Tools and Problems in Partial Differential Equations [1 ed.] 9783030502836, 9783030502843

Table of contents :
Introduction
Acknowledgements
Contents
Part I Tools and Problems
1 Elements of Functional Analysis and Distributions
1.1 Fréchet Spaces
1.2 Elements of Functional Analysis
1.2.1 Fixed Point Theorems
1.2.2 The Banach Isomorphism Theorem
1.2.3 The Closed Graph Theorem
1.2.4 The Banach–Steinhaus Theorem
1.2.5 The Banach–Alaoglu Theorem
1.2.6 The Ascoli Theorem
1.2.7 The Hahn–Banach Theorem
1.2.8 Hilbert Spaces
1.2.9 Spectral Theory of Self-Adjoint Compact Operators
1.2.10 Lp Spaces, 1 ≤p ≤+∞
1.2.11 The Hölder and Young Inequalities
1.2.12 Approximation of the Identity
1.3 Elements of Distribution Theory
1.3.1 Distributions
1.3.2 Tempered Distributions
1.3.3 The Fourier Transform
1.3.4 The Stationary Phase Method
2 Statements of the Problems of Chap.1
3 Functional Spaces
3.1 Sobolev Spaces
3.1.1 Sobolev Spaces on Rd, d ≥1
3.1.1.1 Definition and First Properties
3.1.1.2 Density
3.1.1.3 Operations on Hs(Rd)
3.1.1.4 Sobolev Embeddings
3.1.1.5 Duality
3.1.1.6 Compactness
3.1.1.7 Traces
3.1.1.8 Equivalent Norm
3.1.2 Local Sobolev Spaces Hsloc(Rd)
3.1.3 Sobolev Spaces on an Open Subset of Rd
3.1.3.1 Definition and First Properties
3.1.3.2 Extension
3.1.3.3 Density
3.1.3.4 Poincaré Inequality
3.1.3.5 Sobolev Embedding
3.1.3.6 Duality
3.1.3.7 Compactness
3.1.3.8 Traces
3.1.4 Sobolev Spaces on the Torus
3.2 The Hölder Spaces
3.2.1 Hölder Spaces of Integer Order
3.2.1.1 Definition
3.2.1.2 Properties
3.2.2 Hölder Spaces of Fractional Order
3.2.2.1 Definition
3.2.2.2 Properties
3.3 Characterization of Sobolev and Hölder Spaces in Dyadic Rings
3.3.1 Characterization of Sobolev Spaces
3.3.2 Characterization of Hölder Spaces
3.3.3 The Zygmund Spaces
3.4 Paraproducts
3.5 Some Words on Interpolation
3.6 The Hardy–Littlewood–Sobolev Inequality
4 Statements of the Problems of Chap.3
5 Microlocal Analysis
5.1 Symbol Classes
5.1.1 Definition and First Properties
5.1.2 Examples
5.1.3 Classical Symbols
5.2 Pseudo-Differential Operators
5.2.1 Definition and First Properties
5.2.2 Kernel of a DO
5.2.3 Image of a DO by a Diffeomorphism
5.2.4 Symbolic Calculus
5.2.4.1 Composition
5.2.4.2 Adjoint
5.2.5 Action of the DO on Sobolev Spaces
5.2.6 Garding Inequalities
5.2.6.1 The Weak Inequality
5.2.6.2 The ``Sharp Garding'' Inequality
5.2.6.3 The Fefferman–Phong Inequality
5.3 Invertibility of Elliptic Symbols
5.4 Wave Front Set of a Distribution
5.4.1 Definition and First Properties
5.4.2 Wave Front Set and DO
5.4.3 The Propagation of Singularities Theorem
5.4.3.1 Bicharacteristics
5.4.3.2 The Propagation Theorem
5.5 Paradifferential Calculus
5.5.1 Symbols Classes
5.5.2 Paradifferential Operators
5.5.3 The Symbolic Calculus
5.5.4 Link with the Paraproducts
5.6 Microlocal Defect Measures
6 Statements of the Problems of Chap.5
7 The Classical Equations
7.1 Equations with Analytic Coefficients
7.1.1 The Cauchy–Kovalevski Theorem
7.1.1.1 The Linear Version
7.1.1.2 The Nonlinear Version
7.1.2 The Holmgren Uniqueness Theorem
7.2 The Laplace Equation
7.2.1 The Mean Value Property
7.2.2 Hypoellipticity: Analytic Hypoellipticity
7.2.3 The Maximum Principles
7.2.4 The Harnack Inequality
7.2.5 The Dirichlet Problem
7.2.5.1 Case g=0
7.2.5.2 Case g 0.
7.2.6 Spectral Theory
7.2.6.1 Weyl Law
7.2.6.2 Estimates of the Eigenfunctions
7.3 The Heat Equation
7.3.1 The Maximum Principle
7.3.2 The Cauchy Problem
7.4 The Wave Equation
7.4.1 Homogeneous Cauchy Problem
7.4.1.1 Properties of the Solution
7.4.1.2 Decay at Infinity
7.4.1.3 Finite Speed of Propagation
7.4.1.4 Huygens Principle
7.4.1.5 Influence Domain
7.4.1.6 Conservation of the Energy
7.4.1.7 Strichartz Estimates
7.4.2 Inhomogeneous Cauchy Problem
7.4.2.1 Finite Speed of Propagation
7.4.2.2 Strichartz Estimates
7.4.3 The Mixed Problem
7.5 The Schrödinger Equation
7.5.1 The Cauchy Problem
7.5.1.1 The Homogeneous Cauchy Problem
7.5.2 Properties of the Solution
7.5.2.1 Expression of the Solution
7.5.2.2 Infinite Speed of Propagation
7.5.2.3 Decay at Infinity of the Solution
7.5.2.4 Strichartz Inequality
7.5.2.5 The Inhomogeneous Problem
7.5.2.6 Nonhomogeneous Strichartz Inequality
7.6 The Burgers Equation
7.7 The Euler Equations
7.7.1 The Incompressible Euler Equations
7.7.1.1 Incompressibility
7.7.1.2 The Vorticity
7.7.1.3 Classical Solutions: The Lichtenstein Theorem
7.7.1.4 Weak Solutions: The Yudovitch Theorem
7.7.2 The Compressible Euler Equations
7.8 The Navier–Stokes Equations
7.8.1 Weak Solutions
7.8.2 The Leray Theorem (1934)
7.8.3 Strong Solutions: Theorems of Fujita–Kato and Kato
8 Statements of the Problems of Chap.7
Part II Solutions of the Problems and Classical Results
9 Solutions of the Problems
10 Classical Results
10.1 Some Classical Formulas
10.1.1 The Leibniz Formula
10.1.2 The Taylor Formula with Integral Reminder
10.1.3 The Faa–di-Bruno Formula
10.2 Elements of Integration
10.2.1 Convergence Theorems
10.2.2 Change of Variables in Rd
10.2.3 Polar Coordinates in Rd
10.2.4 The Gauss–Green Formula
10.2.5 Integration on a Graph
10.3 Elements of Differential Calculus
10.4 Elements of Differential Equations
10.4.1 The Precise Cauchy–Lipschitz Theorem
10.4.2 The Cauchy–Arzela–Péano Theorem
10.4.3 Global Theory
10.4.4 The Gronwall Inequality
10.5 Elements of Holomorphic Functions
References
Index

Citation preview

Universitext

Thomas Alazard Claude Zuily

Tools and Problems in Partial Differential Equations

Universitext

Universitext

Series Editors Sheldon Axler San Francisco State University Carles Casacuberta Universitat de Barcelona John Greenlees University of Warwick Angus MacIntyre Queen Mary University of London Kenneth Ribet University of California, Berkeley Claude Sabbah École Polytechnique, CNRS, Université Paris-Saclay, Palaiseau Endre Süli University of Oxford Wojbor A. Woyczy´nski Case Western Reserve University

Universitext is a series of textbooks that presents material from a wide variety of mathematical disciplines at master’s level and beyond. The books, often well classtested by their author, may have an informal, personal even experimental approach to their subject matter. Some of the most successful and established books in the series have evolved through several editions, always following the evolution of teaching curricula, into very polished texts. Thus as research topics trickle down into graduate-level teaching, first textbooks written for new, cutting-edge courses may make their way into Universitext.

More information about this series at http://www.springer.com/series/223

Thomas Alazard • Claude Zuily

Tools and Problems in Partial Differential Equations

Thomas Alazard École Normale Supérieure Paris-Saclay Université Paris-Saclay Gif sur Yvette, France

Claude Zuily Institut de Mathématique d’Orsay Université Paris-Saclay Orsay, France

ISSN 0172-5939 ISSN 2191-6675 (electronic) Universitext ISBN 978-3-030-50283-6 ISBN 978-3-030-50284-3 (eBook) https://doi.org/10.1007/978-3-030-50284-3 Mathematics Subject Classification: 35, 76 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 This work is subject to copyright. All rights are solely and exclusively licensed 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, expressed 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

To our families.

Introduction

The aim of this book is to present, through 65 fully solved long problems, various aspects of the current theory of partial differential equations (PDE). It is intended for graduate students who would like, through practice, to test their understanding of the theory. The book is made of two parts. The first part (Chaps. 1–8) contains some theory (in the odd chapters) and the statements of the problems (in the even chapters). The second part contains the solutions (Chap. 9) and an appendix (Chap. 10), where we recall some basics of the classical analysis. Even though the main purpose of this book is to present these problems and their solutions, for the reader’s convenience, we have recalled some of the main theoretical results concerning each topic. This is why each chapter of problems is preceded by a short introduction recalling without proof the basic facts. However, some comments indicate where one may find the details of the proofs. This makes the book essentially self-contained for the reader. Since the theory of partial differential equations is a very wide subject, it is by no means realistic to hope to describe all the topics in a single volume. Therefore, choices have to be made and we have chosen to focus on a few of them. Let us now describe more precisely the contents of this book. The first chapter is introductory. Some of the essential tools in functional analysis which are commonly used in PDE are recalled. This includes the main theorems concerning Fréchet, Banach, or Hilbert spaces. We have also recalled the main notions in distributions theory, including the analysis of the Fourier transform and the stationary phase formula. The second chapter contains three problems on the subjects discussed in the previous one. The third chapter begins with the description of some of the main function spaces used in PDE, namely Sobolev, Hölder, and Zygmund spaces (including their Littlewood–Paley characterization). Other tools, such as interpolation theory and paraproducts, are also presented.

vii

viii

Introduction

In the fourth chapter, through 15 problems, we discuss various applications such as functional inequalities (product and composition rules in Sobolev spaces, Hardy inequality, etc.) and we introduce other function spaces such as spaces of analytic functions, uniformly local Sobolev spaces, weak Lebesgue spaces, space of bounded mean oscillation, etc. The fifth chapter is concerned with the theory of microlocal analysis. We review the main notions about the pseudo-differential and paradifferential operators, the wave front set, and the microlocal defect measures and we describe some of the main results, such as continuity, Garding inequalities, and propagation of singularities. In the sixth chapter, through 18 problems, we give some applications of these notions, in particular we explore the notions of symbols, hypoelliptic operators, smoothing effect, and Carleman inequalities. The seventh chapter reviews the main classical partial differential equations that is the Laplace, wave, Schrödinger, heat, Burgers, Euler, Navier–Stokes equations, and their main properties are recalled. The last chapter of the first part of the book contains 29 problems on the above equations. For instance, several problems about spectral theory for the Laplace equation, about linear and nonlinear wave and Schrödinger equations are stated. Moreover, other equations such as Monge–Ampère, the mean-curvature, kinetic, and Benjamin–Ono equations are discussed. The second part of the book contains the detailed solutions to all the problems and a chapter gathering several fundamental results concerning the basics of classical analysis, such as Lebesgue integration, differential calculus, differential equations and holomorphic functions.

Acknowledgements We would like to warmly thank the anonymous referees for their excellent work, which helped us to improve the presentation of this book. However, it should be clear that the authors are solely responsible of any mistake remaining. A list of possible corrections will be available at http://talazard.perso. math.cnrs.fr. After this book was proposed to Springer, we had several mail exchanges with Mr Rémi Lodh. We would like to warmly thank him for his suggestions, support, and efficiency. April 2020

Thomas Alazard Claude Zuily

Contents

Part I

Tools and Problems

1

Elements of Functional Analysis and Distributions. . . . . . . . . . . . . . . . . . . . . 1.1 Fréchet Spaces .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2 Elements of Functional Analysis . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.1 Fixed Point Theorems . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.2 The Banach Isomorphism Theorem . . .. . . . . . . . . . . . . . . . . . . . 1.2.3 The Closed Graph Theorem .. . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.4 The Banach–Steinhaus Theorem.. . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.5 The Banach–Alaoglu Theorem . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.6 The Ascoli Theorem .. . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.7 The Hahn–Banach Theorem . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.8 Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.9 Spectral Theory of Self-Adjoint Compact Operators . . . . . 1.2.10 Lp Spaces, 1 ≤ p ≤ +∞ . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.2.11 The Hölder and Young Inequalities .. . .. . . . . . . . . . . . . . . . . . . . 1.2.12 Approximation of the Identity . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.3 Elements of Distribution Theory . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.3.1 Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.3.2 Tempered Distributions.. . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.3.3 The Fourier Transform . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 1.3.4 The Stationary Phase Method .. . . . . . . . .. . . . . . . . . . . . . . . . . . . .

3 3 5 5 5 5 6 6 7 7 7 9 9 10 11 12 12 13 15 16

2

Statements of the Problems of Chap. 1 . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

19

3

Functional Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.1 Sobolev Spaces .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.1.1 Sobolev Spaces on Rd , d ≥ 1 . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . s (Rd ) . . . . . .. . . . . . . . . . . . . . . . . . . . 3.1.2 Local Sobolev Spaces Hloc 3.1.3 Sobolev Spaces on an Open Subset of Rd .. . . . . . . . . . . . . . . . 3.1.4 Sobolev Spaces on the Torus .. . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

25 25 25 28 29 31

ix

x

Contents

3.2

The Hölder Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.2.1 Hölder Spaces of Integer Order .. . . . . . .. . . . . . . . . . . . . . . . . . . . 3.2.2 Hölder Spaces of Fractional Order.. . . .. . . . . . . . . . . . . . . . . . . . Characterization of Sobolev and Hölder Spaces in Dyadic Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 3.3.1 Characterization of Sobolev Spaces . . .. . . . . . . . . . . . . . . . . . . . 3.3.2 Characterization of Hölder Spaces. . . . .. . . . . . . . . . . . . . . . . . . . 3.3.3 The Zygmund Spaces. . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . Paraproducts .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . Some Words on Interpolation .. . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . The Hardy–Littlewood–Sobolev Inequality .. . . .. . . . . . . . . . . . . . . . . . . .

33 34 34 35 35 37 39

4

Statements of the Problems of Chap. 3 . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

41

5

Microlocal Analysis.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.1 Symbol Classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.1.1 Definition and First Properties . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.1.2 Examples .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.1.3 Classical Symbols . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.2 Pseudo-Differential Operators . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.2.1 Definition and First Properties . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.2.2 Kernel of a DO . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.2.3 Image of a DO by a Diffeomorphism . . . . . . . . . . . . . . . . . . . 5.2.4 Symbolic Calculus .. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.2.5 Action of the DO on Sobolev Spaces . . . . . . . . . . . . . . . . . . . 5.2.6 Garding Inequalities . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.3 Invertibility of Elliptic Symbols . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.4 Wave Front Set of a Distribution . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.4.1 Definition and First Properties . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.4.2 Wave Front Set and DO .. . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.4.3 The Propagation of Singularities Theorem .. . . . . . . . . . . . . . . 5.5 Paradifferential Calculus . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.5.1 Symbols Classes . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.5.2 Paradifferential Operators .. . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.5.3 The Symbolic Calculus .. . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.5.4 Link with the Paraproducts .. . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 5.6 Microlocal Defect Measures .. . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

61 61 61 62 62 62 62 63 63 64 65 65 66 66 66 67 67 68 68 69 70 71 71

6

Statements of the Problems of Chap. 5 . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

75

7

The Classical Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.1 Equations with Analytic Coefficients . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.1.1 The Cauchy–Kovalevski Theorem .. . . .. . . . . . . . . . . . . . . . . . . . 7.1.2 The Holmgren Uniqueness Theorem . .. . . . . . . . . . . . . . . . . . . .

101 101 102 103

3.3

3.4 3.5 3.6

31 31 32

Contents

7.2

7.3

7.4

7.5

7.6 7.7

7.8

8

The Laplace Equation .. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.2.1 The Mean Value Property . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.2.2 Hypoellipticity: Analytic Hypoellipticity . . . . . . . . . . . . . . . . . 7.2.3 The Maximum Principles . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.2.4 The Harnack Inequality . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.2.5 The Dirichlet Problem .. . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.2.6 Spectral Theory .. . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . The Heat Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.3.1 The Maximum Principle . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.3.2 The Cauchy Problem . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . The Wave Equation.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.4.1 Homogeneous Cauchy Problem . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.4.2 Inhomogeneous Cauchy Problem .. . . . .. . . . . . . . . . . . . . . . . . . . 7.4.3 The Mixed Problem . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . The Schrödinger Equation . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.5.1 The Cauchy Problem . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.5.2 Properties of the Solution . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . The Burgers Equation .. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . The Euler Equations.. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.7.1 The Incompressible Euler Equations . .. . . . . . . . . . . . . . . . . . . . 7.7.2 The Compressible Euler Equations . . . .. . . . . . . . . . . . . . . . . . . . The Navier–Stokes Equations . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.8.1 Weak Solutions . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.8.2 The Leray Theorem (1934) . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 7.8.3 Strong Solutions: Theorems of Fujita–Kato and Kato .. . .

104 104 104 105 105 106 106 108 108 109 109 109 111 112 113 113 114 116 117 117 121 122 122 122 123

Statements of the Problems of Chap. 7 . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 125

Part II 9

xi

Solutions of the Problems and Classical Results

Solutions of the Problems .. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 181

10 Classical Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.1 Some Classical Formulas . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.1.1 The Leibniz Formula . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.1.2 The Taylor Formula with Integral Reminder . . . . . . . . . . . . . . 10.1.3 The Faa–di-Bruno Formula . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.2 Elements of Integration . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.2.1 Convergence Theorems . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.2.2 Change of Variables in Rd . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.2.3 Polar Coordinates in Rd . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.2.4 The Gauss–Green Formula .. . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.2.5 Integration on a Graph.. . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.3 Elements of Differential Calculus . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

343 343 343 343 344 344 344 345 345 346 347 348

xii

Contents

10.4 Elements of Differential Equations .. . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.4.1 The Precise Cauchy–Lipschitz Theorem . . . . . . . . . . . . . . . . . . 10.4.2 The Cauchy–Arzela–Péano Theorem ... . . . . . . . . . . . . . . . . . . . 10.4.3 Global Theory . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.4.4 The Gronwall Inequality . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 10.5 Elements of Holomorphic Functions . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . .

348 348 349 349 350 350

References .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 353 Index . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . 355

Part I

Tools and Problems

Chapter 1

Elements of Functional Analysis and Distributions

The goal of this chapter is to recall, without proof, the main results in functional analysis: classical theorems about Fréchet, Hilbert, and Banach spaces, as well as fixed point theorems and an introduction to spectral theory. This is complemented by the main definitions in distributions theory, including results about the Fourier transform.

1.1 Fréchet Spaces • Seminorms. Consider a vector space E on C. A seminorm on E is a map p : E → [0, +∞) such that, for all x, y ∈ E and all λ ∈ C, (i)

p(λx) = |λ|p(x),

(ii) p(x + y) ≤ p(x) + p(y).

A family P of seminorms on E is called separating if for every x ∈ E there exists p ∈ P such that p(x) = 0. • Topology. Let P be a family of separating seminorms on a vector space E. For x0 ∈ E, n ∈ N, n ≥ 1 and p ∈ P, set   1 . V(p,n) (x0 ) = x ∈ E : p(x − x0 ) < n Let Vx0 be the collection of all finite intersections of the sets V(p,n) (x0 ) and define a neighborhood of x0 as a set which contains an element of Vx0 . This defines a topology on E. Then (E, P) is called a locally convex space.

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 T. Alazard, C. Zuily, Tools and Problems in Partial Differential Equations, Universitext, https://doi.org/10.1007/978-3-030-50284-3_1

3

4

1 Elements of Functional Analysis and Distributions

– A subset A of E is said to be bounded if, ∀p ∈ P,

∃M > 0 : p(x) ≤ M,

∀x ∈ A.

– If (E, P) and (F, Q) are two locally convex spaces and T : E → F is a linear map then T is continuous if and only if, ∀q ∈ Q, ∃P0 ⊂ P finite, ∃C > 0 : q(T x) ≤ C



p(x), ∀x ∈ E.

p∈P0

– If P = (pj )j ∈N is countable then the topology defined above is metrizable, that is there exists a metric d on E which induces the same topology. Indeed for x, y ∈ E we set, d(x, y) =

+∞  1 pj (x − y) · 2j 1 + pj (x − y) j =0

If moreover the metric space (E, d) is complete then we say that E is a Fréchet space. • Examples. – A Banach space is a Fréchet space. – Let  be an open subset of Rd . Then C 0 () denotes the space of continuous functions on  with complex values, C 1 () denotes the space of differentiable functions whose partial derivatives belong to C 0 (), and, for k ∈ N, k ≥ 2,   ∂u C k () = u ∈ C k−1 () : ∈ C k−1 (), 1 ≤ j ≤ d . ∂xj Recall that  can be written as  = ∪+∞ j =0 Kj , where the Kj ’s are compact and Kj ⊂ Kj +1 . For k ∈ N ∪ {+∞} , u ∈ C k () and j ∈ N set, pj (u) =



sup |∂ α u(x)|

(k < +∞),

|α|≤k x∈Kj

pj (u) =



sup |∂ α u(x)|

(k = +∞).

|α|≤j x∈Kj

Then (C k (), (pj )j ∈N ) is a Fréchet space. A sequence (fn )n∈N ⊂ C k () converges to f for this topology if and only if, for every |α| ≤ k, (∂ α fn )n∈N converges to ∂ α f uniformly on each compact subset of .

1.2 Elements of Functional Analysis

5

1.2 Elements of Functional Analysis In this section we recall several important results of functional analysis used in the study of partial differential equations. Unless expressly stated all the normed spaces considered below are normed spaces on C.

1.2.1 Fixed Point Theorems • The Banach fixed point theorem. Let (X, d) be a complete metric space and consider a map f : X → X satisfying the following condition: There exists k ∈ ]0, 1[ such that, d(f (x), f (y)) ≤ kd(x, y),

∀x, y ∈ X.

Then f has a unique fixed point, that is a point x such that f (x) = x. • The Schauder fixed point theorem. Let E be a Banach space and K ⊂ E be a nonempty convex and compact subset. Then any continuous function f : K → K has a fixed point.

1.2.2 The Banach Isomorphism Theorem • Let E, F be two Banach spaces. If T : E → F is a linear continuous and bijective map then T −1 is continuous.

1.2.3 The Closed Graph Theorem • Let E, F be two Banach spaces and consider a linear map T : E → F . Assume that the set (called the graph of T ), G = {(x, T x) : x ∈ E} is closed in E × F . Then T is continuous. • Notice that the converse is also true. If T is continuous then G is closed. • This result remains true if E and F are Fréchet spaces.

6

1 Elements of Functional Analysis and Distributions

1.2.4 The Banach–Steinhaus Theorem • Let E be a Banach space, F be a normed space, and  be a set. Let (Tλ )λ∈ be a family of linear and continuous maps from E to F . Assume that, ∀x ∈ E, ∃Cx > 0 :

Tλ x F ≤ Cx ,

∀λ ∈ .

Then there exists C > 0 such that, Tλ x F ≤ C x E ,

∀x ∈ E,

∀λ ∈ .

This result can be extended to the case where E is a Fréchet space. • Let (E, (pj )j ∈N ) be a Fréchet space and (F, Q) be a locally convex space. Let (Tλ )λ∈ be a family of linear and continuous maps from E to F . Assume that, ∀x ∈ E, ∀q ∈ Q, ∃Cq,x > 0 :

q(Tλ x) ≤ Cq,x ,

∀λ ∈ .

Then, for any q ∈ Q, there exists a finite set J ⊂ N and C > 0 such that, q(Tλ x) ≤ C



pj (x),

∀x ∈ E, ∀λ ∈ .

j ∈J

1.2.5 The Banach–Alaoglu Theorem • Let E be a normed space and E be its dual, that is the space of linear and continuous maps T : E → C. One can endow E with the strong topology induced by the norm T E = sup x E =1 |T (x)|. • One can also define on E a weaker topology as follows. For x ∈ E and T ∈ E set px (T ) = |T (x)|. Then px is a seminorm on E and the family (px )x∈E defines a topology on E called the weak-star topology. For instance, a sequence (Tn )n∈N in E converges to T ∈ E in this topology if and only if Tn (x) → T (x) in C for every x ∈ E. This is therefore the topology of pointwise convergence. • The Banach–Alaoglu  theorem. Let E be a Banach space. Then the set T ∈ E : T E ≤ 1 is compact in the weak-star topology. • As a Corollary, if E is a separable Banach space then each bounded sequence in E (for the norm) has a convergent subsequence for the weak-star topology.

1.2 Elements of Functional Analysis

7

1.2.6 The Ascoli Theorem • Let X be a compact metric space. Let C 0 (X) be the Banach space of all continuous functions f : X → C endowed with the norm f = supx∈X |f (x)|. Let A ⊂ C 0 (X). Assume that, 1. for every x ∈ X, the set {f (x) : f ∈ A} is bounded in C, that is, ∀x ∈ X,

∃Mx > 0 : |f (x)| ≤ Mx ,

∀f ∈ A.

2. A is equicontinuous, that is, ∀ε > 0, ∀x ∈ X,

∃δ > 0 : d(x, y) ≤ δ ⇒ |f (x) − f (y)| ≤ ε,

∀f ∈ A.

Then A has a compact closure in C 0 (X).

1.2.7 The Hahn–Banach Theorem • Let E be a normed vector space and F be a subspace. Let L : F → C be a : E → C continuous linear form. Then there exists a continuous linear form L   such that L(x) = L(x) for every x ∈ F and L E = L F . As a consequence we have the following result. • Let E be a normed vector space and F be a subspace. Then F is dense in E if and only if, for every L ∈ E , L = 0 on F ⇒ L = 0.

1.2.8 Hilbert Spaces Recall first that a Hilbert space H is a vector space endowed√with a scalar product denoted by (x, y)H and with the corresponding norm x = (x, x)H . • Weak convergence. A sequence (xn )n∈N in H is said to converge weakly to x ∈ H (and write xn x) if limn→+∞ (xn , y)H = (x, y)H for every y ∈ H. – The strong convergence (for the norm) implies the weak convergence. – If xn x and xn → x then (xn )n∈N converges strongly to x. – A sequence (xn )n∈N which converges weakly in H is bounded for the norm, that is, there exists M > 0 such that xn ≤ M for every n ∈ N. – Let (xn )n∈N be a bounded sequence for the norm. Then there exists a subsequence (xσ (n) )n∈N which converges weakly in H . – If (xn )n∈N converges weakly to x in H then x H ≤ lim infn→+∞ xn H .

8

1 Elements of Functional Analysis and Distributions

• A Riesz theorem – Let H be a complex Hilbert space. An antilinear form L on H is a map L : H → C such that for any x, y ∈ H, λ ∈ C, L(x + y) = L(x) + L(y),

L(λx) = λL(x).

– A Riesz theorem. For every antilinear and continuous map L : H → C, there exists a unique x ∈ H such that L(y) = (x, y)H for every y ∈ H . • The Lax–Milgram lemma – Let H be a complex Hilbert space. A sesquilinear form on H × H is a map a : H × H → C such that for every y ∈ H the map x → a(x, y) is linear and for every x ∈ H the map y → a(x, y) is antilinear. – The Lax–Milgram lemma. Let a be a sesquilinear form on H × H. Assume that, (i) a is continuous : ∃C > 0 : |a(x, y)| ≤ C x H y H , (ii) a is coercive : ∃c0 > 0 : Re a(x, x) ≥ c0 x 2H ,

∀x, y ∈ H,

∀x ∈ H.

Then for any L : H → C antilinear there exists a unique x ∈ H such that, a(x, y) = L(y),

∀y ∈ H.

– Examples: (i) a(x, y) = (x, y)H .   (ii) Take H = H 1 (Rd ) = u ∈ L2 (Rd ) : ∂j u ∈ L2 (Rd ), j = 1, . . . , d and for u, v ∈ H set, a(u, v) =

d   d j,k=1 R

 aj k (t)∂j u(t)∂k v(t) dt +

u(t)v(t) dt, Rd

where aj k ∈ L∞ (Rd ), j, k = 1, . . . , d and, Re

d  j,k=1

aj k (t)ζj ζk ≥ c0 |ζ |2 ,

∀t ∈ Rd , ∀ζ ∈ Cd .

1.2 Elements of Functional Analysis

9

1.2.9 Spectral Theory of Self-Adjoint Compact Operators We first recall several definitions. • Compact operators. Let E, F be two normed vector spaces and T ∈ L(E, F ) (the space of continuous linear maps). Then T is said to be compact if the image of the unit ball in E has a compact closure in F . In the particular case where E and F are separable Hilbert spaces, this is equivalent to say that the image by T of a weakly convergent sequence in E is convergent in F for the norm. • Adjoint operator. Let H be a Hilbert space with scalar product (·, ·)H . Let T ∈ L(H, H ). Then there exists a unique operator T ∗ ∈ L(H, H ) such that, (T x, y)H = (x, T ∗ y)H for every x, y ∈ H. It is called the adjoint of T . The operator T is said self-adjoint if T = T ∗ . • Spectrum. Let T ∈ L(H, H ). The spectrum of T is the closed subset of C defined by, σ (T ) = {μ ∈ C : T − μ Id is not invertible} , where Id denotes the identity operator. If T − μ Id is non-injective, which means that there exists x ∈ H, x = 0 such that T x = μx, then μ is called an eigenvalue and x an eigenvector of T . • Spectral theory of compact self-adjoint operators. Let H be a Hilbert space and T ∈ L(H, H ) be a compact self-adjoint operator. Then, 1. the nonzero eigenvalues form a finite subset of R or a sequence (μn )n∈N of real numbers converging to zero when n → +∞, 2. if μn is a nonzero eigenvalue then the space En = Ker (T − μn Id) has finite dimension, 3. H = ⊕En ⊕ Ker T , 4. if H is finite dimensional then σ (T ) = {eigenvalues} and if H has infinite

dimension then σ (T ) = {0} ∪ ∪+∞ n=1 {μn } .

1.2.10 Lp Spaces, 1 ≤ p ≤ +∞ Let (X, T , μ) be a measured space. • If 1 ≤ p < +∞, Lp (X) is the set of (classes of) measurable functions f : X →  C such that X |f (x)|p dμ < +∞. On this space the quantity, 

1

f Lp (X) =

|f (x)| dμ p

p

X

is a norm. Endowed with this norm Lp (X) is a Banach space.

10

1 Elements of Functional Analysis and Distributions

The particular space L2 (X) is a Hilbert space when endowed with the scalar product,  (f, g)L2 (X) =

f (x)g(x) dμ. X

• L∞ (X) is the set of (classes of) measurable functions f : X → C which are essentially bounded, that means that there exists M > 0 such that, μ ({x ∈ X : |f (x)| > M}) = 0. By definition f L∞ (X) is the infimum of such numbers M. It is characterized as follows: μ



x ∈ X : |f (x)| > f L∞ (X)



=0

and

∀α < f L∞ (X) , μ ({x ∈ X : |f (x)| > α}) = 0. Endowed with this norm L∞ (X) is a Banach space. Notice that if f is a continuous and bounded function on an open set  ⊂ Rd then f L∞ () = sup |f (x)|.

• If 1 ≤ p < +∞ the dual of Lp (X) is Lp (X) where p is the conjugate exponent of p defined by, 1 1 + = 1. p p Notice that if p = 1 then p = +∞. Notice also that the dual of L∞ (X) is NOT L1 (X) but a wider space. • Let 1 ≤ p ≤ +∞. Every convergent sequence in Lp (X) has a subsequence which converges almost everywhere.

1.2.11 The Hölder and Young Inequalities • Let (X, T , μ) be a measured space and let 1 ≤ p, q, r ≤ +∞ be real numbers such that, 1 1 1 + = . p q r If u ∈ Lp (X) and v ∈ Lq (X) then uv ∈ Lr (X) and, uv Lr (X) ≤ u Lp (X) v Lq (X) .

1.2 Elements of Functional Analysis

11

• Let 1 ≤ p, q ≤ +∞ and let r ≥ 1 be such that, 1 1 1 = + − 1. r p q If u ∈ Lp (Rd ) and v ∈ Lq (Rd ) then for almost all x in Rd the integral,  (u v)(x) :=

Rd

u(x − y)v(y) dy

is convergent, and u v belongs to Lr (Rd ). Moreover we have, u v Lr (Rd ) ≤ u Lp (Rd ) v Lq (Rd ) .

1.2.12 Approximation of the Identity Let ρ ∈ C ∞ (Rd ) be such that,   supp ρ ⊂ x ∈ Rd : |x| ≤ 1 ,

 ρ ≥ 0,

Rd

ρ(x) dx = 1.

For ε in (0, 1] set, ρε (x) = ε−d ρ

x  ε

.

The family (ρε )ε∈(0,1] is called an approximation of the identity. Then we have, • for any 1 ≤ p < +∞, if u belongs to ∈ Lp (Rd ), then, lim (ρε u) = u

ε→0

in Lp (Rd ).

• If u : Rd → C is a bounded uniformly continuous function, then lim (ρε u) = u

ε→0

in L∞ (Rd ).

• For any 1 ≤ p ≤ +∞, if u belongs to Lp (Rd ), then for almost all x in Rd , lim (ρε u)(x) = u(x).

ε→0

• For any 1 ≤ p ≤ +∞, if u belongs to Lp (Rd ), then for any ε ∈ (0, 1], the function ρε u belongs to C ∞ (Rd ).

12

1 Elements of Functional Analysis and Distributions

1.3 Elements of Distribution Theory • In what follows  will be an open subset of Rd . We shall write K ⊂⊂  if K is a compact subset of . • Recall that the support of a continuous function f :  → C is the set, supp f = {x ∈  : f (x) = 0}. • If x0 ∈  we denote by Vx0 the set of open neighborhoods of x0 .

1.3.1 Distributions • The space C0∞ () – C0∞ () is the vector space of functions f :  → C which are C ∞ on  and whose support is a compact subset of . – There is a topology on this space (called a LF -topology) which is useful for applications. We shall not describe it. We just notice that a sequence (ϕj )j ∈N ⊂ C0∞ () converges to ϕ ∈ C0∞ () for this topology if and only if, (i) ∃K ⊂⊂  : supp ϕj ⊂ K, ∀j ∈ N, (ii) ∀α ∈ Nd , (∂ α ϕj )j converges uniformly to ∂ α ϕ on K.   – If K ⊂⊂  we set C0∞ (K) = ϕ ∈ C0∞ () : supp ϕ ⊂ K . • The space of distributions D () – Definition. D () is the vector space of linear map T : C0∞ () → C satisfying: for any K ⊂⊂ , there exist CK > 0 and k ∈ N such that, | T , ϕ | ≤ CK



sup |∂ α ϕ|,

|α|≤k K

∀ϕ ∈ C0∞ (K).

(1.1)

– If (1.1) is true with the same k ∈ N for any compact K we say that T is of order ≤ k. – The set of distributions of order ≤ k is denoted by D (k) (). – A linear map T : C0∞ () → C is a distribution if and only if for every 

sequence (ϕj )j ∈N converging to zero in C0∞ (), the sequence T , ϕj j ∈N converges to zero in C. – Support. Let T ∈ D (). The support of T is defined as follows. / supp T ⇐⇒ ∃V ∈ Vx0 : T , ϕ = 0 ∀ϕ ∈ C0∞ (V ). x0 ∈ The support is a closed set.

1.3 Elements of Distribution Theory

13

– Distributions with compact support. We denote by E () the set of distributions T on  whose support is a compact set contained in . It can be identified with the set of linear maps T : C ∞ () → C satisfying: for some C > 0, m ∈ N and K ⊂⊂ ,  sup |∂ α ϕ(x)|, ∀ϕ ∈ C ∞ (). | T , ϕ | ≤ C |α|≤m K

– Product by a C ∞ function. If T ∈ D () and a ∈ C ∞ () we define aT ∈ D () by, aT , ϕ = T , aϕ ,

∀ϕ ∈ C0∞ ().

– Differentiation of distributions. If T ∈ D () we define ∂xj T ∈ D () by, 

   ∂xj T , ϕ = − T , ∂xj ϕ ,

∀ϕ ∈ C0∞ ().

Therefore we have ∂ α T , ϕ = (−1)|α| T , ∂ α ϕ for ϕ ∈ C0∞ ().  – For T ∈ D () we have supp T ⊂ {0} if and only if T = |α|≤m cα ∂ α δ0 where δ0 is the Dirac distribution at zero. – Convergence of sequences in D (). A sequence (Tj )j ∈N ⊂ D () converges to T ∈ D () if, lim

j →+∞

  Tj , ϕ = T , ϕ ,

∀ϕ ∈ C0∞ ().

– If (Tj ) → T in D () then for all α ∈ Nd , (∂ α Tj ) → ∂ α T in D ().

– Let C0∞ () the sequence  (Tj )j ∈N ⊂ D (). Assume that for every ϕ ∈

( Tj , ϕ ) is convergent in C. Then there exists T ∈ D () such that (Tj ) → T in D (). – Density. C0∞ () is dense is D (). – Fundamental solution. Let P (D) be a differential operator with constant coefficients. A fundamental solution of P is a distribution E ∈ D (Rd ) such that P (D)E = δ0 (the Dirac distribution at zero).

1.3.2 Tempered Distributions • The Schwartz space S(Rd ) – Definition. S(Rd ) is the vector space of functions u : Rd → C which are C ∞ and satisfy, ∀α, β ∈ Nd , ∃Cα,β > 0 : |x α ∂ β u(x)| ≤ Cα,β , ∀x ∈ Rd .

14

1 Elements of Functional Analysis and Distributions

Examples: C0∞ (Rd ) ⊂ S(Rd ) and u(x) = e−z|x| ∈ S(Rd ) for Re z > 0. d α β d – For α,d β ∈ N pαβ (u) = supx∈Rd |x ∂ u(x)| is a seminorm on S(R ). Then S(R ), (pαβ )α,β∈Nd is a Fréchet space. 2

• The space of tempered distributions S (Rd ) – Definition. S (Rd ) is the vector space of linear maps T : S(Rd ) → C such that,  ∃k,  ∈ N, ∃C > 0 : | T , ϕ | ≤ C pαβ (ϕ), ∀ϕ ∈ S(Rd ). |α|≤k |β|≤

– Density. S(Rd ) is dense in S (Rd ). – Examples. For 1 ≤ p ≤ +∞, Lp (Rd ) ⊂ S (Rd ). – Product. Let f ∈ C ∞ (Rd ) be such that, ∀α ∈ Nd , ∃Cα > 0, ∃mα ∈ R : |∂ α f (x)| ≤ Cα (1 + |x|)mα ,

∀x ∈ Rd

and let T ∈ S (Rd ). Then f T defined by, f T , ϕ = T , f ϕ ,

∀ϕ ∈ S(Rd )

is an element of S (Rd ). – Derivation. Let T ∈ S (Rd ), then for every α ∈ Nd , ∂ α T defined by, 

   ∂ α T , ϕ = (−1)|α| T , ∂ α ϕ ,

∀ϕ ∈ S(Rd )

is an element of S (Rd ). – Convergence of sequences in S (Rd ). A sequence (Tj )j ∈N ⊂ S (Rd ) converges to T ∈ S (Rd ) if, lim

j →+∞



 Tj , ϕ = T , ϕ ,

∀ϕ ∈ S(Rd ).

– If (Tj ) → T in S (Rd ) then for all α ∈ Nd , (∂ α Tj ) → ∂ α T in S (Rd ).

d d – Let  (Tj )j ∈N ⊂ S (R ). Assume that for every ϕ ∈d S(R ) the sequence ( Tj , ϕ ) is convergent in C. Then there exists T ∈ S (R ) such that (Tj ) → T in S (Rd ).

1.3 Elements of Distribution Theory

15

1.3.3 The Fourier Transform • The Fourier transform in S(Rd ). The Fourier transform of ϕ ∈ S(Rd ) is the function defined by,  (F ϕ)(ξ ) =

e−ix·ξ ϕ(x) dx.

F ϕ is often denoted by  ϕ. – The map F is continuous from S(Rd ) to S(Rd ). – The map F is invertible. If we introduce for ψ ∈ S(Rd ), (F −1 ψ)(x) = (2π)−d



(ξ ) dξ, eix·ξ ψ

then F (F −1 ψ) = ψ and F −1 (F ϕ) = ϕ for all ψ, ϕ ∈ S(Rd ). • The Fourier transform in S (Rd ). If T ∈ S (Rd ) then F T defined by, F T , ϕ = T , F ϕ ,

∀ϕ ∈ S(Rd )

belongs to S (Rd ). We define F −1 by an analogous formula. – The maps F , F −1 are continuous from S (Rd ) to S (Rd ). – F (F −1 T ) = T and F −1 (F T ) = T for all T ∈ S (Rd ). – For α ∈ Nd we have, F (∂ α T ) = (iξ )α F (T )

and F (x α T ) = (−Dξ )α (F T ),

where Dξα = Dξα11 . . . Dξαdd with Dξj =

1 ∂ i ∂ξj

.

• Fourier transform of distributions with compact support – If T ∈ E (Rd ) then F T is the C ∞ function on Rd given by,   (F T )(ξ ) = Tx , e−ix·ξ . Moreover, ∃k ∈ N : ∀α ∈ Nd , ∃Cα > 0 : |∂ξα (F T )(ξ )| ≤ Cα (1 + |ξ |)k • Fourier transform in Lebesgue spaces

∀ξ ∈ Rd .

 – If T ∈ L1 (Rd ) then F T is given by the continuous function e−ix·ξ T (x) dx. Moreover (F T )(ξ ) tends to zero when |ξ | → +∞.

16

1 Elements of Functional Analysis and Distributions d

– The map T → (2π)− 2 F T is a bijective isometry from L2 (Rd ) to L2 (Rd ). – More generally, let 1 ≤ p ≤ 2 and p its conjugate, defined by p1 = 1 − p1 .

Then F is continuous from Lp (Rd ) to Lp (Rd ). • Fourier transform and convolution – If T ∈ S (Rd ) and S ∈ E (Rd ) then T S ∈ S (Rd ) and, F (T S) = (F T ) · (F S). – Let p, q be such that, 1 ≤ p ≤ 2,

1 ≤ q ≤ 2,

1 3 1 + ≥ . p q 2

If T ∈ Lp (Rd ) and S ∈ Lq (Rd ) then T S ∈ Lr (Rd ) where and,

1 r

=

1 1 p+q

−1

F (T S) = (F T ) · (F S). For instance one may take p = q = 1 (then r = 1) or p = 1, q = 2 (then r = 2). • Bernstein inequalities. Let p be a real number such that 1 ≤ p ≤ +∞ and let θ ∈ [0, p1 ]. Let r ≥ 1 be defined by 1r = p1 − θ. Let a ∈ Lp (Rd ), R > 0 and assume that,   supp a ⊂ ξ ∈ Rd : |ξ | ≤ R . Then for every α ∈ Nd there exists Cα > 0 independent of a, R such that, ∂ α a Lr (Rd ) ≤ Cα R |α|+dθ a Lp (Rd ) . (See Problem 3.) A case which will be used in the problems is p = 1, θ = 1, r = +∞.

1.3.4 The Stationary Phase Method It is a method which describes the behavior when λ → +∞ of integrals on the form,  I (λ) = eiλϕ(x) a(x) dx , (1.2) Rd

1.3 Elements of Distribution Theory

17

where ϕ is a C ∞ function on Rd with real values and a ∈ C0∞ (Rd ) (ϕ is called the phase, a the amplitude). The asymptotic of I (λ) depends on the behavior of ϕ on the support of a. A point x such that ϕ (x) = 0 is called a stationary (or a critical) point of ϕ. • Nonstationary phase. Assume that: ϕ (x) = 0,

∀x ∈ supp a.

Then, for any N ∈ N, there exists CN > 0 such that, |I (λ)| ≤ CN λ−N ,

∀λ ≥ 1. ◦ 

• Stationary phase. Assume that there exists a unique x0 ∈ supp a such that, ϕ (x0 ) = 0 and ϕ

(x0 ) is nondegenerate, where ϕ

(x0 ) denotes the Hessian matrix of ϕ at x0 namely, (∂xj ∂xk ϕ(x0 ))1≤j,k≤d. Then for any N ∈ N, there exists b0 , . . . , bN in C (depending on a and ϕ), a function RN and a constant CN > 0 such that, for all λ ≥ 1 we have, I (λ) = eiλϕ(x0 )

N 

d

bk λ− 2 −k + RN (λ),

d

|RN (λ)| ≤ CN λ− 2 −N−1 .

(1.3)

k=0

Moreover b0 = √

π (2π)d/2 ei 4 | det ϕ

(x0 )|

sgn ϕ

(x0 )

a(x0),

where sgn is the signature. Comments For a detailed exposition of the classical results in functional analysis we refer to the books by T. Alazard [1], H. Brezis [7], and W. Rudin [24]. Concerning the Distribution theory, including a proof of the stationary phase formula, we refer to the books by L. Hörmander [14] and C. Zuily [32, 33].  

Chapter 2

Statements of the Problems of Chap. 1

This chapter contains problems about functional analysis.

Problem 1

This problem states the basic results which will be used in the study of spectral theory for the harmonic oscillator in Problem 39. In all what follows we shall denote E = E(Rd ) if E = L2 , C0∞ .   Let V = u ∈ L2 : ∇x u ∈ L2 : x u ∈ L2 endowed with the scalar product, (u, v)V = (∇x u, ∇x v)L2 + (x u, x v)L2 ,

1

x = (1 + |x|2 ) 2 ,

1

and with the corresponding norm u V = (u, u)V2 . Then V is a Hilbert space. 1. a) Let Vc = {u ∈ V : supp u is compact} endowed with the norm of V . Show that Vc is dense in V . b) Show that C0∞ is dense in Vc . Hint: Let (θε ) be an approximation of the identity. Use the fact that θε v → v in L2 for every v ∈ L2 when ε → 0. 2. Show that the map u → u from V to L2 is compact. Hint: Let B be the unit ball in V . For u ∈ B write u = χR u+(1−χR )u where χR (x) = χ( Rx ), χ ∈ C0∞ and use the compactness of the map u → u from HK1 to L2 when K is a compact set in Rd .

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 T. Alazard, C. Zuily, Tools and Problems in Partial Differential Equations, Universitext, https://doi.org/10.1007/978-3-030-50284-3_2

19

20

2 Statements of the Problems of Chap. 1

3. a) Let ek be the k t h vector of the canonical basis of Rd . For 0 < |h| ≤ 1 and v ∈ V set Dh v =

1 (v(x − hek ) − v(x)). h

Prove that Dh v ∈ V . b) Let s ∈ R. Prove that for w ∈ H s we have, limh→0 Dh w = −∂k w in H s−1 . (For the definition of the H s space see next chapter.) Hint: Use the Fourier transform, the fact that for θ ∈ R we have |eiθ − 1| ≤ |θ |, |eiθ − 1 − iθ | ≤ 12 |θ |2 and the dominated convergence theorem. c) Prove that, Dh w L2 ≤ ∂k w L2 ,

d

∀w ∈ H 1 ,

Dh g H −1 ≤ (2π ) 2 g L2 ,

∀g ∈ L2 .

Hint: Use the same method as in b). 4. Let L ∈ V . Prove that there exists a unique u ∈ V such that, d   d j =1 R

 ∂j v(x)∂j u(x) dx +

Rd

x v(x) x u(x) dx = L(v),

∀v ∈ V (2.1)

and that there exists C > 0 such that, u V ≤ C L V . 5. In this question we want to give an identification of V . Let   E1 = f1 ∈ S : ξ −1 f1 ∈ L2 ,

  E2 = f2 ∈ S : x−1 f2 ∈ L2 ,

endowed with the norms   f1 E1 =  ξ −1 f1 L2 ,

f2 E2 = x f2 L2 .

  Let E = E1 + E2 := f = f1 + f2 , fj ∈ Ej , j = 1, 2 endowed with the norm, f E =

inf



f =f1 +f2 fj ∈Ej

f1 E1 + f2 E2 .

a) Let fj ∈ Ej , j = 1, 2. Show the map, d

v → L(f1 ,f2 ) (v) = (2π)− 2

 Rd

ξ −1 f1 (ξ ) ξ  v (−ξ ) dξ  +

Rd

x−1 f1 (x) x v(x) dx

2 Statements of the Problems of Chap. 1

21

belongs to V and, ∃ C > 0 : |L(f1 ,f2 ) (v)| ≤ C( f1 E1 + f2 E2 ) v V ,

∀v ∈ V .

(2.2)

b) Show that if v = ϕ ∈ S then L(f1 ,f2 ) (ϕ) = f, ϕS ×S where f = f1 + f2 . c) Deduce that if f1 + f2 = g1 + g2 then L(f1 ,f2 ) = L(g1 ,g2 ) on V . d) For f ∈ E we can therefore set, Lf (v) = (2π )− 2 d

 Rd

ξ −1 f1 (ξ ) ξ  v (−ξ ) dξ +

 Rd

x−1 f2 (x) x v(x) dx.

Show that Lf ∈ V and that there exists C > 0 such that |Lf (v)| ≤ C f E v V . e) We consider the map  : E → V , f → Lf . Show that  is linear, continuous, and injective.

f) We want to show that  is surjective. Let L ∈ V   . Using question 4 show that 2 there exists u ∈ V such that −u + x u, ϕ S ×S = L(ϕ) for all ϕ ∈ S. We set f1 = −u, f2 = x2 u. Show that fj ∈ Ej , j = 1, 2. We set f = f1 + f2 . Show that L = Lf on V . g) Deduce that  : E → V is linear, bijective, continuous, and −1 is continuous and that this identifies the dual of V with the space E.

Problem 2

This problem proves a result used in the study of the wave equation. It uses the stationary phase method. Let d ≥ 2. For λ > 0 and θ ∈ Sd−1 (the unit sphere in Rd )) we consider the integral,  F (λ, θ ) =

eiλθ,ω dω. Sd−1

The goal of this problem is to show that there exists c0 ∈ C such that, F (λ, θ ) =

c0 λ

d−1 2

(1 + o(1)),

when λ → +∞.

Part 1. Preliminaries 1. Show that for any d × d orthogonal matrix A we have, F (λ, Aθ ) = F (λ, θ ).

22

2 Statements of the Problems of Chap. 1

 Hint: Consider the integral = {y∈Rd :|y| 0 such that, uv H s ≤ C u H s v H s ,

∀u, v ∈ H s (Rd ).

• For s ≥ 0, H s (Rd ) ∩ L∞ (Rd ) is an algebra and there exists C > 0 such that, ∀u, v ∈ H s (Rd ) ∩ L∞ (Rd ).

uv H s ≤ C( u L∞ v H s + u H s v L∞ ),

Such an estimate is called a tame estimate because the right-hand side is linear with respect to the H s norm. • Let s ≥ 0 and let f : C → C be a C ∞ function such that f (0) = 0. Then there exists a nondecreasing function F : R+ → R+ such that, ∀u ∈ H s (Rd ) ∩ L∞ (Rd ).

f (u) H s ≤ F ( u L∞ ) u H s ,

(3.3)

In particular (3.3) holds if u ∈ H s (Rd ) for s > d2 . 3.1.1.4 Sobolev Embeddings • We have E (Rd ) ⊂ ∪ H s (Rd ). s∈R

  k Let C→0 (Rd ) be the space of u ∈ C k (Rd ) such that lim|x|→+∞ ∂xα u(x) = 0 for all |α| ≤ k, endowed with the norm |u|k = |α|≤k supRd |∂xα u|. • Let k ∈ N and s ∈ R be such that s >

d 2

+ k. Then,

k H s (Rd ) ⊂ C→0 (Rd ),

with continuous embedding. In particular if s > with continuous embedding.

d 2

we have H s (Rd ) ⊂ L∞ (Rd )

3.1 Sobolev Spaces

27

• Let s = d2 . For all p ∈ [2, +∞[ there exists C > 0 such that, u Lp ≤ C u H s , • Let 0 ≤ s < d2 . For all p such that

1 2

≥

1 p

u Lp ≤ C u H s ,

∀u ∈ H s (Rd ). ≥

−

1 2

s d

there exists C > 0 such that,

∀u ∈ H s (Rd ).

(3.4)

3.1.1.5 Duality u(ξ ) v (−ξ ) belongs to Let s ∈ R and v ∈ H −s (Rd ). If u ∈ H s (Rd ) the function  L1 (Rd ). Indeed,  u(ξ ) v (−ξ ) = ξ s  u(ξ ) ξ −s  v (−ξ ) and the right-hand side is the product of two L2 (Rd ) functions. On the other hand, by the Cauchy–Schwarz inequality we have,   

Rd

   u(ξ ) v (−ξ ) dξ  ≤ u H s v H −s .

(3.5)

Therefore, if v ∈ H −s (Rd ), the map Lv defined by, u −→ Lv (u) = (2π)

−d



 Rd

 u(ξ ) v (−ξ ) dξ =

Rd

 u(ξ ) (F −1 v)(ξ ) dξ

(3.6)

is a continuous linear form on H s (Rd ) (it thus belongs to (H s (Rd )) ) and Lv (H s ) ≤ (2π)−d v H −s . So we have a map, L : H −s (Rd ) → (H s (Rd )) ,

v −→ Lv .

(3.7)

• The map L, defined by (3.7), is linear, bijective, and bicontinuous. It identifies the dual of H s (Rd ) with H −s (Rd ). We write, for u ∈ H s , v ∈ H −s ,  u, vH s ×H −s = Lv (u) = (2π)−d  u(ξ ) v (−ξ ) dξ. Rd

• If v ∈ L2 (Rd ) and u ∈ L2 (Rd ) we have,   −1  u(ξ ) (F v)(ξ ) dξ = Lv (u) = Rd

u(x) v(x) dx. Rd

28

3 Functional Spaces

3.1.1.6 Compactness • Let K be a compact set in Rd and let s, s ∈ R be such that s > s . Denote

HKs = u ∈ H s (Rd ) : supp u ⊂ K . Then the map HKs → H s (Rd ), u → u, is compact.

3.1.1.7 Traces • For all s > 12 , the map u → u(x , 0) from C0∞ (Rd ) to C ∞ (Rd−1 ) can be uniquely extended to a linear, continuous, surjective map γ , from H s (Rd ) to 1 H s− 2 (Rd−1 ).

3.1.1.8 Equivalent Norm • Let 0 < σ < 1. Set,  [[u]]2σ = u 2L2 (Rd ) +

Rd ×Rd

|u(x) − u(y)|2 dx dy. |x − y|d+2σ

Then u H σ ∼ [[u]]σ in the sense that, ∃ 0 < C1 ≤ C2 : C1 u H σ ≤ [[u]]σ ≤ C2 u H σ ,

∀u ∈ H σ (Rd ).

If s = k + σ where k ∈ N and 0 < σ < 1 we have,  [[∂ α u]]σ ∀u ∈ H s (Rd ). u H s ∼ |α|≤k

s 3.1.2 Local Sobolev Spaces Hloc (Rd )

• For s ∈ R we set,  s (Rd ) = u ∈ S (Rd ) : ϕu ∈ H s (Rd ), Hloc

 ∀ϕ ∈ C0∞ (Rd ) .

Its topology is given by the semi-norms pϕ (u) = ϕu H s .

3.1 Sobolev Spaces

29

3.1.3 Sobolev Spaces on an Open Subset of Rd 3.1.3.1 Definition and First Properties • Let  be an open subset of Rd and k ∈ N. We set,   H k () = u ∈ L2 () : ∂ α u ∈ L2 (), |α| ≤ k . It is endowed with the scalar product and the corresponding norm, 

(u, v)H k =

|α|≤k

α

α

∂ u, ∂ v

⎛

L2 ()

,

u H k

=⎝



|α|≤k

⎞1 2

∂

α

u 2L2 () ⎠

. (3.8)

• Endowed with this scalar product, H k () is a Hilbert space. • If k2 ≥ k1 , H k2 () is continuously embedded in H k1 (). • If s = k ∈ N and  = Rd , the spaces H s (Rd ) and H k () coincide and the norms (3.1) and (3.8) are equivalent.

3.1.3.2 Extension • If  has a smooth boundary, then for every k ∈ N, there exists a linear and continuous map Pk : H k () → H k (Rd ) such that Pk u = u in . It is called the extension operator. Consequently, for k > d2 , H k () is an algebra. • Notice that the restriction operator defined by Ru = u| is continuous from H k (Rd ) to H k () for every k ∈ N.

3.1.3.3 Density • If  = Rd and k ≥ 1, C0∞ () is not dense in H k (). We denote by H0k () the closure of C0∞ () in H k () endowed with the norm (3.8). • C0∞ () is the space of restrictions to  of functions in C0∞ (Rd ). It is dense in H k () for any k ∈ N when  has a smooth boundary.

30

3 Functional Spaces

3.1.3.4 Poincaré Inequality Let  be a bounded open subset of Rd with diameter d. Then we have, u L2 () ≤ 2d

d    ∂u   ∂x j =1

j

   

, L2 ()

∀ u ∈ H01 ().

Actually it is sufficient for  to be bounded in one direction for the inequality to hold.

3.1.3.5 Sobolev Embedding • If k > p + d2 and  has a smooth boundary, then H k () is contained in the space C p () which is the space of restrictions to  of functions in C p (Rd ).

3.1.3.6 Duality • The space C0∞ () being, by definition, dense in H0k () the dual of H0k () can be embedded in the space of distributions. By definition we set H −k () = k

H0 () endowed with the dual norm that is, T −k =

sup

|T (ϕ)|.

ϕ∈H0k () ϕ H k =1

3.1.3.7 Compactness • Let  ⊂ Rd be bounded with a smooth boundary. Then for any k > k the

embedding from H k () to H k () is compact.

3.1.3.8 Traces • The map C0∞ ()

∞

→ [C (∂)] , k

 k−1  ∂u  ∂  u → u ∂ ,  , . . . , ∂ν ∂ ∂ν

  u

! ∂

3.2 The Hölder Spaces

31

can be extended as a linear, continuous, and surjective map γ = (γ0 , . . . , γk−1 ) k−1

1

from H k () to  H k−j − 2 (∂). Here ν is the exterior normal to ∂ and j =0

∂ ∂ν

is

the normal derivative.   • Then H0k () = u ∈ H k () : γ u = 0 .

3.1.4 Sobolev Spaces on the Torus Consider the d-dimensional torus Td = Rd /(2πZ)d . A function f : Td → C is a function f : Rd → C such that f (x + 2πej ) = f (x) for any 1 ≤ j ≤ d, where (e1 , . . . , ed ) is the canonical basis of Rd . If f ∈ L1 (Td ) we define its Fourier coefficient by, f(k) =



e−ik·x f (x) dx, (0,2π)d

k ∈ Zd .

Then for s ≥ 0, the Sobolev space H s (Td ) is the space of functions f such that, f 2H s (Td ) :=



(1 + |k|2 )s |f(k)|2 < +∞.

k∈Zd

For s > 0 we define H −s (Td ) as the dual of H s (Td ).

3.2 The Hölder Spaces 3.2.1 Hölder Spaces of Integer Order 3.2.1.1 Definition • Let m ∈ N. We set,   W m,∞ (Rd ) = u ∈ L∞ (Rd ) : ∂ α u ∈ L∞ (Rd ), ∀|α| ≤ m , endowed with the norm, u W m,∞ =

 |α|≤m

∂ α u L∞ .

(3.9)

32

3 Functional Spaces

3.2.1.2 Properties • Endowed with the norm (3.9), W m,∞ (Rd ) is a Banach space. • The space W m,∞ (Rd ) is an algebra and there exists C > 0 such that, uv W m,∞ ≤ C u W m,∞ v W m,∞ ,

∀u, v ∈ W m,∞ (Rd ).

• Notice that W 1,∞ (Rd ) is the space of bounded Lipschitz functions.

3.2.2 Hölder Spaces of Fractional Order Let us introduce a notation. For σ ∈]0, 1[ we set, [u]σ = sup x,y∈Rd x=y

|u(x) − u(y)| . |x − y|σ

3.2.2.1 Definition • Let m ∈ N, 0 < σ < 1 and ρ = m + σ. We set,    [∂ α u]σ < +∞ , W ρ,∞ (Rd ) = u : ∂ α u ∈ L∞ (Rd ), ∀|α| ≤ m and |α|=m

endowed with the norm, u W ρ,∞ (Rd ) =

 |α|≤m

∂ α u L∞ +



[∂ α u]σ .

|α|=m

Notice that these spaces are also denoted in the literature by C m,σ (Rd ).

