then ||fn ∗ g − g||L1 ≤ ||fn ∗ g − fn ∗ h||L1 + ||fn ∗ h − h||L1 + ||g − h||L1 .

(1.12)

We have that ||fn ∗ g − fn ∗ h||L1 = ||fn ∗ (g − h)||L1 ≤ ||fn ||L1 ||g − h||L1 < 4ε (see Problem 1.2.1). Therefore, by plugging previous inequalities into (1.12) we obtain ||fn ∗ g − g||L1 ≤ 4ε + 2ε + 4ε = ε. ♦ Problem 1.2.9 Compute the convolution of f (x) = sin x and g(x) =

1 . 1+x 2

Solution By the definition we need to compute the integral 

∞ −∞

sin y dy. 1 + (x − y)2

We use the residue calculus. We write 

∞ −∞

sin y dy = Im 1 + (x − y)2



∞ −∞

eiy dy. 1 + (x − y)2

To compute the last integral, fix R > 0 and consider γR to be the contour Reit for t ∈ [0, π] (upper half circle); see Fig. 1.1. Suppose R > |y + i|. Then the integrand is holomorphic in the interior of the region bounded by [−R, R] ∪ γR , except at y = x + i, where it has a pole of order 1. By the residue theorem we have that 

R −R

eiy dy + 1 + (x − y)2

 γR

eiy eiy dy = 2πi resy=x+i . 2 1 + (x − y) 1 + (x − y)2

22

1 Preliminaries

γR

−R

R

Fig. 1.1 Contour of the integral in the solution to Problem 1.2.9

iy

e The residue of 1+(x−y) 2 is promptly computed to be ∞, the integral

 γR

−i ix−1 . 2 e

Moreover, as R →

eiy dy 1 + (x − y)2

   eiy  1 −2 for y ∈ γ and R sufficiently tends to 0: we have  1+(x−y) R 2  ≤ 1+(x−y)2 < 2R large (with x fixed). Therefore we obtain 

∞ −∞

eiy dy = πe−1+ix . 1 + (x − y)2

Hence f ∗ g = πe−1 sin y. ♦

1.2.3 Problems Problem 1.2.10 Let f = χ[0,∞) be the function equal to 1 on [0, ∞) and 0 outside. Calculate f ∗ f . Note that f ∈ / L1 (R), but the integral is well defined. Problem 1.2.11 Calculate the convolution of f (x) = e−ax and g(x) = e−bx for real parameters a, b > 0. 2

Problem 1.2.12 Prove that f ∗ g = g ∗ f .

2

1.2 Convolutions

23

Problem 1.2.13 Prove that if f, g, h ∈ L1 (R), then f ∗ (g ∗ h) = (f ∗ g) ∗ h. Problem 1.2.14 Let g ∈ L1 (R). Prove that for any ε > 0 there exists a smooth, compactly supported function h such that ||g − h||L1 (R) < ε. Problem 1.2.15 Generalize the solution of Problem 1.2.2 to show that if f ∈ p q Lp (R) and g ∈ Lq (R) and p1 + q1 = 1, then |f ∗ g(x)| ≤ p1 ||f ||Lp + q1 ||g||Lq . Problem 1.2.16 Use Hölder’s inequality to prove Young’s inequality: for p, q, r ≥ 1 satisfying p1 + q1 = 1 + 1r we have ||f ∗ g||Lr ≤ ||f ||Lp ||g||Lq . Problem 1.2.17 Calculate the convolution of

sin x 1+x 2

and

1 . 1+x 2

Problem 1.2.18 Show that the convolution turns L1 (R) into a commutative algebra. Show that there is no identity in L1 , that is, there is no function g ∈ L1 (R) such that g ∗ f = f for any f ∈ L1 (R). Hint Show that if such g existed, it would have to be zero on any closed interval of R that does not contain zero. Problem 1.2.19 Prove that if f, g ∈ L1 (R), f is k times differentiable and all the derivatives up to the k-th one are bounded on R, then f ∗ g is k times differentiable and for all j = 1, . . . , k: (f ∗ g)(j ) = f (j ) ∗ g. Problem 1.2.20 Generalize the solution to Problem 1.2.6 to show that if fn is given by (1.11) and g is a compactly supported C 1 -function, then fn ∗ g converges to g in the C 1 -norm. Argue that if g is C n -smooth, then the convergence is in C n -norm. Problem 1.2.21 Let fn be given by (1.11) and let g ∈ Lp (R). Prove that fn ∗ g converges to g in Lp -norm. Hint See Problem 1.2.8. Problem 1.2.22 Prove the statement of Problem 1.2.8 in a different way. Namely, show that the statement holds for a characteristic function of an interval, then conclude that it holds for a finite linear combination of such functions and finally pass to a suitable limit. Problem 1.2.23 Suppose hn (x) =

3/2 2 2n √ xe −nx . π

• Prove that if g(x) is a C 1 -smooth function with compact support, then the sequence hn ∗ g(x) converges to g uniformly on R. • Prove that if g ∈ W 1,p (R), then hn ∗ g converges to the weak derivative of g in the Lp -norm (see Sect. 2.3 for introduction to Sobolev spaces)

24

1 Preliminaries

1 Problem 1.2.24 Let f (x) = π(1+x 2 ) . Define fn (x) = nf (nx). Show that fn is an approximate identity in the sense that for any continuous function g with compact support the sequence fn ∗ g converges uniformly to g. Do we have convergence in L1 norm as in Problem 1.2.8?

Problem 1.2.25 ∞ This is a generalization of the previous problem. Suppose h ∈ L1 (R) and −∞ h(x)dx = 1. Set hn (x) = nh(nx). Prove that for any function g that is continuous with compact support we have lim hn ∗ g(x) = g(x)

n→∞

and the convergence is uniform. Hint Repeat the argument used in the solution to Problem 1.2.6. Problem 1.2.26 Find sufficient conditions for a continuous and compactly supported function f : R → R, which guarantee that the sequence fn ∗ g, where fn (x) = n3 f (nx), converges uniformly to g

for any smooth function g with compact support. Problem 1.2.27 Generalize the results on approximate identities to convolutions 2 m m R by√ f (x) = σm e−|x| , where σm is such that in R . Namely, define f : R →m/2 f ( nx). Let g : Rm → R be a measurable R m f (x)dx = 1. Set fn (x) = n function. Prove that • • • •

if g is a bounded continuous, then fn ∗ g converges to g uniformly on Rm ; if g ∈ L1 , then fn ∗ g converges to g in L1 -norm; if g ∈ Lp , then fn ∗ g converges to g in Lp -norm; if g is C k -smooth, bounded and all the derivatives of order less or equal to k are bounded continuous, then fn ∗ g converges to g in the C k -norm.

Problem 1.2.28 Let A ⊂ R be a Lebesgue measurable set with finite measure. Let g be the characteristic function of A, that is, g(x) = 1 if x ∈ A and g(x) = 0 otherwise. Let fn be a sequence of functions as in Problem 1.2.27. Prove that for any x ∈ R we have lim fn ∗ g(x) = dA (x),

n→∞

where dA (x) is the density of A at the point x, that is dA (x) = lim

h→0+

1 (A ∩ [x − y, x + y]) 2y

and 1 denotes the one-dimensional Lebesgue measure.

1.2 Convolutions

25

Problem 1.2.29 Suppose f : R → R is a continuous and integrable function such that f (−x) = f (x), but f ∈ / L2 . Show that f ∗ f (x) is not bounded near x = 0. Problem 1.2.30 Show that if f ∈ L1 (R) and f is uniformly continuous, then |f | is a bounded function on R. Problem 1.2.31 This is a continuation of Problem 1.2.30. Prove that if f ∈ L1 (R) is such that |f (x)| < M for a constant M, then f ∈ Lp (R) for all p > 1, and moreover ||f ||Lp ≤ M p−1 ||f ||L1 . This problem shows, in particular, that a function f satisfying the assumptions of Problem 1.2.29 can be neither uniformly continuous on R nor bounded. Problem 1.2.32 Prove that the equality (*) in Problem 1.2.3 holds. Explain, why it does not have to be true if the assumptions in Problem 1.2.3 are not satisfied. Hint Use Problem 1.2.30 for the first part. Problem 1.2.33 In this problem we construct an example of a function satisfying the assumptions in Problem 1.2.29. Denote by y → Bx,δ,M (y) a continuous function given by ⎧ ⎪ 0, ⎪ ⎪ ⎪ ⎨M(y − (x − δ)), Bx,δ,M (y) = ⎪ −M(y − (x + δ)), ⎪ ⎪ ⎪ ⎩ 0

y ≤ x − δ, y ∈ [x − δ, x], y ∈ [x, x + δ], y ≥ x + δ.

For a sequence an , n ∈ Z, define the function F (y) =

∞  n=−∞

Bn,

1 ,an |n|2 +2

(y).

Show that F is continuous. Prove that  • F ∈ L1 (R) if and only if  |an |/(|n|2 + 2) < ∞; • F ∈ L2 (R) if and only if |an |2 /(|n|2 + 2) < ∞. Find a sample sequence an such that the resulting function F satisfies F ∈ L1 (R), F ∈ / L2 (R). Problem 1.2.34 In analogy to Problem 1.2.33 define a function y → Cx,δ,M (y) to be a continuous piecewise linear function equal to 0 outside of (x − δ, x + δ), linear over the three intervals [x − δ, x − δ/3], [x − δ/3, x + δ/3] and [x + δ/3, x + δ] and attaining value ∓M at x ± δ/3.

26

1 Preliminaries

Find asequence an , n = 1, 2, . . . of positive integers such that the functions F (y) = Bn,1/n2 ,an (y), G(y) = Cn,1/n2 ,an (y) satisfy • F, G ∈ L1 ; • limx→0− F ∗ G(x) > 0; • limx→0+ F ∗ G(x) < 0. Note that F ∗ G(0) = 0 if F, G ∈ L1 .

1.3 Bibliographical Remarks Although integration over surfaces is covered in many textbooks, the general formula (1.4) using determinants of Gram matrix is harder to find. We refer to Section 8.3 of the book of Makarov and Podkorytov [18], or to the book of Evans and Gariepy [9]. A standard resource for various versions of Green’s formula is Appendix C of the Evans’ book on partial differential equations, [8]. As for differential forms, they are discussed in many places, starting from Section 10 of Rudin’s “Principles of mathematical analysis” [19]. One of the excellent resources, although often underestimated, is Arnold’s book on classical mechanics [3], which gives a very intuitive introduction to differential forms. The Hodge star operator as we discussed it is slightly less standard technique. A very detailed description, including the shape of the operator for general Riemannian metric (as well as for manifolds with indefinite metrics, which makes it useful for Poincaré metric for instance) is given in the first volume of “Modern geometry” of Dubrovin et al. [7]. A slightly less detailed, but maybe more readable exposition is given in Chapter 2 of the textbook of Jost [13]. Theory of convolutions is discussed in detail in many textbooks on calculus including [20] and [21] (the first of the two gives a more elementary treatment). There is also an appendix in Evans’ book that presents the rudiments of the theory. The residue calculus used in the solution to Problem 1.2.9 is also commonly used in many places, including [20]. An excellent resource for learning residue calculus is the problem book of Volkovyski˘ı et al. [29].

Chapter 2

Distributions, Sobolev Spaces and the Fourier Transform

In this chapter we introduce the most important tools and notions in modern PDEs theory: we explain the concepts of a distribution and of a weak derivative. For this purpose we present the basics of the theory of distributions, which requires a bit of topology. The great virtue of the distribution theory is that it permits to take derivatives of arbitrary order of locally integrable functions. We are particularly interested in functions whose distributional derivatives are locally integrable. In this way, we come to the theory of Sobolev spaces of functions having weak derivatives integrable with the p-th power. The importance of these spaces stems from the fact that it is relatively easy to find weak solutions there. By definition, weak solutions have fewer (weak) derivatives than the equation stipulates and they satisfy integral identities in place of differential equations. The theory of Sobolev spaces is an important topic, which we treat very carefully. It turns out that the theory of weak solutions is intimately related to the Fourier transform. Since this tool is of independent interest, we begin this chapter with its presentation. The great virtue of the Fourier transform is that it transforms PDEs into ordinary differential equations (ODEs for short) with a parameter.

2.1 The Fourier Transform The Fourier transform is a particularly useful tool in the study of partial differential equations. It allows us to solve partial differential equations by solving certain algebraic or ordinary differential equations.

© Springer Nature Switzerland AG 2019 M. Borodzik et al., Problems on Partial Differential Equations, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-14734-1_2

27

28

2 Distributions, Sobolev Spaces and the Fourier Transform

In this book, the Fourier transform is used in problems in distribution theory (Sect. 2.2) and in Sobolev spaces (Sect. 2.3). Problems illustrating applications to the wave equation and to the estimates of decay of a solution in the presence of a damping term in the equation can be found in Sect. 5.

2.1.1 Theoretical Background The Schwartz Space Definition 2.1.1 Let f : Rn → C be a smooth, complex-valued function on Rn . We say that f is rapidly decreasing (or, often, rapidly decreasing at infinity) on Rn , if for any multiindices α, β, we have   sup x α D β f (x) < ∞.

x∈Rn

The space of rapidly decreasing functions on Rn is called the Schwartz space and is denoted S(Rn ), Sn or simply S. Note that for all m = 0, 1, 2, . . . and x ∈ Rn there holds  C −1 (1 + |x|2 )m ≤ |x α |2 ≤ C(1 + |x|2 )m ,

(2.1)

|α|≤m

with the constant C > 0 dependent only on n and m. Thus we can give an equivalent definition of the Schwartz space: the function f is in the Schwartz space S if and only if for every polynomial P and any multiindex β the function P · D β f is bounded on Rn . In particular, this means that if f ∈ S(Rn ), then all the derivatives D α f (x) tend to zero with |x| → ∞, which explains the name “rapidly decreasing functions.” Obviously, every smooth, compactly supported function on Rn lies in S (i.e., C0∞ (Rn ) ⊂ S(Rn )), but the space S contains functions which are not compactly supported, and they may be positive everywhere (see Problem 2.1.6). The functions in S are integrable with an arbitrary exponent, in particular S ⊂ L2 (Rn ). The topology of the space S is defined by a family of semi-norms pk (f ) =



sup (1 + |x|2)k/2 |D α f (x)| k = 0, 1, 2, . . . ,

n |α|≤k x∈R

i.e., a sequence of functions fn ∈ S converges in S to a function f , if for all k = 0, 1, 2, . . . we have limn→∞ pk (fn − f ) = 0.

2.1 The Fourier Transform

29

The Schwartz space with the distance function given by d(f, g) =

∞ 

2−k

k=0

pk (f − g) 1 + pk (f − g)

(2.2)

is a complete metric space; a complete vector space with topology defined in this way—through a sequence of semi-norms—is often called a Fréchet space. Since compactly supported smooth functions are dense in L2 (Rn ), so is the Schwartz space S.

The Fourier Transform Definition 2.1.2 Assume f ∈ S. The Fourier transform of the function f is the function F (f ) : Rn → C given by the formula F (f )(ξ ) = fˆ(ξ ) =

 Rn

e−2πix·ξ f (x) dx

for ξ ∈ Rn .

(2.3)

Remark 2.1.3 In textbooks one may often find different definitions of the Fourier transform. The most popular are probably F1 (f )(ξ ) = (2π)

−n/2

 Rn

e−ix·ξ f (x) dx

for ξ ∈ Rn

and  F2 (f )(ξ ) =

Rn

e−ix·ξ f (x) dx

for ξ ∈ Rn .

The choice between our definition and the other two is a matter of personal taste, type of applications, and custom; the properties of any of them easily translate to the two other definitions. One can also find in the literature the term Fourier transformation, used to denote the mapping of function spaces that attributes to the function f its Fourier transform fˆ. Note that the definition of the Fourier transform makes sense if we assume only that f is integrable, i.e., f ∈ L1 (Rn ), and thus we can define F (f ) for any f ∈ L1 (Rn ).

30

2 Distributions, Sobolev Spaces and the Fourier Transform

Properties of the Fourier Transform Assume x, y ∈ Rn . We shall use the following notation: ex (y) := e2πix·y = exp{2πi(x1y1 + · · · + xn yn )}, τx f (y) := f (y − x). One can easily check, using Fubini’s theorem, that the Fourier transform satisfies the following algebraic identities. Theorem 2.1.4 Assume f, g ∈ L1 (Rn ) and x ∈ Rn . (a) (b) (c) (d)

−2πix·ξ fˆ(ξ );  (τ x f )(ξ ) = e−x (ξ )fˆ(ξ ) = e  ˆ (ex f )(ξ ) = τx f (ξ );  (f ∗ g)(ξ ) = fˆ(ξ )g(ξ ˆ ). ˆ ) = λn fˆ(λξ ). If λ > 0 and h(x) = f (x/λ), then h(ξ

Remark 2.1.5 Additionally, if f · g ∈ L1 (Rn ), then f · g = fˆ ∗ g. ˆ The following theorem describes the properties of the Fourier transform on the Schwartz space S. Theorem 2.1.6 Assume f, g ∈ S. Then (a) each of the mappings f → Pf,

f → gf,

f → D α f,

where P is a polynomial and α is a multiindex, is a continuous linear mapping from S to S; (b) fˆ ∈ C ∞ (Rn ) and D α fˆ(ξ ) = (−2πi)|α| F (x α f )(ξ ), F (D α f )(ξ ) = (2πi)|α| ξ α fˆ(ξ )

for ξ ∈ Rn ;

(c) it holds 

 Rn

f (x)g(x) ˆ dx =

Rn

fˆ(x)g(x) dx;

(d) the Fourier transform F (f ) = fˆ is a continuous linear mapping from S to S.

2.1 The Fourier Transform

31

Assume f ∈ S and set g(x) = e−|x| . 2

The Fourier transform of g, calculated in Problem 2.1.1, equals g(ξ ˆ )=

 π n/2 

e−π

2 |ξ |2 /

.

Using the identity (c) of Theorem 2.1.6 we see that  π n/2  

Rn

f (x)e−π

2 |x|2 /

 dx =

Rn

2 fˆ(x)e−|x| dx.

 Next, we change the integration variable in the left-hand side, setting x = 2 πy and we take  → 0. By Lebesgue’s Dominated Convergence Theorem we may take on both sides the limit under the integral sign, obtaining  f (0) =

Rn

fˆ(ξ ) dξ.

Finally, identity (a) in Theorem 2.1.4 yields  f (x) =

Rn

e2πix·ξ fˆ(ξ ) dξ,

for f ∈ S.

Definition 2.1.7 Assume f ∈ S. The inverse Fourier transform of the function f is the function given by the formula fˇ(x) = F −1 (f )(x) =

 Rn

e2πix·ξ f (ξ ) dξ

for x ∈ Rn .

(2.4)

Parseval’s Identity and Plancherel’s Theorem Note that ˆ fˆ(x) = f (−x), thus, up to a symmetry with respect to the origin, the Fourier transform on S is equal to its inverse. In particular, it is a bijection of S onto S. Note also that it commutes with complex conjugation f¯ˆ = f.ˆ¯

32

2 Distributions, Sobolev Spaces and the Fourier Transform

Using the identity (c) of Theorem 2.1.6 we obtain the so-called Parseval’s identity: u, v L2 (Rn ) = u, ˆ v ˆ L2 (Rn ) ,

for u, v ∈ S.

(2.5)

This identity shows that the Fourier transform is not only a bijection, but also an isometry of S endowed with the scalar product (and metric) of L2 . As the space S is dense in L2 (Rn ), the Fourier transform can be uniquely extended to an isometry of the whole space L2 (Rn ). Theorem 2.1.8 (Plancherel’s Theorem, see, e.g., Chapter 9 in [20]) There exists a linear isometry T of L2 (Rn ) onto itself, uniquely defined by the condition Tf = fˆ for every f ∈ S. Also, Parseval’s identity (2.5) holds for all u and v in L2 (Rn ).

2.1.2 Worked-Out Problems Problem 2.1.1 Calculate the Fourier transform of the function u : Rn → C, u(x) = e−|x|

2

for any  > 0. Solution Consider first the case of  =  φ(ξ ) := u(ξ ˆ )=

∞ −∞

1 2

e−x

and n = 1. We have then

2 /2

e−2πixξ dx

for ξ ∈ R.

Differentiating φ we get dφ (ξ ) = −2πi dξ  = 2πi



∞

xe−x

−∞ ∞ ∂

−∞

= −(2π)2 ξ



∂x  ∞ −∞

= −(2π)2 ξ φ(ξ ).

2 /2

e−x

e−2πixξ dx

2 /2

e−x



2 /2

e−2πixξ dx

e−2πixξ dx (2.6)

2.1 The Fourier Transform

33

√ Moreover, we know that φ(0) = 2π. The only solution to the differential equation (2.6) with this particular value at 0 is the function φ(ξ ) =

√ 2 2 2πe−2π ξ .

This way we calculated the Fourier transform of the density function of the normal distribution, in the one-dimensional case, with  = 12 . For n ≥ 2 we simply use Fubini’s theorem (note that exp(−|x|2/2) = exp(−x12/2) exp(−x22 /2) · · · exp(−xn2 /2)), obtaining F (e−|x|

2 /2

)(ξ ) = φ(ξ1 ) · · · φ(ξn ) = (2π)n/2 e−2π

2 |ξ |2

.

To deal with an arbitrary  > 0 we calculate: F (e−|x| )(ξ ) = 2



e−|x| e−2πix·ξ dx 2

Rn

= (2)

−n/2



e−|y|

2 /2

Rn

= (π/)n/2 e−π

2 |ξ |2 /

e−2πiy·ξ/

√ 2

dy

.

We proved that the Fourier transform of the density function of the normal distribution is again a density of the normal distribution, albeit rescaled (i.e., with a different variance). ♦ Problem 2.1.2 Calculate the Fourier transform of the function u(x) =

sin x x .

Solution We present a solution that omits a few points, left as exercises for the reader (see Problems 2.1.16 and 2.1.17 below). The problem boils down to calculating the integral   u(ξ ) =

∞

−∞

sin x −2πixξ e dx. x

(2.7)

1 }. We sketch an argument that the integral above is convergent for ξ ∈ R \ {± 2π Filling the details is left as an exercise (Problem 2.1.16 below). To prove the convergence of the above integral, write

e−2πixξ sin x = cos(2πxξ ) sin x − i sin(2πxξ ) sin x. Then cos(2πixξ ) sin x = sin(2πixξ + π2 ) sin x and sin(2πxξ ) sin x can be presented as linear combination of sine and cosine functions, i.e., of periodic functions. We apply the Leibniz criterion.

34

2 Distributions, Sobolev Spaces and the Fourier Transform

To calculate the integral in (2.7) we write sin x = u1 (z) =

ei(1−2πξ )z 2iz

u2 (z) =

ei(−1−2πξ )z . 2iz

eix −e−ix 2i

and set

∞ ∞ Then  u(ξ ) = −∞ (u1 (z) − u2 (z)) dz. Note that we cannot write −∞ (u1 (z) ∞ ∞ −u2 (z)) dz = −∞ u1 (z) dz − −∞ u2 (z) dz, because u1 and u2 are not integrable at 0. We have   u(ξ ) = lim

−ε

ε→0 −R R→∞



R

(u1 (z) − u2 (z)) dz +

(u1 (z) − u2 (z)) dz.

ε

 R −ε Fix ε > 0 and R  0. Our goal is to calculate −R + ε u1 (z) dz and  R −ε u2 (z) dz up to terms vanishing as ε → 0 and R → ∞. We use complex + −R ε integration. 1 Suppose first ξ < − 2π . We integrate the functions uj (z), j = 1, 2 over the left contour in Fig. 2.1. The functions uj (z) are holomorphic inside the region bounded by the contour, so by the Cauchy’s Integral Formula:  γ1 ∪γR ∪γ2 ∪γε

uj (z) dz = 0.

The integral over γ1 ∪ γ2 is precisely the integral we want to calculate. Therefore we need to estimate the integral over γR and γε . Claim We have limR→∞ γR uj (z) dz = 0.

γR γε −R

γ2 −ε ε

−R γ1

γ2 −ε ε

R

γ1

R

γε γR

Fig. 2.1 The contours γε and γε in the solution of Problem 2.1.2

2.1 The Fourier Transform

35

To prove the √claim we present γR as a union of two parts. One part is γR,0 = γR ∩ R × [0, R] and the other part is γR,1 = γR ∩ R × [R, ∞]. It is easy to see that on√γR,0 the function uj (z) is bounded by 1/R and the length of γR,0 is √ less than π R, therefore the integral is bounded from above√by π/ R. The length of γR,1 is bounded by πR, but on γR,1 we have Im z > R, hence ||uj (z)|| ≤ √ −(−1−2πξ ) R e /R, and since (−1 − 2πξ ) > 0, the exponent is always negative. Therefore γR,1 uj (z) converges to zero as R → ∞. This concludes the proof of the claim. We pass to calculating the integral over γε . To obtain a uniform notation choose θ ∈ R and calculate  γε

thus

γε

eiθz dz = z

u1 (z) dz −



γε

γε

1 + iθ z − θ 2 z2 /2 + . . . dz = −π + O(ε), z

(2.8)

u2 (z) dz = O(ε). Consequently,  u(ξ ) = 0 for ξ < −

1 . 2π

An analogous argument, but using the integral over the right contour in Fig. 2.1, reveals that  u(ξ ) = 0 for ξ >

1 . 2π

1 1 If ξ ∈ (− 2π , 2π ), we have to calculate the integral of u1 using the left contour and the integral of u2 using the right contour, because γR u1 dz → 0 and γ u2 dz → R 0. Using (2.8) we obtain

 γ1 ∪γ2

u1 (z) dz = π.

An analogue of (2.8) for the contour γε depicted in Fig. 2.1 (right) is  γε

u2 (z) dz = π + O(ε),

because the contour γε has opposite orientation to that of γε . Therefore, γ1 ∪γ2 u2 (z) dz = −π, hence finally  u(ξ ) =

 −iπ 0

1 1 , 2π ), for ξ ∈ (− 2π 1 1 for ξ ∈ / [− 2π , 2π ].

1 Calculating the value of  u at ξ = ± 2π is left to the reader as Problem 2.1.17. ♦

36

2 Distributions, Sobolev Spaces and the Fourier Transform

The following problem might seem departed from the subject of this section, since it deals with Fourier series, not with the Fourier transform, but it is a preparatory step for constructing the example discussed in Problem 2.1.23.  sin nx Problem 2.1.3 Prove that the series ∞ n=2 ln n represents a function f (x) that is not Lebesgue integrable. Solution It can be seen (see Problem 2.1.21) that tthe series is convergent for all x ∈ R. Suppose f (x) is integrable and let F (t) = 0 f (x)dx. Then F (t) is periodic and absolutely continuous. Therefore F can be expanded into Fourier  series, and, since F (t) = F (−t), the Fourier series for F (t) is of the form ∞ n=0 an cos nt 2π for some sequence an . Integrating 0 F (t) cos nt by parts (which is allowed, since F (t) exists almost everywhere and F = f ), we obtain an = −(n ln n)−1 for n ≥ 2; see Problem 2.1.22. Now F , as an absolutely  continuous function, is a pointwise limit of its Fourier ∞ expansion, thus F (t)= ∞ n=0 an cos nt for all t. In particular, F (0) = n=0 an . ∞ However, the series n=0 an is divergent. ♦ Problem 2.1.4 Find a solution to the problem in R2+ = (−∞, ∞) × (0, ∞),

u = 0 u(x, 0) = g(x)

for x ∈ R,

u(x, y) → 0

(2.9)

for |(x, y)| → ∞.

Solution For the moment, fix y > 0 and apply to u(x, y) the Fourier transform with respect to the x variable.  u(ξ, ˆ y) =

∞ −∞

e−2πixξ u(x, y) dx.

Using the properties of F described in Theorem 2.1.6, we transform the problem (2.9) to −(2πξ )2 u(ξ, ˆ y) +

∂ 2 uˆ (ξ, y) = 0 for (ξ, y) ∈ R2+ , ∂y 2 u(ξ, ˆ y) = g(ξ ˆ ).

Let us now fix ξ ∈ R and denote f (y) = u(ξ, ˆ y). The function f satisfies the ordinary differential equation f

(y) = (2πξ )2 f (y)

2.1 The Fourier Transform

37

with an initial condition f (0) = g(ξ ˆ ). We can solve the equation, using (as a second boundary condition) the fact that u vanishes at infinity. f (y) = u(ξ, ˆ y) = g(ξ ˆ )e−2π|ξ |y

for y > 0.

To determine u, we apply to u(ξ, ˆ y) the inverse Fourier transform (with respect to ξ ):     ˆ )e−2π|ξ |y (x, y) = g ∗ F −1 e−2π|ξ |y (x, y). u(x, y) = F −1 g(ξ Let us keep in mind that the convolution above is taken only with respect to the first variable. We need yet to calculate the inverse Fourier transform of the function e−|ξ |y :    F −1 e−2π|ξ |y (x, y) = 

∞ −∞ ∞

=

e2πixξ e−2π|ξ |y dξ  e2π(ix−y)ξ dξ +

0

0

−∞

e2π(ix+y)ξ dξ

1 1 + 2π(y − ix) 2π(y + ix) y . = 2 π(x + y 2 ) =

Thus the solution of our problem is given by u(x, y) =

y π



∞ −∞

g(z) dz. (x − z)2 + y 2

This solution agrees with our earlier study of harmonic functions: we get the same formula using Green’s function for the half-plane. Note, however, that it is not completely obvious from the formula that u(x, y) is indeed a solution to our problem, in particular that it satisfies the boundary condition. This problem is studied in more detail as Green’s functions are introduced (Sect. 4.1). ♦

38

2 Distributions, Sobolev Spaces and the Fourier Transform

Problem 2.1.5 Find a solution to the following initial problem for the heat equation. ut − u = 0 in (0, ∞) × Rn , g ∈ L2 (Rn ).

u(0, x) = g(x),

Solution Let us fix, for the moment, t ∈ (0, ∞). Applying to u(t, x) the Fourier transform with respect to the spacial coordinate x and using Theorem 2.1.6 we obtain the transformed problem d uˆ ˆ ξ ) = 0, (t, ξ ) + (2πξ )2 u(t, dt u(0, ˆ ξ ) = g(ξ ˆ ). Denote f (t) = u(t, ˆ ξ)

for t ∈ [0, ∞).

The function f satisfies the ordinary differential equation f (t) = −(2πξ )2 f (t) with the initial condition f (0) = g(ξ ˆ ), thus u(t, ˆ ξ ) = f (t) = g(ξ ˆ )e−4π

2t ξ 2

.

To obtain the solution u, we apply to the above the inverse Fourier transform, again with respect to the x variable, treating t as fixed, and using Theorem 2.1.6 (c):   2 2 ˆ )e−4π t ξ = g ∗ F (t, x), u(t, x) = F −1 g(ξ where F (t, x) = F

−1



e

−4π 2 t ξ 2

 (x) =

 

= =

Rn

Rn

e2πix·ξ e−4π

2t ξ 2

e−2πix·ξ e−4π

−|x|2 1 e 4t . n/2 (4πt)

dξ

2t ξ 2

dξ

2.1 The Fourier Transform

39

Finally, we get the following formula for the solution. 1 u(t, x) = (4πt)n/2

 Rn

exp

 −(x − y)2  g(y) dy, 4t

which is a convolution of the initial data with the fundamental solution for the heat equation; compare Proposition 5.3.4. ♦

2.1.3 Problems Problem 2.1.6 Check that f (x) = exp(−|x|2) is in the Schwartz space S(Rn ). Problem 2.1.7 Check that the formula (2.2) indeed defines a metric in the Schwartz space (i.e., d(f, g) ≥ 0, d(f, g) = d(g, f ), d(f, g) + d(g, h) ≥ d(f, h) and d(f, g) = 0 ⇐⇒ f = g for all f, g, h ∈ S). Problem 2.1.8 Show that if φ, ψ ∈ S(Rn ), then their product φ · ψ also lies in S(Rn ). Problem 2.1.9 Show that for any f ∈ S(Rn ) and λ > 0, if fλ (x) = f (x/λ), then fˆλ (ξ ) = λn fˆ(λξ ). Problem 2.1.10 Assume that φ ∈ L1 (Rn ) and D α φ ∈ L1 (Rn ) for all multiindices ˆ ) is a bounded function of α satisfying |α| = k. Prove that κ(ξ ) = (1 + |ξ |k )φ(ξ n ξ ∈R . Problem 2.1.11 Let f ∈ L2 (R). Calculate the Fourier transform of g(x) = f (x)e−2πixa , in terms of the Fourier transform of f , for a ∈ R. Problem 2.1.12 Let A be a symmetric, nonsingular n × n matrix. Calculate the Fourier transform of the function f (x) = e−Ax·x ,

x ∈ Rn ,

where a · b denotes the standard scalar product of vectors a, b ∈ Rn . Problem 2.1.13 Consider the partial differential equation −uxx + uyyyy + u = f

in R2 .

Show that if f ∈ L2 (R2 ), then also the functions u, ux , uyy , uxyy are square integrable (i.e., lie in L2 ).

40

2 Distributions, Sobolev Spaces and the Fourier Transform

Problem 2.1.14 Find a solution to the problem ut = uxx + ux , u(x, 0) = u0 (x),

x ∈ R, t > 0, x ∈ R.

where we additionally know that the initial condition u0 is in L2 (R). Problem 2.1.15 Let u0 be a function in L2 (R). (a) Find a solution to the problem ut = uxx − u, u(x, 0) = u0 (x),

x ∈ R, t > 0, x ∈ R.

(b) Prove that the solution satisfies u(·, t)2 ≤ Ce−t for a positive constant C and all t > 0. The next two problems are related to the problem of computing the Fourier transform of the function sinx x , see Problem 2.1.2 above. ∞ Problem 2.1.16 Explain, why the proof that the integral −∞ sinx x e−2πixξ dx is 1 convergent, given in the solution to Problem 2.1.2, fails precisely when ξ = ± 2π . Is the integral convergent for these values of ξ ? Problem 2.1.17 Let  u(ξ ) be the Fourier transform of

sin x x .

1 Compute  u(± 2π ).

Problem 2.1.18 Calculate the Fourier transform of the function

sin2 x . x2

Problem 2.1.19 Calculate the Fourier transform of the function

sin3 x . x3

Problem 2.1.20 Suppose that f ∈ L1 (R) is bounded. Write f for the Fourier transform. Prove that lim f(ξ ) = 0.

ξ →±∞

Hint Prove it first for f with compact support and then present f as a sum of a function with compact support and a function with small L1 -norm. We come back now to Problem 2.1.3. The next three problems are taken from [10, pp. 70–73].  sin nx Problem 2.1.21 Prove that the series ∞ n=2 ln n is convergent for all x.

2.1 The Fourier Transform

41

Hint Show that the partial sums of criterion.

∞

n=2 sin nx

are bounded and apply Abel’s

 sin nx Problem 2.1.22 In Problem 2.1.3 we took f (x) = n≥2 ln n and F (t) = t 0 f (x)dx. Assuming f (x) is integrable, calculate rigorously the Fourier expansion of F , justifying all steps in the calculations.  inx Problem 2.1.23 Let cn , n ∈ Z be a series of numbers such that ∞ n=−∞ cn e ∞ inx is convergent, but H (x) = n=−∞ cn e is not Lebesgue integrable. Let h be a smooth function supported on [−1/2, 1/2] such that h(0) = 2π and set ∞ 

f (x) =

cn h(x − n).

n=−∞

(a) Argue that limx→±∞ f (x) = 0. (b) Prove that f (x) is also smooth.  = f . Define Supposethat F (x) is a Lebesgue integrable function such that F g(t) = ∞ F (t + 2πm). m=−∞ (c) (d) (e) (f)

Show that g is well defined. π Prove that −π |g(t)|dt < ∞. π 1 −int = c . Show that 2π n −π g(t)e Argue that cn are Fourier coefficients of a Lebesgue integrable function, contradicting the definition of cn . (g) Conclude that f (x) is not a Fourier transform of any Lebesgue integrable function. Problem 2.1.24 Let fn : R → R be a piecewise-linear function equal to 0 outside of [−1 − n1 , 1 + n1 ], equal 1 on [−1, 1] and linear on [−1 − n1 , −1] and [1, 1 + n1 ]. Let fn be the Fourier transforms of fn . Show that lim ||fn ||L1 = ∞.

n→∞

Problem 2.1.25 Let fn be as in Problem 2.1.24. Let nk be an increasing sequence of indices such that ||fnk ||L1 ≥ 3k /k 2 . Define h(x) = k k −2 fnk (x − 2k). Prove that h is a continuous function, h ∈ L1 (R), but its Fourier transform is not Lebesgue integrable. Problem 2.1.26 Prove Theorem 2.1.4. Problem 2.1.27 Let f : R → R be a continuous compactly supported function and let fbe its Fourier transform. Prove that for any R > 0 

R

−R

f(ξ )e2πiξy dξ = f ∗ SR (y),

42

2 Distributions, Sobolev Spaces and the Fourier Transform

where SR (y) =

sin 2πRy . πy

Problem 2.1.28 Prove that if f is a compactly supported continuous function, then 

∞ −∞

f(ξ )e2πiξy dξ = f (y).

Problem 2.1.29 Let a1 , a2 , . . . be a sequence of real numbers such that ai ∈ (0, 1) for i = 1, 2, . . . and ∞ i=1 (1 − ai ) > 0. Construct the fat Cantor set by removing the middle segment of length a1 at the first stage of the construction, then, from the two remaining segments (of length (1 − a1)/2 each) the middle segments of relative length a2 (i.e., of length a2 (1 − a1 )/2), and so on. (a) Prove that the fat Cantor set is a compact set homeomorphic to the standard Cantor set. (b) Calculate the measure of the fat Cantor set and show that it is positive. (c) Let g be the characteristic function of the fat Cantor set. Prove that it is discontinuous on a set of positive measure. (d) Let  g denote the Fourier transform of g. Prove that  h(y) =

∞ −∞

 g (ξ )e2πiξy dy

is equal to g almost everywhere.

2.2 The Theory of Distributions The theory of distributions is an important part of functional analysis and an indispensable tool in the study of partial differential equations. Standard university courses in PDEs rarely cover this topic in sufficient detail and often it is completely omitted by lecturers who prefer to concentrate on other aspects of differential equations.

2.2.1 Theoretical Background Let  be an open subset of Rn . We shall study the space C0∞ () of smooth functions with compact support. This space is contained in many important function spaces (C(), Lp (), ∞ C (), etc.) and we can define many different notions of convergence in C0∞ (),

2.2 The Theory of Distributions

43

inherited from these larger spaces. Probably the most natural is the convergence inherited from C ∞ (): a sequence {φn } of functions φn ∈ C0∞ () converges to a function ψ, if {D α φn } converges almost uniformly to D α ψ for any multiindex α. In this section we shall introduce a slightly different and stronger notion of convergence in C0∞ (). Definition 2.2.1 We say that a sequence {fn } of functions in C0∞ () converges to f ∈ C0∞ (), if • there exists a compact set K ⊂  such that supp fn ⊂ K

for all n

and • for any multiindex α sup |D α (fn ) − D α (f )| −−−→ 0. n→∞

x∈K

Note that the second condition in the above definition means simply that fn , together with all its derivatives, converges to f (and its respective derivatives) uniformly on K. Definition 2.2.2 The space C0∞ () with the above notion of convergence (and the topology defined by it) is called the space of test functions. We denote it by D(), or simply by D. Definition 2.2.3 A distribution is a continuous linear functional on D. We denote the space of all distributions by D . Traditionally, the value of a distribution φ ∈ D

on a test function w ∈ D is denoted by angle brackets: w(φ) = w, φ . Every locally integrable function g ∈ L1loc (Rn ) defines a distribution:  g, φ =

Rn

g(x)φ(x) dx

for φ ∈ D.

(2.10)

A distribution of this kind will be called a regular distribution. In what follows, we shall often identify the function g ∈ L1loc (Rn ) with the regular distribution it defines. Any distribution can be multiplied by a smooth function. Such a product of a smooth function f ∈ C ∞ and a distribution w ∈ D is defined as f w, φ = w, f φ for any φ ∈ D.

44

2 Distributions, Sobolev Spaces and the Fourier Transform

Distributional Derivatives Similarly to multiplication by smooth functions we can define, for any multiindex α, the operation of partial derivation on the space of distributions D α : D (Rn ) → D (Rn ) in the following way: if w ∈ D , then D α w, f = (−1)|α| w, D α f for all f ∈ D. Every test function is, obviously, locally integrable, which gives a natural embedding of D into the space of distributions. Integrating by parts one may easily check that this embedding j : D → D

satisfies j (D α f ) = D α j (f )

for f ∈ D.

This shows that the operator D α defined on the space of distributions is an essential extension of the classical partial derivative D α : the two coincide on the space D, but the space of distributions contains functions that are not differentiable in any classical sense. This motivates the definition of a distributional derivative of a locally integrable function. Definition 2.2.4 Let g be a locally integrable function on Rn . For any multiindex α, the distributional derivative of order α of the function g is the distribution D α g ∈ D

given by D α g, φ = (−1)|α| g, D α φ

for φ ∈ D.

Tempered Distributions Recall that in Sect. 2.1 we introduced the Schwartz space S of rapidly decreasing functions. Let us now consider a continuous linear functional w : S(Rn ) → C. Continuity of w means that there exist m ≥ 0 and C ≥ 0 such that w(f ) ≤ Cpm (f )

for all f ∈ S(Rn ),

2.2 The Theory of Distributions

45

where pm (f ) =



sup (1 + |x|2)m/2 |D α f (x)| m = 0, 1, 2, . . .

n |α|≤m x∈R

are the semi-norms defining the topology of the Schwartz space S. Definition 2.2.5 A tempered distribution is a continuous linear functional on the Schwartz space S(Rn ). The space of all tempered distributions on Rn will be denoted by S (Rn ) or simply S . One can easily check that D(Rn ) embeds continuously in S and S ⊂ D (Rn ), thus tempered distributions are indeed distributions. The space Lp (Rn ), for p ∈ [1, ∞], embeds naturally in the space S , and any w ∈ Lp (Rn ) defines a tempered distribution:  w, f =

Rn

f (x)w(x) dx

for f ∈ S(Rn ).

The same formula gives an embedding of S in S (obviously S ⊂ Lp (Rn )). Similarly, every finite measure μ on Rn defines a tempered distribution  μ, f =

Rn

f (x) dμ(x) for f ∈ S(Rn );

a good example is the Dirac δ0 measure, defining the Dirac δ0 distribution. δ0 , f = f (0). The main reason for introducing tempered distributions is the fact that for these distributions we can define the Fourier transform: as we noted, the Schwartz space S embeds in S , and we can extend F from S to the whole space S . Definition 2.2.6 Let w ∈ S (Rn ) be a tempered distribution. The Fourier transform w  = F (w) ∈ S (Rn ) of the distribution w is the distribution defined as F (w), f = w, F (f )

for all f ∈ S(Rn ).

The Fourier transform maps S to S . In order to see that this definition indeed extends the Fourier transform (defined on S), we should make sure that the two notions coincide on the distributions that are defined by Schwartz functions, i.e., on distributions that are in the image of the embedding j : S → S . Using (c), Theorem 2.1.6, we easily show that F (j (f )) = j (F (f )), which proves that indeed the two notions coincide on S.

46

2 Distributions, Sobolev Spaces and the Fourier Transform

Convolutions Recall that a convolution f ∗ g of two measurable functions f, g : R → R is given by the formula  f ∗ g(x) =

∞ −∞

f (x − y)g(y) dy.

Section 1.2 is devoted to a systematic study of convolutions of functions. Let τx : S → S denote the translation operator τx φ(y) = φ(y − x), and let f˜(x) = f (−x). Definition 2.2.7 A convolution of a test function f ∈ D and a distribution w ∈ D

is the function f ∗ w : Rn → C given by (f ∗ w)(x) = w, τx f˜ for x ∈ Rn . The same formula allows us to define the convolution of a function f ∈ S and a tempered distribution w ∈ S . In both cases, this convolution is a C ∞ function in Rn and there holds D α (f ∗ w) = (D α f ) ∗ w = f ∗ (D α w).

(2.11)

Moreover, if f ∈ S and w is a tempered distribution, then the convolution f ∗ w (identified with the distribution it defines through (2.10)) is again a tempered distribution: f ∗ w ∈ S

for f ∈ S and w ∈ S .

Properties of the Fourier transform with respect to convolutions, described in Theorem 2.1.4, extend naturally to the setting of distributions, in particular we have, for all f ∈ S and w ∈ S , F (f ∗ w) = F (f )F (w).

(2.12)

2.2 The Theory of Distributions

47

Fundamental Solutions Assume P is a polynomial in n variables of non-zero degree and consider the following linear partial differential equation with constant coefficients P (D)u = v.

(2.13)

 αk As usual, we use the convention that if P (x) = m k=0 ak x , where α0 , α1 , . . . , αm are multiindices, then the differential operator P (D) is defined by P (D) =  m αk k=0 ak D . Definition 2.2.8 The fundamental solution of the operator P (D) (or, often, of Eq. (2.13)) is the distribution E ∈ D satisfying P (D)E = δ0 ,

(2.14)

where δ0 stands for the Dirac delta distribution. The importance of fundamental solutions lies in the following fact, which finds applications in constructing solutions to the Laplace equation, see Theorem 4.1.10, and parabolic equations, see Proposition 5.3.4. Theorem 2.2.9 Assume E is the fundamental solution of P (D) and v ∈ D is a test function, i.e., it is smooth and compactly supported. Then u = E ∗ v is a solution of Eq. (2.13). This theorem is proved in the worked-out Problem 2.2.4.

2.2.2 Worked-Out Problems Problem 2.2.1 Prove that the Dirac delta δ0 is indeed a distribution and a tempered distribution. Solution Recall that the δ0 distribution acts on a test function φ by the following formula: δ0 , φ := φ(0). It is obvious that δ0 defines a linear functional on D, and likewise on S. We need to show that this functional is continuous with respect to the convergence both in D and in S. Assume we have a sequence of test functions φn , convergent in D to a function φ. This means that there is a compact set K ⊂ Rn such that ∀n supp φn ⊂ K and φn

48

2 Distributions, Sobolev Spaces and the Fourier Transform

converge uniformly to φ. This implies in particular that δ0 , φn = φn (0) → φ(0) = δ0 , φ , which proves that δ0 ∈ D . Similarly, the convergence of φn to φ in the Schwartz space S implies pointwise convergence of φn to φ, thus again δ0 , φn → δ0 , φ and δ0 ∈ S . ♦ Problem 2.2.2 Calculate the Fourier transform of the Dirac δ0 distribution. Solution Following the definition of the Fourier transform on distributions (Definition 2.2.6), we have for all φ ∈ S  ˆ φ(x) dx = 1, φ . F (δ0 ), φ = δ0 , F (φ) = φ(0) = Rn

Thus δˆ0 ≡ 1. ♦ Problem 2.2.3 Prove that for any function f ∈ S we have f ∗ δ0 = f . Solution By Definition 2.2.7, for any x ∈ Rn f ∗ δ0 (x) = δ0 , τx f˜ = (τx f˜)(0) = f˜(−x) = f (x). ♦ Problem 2.2.4 Show that if E is the fundamental solution of the operator P (D), and v ∈ D, then u = E ∗ v satisfies Eq. (2.13). Solution Since v is a test function, the convolution v ∗ E is well defined and, by (2.11), D α (v ∗ E) = v ∗ D α (E), thus P (D)(v ∗ E) = v ∗ P (D)(E) = v ∗ δ0 , and we proved in Problem 2.2.3 that v ∗ δ0 = v (we know that v, as a test function, lies in the Schwartz space S, so we may apply that result). ♦ Problem 2.2.5 Calculate the distributional derivative H of the Heaviside step function: H (x) =

 0

for x < 0,

1

for x ≥ 0.

2.2 The Theory of Distributions

49

Solution The function H is locally integrable, thus, following the definition of the distributional derivative, for every φ ∈ D we have H , φ = −



∞

−∞

H (x)φ (x) dx = −



∞

φ (x)dx = φ(0) = δ , φ .

0

Thus the distributional derivative of the Heaviside step function is equal to the Dirac δ distribution. ♦ Problem 2.2.6 Calculate the distributional derivative of f (x) = ln |x|. Solution We need to find a distribution w ∈ D (R) such that for all φ ∈ D(R)  w , φ = −

∞ −∞

ln |x|φ (x)dx.

We fix  ∈ (0, 1) and split the integral on the right-hand side into integrals over (−∞, −), (−, 0), (0, ) and (, ∞). Calculating separately the first and the last term and integrating by parts we get 

− −∞







ln |x|φ (x) dx =

∞

−∞

ln |x|φ (x) dx =



−







ln(−x)φ (x) dx = φ(−) ln  −

∞

ln x φ (x) dx = −φ() ln  −

−∞





−

∞ 

1 φ(x) dx, x

1 φ(x) dx, x

while the middle two can be estimated in the following way:    

0 −

  

 ln |x|φ (x) dx  ≤ sup |φ | ·

(−1,1)

0 −

(− ln(−x)) dx = sup |φ | · (1 − ln ) (−1,1)

and similarly    

 0

   

 ln |x|φ (x) dx  ≤ sup |φ | · (− ln(x)) dx = sup |φ | · (1 − ln ),

0

(−1,1)

which shows that  lim



→0 −

ln |x|φ (x) dx = 0.

(−1,1)

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2 Distributions, Sobolev Spaces and the Fourier Transform

Using the above calculations we get 

 ∞   1 1 ln |x|φ (x) dx φ(x) dx + φ() ln  + φ(x) dx + x −∞ x  −    ∞ 1 1 φ(−) − φ(0) φ() − φ(0) φ(x) dx + φ(x) dx + +  ln  x x −  

w , φ = −φ(−) ln  +  =  +

−

−∞ 

−

−

ln |x|φ (x) dx.