3.2.2.2 Properties • Endowed with the norm (3.10), W ρ,∞ (Rd ) is a Banach space.

(3.10)

3.3 Characterization of Sobolev and Hölder Spaces in Dyadic Rings

33

3.3 Characterization of Sobolev and Hölder Spaces in Dyadic Rings For j ∈ N we set,   1 Cj = ξ ∈ Rd : 2j ≤ |ξ | ≤ 2 · 2j . 2 Let ψ ∈ C0∞ (Rd ) be such that,   supp ψ ⊂ ξ ∈ Rd : |ξ | ≤ 1 ,

ψ(ξ ) = 1 if

|ξ | ≤

1 , 2

and set, ξ ϕ(ξ ) = ψ( ) − ψ(ξ ). 2

(3.11)

Then, ψ(ξ ) +

N−1 

ϕ(2−j ξ ) = ψ(2−N ξ ),

∀ξ ∈ Rd , ∀N ≥ 1.

j =0

For u ∈ S (Rd ) we set, −1 u = F −1 (ψ(ξ ) u) ,

  u j u = F −1 ϕ(2−j ξ )

for j ≥ 0,

and by convention we set j u = 0 if j ≤ −2. Then in S (Rd ) we have, u=

+∞ 

j u.

j =−1

It is called the Littlewood–Paley decomposition of u. We will be using in the sequel other rings of the form,   j = ξ ∈ Rd : a2j ≤ |ξ | ≤ b 2j , where 0 < a < b. C

(3.12)

34

3 Functional Spaces

3.3.1 Characterization of Sobolev Spaces • Let s ∈ R. If u ∈ H s (Rd ) we have, ⎛ j u L2 (Rd ) ≤ cj 2

−j s

, j ≥ −1,

⎝

where

+∞ 

⎞1 2

cj2 ⎠

≤ C u H s (Rd ) .

j =−1

j and, • For j ∈ Z, j ≥ −1 let uj ∈ L2 (Rd ) be such that supp  uj ⊂ C ⎛ uj L2 (Rd ) ≤ cj 2−j s ,

where

⎝

+∞ 

⎞1 2

cj2 ⎠ < +∞.

j =−1

Set u =

+∞

j =−1 uj .

Then, ⎛ and u H s (Rd ) ≤ C ⎝

u ∈ H s (Rd )

+∞ 

j =−1

⎞1 2

22j s uj 2L2 (Rd ) ⎠ .

• Notice that if s > 0 we have the same above statements with the ring Cj replaced by a ball B(0, C2j ).

3.3.2 Characterization of Hölder Spaces • Let m ∈ N, 0 < σ < 1 and ρ = m + σ. If u ∈ W ρ,∞ (Rd ) we have, j u L∞ (Rd ) ≤ C2−jρ u W ρ,∞ (Rd ) . j and, • For j ∈ Z, j ≥ −1 let uj ∈ L∞ (Rd ) be such that supp  uj ⊂ C uj L∞ (Rd ) ≤ C2−jρ . Set u =

+∞

j =−1 uj .

Then,

u ∈ W ρ,∞ (Rd )

and

  u W ρ,∞ (Rd ) ≤ C sup 2jρ uj L∞ (Rd ) . j ≥−1

3.4 Paraproducts

35

3.3.3 The Zygmund Spaces • For ρ ∈ R we set, #   jρ = u ∈ S (R ) : sup 2 j u L∞ (Rd ) < +∞ , "

ρ C (Rd )

d

j ≥−1



endowed with the norm u C ρ = supj ≥−1 2jρ j u L∞ (Rd ) . If ρ = m + σ, m ∈ N, 0 < σ < 1 these are the spaces W ρ,∞ (Rd ) defined in the previous paragraph. If ρ = m ∈ N we have W m,∞ (Rd ) ⊂ C m (Rd ), with strict inclusion.

3.4 Paraproducts We keep the notations introduced in the beginning of Sect. 3.3 that is, −1 u = ψ(D)u,

j u = ϕ(2−j D)u,

j ≥ 0,

and we set, S (u) =

−1 

k u = ψ(2− D)u,

(3.13)

k=−1

by (3.12). Notice that S (u) = 0 if  ≤ −1. If f ∈ S (Rd ) we call “spect(f )” (spect. for spectrum) the support of its Fourier transform. According to our choice of the support of ϕ, ψ for every a, u ∈ S (Rd ) we have,   spect(S (a)) ⊂ ξ ∈ Rd : |ξ | ≤ 2 , (3.14)   d j −1 j +1 . ≤ |ξ | ≤ 2 spect(j u) ⊂ ξ ∈ R : 2 In particular if  ≤ j − 2 we have, spect(S (a)) ∩ spect(j (u)) = ∅. • Let a, u ∈ S (Rd ). The paraproduct of a and u is denoted by Ta u and is formally defined by the formula, Ta u =

 j ≥−1

Sj −2 (a)j u.

(3.15)

36

3 Functional Spaces

We shall give below conditions on a and u under which the above series is convergent. In (3.15), according to (3.14), the frequencies of a are smaller than those of u. j . Notice also that the spectrum of Sj −2 (a)j u is contained in a ring C Now if a ∈ L∞ (Rd ) and u ∈ H s (Rd ) the product au, when it is defined, does not belong in general to H s (Rd ). However, we have the following result. • Let s ∈ R and a ∈ L∞ (Rd ). There exists C > 0 depending on d, s such that, Ta u H s (Rd ) ≤ C a L∞ (Rd ) u H s (Rd ) ,

(3.16)

Ta u C∗s (Rd ) ≤ C a L∞ (Rd ) u C∗s (Rd ) .

(3.17)

• Let s ∈ R and a ∈ H β (Rd ). Then there exists C > 0 such that, d , Ta u H s (Rd ) ≤ C a H β (Rd ) u H s (Rd ) , (3.18) 2 d (ii) if β = , Ta u H s (Rd ) ≤ C a d d u H s+ε (Rd ) , ∀ε > 0, H 2 (R ) 2 (3.19)

(i) if β >

(iii) if β
0 and consider a ∈ H α (Rd ), u ∈ d H β (Rd ). Then R(a, u) ∈ H α+β− 2 (Rd ) and there exists C > 0 depending on d, α, β such that, R(a, u)

H

α+β− d 2

(Rd )

≤ C a H α (Rd ) u H β (Rd ) .

Notice that if ε = α − d2 > 0, then the reminder R(a, u) is ε smoother than u. ρ We have the same result if a ∈ C∗ (Rd ) and u ∈ H β (Rd ) with ρ + β > 0. β+ρ The reminder R(a, u) belongs to H (Rd ) and we have, R(a, u) H β+ρ (Rd ) ≤ C a C∗ρ (Rd ) u H β (Rd ) .

3.5 Some Words on Interpolation

• Let α >

d 2

37

and F ∈ C ∞ (C, C). For every a ∈ H α (Rd ) we have d

F (a) − F (0) − TF (a)a ∈ H 2α− 2 (Rd ) and there exists C > 0 depending only on α, d such that, F (a) − F (0) − TF (a) a

H

2α− d 2

(Rd )

≤ C a H α (Rd ) ,

where F denotes the differential of F . Notice that 2α − d2 > α, so again the reminder is smoother than a. • Let α > d2 and ρ = α − d2 . If a ∈ H α (Rd ) and b ∈ H α (Rd ), then for every s ∈ R, Ta ◦ Tb − Tab is continuous from H s (Rd ) to H s+ρ (Rd ), (Ta )∗ − Ta is continuous from H s (Rd ) to H s+ρ (Rd ). Here (Ta )∗ denotes the adjoint of Ta in L2 (Rd ) and a the complex conjugate of a. • We shall introduce later in Chap. 5 a wider class of operators and prove more general results.

3.5 Some Words on Interpolation We work in the category C of normed paces and we shall denote by L(E, F ) where E, F ∈ C, the set of linear and continuous maps from E to F. Let A0 , A1 ∈ C. • Compatible couples. We say that (A0 , A1 ) is a compatible couple if there exists a topological vector space U such that  A0 and A1 are subspaces of U. Then we can consider the sum A0 + A1 = a ∈ U : a = a0 + a1 , aj ∈ Aj and the intersection A0 ∩ A1 . • Intermediate space. It is a space A such that A0 ∩ A1 ⊂ A ⊂ A0 + A1 . • Interpolation spaces. Let A = (A0 , A1 ), B = (B0 , B1 ) be two compatible couples. We say that a couple (A, B) is of interpolation if A, B are intermediate spaces and if, T ∈ L(A0 , B0 ) and T ∈ L(A1 , B1 ) ⇒ T ∈ L(A, B). Here are some examples. • Let (X, T , μ) be a measured space where μ is a σ -finite positive measure. Let 1 ≤ p0 , p1 < +∞,

1 < q0 , q1 ≤ +∞.

38

3 Functional Spaces

Set, A0 = Lp0 (X, dμ),

B0 = Lq0 (X, dμ),

A1 = Lp1 (X, dμ),

B1 = Lq1 (X, dμ),

1−θ θ 1 = + , p p0 p1

1−θ 1 θ = + , q q0 q1

0 < θ < 1.

Then (A = Lp (X, dμ), B = Lq (X, dμ)) is of interpolation. Moreover , θ T L(A,B) ≤ C T 1−θ L(A0 ,B0 ) T L(A1 ,B1 ) .

This is the Riesz–Thorin theorem. A classical application of this theorem is the continuity of the Fourier transform F on the Lp (Rd ) spaces. Indeed we know that F is continuous from L1 (Rd ) to L∞ (Rd ) and from L2 (Rd ) to L2 (Rd ). Taking p0 = 1, q0 = +∞, p1 = q1 = 2 we

see that F is continuous from Lp (Rd ) to Lp (Rd ) where 1 < p < 2 and p1 + p1 = 1. • Let (X, T , μ) be a measured space where μ is a σ -finite positive measure. Let 1 ≤ p ≤ +∞. Let w0 , w1 , w 0 , w 1 be positive μ-measurable functions and, A0 = Lp (X, w0 dμ),

B0 = Lp (X, w 0 dμ),

A1 = Lp (X, w1 dμ),

B1 = Lp (X, w 1 dμ).

Set for 0 < θ < 1, w(x) = w0 (x)1−θ w1 (x)θ ,

w (x) = w 0 (x)1−θ w 1 (x)θ .

Then the couple  dμ)) (A, B) = (Lp (X, w dμ), Lp (X, w is of interpolation. Moreover, θ T L(A,B) ≤ C T 1−θ L(A0 ,B0 ) T L(A1 ,B1 ) .

• Let s0 , σ0 , s1 , σ1 be real numbers. Set A0 = H s0 (Rd ),

B0 = H σ0 (Rd ),

s = (1 − θ )s0 + θ s1 ,

A1 = H s1 (Rd ),

B0 = H σ1 (Rd ),

σ = (1 − θ )σ0 + θ σ1 ,

0 < θ < 1.

3.6 The Hardy–Littlewood–Sobolev Inequality

39

Then the couple, (A = H s (Rd ),

B = H σ (Rd ))

is of interpolation. Moreover, θ T L(A,B) ≤ C T 1−θ L(A0 ,B0 ) T L(A1 ,B1 ) .

Notice that the result about the Sobolev spaces is a consequence of the second result about the Lp spaces. Indeed H s (Rd ) can be described as,   H s (Rd ) = u ∈ S (Rd ) :  u ∈ L2 (Rd , ξ 2s dξ ) ,

1

ξ  = (1 + |ξ |2 ) 2 .

3.6 The Hardy–Littlewood–Sobolev Inequality • Let 1 < p < +∞ and 0 < α < d. This inequality expresses the fact that the convolution by the function |x|−α maps continuously Lp (Rd ) to Lr (Rd ) for 1 α 1 = + − 1. r p d Namely, it states that, |x|−α f Lr (Rd ) ≤ C(α, p, d) f Lp (Rd ) ,

∀f ∈ Lp (Rd ).

(3.21)

• The inequality (3.21) is a generalization of the Young inequality which says that for p, q ≥ 1, 1 1 1 = + −1 r p q

⇒

Lp (Rd ) Lq (Rd ) ⊂ Lr (Rd ).

The inequality (3.21) is not a consequence of the Young inequality because taking q1 = αd or q = αd , the function |x|−α does not belong to Lq (Rd ). However, q the latter belongs to the weak Lq space which is denoted by Lw (Rd ). q q d d • The space Lw (R ): let μ be the Lebesgue measure on R . Then f ∈ Lw (Rd ) if f is measurable and, S(f ) =

sup 0 λ = μ x ∈ Rd : |x| < λ− α q

On the other hand, Lq (Rd ) ⊂ Lw (Rd ) since for 0 < λ < +∞ we have,     |f (x)|q dx ≥ |f (x)|q dx ≥ λq μ x ∈ Rd : |f (x)| > λ . Rd

{x:|f (x)|>λ}

q

• The quantity S(f ) defined in (3.22) is not a norm on Lw (it does not satisfy the triangle inequality). However, the quantity, 1

f Lqw (Rd ) = sup μ(A) q A

−1

 |f (x)| dx,

(3.23)

A

where A is a measurable set of positive and finite measure, is equivalent to S(f ) q and it is a norm on Lw (Rd ) (see Problem 14). • The generalization of the inequality (3.21) can be stated as follows. Let 1 < p, q < +∞ and r be such that, 1 1 1 = + − 1. r p q There exists C = C(p, q, d) > 0 such that, f g Lr (Rd ) ≤ C f Lp (Rd ) g Lqw (Rd ) . Comments For a description of the main functional spaces, the Littlewood– Paley theory, and the paraproducts we refer to H. Bahouri, J.Y. Chemin, and R. Danchin [5] (Chapters 1 and 2) and M. Taylor [31] (Chapter 13). A comprehensive introduction to interpolation theory can be found in J. Bergh and J. Löfström [6]. Different proofs of the Hardy–Littlewood–Sobolev inequality can be found in E. Stein [28] (Chapter VIII, section 4.2) and E. Lieb and M. Loss [19] (Chapter 4).  

Chapter 4

Statements of the Problems of Chap. 3

This chapter contains problems about the main functional spaces.

Problem 4 Let I = (a, b) ⊂ R, s ∈ R, and d ≥ 1. We set H s = H s (Rd ) and we consider a family (u(t, ·))t ∈I ⊂ H s such that, (H1) the map t → u(t, ·) H s is continuous from I to R, (H2) there exists δ > 0 such that t → u(t, ·) is continuous from I to H s−δ . The goal of this problem is to prove that the map t → u(t, ·) is continuous from I to H s . Denote by ·, · the duality bracket between H s and H −s . Let t0 ∈ I and consider a sequence (tn ) ⊂ I with tn → t0 . 1. Let An := u(tn , ·) − u(t0 , ·), ϕ where ϕ ∈ H −s . Show that limn→+∞ An = 0. 2. Conclude.   Hint: Note that if u, v ∈ H s (Rd ) then (u, v)H s = (2π)d u, 2s v H s ×H −s 1

where  = (I − ) 2 .

Problem 5 An Interpolation Result

If I ⊂ R is an interval and s ∈ R we set,   1 1 W s (I ) = f ∈ L2z (I, H s+ 2 (Rd )) : ∂z f ∈ L2z (I, H s− 2 (Rd )) , © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 T. Alazard, C. Zuily, Tools and Problems in Partial Differential Equations, Universitext, https://doi.org/10.1007/978-3-030-50284-3_4

41

42

4 Statements of the Problems of Chap. 3

endowed with its natural norm, f W s (I ) = f

L2z (I,H

s+ 21

(Rd ))

+ ∂z f

L2z (I,H

s− 1 2

(Rd ))

.

Part 1 We assume I = R. We will admit that C0∞ (R × Rd ) is dense in W s (R). 1. Let v ∈ C0∞ (R × Rd ). Prove that the map z → v(z, ·) 2H s (Rd ) is differentiable on R and that, d v(z, ·) 2H s (Rd ) = 2Re (v(z, ·), ∂z v(z, ·))H s (Rd ) . dz 2. Deduce that there exists C, C > 0 such that, 1

sup v(z, ·) H s (Rd ) ≤ C v 2

s+ 1 L2z (R,H 2 (Rd ))

z∈R

1

∂z v 2

L2z (R,H

s− 12

(Rd ))

≤ C v W s (R) .

(4.1)

3. Show that if u ∈ W s (R) then u ∈ C 0 (R, H s (Rd )) ∩ L∞ (R, H s (Rd )) and that an inequality analogue to (4.1) still holds for u. Hint: Use question 2 and the density of C0∞ (R × Rd ) in W s (R). Part 2 u for We assume I = (0, +∞). Show that there exists  u ∈ W s (R) such that u =  z ∈ (0, +∞) and  u W s (R) ∼ u W s ((0,+∞)). 1. Prove that if u ∈ W s ((0, +∞)) then u ∈ (C 0 ∩ L∞ )([0, +∞), H s (Rd )) and that there exists C > 0 (independent of u) such that, sup

u(z, ·) H s ≤ C u W s ((0,+∞)).

z∈(0,+∞)

2. Deduce that this is still true if I = (a, +∞) or (−∞, b), a, b ∈ R. Part 3 We assume I = (−1, 0). Let χj ∈ C ∞ (R), j = 1, 2 be such that, supp χ1 ⊂

$ 1 − ∞, − , 4

% supp χ2 ⊂

3 − , +∞ , 4

χ1 (z) + χ2 (z) = 1 on (−1, 0).

Let u ∈ W (I ). We write u = u1 + u2 where uj (z, ·) = χj (z)u(z, ·), j = 1, 2. 1. Show that u1 ∈ W s ((−1, +∞)) and u2 ∈ W s ((−∞, 0)). 2. Show that u ∈ C 0 ([−1, 0], H s (Rd )) and there exists C > 0 such that, sup u(z, ·) H s (Rd ) ≤ C u W s ((−1,0)).

z∈[−1,0]

4 Statements of the Problems of Chap. 3

43

3. Deduce the same result if I = (a, b), a, b ∈ R.

Problem 6 Spaces of Analytic Functions Let σ > 0 and s ∈ R. We set Sσ = {z = x + iy ∈ C : |y| < σ } . We consider the space   Gσ,s = u ∈ S (R) : eσ |ξ | ξ s  u ∈ L2 (R) ,

1

where ξ  = (1 + |ξ |2 ) 2 ,

endowed with the norm: u L2 (R) . u Gσ,s := eσ |ξ | ξ s  Part 1 1. Prove that for u ∈ Gσ,s and z ∈ Sσ the function ξ → eiz·ξ  u(ξ ) belongs to L1 (R). We set for z ∈ Sσ  U (z) = (2π)−1 eiz·ξ  u(ξ ) dξ. R

2. Show that U is holomorphic in Sσ and that U |y=0 = u. 3. Show that there exists C > 0 such that, sup U (x + iy) Hxs (R) ≤ C u Gσ,s .

|y| 0 fixed we set, V (z) = U (z − t), ϕλ (t) = (U (· + iy) ϕλ ) (x). We want to show that V is a holomorphic function in Sσ . We have ϕλ ∈ S(R) ⊂ H −s (R). Let (χj ) ⊂ C0∞ (R) be such that (χj ) → ϕλ in H −s (R).   2. a) Set Vj (z) = U (z − t), χj (t) . Show that Vj is holomorphic in Sσ . Hint: Let K ⊂ Sσ be compact. Use the fact that Vj (z) = R U (z − t)χj (t) dt for z ∈ K. b) Show that for z ∈ Sσ , |Vj (z) − V (z)| ≤ M0 χj − ϕλ H −s (R) and conclude. 3. Prove that V |y=0 = F |y=0 and deduce that V (z) = F (z) for all z in Sσ . 4. Deduce that Fx (U (· + iy))(ξ ) = e−y·ξ  u(ξ ) and that,  sup

|y| 12 . 1. Prove that there exists C > 0 such that, au Gσ,0 ≤ C a Gσ,s0 u Gσ,0 ,

∀a ∈ Gσ,s0 ,

∀u ∈ Gσ,0 .

2. a) Show that for all s ≥ 0 there exists C > 0 such that, au Gσ,s ≤ C a Gσ,s0 u Gσ,s + a Gσ,s u Gσ,s0 , b) Deduce that for s >

1 2

∀a, u ∈ Gσ,s0 ∩ Gσ,s .

the space Gσ,s is an algebra.

Comments These spaces appear to be useful if one wants to deal with analytic solutions in a microlocal setting.

Problem 7 Uniformly Local Sobolev Spaces

The goal of this problem is to study a family of spaces introduced by Tosio Kato.

4 Statements of the Problems of Chap. 3

45

Part 1 1. Show that there exists a positive function χ ∈ C ∞ (Rd ) such that, (i)

supp χ ⊂ [−1, 1]d ,

(ii)



χq (x) = 1,

∀x ∈ Rd ,

(4.4)

q∈Zd

where χq (x) = χ(x − q). s (Rd ) is the space of elements u ∈ H s (Rd ) such that, For s ∈ R, Hul loc u H s (Rd ) := sup χq u H s (Rd ) < +∞. ul

(4.5)

q∈Zd

s (Rd ) is a Banach space. 2. Show that endowed with the norm (4.5) Hul 3. Give simple examples of functions belonging to L2ul (Rd ) but not to L2 (Rd ).

Part 2 s does not depend on the The goal of this part is to prove that the above definition Hul function χ introduced in (4.4). 1. a) Let ρ > 0. Show that for any α ∈ Nd and all x ∈ Rd we have ∂xα (1 + |x|2)ρ = (1 + |x|2 )ρ−|α| Pα (x) where Pα is a polynomial of degree |α|. b) Deduce that for any α ∈ Nd there exists Cα > 0 such that, |∂xα (1 + |x|2 )ρ | ≤ Cα (1 + |x|2)ρ− 2 |α| , 1

∀x ∈ Rd .

c) Show, by induction on |α|, that for all α ∈ Nd there exists Cα > 0 such that, |∂xα (1 + |x|2 )−ρ | ≤ Cα (1 + |x|2)−ρ− 2 |α| , 1

∀x ∈ Rd .

Hint: Use the fact that (1 + |x|2)ρ (1 + |x|2)−ρ = 1. d) Deduce that for any α ∈ Nd there exists Cα > 0 such that for all a ∈ Rd and all x ∈ Rd we have, |∂xα (1 + |x − a|2 )−ρ | ≤ Cα (1 + |x − a|2)−ρ− 2 |α| . 1

k (x) = χ (x − k). For k, q ∈ Zd and N ∈ N we consider 2. Let χ  ∈ C0∞ (Rd ) and χ the function, k,q (x) =

k − qN x − q

N

χ k ,

1

where y = (1 + |y|2 ) 2 .

Show that for any M ∈ N there exists C = CN,M, χ > 0 independent on k, q such that, k,q H M (Rd ) ≤ C.

46

4 Statements of the Problems of Chap. 3

3. Let s ∈ R and M ∈ N, M ≥ |s|. Using the fact that, 

ξ |s| |θ (ξ )| dξ u H s (Rd ) θ u H s (Rd ) ≤ C and writing,   χ k χq u(x) = k − q−N k,q (x) x − qN ψq (x) χq u(x), where ψ ∈ C0∞ (Rd ) is equal to one on the support of χ, ψq (x) = ψ(x − q) and N > d + 1 show that,   k − q−N sup χq u H s (Rd ) .  χk u H s (Rd ) ≤  χk χq u H s (Rd ) ≤ C q∈Zd

q∈Zd

q∈Zd

4. Conclude. Comments The uniformly local Sobolev spaces have been introduced by T. Kato in [17]. They are useful if one wants to work with solutions having a local Sobolev regularity but do not decrease at infinity. For instance, it contains in a unified setting the Sobolev spaces H s (Rd ) and the periodic Sobolev spaces H s (Td ). It also contains Hölder spaces, as it will be shown in the next problem.

s and C s Problem 8 Link Between Hul ∗

The goal of this problem is to explore the link between the uniformly local Sobolev spaces and the Zygmund spaces. m 1. Prove that for m ∈ N we have, W m,∞ (Rd ) ⊂ Hul (Rd ). m m+ε 2. Let m ∈ N. Prove that for every ε > 0 we have C∗ (Rd ) ⊂ Hul (Rd ).

Let s = m + σ, m ∈ N, 0 < σ < 1. The goal of the following questions is to s show that for δ > 0, C∗s+δ (Rd ) ⊂ Hul (Rd ). 3. a) Let s = m + σ, m ∈ N, 0 < σ < 1 and let ε > 0 be so small that 0 < σ + ε < 1. Then s + ε ∈ / N and C∗s+ε (Rd ) = W s+ε,∞ (Rd ). Prove that,     σ s W σ +ε,∞ (Rd ) ⊂ Hul (Rd ) ⇒ W s+ε,∞ (Rd ) ⊂ Hul (Rd ) . We recall that,  ⎧ ⎪ 2 2 ⎪ χq u H σ (Rd ) ≈ χq u L2 (Rd ) + fq (x, y) dx dy = (1) + (2), ⎪ ⎨ ⎪ |χq (x)u(x) − χq (y)u(y)|2 ⎪ ⎪ ⎩ where, fq (x, y) = . |x − y|d+2σ

4 Statements of the Problems of Chap. 3

47

b) Prove that (1) ≤ C1 u 2L∞ (Rd ) . We write   (2) = fq (x, y) dx dy + |x−y|>1

|x−y| 0. Let u ∈ C0∞ (Rd ) and set uε = χε u. Prove that (uε ) converges to u in H 1 (Rd ) when ε → 0. 2. Prove that C0∞ (Rd \ {0}) is dense in H 1 (Rd ).

48

4 Statements of the Problems of Chap. 3

Part 2 We assume in what follows that d ≥ 2. α Let k ∈ N and P (ξ ) = ξ ∈ Rd , cα ∈ C. We set Pj (ξ ) = |α|≤k cα ξ ,  α and we suppose that P is exactly of order k which, Pk being |α|=j cα ξ homogeneous, reads, (∗)

∃ ω0 ∈ Sd−1 : |Pk (ω0 )| := c0 > 0.

1. Prove that there exists Vω0 a neighborhood of ω0 in Sd−1 and r0 > 0 such that, |P (rω)| ≥

c0 k r , 4

∀ω ∈ Vω0 ,

∀r ≥ r0 .

2. Let U ∈ H −1 (Rd ) be such that supp U ⊂ {0} . Show that U = 0. Let E ⊥ be the orthogonal of E in H 1 (Rd ) and  be the Laplacian in Rd . 3. Let u ∈ E ⊥ . Prove that supp (−u + u) ⊂ {0} . 4. Deduce that −u + u = 0 in S (Rd ) and then that u = 0. 5. Prove that C0∞ (Rd \ {0}) is dense in H 1 (Rd ). 6. Show that, for d ≥ 2, the space H 1 (Rd ) cannot be continuously embedded in the space C 0 (B(0, ε)) where B(0, ε) is the closed ball in Rd with center 0 and radius ε > 0. Part 3 Let d ≥ 2 and let  be an open subset of Rd with smooth boundary such that 0 ∈ . We shall denote by C0∞ ( \ {0}) the set of restrictions to  of functions in C0∞ (Rd \ {0}). 1. Prove that C0∞ ( \ {0}) is dense in H 1 (). Hint: Use the extension and the restriction operators P1 and R (see Sect. 3.1.3 in Chap. 2). Remark In all this problem we can of course replace the point 0 by any point x0 ∈ Rd (or x0 ∈ ).

Problem 10 A Poincaré Inequality

The second part of this problem uses the result proved in Part 3 of Problem 9. In this problem we shall denote by |A| the Lebesgue measure of a measurable set A ⊂ Rd .

4 Statements of the Problems of Chap. 3

49

Part 1 1. Let  be a bounded open subset of Rd . Show that for every α > 0 there exists Cα > 0 such that for every u ∈ H 1 () such that, ( )

| {x ∈  : u(x) = 0} | ≥ α

we have, 

 |u(x)| dx ≤ Cα

|∇u(x)|2 dx.

2



(4.6)



Hint: Argue by contradiction and use the compactness of the map u → u from H 1 () to L2 (). Part 2 The purpose of this part is to see what happens if we relax the condition ( ). 1. Prove by a simple example that the condition (

)

| {x ∈  : u(x) = 0} | = 0

is not sufficient to have (4.6). 2. We assume d = 1,  = (a, b). Show that one can find C > 0 depending only on a, b such that for every u ∈ H 1 ((a, b)) satisfying, ( )

{x ∈ (a, b) : u(x) = 0} = ∅,

we have, 

b



b

|u(x)|2 dx ≤ C

a

|u (x)|2 dx.

a

3. Let d ≥ 2 and let  be a bounded open subset of Rd with smooth boundary. We assume for simplicity that 0 ∈ . Show that the condition ( )

| {x ∈  : u(x) = 0} | > 0

is not sufficient to have (4.6) with a constant C > 0 independent of u. Hint: Use Part 3 of Problem 9.

50

4 Statements of the Problems of Chap. 3

Problem 11 A Composition Rule in Sobolev Space

Part 1 Let p ≥ 1, m ≥ 1 be two integers and let G : Rp → C be a C ∞ function such that, |∂ α G(y)| ≤ M|y|m−|α| ,

∀y ∈ Rp ,

∀|α| ≤ m.

(4.7)

For u = (u1 , . . . , up ) : Rd → Rp and any norm · we shall set u =  p j =1 uj . 1. a) Prove that if m = 1 and d ≥ 1 we have, G(u) H 1 ≤ K u H 1 ,

∀u ∈ (H 1 (Rd ))p .

b) Prove that if m = 2 and 1 ≤ d ≤ 3 we have, G(u) H 2 ≤ K u 2H 2

∀u ∈ (H 2 (Rd ))p .

Hint: Use the Sobolev embeddings. c) Prove, by induction on m ≥ 2, that if 1 ≤ d ≤ 3 we have, G(u) H m ≤ K u m Hm

∀u ∈ (H m (Rd ))p

Hint: Use the fact that ∂xk (G(u)) = (∇y G)(u) · ∂xk u and the induction. Part 2 Let p ≥ 1, m ≥ 1 be two integers and let F : Rp → C be a function in C ∞ (Rp \{0}) homogeneous of degree m, that is: F (λy) = λm F (y),

∀λ ≥ 0,

∀y ∈ Rp .

(4.8)

The goal of this part is to prove that there exists K > 0 such that, for all u = (u1 , . . . , up ) ∈ H m (Rd )p , F (u) ∈ H m (Rd ) and, F (u) H m ≤ K u m Hm , where we assume d ≥ 1 if m = 1 and 1 ≤ d ≤ 3 if m ≥ 2. 1. Prove that for all m ≥ 1 we have, |∂ α F (y)| ≤ ∂ α F L∞ (Sp−1 ) |y|m−|α| ,

∀y ∈ Rp \ {0} ,

where Sp−1 is the unit ball in Rp . We set M =



|α|≤m ∂

|α| ≤ m,

αF L∞ (Sp−1 ) .

4 Statements of the Problems of Chap. 3

51

2. Let χ ∈ C0∞ (Rp ) be such that χ(y) = 1 for |y| ≤ 1, χ(y) = 0 for |y| ≥ 2 and 0 ≤ χ ≤ 1. For  ∈ N,  ≥ 1, define F ∈ C ∞ (Rp ) by, F (y) = (1 − χ (y)) F (y). Prove that for α ∈ Np , |α| ≤ m there exists C = C(χ, α) > 0 such that, |∂ α F (y)| ≤ CM|y|m−|α| ,

∀y ∈ Rp , ∀ ∈ N,  ≥ 1.

3. Prove that there exists K > 0 such that if m = 1, d ≥ 1 or m ≥ 2, 1 ≤ d ≤ 3, F (u) H m ≤ K u m Hm

∀ ≥ 1, ∀u ∈ (H m (Rd ))p .

∞ d m d p 1 d 4. a) Let m ≥ 1,  ϕ ∈ C0 (R ) and u ∈ (H (R )) . Prove that F (u)ϕ ∈ L (R ).   b) Let I = Rd F (u(x))ϕ(x) dx . Prove that there exists K > 0 such that, for all u ∈ H m (Rd )p we have I ≤ K u m H m ϕ H −m . Deduce that F (u) ∈ H m (Rd ) and F (u) H m ≤ K u m Hm .

Problem 12 Let H = 1(0,+∞) be the Heaviside function. Show that for any 0 ≤ s ≤ 1 there exists C > 0 such that for all u ∈ H 1 (R) satisfying u(0) = 0 we have, H u H s (R) ≤ C u H s (R) .

Problem 13 Hardy Inequality

The goal of this part is to show that in dimension d ≥ 3 we have,  Rd

|u(x)|2 dx |x|2

12

d   2 ∂u ≤ | (x)|2 dx d d −2 R ∂xi

! 12 ,

(4.9)

i=1

for all u ∈ C0∞ (Rd ). 1. Let ω ∈ Sd−1 . Prove that,  +∞  +∞ |u(rω)|2 d−1 2 Re r dr = − u(rω) (ω · ∇x u) (rω)r d−2 dr, 2 r d − 2 0 0 (4.10) where ∇x u = (∂xi u)i=1,...,d denotes the gradient of u. 1 d d−2 Hint: Use the fact that r d−3 = d−2 . dr r

52

4 Statements of the Problems of Chap. 3

2. Using the Cauchy–Schwarz inequality and the polar coordinates (r, ω) ∈ (0, +∞) × Sd−1 in Rd deduce the inequality (4.9).

Problem 14 Norm in the Weak Lebesgue Space q

Let q > 1 and μ be the Lebesgue measure on Rd . We denote by Lw (Rd ) the space of Lebesgue measurable functions f : Rd → C such that, S(f ) :=

  1 q λ μ x ∈ Rd : |f (x)| > λ < +∞.

sup 0 λ , μ(A) dλ.

0

b) Deduce that there exists C(q) > 0 such that, f Lqw (Rd ) ≤ C(q) S(f ). q

c) Show that f Lqw (Rd ) is a norm on Lw (Rd ). Problem 15 Functions of Bounded Mean Oscillations We recall that L1loc (Rd ) denotes the space of functions f : Rd → C which are integrable on each compact subset of Rd .

4 Statements of the Problems of Chap. 3

53

  In what follows we shall denote by B a ball in Rd , B = x ∈ Rd : |x − x0 | < r where r > 0, x0 ∈ Rd and by |B| its Lebesgue measure. Moreover, for f ∈ L1loc (Rd ) we set, fB =

1 |B|

 f (x)dx. B

We shall denote by BMO the space of functions f such that, ⎧ 1 d ⎪ ⎨ (i) f ∈ Lloc (R ),  1 ⎪ ⎩ (ii) ∃A > 0 : |f (x) − fB |dx ≤ A, |B| B

for every ball B ⊂ Rd .

1. Show that L∞ (Rd ) ⊂ BMO. 2. Let f ∈ BMO and δ > 0. We denote by fδ the function x → f (δx). Prove that fδ ∈ BMO for all δ > 0.  of functions f : Rd → C such that, 3. We consider the space E ⎧ 1 d ⎪ ⎨ (iii) f ∈ Lloc (R ), ⎪ ⎩ (iv)

∃A > 0 : ∀B

∃ cB > 0 :

1 |B|



|f (x) − cB |dx ≤ A . B

 (take cB = fB ). We have obviously BMO ⊂ E  B be a ball, A and cB satisfying (iv). Prove that |cB − fB | ≤ A . a) Let f ∈ E,  = BMO. b) Deduce that E 4. Our purpose is to prove that if f ∈ BMO then for all ε > 0 the function gε f (x) 1 d defined by gε (x) = (1+|x|) d+ε belongs to L (R ). For k ∈ N we set, Bk =   x ∈ Rd : |x| ≤ 2k . a) Show that it is enough to prove that,  (1)

|x|≥1

|f (x) − fB0 | dx < +∞. |x|d+ε

b) Let f ∈ BMO. Prove that there exists C1 > 0 such that for all k ∈ N,  |f (x) − fBk |dx ≤ C1 2nk A (2) Bk

where A is linked to f by (ii).

54

4 Statements of the Problems of Chap. 3

c) Prove that there exists C2 > 0 such that for all k ∈ N, (3)

|fBk+1 − fBk | ≤ C2 A.

Deduce that there exists C3 > 0 such that for all k ∈ N, (4)

|fBk − fB0 | ≤ kC3 A.

d) Prove that there exists C4 > 0 such that for all k ∈ N,  |f (x) − fB0 |dx ≤ kC4 A2nk . (5) Bk

e) Writing,     d k−1 k x ∈ Rd : |x| ≥ 1 = ∪+∞ x ∈ R : 2 ≤ |x| < 2 k=1 and using question 4d) prove our claim. 5. For a ∈ Rd we denote  by 1B(a,1) the indicator function of the ball B(a, 1) = x ∈ Rd : |x − a| ≤ 1 . a) Let (ak ) ⊂ Rd be a sequence such that limk→+∞ ak = a ∈ Rd . Prove that 1B(ak ,1) → 1B(a,1) almost everywhere. 1 d d b) Let  f ∈ Lloc (R ). Prove that the map  : R → C defined by (a) = B(a,1) f (x)dx is continuous. c) Let f be the function defined on R \ {0} by f (x) = ln |x|. Show that f ∈ L1loc (R). d) Prove that there exists A > 0 such that for all |y0 | ≤ 2 we have,  | ln |y||dy ≤ A. |y−y0 | 0 such that for all |y0 | > 2 we have,  |y−y0 | 0 such that, mu H s ≤ C m L1 u H s .

(4.11)

We shall use the Littlewood–Paley decomposition and denote by C various universal constants depending only on s. The notation A  B means that, A ≤ CB for such a constant C. 1. Prove that m belongs to L1 (R) and that the L∞ -norm of m is controlled by the L1 -norm of m . Deduce that, mu L2  m L1 u L2 . This proves the inequality (4.11) when s = 0. In the following we assume that s ∈ (0, 12 ). 2. Prove that, Tm u H˙ s  m L∞ u H˙ s . 3. Prove that,     j m 2  2− 12 j m  1 L L

 1  and p u L∞  2 2 p p uL2 .

4. Prove that there exists a constant C such that, for all 3 ≤ k, ∞ 2     (j −k u)j m s ≤ C2(2s−1)k m 2L1 u 2H˙ s .  ˙ H

j =k−1

5. In this section we use the convention that  u for any  ≤ −2. Verify that Tu m can be written under the form Tu m =

∞  ∞ 

(j −k u)j m.

k=3 j =k−1

and conclude that Tu m belongs to H˙ s (R) together with the estimate, Tu m H˙ s  m L1 u H˙ s .

56

4 Statements of the Problems of Chap. 3

6. Estimate similarly the reminder um − Tu m − Tm u and conclude the proof of (4.11). 7. The goal of this question is to prove that, in general, there does not exist a constant C > 0 such that, for any u in H 1 (R), mu H 1 ≤ C m L1 u H 1 .

(4.12)

a) Consider the function f : [0, +∞) → R defined by f (x) =

2 arctan(x) π

1 2

Prove that the function f is bounded on [0, +∞), is C ∞ on (0, +∞), that f is nonincreasing, and that f ∈ L1 ((0, +∞)). ∞ b) Let supported in [0, 1] such that  ρ : R → (0, +∞) be a C function ρ(z) dz = 1. Set for ε ∈ (0, 1], ρε (x) = 1ε ρ( xε ) and for x ∈ R, mε (x) = ρε f, where f(x) = 0 for x < 0 and f(x) = f (x) for x ≥ 0. Prove that mε is a C ∞ function on R and that,  mε (x) =

x ε

−∞

ρ(z)f (x − εz) dz.

Deduce that, for any ε > 0, mε is a nondecreasing C ∞ function such that limx→−∞ mε = 0 and limx→+∞ mε = 1. Prove that, sup mε L∞ (R) < +∞. ε∈(0,1]

c) Compute the function m ε and show that it belongs to Lp (R) for all ε ∈ (0, 1] and all p ∈ [1, +∞). Prove that,   sup m ε L1 (R) < +∞.

ε∈(0,1]

4 Statements of the Problems of Chap. 3

57

d) Prove that, 

1

sup ε∈(0,1] 0

m ε (x)2 dx = +∞.

Hint: Show that for all x in [ε, 1], one has m ε (x) ≥ e) Use this family of functions to contradict (4.12).

C √ x

for some C > 0.

Problem 17 Continuity of Paraproducts

The purpose of this problem is to investigate the continuity of a paraproduct Ta when a belongs to a Sobolev space H β (Rd ). More precisely our goal is to prove that if α, β, γ are real numbers such that, (i)

α ≤ γ,

(ii) α ≤ β + γ −

d , 2

(iii) α < γ if β =

d , 2

(4.13)

then there exists a constant C > 0 depending only on α, β, γ , d such that, Ta u H α (Rd ) ≤ C a H β (Rd ) u H γ (Rd ) . We recall that Ta u =



j ≥−1 Sj −2 (a)j u

(4.14)

and we set vj = Sj −2 (a)j u.

1. Prove (4.14) when β > d2 . 2. We assume β = d2 , so α < γ . a) Prove that Sj −2 (a)

H

Hint: Use (3.13). b) Using the fact that

d 2

d +γ −α 2 (Rd )

+γ −α >

d 2

≤ C2j (γ −α) a

d

H 2 (Rd )

.

prove (4.14).

3. We assume β < d2 . a) Prove that vj L2 (Rd ) ≤

j −2  k=−1

k a L∞ (Rd ) j u L2 (Rd ) .

58

4 Statements of the Problems of Chap. 3

b) Prove that k a L∞ (Rd ) ≤ C 2kd χ(2k ·) L2x (Rd ) k a L2 (Rd ) , where χ ∈ S(Rd ). c) Deduce that, ⎛ vj L2 (Rd )

≤C⎝

⎞1

j −2 

2

2

−2kβ

2 χ(2 kd

k

·) 2L2 (Rd ) ⎠

k=−1

⎛ ×⎝

j −2 

⎞1 2

22kβ k a 2L2 (Rd ) ⎠ j u L2 (Rd ) .

k=−1

d) Prove that, d

vj L2 (Rd ) ≤ 2j (α−γ + 2 −β) 2j γ k u L2 (Rd ) a H β (Rd ) , and deduce (4.14). e) Show that (4.14) can be summarized as follows: if a ∈ H β (Rd ) then Ta is of order zero if

β>

d , 2

d −β 2

if β
0

if

β=

d . 2

Problem 18 Products in Sobolev Spaces

This problem uses the result (4.14) in Problem 17. Let d ∈ N, d ≥ 1 and sj ∈ R, j = 1, 2 be such that s1 + s2 > 0. Let s0 be such that, (i)

s0 ≤ sj ,

j = 1, 2,

(ii) s0 ≤ s1 + s2 −

d , 2

where the inequality in (ii) is strict if s1 = d2 or s2 = d2 or s0 = − d2 . The purpose of this problem is to prove that if uj ∈ H sj (Rd ), j = 1, 2 then u1 u2 ∈ H s0 (Rd ) and there exists C > 0 depending only on s1 , s2 , d such that, u1 u2 H s0 (Rd ) ≤ C u1 H s1 (Rd ) u2 H s2 (Rd ) .

(4.15)

4 Statements of the Problems of Chap. 3

59

We write, u1 u2 = Tu1 u2 + Tu2 u1 + R, 

where Ta denotes the paraproduct by a and R =

|j −k|≤2 j u1 k u2 .

Part 1 1. Using (4.14) prove that there exists C > 0 such that, Tu1 u2 H s0 (Rd ) + Tu2 u1 H s0 (Rd ) ≤ C u1 H s1 (Rd ) u2 H s2 (Rd ) .

(4.16)

Part 2 The goal of this part is to prove that there exists C > 0 such that, R

H

s1 +s2 − d 2

(Rd )

≤ C u1 H s1 (Rd ) u2 H s2 (Rd ) .

(4.17)

We have R=



⎛ Rj ,

Rj = ⎝

⎞

j +2 

 k u1 ⎠  j u2 .

k=j −2

j

Notice that the  spectrum of Rj is contained in a ball B(0, C2j ) (not in a ring). So we write, R = p≥−1 p R. 1. a) Show that there exists N0 ∈ N such that, p R =



p Rj .

j ≥p−N0

b) Prove that p Rj = 2pd (F −1 ϕ)(2p ·) Rj and that there exists C1 > 0 such that, p Rj L2 (Rd ) ≤ C1 2

pd 2

Rj L1 (Rd ) .

c) Deduce that there exists C2 > 0 such that, p Rj L2 (Rd ) ≤ C2 2 αj =

pd 2

j +2  k=j −2

2−j (s1 +s2 ) αj

with

2j s1 k u1 L2 (Rd ) 2j s2 j u2 L2 (Rd ) ,

60

4 Statements of the Problems of Chap. 3

and prove that there exists C3 > 0, (dj ) ∈ 2 such that, αj ≤ C3 dj u1 H s1 (Rd ) u2 H s2 (Rd ) . 2. a) We set βp = 2p(s1 +s2 − 2 ) p R L2 (Rd ) . Prove that, d



βp ≤ C2

2(p−j )(s1+s2 ) αj .

j ≥p−N0

b) Deduce that there exists C4 > 0 such that,  p≥−1

c) Deduce (4.17). d) Conclude.

βp2 ≤ C4

 j ≥−1

αj2 .

Chapter 5

Microlocal Analysis

In this chapter we recall, without proof, the definitions and the basic results in microlocal analysis.

5.1 Symbol Classes 5.1.1 Definition and First Properties • For m ∈ R, S m (Rd ) is the set of all complex-valued functions a ∈ C ∞ (Rd × Rd ) such that,     β ∀α, β ∈ Nd , ∃ Cαβ > 0 : ∂xα ∂ξ a(x, ξ ) ≤ Cαβ (1+|ξ |)m−|β| , ∀(x, ξ ) ∈ Rd ×Rd . • This set is a vector space endowed with the semi-norms, pα,β (a) =

• • • •

sup (x,ξ )∈Rd ×Rd

    β (1 + |ξ |)−m+|β| ∂xα ∂ξ a(x, ξ ) ,

α, β ∈ Nd .

(5.1)

We will denote this space by S m if there is no confusion. Then S m is a Fréchet space. Its elements are called “symbols.”

If m ≥ m we have S m ⊂ S m . We set S +∞ = ∪m∈R S m and S −∞ = ∩m∈R S m . If a ∈ S m then ∂xj a ∈ S m and ∂ξj a ∈ S m−1 for all j = 1, . . . , d.

If a ∈ S m and b ∈ S m then ab ∈ S m+m . If F : Ck → C is a C ∞ function and if a1 , . . . , ak ∈ S 0 then F (a1 , . . . , ak ) ∈ S 0 .

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 T. Alazard, C. Zuily, Tools and Problems in Partial Differential Equations, Universitext, https://doi.org/10.1007/978-3-030-50284-3_5

61

62

5 Microlocal Analysis

5.1.2 Examples • If a = a(ξ ) belongs to the Schwartz class S(Rd ), then a ∈ S −∞ (Rd ). • Let m ∈ N and for α ∈ Nd , |α| ≤ m let cα be C ∞ (Rd ) functions all of whose derivatives are bounded. Then a(x, ξ ) = |α|≤m cα (x)ξ α ∈ S m . m

• For m ∈ R, a(ξ ) = (1 + |ξ |2 ) 2 ∈ S m .

5.1.3 Classical Symbols • A symbol a ∈ S m is called “classical” if there exists a sequence (aj )j ∈N ,where aj ∈ S m−j is a homogeneous function in ξ of order m − j for |ξ | ≥ 1 (that is, aj (x, λξ ) = λm−j aj (x, ξ ), |ξ | ≥ 1, λ ≥ 1) such that for all N ∈ N we have, a−

N−1 

aj ∈ S m−N .

j =0



We write a ∼ j aj . • Let aj ∈ S mj where (mj )j ∈N is a strictly decreasing sequence with limj →+∞ mj = −∞. Then there exists a ∈ S m0 such that a ∼ j aj .

5.2 Pseudo-Differential Operators 5.2.1 Definition and First Properties • Let a ∈ S m and A be the operator defined on S(Rd ) by the formula, Au(x) = (2π)−d

 u(ξ ) dξ. eix·ξ a(x, ξ )

A is called a pseudo-differential operator (DO) with symbol a. We denote sometimes A = Op(a). • Such an operator is continuous from S(Rd ) to S(Rd ). It can be extended as a continuous operator from S (Rd ) to S (Rd ). • For u ∈ S (Rd ) we have, sing supp(Au) ⊂ sing supp(u),

(5.2)

5.2 Pseudo-Differential Operators

63

where sing supp, the singular support, is defined as follows: x∈ / sing supp(u) ⇐⇒ u is a C ∞ function in a neighborhood of x. We say then that the DO are pseudo-local. Notice that the analogue of the inclusion (5.2) for the support is false unless A is a differential operator (see Problem 23).

5.2.2 Kernel of a DO • Consider a symbol a ∈ S m and set A = Op(a). There exists a distribution K = K(x, y) ∈ D (Rd × Rd ) such that, for all u, v ∈ C0∞ (Rd ) we have, Au, ϕ = K(x, y), u(y) ⊗ v(x) . • If m < −d we have, K(x, y) = (2π)

−d

 ei(x−y)·ξ a(x, ξ ) dξ,

where the integral converges absolutely.   • The kernel K is C ∞ outside the diagonal D = (x, y) ∈ Rd × Rd : x = y .

5.2.3 Image of a DO by a Diffeomorphism • Let χ be a C ∞ diffeomorphism from Rdx to Rdy all of whose derivatives are bounded. Let A = Op(a) where a ∈ S m (Rdx ). Let B be defined for u ∈ S(Rdy ) by the formula, Bu = A(u ◦ χ) ◦ χ −1 . Then B = Op(b) where b ∈ S m (Rdy ) and, b(χ(x), η) ∼

 1 t

φα (x, η)(∂ξα a) x, χ (x)η , α! d

α∈N

where χ is the differential of χ, tχ is its transpose, and *   + φα (x, η) = Dzα exp i( χ(z) − χ(x) − χ (x)(z − x), η |z=x .

64

5 Microlocal Analysis

• In particular if A is a classical DO with principal symbol am then B is classical with principal symbol, t

bm (χ(x), η) = am x, χ (x)η .

5.2.4 Symbolic Calculus 5.2.4.1 Composition • If a1 ∈ S m1 , a2 ∈ S m2 then Op(a1 ) ◦ Op(a2 ) = Op(b) where b := a1 #a2 ∈ S m1 +m2 is such that, ⎛ ⎞  1 ⎝b − ∂ α a1 ∂xα a2 ⎠ ∈ S m1 +m2 −N , ∀N ≥ 1. i |α| α! ξ |α|≤N−1

• In particular if a1 ∈ S m1 , a2 ∈ S m2 the commutator, [Op(a1 ), Op(a2 )] = Op(a1 ) ◦ Op(a2 ) − Op(a2 ) ◦ Op(a1 ) is a DO of order m1 + m2 − 1 and its symbol b satisfies, b−

1 {a1 , a2 } ∈ S m1 +m2 −2 i

where {·, ·} denotes the Poisson bracket, defined by,

d  ∂f ∂g ∂f ∂g {f, g} = · − ∂ξj ∂xj ∂xj ∂ξj j =1

5.2.4.2 Adjoint Let A be a DO. Its adjoint in the L2 (Rd ) sense is an operator denoted by A∗ and defined by,

(Au, v)L2 (Rd ) = u, A∗ v L2 (Rd ) ,

∀u ∈ S (Rd ), ∀v ∈ S(Rd ).

(5.3)

5.2 Pseudo-Differential Operators

65

• Let a ∈ S m . There exists a ∗ ∈ S m such that if set we A∗ = Op(a ∗ ) then A∗ satisfies (5.3). Moreover, ⎛



⎝a ∗ −

|α|≤N−1

⎞ 1 ∂ α ∂ α a ⎠ ∈ S m−N , i |α| α! ξ x

∀N ≥ 1.

5.2.5 Action of the DO on Sobolev Spaces • Let m ∈ R and s ∈ R. If a ∈ S m then Op(a) is continuous from H s (Rd ) to H s−m (Rd ). Moreover, its operator norm can be bounded by a finite number of semi-norms of the symbol defined by (5.1).

5.2.6 Garding Inequalities The Garding inequalities explore the link between the positivity of a symbol and the positivity of the corresponding operator.

5.2.6.1 The Weak Inequality • Let a ∈ S 2m satisfying, ∃c0 > 0, ∃R > 0 :

Re a(x, ξ ) ≥ c0 |ξ |2m ,

∀(x, ξ ) ∈ Rd × Rd with |ξ | ≥ R.

Then there exists C > 0 such that for all N ∈ N there exists CN > 0 such that, Re (Op(a)u, u)L2 (Rd ) + CN u 2H −N (Rd ) ≥ C u 2H m (Rd ) . 5.2.6.2 The “Sharp Garding” Inequality • Let a ∈ S 2m satisfying: ∃R > 0 :

Re a(x, ξ ) ≥ 0,

∀(x, ξ ) ∈ Rd × Rd , |ξ | ≥ R.

Then ∃C > 0 :

Re (Op(a)u, u)L2 (Rd ) + C u 2

H

m− 1 2 (Rd )

≥ 0.

66

5 Microlocal Analysis

5.2.6.3 The Fefferman–Phong Inequality • Let a ∈ S 2m be a real valued symbol satisfying: a(x, ξ ) ≥ 0,

∀(x, ξ ) ∈ Rd × Rd .

Then ∃C > 0 :

Re (Op(a)u, u)L2 (Rd ) + C u 2H m−1 (Rd ) ≥ 0.

5.3 Invertibility of Elliptic Symbols Let a ∈ S m . We say that a is an elliptic symbol if ∃c0 > 0, ∃R > 0 :

Re a(x, ξ ) ≥ c0 |ξ |2m ,

∀(x, ξ ) ∈ Rd × Rd ,

|ξ | ≥ R.

• If a is an elliptic symbol there exists b ∈ S −m such that, Op(a) ◦ Op(b) − Id ∈ Op(S −∞ ),

Op(b) ◦ Op(a) − Id ∈ Op(S −∞ ).

5.4 Wave Front Set of a Distribution 5.4.1 Definition and First Properties • If ξ0 ∈ Rd \ {0} , a conical neighborhood of ξ0 is a set of the form, ξ0

    ξ ξ0  d  − = ξ ∈ R \ {0} :  < ε , |ξ | |ξ0 | 

ε > 0.

/ WF(u) • Let u ∈ D (Rd ) and (x0 , ξ0 ) ∈ Rd × (Rd \ {0}). We say that (x0 , ξ0 ) ∈ if there exists a neighborhood Vx0 of x0 , a conical neighborhood of ξ0 of ξ0 , a function ϕ ∈ C0∞ (Vx0 ) equal to 1 in a neighborhood of x0 such that, ∃R ≥ 1 : ∀N ∈ N, ∃CN > 0 :

|, ϕ u(ξ )| ≤ CN |ξ |−N ,

∀ξ ∈

ξ0 ,

|ξ | ≥ R.

• WF(u) is called the “wave front set ” of u. It is a closed subset of Rd × (Rd \ {0}). • WF(u) is a conical set, that is (x0 , ξ0 ) ∈ WF(u) ⇒ (x0 , tξ0 ) ∈ WF(u), ∀t > 0.

5.4 Wave Front Set of a Distribution

67

• Let π1 : Rd × Rd \ {0} → Rd be the map (x, ξ ) → x. Then, π1 (WF(u)) = sing supp(u).

5.4.2 Wave Front Set and DO • Let u ∈ D (Rd ) and (x0 , ξ0 ) ∈ Rd × (Rd \ {0}). If (x0 , ξ0 ) ∈ / WF(u) then, ∃ϕ ∈ C0∞ (Rd ), ϕ = 1 near x0 , ∃a ∈ S 0 : a(x0, ξ0 ) = 0 such that , Op(a)(ϕu) ∈ C ∞ (Rd ). • Let u ∈ D (Rd ) and (x0 , ξ0 ) ∈ Rd × (Rd \ {0}). Assume that, ∃ϕ ∈ C0∞ (Rd ), ϕ = 1 near x0 , ∃a ∈ S 0 : a(x0, ξ0 ) = 0 such that Op(a)(ϕu) ∈ C ∞ (Rd ). Then (x0 , ξ0 ) ∈ / WF(u). • Let u ∈ S (Rd ) and a ∈ S m be a classical symbol , a ∼



j ≥0 am−j .

The set

  ! = (x, ξ ) ∈ Rd × (Rd \ {0}) : am (x, ξ ) = 0 is called the characteristic set of a. We have then, WF(u) ⊂ WF(Op(a)u) ∪ !.

5.4.3 The Propagation of Singularities Theorem 5.4.3.1 Bicharacteristics • Let a ∈ S m be a real symbol. Let ρ0 = (x0 , ξ0 ) ∈ Rd × (Rd \ {0}). The bicharacteristic of a issued from the point ρ0 is the parametric curve (x(t), ξ(t)) such that for 1 ≤ j ≤ d, x˙j (t) =

∂a (x(t), ξ(t)), ∂ξj

∂a (x(t), ξ(t)) ξ˙j (t) = − ∂xj

xj (0) = x0,j , ξj (0) = ξ0,j .

68

5 Microlocal Analysis

The Cauchy–Lipschitz theorem implies the existence of a maximal solution of this system. • If we define the Hamiltonian of a as the vector field on Rd × (Rd \ {0}), Ha =

d  ∂a ∂ ∂a ∂ − , ∂ξj ∂xj ∂xi ∂ξj j =1

we see that the bicharacteristics are the integral curves of this field, that is the curves along which the field is tangent at every point. • The symbol a is constant along the bicharacteristic curves. Those where a = 0 are called the null bicharacteristics.

5.4.3.2 The Propagation Theorem • Let a ∈ S m be a real symbol. Let u ∈ S (Rd ). Let ρ0 = (x0 , ξ0 ) ∈ Rd ×(Rd \{0}) and let γρ0 be the maximal null bicharacteristic issued from this point. Assume that γρ0 ∩ WF(P u) = ∅. Then, ρ0 ∈ WF(u) ⇐⇒ γρ0 ⊂ WF(u), / WF(u) ⇐⇒ γρ0 ∩ WF(u) = ∅. ρ0 ∈

5.5 Paradifferential Calculus Paradifferential operators are the generalization of pseudo-differential operators to the case where the symbol has a limited regularity with respect to x. Recall (see Sect. 3.2) that for k ∈ N, W k,∞ (Rd ) denotes the space of functions whose derivatives up to the order k belong to L∞ (Rd ) and for ρ = k +σ, σ ∈ (0, 1), W ρ,∞ (Rd ) is the space of functions whose derivatives up to the order k belong to L∞ (Rd ) and are uniformly Hölder continuous with exponent σ.

5.5.1 Symbols Classes • Given ρ ≥ 0 and m ∈ R, ρm (Rd ) denotes the space of locally bounded functions a(x, ξ ) on Rd × (Rd \ {0}), which are C ∞ with respect to ξ for ξ = 0 and such that, for all α ∈ Nd and all ξ = 0, the function x → ∂ξα a(x, ξ ) belongs

5.5 Paradifferential Calculus

69

to W ρ,∞ (Rd ) and there exists a constant Cα such that for every |ξ | ≥ 12 ,     α ∂ξ a(·, ξ )

W ρ,∞ (Rd )

≤ Cα (1 + |ξ |)m−|α| .

The elements of ρm (Rd ) are called symbols. • For a ∈ ρm (Rd ) we introduce the following semi-norms: Mρm (a) =

    sup (1 + |ξ |)|α|−m ∂ξα a(·, ξ )

sup

W ρ,∞ (Rd )

|α|≤2(d+2)+ρ |ξ |≥1/2

.

(5.4)

5.5.2 Paradifferential Operators • Given a symbol a we define the paradifferential operator Ta by, −d T a u(ξ ) = (2π)

 χ(ξ − η, η) a (ξ − η, η)ψ(η) u(η) dη,

(5.5)

 where  a (θ, ξ ) = e−ix·θ a(x, ξ ) dx is the Fourier transform of a with respect to the first variable and χ, ψ are two fixed C ∞ functions such that, ψ(η) = 0 for |η| ≤

1 , 5

ψ(η) = 1 for |η| ≥

1 , 4

(5.6)

and χ(θ, η) satisfies, for 0 < ε1 < ε2 small enough, χ(θ, η) = 1 if

|θ | ≤ ε1 |η| ,

χ(θ, η) = 0

if

|θ | ≥ ε2 |η| ,

and, ∀(θ, η) :

   α β  ∂θ ∂η χ(θ, η) ≤ Cα,β (1 + |η|)−|α|−|β| .

Notice that in the above integral we have taken the frequencies of a (with respect to its first variable) smaller than those of u since on the support of χ we have |ξ − η| ≤ ε2 |η|.