Then, recalling that lim→0  ln  = 0, we have  w , φ = lim w , φ = lim →0

→0

−

1 φ(x) dx + x

−∞



∞



1 φ(x) dx x





+ 2φ (0) · 0 + 0   −  ∞ 1 1 = lim φ(x) dx + φ(x) dx . →0 x −∞ x  This distribution is known as the principal value of x1 and it is denoted p.v. x1 . Note that the function x1 is not locally integrable near 0 and thus it does not define a regular distribution. ♦

2.2.3 Problems Problem 2.2.7 Calculate the distributional partial derivatives of the function f : R2 \ {0} → R, f (x1 , x2 ) = 

1

.

x12 + x22

Problem 2.2.8 Calculate the second distributional derivative of the function f (x) = ln |x|. Problem 2.2.9 Let f : R2 → R denote the characteristic function of the set {x ∈ R2 : x1 > 0, x2 > 0}. Calculate the distributional partial derivatives of f . Problem 2.2.10 Show that the Dirac delta is not a regular distribution, i.e., there is no locally integrable function f satisfying  δ0 (φ) = φ(0) = for every test function φ ∈ D.

∞

−∞

f (x)φ(x) dx

2.2 The Theory of Distributions

51

Fig. 2.2 The vector field in Problem 2.2.11

Remark 2.2.10 To calculate the distributional divergence of a locally integrable vector field we identify its coordinates with the regular distributions they define and then use the definition (1.3) of the divergence, treating the partial derivatives that appear in (1.3) as distributional derivatives. A vector field has distributional divergence zero if the result is the zero distribution. Problem 2.2.11 Let a vector field v : R2 → R2 be defined as (see Fig. 2.2) ⎧ ⎪ (0, 1) ⎪ ⎪ ⎪ ⎨(1, 0) v(x, y) = ⎪(0, −1) ⎪ ⎪ ⎪ ⎩ (−1, 0)

for x ∈ (−y, y] and y ≥ 0, for x ≥ 0 and y ∈ [−x, x), for x ∈ [y, −y) and y ≤ 0, for x ≤ 0 and y ∈ (x, −x].

Show that the distributional divergence of v is zero. Problem 2.2.12 Show that f (x) = ex does not define (through the formula (2.10)) a tempered distribution, while g(x) = ex cos(ex ) does. Problem 2.2.13 Find the fundamental solution for the operator

∂2 ∂x1 ∂x2

in R2 .

Problem 2.2.14 Find the fundamental solution of the Laplacian  in Rn , for n ≥ 2. Compare with Example 4.1.9. Problem 2.2.15 Show that if v ∈ D(R3 ), then 1 u(x) = − 4π

 R3

v(y) dy |x − y|

is a solution to the Poisson equation u = v in R3 .

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2 Distributions, Sobolev Spaces and the Fourier Transform

2.3 Sobolev Spaces Sobolev spaces play central role in modern theory of partial differential equations. The choice of an appropriate function space in which we look for solutions to a specific partial differential equation is a key step in both proving their existence and to further study of their properties. Sobolev spaces are a natural answer to that problem, because they allow for the use of numerous tools of functional analysis and variational calculus. Good introductory expositions of the theory of Sobolev spaces can be found in textbooks on partial differential equations, including, e.g., the books of Evans [8] and Taylor [26]. Those interested in more intricate properties of these spaces should turn to the books of Leoni [15] and of Adams and Fournier [1]. In this chapter we present different definitions and basic properties of Sobolev spaces, as well as important results on embeddings of Sobolev spaces. Sobolev spaces are omnipresent in Chaps. 4 and 5.

2.3.1 Theoretical Background Sobolev Spaces W m,p In Sect. 2.2 we introduced the notion of a derivative of a distribution. Now, we define a closely related notion of a weak derivative. Definition 2.3.1 Let  be an open subset of Rn , assume f, g ∈ L1loc () and let α be a multiindex. We say that g is an α-weak derivative of f in , if 

g(x)φ(x) dx = (−1)|α| 

 f (x)D α φ(x) dx 

holds for any function φ ∈ C0∞ (). In that case we write g(x) = D α f (x). Remark 2.3.2 If a weak derivative exists, then it is defined uniquely. This can be shown easily using Problem 4.2.7 in Sect. 4.2. If f has a continuous classical α-derivative, one can easily check through integration by parts that this classical derivative coincides with the weak one. This is the reason why we do not need a new symbol for this new kind of differentiation. Do note, however, that the weak derivative (if it exists) is only a locally integrable function, thus it is defined up to a set of measure zero. Recall that a locally integrable function f defines a distribution (called then a regular distribution) wf . We can ask about the relation between the weak derivative of f and the distributional derivative of the regular distribution wf . It is easy to check that whenever the weak derivative exists, the two notions coincide. More precisely, if the weak derivative D α f exists, then the distributional derivative D α of the distribution wf is again a regular distribution, defined by the locally integrable function D α f .

2.3 Sobolev Spaces

53

The notion of a distributional derivative is, however, a more general one. A locally integrable function (identified with the regular distribution it defines) always has a distributional derivative of arbitrary order. At the same time, a weak derivative of such a function might not exist (see Problem 2.3.1). In particular, if f is the characteristic function of a nonempty, bounded, and open set , then even the first weak derivatives of f on Rn do not exist. Let  be an open subset of Rn ; assume m ∈ N and 1 ≤ p < ∞. Definition 2.3.3 The Sobolev space W m,p () is the vector space of functions {u ∈ Lp () : D α u ∈ Lp () for all α such that |α| ≤ m}. In the definition above, by D α u we denote the weak derivative (or, equivalently, distributional derivative) of the function u. In W m,p , we introduce the norm up,m, =



D α u2Lp () .

(2.15)

|α|≤m

Remark 2.3.4 Many textbooks on partial differential equations omit the theory of distributions, and, in particular, the notion of distributional derivatives, using only weak derivatives. Some other, in turn, do not introduce weak derivatives, using only distributional derivatives. Both approaches are equivalent and both have their advantages (and disadvantages). In this book, we introduce both notions; distributions, especially tempered distributions, will prove useful in describing Sobolev spaces by means of the Fourier transform. Before we study Sobolev spaces in greater detail, let us first recall an important and useful property of the Lebesgue spaces Lp (). Theorem 2.3.5 (Riesz–Fischer Theorem) The space Lp () is complete. Moreover, if a sequence {fk } is convergent in Lp (), then there is a subsequence {fkj } which is convergent a.e. in . We should mention the following basic properties of W m,p (): • W m,p () is complete (and thus it is a Banach space). One can easily prove that W m,p () is isomorphic to a closed subspace of Lp () × Lp (, Rn ) ≈ Lp ()n+1 . The latter space is complete by the Riesz– Fischer theorem and a simple observation that a product of complete spaces is complete. Finally, closed subspaces of complete spaces are complete. • Smooth functions are dense in W m,p (), i.e., C ∞ () ∩ W m,p () is a dense subset of W m,p (). In other words, Theorem 2.3.6 (Meyers-Serrin’s Theorem) If u ∈ W m,p (), then for every  > 0, there exists v ∈ C ∞ (), such that u − vp,m, < .

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2 Distributions, Sobolev Spaces and the Fourier Transform

These two properties provide an alternative definition of W m,p () as the completion—in the Sobolev norm (2.15)—of the set ⎧ ⎨ ⎩

 

u ∈ C ∞ () :

|D α u|p dx < ∞

|α|≤m 

⎫ ⎬ ⎭

It turns out that if  = Rn and 1 ≤ p < ∞, then any function in W m,p (Rn ) can be approximated by smooth functions with compact support: one can prove that C0∞ (Rn ) is a dense subset of W m,p (Rn ). This is not the case, however, for an arbitrary open . m,p

Definition 2.3.7 The space W0 () of Sobolev functions with null trace on the boundary is defined as the closure1 of the space of compactly supported smooth functions C0∞ () in the Sobolev space W m,p (). m,p

As we mentioned, W0 () is usually a proper subset of W m,p (). It is good to have in mind the following example: m,p

Example 2.3.8 Let u ∈ W0

(). Define the function u˜ as

u(x) ˜ =

 u(x) for x ∈ , 0

for x ∈ Rn \ .

Then, for any open set 1 ⊃ , we have u˜ ∈ W m,p (1 ). In particular, u˜ ∈ W m,p (Rn ). Note, however, that for an arbitrary function u from W m,p (), the function u˜ constructed as above in general need not have weak derivatives in Rn at all, in particular it need not belong to W m,p (Rn ). This is the case, e.g., when u(x) = 1 for all x ∈ . Then u˜ equals the characteristic function of . If  is an open set with sufficiently regular (e.g., Lipschitz) boundary,2 one can slightly improve the result given by Meyers-Serrin’s Theorem: Theorem 2.3.9 Assume  has Lipschitz boundary, then any function in the Sobolev space W m,p () can be approximated by functions which are continuous, and even smooth up to the boundary.3 In other words, C ∞ ()∩W m,p () is dense in W m,p ().

1 Once

we know that W m,p is complete, we can alternately talk about the completion of a subset of W m,p in the Sobolev norm or about the closure of that subset in W m,p —both constructions lead to the same set. 2 This means that the boundary is locally a graph of a Lipschitz continuous function. 3 A function is smooth up to the boundary of  if it extends to a smooth function on some open V ⊃ .

2.3 Sobolev Spaces

55

The spaces W m,2 () are of particular interest, since they have the structure of a Hilbert space. The scalar product in that case is given by the formula  u, v W m,2 () =



D α u(x)D α v(x) dx.

 |α|≤m

The space W m,2 () is often denoted as H m (). The next section is devoted to another approach to this space, which uses the Fourier transform. Remark 2.3.10 In Sect. 3.3, the Hilbert structure of W 1,2 () plays crucial role, therefore we consistently use there the notation H 1 (). In contrast, some of the tools introduced in Sect. 4.2 generalize to nonlinear problems, in which Sobolev spaces with p = 2 are necessary; we stick to the notation W 1,2 () there. Spaces H s (Rn ) If  = Rn , we can use the Fourier transform to provide an alternate definition of the Sobolev space W m,2 . In this context, that space is often denoted H m . Using Theorem 2.1.6 (property (b)) and Parseval’s Identity (2.5), we get ⎛ u2,m,Rn = ⎝



⎞1/2

 Rn

|ξ α |2 |u(ξ ˆ )|2 dξ ⎠

for u ∈ S.

|α|≤m

This, together with the inequality (2.1), shows that on the Schwartz space S the norm u2,m,Rn and the norm  u∗,m =

1/2 Rn

(1 + |ξ |2 )m |u(ξ ˆ )|2 dξ

(2.16)

are equivalent. Plancherel’s theorem (Theorem 2.1.8) together with the density of S in the space L2 allow us to give an alternate definition of the Sobolev space H m (Rn ) for m ∈ N:  #   1/2

H m (Rn ) = u ∈ L2 (Rn ) :

Rn

(1 + |ξ |2 )m |u(ξ ˆ )|2 dξ

r + n/2, then any function from H s (Rn ) can be modified on a set of measure zero to a C r (Rn ) function. For 0 < s < 1, we can introduce an equivalent norm in H s (Rn ),  u∗,H s =

 Rn

|u|2 dx +

 Rn

Rn

|u(x) − u(y)|2 dx dy |x − y|n+2s

1/2 .

In the same spirit, for any noninteger, positive s > 0, we can use the equivalent norm in H s (Rn ) given by the formula ⎛ u∗,H s = ⎝

  n |α|≤[s] R

|D α u|2 dx +

  n |α|=[s] R



⎞1/2 |D α u(x) − D α u(y)|2 dx dy ⎠ , |x − y|n+2{s} Rn

(2.17) where by [s] and {s} we denote, respectively, the integer and the fractional part of s. Spaces H s () Let  be an open set in Rn . In the previous section we proposed a way to define Sobolev spaces H s (Rn ) for noninteger exponents. Is it possible to do the same for functions defined not on the whole Rn , but only on ? It turns out that indeed, we can define H s (). Our considerations in the previous section provide us with two possible approaches. 1. Use the equivalent norm on H s , given by the formula (2.17) for s ≥ 0. Following this idea, we define the Sobolev space H s (), s ≥ 0, as the space of these functions in L2 () which have weak derivatives up to order k = [s], all these weak derivatives belong to L2 () and such that the following norm  · H s () is finite:

2.3 Sobolev Spaces

57

• for integer s: uH s () =



D α uL2 ()

(2.18)

|α|≤s

(note that this is the standard Sobolev norm in W s,2 ()), • for noninteger s: ⎛ uH s ()

=⎝

⎞1/2    |D α u(x) − D α u(y)|2 |D u| dx + dx dy ⎠ . |x − y|n+2{s}   

 

|α|≤[s]

2

α

|α|=[s]

(2.19) 2. For any s ∈ R, we can define the space Hs () as the space of all restrictions of H s (Rn ) functions to : u ∈ Hs () ⇐⇒ ∃v ∈ H s (Rn ) : v ≡ u

a.e. in .

Theorem 2.3.12 If  is open, bounded, with Lipschitz boundary and s ≥ 0, then the two above definitions of Sobolev spaces coincide, i.e., H s () = Hs (). In the two approaches we avoided a direct application of the Fourier transform, which is well defined for functions given on the whole Rn . On the other hand, any function u ∈ L2 () can be extended to a function in L2 (Rn ), simply by taking u = 0 on Rn \ . Through this extension, we can consider the Fourier transform uˆ of u, in particular we have uˆ ∈ L2 (Rn ). Let s > 0. The Sobolev space of functions with null boundary values can be defined in the standard way: Definition 2.3.13 The space H0s () is defined as the completion of C0∞ () in the norm given by formulas (2.18) (for positive, integer s) and (2.19) (for positive, noninteger s). Lemma 2.3.14 Let  be an arbitrary open set in Rn and assume that m ∈ N. Then, ˆ ) ∈ L2 (Rn )}. H0m () ⊂ {u ∈ L2 () : (1 + |ξ |2 )m/2 u(ξ If  is bounded, with Lipschitz boundary, then the above inclusion is an equality. Theorem 2.3.15 If 0 < s < 12 , then H s () = H0s (). 

Duality and the Spaces W −m,p () Assume 1 ≤ p < ∞ and let p1 + q1 = 1. Any continuous linear functional φ on W m,p () has the following form: for all multiindices α, 0 ≤ |α| ≤ m, there exist

58

2 Distributions, Sobolev Spaces and the Fourier Transform

functions vα ∈ Lq () such that for all u ∈ W m,p ,   D α u(x) vα (x) dx φ(u) = 0≤|α|≤m 

and if 1 < p < ∞, then the functions vα are defined uniquely. In this case, φ(W m,p ) =



vα Lq

0≤|α|≤m

is a norm on the dual space (W m,p ())∗ . m,p For the spaces W0 , with 1 < p < ∞, we have a more convenient description of the dual space. −m,q () consists of all distributions T ∈ D () of Definition 2.3.16  The space W α the form T = 0≤|α|≤m (−1) D α Tvα , where Tvα are regular distributions given by functions vα ∈ Lq ().

This is a Banach space, equipped with the norm T W −m,q = sup{| T , u | : u ∈ W m,p , uW m,p ≤ 1}. The above norm is finite for any regular distribution Tv given by a function v ∈ Lq . Identifying Tv with the function v, we can consider the W −m,q norm on the space Lq ():  sup u(x)v(x) dx. vW −m,p = u∈W m,p uW m,p ≤1



It turns out that the space W −m,p () is the completion of Lq () with respect to the above norm. m,p

Theorem 2.3.17 The space (W0

())∗ is equal to W −m,q ().

Since W0 (Rn ) = W m,p (Rn ), we also have that (W m,p (Rn ))∗ = W −m,q (Rn ). This need not be true for Sobolev spaces defined on an arbitrary open set  and in general we only have an inclusion: (W m,p ())∗ ⊂ W −m,q (). In the case of p = 2, the Fourier transform approach to the Sobolev spaces provides yet another characterization of W −m,2 . It turns out that the dual space to H s (Rn ) is the space H −s (Rn ), in particular W −m,2 (Rn ) is isomorphic to H −m (Rn ). For functions defined on a domain , we have (H0s ())∗ = H −s (), in particular W −1,2 () is isomorphic to H −1 (). One should note that the spaces H s (on a domain or on the whole Rn ) are, for any s, Hilbert spaces, and the Riesz Representation Theorem (see Theorem 3.1.5) provides us with another identification, this time between (H s )∗ and H s . This identification shows that the spaces H −s and H s are isometric. It does not, however, m,p

2.3 Sobolev Spaces

59

mean that they are equal as subspaces of D . In fact, any two separable, infinitedimensional Hilbert spaces are isometric (see, e.g., [4, Chapter 17]).

Trace of a Sobolev Function m,p

Introducing the space W0 (), we called it the Sobolev space with null trace on the boundary. In this context, by trace, we mean restriction. Recall that the Sobolev space W m,p () is defined as a particular subspace of the Lebesgue space Lp (). In particular, functions from Sobolev spaces are defined only up to a set of measure zero. Unless the domain  ⊂ Rn is extremely irregular (and, in what follows, we shall assume it is not), its boundary is a set of n-dimensional Lebesgue measure zero. Is there then any sense in considering restrictions of Sobolev functions to the boundary of ? Obviously, the classical notion of the restriction of a function makes no sense: taking any Sobolev function f on , we can arbitrarily modify its values on ∂ and the function f˜, obtained through that modification, represents the same function in the Sobolev space: the elements of the Sobolev space are not really functions, but equivalence classes of functions defined up to a set of measure zero, just like the elements of Lp spaces. There is, however, a sensible generalization of the notion of restriction that is well defined (and very useful) in the realm of Sobolev spaces. Theorem 2.3.18 Assume  ⊂ Rn is open and bounded, with C 1 boundary. Then there exists a continuous operator T : W 1,p () → Lp (∂) such that for any smooth (C ∞ ) function u :  → R we have T (u) = u|∂ . Definition 2.3.19 Let u ∈ W 1,p (). The function T (u) ∈ Lp (∂) is called the trace of the function u on ∂. The above definition, given through Theorem 2.3.18, seems purely existential and gives no immediate clue how to define it, except the case when u smooth up to the boundary of . However, this is where Theorem 2.3.9 comes into play. Assume u is a discontinuous function in W m,p (), thus a function on which we do not readily know the value of the trace operator T . We know that u can be approximated by smooth functions, which are continuous up to the boundary of , thus there exists a sequence {uk }, such that ∀k uk ∈ C() and uk → u in W m,p (). For each of the functions from the approximating sequence uk , the trace operator is well defined: T (uk ) = uk |∂ . It turns out that these restrictions converge in the space Lp (∂) and the limit does not depend on the choice of the approximating sequence {uk }. We can thus set the value of T (u) to be the limit of T (uk ) = uk |∂ in Lp (checking these claims is essentially the proof of Theorem 2.3.18).

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2 Distributions, Sobolev Spaces and the Fourier Transform

In the same way, to know T (u) for an arbitrary u ∈ W m,p (), we find a sequence {uk } of smooth functions, continuous up to the boundary of , that approximate u in W m,p (); their restrictions to ∂ form a convergent sequence in Lp (∂). The trace of u on ∂ is the limit of that sequence of restrictions. Similarly, we define the trace in the spaces H s (Rn ), for s ∈ R—this time, obviously, not on the boundary of the domain, but on an (n − 1)-dimensional hyperplane. For x ∈ Rn , we can write x = (x , xn ), x = (x1 , . . . , xn−1 ). For smooth, compactly supported functions on Rn , the operator T of the trace on the hypersurface H = {xn = 0} is simply a restriction to H: T : C0∞ (Rn ) → C0∞ (Rn−1 ), (T u)(x ) = u(x , 0). Theorem 2.3.20 Assume s > 12 . The trace operator T : C0∞ (Rn ) → C0∞ (Rn−1 ) can be extended to a continuous linear operator 1

T : H s (Rn ) → H s− 2 (Rn−1 ). In particular, T (u)

H

s− 1 2 (Rn−1 )

≤ CuH s (Rn )

for a constant C > 0, dependent on s and n. Note that for integer s, any function f ∈ H s (Rn ) lies also in H s (Rn+ ) and H = ∂Rn+ . However, if we apply Theorem 2.3.18 to a function f ∈ H s (Rn+ ) = W 2,s (Rn+ ), we only get that the trace of f lies in L2 (Rn−1 ), while Theorem 2.3.20 gives a much stronger result: the trace of f lies in a proper subspace of L2 (Rn−1 ), 1 namely H s− 2 (Rn−1 ) and that if (uk ) is a sequence of smooth, compactly supported functions converging to u in H s (Rn ), then the restrictions of uk to H converge to 1 T (u) not only in L2 (Rn−1 ), but also in H s− 2 (Rn−1 ). For a counterpart of Theorem 2.3.20 for the spaces H s () (under sufficient regularity assumptions on ∂), and for further generalizations we refer the reader to the first volume of Taylor’s book [26] or to the book of Adams and Fournier [1].

Embedding Theorems Many important theorems in mathematics tell us that one space of functions is contained in another one. For example, the fact that a function is differentiable implies that it is continuous, thus we can say that differentiable functions form a

2.3 Sobolev Spaces

61

subset of the space of continuous functions. In this chapter we shall deal with the following general problem: Which function spaces X contain the Sobolev space W 1,p () as their subset?

Recall that the Sobolev space W 1,p is, in simplest words, the space of such functions u, that both u and its first generalized derivative Du belong to the Lebesgue space Lp . Thus, we can reformulate our problem in the following way:   Do the assumptions u ∈ Lp and Du ∈ Lp imply any additional regularity of the function u?

We will answer this question through a long series of exercises. We will start by recalling a few fundamental and useful inequalities, among them Hölder’s inequality, which will be the key tool for our considerations. It is quite surprising how far one can go by using only this inequality and Newton–Leibniz’s Fundamental Theorem of Calculus. Then, we will show the four key embedding results: Poincaré’s, Sobolev’s, Morrey’s and Rellich–Kondrachov’s Theorems. Let us now formulate these theorems, with a commentary.

Poincaré’s Theorem The simplest one of the aforementioned series is Poincaré’s Theorem. It translates to the language of Sobolev spaces the following elementary observation: Assume the function f is differentiable on an open interval (a, b), continuous on [a, b] and that it vanishes at the endpoints a and b. It is clear (one can argue, e.g., using Lagrange’s Mean Value Theorem) that the maximum value of f on [a, b] can be bounded by the maximal rate of growth of f , that is   sup |f (x)| ≤ |b − a| · sup f (x) . x∈[a,b]

x∈(a,b)

In other words, up to a multiplicative constant dependent only on the domain , the supremum norm of a function u vanishing on the boundary of  cannot exceed the supremum norm of its gradient ∇u. The length of the gradient ∇u measures the rate of growth of u and the maximal length of ∇u determines, how much u can increase when we move along an interval of fixed length in . Poincaré’s Theorem (often referred to as Poincaré’s Inequality) translates the above observation to these spaces, in which the size of a function is measured by an integral. It says that the size of a function that vanishes on the boundary of the domain can be estimated by the size of its gradient—as long as we measure both quantities in the norm of the Lebesgue space Lp .

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2 Distributions, Sobolev Spaces and the Fourier Transform

Theorem 2.3.21 (Poincaré’s Theorem) Let  be a bounded domain in Rn . There exists a constant C > 0, dependent on p and , such that the inequality uLp () ≤ C∇uLp ()

(2.20)

1,p

holds for all functions u ∈ W0 (). The proof of this theorem in the particular case n = 2 and p = 2 is included as Problem 2.3.7 in worked-out problems. Generalization of this proof to other values of p and n is the objective of Problem 2.3.23, the dependence of the constant C on parameters is studied in Problem 2.3.24. Another version of the theorem, for functions with not necessarily zero boundary data, is also available. Theorem 2.3.22 Let  be a bounded, connected, open subset of Rn . There exists a constant c, dependent on  and p, such that for every u ∈ W 1,p () &  & & & &u − u& & & 

≤ c∇uLp .

Lp

In the above inequality, the barred integral 1 ||  u, where || is the measure of .



denotes the integral average:



u

=

Sobolev’s Theorem The key to Sobolev’s Theorem is the following Sobolev’s inequality. Theorem 2.3.23 (Sobolev’s Inequality) Let 1 ≤ p < n. There exists a constant C > 0, dependent only on p and n, such that the inequality uLq (Rn ) ≤ CDuLp (Rn ) , where q =

np , holds for all functions u ∈ C01 (Rn ). n−p

The proof of Sobolev’s inequality for n = 2 and p = 1 is given as the worked-out Problem 2.3.8. Proving Sobolev’s inequality for n = 3 and p = 1, which is slightly more difficult, is given as Problem 2.3.28. See Problem 2.3.29 for an idea how to prove Sobolev’s inequality for p > 1 once we have it established for fixed n and p = 1. In Problem 2.3.26, we show that Sobolev’s inequality cannot hold for any q np other than n−p . Combining Sobolev’s inequality with the density of smooth functions in Sobolev spaces we obtain the following theorem. np and assume Theorem 2.3.24 (Sobolev’s Theorem) Let 1 ≤ p < n, q = n−p n  ⊂ R is an open and bounded set with smooth boundary. There exists C > 0,

2.3 Sobolev Spaces

63

dependent only on p and , such that uLq () ≤ CuW 1,p () for all functions u ∈ W 1,p (). Morrey’s Theorem It turns out that for p > n any function u ∈ W 1,p () is in fact continuous (possibly after changing its values on a set of measure zero). Since this is somewhat delicate, let us look at the following example. Example 2.3.25 Dirichlet’s function, which assigns 1 to every rational and 0 to every irrational number becomes continuous after we redefine it (in a proper way) on a set of measure zero, even though in the classical sense it is discontinuous at every point of the real line. When treated as a function in the Sobolev space W 1,p (R), it represents, for any p, the same equivalence class as the zero function. For contrast, the function whose graph is depicted in Fig. 2.3 is discontinuous only in one point, but it cannot be redefined on a set of measure zero in such a way that it becomes continuous. Morrey’s Theorem not only ensures that a function u in W 1,p (), where p > n, is continuous, but it also allows us to estimate the norm of u in a particular space of Hölder continuous functions. A key tool here is Morrey’s inequality. Recall that the Hölder coefficient ' ( |u(x) − u(y)| n [u]C 0,β (Rn ) = sup : x, y ∈ R , x = y |x − y|β serves as a seminorm in the space C 0,β (Rn ). Fig. 2.3 An essentially discontinuous function

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2 Distributions, Sobolev Spaces and the Fourier Transform

Theorem 2.3.26 (Morrey’s Inequality) Assume p > n. There exists a constant C > 0, dependent only on p and n, such that the inequality [u]C 0,β (Rn ) ≤ C∇uLp (Rn ) ,

where β = 1 −

n , p

holds for every function u ∈ C 1 (Rn ) ∩ Lp (Rn ). Morrey’s inequality yields as a consequence, Morrey’s Theorem. Theorem 2.3.27 (Morrey’s Theorem) Assume p > n and let  ⊂ Rn be a bounded, open set with smooth boundary. Then every function u ∈ W 1,p () is Hölder continuous with exponent β = 1 − pn (possibly after redefining it on a set of measure zero). A particular case of this theorem is proved in Problems 2.3.31–2.3.35. Rellich–Kondrachov’s Theorem One of the most useful theorems in the theory of Sobolev spaces, Rellich– Kondrachov’s Theorem4 is, in a way, a generalization of theorems of Bolzano and Weierstrass and of Arzela and Ascoli. Let us recall that, according to the first of these theorems, every sequence of points from a cube in Rn contains a convergent subsequence. The Arzela–Ascoli Theorem provides us with an analogous result in an infinite-dimensional space: assume we have a sequence of functions defined on a compact set. If the functions are uniformly bounded and equicontinuous, then the sequence contains a subsequence that is uniformly convergent. Theorem 2.3.28 (Rellich–Kondrachov’s Theorem, see [8, Theorem 5.7.1]) np Assume  ⊂ Rn is a bounded and open set and let r be such that 1 ≤ r < n−p . 1,p (), contains , bounded in the space W Then every sequence of functions {um }∞ m=1 r a subsequence {umk }∞ k=1 that is convergent in the space L ().

2.3.2 Worked-Out Problems Problem 2.3.1 Prove that the Heaviside step function H : R → R, H (x) =

 0

for x < 0,

1

for x ≥ 0.

does not have a weak derivative.

also Rellich’s Theorem; F. Rellich proved it for p = 2, V.I. Kondrachov (spelled also Kondrashov or Kondrashev) for all other p.

4 Called

2.3 Sobolev Spaces

65

Solution Let {φn }∞ n=1 be a sequence of smooth, compactly supported functions such that   1 1 supp φn ⊂ − , , n n and φn (0) = 1. Suppose that H ∈ L1loc (R) is the weak derivative of H . Then, by the definition of the weak derivative, 

∞ −∞

H (x)φ(x) dx = −



∞

−∞

H (x)φ (x) dx

for any test function φ ∈ C0∞ (R). In our case, for any n ∈ N, we obtain  −

∞

−∞

H (x)φn (x) dx = −



∞ 0

φn (x) dx = φn (0) = 1,

thus for any n ∈ N 

∞ −∞

H (x)φn (x) dx = −1.

Note that for any x = 0 we have limn→∞ φn (x) = 0. Since H ∈ L1loc (R) and φn ∈ C0∞ (R), we can take the limit of the above integral (with n → ∞), passing to the limit under the integral sign. We get 

∞

lim

n→∞ −∞

H (x)φn (x) dx =



∞

−∞

H (x) lim φn (x) dx = 0. n→∞

This gives a contradiction, proving that the Heaviside step function does not have a weak derivative. ♦ Problem 2.3.2 Find all s ∈ R such that the Dirac distribution δ0 is in H s (R2 ). Solution We know that the Fourier transform of δ0 is 1: F (δ0 ) = 1 (worked-out Problem 2.2.2). Thus   2 s 2 (1 + |ξ | ) |F (δ0 )| dξ = (1 + |ξ |2 )s dξ. R2

R2

The last integral is finite for s < −1 and infinite for s ≥ −1. Thus δ0 ∈ H s (R2 ) if and only if s < −1. ♦ Problem 2.3.3 Let F : R → R be a C 1 function with bounded derivative F . Assume that  ⊂ Rn is bounded with Lipschitz boundary and u ∈ W 1,p () for

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2 Distributions, Sobolev Spaces and the Fourier Transform

1 < p < ∞. Prove that the function v(x) := F (u(x)) is in W 1,p () and its first order weak derivatives satisfy vxi = F (u)uxi . Solution First, note that since F is bounded, both F (u)uxi and F (u) belong to the space Lp (). We need yet to confirm that vxi = F (u)uxi . Take any φ ∈ C0∞ (); we have to check if  

F (u)uxi φ dx = −

 

(2.21)

F (u)φxi dx.

Take a sequence {un }∞ n=1 of functions smooth up to the boundary that approximates 1,p u in W (): un ∈ C ∞ (),

W 1,p ()

un −−−−−−−−→ u n→∞

(see Theorem 2.3.9). Obviously, for such smooth un , we can integrate by parts:  

F (un )(un )xi φdx = −

 

(2.22)

F (un )φxi dx.

To conclude, we will show that both the left- and the right-hand sides of (2.22) converge to, respectively, the left- and the right-hand sides of (2.21)—at least after passing to a certain subsequence of the sequence {un }∞ n=1 . The convergence of the right-hand side is easy (recall that F is a Lipschitz mapping):        F (un )φx − F (u)φx , dx  ≤ |φxi ||F (un ) − F (u)| dx i i   







 ≤C

|un − u| dx ≤ C1 

1 |un − u|p dx

p

n→∞

−−−→ 0.



Let us turn our attention to the left-hand side of (2.22).      

 F (un )(un )x φ dx − ≤ F (u)u φ dx |φ(x)||F (un )(un )xi − F (u)uxi | dx xi i        ≤C |F (un )uxi − (un )xi | dx + C |uxi ||F (u) − F (un )| dx. 



(2.23) 1,p (), thus also in Lp (). By the Riesz– We know that {un }∞ n=1 converges to u in W Fischer Theorem 2.3.5, from a sequence convergent in Lp () we may choose a subsequence convergent a.e. in ; let {unk }∞ k=1 be such a subsequence. We have

2.3 Sobolev Spaces

67

k→∞

then F (unk ) −−−→ F (u) a.e. in ; rewriting (2.23) with unk in place of un , we may use the Dominated Convergence Theorem to pass with k to +∞ under the integral signs, proving that      k→∞

 F (un )(un )x φ dx − F (u)uxi φ dx  −−−→ 0. k i  



♦ Problem 2.3.4 Consider the equation ut + (u) = 0 in

(0, T ) × Rn ,

(2.24)

with initial condition u|t =0 = u0

on Rn ,

where u0 ∈ H s (Rn ) for an s > 0. Prove that there exists a solution u belonging to the space C([0, T ]; H s (Rn )). Solution We have to prove that for any  > 0 there exists δ > 0 such that whenever |t1 − t2 | < δ, t1 , t2 ∈ [0, T ], there holds u(t1 , ·) − u(t2 , ·)H s (Rn ) < .

(2.25)

Let us recall that the norm in H s (Rn ) is given by the formula 

1/2 2 s

uH s (Rn ) =

Rn

2

(1 + |ξ | ) |u(ξ ˆ )| dξ

,

thus we can expand the inequality (2.25) to  u(t1 , ·) − u(t2 , ·)2H s (Rn ) =

Rn

(1 + |ξ |2 )s |u(t ˆ 1 , ξ ) − u(t ˆ 2 , ξ )|2 dξ <  2 .

To prove the above inequality, we need some knowledge about the Fourier transform of u. Applying the (spatial) Fourier transform to the differential equation (2.24), we obtain d uˆ + (2π|ξ |)4 u(t, ˆ ξ) = 0 dt with the initial condition u(0, ˆ ξ ) = u0 (ξ ).

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2 Distributions, Sobolev Spaces and the Fourier Transform

Solving the above ordinary differential equation, we obtain u(t, ˆ ξ ) = u0 (ξ )e−(2π|ξ |) t . 4

Then, the Sobolev norm of the difference of the solution at two different time points is  u(t1 , ·) − u(t2 , ·)2H s (Rn ) =

Rn

 4 4 2 (1 + |ξ |2 )s u20 (ξ )e−(2π|ξ |) t1 − e−(2π|ξ |) t2  dξ.

Without loss of generality we may assume that T ≥ t2 > t1 ≥ 0. We shall use the inequality 2  −(2π|ξ |)4t 4 2 4  4 1 e − e−(2π|ξ |) t2  = e−2(2π|ξ |) t1 1 − e−(2π|ξ |) (t2 −t1 )  ≤ 1. By assumption, u0 ∈ H s (Rn ), thus the expression under the integral sign is bounded by an integrable function. We may now use the Dominated Convergence Theorem t1 →t2

to show that u(t1 , ·) − u(t2 , ·)2H s (Rn ) −−−→ 0, thus the inequality u(t1 , ·) − u(t2 , ·)2H s (Rn ) <  2 holds for t2 sufficiently close to t1 . ♦ Problem 2.3.5 Prove that if  ⊂ Rn is an open and bounded set and 1 ≤ q ≤ p, then Lq () ⊂ Lp (). Solution The set  is bounded, therefore the function g(x) ≡ 1 is in Ls () for all s ≥ 1. Assume the function u lies in Lp (). Hölder’s inequality implies that 



 |f (x)| dx =

|f (x)| · 1dx ≤

q



|f (x)|

q



q· pq

q/p 

where the exponent s is chosen in such a way that 1 1 + = 1. s p/q Then, q

1 dx

dx



q

1/s s

f Lq () ≤ f Lp () ||1/s .



,

2.3 Sobolev Spaces

69

Taking both sides of the inequality to the power 1/q we obtain, after simple calculations f Lq () ≤ ||1/q−1/p f Lp () . ♦ Problem 2.3.6 Assume u ∈ W 2,1 (), where  ⊂ R4 is a bounded open set with smooth boundary. Using Sobolev’s Theorem show that u ∈ L2 (). Solution The assumption on u implies that u ∈ L1 () and Du ∈ L1 (), thus u ∈ W 1,1 (). Then, using Sobolev’s Theorem and the fact that 4 4·1 = 4−1 3 we obtain that u ∈ L4/3 (). Now, let vk = Dk u. Then vk ∈ L1 () and Dvk ∈ L1 (), therefore vk ∈ W 1,1 () and reasoning as before we show that vk ∈ L4/3 (), and thus u ∈ W 1,4/3 (). Applying Sobolev’s Theorem one more time we get that u ∈ Lq () for q=

4 · 4/3 4−

4 3

= 2.

♦ Problem 2.3.7 Prove Poincaré’s Theorem 2.3.21 for n = 2 and p = 2. Solution Smooth, compactly supported functions are dense in W01,2 , thus it is enough to prove the inequality for u ∈ C0∞ (). Assume thus that u is a smooth function on , with compact support. Moreover, assume that  is a subset of the rectangle I = [a, b] × [c, d]. By the Fundamental Theorem of Calculus, 

x1

u(x1 , x2 ) =

D1 u(t, x2 )dt, a

thus 

x1

|u(x1 , x2 )| ≤



b

|D1 u(t, x2 )|dt ≤

|Du(t, x2 )|dt.

a

a

Applying Hölder’s (or, more precisely, Schwarz’) Inequality, we get  |u(x1 , x2 )| ≤ a

b

1/2 

b

12 ds a

1/2 |Du(t, x2 )|2 dt

,

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2 Distributions, Sobolev Spaces and the Fourier Transform

which yields  |u(x1 , x2 )| ≤ |b − a| 2

b

|Du(t, x2 )|2 dt.

a

Let us now integrate both sides with respect to x2 from c to d: 

d



d

|u(x1 , x2 )|2 dx2 ≤

c



b

|b − a|

c

a

|D(t, s)|2 dtds = |b − a| · Du2L2 () .

Integrating again, this time with respect to x1 (note that the right-hand side does not depend on x1 ) from a to b we obtain  

|u(x)|2 dx ≤ |b − a|2 · Du2L2 () .

Taking square root of both sides we get the desired inequality. ♦ Problem 2.3.8 Prove Sobolev’s Inequality 2.3.23 for n = 2 and p = 1. Solution As in the previous problem, we begin by applying the Fundamental Theorem of Calculus to u, integrating once with respect to the first variable and once with respect to the second variable. This way we obtain two integral representations of u:  x1  x2 u(x1 , x2 ) = D1 u(s, x2 ) ds and u(x1 , x2 ) = D2 u(x1 , s) ds. −∞

−∞

This yields  |u(x1 , x2 )| ≤

∞

−∞

 |Du(s, x2 )| ds

and

|u(x1, x2 )| ≤

∞ −∞

|Du(x1 , s)| ds.

Both sides of the above inequalities are nonnegative, thus we can multiply them side by side. This gives  |u(x1 , x2 )|2 ≤

∞

−∞

  |Du(s, x2 )| ds

∞ −∞

 |Du(x1 , s)| ds .

Next, we integrate both sides with respect to x1 : 

∞ −∞

 2

|u(x1 , x2 )| dx1 ≤ =

∞

−∞  ∞ −∞

 |Du(s, x2 )| ds

∞



∞

−∞ −∞

|Du(x1 , s)| ds dx1

|Du(s, x2 )| ds DuL1 () ,

2.3 Sobolev Spaces

71

and integrate again, this time with respect to x2 , obtaining: 

∞



∞

−∞ −∞

 |u(x1 , x2 )|2 dx1 dx2 ≤ DuL1 ()

∞



∞

−∞ −∞

|Du(s, x2 )| ds dx2 ,

which means that u2L2 () ≤ Du2L1 () . Taking the square root of both sides yields Sobolev’s Inequality. ♦

2.3.3 Problems Problem 2.3.9 Prove that for n > 1 the unbounded function   1 u(x) = log log 1 + 2 |x| belongs to W 1,n (B(0, 1)), where by B(0, 1) ⊂ Rn we denote the unit ball. Problem 2.3.10 Assume n > 1 and let B(0, 1) denote, as above, the unit ball in Rn . Find a function ) W 1,p (B(0, 1)), v∈ 1≤p≤n

that is discontinuous in every nonempty open subset of B(0, 1). Hint Use the function u from the previous problem and consider the series ∞ 

2−N u(x − aN )

N=1

where {aN }∞ N=1 is a sequence enumerating all these points in B(0, 1) which have all coordinates rational. Problem 2.3.11 Prove that if u ∈ W 1,p (0, 1) and p ≥ 1, then u has a representative which is absolutely continuous in (0, 1). Problem 2.3.12 Without referring directly to Morrey’s Inequality, prove that if u ∈ W 1,p (0, 1) and p > 1, then u satisfies |u(x) − u(y)| ≤ |x − y|

1− p1



1

p

|u |

 p1

0

for almost every x, y ∈ (0, 1); in particular, it has a representative which is a Hölder continuous function on (0, 1).

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2 Distributions, Sobolev Spaces and the Fourier Transform

Problem 2.3.13 Give an example of a function u ∈ H 1/2(R) which is not in L∞ (R). Problem 2.3.14 Assume  ⊂ Rn is connected and u ∈ W 1,p () satisfies ∇u = 0 a.e. in . Prove that u is constant a.e. in . Problem 2.3.15 Assume  ⊂ Rn is bounded. Prove that if u ∈ W 1,p (), then the functions |u|, u+ = max(u, 0), u− = min(u, 0) are in W 1,p (), as well. Problem 2.3.16 Prove that if u ∈ W 1,p (), then ∇u = 0 a.e. in the set {u = 0}. Problem 2.3.17 Assume  ⊂ Rn is bounded. Prove that the inequality 



 1 

|∇u| dx ≤ C 2



2

2

|D u| dx

u dx 

1 2

2

2



holds for any function u ∈ H 2 () ∩ H01 (). Hint Use Green’s First Identity (1.9). Problem 2.3.18 Assume  ⊂ Rn is a bounded open set with smooth boundary. T

Prove that if X = Lp (), then there is no continuous linear operator X − → Lp (∂) such that for any f continuous up to the boundary of  (i.e., for f ∈ C() ⊂ X) we have T (f ) = f |∂ . In other words, prove that one cannot define a continuous trace operator on Lp (). Problem 2.3.19 Prove Young’s inequality |ab| ≤

|a|p |b|q + , p q

where a, b ∈ R, p ≥ 1, q ≥ 1 and p−1 + q −1 = 1. Hint Use concavity of the logarithm function. Problem 2.3.20 Prove Hölder’s inequality: Assume 1 ≤ p, q ≤ ∞ and p−1 + q −1 = 1. If f ∈ Lp () and g ∈ Lq (), then the function fg is in L1 () and the following inequality holds: fgL1 () ≤ f Lp () gLq () . Problem 2.3.21 Assume f ∈ Lr () ∩ Ls () for some r, s ≥ 1. Prove the following interpolation inequality: f Lp ≤ f θLr () f 1−θ Ls ()

2.3 Sobolev Spaces

73

for all p, θ satisfying 0 ≤ θ ≤ 1 and θ 1−θ 1 = + . p r s Problem 2.3.22 Give an example of a function f ∈ L2 (Rn ) which is neither in L3 nor in L1 . Problem 2.3.23 Prove Poincaré’s Theorem 2.3.21 for (a) n = 2 and p ≥ 1, (b) n = 3 and p ≥ 1. Problem 2.3.24 Assume  ⊂ R2 is (a) a disk of radius d, (b) an ellipse with axes a and b. In both cases find the constant C from Poincaré’s Theorem 2.3.21. Does this constant depend on p? Problem 2.3.25 Assume  = R × (−1, 1). Does the Poincaré Inequality uL2 () ≤ CDuL2 () hold for every function u ∈ W01,2 ()? Problem 2.3.26 Let u be a smooth, compactly supported function on Rn . We define uλ = u(λx). Analyze the behavior of the norms uλ Lq (Rn ) and Duλ Lp (Rn ) with respect to λ. np Conclude that if q = n−p , then there is no constant C (dependent on p and n only) such that the inequality f Lq (Rn ) ≤ CDf Lp (Rn ) holds for every smooth, compactly supported function f on Rn . Problem 2.3.27 Consider the family of functions fk : R → R fk (x) =

⎧ ⎨

0 if |x| > k + 1, 1 if |x| < k, ⎩ k + 1 − |x| if k ≤ |x| ≤ k + 1.

Does the weak derivative Dfk of fk exist for n = 0, 1, 2, . . .? Does there exist a constant C and p, q ≥ 1 such that for all k ∈ {0, 1, 2, . . .} we have fk Lp (Rn ) ≤ CDfk Lq (Rn ) ?

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2 Distributions, Sobolev Spaces and the Fourier Transform

Problem 2.3.28 Prove Sobolev’s Inequality for n = 3 and p = 1. Hint Begin with integral representations (cf. worked-out Problem 2.3.8, this time we obtain three different representations of u, depending on D1 u, D2 u, and D3 u). Find an integral estimate of |u|3/2 by a triple product of roots of appropriate integrals. Finally, integrate both sides of the estimate three times, with respect to x1 , x2 , and x3 , using Hölder’s Inequality on the way (this is the main, and essentially the only difference between the cases of n = 2 and n = 3). Problem 2.3.29 Assuming that Sobolev’s Inequality holds for all n > 1 and p = 1, prove that it holds for all n > 1 i p > 1, as well. Hint Consider the function v = |u|γ , with γ suitably chosen. Problem 2.3.30 The function f is given by the formula 

x

f (x) =

g(s) ds, 0

for a g ∈ L2 (0, 1). Prove that f is Hölder continuous with exponent 12 . Problem 2.3.31 Prove that there exists a constant C (dependent on p only), such that the inequality  B(x,r)

|Du(z)| dz ≤ CDuLp (R2 ) , |z − x|

where B(x, r) ⊂ R2 denotes a ball centered at x and of radius r, holds for any p > 2, r > 0 and u ∈ W 1,p (R2 ). Problem 2.3.32 Assume u ∈ C 1 (R2 ) ∩ W 1,p (R2 ) for a p > 2. Consider two concentric disks: B(x, R), of radius R, and a smaller B(x, r), of radius r < R. Let x = (x1 , x2 ), h = (cos ϕ, sin ϕ) and y = x +rh = (x1 +r cos ϕ, x2 +r sin ϕ). Using the Fundamental Theorem of Calculus, prove that 

r

|u(x + rh) − u(x)| ≤

|Du(x + sh)| ds.

0

Deduce that 

 |u(y) − u(x)| dσ ≤ r ∂B(x,r)

B(x,r)

|Du(z)| dz |z − x|

and  |u(y) − u(x)| dy ≤ B(x,R)

R2 2

 B(x,R)

|Du(z)| dz. |z − x|

2.3 Sobolev Spaces

75

Problem 2.3.33 Using Problem 2.3.32 and the inequality     1 1 |u(y)| + |u(x) − u(y)| dy |u(x)| dy ≤ |u(x)| = πr 2 B(x,r) πr 2 B(x,r) prove that (under the assumptions of Problem 2.3.32) sup |u(x)| ≤ CuW 1,p (R2 ) .

x∈R2

Problem 2.3.34 Let x, y ∈ R2 and denote r = |x − y|. Using the final inequality from Problem 2.3.32 together with the inequality    1 |u(x) − u(y)| ≤ |u(x) − u(z)| + |u(z) − u(y)| dz πr 2 B(x,r)   1 1 ≤ |u(x) − u(z)| dz + |u(z) − u(y)| dz πr 2 B(x,r) πr 2 B(y,2r) prove that, under the assumptions of Problem 2.3.32, the following inequality holds sup |u(x) − u(y)| ≤ C|x − y|1−2/p uW 1,p (R2 ) .

x,y∈R2

Problem 2.3.35 Assume  ⊂ R2 is a bounded open set with smooth boundary. Using the inequalities derived in Problems 2.3.33–2.3.34, show that any function in 1,p W0 (), p > 2, can be modified on a set of measure zero in such a way that the resulting function is Hölder continuous on , with exponent α = 1 − p2 . Problem 2.3.36 Consider a sequence of functions fm ∈ L2 (0, π): fm (x) =

∞ 

am,n sin(nx),

where

n=1

∞ 

2 am,n < ∞.

n=1

Prove that if there exists a constant C > 0 such that for every m fm L2 (0,2π) < C, then the sequence {fm } has a subsequence {fmk } convergent in L2 (0, π). Problem 2.3.37 Assume  ⊂ Rn is an open, bounded domain with smooth boundary. Prove, using Rellich–Kondrachov’s Theorem, that there exists a constant C such that the inequality u − u Lp () ≤ CDuLp () , where u =



1 ||  u(x)dx,

holds for all functions u ∈ W 1,p ().

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2 Distributions, Sobolev Spaces and the Fourier Transform

Problem 2.3.38 Prove that under the assumptions of Problem 2.3.37 there exists a constant C, independent of r, such that u − uB(x,r)Lp (B(x,r)) ≤ CrDuLp (B(x,r)). Problem 2.3.39 Prove that if  is an open subset of Rn and u ∈ W m,2 () for all m = 1, 2, . . ., then u is, up to a modification on a set of measure zero, a C ∞ function. Hint Begin by showing that it suffices to solve the problem for compactly supported u. Then use Sobolev’s Inequality for subsequent derivatives of u, remembering that q=

np >p n−p

if 1 ≤ p < n.

For every multiindex α, after a finite number of such steps, one can apply Morrey’s Theorem to D α u, proving it is continuous. Problem 2.3.40 Let us suppose that  ⊂ RN is open and nonempty. Show that for all k ∈ N and 1 ≤ p ≤ ∞ we have dim W k,p () = ∞. Problem 2.3.41 Let us suppose that  ⊂ RN is open and bounded, p ∈ [1, ∞). Show that the function 1,p

W0 () $ u → ∇uLp (;RN ) 1,p

defines a norm on the space W0 (), which is equivalent to the standard one, i.e., there is Cp > 0 such that 1 ∇uLp (;RN ) ≤ uW 1,p () ≤ Cp ∇uLp (;RN ) Cp 1,p

for all u ∈ W0 ().

2.4 Bibliographical Remarks The reader interested in deeper study of the Fourier transform should refer to the excellent books of Rudin [20, 21], Strichartz [24], and Stein and Shakarchi [23]. Applications to partial differential equations are studied in detail in the books of Evans [8] and Taylor [26]. The topics in distribution theory covered in this chapter address only a fraction of the general theory, students interested in a more comprehensive approach should refer, e.g., to the books of Rudin [21], Strichartz [24], Blanchard and Brüning [4] or

2.4 Bibliographical Remarks

77

to a slightly forgotten textbook of Treves [28]. Applications of distribution theory to partial differential equations are covered, e.g., in volume I of Taylor’s book [26]. There is a lot of literature concerning Sobolev spaces. The basic properties are addressed in the book of Evans [8]. A more detailed approach can be found in the book of Evans and Gariepy [9] and in the excellent textbook of Leoni [15]. Probably the most comprehensive source is the book of Adams and Fournier [1].

Chapter 3

Common Methods

We will present here the theoretical background from functional analysis, useful when approaching different types of problems. Having done that, we will discuss the Galerkin method of finding weak solutions to linear PDE, which is based on constructing a sequence of approximating solutions. This plan raises a number of interrelated questions: 1. What is the notion of a solution to the original PDE, which is suitable for the limiting process? 2. What is the appropriate convergence? Section 2.3 suggests that the theory of Sobolev spaces will play a role, but the dimension of W k,p () is infinite. A standard course in linear algebra covers solving finite systems of linear equations. Likewise, an introductory course on ordinary differential equations deals with equations in RN . We will have to learn how to present W k,p () as a limit of finite dimensional spaces which are isomorphic to RN . We also have to devise a way to reduce a linear PDE to a linear system of algebraic equations or to a system of ordinary differential equations (ODEs for short), depending on the nature of the problem. Finally, we will learn how to pass to a limit with the sequence of approximate solutions. This is a way of obtaining a solution of the original problem in W k,p . The question of a proper notion of convergence is challenging. The need to address this question is the reason for developing the method of weak compactness. In separable Hilbert spaces we will see a result that looks familiar to everyone acquainted with the Bolzano–Weierstrass Theorem in RN : if a sequence {xn }∞ n=1 is bounded in H , then we can always extract a weakly convergent subsequence. This idea is carefully explained in Sect. 3.1. Let us stress that in a general Banach space, it is not true (contrary to the case RN ) that we can extract a convergent subsequence from a bounded one. This weak compactness is the basis of the Galerkin method, see Sect. 3.3, which is quite versatile. It is useful also when we discuss the variable separation method, © Springer Nature Switzerland AG 2019 M. Borodzik et al., Problems on Partial Differential Equations, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-14734-1_3

79

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Sect. 3.2, which is restricted to regions having the structure of a Cartesian product, e.g., a rectangle. Since dealing with a separable Hilbert space is of great advantage, we are able to present an elementary proof of the Banach–Alaoglu Theorem, see Theorem 3.1.8. This theorem is the basis of the present chapter. After discussing the issue of weak convergence, we may return to the first question we raised, about the notion of a solution. Having constructed u as a weak limit of approximations, we should not expect that u, our candidate for a solution, has all the classical derivatives stipulated by the equation. We should settle for less: for a solution, which has weak, but not necessarily classical derivatives and which is fewer times weakly differentiable than the equation stipulates. Thus, we are forced to introduce in this chapter the notion of a weak solution, which is studied in detail in the next chapters. At the end of this chapter, we also present an application of the Galerkin method to nonlinear problems. It turns out that the framework based on Hilbert spaces is not sufficient and we have to deal with more general Banach spaces. For this reason, we develop weak convergence in Banach spaces, which is done in the section devoted to the theoretical background, see Sect. 3.1.1. The paragraph on dual spaces of functionals facilitates the reading of section on the Lax–Milgram Lemma in Chap. 4, see Theorem 4.2.4. However, we suggest the reader to return to Sect. 3.1.1 after reading Chap. 4. The reader may also skip the part devoted to the weak convergence in a general Banach space in the first reading.

3.1 Weak Convergence 3.1.1 Theoretical Background Weak Convergence in Hilbert Spaces We present basic facts from functional analysis that are necessary to define the weak convergence in Hilbert spaces. From the point of view of linear PDEs developed here, it is sufficient to restrict our attention to separable Hilbert spaces. As a matter of fact nonseparable Hilbert spaces are useful in physics, but they are quite far from any mainstream introductory PDE course. However, when we consider a nonlinear problem in Sect. 3.3.2, then a need to consider a weak convergence in more general Banach spaces arises. We devote two paragraphs to this issue at the end of the present section. For the sake of simplicity, we will restrict our attention to Hilbert spaces over the real numbers. Of course, one might consider their complex numbers counterparts, but we will not do this here. The main examples, which we have in mind, are L2 (),

H01 (),

H m ().

3.1 Weak Convergence

81

We recall that we defined the Sobolev spaces, H m (), in Sect. 2.3. Sometimes the space l 2 = {{xn }∞ n=1 : xn ∈ R,

∞ 

|xn |2 < ∞}

n=0

turns out to be useful. It may be briefly defined as follows: l 2 := L2 (N, μ), where μ is a counting measure, i.e. μ({k}) = 1, k ∈ N. We begin with fundamental definitions. Definition 3.1.1 Let us suppose that H is a linear space over the field of real numbers equipped with a bilinear form ·, · : H × H → R. This form is called a scalar product provided that the following conditions hold: (i) (ii) (iii) (iv)

x, x = 0 if and only if x = 0; x, y = y, x for all x, y ∈ H ; λx, y = λ x, y for all x, y ∈ H and λ ∈ R; x + y, z = x, z + y, z for all x, y, z ∈ H .

Let us define the following function on H , x := x, x 1/2 ,

x ∈ H.

(3.1)

The inequality below is called the Cauchy–Schwarz inequality and it is relatively easy to prove, | x, y | ≤ xy,

x, y ∈ H.

(3.2)

Once we establish (3.2), then it is not difficult to see that  ·  defined above is a norm. Definition 3.1.2 We say that a linear space H , equipped with a norm given by formula (3.1), is a Hilbert space, provided that the normed space (H,  · ) is complete, i.e. every Cauchy sequence has a limit. Definition 3.1.3 Let us suppose that H is a Hilbert space. We say that a linear functional h∗ : H → R is bounded provided that h∗  := sup{(h∗ , h) : h ∈ H, h ≤ 1} is finite. The space of all bounded linear functionals is denoted by symbol H ∗ . We call it the dual of H .

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Remark 3.1.4 We note that the continuity of a linear mapping is equivalent to its boundedness, see Problem 3.1.8. We notice that h∗ , defined above, is a norm in H ∗ . Moreover, H ∗ with this norm is a Banach space, too. If H is a Hilbert space, then its dual space H ∗ may be identified by a linear isometry with H in a natural way. This is formulated in the theorem below (e.g., see [21]), its proof is the content of Problem 3.1.7. Theorem 3.1.5 (Riesz Functional Representation Theorem) Let us suppose that h∗ ∈ H ∗ is a continuous linear functional on a Hilbert space H . Then, there is an element h ∈ H , uniquely determined by h∗ such that (h∗ , g) = g, h

for all g ∈ H.

and h = h∗ . Here, h∗  is the norm of a linear mapping while h is a norm of a vector. Moreover, the mapping h∗ → h is a linear isometric isomorphism. In particular, h∗ , g ∗ H ∗ := h, g H defines an inner product in H ∗ . After these preparations, we may introduce the main definition of this section. Definition 3.1.6 We say that a sequence {hn }∞ n=1 ⊂ H converges weakly to h ∈ H if and only if hn , g → h, g

for all g ∈ H.

The weak convergence is usually denoted in the following way: hn  h

in H.

Obviously, the weak convergence follows from the strong (i.e., the norm) convergence, i.e. we have the implication, hn − h → 0

⇒

hn  h.

However, the converse does not hold, as shown by a simple example. Example 3.1.7 We consider the following sequence of functions from the Hilbert space L2 (R), ' un (x) =

1 for x ∈ [n, n + 1), 0 for x ∈ R \ [n, n + 1),

for n = 0, 1, . . .

We notice that un L2 (R) = 1 and un − um L2 (R) =

√ 2(1 − δmn ),

3.1 Weak Convergence

83

where δmn is the Kronecker delta. As a result, the sequence {un } does not contain a strongly convergent subsequence. On the other hand, for any function v ∈ L2 (R), we have  un , v =

 R

n+1

un (x)v(x)dx =

v(x)dx. n

Hence, if we use the Cauchy–Schwarz inequality (3.2), we obtain 

n+1

| un , v | ≤

1/2 → 0,

v 2 dx

n

because the integral R v 2 dx is finite. Thus, according to Definition 3.1.6, we see that the sequence un weakly tends to zero in L2 (R), i.e., un  0

in L2 (R).

Since the weak convergence is different from both the strong convergence and the pointwise convergence of functions, we must use this notion very carefully in nonlinear equations. It may happen that the sequence un converges weakly to u, while the sequence u2n need not converge to u2 , even if we assume that this sequence has a limit, see Problem 3.1.14. We should emphasize that the notion of weak convergence is defined for a broader class of Banach spaces, see [21] and Sect. 3.3.2 below. However, most of the time, the analysis in Hilbert spaces is sufficient. Below we present a result, which is the main motivation for the introduction of the weak convergence to the theory of PDEs. It is a simple version of the general Banach-Alaoglu Theorem, see Theorem 3.1.16. The proof of the latter exploits the properties of product topologies (see [21]). However, in the context of separable Hilbert spaces we can offer an elementary proof, which is given below. This result will make the presentation of the core of the Galerkin method complete, see the next section for details. However, in order to claim that the method we are presenting is successful, we have to show that u, the weak limit of approximate solutions un , is a solution in an appropriate sense. This issue is resolved with the help of the theory of distributions and Sobolev spaces. Namely, we define weak solutions, see next section on the Galerkin method for details, and u turns out to be a weak solution. Theorem 3.1.8 Let H be a separable Hilbert space and sequence {un }∞ n=0 ⊂ H be bounded, i.e., sup un  ≤ M.

(3.3)

n

Then, there exists u ∈ H and a subsequence {unk } such that unk  u

in H

and

u ≤ M.

We recall first the notion of a basis in a Hilbert space.

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3 Common Methods

Definition 3.1.9 We say that an orthonormal set of vectors {ek }∞ k=0 is a basis of a separable Hilbert space H if it is maximal with respect to the set inclusion. Remark 3.1.10 We recall that an orthonormal set of vectors {ek }∞ k=0 is a basis of a separable Hilbert space H , if and only if each vector v may be uniquely represented  k e . In this case a k = (v, e ), k ∈ N. as v = ∞ a k k k=0 Proof of Theorem 3.1.8 Step 0 Let {wk }k∈N be an orthonormal basis in H . Then, each element un is the sum of a series convergent in H un =

∞ 

ank wk ≡ lim

N→∞

k=0

N 

ank wk ,

k=0

where the sequence of coefficients {ank }k∈N is given by formula ank = un , wk ,

k ∈ N.

In other words, the upper index denotes the coordinate number in the orthonormal ∞ basis {wk }∞ k=0 . Since the set {wk }k=0 is an orthonormal basis, we deduce that for all n ∈ N sequences an = (an0 , an1 , . . . , ank , . . .)  k belong to space l 2 . Indeed, if v = ∞ k=0 a wk , then the condition wk , wl = δkl immediately implies the equality of the norms, v2 = lim  N→∞

= lim

N→∞

N 

N→∞

k=0 N 

N 

a k wk 2 = lim

|a k |2 =

k=0

∞ 

k,j =0

|a k |2

k=0

= a2l 2 . Hence, un  = 2

∞  k=0

|ank |2 .

a k wk , a j wj

3.1 Weak Convergence

85

Step 1 Since the sequence {un }∞ n=0 is bounded, from the Cauchy–Schwarz inequality (3.2) and (3.3), we obtain a uniform estimate of sequence {anm }n∈N of coefficients for a fixed m ∈ N, |anm | = | un , wm | ≤ M

for n = 1, 2, . . .

∞ Thus, we may choose a subsequence {unml }∞ l=0 ⊂ {un }n=0 , such that

vlm := unml

lim a mm l→∞ nl

and

= a∗m ∈ R.

∞ The upper index m in {nm l }l=0 indicates that this sequence was chosen for a given m. We have to make this choice carefully.

Step 2 (The Diagonal Method) We construct the desired subsequence {unk }∞ k=0 . We proceed inductively. For m = 0 we consider the selected subsequence, {vl0 }∞ l=0 . m+1 ∞ m = u m , we extract from it a subsequence {v } . Since v Next, given {vlm }∞ nl l l=0 l=0 l

m+1 m ∞ we select {nm+1 }∞ = nm m l l=0 an increasing subsequence from {nl }l=0 such that n0 and

lim a m+1 m+1 l→∞ nl

= a∗m+1 .

We claim that the subsequence vll = unl l

has the desired properties. By the above construction, we can see that for a given k sequence {vll }l>k is a subsequence of {vlk }l>k . Hence, for each k ankl → a∗k

l → ∞,

for

l

i.e., vll , wk = ankl → a∗k

for

l

l → ∞.

Step 3 (A Candidate for a Limit) In order to prove that the sequence we have selected has a weak limit, we must first check that our candidate u=

∞  k=0

a∗k wk

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3 Common Methods

is an element of H . Let us consider a partial sum uN =

N 

a∗k wk for N ∈ N. We

k=0

notice that uN belongs to HN , a finite dimensional subspace of H , given by HN = span {w0 , . . . , wN } and (see also the proof of (3.15)) uN  = sup{ uN , φ : φ ∈ H, φ ≤ 1}. Moreover, we have sup uN , φ = sup uN , PN φ = sup lim vll , PN φ ,

φ≤1

φ≤1 l→∞

φ≤1

where PN denotes the orthogonal projection onto HN . We leave the justification of the last equality to the reader. Therefore,   N  k 2 1/2 a  = uN  ≤ sup lim vll PN φ ≤ M. ∗ φ≤1 l→∞

k=0

We conclude that the series

N 

|a∗k |2 converges. As a result,

k=0

u∈H

and u ≤ M.

Step 4 (Weak Convergence) We must show that u is a weak limit of {vll }. We assume that h ∈ H and ε > 0. Then, vll − u, h = vll − u, PN h + vll − u, (I d − PN )h . Let us choose N so large that (I d − PN )h ≤

ε . 4M

This yields the following estimate of the second term: | vll − u, (I d − PN )h | ≤ 2M

ε ε = . 4M 2

After having fixed N, we may take l0 so large that for each l > l0 the inequality (alm − a∗m )2 ≤

ε2 4(N + 1)h2

3.1 Weak Convergence

87

holds for all m = 0, 1, . . . , N. Then, | vll − u, PN h | = | PN (vll − u), h | ≤ PN (vll − u)h +1/2 *N  m m 2 = (al − a∗ ) h k=0

 ≤ (N + 1)

ε2 4(N + 1)h2

1/2 h = ε/2.

Thus, we showed that for any ε > 0 lim | vll − u, h | ≤ ε.

l→∞

Since ε is arbitrary, we conclude that lim | vll − u, h | ≤ 0,

l→∞

i.e.,

unl  u. l

& %

Theorem 3.1.8 has been proved.

Dual Spaces The weak convergence in Hilbert spaces is probably easiest to explain in the setting of 2 , the space of square summable sequences. Let us suppose that a sequence 2 2 {xn }∞ n=1 ⊂  converges weakly to x∞ , i.e., for all vectors e ∈  we have the limit, lim xn , e = x∞ , e .

n→∞

(3.4)

Thus, if we define the vector ek componentwise, as eki = δik (1 at the i-th coordinate, otherwise zero), applying (3.4) to ek in place of e yields lim x k n→∞ n

k = x∞ ,

for all natural k. In other words, the weak convergence of a sequence implies the convergence of each of its coordinates. We check in Problem 3.1.9 that in case of the Hilbert space 2 it is sufficient to take ek in the definition of the weak convergence. That is, the weak convergence is equivalent to coordinate-wise convergence. If we introduce an orthonormal basis in a separable Hilbert space H , then we can identify H with 2 . Hence, the same statement is true for H , see Problem 3.1.1.

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However, such a statement as above is far from obvious in a general Banach space. A special, but important example are the Lebesgue spaces Lp (), where  is an open subset of RN and p ∈ [1, ∞]. Considering them is stipulated by nonlinear problems. Let us present now another aspect of the weak convergence. Let us observe that if H is a real Hilbert space and e is any of its elements, then H $ v → e, v ∈ R is a linear continuous mapping. Let us consider a specific example of H , the Sobolev space H01 (0, 1), i.e., the Sobolev space over the interval (0, 1), then from the Morrey Theorem, see Theorem 2.3.27, we know that every element v of this space has a continuous representative. As a result, if we fix x0 ∈ (0, 1), then H01 (0, 1) $ v → v(x0 ) =: (φx0 , v) ∈ R

(3.5)

is a linear functional. We will check that it is continuous. Indeed, if vn is a sequence of elements of H01 (0, 1) converging to v in the norm of H01 (0, 1), then |vn (x0 ) − v(x0 )| ≤ max |vn (x) − v(x)| = vn − vC([0,1]) x∈[0,1]

≤ Cvn − vH 1 (0,1). 0

At the first glance, the functional φx0 does not look like an inner product with an element of H01 (0, 1). The Riesz Representation Theorem, Theorem 3.1.5, tells us that φx0 must be given by an inner product. In other words, the functional φx0 we defined above is given by an inner product with g, an element of H01 (0, 1). We will find it. Element g must satisfy 

1

v(x0 ) = 0



1

vg dx +

v g dx,

g(0) = 0 = g(1).

(3.6)

0

In the right-hand side, we recognize the weak form of the operator g − gxx , (see Remark 3.1.18 and also Sect. 4.2 and Theorem 4.2.4). Thus, Eq. (3.6) becomes g − gxx = φx0 . Since φx0 is a continuous functional, then the Lax-Milgram Lemma, see Theorem 4.2.4, guarantees existence of a weak solutions such that φx0  = gH 1 (0,1). 0 We shall see in Problem 4.2.16 that the assumptions of Lax-Milgram Lemma are indeed satisfied here. This example shows that functionals over a Hilbert space H may not look as expected. This problem is even more pronounced for general Banach spaces. Thus, we introduce the linear space of linear continuous functionals over a Banach space

3.1 Weak Convergence

89

X, it is denoted by X∗ . It is called the dual space of X. This space is equipped with the operator norm, which takes the following form: ϕX∗ := sup{|(ϕ, v)| : v ∈ X, v ≤ 1}.

(3.7)

In fact X∗ is a Banach space too, see Problem 3.1.23. Let us present a few examples of dual spaces. Let us take first X = Lp () and p ∈ [1, ∞). We consider any element g ∈ Lq (), where q ∈ (1, ∞] and p1 + q1 = 1, where q = ∞ if p = 1. We claim that the linear functional φg : X → R, given by  (φg , v) =

v(x)g(x) dx, 

is continuous. By linearity of φg , it is sufficient to check that φg is bounded, see Problem 3.1.8. By the Hölder inequality, we have |(φg , v)| ≤ vLp gLq . The converse statement is also true. Theorem 3.1.11 Let us suppose that X = Lp (), where p ∈ [1, ∞). Then, (a) X∗ = Lq (), where p ∈ (1, ∞) and (b) (L1 ())∗ = L∞ ().

1 p

+

1 q

= 1;

Part (b) of the theorem above shows that spaces X and X∗ may be quite different. Indeed, the space L1 () is separable, while L∞ () is not, see Problem 3.1.24. Interestingly, the space of continuous functions on Rn vanishing at infinity is dense in L1 (Rn ), while it is a proper closed subspace of L∞ (Rn ). Since we introduced the dual space, we could ask a simple question: what is the dual of the dual space (called sometimes the bi-dual space)? In other words, if we know X and X∗ , we can find X∗∗ := (X∗ )∗ . Sometimes the answer is well-known, as in the case of Lebesgue spaces Lp () with p ∈ (1, ∞). Namely, see [5, Section 4.3], (Lp ())∗∗ = (Lq ())∗ = Lp (), where p1 + q1 = 1. However, computing the bi-dual of L1 () is more difficult. If g is in L1 (), then the following mapping L∞ () $ u →

 u(x)g(x) dx 

is a continuous functional on L∞ (). However, it turns out that the space (L∞ ())∗ is much bigger than L1 ().

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In order to clarify the connection between X and X∗∗ we introduce a canonical and natural embedding of X into X∗∗ , κ : X → X∗∗ . If v ∈ X, then κ(v) is a linear functional acting on X∗ , defined by (κ(v), x) = (v, x).

(3.8)

The operator κ is used in the next definition. Definition 3.1.12 We say that a Banach space is reflexive if κ(X) = X∗∗ We showed above that (Lp ())∗∗ , p ∈ (1, ∞) is isometrically isomorphic to In fact, a stronger statement is true: this isometric isomorphism is given by the above embedding κ. Lp ().

Theorem 3.1.13 The Lebesgue spaces Lp () are reflexive for p ∈ (1, ∞). Weak and Weak∗ Convergence in Banach Spaces Here, we show further usefulness of the notion of the dual space to define weak convergence in Banach spaces. Here is the definition: Definition 3.1.14 We say that a sequence {xn }∞ n=1 ⊂ X, where X is a Banach space, converges weakly to x0 ∈ X if for every functional φ ∈ X∗ we have lim (φ, xn ) = (φ, x0 ).

n→∞

(3.9)

In order to explain the purpose of introducing this notion we recall the classical Bolzano–Weierstrass Theorem, stating that we may extract a convergent subsequence from a bounded sequence in RN . This theorem fails in infinite-dimensional Banach spaces. In order to ameliorate this situation, we introduce the notion of weak convergence, see Theorem 3.1.16. However, we have to find the right topological framework. This is why we rewrite (3.9) from a more general point of view. We may say that for any neighborhood of φ(x0 ) ≡ (φ, x0 ), U , which might be a ball with radius ε, i.e. U = (φ(x0 ) − ε, φ(x0 ) + ε) almost all elements xn belong to the pre-image of U , φ −1 (U ) =: V (φ, ε, x0 ), (i.e. there is N ∈ N so that for k > N we have xk ∈ φ −1 (U )). Since we assumed that φ is a continuous functional, we infer that V (φ, ε, x0 ) ≡ {x ∈ X : |φ(x − x0 )| < ε}

(3.10)

is an open subset of X. However, in general, the collection V = {V (φ, ε, x0 ) ⊂ X : φ ∈ X∗ , ε > 0, x0 ∈ X} does not exhaust all open sets in the norm topology of X. The sets V (φ, ε, x0 ) for varying φ, ε, and x0 play the role of the open balls in the norm topology, where all open sets are sums of open balls they contain. We express this property of the family V by saying that V is a basis of new topology, see

3.1 Weak Convergence

91

Problem 3.1.26. In general, the new topology contains fewer sets than the original norm topology. This is why it is called a weak topology, see Problem 3.1.27. Actually, in order to proceed, we need to generalize the newly introduced notion of the weak topology. Let us suppose that  ⊂ X∗ is any collection of continuous functionals, possibly  is smaller than X∗ . In this case the family {V (φ, ε, x0 ) ⊂ X : φ ∈ , ε > 0, x0 ∈ X} is a basis of a new topology, denoted by σ (X, ), which may be even weaker than the weak one. In these terms, the weak topology is σ (X, X∗ ). Interestingly, if the considered Banach space X happens to be a dual space, X = Y ∗ , then besides  = X∗ = Y ∗∗ there is another natural example of , that is  = κ(X). This choice of  leads to a topology σ (X∗ , X), which is called the weak∗ -topology. If X is not reflexive, then we obtain essentially a new topology σ (X∗ , X∗∗ ) = σ (X∗ , X) and a new notion of weak∗ convergence: ∗ ∗ Definition 3.1.15 We say that a sequence {φn }∞ n=1 ⊂ X , where X is a dual of a ∗ ∗ Banach space, converges weakly to φ0 ∈ X if for every element x ∈ X we have

lim (φn , x) = (φ0 , x).

n→∞

(3.11)

Here comes the main result of this theory, which generalizes Theorem 3.1.8 to the Banach space setting, see [21] for a proof. Theorem 3.1.16 (Banach, Alaoglu) If B is a closed ball in X∗ , then it is compact in the σ (X∗ , X)-topology. At this moment, we would like to recall that a subset K of a general topological space X is called compact, if we can extract a finite subcovering from an open covering of K. If X happens to be a metric space, then compactness of K is equivalent to the ability of selecting a convergence subsequence from any sequence of points from K. Such equivalence is not true in general. However, when X is separable, like Lp () for 1 ≤ p < ∞, then the following result holds: see [21, Theorem 3.16]. ∗ Theorem 3.1.17 Let us suppose that X is separable. If the sequence {xn }∞ n=0 ⊂ X ∞ ∈ X ∗ , such that {x }∞ is bounded, then there is a subsequence {xnk }∞ and x nk k=0 k=0 converges to x ∞ in the weak∗ topology.

Banach Space-Valued Functions We consider here and in Chap. 5 functions defined over the real line with values in a Banach space X. A simple case is C([0, T ]; X) = {f : [0, T ] → X : f is continuous},

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3 Common Methods

(continuity may be understood according to the usual ε − δ definition). We equip C([0, T ]; X) with a natural norm, f  = sup{f (t)X : t ∈ [0, T ]},

f ∈ C([0, T ]; X).

The space C([0, T ]; X) with this norm is a Banach space, see Problem 3.1.37. Occasionally, we need to know when a function f : [0, T ] → X is integrable, where X is a Banach space. The key issue is that f has to be strongly measurable, i.e., it should be a limit of simple functions, lim sk (t) − f (t)X = 0

k→∞

where sk (t) =

Nk 

for a.e. t ∈ [0, T ],

χE k (t)uki , Eik ⊂ [0, T ] are measurable and uki ∈ X. The Pettis i

i=1

Theorem characterizes this type of measurability. Finally, the Bochner Theorem explains, when such an f is integrable. More details are to be found [8, Appendix E.5]. The proofs of both theorems can be found in [30, Chapter V, Sections 4–5]. After settling the measurability issue, the definition of Lp (0, T ; X) is natural. If p ∈ [1, ∞), then Lp (0, T ; X) = {f : [0, T ] → X : f is strongly measurable,

 T 0

p

f (s)X ds < ∞}

and L∞ (0, T ; X) = {f : [0, T ] → X : f

is strongly measurable, ess supt ∈(0,T ) f (s)X < ∞}.

We also know that L2 (0, T ; H ) is a Hilbert space with the natural inner product,  f, g L2 (0,T ;H ) =

T

f (s), g(s) H ds.

0

Hence, we note that (L2 (0, T ; H ))∗ = L2 (0, T ; H ), due to the Riesz Representation Theorem (see Theorem 3.1.5). The norms in Lp (0, T ; X) are defined naturally, f Lp (0,T ;X) =

* T 0

+1/p p f (s)X ds

, 1 ≤ p < ∞, f L∞ (0,T ;X) = ess

supt∈(0,T ) f (s)X .

3.1 Weak Convergence

93

3.1.2 Worked-Out Problems Problems related to the theory of Hilbert spaces appear also in Sect. 4.2. Problem 3.1.1 Let us suppose that {ek }∞ k=1 is an orthonormal basis of a Hilbert space H (see Remark 3.1.10). Show that a sequence vn converges weakly to v if and only if lim vn , ek = v, ek

for all k = 1, 2, . . . .

n→∞

Solution We have to show the following implication. If {vn }∞ n=0 ⊂ H is a sequence such that there is v ∈ H such that for all k ∈ N we have lim vn , ek = v, ek ,

n→∞

then this implies that for all e ∈ H we have lim vn , e = v, e .

(3.12)

n→∞

First of all, we notice (please complete the details) that there is a constant M > 0 such that for all n ∈ N we have vn  ≤ M.

(3.13)

Using the given basis of H , we can write e as e=

∞ 

a k ek .

k=1

For a given ε > 0, we can find N ∈ N, such that &2 & ∞ & &  ε2 & & . a k ek & < & & & 16(v2 + M 2 ) k=N+1

Then, we estimate using the triangle inequality, | vn , e − v, e | ≤ | vn − v,

N  k=1

≤ | vn − v,

N 

ak ek | + | vn − v,

∞ 

ak ek |

k=N+1

ε ak ek | + vn − v

. 4 v2 + M 2 k=1

94

3 Common Methods

Estimate (3.13) yields ε ε ε ≤ (vn  + v)

≤ . vn − v

2 2 2 2 2 4 v + M 4 v + M We can also find N1 such that for n > N1 we have | vn − v, ek | ≤

ε . 2eN

Combining these inequalities, we gather | vn , e − v, e | ≤

N  ε ε|ak | + ≤ ε. 2eN 2 k=1

Thus (3.12) follows. ♦ Problem 3.1.2 Let us suppose that H is a Hilbert space and the sequence {un }∞ n=0 ⊂ H converges weakly to u. Show that if lim sup un  ≤ u, n→∞

then un − u goes to 0 as n → ∞. Solution We notice that 0 ≤ un − u2 = un 2 − 2 un , u + u2 . Thus, lim sup un − u2 = u2 − 2 lim un , u + lim sup un 2 ≤ u2 − 2u2 + u2 = 0. n→∞

n→∞

n→∞

Here, we used that un converges weakly to u. ♦ We also present a similar example in the case of a Banach space, see below. Problem 3.1.3 Let us suppose that a sequence {xn }∞ n=1 of elements of a Banach space Xconverges weakly to x0 ∈ X. Show that lim inf xn  ≥ x0 . n→∞

Solution Let us take ϕ an element of X∗ such that ϕ ≤ 1 and x0  = (ϕ, x0 ). Its existence is guaranteed by the Hahn–Banach Theorem, see [21]. Then, x0  = lim (ϕ, xn ) ≤ lim inf xn  n→∞

as desired. ♦

n→∞

3.1 Weak Convergence

95

In the following problem we assume the existence of an object called a weak solution to a differential equation. The point is to find an estimate on the solution in terms of the problem data. Problem 3.1.4 Let us suppose that u ∈ H01 (0, 1) is a unique weak solution, see the remark below, to u − uxx = φx0 , x ∈ (0, 1), u=0 x = 0 and x = 1, where φx0 is given by (3.5). Check that uH 1 = φx0 , where φx0  is the norm in 0

the space dual to H01 .

Remark 3.1.18 We say that u ∈ H01 (0, 1) is a weak solution to the above problem if and only if for all functions ϕ ∈ H01 (0, 1) the following identity holds, see also Problem 4.2.16, 

1 0

(uϕ + u ϕ ) dx = (φx0 , ϕ).

(3.14)

We will talk about weak solutions of various equations in the next chapters. Solution We notice that vH = sup{ v, ϕ : ϕ ∈ H, ϕH ≤ 1}

(3.15)

for any Hilbert space H . Indeed, by the Cauchy–Schwarz inequality we have | v, ϕ | ≤ vϕ ≤ v, when ϕ ≤ 1. However, we can take ϕ = v/v, then v, ϕ = v. Hence, our claim follows. Thus, if we take supremum in (3.14), then due to the norm definition in H01 , the left-hand side becomes just uH 1 . On the right-hand side we have just the definition 0

of the norm of a functional over H01 . ♦

3.1.3 Problems Problem 3.1.5 Prove that the norm in any Hilbert space H satisfies the parallelogram law: x + y2 + x − y2 = 2x2 + 2y2

for all x, y ∈ H.

(3.16)

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3 Common Methods

Problem 3.1.6 Prove the converse to Problem 3.1.5. Suppose V , || · || is a Banach space and for all x, y ∈ V the parallelogram law (3.16) is satisfied. Prove that there exists a scalar product x, y → x, y on V such that for any x ∈ V we have ||x||2 = x, x . Problem 3.1.7 Prove the Riesz Representation Theorem: Assume H is a Hilbert space with scalar product · , · . For every continuous linear functional L on H there exists a unique v ∈ H such that for all u ∈ H L(u) = u , v . Problem 3.1.8 Let us suppose that (X,  · X ) and (Y,  · Y ) are normed linear spaces and a mapping L : X → Y is linear. Show that the following statements are equivalent: (a) L is bounded; (b) L is continuous; (c) L is continuous at x = 0. Problem 3.1.9 Show that in the case of the space H = 2 the weak convergence is equivalent to the coordinate-wise convergence. In other words, please reprove worked-out Problem 3.1.1 in the particular situation, when H = 2 with ek given below ' 1 if i = k, eki = 0 otherwise. Problem 3.1.10 Let un : (0, 1) → R

and un (x) = sin nx.

Prove that un  0 in L2 (0, 1) 2 and that the sequence {sin2 nx}∞ n=1 does not converge to zero in L (0, 1) (as n → ∞).

Problem 3.1.11 Let {un }∞ n=0 ⊂ H , where H is a Hilbert space. We assume that un  u in H and un  → u. Show that un → u strongly in H , i.e., un − u → 0. 2 Problem 3.1.12 Let us suppose that {un }∞ n=0 ⊂ L (0, 1) converges weakly to zero. ∞ Does this imply that {un }n=0 converge to 0 a.e.?

3.1 Weak Convergence

97

2 Problem 3.1.13 Let us suppose that {un }∞ n=0 ⊂ L (0, 1) converges strongly to ∞ zero. Does the sequence {un }n=0 converge pointwise to 0 a.e.? Shall the answer change if we limit ourselves to a subsequence?

Hint Consider u2m +i (x) = χ(i/2m ,(i+1)/2m) (x) for i = 0, . . . , 2m−1 and m ∈ N. Problem 3.1.14 Let us suppose that un ∈ H 1 (0, 1). We know that un  u in L2 (0, 1),

hence

sup un  ≤ M. n

Is it true that u2n  u2 in L2 (0, 1)? Hint It is easier to consider u = 0. Problem 3.1.15 We know that bn  b

in

L2 (0, 1) and sup un H 1 (0,1) ≤ M. n

Calculate 

1

lim

n→∞ 0

bn un dx.

Problem 3.1.16 Let P be a polynomial, un ∈ W 1,1 (0, 1) and sup un W 1,1 (0,1) ≤ M. n

Does sequence {P (un )}n∈N have a weakly convergent subsequence in L2 (0, 1)? Problem 3.1.17 Let us suppose that H is a Hilbert space and a sequence {un }∞ n=0 ⊂ H converges weakly to u. Show that lim inf un  ≥ u. n→∞

Problem 3.1.18 Set H = H01 (). Let f ∈ L2 (), g ∈ L2 (, Rn ). Show that  : H → R given by  (v) =

v(x)f (x) dx − 

is an element of H ∗ .

n   i=1

gi (x) 

∂v(x) dx, ∂xi

v∈H

98

3 Common Methods

Problem 3.1.19 Set X = W 1,p (), p ∈ (1, ∞). Suppose that f ∈ Lq (), g ∈ Lq (, Rn ), p1 + q1 = 1. Set  (v) =

v(x)f (x) dx − 

n   i=1

gi (x) 

∂v(x) dx, ∂xi

v ∈ X.

Show that  ∈ X∗ . 1,p

Problem 3.1.20 Set X = W0 (),  is a region in Rn , p ∈ (n, ∞). Suppose that f ∈ Lr (), where 1r + n1 − p1 = 1 and g ∈ Lq (, Rn ), p1 + q1 = 1. Set  (v) =

v(x)f (x) dx − 

n   i=1

gi (x) 

∂v(x) dx, ∂xi

v ∈ X.

Show that  ∈ X∗ . Comment on the difference between this one and the previous problem. Problem 3.1.21 Let us suppose that g ∈ L2 (RN−1 ). If v ∈ H 1 (RN ), then we define  (φg , v) = g(x )v(x , 0) dx . RN−1

Show that φg is a well-defined continuous functional over H 1 (RN ). Problem 3.1.22 Let  = B(0, 1) ⊂ RN and D = {x ∈  : xN = 0}. We set H := H 1,2() and  (v) = v(x , 0) dx , D

where x = (x , xN ). Show that  ∈ H ∗ . Problem 3.1.23 Check that X∗ with the norm defined by (3.7) is indeed a Banach space, provided that X is a normed space. Problem 3.1.24 Show that L∞ () is not separable. Hint Notice that if E1 , E2 ⊂  are measurable such that |E1 |, |E2 |, |E1 E2 | > 0, then χE1 − χE2 L∞ = 1. Problem 3.1.25 Check that κ defined in (3.8) is a linear, continuous immersion, i.e., the kernel of κ is trivial.

3.1 Weak Convergence

99

Problem 3.1.26 Let us suppose that sets V (φi , ε, x0 ) are as defined in (3.10). Show that the family of intersections N )

N ∈ N, φi ∈ X∗ , ε > 0, x0 ∈ X,

V (φi , ε, x0 ),

i=1

is the basis of the new topology in X, i.e. every open set in the weak topology in X is a sum of a family of such intersections. Problem 3.1.27 Give an example of a Banach space and an open set, which is not open in the weak topology. Problem 3.1.28 Let us suppose that a sequence {xn }∞ n=1 of elements of a Banach space converges weakly to x0 ∈ X. If lim sup xn  ≤ x0 , n→∞

then xn − x0  → 0 as n → ∞. 1,p (), where  ⊂ Problem 3.1.29 Let us suppose that a sequence {un }∞ n=0 ⊂ W Rd is open, converges weakly to u0 and A ∈ L∞ (, M(n × n)), where M(n × n) is the linear space of n × n matrices. Then,

A∇un  A∇u

in Lp .

1,p (), where  ⊂ Problem 3.1.30 Let us suppose that a sequence {un }∞ n=0 ⊂ W d R is open and bounded with a smooth boundary, converges weakly to u0 . For which q ∈ [1, ∞) this sequence converges to u0 in the Lq -norm? 1,p (), where  ⊂ Problem 3.1.31 Let us suppose that a sequence {un }∞ n=0 ⊂ W d ∞ ∞ R is open, converges weakly to u0 . Moreover, {An }n=0 ⊂ L (, M(d × d)) is a sequence of matrices converging to A in the L∞ (, M(d × d)) norm. Show that

An ∇un  A∇u

in Lp .

Problem 3.1.32 Let us suppose that  ⊂ Rd is open and bounded with a smooth boundary and p ∈ (1, ∞). We assume that function F : W 1,p () → R, is given by the following formula:  F (u) = f (∇u) dx, 

where f is convex and 0 ≤ f (ξ ) ≤ C(|ξ |2 + 1) and C > 0. Show that if un  u0 in W 1,p (), then lim inf F (un ) ≥ F (u0 ). n→∞

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3 Common Methods

Hint If f is convex, then f is a supremum of a family of linear functions. Problem 3.1.33 Let us suppose that ⊂ Rd is open and bounded with a smooth   ∗ 1 −1,2 boundary and f ∈ W () := H0 () . We define F : H01 () → R by the formula,  1 F (u) = |∇u|2 + f, u L2 . 2  Show that if un  u0 , then lim inf F (un ) ≥ F (u0 ). n→∞

Problem 3.1.34 Let us suppose that  and F are as above. We consider a sequence 1 {un }∞ n=0 ⊂ H0 () such that lim F (un ) = inf{F (v) : v ∈ H01 }.

n→∞

(a) Show that there exists M > 0 such that for all n ∈ N we have un H 1 ≤ M. 0

(b) Show that there is u0 ∈ H01 , such that un  u0 and F (u0 ) = inf{F (v) : v ∈ H01 }. (c) Show that u0 found in part (b) is a weak solution to the problem −u + f = 0 in , u=0 on ∂. −1,2 Problem 3.1.35 Let us suppose that  is as in Problem 3.1.33, {fn }∞ n=0 ⊂ W 1 ∗ converges weakly to f and un ∈ H0 () are weak solutions to

−un + fn = 0 in , on ∂. un = 0 ∗ Assume moreover that {un }∞ n=0 converges weakly to u. Show that

−u + f = 0 in , u=0 on ∂.

3.2 The Separation of Variables Method

101

−1,2 Problem 3.1.36 Let us suppose that  is as in Problem 3.1.33, {fn }∞ n=0 ⊂ W converges weakly∗ to f and un ∈ H01 () are weak solutions to

−un + fn = 0 in , un = 0 on ∂. ∗ Show that {un }∞ n=0 converges weakly to u, which is a weak solution to

−u + f = 0 in , u=0 on ∂. Problem 3.1.37 Show that if X is a Banach space so are the spaces C([0, T ]; X), Lp ([0, T ]; X), where p ≥ 1 or p = ∞.

3.2 The Separation of Variables Method 3.2.1 Theoretical Background In this section we present the variable separation method, which is also called the Fourier method. We will use these two names interchangeably. It may be used to show solutions to specific PDEs by means of a closed formula. We shall focus our attention on hyperbolic and parabolic problems set up on a bounded interval [0, l] ⊂ R. However, this method may be also applied to elliptic problems on rectangles or even balls. Our prerequisite is the knowledge of the notion of the Fourier series and of the ordinary differential equations. These two theories meet when we study the eigenvalue problem. The simplest one considered here has the form, d 2u dx 2

= λu, for x ∈ (0, l),

u(x) = 0, for x ∈ {0, l}. The set of solutions is formed by the trigonometric functions, Xk (x) = sin k( πl x), 2 k = 1, 2, . . .. We know from calculus, see [20], that every ∞ function u from L (0, l) can be represented as a Fourier series, i.e. u(x) = k=1 ak Xk (x). However, the above problem is a special case of the following Sturm–Liouville problem, d du dx (p dx ) −

qu = λu, for x ∈ (0, l),

u(x) = 0,

for x ∈ {0, l},

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3 Common Methods

where p, q > 0 and p ∈ C 1 ([0, l]), q ∈ C([0, l]), λ ∈ R. The solutions to this problem need not be given explicitly, hence we do not pay much attention to them. Nonetheless, the eigenfunctions form an orthonormal basis of L2 (0, l). This can be deduced by invoking the Hilbert–Schmidt Theorem, see [22, Section 3.2]. The above boundary value problems should be considered as one-dimensional elliptic equations. To learn more about elliptic problem, please see Chap. 4. Here, we also face the problem of interpreting the formulas that we come up with. If we make strong assumptions on the data, then it is likely that the formulas will give us sufficiently regular functions which may be inserted into the equation and make it hold pointwise. However, we shall frequently see that our formulas give a well-defined object, which does not enjoy regularity stipulated by the equations, but belong to a Sobolev space. In this way we will encounter weak solutions in this chapter. We will discuss it in detail in the due time.

Hyperbolic Problems Let us start with the equation of a vibrating string with length l and fixed at its ends: ut t = uxx , x ∈ (0, l), t > 0, u(0, t) = u(l, t) = 0, t ≥ 0, u(x, 0) = u0 (x), x ∈ (0, l), ut (x, 0) = u1 (x), x ∈ (0, l).

(3.17)

We assume that the initial conditions u0 and u1 belong to the space L2 (0, l). In order to solve the above problem, we shall use the Fourier method, known also as the method of separation of variables. It means that we look for a solution in a form of u(x, t) = X(x)T (t).

(3.18)

After inserting this expression into Eq. (3.17) and dividing both sides by X(x)T (t), we get X

(x) T

(t) = . X(x) T (t)

(3.19)

Since the left-hand side of the above equation depends only on variable x, and the right-hand side only on variable t, then both sides must be equal to the same constant. Let us call it −λ. Thus, we get − X

(x) = λX(x) X(0) = X(l) = 0.

for x ∈ (0, l),

(3.20)

3.2 The Separation of Variables Method

103

The boundary conditions for the function X result from the required form of the solution, see (3.18), and from boundary conditions for the function u set in Eq. (3.17). Solutions to (3.20) are the eigenvalues λn =

n2 π 2 , l2

n = 1, 2, 3, . . .

and the corresponding eigenfunctions Xn (x) = sin

 nπx  . l

The task of finding eigenvalues and eigenfunctions belongs to the field of ordinary differential equations. We suggest to consider separately λ negative, λ = 0 and λ positive. In the particular example we study, the boundary condition will be satisfied only in the latter case. For each n we find functions Tn (t) satisfying the equation Tn

(t) = −λn Tn (t). Hence,



Tn (t) = An sin( λn t) + Bn cos( λn t). Let us note that for each n = 1, 2, . . . the function Xn (x)Tn (t) satisfies Eq. (3.17) and the expected boundary conditions. Nevertheless, in order to find a solution satisfying the initial conditions, we usually have to consider the whole series u(x, t) =

∞       

 An sin λn t + Bn cos λn t sin λn x .

(3.21)

n=1

It remains to find constants An and Bn . The function u must satisfy the conditions u(x, 0) =

∞ 

Bn sin



 λn x = u0 (x)

n=1

and ut (x, 0) =

∞ 

An λn sin



 λn x = u1 (x).

n=1

We should check that the above expansions with respect to sine functions are correct.

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3 Common Methods

We remind that each function belonging to space L2 (−l, l) may be expanded in a Fourier series, i.e. with respect to functions '

( 

 1  1 1 

, sin λn x , cos λn x , 2l l l n=1,1,2,...

x ∈ (−l, l),

which form an orthonormal basis of L2 (−l, l). This is a well-known fact, see [20]. Our data, u0 , u1 , are defined on (0, l) and we may extend them to (−l, l) by reflection to an odd or even function, depending upon the boundary conditions we consider. For example, if u0 is continuous and vanishes at x = 0, then the odd reflection yields a continuous function. If we deal with u0 ∈ C 1 and satisfying the 1 0 condition du dx (0) = 0, then the even extension yields a C function. No matter what kind of extension we consider, if u0 , u1 ∈ L2 (0, l), then their extensions are in L2 (−l, l). In the case of our problem, we extend the data u0 , u1 to odd functions, denoted by u˜ 0 , u˜ 1 . When we compute the coefficients of expansions of u˜ 0 and u˜ 1 in a Fourier series, we notice that the following integrals vanish 

l

−l



 u˜ i cos λn x dx = 0,

n = 1, 2, . . . ,

i = 0, 1.

As a result, we expand u0 and u1 in the sine series, i.e. the series with respect to functions {Xn }n=1,2,... . Thus, coefficients Bn and An , n = 1, 2, . . . , are given by the formulas below 1 Bn = l



l

u0 (x)Xn (x) dx

(3.22)

0

and 1 An = √ λn l



l

u1 (x)Xn (x) dx.

(3.23)

0

As a result, we found a formula for solving problem (3.19). It remains to decide the type of convergence of series (3.21). Certainly, if u0 , u1 ∈ L2 (0, l), the u given by (3.21) converges in the L2 norm. But we would prefer to know that u is either a classical or a weak solution. In other words, is it possible to differentiate the series term by term? In order to make u a classical solution of problem (3.17), we have to know that the series (3.21) and the series of the second derivatives with respect to variables x and t converge uniformly. Therefore, we need to specify assumptions on functions u0 and u1 .

3.2 The Separation of Variables Method

105

For instance, if u0 ∈ C 4 [0, l] and u1 ∈ C 3 [0, l], satisfy the compatibility conditions (cf. Problem 5.2.14), i.e., u0 (0) = u0 (l) = u

0 (0) = u

0 (l) = 0, u1 (0) = u1 (l) = 0, then function u of the form (3.21) is a solution of problem (3.17). We will sketch the computations, because they can be used in different contexts. The definitions of Bn and Xn yield Bn =

1 λn l

 0

l

u0 (x)Xn

(x) dx =

1 λn l



l 0

u

0 (x)Xn (x) dx,

(3.24)

where the integration by parts and the compatibility condition gave the second equality. If u0 ∈ C 4 [0, l] and it satisfies the compatibility conditions, then from (3.23), by the necessary integration by parts, we obtain |Bn | ≤

const n4

A similar argument yields the following estimate for An , provided that u1 ∈ C 3 [0, l] and it satisfies the compatibility conditions, |An | ≤

const . n4

Differentiating series (3.21) twice, we note that this series is uniformly convergent together with its first and second derivatives. This is why function u is of class C 2 and it is a classical solution of problem (3.17). If we compare the above assumptions required to get classical solution with the d’Alembert formula, see (5.31), where u0 ∈ C 2 and u1 ∈ C 1 suffices, then we see that our approach is far from optimal, in terms of regularity of solutions. Comments on the Regularity of u Given by Formula (3.21) Let us assume that u0 ∈ C 2 [0, l] and u0 (0) = 0 = u0 (l) (resp. u0 ∈ H 2 ∩ H01 (0, l)), then we can compute from (3.24) and the definition of An , Bn that |An |,

|Bn | ≤

const . n2

However, this is not sufficient to claim that u is a C 2 -function, but we can show (see Problem 3.2.23) u ∈ L∞ (0, T ; H 2 ∩ H01 (0, l))

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3 Common Methods

and the weak derivative ut t exists and ut t ∈ L∞ (0, T ; L2 (0, l)). Moreover, ut t = uxx as two functions from L2 ((0, l) × (0, T )). We do not discuss here the initial conditions. If we further relax the regularity assumption on the data, i.e., we assume u0 ∈ H01 (0, l) and u1 ∈ L2 (0, l), then we notice that Formula (3.21) yields a function u in L∞ (0, T ; H01 (0, l)) and ut ∈ L∞ (0, T ; L2 (0, l)), see Problems 3.2.24 and 3.2.25. This function is a weak solution of problem (3.17). The meaning of this term is explained in Definition 5.2.2. We refer the reader to Sect. 5.2 for more details.

Hyperbolic Problems: A Non-homogeneous Case Let us consider the problem of a finite string and a given source term f . For instance, let us consider boundary conditions, which are different from the ones previously studied, ut t − uxx = f

in (0, l) × (0, ∞)

ux (0, t) = u(l, t) = 0,

(3.25)

t ≥ 0,

u(x, 0) = u0 (x),

x ∈ (0, l),

ut (x, 0) = u1 (x),

x ∈ (0, l).