70

5 Microlocal Analysis

5.5.3 The Symbolic Calculus Here are the main features of the symbolic calculus for paradifferential operators. Let m ∈ R. An operator T is said to be of order m if, for all μ ∈ R, it is bounded from H μ (Rd ) to H μ−m (Rd ). • If a ∈ 0m (Rd ), then Ta is of order m. Moreover, for all μ ∈ R there exists a constant K such that, Ta H μ →H μ−m ≤ KM0m (a). • Let m ∈ R and ρ > 0. If a ∈ order m + m − ρ where, a#b =

m d ρ (R ), b

 |α| 0. If a ∈ ρm (Rd ) we denote by (Ta )∗ the adjoint operator of Ta . Then (Ta )∗ − Ta ∗ is of order m − ρ where, a∗ =

 |α| 0 we set,  Sρm = a ∈ C ∞ (R2d ) : ∀α, β ∈ Nd ∃Cαβ > 0 :  β |∂xα ∂ξ a(x, ξ )| ≤ Cαβ (1 + |ξ |)m−ρ|β| , ∀(x, ξ ) ∈ R2d .

1. Assume m < 0, ρ > 1. Let a ∈ Sρm . a)

Prove, by induction on k, that for any α, β ∈ Nd there exists Cαβk > 0 such that, (Hk )

β

|∂xα ∂ξ a(x, ξ )| ≤ Cαβk (1 + |ξ |)m−ρ|β|−k(ρ−1) ,

∀(x, ξ ) ∈ R2d .

Hint: Use the fact that a function is the integral of its derivative. b) Deduce that Sρm = S −∞ . 2. a) Let m ∈ N, m ≥ 1 and let p(ξ ) be a polynomial of degree m − 1. Set ρ = 1 + ε, ε > 0. Prove that if ε(m − 1) ≤ 1 we have p ∈ Sρm but p ∈ / S −∞ .

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 T. Alazard, C. Zuily, Tools and Problems in Partial Differential Equations, Universitext, https://doi.org/10.1007/978-3-030-50284-3_6

75

76

6 Statements of the Problems of Chap. 5

b) Give a simple example with m = 0 showing that the result of question 1b) is false in this case.

Problem 20

Let a ∈ S 1 be a real valued symbol satisfying, ∃c > 0 : a(x, ξ ) ≥ c ξ  ,

∀(x, ξ ) ∈ Rd × Rd ,

1

ξ  = (1 + |ξ |2 ) 2 .

Let σ > 0. Prove that for any α, β ∈ Nd we have, (∂xα ∂ξ e−σ a )(x, ξ ) = β

|α|+|β| 

σ j dj,αβ (x, ξ )e−σ a ,

dj,αβ ∈ S j −|β| .

j =0

Problem 21 A Commutator Estimate

If P is a first order differential operator with constant coefficients, then P (f u) − f P u L2 ≤ ∇f L∞ u L2 . The goal of this problem is to prove a similar result when P is replaced by a pseudo differential operator with constant coefficients and L2 is replaced by a Sobolev space Hσ. More precisely, consider real numbers σ0 and m such that σ0 > 1 + d2 and m ∈ [1, σ0 ]. Let p = p(ξ ) ∈ S m and P = Op(p). We want to prove that, for all σ ∈ [−σ0 + m, 0] ∪ [m − 1, σ0 − 1] there exists K > 0 such that, P (f u) − f P u H σ −(m−1) ≤ K f H σ0 u H σ . for all f, u ∈ S(Rd ). Part 1 Assume σ ∈ [m − 1, σ0 − 1]. 1. Set v := P (f u) − f P u with f, u in S(Rd ). Verify that v = vL + vH with,  vL (ξ ) =

|ξ −η|≤ 21 |η|

 v, H (ξ ) =

|ξ −η|> 12 |η|



p(ξ ) − p(η) f(ξ − η) u(η) dη,



p(ξ ) − p(η) f(ξ − η) u(η) dη.

(6.1)

6 Statements of the Problems of Chap. 5

77

2. Verify that there exists C such that for all (ξ, η) satisfying |ξ − η| ≤ have,

1 2

|η| we

|p(ξ ) − p(η)| ≤ C |ξ − η| ηm−1 . 3. a) Prove that, ξ σ −m+1 |vL (ξ )| ≤ K



  u(η)| dη. |ξ − η|f(ξ − η) ησ |

b) Deduce that, vL H σ −m+1 ≤ C f H σ0 u H σ . 4. Prove that vH (ξ )| ≤ K ξ σ −m+1 |,



  ησ ξ − ησ0 f(ξ − η) σ −1 | u(η)| dη, η 0

and deduce that vH H σ −m+1 ≤ C f H σ0 u H σ , and conclude that the estimate (6.1) holds for σ ∈ [m − 1, σ0 − 1]. Part 2 Assume σ ∈ [−σ0 + m, 0]. 1. Define the operator C by Cu = P (f u) − f P u. By using the previous result for σ ∈ [−1, σ0 ] and a duality argument, show that C ∗ ∈ L(H −σ +m−1 , H −σ ) with operator norm bounded by K f H σ0 . Deduce that the estimate (6.1) holds for σ ∈ [−σ0 + m, 0).

Problem 22

The goal of this problem is to study the commutator between a function in the Hölder space W ν,∞ (R) where 0 < ν < 1 and the Hilbert transform H. Recall that H is the operator acting on tempered distribution u by,  ) := ξ  Hu(ξ u(ξ ). i |ξ | We will use the paradifferential calculus. In particular, given a function f = f (x), we denote by Tf the paraproduct by f .

78

6 Statements of the Problems of Chap. 5

1. Given two functions f and g verify that, fg − Tf g =



(Sj +2 g)j f.

j ≥−1

Deduce that, for all real numbers 0 < θ < ν < 1, there is a constant C such that, for all g ∈ H −θ (R), we have,   fg − Tf g 

L2

≤ C f W ν,∞ g H −θ .

2. Let q = q(ξ ) be a symbol in S 0 and Q = Op(q). Prove that, for 0 < θ < ν < 1, there exists a constant K such that for all f ∈ W ν,∞ (R), and all u ∈ H −θ (R) we have, Q(f u) − f Qu L2 ≤ K f W ν,∞ u H −θ . 3. Deduce that, for 0 < θ < ν < 1, there exists a constant K such that for all f ∈ W ν,∞ (R), and all u in H −θ (R), H(f u) − f Hu L2 ≤ K f W ν,∞ u H −θ .

Problem 23 Local DO Are Differential

The purpose of this problem is to show that the only local pseudodifferential operators are the differential operators. Part 1 We consider in this part pseudo-differential operators whose symbols belong to the S m classes. We say that an operator is local if it satisfies, supp P u ⊂ supp u,

∀u ∈ S(Rd ).

1. Prove that a differential operator with C ∞ -bounded coefficients is local. 2. Prove that if P and Q are local operators the same is true for P Q and [P , Q] = P Q − QP . 3. Prove that if P is local the same is true for its adjoint P ∗ . 4. a) Prove that if P ∗ P ≡ 0 then P ≡ 0. b) Let  ≥ 2. Show that if (P ∗ P ) ≡ 0 then (P ∗ P )−1 ≡ 0. Deduce then that P ≡ 0.

6 Statements of the Problems of Chap. 5

79

Part 2 Let a ∈ S m and A = Op(a). The goal of this part is to prove that if A is local then A is a differential operator. 1. Let u ∈ S(Rd ) and x0 ∈ Rd . Show that there exists a sequence (uk )k∈N ⊂ S(Rd ), whose elements vanish identically in a neighborhood of x0 , which converges to u in L2 (Rd ). 2. Assume m < − d2 . Show that there exists C > 0 such that, sup |Av(x)| ≤ C v L2 (Rd ) ,

∀v ∈ S(Rd ).

x∈Rd

3. Assume that A is local and m < − d2 . Let x0 ∈ Rd . Show that Au(x0) = 0 for every u ∈ S(Rd ). Deduce that A ≡ 0. 4. Assume that A is local and m < 0. Show that A ≡ 0. Hint: We may assume that − d2 ≤ m < 0. Let  ∈ N such that 2m < − d2 . Consider the operator (A∗ A) . Notation: If P and Q are two operators we set (ad P )(Q) = [P , Q] = P Q − QP . 5. Let k ∈ N∗ . Let a ∈ S m where m < k. Let g1 , . . . , gk be C ∞ (Rd )-bounded. Prove that (ad g1 )(ad g2 ) · · · (ad gk )(A) = 0. 6. Let a ∈ S m , w ∈ C0∞ (Rd ) and g ∈ C0∞ (Rd ) be such that g(x0 ) = 0. Show, by induction on  ≥ 1 that for x near x0 we have, A(g  w)(x) = (−1) (ad g) (A)w(x) + O(|x − x0 |). 7. Let k ∈ N∗ , x0 ∈ Rd and u ∈ C0∞ (Rd ). Assume that ∂xα u(x0 ) = 0, |α| ≤ k−1.  a) Show that u(x) = |β|=k (x − x0 )β vβ (x) where vβ ∈ C ∞ (Rd ). b) Let θ ∈ C0∞ (Rd ), θ = 1 on supp u. Noticing that θ k+1 u = u show that, Au(x) =



(−1)|β| (ad f1 )β1 (ad f2 )β2 · · · (ad fd )βd (A)(θvβ )+O(|x−x0 |),

|β|=k

where fj = (xj − x0,j )θ. Hint: Use question 6. c) Let a ∈ S m be such that A = Op(A) is local. Let k > m. Show that Au(x0 ) = 0. 8. Let u ∈ C0∞ (Rd ). Setting w(x) = u(x) −

 |α|≤k−1

1 (x − x0 )α χ(x)∂xα u(x0 ), α!

80

6 Statements of the Problems of Chap. 5

where χ ∈ C0∞ (Rd ) is equal to 1 near x0 , show that 

Au(x0 ) =

cα ∂xα u(x0 ),

|α|≤k−1

and conclude.

s (Rd ) Spaces Problem 24 Continuity of the DO in the Hul

This problem follows Problem 7. Let P be pseudo-differential operator with symbol p ∈ S m (Rd ) and s ∈ R. The s+m s (Rd ). goal of this problem is to show that P is continuous from Hul (Rd ) to Hul The continuity of the pseudo-differentials operators on the classical Sobolev spaces H s (Rd ) will be assumed. We keep the notations of Problem 7. We write for k ∈ Zd (with |x| = sup1≤j ≤d |xj |), χk P u =



χk P χq u +

|k−q|≤2



χk P χq u := Ak +

|k−q|≥3



Bk,q .

(6.2)

|k−q|≥3

1. Show that there exists a positive constant C independent of k, q such that, Ak H s (Rd ) ≤ C u H s+m (Rd ) , ul

s+m d ∀u ∈ Hul (R ).

The following questions will be devoted to prove the estimate, Bk,q H s (Rd ) ≤

C u H s+m (Rd ) . ul |k − q|d+1

 2. Let n0 ∈ N, n0 ≥ s. Then Bk,q H s (Rd ) ≤ |α|≤n0 ∂xα Bk,q L2 (Rd ) . Show that there exists C > 0 such that for all k, q ∈ Zd and all α ∈ Nd we have, ∂xα Bk,q L2 (Rd ) ≤ C ∂xα Bk,q L∞ (Rd ) . 3. Prove that for |α| ≤ n0 , ∂xα Bkq (x) is a finite linear combination of terms of the form,  Ckqα (x) =: ∂xα1 χk (x) eix·ξ aα2 (x, ξ )χ |α1 | + |α2 | ≤ |α|, q u(ξ ) dξ, where, aα2 ∈ S m+|α2 | .

6 Statements of the Problems of Chap. 5

81

4. Let θ ∈ C0∞ (Rd ) be such that supp θ ⊂ {ξ : |ξ | ≤ 1} . Show that,  Ckqα (x) = lim

ε→0

ε ε (x, y)χq u(y) dy := lim Ckqα (x), where, Kkqα ε→0



ε Kkqα (x, y) = ∂xα1 χk (x) χq (y)

ei(x−y)·ξ aα2 (x, ξ )θ (εξ ) dξ,

   = 1 on the support of χ.  ⊂ ξ : |ξ | ≤ 32 , χ where χ  ∈ C0∞ (Rd ), supp χ 5. Let n1 ∈ N, n1 ≥ −(m + s). Show that, ε ε |Ckqα (x)| ≤ Kkqα (x, ·) H n1 (Rd ) u H s+m (Rd ) . ul

6. Show that there exists C > 0 independent of k, q, α, ε such that, ε Kkqα (x, ·) H n1 (Rd ) ≤ C



ε ∂yβ Kkqα (x, ·) L∞ (Rd ) .

|β|≤n1 β

ε (x, y) is a finite linear combination of terms of the form, 7. Show that ∂y Kkqα

 ε Kkqαβ (x, y) = ∂xα1 χk (x)∂yβ1 χ q (y)

ei(x−y)·ξ bα2 ,β2 (x, ξ )θ (εξ ) dξ,

where bα2 ,β2 ∈ S m+|α2 |+|β2 | and |α1 | + |α2 | ≤ |α|, β1 + β2 = β. 8. Show that if |k − q| ≥ 3 we have |x − y| ≥ 16 |k − q| on the support of ε Kkqαβ (x, y). d x−y 9. Let M(x) = |x−y| 2 and L = M(x) · ∇ξ = j =1 Mj (x)∂ξj .  γ γ N i(x−y)·ξ = a) Show that for N ∈ N, LN = |γ |=N cγ M ∂ξ and that L e ei(x−y)·ξ . b) Integrating by parts with LN with N large, in the integral giving ε Kkqαβ (x, y), show that there exists C > 0 independent of x, y, k, q, α, β, ε such that, ε |Kkqαβ (x, y)| ≤

10. Conclude.

Problem 25

This problem uses Problem 20.

C u H s+m (Rd ) , ul |k − q|d+1

∀x, y ∈ Rd .

82

6 Statements of the Problems of Chap. 5

Let I = (0, 1), d ≥ 1, and μ < 12 . Let a = a(ξ ) ∈ S 1 be a real symbol such that ∃c0 > 0 : a(ξ ) ≥ c0 ξ  ,

∀ξ ∈ Rd ,

1

ξ  = (1 + |ξ |2 ) 2 .

The goal of this problem is to show that for every ρ ∈ R the map u → e−za(Dx ) u is ρ ρ+μ continuous from C∗ (Rd ) to L2z (I, C∗ (Rd )).

|α| We recall (see Problem 20) that, ∂ξα e−za(ξ ) = j =0 zj qj α (ξ )e−za(ξ ) , where qj α ∈ S j −|α| . Let μ < 12 and p(ξ ) = zμ ξ μ e−za(ξ ). 1. Show that for every α ∈ Nd there exists Cα > 0 such that, |∂ξα p(ξ )| ≤ Cα zμ ξ μ−|α| e− 2 za(ξ ), 1

∀ξ ∈ Rd .

p (x)| ≤ Cd , ∀|x| ≥ 1. 2. Deduce that |x|d+1| 1 3. a) We set z, η = (z2 + |η|2 ) 2 . Prove that,  c0 (x)| ≤ Cα z|α|−d z, ημ−|α| e− 2 z,η dη. |x α p b) Prove that for |α| = d and |α| = d − 1 there exists M > 0 such that,  c0 z, ημ−|α| e− 2 z,η dη ≤ M. c) Prove that for every ε ∈]0, 1] we have, p(x)| ≤ Cε z−ε , |x|d−ε |

∀|x| ≤ 1.

d) Deduce that for every ε ∈]0, 1] there exists Cε > 0 such that,  p L1 (Rd ) ≤ Cε z−ε . 4. Prove that for every ε ∈]0, 1] there exists Cε > 0 such that, j e−za(Dx ) u L∞ (Rd ) ≤ Cε z−μ−ε 2−j (μ+ρ) u C∗ρ . 5. Conclude.

Problem 26 Let α ∈]0, +∞[ with α = 1 and let S(t) = e−it |Dx | be the operator defined on α S (Rd ) by S(t)u = F −1 (e−it |ξ |  u). α

6 Statements of the Problems of Chap. 5

83

The goal of the first two parts problem is to prove the following result. Let s, σ ∈ R. Assume that there exists t0 = 0 such that S(t0 ) is continuous from C∗σ (Rd ) to C∗s (Rd ). Then s≤σ−

dα . 2

(6.3)

In a third part we shall prove that this result does not hold for α = 1. Part 1. Preliminaries Let K ∈ C 0 (Rd ) ∩ L1 (Rd ). Let T be the operator defined by,  T u(x) =

K(x − y)u(y)dy.

1. Show that T is continuous from L∞ (Rd ) to L∞ (Rd ) and that,  ∞ ∞ |K(z)| dz. T L →L = Rd

2. For λ ∈ R let M(λ) = (mj k (λ))1≤j,k≤d , where mj k (λ) = δj k + λωj ωk , ω ∈ Sd−1 , be a d × d matrix. Set F (λ) = det M(λ). We want to show that F (λ) = 1 + λ. We have trivially F (0) = 1. a) Denote by i (λ) the i t h line of the matrix M(λ). Noticing that, F (λ) =

d 

det(1 (λ), . . . ,  k (λ), . . . , d (λ)),

k=1

show that F (0) = 1. b) Noticing that i (λ) is linear with respect to λ show that F

(λ) ≡ 0 and conclude. Part 2 We assume first that t0 = −1. 1. Let u ∈ L∞ (Rd ). Consider the canonical decomposition of u (see Sect. 3.3), u = −1 u +

+∞ 

j u.

j =0

Show that there exist C, C > 0 such that for all j ∈ N we have, j u ∈ C∗σ (Rd ) and, j u C∗σ (Rd ) ≤ C2j σ j u L∞ (Rd ) ≤ C 2j σ u L∞ (Rd ) .

84

6 Statements of the Problems of Chap. 5

2. Show that there exists C > 0 such that, ∀u ∈ L∞ (Rd ),

2j s S(−1)j j u L∞ (Rd ) ≤ C2j σ u L∞ (Rd )

∀j ∈ N.

3. We fix j and we set, Tj = S(−1)j j , h = 2−j . a) Check that,  Tj u(x) =

Kh (x − y)u(y)dy,

Rd

where, Kh (z) = (2πh)

−d



i

1−α |η|α )

e h (z·η+h

ϕ 2 (η)dη.

Rd

b) Show that Kh ∈ C 0 (Rd ) ∩ L1 (Rd ). c) Deduce that,  Tj

L∞ →L∞

=h

d(1−α) Rd

h (s)| ds, |K

with, h (s) = (2πh)−d K



−α φ(s,η)

eih

ϕ 2 (η)dη,

Rd

φ(s, η) = s · η + |η|α .

4. Assume that, |s| ≤

1 α . 2 2|1−α|

Using the vector field L=

d hα 1  ∂φ ∂ , i |∂η φ|2 ∂ηk ∂ηk k=1

show that, h (s)| ≤ CN hN , |K

∀N ∈ N.

5. Assume now that |s| ≥ 21+|α−1| α. Check that on the support of ϕ we have,     ∂φ  (s, η) ≥ |s| − α|η|α−1 ≥ 1 |s|.   ∂η 2

6 Statements of the Problems of Chap. 5

85

Using the same vector field show that, h (s)| ≤ CN |s|−N hN , |K

∀N ∈ N.

6. Assume now that, 1 α ≤ |s| ≤ 21+|α−1| α. 2 2|1−α| 2−α

a) Check that the phase φ has a critical point given by ηc = cα s|s| α−1 , and that ∂ 2φ (s, η) = α|η|α−2 mj k ∂ηj ∂ηk

mj k = δj k − (α − 2)ωj ωk , 

b) Prove that the determinant of the matrix

∂2φ ∂ηj ∂ηk

ω=

η . |η|

 is different from zero and (α−2)d

that its value at the critical point ηc is equal to Cα,d |s| (α−1) . c) Check that the stationary phase formula implies that there exists Cα,d > 0 such that for h small enough we have, " h (s) = Cα,d h−d h K

αd 2

−α φ(s,η

eih

|s|

c)

(α−2)d 2(α−1)

# ϕ 2 (ηc ) + O(hα ) .

7. Deduce from the previous questions that for h = 2−j small enough we have, Tj L∞ →L∞ ≥ Chd(1−α)h−d h

dα 2

− CN hN ≥ C h−

dα 2

= C 2j

dα 2

.

8. Conclude. 9. We assume that there exists t0 = 0 such that S(t0 ) is continuous from C∗σ (Rd ) to C∗s (Rd ). a) If t0 < 0 show that j S(−1)u(x) = λj S(t0 )u 1 (λx) where λα = −t0 , λj v = F −1 (ϕ(2−j λξ ) v ), u 1 (x) = u( xλ ).

λ

λ

b) If t0 > 0 show that j S(−t0 )u = j S(t0 )u. c) Deduce that s ≤ σ − dα 2 . Part 3 The purpose of this part is to show that the inequality (6.3) is false when α = 1. We assume for simplicity that d = 3. For u0 ∈ S (R3 ), t ≥ 0 and x ∈ R3 , set u(t, x) = (e−it |Dx | u0 )(x). 1. Show that u is a solution of the problem, u = 0,

in (0, +∞) × R3 ,

u|t =0 = u0

∂t u|t =0 = −i|Dx |u0 ,

86

6 Statements of the Problems of Chap. 5

where  = ∂t2 − x . 2. Using the formula (7.10) in Chap. 4 prove that if u0 ∈ C∗σ (R3 ) then the function x → u(t0 , x) belongs to C∗σ −1 (R3 ) and conclude. Problem 27 A Carleman Inequality Notations If f ∈ C ∞ (Rt × Rdx ) we denote by f (t) the function x → f (t, x). On the other hand, for T > 0 we set,   ET = u ∈ C ∞ (R, S(Rd )) : supp u ⊂ (0, T ) . Notice that for u ∈ ET we have u(0, ·) = u(T , ·) = 0. Part 1 Let a ∈ S 1 (Rd ) be a real symbol. We set A = Op(a) and we consider the operator P = Dt + A, Dt = 1i ∂t∂ . k

1. Let k ≥ 1, T > 0 and u ∈ C ∞ (R × Rd ). Set v = e 2 (t −T ) u. Show that, 2 e 2 (t −T ) P u = Pv,

P = X + Y,

k

2

X = Dt + A,

Y = ik(t − T ).

2. a) Justify the fact that A∗ = A + B0 , where B0 ∈ Op(S 0 ). b) Prove that there exists T0 > 0 small enough such that for all T ≤ T0 , 

T

2Re

(Xv(t), Y v(t)) L2 (Rd )

0

k dt ≥ 2



T 0

v(t) 2L2 (Rd ) dt,

∀v ∈ ET .

3. Deduce that there exists C > 0 and T0 > 0 such that for all T ≤ T0 , 

T

k 0

ek(t −T ) u(t) 2L2 (Rd ) dt ≤ C 2



T 0

ek(t −T ) P u(t) 2L2 (Rd ) dt, 2

∀v ∈ ET .

1

4. a) Let a(ξ ) = (1 + |ξ |2 ) 2 and A = Op(a). Show that A2 = −x + I d. b) What is the operator Q = (Dt − A)(Dt + A)? c) Noticing that (Dt ± A)u ∈ ET if u ∈ ET , show that there exist positive constants C, T0 , k0 such that for T ≤ T0 , k ≥ k0 and for all u ∈ ET , we have, 

T

k2 0

ek(t −T ) u(t) 2L2 (Rd ) dt ≤ C

where  = ∂t2 −

2

d

2 j =1 ∂xj



T 0

ek(t −T ) u(t) 2L2 (Rd ) dt,

is the wave operator.

2

6 Statements of the Problems of Chap. 5

87

Part 2 Let a, b ∈ S 1 (Rd ) be two real symbols. We assume that b is elliptic in the following sense: ∃ c0 > 0 : b(ξ ) ≥ c0 (1 + |ξ |),

∀ξ ∈ Rd .

Set B = Op(b) and consider the operator P = Dt + A + iB, k

Dt =

1 ∂ i ∂t .

1. For u ∈ C ∞ (R × Rd ) set v = e 2 (t −T ) u. Show that 2 v, e 2 (t −T ) P u = P k

2

P = X + iY,

X = Dt + A,

Y = iB + ik(t − T ).

2. Let r > 0. Prove that there exists T0 > 0 such that for T ≤ T0 and   all v ∈ C ∞ (R × Rd ) with supp v ⊂ (t, x) ∈ R × Rd : 0 ≤ t ≤ T , |x| ≤ r we have, 

T

2Re 0

k (1) ≥ 2

(Xv(t), Y v(t)) L2 (Rd ) dt = (1) + (2) + (3), 

T 0

 v(t) 2L2 (Rd ) dt,



T

(2) = 2Re 0

(Av(t), iBv(t))L2 (Rd ) dt,

T

(3) = 2Re

(Dt v(t), iBv(t)) L2 (Rd ) dt.

0

3. a) Set C = A∗ B. Show that C ∈ Op(S 2 ) and that its symbol c satisfies c = c2 + r1 where c2 is real and r1 ∈ S 1 . Deduce that C − C ∗ ∈ Op(S 1 ). b) Show that there exists C > 0 such that, 

T

|(2)| ≤ C 0

v(t) H 1 (Rd ) v(t) L2 (Rd ) dt.

4. a) Show that there exists C > 0 such that, 

T

|(3)| ≤ C 0

v(t) L2 (Rd ) Dt v(t) L2 (Rd ) dt.

b) Writing Dt = Dt + A − A, deduce that there exists C > 0 and for any ε > 0 there exists Cε > 0 such that, 

T

|(3)| ≤ Cε 0

 Xv(t) 2L2 (Rd ) dt + Cε  +C 0

T 0

T

v(t) 2L2 (Rd ) dt

v(t) L2 (Rd ) v(t) H 1 (Rd ) dt.

88

6 Statements of the Problems of Chap. 5

5. Show that, 

T

Pv(t) 2L2 (Rd ) dt ≥

0

+

k 2



T 0

1 2



T 0

 Xv(t) 2L2 (Rd ) dt + 

v(t) 2L2 (Rd ) dt − C

T 0

T

0

Y v(t) 2L2 (Rd ) dt

v(t) L2 (Rd ) v(t) H 1 (Rd ) dt.

6. a) Using the ellipticity of B show that there exists C > 0 such that,

v(t) H 1 (Rd ) ≤ C Y v(t) L2 (Rd ) dt + (1 + kT ) v(t) L2 (Rd ) .

(6.4)

b) Deduce that for any ε > 0 there exists Cε > 0 such that, 



T

v(t) L2 (Rd ) v(t) H 1 (Rd ) dt ≤ ε

0

T

0

Y v(t) 2L2 (Rd ) dt 

+ Cε (1 + kT ) 0

T

v(t) 2L2 (Rd ) .

7. Prove that there exists C > 0 and T0 > 0 such that for all T ≤ T0 , 

T 0

 Xv(t) 2L2 (Rd ) dt

T

+ 0

 Y v(t) 2L2 (Rd ) dt

T

+k 0

 ≤C

0

T

v(t) 2L2 (Rd ) Pv(t) 2L2 (Rd ) dt.

8. a) Using (6.4) show that there exists C > 0 such that, 

T 0

 v(t) 2H 1 (Rd ) dt

T

≤ C(1 + kT ) 2

0

Pv(t) 2L2 (Rd ) dt.

b) Deduce that, 

T 0

 Dt v(t) 2L2 (Rd ) dt ≤ C(1 + kT 2 )

T

T 0

Pv(t) 2L2 (Rd ) dt.

c) Set I = 0 ek(t −T ) P u(t) 2L2 (Rd ) dt. Show that there exists C > 0 such that for all u ∈ ET , 

T

k 0



T 0

2

ek(t −T ) u(t) 2L2 (Rd ) dt ≤ CI,

ek(t −T )

2

2

  u(t) 2H 1 (Rd ) + Dt u(t) 2L2 (Rd ) dt ≤ C(1 + kT 2 )I.

6 Statements of the Problems of Chap. 5

89

1

9. Let a = 0, b(ξ ) = (1 + |ξ |2 ) 2 , B = Op(b). a) What is the operator Q = (Dt − iB)(Dt + iB)? b) Using the estimates proved in question 8b) show that there exist positive constants C, T0 , k0 such that for all T ≤ T0 , k ≥ k0 and all u ∈ ET we have, 

T 0

ek(t −T )

2

  u(t) 2H 1 (Rd ) + Dt u(t) 2L2 (Rd ) dt 1 ≤ C( + T 2 ) k

where Q = ∂t2 +

d

2 j =1 ∂xj



T 0

ek(t −T ) Qu(t) 2L2 (Rd ) dt, 2

is the Laplacian.

10. Deduce that we have the same estimate as in question 9b) for the operator L = Q + c0 (t, x)∂t +

d 

cj (t, x)∂xj + d(t, x),

j =1

where c0 , cj , d belong to L∞ ((0, T ) × Rd ). Comments Carleman inequalities are L2 estimates with weights of the form ekϕ . They have been introduced by T. Carleman in 1939 (see his article [8]). These inequalities are a basic tool for many subjects in PDE, like uniqueness in the Cauchy problem, control theory, inverse problems. A comprehensive treatment of this tool can be found in the book [18] by N. Lerner.  

Problem 28 Smoothing for the Heat Equation

This problem uses Problem 20. Let T > 0 and a ∈ S 1 be a real symbol satisfying, ∃c > 0 : a(x, ξ ) ≥ c ξ 

∀(x, ξ ) ∈ Rd × Rd ,

1

ξ  = (1 + |ξ |2 ) 2 .

The goal of this problem is to prove that if u ∈ C 0 ([0, T ], L2 (Rd )) is a solution of the problem, ∂t u + Op(a)u = f ∈ C 0 ([0, T ], L2 (Rd )),

u(0) = 0,

then for all ε > 0 the function u(T ) = u|t =T belongs to H 1−ε (Rd ).

(6.5)

90

6 Statements of the Problems of Chap. 5

Let σ > 0. We recall (see Problem 20) that for all α, β ∈ Nd we have, (∂xα ∂ξ e−σ a )(x, ξ ) = β

|α|+|β| 

σ j dj αβ (x, ξ )e−σ a ,

dj αβ ∈ S j −|β| .

j =0

1. For t ∈ [0, T ) set et (x, ξ ) = e(t −T )a(x,ξ ). Show that for all m ∈ [0, +∞[ and all α, β ∈ Nd there exists Cm,αβ > 0 such that, β

|∂xα ∂ξ et (x, ξ )| ≤

Cm,αβ (1 + |ξ |)−m−|β| , (T − t)m

∀t ∈ [0, T ), ∀(x, ξ ) ∈ Rd × Rd .

2. Deduce that for all ε > 0 there exists K > 0 such that for all v ∈ L2 (Rd ) and all t ∈ [0, T ) we have, Op(et )v H 1−ε (Rd ) ≤

K v L2 (Rd ) , (T − t)1−ε

(Op(∂t et ) − Op(et ) Op(a)) v H 1−ε (Rd ) ≤

K v L2 (Rd ) . (T − t)1−ε

3. Let u be the solution of (6.5). Show that  u(T ) =

T



T

(Op(∂t et ) − Op(et ) Op(a)) u(t) dt +

0

Op(et )f (t) dt. 0

4. Conclude.

Problem 29 A Characterization of Ellipticity

Part 1 We introduce the space   1 ϕ (η) = 0 if |η| ≥ 1 . E = ϕ ∈ S(Rd ) :  ϕ ∈ C0∞ (Rd ),  ϕ (η) = 1 if |η| ≤ ,  2 Let m > 0. We consider a family (rλ )λ≥1 of C ∞ functions on Rd × Rd such that: there exists λ0 ≥ 1 such that for all α ∈ Nd there exists Cα > 0 such that, 1

|Dηα rλ (y, η)| ≤ Cα λm− 2 (1 + |y|),

∀y ∈ Rd ,

∀|η| ≤ 1,

∀λ ≥ λ0 .

(6.6)

6 Statements of the Problems of Chap. 5

91

We consider Rλ = Op(rλ ) the DO with symbol rλ defined on E. Show that for any ϕ ∈ E there exists λ0 > 0, C(ϕ) > 0 such that, 1

Rλ ϕ L2 (Rd ) ≤ C(ϕ)λm− 2 ,

∀λ ≥ λ0 .

Hint: Prove a pointwise estimate of yjN (Rλ ϕ)(y) for all N ∈ N. Part 2  Let m > 0 and p ∈ S m be a classical symbol, which means that p ∼ j ≥0 pm−j where pm−j is homogeneous of degree m − j . Assume in addition that P = Op(p) satisfies the following condition: there exist two constants 0 ≤ δ < 12 and C > 0 such that

u H m−δ (Rd ) ≤ C P u L2 (Rd ) + u L2 (Rd ) ,

∀u ∈ C0∞ (Rd ).

(6.7)

The goal of this part is to prove that P is elliptic, which means that ∃c > 0 :

|pm (x, ξ )| ≥ c|ξ |m ,

∀x ∈ Rd , ∀ξ = 0.

Let x 0 ∈ Rd , ξ 0 ∈ Sd−1 and let ϕ ∈ E be such that ϕ L2 (Rd ) = 1. For λ ≥ 1 we set, d

0

1

uλ (x) = λ 4 eiλx·ξ ϕ(λ 2 (x − x 0 )).

1. Compute the Fourier transform of uλ in terms of that of ϕ. 2. Deduce that there exists c0 > 0 such that uλ H m−δ (Rd ) ≥ c0 λm−δ . 3. Show that, 

 1 d 0 e−iλx·ξ P uλ (x) = λ 4 Qλ (y, Dy )ϕ λ 2 (x − x 0 ) , where Qλ has symbol   1 1 qλ (y, η) = p x 0 + λ− 2 y, λξ 0 + λ 2 η . 4. Show that qλ (y, η) = p(x 0 , λξ 0 ) + rλ (y, η), where rλ satisfies the estimates (6.6). Hint: Use Taylor formula with integral reminder at the order 1.

92

6 Statements of the Problems of Chap. 5

5. Using the inequality (6.7) and the result of Part I show that, ∃c1 > 0, ∃λ0 ≥ 1 : |p(x 0 , λξ 0 )| ≥ c1 λm−δ ,

∀λ ≥ λ0 .

6. Using the fact that, p being classical, p − pm ∈ S m−1 deduce that, ∃c2 > 0, ∃λ1 ≥ 1 : |pm (x 0 , λξ 0 )| ≥ c2 λm−δ ,

∀λ ≥ λ1 .

7. Show eventually that, ∃c3 > 0 : |pm (x, ξ )| ≥ c3 |ξ |m ,

∀x ∈ Rd , ∀ξ = 0.

Comments 1. The converse statement of this problem is true. If P is elliptic in the sense given in the beginning of Part 2 then the estimate (6.7) is true with δ = 0 (therefore with all 0 ≤ δ < 12 ). This is proved for instance in [1, 3, 30]. 2. The result of the problem is false if δ = 12 since then P is no more necessarily elliptic. There are characterizations of the operators satisfying (6.7) with 12 ≤ δ < 1, 1 ≤ δ < 32 and so on. These operators are called “subelliptic.” Their characterization is more difficult and can be found in Chapter XXVII of the book by L. Hörmander [15].  

Problem 30 Local Solvability of Real Vector Fields

Part 1 We consider, in a neighborhood of zero in Rd , d ≥ 2 in which the coordinates are denoted by (t, x), t ∈ R, x ∈ Rd−1 , a differential operator of order one of the form,  ∂ ∂ + aj (t, x) , ∂t ∂xj d

X=

j =2

where the coefficients aj are C 1 real valued functions. 1. We consider the differential system, t˙(s) = 1,

t (0) = 0,

x˙j (s) = aj (t (s), x(s)),

xj (0) = yj ,

2 ≤ j ≤ d.

Prove that for all y = (y2 , . . . , yd ) in a neighborhood of zero in Rd−1 this system has a unique solution, denoted by ϕ(s, y) = (s, x2 (s, y), . . . , xd (s, y)), which is a C 1 function of all its arguments.

6 Statements of the Problems of Chap. 5

93

2. a) Prove that the Jacobian of ϕ does not vanish at (0, 0). b) Deduce that there exists open neighborhoods U of (t, x) = (0, 0) and V of (s, y) = (0, 0) such that ϕ is a diffeomorphism from V to U. c) For u ∈ C 1 (U ) we set  u(s, y) = u(ϕ(s, y)). Show that, ∂ u (s, y) = (Xu)(ϕ(s, y)). ∂s 3. Let b and f be two C 1 functions in U with values in C and let u0 be a C 1 function in a neighborhood of x = 0. Show that the problem Xu + bu = f,

u|t =0 = u0 (x)

has a unique C 1 solution u in U . Part 2 Let x 0 ∈ Rd and let V be a neighborhood of x 0 . Let bj ∈ C 1 (V , R), 1 ≤ j ≤ d and  c, g ∈ C 1 (V , C). We assume that dj=1 |bj (x 0 )| = 0 and set Y =

d  j =1

bj (x)

∂ . ∂xj

For j0 ∈ {1, . . . , d} we set,   V 0 = (x1 , . . . , xj0 −1 , xj0 +1 , . . . , xd ) : (x1 , . . . , xj0 −1 , xj00 , xj0 +1 , . . . , xd ) ∈ V .

1. Prove that there exists j0 ∈ {1, . . . , d} such that for all u0 ∈ C 1 (Vx0 ) the problem, Y u + cu = g,

u|{xj

0

=xj0 } 0

= u0 (x1 , . . . , xj0 −1 , xj0 +1 , . . . , xd )

has a unique C 1 solution u in a neighborhood W ⊂ V of x 0 .

Problem 31 A Necessary Condition for Hypoellipticity

Let  be an open subset of Rd and let, P (x, D) =

d  j,k=1

aj k (x)Dj Dk +

d  k=1

bk (x)Dk + c(x),

Dj =

1 ∂ i ∂xj

(6.8)

94

6 Statements of the Problems of Chap. 5

be a second order differential operator with C ∞ coefficients in  such that for all x ∈  the matrix A(x) = (aj k (x)) is symmetric with real coefficients. Let p(x, ξ ) =

d 

aj k (x)ξj ξk = A(x)ξ, ξ 

j,k=1

be the principal symbol of P . We assume that, (H ) ∃(x , ξ ) ∈  × R \ {0} : 0

0

d

p(x , ξ ) = 0, 0

0

 d    ∂p 0 0     ∂ξ (x , ξ ) = 0. =1



We will denote by Vx 0 the set of open neighborhoods of x 0 in . We will admit that (H ) implies, ( )

∃V ∈ Vx 0 ,

∃ϕ ∈ C ∞ (V ) : p(x, gradϕ(x))= 0, ∀x ∈ V ,

gradϕ(x 0 )= ξ 0 .

Part 1 Our goal is to prove, by contradiction, that P is not hypoelliptic in . 1. Give an example of operator on the form (6.8) satisfying (H ). We shall assume in what follows that P is hypoelliptic.   0 (V ) : P u ∈ C ∞ (V ) Let V ∈ Vx 0 . We consider the space E = u ∈ C  endowed with the semi-norms pK,N (u) = supK |u| + |α|=N supK |D α P u| where K is a compact subset of V and N ∈ N. 2. Using the closed graph theorem prove that there exists a compact K ⊂ V , N0 ∈ N and C > 0 such that, ⎛ ⎞  sup |D α P u|⎠ , ∀u ∈ C ∞ (V ). (6.9) |grad u(x 0 )| ≤ C ⎝sup |u| + K

|α|≤N0 K

3. Let ϕ ∈ C ∞ (V ) be given by ( ) and w ∈ C ∞ (V ). Show that, P (eiλϕ w) = eiλϕ (−iλLw + P w) where L=

d  ∂p ∂ (x, gradϕ(x)) + b, b ∈ C ∞ (V ). ∂ξk ∂xk k=1

6 Statements of the Problems of Chap. 5

95

4. a) Prove that for all α ∈ Nd and f ∈ C ∞ (V ) there exist cαk ∈ C ∞ (V ), 0 ≤ k ≤ |α| such that, α

∂ (e

iλϕ

f) =

|α| 

cαk λk eiλϕ .

k=0

b) Deduce that if K ⊂ V is compact and α ∈ Nd there exists CK,α > 0 such that, sup |∂ α (eiλϕ f )| ≤ CK,α λ|α| ,

∀λ ≥ 1.

K

5. From (H ) and ( ) there exists 0 such that ∂ξ∂p (x 0 , gradϕ(x 0 )) = 0. Let N0 be 0 defined in question 2. Show that the problems, Lw0 = 0, w0 |x

0

=x0

0

= 1, Lwj +1 = −iP wj , wj +1 |x

0

=x0

= 0, 0 ≤ j ≤ N0 − 1

0

have C ∞ solutions w0 , w1 , . . . , wN0 in V1 ∈ Vx 0 , V1 ⊂ V . Hint:  Use Problem 30. 0 −j iλϕ w , 6. Set v = N λ ≥ 1. j j =0 λ e a) Show that P v = λ−N0 eiλϕ P wN0 . b) Deduce that if K ⊂ V1 is compact there exists CK > 0 such that sup |∂ α P v| ≤ CK λ|α|−N0 K

and

sup |v| ≤ CK . K

c) Prove that there exists C > 0 such that |gradv(x0 )| ≥ Cλ for λ large enough and deduce a contradiction. Part 2 Let P be a differential operator of the form (6.8) and x 0 ∈ . Assume that there exists ξ 1 , ξ 2 ∈ Rd \ {0} such that,   A(x 0 )ξ 1 , ξ 1 > 0,



 A(x 0 )ξ 2 , ξ 2 < 0.

(6.10)

The goal is to prove that there exists ξ 0 ∈ Rd \ {0} such that (H ) is satisfied. 1. Show that the matrix A(x 0 ) has d real eigenvalues λ1 , . . . , λd , not all vanishing and that it has at least one strictly positive eigenvalue λj0 and at least one strictly negative eigenvalue λj1 . 2. Set D = diag(λj ) and q(η) = Dη, η . Show that there exists η0 ∈ Rd \ {0}   ∂q 0  such that q(η0) = 0 and d=1  ∂η (η ) = 0. 

96

6 Statements of the Problems of Chap. 5

3. Deduce that there exists ξ 0 ∈ Rd \ {0} such that, 0

0

p(x , ξ ) = 0 and

 d    ∂p 0 0     ∂ξ (x , ξ ) = 0. =1



4. State the conclusion of Parts 1 and 2 as a theorem. Comments The result proved in this problem appears in the paper [13] by L. Hörmander. It was the motivation for this author to study the hypoellipticity of second order operators whose principal symbol does not change sign, namely the famous  operators “sum of squares”: P = rj =1 Xj2 + X0 + c.   Problem 32 Wave Front Set of an Indicator Function Let d ≥ 2. If x = (x1 , . . . , xd ) ∈ Rd we set x = (x1 , . . . , xd−1 ), so x = (x , xd ). We consider the function u ∈ L∞ (Rd ) defined by, u(x) = 1 if xd > 0,

u(x) = 0 if xd < 0.

  1. Show that WF(u) ⊂ (x, ξ ) ∈ Rd × (Rd \ {0}) : xd = 0 . 2. a) Let ξ0 ∈ Rd \ {0} be such that ξ0 = 0. Show that there exist ε > 0 and c0 > 0 such that,     ξ ξ0  d  − ∀ξ ∈ ξ0 =: ξ ∈ R \ {0} :  0 such that, 

  d k Ck 1   , ≤   dη 1 + η2  (1 + η)k+2

∀η ≥ 0.

2. a) Prove that for all N ∈ N∗ there exist aj,N ∈ C, 0 ≤ j ≤ N − 1 such that for x = 0 we have u(x) =

N−1  j =0

aj,N i N−1 + xj x N−1





+∞

e

ixη

0

d dη

N−1

1 1 + η2

dη.

b) Deduce that W F (u) ⊂ {(0, ξ ) : ξ ∈ R \ {0}}. 3. Let ϕ ∈ C0∞ (R) be such that ϕ(x) = 1 if |x| ≤ 12 and ϕ(x) = 0 if |x| ≥ 1.  +∞ 1 a) Show that ϕ , u(ξ ) = 0  ϕ (ξ − η) 1+η 2 dη. b) Show that ϕ , u(ξ ) = O(|ξ |−N ), W F (u) ⊂ {(0, ξ ) : ξ > 0} .

∀N ∈ N,

ξ → −∞. Deduce that

4. Let ψ ∈ C0∞ (R) with values in (0 + ∞) be even and such that Set ϕ = ψ ψ.



ψ(y)2 dy = 1.

a) Prove that ϕ(0) = 1, ϕ (0) = 0,  ϕ ≥ 0,  ϕ (0)  > 0. 1 b) Using question 3a) and noticing that ϕ , u(ξ ) ≥ A  ϕ (ξ − η) 1+η 2 dη where A = {η ∈ (0, +∞) : |ξ − η| ≤ ε} , ε > 0 small to be chosen, show that there , u(ξ ) ≥ C for any ξ > 0 large enough. exists C > 0 such that ξ 2 ϕ c) Let ϕ1 ∈ C0∞ (R) be such that ϕ1 = 1 in a neighborhood of ϕ. Set θ = ϕ − ϕ1 . Show that θ (x) = x 2 θ1 (x) where θ1 ∈ C0∞ (R). Using questions 3a) and 1b) show that, |θ, u(ξ )| ≤ |( θ1 ) (ξ )| + C



+∞ 0

|θ1 (ξ − η)|

1 dη. (1 + η)4

98

6 Statements of the Problems of Chap. 5

Deduce that for ε > 0 small enough there exists C > 0 such that, ξ 3−ε |θ, u(ξ )| ≤ C,

∀ξ ≥ 1.

d) Deduce that there exists C > 0 such that ξ 2 ϕ 1 u(ξ ) ≥ C for ξ > 0 large enough, and then that W F (u) = {(0, ξ ) : ξ > 0} .

Problem 34 A Problem on the Wave Front Set Let u ∈ D (R2 ) be such that x1 u ∈ C ∞ (R2 ), x2 u ∈ C ∞ (R2 ) but u ∈ / C ∞ (R2 ). We want to determine the wave front set of u.   1. Show that W F (u) ⊂ (x, ξ ) ∈ R2 × R2 \ {0} : x = 0 . / 2. We assume in what follows that there exists ξ0 ∈ R2 \ {0} such that (0, ξ0 ) ∈ W F (u). We set ξ0 = λω0 , λ = 0, ω0 ∈ S1 . Let ϕ ∈ C0∞ (R2 ) be equal to 1 near zero. a) Show that

∂ ∂ξj

(, ϕ u) ∈ S(R2 ), j = 1, 2.

b) Let ξ = λω, λ > 0, ω ∈ S1 be such that the angle θ = (ω, ω0 ) is strictly less than π. Show that there exists c > 0 such that |tω + (1 − t)ω0 | ≥ c. Hint: Compute the minimum of the function t → |tω + (1 − t)ω0 |2 from [0, 1] to R. c) Applying the Taylor formula at the order one to the function ϕ(λω ) where ω is close to ω, show that (0, ξ ) ∈ / W F (u). d) Deduce that u is C ∞ in a neighborhood of zero and determine W F (u). 3. Give an alternative proof of this result using the theorem of propagation of singularities for the operator P = x1 +λx2 ( or P = μx1 +x2 ) for an appropriate λ (or μ) in R.

Problem 35 On Propagation of Singularities

This problem is an application of the theorem on propagation of singularities for a concrete operator. We denote by P the differential operator in R2 given by, P =x

∂2 ∂2 − ∂y 2 ∂x 2

  and let p(x, y, ξ, η) = ξ 2 −xη2 be its symbol. We set H = (x, y) ∈ R2 ) : x < 0 .

6 Statements of the Problems of Chap. 5

99

1. Let u ∈ D (R2 ) be such that P u ∈ C ∞ (R2 ). Prove that u ∈ C ∞ (H ). 2. Give the equations of the bicharacteristics of p. Solve these equations. ∞ 2 3. Let v ∈ D (R2 ) be such that if v is C ∞ in a  that P v2 ∈ C (R  ). Prove ∞ neighborhood of the set (x, y) ∈ R : x = 0 then v ∈ C (R2 ).

Problem 36 A Microlocal Defect Measure For ξ0 ∈ Sd−1 and k ≥ 1 we set uk (x) = b(x)eikx·ξ0 where b ∈ C0∞ (Rd ). 1. Prove that the sequence (uk )k∈N∗ is bounded in L2loc (Rd ) and converges to zero weakly in the sense of (5.13). The purpose of this problem is to show that this sequence has a defect measure μ given by μ = |b(x)|2dx ⊗ δ ξ0 . |ξ0 |

Let A be a pseudo-differential operator whose symbol a has compact support in x and is homogeneous of degree zero for |ξ | ≥ 1. Let χ be in C0∞ (Rd , R) equal to one on the support in x of the symbol of A. Set K = supp χ and, Ik = (A(χuk ), χuk )L2 (Rd ) . 2. Prove that we have,  Ik =

Rd ×Rd

fk (x, η) dx dη,

where,

, fk (x, η) = (2π)−d eix·η a(x, η + kξ0 )χb(η)χ(x)b(x). We set  Jk =

 |η|≥ 12 k

fk (x, η) dx dη,

Hk =

|η|≤ 12 k

fk (x, η) dx dη.

3. a) Prove that limk→+∞ Jk = 0. b) Prove that when k ≥ 2 and |η| ≤ 12 k we have,  η 1 a0 (x, η + kξ0 ) = a0 x, ξ0 + = a(x, ξ0 ) + ck (x, η), k k where ck is such that |ck (x, η)| ≤ C|η| where C is independent of k. c) Prove that when k → +∞ we have,  −d , Hk = (2π) eix·η a(x, ξ0 )χb(η)χ(x)b(x) dη dx + o(1). |η|≤ 21 k

100

6 Statements of the Problems of Chap. 5

d) Prove that when k → +∞ we have,  −d , (2π) eix·η a(x, ξ0 )χb(η)χ(x)b(x) dη dx |η|≤ 12 k

 =

Rd

a(x, ξ0 )|b(x)|2 dx + o(1).

e) Conclude. 4. The goal of this question is to prove that the same result holds for uk (x) = b(x)eikx·ξ0 where b ∈ L2loc (Rd ). Let χ1 ∈ C0∞ (Rd ) be equal to one on a neighborhood of supp χ. We set  b = χ1 b ∈ L2 (Rd ). There exists (bn )n∈N ⊂ ∞ d 2 C0 (R ) converging to  b in L (Rd ). Set,   beikx·ξ0 ), χ  beikx·ξ0 Ik = A(χ(

L2

 −

Rd

| b(x)|2 a(x, ξ0 )|b(x)|2 dx

and write Ik = (1) + (2) + (3) + (4) where,   (1) = A(χ( b − bn )eikx·ξ0 ), χ  beikx·ξ0 2 , L   (2) = A(χbn eikx·ξ0 ), χ( b − bn )eikx·ξ0 ) 2 , L    |bn (x)|2 a(x, ξ0 ) dx, (3) = A(χbn eikx·ξ0 ), χbn eikx·ξ0 2 − L

 (4) = −

Rd

| b(x)|2 a(x, ξ0 )|b(x)|2 dx +

Rd



Rd

|bn (x)|2 a(x, ξ0 ) dx.

a) Prove that there exists a constant C0 > 0 independent of k, n such that, |(1)| + |(2)| + |(4)| ≤ C( bn L2 +  b L2 ) bn −  b L2 . b) Prove that limk→+∞ Ik = 0 and conclude.

Chapter 7

The Classical Equations

In this chapter, we review some of the main properties of the solutions to the classical partial differential equations. This includes results about equations with analytic coefficients, the Laplace equation, the wave equation, the heat equation, the Schrödinger equation, the Burgers equation, and also the Euler and Navier–Stokes equations in fluid dynamics.

7.1 Equations with Analytic Coefficients First of all let us recall some definitions concerning the concept of analyticity. • Let  ⊂ Rd be open. A function f :  → C is said to be real analytic in  if every point x0 ∈  has a neighborhood Vx0 in which one can write, f (x) =



aα (x − x0 )α

(aα ∈ C),

α∈Nd

where the series is absolutely convergent. • In Cd denote by z = (z1 , . . . , zd ) the variable, where zj = xj + iyj , xj , yj ∈ R. Let O ⊂ Cd be open. A function f : O → C is said to be holomorphic if it is C 1 with respect to the real variables (xj , yj ), 1 ≤ j ≤ d and satisfies, ∂j f = 0

in O

for j = 1, . . . , d,

where ∂ j =

1 2



∂ ∂ +i ∂xj ∂yj

.

• There is of course a tight link between these two notions. If f is real analytic in x0 of x0 in Cd and a neighborhood Vx0 of x0 in Rd one can find a neighborhood V

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 T. Alazard, C. Zuily, Tools and Problems in Partial Differential Equations, Universitext, https://doi.org/10.1007/978-3-030-50284-3_7

101

102

7 The Classical Equations

x0 such that f = f on Vx0 . On the other hand, if a holomorphic function f in V x0 the restriction of fto V x0 ∩ Rd is real analytic. f is holomorphic in V 1 • If ϕ is a C real valued function in a neighborhood Vx0 of x0 ∈ Rd we set   S = x ∈ Vx0 : ϕ(x) = ϕ(x0 ) ,

(7.1)

and we assume that dϕ(x) = 0 on S, where dϕ is the differential of ϕ. Then S is a C 1 hypersurface. ∂ We shall denote by n the unit normal to S and by ∂n the normal derivative.

7.1.1 The Cauchy–Kovalevski Theorem The Cauchy–Kovalevski theorem concerns the question of existence of local solutions to the Cauchy problem for partial differential equations in the framework of real analytic or holomorphic functions. We assume in this paragraph that the function ϕ appearing in (7.1) is real analytic.

7.1.1.1 The Linear Version  • Let m ∈ N, m ≥ 1 and let P = |α|≤m aα (x)D α be a linear differential operator in of x0 with real analytic coefficients and principal symbol pm = a neighborhood α . Assume that S is non-characteristic for P that is, a (x)ξ |α|=m α pm (x, dϕ(x)) = 0 ,

∀x ∈ S.

(7.2)

Then for every real analytic functions u0 , . . . , um−1 on S and every real analytic function f near x0 the problem, P u = f,

∂ ∂n

j ⏐ ⏐ u⏐ ⏐ = uj ,

0 ≤ j ≤ m − 1,

S

has a unique real analytic solution u in a neighborhood of x0 . • Notice that if we make a real analytic change of coordinates from x to (t, y) ∈ R × Rd−1 in which x0 becomes (0, y0 ) and S becomes {t = 0} (which is always possible locally) then condition (7.2) ensures that the coefficient of Dtm in the operator P in the new coordinates does not vanish at (0, y0 ).

7.1.1.2 The Nonlinear Version   Let m ∈ N, m ≥ 1. We set N = " α ∈ Nd : |α| ≤ m and p = (pα )|α|≤m ∈ RN .

7.1 Equations with Analytic Coefficients

103

Then we have the following result. • Let p0 ∈ RN and let F : Vx0 × RN → R be a real analytic function in a neighborhood of (x0 , p0 ) in Rd × RN . Assume that, (i)

F (x0 , p0 ) = 0,

(ii)

 ∂F (x0 , p0 )(dϕ(x0 ))α = 0. ∂pα

|α|=m

u(x0 ) = pα0 Let  u be any real valued real analytic function near x0 such that ∂ α for |α| ≤ m. Then the Cauchy problem, F (x, (∂ u(x)) = 0, α

∂ ∂n

j ⏐ j ⏐ ⏐ ⏐ ∂ ⏐ u⏐ =  u⏐ ⏐ , ∂n S S

0 ≤ j ≤ m − 1,

has a unique real valued real analytic solution u near x0 . Comments i) If we want to consider non-real solutions one has to assume holomorphy for F and then the same result holds where real analytic is replaced by holomorphic. ii) Nevertheless if F is a polynomial with respect to the variable p then the above result still holds in the category of real analytic functions even for non-real data. iii) A short abstract proof of the nonlinear Cauchy-Kovalevski theorem is given by T. Nishida in [23].  

7.1.2 The Holmgren Uniqueness Theorem We assume in this paragraph that the function ϕ appearing in (7.1) is C 1 .  α • Let P = |α|≤m aα (x)D , m ≥ 1, be a linear differential operator with real analytic coefficients in Vx0 . Assume that, pm (x0 , dϕ(x0 )) = 0, where pm is the principal symbol of P . Then there exists a neighborhood Wx0 ⊂ Vx0 of x0 such that if u ∈ D (Vx0 ) satisfies, P u = 0 in Vx0 , then u = 0 in Wx0 .

u = 0 when ϕ(x) < ϕ(x0 ),

104

7 The Classical Equations

• The above result implies the same conclusion when ϕ ∈ C m and u is a C m solution of the problem, P u = 0 in Vx0 ,

∂ ∂n

j ⏐ ⏐ u⏐ ⏐ = 0,

0 ≤ j ≤ m − 1.

S

Comments A proof of the Holmgren Theorem can be found in Chapter 3 of the book by F. John [16].  

7.2 The Laplace Equation The Laplace operator (or the “Laplacian”) is given by =

d  ∂2 . ∂xj2 j =1

7.2.1 The Mean Value Property Recall that a harmonic function is a solution of u = 0. Let  be an open subset of Rd and u ∈ C 2 (). Then u is harmonic if and only if u satisfies the mean value property: for any x ∈  and any r > 0 such that B(x, r) ⊂ ,  1 u(y) dy. u(x) = |B(x, r)| B(x,r) where |B| denotes the Lebesgue measure of the set B.

7.2.2 Hypoellipticity: Analytic Hypoellipticity Recall that a fundamental solution of a constant coefficient differential operator P is a distribution E ∈ D (Rd ) satisfying P E = δ0 . • A fundamental solution of the Laplacian is given by, E=

1 Log |x| if d = 2, 2π

E=

|x|2−d |Sd−1 |(2 − d)

if d ≥ 3.

7.2 The Laplace Equation

105

• The Laplacian is hypoelliptic (resp. analytic hypoelliptic). This means the following. Let u be a distribution in an open set  ⊂ Rd . For every open set ω ⊂  if u belongs to C ∞ (ω) (resp. is real analytic in ω) then u belongs to C ∞ (ω) (resp. is real analytic in ω). These properties are shared by all elliptic operators with C ∞ (resp. real analytic) coefficients.

7.2.3 The Maximum Principles • The weak maximum principle. Let  be a bounded open subset of Rd . Let u be a real valued function belonging to C 0 () ∩ C 2 () such that u(x) ≥ 0 (resp. u(x) ≤ 0) for all x ∈ . Then for all x ∈  we have, u(x) ≤ sup u(y),

(resp. u(x) ≥ inf u(y)). y∈∂

y∈∂

• The strong maximum principle. Let  be a bounded and connected open subset of Rd . Let u be a real valued function belonging to C 0 () ∩ C 2 () such that u(x) ≥ 0 (resp. u(x) ≤ 0) for all x ∈ . Then if u reaches its maximum (resp. its minimum) at a point x ∈  then u is constant.   • The Hopf Lemma. Let x0 ∈ Rd and R > 0. Set B = x ∈ Rd : |x − x0 | < R . ∂ Denote by ∂B its boundary, by n the unitary normal to the boundary and by ∂n the normal derivative. Let u be a C 2 function in a neighborhood of B such that u(x) ≥ 0, ∀x ∈ B. Assume that there exists a point y ∈ ∂B such that u(y) > u(x) for all x ∈ B. Then ∂u (y) > 0. ∂n

7.2.4 The Harnack Inequality • Let  be an open subset of Rd and let u be a nonnegative C 2 () solution of u = 0. Then for any bounded open subset O ⊂  there exists C > 0 depending only on d, O,  such that, sup u ≤ C inf u. O

O

106

7 The Classical Equations

7.2.5 The Dirichlet Problem A relevant problem for the Laplacian is the Dirichlet problem. Roughly speaking the question is the following. Given an open subset  in Rd , whose boundary is denoted by , given a distribution f on  and u0 defined on , solve the problem, u = f in ,

u = g on .

(7.3)

Of course the spaces where f and u0 are to be taken should be specified. Here are some classical results. 7.2.5.1 Case g = 0 • Let λ ≥ 0 and let  be an open subset of Rd (λ > 0 if  is not bounded). Then the operator P = − + λ is a linear isomorphism from H01 () on H −1 (). • Assume that  is a C ∞ open set and let λ ≥ 0 (λ > 0 if  is not bounded). For any k ∈ N, the operator − + λ is a linear isomorphism from H k+2 () ∩ H01 () on H k (). • Under the above conditions, − + λ is a linear isomorphism from ∩ H k () ∩ k∈N

H01 () to ∩ H k (). k∈N

• If  is a C ∞ bounded open set we have ∩ H k () = C ∞ (). Otherwise we have C0∞ () ⊂ ∩ H k () ⊂ C ∞ ().

k∈N

k∈N

7.2.5.2 Case g ≡ 0. • Let  be a C 1 open subset of Rd . Let λ ≥ 0 (λ > 0 if  is not bounded). Let 1 f ∈ H −1 (), g ∈ H 2 (∂). There exists a unique u ∈ H 1 () such that, (− + λ) u = f in D (),

γ0 u = g on .