Assume that function f : (0, l) × (0, ∞) → R may be expanded in a series with respect to the eigenfunctions of the following problem: − Xn

(x) = λn Xn (x),

(3.26)

Xn (0) = Xn (l) = 0,

(3.27)

and that it satisfies the consistency condition, f (0, t) = f (l, t) = 0

t > 0.

Following the previous example, we look for a solution u in the form of a series u(x, t) =

∞  n=1

Tn (t)Xn (x),

(3.28)

3.2 The Separation of Variables Method

107

where Xn are the eigenfunctions of problem (3.26). Thus,

Xn (x) = cos( λn x), where λn =

n2 π 2 , l2

n = 1, 2, . . .

In order to find the functions Tn (t), we shall expand the right-hand side of Eq. (3.25), i.e. the function f. f (t, x) =

∞ 

Fn (t)Xn (x)

(3.29)

n=1

in a series of Xn (x), 0 < n ∈ N. Formally inserting u and f , defined as the series (3.28) and (3.29) into Eq. (3.25), gives us (after differentiating formally the series term by term) ∞ 

(Tn

(t) + λn Tn (t))Xn (x) =

n=0

∞ 

Fn (t)Xn (x).

n=0

Thus, u is a solution, provided that the functions Tn (t) satisfy the ordinary differential equations Tn

(t) + λn Tn (t) = Fn (t), Tn (0) = un0 , Tn (0) = un1 ,

n = 1, 2, . . . ,

where un0 and un1 are the coefficients of expansions of u0 (·) and u1 (·) in the basis 2 {Xn (·)}∞ n=1 of L (0, l). Parabolic Problems Let us consider the problem of heat conduction in a homogeneous bar, ut = uxx , x ∈ (0, l), t > 0, u(0, t) = u(l, t) = 0, t ≥ 0, u(x, 0) = u0 (x), x ∈ (0, l),

(3.30)

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3 Common Methods

for u0 (x) ∈ L2 (0, l). Since this equation is considered in a rectangular region, we may seek solutions in the analogous form as for a vibrating string, i.e. u(x, t) =

∞ 

An e−λn t sin( λn x),

n = 1, 2, . . .

(3.31)

n=1

where λn =

n2 π 2 , l2

n = 1, 2, . . . .

Before we come to the calculations of the coefficients, we note that the parabolic problems are studied in a greater detail in Sect. 5.3. We note that despite the formal similarities between formulas, there are deep differences between solutions to hyperbolic and parabolic equations. After these general remarks, we come back to the problem of finding An ’s. The coefficients An are calculated from the initial condition. These are simply the coefficients of the expansion of function u0 into a series with respect to √ {sin( λn x)}n=1,2,... : u(x, 0) = u0 (x) =

∞ 

An sin( λn x).

n=1

We assume that u0 ∈ L2 (0, l), thus the coefficients An are bounded independently of n. As a result, the coefficients An e−λn t in series (3.31) decay exponentially fast because of factors e−λn t . This implies that function u is smooth in set (0, l) × (0, ∞). Another problem, which we do not address here, is the regularity of function u at the time instant t = 0. Certainly, this depends on the initial data.

3.2.2 Worked-Out Problems We offer one worked-out example in each category: homogeneous hyperbolic equations, hyperbolic equations with forcing and parabolic equations. Problem 3.2.1 Solve the following problem by using the Fourier method, (x, t) ∈ (0, π) × (0, T ), ut t = uxx + 2ux − u, u(0, t) = 0 = u(π,√t), t ∈ (0, T ), √ u(x, 0) = e−x sin( 2x), ut (x, 0) = e−x sin( 6x)

3.2 The Separation of Variables Method

109

Remark 3.2.1 This problem serves as a warning that not every eigenvalue problem encountered has an orthogonal system of eigenfunctions. Since the operator Lu = uxx + 2ux − u is not self-adjoint on any domain in L2 (0, π), the orthogonality of the eigenfunctions is lost. We need other tools, not mentioned here, to study the completeness of the eigenfunction system we obtain. However, due to the simplicity of the initial and boundary data, we can find an explicit solution. Solution First, we study the eigenvalue problem, λu = uxx + 2ux − u,

x ∈ (0, π),

u(0) = 0 = ux (π).

(3.32)

For this purpose, we write the characteristic equation of the above ordinary differential equation, r 2 + 2r − 1 − λ = 0.

(3.33)

Its discriminant is  = 4(2 + λ). If  ≥ 0, then the roots of (3.33) are real. Then, it is easy to check that there is no solution to (3.32). The reader is encouraged to do so. However, if  < 0, then the roots of (3.33) are complex,



r1 (λ) = −1 − |2 + λ|i, r2 (λ) = −1 + |2 + λ|i. Real solutions to (3.32) have the following form:



uλ (x) = C1 e−x sin( |2 + λ|x) + C2 e−x cos( |2 + λ|x). We now check, when uλ satisfies the boundary conditions. We note that at x = 0 we have 0 = uλ (0) = C1 · 0 + C2 . Hence, C2 = 0. However, at x = π we have

0 = C1 e−π sin( |2 + λ|π). Thus, we have non-trivial solutions if and only if number. In other words, 0 > λk = −2 − k 2 ,

√ |2 + λ| is a positive natural

k ∈ N,

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3 Common Methods

so the set of eigenfunctions consists of

Xk (x) = e−x sin( 2 + k 2 x),

k ∈ N.

We see that the eigenfunctions appearing in the Fourier method need not be trigonometric functions. We seek solutions u in the linear space spanned by the eigenfunctions {Xk }∞ k=0 . So, if we consider the product uk = Xk (x)Tk (t), then Tk must solve the equation Tk

= −(2 + k 2 )Tk ,

k = 0, 2,

(3.34)

with the initial conditions T0 (0) = 1,

T0 (0) = 0,

T2 (0) = 0,

T2 (0) = 1.

We note that Tj (0) = Tj (0) for j = 0, 2, hence by the theory of ordinary differential equations we deduce that Tj ≡ 0 for j = 1, 3, 4, . . .. In other words, we are interested in just two equations, among infinitely many, which are given by (3.34). The solutions are √ √ √ 6 sin( 6t). T0 (t) = cos( 2t), T2 (t) = 6 Finally, the solution to our problem is √ √ u(x, t) = cos( 2t)e−x sin( 2x) +

√ √ √ 6 sin( 6t)e−x sin( 6x). 6

♦ Problem 3.2.2 Solve the following problem with the help of the Fourier method, i.e., the variable separation method, π ut t = uxx + sin( x) cos t, 2

(x, t) ∈ (0, 1) × (0, T ),

u(0, t) = 0 = ux (1, t), u(x, 0) = x 2 − x,

t ∈ (0, T ), ut (x, 0) = 0,

x ∈ (0, 1).

Solution First, we study the eigenvalue problem λu = uxx

x ∈ (0, 1),

u(0) = 0 = ux (1).

(3.35)

We note that for λ ≥ 0 there are no√solutions to this √ problem. If we consider λ < 0, then the function uλ (x) = C1 sin( |λ|x) + C2 cos( |λ|x) satisfies λuλ = (uλ )xx .

3.2 The Separation of Variables Method

111

We have to find λ’s such that uλ fulfill the boundary conditions. At x = 0, we have 0 = uλ (0) = C1 · 0 + C2 cos 0 = C2 . At the other endpoint of interval (0, 1), we have 0=



d uλ (1) = C1 |λ| cos |λ|. dx

Thus, we immediately conclude that 1 λk = −π 2 (k + )2 , 2

k ∈ N.

Hence, the eigenfunctions are {sin(π(k + 12 )x)}∞ k=0 . Each of these functions has the √ 2 L norm equal to 2. Finally, we have the following orthonormal system: √ 1 2 sin(π(k + )x), ϕk (x) = 2 2

k ∈ N.

(3.36)

We have to expand the initial condition u0 (x) = x 2 − x in a trigonometric series. We note that √ √  1  1 1 2 2 1 d cos(π(k + )x) dx x sin(π(k + )x) dx = − x 1 2 0 2 2 2π(k + 2 ) 0 dx √ √  1 1 x=1 2 2 1 = )x) dx − x cos(π(k + )x) cos(π(k + . x=0 2 2 2 2π(k + 12 ) 0 √ 2 . = 2 2π (k + 12 )2 On the way we used the integration by parts. If we continue our computations along the lines indicated above, we will reach the following conclusion: * + √ √  1 1 2 2 1 2− x 2 sin(π(k + )x) dx = . 2 0 2 2π(k + 12 )2 π 2 (k + 12 ) Hence, 

1

ak = 0

(x 2 − x)ϕk (x) dx =

√ 2 (2π − 1)(k + 12 )π − 1 . 2 π 3 (k + 12 )3

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3 Common Methods

As a result, x2 − x =

∞ 

x ∈ (0, 1).

ak ϕk (x),

k=0

Expanding the forcing term, sin( π2 x) cos t, in a series with respect to the orthonormal system (3.36) is easy, f (x, t) =

√ 2ϕ0 (x) cos t.

Finally, we have to solve the family of ODEs, forming the solutions uk (x, t) = ϕk (x)Tk (t), T0

= λ0 T0 +

√ 2 cos t,

Tk

= λk Tk ,

a00 = a0 ,

ak0 = ak ,

a01 = 0,

ak1 = 0.

The solutions are

T0 (t) = A cos t + (a0 − A) cos( |λ0 |t), where A =

√ 4 2 . π 2 −4

Furthermore,

Tk (t) = ak cos( |λk |t),

k = 1, 2, . . .

Finally, ∞ 



u(x, t) = A cos tϕ0 (x) + (a0 − A) cos( |λ0 |t)ϕ0 (x) + ak cos( |λk |t)ϕk (x) k=1

is the solution to our problem. Uniqueness is shown by different methods, see Sect. 5.2. ♦ Problem 3.2.3 Solve the following problem with the help of the Fourier method, ut = uxx − u, (x, t) ∈ (0, π) × (0, T ), u(0, t) = 0 = u(π, t), t ∈ (0, T ), x ∈ (0, π). u(x, 0) = x − π2 , Investigate the continuity of the obtained solution at t = 0. Is the solution smooth in (0, π) × (0, T )?

3.2 The Separation of Variables Method

113

Solution We set up the eigenvalue problem λu = uxx − u,

x ∈ (0, π),

u(0) = 0 = u(π).

Solutions to these equations are



uλ (x) = C1 cos( |λ + 1|x) + C2 sin( |λ + 1|x). √ |λk + 1| = k, 0 < k ∈ N, i.e.,

The boundary conditions give us

λk = 1 − k 2 π 2 ,

k = 1, 2, . . . .

The corresponding eigenfunctions, after normalization, are √

2 ϕk (x) = √ sin(kx), π

k = 1, 2, . . .

and they form an orthonormal system. We expand the initial condition in a Fourier series, repeating the computations as in the previous problems. We can see that u0 (x) = 2

∞  (−1)k+1 − 1

k

k=1

sin(πkx).

Since the solutions to Tk (t) = λk T (t),

Tk (0) = ak ≡ 2

(−1)k+1 − 1 , Tk (0) = 0 k

are Tk (t) = ak et

√

k 2 π 2 −1

,

k = 1, 2, . . . ,

we deduce that u is given by the formula, u(x, t) = 2

∞  (−1)k+1 − 1 t √k 2 π 2 −1 e sin(πkx). k k=1

♦ Remark 3.2.2 We note that the Fourier series is zero at x = 0 or x = π, hence it does not converge to the initial datum there. As a result, the solution is not continuous at t = 0, in general.

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3 Common Methods

3.2.3 Problems Let us recall that we use the terms “the Fourier method” and “the variable separation method” interchangeably.

Hyperbolic Equations Problem 3.2.4 Solve the following problem by the Fourier method ut t = uxx + 1, in (0, π) × R+ , u(0, t) = u(π, t) = 0, for t > 0, u(x, 0) = 0, x ∈ (0, π), ut (x, 0) = 0, x ∈ (0, π). Problem 3.2.5 Solve the following problem by the method of separation of variables ut t = uxx + t cos x, in (0, π/2) × R+ , ux (0, t) = u(π/2, t) = 0, for t > 0, u(x, 0) = 0, x ∈ (0, π/2), ut (x, 0) = cos x, x ∈ (0, π/2). Problem 3.2.6 Solve the following problem using the Fourier method ut t = uxx , in (0, π) × R+ , u(0, t) = u(π, t) = 0, for t > 0, π π u(x, 0) = − |x − |, x ∈ (0, π), 2 2 ut (x, 0) = χ[ π , 3π ] , x ∈ (0, π). 4

4

Problem 3.2.7 Solve the following problem using the Fourier method ut t = uxx + sin t, in (0, π) × R+ , u(0, t) = u(π, t) = 0, for t > 0, u(x, 0) = sin x, x ∈ (0, π), ut (x, 0) = x(x − π), x ∈ (0, π).

3.2 The Separation of Variables Method

115

Problem 3.2.8 Solve the following problem using the Fourier method ut t = uxx + cos(3x), in (0, π) × R+ , ux (0, t) = ux (π, t) = 0, for t > 0, u(x, 0) = 2 cos x, x ∈ (0, π), ut (x, 0) = 5 cos(2x), x ∈ (0, π). Problem 3.2.9 Solve the following problem using the Fourier method ut t = uxx + sin(5x), in (0, π/2) × R+ , π u(0, t) = ux ( , t) = 0, for t > 0, 2 u(x, 0) = 2 sin x, , x ∈ (0, π/2), ut (x, 0) = 4 sin(3x), x ∈ (0, π/2). Problem 3.2.10 Solve the following problem using the Fourier method ut t = uxx , in (0, π) × R+ , u(0, t) = t, u(π, t) = 0, for t > 0, u(x, 0) = 2 sin x, x ∈ (0, π/2), ut (x, 0) = 4 sin(3x), x ∈ (0, π/2). Problem 3.2.11 Let us suppose that u = u(x, y, t) and Q ∈ (0, π)×(0, 2π). Solve the following problem using the Fourier method ut t = uxx + uyy , in Q × R+ , u(x, y, t) = ut (x, y, t) = 0, for t > 0 and (x, y) ∈ ∂Q, u(x, y, 0) = sin x cos(y/2), (x, y) ∈ Q, ut (x, y, 0) = 0, (x, y) ∈ Q. Problem 3.2.12 Let u = u(x, y, t) and Q ∈ (0, π) × (0, 2π). Solve the following problem using the Fourier method ut t = uxx + uyy + xy, in Q × R+ , u(x, y, t) = ut (x, y, t) = 0, for t > 0 and (x, y) ∈ ∂Q, u(x, y, 0) = 0, (x, y) ∈ Q, ut (x, y, 0) = 0, (x, y) ∈ Q.

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3 Common Methods

Problem 3.2.13 Solve the following problem using the Fourier method ut t + ut = uxx , in (0, 1) × R+ , u(0, t) = u(1, t) = 0, for t > 0, u(x, 0) = cos(2πx), x ∈ (0, 1), ut (x, 0) = sin(πx), x ∈ (0, 1). Problem 3.2.14 Solve the following problem using the Fourier method ut t + ut = uxx + 1, in (0, π) × R+ , u(0, t) = u(π, t) = 0, for t > 0, u(x, 0) = 0, x ∈ (0, π), ut (x, 0) = 0, x ∈ (0, π).

Parabolic Equations Problem 3.2.15 Solve the following problem using the Fourier method ut = uxx , in (0, π) × R+ , u(0, t) = u(π, t) = 0, for t > 0, u(x, 0) = χ[ π , 3π ] , x ∈ (0, π). 4

4

Problem 3.2.16 Solve the following problem by the method of separation of variables ut = uxx , in (0, π) × R+ , u(0, t) = 0, u(π, t) = sin t, for t > 0, u(x, 0) = 4 sin(3x), for x ∈ (0, π). Problem 3.2.17 Solve the following problem by the method of separation of variables ut = uxx + tx 2 , in (0, π) × R+ , ux (0, t) = ux (π, t) = 0, for t > 0, u(x, 0) = cos x, x ∈ (0, π). Show that the solution of the above problem is smooth in (0, π) × [T , ∞) for each T > 0.

3.2 The Separation of Variables Method

117

Problem 3.2.18 Solve the following problem by the method of separation of variables ut = uxx + e−t , in (0, 1) × R+ , u(0, t) = u(1, t) = 0, for t > 0, u(x, 0) = 0, x ∈ (0, 1). Investigate the behavior of solutions as t → ∞. Problem 3.2.19 Solve the following problem by the method of separation of variables ut = uxx − u + sin x, in (0, π) × R+ , u(0, t) = u(π, t) = 0, for t > 0, u(x, 0) = sin(5x), x ∈ (0, π). Problem 3.2.20 Show that the solution of problem ut = uxx − u, in (0, π) × R+ , u(0, t) = u(π, t) = 0, for t > 0, u(x, 0) = u0 (x), x ∈ (0, π), obtained by Fourier method is a smooth function (0, π)×(T , ∞) for u0 ∈ C 1 (0, π). Problem 3.2.21 Let us consider problem x 1 sin( ), in (0, π) × R+ , 4 2 u(0, t) = u(π, t) = 0, for t > 0, x u(x, 0) = sin( ) + sin x, x ∈ (0, π). 2 ut = uxx +

What is the limit lim u(·, t) and in which topology the limit exists, if u is the t →∞ solution obtained by the method of separation of variables? Problem 3.2.22 Let u0 ∈ L2 (0, π). Show that the solution of the problem ut = uxx , in (0, π) × R+ , u(0, t) = u(π, t) = 0, for t > 0, u(x, 0) = u0 (x), x ∈ (0, π).

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3 Common Methods

obtained by the Fourier method satisfies the estimate u(·, t)L2 ≤ e−t u0 L2 . Problem 3.2.23 Suppose that u0 ∈ H 2 ∩ H01 (0, l) and u1 ∈ H 1 (0, l) and u is given by (3.21). Show that: (a) for almost every t ∈ (0, ∞) function u(·, t) is a well-defined element of L2 (0, l); (b) for almost every t ∈ (0, ∞) function u(·, t) is a well-defined element of H 2 (0, l); (c) for almost every t ∈ (0, ∞) function u(·, t) is a well-defined element of H 2 ∩ H01 (0, l); (d) there is a positive number M, such that u(·, t)H 2 ≤ M for a.e. t ∈ [0, ∞), M depends on u0 and u1 ; (e) for almost every t ∈ (0, ∞) function ut t (·, t) is a well-defined element of L2 (0, l) Problem 3.2.24 Suppose that u0 ∈ H 2 ∩ H01 (0, l) and u1 ∈ H 1 (0, l) and u is given by (3.21). Show that u is a weak solution of (3.17) in the sense of Definition 5.2.2. Problem 3.2.25 Suppose that u0 ∈ H01 (0, l) and u1 ∈ L2 (0, l) and u is given by 3.21. Show that (a) for almost every t ∈ [0, ∞) function u(·, t) is a well-defined element of H 1 (0, l); (b) u is a weak solution to 3.17 in the sense of Definition 5.2.2.

3.3 Galerkin Method 3.3.1 Theoretical Background Partial differential equations are very important for applied mathematics and mathematical analysis, because many models of physical phenomena of industrial processes are written in the language of differential equations. However, finding a classical (i.e., differentiable enough times) solution is usually very difficult. The discussion on the regularity of the function given by formula (3.21) gave us a taste of it. Despite these shortcomings, we need an effective method of constructing solutions, possibly less regular, than the equation stipulates, as well as forming a tool capable of yielding solutions numerically. The Sobolev spaces, discussed in Sect. 2.3, allow the use of the powerful toolbox of functional analysis. In this section we introduce a notion of weak solutions and we present one of possible methods of their construction, related to the theory of separable Hilbert spaces, namely the Galerkin method (see, e.g., books by Evans and Temam [8, 27]). This may be applied to elliptic equations, as well as to a large class of evolution equations. We want to present here a general concept of this method and two examples, so that the reader can have a good start for independent generalizations

3.3 Galerkin Method

119

of more complicated systems. It is worth stressing the link between this theory and numerical analysis, in particular the method of finite element. For further discussion on weak solutions, on methods of proving their existence, and on their properties, we refer the reader to the next chapters. We shall proceed according to the following scheme: The Recipe, i.e. the Foundation of the Galerkin Method Step I Find a suitable functional space, H , and its carefully chosen basis {ek }∞ k=1 . Hence, if we set HN = span {e1 , . . . , eN }, then for any v ∈ H there is a sequence {vN }∞ N=1 such that vN ∈ HN and vN → v in the norm of H . We will construct solutions in space H . Step II Construct in HN a finite dimensional approximation of an equation we study. Prove existence of an approximate solution in HN . Step III Find uniform estimates on solutions of the approximate systems in HN . Step IV Find a weak limit of a sequence of approximate solutions and show that they satisfy the original equation. Most frequently we can do this only in a weak sense. Step V Examine additional properties of solutions. Sometimes Galerkin method enables us to prove a higher (than expected) regularity of obtained solution.

3.3.2 Worked-Out Problems We present here two applications of the theory we developed above. In the first example we consider a linear elliptic boundary value problem, while the second equation is nonlinear. This is the only section apart from Sect. 5.1 where we deal with a nonlinear problem. The First Example: A Linear Elliptic Equation In order to illustrate the method, we will first present an application of the Galerkin method to a simple elliptic equation with the Dirichlet boundary condition,  −div (k(x)∇u + B(x) · ∇u = f u=0

in , on ∂.

(3.37)

We have not yet made our assumptions precise. We shall formulate them later in the process of constructing a solution. Step I Definitely, we look for solutions with relaxed regularity requirements, i.e. weak solutions of (3.37). We have already mentioned that we expect only weak convergence of the sequence of approximate solutions, uN , i.e., lim uN , φ = u, φ

N→∞

∀φ ∈ H.

(3.38)

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3 Common Methods

Let us stress that the space H has to be determined yet. However, our main intuition is that H should be a separable Hilbert space, containing D = C0∞ () as a dense subset. In general, it is sufficient that the identity (3.38) holds for all φ from a dense subset of H . Verifying (3.38) usually comes very naturally. Now, we have to find a way of relaxing regularity of solutions. For this purpose we assume for a moment that u is a smooth solution to (3.37). We multiply Eq. (3.37) by a function φ ∈ D. After integration by parts we obtain k(x)∇u, ∇φ L2 (;Rn ) + B(x) · ∇u, φ L2 () = (f, φ) for each function φ ∈ C0∞ ().

(3.39) We choose the notation of the right-hand side so that it works to our advantage. We write (f, φ) stressing that f ∈ H ∗ acts on a test function φ ∈ D ⊂ H . However, if the reader prefers a simpler statement, then they should think about f as an element of L2 () and use the inner product, f, φ L2 () , instead of pairing (f, φ). We wish that a solution be a test function, because this simplifies computations. Thus, (3.39) suggests that ∇u should be square integrable, as well as ∇φ. We recall that the closure of D in the H 1 norm is H01 (), where the subscript 0 tells us that u has a zero trace, i.e. such u satisfies the conditions (3.372). Thus, we come to a conclusion that a solution and the test functions should belong to H01 (). This discussion prompts the following definition. Definition 3.3.1 We shall say that u ∈ H01 () is a weak solution to (3.37), provided that the integral identity (3.39) holds. This definition indicates that H01 () may be a Hilbert space H we are looking for. It is a closed subspace of H 1 () consisting of functions with the zero trace, see the paragraph Trace of a Sobolev function in the previous chapter. The Hilbert space H01 () is separable. Let us take {wk }∞ k=1 any (not necessarily 1 orthogonal) basis of H0 (), then H01 () = span{w1 , . . . , wn , . . .}

H 1 ()

.

Due to the above fact, we may define an increasing sequence of finite dimensional spaces, V N = span{w1 , . . . , wN }

for N ∈ N,

whose sum is H01 (). We look for an approximate solution in the following form, u = N

N 

dkN wk .

(3.40)

k=1

We would like to stress that, in general, the coefficients dkN depend on N in an essential way and this dependence cannot be ignored.

3.3 Galerkin Method

121

Step II We assume that uN satisfies the following finite dimensional version of Problem (3.39) k(x)∇uN , ∇φ N L2 (;Rn ) + B(x) · ∇uN , φ N L2 () = (f, φ N )

(3.41)

for all φ N ∈ V N . System (3.41) reduces to N equations k(x)∇uN , ∇wk L2 (;Rn ) + B(x)·∇uN , wk L2 () = (f, wk )

for k = 1, . . . , N.

Now, we must show the existence of solutions of (3.41). Let us recall that we have not stated any assumption on functions f and B appearing in our problem, yet. We did it on purpose, so that we could understand the need and the goal of a particular hypothesis. If uN is to have the form (3.40), then we notice that our problem is reduced to a system of algebraic equations, N 

aj k djN = fk

for k = 1, . . . , N,

(3.42)

j =1

where aj k = k(x)∇wk , ∇wj L2 (;Rn ) + B(x) · ∇wk , wj L2 () and fk = (f, wk ), j, k = 1, . . . , N. Since the original Problem (3.37) is elliptic, see also Chap. 4, then it suffices to show that matrix {aj k }N j,k=1 is positive definite, i.e., there exists a constant γ > 0 such that N 

aj k ξ j ξ k ≥ γ |ξ |2

for each ξ ∈ RN .

(3.43)

j,k=1

Then, obviously {aj k }N j,k=1 is invertible and after solving system (3.42), we get N coefficients dk , enabling us to define the approximate solution uN by formulas (3.40).  k Let XN = N k=1 ξ wk . Then, N 

aj k ξ j ξ k = k(x)∇XN , ∇XN L2 (;Rn ) + B(x) · ∇XN , XN L2 () = AN .

j,k=1

Let us observe that if div B = 0,

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3 Common Methods

then  1 B(x) · ∇(ψ 2 ) dx 2     1 div B ψ 2 dx , because div B = 0, (3.44) = 2   1 = n · B(x)ψ 2 dσ due to Gauss’ law, see Theorem 1.1.2, 2 ∂

B(x) · ∇ψ, ψ L2 () =

= 0.

The latter equality holds, because of ψ = 0 on the boundary of . Moreover, if k(x) ≥ k∗ > 0

for all x ∈ Rn

and we insert XN in place of ψ in (3.44), then AN ≥ k∗ ∇XN 2L2 (;Rn ) .

(3.45)

We see that (3.45) implies (3.43), due to the fact that all norms in a finite dimensional Banach space V N are equivalent. Thus, we conclude that the matrix {aj k }N j,k=1 is N strictly positive definite, which implies the existence of a vector dk for any f ∈ H −1 (). Step III The next goal is to find a common estimate for solutions of approximate systems. We know that they satisfy k(x)∇uN , ∇φ N L2 (;Rn ) + B(x) · ∇uN , φ N L2 () = (f, φ N ) for all φ N ∈ V N . However, uN ∈ V N , hence we may take φ N = uN . Thus, we get the equality k(x)∇uN , ∇uN L2 (;Rn ) + B(x) · ∇uN , uN L2 () = (f, uN ). Let us notice that k(x)∇uN , ∇uN L2 (;Rn ) ≥ k∗ ∇uN 2L2 (;Rn ) , because k(x) ≥ k∗ . Moreover, (3.44) implies that B(x) · ∇uN , uN L2 () = 0. The ultimate element of this puzzle is the estimate |(f, uN )| ≤ f H −1 uN H 1 , 0

3.3 Galerkin Method

123

which follows from f ∈ H −1 (). Therefore, we obtain k∗ ∇uN 2L2 (;Rn ) ≤ f H −1 uN H 1 . 0

Since u → ∇uN L2 (;Rn ) defines a norm on H01 (), which is equivalent to the standard one (see Problem 2.3.41), then we reach uN H 1 ≤ 0

C2 f H −1 , k∗

(3.46)

where the right-hand side does not depend on N. This is the required estimate. Step IV In a linear case this step is usually easy, but we need to remember that for nonlinear equations it usually becomes nontrivial. We look for a limit of a sequence of approximate solutions uN , so that we could later show that it is a desired solution of our original equation. Besides, we are supposed to determine the limiting behavior of all elements of the equation. Applying Theorem 3.1.8, we quickly find a subsequence uNk , weakly convergent to an element u∗ in space H01 (). Without the loss of generality, we may assume that the whole sequence uN converges, i.e., uN  u∗ weakly in H01 (). Besides, up to selecting another subsequence, we also have (see Problem 3.1.29) k(x)∇uN  k(x)∇u∗

weakly in L2 ()

(3.47)

and B(x) · ∇uN  B(x) · ∇u∗

weakly in L2 ().

(3.48)

We will show that u∗ is a weak solution, i.e. (3.39) holds. For ψ, φ ∈ H01 (), we set L(ψ, φ) = k(x)∇u∗ , ∇φ L2 (;Rn ) + B(x) · ∇u∗ , φ L2 () . We want to check that for all φ ∈ H01 (), L(u∗ , φ) − (f, φ) = 0.

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3 Common Methods

We have |L(u∗ , φ) − (f, φ)| = | lim L(u∗ − uN + uN , φ − φ N + φ N ) − (f − f N + f N , φ − φ N + φ N )| N→∞

= | lim

N→∞



L(u∗ − uN , φ) + L(uN , φ − φ N ) + [L(uN , φ N ) − (f N , φ N )]

  + (f − f N , φ) + (f N , φ − φ N ) 

= | lim (K1 + K2 + K3 + K4 + K5 )|, N→∞

where φ N (respectively, f N ) is the orthogonal projection of φ (respectively, f ) on subspace V N (respectively, (V N )∗ ). Since (f, φN ) = (fN , φN ), due to properties of the projection, then uN is not only the solution of considered above (in the second step) approximative equation (3.41) with the right side f , but also of the same equation with the right side f N . This immediately yields K3 = L(uN , φ N ) − (f N , φ N ) = 0

for all x ∈ Rn .

Now, we estimate terms K2 , K4 , and K5 . Let us suppose that k(x) ≤ k ∗

and

|B(x)| ≤ b∗ .

φ − φ N H 1 ≤ ε

k∗ f H −1 k ∗ + b∗

Next, we take N so large that

and f − f N H −1 ≤

ε . φH 1

Then, |K2 | ≤ | k(x)∇uN , ∇φ − ∇φ N L2 (;Rn ) | + | B(x)∇u∗ , φ − φ N L2 () | ≤ k ∗ ∇uN L2 (;Rn ) φ − φ N H 1 + b ∗ ∇u∗ L2 (;Rn ) φ − φ N L2 (;Rn ) ≤ (f H −1 +

b∗ f H −1 )φ − φ N L2 () ≤ ε. k∗

Let us also observe that |K4 | ≤ |(f − f N , φ)| ≤ ε

and

|K5 | ≤ |(f N , φ − φ N )| ≤ ε.

3.3 Galerkin Method

125

Estimating K1 is the key to success. We notice that for a fixed φ ∈ H01 () the function H01 () $ ψ → L(ψ, φ) = ∇ψ, ∇φ L2 (;Rn ) + B(x) · ∇ψ, φ L2 () is a bounded functional on H01 (). Thus, taking sufficiently large N, due to the weak convergence, (3.47) and (3.48), we obtain |L(uN − u∗ , φ)| ≤ ε

for suitably large N.

Summing up the estimates of Ki ’s yields |L(u∗ , φ) − (f, φ)| ≤ 4ε . Since ε was arbitrary, we reach (3.39), which means that u∗ is indeed a weak solution. We are going also to show that u∗ is the only solution. Let us assume that for a certain f , we have two solutions u1 and u2 . Due to the linearity of the problem, we get k∇(u1 − u2 ), ∇φ L2 (;Rn ) + B(x) · ∇(u1 − u2 ), φ L2 () = 0 for any φ ∈ H01 (). In particular, we may put φ = u1 − u2 . This leads to equality ∇k(u1 − u2 ), ∇(u1 − u2 ) L2 (;Rn ) + B(x) · ∇(u1 − u2 ), (u1 − u2 ) L2 () = 0. (3.49) Now, we note that (3.44) implies k∗ ∇(u1 − u2 )L2 (;Rn ) ≥ k∇(u1 − u2 )L2 (;Rn ) = 0, that is, after taking into account the boundary conditions, u1 ≡ u2 . Summing up, we proved the following statement: Theorem 3.3.2 Let k ∈ L∞ () be such a function that k∗ ≤ k(x) ≤ k ∗

for all x ∈ R n

for certain positive numbers k ∗ , k∗ and B ∈ L∞ () is a vector field such that div B = 0 in D (). If f ∈ H −1 (), then there exists exactly one weak solution u ∈ H01 (), satisfying system (3.37) in the sense of (3.39). The Second Example: The Heat Equation with a Nonlinear Force The main goal of this subsection is to show the reader that the Galerkin method is an adequate tool to construct solutions for nonlinear problems. It appears, however,

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3 Common Methods

that even the simplest nonlinearity leads to serious technical complications. This shall be obvious for every reader who follows carefully the subsequent calculations and independently furnishes numerous computational details. In fact, we strongly suggest that the reader first get acquainted with Sect. 5.3. Let us consider the following example: ut − u = u2 u=0 u|t =0 = u0

in  × (0, T ), on ∂ × (0, T ), on .

(3.50)

Let us focus on a case when  is a region with a smooth boundary in R3 . Below, we define a weak solution. Definition 3.3.3 We say that u ∈ L2 (0, T ; H01()) is a weak solution to (3.50) if and only if the following integral identity −

 T 0



uφt dx dt +

 T 0



∇u · ∇φ dx dt =

 T 0



 u2 φ dx dt +



u0 φ(x, 0) dx

holds for any function φ ∈ C ∞ ( × [0, T )) such that φ = 0 on ∂ × (0, T ), as well as φ = 0 on  × {T }. We proceed according to our plan delineated earlier. Step I Evolutionary problems have slightly different features than elliptic ones. Namely, they require an appropriate treatment of the time derivative. In our relatively simple case, we shall limit ourselves again to space H01 () with a basis {wi }∞ i=1 . We will look for approximate solutions in the following form: uN (x, t) =

N 

bkN (t)wk (x).

(3.51)

k=1

Finding suitable functions bkN : [0, TN ) → R is a part of the problem. Step II In order to find suitable approximations, let us consider a problem on a finite dimensional subspace. We require that uN satisfy the equation N N N N 2 N uN t , ψ L2 () + ∇u , ∇ψ L2 (;Rn ) = (u ) , ψ L2 () ,

(3.52)

on [0, T ) in a distributional sense. This problem is augmented with initial conditions bkN (0) = (u0 , wk ), i.e. we want that the above identity be true for every function ψ N ∈ C ∞ ([0, T ), V N ). As before, V N denotes the N-dimensional space spanned by w1 , . . . , wN .

3.3 Galerkin Method

127

Without the loss of generality, after having applied the Gramm–Schmidt orthogonalization process, we may assume that the basis is orthogonal, i.e., wk , wl L2 () = δkl . Inserting ψ N = wk for k = 1, . . . , N, into (3.52), we get the following system of ordinary differential equations, d N b (t) + ∇uN , ∇wk L2 (;Rn ) = (uN )2 , wk L2 () , dt k

k = 1, . . . , N.

The basic theorem from the ordinary differential equations theory implies the existence of a solution, which is local in time, i.e. it is defined on [0, TN ], where generally speaking, TN may depend upon N in an essential way. Step III Observe that Step II yields only the existence of approximate solutions. In order to pass to the limit with N, N → ∞, we must find a common time interval, on which the solutions uN are defined. For this purpose, we shall use energy estimates. Let us take (3.52); we know that this equation may be tested by solution uN . We know also that for a fixed integer N the number TN is non-zero. Thus, we get 1 d N 2 u L2 + ∇uN 2L2 = 2 dt

 (uN )3 dx. 

Let us assume for a moment that the solution uN is defined on the interval [0, T ). Then, integration of both sides of the above identity with respect to t ∈ [0, τ ) yields  u

N

(τ )2L2

+2

τ

 ∇u

0

N

(s)2L2

ds = 2 0

τ

 

(uN )3 dxds + u0 2L2 .

(3.53)

We recall (see Problem 2.3.41) that there is C2 > 0 such that for all v ∈ H01 (), we have C2 ∇vL2 ≥ vH 1 . We use this inequality for uN (s) in place of v to estimate the left-hand side of (3.53) from below, then we take supremum over [0, T ). Finally, we reach the following estimate  sup uN 2L2 +

0≤t ≤T

T 0

uN 2H 1 () dt ≤ C2 uN 3L3 (×(0,T )) + C2 u0 2L2 .

(3.54)

Let us remind that dim  = 3. Recalling Problem 3.3.5, providing the so-called parabolic embedding, yields +

* u L10/3 (×(0,T )) ≤ C N

sup u L2 + u L2 (0,T ;H 1 ()) , N

0≤t ≤T

N

(3.55)

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3 Common Methods

where the constant does not depend on T . We introduce * XN (T ) =

 sup

0≤t ≤T

uN 2L2

T

+ 0

+1/2 uN 2H 1 () dt

.

(3.56)

Then, applying (3.54), (3.55), and Hölder inequality, we obtain 3 (T ) + Cu0 2L2 , XN (T )2 ≤ CT 1/10XN 3 3 3 + 10 + 10 + if we remember that 1 = 10 This leads to the following estimate

XN (T ) ≤ 0 (T , u0 L2 )

(3.57)

1 10 .

for all N = 1, 2, . . . and T < Tmax .

(3.58)

The reader may kindly consider how this should be done. The number Tmax = Tmax (u0 L2 ) depends only on the initial data norm and it converges to zero when this norm tends to infinity. However, we see that this estimate does not depend anymore on N. As a result, we get a uniform (with respect to N) estimate on approximate solutions and also a common bound from below on times TN . Step IV Due to the uniform estimates on functions uN and numbers XN (T ), we may try to find a limit of uN and define a solution. In particular, we note that we may deduce from estimate (3.58) the existence of such u, that uN  u weakly in L2 (0, T ; H 1 ()), ∗

uN  u weakly in L∞ (0, T ; L2 ()).

(3.59) (3.60)

It is sufficient to control the left-hand side of the equation but problems occur on the right-hand side. The nonlinearity requires additional information about convergence. Mere weak convergence is no longer sufficient. We must have more detailed information about the sequence uN . As we have at our disposal the norms estimate in H 1 (), we can get a strong convergence, due to the Rellich Theorem, see Theorem 2.3.28, but only in the spatial directions. Thus, we need extra information about the time regularity. In the beginning, we noticed that due to (3.57) and (3.55) (uN )2  ξ weakly in L5/3 ( × (0, T )) .

(3.61)

Thus, the limit u of the sequence uN satisfies the equation − u, φt L2 + ∇u, ∇φ L2 = ξ, φ L2 in the sense of distributions on [0, T ).

(3.62)

3.3 Galerkin Method

129

We want to show that ut ∈ L2 (0, T ; H −1 ()). For this purpose, we will focus on the nonlinear expression and we will show that (uN )2 ∈ L2 (0, T ; H −1 ()). In order to do this it suffices to prove that (uN )2 ∈ L2 (0, T ; L6/5 ()), because it is easy to check (the reader is encouraged to do so) that any element of L6/5 () defines in a natural way a continuous functional over H01 (). Thus, (uN )2 L2 (0,T ;L6/5 ()) ≤ uN L∞ (0,T ;L2 ()) uN L2 (0,T ;L3 ()) , because 56 = 12 + 13 and in dimension n = 3, we have H 1 () ⊂ L6 () ⊂ L3 (). In this way, having in mind estimates on XN (T ), obtained in the previous step, we reach inequality (uN )2 L2 (0,T ;H −1 ()) ≤ 1 (T , u0 )2 ,

(3.63)

Thus, we conclude that ξ ∈ L2 (0, T ; H −1 ()). Summing up (3.59) and (3.63), we get from (3.62) information about ut and uN t . Namely, 2 −1 ()). uN  u in L2 (0, T ; H 1()) and uN t  ut in L (0, T ; H

Then, by Lions–Aubin Theorem, see Problem 3.3.6, we obtain strongly in L2 ( × (0, T )).

uN → u

(3.64)

Taking into account (3.61) and (3.64), we assert that ξ = u2 . Thus, u satisfies Eq. (3.50) in the sense of Definition 3.3.3. Step V In our example we may obtain exact information about regularity of a solution. Observe that we can test the approximate solution with uN t . We get then  

2 (uN t ) dx +

1 d 2 dt



 |∇uN |2 dx = 



(uN )2 uN t dx.

(3.65)

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3 Common Methods

Let us introduce 

T

IN (T ) := 0

2 uN t L2 dt,

JN (T ) :=

 T 0

(uN )4 dxdt, 

as well as SN (T ) := sup ∇uN (·, t)2L2 () . 0≤t ≤T

We apply the Schwarz inequality on the right-hand side of (3.65), then we integrate both sides with respect to t ∈ [0, T ], where T ≤ T and we see that IN , JN , and SN satisfy the following inequality:   IN (T ) + SN (T ) ≤ C IN (T )1/2 JN (T )1/2 + ∇u0 2L2 .

(3.66)

Since I 1/2 J 1/2 ≤ εI + J /(4ε) for all I, J ≥ 0 and ε > 0, then it follows from (3.66) that   IN (T ) + SN (T ) ≤ C JN (T ) + ∇u0 2L2 .

(3.67)

The constant C is obviously different than previously. Let us show how to get rid of an inconvenient term JN (T ) from (3.67). Let us see that  

1/2 



 (uN )4 dx =

|uN |3 |uN |dx ≤ 

1/2

(uN )6 dx

(uN )2 dx



.



Thus, integrating over [0, T ] and keeping in mind definition (3.56) of XN (T ) and estimate (3.58), that we reached in Step III, we obtain 

T

JN (T ) = 0





T

(uN )4 dx dt ≤ 

0

uN 3L6 () uN L2 () dt 

≤ sup u(·, t)L6 () 0≤t ≤T

T 0

u2L6 () uL2 () dt

≤ sup u(·, t)L6 () XN (T ) 0≤t ≤T



T

0

u2L6 () dt

≤ C sup u(·, t)L6 () XN (T )3 . 0≤t ≤T

In the latter line, we benefited from the embedding H 1 () ⊂ L6 () and definition of XN (T ).

3.3 Galerkin Method

131

Now, we insert the above estimate of integral JN (T ) to (3.67). Next, let us notice that due to embedding we have H01 () ⊂ L6 () sup u(·, t)L6 () ≤ C sup ∇uN (·, t)L2 () = CSN (T )1/2 ,

0≤t ≤T

0≤t ≤T

which leads to the following inequality   IN (T ) + SN (T ) ≤ C SN (T )1/2XN (T )3 + ∇u0 2L2 . Now, it remains only to get rid of SN (T )1/2 from the right-hand side. In order to do this we perform a similar trick as before, when we were eliminating IN (T )1/2 from the right-hand side of (3.66). Finally, we get  sup ∇uN 2L2 +

0≤t ≤T

T 0

2 2 uN t L2 dt ≤ 2 (T , ∇u0 L2 ) ,

(3.68)

where 2 (T , ∇u0 L2 ) is a quantity bounding all XN (T ), N = 1, 2, . . .. We remember that due to (3.58) such a quantity depends on the datum u0 and (suitably chosen) time T , and 2 (T , ∇u0 L2 ) exists indeed. In particular, we may get such a subsequence that uN t  ut

weakly in L2 ( × (0, T )).

Therefore, our weak solution satisfies equation u = ut − u2 ∈ L2 ( × (0, T )), in the distribution sense. This, due to Problem 3.3.9 and estimate (3.68), yields the inequality ∇ 2 uL2 (×(0,T )) ≤ 3 (T , ∇u0 L2 ). Thus, we reach the following theorem: Theorem 3.3.4 Let u0 ∈ L2 (). Then, there is a time instant T > 0 such that there exists a weak solution of (3.50), which conforms to Definition 3.3.3 and the condition u ∈ L∞ (0, T ; L2 ()) ∩ L2 (0, T ; H01 ()). Moreover, if u0 ∈ H 1 (), then u ∈ L2 (0, T ; H 2 ()) ∩ H 1 (0, T ; L2 ()).

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3 Common Methods

3.3.3 Problems Problem 3.3.1 Use the Galerkin method to prove the Lax–Milgram Theorem, see Theorem 4.2.4, for separable Hilbert spaces. Problem 3.3.2 Let us suppose f ∈ L2 (Rn ). Show the existence of distributional solutions of the following equation −div (1 + |x|3 )∇u + u = f

in Rn .

Remark 3.3.5 By the distribution solution of an equation we understand a function u ∈ H 1 (Rn ), which satisfies the following integral identity 

 Rn



(1 + |x| )∇u · ∇φ + uφ dx = 3

 Rn

f φ dx

for all φ ∈ C0∞ (Rn ). Problem 3.3.3 Investigate the conditions for f and n ∈ N, such that the problem −u + u3 = f w , u=0 on ∂ has a unique weak solution. We assume that  is a region in Rn . Hint Use Problem 3.3.10. Problem 3.3.4 Investigate the conditions for f and n ∈ N, such that the problem −div (1 + u2 )∇u + u = f in , u=0 on ∂. has a unique weak solution. We assume that  is a region in Rn . Hint Use Problem 3.3.10. Problem 3.3.5 Let  ⊂ R3 and T > 0. Show that if u ∈ L∞ (0, T ; L2 ()) ∩ L2 (0, T ; H01 ()), then u ∈ L10/3( × (0, T )). Estimate the norm of the embedding operator.

3.3 Galerkin Method

133

2 Problem 3.3.6 (Lions–Aubin Theorem) Let us suppose {un }∞ n=1 ⊂ L (0, T ; H01 ()) is a sequence converging weakly to u. We also assume that the sequence of t-derivatives converges weakly in L2 (0, T ; H −1 ()). Prove that the sequence un converges to u in the norm of L2 ( × (0, T )).

Hint This problem is difficult. The solution is based on an application of the ArzelaAscoli Theorem. Problem 3.3.7 Show by the Galerkin method that there exists a solution to ut t − u = f u=0

in  × (0, T ), on ∂ × (0, T ),

u|t =0 = u0 ,

ut |t =0 = u1

on .

Finding a space sufficient for existence of weak solutions is a part of the problem. Problem 3.3.8 Let u0 ∈ L2 (Rn ). Show the existence of weak solutions to ut +  2 u + u = 0 u|t =0 = u0

in Rn × (0, T ), on Rn .

Define weak solutions and determine the function space, where weak solutions live. Problem 3.3.9 Let f ∈ L2 (), where  ⊂ Rn is a bounded region in Rn with a smooth boundary. Show that if u is a weak solution to u = f

in ,

u=0

on ∂,

then u ∈ H 2 () and uH 2 () ≤ Cf L2 () . Hint Use the Fourier transform and Plancherel Theorem. Problem 3.3.10 Let P : RN → RN be a continuous function. Let us assume that for all |X| ≥ r the following condition holds: P (X) · X ≥ 1 . Show that there exists a point X0 ∈ RN such that |X0 | ≤ r and P (X0 ) = 0.

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3 Common Methods

3.4 Bibliographical Remarks The theory of weak convergence and weak topology is the key to developing existence results based on approximation. Many functional analysis textbooks, like [21], can be used for further studies. However, texts aiming at PDE and Sobolev spaces are more appropriate, the book by Brezis, [5], seems a better choice. The method of variable separation (or the Fourier method) is very simple and it appears in many texts devoted to PDEs, e.g. [8]. On the other hand, the Galerkin method, which is easy to expose requires more effort to justify it, thus it is more likely to appear in more advanced books dealing with nonlinear problems. The book by Evans [8], serves as an example.

Chapter 4

Elliptic Equations

Elliptic equations describe the steady states of parabolic or hyperbolic problems. However, they share properties only with parabolic equations. The elliptic equations are treated differently than evolution problems, since they lack the time variable. We pay a lot of attention to harmonic functions, which are intimately related to holomorphic functions in the case of two-dimensional problems. We introduce classical approach to the Laplace and Poisson equations, which is based on the notion of the Green function. However, the most important part of this chapter is devoted to the theory of existence and uniqueness of weak solutions to linear elliptic boundary value problems. It is relatively easy to build with the use of the Lax–Milgram Lemma (cf. Theorem 4.2.4) and the theory of Sobolev spaces (see Sect. 2.3). We also study the properties of solutions, e.g., the maximum principle for classical solutions. We address the issue of regularity of weak solutions to elliptic equations. It is relatively easy in case of the Laplace problem. We show that weak solutions are in fact classical solutions, which justifies the development of the modern theory.

4.1 Classical Theory of Harmonic Functions 4.1.1 Theoretical Background Definitions Let  ⊂ Rn be an open set. A function f :  → R is called harmonic if it is twice differentiable and at each point of  it satisfies f = 0,

© Springer Nature Switzerland AG 2019 M. Borodzik et al., Problems on Partial Differential Equations, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-14734-1_4

135

136

4 Elliptic Equations

where  is the Laplace operator: f =

n  ∂ 2f k=1

∂xk2

.

(4.1)

It is also possible to define harmonic functions on Riemannian manifolds. A standard definition begins with the formula f = div ∇f and both div and ∇ can be defined on any Riemannian manifold. See also Problem 1.1.9. We will now gather results on harmonic functions. We begin with a regularity theorem. Theorem 4.1.1 (Regularity of Harmonic Functions) If f is harmonic in , then f is real analytic in . The reader is encouraged to compare this result with Weyl’s Lemma (Theorem 4.2.6). The following result can be deduced from Green’s formulae, see Problem 1.1.11. Theorem 4.1.2 (Mean Value Theorem for Harmonic Functions) Suppose f is harmonic and the ball B(x, r) is contained in . Then B(x,r) f (y) dy S(x,r) f (y) dσ f (x) = = . B(x,r) 1 dy S(x,r) 1 dσ Here



S(x,r) 1 dσ

is, obviously, the (n − 1)-dimensional area of the sphere S(x, r).