7.2.6 Spectral Theory • Let  be a bounded open subset of Rd . The problem − u = λu,

u ∈ H01 ()

(7.4)

7.2 The Laplace Equation

107

is solvable with u = 0 if and only if λ belongs to a sequence (λj )j ∈N∗ of strictly positive real numbers (called eigenvalues) which tends to +∞ when j → +∞. We can order them as, 0 < λ1 ≤ λ2 ≤ · · · ≤ λj ≤ · · · where each eigenvalue is repeated according to its finite multiplicity. • There exists an orthonormal basis (ej )j ≥1 of L2 () such that, − ej = λj ej ,

ej ∈ H01 ().

(7.5)

The ej s are called eigenfunctions. The eigenspace corresponding to each eigenvalue is finite dimensional. By the hypoellipticity of the Laplacian we have ej ∈ C ∞ () and, if  has a ∞ C boundary, we have ej ∈ C ∞ ().  • (ej ) is also an orthogonal basis of H01 () and we have ej H 1 () = λj . More k

0

generally, for every k ∈ N we have ej H k () ≤ Cλj2 . 7.2.6.1 Weyl Law Let  be a bounded open subset of Rd with C ∞ boundary. For λ > 0 set,   N(λ) = # j ≥ 1 : λj ≤ λ . Then we have, d

N(λ) ∼ (2π)d Cd vol ()λ 2 ,

λ → +∞,

(7.6)

where vol () is the Lebesgue measure of  and Cd is the Lebesgue measure of the unit ball in Rd . • As a corollary we have, λj ∼

(2π)2 [Cd vol ()]

2

2 d

jd,

j → +∞.

7.2.6.2 Estimates of the Eigenfunctions • Let (ej )j ∈N∗ be an orthonormal basis of L2 () made of eigenfunctions. Then by the estimates of the H k norms of ej and the Sobolev embedding one can see d

that ej L∞ () ≤ Cλj4

+ε

, ε > 0. However, this is not the best result. Indeed we

108

7 The Classical Equations

have the following better bound: ∃C > 0 : ∀j ≥ 1,

d−1

ej L∞ () ≤ Cλj 4 ,

(7.7)

and these estimates are optimal in general. (See Problem 39 for the case where  is the unit a ball in R3 and Problem 40 for the case of the sphere.) • If we call Ej the eigenspace corresponding to the eigenvalue λj then there exists C > 0 independent of j such that, d−1

dim Ej ≤ Cλj 2 . (See Problem 39.) Comments A comprehensive introduction to the study of the Laplace equation can be found in Chapters 1–9 of D. Gilbarg and N.S. Trudinger [11] and in L.C. Evans [9]. The spectral theory for the Laplace operator, including a proof of the Weyl law, can be found in Chapter 13 of the book by C. Zuily [33]. More advanced results concerning the spectral theory can be found in the book by C.D. Sogge [27].  

7.3 The Heat Equation The heat operator in Rt ×Rdx is given by P =

−x where x is the Laplacian.

∂ ∂t

• A fundamental solution of this operator is given by the function, E(t, x) =

H (t) (4πt)

d 2

e−

|x|2 4t

,

where H (t) is the Heaviside function, H (t) = 1 if t > 0, H (t) = 0 if t ≤ 0. • The heat operator is hypoelliptic.

7.3.1 The Maximum Principle • Let  be an open bounded subset of Rdx and T > 0. Set Q =  × (0, T ) and denote by ∂Q the boundary of Q. If u ∈ C 2 (Q) ∩ C 0 (Q) is such that ∂u ∂t − x u ≤ 0 in Q then, sup u = sup u. Q

∂Q

7.4 The Wave Equation

109

7.3.2 The Cauchy Problem Roughly speaking, when  = Rd , the problem is the following. Given f defined in Q = (0, T ) × , u0 defined in  and g defined on (0, T ) × ∂ (belonging to some spaces) find u defined on Q satisfying, ∂u − x u = f, ∂t

u|t =0 = u0 ,

γ0 u(t, ·) = g(t, ·), t ∈ (0, T ),

(7.8)

where γ0 denotes the trace on the boundary of  (when  = Rd we remove this last condition on γ0 u). For instance if g = 0 we require u(t, ·) ∈ H01 (). Here is an example of result which holds. Let  be an open bounded subset of Rdx and T > 0. Set Q =  × (0, T ). • Let f ∈ L2 (Q), u0 ∈ L2 (), g = 0. Then the Cauchy problem (7.8) has a unique solution u in C 0 ([0, T ], L2 ()) ∩ L2 ((0, T ), H01 ()). • Assume  bounded and let (en )n∈N , en ∈ H01 (), be an orthonormal basis on L2 () consisting of eigenfunctions of the Laplacian, which means −en = λn en where λn > 0. Then the solution in the above result has the following form. Write,    un (t)en (x), u0 = u0n en (x), f (t, x) = fn (t)en (x). u(t, x) = n≥0

n≥0

n≥0

Then, un (t) = e−λn t u0n +



t

e−λn (t −s)fn (s) ds.

0

7.4 The Wave Equation  The wave operator on Rt × Rdx is given by  = ∂t2 − dj=1 ∂x 2 . j A relevant problem for this operator is the Cauchy problem.

7.4.1 Homogeneous Cauchy Problem Given u0 , u1 defined on Rd (in convenient spaces) we have to solve, u = 0,

u|t =0 = u0 ,

∂t u|t =0 = u1 .

(7.9)

110

7 The Classical Equations

• In any dimension the solution is given by the explicit formula,

√ √  sin t − u(t, x) = cos t − u0 + u1 . √ − • In dimension d = 3 the above formula simplifies to, u(t, x) =

t 4π

 S2

u1 (x − tω) dω +

1 4π

 S2

u0 (x − tω) dω

3  t  ∂u0 + (x − tω)ωi dω, 4π S2 ∂xi

(7.10)

i=1

where S2 is the unit sphere in R3 .

7.4.1.1 Properties of the Solution 7.4.1.2 Decay at Infinity • If u0 , u1 belong to C0∞ (Rd ) there exists C = C(u0 , u1 ) such that for t ≥ 1, sup |u(t, x)| ≤ x∈Rd

C t

d−1 2

.

• In dimension d = 3 we have the following more precise estimate, ⎞ ⎛ 3   1 ⎝   ∂u1  sup |u(t, x)| ≤ + ∂ α u0 L1 (R3 ) ⎠ ,   4πt ∂xi L1 (R3 ) x∈R3 i=1

t ≥ 1.

1≤|α|≤2

7.4.1.3 Finite Speed of Propagation • If u0 , u1 ∈ C0∞ (Rd ) are such that supp u0 and supp u1 are included in the set     x ∈ Rd , |x| ≤ R then for all t > 0, supp u(t, ·) ⊂ x ∈ Rd : |x| ≤ R + t .

7.4.1.4 Huygens Principle • Let d be odd and consider initial data u0 , u1 ∈ C0∞ (Rd ) such that supp u0   and supp u1 are contained in the set x ∈ Rd : |x| ≤ R . Then u vanishes in {(t, x) : t > R, |x| < t − R}.

7.4 The Wave Equation

111

7.4.1.5 Influence Domain • Let (t0 , x0 ) ∈ R+ × Rd . The value of the solution u at (t0 , x0 ) depends only on the values of the data u0 , u1 on the intersection of the hyperplane t = 0 with the backward cone C(t0 , x0 ) = {(t, x) : |x − x0 | < t0 − t, t < t0 }.

7.4.1.6 Conservation of the Energy • Let u0 , u1 ∈ C0∞ (Rd ) and let u be the solution of (7.9). For t ∈ R set, 1 2

E(t) = E(0) = where |∇x v|2 =

d



1 2

 Rd



Rd



 ∂v 2 i=1 ∂xi .



2  ∂u  2 (t, x) + |∇ u(t, x)| dx,   x ∂t   |u1 (x)|2 + |∇x u0 (x)|2 dx, Then for all t ∈ R we have E(t) = E(0).

The quantity E(t) is called the energy of the solution at time t; the above property says that the energy is conserved and therefore is equal to the energy of the data.

7.4.1.7 Strichartz Estimates • Let T > 0, s ≥ 0 and d ≥ 2. There exists C > 0 such that for every solution of (7.9) we have,   u Lq ((0,T ),Lr (Rd )) ≤ C u0 H˙ s (Rd ) + u1 H˙ s−1 (Rd ) ,

(7.11)

whenever the following conditions are satisfied, r < +∞,

d −1 2 d −1 + ≤ , q r 2

1 d d + = − s. q r 2

7.4.2 Inhomogeneous Cauchy Problem This is the problem u = f,

u|t =0 = u0 ,

∂t u|t =0 = u1 .

(7.12)

112

7 The Classical Equations

• In dimension d = 3 the solution is given by the formula, u(t, x) =

t 4π

 S2

u1 (x − tω) dω+

1 4π

 S2

u0 (x − tω) dω+

 t  3  t  ∂u0 1 + (x − tω)ωi dω + τ f (t−τ, x−τ ω) dω dτ. 4π 4π 0 S2 ∂xi S2 i=1

7.4.2.1 Finite Speed of Propagation   • If for j = 0, 1 we have uj ∈ C0∞ (Rd ) with supp uj ⊂ x ∈ Rd , |x| ≤ R and if   for t > 0 supp f (t, ·) ⊂ x ∈ Rd , |x| ≤ R + t then for all t > 0, supp u(t, ·) ⊂   x ∈ Rd : |x| ≤ R + t .

7.4.2.2 Strichartz Estimates • We also have an inhomogeneous Strichartz estimate where in the right-hand side of (7.11) we add, f La ((0,T ),Lb (Rd )) , where, b < +∞,

d −1 2 d −1 + ≤ , b a 2

1 d d + − 2 = − s. a b 2

Here a , b denote the conjugate exponents of a, b. Notice that the pair (a, b) is completely independent of (q, r) which appears in the left-hand side of (7.11).

7.4.3 The Mixed Problem When the variable x lies in a bounded open set  ⊂ Rd , then one of the relevant problem for the wave equation is the Cauchy–Dirichlet problem, also called a mixed problem. The homogeneous one can be stated as follows. Given 0 < T ≤ +∞, u0 , u1 defined on  and g defined on (−T , T ) × ∂, find u satisfying, u = 0 in (−T , T ) × ,

u|t =0 = u0 ,

∂t u|t =0 = u1 ,

u|(−T ,T )×∂ = g. (7.13)

7.5 The Schrödinger Equation

113

Here are some classical results on this problem. Set I = (−T , T ). • For all (u0 , u1 ) ∈ L2 () × H −1 (), g ∈ L2loc (I × ∂) the problem (7.13) has a unique solution u ∈ C 0 (I, L2 ()) ∩ C 1 (I, H −1 ()). • If (u0 , u1 ) ∈ H01 ()×L2 () and g = 0 the problem (7.13) has a unique solution u ∈ C 0 (I, H01 ()) ∩ C 1 (I, L2 ()). Comments For the basic results concerning the wave equation we refer to the books by S. Alinhac [2] and by F. John [16].  

7.5 The Schrödinger Equation The Schrödinger operator on Rt × Rdx is given by P = i∂t + x where x = d 2 j =1 ∂xj .

7.5.1 The Cauchy Problem 7.5.1.1 The Homogeneous Cauchy Problem We have to solve the problem i∂t u + x u = 0,

u|t =0 = u0 .

(7.14)

• If u0 ∈ S (Rd ) the problem (7.14) has a unique solution u ∈ S (R × Rd ). • If u0 ∈ S(Rd ) the solution u belongs to C ∞ (R, S(Rd )). It is given by the formula,  2 −d u(t, x) = (2π) u0 (ξ ) dξ . (7.15) eix·ξ −it |ξ |  We set u(t, ·) = S(t)u0 (·). • If u0 ∈ H s (Rd ), where s ∈ R, the solution u = S(t)u0 belongs to C 0 (R, H s (Rd )). For k ∈ N, (∂tk u) belongs to C 0 (R, H s−2k (Rd )) and, S(t)u0 H s = u0 H s

∀t ∈ R,

∂tk S(t)u0 H s−2k ≤ u0 H s

∀ t ∈ R , ∀ k ∈ N∗ .

(7.16)

114

7 The Classical Equations

7.5.2 Properties of the Solution 7.5.2.1 Expression of the Solution • For u0 ∈ S (Rd ) the unique solution u = S(t)u0 of the problem (7.14) can be written for t = 0, % 2 $   2 |·| · 1 −id π4 sgn t i |·|4t , (7.17) F ei 4t u0 S(t)u0 (·) = e e (4π|t|)d/2 2t where sgn t denotes the sign of t.

7.5.2.2 Infinite Speed of Propagation By this we mean the following fact: the smoothness of the solution at a time t = 0 does not depend on the smoothness of the initial data u0 but depends on its behavior at infinity. This can be made more precise but we shall give here only two examples describing this phenomena. • If u0 ∈ E (Rd ) and S(t)u0 is the solution of problem (7.14) then, for all t = 0, u(t, ·) ∈ C ∞ (Rd ). 2 • Let u0 (x) = e−i λ|x| , where λ > 0. Then d π 1 S u0 = (4πλ) 2 e−in 4 δ0 . 4λ Summarizing, an irregular datum but vanishing at infinity gives rise to a C ∞ solution for t = 0 while a perfectly smooth datum but oscillating at infinity gives rise to a singular solution.

7.5.2.3 Decay at Infinity of the Solution • If u0 ∈ L1 (Rd ), the solution S(t)u0 of the problem (7.14) satisfies S(t)u0 L∞ (Rd ) ≤

d g L1 |t|− 2 , for t = 0 . d/2 (4π)

7.5.2.4 Strichartz Inequality Let (q, r) be two real numbers. We say that (q, r) is admissible if, 2 d d = − q 2 r

with

2 ≤ q ≤ +∞ if d ≥ 3

and

2 < q ≤ +∞ if d = 2. (7.18)

7.5 The Schrödinger Equation

115

• Let (q, r) be an admissible couple. There exists C > 0 such that, S(t)u0 Lq (R,Lr (Rd )) ≤ C u0 L2 (Rd ) ,

∀u0 ∈ L2 (Rd ).

(7.19)

Remark (i) The above condition on (q, r) comes from the scale invariance of the Schrödinger equation. Indeed if u is a solution of (7.14) with datum u0 , then for all λ > 0 the function uλ (t, x) = u(λ2 t, λx) is also a solution of (7.14) with datum uλ0 (x) = u0 (λx). The estimate (7.19) should therefore be satisfied by uλ for all λ > 0 with the same constant C. This is possible only if (7.18) holds. (ii) If d = 2 the above inequality with q = 2, r = +∞ is false.  

7.5.2.5 The Inhomogeneous Problem We have to solve the problem i∂t u + x u = f,

u|t =0 = u0 .

(7.20)

The solution of this problem is given by u(t, ·) = S(t)u0 (·) +

1 i



t

S(t − s)[f (s, ·)] ds.

0

7.5.2.6 Nonhomogeneous Strichartz Inequality • Let (qj , rj ), j = 1, 2, be two admissible couples and (q2 , r2 ) be the conjugate exponents of (q2 , r2 ). Then there exists C > 0 such that for any f ∈

Lq2 (R, Lr2 (Rd )) we have,  t     S(t − s)[f (s, ·)] ds    0

Lq1 (R,Lr1 (Rd ))

≤ C f

q

r

L 2 (R,L 2 (Rd ))

.

It is important to notice that the couples (q1 , r1 ) and (q2 , r2 ) are independent. Comments For the Strichartz estimates and applications to the nonlinear wave and Schrödinger equations we refer to the book by T. Tao [29].  

116

7 The Classical Equations

7.6 The Burgers Equation This is a quasi-linear first order partial differential equation which appears in the study of various domains of applied mathematics. It has been introduced by J. M. Burgers in 1948. We will be interested in solving the Cauchy problem for this equation, that is, given a real valued function u0 find u satisfying, ∂t u + u ∂x u = 0,

(t, x) ∈ R × R,

u|t =0 = u0 .

(7.21)

• Case where u0 ∈ C 1 (R) – Let u0 ∈ C 1 (R) be real valued satisfying infx∈R u 0 > −∞ where u 0 = ∂x u0 . Set, T ∗ = +∞ if inf u 0 ≥ 0 x∈R

– – – –

and

T∗ =

1 if inf u 0 < 0. x∈R − infx∈R u 0

Then the problem (7.21) has a unique solution u ∈ C 1 ([0, T ∗ ) × R). In particular if u 0 ∈ L∞ (R) and T > 0 is such that T u 0 L∞ (R) < 1 then T < T ∗ . Thus the solution exists on the time interval [0, T ]. Notice that if u0 ∈ C01 (R) (that is has compact support), u0 ≡ 0, then we have necessarily T ∗ < +∞. If u0 ∈ C k (R), k ≥ 2 with infx∈R u 0 > −∞ then the above solution belongs to C k ([0, T ∗ ) × R). One can show that, when it is finite, T ∗ is the right maximal time of existence. For instance assume u0 ∈ C 2 (R). Let x0 be a point where u 0 (x0 ) < 0 then we can show that limt → 1 |∂x u(t, x0 )| = +∞. However, notice that the −u (x0 ) 0

function u itself remains bounded. • Case where u0 ∈ H 2 (R) – Let u0 ∈ H 2 (R) ⊂ W 1,∞ (R) and let T > 0 be such that T u 0 L∞ (R) < 1. We can apply the previous result and infer that (7.21) has a unique C 1 solution. Then this solution belongs to C 0 ([0, T ], H 2 (R)). – More generally if u0 ∈ H k (R) for k ≥ 2 then the solution belongs to C 0 ([0, T ], H k (R)). (See Problem 65 for the above claims.) – Continuity of the flow. The map u0 → u is continuous from H 2 (R) to C 0 ([0, T ], H 2 (R)). – Nonuniform continuity of the flow. The map u0 → u from H 2 (R) to C 0 ([0, T ], H 2 (R)) is not uniformly continuous (see Problem 65).

7.7 The Euler Equations

117

• Case where u0 ∈ L∞ (R). When the data is poorly regular (in particular discontinuous) we have to weaken the notion of solution. We write the Burgers problem as, 1 ∂t u + ∂x (u2 ) = 0, 2

u|t =0 = u0 .

(7.22)

Let I = (0, +∞). We shall say that u is a weak solution of (7.22) if for all ϕ ∈ C01 ([0, +∞) × R) we have, 

 1 u0 (x)ϕ(0, x) dx = 0. u∂t ϕ + u2 ∂x ϕ (t, x) dx dt + 2 I ×R R

Then we have the following result. – Let u0 ∈ L∞ (R). Then the problem (7.22) has a unique weak solution u ∈ L∞ (I × R) satisfying, (i) u L∞ (I ×R) ≤ u0 L∞ (R) , (ii)

there exists E > 0 such that for every a > 0, t ∈ I, x ∈ R, E u(t, x + a) − u(t, x) < . a t

– Condition (ii) above is called the entropy condition. This condition permits to select the physically acceptable solution and is essential for the uniqueness. The solution u above is called the entropy solution. – Assume that u0 ∈ L∞ (R) has compact support, then the entropy solution satisfies, u(t, ·) L∞ (R) ≤ Ct − 2 , 1

t ≥ 1.

Comments For further information on the Burgers equation we refer the reader to the book by J. Smoller [25], Chapters 15 and 16.  

7.7 The Euler Equations 7.7.1 The Incompressible Euler Equations These are equations which describe the dynamic of a fluid without viscosity submitted to exterior forces. If we denote by X(t) = (x1 (t), . . . , xd (t)) ∈ Rd a

118

7 The Classical Equations

particle of fluid, its velocity at time t at this point is given by, dX ˙ (t). v(t, X(t)) = X(t) := dt

(7.23)

It is a vector in Rd . The Euler equations are the mathematical expression of the Newton law which says that the acceleration is proportional to the forces which act ¨ on the particle. In our case the acceleration is given by X(t). According to (7.23) d ∂vj ∂vj we have x¨j (t) = ∂t + k=1 x˙k (t) ∂xk = F. Using again (7.23) we obtain, ∂vj  ∂vj + vk = F, ∂t ∂xk d

j = 1, . . . , d.

(7.24)

k=1

In the case of basic fluid dynamics, the force which acts comes from the gradient of pressure, so that the Euler system reads, ∂vj  ∂vj ∂P + vk =− . ∂t ∂xk ∂xj d

k=1

We can write these equations in a shorter manner. Set we have, ∂v + (v · ∇x )v = −∇x P . ∂t

d

∂ k=1 vk ∂xk

= v · ∇x , then

(7.25)

7.7.1.1 Incompressibility It is a physical notion which in fluid dynamics expresses the fact that over time the volume of the fluid remains constant under the action of the forces which act on it. This fact can be expressed mathematically. The incompressibility is equivalent to div v =:

d  ∂vj j =1

∂xj

= 0.

(7.26)

(See Problem 59.) In the incompressible case, the pressure P appearing in the equations (7.25) is determined by the velocity v. Indeed let us take the divergence of both members ∂ of (7.25). Since div ∂v ∂t = ∂t (div v) = 0, using the fact that div ∇x =  = d ∂2 2 (the Laplacian) we obtain −P = div (v · ∇x )v, which can be written as j =1 ∂xj

7.7 The Euler Equations

119

−P = −1 div ((v · ∇x )v) , so that the equations (7.25) are equivalent to ∂v + (v · ∇x )v = ∇x −1 div ((v · ∇x )v) . ∂t

(7.27)

Thus the Cauchy problem for the incompressible Euler equation consists in finding (v, P ) satisfying, ⎧ ∂v ⎪ + (v · ∇x )v = −∇x P , ⎪ ⎪ ⎨ ∂t (7.28)

div v = 0, ⎪ ⎪ ⎪ ⎩ v|t =0 = v0 .

 Define div(v ⊗ v) as the vector whose i t h coordinates is dj=1 ∂xj (vi vj ). Then it is easy to see that the system (7.28) is equivalent to the system, ⎧ ∂v ⎪ + div (v ⊗ v) = −∇x P , ⎪ ⎪ ⎨ ∂t (7.29)

div v = 0, ⎪ ⎪ ⎪ ⎩ v|t =0 = v0 . 7.7.1.2 The Vorticity If v is the velocity, the vorticity, is defined by the skew-symmetric matrix ω = (ωij )1≤i,j ≤d ,

ωij = ∂xi vj − ∂xj vi ,

1 ≤ i, j ≤ d.

• Notice that a fluid for which ω = 0 is called irrotational. If v is a solution of the Euler system (7.28) such that ω|t =0 = 0 then for every t ≥ 0 we have ω(t, ·) = 0. That means that the irrotationality is propagated by the flow. (See Problem 60.) • If d = 3, ω may be identified with the vector ∇ × v. Then, if v is a sufficiently regular solution of the Euler system (7.28), the vorticity satisfies the vectorial equation, ∂t ω + (v · ∇x )ω = (ω · ∇x )v,

(7.30)

 where u · ∇x = 3i=1 ui ∂xi . • If d = 2 then the vorticity may be identified with the scalar function ω = ∂x1 v2 − ∂x2 v1 . In this case the vorticity equation reduces to ∂t ω + (v · ∇x )ω = 0.

(7.31)

120

7 The Classical Equations

Moreover, (still when d = 2) since v satisfies the incompressibility condition div v = 0 one can formally recover v from ω as follows. There exists a function ψ such that v = ∇ ⊥ ψ = t (−∂x2 ψ, ∂x1 ψ). Taking 1 the curl of both members we find that x ψ = ω. Let E = 2π Ln |x| be the 2 fundamental solution of the Laplacian in R . Then if ω decays sufficiently at infinity we get, ψ(t, x) = (E ω(t, ·))(x), which implies that,  v(t, x) = (K ω(t, ·))(x) =

R2

K(x − y)ω(t, y) dy,

  x2 x1 1 t − |x| where K(x) = ∇x⊥ E(x) = 2π 2 , |x|2 . This is the so-called Biot–Savart law. • Conversely if v, ω satisfy the vorticity equation (7.31) and div v = 0 then, curl (∂t v + (v · ∇x )v) = ∂t ω + (v · ∇x )ω = 0. Therefore, there exists P such that, ∂t v + v · ∇x v = −∇x P . Thus (v, P ) satisfies the system (7.28). Moreover, taking the divergence of both members we find that P is a solution of the equation, 2 

−x P =

∂xi vj ∂xj vi .

i,j =1

Concerning the existence of solutions for the system (7.28) there is a huge literature and thousands of papers. We just give here a few examples of such results.

7.7.1.3 Classical Solutions: The Lichtenstein Theorem • Let d ≥ 2 and 0 < α < 1. Set ρ = 1 + α. Let v0 ∈ W ρ,∞ (Rd , Rd ) be such that div v0 = 0. There exists a unique T ∗ > 0 and a unique solution v of (7.28) on ∗ ρ,∞ (Rd , Rd )). Moreover, [0, T ∗ ) × Rd belonging to the space L∞ loc ([0, T ), W T ∗ < +∞ ⇒



T∗ 0

v(t, ·) W ρ,∞ dt = +∞.

7.7 The Euler Equations

121

• Notice that under the above hypothesis on v0 the map v0 → v need not be continuous from W ρ,∞ (Rd , Rd ) to L∞ ([0, T ], W ρ,∞ (Rd , Rd )) for T < T ∗ (see Problem 64).

7.7.1.4 Weak Solutions: The Yudovitch Theorem It is a classical result on the existence and uniqueness of a weak solution for the vorticity equation in dimension d = 2. • Set I = (0, +∞). Let ω0 ∈ L1 (R2 ) ∩ L∞ (R2 ). There exists a unique couple (ω, v) with ω ∈ L∞ (I, L1 (R2 ) ∩ L∞ (R2 )), v = K ω satisfying, 

 I ×R2

ω(t, x) (∂t ϕ + (v · ∇x )ϕ) (t, x) dt dx +

R2

ϕ(0, x)ω0 (x) dx = 0,

for every ϕ ∈ C01 ([0, +∞) × R2 ). • Notice that by the above discussion this gives a result on existence of a weak solution for the Euler system (7.29) in dimension d = 2. Let v0 be a vector field such that div v0 = 0 which belongs to L2 (R2 , R2 ). Assume moreover that ω0 = curl v0 ∈ L1 (R2 ) ∩ L∞ (R2 ). Then there exists a unique weak solution v of problem (7.29) belonging to C 0 ([0, +∞), L2 (R2 , R2 )) such that ω ∈ L∞ (I × R2 ) ∩ L∞ (I, L1 (R2 )).

7.7.2 The Compressible Euler Equations These are also classical equations. We have to introduce another quantity, the fluid density, which will change with time. If we denote by ρ this density, the compressible Euler equations are, ⎛ ⎞ d d  ∂vj ∂ρ  ∂ρ ⎠ ρ = 0, + vj +⎝ ∂t ∂xj ∂xj ⎛

j =1

ρ|t =0 = ρ0 ,

j =1

⎞ d γ  ∂v ∂v k k ⎠ + ∂ρ = 0, + vj ρ⎝ ∂t ∂xj ∂xk

vk |t =0 = v0k ,

k = 1, . . . , d,

γ > 0.

j =1

(7.32)

122

7 The Classical Equations

7.8 The Navier–Stokes Equations These are partial differential equations describing the dynamics of viscous fluids. As for the Euler equations u denotes the velocity of a particle of fluid, P denotes the pressure, and here we add a parameter μ which stands for the viscosity of the media. The Cauchy problem for these equations consists in finding u = (u1 , . . . , ud ) and P solutions of, ⎧ ∂u ⎪ + div (u ⊗ u) − μx u = −∇x P , ⎪ ⎪ ⎨ ∂t (7.33)

div u = 0, ⎪ ⎪ ⎪ ⎩ u|t =0 = u0 . Recall that div(u ⊗ u) is the vector whose i t h coordinate is

d

j =1 ∂xj (ui uj ).

7.8.1 Weak Solutions Let 0 < T ≤ +∞ and I = (0, T ). We introduce the space,   1,1 Wloc = u ∈ L1loc (I × Rd , Rd ) : ∂xk u ∈ L1loc (I × Rd , Rd ), 1 ≤ k ≤ d . 1,1 is called a weak solution of (7.33) on I × Rd if div u = 0 and, Then u ∈ Wloc

 I ×Rd

u(t, x) · ∂t φ(t, x) dx dt +

 −μ

I ×Rd

d   i,j =1

∇x u(t, x) · ∇x φ(t, x) dx dt −

I ×Rd

ui (t, x)uj (t, x)∂xj φi (t, x) dx dt

 Rd

u0 (x) · φ(0, x) dx = 0,

(7.34)

for every φ = (φ1 , . . . , φd ) ∈ C0∞ ([0, T ) × Rd , Rd ) such that div φ = 0. As for the Euler equations there are thousands of papers dealing with these equations. We will just mention some of the most famous results on it.

7.8.2 The Leray Theorem (1934) • Let I = (0, +∞) and d ≥ 2. Consider a vector field u0 ∈ L2 (Rd , Rd ) such that div u0 = 0. Then the Cauchy problem for the Navier–Stokes equation (7.33)

7.8 The Navier–Stokes Equations

123

has a weak solution u ∈ L∞ (I, L2 (Rd , Rd )) with ∇x u ∈ L2 (I, L2 (Rd , Rd )). Moreover, this solution satisfies,  u(t, ·) 2L2

+ 2μ

t

0

∇x u(s, ·) 2L2 ds ≤ u0 2L2 .

• If d = 2 the solution belongs to C 0 (I, L2 (Rd , Rd )) and is unique in the set,   u ∈ L∞ (I, L2 (Rd , Rd )) : ∇x u ∈ L2 (I, L2 (Rd , Rd )) and div u = 0 . Moreover, it satisfies, for all time t > 0, the equality,  u(t, ·) 2L2 + 2μ

0

t

∇x u(s, ·) 2L2 ds = u0 2L2 .

• Notice that when d ≥ 3 the uniqueness of the weak solution above is, since 1934, an open problem. Moreover, since there are other results showing the uniqueness of the solution in spaces of more regular functions, the smoothness of these weak solutions is as well an open problem.

7.8.3 Strong Solutions: Theorems of Fujita–Kato and Kato We restrict ourselves to the physical dimension d = 3. • Fujita–Kato (1964): let u0 ∈ H˙ 2 (R3 , R3 ) be such that div u0 = 0. There exists a maximal time T ∗ > 0 such that the problem (7.33) has a unique solution u such that,     1 3 u ∈ C 0 [0, T ∗ ), H˙ 2 (R3 , R3 ) ∩ L2loc (0, T ∗ ), H˙ 2 (R3 , R3 ) . 1

Moreover, there exists ε0 > 0 such that if u0

1

H˙ 2 (R3 ,R3 )

≤ ε0 then T ∗ = +∞.

• Kato (1984): Let u0 ∈ L3 (R3 , R3 ) be such that div u0 = 0. Then there exists a maximal time T ∗ > 0 such that the problem (7.33) has a unique solution u such that,   1 and t 2 u(t) L∞ (R3 ,R3 ) ∈ L∞ ((0, T ∗ )). u ∈ C 0 [0, T ∗ ), L3 (R3 , R3 ) Moreover, there exists ε0 > 0 such that if u0 L3 (R3 ,R3 ) ≤ ε0 , then T ∗ = +∞. Comments Concerning the equations involved in fluid mechanics we refer to the book by H. Bahouri, J.-Y. Chemin, and R. Danchin [5].  

Chapter 8

Statements of the Problems of Chap. 7

This chapter contains problems about the classical partial differential equations.

Problem 37 Monge–Ampère Equation

Part 1 In R2 , where the variable is denoted by (x1 , x2 ), if u is a C 2 function we shall set uj k = ∂xj ∂xk u. The purpose of this part is to show that, using the Cauchy– Kovalevski theorem, for every given real analytic and real valued functions K, g, h near the origin one can solve locally the Cauchy problem, u11 u22 − u212 = K(x1 , x2 ),

u|x2 =0 = g(x1 ),

∂u |x =0 = h(x1 ) ∂x2 2

(8.1)

in the analytic framework, under the hypothesis g

(0) = 0. 1. Show that there exists a constant A ∈ C such that if we set, 1  u(x1 , x2 ) = g(x1 ) + h(x1 )x2 + Ax22 , ϕ(x1 , x2 ) = x2 , 2

p0 =  u(0, 0), (∂xj  u(0, 0))j =1,2, (∂xj ∂xk  u(0, 0))j,k=1,2 , then all the conditions in the nonlinear Cauchy–Kovalevski theorem are satisfied. 2. Conclude. Part 2 We want to extend this result to higher dimensions.

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 T. Alazard, C. Zuily, Tools and Problems in Partial Differential Equations, Universitext, https://doi.org/10.1007/978-3-030-50284-3_8

125

126

8 Statements of the Problems of Chap. 7

If d ≥ 3 we denote by (x , xd ) ∈ Rd−1 × R the variable in Rd . For u ∈ C 2 we denote by Hess(u) = (∂xj ∂xk u)1≤j,k≤d its Hessian matrix. Given real analytic and real valued functions K, g, h near the origin we consider the problem, u|xd =0 = g(x ),

det(Hess(u)) = K,

∂u |x =0 = h(x ), ∂xd d

(8.2)

under the condition det(Hess(g)(0)) = 0,

(8.3)

where Hess(g)(0) = (∂xj ∂xk g(0))1≤j,k≤d−1 is the Hessian matrix of g at zero. 1. Case 1. Assume that in a neighborhood of zero in Rd−1 we have, g(x ) = a0 +

d−1  j =1

aj xj +

d−1 

λj xj2 + O(|x |3 ),

j =1

where λj = 0 for j = 1, . . . , d − 1. Show that one can find A ∈ C such that if we set, 1  u(x) = g(x ) + h(x )xd + Axd2 , ϕ(x) = xd , 2

p0 =  u(0), (∂xj  u(0))j =1,...,d , (∂xj ∂xk  u(0))j,k=1,...,d , then we can apply the nonlinear Cauchy–Kovalevski theorem. 2. Case 2. Assume (8.3) true. For k ∈ N, k ≥ 2, denote by O(k) the set or k × k orthogonal matrices. a) Prove that one can find M = (m j k ) ∈ O(d − 1) such that M (Hess(g)(0)) M = diag(λj ) ,

t

where λj = 0 for j = 1, . . . , d − 1 and tM is the transposed matrix of M . b) Set, mj k = m j k , for 1 ≤ j, k ≤ d − 1, mj d = mdj = 0, for 1 ≤ j ≤ d − 1, mdd = 1.

Prove that M = (mj k ) ∈ O(d) and that My = (M y , yd ) if y = (y , yd ). c) If u ∈ C 2 (Rd ) we set v(y) = u(My). Prove that, Hess(v)(y) = tMHess(u)(My)M.

8 Statements of the Problems of Chap. 7

127

Hint: Apply the Taylor formula at the order two to v(y + hy0 ) and u(My + hMy0 ) where y, y0 ∈ Rd and h ∈ R is small. d) Deduce that det (Hess(u)(My)) = det (Hess(v)(y)) . e) We set  g (y ) = g(M y ). Prove that in a neighborhood of the origin of Rd−1 we have,  g (y ) = a0 +

d−1 

1 λj yj2 + O(|y |3 ). 2 d−1

aj yj +

j =1

j =1

f) Applying the result proved in Case 1 show that the problem (8.2) has a unique real analytic solution in a neighborhood of the origin. Comments The Monge–Ampère equations appear in geometry since, for instance, the Gauss curvature of a surface in R3 given by a graph z = u(x, y) is given by the formula K =

uxx uyy −u2xy

1

(1+u2x +u2y ) 2

. They appear also in recent works concerning the theory  

of optimal transportation (see the Lecture Notes [4]).

Problem 38 An Application of the Harnack Inequality For n ≥ 1, y 0 ∈ Rn and ρ > 0 we shall denote in what follows,   n 0 0 Bn (y , ρ) = y ∈ R : |y − y | := max |yj − yj | ≤ ρ 0

1≤j ≤n

the closed ball with center y 0 and radius ρ. 1. Let  be an open subset of Rn and u be a C 2 nonnegative harmonic function in . Let X0 ∈  and r > 0 be such that Bn (X0 , r) ⊂ . The purpose of this question is to prove that there exists a constant M depending only on the dimension n such that, sup Bn (X0 , r2 )

u≤M

inf

Bn (X0 , 2r )

u.

(8.4)

a) Let X1 , X2 ∈ Bn (X0 , 2r ). Using the fact that Bn (X2 , r) ⊂ Bn (X1 , 2r) and the mean value formula prove that there exists C = C(n) > 0 such that, u(X1 ) ≥ Cu(X2 ). b) Deduce (8.4).

128

8 Statements of the Problems of Chap. 7

2. Let ω ⊂ Rd , d ≥ 1, be open and let ϕ ∈ C 2 (ω) ∩ C 0 (ω), ϕ ≡ 0, be a solution of the problem, ϕ|∂ω = 0,

− ϕ = λϕ in ω,

λ > 0.

(8.5)

√

a) We set for t ∈ R, ψ(t, x) = et λ ϕ(x). Show that ψ is a C 2 harmonic function in Rt × ω. Let Z(ϕ) = {x ∈ ω : ϕ(x) = 0} . Let x0 ∈ ω be such that x0 ∈ / Z(ϕ) and let r > 0 be such that Bd (x0 , 4r)∩Z(ϕ) = ∅. Changing ϕ to −ϕ, if necessary, we may assume that ϕ(x) > 0 for x ∈ Bd (x0 , 4r). r r Let X0 = ( 3r 2 , x0 ) so that (r, 2r) × Bd (x0 , 2 ) = Bd+1 (X0 , 2 ). b) Prove that, sup Bd+1 (X0 , r2 )

ψ ≥ er

√ λ

inf

Bd+1 (X0 , 2r )

ψ.

c) Using question 1 with n = d +1 prove that there exists a constant C depending only on the dimension d such that r ≤ √C , which is equivalent to say that, λ

C r > √ ⇒ Bd (x0 , 4r) ∩ Z(ϕ) = ∅. λ 3. Using the conclusion of question 2c) prove by contradiction that there exists a constant M depending only on d such that, ∀λ > 0,

∀x ∈ ω,

M d(x, Z(ϕ)) ≤ √ , λ

(8.6)

where d(x, Z(ϕ)) = infz∈Z(ϕ) |x − z|.

Problem 39

Part 1 The goal of this part is to show that the estimate (7.7) is optimal in the case where  is a ball in R3 .  Let B = x ∈ R3 : |x| < 1 . For k ∈ Z, k = 0, we set uk (x) = sin(kπ|x|) . |x| 1. Show that uk is an eigenfunction of the operator − for the Dirichlet problem in the ball B. 2. Compute the L2 (B) norm of uk . Compute the L∞ (B) norm of uk . 3. Conclude.

8 Statements of the Problems of Chap. 7

129

Part 2 In this part we give an application of the estimate (7.7). Let  be a bounded open subset of Rd with C ∞ boundary. Let λ be an eigenvalue of the Dirichlet problem for the Laplacian and let V be the eigenspace corresponding to λ endowed with the L2 () norm. We recall (see Sect. 7.2.6) that V ⊂ C ∞ () and that m := dim V < +∞. Our goal is to show that there exists C > 0 independent of λ such that, m ≤ Cλ

d−1 2

(8.7)

.

We shall use for that the estimate (7.7). 1. Let (v1 , . . . , vm ) be an orthonormal basis of V . For y ∈  we set, uy (x) =

m 

vi (y)vi (x),

a(y) = uy (y) =

i=1

m 

|vi (y)|2 .

i=1

Prove that there exists y0 in  such that, m ≤ || a(y0), where || denotes the Lebesgue measure of . 2. Compute uy0 2L2 () and show that uy0 L∞ () ≥ a(y0 ). 3. Deduce that, u L∞ () ≥ ||− 2 m 2 . 1

sup

1

u∈V u L2 () =1

4. Using the estimate (7.7) prove (8.7).

Problem 40 An Estimate of the Eigenfunctions of the Laplacian on the Sphere For d ≥ 2 let Sd be the unit sphere of Rd+1 and Rd+1 =

d+1

∂2 j =1 ∂x 2 . j

The operator ω on Sd (the Laplace–Beltrami operator) is such that for f ∈ C 2 (Rd+1 ), x = rω, r > 0, ω ∈ Sd and F (r, ω) = f (rω), (Rd+1 f )(rω) =

1 ∂ 2F d ∂F + +  F (r, ω). ω r ∂r ∂r 2 r2

(8.8)

130

8 Statements of the Problems of Chap. 7

1. For j ∈ N and ω = (ω1 , . . . , ωd+1 ) ∈ Sd we set ej (ω) = (ω1 + iω2 )j . Prove that we have, −ω ej = j (j + d − 1)ej . Hint: Use the holomorphy of the function f (x) = (x1 + ix2)j in C. 2. Show that there exists c0 > 0 such that, lim j

j →+∞

d−1 2

ej 2L2 (S d ) = c0 .

  Hint: Write Sd = ω = (ω1 , ω2 , ω ) ∈ R × R × Rd−2 : |ω| = 1 , parametrize 1

1

Sd by |ω | ≤ 1, ω1 = (1−|ω |2 ) 2 cos α, ω2 = (1−|ω |2 ) 2 sin α, 0 < α < 2π, then set ω = ρ $, 0 < ρ < 1, $ ∈ Sd−2 and eventually ρ = √t j . 3. Set uj =

ej ej L2 (Sd ) .

Deduce that there exists c1 > 0 such that, uj L∞ (Sd ) ∼ c1 j

d−1 4

j → +∞.

Problem 41 Spectral Theory for the Harmonic Oscillator

This problem uses heavily the results of Problem 1. The goal of this problem is to study in Rd , d ≥ 1 the harmonic oscillator, P = − + |x|2 and its spectral theory. In all what follows we shall denote E = E(Rd ) if E = L2 , H s , C0∞ , S, S .   Let V = u ∈ L2 : ∇x u ∈ L2 : x u ∈ L2 endowed with the scalar product, (u, v)V = (∇x u, ∇x v)L2 + (x u, x v)L2 ,

1

x = (1 + |x|2 ) 2 ,

1

and with the corresponding norm u V = (u, u)V2 . Then V is a Hilbert space. We shall denote by V its dual. Part 1 Study of the operator P = − + x2 . 1. Prove that the operator P = − + x2 sends V to V , that it is bijective, continuous and that P −1 is continuous that is, ∃C > 0 : w V ≤ C (− + x2 )w V ,

∀w ∈ V .

(8.9)

8 Statements of the Problems of Chap. 7

131

Let f ∈ L2 ⊂ V . Our aim is to show that, 

 u ∈ V , −u + x2 u = f ⇒ u ∈ H 2 , x ∇x u ∈ L2 , x2 u ∈ L2 . (8.10)

Let ek be the k t h vector of the canonical basis of Rd . For 0 < |h| ≤ 1 and v ∈ V set Dh v = h1 (v(x − hek ) − v(x)). 2. a) Prove that we have,

(∇x D−h u, ∇x v)L2 + (x D−h u, x v)L2 = D−h f , v L2  − (2yk + h)u(y + hek )v(y) dy. Rd

b) Taking v = D−h u ∈ V prove that there exists C > 0 independent of u, h such that,

∇x D−h u L2 + x D−h u L2 ≤ C f L2 + x u L2 . c) Prove that, u ∈ L2 , x ∇x u ∈ L2 , x2 u ∈ L2 and deduce (8.10). Hint: Use thefact that a ball in L2 isweakly compact and Problem 1. d) We set D = u ∈ H 2 : x2 u ∈ L2 endowed with the norm u D = u H 2 + x2 u L2 . Prove that the operator P = − + x2 is an isomorphism from D to L2 . 3. We consider the operator P −1 from L2 to L2 as follows. P −1 : L2 → V → V → L2 ,

u → u → P −1 u → P −1 u.

Prove that Ker (P −1 )L2 →L2 = {0} , that P −1 (L2 ) = D and that P −1 is a positive compact and self-adjoint operator. According to the general theory (see Chap. 1, Sect. 1.2.9) σ (P −1 ) ( the spectrum of P −1 ) consists in {0} and a sequence of positive eigenvalues (μn ) −1 − μ I d). which tends to zero when n → +∞ and H = ⊕+∞ n n=1 Ker(P Therefore, there exists an orthonormal basis (en )n∈N of L2 such that P −1 en = μn e n . 4. Show that en ∈ D and that (− + |x|2 )en = λn en where λn = μ1n − 1. 5. a) Let u ∈ D, u ≡ 0, be a solution of the equation −u + |x|2u = λu where λ ≥ 0. Prove that for every k ∈ N, k ≥ 2 we have, u ∈ H k , x u ∈ k−1 H , x2 u ∈ H k−2 . Hint: Use (8.10) proved in question 2 and an induction on k. b) Prove that, ∇x u 2L2 + |x|u 2L2 = λ u 2L2 .

132

8 Statements of the Problems of Chap. 7

c) Prove that, d u 2L2 ≤ ∇x u 2L2 + |x|u 2L2 .

d) 6. a) b) c) 7. a)

b)

 Hint: Compute the quantity 2Re Rd ∂xj u(x)xj u(x) dx. Deduce that λ ≥ d. Prove that the family (en )n∈N is orthogonal in V and that en 2V = λn + 1.     Prove that V = u = n∈N un en : n∈N λn |un |2 < +∞ .     Prove that D = u = n∈N un en : n∈N λ2n |un |2 < +∞ . Let λ = λn for every n ∈ N. Prove that the operator − + |x|2 − λ : D → L2 is invertible.   n Hint: To prove the surjectivity if f = n fn en ∈ L2 set u = n λnf−λ en . 2 Deduce that if we define the spectrum of − +   |x| as the set S = λ ∈ C : − + |x|2 − λ : D → L2 is not invertible then S = ∪n∈N {λn } .

Part 2 The goal of this part is to determine explicitly the spectrum of H = − + |x|2 . Case 1. d = 1 We shall set in what follows, L+ = −

d + x, dx

L− =

d + x, dx

H =−

d2 + x2. dx 2

1. Prove the following identities, H = L+ L− + 1,

L− (L+ )n = (L+ )n L− + 2n(L+ )n−1 ,

n ∈ N, n ≥ 1.

Hint: Use an induction. 2. a) Let φ0 (x) = exp(− 12 x 2 ) ∈ S(R). Prove that Hφ0 = φ0 . b) We set φn = (L+ )n φ0 . Prove that φn (x) = Pn (x)exp(− 12 x 2 ) where Pn is a polynomial of order n whose coefficient of x n is 2n . c) Prove that Hφn = (2n + 1)φn for all n ∈ N. d) Prove that the family (φn ) is orthogonal in L2 . Hint: Prove by induction on n ≥ 1 that for all m < n we have (φm , φn )L2 = 0 and use question 1. e) Prove that the family (φn ) is an orthonormal basis of L2 . Hint: Prove that 

. / 2 2 n − x2 f ∈ L and (f, φn )L2 = 0∀n ∈ N ⇒ x f (x)e dx = 0, ∀n ∈ N ⇒ f = 0.

8 Statements of the Problems of Chap. 7

133

Let en = φφn n 2 . Then (en ) is an orthonormal basis of L2 and Hen = L (2n + 1)en . 3. Let u ∈ D, u ≡ 0, λ ∈ R such that Hu = λu. We want to prove that there exists k ∈ N such that λ = 2k + 1.  a) Prove that u = n∈N (u, en )L2 en where the series converges in L2 . b) Prove that (u, (H − λ)ϕ)L2 = 0 for all ϕ ∈ D. c) Taking ϕ = ek , k ∈ N prove that there exists k ∈ N such that λ = 2k + 1. 4. For any k ∈ N let Ek = Ker (H − (2k + 1)). We want to show that we have dim Ek = 1. Let k ∈ N be fixed. Let u ∈ D be a nontrivial solution of (H −(2k +1))u = 0. Let n ∈ N, n = k. Show that (u, en )L2 = 0. Conclude. Case 2. d ≥ 2 1. Extend the results of questions 1 to 4 to higher dimensions. Hint: For j = 1, . . . , d set L+ j =−

∂ + xj , ∂xj

L− j =

∂ + xj . ∂xj

 αk Introduce also φ0 (x) = exp(− 12 |x|2) and, for α ∈ Nd , set φα = dk=1 (L+ k ) φ0 . Prove that if H = − + |x|2 then Hφα = (2|α| + d)φα . Eventually prove that the multiplicity of λ = 2n + d is equal to |α|=n 1. Problem 42 Ground State for the Hydrogen Atom

This problem uses Problem 13. For ψ ∈ H 1 (R3 ) we set,    ψ 2   F (ψ) = ∇x ψ 2L2 (R3 ) −  1   |x| 2  2

,

(8.11)

L (R3 )

  E = inf F (ψ) : ψ ∈ H 1 (R3 ), ψ L2 (R3 ) = 1 ≥ −∞.

(8.12)

The goal of this problem is to show that E = − 14 , that the infimum is reached  = √1 e− 12 |x| and that the operator − − 1 − E from H 2 (R3 ) to L2 (R3 ) for ψ |x| 8π  has a kernel of dimension one generated by ψ. This is linked with the study of the hydrogen atom in quantum mechanics. Unless otherwise stated we shall denote in what follows X = X(R3 ) if X = 2 L , H k , C0∞ , D .

134

8 Statements of the Problems of Chap. 7

Part 0 ψ 1. Prove that if ψ ∈ H 1 then |x| ∈ L2 . Prove also that   1 1  ψ   1  ≤ C ψ 2 2 ∇x ψ 2 2 .   L L |x| 2

ψ 1

|x| 2

∈ L2 and that we have

L2

Hint: Use the inequality of Hardy (4.9) (see Problem 13).

Part 1. Existence of a Minimizer An element u ∈ H 1 will be called a minimizer if u L2 = 1 and F (u) = E. 1. Prove that E < 0. 3 Hint: Consider ψ(x) = ε 2 χ(εx) where χ ∈ C0∞ . 2. a) Prove that, ∃(ψn )n∈N ⊂ H 1 with ψn L2 = 1 such that

lim F (ψn ) = E.

n→+∞

b) Prove that the sequence (ψn ) is bounded in H 1 . c) Deduce that there is a subsequence (ψσ (n) ) which converges weakly in L2 to . Deduce that,  and (∇x ψσ (n) ) converges weakly in L2 to ∇x ψ ψ  L2 ≤ 1, ψ

 2 2 ≤ lim inf ∇x ψσ (n) 2 2 . ∇x ψ L L n→+∞

   ψσ (n) −ψ   d) Prove that limn→+∞  1   |x| 2

3. a) b) c) 4. a)

b)

= 0.

L2

Hint: Cut the integral into {|x| ≤ R} and {|x| > R} and use questions a), b), and c). ) ≤ E. Prove that F (ψ Hint: Use questions 2c) and d). ) ≥ E ψ  2 2 . Prove that F (ψ L  is a minimizer. Deduce that ψ Let ψ ∈ H 1 and (ϕn ) ⊂ H 2 such that (ϕn ) → ψ in H 1 . Prove that limn→+∞ F (ϕn ) = F (ψ). Hint: Use question 1.   For j = 1, 2 we set Ej = inf F (ψ) : ψ ∈ H j , ψ L2 = 1 . Prove that E1 = E2 .

Part 2 A minimizer is an eigenfunction of H = − −  be a minimizer. Let ψ

1 |x|

with eigenvalue E.

 ±εϕ. Using the fact that F (f ) ≥ E f 2 2 1. a) Let ε > 0 and ϕ ∈ C0∞ . Set f = ψ L prove that, 1 0  ψ , ϕ = 0, − − Eψ Re −ψ |x|

8 Statements of the Problems of Chap. 7

135

where ·, · denotes the bracket between D and C0∞ .

  b) Changing ϕ to iϕ prove  that, H ψ= E ψ in D . ψ 2 .  ∈ D = u ∈ H2 : c) Prove that ψ |x| ∈ L 2. Let Ker (H − E) = {u ∈ D : (H − E)u = 0} . Prove that if u ∈ Ker (H − E), u ≡ 0 then u u 2 is a minimizer. L

Part 3. Properties of the Minimizers  be a minimizer. We show that ψ (x) = 0 for all x ∈ R3 . Let ψ

(8.13)

1. Let u ∈ H 1 be a real valued function. Prove that |u| ∈ H 1 and that ∇x |u| = u 1{x:u(x)=0} |u| ∇x u. √ Hint: Consider for ε > 0, uε = u2 + ε2 − ε.  is a minimizer then |ψ|  is also a minimizer. 2. a) Prove that if ψ  ∈ D and, b) Deduce that |ψ|   = |ψ| , (− + ω2 )|ψ| |x|

where E = −ω2 = 0.

c) Prove that, 1  |ψ(x)| = 4π

 R3

 e−ω|x−y| |ψ(y)| dy. |x − y| |y|

−ω|x|

1 e 2 Hint: Let G = 4π |x| . Then (− + ω )G = δ0 .  d) Prove that |ψ(x)| > 0 for every x ∈ R3 .

We show that

| ∈ L2 ∩ L∞ , |x|k |ψ

∀k ∈ N.

(8.14)

 ∈ L2 ∩ L∞ for all k ∈ N. 3. Prove by induction on k ∈ N that |x|k |ψ| Hint: Use question c) and L1 L2 ⊂ L2 , L2 L2 ⊂ L∞ . We show that dim Ker(H − E) = 1.

(8.15)

4. a) Prove that if u ∈ Ker(H − E) then u ≡ 0 or u = 0 everywhere. b) Let u1 , u2 ∈ Ker(H − E) with u2 ≡ 0. Let x0 be such that u2 (x0 ) = 0. Let 0) u = u1 − uu12 (x (x0 ) u2 . Prove that u ≡ 0 and conclude.  > 0 then ψ  is radial. We show that if ψ

(8.16)

136

8 Statements of the Problems of Chap. 7

A (x) = ψ (Ax). Show that ψ A ∈ 5. Let A be a d × d orthogonal matrix. Set ψ  is radial, that means ψ (x) = u(|x|). Ker (H − E) and deduce that ψ A (x) = (ψ )(Ax). Hint: Use the fact that ψ Part 4  The goal of this part is to determine E and ψ. Let θ ∈ D be a solution of Hθ = λθ, with λ = −μ2 , μ > 0 such that, θ > 0,

|x|k θ ∈ L2 ∩ L∞

∀k ∈ N,

θ is radial.

1. Prove that F (θ ) = λ θ 2L2 where F is defined in (8.11). 2. a) We set |x| = r and θ (x) = u(r). Prove that u is continuous on [0, +∞) and satisfies the equation −u

(r) − 2r u − 1r u(r) = λu(r) in D ((0, +∞)). b) Let v(r) = ru(r). Prove that r k v ∈ L∞ ((0, +∞)) for all k ∈ N. Deduce that v ∈ L1 ((0, +∞)). c) Prove that v satisfies −v

(r) − 1r v(r) = λv(r) in D ((0, +∞)). 3. a) Set w(r) = 1{r>0} v(r) ∈ L1 (R) and T = −rw

− w − λrw ∈ S (R). Show  (k) that T = N k=0 ck δ0 , ck ∈ C. ∞ b) Let ϕ ∈ C0 (R). For ε > 0 set ϕε (r) = ϕ( rε ). Prove that limε→0 T , ϕε  = 0. Hint: Use the fact that v ∈ L∞ ((0, +∞)) ∩ C 0 ([0, +∞)) and v(0) = 0. c) Using question a) prove that T = 0.  +∞ 4. a) Let f (ξ ) = w (ξ ) = 0 e−irξ v(r) dr. Prove that f is a C ∞ function on R which can be extended as a holomorphic function (still denoted by f ) in the set {z ∈ C : Im z < 0} . b) Show that f satisfies the differential equation, (ξ 2 + μ2 )f (ξ ) + (2ξ + i)f (ξ ) = 0,

ξ ∈ R,

λ = −μ2 , μ > 0.

5. a) Let G be a holomorphic function in the open set {z ∈ C : Im z < 0} and continuous in the set {z ∈ C : Im z ≤ 0.} Assume that G|Im z=0 = 0. Prove that G = 0. Hint: Use the Holmgren theorem and the principle of isolated zeros. b) Deduce that, (z2 + μ2 )f (z) + (2z + i)f (z) = 0,

z ∈ C, Im z < 0.

 (z) dz. Show 6. a) Let ε > 0 small and B = {z ∈ C : |z + iμ| ≤ ε} . Set I = B ff (z) that I is an integer n ∈ N. 1 1 b) Using the residue formula prove that I = 2μ − 1 and then that μ = 2(n+1) . c) Deduce that the minimum energy E is equal to − 14 . d) Using the form of

f f

deduce that f (z) =

C . (z− 2i )2

8 Statements of the Problems of Chap. 7

7. Recall that

C (ξ − 2i )2

137

= f (ξ ) = w (ξ ) so w(r) =

Fξ−1 →r



C (ξ − 2i )2

since w and f

are in L1 (R). Our goal is to compute w. ϕ and deduce F (rϕ). a) Let ϕ(r) = 1r>0 e− 2 ∈ L1 (R). Compute  r b) Prove that w(r) = 1r>0 Cre− 2 and then that the set of minimizers is generated (x) = √1 e− 12 |x| . by ψ r

8π

Comments We would like to warmly thank our colleague P. Gérard for sharing his notes on the subject of this problem.  

Problem 43 On the Dirichlet–Neuman Operator

This problem uses Problem 5. The goal of this problem is to define the Dirichlet–Neuman operator in a particular geometry and to prove its continuity in the Sobolev spaces. Notations Let h > 0 and η ∈ W 1,∞ (R) be such that η L∞ (R) ≤ h2 . We set,    = (x, y) ∈ R2 : −h < y < η(x) ,    = (x, z) ∈ R2 : −h < z < 0 . 

  ! = (x, y) ∈ R2 : y = η(x) ,

We shall denote by C ∞,0 () the space of C ∞ functions on  which vanish near ! and by H 1,0() the closure of C ∞,0 () in H 1 (), endowed with the H 1 () norm. We will admit that,   H 1,0() = u ∈ H 1 () : γ0 (u) = 0 , where γ0 denotes the trace on !. C0∞ () will denote the space of C ∞ functions with compact support in . We will set eventually xy =

∂2 ∂x 2

+

∂2 . ∂y 2

Part 1

  Case η = 0. Assume η ≡ 0, that is,  = (x, y) ∈ R2 : −h < y < 0 . Let ψ ∈ S(R). We consider the problem, xy u = 0 in ,

u|y=0 = ψ,

(∂y u)|y=−h = 0.

(8.17)

138

8 Statements of the Problems of Chap. 7

1. Prove that this problem has at most one solution belonging to C 2 ([−h, 0], H s (R)) for all s ∈ R.

Hint: If u1 , u2 are two solutions compute − xy u, u L2 () where u = u1 − u2 . 1 t x |) −t 2. a) Show that u = ch((y+h)|D ch(h|Dx |) ψ, where ch(t) = 2 (e + e ) is the hyperbolic cosine, is the unique solution of (8.17) in C 2 ([−h, 0], H s (R)) for every s ∈ R. Hint: Use the Fourier transform in x. b) Set G(0)ψ = (∂y u)|y=0 . Deduce that G(0)ψ = a(Dx )ψ where a(ξ ) = |ξ |th (h|ξ |) and th denotes the hyperbolic tangent. c) Show that a ∈ S 1 and that for all s ∈ R we have, G(0)ψ

H

s− 1 2 (R)

≤ ψ

H

s+ 1 2 (R)

1

∀ψ ∈ H s+ 2 (R).

,

Part 2 Case η ≡ 0. 1,0 (). u L2 () ≤ 3h 2 ∂y u L2 () , for all u ∈ H 1  let ψ(x,  z) = χ(z)ez|Dx | ψ(x) where χ ∈ 2. a) Let ψ ∈ H 2 (R). For (x, z) ∈  C ∞ , χ(z) = 1 if z ≥ − h4 , χ(z) = 0 if z ≤ − h3 . |z=0 = ψ, ψ |z=−h = 0 and that there exists C > 0 such Show that ψ that,

1. Prove the inequality,

 H 1 ( ψ ) ≤ C ψ

1

H 2 (R)

.

b) Deduce that there exists ψ ∈ H 1 () such that, ψ|! = ψ,

5 ψ(x, y) = 0 if y ≤ − h, and, 6

∃ C > 0 : ψ H 1 () ≤ C(1 + η W 1,∞ (R) ) ψ

1

H 2 (R)

.

3. We consider the sesquilinear form, 

 ∂x u(x, y)∂x u(x, y) dy dx +

a(u, v) = 

∂y u(x, y)∂y u(x, y) dy dx. 

Show that it is coercive on H 1,0(). 1 4. a) Let ψ ∈ H 2 (R). Prove that there exists a unique u ∈ H 1,0() such that, 

   −xy u, ϕ = xy ψ, ϕ

∀ϕ ∈ C0∞ ().

8 Statements of the Problems of Chap. 7

139

b) Let φ = u + ψ. Prove that φ is the unique solution of the problem,   −xy u, ϕ = 0 ∀ϕ ∈ C0∞ (),

γ0 (φ) = ψ on !

(8.18)

and that there exists C > 0 such that, ∂x φ L2 () + ∂y φ L2 () ≤ C(1 + η W 1,∞ (R) ) ψ

1

H 2 (R)

.