One of the consequences of the mean value theorem is the following result. Theorem 4.1.3 (Maximum Principle for Harmonic Functions) A nonconstant harmonic function does not have local maxima (or minima) in . In particular, if  is bounded, then sup{f (x) : x ∈ } = sup{f (x) : x ∈ ∂}. Motivated by the mean value theorem, we introduce the following definition. Definition 4.1.4 Let  ⊂ Rn be an open subset. • A locally integrable function f :  → R (that is, integrable over any compact subset of ) has the mean value property for balls, if for all x ∈  and for all r > 0 such that B(x, r) ⊂ , we have B(x,r) f (y) dy f (x) = . B(x,r) 1 dy

4.1 Classical Theory of Harmonic Functions

137

• A continuous function f :  → R has the mean value property for spheres if for all x ∈ , r > 0 such that B(x, r) ⊂ , we have S(x,r) f (y) dσ f (x) = . S(x,r) 1 dσ The mean value properties for spheres and for balls are equivalent for continuous functions. We will discuss these definitions in a series of problems starting with Problem 4.1.11. One of the core problems in the theory of harmonic functions is to find non-trivial solutions to the Laplace equation u = 0. A precise formulation of this problem is called the Dirichlet problem, which we now define. Definition 4.1.5 Let  ⊂ Rn be an open subset with the boundary of class C 1 . Let g : ∂ → R be a continuous function. The Dirichlet problem is the problem of finding a C 2 function f on , continuous up to the boundary, such that f |∂ = g, and f = 0 on . The Poisson problem is a natural generalization. Definition 4.1.6 Let  ⊂ Rn be as above, let g : ∂ → R be continuous and let h :  → R be continuous. The Poisson problem is the problem of finding a C 2 smooth function on , continuous on the boundary, and such that f |∂ = g and − f = h on .

Green’s Functions Both the Dirichlet problem and the Poisson problem can be solved if one knows the so-called Green’s function for the given open set . Definition 4.1.7 Let  ⊂ Rn be an open set. Green’s function for  (with Dirichlet boundary conditions) is a function ¯ × ¯ → R, :  such that (a) for x ∈  and y ∈ ∂ we have (x, y) = 0; (b) for x ∈  we have −y  = δx , where y denotes the distributional Laplace operator (see Sect. 2.2) with respect to the variable y and δx is the Dirac distribution centered at x. Remark 4.1.8 In the definition of Green’s function the derivatives of  when y = x are considered in distributive sense.

138

4 Elliptic Equations

Example 4.1.9 For  = Rn , Green’s function can be obtained via Fourier transform; compare Problem 2.1.4. We obtain the following formula. ⎧ 1 ⎪ ⎨− ln |x − y| 2π (x, y) = 1 1 ⎪ ⎩ n(n − 2)α(n) |x − y|n−2

n = 2, n > 2.

Here α(n) denotes the volume of the unit n-dimensional ball. The importance of Green’s functions can be seen in the following result. Theorem 4.1.10 Let  be an open set in Rn , whose boundary is C 1 smooth and let h :  → R and g : ∂ → R be continuous. If  is Green’s function for , then the function f given by 

∂ g(y) dσ (y) + f (x) = − ∂ n y ∂

 h(y)(x, y) dy

(4.2)



solves the Poisson problem for functions g and h. Here, ∂ ∂ny denotes the directional derivative of  in the direction of the normal vector n, and the derivative is taken with respect to the variable y. Putting h = 0 in (4.2), we obtain a solution to the Dirichlet problem.  f (x) = −

g(y) ∂

∂ dσ (y). ∂ n y

(4.3)

We will need one more useful notion. Definition 4.1.11 The function P (x, y) :  × ∂ → R, defined by P (x, y) =

∂ (x, y), ∂ n y

is called the Poisson kernel for . The formula (4.3) written as  f (x) = − g(y)P (x, y) dσ (y). ∂

is called the Poisson formula for . The following electrostatic interpretation might be particularly useful in determining Green’s functions for many open sets. Let  be Green’s function for Rn . For fixed x ∈ Rn , y → (x, y) is the electrostatic potential determined by a unit charge placed at x. Now, if we are given an open set  ⊂ Rn , the potential does not necessarily vanish on ∂. It is often possible to compensate the potential by placing

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139

other charges outside of , so that the total potential equals zero on ∂. We refer to Problem 4.1.2 for an instructive example.

Holomorphic Functions and Harmonic Functions For n = 2, that is, when one considers functions on open subsets of R2 , there are deep and beautiful connections between harmonic functions and holomorphic functions. Let us recall the definition: Definition 4.1.12 Let U ⊂ C. A function F : U → C is called holomorphic if it is continuous and for all z0 ∈ U one of the following equivalent conditions is satisfied. • There exists a finite limit

lim

w→0,w∈C

F (z0 +w)−F (z0 ) . w

  a −b . b a • If we write z = x + iy and F (z) = u(x, y) + iv(x, y), then the following equations, known as the Cauchy–Riemann equations are satisfied

• The derivative DF , regarded as a real 2 × 2 matrix, has the form

∂v ∂u ∂u ∂u = and =− . ∂x ∂y ∂y ∂x • The differential form F (z)dz = F (z)dx + iF (z)dy is closed. Holomorphic functions have numerous special properties, many of them can be deduced from the Cauchy Integral Formula. Below we state this formula in a simplified version. A more general statement can be found in [20]. Theorem 4.1.13 (Cauchy Integral Formula for a Disk) If f : U → C is holomorphic and B(z0 , r) ⊂ U , then  ∂B(z0 ,r)

f (z) dz = 2πif (z0 ). z − z0

The integral on the left-hand side is understood either as an integration over path f (z) (see, e.g., [20, Section 10.8]), or, equivalently, as an integral of a 1-form z−z dz. 0 Parameterizing the boundary ∂B(z0 , r) by z = z0 + reit , t ∈ [0, 2π], we obtain 

2π

2πif (z0 ) = 0

f (z0 + reit ) it ire dt = i reit

 0

2π

f (z0 + reit )dt =

i r

 f (z) dσ. ∂B(z0 ,r)

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4 Elliptic Equations

In the above formula the integral on the right-hand side is an integral with respect to the measure on the circle (in the sense of Sect. 1.1.1). The Cauchy Integral Formula (Theorem 4.1.13) implies then 1 2πr

 f (z) dσ = f (z0 ).

(4.4)

∂B(z0 ,r)

In other words, a holomorphic function has the mean value property for spheres.

Subharmonic Functions Let us recall the following notion. Definition 4.1.14 Let  ⊂ Rn be an open subset. A function f :  → R ∪ {−∞} is called upper semicontinuous if for any a ∈ R the set {x ⊂  : f (x) < a} is open in . Equivalently, f is upper semicontinuous if for every x0 ∈  lim sup f (x) ≤ f (x0 ). x→x0

We give now the definition of a subharmonic function. The properties of subharmonic functions will be studied in a series of problems in Sect. 4.1.47. Definition 4.1.15 An upper semicontinuous function f :  → R ∪ {−∞} is called subharmonic if for every x ∈  and r > 0, such that B(x, r) ⊂ , and for every harmonic function h : B(x, r) → R continuous on B(x, r), the condition ∀x ∈ ∂B(x, r) : f (x) ≤ h(x) implies the condition ∀x ∈ B(x, r) : f (x) ≤ h(x). As a reader might have already noticed, if in Definition 4.1.15 we replace the condition that h is harmonic with the condition that h is linear, we obtain the definition of a convex function. Remark 4.1.16 In Problem 4.1.56 we show that a C 2 -smooth function u is subharmonic if and only if it satisfies u ≥ 0.

General Elliptic Operators Having discussed harmonic functions, we pass to the classical theory of more general elliptic operators. We focus on maximum principles for these operators.

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141

To begin with, consider an elliptic operator in nondivergence form Lu = −

n 

aij uxi xj +

i,j =1

n 

b i ux i ,

i=1

where aij and bi are continuous functions and the matrix {aij } is positive definite. Theorem 4.1.17 (Weak Maximum Principle) Let u ∈ C 2 () ∩ C 0 (). If Lu ≤ 0

in ,

If Lu ≥ 0

in ,

then then

max u = max u. ∂



min u = min u. 

∂

The stronger version of the principle may be proved under the additional assumption that  is a connected set. Theorem 4.1.18 (Strong Maximum Principle) Assume  is an open, bounded, and connected set. Let u ∈ C 2 () ∩ C 0 (). If Lu ≤ 0

in ,

and u attains its maximum over  at an interior point, then u is constant within . Similarly, if Lu ≥ 0

in ,

and u attains its minimum over  at an interior point, then u is constant within . The notions of subharmonic and superharmonic functions can be generalized to the case of general elliptic operators in the following way. Definition 4.1.19 If a function u satisfies the differential inequality Lu ≤ 0

in ,

we call it a subsolution for the operator L. Accordingly, if u satisfies Lu ≥ 0

in ,

we call it a supersolution. An essential tool in the proof of the strong maximum principle is the following result.

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4 Elliptic Equations

Lemma 4.1.20 (The Hopf Lemma) Let u ∈ C 2 () ∩ C 1 (). Suppose that the following conditions are satisfied: 1. Lu ≤ 0 in ; 2. there is a point x0 ∈ ∂ such that u(x0 ) > u(x)

for all x ∈ ;

3. there exists an open ball B ⊂  such that x0 ∈ ∂B. Then ∂u (x0 ) > 0, ∂ n where n is the outer unit normal vector to B at x0 . The above theorems may be easily generalized to the case when the elliptic operator has a zero-order term, provided the zero-order coefficient is nonnegative, i.e., Lu = −

n  i,j =1

aij uxi xj +

n 

bi uxi + cu

with c ≥ 0.

i=1

Theorem 4.1.21 (Weak Maximum Principle for c ≥ 0) Let u ∈ C 2 () ∩ C 0 () and c ≥ 0. • If Lu ≤ 0 in , then max u(z) ≤ max max(u(w), 0). z∈

w∈∂

• If Lu ≥ 0 in , then min u ≥ − max (− min(u(w), 0)). z∈

w∈∂

Theorem 4.1.22 (Strong Maximum Principle for c ≥ 0) Assume  is an open, bounded, and connected set and c ≥ 0. Let u ∈ C 2 () ∩ C 0 (). If Lu ≤ 0

in ,

and u attains nonnegative maximum over  at an interior point, then u is constant within . Similarly, if Lu ≥ 0

in ,

and u attains nonpositive minimum over  at an interior point, then u is constant within .

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143

Lemma 4.1.23 (The Hopf Lemma for c ≥ 0) Let u ∈ C 2 () ∩ C 1 () and c ≥ 0. Suppose that the following conditions are satisfied: 1. Lu ≤ 0 in ; 2. there is a point x0 ∈ ∂ such that u(x0 ) ≥ 0 and u(x0 ) > u(x)

for all x ∈ ;

3. there exists an open ball B ⊂  such that x0 ∈ ∂B. Then ∂u (x0 ) > 0, ∂ n where n is the outer unit normal vector to B at x0 .

4.1.2 Worked-Out Problems The Laplace Operator in Different Coordinate Systems Knowledge of the form of the Laplace operator in other coordinate systems (particularly, in spherical coordinates) can be quite useful not only in partial differential equations, but also, for instance, in quantum mechanics. The problems below are about this variable change. In Problem 4.1.1 we show an algorithm based on the Hodge star operator that allows to pass to the spherical coordinate system in a quick way. Problem 4.1.1 Find the Laplace operator in spherical coordinates in R4 given by x = r cos α cos β cos γ y = r cos α cos β sin γ z = r cos α sin β u = r sin α. Solution We are going to perform the calculations using the Hodge star operator. It is easy to check that F = div ∇F = ∗d ∗ dF , where d is the exterior derivative and ∗ is the Hodge operator. Our goal is to determine the Hodge operator in the coordinates r, α, β, γ . The key property is the knowledge of the Hodge operator in an oriented orthonormal basis.

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4 Elliptic Equations

First, note that the two bases of the tangent space to R4 at a given point of R4 : ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ , , , and , , , ∂x ∂y ∂z ∂u ∂r ∂α ∂β ∂γ are related in the following way: ∂ ∂ ∂ = cos α cos β cos γ + cos α cos β sin γ ∂r ∂x ∂y ∂ ∂ + sin α , ∂z ∂u ∂ ∂ ∂ = −r sin α cos β cos γ − r sin α cos β sin γ ∂α ∂x ∂y + cos α sin β

∂ ∂ + r cos α , ∂z ∂u ∂ ∂ ∂ = −r cos α sin β cos γ − r cos α sin β sin γ ∂β ∂x ∂y − r sin α sin β

+ r cos α cos β

∂ , ∂z

∂ ∂ ∂ = −r cos α cos β sin γ + r cos α cos β cos γ . ∂γ ∂x ∂y As the vectors

∂ ∂ ∂ ∂ ∂x , ∂y , ∂z , ∂u

form an oriented orthonormal basis of the tangent

space, one checks that the vectors length is, respectively, equal to

∂ ∂ ∂ ∂ ∂r , ∂α , ∂β , ∂γ

are pairwise orthogonal and their

1, r, r cos α, r cos α cos β. It follows that the following normalized vectors 1 ∂ 1 ∂ ∂ 1 ∂ , , , ∂r r ∂α r cos α ∂β r cos α cos β ∂γ form an oriented orthonormal basis of the tangent space. Passing to the dual, we obtain that the differential forms e1 = dr, e2 = rdα, e3 = r cos αdβ, e4 = r cos α cos βdγ ,

4.1 Classical Theory of Harmonic Functions

145

in this order, form an oriented orthonormal basis of the cotangent space. This implies that ∗ e1 = e2 ∧e3 ∧e4 , ∗ e2 = −e3 ∧e4 ∧e1 , ∗ e3 = e4 ∧e1 ∧e2 , ∗ e4 = −e1 ∧e2 ∧e3 and ∗(e1 ∧ e2 ∧ e3 ∧ e4 ) = 1 (we list only the action of the Hodge star operator on the forms that we need). Going back to dr, dα, dβ, and dγ , we obtain ∗ dr = r 3 cos2 α cos βdα ∧ dβ ∧ dγ , ∗ dα = −r cos2 α cos βdβ ∧ dγ ∧ dr, ∗ dβ = r cos βdγ ∧ dr ∧ dα, ∗ dγ = − ∗(dr ∧ dα ∧ dβ ∧ dγ ) =

r dr ∧ dα ∧ dβ, cos β 1

r 3 cos2 α cos β

.

We are now close to the end. We have dF =

∂F ∂F ∂F ∂F dr + dα + dβ + dγ , ∂r ∂α ∂β ∂γ

so ∗ dF =

∂F 3 r cos2 α cos βdα ∧ dβ ∧ dγ − ∂r ∂F − r cos2 α cos βdβ ∧ dγ ∧ dr+ ∂α ∂F + r cos βdγ ∧ dr ∧ dα− ∂β −

∂F r dr ∧ dα ∧ dβ. ∂γ cos β

The only place where more complicated calculations appear is the computation of d ∗ dF :  2 ∂ F 3 ∂F 2 d ∗ dF = 3r cos2 α cos β + r cos2 α cos β + 2 ∂r ∂r +

∂F ∂ 2F 2r sin α cos α cos β+ r cos2 α cos β − 2 ∂α ∂α

∂ 2F ∂F r sin β+ r cos β − ∂β 2 ∂β  r ∂ 2F + dr ∧ dα ∧ dβ ∧ dγ . cos β ∂γ 2

+

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4 Elliptic Equations

Finally, F = ∗ d ∗ dF =

∂ 2F ∂F 1 ∂ 2F 1 1 ∂ 2F + + + 2 + ∂r 2 r 2 ∂α 2 r 2 cos2 α ∂β 2 r cos2 α cos2 β ∂γ 2 +

2 tan α ∂F tan β ∂F 3 ∂F − − 2 . 2 r ∂r ∂α r r cos2 α ∂β

& %

Remark 4.1.24 The above method was efficient, because the matrix DF of the base F change (r, α, β, γ ) → (x, y, z, u) has the property that DF T · DF is diagonal. ∂ ∂ ∂ ∂ Therefore, the vectors ∂r , ∂α , ∂β , and ∂γ are orthogonal and the Hodge star operator has rather simple form.

Green’s Functions Problem 4.1.2 Determine Green’s function for the open set {(x1 , x2 , x3 ) ∈ R3 : x1 > 0}. 1 |x −y|−1, which is Green’s Solution Our first attempt is the function 0 (x, y) = 4π 3 function for R . However 0 does not vanish when y1 = 0. The motivation for the answer comes from physics, namely, the interpretation of Green’s function as the electric potential. The function y → 0 (x, y) is the potential associated with a unit charge placed at x. If we want the potential to be 0 on the plane y1 = 0, we need to place another charge, symmetrically to x with respect to y1 = 0 and give it the opposite sign. Therefore, our educated guess for Green’s function of the half space is

G(x, y) = 0 ((x1 , x2 , x3 ), y) − 0 ((−x1 , x2 , x3 ), y). We leave it to the reader to check that G is indeed Green’s function for the halfspace. & % Problem 4.1.3 Find Green’s function of the unit ball B(0, 1) ⊂ R3 . Solution We will act in a similar way as above, i.e., we are going to place another charge so that the sum of the potentials cancels on the boundary of B(0, 1). Now the symmetry is not a reflection, but the inversion with respect to the unit sphere, that is, the map x → i(x) =

x . |x|2

4.1 Classical Theory of Harmonic Functions

147

We leave it to the reader to check that the function G(x, y) = 0 (x, y) − 0 (i(x)|x|, y|x|), where 0 =

1 4π |x

− y|−1 , is Green’s function for the unit ball.

& %

Problem 4.1.4 Deduce the following Poisson formula for the unit ball B(0, 1) in R2 :   it  π e +x f (eit ) dt, Re it (4.5) u(x) = e − x −π where x ∈ B(0, 1) and f : ∂B(0, 1) → R is the function in the boundary condition. Then the function u given by (4.5) is harmonic inside the disk and its restriction to the boundary is f . Solution The integral on the right-hand side of (4.5) is equal to 1 2π





y+x Re y −x S(0,1)

 f (y) dy,

so we need to show that the normal derivative of Green’s function is equal to   y+x 1 Re , 2π y−x where the division in the parenthesis is in the sense of complex numbers. By analogous arguments as in Problem 4.1.3, Green’s function is equal 1 x (− ln |x − y| + ln | 2 − y| − ln |x|), 2π |x| and the last term will be zero when differentiated over y. At |y| = 1 and x fixed, the normal derivative of ln |x − y| with respect to y is equal to (x1 − y1 )y1 + (x2 − y2 )y2 . |x − y| Here, we write x = (x1 , x2 ), y = (y1 , y2 ). The normal derivative of Green’s function at y ∈ ∂B(0, 1) equals (y1 − x1 /|x|2 )y1 + (y2 − x2 /|x|2)y2 (y1 − x1 )y1 + (y2 − x2 )y2 − , 2 |x − y| | |x|x 2 − y|2

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4 Elliptic Equations

As |y| = 1, we have |y − x| = |y −

x | · |x|, |x|2

so the last formula becomes

(y1 − x1 )y1 + (y2 − x2 )y2 − (|x|2y1 − x1 )y1 − (|x|2 y2 − x2 )y2 = |y − x|2 (y 2 + y22 )(1 − |x|2 ) . = 1 |y − x|2

(4.6)

Now, writing y = y1 + iy2 and x = x1 + ix2 , we get |y|2 (1 − |x|2) = |y|2 − |x|2 = Re(y +x)(y¯ − x), ¯ so (4.6) becomes Re y+x y−x . From this and (4.3), the formula follows readily. & %

Harmonic and Holomorphic Functions Problem 4.1.5 Using Stokes’ Theorem (Theorem 1.1.6) prove the Cauchy Integral Formula for a disk (Theorem 4.1.13). Solution Take z0 and ε > 0 small enough. Set  = B(z0 , r) \ B(z0 , ε); clearly f (z)  ⊂ U (we use the notation from Theorem 4.1.13). Let ω = z−z dz. As f is 0 holomorphic, dω = 0. Hence, by Stokes’ formula 

 ω= ∂

dω = 0, 

that is 

 ω= ∂B(z0 ,r)

ω. ∂B(z0 ,ε)

To estimate the last integral, we note that for any δ > 0, there exists an ε > 0 such that if |z − z0 | ≤ ε, then |f (z) − f (z0 )| < δ. Using this, we obtain    

∂B(z0 ,ε)

      f (z) f (z) f (z0 )    − 2πif (z0 ) =  dz − dz ≤ z − z0 ∂B(z0 ,ε) z − z0 ∂B(z0 ,ε) z − z0   δ |f (z) − f (z0 )| δ dσ ≤ = 2πε = 2πδ, |z − z | ε ε 0 ∂B(z0 ,ε) ∂B(z0 ,ε)

where we have used the fact that |



f (z)dz| ≤



γ |f (z)| dσ . f (z) This estimate shows that the difference between ∂B(z0 ,r) z−z dz and 2πif (z0 ) 0 is bounded by 2πδ, but δ is arbitrary. This proves Theorem 4.1.13. & % γ

4.1 Classical Theory of Harmonic Functions

149

Problem 4.1.6 Assume  = {(x, y) : x 2 + y 2 < 1} is a unit disk in R2 . Assume u ∈ C 2 () ∩ C() is a classical solution of the Dirichlet problem u = x 2

in ,

u|∂ = x 2 + y 2 − 5xy + 2. Calculate the maximal value of u in . Solution Since −u ≤ 0 in , we know that u satisfies the maximum principle (Theorem 4.1.17) and therefore max u = max u = max (3 − 5xy). x 2 +y 2 =1

∂



An easy calculation shows that max u = 

11 . 2

♦

4.1.3 Problems Harmonic Functions: Basic Properties Problem 4.1.7 Prove that if  ⊂ Rn is open and f :  → R is C 3 and harmonic, then for every vector v ∈ Rn the directional derivative v · ∇f is harmonic. Problem 4.1.8 Let n > 2. Use the definition of a harmonic function to show that 1−n/2  u(x1 , . . . , xn ) = x12 + · · · + xn2 is harmonic in Rn \ {0}. Problem 4.1.9 Prove that the function u : R2 \ {0} → R given by     u(x, y) = ln x 2 + y 2  is harmonic.

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4 Elliptic Equations

For the next problem, it might be useful to know how to write the Laplace operator in spherical coordinates (even though the full formula is not needed). We refer to Problem 4.1.1 above for an algorithm. Problem 4.1.10 Find all functions f : Rn \ {0} → R which are harmonic and spherically  symmetric, that is, f (x1 , . . . , xn ) = u(r) for a function u : (0, ∞) → R, where r =

x12 + . . . + xn2 , when

(a) n = 2, (b) n > 2. Compare the results with Problems 4.1.8 and 4.1.9. Problem 4.1.11 Suppose u :  → R is a C 2 smooth function, where  ⊂ Rn is an open set. Show that u is harmonic if and only if it has the mean value property (either for balls or for spheres). Problem 4.1.12 Show that if u is a continuous function, then it has the mean value property for spheres if and only if it has the mean value property for balls. Problem 4.1.13 Let f : Rn → R be a continuous function having the mean value property. Let η : R≥0 → R be an integrable function such that Rn η(|x|) dx = 1. Show that  f (y)η(|x − y|) dy = f (x). Rn

Hint One way to show this is by means of the mean value property for balls, i.e., to show the result when η is a characteristic function, then when η is a finite sum of characteristic functions and finally to pass to a suitable limit. Another approach is an application of the mean value property for spheres and the Fubini theorem. Problem 4.1.14 Show that a continuous function f :  → R, which has the mean value property, is necessarily C ∞ smooth. Hint Consider functions η from Problem 4.1.13 above, which are C ∞ smooth and have small support. In particular, we know that continuity plus the mean value property imply harmonicity. Problem 4.1.15 Let un be a sequence of harmonic functions on some open set U ⊂ Rn . Assume that un converges to a function u uniformly on all compact subsets of U . Prove that u is harmonic. Problem 4.1.16 Prove Liouville’s Theorem: a bounded harmonic function on Rn is necessarily constant.

4.1 Classical Theory of Harmonic Functions

151

Problem 4.1.17 Suppose that B = B(0, 3r) is a ball of radius 3r > 0 and u : B → R is a harmonic function that takes only nonnegative values for all x ∈ B(0, r). Prove that for all x, y ∈ B(0, r) we have u(x) ≤ 33n u(y). Hint Connect x and y by a segment and consider the points w, z lying on 1/3 and 2/3 of the segment. Use the mean value theorem to show that u(x) ≤ 3n u(w), u(w) ≤ 3n u(z), and u(z) ≤ 3n u(y). Problem 4.1.18 Use the above problem to show the following Harnack inequality. For any bounded open set  ⊂ Rn and any open subset U ⊂  such that U ⊂  there exists a constant C > 0 such that if u :  → R is harmonic and u(x) ≥ 0, then sup u(x) ≤ C inf u(x). x∈U

x∈U

Problem 4.1.19 Determine whether there exist nonconstant harmonic functions on Rn which are bounded from above. The next result is known as the removal of singularities. Problem 4.1.20 Suppose U ⊂ Rn , n > 1 is an open subset and x0 ∈ U . Let F : U \ {x0} → R be a bounded harmonic function. Prove that F extends to a harmonic function across the point x0 . Problem 4.1.21 Characterize harmonic functions on R1 . Find Green’s function for the interval [a, b]. Problem 4.1.22 Prove that harmonic functions whose all critical points (i.e., points where the gradient vanishes) are nondegenerate (the Hessian matrix has non-zero determinant at that point) cannot have a definite Hessian matrix. This implies that such harmonic functions cannot have local maxima or local minima. Remark 4.1.25 The above problem casts some light on the maximum principle, but it does not prove it in general, because we assume that the Hessian is nondegenerate. It is possible to prove the maximum principle by perturbing locally a harmonic function by a linear function to get a harmonic function with nondegenerate critical points, but this requires additional work. A proof of the Maximum Principle through the mean value theorem is simpler and more elegant. Problem 4.1.23 Consider a flat torus S 1 × S 1 with coordinates x, y such that the lengths of the two circles S 1 × {y0 } and {x0 } × S 1 are a and b, respectively. Find the eigenvalues of the Laplace operator  = ∂xx + ∂yy . Discuss the dependence on a and b.

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4 Elliptic Equations

Problem 4.1.24 Find the eigenvalues of the Laplace operator on the rectangle (0, a)×(0, b), where we restrict our attention to functions vanishing on the boundary of the rectangle.

The Laplace Operator in Different Coordinate Systems Problem 4.1.25 Write the two-dimensional Laplace operator in the radial coordinate system (r, φ), where x = r cos φ and y = r sin φ. Problem 4.1.26 Determine the form of the three-dimensional Laplace operator in spherical coordinates (r, ψ, φ), x = r sin φ sin ψ, y = r sin φ cos ψ, z = r cos φ. Problem 4.1.27 Find the form of the Laplace operator in cylindrical coordinates (r, φ, z), where x = r sin φ, y = r cos φ. Problem 4.1.28 Let R > r0 > 0. Prove that the formula x = (R + r cos φ) cos θ y = (R + r cos φ) sin θ z = r sin φ for r ∈ (r0 , R) gives a local coordinate system on an open subset of R3 (this subset is a drilled torus) and determine the Laplace operator  in coordinates r, φ, θ .

Green’s Functions Problem 4.1.29 Find Green’s function for {(x1 , x2 , x3 , x4 ) ∈ R4 , 0 < x1 < 1}. Problem 4.1.30 Find Green’s function for the sector {(x1 , x2 ) : x2 > 0, x1 > 0, x1 < αx2 }, where α = tan πk , k ∈ N. Problem 4.1.31 Show that Green’s function for the first quadrant of the plane tends to zero uniformly over compact sets with respect to x as |y| → ∞. Problem 4.1.32 Find Green’s function for Neumann’s conditions (that is, the condition G(x, y) = 0 for y on the boundary is replaced by the condition ∂∂G ny (x, y) = 0 for y on the boundary of the open set) for the sector {(x1, x2 ) : x2 > 0, x1 > 0, x1 < βx2 }, where β = tan 2π k , k ∈ N.

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153

Problem 4.1.33 Find Green’s function for  = R2 \ B(0, 1). Problem 4.1.34 Find Green’s function for {(x1 , x2 , x3 ) ∈ R3 , 0 < x1 < 1}. Problem 4.1.35 Find Green’s function for the sector {(x1 , x2 ) : x2 > 0, x1 > 0, x1 < αx2 }, where 0 < α < ∞. This is substantially harder than Problems 4.1.30 and 4.1.32. Problem 4.1.36 Find Green’s function for the annulus {(x1 , x2 ) ∈ R2 : 1/4 < x12 + x22 < 1}. Problem 4.1.37 Find the Fourier expansion of Green’s function for the square (0, 1) × (0, 1). Show that the Green’s function is in L2 . In which dimensions n Green’s function for the n-dimensional hypercube [0, 1]n is in L2 ?

Harmonic and Holomorphic Functions Problem 4.1.38 Prove the equivalence of the four conditions in Definition 4.1.12. Problem 4.1.39 Prove that the real part of a holomorphic function is harmonic. Problem 4.1.40 Suppose U ⊂ R2 is an open set and f : U → C is C 2 . Assume that f = 0 and (f 2 ) = 0. Prove that either f or f is holomorphic. Problem 4.1.41 Show that any harmonic function on a simply connected open subset of R2 is a real part of a holomorphic function. Hint Use the fact that on a simply connected space a closed form is exact. Problem 4.1.42 Let  = {z ∈ C : 0 < |z| < 1}. Let h :  → R be given by h(z) = log |z|. (a) Show that h is harmonic. (b) Show that h cannot be a real part of any holomorphic function on . Problem 4.1.43 Deduce the mean value theorem for harmonic functions in R2 from an analogous result for holomorphic functions. Problem 4.1.44 Deduce the Poisson formula for the unit disk (Problem 4.1.4) from the Cauchy Integral Formula. Problem 4.1.45 Let g : ∂B(0, 1) → C be of class L2 . Show that the following two conditions are equivalent.

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4 Elliptic Equations

(a) The Fourier expansion of g contains only positive terms, in other words, for any m ∈ Z, m < 0 we have  g(φ)φ m dφ = 0. S1

(b) There exists a holomorphic function f (z) : B(0, 1) limr→1 f (reiφ ) = g(φ).

→

C such that

Problem 4.1.46 (Maximum Principle for Holomorphic Functions) Assume  ⊂ C is an open, bounded, connected subset of a complex plane, with smooth boundary. Assume f :  → C is a holomorphic function in , which is continuous up to the boundary. Prove that max |f (z)| ≤ max |f (z)|. z∈

z∈∂

Subharmonic Functions Problem 4.1.47 Let  ⊂ Rn be an open subset and f :  → Rn be an upper semicontinuous function bounded from above. Prove that there exists a sequence fj of continuous functions on , bounded from above and such that for every x ∈ , we have fj (x) ' f (x). Hint Consider fj (x) = supy∈ f (y) − j |x − y| for j = 1, 2, . . .. Remark 4.1.26 The symbol fj (x) ' f (x) denotes that the sequence fj (x) is nonincreasing and, if f (x) > −∞, the sequence fj (x) converges to f (x). If f (x) = −∞, we require that limj →∞ fj (x) = −∞. Problem 4.1.48 Suppose fj (x) is a pointwise convergent sequence of continuous functions such that fj (x) ≥ fj +1 (x). Is the limit function upper semicontinuous? Problem 4.1.49 Let f :  → R be upper semicontinuous and K ⊂  be a compact subset. Show that f attains its global maximum on K, but it does not have to attain its global minimum. Problem 4.1.50 Show that an upper semicontinuous function f :  → R ∪ {−∞} is subharmonic if and only if for any x ∈  and r > 0, such that B(x, r) ⊂ , we have B(x,r) f (y) dy f (x) ≤ . B(x,r) dy Problem 4.1.51 Show that a subharmonic function in an open set does not have local maxima.

4.1 Classical Theory of Harmonic Functions

155

Problem 4.1.52 Let f, g be two subharmonic functions. Are min(f, g) and max(f, g) subharmonic? Problem 4.1.53 Show that (x, y, z) → log(x 2 + y 2 + z2 ) is subharmonic on R3 . Problem 4.1.54 Show that if f is subharmonic and φ : R → R is convex increasing, then φ ◦ f is subharmonic. Problem 4.1.55 Show that a subharmonic function on an interval in R is convex. Problem 4.1.56 Show that a C 2 smooth subharmonic function f satisfies f ≥ 0. Problem 4.1.57 Show that a pointwise limit of a decreasing sequence of subharmonic functions is subharmonic. Problem 4.1.58 Let f :  → C,  ⊂ C be holomorphic. Show that for p ≥ 1, the function |f |p is subharmonic. Problem 4.1.59 Assume  = {(x, y) : x 2 + y 2 < 1} is a unit disk in R2 . Assume u ∈ C 2 () ∩ C() is a classical solution of the Dirichlet problem −u − 2ux = −6u2

in ,

u|∂ (x, y) = x 4 + y 2 − x 2 y. Calculate the maximal value of u in . Problem 4.1.60 Let  ⊂ Rn be an open, bounded, connected set with smooth boundary. Show that if u ∈ C 2 () ∩ C 1 () is a solution to the Neumann problem u = 0 in , ∂u = 0 on ∂, ∂ n then u is a constant function. Problem 4.1.61 (Serrin’s Lemma) Let  be an open, bounded, connected subset of Rn with smooth (of class C 2 ) boundary. Assume that u ∈ C 2 () is a nonnegative function and that it is a solution of the equation −u(x) + c(x)u(x) = 0

in ,

where c ∈ L∞ () (no sign condition on c). Show that then either u is strictly positive in  or u ≡ 0. Moreover, if x0 ∈ ∂ is such that u(x0 ) = 0 and u is strictly positive in , then ∂u (x0 ) < 0, ∂ n where n is a unit outward normal vector to the boundary at x0 .

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4 Elliptic Equations

Problem 4.1.62 Let  ⊂ R2 be an open, bounded, connected set with smooth boundary. Let C(∂) denote the Banach space of all continuous functions on ∂. Denote further X = {f ∈ C(∂) : there exists a classical harmonic function u on  s.t. u|∂ = f }.

Suppose φ ∈ X and h ∈ C 2 ()∩C() is a harmonic function with boundary values equal to f . Suppose x0 ∈  is arbitrary. Show that the mapping Lx0 defined as Lx0 (φ) = h(x0 ) is a continuous, linear functional on X. Calculate the norm of L.

4.2 Weak Solutions 4.2.1 Theoretical Background This chapter deals with weak solutions to linear elliptic problems. They appeared previously in Sect. 3.3, here we want to discuss them in more detail. We are interested in partial differential equations of the form −

n 

(aij (x)u(x)xi )xj +

i,j =1

n 

bi (x)u(x)xi + c(x)u(x) = f (x)

(4.7)

i=1

considered in some open subset  of Rn , and in solutions to these equations, satisfying different kinds of boundary conditions. As mentioned in Sect. 3.3, finding a classical (i.e. twice differentiable) solution to (4.7) is usually very difficult, if not downright impossible. To facilitate the task of finding a solution, we shall do two things. First, we shall look for a solution, which is only weakly differentiable (and lies in an appropriate Sobolev space, see Sect. 2.3). Second, we will reformulate Eq. (4.7) into another form (the so-called weak formulation).

Physical Motivation Before we proceed with formal introduction of weak solutions, let us start with main motivation, which comes from the observation of physical systems. They can be described very efficiently by differential equations and, at the same time, they follow certain general principles, saying that some physical quantities (work, distance covered, energy) are, in the evolution of the system, minimized.

4.2 Weak Solutions

157

To illustrate this, consider a particle of mass m, moving along a line, which at the time t = 0 is at the point a, and at the time t = 1—at the point b. Assume also that no external forces act on the particle. Hamilton’s principle says that the position x(t) ¯ of the particle at the time t ∈ [0, 1] is such that the integral from t = 0 to t = 1 of the kinetic energy of the particle is smallest possible. More formally, the functional 

 2   (s) ds,  dt 

1  dx

I (x) = 0

defined on all smooth functions of the x(t) satisfying x(0) = a and x(1) = b, attains its minimal value on the function x. ¯ Assume now that the value of I is indeed smallest at the function x. ¯ Let us perturb x¯ with a smooth function η, vanishing at the endpoints of the interval [0,1]. For fixed x¯ and η, we consider the quadratic function of the z variable: F (z) = I (x¯ + zη) = A + 2Bz + Cz2 , where  1  d x(s) ¯ 2

 A= 0

   dt  ds, B =



1 0

dη(s) d x(s) ¯ ds and C = dt dt

1  dη(s) 2

 0

   dt  ds.

This function attains its minimum at z = 0 (because the value of I is minimal at u), ¯ thus F (0) vanishes; this is a simple, quadratic function and we can easily calculate this derivative, which immediately yields 0 = F (0) = 2B. Integrating the expression for B by parts we get  2B = 2 0

1

dη(s) d x(s) ¯ ds = −2 dt dt



1

0

d 2 x(s) ¯ η(s)ds dt 2

(we use the fact that η vanishes at 0 and 1). Finally, 

1 0

d 2 x(s) ¯ η(s)ds = 0 dt 2

for every smooth function η : [0, 1] → R vanishing at 0 and 1. This implies that ¯ d 2 x(t) =0 dt 2

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4 Elliptic Equations

for every t ∈ (0, 1) (see Problem 4.2.7). The conclusion is the following: if a particle moves in such a way as to minimize the integral of its kinetic energy, then its acceleration is zero. This is in accordance with Newton’s Second Law of Motion. We can express this observation in one more way: a smooth function minimizing the integral 

     dt  ds

1  dx(s) 2 0

in the space of all smooth functions x : [0, 1] → R satisfying the boundary conditions x(0) = a, x(1) = b, solves also the differential equation d 2 x(s) =0 dt 2 with boundary values x(0) = a, x(1) = b. An n-dimensional counterpart of the first derivative of a function is its gradient, while the Laplacian in one dimension is just the second derivative. In this sense the following theorem can be understood as an n-dimensional generalization of the previous result (compare with Problem 4.2.15). Theorem 4.2.1 (Dirichlet’s Principle) Assume  is a bounded domain in Rn with smooth boundary ∂. Consider a functional I , defined on the set K = {u : u ∈ C 2 () ∩ C 1 (), u = φ on ∂} by the formula  |∇u|2 dx.

I (u) = 

Assume u ∈ K. Then the following conditions are equivalent: (i) u  is a solution to the Laplace equation u = 0, (ii) ∇u(x) · ∇ϕ(x) dx = 0 for every ϕ ∈ C0∞ (), 

(iii) I (u) ≤ I (v) for every function v ∈ K. The proof of Dirichlet’s Principle is divided into Problems 4.2.8–4.2.10.

Weak Formulation Theorem 4.2.1 explains origins of weak formulation. In certain cases a partial differential equation may be interpreted as the Euler–Lagrange equation of a variational functional. We will not discuss deeply variational approach to problem of solving PDEs and we will restrict ourselves only to the case of Hilbert spaces.

4.2 Weak Solutions

159

However, it is important to know that weak formulation may be derived also to equations which do not necessary have variational origin. In order to derive the weak formulation of Eq. (4.7), we repeat the same procedure as in Sect. 3.3 (here we deal with more general form of the equation). We multiply (4.7) on both sides by a smooth function φ ∈ C ∞ () and then we integrate both sides of the equation over . Finally, we integrate the terms with the second order derivatives of u by parts, obtaining 

n   i,j =1

 (aij (x)uxi )φxj dx +

+

∂ i,j =1

  n 

n 

i=1

(aij (x)uxi )φnj dσ

  bi (x)uxi + c(x)u φ dx = f φ dx.

(4.8)



As a result, we do not have any terms with second order derivatives of u any more. As long as we neglect possible difficulties with the integral over ∂ (dubbed the boundary term), and if all the integrands are integrable, this equation makes sense for functions with only first weak derivatives. Let us observe that in the case of Laplace’s equation and under assumption φ ∈ C0∞ () we obtain exactly the statement (ii) of Theorem 4.2.1. The idea expressed in the theorem states that, for functions which are sufficiently regular, Eqs. (4.7) and (4.8) are equivalent. On the other hand, the expression (4.8) may be satisfied for more general class of functions. In order not to lose too much original information on the problem, to deal with the boundary term and to formulate the definition of the weak solution of Eq. (4.7) properly, we must take into account: • regularity of the coefficients of the differential operator, i.e. coefficients aij , bi , and c, as well as regularity of the data f appearing in (4.8); • boundary conditions initially imposed on the solution u. • density of smooth functions (in some cases, smooth functions with compact support) in desired function space in which we will seek weak solutions. Since the weak formulation depends both on the equation and on the boundary conditions, we should rather talk of weak formulation of a problem (i.e., an equation augmented with initial and/or boundary conditions) than of an equation. When choosing the appropriate Sobolev space in which we look for the weak solution, we take into account the first point of the above list (integrability conditions of the function u and its gradient ∇u are important). However, sometimes the choice of the correct space requires broader point of view. This will be discussed in the next subsection—see the discussion of Definition 4.2.8. The second of the above points leads to a reformulation of the boundary condition in the sense of traces of Sobolev functions (discussed in Sect. 2.3). The last requirement is important since it enables us to enlarge also the class of available test functions φ. If it is satisfied, we can use a weak solution itself as the test function. This technique is used often to derive the

160

4 Elliptic Equations

so-called a priori estimates for weak solutions, which in turn are crucial to show their further regularity. For these reasons we must stress that the general definition of a weak solution does not exist. It is always tailored for the specific equation. One example was given in Sect. 3.3, see Definition 3.3.1. You will see more examples further in this section. We emphasize that solving Problems 4.2.15–4.2.31 requires specifying the appropriate Sobolev space. Weak solutions are usually much easier to find than classical solutions discussed in Sect. 4.1. This is true already for linear problems like (4.7), with coefficients a, b, and c depending on x, but this approach (and the notion of a weak solution) extends also to the scope of nonlinear equations, which, in general, are much more difficult. The Sobolev spaces, in which we look for solutions of (4.8), allow the use of the whole powerful toolbox of functional analysis. A list of methods which can be applicable include: • • • •

direct method of the Calculus of Variations in Banach spaces; fixed point theorems in Banach spaces; methods of separable Hilbert spaces (the Galerkin method discussed in Sect. 3.3); methods based on duality of Hilbert spaces;

In this book we do not discuss the first two approaches listed above (an interested reader may look, e.g., at the book of Evans [8] or Dacorogna [6]). In the case when the underlying Sobolev space is a Hilbert space, the third and the fourth approach are the most fruitful. The Galerkin method was presented in Sect. 3.3. We present below two other tools based on the theory of Hilbert spaces.

Variational Approach on Hilbert Spaces Dirichlet’s principle suggests that the minimization of certain functionals might play an important role in the study of differential equations. In order to study a larger class of problems, staying away from too abstract and technical level, we shall restrict ourselves to a particular class of functionals on a Hilbert space. Let H be a Hilbert space, endowed with a scalar product (·, ·) and a norm  · . In this space we consider the functional I (u) = u2 + 2L(u), where L is a fixed continuous linear functional on H . Such functional can be understood as an infinite-dimensional counterpart of the quadratic function f (x) = x 2 + 2bx. The function f obviously attains a minimum on R, at the point x, ¯ which is a solution of the equation x + b = 0. The next theorem states that a similar result holds for the functional I .

4.2 Weak Solutions

161

Theorem 4.2.2 The functional I defined as above attains its minimum at exactly one point u¯ ∈ H . The proof of this theorem is split into Problems 4.2.11–4.2.13. First, in Problem 4.2.11 we prove that the functional I is bounded from below, so the infimum of values of I on H is finite, and if M = infu∈H I (u), we may take a sequence un ∈ H such that limn→∞ I (un ) = M. Next, in Problem 4.2.12 we prove that (un ) is a Cauchy sequence, and thus it converges in H to some u. ¯ One easily checks that I (u) ¯ = M. Finally, in Problem 4.2.13 we prove the uniqueness of the minimum. The next theorem shows the relation between finding a minimum of the functional I with solving certain equation in the Hilbert space H . Theorem 4.2.3 Let the functional I (u) = u2 + 2L(u) be as in Theorem 4.2.2. If I attains its minimum at u¯ ∈ H , then u¯ satisfies the equation (u¯ , v) + L(v) = 0

∀v ∈ H.

(4.9)

The proof of the above theorem is the subject of Problem 4.2.14. Lax–Milgram’s Lemma In the case our underlying space of weak solution is the Hilbert space W 1,2 () or W01,2 () but the weak formulation (4.8) need not be written in the form (4.9) for some functional L, we have another one method available. We can think of the left-hand side of (4.8) as of a bilinear form, defined on the Hilbert space and we can apply the Lax–Milgram Lemma. The Lemma is generalization of the Riesz Representation Theorem (see Theorem 3.1.5) in the sense that our bilinear form need not to be symmetric and therefore does not necessarily define a scalar product on the Hilbert space. Theorem 4.2.4 (Lax-Milgram’s Lemma) Assume B : H × H → R is a bilinear form on a Hilbert space H , satisfying the following conditions. (i) B is bounded, i.e. there exists c > 0 such that for all u, φ ∈ H |B[u, φ]| ≤ cuH φH (ii) B is coercive, i.e. there exists d > 0 such that for all u ∈ H B[u, u] ≥ du2H , and let f be a continuous linear functional on H . Then there exists a unique u ∈ H such that the identity B[u, φ] = f (φ) holds for all φ ∈ H .

162

4 Elliptic Equations

Remark 4.2.5 Some authors prefer to call forms satisfying (ii) elliptic rather than coercive.

Regularity of Weak Solutions Once we prove the existence of some weak solution u, we may then ask, if, perhaps, u is more regular than a general Sobolev function, or if it is, in fact, a C 2 () ∩ C() function. If this is the case, then we can easily prove, again integrating by parts, that u solves the original problem (4.7). On the other hand, the functions in a Sobolev space might be discontinuous. If n ≥ 2 and u ∈ W 1,2 (), there is no guarantee that u is continuous (see Problem 2.3.9). Thus, a solution obtained through application of the Galerkin method, Lax–Milgram’s Lemma, or minimization of a functional on a Hilbert space, could be—at least a priori—not differentiable, or even discontinuous, possibly everywhere in its domain, like the function described in Problem 2.3.10. Is it possible, for some equations, to prove that their weak solutions are in fact classical solutions, differentiable enough times to have all the derivatives that appear in the equation continuous? In general it is not true. This loss of smoothness is, e.g., a common feature of hyperbolic equations, see Sect. 5.2. In simple cases we can answer the question of regularity of weak solutions more or less completely. Theorem 4.2.6 (Weyl’s Lemma, Version I) Assume  ⊂ Rn is open. If a function u ∈ W 1,2 () is a weak solution of Laplace’s equation in , i.e.  ∇u · ∇ϕ dx = 0 

for every ϕ ∈ C0∞ (),

then u ∈ C ∞ () and u(x) = 0 at every point x ∈ . Let us remark here that the theorem known as Weyl’s lemma is in fact stronger: it says that not only every weak, but also every distributional solution of Laplace’s equation is smooth. The next theorem explains that notion. Theorem 4.2.7 (Weyl’s Lemma, Version II) Assume  ⊂ Rn is open. If a distribution u ∈ D () is a distributional solution to Laplace’s equation, i.e. it satisfies the equation u, ϕ = 0

for every ϕ ∈ C0∞ (),

then u ∈ C ∞ () and u = 0 at every point x ∈ . In the above theorem, by saying that a distribution u is a C ∞ function we mean that u is a regular distribution, given by a smooth function. One easily checks that a function v is a weak solution of Laplace’s equation in  if and only if v ∈ W 1,2 () and the distribution φ →  vφ (i.e., the regular distribution defined by v) is a distributional solution to Laplace’s equation (we refer to Sect. 2.2 for the brief introduction to the theory of distributions).

4.2 Weak Solutions

163

Similar results can be proved for weak solutions to elliptic and parabolic equations; the interested reader can look for more information in Evans’ textbook [8] (e.g., Theorem 3, Section 6.3, Theorem 7, Section 7.1, and also Section 8.3, dealing with nonlinear equations). There are several ways of proving Theorem 4.2.6. We will show two of them, leading the reader through a series of problems (see Problems 4.2.33, 4.2.34 and 4.2.38). Let us finish the section with the following observation: It might happen that we do not succeed in proving regularity of weak solutions. Still, considering a solution, which is not continuous, or, in fact, need not be even a function (it might be a measure or a distribution), may (and does) lead to new, interesting applications of partial differential equations.

4.2.2 Worked-Out Problems Now, we shall concentrate on specific boundary value problems and we shall see how all that theory works. Let us focus on the Dirichlet problem for Poisson’s equation. In Sect. 4.1 we solved the Poisson equation on the unit ball B(0, 1) ⊂ R2 (see Problem 4.1.4). Now we will show two ways of proving the existence of a unique weak solution on an arbitrary smooth domain in Rn . Problem 4.2.1 Assume  ⊂ Rn is a bounded domain with C 1 boundary. Consider the problem −u = f u=0

in ,

(4.10)

on ∂,

where f ∈ L2 (). Show that there exists a unique weak solution to (4.10). Solution (Variational Approach) The precise definition of a weak solution depends, as we mentioned in the introduction, on the equation and on the boundary conditions imposed. We multiply both sides of the equation by φ and integrate over . We obtain    ∇u · ∇φ dx + φ∇u · n dσ = f φ dx, 

∂



where n stands for the unit normal vector to ∂, pointing outwards of . Let us now look closer at the term with the integral over the boundary of . We are looking for solutions with null trace on ∂. We take this into account asking that u belongs to the Hilbert space W01,2 (). Then, since C0∞ () functions are dense in W01,2 (), and

164

4 Elliptic Equations

eventually we want φ to belong to the same Hilbert space as u, we restrict ourselves to φ ∈ C0∞ () and the whole integral over boundary vanishes. We obtain 

 ∇u(x) · ∇ϕ(x)dx =

f (x)ϕ(x)dx.