5. Prove that the map, (x, z) → (x, ρ(x, z)),

where

ρ(x, z) =

1 (z + h)η(x) + z, h

 to . is a diffeomorphism from   6. Let φ(x, z) = φ(x, ρ(x, z)). Prove that  φ is the solution of problem, ), φ = 0 in D ( (21 + 22 )

 φ |z=0 = ψ,

where 1 =

1 ∂z , ∂z ρ

2 = ∂x −

∂x ρ ∂z . ∂z ρ

7. Let U = (1 − (∂x ρ)2 ) φ. Show that, ∂z U = −∂x ((∂z ρ)2 U ) . 8. a) Prove that there exists C > 0 such that, U L2 ((−h,0),L2(R))+ ∂z U L2 ((−h,0),H −1(R)) ≤ C(1+ η W 1,∞ (R) ) ψ

1

H 2 (R)

.

b) Deduce from Problem 5 that U is continuous on [−h, 0] with values in 1 H − 2 (R) and that there exists C > 0 such that, U |z=0

H

− 21

(R)

≤ C(1 + η W 1,∞ (R) ) ψ

9. Show that U |z=0 = where

∂ ∂n |!

1

H 2 (R)

2

1 + (∂x

is the normal derivative on !.

η)2

∂φ ∂n

|! ,

.

140

8 Statements of the Problems of Chap. 7

Comments The  map  which to ψ (the Dirichlet data of the problem (8.18)) gives  1 + (∂x η)2 ∂φ ∂n |! (the normalized Neumann data of the same problem) is called the Dirichlet–Neuman operator. It is denoted by G(η). Therefore, we have shown that G(η)ψ

H

− 21

≤ C(1 + η W 1,∞ (R) ) ψ

(R)

1

H 2 (R)

,

1

for all ψ ∈ H 2 (R). This operator plays a role in establishing the water wave equations.  

Problem 44 Let u ∈ H 1 (R3 ) be a solution of the equation, − u + u = u3 , where  =

3

∂2 j =1 ∂x 2 j

(8.19)

is the Laplacian.

The goal of the problem is to prove that u decays exponentially at infinity. Part 1 1. Using the Sobolev embedding show that u3 ∈ L2 (R3 ). Deduce that u ∈ H 2 (R3 ). 2. Using (3.3) prove by induction on k that u ∈ H k (R3 ) for every k ≥ 2. Part 2. Preliminaries e−|x| 1. Compute the Fourier transform of the L1 (R3 ) function E(x) = 4π|x| .  is radial, for fixed ξ use an orthogonal matrix A Hint: Use the fact that E sending ξ to (|ξ |, 0, . . . , 0) then use the polar coordinates in R3 . 2. Deduce that E is the unique solution in S (R3 ) of the equation E − E = δ0 where δ0 is the Dirac distribution at zero. Part 3 Let u ∈ H 1 (R3 ) be a solution of (8.19). 1. Show that we have,  u(x) = (E u3 )(x) =

f (x, y) dy, R3

For n ∈ N large, set ωn supx∈ωn |u(x)|. 2. Prove that limn→+∞ An = 0.

=



f (x, y) =

e−|x−y| u(y)3 . 4π|x − y|

x ∈ R3 : 2n ≤ |x| ≤ 2n+1



and An

=

8 Statements of the Problems of Chap. 7

141

3. Prove that there exists C = C(u) > 0 independent of n such that for x ∈ ωn ,  I1 := 

|y|≤2n−1

I3 :=

f (x, y) dy ≤ Ce−2

y∈ωn−1

I5 := y∈ωn+1

 , I2 := 

f (x, y) dy ≤ 

n−1

CA3n−1 ,

|y|≥2n+2

I4 := y∈ωn

f (x, y) dy ≤ Ce−2

n−1

,

f (x, y) dy ≤ CA3n ,

f (x, y) dy ≤ CA3n+1 ,

and deduce that, An ≤ C(e−2

n−1

+ A3n−1 + A3n + A3n+1 ).

(8.20)

4. Prove that there exists C > 0, N0 ∈ N such that for N ≥ N0 we have, +∞ 

An ≤ C (e−2

N−2

+ A3N−1 ).

n=N

N+M Hint: Using (8.20) estimate n=N An ; then take N large enough and use question 2. Then let M go to +∞. N−2 + AN . Show that there exists N1 ∈ N such that BN+1 ≤ 5. a) Set BN = e−2 (1 + C )BN2 for N ≥ N1 and deduce that for N ≥ N1 we have, BN ≤

2N−N1 1 

2 (1 + C ) B N 1 1 + C

b) Prove that if N1 is large enough one can find K > 0 and δ > 0 such that N BN ≤ Ke−δ2 for N ≥ N1 . c) Prove that there exist C > 0, R > 0, δ1 > 0 such that |u(x)| ≤ Ce−δ1 |x| for |x| ≥ R.

Problem 45 On the Smoothing Effect for the Heat Equation

The goal of this problem is to study the smoothing effect for the heat equation. Consider the Cauchy problem, "

∂t u − u = 0, u|t =0 = g.

(8.21)

142

8 Statements of the Problems of Chap. 7

We first prove a classical result: if g ∈ H s (Rd ) then u ∈ L2 ((0, T ); H s+1(Rd )) for any T > 0. We then prove a refined version which, in dimension d = 2 for instance, states that if g ∈ L2 (R2 ) then u belongs to L2 ((0, T ); L∞ (R2 )). Part 1 Let d ≥ 1 and s ∈ R. Given T > 0 and g in H s (Rd ), we denote by u the unique function in C 0 ([0, T ]; H s (Rd )) ∩ C 1 ([0, T ]; H s−2(Rd )) solution to the Cauchy problem (8.21). 1. Given an integer  ∈ N∗ , introduce the Fourier multiplier P defined by, P  f (ξ ) = f(ξ )

P  f (ξ ) = 0

if |ξ | ≤ ,

if |ξ | > ,

 where  v (ξ ) = v(y)e−iyξ dy is the Fourier transform of v. Define u by u (t, ·) = P (u(t, ·)). Show that u ∈ C 1 ([0, T ], H s (Rd )) and that, for any time t ∈ (0, T ), 1 d u (t, ·) 2H s + ∇u (t, ·) 2H s = 0. 2 dt 2. Deduce that u belongs to L2 ((0, T ), H s+1 (Rd )). Part 2 d We now take g ∈ H 2 −1 (Rd ) and we denote by u the unique solution of (8.21). It follows from the previous part that, u

d

L2 ((0,T ),H 2 (Rd ))

≤ C g

H

d −1 2 (Rd )

.

The goal of this part is to improve this to, u L2 ((0,T ),L∞ (Rd )) ≤ C g

H

d −1 2 (Rd )

.

(8.22)

 1. Consider the Littlewood–Paley decomposition in Rd , I = j ≥−1 j . Set (j u)(t, ·) = j (u(t, ·)). Prove that j u is the unique solution of the problem, (∂t − )j u = 0,

(j u)|t =0 = j g.

2. Show that there exists C > 0 such that, for any time t in [0, T ] and any j ≥ −1,   j u(t, ·)

L∞

≤C

 j ≥−1

 d 2j 2 j u(t, ·)L2 .

Deduce that, for some c > 0, u(t, ·) L∞ ≤ C

 j ≥−1

 d 2j  2j 2 e−ct 2 j g L2 .

8 Statements of the Problems of Chap. 7

143

3. a) Consider two sequences (fn )n∈Z ∈ 1 (Z) and (gn )n∈Z ∈ 2 (Z). Prove that the sequence (hn )n∈Z defined by, hn =



fn−k gk

k∈Z

belongs to 2 (Z). b) Consider two sequences (aj ) and (ak ) indexed by j, k = −1, 0, 1, . . .. Prove that, +∞ ∞  ∞   aj ak ≤ an2 . 2|k−j |

j =−1 k=−1 

c) Set, aj = 2

j

d 2 −1



n=−1

  j g  2 . Prove that, L u 2L2 ((0,T );L∞ ) ≤

∞  ∞  j =−1 k=−1

aj ak , c2|k−j |

and then deduce (8.22).

Problem 46 Entropy Notions and the Heat Equation

In this problem, we are interested in finding quantities which are nonincreasing along the flow of the heat equation. We shall consider functions h = h(t, x) : [0, +∞) × Rd → (0, +∞) which are 2π-periodic with respect to xj for any 1 ≤ j ≤ d, belong to C ∞ ([0, +∞) × Rd ) and satisfy the heat equation, ∂t h − h = 0, where  =

d

2 j =1 ∂j ,

∂j =

∂ ∂xj

in (0, +∞) × Rd ,

.

Notations We shall set in what follows, ∇h = (∂j h)1≤j ≤d and, ∇ 2 h = (∂i ∂j h)1≤i,j ≤d ,

∇h ⊗ ∇h = ((∂i h)(∂j h))1≤i,j ≤d ,

d    (aij )2 = aij2 . i,j =1

144

8 Statements of the Problems of Chap. 7

1. Show that, (∂t − )Log h =

|∇h|2 , h2

|∇h|2 , h  2   ∇ h ∇h ⊗ ∇h 2 |∇h|2   . (∂t − ) = −2h  −  h h h2 (∂t − )(h Log h) = −

2. Introduce the function t → H (t), known as the Boltzmann’s entropy, defined by,  H (t) =

[0,2π]d

h(t, x)Log (h(t, x)) dx.

Prove that the function H decays in a convex manner that is, d H ≤ 0 and dt

d2 H ≥ 0. dt 2

(The quantity H is also known as the Gibbs’ entropy or also as the Shannon’s entropy.) 3. In this question, we further assume that the initial data h0 (x) = h(0, x) is a probability density which means that,  [0,2π]d

h0 (x) dx = 1.

 a) Verify that [0,2π]d h(t, x) dx = 1 for all time. b) Introduce the functions t → F (t) and t → J (t) defined by,  F (t) =

[0,2π]d

 J (t) =

[0,2π]d

|∇h(t, x)|2 dx, h(t, x)   2  ∇ h(t, x) ∇h(t, x) ⊗ ∇h(t, x) 2  dx.  − h(t, x)   h(t, x) h(t, x)2

Show that, d F + 2J = 0. dt (The quantity F is called the Fisher’s information).

8 Statements of the Problems of Chap. 7

145

c) Given a real number λ introduce the quantity, A(λ) =



d   d i,j =1 [0,2π]

h

∂ij h (∂i h)(∂j h) − + λδij h h2

2 dx,

where δij = 1 when i = j and 0 when i = j . Verify that, A(λ) = J − 2λF + λ2 d. Then by choosing λ appropriately deduce that, J ≥

1 2 F . d

d) Consider the function t → N(t) defined by,

2 N(t) = exp − H (t) . d (This function, introduced by Shannon, is called the entropy power). Prove that the function t → N(t) is concave that is, d2 N ≤ 0. dt 2 e) Let u : [0, +∞) → (0, +∞) be is a C 1 function satisfying for some K > 0 the inequality, ∂t u + Ku2 ≤ 0. Prove that, u(t) ≤

u(0) . 1 + Ku(0)t

Hint: Consider the function 1/u. f) Conclude that there exists a constant C > 0 such that, for any time t ≥ 1,  [0,2π]d

|∇h(t, x)|2 C dx ≤ . h(t, x) t

Problem 47 Lyapunov Functions for the Mean-Curvature Equation

146

8 Statements of the Problems of Chap. 7

In this problem, we consider functions h = h(t, x) : [0, +∞) × Rd → R which are 2π-periodic with respect to xj for any 1 ≤ j ≤ d, belong to C ∞ ([0, +∞) × Rd ) and are solutions to the mean-curvature equation, ∂t h +

2

1 + |∇h|2 κ = 0

where κ = − div 

∇h 1 + |∇h|2

! .

Here ∇ and div denote the derivatives and the divergence with respect to the spatial variable x = (x1 , . . . , xd ). We are interested in finding quantities which are nonincreasing along the flow of this equation. 1. Show that, d dt



2

[0,2π]d

 1 + |∇h|2 dx =

[0,2π]d

(∂t h)κ dx.

Deduce that d dt

 [0,2π]d

2 1 + |∇h|2 dx ≤ 0.

2. In the rest of the problem we assume that the space dimension d is equal to 1 and we use the notations, hx = ∂x h,

hxx = ∂xx h.

Verify that, ∂t h +

2

1 + h2x κ = ∂t h −

∂xx h , 1 + h2x

and deduce that, d dt



2π 0

h2x dx ≤ 0.

3. a) Verify that, ∂t h +

2

1 + h2x κ = ∂t h − ∂x arctan(∂x h).

b) Prove that, d dt



2π 0

h2 dx ≤ 0.

8 Statements of the Problems of Chap. 7

147

c) Compute ∂t (hx arctan(hx )) and then prove that, d2 dt 2



2π

h2 dx ≥ 0.

0

(Together with the previous result this shows that convex manner.) d) Prove that h˙ = ∂t h satisfies the equation, ∂t h˙ = −∂x



 2π 0

h2 dx decays in a

h˙ x . 1 + h2x

e) Prove that, d dt



2π

(∂t h)2 dx ≤ 0,

0

and then deduce that, d dt



2π 0

(1 + h2x )κ 2 dx ≤ 0.

Problem 48 The Boussinesq Equation In this problem, we consider functions h = h(t, x) : [0, +∞) × Rd → (0, +∞) which are 2π-periodic with respect to xj for any 1 ≤ j ≤ d, belong to C ∞ ([0, +∞) × Rd ) and are solutions to the following Boussinesq’s equation, ∂t h − div(h∇h) = 0, where ∇ and div denote the derivatives and the divergence with respect to the spatial variable x = (x1 , . . . , xd ). As in the previous problems we are interested in finding quantities which are nonincreasing along the flow of this equation. Part 1. A Functional Inequality The goal of this part is to prove a Sobolev type inequality which will be used to study the Boussinesq equation. We want to prove that, for any d ≥ 1 and any positive function θ ∈ C ∞ (Rd ) which is 2π-periodic with respect to xj for any 1 ≤ j ≤ d we have,  [0,2π]d

 1/2 4 ∇θ  dx ≤ 9 16

 [0,2π]d

(θ )2 dx.

148

8 Statements of the Problems of Chap. 7

1. Verify that, 

 [0,2π]d

(θ )2 dx =

[0,2π]d

   2 2 ∇ θ  dx,

where ∇ 2 θ = (∂j ∂k θ )1≤j,k≤d . 2. Show that,  I := 16

[0,2π]d

 1/2 4 ∇θ  dx,

satisfies,  I =−

[0,2π]d

  ∇θ −1 · ∇θ |∇θ |2 dx.

3. Deduce that I can be written under the form,   −1 2 θ θ |∇θ | dx + 2 I= [0,2π]d

[0,2π]d

θ −1 [(∇θ · ∇)∇θ ] · ∇θ dx.

  4. Prove that, |(∇θ · ∇)∇θ | ≤ |∇θ | ∇ 2 θ  and deduce that,  I ≤ 3 I 1/2

1/2 [0,2π]d

(θ )2 dx

.

Conclude the proof. Part 2. Lyapunov Functionals Assume that h is a smooth positive solution to ∂t h − div(h∇h) = 0 which is 2πperiodic with respect to xj for any 1 ≤ j ≤ d. 1. Let m ≥ 0. Show that,   d m+1 h dx + m(m + 1) hm |∇h|2 dx = 0. d dt [0,2π]d [0,2π] 2. Show that, d dt

 [0,2π]d

h2 |∇h|2 dx ≤ 0.

Hint: Multiply the equation by ∂t (h2 ) and integrate by parts.

8 Statements of the Problems of Chap. 7

149

3. The goal of this question is to prove that, m ∈ [0, (1 +

√ 7)/2]

⇒



d dt

[0,2π]d

hm |∇h|2 dx ≤ 0.

a) Check that,   d m 2 h |∇h| dx = mhm−1 div(h∇h)|∇h|2 dx dt [0,2π]d [0,2π]d  + 2hm ∇h · ∇div(h∇h) dx. [0,2π]d

b) Prove that, d dt

 hm |∇h|2 dx =

[0,2π]d



+m

[0,2π]d

m2 + 1 2



hm h|∇h|2 dx − 2

[0,2π]d

hm−1 |∇h|4 dx



[0,2π]d

  2 div h(m+1)/2 ∇h dx.

Hint: Use the identity,   2 (m − 1)2 div(hm ∇h) div(h∇h) = div h(m+1)/2∇h hm−1 |∇h|4 . − 4 c) Prove that,  (m + 1)

[0,2π]d

 hm |∇h|2 h dx = −  +

Hint: Integrate by parts twice. Deduce that,  hm |∇h|2 h dx = − [0,2π]d

1 m+1

[0,2π]d

[0,2π]d

hm+1 (h)2 dx   div hm+1 ∇h h dx.

 [0,2π]d

hm+1 (h)2 dx

 2   1 dx div h(m+1)/2 ∇h + m + 1 [0,2π]d  m+1 hm−1 |∇h|4 dx. − d 4 [0,2π]

150

8 Statements of the Problems of Chap. 7

Hint: Use the identity,    2 (m + 1)2  hm−1 |∇h|4 . − div hm+1 ∇h h = div h(m+1)/2 ∇h 4 d) Deduce that, d dt



m2 − m + 2 h |∇h| dx = 4

[0,2π]d

m − m+1





2

m

m+1

[0,2π]d

h

m+2 (h) dx− m+1

[0,2π]d

hm−1 |∇h|4 dx



2

[0,2π]d

  2 div h(m+1)/2∇h dx.

e) Use the inequality proved in Part 1 to show that, 



[0,2π]d



(m+1)/2

div h

 2 (m + 3)2 ∇h dx ≥ hm−1 |∇h|4 dx. 36 [0,2π]d

f) Deduce that for any m ≥ 0, d dt

 [0,2π]d

hm |∇h|2 dx + Im ≤ 0,

with, m Im = m+1



 m+1

[0,2π]d

h

(h) dx + Cm 2

[0,2π]d

hm−1 |∇h|4 dx,

where m + 2 (m + 3)2 m2 − m + 2 · − . m+1 36 4 √ g) Verify that Cm ≥ 0 for m ∈ [0, (1 + 7)/2]. Cm =

4. Conclude from the previous questions that the square of the L2 -norm decays in a convex manner that is, d dt

 [0,2π]d

h2 dx ≤ 0 and

d2 dt 2

 [0,2π]d

h2 dx ≥ 0.

5. We now introduce the Boltzmann’s entropy (already considered in Problem 46) that is the function t → H (t) defined by,  H (t) =

[0,2π]n

h(t, x)Log (h(t, x)) dx.

8 Statements of the Problems of Chap. 7

151

Prove that H decays in a convex manner that is, d H ≤ 0 and dt

d2 H ≥ 0. dt 2

Problem 49 A Morawetz’ Inequality

This problem uses Problem 13. Let  = ∂t2 − 3i=1 ∂x2i be the wave operator in R × R3 and set I = (0, +∞). Let f, g ∈ C0∞ (R3 ). We recall that the unique solution of the problem, u = 0 in I × R3 ,

u|t =0 = f,

∂t u|t =0 = g

can be written as u = u1 + u2 + u3 , where u1 (t, x) = u2 (t, x) =

1 4π

 S2

f (x − tω) dω,

u3 (t, x) =

t 4π

 S2

g(x − tω) dω,

3  ∂f t  (x − tω)ωi dω. 4π S2 ∂xi i=1

1. Using the Cauchy–Schwarz inequality and the polar coordinates in R3 prove that there exist constants Cj > 0 (independent of f, g) such that for all x ∈ R3 , u1 (·, x) L2 (I ) ≤ C1 g L2 (R3 ) ,    f (x − ·)    u2 (·, x) L2 (I ) ≤ C2  , | · | L2 (R3 ) u3 (·, x) L2 (I ) ≤ C3

3 

(8.23) (8.24)

! 12 ∂xi f 2L2 (R3 )

(8.25)

.

i=1

2. Deduce that there exists C > 0 (independent of f, g) such that, ⎛ ⎜ sup u(·, x) L2 (I ) ≤ C ⎝ g L2 (R3 ) +

x∈R3

3 

! 12 ∂xi f 2L2 (R3 )

i=1

Hint: Use the Hardy inequality (4.9) (see Problem 13).

Problem 50 A Nonlinear Wave Equation

⎞ ⎟ ⎠.

152

8 Statements of the Problems of Chap. 7

 In what follows we will denote  = ∂t2 − 3j =1 ∂x2j the wave operator in R × R3  and |∇x u|2 = 3j =1 (∂xj u)2 . The goal of this problem is to study the existence of a solution to the problem u = |∇x u|2 − (∂t u)2 ,

u|t =0 = 0,

∂t u|t =0 = g(x),

(8.26)

where g will be specified in the sequel. Part 1 1. Let u be a smooth solution of (8.26). Set v = eu . Of which problem v is it solution?  t 2. Show that v = 1 + w where w(t, x) = 4π S2 g(x − tω) dω. We assume in what follows that g ∈ L∞ (R3 ), ∇x g ∈ L1 (R3 ) and that lim|x|→+∞ g = 0.  +∞ d 3. Writing g(x − tω) = − t dt [g(x − sω)] ds show that, |w(t, x)| ≤

1 ∇x g L1 (R3 ) , 4πt

 where ∇x g L1 (R3 ) = 3j =1 ∂xj g L1 (R3 ) . 1 ∇x g L1 (R3 ) . 4. a) Prove that |w(t, x)| < 1 if t > 4π b) Prove that, |w(t, x)| ≤

1 ∇x g L1 (R3 ) g L∞ (R3 ) 4π

if

t≤

1 ∇x g L1 (R3 ) . 4π

5. Assume that ∇x g L1 (R3 ) g L∞ (R3 ) < 4π. Show that the problem (8.26) has a unique solution defined on R × R3 . Part 2 We consider the case where lim|x|→+∞ g = 0, but ∇x g L1 (R3 ) g L∞ (R3 ) > 4π. Let δ > 0 be small. Let ϕ0 be the piecewise C 1 function on [0, +∞) defined by, ϕ0 (t) = 1

for t ∈ [0, 1],

1+δ 1 ϕ0 (t) = − t + δ δ

for t ∈ [1, 1 + δ],

ϕ0 (t) = 0

for t ≥ 1 + δ.

Let ε > 0 be small and set, λ=

1+ε , 1−ε

g(x) = −ϕ0

|x| . λ

8 Statements of the Problems of Chap. 7

153

1. Let |x0 | ≤ ελ and 1 < t0 < (1 − ε)λ = (1 + ε). Show that w(t0 , x0 ) = −t0 . Deduce that v(t0 , x0 ) < 0, and then that the problem (8.26) has no globally defined solution on R × R3 . We want now to give a lower bound for the quantity ∇x g L1 (R3 ) g L∞ (R3 ) . 2. Using polar coordinates show that ∇x g L1 (R3 ) g L∞ (R3 )



δ2 ≥ 4π 1 + δ + . 3

Part 3 Give a very simple example of function g, which does not tend to zero at infinity, for which the problem (8.26) has no globally defined solution on R × R3 . Comments The equation considered in this problem is the very beginning example of equations of the type u = f (∂t u, ∇x u) fully studied by S. Klainerman in a series of papers (see the book by C. Sogge [26]).  

Problem 51 On the Strichartz Estimate for the Wave Equation Consider the Cauchy problem for the wave equation in Rt × Rdx , u = 0,

u|t =0 = 0,

∂t u|t =0 = u1 .

(8.27)

Part 1 The goal of this part is to show that in dimension d = 3 the end point Strichartz estimate does not hold. More precisely we want to show that for every T > 0 there is no positive constant C such that, u L2 ((0,T ),L∞ (R3 )) ≤ C u1 L2 (R3 ) .

(8.28)

In R3 we denote by B(x0 , r) the Euclidian ball of center x0 and radius r and by e1 the point (1, 0, 0). Let A = B(2e1 , 2) \ B(e1 , 1) and 1A (x) be its indicator function.   1. a) Show that A = x ∈ R3 : 14 |x|2 < x1 ≤ 12 |x|2 . b) Deduce that in the spherical coordinates (ρ, θ1 , θ2 ) ∈ (0, +∞) × (0, π) × (0, 2π) we have,  π A ⊂ (ρ, θ1 , θ2 ) ∈ (0, 4) × (0, ) × (0, 2π) : 2   ρ   ρ arccos min( , 1) ≤ θ1 ≤ arccos . 2 4

154

8 Statements of the Problems of Chap. 7

Hint: We recall the spherical coordinates: x1 = ρ cos θ1 , x2 = ρ cos θ1 cos θ2 , x3 = ρ cos θ1 sin θ2 , dx = ρ 2 sin θ1 dρ dθ1 dθ2 . c) Let gA : R3 \ {0} → R be defined by, gA (x) =

1A (x) , |x|2(1 + |ln |x||)α

1 < α ≤ 1. 2

(8.29)

Show that gA ∈ L2 (R3 ). In what follows we shall take in (8.27) u1 = gA and we recall then that the solution u can be written as,  t u(t, x) = gA (x − tω) dω. (8.30) 4π S2   2. We set for k ∈ N, Ak = A ∩ x ∈ R3 : 2−k+1 < |x| ≤ 2−k+2 . a) Show that u(t, te1 ) =

+∞ t  mk , 4π

 where mk =

k=0

S2

gAk (te1 − tω) dω.

b) Show that for 1 < t < 2 and ω ∈ S2 we have, te1 − tω ∈ Ak ⇐⇒ 2−2k+1 < t 2 (1 − ω1 ) ≤ 2−2k+3 . c) Show that there exists c0 > 0 such that for large k we have,  Ik :=

{

ω∈S2 :t e1 −t ω∈Ak

}

dω = c0 t −2 2−2k .

Hint: Use the spherical coordinates recalled in question 1b) with ρ = 1. d) Deduce that there exists c1 > 0 such that, mk ≥ c1 t −2 /(1 + k)α . 3. Show that (8.28) does not hold for this u. Part 2 The goal of this part is to show that for any s ≥ 0 there is no C > 0 such that, u L2 (R+ ,L∞ (R3 )) ≤ C u1 H s (R3 ) , for every u1 ∈ H s (R3 ), where u is the solution of (8.27).

(8.31)

8 Statements of the Problems of Chap. 7

155

Let gA ∈ L2 (R3 ) be the function defined in (8.29). Let (gn ) ∈ S(R3 ) be a sequence which converges to gA in L2 (R3 ) and denote by u and un the corresponding solutions of (8.27) given by (8.30). 1. Our first goal is to prove that (un ) cannot be bounded in L2 (R+ , L∞ (R3 )). We argue by contradiction. Assume that there exists M > 0 such that un L2 (R+ ,L∞ (R3 )) ≤ M. a) Show that un − u L∞ (R3 , L2 (R)) ≤ C gn − gA L2 (R3 ) . t

x

b) Using the Banach–Alaoglu theorem show that there is a subsequence (uσ (n) ) and v ∈ L2 (R+ , L∞ (R3 )) such that,   uσ (n) , ϕ → v, ϕ ,

∀ϕ ∈ L2 (R+ , L1 (R3 )).

c) Prove that v = u and using Part 1 show a contradiction. d) Deduce that there exists a subsequence unk tending to +∞ in L2 (R+ , L∞ (R3 )) when nk goes to +∞. 2. For λ > 0 we set gnλk (x) = λ− 2 gnk ( xλ ) and we denote by uλnk the corresponding solution of (8.27) with u0 = 0, u1 = gnλk . a) Prove that 3

(i)

gnλk L2 (R3 ) = gnk L2 (R3 ) ,

(ii)

uλnk L2 (R+ ,L∞ (R3 )) = unk L2 (R+ ,L∞ (R3 )) ,

(iii)

3

lim gnλk H s (R3 ) = (2π) 2 gnk L2 (R3 ) .

λ→+∞

b) Prove that (8.31) cannot hold with a fixed C.

Problem 52 For u0 ∈ S (Rd ) we denote by S(t)u0 the solution of the problem, i∂t u + u = 0,

u|t =0 = u0 .

156

8 Statements of the Problems of Chap. 7

1. Let 2 ≤ p ≤ +∞ and p its conjugate defined by

1 p

+

Prove that there exists C > 0 such that for all u0 ∈ have, S(t)u0 Lp (Rd ) ≤ C|t|

−d( 21 − p1 )

1 p = 1.

Lp (Rd )

and all t = 0 we

u0 Lp (Rd ) .

2d 2. Assume d ≥ 2 and 2 < p < d−2 . Let u0 ∈ H 1 (Rd ). Using the density of ∞ d 1 d C0 (R ) in H (R ), the Sobolev embedding and question 1 show that,

lim S(t)u0 Lp (Rd ) = 0.

t →+∞

Problem 53 On the Smoothing Effect for the Schrödinger Equation Let u0 ∈ C0∞ (Rd ). We know that the problem, i∂t u + u = 0,

u|t =0 = u0 ,

where  =

d  j =1

∂x2j ,

(8.32)

has a unique solution u ∈ C 1 (R, L2 (Rd )) such that u(t) L2 (Rd ) = u0 L2 (Rd ) . The purpose of this problem is to show that, when d = 1, for any σ > 12 we have x−σ u ∈ L2 ((0, T ), H 2 (R)). 1

1. Let f, g ∈ C 1 (R, L2 (Rd )). Show that the map t → (f (t), g(t))L2 (Rd ) is C 1 on R and that, ∂t (f (t), g(t))L2 (Rd ) = (∂t f (t), g(t))L2 (Rd ) + (f (t), ∂t g(t))L2 (Rd ) . 2. Let B ∈ Op(S 0 ) and let u be the solution of the problem (8.32). Show that, ∂t (Bu(t), u(t))L2 (Rd ) = (i[B, ]u(t), u(t))L2 (Rd ) . 3. Deduce that for all T > 0 there exists C > 0 such that,    

T 0

  (i[B, ]u(t), u(t))L2 (Rd ) dt  ≤ C u0 2L2 (Rd ) .

We assume in what follows that d = 1.

8 Statements of the Problems of Chap. 7

157

 x dy 1 2 21 Let b = − 12 ξξ  0 y 2σ dy where σ > 2 and a = (1 + a ) . Let B = Op(b). 4. a) Show that b ∈ S 0 . b) Prove that i[B, ] = − x−2σ −1 ∂x2 + R where R ∈ Op(S 0 ) and  = 1 (I d − ∂x2 ) 2 . c) Deduce that for all T > 0 there exists C > 0 such that, 

T 0

∂x − 2 x−σ u(t) 2L2 (R) dt ≤ C u0 2L2 (R), 1

and then that, 

T 0

x−σ u(t) 2

1

H 2 (R)

dt ≤ C(1 + T ) u0 2L2 (R) .

(8.33)

5. Let u0 ∈ L2 (R) and (ϕk ) ⊂ C0∞ (R) be a sequence which converges to u0 in L2 (R). Denote by u (resp. uk ) the solution of the problem (8.32) with datum u0 (resp. ϕk ). a) Prove that (x−σ uk ) converges to an element v in L2 ((0, T ), H 2 (R)). b) Prove that the sequence (uk ) converges to u in L2 ((0, T ), L2 (R)) and deduce that it converges to u in D ((0, T ) × R). 1 c) Deduce that u satisfies x−σ u ∈ L2 ((0, T ), H 2 (R)) together with the inequality (8.33). 1

Problem 54 Let d ≥ 1, s > d/2 and let f ∈ C ∞ (C, C) be such that f (0) = 0. Our goal is to study the existence of a solution to the problem, i∂t u + u = f (u),

u|t =0 = u0 ,

(8.34)

when u0 ∈ H s (Rd ). We shall use the following result (see (3.3)): (i) if u ∈ H s (Rd ) we have f (u) ∈ H s (Rd ) and (ii) there exists F : R+ → R+ nondecreasing such that

f (u) H s (Rd ) ≤ F u H s (Rd ) u H s (Rd ) . Notations • We shall denote by u(t) the function x → u(t, x) and we shall set H s = H s (Rd ).

158

8 Statements of the Problems of Chap. 7

  2 • S(t) will denote the Schrödinger propagator defined by S(t)f =F −1 e−it |ξ | f . • We recall that for a ∈ R and f ∈ S (Rd ) S(t + a)f satisfies, (i∂t + ) (S(t + a)f ) = 0,

(S(t + a)f ) |t =0 = S(a)f.

Part 1 We study here the existence problem. We fix u0 ∈ H s with s > d/2. Let A be the operator u → A(u) where, 

t

A(u)(t) = S(t)u0 − i

S(t − σ )[f (u(σ )] dσ.

0

We set R = 2 u0 H s , I = (−T , T ) and,   B = u ∈ L∞ (I, H s ) ∩ C 0 (I, H s ) : u L∞ (I,H s ) ≤ R . endowed with the L∞ (I, H s ) norm. 1. Show that if T is small enough with respect to u0 H s the operator A sends B to B. 2. Show that if T is small enough with respect to u0 H s the operator A is contractive from B to B. 3. Deduce that the equation u = A(u) has a solution u ∈ B. 4. Using the linear theory show that this solution u satisfies (8.34). Part 2 We study here the uniqueness in the space L∞ (I, H s ) ∩ C 0 (I, H s ). Let u1 , u2 be two solutions in L∞ (I, H s )∩C 0 (I, H s ) of the equation u = A(u). We consider the set,   F = t ∈ I : u1 (t) = u2 (t) in H s . 1. Show that F is a nonempty closed subset of I. 2. We want to show that F is open in I . a) Let t0 ∈ F. Set vj (t) = uj (t + t0 ), j = 1, 2. Show that vj satisfies, 

t

vj (t) = S(t)uj (t0 ) − i

S(t − σ )[f (vj (σ )] dσ ,

t ∈ (−T − t0 , T − t0 ).

0

Hint: Use the fact that S(a)S(b) = S(a + b) for a, b ∈ R. b) Let ε > 0 be such that (t0 − ε, t0 + ε) ⊂ I. Using question a) show that,  v1 (t) − v2 (t) H s ≤

ε −ε

f (v1 (σ )) − f (v2 (σ )) H s dσ,

t ∈ (−ε, ε).

8 Statements of the Problems of Chap. 7

159

c) Deduce that there exists ε > 0 such that (t0 − ε, t0 + ε) ⊂ F and conclude. Part 3 We show here that if T is small enough, the map u0 → u is locally Lipschitz from H s to L∞ (I, H s ). 1. Let u0 , v0 ∈ H s be such that u0 H s ≤ R2 , v0 H s ≤ R2 and let u, v be the corresponding solutions. Prove that there exist T0 > 0 and C > 0 such that for T ≤ T0 we have, u(t) − v(t) H s ≤ C u0 − v0 H s ,

∀t ∈ (−T , T ).

Problem 55 Infinite Speed Propagation for the Schrödinger Equation Let P = i∂t + x be the Schrödinger operator and for j = 1, . . . , d, set Lj = xj + 2it∂xj . We consider the Cauchy problem, P u = 0,

u|t =0 = g.

1. Prove that [P , Lj ] = 0. α 2. Set Lα = Lα1 1 · · · Ld d . Using the fact that [P , Lα ] = 0 prove that if g ∈ L2 (Rd ) α is such that x g ∈ L2 (Rd ) for all α ∈ Nd then for t = 0, the function x → u(t, x) belongs to C ∞ (Rd ).

Problem 56 Soliton for the Benjamin–Ono Equation 1. Compute the Fourier transform of the function f (ξ ) = e−|ξ | , ξ ∈ R. 2. Deduce the Fourier transform of the function g(ξ ) = |ξ |e−|ξ | . Hint: We may consider for λ > 0 the functions gλ (ξ ) = |ξ |e−λ|ξ | . 1 3. Set for x ∈ R, Q(x) = 1+x 2 and for c ∈ R \ {0} , u(t, x) = 4cQ(c(x + ct)),

t ∈ R, x ∈ R.

a) Show that Q belongs to H s (R) for any real number s. b) We define the operator |Dx | on S (R) by

|Dx |T = F −1 |ξ |T .

160

8 Statements of the Problems of Chap. 7

Show that 1 |Dx |u = −cu + u2 . 2 c) Deduce that u satisfies the Benjamin–Ono equation, ∂t u + ∂x |Dx |u =

1 ∂x (u2 ). 2

Problem 57 Averaging Lemma

Let (, T , μ) be a measured set such that μ() < +∞. We consider a family (pω (ξ ))ω∈ which is L∞ in ω with values in the set of homogeneous symbols of degree 1 on Rd . Part 1 We assume that there exists δ ∈ (0, 1) and c0 > 0 such that, ∀θ ∈ Sd−1 , ∀a > 0,

μ {ω ∈  : |pω (θ )| ≤ a} ≤ c0 a 2δ .

( )

1. Let u ∈ L2 (Rd × , dx dμ). Show that the function ω → u(x, ω) belongs to L1 (, dμ) for almost all x ∈ Rd . Set,  U (x) = u(x, ω) dμ(ω). 

The goal of this part is to show that ( ) implies that there exists C > 0 such that,

U H δ (Rd ) ≤ C u L2 (Rd ×,dx dμ) + Op(pω )u L2 (Rd ×,dx dμ) ,

(

)

for all u ∈ L2 (Rd × , dx dμ) such that Op(pω )u ∈ L2 (Rd × , dx dμ). 2. a) Prove that U ∈ L2 (Rd ). (ξ ) = b) Prove that U u(ξ, ω)dμ(ω) (where  u is the Fourier transform of u  with respect to x) and that (ξ )|2 ≤ C |U

 | u(ξ, ω)|2 dμ(ω). 

8 Statements of the Problems of Chap. 7

161

For ξ ∈ Rd , ξ = 0 we set,  I1 (ξ ) =

 |pω (ξ )|≤1

 u(ξ, ω) dμ(ω),

I2 (ξ ) =

|pω (ξ )|≥1

3. Show that there exists C > 0 such that,  C |I1 (ξ )|2 ≤ | u(ξ, ω)|2 dμ(ω), |ξ |2δ 

 u(ξ, ω) dμ(ω).

∀ξ = 0.

4. Prove that for ξ = 0 we have, 1 |I2 (ξ )| ≤ 2 |ξ |



2

1 |pω (θ)|≥ |ξ|

dμ(ω) |pω (θ )|2

! 

|pω (ξ ) u(ξ, ω)|2 dμ(ω) , 

where θ = |ξξ | ∈ Sd−1 . 5. We fix θ ∈ Sd−1 . Let ν be the measure on R+ which is the image of the measure μ by the map  → R+ , ω → |pω (θ )|. a) How does condition ( ) translates on ν? b) Show that, 1 |ξ |2

 1 |pω (θ)|≥ |ξ|

dμ(ω) 1 = |pω (θ )|2 |ξ |2



+∞ 1 |ξ|

dν(t) 1 = lim N→+∞ |ξ |2 t2 :=

6. Writing JN (ξ ) = C > 0 such that,

1 |ξ |2

N−1  k=1

k+1 |ξ| k |ξ|

dν(t ) t2

JN (ξ ) ≤



N |ξ| 1 |ξ|

dν(t) , t2

lim JN (ξ ).

N→+∞

and using ( ) prove that there exists

C . |ξ |2δ

7. Deduce (

). Part 2 We study the converse. We assume (

) satisfied. We recall that U (x) =  u(x, ω) dμ(ω). 1. Let g : Rd → R be a continuous function such that,  ∀ψ ∈ S(Rd ),

Rd

 )|2 dξ ≤ 0. g(ξ )|ψ(ξ

162

8 Statements of the Problems of Chap. 7

Show that g(ξ ) ≤ 0 for all ξ ∈ Rd . 2. Applying (

) to u(x, ω) = ψ(x)ϕ(ω) where ψ ∈ S(Rd ) and ϕ ∈ L2 () is well chosen, using the Fourier transform and the previous question prove that ( ) is satisfied.

Problem 58 Kinetic Equations

Part 1 The last question of Part 2 of this problem uses Part 1 in Problem 57. For t ∈ R, x ∈ Rd , v ∈ Rd , d ≥ 1, we consider the problem, ∂f (t, x, v) + v · ∇x f (t, x, v) = 0, ∂t

f (0, x, v) = f 0 (x, v).

(8.35)

1. Using the method of characteristics prove that for f 0 ∈ C 1 (Rdx × Rdv ) this problem has a unique solution f ∈ C 1 (Rt × Rdx × Rdv ). 2. Prove that if in addition f 0 ∈ Lp (Rdx × Rdv ) then for t ∈ R we have f (t) ∈ Lp (Rdx × Rdv ), where f (t) denotes the function (x, v) → f (t, x, v), and, f (t) Lp (Rdx ×Rdv ) = f 0 Lp (Rdx ×Rdv ) . If for every (t, x) ∈ Rt × Rdx the function v → f (t, x, v) belongs to L1 (Rd ) we define the macroscopic density by,  ρ(t, x) =

f (t, x, v) dv. Rd

3. Assume that f 0 ∈ L1 (Rdx , L∞ (Rdv )). Prove that we have for t = 0, |ρ(t, x)| ≤

1 0 f L1 (Rdx ,L∞ (Rdv )) . td

4. In this question we want to prove a kind of Strichartz estimate for the macroscopic density. Let a, q, r be positive real numbers such that, d , 1 1.



2 1 =d 1− , q r

1≤a=

2d 2r < . r +1 2d − 1

8 Statements of the Problems of Chap. 7

163

a) Let  ∈ C0∞ (Rt × Rdx ). Prove the estimate,     

R Rd

  ρ(t, x)(t, x) dx dt       ≤ f 0 La (Rdx ×Rdv )  |(t, x + vt)| dt   R

,

La (Rdx ×Rdv )

where a is the conjugate exponent of a defined by a1 + a1 = 1.  2 b) We set I :=  R |(t, x + vt)| dt La (Rd ×Rd ) . Prove that, x

v

 

2      |(t, x + vt)| dt  I =  R 

,

Lr (Rdx ×Rdv )

where r is the conjugate exponent of r. c) Prove that, 



I≤ R

(t, ·) Lr (Rd )

R

!

1 d

|t − s| r

(s, ·) Lr (Rd ) ds

dt.

d) Deduce that,    1    I ≤  Lq (Rt ,Lr (Rd ))  d || x   |t| r

.

Lq (Rt ,Lr (Rdx ))

e) Using the Hardy–Littlewood–Sobolev inequality prove that one can find C > 0 such that, I ≤ C  2Lq (R ,Lr (Rd )) . t

x

Hint: Show that rd < 1 and that if b is defined by f) Using the previous questions deduce that,     

R Rd

1 q

=

1 b

+ rd − 1 then b = q .

  ρ(t, x)(t, x) dx dt  ≤ f 0 La (Rdx ×Rdv )  Lq (Rt ,Lr (Rd )) . x

5. Prove that there exists C(d) > 0 such that, ρ Lq (Rt ,Lr (Rdx )) ≤ C(d) f 0 La (Rdx ×Rdv ) .

164

8 Statements of the Problems of Chap. 7

Part 2 In this part we want to apply the result obtained in Part 1 of Problem 57 to the following situation.   Let M > 0 and B(0, M) = v ∈ Rd : |v| ≤ M . Let χ ∈ C0∞ (Rdv ) with supp χ ⊂ B(0, M). We shall take,  = Rd ,

μ = χ(v) dv,

pv (τ, ξ ) = τ + v · ξ.

For a > 0 and (τ, ξ ) ∈ Sd−1 we set,   a = v ∈ Rd : |v| ≤ M, |τ + v · ξ | ≤ a . Eτ,ξ 1. Let O be a d × d orthogonal matrix and,   a a O(Eτ,ξ . ) = w ∈ Rd : w = Ov, v ∈ Eτ,ξ Prove that,   a O(Eτ,ξ ) = w ∈ Rd : |w| ≤ M, |τ + w, Oξ  | ≤ a a a μ(Eτ,ξ ) = μ1 (OEτ,ξ ),

and,

where μ1 = χ(O −1 w) dw.

2. Prove that one can find a d × d orthogonal matrix O such that,   a μ(Eτ,ξ ) = μ1 w ∈ Rd : |w| ≤ M, |τ + |ξ |wd | ≤ a . 3. Let a > 12 . Prove that for any (τ, ξ ) ∈ Sd we have, a μ(Eτ,ξ ) ≤ 2μ(B(0, M))a.

4. Let 0 < a ≤ 12 . a) Prove that one can find ε0 depending only on M such that if (τ, ξ ) ∈ Sd and |ξ | ≤ ε0 then |τ + v · ξ | > a. b) Let |ξ | > ε0 . Prove then that there exists a constant C(d, M) > 0 such that, a )≤ μ(Eτ,ξ

Hint: Use question 2.

C(d, M) a. ε0

8 Statements of the Problems of Chap. 7

165

5. Deduce from Part 1 of Problem 57 that if f ∈ L2 (Rt × Rdx × Rdv , dt dx dv) is a solution of the equation, ∂f (t, x, v) + v · ∇x f (t, x, v) = 0 ∂t then for every χ ∈ C0∞ (Rdv ) we have

 Rd

1

χ(v)f (t, x, v) dv ∈ H 2 (Rt × Rdx ).

Problem 59 Incompressibility

The purpose of this problem is to give a mathematical interpretation of incompressibility in hydrodynamics. Part 1 Let Md (R) be the algebra of d × d matrices with real coefficients. Let I ⊂ R be an open interval, A ∈ C 0 (I, Md (R)), B ∈ C 1 (I, Md (R)), t0 ∈ I satisfying, dB (t) = A(t)B(t), dt

t ∈ I,

(8.36)

Let (t) = det B(t) and let Tr (A(t)) be the trace. The goal of this part is to prove that, d (t) = (t)Tr (A(t)) dt



t

and therefore, (t) = (t0 )exp

Tr (A(s)) ds .

t0

Let B(t) = (bij (t))1≤i,j ≤d and let i (t) be the i t h line of the matrix B(t), so that, 

d (t) = det 1 (t), . . . ,  i (t), . . . , d (t) . dt d

i=1

1. Show that,  dbj k d (t) = (t)cof(bj k (t)), dt dt d

i=1

where cof(bj k ) is the cofactor of bj k .

166

8 Statements of the Problems of Chap. 7

2. We set adj(B(t)) = t(cof(bik (t))) the adjoint matrix of B(t). Show that, Tr

d dB (t)adj(B(t)) = (t). dt dt

3. Conclude. Let v ∈ C 1 (Rt × Rdx , Rd ), T > 0 and for t ∈ [0, T ) let X(t) be the solution of the problem, ˙ X(t) = v(t, X(t)),

X(0) = y ∈ Rd .

(8.37)

We shall set X(t) = X(t, y) = t (y). Let 0 ⊂ Rd be a bounded open set and t = t (0 ). Let μ be the Lebesgue measure. Our goal is to show that the equality μ(t ) = μ(t ) for all t, t ∈ [0, T ) is equivalent the fact that (divx v)(t, x) = 0 for all (t, x) ∈ [0, T ) × Rd where to d divx v = j =1 ∂xj vj . Part 2 We assume (divx v)(t, x) = 0 for all (t, x) ∈ [0, T ) × Rd . 1. a) Show that the map t : 0 → t is injective for all t ∈ [0, T ). Hint: Assume X(t0 , y) = X(t0 , y ); show that,   E = t ∈ [0, T ) : X(t, y) = X(t, y ) is a nonempty open and closed subset of [0, T ). b) Compute its Jacobian determinant (t, y).   ∂X Hint: Using (8.37) show that the matrix ∂ykj (t, y) satisfies a differential equation. c) Deduce that t is a C 1 -diffeomorphism from Rd to Rd .   2. a) For t ∈ [0, T ) set V (t) = μ(t ) = t dx. Show that V (t) = 0 (t, y) dy. b) Deduce that,  V (t) = (divx v)(t, X(t, y))(t, y) dy. 0

c) Conclude. Part 3 We assume that there exists (t0 , x0 ) ∈ [0, T ) × Rd such that (divx v)(t0 , x0 ) = 0. Let y0 be the unique point in Rd such that X(t0 , y0 ) = x0 . Then there exists 0 a neighborhood of y0 such that, (divx v)(t0 , X(t0 , y)) = 0,

∀y ∈ 0 .

8 Statements of the Problems of Chap. 7

167

Set t = t (0 ). 1. Prove that there exists t, t ∈ [0, T ) such that μ(t ) = μ(t ).

Problem 60 Propagation of the Irrotationality

This problem uses Problem 59. Let v ∈ C 2 (R × Rd , Rd ) be a solution of the Euler equations on [0, T ) × Rd , ∂s v + (v · ∇x )v = −∇x P ,

divx v = 0,

v|s=0 = v0 .

(8.38)

We recall (see Problem 59) that if X(s, y) is the solution on [0, T ) of the problem dX (s, y) = v(s, X(s, y)), ds

X(0, y) = y ∈ Rd ,

then for all s ∈ [0, T ), the map y → X(s, y) is a diffeomorphism from Rd to Rd . Set, (curl v)(s, x) =: A(s, x) = (ωij (s, x))1≤,i,j ≤d ,

ωij = ∂i vj (s, x) − ∂j vi (s, x).

Our goal is to show that if (curl v0 )(·) = 0 then (curl v)(s, ·) = 0 for all s ∈ [0, T ). 1. Using (8.38) show that for all 1 ≤ i < j ≤ d we have, ∂s ωij + (v · ∇x ) ωij +

d 

∂i vk ∂k vj − ∂j vk ∂k vi = 0.

(8.39)

k=1

2. Prove that, ∂i vk ∂k vj − ∂j vk ∂k vi = ∂i vk ωkj + ∂j vk ωki . 3. Denote by B(s, ·) the matrix (∂i vj (s, ·))1≤i,j ≤d . Show that A(s, ·) satisfies the matrix differential equation, dA (s, ·) + (v · ∇x )A(s, ·) + B(s, ·)A(s, ·) − A(s, ·)tB(s, ·) = 0. ds 4. a) Set C(s) = A(s, X(s, ·)). Show that C(s) satisfies the differential equation, dC (s) + B(s, X(s, ·))C(s) − C(s)tB(s, X(s, ·)) = 0. ds

168

8 Statements of the Problems of Chap. 7

b) Denoting by · any matrix norm deduce that,  C(s) ≤ C(0) + 2

s

B(σ, X(σ, ·)) C(σ ) dσ.

0

5. Conclude.

Problem 61 If v = (vj )1≤j ≤3 is a vector field on R3 and E is a functional space we shall say that  v ∈ E if vj ∈ E, 1 ≤ j ≤ 3. If v ∈ C 1 (R3 ) we shall denote div v = 3j =1 ∂j vj its divergence. 1. Let f ∈ S(R3 ). Show that there exists v ∈ H +∞ (R3 ) = ∩s∈R H s (R3 ) such that div v = f. ξ Hint: Consider the quantity |ξ j|2 f.

2. a) Choosing f appropriately, show that there exists v ∈ H +∞ (R3 ) such that,  R3

div v(x) dx = 0.

b) Deduce that there exists v such that div v ∈ L1 (R3 ) but there exists j ∈ {1, 2, 3} such that ∂j vj ∈ / L1 (R3 ). ξ 3. We set for j = 1, 2, 3, ρj (ξ ) = |ξj|2 ∈ L2 (R3 ). Let Tj be the operator defined on S(R3 ) by



Tj u = F −1 ρj (ξ ) u . Prove that for σ ∈ R there exists C > 0 such that, Tj u H σ +1 (R3 ) ≤ C u L1 (R3 ) + C u H σ (R3 ) ,

∀u ∈ S(R3 ).

Problem 62 Nonuniform Continuity of the Euler Flow

Our goal is to show that the flow of the incompressible Euler system on the torus T2 = R2 /(2πZ)2 (flow which is continuous) is not uniformly continuous in any H s for s ∈ R.

8 Statements of the Problems of Chap. 7

169

Notations A function f : T2 → C is a function f : R2 → C which is 2π-periodic in (x1 , x2 ) that is f (x1 + 2π, x2 ) = f (x1 , x2 ) = f (x1 , x2 + 2π). To f ∈ L1 (T2 ) we associate its Fourier coefficient,  f(k) = e−ik·x f (x) dx, k ∈ Z2 , (0,2π)2

For s ∈ R we define then H s (T2 ) as the space of f such that, f 2H s (T2 ) :=



(1 + |k|2 )s |f(k)|2 < +∞.

k∈Z2

Eventually if U = (u1 , u2 ) we set, U 2(H s (T2 ))2 = u1 2H s (T2 ) + u2 2H s (T2 ) . The Euler system on U = (u1 , u2 ) is, ∂U + (U · ∇x ) U − ∇x −1 div ((U · ∇x ) U ) = 0, ∂t div U = 0,

t > 0, x ∈ T2 ,

(8.40)

U |t =0 = U , 0

v ). where −1 v = F −1 ( |ξ1|2  The goal of the following questions is to construct two sequences of data (Un0 ), (Vn0 ) such that, lim Un0 − Vn0 (H s (T2 ))2 = 0,

n→+∞

(8.41)

but, for t > 0 arbitrary small and for n large enough, Un (t, ·) − Vn (t, ·) (H s (T2 ))2 ≥ c0 (t) > 0,

(8.42)

where Un , Vn denote the solutions of the system (8.40) with data Un0 , Vn0 at t = 0. 1. Let s ∈ R. Compute the norms in H s (T2 ) of the functions f (x1 , x2 ) = C ∈ C, f (x1 , x2 ) = sin(n x1 ), f (x1 , x2 ) = sin(n x2 ) where n ∈ Z. For n ∈ N and ω ∈ R we set Un,ω = (u1 , u2 ) where, u1 (t, x)=ωn−1 + n−s cos(nx2 − ωt),

u2 (t, x) = ωn−1 + n−s cos(nx1 − ωt). (8.43)

2. a) Compute A = div Un,ω ,

B=



∂ Un,ω + Un,ω · ∇x Un,ω ∂t

170

8 Statements of the Problems of Chap. 7

and

C = div (Un,ω · ∇x )Un,ω . b) Set wn = sin(nx1 − ωt) sin(nx2 − ωt). Compute 



1 w 2n2 n

 .

c) Let Q be a periodic solution of equation Q = wn and let R = Q + 2n12 wn . Prove that R = 0. 1  1  ik·x show that R(x) =  d) Writing R(x) = (2π) R(0) and 2 k∈Z2 R(k)e (2π)2

therefore that Q = −1 wn = − 2n12 wn + Cn , Cn ∈ C.

e) Compute ∇x −1 div (Un,ω · ∇x )Un,ω . f) Deduce that Un,ω is a solution of (8.40). Let,   0 = Un,ω |t =0 = ωn−1 + n−s cos(nx2 ), ωn−1 + n−s cos(nx1 ) , Un,ω (8.44) and 0 Un0 = Un,1 ,

0 Vn0 = Un,−1 .

3. Show that (8.41) is satisfied. 4. Let Un and Vn be the solutions of system (8.40) corresponding to these data. Show that, Un (t, ·) − Vn (t, ·) (H s (T2 ))2 ≥ C3 | sin(t)| −

C4 n

and conclude. Comments This problem is inspired by an article by A. Himonas and G. Misiolek [12].  

Problem 63 Finite Time Blow-Up for the Euler Equations Let T > 0. We consider the system on [0, T ) × R3 , ⎞ ⎛ 3 3  ∂vj ∂ρ  ∂ρ ⎠ ρ = 0, + vj +⎝ ∂t ∂xj ∂xj j =1

⎞ 3  ∂v ∂v k k ⎠ + 3ρ 2 ∂ρ = 0, + vj ρ⎝ ∂t ∂xj ∂xk ⎛

ρ|t =0 = ρ0 ,

(8.45)

j =1

j =1

vk |t =0 = v0k ,

k = 1, 2, 3.

(8.46)

8 Statements of the Problems of Chap. 7

171

This system corresponds to the compressible Euler equations in the case where P = ρ 3 . Set v = (v1 , v2 , v3 ). We shall say that (ρ, v) is a smooth solution if, (C0 )

ρ ≥ 0,

(C1 )

ρ, vk ∈ C 1 ([0, T ) × R3 , R),

(C2 )

ρ satisfies (8.45),

and

∂vk  ∂vk + vj =0 ∂t ∂xj

k = 1, 2, 3,

v satisfies (8.46)

if ρ(t, x) = 0

3

if ρ(t, x) = 0.

j =1

Such solutions exist. We consider in what follows a smooth solution. Part 1   For R > 0, let B(0, R) = x ∈ R3 : |x| < R . The goal is to prove that, supp ρ0 ∪ supp v0 ⊂ B(0, R) ⇒ supp ρ(t)∪ supp v(t) ⊂ B(0, R),

∀t ∈ [0, T ). (8.47)

1. a) Let |x0 | > R and T0 < T . Set, ⎛ M = sup ⎝|∇x ρ(t, x0 )| + t ∈[0,T0 ]

3 

⎞ |∇x vj (t, x0 )|⎠ .

j =1

Prove that there exists C > 0 independent of x0 such that for t ∈ [0, T0 ], |∂t ρ(t, x0 )| +

3 

|∂t vj (t, x0 )| ≤ CM(|ρ(t, x0 )| +

j =1

3 

|vj (t, x0 )|).

j =1

 b) Set u(t) = ρ(t, x0 )2 + 3j =1 vj (t, x0 )2 . Deduce that there exists C > 0 independent of x0 such that for t ∈ [0, T0 ], |∂t u(t)| ≤ Cu(t). 2. Prove (8.47). Part 2 We assume here that ρ0 and v0 have compact support and that ρ0 ≡ 0. Our goal is to prove that the solution (ρ(t), v(t)) cannot be global, that is T < +∞.

172

8 Statements of the Problems of Chap. 7

1. Prove, from the equations (8.45), (8.46), the identities, ∂ρ  ∂ + (ρvj ) = 0, ∂t ∂xj 3

j =1

 ∂ ∂ρ ∂ (ρvk ) + (ρvj vk ) + 3ρ 2 = 0, ∂t ∂xj ∂xk 3

1 ≤ k ≤ 3,

j =1

3    ∂  ∂  ρ|v|2 + ρ 3 + vj (ρ|v|2 + 3ρ 3 ) = 0, ∂t ∂xj

|v|2 =

j =1

3  (vj )2 . j =1

(8.48) 2. Prove that the quantity,  E=

R3

  ρ(t, x)|v(t, x)|2 + ρ(t, x)3 dx

is independent of t and that E > 0. 3. Prove that the quantity R3 ρ(t, x) dx is independent of t. We set in what follows,  H (t) = |x|2 ρ(t, x) dx. R3

4. Prove that there exists M > 0 such that H (t) ≤ M, 5. Prove that H is C 1 on (0, T ) and that, H (t) = 2

3   3 k=1 R

∀t ∈ [0, T ).

ρ(t, x)xk vk (t, x) dx.

6. a) Prove that H is C 1 on (0, T ) and that,    ρ(t, x)|v(t, x)|2 + 3ρ 3 (t, x) dx. H

(t) = R3

b) Deduce that H

(t) ≥ E, ∀t ∈ [0, T ). 7. Prove that H (t) ≥ H (0) + tH (0) + 12 Et 2 . 8. Deduce that T is necessarily bounded. Comments This problem is inspired by an article by T. Makino, S. Ukai, and S. Kawashima [20].  

Problem 64 On the Continuity of the Flow Map of the Incompressible Euler Equation

8 Statements of the Problems of Chap. 7

173

Consider the Cauchy problem for the incompressible Euler equations: ⎧ ⎪ ⎨ ∂u + (u · ∇)u = −∇P , ∂t ⎪ ⎩ u| t =0 = u0 .

div u = 0,

(8.49)

Let 0 < α < 1. Set ρ = 1 + α. The purpose of this problem is to show that when u0 ∈ W ρ,∞ (R3 , R3 ) although there exists for some T > 0 a unique solution u ∈ L∞ (([0, T ], W ρ,∞ (R3 , R3 )), the map u0 → u is not necessarily continuous in these spaces. Let T > 0 and let f, g, h be three functions belonging to W ρ,∞ (R, R). For t ∈ [0, T ] and x ∈ R3 we set, u(t, x) = (f (x2 ), 0, h(x1 − tf (x2 ))),

v(t, x) = (g(x2 ), 0, h(x1 − tg(x2 ))).

1. Show that u, v ∈ L∞ ([0, T ], W ρ,∞ (R3 , R3 )). 2. Show that u and v are two solutions of (8.49) with an appropriate P . 3. Given δ > 0 show that we can find f, g such that u0 − v0 W ρ,∞ ≤ δ and f (s) = g(s) for s ∈ R. We set a = T max( f L∞ , g L∞ ) and we choose h such that, h (s) = |s|α

for

− 2a ≤ s ≤ 2a.

4. Prove that, A(t) := u(t, ·) − v(t, ·) W ρ,∞ ≥ h (· − tf (·)) − h (· − tg(·)) W α,∞ . 5. Deduce that, A(t) ≥ B(t) where, B(t) is equal to, sup E

||x1 − tf (x2 )|α − |x1 − tg(x2 )|α − |y1 − tf (y2 )|α + |y1 − tg(y2 )|α )| , |x1 − y1 |α

where E = {(x, y) ∈ [−a, a]2 : x1 = y1 }. 6. Evaluating the right-hand side at x2 = y2 = c with −a < c < a and then at x1 = tg(c), y1 = tf (c) prove that A(t) ≥ 2. 7. Conclude. Comments This problem is inspired by a paper by G. Misiolek and T. Yoneda [22].  