(4.11)



Every classical solution to Poisson’s equation satisfies the above integral identity for an arbitrary smooth function ϕ ∈ C0∞ (). To show this, one should proceed as in the proof of the Dirichlet Principle, Theorem 4.2.1; the crucial step is to use Problem 4.2.7 as a tool. This leads to the following definition (compare it to Definition 3.3.1 in Sect. 3.3). Definition 4.2.8 A function u ∈ W01,2 () is called a weak solution to Poisson’s equation −u = f

in ,

with boundary condition u = 0 on ∂ if the following identity holds for every ϕ ∈ W01,2 (): 

 ∇u(x) · ∇ϕ(x)dx = 

f (x)ϕ(x)dx.

(4.12)



At this point a reader may wonder why we are not requiring the weak solution to belong to, say, the space W 1,1 ()? The reasons for the choice of W 1,2 () are twofold: first, this is a Hilbert space, and such choice enables us to use Hilbert space methods. Second, C0∞ () functions are dense in W01,2 () (see Sect. 2.3) and one easily checks that if (4.12) holds for all ϕ ∈ C0∞ (), then it holds for all ϕ ∈ W01,2 (), as well. Therefore, once we prove a weak solution exists, we will be allowed to use it as a test function—this is the start point of many regularity estimates, see Problem 4.2.6 further. Let us consider the functional   I (u) = |∇u(x)|2dx + 2 f (x)u(x) dx. 



It has the form we considered earlier: I (u) = (u, u) + 2L(u). Indeed, Poincaré’s inequality (see Theorem 2.3.21 in Sect. 2.3) implies that the formula  ∇u(x) · ∇v(x) dx. (u, v) = 

4.2 Weak Solutions

165

defines a scalar product in the Hilbert space W01,2 (). Moreover, the functional  L(u) =

f (x)u(x) dx 

is obviously linear and continuous in that space, since it is bounded:  |L(u)| = | 

f (x)u(x) dx| ≤ f L2 () uL2 () ≤ CuW 1,2 () . 0

Therefore, by Theorem 4.2.2, I attains minimum at exactly one point u ∈ W01,2 (), and by Theorem 4.2.3, this u satisfies the equation ∀ v ∈ W01,2 (),

(u, v) + L(u) = 0

thus u is a weak solution to Poisson’s equation. ♦ Solution (Lax–Milgram Approach)  B[u, φ] =

∇u · ∇φ 

is a bilinear form on H = W01,2 (). We can thus rewrite the identity (4.11) as B[u, φ] = (f, φ), where the functional f is given by  (f, φ) =

f φ dx 

(this is, of course, a slight abuse of notation, but it is in accordance with identifying a regular distribution with a function it is defined by, to what the reader should be already accustomed). To apply Lax–Milgram’s Lemma, we need only to check that the bilinear form B is bounded and coercive and that f is a continuous linear functional. Boundedness follows from the Cauchy–Schwarz inequality:    1/2  1/2   |B[u, φ]| =  ∇u · ∇φ dx  ≤ |∇u|2 dx |∇φ|2 dx ; 





the right-hand side can be obviously estimated by uW 1,2 () φW 1,2 () . 0

0

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4 Elliptic Equations

To prove coercivity, we must estimate the form  |∇u|2 dx

B[u, u] = 

from below, by the full W01,2 ()-norm of u. Such an estimate is less obvious, but it does follow from Poincaré’s inequality (Theorem 2.3.21). We have 

 |∇u|2 dx ≥ min 

 1 1 , u2 1,2 , W0 () 2 2c

where c is the constant in Poincaré’s inequality. This proves coercivity of B. As for the right-hand side of (4.11), we check as in variational approach that the functional f is linear and continuous of H . ♦ Remark 4.2.9 Note that in the above reasoning, to prove that the problem (4.10) has a weak solution, we do not need to assume that the function f is square integrable. In our reasoning we used only the fact that φ ∈ L2 (), while we assumed higher regularity for it (φ ∈ W01,2 ()). Therefore the whole proof would go without changes if f belonged to the space of continuous linear functionals on W01,2 (), i.e. to the space W −1,2 () (see Sect. 2.3). Then, the right-hand side of (4.11) should be written simply as f (φ). Problem 4.2.2 Consider the problem − u = f 

∂u =0 ∂ n

in ,

(4.13)

on ∂,

u = 0, 

with f ∈ L2 (). Find the correct weak formulation of the problem and prove the existence of a weak solution. Solution First, note that the problem does not have a unique solution without the condition that the integral of u over  vanishes. Indeed, if the solution exists at all, then any function that differs from it by a constant again solves the problem. Therefore, the integral condition allows us to choose one solution out of infinitely many, differing by a constant. Let us begin with the weak formulation. Proceeding exactly as in the previous problem, we obtain the identity 





∇u · ∇φ dx + 

φ∇u · n dσ = ∂

f φ dx, 

4.2 Weak Solutions

167

where n stands for the unit normal vector to ∂, pointing outwards of . This time, however, the integral over the boundary vanishes for a different reason: the normal derivative of the function u vanishes on the boundary. Thus the weak formulation of the problem (4.13) takes the form, 

 ∇u · ∇φ dx = 

f φ dx.

(4.14)



The bilinear form B is exactly as in the previous problem:  ∇u · ∇φ dx.

B[u, φ] = 

The domain of B is a different Hilbert space this time. We still ask that both u and φ lie in the Sobolev space W 1,2 (). We do not impose, however, the condition that their trace is zero on ∂, because our boundary condition is different. Moreover, we cannot set the condition that ∂u ∂ n |∂ vanishes along ∂ in the definition of the Hilbert space H , because this condition is ill posed and meaningless for functions in W 1,2 (). The trace theorem, even in its strongest forms, does not allow us to 1,2 . We can, however, include the define ∂u ∂ n |∂ , if all we know about u is that u ∈ W condition that the integral of u over  vanishes. Altogether, we set H =W

1,2

$  % () ∩ u : u=0 . 

In fact, if we assume that ∂ is smooth, we can trace back our steps from ¯ Problem 4.2.2 to its weak formulation (4.14) and prove that every if u ∈ C 1 () ∂u solves (4.14), then ∂ n = 0 on ∂ (doing it rigorously requires a variant of Problem 4.2.7, with ∂ in place of ). In this sense the weak formulation encodes the boundary condition: it must be satisfied by any sufficiently regular weak solution. We prove the boundedness of the form B exactly as in the previous problem, while coercivity follows from the variant of Poincaré’s inequality given in Theorem 2.3.22. ♦ To complete our study of different boundary conditions let us consider now a problem with the Robin boundary condition, (u + ∂u ∂ n )|∂ = 0. Problem 4.2.3 Find the weak formulation and prove the existence of weak solutions for the problem −u + u = f

in ,

∂u + u = 0 on ∂, ∂ n

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4 Elliptic Equations

Solution The weak formulation takes the form     ∇u · ∇φ dx − φ∇u · n dσ + uφ dx = f φ dx. 

∂



(4.15)



Note first that in the boundary term (i.e., the integral over ∂) we cannot leave the normal derivative of the function u, but we need to exchange it for −u, according to the boundary condition. Otherwise the bilinear form B given by the left-hand side of (4.15) would not be well defined for functions in the space W 1,2 (), in which we are looking for solutions. We cannot include that condition in the definition of the Hilbert space, either, because it contains a derivative of u at points of the boundary. Thus H = W 1,2 (). The bilinear form takes the form    B[u, φ] = ∇u · ∇φ dx + uφ dσ + uφ dx. 

∂



Now, its boundedness follows from the trace theorem, which gives us the estimate  ∂

uφ dσ ≤ cuW 1,2 () φW 1,2 () .

This time, coercivity of B is obvious: 





|∇u|2 dx +

B[u, u] = 

u2 dx + 

∂

u2 dσ ≥ u2W 1,2 () .

♦ All the problems considered until now had Laplacian as the differential operator in the equation. Let us now look at a more general example. Problem 4.2.4 Assume A is a symmetric, positive definite 2 × 2 matrix and  is an open and bounded subset of R2 . Let f ∈ L2 (). Prove that the problem −div(A∇u) = f

in ,

u=0

on ∂

has a unique weak solution. Solution We are looking for weak solutions in the space W01,2 (). The bilinear form in the weak formulation is given by  (A∇u) · ∇v dx.

B[u, v] = 

4.2 Weak Solutions

169

To prove its coercivity, recall that if the matrix A is positive definite, then there exists a positive constant λ such that 2 

ai,j ξi ξj ≥ λ|ξ |2

i,j =1

for all ξ = (ξ1 , ξ2 ) ∈ R2 . This in turn means that  

(A∇u) · ∇u dx ≥ λ∇u2L2 ().

The above inequality, together with Poincaré’s inequality (Theorem 2.3.21), implies the coercivity of B. The boundedness of B is fairly easy to show. ♦ Now, let us turn our attention to regularity of solutions. Let us begin with an easy observation: if the solution we find is sufficiently smooth, then we do not need to distinguish whether it is a weak or a classical solution, as these two notions coincide in that case. Problem 4.2.5 Assume  ⊂ Rn is an open set and u ∈ W 1,2 () ∩ C 2 () is a weak solution to the Laplace equation in , i.e.  ∇u · ∇ϕ dx = 0 

for all ϕ ∈ C0∞ (),

then u(x) = 0 at every point x ∈ . Sketch of a Solution Using the assumption that u ∈ C 2 , integrate by parts once, then apply the result of Problem 4.2.7.♦ Problem 4.2.6 (Cacciopoli’s Inequality for Laplace’s Equation) Assume  ⊂ Rn is an open set and u ∈ W 1,2 () is a weak solution to Laplace’s equation u = 0. For a ∈ , let Br = B(a, r) and BR = B(a, R) be two concentric balls with radii satisfying 0 < r < R < dist(a, ∂). Prove that there exists a constant C, independent of R and r, such that the following Cacciopoli’s inequality holds: 

C |∇u| dx ≤ (R − r)2

 |u|2 dx.

2

Br

(4.16)

BR

Sketch of a Solution The weak formulation of Laplace’s equation  ∇u · ∇φ = 0 

is satisfied for all φ ∈ W01,2 ().

(4.17)

170

4 Elliptic Equations

Let ζ be a smooth, compactly supported function on , satisfying the following conditions: • 0 ≤ ζ (x) ≤ 1 for all x ∈ , • ζ ≡ 1 on Br , ζ ≡ 0 outside BR , C . • |∇ζ | ≤ R−r A function ζ satisfying the above three conditions exists for any C > 1. Such ζ are often called cut-off functions. Take φ = ζ 2 u in (4.17):     ∇u · ∇φ = ∇u · ζ 2 ∇u + 2ζ u∇ζ = 0. 



Then 





|∇u| ≤

ζ ∇u · ∇u = −2

2

Br

ζ u∇u · ∇ζ.

2



(4.18)



Use Schwarz’ inequality and the properties of ζ to estimate the right-hand side of (4.18) by C R−r



1/2 

1/2

|u|2 BR

|∇u|2

.

BR

This immediately gives (4.16).♦

4.2.3 Problems Problems related to the theory of Hilbert spaces can be found in Sect. 3.1. Problem 4.2.7 Assume  is an open subset of Rn . Prove that if f ∈ L1 () and  fϕ = 0 

for every ϕ ∈ C0∞ (), then f = 0 almost everywhere. Problem 4.2.8 Prove the implication (i) ⇒ (ii) in Theorem 4.2.1. Problem 4.2.9 Prove the implication (ii) ⇒ (iii) in Theorem 4.2.1. Problem 4.2.10 Prove the implication (iii) ⇒ (i) in Theorem 4.2.1. Problem 4.2.11 Prove that the functional I considered in Theorem 4.2.2 is bounded from below.

4.2 Weak Solutions

171

Hint A linear functional L is continuous if and only if there exists a constant C such that for any x we have L(x) ≤ Cx. Problem 4.2.12 Since the functional I considered in Theorem 4.2.2 is bounded from below, the infimum of values of I on H is finite. If M = infu∈H I (u), we may take a sequence un ∈ H such that limn→∞ I (un ) = M. Using the parallelogram law prove that the sequence un is a Cauchy sequence. Problem 4.2.13 Prove that the functional I considered in Theorem 4.2.2 attains its minimal value at exactly one point u. ¯ Problem 4.2.14 Prove Theorem 4.2.3. Problem 4.2.15 Find the weak formulation of the equation u = 0

in ,

with boundary condition u=ψ

on ∂,

where ψ is equal to the trace of a function  ∈ W 1,2 (). Next, prove the existence of a unique weak solution. Problem 4.2.16 (a) Find the weak formulation of the equation −u + u = f

in ,

with boundary condition u=0

on ∂,

where f ∈ L2 (). Next, prove the existence of a unique weak solution. (b) Let us consider the same equation, but now f ∈ W −1,2 (). Show existence and uniqueness of solutions. Problem 4.2.17 Let  = B(0, 2) ⊂ R2 . Determine values of the parameter k ≥ 0 for which there exists a unique weak solution to the following problem −u +

1 x1 u= 120 |x|k u=0

in , on ∂.

172

4 Elliptic Equations

Problem 4.2.18 Assume that  ⊂ R2 is a bounded domain and f ∈ L2 (). Prove that there exists a unique weak solution to the problem −uxx − 4uyy + ux + uy + 5u = f u=0

in , on ∂.

Problem 4.2.19 Assume that  ⊂ R2 is a bounded domain and f ∈ L2 (). Prove that there exists a unique weak solution to the problem −uxx − 4uyy + ux + uy + 5u = f

in ,

u = 2 on ∂. Problem 4.2.20 Assume  ⊂ R2 to be a bounded domain. Prove that there exists a unique weak solution to the problem −uxx − 2uxy − 3uyy + ux = x

in ,

u = 0 on ∂. Problem 4.2.21 Assume  = B(0, 1) ⊂ R2 . Prove that for any α < α0 = 1 there exists a unique weak solution to the following problem 2uxx − 2uxy + uyy =

1 |x|α

u=0

in , on ∂.

Can you find α0 > 1 with the same property? Problem 4.2.22 Assume  = B(0, 1) ⊂ R3 . Prove that for any α > α0 = −3/2 there exists a unique weak solution to the following problem uxx − 2uxy + 2uyy + uzz = |x|α ln |x| in , u=0

on ∂.

Find α0 < −3/2 with the same property. Problem 4.2.23 Assume  = B(0, 1) ⊂ R2 . Prove that for |k| < 4 there exists a unique weak solution to the following problem −div

   1k ∇u = (ln |x|)x1 04 u=0

in , on ∂.

4.2 Weak Solutions

173

Problem 4.2.24 Assume  = B(0, 2) ⊂ R2 . Prove that there exists a unique weak solution to the following problem    1 12 −div ∇u = 03 |x|   12 n · ∇u = 0 03  u = 0.

in , on ∂,



Problem 4.2.25 Assume  = B(0, 2) ⊂ Rn for n > 1. Determine values of a ∈ R for which there exists a unique weak solution to the following problem −u =

n−1 +a |x|

∂u =0 ∂ n



in , on ∂,

u dx = 0. 

Problem 4.2.26 Let δ0 denote the Dirac delta distribution. Prove that the following problem −uxxxx = 1 − δ0

on (−1, 1),

u(−1) = u(1) = 0, ux (−1) = ux (1) = 0. admits a unique weak solution. Problem 4.2.27 Let  = B(0, R) ⊂ R2 and let f ∈ L2n/(n+2) (). Determine values of α ∈ R for which the following problem −u − α|x|2 u = f u=0 admits a unique weak solution.

in , on ∂

174

4 Elliptic Equations

Problem 4.2.28 Let  = B(0, 1) − B(0, 1/2) be an annulus in R2 . The boundary of  consists of two circles, let 1 denote the outer and 2 —the inner one. Consider the problem −u + 2u = ln |x| in , u=0

on

1,

∂u =0 ∂ n

on

2.

Does the problem admit a unique weak solution? Problem 4.2.29 Assume  ⊂ Rn is a bounded domain. Let f ∈ L2 (R). Prove the existence of a unique weak solution to the problem −u + u = f ∂u +u=0 ∂ n

in  on ∂.

Problem 4.2.30 Assume  ⊂ R2 is a bounded domain. Let f ∈ Lp (R) for some p ∈ [1, ∞). For which values of p there exists a unique weak solution to the problem −u + u = f ∂u =0 ∂ n

in  on ∂ ?

Problem 4.2.31 Assume  ⊂ Rn to be a bounded domain. Let {kj (x)}j ⊂ L∞ be a sequence of functions such that, for every j = 1, 2, 3, . . ., they are positive a.e. and kj → k in L∞ () when j → ∞. Assume that the function uj is a weak solution to the problem   − div (kj (x) + 1)∇uj + uj = 1 uj = 0

in , on ∂,

for j = 1, 2, 3, . . .. Prove that the sequence {uj } converges in W 1,2 () to a function u, which is a weak solution to the problem − div ((k(x) + 1)∇u) + u = 1 u=0

in , on ∂.

4.2 Weak Solutions

175

Problem 4.2.32 Assume  = B(0, 1) ⊂ Rn . Consider a family of problems − uj +

1 uj = fj j uj = 0

in 

(4.19)

on ∂,

where fj (x) ∈ L2 () for j = 1, 2, 3, . . .. Assume that fj → f in L2 () when j → ∞ and that, for j = 1, 2, 3, . . ., the function uj is a weak solution to (4.19). Prove that the sequence (uj ) converges in W 1,2 () to a function u, which is a weak solution to the problem −u = f u=0

in  on ∂.

Remark 4.2.10 Let us recall the notion of approximate identity (c.f. Sect. 1.2, Problems 1.2.25 and 1.2.27). Let ψ be a fixed, nonnegative, smooth function on Rn , with support in the unit ball B(0, 1) and such that Rn ψ = 1. We set, for ε > 0, ψε (x) = ε−n ψ(x/ε).

(4.20)

The family of functions {ψε : ε > 0} is referred to as approximate identity. In Sect. 1.2, we considered as an approximate identity a sequence of functions, corresponding to the choice ε = 1/k for k = 1, 2, . . . in the definition of ψε . In the same way as in Problem 1.2.27 one can show that • whenever u is locally integrable on Rn , the function uε = u ∗ ψε is smooth, • if u is bounded and continuous, then u ∗ ψε → u uniformly with ε → 0+ , Lp

• if u ∈ Lp , for p ≥ 1, then u ∗ ψε −→ u with ε → 0+ . Problem 4.2.33 Assume u ∈ W 1,2 () is a weak solution of Laplace’s equation in . Prove that the convolution uε = u ∗ ψε , with ψε as in (4.20), is well defined at every point of the set ε = {x ∈  : dist (x, ∂) > ε} and satisfies in this set Laplace’s equation: uε = 0 (in the classical sense). Hint Prove that uε is a weak solution of Laplace’s equation in ε and use the result proved in Problem 4.2.5. Problem 4.2.34 Show that the result proved in the previous problem, together with the properties of convolutions and harmonic functions, given in Problems 1.2.19 and 4.1.13, implies Theorem 4.2.6 as an immediate corollary. Problem 4.2.35 Show that in the assumptions of Theorem 4.2.6 it suffices to assume u ∈ W 1,1 ().

176

4 Elliptic Equations

Problem 4.2.36 Generalize the result proved in Problem 4.2.5 to linear elliptic equations with variable coefficients. One of the methods of proving smoothness of solutions to parabolic and elliptic equations is based on the idea to prove that the solution u has in fact more weak derivatives than we a priori assume, and, additionally, that (some or all) derivatives of u are weak solutions to similar equations. More elaborate examples of this approach can be found in Sect. 3.3, devoted to Galerkin’s method. The next problem provides a very simple, one-dimensional example of that approach. Problem 4.2.37 Assume u ∈ W01,2 ((a, b)) is a weak solution of the equation −u

= λu on the interval (a, b) ⊂ R, with Dirichlet boundary conditions u(a) = u(b) = 0. In other words, we assume that 

b a





b

u ϕ dx = λ

uϕ dx a

for all ϕ ∈ C0∞ ((a, b)).

Prove, using only the notion of the weak derivative, that u has weak derivatives of arbitrary order in the space L2 ((a, b)). Deduce that u is a C ∞ function. Hint See Problems 2.3.12 and 2.3.39. Problem 4.2.38 (Convergence of Derivatives in L2 ) Assume  ⊂ Rn is an open set and u ∈ W 1,2 () is a weak solution of Laplace’s equation u = 0 in . Consider the functions uε = u ∗ ψε , with ψε as in (4.20). Prove that for every compact K ⊂  and every multiindex α there exists N > 0 such that the sequence {D α u1/j }∞ j =N is a Cauchy sequence in L2 (K). Using completeness of the space W m,2 , conclude that u is a smooth function. Hint Are the derivatives of uε again solutions to Laplace’s equation? In what sense? Having answered these questions, use the previous problem. (Every compact K can be covered by a finite number of small balls of radius r > 0, such that also slightly larger, concentric balls are still contained in .) At the end, use Problem 2.3.39. The approach presented in Problem 4.2.38 leads to another proof of Theorem 4.2.6. An advantage of this method is that we do not have to rely on a very special property of harmonic functions, given in Problem 4.1.13 (that a convolution of a harmonic function with a radial function is again harmonic)—we only use the linearity of the convolution and ellipticity of the equation. Therefore, this method can be applied to equations other than Laplace’s equation, as well.

4.3 Bibliographical Remarks

177

Problem 4.2.39 (Smoothness of Eigenfunctions of the Laplacian) Assume  ⊂ Rn is a bounded domain. Use methods from Problem 4.2.38 to prove that weak solutions u ∈ W01,2 () of the problem −u = λu, where λ ∈ R is a given constant are C ∞ smooth. Problem 4.2.40 Let  ⊂ Rn be an open, bounded, connected set with smooth boundary. Consider a weak solution of the Dirichlet problem −u = 0 in , u=g

on ∂,

where g ∈ C 0 (∂). Show that u ≥ minx∈∂ g(x) a.e. in .

4.3 Bibliographical Remarks Harmonic functions are covered in many places, including Chapter 2.2 of the Evans book [8]. Chapter 9 of the book of Lieb and Loss [16] gives introduction to harmonic functions as well as subharmonic functions. Connections of harmonic functions and holomorphic functions can be found in “Real and complex analysis” of Rudin [20], especially Chapters 10 and 11. The theory of weak solutions to elliptic equations is well covered in Chapters 5 and 6 of the Evans book [8]. One may also look at the book of Taylor [26]. For more information of the variational methods in Banach spaces, we refer the reader to the book of Dacorogna [6]. A full answer to the question of regularity exceeds by far the scope of this book. The problem, formulated for second order elliptic equations, was the central part of David Hilbert’s XIX problem (asked in 1900), and answering it took almost 60 years. The answer came through long, multi-step effort of many famous mathematicians, among them Sobolev, Morrey, Weyl, De Giorgi, Nash, and Moser; a full list would be much, much longer. For more information, we refer the reader to the book of Evans [8] and Taylor [26].

Chapter 5

Evolution Equations

This chapter is quite heterogeneous. However, the unifying theme is time. Here, we deal with equations describing objects (e.g., a vibrating string or a heat conducting rod), which change in time. Sections 5.1 and 5.2 are closely related, because a firstorder equation may be viewed as a special case of hyperbolic problem. For these problems it makes sense to talk about reversing time: we do this in any transport task; also, if we watch a movie showing a vibrating string, it is hard to say if we are watching it the way it was taped or with time reversed. Thus, we can expect that spatial regularity is (more or less) the same as the initial conditions. Problems in Sect. 5.3 behave differently. Namely, we see that the heat equation immediately smooths out the initial data. As a result, we cannot reverse the time in the equation. Nonetheless, there are tools, which can be used for both types, i.e., hyperbolic and parabolic problems. This is apparent when we try to derive estimates on solutions to time dependent problems or to prove uniqueness of solutions.

5.1 First-Order Equations and the Method of Characteristics 5.1.1 Theoretical Background Consider a partial differential equation of the first order F (x, u(x), ∇u(x)) = 0

for x ∈ ,

(5.1)

with F :  × R × Rn → R

a given smooth function.

© Springer Nature Switzerland AG 2019 M. Borodzik et al., Problems on Partial Differential Equations, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-14734-1_5

179

180

5 Evolution Equations

We assume that  is an open subset of Rn , with smooth boundary. We impose on a function u a boundary condition, fixing the values of u on the boundary ∂ of , or on some part of the boundary: u(x) = g(x)

for x ∈

⊆ ∂.

(5.2)

We shall study a particular method of solving equations of this type, known as the method of characteristics. The key idea of that method is that we find the value of u at a point x ∈  by finding a curve γ in , which connects the point x with some point xo ∈ (where the value of u is fixed by the boundary condition). The curve is chosen in such a way that we can calculate the values of u along it; we find such curves by reducing (5.1) to a certain system of ordinary differential equations. This way, the graph of the function u is woven, with the graphs of u over the curves γ , obtained for different points x, playing the role of thread. Let us start first with a simple example. Assume that the function F in (5.1) is linear and homogeneous with respect to the third variable and that it does not depend on z. In other words, we have F (x, z, p) = a(x) · p, with a :  → Rn a fixed smooth function. Equation (5.1) takes now the form a(x) · ∇u(x) = a1 (x)ux1 (x) + . . . + an (x)uxn (x) = 0.

(5.3)

This means that the gradient vector ∇u(x) is at every point x ∈  perpendicular to the given vector a(x). We know, however, that ∇u(x) is also perpendicular to the level set of u at the point x. This means that the vector a(x) is tangent to the level set of u passing through x, therefore, the trajectories (integral curves) of the vector field a lie in level sets of u (in particular, the value of u is constant on every trajectory). We can now find the level sets of u indirectly, by weaving them from the trajectories of the vector field a starting from points on the boundary of . To this end we solve the system of ordinary differential equations ⎧ x˙1 (s) = a1 (x(s)), ⎪ ⎪ ⎪ ⎨ x˙2 (s) = a2 (x(s)), .. ⎪ ⎪ . ⎪ ⎩ x˙ n (s) = an (x(s)) (here x(s) parameterizes the trajectory of the vector field a). We augment this system with initial data (which we derive from the boundary condition (5.2) imposed on u). If the coordinate functions ai of the vector field a are of the class C 1 , this system has a unique solution. By finding this solution we obtain a trajectory x(s) of the vector field a. We know that this trajectory lies on a level set of u, thus u(x(s)) = const. = u(x(0)), where x(0) ∈ ∂. This way we know the value of u

5.1 First-Order Equations and the Method of Characteristics

181

at all points x(s), which allows us to find an either a direct or at least an implicit formula for u at an arbitrary point x ∈ . The curve x(s) is called the characteristic curve (or, shortly, the characteristic) of Eq. (5.3). It consists of the first n coordinates of a more general, (2n + 1)dimensional curve called the characteristic strip, hence the name of the method. We shall now concentrate on finding characteristic curves and strips for an arbitrary function F , in other words—on the ways of reducing (5.1) to a certain system of ordinary differential equations.

The Characteristic System Assume the function u is a C 2 solution of the problem (5.1) with the boundary condition (5.2) and let the curve γ be parameterized as x(s) = (x1 (s), . . . , xn (s)) for s from some interval I in the real line. Let z : I → R be the function which assigns to s ∈ I the value of the function u at x(s), i.e., z(s) = u(x(s)) and let p(s) = (p1 (s), . . . , pn (s)) be the function attributing to s the value of the gradient of u at x(s), that is p(s) = ∇u(x(s)), pi (s) = uxi (x(s))

where for i = 1, . . . , n.

The derivatives of the functions x, z, p with respect to the s variable will be denoted by a dot: x, ˙ z˙ , p. ˙ The function F (x, z, p) is a function of 2n + 1 unknown real parameters (n coordinates of x, one of z and again n coordinates of p). Its gradient consists of the derivatives with respect to the first n coordinates, denoted by Fx , the derivative with respect to the (n + 1)-st coordinate Fz , and the derivative with respect to the last n coordinates, Fp ; in other words Fx (x, z, p) = Dx F (x, z, p) = (Fx1 (x, z, p), . . . , Fxn (x, z, p)), Fz (x, z, p) = Dz F (x, z, p), Fp (x, z, p) = Dp F (x, z, p) = (Fp1 (x, z, p), . . . , Fpn (x, z, p)). Differentiating the functions z and p with respect to the parameter s we obtain z˙ (s) = ∇u(x(s)) · x(s) ˙ = p(s) · x(s) ˙

(5.4)

182

5 Evolution Equations

and p˙ i (s) = Dx (uxi )(x(s)) · x(s) ˙ =

n  ∂ 2u (x(s)) x˙ j (s), ∂xj ∂xi

(5.5)

j =1

for i = 1, . . . , n. Note that the second-order derivatives of u appear in the last equation. Let us differentiate also the original equation, (5.1), with respect to coordinates of x. We obtain Fxi (x, u, ∇u) + Fz (x, u, ∇u) uxi (x) +

n 

Fpj (x, u, ∇u)

j =1

∂ 2u (x) = 0 ∂xi ∂xj

(5.6)

for i = 1, . . . , n. Now, let γ be a curve in the set , parameterized by x(s), with x(s) satisfying the equation x(s) ˙ = Fp (x(s), z(s), p(s)).

(5.7)

(To be more precise, this is a system of equations, since x(s) ˙ has n coordinates, but we will avoid that distinction unless necessary.) Equation (5.4) takes the form z˙ (s) = p(s) · Fp (x(s), z(s), p(s)).

(5.8)

We can transform Eq. (5.5) using (5.6), (5.7) and the symmetry of the second-order derivative of u (we can do it, because we assumed that the solution u is of class C 2 ). This leads to p˙ i (s) =

n  ∂ 2u (x(s)) Fpj (x(s), z(s), p(s)) ∂xj ∂xi j =1

(5.9)

= −Fxi (x(s), z(s), p(s)) − Fz (x(s), z(s), p(s)) pi (x(s)). Taking together (5.7)–(5.9) we obtain a system of 2n + 1 ordinary differential equations ⎧ ⎨ x˙ = Fp , z˙ = p · Fp , ⎩ p˙ = −Fx − Fz p.

(5.10)

To simplify the notation, we have not written in the above system the arguments of the functions involved, but one should keep in mind that the functions x, z, p and their derivatives depend on the parameter s, while F depends on a vector (x, z, p) ∈ R2n+1 .

5.1 First-Order Equations and the Method of Characteristics

183

Definition 5.1.1 The system (5.10) is called the characteristic system of Eq. (5.1). We should remember that the derivation of the system (5.10) required the assumption that the solution u of Eq. (5.1) is of class C 2 (). The key idea in the reasoning was assuming that the tangent vector to the curve γ satisfies Eq. (5.7). Definition 5.1.2 A curve x(s), obtained as part of the solution of the characteristic system (5.10), is called a characteristic, or a characteristic curve of Eq. (5.1), F (x, u(x), ∇u(x)) = 0. A full solution (x(s), z(s), p(s)) of the system (5.10), representing a curve in a (2n + 1)-dimensional space, will be called the characteristic strip of Eq. (5.1). Remark 5.1.3 There is certain terminological confusion in the literature: in some textbooks (e.g., in the book of Evans [8]) the name characteristic is reserved for what we call the characteristic strip, while the curve x(s), which is the projection of (x(s), z(s), p(s)) onto the first n coordinates, is called the characteristic projection. We stick to the terminology used in the book of John [12]. In the important case when the function F in Eq. (5.1) is linear (or, more generally, affine) with respect to the p coordinate, the characteristic system takes particularly simple form. Example 5.1.4 Assume the function F is an affine function of the second and third coordinate, i.e., F (x, z, p) = a(x) z + b(x) · p + c(x), where a :  ⊂ Rn → R, b :  ⊂ Rn → Rn , and c :  → R are given smooth functions. Then Eq. (5.1) takes the form b1 (x) ux1 (x) + . . . + bn (x) uxn (x) + a(x) u(x) + c(x) = 0 for x ∈ .

(5.11)

The first n + 1 equations of the characteristic system (5.10) are (by Eq. (5.11)) '

x(s) ˙ = b(x(s)), z˙ (s) = b(x(s)) · p(s) = −a(x(s)) z(s) − c(x(s)).

(5.12)

This way we obtain a system of equations on coordinates x and z, not depending of p any more. To determine the characteristic curve x(s), it suffices to solve this simplified form (5.12) of the characteristic system. Example 5.1.5 Assume now that the left-hand side of (5.1) is affine with respect to ∇u, that is, F is affine only with respect to the variable p, F (x, z, p) = a(x, z) · p + c(x, z),

184

5 Evolution Equations

where a :  × R → Rn and c :  × R → R are given smooth functions. Equation (5.1) has the form a1 (x, u) ux1 (x) + . . . + an (x, u) uxn (x) + c(x, u) = 0 for x ∈ .

(5.13)

Writing out, as in the previous example, first n + 1 equations of the characteristic system, we obtain (using (5.13)) '

x(s) ˙ = a(x(s), z(s)), z˙ (s) = a(x(s), z(s)) · p(s) = −c(x(s), z(s)).

(5.14)

Again, the right-hand sides do not depend on the variable p; it suffices to solve this simplified characteristic system (5.14) to find characteristic curves of the system (5.13).

Admissible Boundary Conditions In order to have a unique solution to the characteristic system (5.10), we need to impose appropriate initial conditions. Assume now that xo is an arbitrary point of the set ⊂ ∂, in which the boundary conditions (5.2) are set. In what follows, we assume for simplicity that the set near xo is contained in the hyperplane {xn = 0}, while the set  (or at least its intersection with a neighborhood of xo ) is contained in the half-space {x ∈ Rn : xn > 0}. One can transform our original problem to this setting with a diffeomorphism. We want to fix initial conditions for the characteristic system (5.10), to find the characteristic curve originating at xo . We take thus x(0) = xo , z(0) = zo , p(0) = po . Let us now determine the appropriate values of zo and po . The boundary values of u along are, by (5.2), equal to g, thus z(0) = zo = g(xo ).

(5.15)

It is not obvious, however, what should be the value of po , since the boundary conditions do not give an immediate clue about the values of the gradient of u at the boundary.

5.1 First-Order Equations and the Method of Characteristics

However, we assume that condition (5.2) as

185

⊆ {xn = 0}, which allows us to write the boundary

u(x1 , . . . , xn−1 , 0) = g(x1 , . . . , xn−1 , 0). Differentiating the above equality with respect to xi for i = 1, 2, . . . , n − 1 (that is in directions tangent to ) we obtain uxi (x1 , . . . , xn−1 , 0) = gxi (x1 , . . . , xn−1 , 0)

for i = 1, . . . , n − 1.

We set thus pi (0) = gxi (xo )

for i = 1, . . . , n − 1.

(5.16)

We obtained the initial conditions for the first (n−1) coordinates of the vector p(0). We need one more condition, which will allow us to determine the final coordinate pn (0). For that, we shall use the original equation (5.1). Inserting the values of x(0), z(0) and pi (0) for i = 1, . . . , n − 1 into it we get F (xo , g(xo ), gx1 (xo ), . . . , gxn−1 (xo ), pn (0)) = 0.

(5.17)

In the above equation pn (0) is the only unknown; by solving that equation we find the last initial condition for the characteristic system. Note that in general Eq. (5.17) is nonlinear, which means that a vector po = p(0) satisfying conditions (5.16) and (5.17) might be nonunique, or it might not exist at all. Equations (5.15)–(5.17) are called compatibility conditions. Definition 5.1.6 We say that a vector (xo , zo , po ) ∈ R2n+1 of initial conditions is admissible for the characteristic system (5.10) of a boundary value problem (5.1)– (5.2): F (x, u, ∇u) = 0 in , u=g if xo ∈

in

⊆ ∂,

and the compatibility conditions (5.15)–(5.17) are satisfied, i.e. zo = g(xo ), po = (gx1 (xo ), . . . , gxn−1 (xo ), pn (0)),

and F (xo , zo , po ) = 0.

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5 Evolution Equations

Noncharacteristic Initial Conditions In the previous subsection we learned how to determine the initial conditions for the characteristic system (5.10) in order to find the characteristic curve x(s) which crosses at a point xo and ultimately to find the values of u at the points of the characteristic. To solve the problem (5.1)–(5.2) locally in a neighborhood of a point xo ∈ , we must be able to solve the characteristic system for initial values near xo . In other words, we need to check if the compatibility conditions hold not only for the vector (xo , zo , po ), but also if the point xo is slightly perturbed, i.e., we need to find, for all y ∈ sufficiently close to xo , such q(y) that the vector (y, g(y), q(y)) is admissible. The following lemma holds: Lemma 5.1.7 Assume the vector (xo , zo , po ) of initial values is admissible. Assume moreover that the partial derivative of F with respect to the last coordinate of the vector p ∈ Rn does not vanish at the point (xo , zo , po ): Fpn (xo , zo , po ) = 0. Then there exists a neighborhood U of the point xo ∈ such that for every point y = (y1 , . . . , yn−1 , 0) ∈ U ∩ the following system of n equations for the unknown variables b1 , . . . , bn ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

b1 = gx1 (y), .. . bn−1 = gxn−1 (y), F (y, g(y), b1 , . . . , bn−1 , bn ) = 0

has a unique solution. The above lemma lets us define a function q:U∩ qi (y) = bi

→ Rn , for i = 1, . . . , n.

The function q satisfies q(xo ) = po ; it attributes to a point y ∈ U ∩ such a value q(y) that the vector (y, g(y), q(y)) is an admissible vector of initial values for the system (5.10). This motivates the following definition. Definition 5.1.8 We say that a vector (xo , zo , po ) ∈ R2n+1 of initial values is noncharacteristic, if Fpn (xo , zo , po ) = 0, where F is as in (5.1).

5.1 First-Order Equations and the Method of Characteristics

187

Remark 5.1.9 In the general case, when the part of the boundary is not contained in the hypersurface {xn = 0} in a neighborhood of xo , the condition for the initial values vector to be noncharacteristic is Dp F (xo , zo , po ) · ν(xo ) = 0, where ν(xo ) denotes the external normal vector to the boundary point xo .

⊆ ∂ at the

Local Solutions Finally, we move to the key objective of this section, that is to the constructions of a (local) solution to the boundary value problem (5.1)–(5.2). By a local solution we mean a solution which is valid in a neighborhood of , but not necessarily on the whole . Assume the vector (xo , zo , po ) ∈ R2n+1 of initial values is admissible and noncharacteristic. According to Lemma 5.1.7, there is a function q : Rn → Rn such that po = q(xo ) and for every y ∈ from a neighborhood of xo the vector (y, g(y), q(y)) of initial values is admissible. For every such point y = (y1 , . . . , yn−1 , 0) we solve the characteristic system (5.10) ⎧ ⎨ x˙ = Fp , z˙ = p · Fp , ⎩ p˙ = −Fx − Fz p with initial values x(0) = y, z(0) = g(y), p(0) = q(y). This way we obtain a characteristic strip, that is, the curve (x(s), z(s), p(s)) in the space R2n+1 . Clearly, it depends on the vector of initial values, which in turn is determined by the initial point y ∈ . To emphasize this dependence, we shall sometimes write y explicitly as the parameter of the characteristic strip, namely: (x(y, s), z(y, s), p(y, s)). Recall that we assumed, for simplicity, that ⊂ {x ∈ Rn : xn = 0} and  (or at least its intersection with a neighborhood of xo ) is contained in {x ∈ Rn : xn > 0}. The following lemma holds.

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5 Evolution Equations

Lemma 5.1.10 Assume (xo , zo , po ) is a vector of admissible and noncharacteristic initial values. Then there exist • an open interval I ⊂ R, with 0 ∈ I , • a neighborhood W ⊆ ⊂ Rn−1 of the point xo in • a neighborhood V ⊂ Rn of the point xo in Rn ,

and

such that for every v ∈ V there exists a unique pair (y, s) ∈ W × I satisfying v = x(y, s), where x(y, ·) denotes the characteristic curve originating at the point y ∈ W . Moreover, the mappings v → s(v), v → y(v) are of class C 2 . By the above lemma, there is a unique characteristic passing through a point v ∈ V , which originates at a point y = y(v) ∈ W with v = x(y(v), s(v)) (the first parameter denotes the initial point, the second one parameterizes the characteristic curve). We can thus define a function u which attributes to a point v ∈ V the part z of the solution to the characteristic system (5.10) with initial values (y(v), g(y(v)), q(y(v))): u(v) = z(y(v), s(v)).

(5.18)

Theorem 5.1.11 Assume (xo , zo , po ) is a vector of admissible, noncharacteristic initial values. The function u given by the formula (5.18) is of class C 2 . Moreover, it solves the partial differential equation F (x, u(x), ∇u(x)) = 0

for x ∈ V ⊂ 

with the boundary condition u(xo ) = g(xo )

for xo ∈ W ⊂

(the sets W and V are as in Lemma 5.1.10). One should note that the above theorem deals with local solutions, that is solutions which are well defined in a neighborhood (possibly very small) of boundary points which lie in and which give admissible and noncharacteristic initial data for the characteristic system. To prove the existence of global solutions, i.e., solutions defined in the set , we need to check (using other tools) if these local solutions can be extended to the whole .

5.1 First-Order Equations and the Method of Characteristics

189

In practice, solving the characteristic system might be a difficult task, although there are cases (e.g., when F is affine) when it is rather simple. In the general case, a numerical solution might be the only option. We shall now show some examples and see how the method of characteristics works in particular, practical cases.

5.1.2 Worked-Out Problems To make the solutions easier to follow, we recall the general form of the characteristic system. ⎧ ⎨ x˙ = Fp , z˙ = p · Fp , ⎩ p˙ = −Fx − Fz p. The first four examples deal with a quasilinear equation, i.e., an equation in which F is a linear function of the p variable. In this case the characteristic system becomes simpler than the above general form: to make it complete, we do not need to include the equation for p. ˙ However, even in this case Theorem 5.1.11 provides only local solutions to the characteristic system, and once we find a solution, we need to check on what set it is well defined. Problem 5.1.1 Find a solution to the equation x ux + y uy = 2

in the set  = {(x, y) ∈ R2 : x > 0},

(5.19)

satisfying the condition u(1, y) = 3y 2 .

(5.20)

Solution To be precise, this problem is not exactly a boundary value problem, because the line (1, y) ∈ R2 is not the boundary of the domain . Nevertheless, we can consider two domains, 1 = {(x, y) ∈ R2 : x > 1} and 2 = {(x, y) ∈ R2 : x ∈ (0, 1)}. Then the condition (5.20) is imposed on the whole boundary of 1 and on a piece of the boundary of 2 . Assume (x, y) ∈ 1 . The function F takes the form F (x, y, z, p) = x p1 + y p2 − 2. This is a linear function in the variable p = (p1 , p2 ), thus in the characteristic system we can omit the equation for p(s). ˙ The system becomes ⎧ ˙ = x(s), ⎨ x(s) y(s) ˙ = y(s), ⎩ z˙ (s) = x(s) p1 (s) + y(s) p2 (s) = 2,

190

5 Evolution Equations

with the initial condition x(0) = 1, y(0) = α, z(0) = 3α 2 . The parameter α determines for the position of the initial point at the boundary of the domain. Solving this simple system yields x(s, α) = es , y(s, α) = αes , z(s, α) = 2s + 3α 2 . The characteristic curve starting at the point (1, α) is parameterized by (x(s, α), y(s, α)) = (es , αes ). Since u(x(s, α), y(s, α)) = z(s, α) = 2s + 3α 2 , by taking an inverse of the parametrization (which is possible both in the domain 1 and in 2 ) we obtain a formula for u: u(x, y) = 2 ln x + 3

y2 x2

for (x, y) ∈ 1 .

We can easily check that this formula gives a valid solution in 2 as well, and that u is a smooth function on the whole . ♦ Problem 5.1.2 Find a solution to the boundary value problem x1 ux1 + (x1 + x2 ) ux2 + (x1 + x3 ) ux3 = x1 + x2 + x3 u(x1 , x2 , 0) = x1 − x2

for (x1 , x2 , x3 ) ∈  = R × R × R+ , for x1 , x2 ∈ R.

Solution The function F defining the problem has the form F (x, z, p) = x1 p1 + (x1 + x2 ) p2 + (x1 + x3 ) p3 − x1 − x2 − x3 .

5.1 First-Order Equations and the Method of Characteristics

191

As before, F is linear in the p variable, and the characteristic system takes the simpler form ⎧ x˙1 (s) ⎪ ⎪ ⎨ x˙2 (s) ⎪ x˙ (s) ⎪ ⎩ 3 z˙ (s)

= x1 (s, ) = x1 (s) + x2 (s), = x1 (s) + x3 (s), = x1 (s) + x2 (s) + x3 (s).

Using the boundary condition we add initial conditions to the characteristic system: x1 (0) = α, x2 (0) = β, x3 (0) = 0, z(0) = α − β, where α, β are arbitrary real numbers. The solution to the above system is x1 (s, α, β) = αes , x2 (s, α, β) = βes + αses , x3 (s, α, β) = αses , z(s, α, β) = (β − α)(es − 2) + 2αses . Assume now that (x1 , x2 , x3 ) ∈ 1 = R × R × R+ . We want to find a characteristic curve passing through that point. If x1 = 0, then α = 0 and we obtain s=

x3 , x1

α = x1 e−x3 /x1 , β = (x2 − x3 )e−x3 /x1 . The function u is thus given by the formula u(x1 , x2 , x3 ) = (x2 +x3 −x1 )+2(x1 −x2 +x3 )e−x3 /x1

for x1 = 0, x2 ∈ R, x3 ≥ 0. (5.21)

If x1 = 0, then α = 0 and the initial data is characteristic, because Fp3 (x(0), z(0), p(0)) = x1 (0) + x3 (0) = 0.

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5 Evolution Equations

In this case, Theorem 5.1.11 on the existence of local solution does not tell us anything about solutions passing above the line (0, x2 , 0) ⊂ ∂. We obtain two pieces of the solution: one defined on R+ × R × R+ ⊂ , and the other on R− × R × R+ ⊂ . Can these two pieces be glued together to a solution defined on the whole of ? A look at the formula for u (5.21) shows that the solution cannot be extended continuously to the point (0, x2 , x3 ) if x3 = 0, because the limit of u when we approach that point with x1 > 0 differs from the analogous limit with x1 < 0. We can also check that the solution u has no limit at any of the boundary points of the form (0, x2 , 0). Ultimately, we find a solution of the initial equation, valid in the set  1 = R+ × R × R+ , with the boundary data given on a part of the boundary u(x1 , x2 , 0) = x1 − x2

on

⊂ ∂1 ,

where = {(x1 , x2 , x3 ) : x1 > 0, x2 ∈ R, x3 = 0}, ∂1 =

∪ {(x1 , x2 , x3 ) : x1 = 0, x2 ∈ R, x3 ≥ 0}.

In the same way we show that our problem admits a solution in  2 = R− × R × R+ , with initial data given on a piece

of the boundary of 2 ,

= {(x1 , x2 , x3 ) : x1 < 0, x2 ∈ R, x3 = 0}. ♦ Problem 5.1.3 Find a solution of the problem u ux − uy = u − 1

for (x, y) ∈  = R2

satisfying the condition u(x, x) = 2x. Solution Again, the equation in our problem is quasilinear. The values of the function u are prescribed on the line y = x, which divides the plane R2 into two half-planes. Formally, we should solve the problem separately in each of them, but, in fact, the calculations are exactly the same.

5.1 First-Order Equations and the Method of Characteristics

193

The equation has the form F (x, y, u, ∇u) = 0, with F (x, y, z, p) = z p1 − p2 − z + 1. Assume (x, y) = (x(s), y(s)), z(s) = u(x(s), y(s)), p(s) = (ux (x(s), y(s)), uy (x(s), y(s)) = (p1 (s), p2 (s)). We can write out the characteristic system x(s) ˙ = z(s), y(s) ˙ = −1, z˙ (s) = (p1 (s), p2 (s)) · (z, −1) = zp1 − p2 = z − 1. As in the previous example, we do not need the equation for p(s). ˙ We derive the initial conditions for the above system from the initial data ⎧ ⎪ ˙ = α, ⎪ ⎨x(0) y(0) ˙ = α, ⎪ ⎪ ⎩z˙ (0) = 2α. We solve the system, obtaining z(s) = Aes + 1, y(s) = −s + B, x(s) = Aes + s + C, which, taking into account the initial conditions, yields z(s, α) = (2α − 1)es + 1, y(s, α) = −s + α, x(s, α) = (2α − 1)es + s + 1 − α. Finally u(x, y) = z(x(s, α), y(s, α)) = (2α − 1)es + 1 = x + y. ♦

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5 Evolution Equations

Problem 5.1.4 (Nonhomogeneous Burgers’ Equation) Solve the boundary value problem ut + u ux = 1

for (x, t) ∈ R × R+ ,

u(x, 0) = kx

for x ∈ R,

with k ∈ R a given parameter. Solution The function F has the form F (x, t, z, p) = p2 + zp1 − 1. As before, this function is linear in p, thus the equation in quasilinear. The characteristic system has the form ⎧ ⎪ ⎪ ⎨t˙(s) = 1, x(s) ˙ = z(s), ⎪ ⎪ ⎩z˙ (s) = 1, with initial conditions ⎧ ⎪ ⎪t˙(0) = 0, ⎨

x(0) ˙ = α, ⎪ ⎪ ⎩z˙ (0) = kα, and we can solve it, obtaining t (s, α) = s, x(s, α) =

1 2 s + kαs + α, 2

z(s, α) = s + kα. The parameter s is equal to t, and we can easily calculate α from the coordinates of the point (x, t): α=

x − 12 t 2 . kt + 1

Finally, the solution u is given by the formula u(x, t) = t + k

x − 12 t 2 . kt + 1

The solution depends, obviously, on the parameter k.