Problem 65 On the Cauchy Problem for the Burgers Equation

174

8 Statements of the Problems of Chap. 7

Our goal here is to study the Cauchy problem for the Burgers equation, ∂t u + u ∂x u = 0,

u|t =0 = u0 ,

t ≥ 0, x ∈ R,

(8.50)

where u0 is a real valued function. Part 1. Necessary Condition, Data in C 1 (R) In this part we assume the existence of a real valued solution u which is C 1 on [0, +∞) × R. 1. Consider the characteristic of the equation (8.50) (t (s), x(s)) starting at (0, y) defined by, t˙(s) = 1,

t (0) = 0,

x(s) ˙ = u(s, x(s))

x(0) = y ∈ R.

We have obviously t (s) = s. Prove that, as long as x(s) exists, the solution u is constant on the characteristic. Deduce that x(s) = y + su0 (y) and therefore that x(s) exists for all s ≥ 0. 2. Prove that we have necessarily u(s, y + su0 (y)) = u0 (y). Part 2. Sufficient Condition, Data in C 1 (R) In this part assuming that u0 ∈ C 1 (R), and infy∈R u 0 > −∞, we want to solve the problem (8.50) using the results in Part 1. 1. Let F (t, y) = y + tu0 (y). Prove that for t ≥ 0, inf u 0 ≥ 0, ⇒

y∈R

inf

y∈R

u 0



< 0 ⇒

∂F (t, y) > 0, ∂y

∀(t, y) ∈ [0, +∞) × R ,

∂F 1 ,y ∈ R . (t, y) > 0 for 0 ≤ t < ∂y − infy∈R u 0

2. a) We set T ∗ = +∞ if infR u 0 ≥ 0 and T ∗ =

1 − infR u 0

if infR u 0 < 0. Prove that

there exists a unique map κ ∈ C 1 ([0, T ∗ ) × R) such that, for all t ∈ [0, T ∗ ), y + tu0 (y) = x ⇐⇒ y = κ(t, x),

∀x, y ∈ R.

b) Prove that if T u 0 L∞ (R) < 1 then T < T ∗ . 3. Using question 2 in Part 1. show that the problem (8.50) has a unique solution u ∈ C 1 ([0, T ∗ ) × R). 4. Prove that if u0 has compact support then necessarily T ∗ < +∞. Hint: Use a contradiction argument. 5. We assume t ∈ [0, T ] where T u 0 L∞ (R) < 1. Prove that, (∂x κ)(t, x) =

1 , 1 + tu 0 (κ(t, x))

(∂t κ)(t, x) =

−u0 (κ(t, x)) . 1 + tu 0 (κ(t, x))

8 Statements of the Problems of Chap. 7

175

Part 3. Maximal Time of Existence The purpose of this part is to show that when T ∗ is finite it is the right maximal time of existence. Assume u0 ∈ C 2 (R) with infx∈R u 0 < 0 and let u ∈ C 2 ([0, T ∗ ) × R) be the solution found in Part 2. 1. a) For x0 ∈ R such that u 0 (x0 ) < 0 set Tx∗0 = set q(t) = (∂x u)(t, x0 + tu0 (x0 )). Then q is a solution of the problem, q (t) = −q(t)2 ,

1 and for 0 ≤ t < −u 0 (x0 ) C 1 function. Prove that q

Tx∗0 is a

q(0) = u 0 (x0 ).

b) Deduce that limt →Tx∗ q(t) = −∞ and the maximal time of existence as a C 1 0

function of the solution of the problem (8.50) is T ∗ =

1 −infx∈R u 0 .

Unless otherwise stated we shall denote in what follows X = X(R) if X = L2 , L∞ , H k , C0∞ . Part 4. Data in H 2 The goal of this part is to show that if u0 ∈ H 2 and T is such that T u 0 L∞ < 1 then the solution found in Part 2 belongs to C 0 ([0, T ], H 2 ). 1. Let u0 ∈ H 2 and T be such that T u 0 L∞ < 1. Prove that we may apply the result in Part 2 and that the problem (8.50) has a unique solution u ∈ C 1 ([0, T ] × R) given by u(t, x) = u0 (κ(t, x)). 2. Prove that for fixed t ∈ [0, T ] the function x → u(t, x) belongs to H 1 . 3. Prove that for fixed t ∈ [0, T ] the function x → ∂x2 u(t, x) belongs to L2 . The goal of the following questions is to prove the continuity. Let t0 ∈ [0, T ] and (tn ) ⊂ [0, T ] such that limn→+∞ tn = t0 . Let ϕ ∈ C0∞ with supp ϕ ⊂ {x : |x| ≤ R0 } .  We set In = R |ϕ(κ(tn , x)) − ϕ(κ(t0 , x))|2 dx and we want to prove that limn→+∞ In = 0. 4. a) Set R = R0 + (t0 + 1) u0 L∞ (R) and take n ≥ n0 so large that |tn − t0 | ≤ 1 for n ≥ n0 . Prove that,  In = |ϕ(κ(tn , x)) − ϕ(κ(t0 , x))|2 dx. |x|≤R

Hint: Prove that if |x| ≥ R we have, |κ(tn , x)| ≥ R0 and |κ(t0 , x)| ≥ R0 . b) We set BR = {x : |x| ≤ R} and we denote by |BR | its volume. Prove that for n ≥ n0 we have, In ≤ |BR | ϕ L∞ and conclude.

u0 L∞ |tn − t0 | 1 − T u 0 L∞

176

8 Statements of the Problems of Chap. 7

 5. Let v ∈ L2 and Jn = R |v(κ(tn , x)) − v(κ(t0 , x))|2 dx. Prove that limn→+∞ Jn = 0. Hint: Take (ϕk ) ⊂ C0∞ such that (ϕk ) → v in L2 and use question 4b) above. 6. Let f = f (t, x) be a bounded continuous function on [0, T ] × R and v ∈ L2 . Prove that the function w(t, x) = f (t, x)v(κ(t, x)) belongs to C 0 ([0, T ), L2 ). 7. Let u be the solution considered in question 1. Prove that ∂xk u ∈ C 0 ([0, T ), L2 ) for k = 0, 1, 2. Part 5. Nonuniform Continuity of the Flow The goal of this part is to show that, for T > 0 fixed, the map u0 → u from H 2 (R) to C([0, T ], H 2 (R)) is not uniformly continuous. 1. Prove that it is sufficient to find ε0 > 0 and to construct two sequences of data (u0n ), (vn0 ) ⊂ H 2 such that limn→+∞ u0n − vn0 H 2 = 0 and if (un ), (vn ) denote the corresponding solutions of (8.50) then supt ∈[0,T ] un (t) − vn (t) H 2 > ε0 for n ≥ N. Let χ ∈ C0∞ , χ(x) = 1 if |x| ≤ 12 , χ(x) = 0 if |x| ≥ 1. Let (λn ), (εn ) be two sequences of positive real numbers such that, lim λn = +∞,

n→+∞

lim εn = 0,

n→+∞

lim λn εn = +∞.

n→+∞

(8.51)

Set −3

u0n (x) = λn 2 χ(λn x),

vn0 (x) = u0n (x) + εn χ(x).

(8.52)

Then obviously, lim u0n − vn0 H 2 = 0.

(8.53)

n→+∞ −1

−1

2. a) Prove that (u0n ) L∞ ≤ Cλn 2 and (vn0 ) L∞ ≤ C(λn 2 + εn ). We fix n0 so large that T (u0n ) L∞ ≤ 12 , T (vn0 ) L∞ ≤ 12 for n ≥ n0 . b) Deduce that for n ≥ n0 the solutions un , vn of (8.50) with data u0n , vn0 exist on [0, T ]. 3. Using question 2 Part 2 we set, y + tu0n (y) = x ⇔ y = κ1 (t, x),

y + tvn0 (y) = x ⇔ y = κ2 (t, x).

Using question 6a) Part 2 prove that there exists C > 0 such that, for j = 1, 2, for n ≥ n0 , and all t ∈ [0, T ] we have, 2 ≤ |∂x κj (t, x)| ≤ 2, 3

1

|∂x2 κj (t, x)| ≤ Ct (λn2 |χ

(λn κj (t, x))| + 1).

(8.54)

8 Statements of the Problems of Chap. 7

177

4. Prove that there exists C > 0 such that for all n ≥ n0 and all t ∈ [0, T ] we have, (u0n )

(κ1 (t, ·)) L2 ≥ C.

(8.55)

5. Recall that from Part 1 the solutions un , vn of (8.50) can be written as, un (t, x) = u0n (κ1 (t, x)),

vn (t, x) = vn0 (κ2 (t, x)).

Then we write, vn (t, x) − un (t, x) = An (t, x) + Bn (t, x), An (t, x) = (vn0 − u0n )(κ2 (t, x)) = εn χ(κ2 (t, x)),

(8.56)

Bn (t, x) = u0n (κ2 (t, x)) − u0n (κ1 (t, x)). Prove that limn→+∞ An (t, ·) H 2 = 0. 6. Prove that for j = 1, 2 we have limn→+∞ u0n (κj (t, ·)) H 1 = 0. Deduce that, . / Bn (t, ·) H 2 = ∂x2 u0n (κ2 (t, ·)) − u0n (κ1 (t, ·)) L2 + o(1), n → +∞. 7. a) Using (8.54) prove that for j = 1, 2,     lim (u0n ) (κj (t, ·))∂x2 κj (t, ·)

n→+∞

L2

= 0.

b) Deduce from (8.56) and questions 5, 6, 7a) that when n → +∞ we have, vn (t, ·) − un (t, ·) H 2 = fn (t, ·) − gn (t, ·) L2 + o(1) where, fn (t, x) = (u0n )

(κ2 (t, x)) (∂x κ2 (t, x))2 , gn (t, x) = (u0n )

(κ1 (t, ·)) (∂x κ1 (t, x))2 . Notice that, by construction, we have supp gn (t, ·) ⊂ {x : λn |κ1 (t, x)| ≤ 1} . c) Let x ∈ supp g(t, ·). Writing, κ2 (t, y + tu0n (y)) = κ2 (t, y + tvn0 (y)) + t (u0n (y) − vn0 (y))∂x κ2 (t, x ∗ ) prove that, κ2 (t, y + tu0n (y)) = y + t (u0n (y) − vn0 (y))∂x κ2 (t, x ∗ ) = y − tεn ∂x κ2 (t, x ∗ ).

178

8 Statements of the Problems of Chap. 7

Setting y + tu0n (y) = x deduce that for t > 0 and n large enough we have, λn |κ2 (t, x)| ≥ 2 and therefore that x ∈ / supp f (t, ·). d) Let α, β ∈ L2 be such that (supp α) ∩ (supp β) = ∅. Prove that α − β L2 ≥ √1 ( α L2 + β L2 ). Deduce that when n → +∞ we have, 2

vn (t, ·) − un (t, ·) H 2 ≥ C (u0n )

(κ1 (t, ·)) (∂x κ1 (t, ·))2 L2 + o(1). e) Using (8.54) and (8.55) prove that there exists ε0 > 0 such that for n large enough we have supt ∈[0,T ] vn (t, ·) − un (t, ·) H 2 ≥ ε0 .

Part II

Solutions of the Problems and Classical Results

Chapter 9

Solutions of the Problems

This chapter is devoted to the detailed solutions of the problems stated in the previous chapters.

Solution 1 1. a) Let χ ∈ C0∞ , χ(x) = 1 for |x| ≤ 1 and 0 ≤ χ ≤ 1. Set for k ≥ 1, χk (x) = χ( xk ). If u ∈ V set uk = χk u. Then,  x  |2 x2 |u(x)|2 dx. x (uk − u) 2L2 = |1 − χ d k R The right-hand side tends to zero by the Lebesgue dominated convergence; indeed, for fixed

x the interior of the integral tends to zero when k → +∞ and |1 − χ xk |2 x2 |u(x)|2 ≤ x2 |u(x)|2 ∈ L1 . Eventually we have, ∇x (uk − u) L2 ≤

1 ∇x χ L∞ u L2 + (1 − χk )∇x u L2 . k

When k → +∞ the first term in the right-hand side tends to zero obviously and the second one tends to zero as well by the Lebesgue dominated convergence as above.  b) Let θ ∈ C0∞ , supp θ ⊂ {x : |x| ≤ 1} , θ ≥ 0, Rd θ (x) dx = 1. For ε > 0 set

θε (x) = ε−d θ xε . For u ∈ Vc with supp u ⊂ {x : |x| ≤ R} set uε = θε u. Then uε ∈ C ∞ (Rd ) and supp uε ⊂ supp u + B(0, ε) ⊂ B(0, R + 1) if ε ≤ 1. It follows that, x (uε − u) L2 ≤ R + 1 uε − u L2 , and the right-hand side tends to zero when ε → 0 using the Hint. © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 T. Alazard, C. Zuily, Tools and Problems in Partial Differential Equations, Universitext, https://doi.org/10.1007/978-3-030-50284-3_9

181

182

9 Solutions of the Problems

On the other hand, since ∇x uε = θε ∇x u we have, uε − u 2H 1 = uε − u 2L2 + θε ∇x u − ∇x u 2L2 , and the right-hand side tends to zero when ε → 0 using the Hint. 2. Let B be the unit ball in V . Let (un ) be a sequence in B. We have to show that there is a subsequence (uσ (n) ) which converges in L2 (Rd ). Let ε > 0. Let R0 ≥ 1 be such that R10 ≤ 2ε . Let χ ∈ C0∞ , χ(x) = 1 if |x| ≤ 1, χ(x) = 0 if   |x| ≥ 2, 0 ≤ χ ≤ 1. Set χR0 (x) = χ Rx0 and write un = χR0 un + (1 − χR0 )un . We have, χR0 un H 1 ≤ χR0 un L2 +

1 ∇x χ L∞ un L2 + χR0 ∇x un L2 R0

≤ (1 + ∇x χ L∞ ) un L2 + ∇x un L2 ≤ (1 + ∇x χ L∞ ) un H 1 ≤ (1 + ∇x χ L∞ ). Therefore, (χR0 un ) belongs to a bounded set of HK1 where K is the compact   x ∈ Rd : |x| ≤ 2R0 . Since the map u → u from HK1 to L2 is compact we deduce that one can find a subsequence (χR0 uσ (n) ) which converges in L2 to some v ∈ L2 , that is, there exists N ∈ N such that for n ≥ N we have, χR0 uσ (n) − v L2 ≤

ε . 2

(9.1)

Now since 1 − χR0 = 0 for |x| ≤ R0 and 0 ≤ 1 − χR0 ≤ 1 we have,  (1 − χR0 )uσ (n) 2L2 ≤  ≤ ≤

|x|≥R0

|x|≥R0

|uσ (n) (x)|2 dx 1 x2 |uσ (n) (x)|2 dx |x|2

1 1 ε2 2 x , u ≤ ≤ σ (n) L2 4 R02 R02

since (un ) ⊂ B and R0 ≥ 2ε . It follows from (9.1) that uσ (n) − v L2 ≤ ε for n ≥ N. 3. a) The function v(x − hek ) belongs obviously to H 1 and the function x v(x − hek ) belongs to L2 since x v(x − hek ) L2x = y + hek  v L2y and y + hek  ≤ 2 y . Therefore, Dh v ∈ V . b) Since F (w(x − hek ))(ξ ) = e−ihξk w (ξ ) we have,  Dh w +

∂xk w 2H s−1

=

Rd

ξ 

  

2s−2  e

−ihξk

2 − 1 − (−ihξk )  w(ξ )|2 dξ.  | h

9 Solutions of the Problems

183

For fixed ξ the function in the interior of the integral tends to zero with h since, using the Hint with θ = −ihξk we have,   −ihξ k − 1 − (−ihξ )  e 1 2 k    ≤ 2 hξk .  h Moreover, we have, again by the Hint,   

2s−2  e

ξ 

−ihξk

−1

h

2  + iξk  | w(ξ )|2 ≤ 4ξk2 ξ 2s−2 | w(ξ )|2 ∈ L1 (Rd )

since w ∈ H s . By the Lebesgue dominated convergence theorem we have limh→0 Dh w + ∂xk w 2H s−1 = 0. c) Since for θ ∈ R, |eiθ − 1| ≤ |θ | we have, 2  −ihξ k − 1 e dξ  |  w (ξ )|2   h (2π )d  dξ ≤ ξk2 | w (ξ )|2 = ∂xk w 2L2 . (2π )d Rd

 Dh w 2L2

=

Rd

By the same way,  Dh g 2H −1

=

Rd

2  −ihξ k − 1 e  ξ −2 |  g (ξ )|2 dξ   h  ≤

Rd

ξk2 ξ 2

| g (ξ )|2 dξ ≤ (2π)d g 2L2 .

4. The left-hand side of (2.1) is the scalar product on V . Since f ∈ V the Riesz theorem gives the answer. If we take v ∈ C0∞ we see that the unique solution u ∈ V of (2.1) solves the equation −u + x2 u = f in the space of distributions. Moreover, taking u = v ∈ V and using the fact that L ∈ V we obtain, u 2V ≤ C L V u V , from which we deduce the desired inequality. 5. a) This follows immediately from the Cauchy–Schwarz inequality applied to the two integrals. b) If v = ϕ ∈ S then,   d  − d2 ξ −1 f1 (ξ ) ξ   ϕ (−ξ ) dξ = (2π)− 2 f1 ,  ϕ (−ξ ) S ×S (2π) Rd

= f1 , ϕS ×S ,

184

9 Solutions of the Problems

and 

x−1 f2 (x) x ϕ(x) dx = f2 , ϕS ×S ,

Rd

so L(f1 ,f2 ) (ϕ) = f, ϕS ×S . c) It follows from question b) that L(f1 ,f2 ) (ϕ) = L(g1 ,g2 ) (ϕ) for all ϕ ∈ S(Rd ). By question a) these are two linear and continuous form on V which coincide on S(Rd ) which is dense in V by question 1. Therefore, they coincide on V . d) This follows from the inequality (2.2) proved in question a), taking the infimum on all decompositions f = f1 + f2 and using the definition of the norm in E. e)  is obviously linear. Its continuity follows from the previous question and the fact that Lf V = sup v E =1 |Lf (v)|. Eventually it is injective since, if Lf = 0 then by question b) we have, 0 = Lf (ϕ) = f, ϕS ×S for all ϕ ∈ S, so f = 0 in S . f) This is the distributional formulation of the equality (2.1). Now, since u ∈ V we have, ξ −1 f1 = ξ −1 |ξ |2 u ∈ L2 and x−1 f2 = x−1 x2 u ∈ L2 . Moreover, L(ϕ) = f1 + f2 , ϕ = Lf (ϕ) for all ϕ ∈ S. Then L and Lf are two linear and continuous forms on V which coincide on the dense subspace S; therefore, they coincide on V . g) E and V are Banach spaces.  is linear, bijective, and continuous. Therefore, −1 is continuous.

Solution 2

Part 1 1. Let A be an orthogonal matrix and denote by B its transposed matrix. Set z = By in the integral in the middle below; since |y| = |z| and dy = dz we can write, 

 |y|≤r

eiλAθ,y dy =

 |y|≤r

eiλθ,By dy =

|z|≤r

eiλθ·z dz.

Using the polar coordinates, y = ρω, the equality between the first and the last integrals can be written as, 

r 0



 Sd−1

r

eiλρAθ,ω dω dr = 0

 eiλρθ,ω dω dr. Sd−1

Differentiating both members with respect to r and taking r = 1 we obtain, 

 F (λ, Aθ ) =

Sd−1

eiλAθ,ω dω =

Sd−1

eiλθ,ω dω = F (λ, θ ).

9 Solutions of the Problems

185

2. There exists an orthogonal matrix A such that Aθ = (0 . . . , 0, 1). It follows from the previous question that,  F (λ, θ ) =

eiλ ωd dω. Sd−1

If ω = (ω , ωd ) ∈ Sd−1 where ω = (ω1 , . . . ωd−1 ) we have ωd2 + |ω |2 = 1. Therefore,   2 d−1 d

2 S = E+ ∪ E− , E± = ω ∈ R : |ω | ≤ 1, ωd = ± 1 − |ω | .  It can be parametrized by |ω | ≤ 1, ωd = ± 1 − |ω |2 .  Now recall that the surface measure on a graph ωd = ψ(ω ) is given by   1 + |∇ω ψ|2 dω . Here ψ(ω ) = ± 1 − |ω |2 . It follows that we have, ωj |ω |2 ∂ωj ψ(ω ) = ∓  , |∇ω ψ|2 = , 1 − |ω |2 1 − |ω |2

2

1 1 + |∇ω ψ|2 =  . 1 − |ω |2

(9.2) Summing up we can write,  F (λ, θ ) =

|ω | 0

∃j0 : ∀j ≥ j0

χq uj − χq u H s (Rd ) ≤ ε,

∀q ∈ Zd ,

s (Rd ). which proves that uj → u in Hul 3. The constant functions belong to L2ul (Rd ) but of course not to L2 (Rd ). Indeed let u(x) = 1 for all x ∈ Rd . Then χq u L2 (Rd ) = χ(· − q) L2 (Rd ) = χ L2 (Rd ) , so supq∈Zd χq u L2 (Rd ) = χ L2 (Rd ) < +∞.

196

9 Solutions of the Problems

Part 2 1. a) We use an induction on |α| = k. The claim is true for k = 0. Assume it is γ true for k ≥ 0 and let |γ | = k + 1. Then ∂x = ∂xj ∂xα where |α| = k. Set u(x) = 1 + |x|2. We have, γ

∂x u(x)ρ = ∂xj [Pα (x)u(x)ρ−|α| ], = u(x)ρ−|α|−1 [(∂xj Pα )(x)u(x) + 2(ρ − |α|)xj Pα (x)], = u(x)ρ−|γ | Pγ (x), where Pγ is a polynomial of order |γ |. 1

b) Use a) and the fact that |Pα (x)| ≤ Cα (1 + |x|)|α| ≤ Cα (1 + |x|2 ) 2 |α| . c) By induction on |α| = k. The claim is true for k = 0. Assume it is true for k ≥ 0 and let |γ | = k + 1. Differentiating the inequality given in the Hint and using the Leibniz formula we obtain, γ γ γ ∂x 1 [u(x)ρ ]∂x 2 [u(x)−ρ ]. γ 1 =γ



u(x)ρ ∂x [u(x)−ρ ] = − γ

γ1 +γ2 γ1 =0

Since |γ2 | ≤ k, by induction and by question a), each term of the sum in 1 1 1 the right-hand side is bounded by Cu(x)ρ− 2 |γ1 | u(x)−ρ− 2 |γ2 | = u(x)− 2 |γ | . ρ Dividing both members by u(x) we obtain our claim at the rank k + 1. d) This is obvious since ∂xα [f (x − a)] = (∂xα f )(x − a). 2. By the Leibniz formula for |α| ≤ M we have, ∂xα k,q (x)

= k − q

N



α α1 1 ∂x ∂xα2 χ (x − k). N α x − q 1 =α

 α1 +α1

Using the result in question 1d) with ρ = |∂xα k,q (x)| ≤ Cα k − qN

N 2

 α1 +α1 =α

we get, 1 |∂xα2 χ (x − k)|. x − qN+|α1 |

On the support of χ k we have |x −k| ≤ A. Assume |k −q| ≥ 2A. Then |x −q| = |x − k + k − q| ≥ |k − q| − |x − k| ≥ 12 |k − q|. Since |α1 | ≥ 0 we get, |∂xα k,q (x)| ≤ Cα,N

 |γ |≤M

γ

|∂x χ (x − k)|.

9 Solutions of the Problems

197

If now |k − q| < 2A we have, |∂xα k,q (x)| ≤ Cα,N,A



γ

|∂x χ (x − k)|.

|γ |≤M

It follows that, χ H M (Rd ) , ∂xα k,q H M (Rd ) ≤ CM,N,A 

∀k, q ∈ Zd .

3. By the definition of k,q , and since ψq χq = χq we can write,   χ k χq u(x) = k − q−N k,q (x) x − qN ψq (x) χq u(x).  Fix N > d + 1. Since u = q∈Zd χq u, using the result in the previous question and the fact that the function x − qN ψq is C ∞ − bounded independently of q we can write,   k − q−N sup χq u H s (Rd ) .  χk u H s (Rd ) ≤  χk χq u H s (Rd ) ≤ C q∈Zd

4. Since the sum

 q∈Zd

q∈Zd

q∈Zd

k − q−N is finite, we have,

χk u H s (Rd ) ≤ sup χq u H s (Rd ) sup 

k∈Zd

q∈Zd

for all χ  ∈ C0∞ (Rd ). Exchanging the roles of χ  and χ we see that the fact that supq∈Zd χq u H s (Rd ) is finite is independent of the function χ.

Solution 8 1  2 2 α 1. Let u ∈ W m,∞ (Rd ). We have χq u H m (Rd ) = . |α|≤m ∂ (χq u) L2 (Rd ) Now, by the Leibniz formula, we have for |α| ≤ m, ∂ α (χq u) L2 ≤

α (∂ α1 χq )(∂ α2 u) L2 α 1 =α

 α1 +α2

≤ Cα



∂ α1 χq L2 ∂ α2 u L∞

α1 +α2 =α

≤ C u W m,∞ sup ∂ α χ L2 , |α|≤m

198

9 Solutions of the Problems

since (∂ α χq )(x) = (∂ α χ)(x − q). So sup χq u H m (Rd ) ≤ Cm u W m,∞ (Rd ) .

q∈Zd

2. If ε ∈ (0, 1) then we have by question 1, m d C∗m+ε (Rd ) = W m+ε,∞ (Rd ) ⊂ W m,∞ (Rd ) ⊂ Hul (R ).

If ε ≥ 1 we have C∗m+ε (Rd ) ⊂ C∗m+ε (Rd ) where ε < 1. 3. a) Since s = m+σ , if u ∈ W s+ε,∞ (Rd ) we have, ∂ α u ∈ W σ +ε,∞ (Rd ), |α| ≤ m σ (Rd ), |α| ≤ m and therefore u ∈ H s (Rd ). so ∂ α u ∈ Hul ul b) We write, χq u L2 (Rd ) ≤ χq L2 (Rd ) u L∞ (Rd ) = χ L2 (Rd ) u L∞ (Rd ) . c) Since |χq (x)u(x) − χq (y)u(y)|2 ≤ 2(|χq (x)u(x)|2 + |χq (y)u(y)|2) we have, 

 (3) ≤ C

|χq (x)u(x)|2

|x−y|>1

dy |x − y|d+2σ

dx ≤ C u 2L∞ (Rd ) .

d) In the set {|x − y| < 1} if |y − q| ≥ 2 we have |x − q| ≥ 1 so fq (x, y) = 0. e) This follows from the equality,

χq (x)u(x) − χq (y)u(y) = χq (x) (u(x) − u(y)) + χq (x) − χq (y) u(y). f) Since |u(x) − u(y)| ≤ |x − y|σ +ε u W σ +ε (Rd ) we have, 



 gq (x, y) dx dy ≤ 

Rd

|χq (x)|2

|x−y| 0 such that |Pj (rω)| = r j |Pj (ω)| ≤ Cj r j , ∀ω ∈ Vω0 . Then, for ω ∈ Vω0 there exists r0 > 0 such that, |P (rω)| ≥ |Pk (rω)| −

k−1  j =0

c0 k  c0 r − Cj r j ≥ r k , 2 4 k−1

|Pj (rω)| ≥

j =0

r ≥ r0 .

200

9 Solutions of the Problems

  (α) α (ξ ) = 2. If supp U ⊂ {0} we have U = |α|≤k cα δ0 . Then U |α|≤k cα ξ =: P (ξ ) where P is exactly of order k ∈ N. By hypothesis we should have, (ξ ) = (1 + |ξ |2 )− 2 P (ξ ) ∈ L2 (Rd ). (1 + |ξ |2 )− 2 U 1

1

But this is impossible if P does not vanish identically. Indeed, by the previous question we have for R & r0 > 0, 

R r0



(1 + r 2 )−1 |P (rω)|2 r d−1 dr dω ≥ Vω0

c02 |Vω0 | 16



R

(1 + r 2 )−1 r d−1+2k dr

r0

and the right-hand side tends to +∞ if R → +∞, since d ≥ 2 and k ≥ 0 imply d − 3 + 2k ≥ −1. 3. If u ∈ E ⊥ we have, (u, v)H 1 = (u, v)L2 +

d  j =1

∂j u, ∂j v

L2

=0

∀v ∈ E.

Taking v ∈ C0∞ (Rd \{0}) ⊂ E we deduce that U = −u+u = 0 in D (Rd \{0}). 4. Therefore, we have supp U ⊂ {0} and U ∈ H −1 (Rd ) since u ∈ H 1 (Rd ). We deduce from Part 1 that U = 0 in Rd and by Fourier transform that u = 0 since,  = 0 = (1 + |ξ |2 ) U u. 5. It follows that we have E ⊥ = {0} , so H 1 (Rd ) = E ⊥⊥ = E = E since, by definition, E is closed in H 1 (Rd ). 6. Assume this is true. Let u ∈ C0∞ (Rd ) ⊂ H 1 (Rd ) be such that u(0) = 1 and let (ϕj ) ⊂ C0∞ (Rd \ {0}) which converges to u in H 1 (Rd ). Then (ϕj ) would converge to u in C 0 (B(0, ε)), therefore at each point. Since for all j ∈ N we have ϕj (0) = 0 we would necessarily have u(0) = 0 which is false. Part 3 1. There exists a continuous map P1 : H 1 () → H 1 (Rd ) such that P1 u = u on . Moreover, the restriction map R : v → Rv = v| is continuous from H 1 (Rd ) to H 1 (). Let u ∈ H 1 (). Then P1 u ∈ H 1 (Rd ). By Part 2 there exists a sequence (ψk ) ⊂ C0∞ (Rd \ {0}) such that (ψk ) → P1 u in H 1 (Rd ). Set ϕk = Rψk . Then by definition (ϕk ) ⊂ C0∞ ( \ {0}). Now since R(ψk − P1 u) = ϕk − u we can write, ϕk − u H 1 () = R(ψk − P1 u) H 1 () ≤ C ψk − P1 u H 1 (Rd ) → 0.

9 Solutions of the Problems

201

Solution 10

Part 1 1. If this is not true there exists α0 > 0 such that for every C  > 0 one can find u ∈ H 1 () such that | {x ∈  : u(x) = 0} | ≥ α0 but  |u(x)|2 dx >  C  |∇u(x)|2 dx. Taking C = k ≥ 1 one can find uk ∈ H 1 () such that | {x ∈  : uk (x) = 0} | ≥ α0 with, 

 |uk (x)| dx = 1,

|∇uk (x)|2 dx
3/2 ≥ d2 , H m (Rd ) is an algebra and H m (Rd ) ⊂ L∞ (Rd ). We have, G(u) H m+1 ≤ C( (G(u) L2 + ∇x (G(u)) H m ). First of all we have, m+1 G(u) L2 ≤ M um+1 L2 ≤ K u m L∞ u L2 ≤ u H m+1 .

Then,     ∇x (G(u)) H m = (∇y G)(u)∇x uH m ≤ K (∇y G)(u)H m ∇x u H m . By assumption G ∈ C ∞ (Rp ) and |∂ α ∇y G(y)| ≤ M|y|m+1−|α|−1 = M|y|m−|α| for all y ∈ Rp and |α| ≤ m. Hence the function ∇y G satisfies the estimates (4.7) at the order m. Therefore, we can apply the induction assumption and we obtain,   (∇y G)(u) m ≤ K u m m . H H Therefore, G(u) H m+1 ≤ K u m+1 . H m+1 Part 2 1. By assumption, F ∈ C ∞ (Rp \ {0}). Differentiating the equality (4.8) |α| times, for |α| ≤ m, we find (∂ α F )(λy) = λm−|α| (∂ α F )(y) for y ∈ Rp \ {0} . Let 1 y ∈ Rp \ {0} . Using this equality with λ = |y| we obtain,    y  |(∂ α F )(y)| = |y|m−|α| (∂ α F ) ≤ ∂ α F L∞ (Sp−1 ) |y|m−|α| . |y| 

204

9 Solutions of the Problems

2. Using the Leibniz formula we can write, ∂ α F (y) = (1 − χ(y))(∂ α F )(y) −

 α |β| (∂ β χ)(y)(∂ α−β F )(y). β

β≤α β=0

Using question 1 we obtain for all y ∈ Rp , |∂ α F (y) ≤ M|y|m−|α| +

 α β≤α β=0

≤ M|y|m−|α| +

β

 α

β≤α β=0

β

|β| |(∂ β χ)(y)|M|y|m−|α|+|β|,

supz∈Rd (|z||β| |(∂ β χ)(z)|)M|y|m−|α|

≤ C(α, χ)M|y|m−|α| . 3. This follows immediately from question 1a) Part 1 since F satisfies the uniform estimates (4.7). 4. a) It follows from the definition (4.8) that F (0) = 0. Therefore, if u(x) = 0 we have F (u(x)) = 0 and if u(x) = 0 we have by question 1, |F (u(x)| ≤ M|u(x)|m . If m = 1, d ≥ 1 and u ∈ H 1 (Rd ) we have |F (u(x))ϕ(x)| ≤ M|u(x)ϕ(x)| ∈ L1 (Rd ) by the Cauchy–Schwarz inequality. If m ≥ 2, 1 ≤ d ≤ 3 and u ∈ H m (Rd ) ⊂ L∞ (Rd ) we have, 2 1 d |F (u(x))ϕ(x)| ≤ M|u(x)|m |ϕ(x)| ≤ CM u m−2 L∞ |u(x)| |ϕ(x)| ∈ L (R ).

b) By the Lebesgue dominated convergence theorem we have, 

 Rd

F (u(x))ϕ(x) dx = lim

→+∞ Rd

F (u(x))ϕ(x) dx.

On the other hand, by question 3 we have,    

Rd

  F (u(x))ϕ(x) dx  ≤ F (u) H m ϕ H −m ≤ K u m H m ϕ H −m .

It follows that,    

Rd

  F (u(x))ϕ(x) dx  ≤ K u m H m ϕ H −m ,

which implies F (u) ∈ H m (Rd ) together with F (u) H m ≤ K u m Hm.

9 Solutions of the Problems

205

Solution 12 We have trivially H u L2 (R) ≤ u L2 (R) , so the map u → H u is continuous from L2 (R) to itself. Now if u ∈ H 1 (R) then u is continuous and we have, ∂x (H u) = uδ0 + H ∂x u = u(0)δ0 + H ∂x u = H ∂x u so, ∂x (H u) L2 (R) = H ∂x u L2 (R) ≤ ∂x u L2 (R) ≤ u H 1 (R) . This shows that the map u → H u is continuous from H 1 (R) to itself. By interpolation, for 0 < s < 1, the map u → H u is continuous from H s (R) to itself and there exists C > 0 such that H u H s (R) ≤ C u H s (R) . Solution 13 1. Set J =

 +∞ 0

1 J = d −2

 0

|u(rω)|2 d−1 r dr. r2 +∞

Integrating by parts we can write,

1 d d−2 |u(rω)| dr = − r dr n−2



+∞

2

0

 d  |u(rω)|2 r d−2 dr, dr

since the function |u(rω)|2 r d−2 vanishes at r = 0 and r = +∞. Now, " #   d   d  d ωi (∂xi u)(rω) , |u(rω)|2 = 2Re u(rω) u(rω) = 2Re u(rω) dr dr i=1

which proves the equality (4.10). 2. Integrating both members of the equality (4.10) on Sd−1 we can write,  I ×Sd−1

|u(rω)|2 d−1 r dr dω r2 =−

2 Re d−2

 I ×Sd−1

d−1 u(rω) d−1 r 2 (ω · ∇x u) (rω)r 2 dr dω, r

where I = (0, +∞). By the Cauchy–Schwarz inequality we obtain,  I ×Sd−1

|u(rω)|2 d−1 2 r dr dω ≤ 2 r d −2   ·

  I ×Sd−1

|u(rω)|2 d−1 r dr dω r2

12

1

|ω · ∇x u(rω)| r

2 d−1

I ×Sd−1

2

dr dω

.

206

9 Solutions of the Problems

Using the fact that |ω · ∇x u| ≤ |∇x u| (since |ω| = 1), then setting x = rω, which implies that dx = r d−1 dr dω, we obtain,  Rd

2 |u(x)|2 dx ≤ |x|2 d −2

 Rd

|u(x)|2 dx |x|2

12 

1 2

Rd

|∇x u(x)| dx

2

.

We may of course assume that u does not vanish identically, which allows us to 1  2 2 dx . The desired result follows. divide both members by Rd |u(x)| 2 |x| Solution 14   1. Take A = x ∈ Rd : |f (x)| > λ which is a measurable set of finite measure. We have for all 0 < λ < +∞,   1  −1 |f (x)| dx, f Lqw (Rd ) ≥ (μ x ∈ Rd : |f (x)| > λ ) q {x∈Rd :|f (x)|>λ}   1 ≥ λ(μ x ∈ Rd : |f (x)| > λ ) q , so S(f ) ≤ f Lqw (Rd ) . 2. a) Set g(x) = 1A (x)|f (x)|. Using the Fubini theorem we can write, 

 |f (x)| dx =

R3

A

 =

g(x) dx =

!

g(x)

dλ R3

{x:g(x)>λ}

dx,

0

+∞ 

0







dx

 dλ =

+∞

  μ x ∈ Rd : |g(x)| > λ dλ,

0

+∞

μ {x ∈ A : |f (x)| > λ} dλ,

= 0

 =

+∞

μ 

   x ∈ Rd : |f (x)| > λ ∩ A dλ,

0

≤

+∞

    min μ x ∈ Rd : |f (x)| > λ , μ(A) dλ.

0

  b) By hypothesis we have, μ x ∈ Rd : |f (x)| > λ ≤ λ−q S(f )q . Therefore, 

 |f (x)| dx ≤ A

0

+∞



min λ−q S(f )q , μ(A) dλ.

9 Solutions of the Problems

207

Set λ0 = S(f )μ(A)



and h(λ) = min λ−q S(f )q , μ(A) . We have,

− q1





λ0

|f (x)| dx ≤



0

A

+∞

h(λ) dλ +

h(λ) dλ = I1 + I2 .

λ0

In I1 we have μ(A) ≤ λ−q S(f )q so that, 

λ0

I1 =

μ(A) dλ = μ(A)

1− q1

S(f ).

0

In I2 we have μ(A) ≥ λ−q S(f )q so that, since q > 1,  I2 = S(f )q

+∞

λ−q dλ = S(f )q

λ0

1−q

λ0 1 1− 1 = μ(A) q S(f ). q −1 q −1

It follows that, μ(A)

1 q −1

 |f (x)| dx ≤ A

q S(f ). q −1

q

c) The fact that f Lqw (Rd ) is a norm on Lw (Rd ) is easy. Solution 15 1. If f ∈ L∞ (Rd ) we have |fB | ≤



1 |B| B

|f (x)|dx ≤

1 ∞ |B| |B| f L

≤ f L∞ and

|f (x) − fB | ≤ |f (x)| + |fB | ≤ 2 f L∞ . It follows that, 1 |B|

 |f (x) − fB |dx ≤ 2 f L∞ B

1 |B| ≤ 2 f L∞ . |B|

2. We have, (fδ )B(x0 ,r) =

1 |B(x0 , r)|

 |x−x0 | 1 there exist ε0 > 0 such that |x0 − a| > 1 + ε0 and k0 such that |ak − a| ≤ ε20 for k ≥ k0 . Then, |x0 − ak | ≥ |x0 − a| − |ak − a| ≥ 1 + ε20 . Therefore, 1B(ak ,1) (x0 ) = 0   for all k ≥ k0 and 1B(a1)(x0 ) = 0. Eventually the set x ∈ Rd : |x − a| = 1 has Lebesgue measure zero. b) Let a ∈ Rd and let (ak ) be a sequence which converges to a. Set fk (x) = 1B(ak ,1) f (x). From question a) the sequence (fk ) converges to the function 1B(a,1)f (x) almost everywhere. On the other hand, if k is large enough we have, B(ak , 1) ⊂ B(a, 2) so that |fk (x)| ≤ 1B(a,2)(x)|f (x)| ∈ L1 (Rd ) since f ∈ L1loc (Rd ). We conclude by the Lebesgue dominated convergence.

210

9 Solutions of the Problems 1

c) There exists ε0 > 0 such that for |x| ≤ ε0 we have |x| 2 | ln |x|| ≤ 1. It follows that,   dx | ln |x|| dx ≤ < +∞. 1 |x|≤ε0 |x|≤ε0 |x| 2 On the other hand, the function ln |x| is continuous so locally integrable on the set {x : |x| > ε0 } .  d) Using the notation in question 5b) we have, |y−y0 | 2 and |y − y0 | < 1 we have, |y| > |y0 | − |y − y0 | > 1 so that t|y| + (1 − t)|y0 | > 1. Moreover, ||y| − |y0 || ≤ |y − y0 |. Summing up we obtain, | ln |y| − ln |y0 || ≤ |y − y0 .|. It follows that   | ln |y| − ln |y0 || dy ≤ |y − y0 | dy = 1. |y−y0 | 0. Now ρε ∈ Lp (R) for all p ∈ [1, +∞) and by question a) g ∈ L1 (R). Therefore, by the Young inequality we have ρε g ∈ Lp (R). In particular,   m 

ε L1 (R)

≤ ρε L1 (R) g L1 (R) ≤ g L1 (R) .

(9.12)

d) We now want to prove that the functions m ε are not bounded in L2 . To do so, we are going to prove the pointwise lower bound, ∀x ∈ [ε, 1],

C m ε (x) ≥ √ , x

9 Solutions of the Problems

215

for some positive constant C. Consider x ∈ [ε, 1]. Then xε ≥ 1. Since supp ρ ⊂ [0, 1] we have from the formula proved in question c) that, m ε (x)



1

=

ρ(z)f (x − εz) dz.

0

By question a) we know that f is nonincreasing; therefore, f (x−εz) ≥ f (x) so, m ε (x)



≥ f (x) 0

1

C ρ(z) dz = f (x) ≥ √ x

for x ∈ [ε, 1]. This implies that, 

1 0

m ε (x)2 dx ≥ C |log(ε)| .

(9.13)

e) We are now in a position to conclude the proof. By contradiction, let us assume that there exists a constant C > 0 such that, for all ε ∈ (0, 1] and all u in H 1 (R), mε u H 1 ≤ C m ε L1 u H 1 .

(9.14)

We fix a C ∞ function u supported in [−2, 2] and satisfying, u(x) = 1 for x ∈ [−1, 1]. Then the right-hand side of (9.14) is bounded by C u H 1 . Since by question b) we have, mε L∞ (R) ≤ 1 we obtain, mε u L2 ≤ mε L∞ u L2 ≤ u L2 .

(9.15)

mε u L2 ≤ mε L∞ u L2 ≤ u L2 .

(9.16)

Similarly,

Therefore, if (9.14) holds we would have, m ε u L2 ≤ C u H 1 . This is impossible. Indeed since u equals 1 on [0, 1], we have, using (9.13),   1 m ε u L2 (R) ≥ m ε L2 ([0,1]) ≥ C |log(ε)| 2 → +∞ when ε → 0. .

216

9 Solutions of the Problems

Solution 17

1. Since β > can write,

d 2

we have H β (Rd ) ⊂ L∞ (Rd ) with continuous embedding. So we

2j α vj L2 ≤ 2j α Sj −2 (a) L∞ j u L2 ≤ C a L∞ 2j α j u L2 , ≤ C a H β 2j (α−γ )cj u H γ ≤ Ccj a H β u H γ ,  since α ≤ γ , where j cj2 < +∞. It follows that Ta u belongs to H α (Rd ) and that we have the inequality (4.14). 2. Since γ − α > 0 we have γ − α + d2 > d2 . It follows that, 2j α vj L2 ≤ 2j α Sj −2 (a) L∞ j u L2 , ≤ 2j α Sj −2 (a)

d

H γ −α+ 2

j u L2 .

We know (see (3.13)) that Sj −2 (a) = ψ(2−j +4 D)a. It follows that, Sj −2 (a)

H

γ −α+ d2

≤ C2j (γ −α) a

d 2

H

.

Therefore, we have, 2j α vj L2 ≤ C2j α 2j (γ −α)cj 2−j γ a

H

≤ Ccj a

H

d 2

d 2

u H γ ,

u H γ ,

 where j cj2 < +∞. This proves (4.14) in this case. 3. In this case we have α ≤ β + γ − d2 . We write as before, vj L2 ≤ Sj −2 (a) L∞ j u L2 . Now we have Sj −2 (a) L∞ ≤

j −3 

k a L∞ .

k=−1

j and equal to one on the support of Let ϕ1 ∈ C0∞ (Rd ) be supported in a ring C ϕ. Then, −k −k −k   k a(ξ ) = ϕ(2 ξ )a = ϕ1 (2 ξ )ϕ(2 ξ )a,

9 Solutions of the Problems

217

from which we deduce, taking the inverse Fourier transform of both members, k a = 2kd ϕ1 (2k ·) k a. Therefore, we have, k a L∞ ≤ 2kd ϕ1 (2k ·) L2 k a L2 , so using the above inequality and the Cauchy–Schwarz inequality we get, ⎛ vj L2 ≤ ⎝

j −3 

⎞ 2 ϕ1 L2 k a L2 ⎠ j u L2 , kd 2

k=−1

⎛ ≤ ϕ1 L2 ⎝

j −3 

⎞1 ⎛ 2

2kd 2−2kβ ⎠ ⎝

k=−1

⎛ ≤ C ϕ1 L2 ⎝

j −3 

j −3 

⎞1 2

22kβ k a 2L2 ⎠ j u L2 ,

k=−1

⎞1 2

2

−2k(β− d2 )

⎠ a H β j u L2 .

k=−1

Since β −

d 2

d

d

< 0 we have 2−2k(β− 2 ) ≤ C2−2j (β− 2 ) for k ≤ j − 3 so that, d

vj L2 ≤ C2−j (β− 2 ) a H β j u L2 , therefore, d

2j α vj L2 ≤ C2−j (β−α+γ − 2 ) a H β 2j γ j u L2 , d

≤ C2−j (β−α+γ − 2 ) cj a H β u H γ ,  where j cj2 < +∞. Since β − α + γ − d2 ≥ 0 this proves (4.14). 4. Let us interpretate (4.14). If β > d2 we can take γ = α in (4.13). Thus Ta sends H α (Rd ) to H α (Rd ). It is therefore of order zero. If β = d2 we can take, by (4.13), (iii) γ = α + ε for every ε > 0. Thus Ta is of order ε. Eventually if β < d2 d

then we can take γ = α − β + d2 , thus Ta sends H α (Rd ) to H α−β+ 2 (Rd ). It is therefore of order d2 − β < 0.

218

9 Solutions of the Problems

Solution 18

Part 1 1. To estimate the first term in the left-hand side of (4.16) we want to use (4.14) with α = s0 , β = s1 , γ = s2 , a = u1 , u = u2 . Let us check (4.13). Using the hypotheses (i), (ii) we have, α ≤ γ , α ≤ β + γ − d2 and if β = s1 = d2 then α = s0 < s2 = γ . Therefore, (4.14) can be applied and gives the result. The estimate of the second term in (4.16) is symmetric. Part 2 −p , . The support of R ,j is contained in a ball of  1. a) We have  p Rj = ϕ(2 ξ )R j   j radius C2 and that of ϕ(2−p ξ ) is contained in ξ : 2p−1 ≤ |ξ | ≤ 2p+1 . Therefore, if C2j < 2p−1 that is if p − j > Ln(2C) Ln2 , we .have /p Rj = 0.  + 1. Therefore, we have p R = j ≥p−N0 p Rj where N0 = Ln(2C) Ln2

b) We have p Rj = F −1 (ϕ(2−p ξ )) Rj = 2pd (F −1 ϕ)(2p ·) Rj . By the Young inequality on convolutions we have, p Rj L2 (Rd ) ≤ 2pd (F −1 ϕ)(2p ·) L2 (Rd ) Rj L1 (Rd ) , ≤2

pd 2

F −1 ϕ L2 (Rd ) Rj L1 (Rd ) .

c) The first inequality follows from the fact that, j +2 

Rj L1 (Rd ) ≤

k u1 L2 (Rd ) j u2 L2 (Rd ) .

k=j −2

Moreover, we have, with (ck ), (dj ) ∈ 2 , j +2 

αj ≤ C

ck u1 H s1 (Rd ) dj u2 H s2 (Rd ) ,

k=j −2

and |j − k| ≤ 2 in the sum. Therefore, by Hölder inequality we get, ⎛ αj ≤ C ⎝

⎞



ck2 ⎠ dj u1 H s1 (Rd ) u2 H s2 (Rd ) .

k≥−1

2. a) From the questions 1a) and c) we have, d

βp ≤ C2 2p(s1+s2 − 2 )

 j ≥p−N0

2

pd 2

2−j (s1 +s2 ) αj = C2

 j ≥p−N0

2(p−j )(s1+s2 ) αj .

9 Solutions of the Problems

219

b) Using the Hölder inequality we can write, ⎛



βp2 ≤ C2 ⎝

⎞⎛ 2(p−j )(s1+s2 ) ⎠ ⎝

j ≥p−N0

⎞



2(p−j )(s1+s2 ) αj2 ⎠ .

j ≥p−N0

Now we have, 

2

(p−j )(s1 +s2 )

+∞ 

=

j ≥p−N0

2−k(s1 +s2 ) < +∞

k=−N0

since s1 + s2 > 0. Therefore, we get, 

βp2 ≤ C2





2(p−j )(s1 +s2 ) αj2 =

p≥−1 j ≥p−N0

p≥−1

≤ C3

 j ≥−1

⎛ ⎝

j +1 



⎛ ⎝

j ≥−1

⎞

2−k(s1 +s2 ) ⎠ αj2 ≤ C4



j +N0



⎞ 2(p−j )(s1 +s2 ) ⎠ αj2 ,

p=−1

αj2 .

j ≥−1

k=N0

c) Using the definition of βp it follows from question 1c) that, 

⎛ 2

2p(s1 +s2 − d2 )

p R 2L2 (Rd ) ≤ C5 ⎝



j ≥−1

p≥−1

⎞ dj2 ⎠ u1 2H s1 (Rd ) u2 2H s2 (Rd ) .

This proves (4.17). d) Using (4.16) and (4.17) and the fact that s0 ≤ s1 + s2 − u1 u2 ∈ H s0 (Rd ) and that we have the estimate (4.15).

d 2

we deduce that

Solution 19 β

1. a) Set aαβ = ∂xα ∂ξ a. (H0 ) is true by hypothesis. Assume (Hk ) satisfied and let us prove (Hk+1 ). We have then |aαβ (x, ξ )| ≤ C(1 + |ξ |)m−ρ|β|−k(ρ−1) . Since m − ρ|β| − k(ρ − 1) < 0 we have limξ1 →±∞ aαβ (x, ξ1 , ξ ) = 0. Therefore, for ξ1 > 0,  aαβ (x, ξ ) = −

+∞

ξ1

∂ξ1 aαβ (x, θ, ξ ) dθ.

220

9 Solutions of the Problems

It follows from (Hk ), since inside the integral a is differentiated |β| + 1 times with respect to ξ that, 

+∞

|aαβ (x, ξ )| ≤ C

(1 + θ + |ξ |)m−ρ(|β|+1)−k(ρ−1) dθ.

(9.17)

ξ1

Let M = m − ρ(|β| + 1) − k(ρ − 1). Then for |β| ≥ 0 and k ≥ 0 we have M + 1 ≤ m − ρ + 1 < 0. Therefore, we can integrate the right-hand side of (9.17) and we obtain, |aαβ (x, ξ )| ≤ C(1 + ξ1 + |ξ |)M+1 . We have just to notice that M + 1 = m − ρ|β| − (k + 1)(ρ − 1). For ξ1 < 0 we write,  aαβ (x, ξ ) =

ξ1 −∞

∂ξ1 aαβ (x, θ, ξ ) dθ

and we argue as above. This proves (Hk+1 ). b) If N ∈ N let us take k ∈ N such that m − ρ|β| − k(ρ − 1) < −N which is possible since ρ − 1 > 0. Then |aαβ (x, ξ )| ≤ C(1 + |ξ |)−N . So a ∈ S −∞ . Conversely it is obvious that S −∞ ⊂ Sρm . β

2. a) Since p is of order m − 1 we have ∂ξ p = 0 if |β| ≥ m. Now for |β| ≤ β

m − 1 we have |∂ξ p(ξ )| ≤ C(1 + |ξ |)m−1−|β| . We have just to notice that −(1 + |β|) ≤ −ρ|β|, which is equivalent to (ρ − 1)|β| ≤ 1 or to ε|β| ≤ 1, which is implied by our assumption ε(m − 1) ≤ 1. However, p ∈ / S −∞ . ∞ d 0 b) Take a(x, ξ ) = b(x) where b ∈ C0 (R ). Then a ∈ Sρ since for β = 0 we / S −∞ . have |∂xα b(x)| ≤ Cα and for β = 0, ∂xα ∂ξ b(x) = 0. However, a ∈ β

Solution 20 We make an induction on |α| + |β| = . The claim is true for  = 0. Assume it is  α∂β α β α β | =  + 1. We have the ∂x true for  and let | α | + |β ξ = ∂xi ∂x ∂ξ or ∂ξi ∂x ∂ξ where |α| + |β| = . Let ∂ = ∂xi or ∂ξi . By the induction we have, ⎛  β

α ∂x ∂ξ e−σ a = ⎝∂d0,αβ +

 

⎞ σ j (∂dj,αβ − (∂a)dj −1,αβ ) + σ +1 (∂a)d,αβ ⎠ e−σ a .

j =1

Therefore, d0,  = ∂d0,αβ , dj,  = ∂dj,αβ − (∂a)dj −1,αβ , for 1 ≤ j ≤ αβ αβ , d+1, = (∂a)d .  ,αβ αβ

9 Solutions of the Problems

221

−|β| , j −|β| + S 1 ×  = β. We have, d0, Case 1. ∂ = ∂xi . Then β dj, ∈ S ∈ S αβ αβ 1 −|β| ⊂ S +1−|β| . S j −1−|β| ⊂ S j −|β| , eventually, d+1, ∈ S ×S αβ −(|β|+1) ,  = |β| + 1. We have, d0, Case 2. ∂ = ∂ξi . Then |β| dj,  ∈ S  ∈ αβ αβ j −|β|−1 0 j −1−|β| j −(|β|+1) 0 −|β| S + S ×S ⊂ S , eventually, d+1, ⊂  ∈ S × S αβ S +1−(|β|+1) .

Solution 21

Part 1 1. The Fourier transform of v is given by,   v (ξ ) =



p(ξ ) − p(η) f(ξ − η) u(η) dη,

which immediately implies the desired result by splitting the integral into two parts. 1 2. We can write, p(ξ ) − p(η) = 0 (∇ξ p)(tξ + (1 − t)η)(ξ − η) dt, so using the definition of S m we obtain, |p(ξ ) − p(η)| ≤ C tξ + (1 − t)ηm−1 |ξ − η|. If |ξ − η| ≤ 12 |η| then tξ + (1 − t)ηm−1 ≤ C ηm−1 which gives the result. 3. a) If |ξ − η| ≤ 12 |η|, then |ξ | ≤ |ξ −η|+|η| ≤ 32 |η|. Therefore, since σ −m+1 ≥ 0 we have, ξ σ −m+1 ≤ C ησ −m+1 . So using question 2 we can write, ξ σ −m+1 |vL (ξ )| ≤ K



  u(η)| dη. |ξ − η|f(ξ − η) ησ |

b) By the Young inequality we have, h g L2 ≤ h L1 g L2 . We deduce from question a), taking the L2 norm of both members that, vL H σ −m+1 ≤  ζ  |f(ζ )| dζ u H σ . Now using the Cauchy–Schwarz inequality and the fact that 2(σ0 − 1) > d we can write,   ζ  |f(ζ )| dζ = ζ −(σ0 −1) ζ (σ0 −1) ζ  |f(ζ )| dζ ≤ C f H σ0 . 4. If |ξ − η| ≥ 12 |η| we obtain |η| ≤ 2|ξ − η| and |ξ | ≤ 3|ξ − η|. It follows that for t ∈ [0, 1] we have tξ + (1 − t)η ≤ C ξ − η . Since m ≥ 1 we deduce that |p(ξ ) − p(η)| ≤ C ξ − ηm . On the other hand, since σ ≥ m − 1 we also have ξ σ −m+1 ≤ C ξ − ησ −m+1 . Therefore,  ξ σ −m+1 |, vH (ξ )| ≤ C ξ − ησ +1−σ0 ξ − ησ0 |f(ξ − η)|| u(η)| dη.

222

9 Solutions of the Problems

Since σ +1−σ0 ≤ 0 and |ξ −η| ≥ 12 |η| we have, ξ − ησ +1−σ0 ≤ C ησ +1−σ0 . It follows that,  ξ σ −m+1 |, vH (ξ )| ≤ C ξ − ησ0 |f(ξ − η)| ησ +1−σ0 | u(η)| dη. Using the Young inequality as in question 3b) we deduce that,  vH H σ −m+1 ≤ C f

H σ0

1 ησ0 −1

ησ | u(η)| dη.

Using the Cauchy–Schwarz inequality and the fact that 2(σ0 − 1) > d we obtain the desired inequality. Part 2 1. At this stage, we have proved (6.1) for σ ∈ [m − 1, σ0 − 1]. We now consider the case σ ∈ [−σ0 + m, 0]. The two key observations are that μ := −σ + m − 1 ∈ [m − 1, σ0 − 1],

μ − m + 1 = −σ,

and that the adjoint of the operator C given by Cu = P (f u) − f P u is a commutator of the same form. Consequently, the result of the previous question implies that,  ∗    C ϕ  −σ = C ∗ ϕ  μ−m+1 ≤ K f H σ0 ϕ H μ = K f H σ ϕ −σ +m−1 . H 0 H H This proves that C ∗ ∈ L(H −σ +m−1 , H −σ ) with operator norm bounded by K f H σ0 , which in turn implies the desired result by a duality argument.

Solution 22 In this proof, we shall denote by T X→Y the operator norm of a bounded linear operator T : X → Y . 1. Observe that, f g − Tf g =



(j f )(p g) −

j,p≥−1

=







p≥−1 −1≤j ≤p−3

(p g)(j f ) =

j ≥−1 −1≤p≤j +2



(j f )(p g) = 

j ≥−1

(j f )(p g)

p≥−1 j ≥p−2

Sj +3 (g)j f.

9 Solutions of the Problems

223

Therefore, one can write,        Sj +2 (g) 2 j f  ∞ . fg − Tf g  2 ≤ L L L j ≥−1

Now recall that, according to (3.12), the support of the Fourier transform of Sj +3 (g) is contained in a ball of radius 2j +3 . It follows that,   Sj +3 (g)

L2

≤ C 2j θ g H −θ .

Then, using the characterization of Hölder spaces in terms of Littlewood–Paley decomposition and the assumption θ < ν we can write,    fg − Tf g  2 ≤ C 2θj g H −θ 2−j ν f W ν,∞ ≤ C g H −θ f W ν,∞ . L j ≥−1

2. Denote by f & the multiplication operator u → f u. We have, *

+ * + Q, f & = Q, Tf + Q(f & − Tf ) − (f & − Tf )Q.

(9.18)

Notice that Q = Tq since the symbol q does not depend on x. Consequently, the composition rule for paradifferential operators implies that, ∀σ ∈ R,

* +  Q, Tf 

H σ →H σ +ν

≤ c(ν, σ ) f W ν,∞ .

(9.19)

≤ C f W ν,∞ .

(9.20)

In particular, * +  Q, Tf 

H −ν →L2

With regards to the two other terms in the right-hand side of (9.18) we claim that,     Q(f & − Tf ) −θ 2 + (f & − Tf )Q −θ 2  f W ν,∞ . H →L H →L

(9.21)

This follows from the estimate proved in the first question and the fact that Q is bounded from H σ to itself for any σ ∈ R. 3. Denote by H0 the Fourier multiplier with symbol −i(1 − (ξ ))ξ/ |ξ | where  ∈ C0∞ (R) is such that (ξ ) = 1 on a neighborhood of the origin. With these + * notations, one can rewrite the commutator H, f & as, *

+ * + H, f & = H, Tf + H(f & − Tf ) − (f & − Tf )H, * + = H0 , Tf + (H − H0 )Tf − Tf (H − H0 ), + H(f & − Tf ) − (f & − Tf )H.

(9.22)

224

9 Solutions of the Problems

We want to prove that the five operators in the right-hand side of (9.22) are bounded from H −θ to L2 with operator norm controlled by f W ν,∞ . Notice that the first three terms have been estimated in the second question (one can use the estimates (9.20) and (9.21) since −i(1 − (ξ ))ξ/ |ξ | is a smooth symbol of order 0). So it remains only to study the last two terms. To do so, we notice that H − H0 is a smoothing operator (that is an operator bounded from H σ to H σ +t for any real numbers σ, t ∈ R). On the other hand, recall the following classical estimate for paraproducts: ∀σ ∈ R,

  Tf  σ ≤ c(σ ) f L∞ . H →H σ

We infer that,     (H − H0 )Tf  −θ 2 ≤ H − H0 −θ 2 Tf  −θ ≤ C f L∞ , H →L H →L H →H −θ     Tf (H − H0 ) −θ 2 ≤ Tf  2 2 H − H0 −θ 2 ≤ C f L∞ . H →L H →L L →L This completes the proof.

Solution 23

Part 1 In what follows we shall denote E = E(Rd ) if E = Lp , H s , C0∞ , S, S . . . 1. This follows from the fact that supp (∂xj u) ⊂ supp u and supp(au) ⊂ supp a ∩ supp u. 2. By definition we have supp (P Qu) ⊂ supp (Qu) ⊂ supp u. The same is true for the bracket. 3. We show that (supp u)c ⊂ (supp P ∗ u)c . Let x0 ∈ / supp u. There exists a neighborhood Vx0 such that for all ϕ ∈ C0∞ (Vx0 ) we have (u, ϕ)L2 (Rd ) = 0. Let θ ∈ C0∞ (Vx0 ). Since P is local we have supp (P θ ) ⊂ supp θ ⊂ Vx0 . So ϕ = P θ ∈ C0∞ (Vx0 ). Therefore, (u, P θ )L2 = 0, so (P ∗ u, θ )L2 = 0 which proves that x0 ∈ / supp P ∗ u. ∗ 4. a) If P P = 0, for all ϕ ∈ S we have 0 = (P ∗ P ϕ, ϕ)L2 = P ϕ 2L2 . So P ϕ = 0 for all ϕ ∈ S, therefore P = 0. b) If (P ∗ P ) = 0 for all ϕ ∈ S we have, 0 = ((P ∗ P ) ϕ, (P ∗ P )−1 ϕ)L2 = P (P ∗ P )−1 ϕ 2L2 , so P (P ∗ P )−1 ≡ 0. Then, 0 = (P (P ∗ P )−1 ϕ, P (P ∗ P )−2 ϕ)L2 = (P ∗ P )−1 ϕ 2L2 . By induction we deduce that P ∗ P = 0 and using question 4 that P = 0.

9 Solutions of the Problems

225

Part 2 1. Let ϕ ∈ C0∞ be such that ϕ(x) = 1 if |x| ≤ 1, ϕ(x) = 0 if |x| ≥ 2. Set, for k ≥ 1, uk (x) = (1 − ϕ(k(x − x0 ))u(x). This is an element of S which vanishes for |x − x0 | ≤ 1k . Now,  uk − u 2L2 =

|ϕ(k(x − x0 ))u(x)|2 dx.

The right-hand side goes to zero, when k → +∞, by the dominated convergence theorem. Indeed if x = x0 is fixed we have k|x − x0 | → +∞, so |ϕ(k(x − x0 ))u(x)|2 → 0. Moreover, |ϕ(k(x − x0 ))u(x)|2 ≤ ϕ 2L∞ |u(x)|2 ∈ L1 . 2. Since −m > d2 the embedding from H −m to L∞ is continuous. Therefore, there exists C1 > 0 such that supRd |Au(x)| ≤ C1 Au H −m for all u ∈ S. Now A being of order m it is continuous from L2 to H −m . So there exists C2 > 0 such that Au H −m ≤ C2 u L2 for all u ∈ S and the result follows. 3. By the previous question we have, |Auk (x0 ) − Au(x0 )| ≤ C uk − u L2 (Rd ) .

(9.23)

If A is local we have (suppuk )c ⊂ (suppAuk )c . Since uk ≡ 0 near x0 we have x0 ∈ (suppuk )c so that x0 ∈ (suppAuk )c . Therefore, Auk (x0 ) = 0 for k ≥ 1. Since the right-hand side of (9.23) goes to zero, we have Au(x0 ) = 0. This being true for all x0 ∈ Rd and all u ∈ S, therefore for all u ∈ S , we deduce that A ≡ 0. 4. The operator (A∗ A) is local by Part 1 and it is of order 2m < − d2 ; by the previous question we have (A∗ A) ≡ 0. It follows from Part 1 that A ≡ 0. 5. First of all if f is C ∞ -bounded we have f ∈ S 0 . Now if P ∈ Op(S r ) is local then the bracket (ad f )(P ) belongs to Op(S r−1 ) and it is local by Part 1. Applying this to P = A, f = fk then to P = (ad fk )(A), f = fk−1 and so on, eventually to P = (ad f2 ) . . . (ad fk )(A), f = f1 we see that (ad f1 )(ad f2 ) . . . (ad fk )(A) belongs to Op(S m−k ) and it is local. Since m−k < 0 it follows from the previous question that this bracket vanishes. 6. For  = 1 we have A(gw) = [A, g]w + gAw = −(ad g)(A)w + gAw. Since a ∈ S m and w ∈ C0∞ we have Aw ∈ S and |g(x)| ≤ C|x − x0 |. Assume the formula true at the order  ≥ 1. Using the induction with w replaced by gw and the fact (ad P )(Q) = −(ad Q)(P ) we obtain, (−1) A(g  gw) = (ad g) (A)gw+O(|x−x0|) = −(ad g)+1 (A)w+O(|x−x0|). 7. a) This follows from the Taylor formula with integral reminder. We have, vβ (x) =

k β!



1 0

(1 − t)k−1 (∂xβ u)(tx + (1 − t)x0 ) dt.

It is obviously a C ∞ function on Rd .

226

9 Solutions of the Problems

b) We have θ k+1 u = u so that using the previous question we get, Au(x) =





A (θ (x)(x − x0 ))β (θ vβ )(x) .

|β|=k

Setting for 1 ≤ j ≤ d, fj (x) = θ (x)(xj − x0,j ) we have, Au(x) =

  β β β A f1 1 f2 2 . . . fk k (θ vβ ) .

 |β|=k

Using question 6 with successively g = f1 , . . . , g = fd and w = θ vβ we obtain,  Au(x) = (−1)|β| (ad f1 )β1 . . . (ad fd )βd (A)(θ vβ ) + O(|x − x0 |). |β|=k

c) Since A is local and m < k we can use question 5 and deduce that, Au(x) = O(|x − x0 |). So Au(x0 ) = 0.  β 1 8. Let u ∈ C0∞ and w(x) = u(x) − |β|≤k−1 β! χ(x)(x − x0 )β ∂x u(x0 ) where χ ∈ C0∞ , χ(x) = 1 near x0 . Then w ∈ C0∞ and ∂xα w(x0 ) = 0 for |α| ≤ k − 1. By the previous question we have Aw(x0 ) = 0, so, Au(x0 ) =

 |β|≤k−1

1 A χ(x)(x − x0 )β ∂xβ u(x0 ). β!

This being true for all x0 ∈ Rd we have Au(x) = Therefore, it is a differential operator of order ≤ k − 1.



β |β|≤k−1 aβ (x)∂x u(x).

Solution 24

1. We have, 

Ak H s ≤ C

χk P χq u H s .

|q−k|≤2

χ (ξ )| we have, From (3.2) and the fact that |χk (ξ )| = |  χk v H t ≤ Ct,d

ξ |t | | χ (ξ )| dξ v H t .

9 Solutions of the Problems

227

Using the continuity of the pseudo-differential operators in the usual Sobolev space we deduce that, 

Ak H s ≤ C1 (s, d, χ)

P χq u H s

|q−k|≤2

≤ C2 (s, d, χ, p, m)



χq u H s+m

|q−k|≤2

⎛



≤ C3 (s, d, χ, p, m) ⎝

|q−k|≤2

⎞ 1⎠ u H s+m . ul

  For fixed k, the cardinal of the set q ∈ Zd : |k − q| ≤ 2 is finite and depends only on the dimension d. Therefore, we have, Ak H s ≤ C u H s+m , ul

where C is independent of k. 2. The support of ∂ α Bk,q is included in the support of χk . Let χ  ∈ C0∞ (Rd ) be α α equal to 1 on the support of χ. Then we have χ k ∂ Bk,q = ∂ Bk,q and therefore, χk ∂ α Bk,q L2 ≤  χk L2 ∂ α Bk,q L∞ =  χ L2 ∂ α Bk,q L∞ . ∂ α Bk,q L2 =  3. This follows from the expression of the DO and from the Leibniz formula. Indeed we have, with some constants cα1 ,α2 ,α3 , ∂xα Bkq (x)





=

cα1 ,α2 ,α3 ∂xα1 χk (x)

eix·ξ ξ α2 ∂xα3 p(x, ξ )χ q u(ξ ) dξ,

α=α1 +α2 +α3

and aα2 (x, ξ ) =: ξ α2 ∂x 3 p(x, ξ ) ∈ S m+|α2 | . 4. From the Lebesgue dominated convergence theorem and the Fubini theorem we can write,  α1 Ckqα (x) = lim ∂x χk (x) eix·ξ aα2 (x, ξ )θ (ε ξ )χ q u(ξ ) dξ, α

ε→0

= lim ∂xα1 χk (x) ε→0



 (χq u)(y)

ei(x−y)·ξ aα2 (x, ξ )θ (ε ξ ) dξ

dy.

5. We have, ε ε |Ckqα (x)| ≤ Kkqα (x, ·) H −(s+m) χq u H s+m (Rd ) ≤ Kkqα (x, ·) H n1 u H s+m . ul

228

9 Solutions of the Problems

6. This follows again from the Leibniz formula. We have, 

ε (x, y) = ∂yβ Kkqα

β1 +β2 =β

β α1 ∂ χk (x)∂yβ1 χ q (y) β1 x  ·

ei(x−y)·ξ (−iξ )β2 aα2 (x, ξ )θ (ε ξ ) dξ ,

and bα2 ,β2 (x, ξ ) = (−iξ )β2 aα2 (x, ξ ) ∈ S m+|α2 |+|β2 | . ε (x, y) we have |x − k| ≤ 1, |y − q| ≤ 7. On the support of Kkqα

3 2

so that,

|x −y| = |x −k+k−q +q −y| ≥ |k−q|−|x −k|−|q −y| ≥ |k−q|−

5 1 ≥ |k−q|. 2 6

8. Obvious. 9. a) Easy by induction on N. b) Integrating by parts and using the Leibniz formula we find, ε |Kkqαβ (x, y)| ≤ |∂xα1 χk (x)∂yβ1 χ q (y)|·





dγ ,γ1 M

|γ |

|γ |=N γ1 +γ2 =γ



ε|γ1 | |θ1 (ε ξ )||∂ξ 2 bα2 ,β2 (x, ξ )| dξ, γ

where θ1 ∈ C0∞ (Rd ). Now we have, M |γ | (x, y) ≤

C , |x − y|N

|x − y| ≥

1 |k − q|, 6

|∂xα1 χk (x)∂yβ1 χ q (y)| ≤ C,

where C is independent of k, q, so that we can write, ε |Kkqαβ (x, y)| ≤



CN |k − q|N

(ε ξ )|γ1 | |θ1 (ε ξ ) ξ −|γ1 | |∂ξ 2 bα2 ,β2 (x, ξ )| dξ, γ

and ξ −|γ1 | ∂ξ 2 bα2 ,β2 ∈ S m+n0 +n1 −N since |α2 | ≤ n0 , |β2 | ≤ n1 and |γ1 | + |γ2 | = N. Then it is sufficient to bound ε ξ )|γ1 | |θ1 (ε ξ ) by a constant independent of ε and to take m + n0 + n1 − N ≤ −(d + 1), or equivalently N ≥ m+n0 +n1 +d +1, to ensure that the integral is absolutely convergent. 10. It follows from the previous questions that, γ

 |k−q|≥3

⎛ Bkq H s ≤ C ⎝



|k−q|≥3

⎞ 1 ⎠ u H s+m ≤ C u H s+m , |k − q|N

9 Solutions of the Problems

229

 since, when N > d + 1 the series |p|≥3 |p|1N is convergent. Using question 1 this proves the continuity of the DO in the uniformly local Sobolev spaces.

Solution 25

1. Using the equality given in the statement we have, ∂ξα p(ξ )

=z

 |α2 |  α α1 μ ∂ξ ξ  zj qj α2 (ξ )e−za(ξ ). α 1 =α



μ

α1 +α2

j =0

It follows that, |∂ξα p(ξ )|

≤ Cαμ z

μ



ξ 

μ−|α1 |

α1 +α2 =α

|α2 | 

zj ξ j −|α2 | e−za(ξ ),

j =0 1

≤ Cαμ zμ ξ μ−|α| e− 2 za(ξ )

|α2 | 



1

(z ξ )j e− 2 za(ξ ).

α1 +α2 =α j =0 1

Eventually since a(ξ ) ≥ c0 ξ  we have (z ξ )j e− 2 za(ξ ) ≤ Cj . 2. From question 1 we have, 1

ξ −|α| . |∂ξα p(ξ )| ≤ Cαμ ξ −|α| (z ξ )μ e− 2 za(ξ ) ≤ Cαμ

Taking |α| = d + 1, we obtain, α

α (x)| = |∂ |x α p ξ p(x)| ≤ ∂ξ p L1 (Rd ) ≤ Cαμ



dξ ξ d+1

≤ Cαμ .

(9.24)

 To conclude we have just to notice that |x|d+1 ≤ Cd dk=0 |xk |d+1 and to take α = (0, . . . , d + 1, . . . , 0) in the above estimate. 1 3. a) We have z ξ  = (z2 + z2 |ξ |2 ) 2 so that using (9.24) and the estimate in question 1 we obtain, (x)| ≤ Cαμ z|α| |x α p

 (z2 + z2 |ξ |2 )

μ−|α| 2

c0

e− 2 (z

1 2 +z2 |ξ |2 ) 2

Set zξ = η so dξ = z−d dη. Then, (x)| ≤ Cαμ z|α|−d |x α p



c0

z, ημ−|α| e− 2 z,η dη.

dξ.

230

9 Solutions of the Problems

b) Since z ∈ (0, 1) we have |η| ≤ z, η ≤ η . Then,   c0 c0 z, ημ−|α| e− 2 z,η dη ≤ C e− 3 |η| dη := M0 < +∞. |η|≥1

|η|≥1

Now if |α| ≥ 1 we have,  |η|≤1

z, η

μ−|α| −

e

c0 2 z,η

 dη ≤

|η|≤1

since |α| = d or d − 1. If α = 0 we have,   c0 μ z, ημ e− 2 z,η dη ≤ 2 2 |η|≤1

dη = M1 < +∞, |η||α|−μ

|η|≤1

dη = M2 < +∞.

c) It follows from the above results that, |x|d | p(x)| ≤ M2 ,

|x|d−1 | p(x)| ≤ M3 z−1 .



1−ε d−1

ε |x| | p(x)| = |x|d | p (x)| p (x)| and the Using the equality |x|d−ε | above estimates we obtain the desired result. d) This follows from the fact that using questions 2 and 3 we have, | p (x)| ≤

C , |x|d+1

|x| ≥ 1,

| p(x)| ≤ C

z−ε , |x|d−ε

|x| ≤ 1.

4. Set A = j (e−za(Dx ) u) L∞ . We have, A = z−μ zμ Dx μ e−za(Dx ) Dx −μ j u L∞ = z−μ p(Dx ) Dx −μ j u L∞ ,

p Dx −μ j u L∞ ≤ Cz−μ  p L1 Dx −μ j u L∞ , = z−μ  ≤ Cε z−μ−ε 2−j (ρ+μ) u C∗ρ . 5. Choose ε > 0 such that μ + ε < 12 . Then, 

!2

1

sup 2

0

j ≥−1

j (ρ+μ)

j (e

which proves the result.

−za(Dx )

u) L∞ (Rd )

 dz ≤ Cε2

1 0

dz z2(μ+ε)

u 2C ρ , ∗

9 Solutions of the Problems

231

Solution 26

Part 1 1. We have,  |T u(x)| ≤

Rd

|K(z)| dz u L∞ (Rd ) ,

∀x ∈ Rd .

 So T is continuous from L∞ to itself and T L∞ →L∞ ≤ Rd |K(z)| dz. Now set   A = z ∈ Rd : K(z) = 0 . If μ(A) = 0 we have K ≡ 0, T ≡ 0 (since K is  continuous). If μ(A) = 0 let x0 ∈ Rd be fixed. Then T u(x0 ) = {x0 −y∈A} K(x0 −

K(x0 −y) if x0 − y ∈ A and u0 (y) = 0 y)u(y) dy. Let u0 be such that u0 (y) = |K(x 0 −y)| ∞ otherwise. Then u0 ∈ L , u0 L∞ (Rd ) = 1 and,

 T u0 (x0 ) =

Therefore,

T u0 L∞ (Rd ) u0 L∞ (Rd )

 {x0 −y∈A}

≥

 Rd

|K(x0 − y)| dy =

Rd

|K(z)| dz.

|K(z)| dz, so that T L∞ →L∞ ≥

 Rd

|K(z)| dz

and we have the equality. 2. a) If i = k we have i (0) = (0, . . . , 1, . . . , 0) and  k (λ) = (0, . . . , ωk2 , . . . , 0), which shows that det(1 (0), . . . ,  k (0), . . . , d (0)) = ωk2 . Therefore, F (0) = d 2 k=1 ωk = 1. b) Set Gk (λ) = det(1 (λ) . . . ,  k (λ) . . . d (λ)). Since 

k (λ) ≡ 0 we have G k (λ) =



det(1 (λ), . . . ,  j (λ), . . . ,  k (λ), . . . , d (λ)).

j =k

Now  j (λ) = ωj (ω1 , . . . , ωd ) and  k (λ) = ωk (ω1 , . . . , ωd ). Therefore, each determinant in the sum vanishes; then G k (λ) ≡ 0 and F

(λ) = d

k=1 Gk (λ) ≡ 0. It follows that we have F (λ) = 1 + λ. Part 2 1. Indeed we have j u C∗σ = supk≥−1 (2kσ k j u L∞ ). On the other hand, due to the support of ϕ, we have k j = 0 if |k − j | > 2, so,  j u C∗σ = max 2(j −1)σ j −1 j u L∞ , 2j σ j j u L∞ ,

 2(j +1)σ j +1 j u L∞ .

232

9 Solutions of the Problems

We just have to notice that j is uniformly bounded on L∞ to see that there exists C > 0 such that, j u C∗σ ≤ C2j σ j u L∞ ≤ C 2j σ u L∞ . 2. Since j commutes with S(−1), using the definition of C∗s , the assumption, and the previous question we obtain, 2s j S(−1)j u L∞ ≤ S(−1)j u C∗s ≤ C j u C∗σ , ≤ C 2j σ u L∞ , for all u ∈ L∞ (Rd ). 3. a) We have Tj u(x) = (2π)−d



ei[(x−y)·ξ +|ξ | ] ϕ 2 (2−j ξ )u(y) dy dξ, α

  −d i[(x−y)·ξ +|ξ |α ] 2 −j = (2π) ϕ (2 ξ ) dξ u(y) dy. e Set Kh (z) = (2π)−d We obtain,



ei[z·ξ +|ξ | ] ϕ 2 (2−j ξ ) dξ then η = hξ in the integral. α

Kh (z) = (2πh)

−d



e h [z·η+h i

1−α |η|α ]

ϕ 2 (η) dη.

b) We see that  −α α z . Kh = h−d F −1 eih |η| ϕ 2 (η) h   Since the support of ϕ is contained in 12 ≤ |η| ≤ 2 the function −α

eih |η| ϕ 2 (η) belongs to C0∞ (Rd ). Its inverse Fourier transform belongs to C 0 ∩ L1 . c) It follows from Part 1 that Tj is continuous from L∞ (Rd ) to itself and setting z = h1−α s that,   d(1−α) ∞ ∞ Tj L →L = |Kh (z)| dz = h |Kh (h1−α s)| ds, α

 = hd(1−α)

h (s)| ds, |K

where, h (s) = (2πh)−d K



−α φ(s,η)

eih

ϕ 2 (η) dη,

φ(s, η) = s · η + |η|α .

9 Solutions of the Problems

4. Recall that suppϕ ⊂ the integral we have,

233



1 2

 ≤ |η| ≤ 2 . We have

∂φ ∂η (s, η)

= s + α|η|α−2 η, so in

∂φ α 1 α . (s, η) ≥ α|η|α−1 − |s| ≥ |1−α| − |s| ≥ ∂η 2 2 2|1−α| We consider then the operator, L=

d

hα 1  ∂φ ∂ := hα L0 s, η, ∂η 2 i |∂η φ| ∂ηk ∂ηk k=1

−α

−α

−α

−α

ih φ(s,η) = e ih φ(s,η) for all It satisfies Leih φ(s,η) = eih φ(s,η) so hαN LN 0 e ∞ d N ∈ N. Since ϕ ∈ C0 (R ) we can integrate by parts and we obtain,

h (s) = hαN K



−α φ(s,η)

eih

(tL0 )N ϕ(η) dη,

where tL0 denotes the transpose operator of L0 . It follows that, h (s)| ≤ hαN |K

 |(tL0 )N ϕ(η)| dη.

5. Since |η|α−1 ≤ |η||α−1| ≤ 2|α−1| on the support of ϕ we have, |∂η φ(s, η)| ≥ |s| − α|η|α−1 ≥ |s| − α2|α−1| ≥

1 |s|. 2

(9.25)

Taking the same vector field as in question 4, we obtain by the same way, h (s)| ≤ hαN |K

 |(tL0 )N ϕ(η)| dη.

C Now in this region according to (9.25) the coefficients of L0 are bounded by |s| . αN −N  Then we have |Kh (s)| ≤ CN h |s| . 6. a) In this region, ∂η φ(s, η) = s + αη|η|α−2 = 0 implies that |s| = α|η|α−1 so   1   α−2 2−α α−1 α−1 |η| = |s| where s = −αη |s| and eventually η = cα s|s| α−1 . α α On the other hand,

 ∂ 2φ ∂  sj + αηj |η|α−2 = αδj k |η|α−2 + α(α − 2)ηj ηk |η|α−4 , = ∂ηj ∂ηk ∂ηk η . = α|η|α−2 (δj k + (α − 2)ωj ωk ), ω = |η|

234

9 Solutions of the Problems

b) Using question 2 of Part 1 we obtain,

∂ 2φ det ∂ηj ∂ηk

= (α|η|α−2 )d det(δj k +(α−2)ωj ωk ) = (α−1)(α|η|α−2 )d = 0

since α = 1. c) Recall that, 

h (s) = (2πh)−d K

−α φ(s,η)

eih

ϕ 2 (η)dη,

Rd

φ(s, η) = s · η + |η|α .

We apply the stationary phase formula. The large parameter is λ = h−α . Since 1 |ηc | = Cα |s| α−1 the square root of the determinant of the Hessian matrix at (α−2)d

the critical point is equal to cdα |s| 2(α−1) . The formula then gives, " h (s) = Cα,d h K

−d

h

αd 2

#

−αφ(s,ηc )

eih

ϕ (ηc ) + O(h ) . 2

(α−2)d

|s| 2(α−1)

α

(9.26)

7. We have proved in question 3 that,  Tj L∞ →L∞ = h

d(1−α)

h (s)| ds. |K

α h (s)| ≤ we have proved in question 4 that |K In the region where |s| ≤ 12 2|1−α| N  CN h for all N ∈ N. The integral of Kh on this domain gives a contribution O(hN ). In the region where |s| ≥ 21+|α−1| α we have proved in question 5 that h (s)| ≤ CN hN |s|−N . The integral of K h on this domain gives a contribution |K N O(h ). Eventually in the intermediate region we have the equality (9.26) so the h on this domain gives a contribution which is bounded below by integral of K dα −d Ch h 2 . Therefore,

Tj L∞ →L∞ ≥ Chd(1−α)h−d h

dα 2

− CN hN ≥ C h−

dα 2

= C2j

dα 2

.

8. From the definition of the operator norm, for all ε > 0 there exists u0 ∈ L∞ (Rd ) nonvanishing identically such that, Tj u0 L∞ ≥ (C2j

dα 2

− ε) u0 L∞ .

Taking ε small enough and using question 2 we obtain, C2j σ u0 L∞ ≥ 2j s Tj u0 L∞ ≥ C2j s 2j This proves that we must have σ ≥ s +

dα 2 .

dα 2

u0 L∞ .

9 Solutions of the Problems

235

9. a) We have, setting ξ = λη, λ > 0 in the integral, j S(−1)u(x) = (2π)−d



= λd (2π)−d

u(ξ ) dξ, eix·ξ ei|ξ | ϕ(2−j ξ ) α

 eiλx·η eiλ

α |η|α

ϕ(2−j λη) u(λη) dη

= λj S(t0 )u 1 (λx). λ

since  u(λη) = λ−d u ,1 (η). λ

b)

j S(−t0 )u(x) = (2π)

−d

= (2π)−d

 

u(ξ ) dξ, e−ix·ξ e−it0 |ξ | ϕ(2−j ξ ) α

u(−η) dξ = j S(t0 )u(x), eix·η e−it0 |η| ϕ(2−j η) α

 since  u(−η) = u(η). c) The questions a) and b) show that if S(t0 ) is continuous from C∗σ to C∗s the same holds for S(−1); so we have s ≤ σ − dα 2 . Part 3 1. It is easy to see that u satisfies (i∂t − |Dx |)u = 0, u|t =0 = u0 . Then ∂t u|t =0 = −i|Dx |u0 . It is sufficient to apply the operator i∂t + |Dx | to the equation and to notice that (i∂t + |Dx |)(i∂t − |Dx |) = −. 2. Following the Hint we deduce that for t0 = 0, t0 u(t0 , x) = −i 4π



1 (|Dx |u0 )(x − t0 ω) dω + 2 4π S +

 S2

u0 (x − t0 ω) dω

3  t0  ∂u0 (x − t0 ω)ωi dω. 4π S2 ∂xi i=1

(9.27) 0 Let v0 be one of the functions |Dx |u0 , u0 , ∂u ∂xi . Noticing that,

j [v0 (x − t0 ω)] = (j v0 )(x − t0 ω), we have, j [v0 (· − t0 ω)] L∞ (R3x ) ≤ j v0 L∞ (R3 ) . Assuming that u0 ∈ C∗σ (R3 ) and applying the operator j to both members of (9.27) we deduce that, u(t0 , ·) C σ −1 (R3 ) ≤ C u0 C∗σ (R3 ) . ∗

236

9 Solutions of the Problems

Therefore, the operator e−it0 |Dx | is continuous from C∗σ (R3 ) to C∗ (R3 ) where ρ = σ − 1 and we do not have ρ ≤ σ − 32 . ρ

Solution 27

Part 1 In what follows we shall denote L2 = L2 (Rd ), H 1 = H 1 (Rd ) and by v(t) the function Rd → C : x → v(t, x). 1. We have u = e− 2 (t −T ) v so, k

2

k k 2 Dt u + Au = e− 2 (t −T ) (Dt v + Av − (t − T )v). i

2. a) This follows from the fact that modulo S 0 the symbol of A∗ is a = a since a is real.  b) Since Dt = 1i ∂t and (f, g)L2 = f (x)g(x) dx we have, (Xv(t), Y v(t)) L2 = −k(t − T ) (∂t v(t), v(t))L2 − ik(t − T ) (Av(t), v(t))L2 , =: f (t) + g(t). We have 2Re f (t) = −k(t − T )∂t ( v(t) 2L2 ) so integrating by parts and using the fact that v(0) = v(T ) = 0 we obtain, 

T

2Re

 f (t) dt = k

0

T

0

v(t) 2L2 dt.

Now 2Re g(t) = 2k(t − T )Im (Av(t), v(t))L2 . Since the symbol a of A is real we have A∗ = A − B0 where B0 ∈ Op(S 0 ). Then (Av(t), v(t))L2 = (v(t), Av(t)) L2 + (B0 v(t), v(t)) L2 which shows that 2i Im (Av(t), v(t))L2 = (B0 v(t), v(t)) L2 . The operators with symbols in S 0 being continuous on L2 we have,    2Re 

T 0

   g(t) dt  ≤ Ck

T 0

 (T − t) v(t) 2L2 ≤ CkT

T 0

v(t) 2L2 .

Let T0 > 0 be such that CT0 ≤ 12 . For T ≤ T0 we obtain, 

T

2Re 0

(Xv(t), Y v(t)) L2 dt ≥

k 2



T 0

v(t) 2L2 dt.

9 Solutions of the Problems

237

3. According to question 1 and since P = X + Y we obtain, 

T 0

ek(t −T ) P u(t) 2L2 dt = 2

 

T 0

+ 0

T

Pv(t) 2L2 dt =



T 0

Xv(t) 2L2 dt

Y v(t) 2L2 dt + 2Re (Xv(t), Y v(t)) L2 .

It follows from the previous question that for all v ∈ ET we have, 

T

e

k

k(t −T )2

0

 u(t) 2L2

T

dt ≤ 0

ek(t −T ) P u(t) 2L2 dt. 2

4. a) For u ∈ S we have F (A2 u) = ξ  F (Au) = ξ 2  u = (1 + |ξ |2 ) u = F ((I d − x )u). b) We have Q = (Dt −A)(Dt +A) = Dt2 −A2 = −∂t2 +x −I d = −(−I d). c) Using question 3 successively with P = Dt ± A we obtain, 

T

k2 0

ek(t −T ) u(t) 2L2 dt ≤ C1 k 2



T

≤ C2 0

 0

T

ek(t −T ) (Dt + A)u(t) 2L2 dt, 2

0

ek(t −T ) (Dt − A)(Dt + A)u(t) 2L2 dt, 2

T

≤ C3



ek(t −T ) u(t) 2L2 dt + 2

 0

T

2 ek(t −T ) u(t) 2L2 dt .

Taking k ≥ k0 , where k02 ≥ 2C3 we get, k 2 − C3 ≥ 12 k 2 . Therefore, we can absorb the last term of the right-hand side by the left-hand side and we obtain the desired estimate. Part 2 1. Same proof as in question 1 Part 1. T 2. The term (1) corresponds to the integral 2Re 0 (Dt v(t), ik(t − T )v(t))L2 dt T which by question 2 of Part 1 is bounded below by k2 0 v(t) 2L2 dt. 3. a) This follows from the general theory ; A∗ ∈ Op(S 1 ), B ∈ Op(S 1 ) so A∗ B ∈ Op(S 2 ). Moreover, c = a ∗ b + r1 , r1 ∈ S 1 . Since a ∗ = a + r2 , r2 ∈ S 1 we have c = ab + r3 , r3 ∈ S 1 and ab is real. So, again by the general theory, we have C − C ∗ ∈ Op(S 1 ). T T b) We have (2) = Im 0 (Av(t), Bv(t)) L2 dt = Im 0 (v(t), Cv(t)) L2 dt. It follows from question a) that, (v(t), Cv(t)) L2 = (Cv(t), v(t)) L2 + (Rv(t), v(t)) L2 ,

R ∈ Op(S 1 ).

238

9 Solutions of the Problems

Using the fact that z − z = 2i Im z, the Cauchy–Schwarz inequality and the continuity of R from H 1 to L2 we obtain, |Im (v(t), Cv(t)) L2 | ≤ C1 v(t) H 1 v(t) L2 . It follows that, 

T

|(2)| ≤ C2 0

v(t) H 1 v(t) L2 dt.

T 4. a) We have (3) = −2Re 0 (∂t v(t), Bv(t)) L2 dt. T * + Since v(0) = v(T ) = 0 we have 0 ∂t (v(t), Bv(t)) L2 dt = 0. Therefore, 

T 0

 (∂t v(t), Bv(t)) L2 dt +

T 0

(v(t), B ∂t v(t))L2 dt = 0,

since ∂t and B commute. Now since the symbol of B is real we have B ∗ = B + R where R ∈ Op(S 0 ). Therefore, 

T

0

 (∂t v(t), Bv(t)) L2 dt +

T 0

(Bv(t), ∂t v(t))L2 dt, 

=−

T

(Rv(t), ∂t v(t))L2 dt,

0

which implies that, 

T

2Re 0

 (∂t v(t), Bv(t)) L2 dt = −

T

(Rv(t), ∂t v(t))L2 dt.

0

Since R is continuous from L2 to L2 we obtain, 

T

|(3)| ≤ 0

v(t) L2 ∂t v(t) L2 dt.

b) We have Dt = Dt + A − A = X − A so that, 

T

|(3)| ≤ 0

 v(t) L2 Xv(t) L2 dt +

0

T

v(t) L2 Av(t) L2 dt.

We use the fact that for all positive numbers a, b, ε, ab ≤ εa 2 + the fact that A is continuous from H 1 to L2 .

1 2 4ε b

and

9 Solutions of the Problems

239

5. This follows from the fact that, 

T 0

Pv(t) 2L2 =



T 0

  Xv(t) 2L2 + Y v(t) 2L2 dt 

T

+ 2Re

(Xv(t), Y v(t)) L2 dt

0

and from questions 2, 3, and 4 taking ε small enough. 6. a) Since B ∈ Op(S 1 ) is elliptic there exists C > 0 such that, v(t) H 1 ≤ C( Bv(t) L2 + v(t) L2 ). The inequality to be proved follows then from the fact that Y = iB + ik(t − T ). b) From question a) we have, 

T 0

 v(t) H 1 v(t) L2 dt ≤C

T 0

Y v(t) L2 v(t) L2 dt 

T

+ C(1 + kT )

v(t) 2L2 dt.

0

Then we have just to use the inequality ab ≤ εa 2 + a, b, ε. 7. Set 

T

F = 0

 Xv(t) 2L2

T

dt + 0

Y v(t) 2L2

k dt + 2



1 2 4ε b

T 0

for all positive

v(t) 2L2 dt.

We have shown in question 5 that, 

T

F ≤C 0

Pv(t) 2L2 dt +



T 0

v(t) H 1 v(t) L2 dt .

Using question 6b) we obtain, 

T

F ≤C 0

v(t) 2 2 dt + C ε P L

 0

T

Y v(t) 2L2 dt + C

(1 + kT )



T 0

v(t) 2L2 dt.

Taking ε > 0 small enough so that C ε ≤ 12 , k large enough and T small enough so that C

(1 + kT ) ≤ k4 , we can absorb the two last terms in the righthand side by F and we obtain, 

T 0



  Xv(t) 2L2 + Y v(t) 2L2 + k v(t) 2L2 dt ≤ C1

T 0

Pv(t) 2L2 dt.

240

9 Solutions of the Problems

8. a) From question 6a) we have,   v(t) 2H 1 ≤ C Y v(t) 2L2 + (1 + k 2 T 2 ) v(t) 2L2 . Using question 7 we obtain, if k is large enough, 

T 0

 Y v(t) 2L2 dt + (1 + k 2 T 2 )



T

v(t) 2L2 dt ≤ C(1 + kT 2 )

0

T 0

Pv(t) 2L2 dt,

(9.28) which proves the result. b) We have Dt v = Xv − Av so that this inequality follows from a) and from question 7, since A is continuous from H 1 to L2 . T c) We have I = 0 Pv(t) 2L2 dt so that the first inequality to be proved follows from question 7. To prove the second one we notice that k

e 2 (t −T ) ∂t u(t) = ∂t v(t) − k(t − T )v, 2

so that, 

T

e 0

k(t−T )2

 ∂t u(t) 2L2

dt ≤ C 0

T

 ∂t v(t) 2L2

dt + k T 2

2 0

T

v(t) 2L2

dt .

We have just to use (9.28) and question 8b). 9. a) Since B 2 = I d − x , we have Q = Dt2 + B 2 = −∂t2 − x + I d = −x,t + I d. b) Set  J =

T

ek(t −T )

0

2



 u(t) 2H 1 (Rd ) + Dt u(t) 2L2 (Rd ) dt.

It follows from questions 8b) and 9a) that, 

T

J ≤ C(1 + kT 2 ) 0



T 0

ek(t −T ) (Dt + iB)u(t) 2L2 (Rd ) dt, 2

ek(t −T ) (Dt + iB)u(t) 2L2 (Rd ) dt ≤ 2

C k



T 0

ek(t −T ) Qu(t) 2L2 (Rd ) dt. 2

The inequality to be proved follows from the above two inequalities.

9 Solutions of the Problems

241

10. Since cj , d ∈ L∞ ((0, T ) × Rd ) we have, setting x0 = t, 

T 0

ek(t −T ) Lu(t) 2L2 (Rd ) dt ≤ 2

⎛

+C⎝

d  

T

j =0 0



T 0

ek(t −T ) Qu(t) 2L2 (Rd ) dt 2

ek(t −T ) ∂xj u(t) 2L2 (Rd ) dt + 2

⎞



T 0

ek(t −T ) u(t) 2L2 (Rd ) dt ⎠ . 2

We obtain the desired result by using the inequality in question 9b) and taking 1 2 k + T small enough.

Solution 28 1. By the recalled formula and the fact that 1 + |ξ | ≤ Ca, we can write, setting β σ = T − t and I = |∂xα ∂ξ e−σ a |, I ≤

|α|+|β| 

Cj αβ σ j (1 + |ξ |)j −|β| e−σ a ≤ Cαβ

j =0

≤ Cαβ

|α|+|β| 

[(a σ )j e−σ a ](1 + |ξ |)−|β| ,

j =0

|α|+|β| 

[(a σ )j +m e−σ a ]

j =0

Cαβ 1 −|β| (1 + |ξ |) ≤ (1 + |ξ |)−m−|β| , (σ a)m σm

since for all ρ ∈ (0, +∞) and t > 0 we have t ρ e−t ≤ Cρ . 2. Taking m = 1 − ε in the previous question we see that the symbol et belongs C to S −(1−ε) and that all its semi-norms are bounded by (T −tαβ)1−ε . Therefore, the operator Op(et ) is continuous from L2 (Rd ) in H 1−ε (Rd ) and we have, Op(et )v H 1−ε ≤

C v L2 , (T − t)1−ε

∀v ∈ L2 (Rd ).

Then we have ∂t et = et a ∈ S −1−ε × S 1 ⊂ S −ε and its semi-norms are bounded C by (T −tαβ)1−ε . On the other hand, by the symbolic calculus, Op(et ) Op(a) ∈ Op(S −(1−ε) × S 1 ) ⊂ Op(S −ε ) and its semi-norms are bounded by Eventually, again by the symbolic calculus,

Cαβ . (T −t )1−ε

Op(∂t et ) − Op(et ) Op(a) = Op(et a) − Op(et ) Op(a) ∈ S −(1−ε) ,

242

9 Solutions of the Problems Cαβ . (T −t )1−ε

and its semi-norms are bounded by previous estimate we have,

By the same argument as in the

Op(∂t et ) − Op(et ) Op(a))v H 1−ε ≤

K v L2 . (T − t)1−ε

3. Since ∂t u = − Op(a)u + f one can write, ∂t (Op(et )u(t)) = Op(∂t et )u(t) + Op(et )∂t u(t), = (Op(∂t et ) − Op(et ) Op(a)) u(t) + Op(et )f (t). Since u(0) = 0 and eT = 1 integrating this inequality between 0 and T we obtain, 

T

u(T ) =



T

(Op(∂t et ) − Op(et ) Op(a)) u(t) dt +

0

Op(et )f (t) dt. 0

4. Using question 3 we can write,  u(T ) H 1−ε ≤ C

T 0

!

dt (T − t)1−ε

sup u(t) L2 + sup f (t) L2 ,

t ∈[0,T ]

t ∈[0,T ]

which proves that u(T ) ∈ H 1−ε (Rd ).

Solution 29

Part 1 Integrating by parts, and using the fact that  ϕ has compact support in {η : |η| ≤ 1} we obtain,  N −d yj (Rλ ϕ)(y) = (2π) ϕ (η) dη, yjN eiy·η rλ (y, η) = (2π)

−d

= (2π)−d

 ϕ (η) dη, (DηNj eiy·η )rλ (y, η)  N1 +N2 =N



N N1

 eiy·η DηNj1 rλ (y, η)DηNj2  ϕ (η) dη.

9 Solutions of the Problems

243

It follows from the estimates on rλ that for all N ∈ N, all y ∈ Rd and λ ≥ 1 we have, 1

|yj |N |(Rλ ϕ)(y)| ≤ C(ϕ)λm− 2 (1 + |y|).   Since (1 + dj=1 yj2 )k ≤ C(1 + dj=1 yj2k ) we deduce from the above inequality that for all k ∈ N we have, (1 + |y|) . (1 + |y|2)k

1

|(Rλ ϕ)(y)| ≤ C (ϕ)λm− 2

If 2(2k − 1) > d, the right-hand side belongs to L2 (Rd ). Part 2 1 1. We have, setting y = λ 2 (x − x 0 ), 

d

uλ (ξ ) = λ 4

e−ix·(ξ −λξ ) ϕ(λ 2 (x − x 0 )) dx, 1

0

d

0 ·(ξ −λξ 0 )

d

0 ·(ξ −λξ 0 )

= λ− 4 e−ix = λ− 4 e−ix



e−iλ

− 21

y·(ξ −λξ 0 )

ϕ(y) dy,

1

ϕ (λ− 2 (ξ − λξ 0 )). 

1

d

2. Using question 1 and setting η = λ− 2 (ξ − λξ 0 ), so dξ = λ 2 dη, we obtain, uλ 2H m−δ

− d2

=λ



1

ϕ (λ− 2 (ξ − λξ 0 ))|2 dξ, (1 + |ξ |2 )m−δ |

 =

1

(1 + |λξ 0 + λ 2 η|2 )m−δ | ϕ (η)|2 dη.

On the support of  ϕ we have |η| ≤ 1 so that, since |ξ 0 | = 1 we have, 1

1

1

|λξ 0 + λ 2 η| ≥ λ − λ 2 |η| ≥ λ − λ 2 ≥

1 λ if λ ≥ 4. 2

It follows that,  uλ 2H m−δ ≥ cλ2(m−δ)

| ϕ (η)|2 dη.

244

9 Solutions of the Problems

3. Using question 1 one can write, d

P uλ (x) = (2π)−d λ− 4 = eiλx

0 ·ξ 0



eix·ξ e−ix

λ− 4 (2π)−d d

0 ·(ξ −λξ 0 )

 ei(x−x

0 )·ξ

1

p(x, ξ ) ϕ (λ− 2 (ξ − λξ 0 )) dξ,

p(x, ξ ) ϕ (λ− 2 (ξ − λξ 0 )) dξ. 1

Setting η = λ− 2 (ξ − λξ 0 ), so dξ = λ 2 dη we obtain, 1

P uλ (x) = e

d

iλx·ξ 0

d 4

λ (2π)

−d

 eiλ

1 2 (x−x 0 )·η

1

p(x, λξ 0 + λ 2 η) ϕ (η) dη.

Set, Qλ (y, Dy )ϕ(y) = (2π)

−d



1

1

ϕ (η) dη, eiy·η p(x 0 + λ− 2 y, λξ 0 + λ 2 η)

then, 1

n

e−iλx·ξ P uλ (x) = λ 4 (Qλ (y, Dy )ϕ)(λ 2 (x − x 0 )). 0

4. Let qλ (y, η) = p(x 0 +λ− 2 y, λξ 0 +λ 2 η). Using the Taylor formula with integral reminder at the order 1 we can write, 1

1

qλ (y, η) = p(x0 , λξ 0 ) + rλ (y, η), rλ = rλ1 + rλ2 ,  1 1 1 1 − 12 (y · ∂x p)(Xt ) dt, Xt = (x0 + tλ− 2 y, λξ 0 + tλ 2 η), rλ (y, η) = λ 1

rλ2 = λ 2



0

1

(η · ∂ξ p)(Xt ) dt.

0

Let us show that rλ satisfies the estimates in Part 1. We have, ∂ηα rλ1 (y, η) = λ− 2 1



1

0

(tλ 2 )|α| y · (∂ξα ∂x p)(Xt ) dt. 1

Since p ∈ S m we obtain, |∂ηα rλ1 (y, η)| ≤ λ− 2 λ 1

|α| 2

 |y| 0

1

1

(1 + |λξ 0 + tλ 2 η|)m−|α| dt.

9 Solutions of the Problems

245

Now, since |ξ 0 | = 1, |η| ≤ 1 and 0 ≤ t ≤ 1 we have, 1

1

1 + |λξ 0 + tλ 2 η| ≤ 1 + λ + λ 2 ≤ 3λ 1

if λ ≥ 1,

1

1

1 + |λξ 0 + tλ 2 η| ≥ λ − tλ 2 |η| ≥ λ − λ 2 ≥

(9.29)

1 λ, 2

if λ ≥ 4.

It follows that, |∂ηα rλ1 (y, η)| ≤ Cλm− 2 λ− 1

|α| 2

1

|y| ≤ Cλm− 2 |y|.

Now by the Leibniz formula we have, ∂ηα rλ2 (y, η)

=λ

1 2



1 0

1

(tλ 2 )|α| η · (∂ξα ∂ξ p)(Xt ) dt + Cλ

1 2





|β|≤|α|−1 0

1

1

(tλ 2 )|β| (∂ξ ∂ξ p)(Xt ) dt. β

Thanks to the fact that p ∈ S m and to (9.29), the first term above can be bounded |α| 1 |α| 1 1 by Cλ 2 λ 2 λm−|α|−1 or by λm− 2 λ− 2 thus by Cλm− 2 . The second term can  1 |β| 1 be bounded, in a similar way, by C |β|≤|α|−1 λ 2 λ 2 λm−|β|−1 or by C λm− 2 . Gathering the estimates obtained for rλ1 and rλ2 we obtain eventually, 1

|∂ηα rλ (y, η)| ≤ Cλm− 2 (1 + |y|). 5. It follows from questions 3 and 4 that 1

n

1

n

e−iλx·ξ P uλ (x) = λ 4 p(x 0 , λξ 0 )ϕ(λ 2 (x − x 0 )) + λ 4 (Rλ (y, Dy )ϕ)(λ 2 (x − x 0 )). (9.30) 0

Since f (λ 2 (x − x 0 ) L2 = λ− 4 f L2 we deduce from (9.30) and from the result in Part 1 that, 1

d

1

P uλ L2 ≤ |p(x 0 , λξ 0 )| ϕ L2 + C(ϕ)λm− 2 . Using the fact that ϕ L2 = 1, the estimate (6.7) and question 2 we can write, 1

c0 λm−δ ≤ C|p(x 0 , λξ 0 )| + C(ϕ)λm− 2 .

246

9 Solutions of the Problems 1

Since δ < 12 there exists λ0 > 0 such that for λ ≥ λ0 we have c0 λm− 2 − C(ϕ)λm−δ ≥ c20 λm−δ . It follows that, C|p(x 0 , λξ 0 )| ≥

c0 m−δ λ , 2

∀λ ≥ λ0 .

6. Since p − pm ∈ S m−1 and λ ≥ 1 we have |(p − pm )(x 0 , λξ 0 )| ≤ C(1 + λ)m−1 ≤ C1 λm−1 . It follows from the previous question and from the fact that δ < 12 that there exists λ3 > 0 such that for λ ≥ λ3 we have, |pm (x 0 , λξ 0 )| ≥ |p(x 0 , λξ 0 )| − |(p − pm )(x 0 , λξ 0 )| ≥ C2 λm−δ − C1 λm−1 ≥

1 C2 λm−δ . 2

7. Since pm is homogeneous of degree m, taking λ = λ3 we obtain from the previous question, |pm (x 0 , ξ 0 )| ≥

1 1 C2 λ−m λm−δ = C2 λ−δ . 2 2

Let x ∈ Rd and ξ ∈ Rd , ξ = 0. We can take in the above inequality ξ 0 = Sd−1 . Using again the homogeneity of pm we obtain, |pm (x, ξ )| ≥

ξ |ξ |

∈

1 C2 λ−δ |ξ |m . 2

Solution 30

Part 1 1. The functions aj being real and C 1 with respect to all variables, the Cauchy– Lipschitz theorem ensures that this system has a unique solution in a neighborhood of (t, y) = (0, 0). Moreover, the solution has a C 1 dependence with respect to the initial data. 2. a) We have ∂x1 (0, 0) = 1, ∂s

∂x1 (0, 0) = 0, ∂yj

∂xj (0, 0) = x˙ j (0, 0) = aj (0, 0), ∂s It follows that det ∂ϕ(s,y) ∂(s,y) = 1.

∂xj (0, 0) = δj k , ∂yk

j, k ≥ 2.

9 Solutions of the Problems

247

b) This is a consequence of the inverse function theorem. c) We have ∂ ∂ (u(ϕ(s, y)) = (u(s, x2 (s, y), . . . , xd (s, y)), ∂s ∂s  ∂u ∂u (ϕ(s, y))x˙1 (s, y) + (ϕ(s, y))x˙j (s, y), ∂x1 ∂xj d

=

j =2

 ∂u ∂u (ϕ(s, y)) + aj (ϕ(s, y)) (ϕ(s, y)) = Xu(ϕ(s, y)). ∂x1 ∂xj d

=

j =2

3. Set  b(s, y) = b(ϕ(s, y)), f(s, y) = f (ϕ(s, y)). Then the equation Xu(t, x) + b(t, x)u(t, x) = f (t, x) is equivalent to, ∂ u (s, y) +  b(s, y) u(s, y) = f(s, y). ∂s

(9.31)

Now since ϕ(0, y) = (0, y) we have  u(0, y) = u(0, y) = u0 (y). The equation (9.31) can be solved easily. Indeed it is equivalent to, d ds

 s

 s

  exp b(σ, y) dσ  u(s, y) = exp b(σ, y) dσ f(s, y), 0

0

which according to the initial data provides a unique solution given by,

 s   b(σ, y) dσ u0 (y)+  u(s, y) = exp − 0



s

exp

0

σ

 b(z, y) dz f(σ, y) ds.

s

This solution is C 1 with respect to (s, y) in V so u is C 1 for (t, x) ∈ U. Part 2 1. Setting  bj (x) = bj (x + x 0 ) likewise for c, g, u, we are led to the case where x 0 = 0. By hypothesis there exists j0 such that  bj0 (0) = 0. Then there exists a neighborhood U1 ⊂ U of 0 such that bj0 (x) = 0 for all x ∈ V1 . On the other hand, we can change the indices so that j0 = 1 (setting t = z1 = xj0 , zk = xk if k = j0 .). In the new coordinates Y =  bj0 (t, z) ∂t∂ + d  b g ∂ c  so that setting bj = j , c =  ,g =  the initial problem k=2 bk (t, z)  bj0

∂zk

can be written as,

 bj0

! d  ∂ ∂  u + c + bj (t, z) u = g, ∂t ∂zk k=2

 bj0

 u|t =0 =  u0 .

248

9 Solutions of the Problems

Then we apply the result in Part 1. It follows that this problem has a C 1 unique solution  u in a neighborhood W of 0.

Solution 31

Part 1   1. In Rd , let P = D12 − dj=2 Dj2 be the wave operator. Then p(ξ ) = ξ12 − nj=2 ξj2 ∂p and ξ 0 = (ξ10 = 1, ω0 ), where ω0 ∈ Sd−2 , satisfies p(ξ 0 ) = 0, ∂ξ (ξ 0 ) = 0. 1 ∞ ∞ 2. If P is hypoelliptic we have E ⊂ C (V ). The spaces E and C (V ) are Fréchet spaces. Let us consider the map E → C ∞ (V ) : u → u. It has a closed graph. Indeed, if a sequence (un , un ) converges to (u, v) in E × C ∞ (V ) then (un ) converges to u in E so in C 0 (V ) and (un ) converges to u in C ∞ (V ) so in C 0 (V ). It follows that u = v. It follows from the closed graph theorem ∞ that this map is continuous. αSince the semi-norms of C (V ) are of the form qF,M (u) = |α|≤M supF |D u| where F is a compact subset of V and M ∈ N, taking for F a compact containing the point x 0 there exists a compact K in V and N0 ∈ N such that, ⎛ ⎞   |gradu(x 0 )| ≤ C sup |D α u| ≤ C ⎝sup |u| + sup |D α P u|⎠ |α|≤1 F

K

|α|≤N0 K

for all u in E which contains C ∞ (V ). 3. We have Dk (eiλϕ w) = eiλϕ (Dk w + iλ(Dk ϕ)w). On the other hand, we have,  Dj Dk (eiλϕ w) = eiλϕ Dj Dk w + iλ(Dj ϕ)Dk w + iλ(Dk ϕ)Dj w  + iλ(Dj Dk ϕ)w − λ2 (Dk ϕ)(Dj ϕ)w . We deduce that, e−iλϕ P (eiλϕ w) = λ2 p(x, gradϕ(x))w − 2iλ

d  j,k=1

aj k (x)(∂j ϕ)∂k w + iλ((P − c)ϕ)w + P w.

9 Solutions of the Problems

249

By ( ) we have p(x, gradϕ(x)) = 0. Moreover, we have follows that, e

−iλϕ

P (e

iλϕ

∂p ∂ξk

=2

d

j =1 aj k ξj .

It

! d  ∂p w) = −iλ (x, gradϕ(x))∂k w − ((P − c)ϕ)w + P w. ∂ξk k=1

We have just to set b = −(P − c)ϕ. 4. a) We argue by induction on |α| = n. The formula is true if n = 0. Assume it is true for n and let |β| = n + 1. Then ∂ β = ∂ ∂ α where |α| = n. Then we have, β

∂ (e

iλϕ

f ) = ∂ ∂ (e α

iλϕ

f) =

n 

(∂ ckα )λ + i k

k=0

=

n+1 

n 

! k+1

ckα (∂ ϕ)λ

eiλϕ ,

k=0

cjβ λj eiλϕ .

j =0

b) Since λ ≥ 1 we have λk ≤ λ|α| if k ≤ |α|, so that, sup |∂ (e α

iλϕ

|α|

f )| ≤ λ

K

|α| 

sup |ckα |.

k=0 K

5. By (H ), ( ) and Problem 30, these problems have C ∞ solutions in a neighborhood V1 ⊂ V of x 0 . 6. a) We have, e−iλϕ P v = e−iλϕ

N0 

λ−j P (eiλϕ wj ) =

j =0

= −iλLw0 − i N 0 −1 j =0

λ−j (−iλLwj + P wj ),

j =0 N0 

λ−(j −1) Lwj +

j =1

= −i

N0 

λ−j Lwj +1 +

N0 

λ−j P wj ,

j =0 N 0 −1

λ−j P wj + λ−N0 P wN0 = λ−N0 P wN0 ,

j =0

since Lw0 = 0 and Lwj +1 = −iP wj for 0 ≤ j ≤ N0 − 1. N0 b) Since ϕ is real and λ ≥ 1, we have supK |v| ≤ j =0 supK |wj | ≤ CK and α we deduce from questions 4b) and 6a) that supK |∂ P v| ≤ Cλ|α|−N0 .

250

9 Solutions of the Problems

c) Since by ( ) we have grad ϕ(x0 ) = ξ 0 = 0 we have, e−iλϕ(x0 ) grad v(x0 ) =

N0 

  λ−j iλwj (x0 )ξ 0 + grad wj (x0 ) ,

j =0

= iλw0 (x0 )ξ 0 + i

N 0 −1

λ−j wj +1 (x0 )ξ 0 +

j =0

N0 

λ−j gradwj (x0 ).

j =0

N0 −1 −j N0 −j Since λ ≥ 1 we have |i j =0 λ wj +1 (x0 )ξ 0 + j =0 λ grad wj (x0 )| ≤ C. On the other hand, by construction we have w0 (x0 ) = 1. It follows that, |grad v(x0 )| ≥ λ|ξ 0 | − C ≥

1 0 λ|ξ |, 2

if λ is large enough. Then we use the inequality (6.9). By question 6b) the right-hand side is bounded while by question 6c) the left-hand side tends to +∞ when λ goes to +∞. Therefore, we have a contradiction. This implies that P cannot be hypoelliptic. Part 2 1. The matrix A(x0 ) being real and symmetric it has d real eigenvalues λ1 , . . . , λd . Moreover, there exists an orthogonal matrix O such that OA(x0 )O = D where D = diag(λj ). Therefore, setting η = Oξ, we have, p(x0 , ξ ) = A(x0 )ξ, ξ  = DOξ, Oξ  = Dη, η =

d 

λj ηj2 := q(η).

j =1

(9.32) If all the λj are equal to zero then p(x0 , ξ ) = 0 for all ξ ∈ Rd which contradicts condition (6.10) in the statement. Now if all the eigenvalues are ≤ 0 (resp. ≥ 0) then p(x0 , ξ ) ≤ 0 (resp. p(x0 , ξ ) ≥ 0) for all ξ ∈ Rd by (9.32); this contradicts (6.10) in the statement. Thus there exists λj0 > 0 and λj1 < 0. We may assume that j0 < j1 . 7 2. Let η0 = (ηj0 ) where ηj00 = 1, ηj01 =

λj0 −λj1

and ηj0 = 0 otherwise. Then,

q(η0 ) = λj0 (ηj00 )2 + λj1 (ηj01 )2 = λj0 + λj1 ∂q 0 ∂ηj0 = 2λj0 ηj0 = ξ 0 = O−1 η0 . By

Now,

λj0 = 0. −λj1

2λj0 = 0.

3. Set (9.32) we have p(x0 , ξ 0 ) = 0. On the other hand, 0 0 dξ p(x0 , ξ ) = dη q(η ) ◦ O. So dξ p(x0 , ξ 0 ) = 0.

9 Solutions of the Problems

251

4. We can state the following theorem. Theorem Let P be an operator of the form (6.8) in the statement, where the coefficients aj k are real. If P is hypoelliptic in  then for any point x 0 ∈  we have,     either p(x0 , ξ ) ≥ 0, ∀ξ ∈ Rd or p(x0 , ξ ) ≤ 0, ∀ξ ∈ Rd . Indeed, otherwise the condition (6.10) in the statement would be satisfied, which ensures by Part 2 that the condition (H ) is satisfied, which by Part 1 implies that P is not hypoelliptic.

Solution 32 1. Let x ∈ Rd be such that xd = 0. There exists a neighborhood V of this point in which u is equal to 0 or 1 so is C ∞ in V . It follows that for ξ ∈ Rd \ {0} we have (x, ξ ) ∈ / WF(u). |ξ | |ξ | < 1. Let ε > 0 be such that α := |ξ0,d + ε < 1 and 2. a) If ξ0 = 0 we have |ξ0,d 0|  0|     ξ  ξd ξ0,d  ξ0  d {0} =: ξ ∈ R \ : − we have − < ε . If ξ ∈    ξ0 ξ0 |ξ | |ξ0 | |ξ | |ξ0 |  < ε so |ξd − ξ0,d |ξ|ξ0|| | < ε|ξ |. Then,

   

 |ξ |   |ξ0,d ||ξ |  |ξ0,d |  |ξd | ≤ ξd − ξ0,d + ≤ ε+ |ξ | = α|ξ |. |ξ0 |   |ξ0 |  |ξ0 | It follows that, |ξ | ≤ |ξd | + |ξ | ≤ α|ξ | + |ξ | so (1 − α)|ξ | ≤ |ξ |. We take c0 = 1 − α. b) We have, |ξ | ≤ |ξ | ≤ c10 |ξ |. c) We have,  ϕ ,u(ξ ) =

e

−ix·ξ



+∞ 

ϕ(x)u(x) dx = 0

e−ix·ξ ϕ(x) dx. Rd−1

We use the fact that (I d − x )N e−ix·ξ = (1 + |ξ |2 )N e−ix·ξ where x = d−1 ∂ 2 ∞ d 2 . Since ϕ ∈ C0 (R ) we can, setting P = I d − x , integrate by j =1 ∂xj

parts and write, (1 + |ξ |2 )N ϕ , u(ξ ) =

 

+∞  0

= 0

+∞ 

(P N e−ix·ξ )ϕ(x) dx, Rd−1

e−ix·ξ (P N ϕ)(x) dx. Rd−1

252

9 Solutions of the Problems

It follows that, (1 + |ξ |2 )N |, ϕ u(ξ )| ≤

+∞ 

 0

Rd−1

|(P N ϕ)(x)| dx,

and by question 2b), (1 + |ξ |2 )N |, ϕ u(ξ )| ≤ CN for ξ ∈ ξ0 , which  shows that (x0 , ξ0 ) ∈ / WF(u). Therefore, WF(u) ⊂ (x, ξ ) : xd = 0, ξ = 0 . 3. Let V be a neighborhood of the point (x0 , 0). Take ϕ ∈ C0∞ (V ), ϕ = 1 

near (x0 , 0) with ϕ(x , 0) dx = c = 0; (we can take for instance ϕ(x) = ϕ1 (x )ϕ2 (xd )). Set I = i ξ0,d ϕ , u(0, ξ0,d ). We can write, 

 I=  =

Rd−1

Rd−1

+∞ 0

  −∂xd e−ixd ξ0,d ϕ(x , xd ) dxd dx

ϕ(x , 0) dx +



 Rd−1

+∞ 0

e−ixd ξ0,d (∂xd ϕ)(x , xd ) dxd dx , = c + A.