5.1 First-Order Equations and the Method of Characteristics

195

Note that also the domain of the solution depends on the value of k. A characteristic curve starting at a point (α, 0) is parameterized by t (s, α) = s, x(s, α) =

1 2 s + kαs + α. 2

For k ≥ 0, characteristic curves do not intersect in R × R+ . This allows us to construct a solution, which is well defined in the whole half-plane R × R+ . For k < 0, the characteristic curves intersect at the point ( 2k12 , − 1k ), so the inverse of the parametrization is not well defined. The interval I in Theorem 5.1.11 is equal to (0, − 1k ), and the solution of our boundary value problem blows up to infinity as we approach the line R × {− 1k }, thus it is not well defined in the whole half-space. ♦ Problem 5.1.5 (The Eikonal Equation) Let B(0, 1) denote an open unit disk in R2 . Solve the boundary value problem u2x + u2y = 1 for (x, y) ∈ R2 \ B(0, 1),

(5.22)

u(x, y) = 0 for x 2 + y 2 = 1. Solution This is a fully nonlinear equation, with the function F taking the form F (x, y, z, p) = p12 + p22 − 1

for p = (p1 , p2 ).

We write the characteristic system x(s) ˙ = 2p1 (s), y(s) ˙ = 2p2 (s), z˙ (s) = p1 Fp1 + p2 Fp2 = 2p12 + 2p22 = 2, p˙ 1 (s) = −Fx − Fz p1 = 0, p˙ 2 (s) = −Fy − Fz p2 = 0. Next, we need to determine the initial conditions for that system. We parameterize the unit circle by arc-length [0, 2π) $ α → (cos α, sin α)

196

5 Evolution Equations

obtaining x(0) = cos α, y(0) = sin α, z(0) = 0. We need yet to find the initial conditions for p1 and p2 . Recall that p1 (s) =

∂u (x(s), y(s)), ∂x

p1 (0) =

∂u (cos α, sin α). ∂x

thus

Differentiating both sides of the equation u(cos α, sin α) = 0 with respect to α we get ∂u ∂u (cos α, sin α)(− sin α) + (cos α, sin α) cos α = 0, ∂x ∂x thus −p1 (0) sin α + p2 (0) cos α = 0. Moreover, directly from Eq. (5.22), p12 (0) + p22 (0) = 1. This gives us two possible sets of initial conditions for the vector p:  p1 (0)

= cos α,

p2 (0)

= sin α,

or

 p1 (0) = − cos α, p2 (0) = − sin α.

Let us check if this set of initial data is noncharacteristic, i.e., satisfies the condition 5.1.8. Note that parametrization of the unit circle by arc length corresponds to introducing radial coordinates (α, r) in R2 . This coordinate system “straightens” the boundary of : the unit circle {x 2 + y 2 = 1} is replaced by the interval {α ∈ [0, 2π); r = 1}. However, it is easier to work in the original, Euclidean coordinates (x, y). Since the boundary of  is not flat in these coordinates, the condition for the data to be noncharacteristic reads Fp (x(0), y(0), z(0), p(0)) · ν = 0,

5.1 First-Order Equations and the Method of Characteristics

197

where ν = (n1 , n2 ) is the external normal vector to the boundary of  at the point (x(0), y(0)). In our case, n1 = − cos α, n2 = − sin α. We have thus (2p1 (0), 2p2 (0)) · (n1 , n2 ) = −2(± cos α, ± sin α) · (cos α, sin α) = ∓2, which means that both sets of initial data we obtained are noncharacteristic. We shall work with the set x(0) = cos α, y(0) = sin α, z(0) = 0, p1 (0) = cos α, p2 (0) = sin α, the solution with the other set is analogous. Solving the characteristic system we obtain x(s, α) = (2s + 1) cos α, y(s, α) = (2s + 1) sin α, z(s, α) = 2s, p1 (s, α) = cos α, p2 (s, α) = sin α. Assume now (x, y) ∈ . By finding the inverse to the parameterization of the characteristic curve passing through that point we obtain the relation 2s + 1 =



x 2 + y 2.

This gives us a formula for the solution of our equation. u(x, y) =



x 2 + y 2 − 1.

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5 Evolution Equations

The other set of initial data leads to u(x, y) = 1 −



x 2 + y 2.

We see, in particular, that the solution to this equation is nonunique.

♦

5.1.3 Problems Problem 5.1.6 Find a solution to the boundary value problem −yux + xuy = 1 for (x, y) ∈ R+ × R+ , u(x, 0) = x

for x > 0.

Problem 5.1.7 Find a solution to the boundary value problem y ux − x uy = u u(x, 0) = x

in  = {(x, y) ∈ R2 : x ≥ 0}, for x ≥ 0.

Does the problem have a solution defined on the whole R2 ? If so, is it unique? Problem 5.1.8 Find a solution to the problem x 2 ux + y 2 uy = (x + y)u for (x, y) ∈ R+ × R+ , x u(x, ) = 1 for x > 0. 2 In Problems 5.1.9–5.1.19 the domain of the solution is not specified. You are expected to find the maximal possible domain of the solution and discuss the problems with extending it further. Problem 5.1.9 Determine a solution to the problem ux − xuy = u which satisfies the condition u(x, 0) = x. Problem 5.1.10 Determine a solution to the problem xux + 2yuy + uz = 3u

5.1 First-Order Equations and the Method of Characteristics

which satisfies the condition u(x, y, 0) = x − y. Problem 5.1.11 Determine a solution to the problem uy − uux + 2u = 0 which satisfies the condition u(x, 0) = x. Problem 5.1.12 Determine a solution to the problem uux + uy = 1 which satisfies the condition u(x, x) =

1 x. 2

Problem 5.1.13 Determine a solution to the problem uy + 3ux = −u2 which satisfies the condition u(x, 0) = x 2 . Problem 5.1.14 Determine a solution to the problem ux + uy = u2 which satisfies the condition u(x, −x) = x. Problem 5.1.15 Determine a solution to the problem ux + uy = u2 which satisfies the condition u(x, 0) = g(x), where g is a given continuous function.

199

200

5 Evolution Equations

Problem 5.1.16 Determine a solution to the problem x(y 2 + u)ux − y(x 2 + u)uy = (x 2 − y 2 )u which satisfies the condition u(x, −x) = 1. Problem 5.1.17 Determine a solution to the problem u2x − u2y = xu + y which satisfies the condition u(x, 0) = −x. Problem 5.1.18 Determine a solution to the problem u2x − u2y = xu2 which satisfies the condition u(x, x) = 1. Problem 5.1.19 Determine a solution to the problem ux uy = u

w  = {(x, y) ∈ R2 : x > 0}

which satisfies the condition u(0, y) = y 2 . Problem 5.1.20 Let B = B(0, 1) denote an open disc in R2 . Assume u ∈ C 1 (B) is a solution to the equation a(x, y)ux + b(x, y)uy + u = 0, where a and b are given continuous functions. Show that if a(x, y)x + b(x, y)y > 0 for (x, y) ∈ ∂B, then u ≡ 0.

5.1 First-Order Equations and the Method of Characteristics

201

Problem 5.1.21 Consider the problem  ut +

u2 2

 ≡ ut + uux = 0

for (x, t) ∈ R × R+ ,

x

u(x, 0) = u0 (x). 1. Set u0 (x) = arctg x and let u be a solution to the problem given above. Show that for fixed x ∈ R there holds u(x, t) −−−→ 0. t →∞

Is the convergence uniform with respect to x? 2. Investigate, for arbitrary u0 , the behavior of a solution with t → ∞. Problem 5.1.22 Consider the problem x 3 ux + yuy = c(x, y)u in R2 , u(cos α, sin α) = g(α)

for α ∈ (−π, π],

where g is a continuous function on the unit in R2 and c ∈ C 1 (R2 ). 1. Find a formula for a solution u ∈ C 1 (R2 \ (0, 0)) in R2 \ (0, 0). 2. Assume that c(0, 0) < 0. Show that the function u can be extended to a solution in the class C 1 (R2 ) iff g ≡ 0 (and then u ≡ 0). 3. Assume that c(0, 0) > 0. Show that u can continuously be extended to be a solution onto the whole R2 . 4. Assume that c(0, 0) = 0. Show that in general, for arbitrary c and arbitrary g, one cannot extend u to a continuous solution onto the whole R2 . 5. Let c(x, y) = y and assume that g(0) = g(π) =

π 1 π g( ) = eg(− ) = K e 2 2

for some K ∈ R. Show that lim u(t cos α, t sin α) = K

t →0+

∀α ∈ (−π, π].

Is it possible to extend u to a solution continuous on R2 ? 6. Let c(x, y) = x 2 and assume that π π g( ) = g(− ) = 0. 2 2 Is it possible to extend u to a solution continuous on R2 ?

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5 Evolution Equations

Problem 5.1.23 Consider the problem ut + h(u)ux = 0 in  = {(x, t) : x ∈ R, t > 0} u(x, 0) = g(x)

for x ∈ R,

where h ∈ C ∞ (R) and g ∈ C 0 (R). 1. Solve the characteristic system for the problem. 2. Assume that g ∈ C 1 (R). Formulate conditions on functions h and g that ensure the existence of a unique solution to the problem in R × [0, T ] for T sufficiently small. Find the maximal value Tmax for T . Show that the condition indeed gives a unique solution u ∈ C 1 (R × [0, Tmax ]). 3. Assume that  1 for x < 0, g(x) = 0 for x ≥ 0. Formulate a condition on the function h which ensures that characteristic curves (x(s), t (s)) do not intersect.

5.2 Hyperbolic Problems In this section we begin our work by studying the wave equation, which is probably the most important example of a second-order hyperbolic problem. This equation has been first introduced in the case of one spatial dimension by d’Alembert, to describe the motion of a vibrating string; the multidimensional case was derived by Euler, to describe acoustic waves, and hence the name. The existence of solutions to this equation in the form of a coherent signal traveling with a finite speed is the hallmark of hyperbolic equations. This trait is in a sharp contrast with parabolic equations (discussed in next section), which tend to smear out data. The worked-out problems related to the wave equation are in Sect. 5.2.2 while Sect. 5.2.3 contains exercises. We use here the results on the first-order equations from Sect. 5.1. Generalizations of the wave equation are dealt with in problems in Sect. 5.2.3. The energy estimates, which lead to uniqueness theorems, are so important that they are separated into an independent subsection, see Sects. 5.2.2 and 5.2.3. We also study the first-order hyperbolic linear systems. There is a small intersection of this section with Sect. 5.1 on the first-order equations. We present applications of the Fourier transform to the linear hyperbolic equations with constant coefficients in Sects. 5.2.3 and 5.2.2. The material presented here depends on Sect. 2.3 (Sobolev spaces), on Sect. 5.1 (the first-order equations), and Sect. 3.2 (on the technique of separation of variables).

5.2 Hyperbolic Problems

203

5.2.1 Theoretical Background A key property of hyperbolic equations or systems is that they describe phenomena where a disturbance, e.g. a light signal, travels through a medium with a finite speed. Moreover, the number of directions of propagation is “maximal,” e.g. it is equal to the space dimension in the case of second-order problems. A genuine representative of a hyperbolic second-order equation1 is the wave equation, 1 ut t − u = 0, c2

(x, t) ∈  × R+ ,

 ⊂ Rn ,

(5.23)

augmented with boundary conditions and initial data. A particular case of (5.23) occurs when n = 1. This is the equation of a vibrating string of length l, 1 ut t − uxx = 0, c2

(x, t) ∈ (0, l) × R+ .

(5.24)

In this example u(x, t) is the displacement from the equilibrium position u ≡ 0, at x and at the time instance t. Then, ut t is the acceleration. A derivation of Eq. (5.23) can be found in [8, Chapter 2.4]. From the physical point of view, it is obvious that we should augment (5.24), and likewise the general case (5.23), with the initial string location u(x, 0) = u0 (x) and its initial velocity ut (x, 0) = u1 (x). Now, we have to determine how we are going to hold the string: i.e. whether its ends are fixed, u(0, t) = 0 = u(l, t) for t > 0, or maybe one of its ends is free, for instance ux (0, t) = 0. In the above equations c > 0 is a material constant, called the speed of sound (when we study sound waves) or the speed of light, when (5.23) describes the electromagnetic wave movement in a homogeneous and isotropic region. The region  may be arbitrary. In the simplest case  = Rn , there is no need to impose any boundary conditions. A natural generalization of (5.23) is obtained by admitting a nonhomogeneous medium. In this case the parameters of the equation depend on the space and time variables. We can study anisotropic problems, to complicate things further. Thus, in that most general situation, we get 1 ut t − Lu + gut = f , (x, t) ∈  × R+ , c2 u(x, 0) = u0 (x), ut (x, 0) = u1 (x), x ∈ ,

(5.25)

augmented with appropriate boundary conditions on ∂, where L is a general elliptic operator of the second order in the divergence form, depending on the space

1 In

fact we will only deal with such equations.

204

5 Evolution Equations

variables x and time t. Namely, if Lu =

n 

(aij (x, t)uxi )xj +

i,j =1

n 

bi (x, t)uxi + c(x, t)u,

i=0

where x0 = t and aij = aj i , then we assume that there exists such a number θ > 0, that for each ξ ∈ Rn , n 

aij (x, t)ξi ξj ≥ θ |ξ |2 .

(5.26)

i,j =1

Definition 5.2.1 We shall say that Eq. (5.25) is uniformly hyperbolic, provided that the condition (5.26) is satisfied for all x ∈  and t > 0 and for a positive θ , independent of (x, t). Some lower order terms in (5.25) have physical interpretation, for instance if g ≥ 0, then gut is damping, i.e. it is the resistance proportional to velocity. The term f on the right-hand side is the external force. Our first task is to solve (5.25). In the general case, the Galerkin method can be used here, see Evans’ book [8, Chapter 7.2] and Sect. 3.3 of this text. Naturally, this approach leads to a weak solution. This notion was first introduced in Sect. 3.2, in the context of the Fourier method applied to hyperbolic problems. A thorough discussing of this notion for elliptic problems was presented in Sect. 4.2. We will present it, in the case of the Dirichlet boundary conditions. Definition 5.2.2 Let us suppose that the region  has a smooth boundary. We will say that a function u ∈ H 1 ( × R+ ), satisfying conditions u|∂×(0,T ) = 0, u|×{0} = u0 ∈ H 1 (), is a weak solution to (5.25) with a homogeneous Dirichlet condition, if for each test function ϕ ∈ H 1 ( × R+ ), ϕ|∂×(0,T ) = 0 and ϕ(·, T ) = 0, the following integral identity is satisfied:  0

T



 

(ut ϕt − aij uxi ϕxj − L0 uϕ) dxdt =

where u1 ∈ L2 () and L0 u =

n

 u1 (x)ϕ(x, 0) dx −

i=0 bi (x, t)uxi

+ c(x, t)u.

 f ϕ dxdt,

0



T



(5.27)

5.2 Hyperbolic Problems

205

We refer an interested reader to Theorems 2.3.18 and 2.3.20 from Sect. 2.3 for information about Sobolev spaces and for an explanation how we should understand the restriction of u ∈ H 1 ( × R+ ) to  × {t = 0}. In Sect. 3.3, we introduced the Galerkin method and we saw its application to elliptic and parabolic equations. A particular example of this method is the Fourier method, namely the method of separation of variables; this topic is discussed in Sect. 3.2. For the wave equation (5.23), which is a particular case of (5.25), there are tools adapted to this problem. These include the method of spherical means (Problems 5.2.17, 5.2.18) for dimension n odd and the methods of descent (Problem 5.2.19) in case of n even. Thus, we obtain Kirchhoff’s formulas, when n = 3, and Poisson’s formulas for n = 2. These mentioned formulas lead to classical or weak solutions depending on the smoothness of the data. It is worth noticing that they require the assumption of much higher smoothness of the initial and boundary data to yield C 2 -solutions than we might expect. Now, we will analyze a simple hyperbolic system of equations of the first order. It is convenient to write them down in the matrix form, ut +

n 

Ai (x, t)uxi = f

in  × (0, T ),

(5.28)

i=1

where u(x, t) is a vector in RN , therefore Ai (x, t), i = 1, . . . , n are N by N matrices depending on x and t. We assume that  is a region in Rn with a smooth boundary. We stress that n and N need not be related. Certainly, system (5.28) is complemented with initial conditions and, in some circumstances, with boundary conditions. Definition 5.2.3 We say that the system (5.28) is hyperbolic if for ξ ∈ RN each n and (x, t) ∈  × R+ all the eigenvalues of the matrix A(x, t, ξ ) = i=1 Ai (x, t)ξi are real. If all the eigenvalues of A(x, t, ξ ) are different, then we say that system (5.28) is strictly hyperbolic. The origin of this definition and its relation to a finite speed of signal propagation will be explained in Problems 5.2.38–5.2.41, where plane waves will be introduced. This approach makes it easier to understand the idea of the Fourier synthesis. We concentrate mainly on particular cases where the matrices Ai are constant. Let us note that a single equation ut + aux = f is a hyperbolic system of equations. Let us say more about solutions to (5.28) in a particular form, ut + Aux = 0, when the matrix A is diagonalizable. Let us note that the one-dimensional wave equation may be reduced to such a system.

206

5 Evolution Equations

In a hyperbolic system with constant coefficients we may easily apply a method based on the Fourier transform, see Sects. 5.2.3 and 5.2.2. The Fourier transform of an integrable function was defined in Sect. 2.1. In the case of variable coefficients this method appears difficult and despite its huge theoretical importance we shall not develop it here. It may also be applied to a single hyperbolic equation—see Sects. 5.2.3 and 5.2.2. We have already mentioned that the key property of hyperbolic problems is that the signal propagates with a finite speed. This is strongly connected to the problem of uniqueness of solutions. This problem may be approached in two ways. The first one applies to weak solutions and is related to the Gronwall inequality (see Problem 5.2.7). The other method, more suitable for classical solutions, is based on energy inequalities, leading to the concept of the cone of influence (another group of problems in Sect. 5.2.3, see Problems 5.2.43–5.2.46). In Problems 5.2.47 and 5.2.49, the concept of a characteristic surface appears in a natural way.

Duhamel’s Formula Separately, we will pay attention to inhomogeneous problems, when function f , appearing on the formula right side (5.25) and (5.28), is different than zero. We present a formal derivation below, while its rigorous treatment is done in the problem sections, see Problem 5.2.21 and formulas (5.50) and (5.51), cf. Problems 5.2.3. Let us suppose that S : Rn × Rn × R+ → R is a solution to the wave equation, ∂ 2S  ∂ 2S − =0 ∂t 2 ∂xi2 n

for (t, x) ∈ (s, ∞) × Rn ,

i=1

S(s, s, x) = 0,

∂ S(s, s, x) = f (s, x) ∂t

for x ∈ Rn .

Then, the function 

t

u(t, x) =

S(t, s, x) ds 0

is a solution of the following problem, ∂ 2u  ∂ 2u − =f ∂t 2 ∂xi2 n

for (t, x) ∈ (0, ∞) × Rn ,

i=1

u(0, x) = 0,

∂ u(0, x) = 0, ∂t

for x ∈ Rn .

5.2 Hyperbolic Problems

207

5.2.2 Worked-Out Problems The Wave Equation Problem 5.2.1 Solve the following problem ∂ 2u 1 ∂ 2u (x, t) − (x, t) = 0, c2 ∂t 2 ∂x 2

in R × R+ ,

u(x, 0) = u0 (x),

for x ∈ R,

∂u (x, 0) = u1 (x), ∂t

for x ∈ R,

(5.29)

where the functions u0 and u1 are given. Examine the smoothness of u depending on that of u0 and u1 . Solution We are at the stage of developing ideas, for this purpose we assume that u is a classical solution to (5.29) and our goal is to find out how does it look like. Note that after a change of variables, t = ct, we might assume without the loss of generality that the constant c in Eq. (5.29) equals 1, but we will not do this. Indeed, ∂u ∂t

∂u ∂u =

= c . ∂t ∂t ∂t ∂t However, we will use, for now, the original variable t. We may indicate three possible methods of solving (5.29): (a) We use a “difference of squares” factorization identity 

∂2 1 ∂2 − c2 ∂t 2 ∂x 2



 u=

∂ 1 ∂ − c ∂t ∂x

   ∂ 1 ∂ + ◦ u. c ∂t ∂x

Since differentiations with respect to t and x commute (we assume that u is of class C 2 ), the above formula may be easily verified. Therefore, it is easy to see that u = ϕ + ψ is a solution, provided that 

∂ 1 ∂ − c ∂t ∂x



 ψ =0

and

∂ 1 ∂ + c ∂t ∂x

 ϕ = 0.

We examine each of the above first-order equations by using the method of characteristics, introduced in Sect. 5.1. (b) We reduce (5.29) to a system of first order, see Problem 5.2.25. (c) We introduce new variables, η = x + ct,

ξ = x − ct.

208

5 Evolution Equations

Then, 1 ∂u ∂u ∂u = − , c ∂t ∂η ∂ξ

∂u ∂u ∂u = + . ∂x ∂η ∂ξ

Thus, Eq. (5.29) in new variables will take the following form: ∂ ∂ 2u = −4 0 = −4 ∂η∂ξ ∂η



We conclude from this formula that

∂u ∂ξ

∂u ∂ξ





∂ = −4 ∂ξ

 ∂u . ∂η

(5.30)

does not depend on η, i.e.

∂u = f (ξ ). ∂ξ Next, we see that  u=

ξ

f (s) ds + ϕ(η) = ϕ(x + ct) + ψ(x − ct).

Inserting this formula into the initial conditions yields a system of equations, '

ϕ(x) + ψ(x) = u0 (x), cϕ (x) − cψ (x) = u1 (x).

Differentiating the first of these equations, we obtain a linear system of equations for the first derivatives ϕ and ψ . Solving it, we obtain u(x, t) =

1 1 (u0 (x − ct) + u0 (x + ct)) + 2 2c



x+ct

u1 (s) ds.

(5.31)

x−ct

The above solution is known as d’Alembert’s formula. Let us pass to the regularity analysis of (5.31). Since we assumed that u is in C 2 (up to the boundary), then taking t = 0 implies that u0 is in C 2 . Moreover, taking x = 0 yields u1 ∈ C 1 . Obviously, if u0 ∈ C 2 (R),

u1 ∈ C 1 (R),

then formula (5.31) yields a classical solution. On the other hand, if u0 ∈ H 1 (R),

u1 ∈ L2 (R),

then we claim that d’Alembert’s formula yields not only a function u in H 1 (R × R+ ), but also a weak solution in the sense of Definition 5.2.2. Note that no explicit boundary condition is present in that case.

5.2 Hyperbolic Problems

209

We have to show that u in H 1 (R × R+ ), in order to see this we have to check that  x+a ∂ u1 (s) ds = u1 (x + a) − u1 (x − a), ∂x x−a where the derivative is understood in the weak sense, i.e.  R

∂ϕ ∂x



x+a

 u1 (s) dsdy = −

x−a

R

for all ϕ ∈ C0∞ (R).

ϕ(y)(u1 (y+a)−u1 (y−a)) dy

(5.32)

Of course, if u1 is continuous and vanishing for |x| > R, where R > 0, then (5.32) holds. Since the space C0 (R) is dense in L2 (R), then there is a sequence {un1 }∞ n=1 ⊂ C0 (R) converging to u1 for which (5.32) is satisfied. We will conclude validity of (5.32) for any u1 ∈ L2 (R) after passage to the limit. Now, we check that u is a weak solution. Since u ∈ H 1 (R × R+ ), then by the Trace Theorem, see Theorem 2.3.18, we may compute the trace of u at t = 0, we can see that u(x, 0) = u0 (x). We have to see that (5.27) holds. Let us suppose n ∞ n 2 1 1 2 that {un0 }∞ n=1 ⊂ C ∩ H (R) (respectively, {u1 }n=1 ⊂ C ∩ L (R)) and u0 → n u0 in H 1 (R) (respectively, u1 → u1 in L2 (R)). Then, the d’Alembert formula corresponding to un0 , un1 yields un , a classical solution. Hence, if we take a test function ϕ ∈ H 1 (R × R) vanishing for t > T , then the integration by parts (see also Problem 1.1.25) yields 

T 0



 R

(unt ϕt − unx ϕx ) dxdt =  =

T 0

R



 R

(unxx − untt ) dxdt +

R

unt (x, 0)ϕ(x, 0) dx

un1 (x)ϕ(x, 0) dx.

Taking the limit as n goes to infinity yields (5.27) for the one-dimensional wave equation, i.e. u is a weak solution. ♦ Problem 5.2.2 Solve the following equation ∂ 2u 1 ∂ 2u (x, t) − (x, t) = 0 c2 ∂t 2 ∂x 2

in R+ × R+ ,

u(0, t) = 0

for t > 0,

u(x, 0) = u0 (x),

∂u (x, 0) = u1 (x) ∂t

(5.33)

for x > 0,

where u0 ∈ C 2 (R+ ) and u1 ∈ C 1 (R+ ) are given functions. Analyze the smoothness of the solution.

210

5 Evolution Equations

An Outline of a Solution Let us apply the reflection method. We shall extend u onto R × R+ by the following formula ' u(x, ˜ t) =

u(x, t) for x ≥ 0, −u(−x, t) for x < 0.

Let us note that if u0 (0) = 0, then u˜ is of class C 1 . Since u ∈ C 1 , the boundary condition leads to u1 (0) = 0. This way we obtain additional conditions u0 (0) = 0,

u1 (0) = 0,

which are called compatibility conditions (i.e. the boundary and initial data should be consistent). They are necessary conditions for the smoothness of solutions. Note also that u˜ satisfies (5.33) in R × R+ , hence we may use the formula (5.31). Thus, we have a formula for u; we should add that examination of smoothness of u requires analyzing the solution along the line x = ct. As we shall see later, this is a particular line, see Sect. 5.2.43, the definition of set C, formula (5.60).

Duhamel’s Formula Problem 5.2.3 Solve the equation ∂ 2u ∂ 2u (t, x) − 2 (t, x) = x 2 2 ∂t ∂x

for (t, x) ∈ (0, ∞) × (0, l),

u(t, 0) = 0 = u(t, l), u(0, x) = 0,

for t > 0,

∂ u(0, x) = 0, ∂t

for x ∈ (0, l).

Sketch of a Solution In the classroom practice, it is often easier to guess a particular solution of an inhomogeneous equation. Note that it is easy to solve the equation, vxx + x 2 = 0. 1 4 We obtain v(x) = 12 x + ax + b. We can choose a and b in such a way that v satisfies the boundary conditions i.e.

v(x) =

l3 1 4 x − x. 12 12

5.2 Hyperbolic Problems

211

Then, the function w = u − v satisfies the equation ∂ 2w ∂ 2w (t, x) − (t, x) = 0 ∂t 2 ∂x 2

for (t, x) ∈ (0, ∞) × (0, l)

w(t, 0) = 0 = w(t, l), u(0, x) = −v(x),

for t > 0,

∂ w(0, x) = 0 ∂t

for x ∈ (0, l).

We may apply any of the methods described in Problem 5.2.1 to solve this equation. ♦

Hyperbolic Systems of the First Order Problem 5.2.4 Let us assume that A is a fixed two-by-two matrix, which is diagonalizable. Solve the system of equations, ∂u ∂t (t, x) +

A ∂u ∂x (t, x) = 0, for (t, x) ∈ R+ × R, for x ∈ R, u(0, x) = u0 (x),

(5.34)

where the unknown function u maps R+ × R to R2 . What can be said about its solvability assuming that u0 ∈ C 1 (R; R2 )? Sketch of a Solution Since we assumed that the matrix is diagonalizable, there exist such matrices S and D = diag (λ1 , λ2 ), that SAS −1 = D. Let us introduce a new variable v = Su. We multiply (5.34) by S from the left. We get then vt = Sut = −SAS −1 Sux = −Dvx .

(5.35)

This equation is augmented with initial conditions v0 = Su0 . After rewriting the above system in coordinates of v = (v1 , v2 ), we obtain vt1 (t, x) + λ1 vx1 (t, x) = 0, vt2 (t, x) + λ2 vx2 (t, x) = 0, v 1 (0, x) = v01 (x), v 2 (0, x) = v02 (x).

212

5 Evolution Equations

Above equations are not coupled, thus they may be solved independently (see Sect. 5.1). Using the formula v = Su we find u. Why is u of class C 1 ? ♦

The Use of the Fourier Transform Problem 5.2.5 Let us assume that u0 ∈ H 2 (Rn ) and u1 ∈ H 1 (R). Find a formula for the solution of the equation = uxx (x, t) in Rn × (0, ∞), u(x, 0) = u0 (x), ut (x, 0) = u1 (x) for x ∈ Rn . 1 u (x, t) c2 t t

(5.36)

Examine the smoothness of the constructed solution. Solution Let us start by recalling the definition of the Fourier transform uˆ : Rn → R of the function u, denoted also as F u (see Sect. 2.1). If u ∈ L1 (Rn ), we set  u(ξ ˆ )= e−2πiξ ·x u(x) dx. (5.37) Rn

As a matter of fact, in Sect. 2.1 we were using such a definition for f ∈ S, but it is valid if the integral of the right-hand side exists, i.e. for f ∈ L1 . Parseval’s Identity (see (2.5) in Sect. 2.1) shows that if u ∈ L1 (Rn ) ∩ L2 (Rn ), the Fourier transform is an isometry (in the L2 norm), i.e. uL2 (Rn ) = u ˆ L2 (Rn ) . This fact and the density of the space L1 (Rn ) ∩ L2 (Rn ) in L2 (Rn ) is sufficient to define the transform on L2 (Rn ), see [16, 24, 25]. By Definition 2.2.6, the Fourier transform F is well-defined on tempered distributions. Thus, the application of F to both sides of (5.36) yields an ordinary differential equation with respect to t, with a parameter ξ , uˆ t t = −4c2π 2 |ξ |2 u, ˆ

u(ξ, ˆ 0) = uˆ 0 (ξ ),

uˆ t (ξ, 0) = uˆ 1 (ξ ).

(5.38)

Thus, we immediately obtain that u(ξ, ˆ t) = C1 exp(2πciξ t) + C2 exp(−2πciξ t). Then, we determine the values of C1 and C2 from the initial conditions on (5.38) u(ξ, ˆ t) = uˆ 0 (ξ ) cos(2πctξ ) + uˆ 1 (ξ )

sin(2πctξ ) . 2πc|ξ |

(5.39)

This formula shows that (1 + |ξ |2 )u(ξ, ˆ t) ∈ C([0, T ); L2 (Rn )), in other words, 2 n u ∈ C([0, T ); H (R )) and uˆ t (ξ, t) ∈ C([0, T ); L2 (Rn )), provided that u0 ∈ H 1 , u1 ∈ L2 . We have obtained a weak solution, which enjoys additional regularity.

5.2 Hyperbolic Problems

213

We may obtain a closed formula for u(x, t) by applying the inverse Fourier transform. ♦

Energy Estimates and Uniqueness Problem 5.2.6 Let us assume that A is a real, symmetric, and positively defined N × N matrix. Assume that u is a smooth solution of the problem ut t (x, t) − div (A∇u(x, t)) = 0 in  × (0, +∞), u(x, t) = 0 on ∂ × (0, +∞), ut (x, 0) = u1 (x). u(x, 0) = u0 (x),

(5.40)

Prove that for each t > 0 the following equality holds 

 

 u2t (x, t) + A∇u(x, t) · ∇u(x, t) dx =

 

 2  u1 (x) + A∇u0 (x) · ∇u0 (x) dx. (5.41)

Deduce the uniqueness of solutions. Solution Let us denote the left-hand side of formula (5.41) by I (t). Since u is assumed to be a smooth function and we can differentiate the formula for I with respect to t under the integral sign, we conclude that I is a smooth function of variable t. The differentiation of I yields dI (t) = 2 dt

 (ut t ut + A∇u · ∇ut ) dx. 

Integrating the second term by parts, combined with the observation that ut = 0 on ∂ × (0, +∞), results in  dI (t) = 2 (ut t − div A∇u)ut dx = 0. dt  Here we used the fact that u is a solution to (5.40). In addition, I is a continuous function on [0, ∞), consequently I (t) = I (0) for all t > 0, hence (5.41) holds. Let us assume that we have two solutions u1 and u2 of Eq. (5.40). Let u be their difference, i.e. u = u2 − u1 . Then, u satisfies (5.40) with zero initial conditions and I (0) = 0. Thus, for all t > 0, we have  2 (u2t + A∇u · ∇u) dx = 0. 

214

5 Evolution Equations

If A is positively definite, it implies that for all t ≥ 0 we have u2t = 0 and A∇u · ∇u = 0. Using again the positive definiteness of A, we obtain that ∇u = 0

and ut = 0.

Hence u must be constant, and consequently it equals 0. Finally, 0 = u2 − u1 .

♦

Problem 5.2.7 Let us assume that A is a real, symmetric, and positive definite N by N matrix. Assume also that f is a smooth function and b ∈ R. Prove that, regardless of the sign of constant b, the problem ut t − div A∇u + bu = f in  × (0, +∞), u=0 on ∂ × (0, +∞), ut (x, 0) = u1 (x), if x ∈ , u(x, 0) = u0 (x),

(5.42)

has at most one smooth solution. Solution Let us assume that we have two solutions u1 and u2 of Eq. (5.42). The function u = u2 − u1 solves then the following equation ut t − div A∇u + bu = 0 in  × (0, +∞), u=0 on ∂ × (0, +∞), u(x, 0) = ut (x, 0) = 0. in 

(5.43)

We multiply (5.43) by ut and integrate the result with respect to x over . We get 1 d 2 dt



 

u2t dx −

 ut div A∇u dx = −b



uut dx. 

Let us integrate by parts the second term on the left-hand side and then integrate the result with respect to time t over [0, T ]. We obtain 

 

u2t (x, T ) dx +

 A∇u(x, T ) · ∇u(x, T ) dx ≤ |b|

T



0





(u2 + u2t )(x, t) dxdt.

While transforming the left-hand side, we used the identity, d (A∇u · ∇u) = 2A∇u · ∇ut , dt which is true for all constant symmetric matrices. Then, we applied the inequality 2ab ≤ a 2 + b2 to estimate the right-hand side. Let us note that by the boundary conditions and Poincaré’s inequality, we have 

 u2 dx ≤ C() 

|∇u|2 dx. 

5.2 Hyperbolic Problems

215

Moreover, positive definiteness of the matrix A yields the following estimate A∇u · ∇u ≥ μ|∇u|2 . If we set  a(t) = 

(u2t (x, t) + |∇u|2 (x, t)) dx,

we get the following inequality a(T ) ≤

max{|b|, C()} min{1, μ}



T

(5.44)

a(s) ds. 0

Applying Gronwall’s inequality to (5.44) results in 

max{|b|, C()} T a(T ) ≤ a(0) exp min{1, μ}

 .

Since a(0) = 0, then for all T ∈ (0, +∞) we get a(T ) = 0. Thus, we infer that 0 ≡ u = u2 − u1 . ♦

5.2.3 Problems We present problems divided into sections corresponding to the background, which we exposed earlier. We note that one may use the Fourier method to obtain solutions, but in such a case analysis of the regularity of solutions is more difficult. The tools presented in this section are more appropriate. Frequently, we do not specify what kind of solutions we have in mind. Only the regularity of the data suggests that. However, the reader is asked to determine if the given formula yields a classical or a weak solution.

The Wave Equation Problem 5.2.8 Solve the following problem with homogeneous Neumann boundary conditions. 2 1 ∂2u − ∂∂xu2 u = 0 c2 ∂t 2 ∂u ∂x (0, t) = 0 u(x, 0) = u0 (x), ∂u ∂t (x, 0)

in R+ × R+ , for t > 0, = u1 (x) for x > 0,

where u0 ∈ C 2 i u1 ∈ C 1 are given. Analyze the smoothness of the solution.

(5.45)

216

5 Evolution Equations

Hint Apply the method of reflection. Problem 5.2.9 Carry out detailed calculations for the solutions of equation (5.45), when: u0 (x) = χ[0,π] (x) sin x, u1 (x) = 0. Is the solution of class C 1 , C 2 ? u0 (x) = χ[0,π] (x) sin2 x, u1 (x) = 0. Is the solution of class C 1 , C 2 ? u0 (x) = χ[0,π] (x) cos x, u1 (x) = 0. Is the solution continuous? u0 (x) = 0, u1 (x) = sin(x)χ[0,π] (x). u0 (x) = 0, u1 (x) = sin2 (x)χ[0,π] (x). u0 (x) is given by the formula from point (a), (b), or (c), while u1 (x) is taken from point (d) or (e) (please consider as many combinations as possible). (g) u0 (x) = 0, u1 (x) = χ[0, 1 π] (x) cos2 x.

(a) (b) (c) (d) (e) (f)

2

Problem 5.2.10 Solve the following equation 1 ∂2u c2 ∂t 2

2

∂ u − ∂x in (0, l) × R+ , 2u = 0 u(0, t) = 0 = u(l, t) for t > 0, (x, 0) = u (x) for x ∈ (0, l), u(x, 0) = u0 (x), ∂u 1 ∂t

(5.46)

where u0 ∈ C 2 and u1 ∈ C 1 are given. Analyze the smoothness of the solution. Problem 5.2.11 Solve the following problem, 2 1 ∂2u − ∂∂xu2 u = 0 c2 ∂t 2 ∂u ∂u ∂x (0, t) = 0 = ∂x (l, t) u(x, 0) = u0 (x), ∂u ∂t (x, 0)

in (0, l) × R+ , for t > 0, = u1 (x) for x ∈ (0, l),

where u0 ∈ C 2 and u1 ∈ C 1 are given. Analyze the smoothness of the solution. Problem 5.2.12 Let us assume u ∈ C 2 ((0, l) × (0, +∞)) ∩ C([0, l] × [0, +∞)). Show that u is the solution of equation 1 ∂ 2u ∂ 2u − 2u = 0 c2 ∂t 2 ∂x

in (0, l) × R+

only if u for all x, t, ξ , and η satisfy the following difference equation, u(x, t) − u(x + cξ, t + ξ ) − u(x − cη, t + η) + u(x + cξ − cη, t + ξ + η) = 0. Problem 5.2.13 Give a geometrical interpretation of the difference equation from the above problem. Use it for constructing the solution of equation (5.46). Problem 5.2.14 Let us assume that u belongs to C 2 ([0, l]×[0, ∞) and is a solution of equation (5.46). The point is that u is in C 2 up to the boundary.

5.2 Hyperbolic Problems

217

Show that u0 (0) = 0, u1 (0) = 0, c2 u

0 (0) = 0, u0 (l) = 0, u1 (l) = 0, c2 u

0 (l) = 0.

(5.47)

Remark 5.2.4 Such conditions as (5.47) are called compatibility conditions. Problem 5.2.15 Let us assume that compatibility conditions (5.47) are satisfied. Show that there exists a solution of class C 2 of problem (5.46). Problem 5.2.16 Solve the following problem ∂2u ∂t 2

− u = 0 u(0, x) = u0 (x),

∂u ∂t (0, x)

in R+ × R3 , = u1 (x) for ∈ R3 ,

where given u0 and u1 are spherically symmetrical, i.e. they depend only on |x| and u0 , u1 ∈ C 2 (0, ∞). What else should we assume about u0 and u1 , so that the resulting formula defines functions of class C 2 on R+ × R3 ? Hint (ru)rr = urr + 2rur . Problem 5.2.17 Let us assume that u is a function of class C 2 in Rn × R+ . Let us define a spherical mean of function u over the sphere centered at x and radius R by the formula  1 Mu (x, R) = n−1 u(y) dσ (y). |S (0, R)| S n−1 (x,R) Show that if u is a solution to equation ut t = u

in Rn × R+ ,

(5.48)

then Mu (x, ·) is a solution of equation vt t = vrr +

n−1 vr r

in R × R+ .

(5.49)

Remark 5.2.5 We need to define Mu for R = 0, by setting Mu (x, 0) = u(x), as well as for negative radii, by means of formula Mu (x, R) = Mu (x, −R). Problem 5.2.18 Solve Eq. (5.49) with initial conditions v(R, 0) = Mu0 (x, R),

vt (R, 0) = Mu1 (x, R).

218

5 Evolution Equations

Assuming that n = 3, u0 is of class C 3 (R3 ), while u1 ∈ C 2 (R3 ), derive the Kirchhoff formula for solutions of the wave equation in R3 × R+ , 1 ∂ u(x, t) = 4π ∂t

  t



1 t u0 (x + ty) dσ (y) + 2 4π S (0,1)

 S 2 (0,1)

u1 (x + ty) dσ (y).

Hint First, introduce a new variable w(R, t) = R n−1 v(R, t). Problem 5.2.19 Let us assume that n = 2, u0 is of class C 3 (R2 ), while u1 ∈ C 2 (R2 ). Applying the Kirchhoff formula, derive the Poisson formula for solutions to the wave equation in R2 × R+ , 1 ∂ u(x, t) = 2π ∂t

  t B(0,1)

  u0 (x + ty) u1 (x + ty) t dy + dy. (1 − |y|2)1/2 2π B(0,1) (1 − |y|2 )1/2

Hint Apply a substitution method, i.e., note that solutions of a two-dimensional problem are solutions of the three-dimensional problem but they do not depend on the third variable. If so, we should carry out integration with respect to the third variable in Kirchhoff’s formula. Problem 5.2.20 Let u be a solution to the following problem, ∂2u ∂t 2

2

∂ u − c2 ∂x for (t, x) ∈ R+ × R, 2 = 0 2 ∂u −x for x ∈ R. u(0, x) = 0, ∂t (0, x) = −xe

Show that u is a bounded function satisfying the following estimate, |u(t, x)| ≤

1 . 4c

Moreover, this bound cannot be improved, (i.e. it does not hold if we assume a smaller constant on the right-hand side). Problem 5.2.21 Assume that S(t, s, x), where t > s > 0, x ∈ Rn , n ≤ 3, is of class C 2 with respect to t and x and it is a solution of the following problem, ∂ 2S  ∂ 2S − =0 ∂t 2 ∂xi2 n

for (t, x) ∈ (s, ∞) × Rn ,

i=1

S(s, s, x) = 0,

∂ S(s, s, x) = f (s, x) ∂t

where f is a function of class C 2 and s is a given real parameter.

for x ∈ Rn ,

5.2 Hyperbolic Problems

219

(a) Show that function (t, s, x) → S(t, s, x) is continuous. (b) Show that function 

t

u(t, x) =

S(t, s, x) ds

(5.50)

0

is a solution of the following problem, where f is as above, ∂2u ∂t 2

−

n

u(0, x)

∂2u i=1 ∂x 2 = f i = 0, ∂t∂ u(0, x)

for (t, x) ∈ (0, ∞) × Rn , = 0, for x ∈ Rn .

(5.51)

(c) Solve problem (5.51), when initial conditions do not equal zero, derive a general formula. Remark 5.2.6 Formula (5.50) is called the Duhamel formula. Problem 5.2.22 Show by means of the Fourier transform an explicit form of the Duhamel formula for wave equation in Rn , ut t − u = f

(x, t) ∈ Rn × R+

if u(x, 0) = ut (x, 0) = 0. Problem 5.2.23 Solve equation ∂2u ∂t 2

2

∂ u − ∂x (t, x) ∈ (0, ∞) × (0, l), 2 = f u(t, 0) = 0 = u(t, l), t > 0, u(0, x) = 0, ∂t∂ u(0, x) = 0, x ∈ (0, l),

when (a) (b) (c) (d)

f (x, t) = 1, f (x, t) = sin( πl x), f (x, t) = t sin( πl x), f (x, t) = g(t)h(x), where g and h are smooth.

Hyperbolic Systems of the First Order Problem 5.2.24 Let us assume that A is a given, real two-by-two matrix, such that det A < 0. Let us consider the initial value problem, + A ∂u ∂x + u = 0 for (t, x) ∈ R+ × R, u(0, x) = u0 (x), ∂u ∂t

220

5 Evolution Equations

where u(t, x) ∈ R2 . Show the existence of solutions of class C 1 , assuming that u0 ∈ C 1 (R; R2 ). Hint Consider substitution v(t, x) = exp(λt) u(t, x) for an appropriate number λ. Problem 5.2.25 Write down a one-dimensional wave equation in the form of a system of equations. Notice that it is hyperbolic. Derive d’Alembert’s formula (5.31) by means of Problem 5.2.4. Hint Introduce new variables v1 = ux , v2 = ut , write a system of equations for (v1 , v2 ). Problem 5.2.26 Solve the following system of equations + A ∂u for (t, x) ∈ R+ × R, ∂x = 0 u(0, x) = u0 (x), ∂u ∂t

when u0 ∈ C 1 as well as (a) 

 21 A= ; 11 (b)  A=

 1a , 02

where a is a real number. What shall we assume about a to make the system hyperbolic? Problem 5.2.27 Solve a system of equations + A ∂u ∂x = 0 for (t, x) ∈ R+ × R, u(0, x) = u0 (x) for x ∈ R, ∂u ∂t

when u0 ∈ C 1 and  A=

 1x . 02

Hint We may apply a method from previous problems. Substitution (5.35) leads to two coupled equations of the form wt + μwx = f .

5.2 Hyperbolic Problems

221

Remark 5.2.7 We may show the uniqueness of solutions of hyperbolic systems of equations, by applying, for instance, the methods similar to those used below in Sect. 5.2.3, in Problems 5.2.44–5.2.46.

Hyperbolic Equations of the Second Order, Considered More Generally Problem 5.2.28 Let us assume that  = (0, 1). Let us analyze ut t + ut = (aux )x + f in  × (0, T ), u|∂×(0,T ) = 0, u(x, 0) = u0 (x),

(5.52) ut (x, 0) = u1 (x).

where a is a smooth function, such that a(x) ≥ δ > 0. Let us assume that u0 , u1 ∈ C0∞ (). State and show a theorem about the existence and uniqueness of solutions to problem (5.52), when (a) f ≡ 0, (b) f ≡ f (t) belongs to space C 1 ([0, T ), H 2 () ∩ H01 ()). Problem 5.2.29 Assume that  is a bounded region in Rn , n ≥ 1, with a smooth boundary. Let us consider equation ut t + ut = u + f in  × (0, T ), u|∂×(0,T ) = 0, u(x, 0) = u0 (x), ut (x, 0) = u1 (x).

(5.53)

(a) Show the existence and uniqueness of solutions to problem (5.53) with suitable assumptions on u0 and u1 , provided that f ≡ 0. (b) Show the existence and uniqueness of solutions of problem (5.53) with appropriate assumptions on u0 , u1 , as long as f ≡ f (t) satisfies f ∈ C 1 ([0, T ), H 2 () ∩ H01 ()). (c) (difficult) Show that there is T > 0 such that there exists a unique solution to (5.53) on interval [0, T ), if f (x, t) = −u2 . Find suitable assumptions on u0 and u1 . (d) (difficult) Is T from the (c) above finite?

222

5 Evolution Equations

The Use of the Fourier Transform Problem 5.2.30 Using methods from problem (5.36), derive a formula for the Fourier transform of the following equation − uxx = f in Rn × (0, ∞), u(x, 0) = u0 (x), ut (x, 0) = u1 (x), 1 u c2 t t

(5.54)

where u0 ∈ H 2 (Rn ), u1 ∈ H 1 (R) and f ∈ C([0, T ]; H 1(R)). Show that the obtained solution u belongs to space   C 2 (0, ∞); L2 (Rn )) ∩ C 0 ([0, ∞); H 2(Rn ) . Problem 5.2.31 Let us assume that u0 , u1 ∈ S(R). Find a solution of equation ut t + ut = uxx in R × (0, ∞), u(x, 0) = u0 (x) for x ∈ R, ut (x, 0) = u1 (x) for x ∈ R.

(5.55)

Problem 5.2.32 Let us assume that u0 ∈ L2 (R), u1 ∈ H 1 (R). Show that formula obtained in the above problem defines a weak solution of equation (5.55). Remark 5.2.8 Defining a weak solution is a part of this problem. Problem 5.2.33 Show that there exists at most one weak solution of equation (5.55). Problem 5.2.34 Let us assume that u0 ∈ L2 (R), u1 ∈ H 1 (R) and u is a weak solution to equation (5.55). Find the rate of decay of u in the L2 -norm, i.e. find α > 0 such that u(·, t)2 ≤ Ct −α . Problem 5.2.35 Let us assume that u0 ∈ L2 (R), u1 ∈ H 1 (R) and u is a weak solution of equation (5.54). Show that for each t > 0 the following equality holds:  Rn

(

1 |ut (x, t)|2 + |∇u(x, t)|2 ) dx = c2

 Rn

(

1 |u1 (x)|2 + |∇u0 (x)|2 ) dx. c2

Hint Establish first the claim for smooth solutions. Problem 5.2.36 Solve equation in Rn × R+ , ut t = u − u u(x, 0) = u0 (x), ut (x, 0) = u1 (x).

(5.56)

5.2 Hyperbolic Problems

223

Show that if initial data u0 belongs to space H 1 (Rn ), while u1 ∈ L2 (Rn ), then there exists a constant C depending on u0 1,2 , such that u(·, t) − u0 2 ≤ Ct. Problem 5.2.37 Let us assume that u is a solution of equation (5.56). Show that there exists a constant C, such that for k = 0, 1 and t ≥ 1 the following estimate holds: u(·, t)k,2 ≤ Ce−t . Problem 5.2.38 Let us assume that A is a real, symmetric, and nonsingular twoby-two matrix (the entries are real). Show that if u0 ∈ H 1 (R; R2 ), then function u, which is a solution of system ut + Aux = 0 in R × (0, ∞) u(x, 0) = u0 (x), fulfills the condition u(·, t)L2 (R) = u0 L2 (R) . Problem 5.2.39 Let us assume that A is a real, symmetric, and nonsingular twoby-two matrix, v is an eigenvector, and λ is the corresponding eigenvalue. We set w(x, t) = ei(x−λt )v. Show that w is a solution of a system of equations ut + Aux = 0,

(x, t) ∈ R × R+ ,

where u = (u1 , u2 ) ∈ R2 . Comment Functions w are called plane waves. We may say that formulas obtained by means of the Fourier transform are a generalization of the sum of plane waves. Problem 5.2.40 Let us assume that Ai , i = 1, . . . , N are real, symmetric N by N matrices and system ut +

N  i=1

A i ux i = 0

(5.57)

224

5 Evolution Equations

N is strictly hyperbolic, N see Definition 5.2.3. Let ξ ∈ R while v is an eigenvector of matrix A(ξ ) = i=1 Ai ξi and λ the corresponding eigenvalue. Let us set

w(x, t) = ei(ξ ·x−λt )v.

(5.58)

Show that, w is a solution of system (5.57). Problem 5.2.41 Let us assume that Ai , i = 1, . . . , N and B are real, symmetric N by N matrices, and system ut +

N 

Ai uxi + Bu = 0

(5.59)

i=1

is strictly hyperbolic. Let vector ξ ∈ RN be given. What kind of conditions must be imposed on vector v and a complex number λ, so that the plane wave, given by formula (5.58), is a solution to (5.59)? Prove that Re λ > 0. Find a corresponding estimate of the rate of decay of w, when t tends to infinity. Problem 5.2.42 Let us assume that A is a real, symmetric, and positively defined N by N matrix. Let v ∈ RN be an eigenvector of A, |v|2 = 1 and λ2 is the corresponding eigenvalue. Prove that plane waves w− (x, t) = ei(v·x−λt ),

w+ (x, t) = ei(v·x+λt )

are solutions of equation ut t − div A∇u = 0.

Energy Estimates and Uniqueness Problem 5.2.43 Assume that u ∈ C 2 (Rn × R). Let us introduce the following notation: if (x0 , t0 ) ∈ Rn × R, t1 < t0 , then we set C(x0 , t0 ) := {(x, t) ∈ Rn × R : |x − x0 | ≤ |t0 − t|}, D(x0 , t1 , t0 ) := (Rn × [0, t1 ]) ∩ C(x0 , t0 ).

(5.60)

Let ∇ denote the gradient with respect to the space variables, i.e. Du = (∇u, ut ), and besides, |Du|2 = |∇u|2 + |ut |2 . (a) We set X = (X1 , . . . , Xn , Xn+1 ), where Xi = 2ut uxi ,

i = 1, . . . , n,

Xn+1 = −(|∇u|2 + |ut |2 ).

5.2 Hyperbolic Problems

225

Check that −2ut (ut t − u) = div X, where div X =

n  ∂Xi i=1

∂xi

+

∂ Xn+1 . ∂t

(b) Prove that if (x0 , t0 ) ∈ Rn × R, t1 < t0 , then    2 div X dxdt ≤ |Du| dx − D(x0 ,t1 ,t0 )

B(x0 ,t0 )×{0}

B(x0 ,t0 −t1 )×{t1 }

|Du|2 dx.

Problem 5.2.44 Let us assume that u ∈ C 2 satisfies the equation ut t = u u(x, 0) = u0 (x),

in Rn × R+ , ut (x, 0) = u1 (x) for x ∈ Rn .

(5.61)

(a) In addition, for certain x0 ∈ Rn and t0 > 0 functions u0 and u1 identically vanish in the ball B(x0 , t0 ). Prove that u vanishes in cone C(x0 , t0 ) ∩ {0 ≤ t ≤ t0 }. Which condition is sufficient to prove an analogue statement in the case a coefficient c2 = 1 appears in front of u? (b) Let us assume that v1 , v2 ∈ C 2 satisfy Eq. (5.61) and v1 (x, 0) = v2 (x, 0),

∂ ∂ v1 (x, 0) = v2 (x, 0) ∂t ∂t

in ball B(x0 , t0 ). Show that v1 = v2 in cone C(x0 , t0 ) ∩ {0 ≤ t ≤ t0 }. Problem 5.2.45 Prove the same claim as in the above problem assuming that v1 and v2 are weak solutions. Remark 5.2.9 Cone C(x0 , t0 ) ∩ {0 ≤ t ≤ t0 } is called a cone of dependence. Problem 5.2.46 Let us assume that A is a constant, positively defined symmetric matrix and let us use the notation of Problem 5.2.43. Namely, let us set C(A, x0 , t0 ) := {(x, t) ∈ Rn × R : A(x − x0 ) · (x − x0 ) ≤ |t0 − t|2 }, D(A, x0 , t1 , t0 ) := (Rn × [0, t1 ]) ∩ C(A, x0 , t0 ). Now, let A be a diagonal matrix A = diag (λ1 , . . . , λn ), where λi > 0 for all i = 1, . . . , n. Let us assume that u ∈ C 2 is a solution of problem ut t = div (A∇u), u(x, 0) = u0 (x),

∂u ∂t (x, 0)

x ∈ Rn , t > 0, = u1 (x), x ∈ Rn .

(5.62)

226

5 Evolution Equations

Prove that if u0 and u1 equal zero in the intersection of the cone of dependence with a hyperplane {t = 0}, then u = 0 in D(A, x0 , t1 , t0 ) for any t1 < t0 . The meaning of sets C(A, x0 , t0 ) will be presented below Definition 5.2.10 (See Evans [8, 4.6.1.b]) We shall say that a smooth hypersurface M ⊂ Rn+1 with a field of normal vectors ν = (ν1 , . . . , νn+1 ) is a surface characteristic for equation n+1 

aij (x)uxi xj +

i,j =1

n+1 

bi (x)uxi + cu = f

in ,

i=1

if n+1 

aij (x)νi (x)νj (x) = 0

i,j =1

for each x ∈  ∩ M. Notice that if M is a level set of function ϕ and ∇ϕ = 0 on M, then in above definition, we may take ν = ∇ϕ. The notion of a characteristic surface may be introduced for nonlinear equations and also for systems of equation. Problem 5.2.47 Let C(x0 , t0 ) ⊂ Rn+1 be a set introduced in Problem 5.2.43. Prove that C(x0 , t0 ) is a characteristic surface for wave equation 1 ut t − u = 0 in Rn+1 . c2

(5.63)

Problem 5.2.48 Let C(A, x0 , t0 ) be a set introduced in Problem 5.2.46. Prove that C(A, x0 , t0 ) is a characteristic surface for equation ut t − div (A∇u) = 0

in Rn+1 .

Problem 5.2.49 Let us assume that ϕ : Rn+1 → R is smooth ∇ϕ = 0 on M = {ϕ(x, t) = 0}. Let us assume that  ⊂ Rn+1 is a bounded region with a smooth boundary. Let us introduce sets + = {(x, t) ∈  : ϕ(x, t) > 0},

− = {(x, t) ∈  : ϕ(x, t) < 0}. +

−

Let us suppose that function u :  → R belongs to space C()∩C 2 ( )∩C 2 ( ) and it is a weak solution to (5.63).

5.3 Parabolic Equations

227

(a) Show that M is a characteristic surface. (b) If (x0 , t0 ) ∈ M, then we set [[∇u]](x0, t0 ) =

lim

∇u(x, t) −

(x,t )∈+ (x,t )→(x0,t0 )

lim

∇u(x, t).

(x,t )∈− (x,t )→(x0,t0 )

Show that [[∇u]](x0, t0 ) is a vector perpendicular to M.

5.3 Parabolic Equations This section is devoted to the heat equation and to a lesser extent to general linear parabolic problems. We begin with the classical solutions. We discuss the maximum principle, energy decay, forward uniqueness and, in some special cases, backward uniqueness. We also show that the heat equation instantly regularizes initial conditions, see Proposition 5.3.4, which we contrast with the behavior of hyperbolic equations, discussed in the previous section. We also present an introduction to the theory of weak solutions. Proposition 5.3.4 nicely links this theory with the classical solutions. We present problems on regularizing effects of equations and the continuous dependence of solutions upon the data. We remind the reader that the parabolic equations were presented in other parts of the book. The Fourier method and its applications to parabolic equations are described in Sect. 3.2. Section 3.3.2 shows applications of the Galerkin method for parabolic equations. Fourier transform also provides a way of solving parabolic equations in Rn , cf. 2.1.5, 2.1.14 and 2.1.15.

5.3.1 Theoretical Background Heat Equation Our study of parabolic equations begins with the discussion of the heat equation on Rn . Later we move on to general linear parabolic equations of order 2. Definition 5.3.1 The heat equation on Rn is the equation ∂t u = u, u(x, 0) = u0 (x), where u : Rn × [0, ∞) → R and  is the Laplace operator. Remark 5.3.2 Unless specified otherwise, the function u is assumed to be continuous on Rn × [0, ∞) and C 2,1 -smooth in Rn × (0, ∞); the meaning of C 2,1 is that the second order derivatives of u with respect to the variables xi exist and are continuous, and ∂t u exists and is a continuous function.

228

5 Evolution Equations

There is a class of solutions to the heat equation having form u(x, t) = t −α v(t −β x). A special case of this class, satisfying also rotational symmetry in the x variables is called the fundamental solution. Proposition 5.3.3 The function given by  (x, t) =

2 1 e−|x| /4t (4πt )n/2

t > 0,

0

t0 is an approximate identity, see Sect. 1.2 and Remark 4.2.10. The fundamental solution allows us to give a formula for the solution to the heat equation for a general initial condition. Proposition 5.3.4 If a bounded continuous function u0 : Rn → R is given, then the function  u(x, t) := u0 ∗ (x, t) =

Rn

u0 (x − y)(y, t) dy

(5.65)

is a continuous in Rn × [0, ∞), smooth for t > 0 and it satisfies '

∂t u(x, t) = u(x, t) (x, t) ∈ Rn × (0, ∞), u(x, 0) = u0 (x) x ∈ Rn .

Definition 5.3.5 A nonhomogeneous heat equation on Rn is given by '

∂t u(x, t) − u(x, t) = f (x, t) u(x, 0) = u0 (x).

(5.66)

This nonhomogeneous problem has the following solution. The method for obtaining this solution is essentially the parameter variation formula. Proposition 5.3.6 If f ∈ L1 (Rn × [0, T ]), then the function u(x, t) = u0 ∗  +

 t 0

Rn

(x − y, t − s)f (y, s) dyds,

(5.67)

defined for (x, t) ∈ Rn × [0, T ], is a solution of the nonhomogeneous heat equation (5.66).

5.3 Parabolic Equations

229

Formula (5.67) is derived in Problem 5.3.5 for f ∈ L2 (Rn × R+ ). It is also a source of additional information about the smoothing effect of the heat equation, see Problems 5.3.5, 5.3.6, 5.3.40. Propositions 5.3.4 and 5.3.6 provide a solution to the homogeneous, respectively nonhomogeneous, heat equation. We will now discuss the uniqueness problem. By linearity, the uniqueness of a homogeneous problem is equivalent to the uniqueness of the nonhomogeneous problem. In fact, two solutions to the same nonhomogeneous equation differ by a solution of the homogeneous equation. One of the methods of studying the uniqueness is the maximum principle. Since the nature of the heat equation and of the Laplace equation are a bit different, we cannot expect that the maximum principle for the heat equation will have the same formulation as a similarly called result for the Laplace operator. We begin with introducing a piece of notation. Definition 5.3.7 For (x, t) ∈ Rn × (0, ∞) and given radius r > 0 we define the energy ball: ' ( 1 E(x, t, r) = (y, s) ∈ Rn × (0, ∞) : s ≤ t, (x − y, t − s) ≥ n . r We have the following result. Theorem 5.3.8 (Mean Value Theorem for Heat Equation; see [8, Section 2.3.2]) Suppose  ⊂ Rn × R and u is any C 2 -smooth function on  satisfying ∂t u = u. Then for any (x, t) ∈  and r > 0 such that E(x, t, r) ⊂  we have  1 |x − y|2 u(x, t) = n u(y, s) dyds. 4r E (x,t,r) (t − s)2 From the mean value theorem we can deduce a maximum principle for parabolic equations. To this end we introduce convenient terminology. Definition 5.3.9 Let  ⊂ Rn and choose T > 0. A parabolic boundary of  × T is the set ∂par T defined as ∂par T = ( × {0}) ∪ (∂ × [0, T ]) . See Fig. 5.1. Theorem 5.3.10 (Maximum Principle; see [8, Section 2.3.3]) Let  ⊂ Rn be a bounded set and suppose u :  × [0, ∞) → R is of class C 2 in  × (0, ∞) and is continuous in the closure  × [0, ∞). If u satisfies ∂t u = u, then for any T > 0 we have max

(x,t )∈×[0,T ]

u(x, t) =

max

(x,t )∈∂par T

u(x, t).

230

5 Evolution Equations

R

∂Ω × [0, T ]

Ω × [0, T ]

∂Ω × [0, T ]

Ω × {0}

Rn

Fig. 5.1 The parabolic boundary (the thick part). The dotted line represents the part of ∂( × [0, T ]) that does not belong to the parabolic boundary

Furthermore, if  is connected and there is a point (x0 , t0 ) ∈ / ∂par T such that max

(x,t )∈×[0,T ]

u(x, t) = u(x0 , t0 ),

then u is constant in  × (0, T ]. The maximum principle guarantees the uniqueness of solution to the heat equation for bounded domains: the difference of two solutions with the same boundary conditions is a solution with zero boundary conditions and such a solution must be zero by the maximum principle. An analogue of the maximum principle for Rn is true under an extra assumption. Theorem 5.3.11 (Maximum Principle for Bounded Solutions in Rn ) Suppose u : Rn × [0, T ] → R is bounded continuous and C 2 -differentiable. If u solves the heat equation, then for any t0 ∈ [0, T ], we have sup

(x,t )∈Rn×[0,t0 ]

u(x, t) = sup u(x, 0). x∈Rn

Theorem 5.3.12 (Maximum Principle for Solutions in Rn with Controlled Growth Rate) Suppose u : Rn × [0, ∞) → R satisfies heat equation and for some 2 T > 0 and a, A > 0, we have u(x, t) ≤ Aea|x| , whenever (x, t) ∈ Rn × (0, T ). Then, sup

Rn ×[0,T ]

u(x, t) = sup u(x, 0). x∈Rn

Theorem 5.3.13 Suppose u0 (x) is a continuous function satisfying |u0 (x)| ≤ 2 Aea|x| for some constants a, A > 0. The equation ∂t u = u, u(x, 0) = u0 (x) for x ∈ Rn , t ≥ 0 has at most one solution in the class of functions satisfying 2 |u(x, t)| ≤ Aea|x| for x ∈ Rn , t > 0.

5.3 Parabolic Equations

231

The condition on the bounded growth of u might seem artificial. There is, however, an example due to Tikhonov of a non-zero function u(x, t) satisfying ∂t u = u, u(x, 0) = 0; see Problem 5.3.19. Further properties of the heat equation, as well as a different approach to the proof of maximum principle are presented in the Problems part.

Linear Second-Order Parabolic Equations Suppose  ⊂ Rn is an open subset. Consider a second-order linear parabolic problem given by ⎧ ⎪ ⎪ ⎨∂t u + Lu = f u(x, 0) = u0 (x) ⎪ ⎪ ⎩u(y, t) = g(y, t)

in  × R+ , x ∈ ,

(5.68)

y ∈ ∂ × R+ ,

where f , u0 , and g are known functions (we will usually assume that they are smooth), and L is a second-order linear elliptic operator, that is, Lu = −

n 

aij (x, t)∂xi ∂xj u +

i,j =1

n 

bk (x, t)∂xk u + c(x, t)u

(5.69)

k=1

and aij , bk , c are functions on  × [0, ∞). Ellipticity means that the matrix {aij (x, t)}ni,j =1 is positive definite for all x, t. Usually a stronger condition is imposed, namely uniform ellipticity. We recall the definition. Definition 5.3.14 An operator L as in (5.69) is called uniformly elliptic if there are constants " > θ > 0 such that for any ξ ∈ Rn and (x, t) ∈  × [0, ∞), we have θ

n  i=1

ξi2 ≤

n  i,j =1

aij (x, t)ξi ξj ≤ "

n 

ξi2 ,

(5.70)

i=1

One can compare Definition 5.3.14 with the hypotheses of Theorem 4.2.4 (LaxMilgram Lemma) and Definition 5.2.1.

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5 Evolution Equations

Remark 5.3.15 While many results for second-order linear parabolic equations hold under rather mild assumptions on functions aij , bk and c, in this chapter we assume that aij , bk , and c are continuous functions. A substantial part of the theory for general linear parabolic equations of order two is a direct generalization of the properties of the heat equation. The classical part of the theory is built on Harnack’s inequality and maximum principle. We state now Harnack’s inequality (Theorem 5.3.16). For the proof, we refer to [8, Chapter 7, Theorem 10]. We assume that L is an operator given by (5.69) and it is uniformly elliptic. Theorem 5.3.16 (Parabolic Harnack Inequality) Suppose u ∈ C 2 ( × [0, T ]) satisfies ∂t u + Lu = 0 and u ≥ 0 on  × [0, T ]. Then for any open subset V ⊂  such that V ⊂  and t1 , t2 such that 0 < t1 < t2 < T , there exists a constant C such that sup u(x, t1 ) ≤ C inf u(x, t2 ). x∈V

x∈V

The constant C depends on V , , t1 , t2 , but it does not depend on u. One of the consequences of Harnack’s inequality is the strong maximum principle for parabolic equations. We phrase it in the following way (see Remark 5.3.15 for assumptions on regularity of aij , bk , and c). Theorem 5.3.17 (Strong Maximum Principle) Suppose that  ⊂ Rn is a bounded connected open set and u is a C 2 -smooth function satisfying ∂t u + Lu ≤ 0 in  × [0, T ]. Assume that c is nonnegative on the whole of  × [0, T ]. If u attains maximum at a point (x0 , t0 ) ∈  × (0, T ], then u is constant on  × [0, t0 ].

Weak Solutions Following the previous sections, we present basic information about weak solutions to (5.68). Our motivation is as usual, we look for candidates with relaxed regularity assumptions. The Gauss formula will help us. Definition 5.3.18 We shall say that a function u ∈ L2 (0, T ; H01()) satisfying du 2 −1 2 −1 ()) if dt ∈ L (0, T ; H ()) is a weak solution to (5.68) with f ∈ L (0, T ; H

(

du , ϕ) + dt

⎛



⎝ 

n 

ij =1

∂xi u∂xj ϕ +

n  k=1

⎞ bk ∂xk uϕ + cuv ⎠ dx = (f, ϕ),

5.3 Parabolic Equations

233

where (·, ·) denotes the pairing between H01 and H −1 . Moreover, we require that u(·, 0) = u0 . We note that the conditions imposed on u in this definition imply that u ∈ C([0, T ]; L2 ()), hence u(0) is well-defined. In the problem sections, we consider only f ∈ L2 (0, T ; L2 ()), which is a proper subspace of L2 (0, T ; H −1()).

5.3.2 Worked-Out Problems Problem 5.3.1 Let  ⊂ Rn be a bounded subset with smooth boundary. Assume that u : ×[0, ∞) → R is a smooth function satisfying ∂t u = u, u(x, 0) = u0 (x) and u(y, t) = 0 for y ∈ ∂. Define the energy of a solution by  E0 (t) =

u(x, t)2 dx. 

Prove that

d dt E0 (t)

≤ 0.

Solution We have E0 (t)

 =2

u∂t udx. 

As u satisfies the heat equation, we have 

 u∂t udx = 2

2 

uudx. 

By Green’s First Formula (see (1.9)) we have 

 ∂u u 2 uu = −2 |∇u| + 2 , ∂ν   ∂ 





2

where ∂u ∂ν is the normal derivative. By the assumptions u vanishes on the boundary, hence we obtain  E0 (t) = −2 |∇u|2 ≤ 0. 

♦

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5 Evolution Equations

R

∂Ω × [0, T ]

Ω × {T }

Ω × [0, T ]

∂Ω × [0, T ]

Rn

Fig. 5.2 Problem 5.3.2. Two functions u and , u are assumed to be equal on the thick part of the boundary. Backward uniqueness implies that they are equal on the whole of  × [0, T ]

Problem 5.3.2 Use the energy bound to prove backward uniqueness. More specifically, suppose that  is a bounded set with smooth boundary and u,, u satisfy the nonhomogeneous heat equation ∂t u − u = f and ∂t , u − , u = f . Prove that if u and , u agree on ∂ × [0, T ] for some T and u(x, T ) = , u(x, T ) for all x ∈ , then u ≡, u on  × [0, T ]; see Fig. 5.2. Solution The difference w = u−, u satisfies the equation ∂t w = w and w|∂ = 0. We also have w(x, T ) = 0 for x ∈  and we need to show that w ≡ 0 in  ×[0, T ]. Write E0 (t) =  w(x, t)2 dx. We have E0 (T ) = 0 and u ≡ , u if E0 (t) ≡ 0 for t ∈ [0, T ]. Note that E0 (t) is a continuous function. Now, log E0 (t) is a convex function (see Problem 5.3.22 below). The problem boils down to showing that if log E0 is convex and E0 (T ) = 0, then E0 (t) = 0 for t ≤ T . Suppose conversely, namely that E0 (t) > 0 for t ∈ (α, β) and assume that β is maximal possible, that is, E0 (β) = 0. Write F (t) = log E0 (t) for t ∈ (α, β). The fact that E0 (β) = 0 translates into limt →β F (t) = −∞. Take some γ ∈ (α, β) and let c = F (γ ). As F is convex, the graph of F lies above the line tangent to the graph of F at γ , in particular we have F (t) ≥ F (γ ) + (t − γ )c. Therefore, it is not possible for F (t) to tend to −∞ as t → β. The contradiction excludes the possibility that E0 (t) > 0 on some interval in [0, T ]. This shows that E0 (t) ≡ 0 for t ∈ [0, T ] and hence w ≡ 0 in  × [0, T ]. ♦ Problem 5.3.3 Let  ⊂ Rn be an open, bounded, connected set with smooth boundary. Show that if u is a classical solution to the Cauchy problem ut − u = 6u in  × [0, T ], u0 (x) = 3, ∂u on ∂, ∂ n = 0 then u is nonnegative. Calculate the maximal value of u on  × [0, 10].

5.3 Parabolic Equations

235

Solution First, observe that the function v(x, t) = e−6t u(x, t) is a solution to the heat equation with zero Neumann boundary condition. Therefore v satisfies the weak maximum principle (see Problem 5.3.37 below), and so v is the unique solution. Indeed, if another solution, say v1 (x, t), exists, then z(x, t) = v(x, t) − v1 (x, t) is a solution to the heat equation with zero Neumann boundary condition and zero initial condition. Thus, again by the weak maximum principle, z(x, t) ≤ 0. On the other hand, an analogous reasoning leads us to a conclusion that z ≥ 0. Therefore v(x, t) = v1 (x, t). One can easily check that v(x, t) = 3 is a solution, thus u(x, t) = 3e6t . Hence u is nonnegative and its maximal value equals to 3e60. ♦ Problem 5.3.4 Let  ⊂ R3 be a unit ball. Assume u is a classical solution to the Cauchy problem ut − u = 0 in  × [0, T ], in , u0 (x) = x 2 + y 2 + z2 u = 1 + te13t (x 2 + y 2 ) on ∂. Calculate the maximal value of u in  × [0, 3]. Solution The parabolic weak maximum principle implies that u attains its maximal values on the parabolic boundary of the cylinder  × [0, 3]. On the set  × {0} (the lower base of the cylinder) the maximal value of u equals to 1. On the wall of the cylinder (the set ∂ × [0, 3]), u is an increasing function of t. Therefore it is enough to find the maximal value of x 2 + y 2 on the unit sphere x 2 + y 2 + z2 = 1. Hence the maximal value of u equals to 1 + 3e39. ♦ Problem 5.3.5 Let us suppose that f ∈ L2 (0, T ; L2 (RN )). (a) Show, using the Fourier transform that there is a unique weak solution to the following problem vt = v + f for (x, t) ∈ RN × R+ , v(x, 0) = v0 (x) for x ∈ RN .

(5.71)

(b) Show that if v0 ≡ 0, then v ∈ L2 (0, T ; H 2 (RN )). (c) Show that if v0 ∈ H 2 (RN ), then v ∈ L2 (0, T ; H 2 (RN )). Solution We begin with (a). We apply the Fourier transform to both sides of (5.71). The result is (see Problem 2.1.5), vˆt (ξ, t) = −4π 2 |ξ |2 v(ξ, ˆ t) + fˆ(ξ, t) t > 0, ξ ∈ RN , v(ξ, ˆ 0) = vˆ0 (ξ ),

ξ ∈ RN .

236

5 Evolution Equations

As a result, we immediately see that the unique solution is given by the following formula, v(ξ, ˆ t) = e−4π

2 |ξ |2 t



t

uˆ 0 (ξ ) +

e−4π

2 |ξ |2 (t −s)

fˆ(ξ, s) ds ≡ w(ξ, ˆ t) + u(ξ, ˆ t).

0

(5.72) Applying the inverse Fourier transform yields formula (5.65). Next, we show (b). We raise both sides of (5.72) to power 2 and integrate it over Rn × (0, T ). By Hölder’s inequality we have  u ˆ 2L2 (0,T ;L2 )

T

=  ≤  ≤ =

 RN

0



RN

RN



T 0

 t 2    e−4π 2 |ξ |2 (t −s)fˆ(ξ, s) ds  dξ dt   

0 t

|fˆ(ξ, s)|2 ds

0 T

|fˆ(ξ, s)|2 ds

0



t

e−8π

2 |ξ |2 (t −s)

dsdtdξ

0



T



t

1 dsdtdξ 0

0

T2 ˆ 2 f L2 (0,T ;L2 ) . 2

Since the Fourier transform is an isometry of L2 , we have uL2 (0,T ;L2 ) ≤ T f L2 (0,T ;L2 ) .

(5.73)

ˆ t) being square We note that u ∈ L2 (0, T ; H 2(RN )) is equivalent to (1 + |ξ |2 )u(ξ, integrable over RN × (0, T ). Keeping in mind (5.73), we see that it suffices to check that u(ξ, ˆ t)|ξ |2 is square integrable. We observe that for a.e. ξ ∈ RN the function u(ξ, ˆ ·), given by the second integral in (5.72), is a convolution of fˆ(ξ, ·) with (ξ, ·), where: (ξ, t) = e−4π

2 |ξ |2 t

χ[0,∞) (t),

t ∈ R, ξ ∈ RN .

Now, we recall the Young convolution inequality from Problem 1.2.16: f ∗ gLr ≤ f Lp gLq . We will use it with f equal fˆ(ξ, ·), p = 2 and g = (ξ, ·), q = 1 and r = 2. In this way, we obtain 

T 0

 t 2     fˆ(ξ, s)e−4π 2 |ξ |2 (t −s) ds  dt ≤   0

T 0

|fˆ(ξ, s)|2 ds



T

e 0

−4π 2 |ξ |2 s

2 ds

.

5.3 Parabolic Equations

237

As a result, we see that 

T

 RN

0



 |ξ | |u(ξ, ˆ t)| dξ dt ≤ 4

2

RN

 =

|ξ |

T

4

|fˆ(ξ, s)|2 ds

0



RN

T



T

e

−4π 2 |ξ |2 s

2 ds

0

  |fˆ(ξ, s)|2 ds |ξ |2

0

T

e−4π

2 |ξ |2 s

2 ds

0

1 ≤ fˆL2 (0,T ;L2 ) , 4π 2 

T

1 . 4π 2 0 Finally, in order to show (c), we examine w defined in (5.72), we note that always wˆ ∈ L2 (0, T ; RN ). Moreover,

where we used |ξ |



T 0

2

e−4π

2 |ξ |2 s

ds ≤



 4

RN

T

2

|ξ | |w(ξ, ˆ t)| dξ dt ≤

 RN

0

|e−4π

2 |ξ |2 t

|ξ |4 |vˆ0 (ξ )|2 dξ dt ≤ T v0 2H 2 .

Since u, w belong to L2 (0, T ; H 2(R)), so does v = w + u.

♦

Problem 5.3.6 Let us suppose that v is a weak solution to (5.77) with f ≡ 0 and v0 ∈ L2 (RN ). Show that vH 2 ≤ Ct −1 for t ∈ (0, 1). How does C depend on u0 ? Solution We use the formula (5.72) for v, we see that v = w, when f = 0. Due to the definition of the Sobolev spaces on RN , it suffices to show that  RN

(1 + |ξ |2 )2 |w(ξ, ˆ t)|2 dξ ≤

C12 . t2

By the definition of wˆ and since (1 + a 2 )2 ≤ 2 + 2a 4, we obtain  RN

(1+|ξ |2 )2 |w(ξ, ˆ t)|2 dξ ≤ 2vˆ0 2L2 +

2 t2

 RN

|ξ |4 t 2 e−8π

2 |ξ |2 t

vˆ0 (ξ ) dξ ≤ vˆ0 2L2

max{1, C1 }2 , 4π 2 t 2

where C1 = sup{y 4 e−y : y ∈ R+ }. Since the Fourier transform is an isometry, we deduce that 2

w(·, t)H 2 ≤ as desired.

max{1, C1 } v0 L2 , 2πt ♦

238

5 Evolution Equations

5.3.3 Problems Problem 5.3.7 Find all solutions of the heat equation on Rn having the form u(x, t) = t α v(t β |x|). Problem 5.3.8 Deduce the maximum principle for the solution to the heat equation (that is, prove Theorem 5.3.10) from the mean value theorem (Theorem 5.3.8). Problem 5.3.9 Prove that Rn (x, t)dx = 1, where  defined in (5.64) is the fundamental solution to the heat equation. Problem 5.3.10 Suppose u :  × [0, ∞) → R is C 2 -smooth and satisfies ∂t u = u. Choose time T > 0 and assume  that v(x) = u(x, T ) has a local maximum at x0 2v and the Hessian D 2 v = ∂x∂i ∂x of v is nondegenerate. Prove that ∂t u(x0 ) < 0. j i,j

Generalize this result to the case, when the Hessian of v is degenerate, but not identically zero. Problem 5.3.11 Deduce the maximum principle (Theorem 5.3.10) for solutions to the heat equation in a bounded open set  ⊂ Rn using Problem 5.3.10. Use the following plan. • Assume that a function u :  × [0, ∞) → R is C 2 -smooth in the interior of its domain and satisfies ∂t u = u. • Take T > 0. Suppose u|×[0,T ] has a global maximum at (x0 , t0 ), where x0 ∈  and t0 ∈ (0, T ). Use Problem 5.3.10. to show that v = u|×{t0 } must have a degenerate maximum at x0 (a local maximum is degenerate if the Hessian has zero determinant). • Show that a linear perturbation u + , where  :  → R is a linear function vanishing at x0 has only nondegenerate local maxima, if  is generic. Here by “generic” we mean that there exists a full measure subset of the set of coefficients of , such that if the coefficients of  belong to that set, u +  has only nondegenerate local maxima. • Show that for a generic linear function  :  → R vanishing at x0 the function u +  has a local maximum either at ∂ × [0, T ] or at  × {0}. In particular there exists a point (x , t ) such that (u + )(x , t ) ≥ u(x0 , t0 ). • Pass to the limit with  converging to 0 = 0 and show that there exists a point (x

, t

) either on ∂ × [0, T ] or on  × {0} such that u(x

, t

) ≥ u(x0 , t0 ). Problem 5.3.12 Show that a solution to the heat equation cannot have local maxima. The next two problems give a proof of Proposition 5.3.4. Problem 5.3.13 Suppose u0 (x) is bounded and continuous and u(x, t) is as in (5.65). Prove that ∂t u(x, t) = u(x, t) for x ∈ Rn , t > 0. Problem 5.3.14 Let u0 , u be as in Problem 5.3.13 above. Show that as t → 0 we have u(x, t) → u0 (x) pointwise. Conclude that u(x, t) is a continuous function on Rn × [0, ∞).

5.3 Parabolic Equations

239

Problem 5.3.15 Assume that u0 : Rn → R is of class C k and all the derivatives up to the k-th order are bounded. Let u(x, t) be as in (5.65). Prove that u(x, t) is a C k -smooth function on Rn × [0, ∞). Generalize to the cases when: • u0 is of class C ∞ ; • u0 is analytic. Problem 5.3.16 Suppose u(x, t) is a solution of the heat equation ∂t u(x, t) = u(x, t) for (x, t) ∈ Rn × (0, ∞) with the initial condition u(x, 0) = u0 (x) ∈ L1 (Rn ). Prove that |u(x, t)| ≤ (4πt)−n/2 ||u0 ||L1 , ||u(·, t)||L1 (Rn ) ≤ ||u0 ||L1 , ||u(·, t)||Lp (Rn ) ≤ (4πt)−(n/2)(1−1/p)||u0 ||L1 for p ∈ (1, ∞). Hint Use formula (5.65). Problem 5.3.17 Draw a sketch of the set E(x, t, r) for n = 1 (see Definition 5.3.7). Prove that E(x, t, r) is a bounded set for general n. Problem 5.3.18 Prove that  |x − y|2 1 = 1. 4r n E (x,t,r) |t − s|2 Problem 5.3.19 Consider the series u(x, t) =

∞  k=0

1 2k d k −1/t 2 x e . (2k)! dt k

(5.74)

(a) Prove that the series (5.74) is uniformly convergent for (x, t) ∈ R × R. Argue that u(x, t) is a C ∞ smooth function. (b) Show that u(x, t) is a solution to ∂t u = ∂x2 u, u(x, 0) = 0. Problem 5.3.20 Suppose  ⊂ Rn is a bounded set. Use the maximum principle Theorem 5.3.10 to show that the problem ⎧ ⎪ ⎪ ⎨∂t u(x, t) = u(x, t) + f (x, t) (x, t) ∈  × (0, ∞), x ∈ , u(x, 0) = u0 (x) ⎪ ⎪ ⎩u(y, t) = h(y, t) y ∈ ∂, t ≥ 0, where h, f, u0 are C 2 -smooth functions, has at most one solution.

240

5 Evolution Equations

Problem 5.3.21 Prove the uniqueness of the solution to the heat equation as in Problem 5.3.20 for  with smooth boundary using the energy decay described in Problem 5.3.1. Problem 5.3.22 Suppose  is a bounded open set with smooth boundary and u satisfies ∂ t u = u, u|∂×[0,∞) = 0. Define the energy as in Problem 5.3.1, that is E0 (t) =  u(x, t)2 dx. Prove that log E0 (t) is a convex function. Hint Prove that E0 (t)2 ≤ E0

(t)E0 (t). Problem 5.3.23 We generalize now the notion of energy from Problem 5.3.1. Let  ⊂ Rn be a bounded open set with smooth boundary. Assume that u :  × [0, ∞) → R is a smooth function satisfying ∂t u = u, u(x, 0) = u0 (x) and u(y, t) = 0 for y ∈ ∂. Define the (first-order) energy of a solution by:  E1 (t) =

|∇u(x, t)|2 dx. 

Prove that

d dt E1 (t)

≤ 0.

Problem 5.3.24 Let us assume that u ∈ C ∞ (Rn × R+ ) ∩ C(Rn × [0, ∞)) is a solution to the heat equation  ∂t u = u u(x, 0) = u0 (x)

in Rn × (0, ∞), for x ∈ Rn .

In addition, we assume that u > 0 in Rn × (0, ∞) and for all t > 0 the function d u(·, t) ln u(·, t) is integrable over Rn . Show that dt Rn u(x, t) ln u(x, t) dx ≤ 0. Problem 5.3.25 Let us assume that u ∈ C ∞ (Rn × R+ ) ∩ C(Rn × [0, ∞)) ∩ C(R+ ; H 1 (Rn )) is a solution to the heat equation  ∂t u = u u(x, 0) = u0 (x)

in Rn × (0, ∞), for x ∈ Rn .

Take a smooth convex function W : R → R such that 0 ≤ W (ξ ) ≤ C|ξ |2 for a d constant C. Show that dt Rn W (u(x, t)) dx ≤ 0. Problem 5.3.26 Suppose  ⊂ Rn and consider the nonlinear equation, called the Cole–Hopf equation. ⎧ ⎪ ⎪ ⎨∂t u − au + b ∇u, ∇u = 0 u(x, 0) = u0 (x) ⎪ ⎪ ⎩u(y, t) = g(y, t) where a, b are real parameters, a > 0.

x∈ y ∈ ∂, t ≥ 0

(5.75)

5.3 Parabolic Equations

241

(a) Prove that u satisfies (5.75) if and only if w = e−2bu/a satisfies the heat equation. (b) Give an explicit formula for a solution to (5.75) for  = Rn . Problem 5.3.27 Consider the viscous Burgers equation  ∂t u − a∂x2 u + u∂x u = 0

in R × (0, ∞)

u(x, 0) = u0 (x),

(5.76)

where a > 0. Give an explicit solution to (5.76). x x Hint Consider w(x, t) = −∞ u(y, t)dy and h(x) = −∞ u0 (y)dy. Problem 5.3.28 Suppose g : [0, ∞) × Rn−1 → R is a bounded and continuous function such that g(0, x2 , . . . , xn ) = 0. Extend g to Rn by the formula g(−x1 , x2 , . . . , xn ) = g(x1 , x2 , . . . , xn ). Show that the function u(x, t) : [0, ∞) × Rn−1 × [0, ∞) → R given by u(x1 , . . . , xn , t) = (·, t) ∗ g solves the heat equation on [0, ∞) × Rn−1 × [0, ∞) with initial value u(x1 , . . . , xn , 0) = g(x1 , . . . , xn ) and with boundary condition u(0, x2 , . . . , xn , t) = 0. Here  is as in (5.64). Problem 5.3.29 Use Problem 5.3.28 to prove uniqueness (in particular, specify conditions for uniqueness) of solutions to the problem ∂t u = u on [0, ∞) × Rn−1 × (0, ∞) with initial value u(x, 0) = u0 (x) for x ∈ (0, ∞) × Rn−1 × {0} and u(0, x2 , . . . , xn , t) = f (x2 , . . . , xn , t) for some function f . Problem 5.3.30 Generalize the ideas of Problem 5.3.28 to give a solution to the heat equation on the strip (0, 1) × Rn × [0, ∞). Problem 5.3.31 Prove the following weak version of the maximum principle for a general linear equation of order 2: Suppose  ⊂ Rn is a bounded open set and u :  × [0, T ] → R is a C 2 -smooth function that satisfies (∂t + L)u(x, t) < 0 for all (x, t) ∈  × [0, T ]. Assume that u < 0 on ∂par T . Then u < 0 on  × [0, T ]. Hint Consider t0 = inf{t ∈ [0, T ] : u(x, t) > 0 for some x ∈ } and look at the maximum of the function x → u(x, t0 ). Problem 5.3.32 Let  be a bounded subset of Rn . Assume that c(x, t) in the definition of the operator L (Eq. (5.68)) is zero on  × [0, T ]. Suppose u is a C 2 -smooth function on  × (0, T ) satisfying ∂t u + Lu ≤ 0. Show that sup (x,t )∈×(0,T )

u(x, t) =

sup (x,t )∈∂par T

u(x, t).

242

5 Evolution Equations

Hint Solve the problem first under a stronger assumption that ∂t u + Lu < 0. Then consider the sequence of functions uε (x, t) = u(x, t) − εt. Problem 5.3.33 Let u = u(x, t) defined on [0, 1] × [0, 10] be a classical solution to the Cauchy problem ut − 3uxx = sin x − 1, u(x, 0) = x 2 + 1, u(0, t) = 0, u(1, t) = 1. Calculate the maximum of u. Problem 5.3.34 Let u = u(x, t) defined on [0, π] × [0, 2] be a classical solution to the Cauchy problem ut − uxx = πu, u(x, 0) = sin x + 1, u(0, t) = u(π, t) = 1. Calculate the maximum and minimum of u. Problem 5.3.35 Let  be an open, bounded, connected subset of Rn . Let g ∈ C ∞ () be a nonconstant, nonnegative function. Assume u is a classical solution to the heat equation with the initial condition equal to g and the Neumann boundary condition on ∂ equal to zero. Show that u(·, t) > 0 for every t > 0. Problem 5.3.36 Let  ⊂ Rn be an open, bounded, connected set with smooth boundary. Consider a weak solution to the Cauchy problem in  × (0, T ), ut − u ≤ 0 u(0, x) = u0 (x) for x ∈ , u=g on , where u0 (x) ∈ L∞ () and g ∈ L∞ (∂). Show that u ≤ max{max g(x), max u0 (x)} x∈∂

almost everywhere in  × (0, T ).

x∈

5.3 Parabolic Equations

243

Problem 5.3.37 Let  ⊂ Rn be an open, bounded, connected set with smooth boundary. Consider a weak solution to the Cauchy problem in  × (0, T ), ut − u = 0 u(0, x) = u0 (x) for x ∈ , ∂u on ∂, ∂ n = 0 where u0 (x) ∈ L∞ (). Show that for a.e. (x, t) ∈  × (0, T ) it holds u(x, t) ≤ sup u0 (x). x∈

Problem 5.3.38 Let  ⊂ Rn be an open, bounded, connected set with smooth boundary. Given functions v ∈ C 1 such that ∇v ∈ L∞ ( × (0, T )), and u0 ∈ L∞ () such that u0 (x) ≥ 0, consider a weak solution to the Cauchy problem ut − u = div(u∇v) in  × (0, T ), u(0, x) = u0 (x) for x ∈ , ∂u ∂v on ∂. ∂ n − u ∂ n = 0 Suppose the weak solution is unique. Show that u ≥ 0. Problem 5.3.39 Let us suppose that  is given by (5.64). Show that  is a fundamental solution to the heat equation, i.e. t −  = δ0

in D (Rn × R).

Problem 5.3.40 Let us consider a weak solution of the following equation ut = u + f (x, t) ∈ RN × R+ , u(x, 0) = u0 (x) x ∈ RN .

(5.77)

with f ≡ 0. Show that: (a) If u0 ∈ H 1 (RN ), then u satisfies the estimate uH 2 ≤ Ct −1/2 for t ∈ (0, 1). How does C depend on u0 ? (b) If u0 ∈ H α (RN ), then u satisfies the estimate uH β ≤ Ct α/2−β/2 for β > α and t ∈ (0, 1). Problem 5.3.41 Let us suppose that un is a weak solution to (5.77) with f = fn ∈ L2 (0, T ; L2 (RN )) and u0 = un0 ∈ H 2 (RN ). Show that if fn → f

in

L2 (0, T ; L2 (RN ))

and

un0 → u0

in

H 2 (RN ),

then un → u in L2 (0, T ; H 2 (RN )) and u is a solution to (5.77) with data f ∈ L2 (0, T ; L2 (RN )) and u0 ∈ H 2 (RN ).

244

5 Evolution Equations

Problem 5.3.42 Let us suppose that un is a weak solution to (5.77) with f ≡ 0 and u0 = un0 ∈ H 1 (RN ). Show that if un0 → u0

in

L2 (RN ),

then un → u in L2 (0, T ; L2 (RN )) and u is a solution to (5.77) with f ≡ 0 and u0 ∈ H 1 (RN ). Show that ∇u(·, t1 )L2 ≥ ∇u(·, t2 )L2 for t1 ≥ t2 . Hint Recall Problem 5.3.25. Problem 5.3.43 Let us suppose that u is a weak solution to (5.77) with f ≡ 0 and u0 ∈ L2 (RN ). Show that u(·, t1 )2L2 ≥ u(·, t2 )2L2

if t1 ≤ t2 .

Problem 5.3.44 Show that the result of the previous problem implies the uniqueness of weak solutions to the heat equation. Problem 5.3.45 Use the Fourier method to state and prove analogues of Prob¯ condition lems 5.3.5, 5.3.6 for Eq. (5.68), where the coefficients belong to C(), 2 2 (5.70) holds, u0 ∈ L () and f ∈ L ( × (0, T )). Problem 5.3.46 Let us assume that Eq. (5.68) is in the divergence form and bk = 0 for k = 1, . . . , n and c = 0, i.e. the operator L has the form Lu = div A∇u, where matrix A satisfies (5.70). Show that there is at most one weak solution to this problem.

5.4 Bibliographical Remarks The method of characteristics is well covered in the textbooks of Evans [8] and John [12]. They are good sources for further study in that topic. In general, problems in Sect. 5.2 are not original, because they commonly illustrate classical issues. Inspiration comes from books listed in the bibliography and from many other sources including our own classroom practice. It is worth to emphasize that in a very well-written book by Alinhac [2], many aspects of hyperbolic equations have been considered. Problem 5.2.12 on a difference equation comes from John’s book [12]. Fourier transform is discussed in detail in [16]; notation used in the latter one is compatible with ours.

5.4 Bibliographical Remarks

245

Classical theory on the heat equation and second-order linear parabolic equation is presented in detail in Evans’ book, [8], see Sections 2.3 and 7.1. The book of Lieberman [17] is another resource with a detailed explanation of various maximum principles. Problem 5.3.19 was originally discovered by Tikhonov and it is discussed in detail in the book of John [12].

Bibliography

1. R. A. Adams and J. J. F. Fournier. Sobolev spaces, volume 140 of Pure and Applied Mathematics (Amsterdam). Elsevier/Academic Press, Amsterdam, second edition, 2003. 2. S. Alinhac. Hyperbolic partial differential equations. Universitext. Springer, Dordrecht, 2009. 3. V. I. Arnold. Mathematical methods of classical mechanics, volume 60 of Graduate Texts in Mathematics. Springer-Verlag, New York, [1989]. Translated from the 1974 Russian original by K. Vogtmann and A. Weinstein, Corrected reprint of the second (1989) edition. 4. Ph. Blanchard and E. Brüning. Mathematical Methods in Physics. Distributions, Hilbert Space Operators, Variational Methods, and Applications in Quantum Physics. Springer Cham, Birkhäuser, second edition, 2015. 5. H. Brezis. Functional analysis, Sobolev spaces and partial differential equations. Universitext. Springer, New York, 2011. 6. B. Dacorogna. Direct methods in the calculus of variations, volume 78 of Applied Mathematical Sciences. Springer, New York, second edition, 2008. 7. B. A. Dubrovin, A. T. Fomenko, and S. P. Novikov. Modern geometry—methods and applications. Part I: The geometry of surfaces, transformation groups, and fields, volume 93 of Graduate Texts in Mathematics. Springer-Verlag, New York, second edition, 1992. 8. L. C. Evans. Partial differential equations, volume 19 of Graduate Studies in Mathematics. American Mathematical Society, Providence, RI, second edition, 2010. 9. L. C. Evans and R. F. Gariepy. Measure theory and fine properties of functions. Textbooks in Mathematics. CRC Press, Boca Raton, FL, revised edition, 2015. 10. B. R. Gelbaum and J. M. H. Olmsted. Counterexamples in analysis. Dover Publications, Inc., Mineola, NY, 2003. Corrected reprint of the second (1965) edition. 11. I. M. Gelfand and S. V. Fomin. Calculus of variations. Revised English edition translated and edited by Richard A. Silverman. Prentice-Hall, Inc., Englewood Cliffs, N.J., 1963. 12. F. John. Partial differential equations, volume 1 of Applied Mathematical Sciences. SpringerVerlag, New York, fourth edition, 1991. 13. J. Jost. Riemannian geometry and geometric analysis. Universitext. Springer, Cham, seventh edition, 2017. 14. S. Krantz. Explorations in Harmonic Analysis: With Applications to Complex Function Theory and the Heisenberg Group Birkhäuser Boston, Springer 2009. 15. G. Leoni. A first course in Sobolev spaces. volume 181 of Graduate Studies in Mathematics. American Mathematical Society, Providence, RI, second edition, 2017. 16. E. H. Lieb and M. Loss. Analysis, volume 14 of Graduate Studies in Mathematics. American Mathematical Society, Providence, RI, second edition, 2001.

© Springer Nature Switzerland AG 2019 M. Borodzik et al., Problems on Partial Differential Equations, Problem Books in Mathematics, https://doi.org/10.1007/978-3-030-14734-1

247

248

Bibliography

17. G. M. Lieberman. Second order parabolic differential equations. World Scientific Publishing Co., Inc., River Edge, NJ, 1996. 18. B. Makarov and A. Podkorytov. Real analysis: measures, integrals and applications. Universitext. Springer, London, 2013. 19. W. Rudin. Principles of mathematical analysis. McGraw-Hill Book Co., New York-AucklandDüsseldorf, third edition, 1976. International Series in Pure and Applied Mathematics. 20. W. Rudin. Real and complex analysis. McGraw-Hill Book Co., New York, third edition, 1987. 21. W. Rudin. Functional analysis. International Series in Pure and Applied Mathematics. McGraw-Hill, Inc., New York, second edition, 1991. 22. B. Simon. Operator theory. A Comprehensive Course in Analysis, Part 4. American Mathematical Society, Providence, RI, 2015. 23. E. Stein and R. Shakarchi. Fourier analysis. An introduction. Volume 1 of Princeton Lectures in Analysis, Princeton University Press, Princeton, NJ, 2003. 24. R. S. Strichartz. A guide to distribution theory and Fourier transforms. Reprint of the 1994 original. World Scientific Publishing Co., Inc., River Edge, NJ, 2003. 25. Z. Szmydt. Fourier transformation and linear differential equations. D. Reidel Publishing Co., Dordrecht-Boston, Mass.; PWN—Polish Scientific Publishers, Warsaw, revised edition, 1977. 26. M. E. Taylor. Partial differential equations Vol. I-III, volume 115–117 of Applied Mathematical Sciences. Springer, New York, second edition, 2011. 27. R. Temam. Navier-Stokes equations, volume 2 of Studies in Mathematics and its Applications. North-Holland Publishing Co., Amsterdam, third edition, 1984. Theory and numerical analysis, With an appendix by F. Thomasset. 28. F. Trèves. Topological vector spaces, distributions and kernels. Dover Publications, Inc., Mineola, NY, 2006 (unabridged replication of the 1967 original). 29. L. I. Volkovyski˘ı, G. L. Lunts, and I. G. Aramanovich. A collection of problems on complex analysis. Dover Publications, Inc., New York, 1991. Reprint of the 1965 English translation. 30. K. Yosida, Functional analysis. Sixth edition. Grundlehren der Mathematischen Wissenschaften, 123. Springer-Verlag, Berlin-New York, 1980.