Now we have,  i ξ0,d A =  =

 Rd−1

Rd−1



+∞ 0 +∞ 0

−(∂xd e−ixd ξ0,d )(∂xd ϕ)(x , xd ) dxd dx , e−ixd ξ0,d (∂x2d ϕ)(x , xd ) dxd dx ,

since (∂xd ϕ)(x , 0) = (∂xd ϕ)(x , +∞) = 0. It follows that |ξ0,d A| ≤ C. For |ξ0,d | → +∞ we can write, ϕ , u(0, ξ0,d ) =



c 1 +O . iξ0,d |ξ0,d |2

We deduce that, |, ϕ u(0, ξ0,d )| ≥

c 1 , 2 |ξ0,d |

for |ξ0,d | large enough. This shows that ϕ ,u does not decay rapidly in any cone

, 0, 0, ξ ) ∈ WF(u). and therefore that (x ξ0 0,d 0

9 Solutions of the Problems

253

Solution 33 1. a) The equality is obviously true for k = 0 with P0 (η) = 1. Assume it is true up to the order k ≥ 0 then we have,

d dη

k+1

1 1 + η2



Pk (η) (1 + η2 )k+1

Pk (η) 2(k + 1)ηPk (η) − , (1 + η2 )k+1 (1 + η2 )k+2

=

d dη

=

(1 + η2 )Pk (η) − 2(k + 1)ηPk (η) , (1 + η2 )k+2

=

and Pk+1 (η) = (1 + η2 )Pk (η) − 2(k + 1)ηPk (η) is a polynomial of degree ≤ k + 1. b) According to a) we can write,

d dη

k+1

1 1 + η2

=

1 (1 + η)k+2 Pk (η) . (1 + η)k+2 (1 + η2 )k+1

k+2

Pk (η) The function vk (η) = (1+η) is continuous on [0, +∞) and tends to a (1+η2 )k+1 finite limit when η goes to +∞. It is therefore bounded by a constant Ck . 1 2. a) Set f (η) = 1+η 2 . The equality is true for N = 1 with a0,N = 0. Assume it is true up to the order N ≥ 1 and let us prove it at the order 1 d ixη so that integrating by parts we obtain, N + 1. We have eixη = ix dη e



+∞

 1 . ixη (N−1) /+∞ 1 +∞ ixη (N) (η) − e f (η) dη, e f 0 ix ix 0  f (N−1) (0) 1 +∞ ixη (N) =− e f (η) dη, − ix ix 0

eixη f (N−1) (η) dη =

0

since from question 1b) we have limη→+∞ |f (N−1) (η)| = 0. Using the induction hypothesis at the order N we can write, u(x) =

N−1  j =0

aj,N i N f (N−1) (0) iN + + xj xN xN



+∞

eixη f (N) (η)) dη,

0

which proves the desired inequality at the order N + 1. b) Let us show that u is C ∞ in R \ {0} . Let p ∈ N. Using question a) with N =  +∞ p + 1 it is sufficient to prove that the function x → 0 eixη f (N−1) (η) dη is C p . But this results from a classical theorem of differentiability of the

254

9 Solutions of the Problems

integrals. Indeed for j ≤ p we have,    d j  CN ηj CN (1 + η)p   ixη (N−1) e f (η)  = |ηj f (N−1) (η)| ≤ ≤   dx  (1 + η)N+1 (1 + η)N+1 ≤

CN ∈ L1 (R), (1 + η)N+1−p

since N + 1 − p ≥ 2. Therefore, the integral is C p for all p ∈ N so it is C ∞ on R. It follows that u is C ∞ on R \ {0} . We deduce that for all x0 = 0 and all ξ ∈ R \ {0} we have, (x0 , ξ ) ∈ / W F (u) that is W F (u) ⊂ {(0, ξ ), ξ = 0} .  −ixξ  +∞ eixη 3. a) We have ϕ ,u(ξ ) = R e ϕ(x) 0 dη dx. The function (x, η) → 1+η2 e−ix(ξ−η) ϕ(x) 1+η2

is in L1 (R × (0, +∞)), so we can use the Fubini theorem and

deduce that, +∞ 

 ϕ ,u(ξ ) = 0

e−ix(ξ −η)ϕ(x) dx

R

1 = 1 + η2



+∞ 0

(ξ − η) ϕ dη. 1 + η2

ϕ (ζ ) is bounded on R by b) Since  ϕ ∈ S(R), for all N ∈ N the function |ζ |N  a constant CN . Then for ξ < 0 and η ≥ 0 we have |ξ | + η = −ξ + η = −(ξ − η) = |ξ − η|. Therefore, 

+∞

|ξ | | ϕ (ξ )| ≤ N



0 +∞

≤ 0

| ϕ (ξ − η)| |ξ | dη ≤ 1 + η2



+∞

N

(|ξ | + η)N

0

| ϕ (ξ − η)| |ξ − η| dη ≤ CN 1 + η2



+∞

N



0

| ϕ (ξ − η)| dη, 1 + η2

1 dη. 1 + η2



4. a) We have ϕ(0) = ψ(y)ψ(−y) dy. So ϕ(0) = ψ(y)2 dy = 1 since ψ is 

even. Now, ϕ (0) = ψ(y)ψ (−y) dy = − ψ(y)ψ (y) dy since ψ is odd,  d  is real and we so ϕ (0) = − 12 dy [ψ(y)2 ] dy = 0. Now, since ψ is even ψ 

2 2 =   2 ≥ 0. Eventually  ϕ (0) = (ψ(0)) ψ(y) dy = 0 since have  ϕ = (ψ) ψ has positive values and L2 norm equal to 1.  1 b) Since  ϕ is positive we have from question 3a), ϕ , u(ξ ) ≥ A  ϕ (ξ − η) 1+η 2 dη. 1 ϕ ) (λζ ) dλ =  ϕ (0) + ζ χ(ζ ) where Now we write  ϕ (ζ ) =  ϕ (0) + ζ 0 ( supR |χ| ≤ C. Then,  ϕ , u(ξ ) ≥  ϕ (0) A

1 dη + 1 + η2

 (ξ − η)χ(ξ − η) A

1 dη = (1) + (2). 1 + η2

9 Solutions of the Problems

255

We have |(2)| ≤ Cε



1 A 1+η2

dη so, 

ϕ , u(ξ ) ≥ ( ϕ (0) − Cε) A

We fix ε > 0 such that Cε = {η : ξ − ε ≤ η ≤ ξ + ε} . Then, 

ξ +ε

ξ −ε

1 ϕ (0). 2

1 dη. 1 + η2

For ξ > 0 large, we have A =

1 2ε dη ≥ . 2 1+η 1 + (ξ + ε)2

Summing up we have, ϕ , u(ξ ) ≥

ε ϕ (0) 1 + (ξ + ε)2

ξ 2ϕ , u(ξ ) ≥ ε  ϕ (0)

so

ξ2 . 1 + (ξ + ε)2

This proves our claim since the right-hand side tends to ε ϕ (0) > 0 when ξ → +∞. To prove that {(0, ξ ) : ξ > 0} ⊂ W F (u) we have to prove a lower bound, similar to that proved above, for all functions ϕ1 ∈ C0∞ equal to 1 near zero (not only for one function !). c) Let ϕ1 be given in the statement. We have θ (0) = ϕ(0) − ϕ1 (0) = 1 − 1 = 0 and θ (0) = ϕ (0) − ϕ1 (0) = 0 − 0 = 0. Since θ ∈ C0∞ (R) the Taylor formula with integral reminder shows that θ (x) = x 2 θ1 (x) and θ1 ∈ C0∞ (R). 2 Then  θ (ξ ) = −( d 2  θ1 )(ξ ). It follows from question 3a) that, dξ

θ, u(ξ ) =



+∞ 0



θ (ξ − η)

+∞

=− 0

d2 dη2

1 dη = − 1 + η2

* +  θ1 (ξ − η)

+∞

 0

d2 1  (ξ − η) θ dη, 1 2 dξ 1 + η2

1 dη. 1 + η2

Integrating two times by parts we obtain, since ( θ1 ) ∈ S(R), , θ u(ξ ) = −( θ1 ) (ξ ) −



+∞

 θ1 (ξ − η)

0

d2 dη2



1 1 + η2

dη.

Then according to question 1b) with k = 2 we have, |θ, u(ξ )| ≤ |( θ1 ) (ξ )| + C1

 0

+∞

| θ1 (ξ − η)|

1 dη. (1 + η)4

256

9 Solutions of the Problems

Let ε > 0 be small. Writing |ξ |3−ε ≤ C2 (|ξ − η|3−ε + |η|3−ε ) we get, u(ξ )| ≤ |ξ |3−ε |( θ1 ) (ξ )| + C3 |ξ |3−ε |θ, +∞

+ C3 0

|ξ |

+∞

|ξ − η|3−ε

0



3−ε



|θ, u(ξ )| ≤ C4 + C5

| θ1 (ξ − η)| dη (1 + η)4

|η|3−ε dη, (1 + η)4

 +∞ 1 1 dη + dη . (1 + η)4 (1 + η)1+ε 0

| θ1 (ξ − η)|



+∞ 0

d) Then we have for ξ ≥ 1, ξ 2 |θ, u(ξ )| ≤

C6 . |ξ |1−ε

We write,

2 u(ξ ). , u(ξ ) − ξ 2 θ, ξ 2 ϕ 1 u(ξ ) = ξ ϕ

It follows from question 4b) that for ξ ≥ 1, ξ 2 ϕ 1 u(ξ ) ≥ C −

C6 ≥ C7 > 0, |ξ |1−ε

if ξ is large enough. We deduce that {(0, ξ ), ξ > 0} ⊂ W F (u).

Solution 34 1. It follows from the hypothesesthat (x12 + x22 )u ∈ C ∞ (R2 ), so u ∈ C ∞ (R2 \ {0})  which implies that W F (u) ⊂ (x, ξ ) ∈ R2 × (R2 \ {0}) : x = 0 . ∂ ϕ u) ∈ S(R2 ), j = 2. a) By hypothesis we have xj ϕu ∈ C0∞ (R2 ), so x j ϕu = i ∂ξj (, 1, 2. b) Let f (t) = |tω + (1 − t)ω0 |2 = t 2 + (1 − t)2 + 2t (1 − t)(ω · ω0 ). We have ω ·ω0 = cos θ where θ ∈ (0, π) is the angle between ω and ω0 . By hypothesis 0 ≤ θ < π. We have f (t) = 2(1−cos θ )t 2 −2(1−cos θ )t +1. If θ = 0 we have f (t) = 1. If 0 < θ < π we have 1 + cos θ > 0 and f (t) = 2t (1 − cos θ )(2t − 1). We see then that f has a minimum at t = 12 and that f ( 12 ) = 12 (1 + cos θ ). Therefore, f (t) ≥ 12 (1 + cos θ ) = c2 > 0. c) Assume that the angle (ω, ω0 ) is strictly smaller than π. There exists ε > 0 such that for all ω ∈ S1 with |ω−ω | < ε the angle (ω , ω0 ) is strictly smaller than π. We write, , u(λω0 ) + ϕ , u(λω ) = ϕ

2  j =1

λ(ωj − ω0,j )



1 0

∂ (, ϕ u)(λ(tω + (1 − t)ω0 )) dt. ∂ξj (9.33)

9 Solutions of the Problems

257

By hypothesis we have |, ϕ u(λω0 )| ≤ CN λ−N for all N ∈ N. Now, since from ∂ question a) we have ∂ξj (, ϕ u) ∈ S(R2 ), for all N ∈ N there exists CN > 0 such that,    ∂ 

 ϕ u)(λ(tω + (1 − t)ω0 )) ≤ CN |λ(tω + (1 − t)ω0 )|−N .  ∂ξ (, j It follows from question b) applied to ω that,    ∂ 

 ϕ u)(λ(tω + (1 − t)ω0 )) ≤ CN c−N λ−N .  ∂ξ (, j

We deduce from formula (9.33) that for all N ∈ N there exists DN > 0 such that, |, ϕ u(λω )| ≤ DN λ−N . Therefore, (0, λω) ∈ / W F (u). d) Since S1 can be covered by three sectors making an angle < π, if there exists ω0 such that (0, λω0 ) ∈ / W F (u), then for all ω ∈ S1 we have, from question c), (0, λω) ∈ / W F (u). Therefore, 0 ∈ / supp sing(u) which means that u is C ∞ in a neighborhood of zero. Since from question 1 we know that u is C ∞ in R2 \ {0} we have u ∈ C ∞ (R2 ) which contradicts our hypothesis. Therefore, all points (0, λω) belong to W F (u), i.e., W F (u) = {0, ξ ), ξ = 0} . A simple example of such a distribution is given by the Dirac mass at the origin δ(0,0). 3. Assume that we show,   ∃ξ 0 = 0 : (0, ξ 0 ) ∈ / W F (u) ⇒ (∀ξ = 0 : (0, ξ ) ∈ / W F (u)) . (9.34) This would imply that u ∈ C ∞ near zero, therefore u ∈ C ∞ (R2 ), which is not true by hypothesis. So (0, ξ 0 ) ∈ W F (u) for every ξ 0 = 0. Let us prove (9.34). Let (0, ξ ) with ξ = 0, ξ = ξ 0 . We may assume that

ξ2 −ξ20 and consider the operator P = x1 + λx2 . Since by ξ1 −ξ10 ∞ 2 C (R ) the bicharacteristic issued from (0, ξ 0 ) does not meet

ξ1 = ξ10 . Set λ =

hypothesis P u ∈ W F (u). Now this bicharacteristic is given for all t ∈ R by x(t) = 0, ξ1 (t) = −t + ξ10 , ξ2 (t) = −λt + ξ20 . Now if t = ξ10 − ξ1 we have ξ1 (t) = ξ1 , ξ2 (t) = ξ2 . Thus (0, ξ ) ∈ / W F (u).

258

9 Solutions of the Problems

Solution 35   1. Let us denote by ! = (x, y, ξ, η) ∈ R2 × (R2 \ {0}) : p(x, y, ξ, η) = 0 =   (x, y, ξ, η) ∈ R2 × (R2 \ {0}) : ξ 2 = xη2 the characteristic set of P . If u ∈ D (R2 ) we know that, WF(u) ⊂ WF(P u) ∪ !. Since P u ∈ C ∞ (R2 ) we have WF(P u) = ∅. The points (x, y, ξ, η) such that x < 0 do not belong to ! so neither to WF(u). This implies that u is C ∞ near these points, so u ∈ C ∞ (H ). 2. Denote by (x(t), y(t), ξ(t), η(t)) the bicharacteristic starting at t = 0 from a point (x0 , y0 , ξ0 , η0 ). Its equations are, x(t) ˙ = 2ξ(t),

y(t) ˙ = −2x(t)η(t),

ξ˙ (t) = η(t)2 ,

η(t) ˙ = 0.

The last equation gives η(t) = η0 , the one before gives ξ(t) = η02 t + ξ0 , the first one implies that x(t) = η02 t 2 + 2ξ0 t + x0 eventually the second one gives y(t) = − 23 η03 t 3 − 2η0 ξ0 t 2 − 2x0 η0 t + y0 . 3. We know from question 1 that u is C ∞ for x < 0. On the other hand, by hypothesis u is C ∞ in a neighborhood of x = 0. It is therefore sufficient to show that u is C ∞ for x > 0. By the propagation of singularities theorem it is sufficient to show that, starting from a point (x0 , y0 , ξ0 , η0 ) where x0 > 0 and (ξ0 , η0 ) is to be chosen, the null bicharacteristic starting at this point meets, at a time t > 0, the line {x = 0} and therefore a neighborhood of this line (where u is C ∞ ). From question 2 we have x(t) = η02 t 2 +2ξ0 t +x0 where ξ02 = x0 η02 . Notice that η0 = 0 otherwise ξ0 = η0 = 0, which is excluded. The reduced discriminant of this second order polynomial in t is ξ02 −x0 η02 = 0 so that we have x(t) = η02 (t − ξ02 )2 . Take ξ0 > 0, then at the time t =

ξ0 η02

we have x(t) = 0.

η0

Solution 36   1. If K is a compact subset of Rd we have K |uk (x)|2 dx = K |b(x)|2 dx < +∞. Moreover, if f ∈ L2c (Rd ) ⊂ L1 (Rd ) we have,  Rd

eikx·ξ0 f (x) dx = f(−kξ0 ),

9 Solutions of the Problems

259

and the right-hand side tends to zero, when k → +∞, since the Fourier transform 0 (Rd ). sends L1 (Rd ) to C→0 2. We have  , − kξ0 ). χu k (ξ ) = e−iyξ χ(y)b(y)eiky·ξ0 dy = χb(ξ Rd

Therefore, if we set (1) = (2π)d A(χuk )(x) we obtain,   , − kξ0 ) dξ, eixξ a(x, ξ )χu k (ξ ) dξ = eixξ a(x, ξ )χb(ξ (1) = Rd

Rd



= eikx·ξ0

, dη. eix·η χb(η) Rd

Since χuk (x) = e−ikx·ξ0 χb(x) it follows that Ik = (2π)−d

 Rd ×Rd

, eix·η a(x, η + kξ0 )χb(η)χ(x)b(x) dη dx.

3. a) For |η| ≥ 12 k we can write, , |fk (x, η)| = (2π)−d |η|−1 |a(x, η + kξ0 )|(|η||χb(η)||χ(x)b(x)|, ≤

C , (|η||χb(η)||χ(x)b(x)|. k

Then the conclusion follows from the fact that the function , (|η||χb(η)||χ(x)b(x)| belongs to L1 (Rd × Rd ). b) If |η| ≤ 12 k we write, η + kξ0 = k2 (2ξ0 + 2k η). Now 2 2 |2ξ0 + η| ≥ 2 − |η| ≥ 2 − 1 = 1. k k Since λ =

k 2

≥ 1 and a is homogeneous of degree zero for |ξ | ≥ 1 we have,



k 2 2 a(x, η + kξ0 ) = a x, (2ξ0 + η) = a x, 2ξ0 + η , 2 k k

 1 2 2t = a(x, 2ξ0 ) + η · ∂η a x, 2ξ0 + η dt k 0 k 1 = a(x, ξ0 ) + ck (x, η). k

260

9 Solutions of the Problems

Since a is a symbol of order zero we have,  |ck (x, η)| ≤ C1 |η|

10

0

2t 2ξ0 + η k

1−1

≤ C2 |η|.

c) It follows that,        ix·η 1 , ck (c, η)χb(η)χ(x)b(x) e dη dx    |η|≤ 1 k  k 2  C , ≤ |η||χb(η)||χ(x)b(x)| dη dx, k Rd ×Rd and the right-hand side tends to zero when k goes to +∞.  , dη dx. We have, d) Set Ak = |η|≥ 1 k eix·η a(x, ξ0 )χb(η)χ(x)b(x) 2

 |Ak | ≤ ≤ It follows that,  (2π)−d

|η|≥ 21 k

C k

|η|≤ 21 k

= (2π)−d  =

Rd

, |η|−1 |a(x, ξ0 )||η||χb(η)||χ(x)b(x)| dη dx,



Rd ×Rd

, |η||χb(η)||χ(x)b(x)| dη dx → 0.

, eix·η a(x, ξ0 )χb(η)χ(x)b(x) dη dx



Rd ×Rd

, eix·η a(x, ξ0 )χb(η)χ(x)b(x) dη dx + o(1),

 a(x, ξ0 )χ(x)b(x) (2π)−d

 =

Rd

, eix·η χb(η) dη Rd

dx + o(1),



a(x, ξ0 )χ(x)2 |b(x)|2 dx + o(1) =

Rd

a(x, ξ0 )|b(x)|2 dx + o(1),

since χ is equal to one on the support in x of a. e) It follows from the previous questions that,  lim (A(χuk ), χuk )L2 (Rd ) =

k→+∞

  since a(x, ξ0 ) = a x, |ξξ00 | .

Rd



ξ0 a x, |b(x)|2 dx, |ξ0 |

9 Solutions of the Problems

261

4. a) The estimates of (1) and (2) follow from the Cauchy–Schwarz inequality and the fact that A ∈ Op S 0 thus is continuous on L2 . Now since a ∈ S 0 it is bounded so we have,  (|bn (x)| + | b(x)|)(|bn (x)) −  b(x)|) dx. |(4)| ≤ C Rd

Using the Cauchy–Schwarz inequality we obtain, b L2 ) bn −  b L2 . |(4)| ≤ C( bn L2 +  b) Let ε > 0. We fix n so large that |(1)| + |(2)| + |(4)| ≤ ε2 . Now n being fixed we can use the previous questions with b = bn ∈ C0∞ (Rd ). It follows that, limk→+∞ (3) = 0 so there exists k0 such that for k ≥ k0 we have |(4)| ≤ 2ε . Therefore, for k ≥ k0 we have |Ik | ≤ ε. To conclude we have just tonotice that b(x)|2a x, |ξξ00 | = since χ1 is equal to one on the support in x of a we have |   |b(x)|2a x, |ξξ00 | .

Solution 37

Part 1 2 − K(x). Since, 1. We have F (x, p) = p11 p22 − p12 0 0 0 p11 = ∂x21  u(0, 0) = g

(0), p22 = ∂x22 u(0, 0) = A, p12 = ∂1 ∂x2 u(0, 0) = h (0),

the condition F (0, p0 ) = 0 is equivalent to Ag

(0) − h (0)2 − K(0, 0) = 0.

2 Therefore, it is sufficient to take A = h (0)g+K(0,0) . On the other hand, since

(0) ϕ(x) = x2 we have ∂x1 ϕ(0, 0) = 0, ∂x2 ϕ(0, 0) = 1. Therefore, 2  ∂F ∂F 0 (0, p0 )(∂xj ϕ(0, 0)∂xk ϕ(0, 0)) = (0, p0 ) = p11 = g

(0) = 0. ∂pj k ∂p22

j,k=1

2. We can therefore apply the Cauchy–Kovalevski theorem. Notice that F is a polynomial in p. Then for every K, every g, h real analytic near the origin with g satisfying the condition g

(0) = 0 the problem (8.1) has a unique real analytic solution u near the origin.

262

9 Solutions of the Problems

Part 2  1. Recall that  u(x) = g(x ) + h(x )xd + 12 Axd2 where, g(x ) = a0 + d−1 j =1 aj xj +

d−1 2

3 0 u(0), (∂xj  u(0))j =1,...,d , (∂xj ∂xk  u(0))j,k=1,...,d . j =1 λj xj +O(|x | ) and p =  Then, if |α| = 0, p00 = g(0); if |α| = 1, pj0 = ∂xj g(0), j = 1 . . . , d − 1, pd0 = h(0), if |α| = 2, pj0k = 0, 1 ≤ j = k ≤ d − 1, pj0d = ∂xj h(0), 1 ≤ j ≤ d − 1,

0 pjj = 2λj , 1 ≤ j ≤ d − 1, 0 pdd = A.

We have F (0, p0 ) = det(pj0k ) − K(0). Developing the determinant with respect 8 to the last line we get, det(pj0k ) = A d−1 j =1 (2λj )+F (λj , ∂xj h(0))1≤j ≤d−1 . Since by condition (8.3) we have λj = 0, j = 1, . . . , d − 1 one can find A such that F (0, p0 ) = 0. On the other hand, since ϕ(x) = xd we have,  ∂F ∂F (0, p0 )(dϕ(0))α = (0, p0 ) = det((pj0k )1≤j,k≤d−1 ), ∂pα ∂pdd

|α|=2

=

d−1 9

(2λj ) = 0.

j =1

Thus the conditions in the Cauchy–Kovalevski theorem are satisfied. Then we have the same conclusion as in Part 1. 2. a) The Hessian matrix of g at zero is a real and symmetric matrix. It is therefore diagonalizable in an orthonormal basis, that means that we can find M ∈ O(d − 1) such that tM Hess(g(0))M = diag(λj ). In particular, since 8 | det M | = 1 it follows from (8.3) that det(Hess(g(0))) = d−1 j =1 λj = 0. t

b) This follows from a simple computation since M M is the (d − 1) × (d − 1) identity matrix. Moreover, since mj d = mdj = 0 for 1 ≤ j ≤ d − 1 and mdd = 1 we have, M(y , yd ) = (M y , yd ). c) Using the Taylor formula we can write in a neighborhood of the origin, v(y + hy0 ) = v(y) + h∇y v(y) · y0 +

h2 Hess(v)(y)y0 , y0  + O(|h|3 ), 2

u(My + hMy0 ) = u(My) + h∇x u(My0 ) · (My0 ) +

h2 Hess(u)(My)My0 , My0  + O(|h|3 ). 2

We have by definition v(y + hy0 ) = u(My + hMy0 ), v(y) = u(My). Then dividing by h and taking h = 0 we get ∇y v(y) · y0 = ∇x u(My0 ) · (My0 ).

9 Solutions of the Problems

263

By the same way we obtain Hess(v)(y)y0 , y0  = Hess(u)(My)My0 , My0  which can be written as,   Hess(v)(y)y0 , y0  = tMHess(u)(My)My0 , y0 . This equality for any y0 ∈ Rd implies that Hess(v)(y) = tMHess(u)(My)M. d) This follows from the fact that | det M| = 1. e) Applying the Taylor formula to g we can write,  1 Hess(g)(0)M y , M y + O(|y |3 ), 2  1 = g(0) + ∇x g(0) · (M y ) + tM (Hess(g))(0)M y , y + O(|y |3 ), 2

 g (y ) = g(0) + ∇x g(0) · (M y ) +

= g(0) + ∇x g(0) · (M y ) +

1 λj yj2 + O(|y |3 ), 2 d−1 j =1

by question 2a). f) Summing up we have proved that the function v(y) = u(My) = u(M y , yd ) satisfies, det(Hess(v)(y)) = K(My),

v|yd =0 =  g (y ),

∂v |y =0 = h(M y ), ∂yd d (9.35)

with,  g (y ) = g(0) + ∇x g(0) · (M y ) +

1 λj yj2 + O(|y |3 ). 2 d−1 j =1

Therefore, we can apply the result obtained in Case 1 and deduce that the problem (9.35) has a unique real analytic solution in a neighborhood of the origin. It follows that u(x) = v(M −1 x) is the unique real analytic solution of problem (8.2).

Solution 38 1. a) We have |X1 − X2 | ≤ |X1 − X0 | + |X0 − X2 | ≤ r. Let Y ∈ Bn (X2 , r) then |Y − X1 | ≤ |Y − X2 | + |X2 − X1 | ≤ 2r. Now by the mean value formula we

264

9 Solutions of the Problems

have, since u ≥ 0,  1 u(Y ) dY ≤ u(Y ) dY, |Bn (X2 , r)| Bn (X1 ,2r) Bn (X2 ,r)  1 |Bn (X1 , 2r)| |Bn (X1 , 2r)| u(X1 ). u(Y ) dY = ≤ |Bn (X2 , r)| |Bn (X1 , 2r)| Bn (X1 ,2r) |Bn (X2 , r)|

1 u(X2 ) = |Bn (X2 , r)|



Now |Bn (X2 , r)| = cn r n and |Bn (X1 , 2r)| = cn (2r)n . It follows that u(X2 ) ≤ 2n u(X1 ). b) Now since u is continuous on the closed ball Bn (X0 , 2r ) we have infBn (X0 , 2r ) u = u(X1 ) and supBn (X0 , r ) u = u(X2 ) for some points 2 X1 , X2 ∈ Bn (X0 , 2r ). Applying the above inequality with these two points we obtain, sup Bn (X0 , r2 ) √

u ≤ 2n

inf

Bn (X0 , 2r )

u.

√

2. a) We have ∂t2 ψ = et λ λϕ = −et λ x ϕ = −x ψ so t,x ψ = 0. r Let X0 = ( 3r 2 , x0 ). We have X = (t, x) ∈ Bd+1 (X0 , 2 ) if and only if

3r r 3r r r max |t − |, |x − x0 | ≤ ⇔ |t − | ≤ , |x − x0 | ≤ , 2 2 2 2 2 which is equivalent to, r X ∈ (r, 2r) × Bd (x0 , ). 2 b) According to the previous question we have, sup Bd+1 (X0 , 2r )

ψ =e

√ 2r λ

sup ϕ ≥ e

√ 2r λ

inf

Bd (x0 , 2r )

Bd (x0 , 2r )

ϕ=e

√ r λ

= er

e

√ λ

!

√ r λ

inf

Bd (x0 , r2 )

inf

Bd+1 (X0 , r2 )

ϕ ,

ψ.

c) Applying the previous question and question 1 in  = Bd+1 (X0 , 4r) to the positive harmonic function ψ we can write, er

√ λ

inf

Bd+1 (X0 , 2r ) √

ψ≤

sup Bd+1 (X0 , 2r )

ψ ≤M

It follows that er λ ≤ M which implies that r ≤ on the dimension d.

inf

Bd+1 (X0 , 2r )

C √ λ

ψ.

where C depends only

9 Solutions of the Problems

265

3. By the conclusion of question 2c) there exists C > 0 (depending only on d) such that, C r > √ ⇒ Bd (x0 , 4r) ∩ Z(ϕ) = ∅. λ

(9.36)

We proceed by contradiction. If the claim (8.6) is not true we have, M ∀M > 0, ∃λ > 0, ∃x ∈ ω : d(x, Z(ϕ)) > √ . λ √ > Take M = 5(C + 1) and the corresponding λ, x. Take r = C+1 λ we have Bd (x, 4r) ∩ Z(ϕ) = ∅. But if z ∈ Z(ϕ) one can write,

|z − x| ≥ d(x, Z(ϕ)) >

(9.37) C √ . By λ

(9.36)

5(C + 1) = 5r √ λ

which gives a contradiction.

Solution 39

Part 1 1. Recall that in the polar coordinates (r, ω) ∈ (0, +∞) × S2 , the Laplacian can ∂2 2 ∂ 1 2 be written as  = ∂r 2 + r ∂r + r 2 ω , where ω is an operator on S (the Laplace–Beltrami operator). The function uk is C ∞ on B and depends only on r. It follows that, uk =

∂2 2 ∂ + 2 ∂r r ∂r

sin(kπr) . r

An elementary computation shows that −uk = (kπ)2 uk . On the other hand, uk |r=1 = sin(kπ) = 0. Thus uk is an eigenfunction corresponding to the eigenvalue (kπ)2 . 2. Using the polar coordinates we get,   uk 2L2 (B)

=

S2

1 0

|S2 | sin (kπr) dr dω = 2



1

2

0

(1 − cos(2kπr)) dr =

|S2 | . 2

Now since | sinx x | ≤ 1 we have |uk (x)| ≤ kπ. On the other hand, uk (0) = kπ. Therefore, uk L∞ (B) = kπ.

266

3. Let ek =

9 Solutions of the Problems uk uk L2 (B) .

2

Then ek is also an eigenfunction such that ek L2 (B) = 1 and

ek L∞ (B) = |S22 | kπ = Ck = Ck improved in that case.

3−1 2

. Therefore, the estimate (7.7) cannot be

Part 2    2 1. We have  a(y) dy = m i=1  |vi (y)| dy = m since vi L2 () = 1. On the other hand, since the function a is continuous on  there exists y0 ∈  such that sup a = a(y0). Therefore, by the above equality we get m ≤  || a(y0). 2 2. Since (vi ) is an orthonormal basis of V we have uy0 2L2 () = m i=1 |vi (y0 )| = a(y0 ). On the other hand, we have uy0 L∞ () ≥ |uy0 (y0 )| = a(y0 ). uy 3. Let U = uy 02 . Then U ∈ V and U L2 () = 1. By the previous question 0 L ()

we have, U L∞ () =

 uy0 L∞ () a(y0) ≥√ = a(y0). uy0 L2 () a(y0)

It follows from question 1 that, sup

1

1

u L∞ () ≥ U L∞ () ≥ ||− 2 m 2 .

u∈V u L2 () =1

4. The estimate (7.7) shows that for u ∈ V we have u L∞ () ≤ Cλ then from question 3 that, ||− 2 m 2 ≤ Cλ 1

1

d−1 4

,

which proves (8.7).

Solution 40 1. If f (x) = (x1 + ix2 )j = zj we have Rd+1 f =

∂2 ∂2 + ∂x12 ∂x22

! f =

1 ∂ ∂ j z = 0. 4 ∂z ∂z

d−1 4

. It follows

9 Solutions of the Problems

267

On the other hand, f (rω) = r j (ω1 + iω2 )j so that, using formula (8.8) in the statement, we obtain for r > 0, 0 = Rd+1 f * + = r j −2 j (j − 1)ej + d j ej + ω ej * + = r j −2 ω ej + j (j + d − 1)ej , which proves our result. 2. We have |ej (ω)|2 = (ω12 + ω22 )j = (1 − |ω |2 )j , so that, using the hint given in the statement, we can write, when d ≥ 3,  ej 2L2 (Sd )

=

 {

ω ∈Rd−1 :|ω |≤1



1

= 2π 0

= 2π|S|

}

2π

(1 − |ω |2 )j (1 − |ω |2 ) 2 dα dω , 1

0

(1 − ρ 2 )j + 2 ρ d−2 dρ d$, 1

Sd−2  √j d−2 0



j + 12 t2 t d−2 dt √ √ . 1− j j j

Since by the Lebesgue dominated convergence theorem we have, √



j

lim

j →+∞ 0



j + 12  +∞ t2 2 t d−2 dt = e−t t d−2 dt := c 1− j 0

we obtain, lim j

j →+∞

d−1 2

ej 2L2 (Sd ) = 2π|Sd−2|c := c0 .

If d = 2 we have ω ∈ (−1, 1) so that,  ej 2L2 (S2 ) = Then we set ω =

√t

j

1



−1 0

2π

(1 − |ω |2 )j (1 − |ω |2 ) 2 dα dω . 1

and we conclude as above.

3. We have ej L∞ (Sd ) = supω∈Sd (1 − |ω |2 )j = 1 so that, uj L∞ (Sd ) =

d−1 1 ∼ c1 j 4 , ej L2 (Sd )

j → +∞.

268

9 Solutions of the Problems

Solution 41

Part 1   1. Recall that we have, V = H −1 + f ∈ S : x−1 f ∈ L2 . Since u ∈ H 1 we have −u ∈ H −1 ; moreover x−1 (x2 u) = x u ∈ L2 − . Therefore, −u + x2 u ∈ V . The operator − + x2 is bijective by the existence and uniqueness result proved in Problem 1 question 3. It is continuous since, (− + x2 )u V ≤ u H −1 + x−1 x2 u L2 , ≤ u H 1 + x u L2 ≤ u V . The inverse operator is therefore continuous by the Banach theorem. Summing up, P = − + x2 is an isomorphism from V to V . 2. a) If f ∈ L2 we deduce from (2.1) that,

(∇x u, ∇x w)L2 + (x u, x w)L2 = f , w L2 ,

∀w ∈ V .

Taking w = Dh v where v ∈ V we obtain,   (∇x u, ∇x Dh v)L2 + x2 u, Dh v

L2



= f , Dh v L2 ,

∀v ∈ V .

(9.38)

Now we have, (∇x u, ∇x Dh v)L2 = (∇x D−h u, ∇x v)L2 (Rd ) ,



f , Dh v

L2



= D−h f , v L2 ,

and,   x2 u, Dh v

L2

  = y2 D−h u, v

L2

 1 + [y + hek 2 − y2 ]u(y + hek ), v 2 . L h

We have, h1 y + hek 2 − y2 = 2yk + h. It follows from (9.38) that,  

(∇x D−h u, ∇x v)L2 + y2 D−h u, v 2 = D−h f , v L2 L  − (2yk + h)u(y + hek )v(y) dy, ∀v ∈ V . Rd

(9.39)

9 Solutions of the Problems

269

b) We can take in (9.39) v = D−h u ∈ V . Using the Cauchy–Schwarz inequality, setting x = y + hek in the integral and using the fact that |2xk − h| ≤ C x, we obtain,

∇x D−h u 2L2 + x D−h u 2L2 ≤ | D−h f , D−h u L2 | + C x u L2 D−h u L2 .

Now using Problem 1 question 3c) we have, x u L2 D−h u L2 (Rd ) ≤ x u L2 ∇x u L2 ≤ u 2V ,

| D−h f , D−h u L2 | ≤ D−h f H −1 D−h u H 1 ≤ C f L2 D−h u H 1 , ≤ ε D−h u 2H 1 + Taking ε =

1 2

C f 2L2 . 4ε

we obtain,

∇x D−h u 2L2 + x D−h u 2L2 ≤ C ∇x D−h u 2L2 + x D−h u 2L2 . c) Let (hn ) be a sequence in R which converges to zero. From the above inequality we see that the sequence (∇x D−hn u) is uniformly bounded in L2 . Therefore, it has a subsequence which converges weakly in L2 , thus in D , to a v ∈ L2 . On the other hand, since ∇x u ∈ L2 Problem 1 question 3b) shows that (D−hn ∇x u) converges to ∂xk ∇x u in H −1 , thus in D . Therefore, ∂xk ∇x u = v ∈ L2 . It follows that u ∈ L2 and the equation shows that x2 u ∈ L2 . d) The operator P is clearly continuous from D to L2 . It is bijective; indeed if f ∈ L2 ⊂ V there exists a unique u ∈ V such that −u + x2 u = f. Using (8.10) we see that u ∈ D. Its inverse is continuous by the Banach isomorphism theorem since D and L2 are Banach spaces. 3. If f ∈ L2 is such that P −1 f = 0 ∈ D then f = 0 by the previous question. Moreover, we have P −1 (L2 ) = D. Now, since a composition of continuous map and compact map is compact it follows from Problem 1 question 2 that P −1 is compact from L2 to itself. On the other hand, by question 2c) P is an isomorphism from D to L2 . Therefore, if f ∈ L2 then P −1 f = u ∈ D and this is equivalent to (− + x2 )u = f. Therefore, we have,     P −1 f, f 2 d = u, (− + x2 )u 2 . L (R )

L

Now for u ∈ H 2 we have (u, −u)L2 = ∇x u 2L2 . Therefore, 

P −1 f, f

 L2

= ∇x u 2L2 + x u 2L2 ≥ 0.

This means that P −1 is a positive operator.

270

9 Solutions of the Problems

By the same argument we see that for f, g ∈ L2 setting P −1 g = v we have,     P −1 f, g 2 = u, (− + x2 )v 2 = (∇x u, ∇x v)L2 + (x u, x v)L2 , L L     −1 2 f, P g 2 = (− + x )u, v 2 = (∇x u, ∇x v)L2 + (x u, x v)L2 . L

L



Therefore, we have P −1 f, g L2 = f, P −1 g L2 , which proves that P −1 is selfadjoint. 4. This is obvious. 5. a) We have u ∈ V and −u + x2 u = (λ + 1)u. Our claim is true when k = 2 since u ∈ D. Now assume the claim true at the order k ≥ 2 and let α ∈ Nd , |α| = k − 1. Then ∂ α u ∈ V .  Indeed, by

the induction, we have ∂ α u ∈ H 1 and x ∂ α u = ∂ α (x u) − β=0 αβ ∂ β x ∂ α−β u ∈ L2 since x u ∈ H k−1 , ∂ β x is bounded and |α − β| ≤ k − 2. Moreover, differentiating the equation we obtain,



 α ∂ β x2 ∂ α−β u =: g. β

(− + x2 )∂ α u = (λ + 1)∂ α u −

β=0

By the induction we have (λ+1)∂ α u ∈ L2 , then for |β| = 1, x ∂ α−β u ∈ L2 , for |β| = 2, ∂ β x2 is bounded, ∂ α−β u ∈ L2 and for |β| ≥ 3, ∂ β x2 = 0, so g ∈ L2 . Therefore, we can apply (8.10) to ∂ α u and deduce that ∂ α u ∈ H 2 and x ∇x ∂ α u ∈ L2 . Therefore, u ∈ H k+1 and x u ∈ H k . Eventually we deduce from the equation that x2 u = u + (λ + 1)u ∈ H k−1 . b) We have,   −u + |x|2 u, u

L2

= λ u 2L2 .

∞ . Therefore, we can integrate by parts in the above Now u ∈ ∩k∈N H k ⊂ C→0 formula and deduce that,

∇x u 2L2 + |x|u 2L2 = λ u 2L2 . c) Integrating by parts we can write, 

 Rd

∂xj u(x)xj u(x) dx = −

 Rd

|u(x)|2 dx −

Rd

xj u(x)∂xj u(x) dx.

Therefore, by the Cauchy–Schwarz inequality,  u 2L2

= −2Re

Rd

∂xj u(x)xj u(x) dx ≤ 2 ∂xj u L2 xj u L2 .

9 Solutions of the Problems

271

Summing in j from 1 to d and using the Cauchy–Schwarz inequality we obtain,

d u 2L2

⎛ ⎞1 ⎛ ⎞1 2 2 d d   2 ⎠ ⎝ 2 ⎠ ⎝ ≤2 ∂xj u L2 xj u L2 = 2 ∇x u L2 |x|u L2 , j =1

≤

∇x u 2L2

j =1

+ |x|u 2L2 .

d) Follows immediately from questions b) and c) since u L2 = 0. 6. a) If j, k ∈ N we have, (ej , ek )V = (∇x ej , ∇x ek )L2 + (x2 ej , ek )L2 = ((− + x2 )ej , ek )L2 , = (λj + 1)(ej , ek )L2 = 0. Therefore, (ej , ek )V = 0 if j = k and ej 2V = (λj + 1).   b) It follows that (u, u)V = j,k uj uk (ej , ek )V = j (λj + 1)|uj |2 . c) We have, u ∈ D ⇔ u ∈ V and (− + x2 )u ∈ L2 ⇔ u =

 n

un en and



λ2n |un |2 < +∞.

n

7. a) Since λ = λn for all n ∈ N the operator − + |x|2 −  λ is injective. To see n that it is surjective let f = n∈N fn en ∈ L2 . Set u = n λnf−λ en . Then by question 6c) u ∈ D and (− + |x|2 − λ)u = f. b) Thus (− + |x|2 − λ) : D → L2 is not invertible if and only if λ = λn for some n ∈ N. Part 2 Case 1. d = 1 1. The first identity is obvious. Let us prove the second by induction on n. If n = 1 we have, (L− L+ − L+ L− )u = (

d d + x)(−u + xu) − (− + x)(u + xu) = dx dx

= (−u

+ u + xu − xu + x 2 u) − (−u

− u − xu + xu + x 2 u) = 2u. Assume the identity true at the order n. Then using the induction we can write, L− (L+ )n+1 = L− (L+ )n L+ = (L+ )n L− L+ + 2n(L+ )n .

272

9 Solutions of the Problems

Using the case n = 1 above we deduce that, L− (L+ )n+1 = (L+ )n+1 L− +2(L+ )n +2n(L+ )n = (L+ )n+1 L− +2(n+1)(L+ )n . 2. a) This follows from the fact that L− φ0 = 0 and question 1. This shows that λ = 1 is an eigenvalue of H. b) We argue by induction on n. The claim is obviously true for n = 0. Assume it is true for n and let us prove it for n + 1. We have 2

d x + x) Pn (x)exp − , dx 2 2 x

= (−Pn (x) + 2xPn (x))exp − , 2

φn+1 (x) = (L+ φn )(x) = (−

and −Pn + 2xPn is a polynomial of order n + 1 whose coefficient of x n+1 is equal to 2n . c) We have, Hφn = (L+ L− + 1)φn = L+ L− (L+ )n φ0 + φn . We deduce from question 1 that, Hφn = (L+ )n+1 L− φ0 + 2n(L+ )n φ0 + φn = (2n + 1)φn , since L− φ0 = 0.



d) Notice that for ϕ, ψ ∈ S(R) we have ϕ, L+ ψ L2 = L− ϕ, ψ L2 . It follows



that for any m ≥ 0 we have ϕ, (L+ )m ψ L2 = (L− )m ϕ, ψ L2 . Therefore, we have for n, m ∈ N with n < m,

(φn , φm )L2 = (L− )m (L+ )n φ0 , φ0 L2 . We shall prove, by induction on n that for every n ≥ 1 and every m < n we have Im,n := (L− )m (L+ )n φ0 , φ0 L2 = 0. If n = 1, m = 0, since L− φ0 = 0 we have,



I0,1 = L+ φ0 , φ0 L2 = φ0 , L− φ0 L2 = 0. Assume that our claim is true at the order n ≥ 1 for all m < n. Let m < n + 1 and consider Im,n+1 . If m = 0 we have,   I0,n+1 = (L+ )n+1 φ0 , φ0

L2

  = φ0 , (L− )n+1 φ0

L2

= 0.

9 Solutions of the Problems

273

If m ≥ 1 we write using question 1,   Im,n+1 = (L− )m−1 L− (L+ )n+1 φ0 , φ0

L2



= (L− )m−1 (L+ )n+1 L− φ0 , φ0



L2

,   + (n + 1) (L− )m−1 (L+ )n φ0 , φ0

L2

= 0,

using L− φ0 = 0 and the induction since m − 1 < n. x2

e) Recall that φj (x) = Pj (x)e− 2 where the polynomials Pj are of exact order j. Then (P0 , . . . , Pn ) form a basis of the space of polynomials of order ≤ n  x2 for every n ∈ N which implies that x j e− 2 f (x) dx = 0 for every j ∈ N. x2

For ξ ∈ R the function x → e−ixξ e− 2 f (x) belongs to L1 (R) since x2

f ∈ L2 (R) and e− 2 ∈ L2 (R). Then we have,  F (ξ ) =

x2

e−ixξ e− 2 f (x) dx =

  +∞ (−ixξ )j − x 2 e 2 f (x) dx. j! j =0

Since,   +∞ (|xξ |)j j =0

j!

x2

e− 2 |f (x)| dx =



x2

e|xξ | e− 2 |f (x)| dx < +∞,

we can apply the Fubini theorem and write, F (ξ ) =

 +∞  (−iξ )j j =0

j!

x2

x j e− 2 f (x) dx = 0,

∀ξ ∈ R.

x2

This shows that the Fourier transform of e− 2 f (x) vanishes identically. This x2

implies that e− 2 f (x) ≡ 0 thus f ≡ 0. 3. a) This follows immediately from the fact that (en ) is an orthonormal basis of L2 . b) This follows from question 3 of Part 2 where we have proved that (Hu, ϕ)L2 = (u, Hϕ)L2 for u, ϕ ∈ D. c) Taking ϕ = ek in question b) we obtain, since Hek = (2k + 1)ek , (u, (2k + 1 − λ)ek )L2 = 0 = (2k + 1 − λ) (u, ek )L2 . If 2k + 1 − λ = 0 for all k ∈ N then (u, ek )L2 = 0 for all k ∈ N which implies that u = 0 which contradicts our assumption u ≡ 0. Therefore, there exists k ∈ N such that λ = 2k + 1. 4. Let k ∈ N be fixed. For any n = k we have (Hu, en )L2 = (2k + 1) (u, en )L2 ,

274

9 Solutions of the Problems

and (Hu, en )L2 = (u, Hen )L2 = (2n + 1) (u, en )L2 . Therefore, 2(k − n) (u, en )L2 = 0. This implies that (u, en )L2 = 0. Now, (en ) being an orthonormal basis of L2 we have  u= (u, en )L2 en = (u, ek )L2 ek . n∈N

This shows that u is proportional to ek . Case 2. d ≥ 2 The proof is straightforward using the path given by questions 1 to 4 above.

Solution 42

Part 0 1. The first claim follows from the Hardy inequality (4.9). For the second we use the Cauchy Schwarz and the Hardy inequalities. We obtain, 

|ψ(x)|2 dx ≤ |x|



1  2

|ψ(x)| dx

2

|ψ(x)|2 dx |x|2

12

≤ C ψ L2 ∇x ψ L2 .

Part 1 1. Using the Hint and setting εx = y, ε > 0 in the integral we obtain,

  |χ(y)|2 dy < 0 F (ψ) = ε ε |∇x χ(y)|2 dy − |y| if ε > 0 is sufficiently small and χ ≡ 0. Thus the infimum E is strictly negative. 2. a) This follows from the definition of an infimum. Indeed in the case where E > −∞ for any n ≥ 1 E + n1 is no more an infimum that is there exists ψn ∈ H 1 with ψn L2 = 1 such that E ≤ F (ψn ) ≤ E + n1 . If E = −∞ then for any n > 0 there exists ψn as above such that F (ψn ) ≤ −n. In all cases F (ψn ) → E. b) We have ψn L2 = 1. On the other hand, since F (ψn ) < 0, for n large enough, using question 1 we obtain, ∇x ψn 2L2

   ψ 2  n  ≤  1  ≤ C ψn L2 ∇x ψn L2 = C ∇x ψn L2 .  |x| 2 

Therefore, ∇x ψn L2 ≤ C.

9 Solutions of the Problems

275

c) Since ψn L2 = 1 there exists a subsequence (ψσ1 (n) ) which converges  in D . Since . Thus (∇x ψσ1 (n) ) converges to ∇x ψ weakly in L2 to ψ ∇x ψσ1 (n) L2 is uniformly bounded (by C) there exists a subsequence (ψσ (n) ) 2

 such that (∇x ψσ (n) ) converges weakly in

L (thus in D ) to v. Then v = ∇x ψ . 2 Now for ϕ ∈ L we have, | ψσ (n) , ϕ L2 | ≤ ϕ L2 . The left-hand side con  



 L2 = sup | ψ,  ϕ 2 | so we obtain, ψ  ϕ 2 | : ϕ L2 = 1 verges to | ψ, L L ≤ 1. The same argument can be applied to ∇x ψσ (n) .  Let ε > 0. Fix R > 0 such that 2 (1 + ψ  2 2) ≤ ε . d) Set θn = ψσ (n) − ψ. R 2 L Then we have,  In =:

{|x|>R}

2 ε |θn (x)|2  2 2) ≤ . dx ≤ ( ψσ (n) 2L2 + ψ L |x| R 2

Now by the Hardy inequality and the fact that (ψn ) is bounded in H 1 by K we have,  Jn =:

{|x|≤R}

|θn (x)|2 dx ≤ θn L2 (|x|≤R) |x|



|θn (x)|2 dx |x|2

12 ,

 H 1 ). ≤ C θn L2 (|x|≤R) ∇x θn L2 ≤ C θn L2 (|x|≤R)(K + ψ Let χ ∈ C0∞ be such that χ(x) = 1 if |x| ≤ 1, χ(x) = 0 if |x| ≥ 2 and 0 ≤ χ ≤ 1. Then,  x    θn L2 (|x|≤R) ≤ χ( )θn  . R L2 Now by question c) we know that (θn ) converges weakly to 0 in H 1 which implies that (χ( Rx )θn ) converges as well weakly to 0 in H 1 . Since the   map u → u from u ∈ H 1 : supp u ⊂ {|x| ≤ 2R} to L2 is compact we deduce that (χ( Rx )θn ) converges strongly (that is for the norm) to zero in L2 . Therefore, there exists N ∈ N such that for n ≥ N, Jn ≤ C θn L2 (|x|≤R)(K +  H 1 ) ≤ ε . Therefore, In + Jn ≤ ε for n ≥ N. ψ 2 3. a) Using questions c) and d) we can write,  2 2 − ) = ∇x ψ F (ψ L



(x)|2 |ψ dx |x|

n→+∞

≤ lim inf ∇x ψσ (n) 2L2 n→+∞



|ψσ (n) |2 dx, n→+∞ |x|

 |ψσ (n) |2 dx = lim inf F (ψσ (n) ) = E. − n→+∞ |x|

≤ lim inf ∇x ψσ (n) 2L2 − lim

276

9 Solutions of the Problems

b) Since θ =

 ψ  2 ψ L

belongs to H 1 and has an L2 norm equal to one we have

F (θ ) ≥ E.  2 2 ) ≥ 0. Since E < 0 we c) It follows from questions a) and b) that E(1 − ψ L 2  2 ≥ 1. Using question 2c) we deduce that ψ  2 2 = 1 and deduce that ψ L L ) = E. from questions 3a) and b) that F (ψ 4. a) When  n → +∞ we have ∇x ϕn − ∇x ψ L2 → 0 and by question 1  ϕn −ψ   |x|  2 → 0 so F (ϕn ) → F (ψ). L

b) Since H 2 ⊂ H 1 we have E1 ≤ E2 . Now let ψ ∈ H 1 with ψ L2 = 1 and (ϕn ) ⊂ H 2 such that ϕn → ψ in 1 H . Let ε ∈ (0, 1). There exists n ∈ N such that ϕn L2 ≥ ψ L2 − ε = 1 − ε > 0 and by question a) F (ϕn ) ≤ F (ψ) + ε. Then we can write, E2 ≤ F

ϕn ϕn L2

≤

1 (F (ψ) + ε). (1 − ε)2

Taking that infimum on all ψ ∈ H 1 with ψ L2 = 1 we obtain, E2 ≤

1 (E1 + ε). (1 − ε)2

Since this holds for all ε > 0 we obtain E2 ≤ E1 . Part 2  = 0 we  ± εϕ and G(f ) = F (f ) − E f 2 2 ≥ 0. Since G(ψ) 1. a) Set f = ψ L obtain,  ± 2εRe

  ψ(x)ϕ(x) dx |x|

  − E ψ (x)ϕ(x) dx ≥ −ε2 G(ϕ).

 ∇x ψ(x) · ∇x ϕ(x) dx −

Dividing both members by 2ε > 0 and letting ε go to zero we obtain,  Re

 ∇x ψ(x) · ∇x ϕ(x) dx −

!    ψ(x)ϕ(x)  dx − E ψ(x)ϕ(x) dx = 0, |x|

which implies that, 1 0   ϕ = 0,  − ψ − E ψ, Re −ψ |x| in the sense of distributions.

9 Solutions of the Problems

277

b) Changing ϕ to iϕ we see that the imaginary part of the quantity above in the brackets is equal to zero. This proves our claim.  ∈ H 1 , question 1 Part 1 shows that ψ ∈ L2 . It follows from the c) Since ψ |x|  ∈ L2 which implies that u ∈ H 2 . Thus ψ  ∈ D. equation that ψ 2. If u ≡ 0 there exists x0 ∈ R3 such that u(x0 ) = 0. Since H 2 (R3 ) ⊂ C 0 , u is continuous, so there exists a neighborhood of x0 in which u does not vanish. Therefore, u L2 = 0. On the other hand, if u ∈ D we have, E u 2L2 Therefore,

 

 u 2 u   ,u = −u − = ∇x u 2L2 −  1  = F (u).  |x| L2 |x| 2 L2 u u L2

(9.40)

is a minimizer.

Part 3 √ 1. Obviously |u| ∈ L2 . Now we have ε ≤ u2 + ε2 ≤ |u| + ε. Therefore, 0 ≤ uε ≤ |u| and for almost all x ∈ R3 limε→0 uε (x) = |u(x)|. It follows from the Lebesgue dominated convergence that 

 lim

ε→0

uε (x)ϕ(x) dx =

|u(x)|ϕ(x) dx,

for all ϕ ∈ C0∞ . This implies that uε → |u| in D . It follows that ∂j uε → ∂j |u| u∂ u in D , 1 ≤ j ≤ 3. Now ∂j uε = √ 2j 2 . It follows that ∂j uε (x) converges to u +ε

u(x) zero if u(x) = 0 and to |u(x)| ∂j u(x) if u(x) = 0. Moreover, for ϕ ∈ C0∞ we have |∂j uε (x)ϕ(x)| ≤ |∂j u(x)ϕ(x)| ∈ L1 . It follows from the Lebesgue dominated convergence that

∂j uε → 1{u(x)=0}

u(x) ∂j u |u(x)|

in D .

u(x) ∂j u ∈ L2 . Therefore, ∂j |u| = 1{u(x)=0} |u(x)| |. This shows || ≤ |∇x ψ | ∈ H 1 and |∇x |ψ 2. a) It follows from question 1 that |ψ    that F (|ψ |) ≤ F (ψ ). Since F (ψ ) is the infimum we must have the equality. It follows from question 1a) in Part 2 that we have the claimed equality. b) This follows from question 1c) in Part 2. −ω|x| c) Let G(x) = e4π|x| . Then G ∈ L1 ∩ L2 and (− + ω2 )G = δ0 (the Dirac distribution at zero). It follows that,

   |)(x), |ψ(x)| = (δ0 |ψ|)(x) = ((− + ω2 )G |ψ|)(x) = (G (− + ω2 )|ψ

  (y)| e −ω|x−y| |ψ |ψ| (x) = dy. = G |x| 4π |x − y| |y|

278

9 Solutions of the Problems

(x0 )| = 0, question c) shows that ψ (y) = 0 d) If there is a point x0 such that |ψ  L2 = 1. Therefore (8.13) almost everywhere, which is impossible since ψ is proved. | ∈ H 2 ⊂ L2 ∩ L∞ . Assume it is true up to the 3. (8.14) is true for k = 0 since |ψ k+1 order k ≥ 0 then since |x| ≤ Ck (|x − y|k+1 + |y|k+1 ) we have, |x|

k+1

| ≤ C |ψ





k −ω|x−y| |ψ (y)|

|x − y| e

|y|

 dy +

e−ω|x−y| k  |y| |ψ (y)| dy . |x − y|

Now on one hand, |x|k e−ω|x| ∈ L1 ∩ L2 ,

(y)| |ψ ∈ L2 , |y|

and on the other hand, by the induction, e−ω|x| ∈ L1 ∩ L2 , |x|

| ∈ L2 . |y|k |ψ

By the Young inequality we have L1 L2 ⊂ L2 and L2 L2 ⊂ L∞ . Therefore,  ∈ L2 ∩ L∞ . |x|k+1 |ψ| 4. a) Let u ∈ Ker(H − E). If u ≡ 0 it follows from question 2. Part 2 that u u 2 is L

a minimizer and from question 2d) Part 3 that |u(x)| > 0 for all x ∈ R3 . b) We have obviously u ∈ Ker(H − E) and u(x0 ) = 0. It follows from question a) above that u ≡ 0. Therefore, u1 = λu2 , λ ∈ C which shows that dim Ker(H − E) = 1. 5. Since |Ax| = |x| we have, A (x) − −ψ

A (x) (Ax) ψ ψ A (x) = (−ψ (Ax) )(Ax) − − Eψ − Eψ |x| |Ax| )(Ax) = 0. = ((H − E)ψ

A L2 = ψ  L2 = 1. Therefore, by question 2 Part 2, ψ A is a Moreover, ψ minimizer. It follows from the previous question that there exists c = 0 such that A = c ψ . Since the L2 norms are equal to one and ψ  > 0 we have c = 1. Thus ψ  is radial. ψ Part 4 1. This follows from (9.40) in question 2 Part 2. 2. a) Since θ ∈ H 2 ⊂ C 0 the function u is continuous on [0, +∞). Now since θ (x) = u(r) is radial we have in dimension d = 3, θ = u

+ 2r u . b) By hypothesis r k u ∈ L∞ ((0, +∞)) for every k ∈ N. This implies the same properties for v and the fact that v ∈ L1 ((0, +∞)).

9 Solutions of the Problems

279

c) This follows from the fact that



1 2 v v

= ru

+ 2u = r u

+ u = r − u − λu = − − λv. r r r 3. a) By the equation satisfied by v we see that T = 0 on (0, +∞) and, by definition of w, on (−∞, 0). Therefore, we have supp T ⊂ {0} and the result follows. b) We have,   T , ϕε  = w, −(rϕε )

− ϕε − λrϕε ,  +∞ 

=− v(r)(rϕε (r)) dr − 0

0

+∞

 r / . r  + λrϕ dr v(r) ϕ ε ε

= I + J.  +∞ Setting r = εs in the integral we see that |J | ≤ ε v L∞ 0 (1 + ε|λ|s)|ϕ(s)| ds. Therefore, J → 0 when ε → 0. On the other hand, setting r = εs in the integral we have, 

+∞

I =− 

0 +∞

=−

%

$ r

 r  2  r  + ϕ v(r) 2 ϕ dr ε ε ε ε

v(εs) sϕ

(s) + 2ϕ (s) ds.

0

Since u is continuous on [0, +∞) and v = ru we have v(0) = 0. So using the dominated convergence theoremwe see that limε→0 I = 0. k −k (k) c) By question a) we have, T , ϕε  = N k=0 (−1) ck ε ϕ (0) and by question b) T , ϕε  → 0 when ε → 0. This implies T = 0. Indeed we show that cN = cN−1 = · · · c0 = 0 as follows. We have εN T , ϕε  → 0 thus cN = 0. Then εN−1 T , ϕε  → 0 which implies cN−1 = 0 and we continue up to c0 . 4. a) The fact that f is C ∞ on R follows from the fact that r k v ∈ L1 (R) for every k ∈ N and the Lebesgue theorem on differentiability of integrals. Now for Im z < 0 and r ∈ (0, +∞) we have |eizr v(r)| = e−Im zr |v(r)| ≤ |v(r)| ∈  +∞ 1 L . Therefore, we can set f (z) = 0 eizr v(r) dr. Again by the Lebesgue theorem we see that f is holomorphic in the set Im z < 0. b) Recall that −rw

− w − λrw = 0 in S (R) and that w  = f . Taking the Fourier transform of both members and using the fact that F (rw

) =

1 d 2 λ d (ξ f ), F (λrw) = − f, i dξ i dξ

we deduce that (ξ 2 + μ2 )f + (2ξ + i)f = 0 in S (R).

280

9 Solutions of the Problems

5. a) Since G is holomorphic in Im z < 0 and vanishes when Im z = 0 the function H (x, y) = G(x + iy) is a solution of the problem, 1 2

∂H =



∂ ∂ +i ∂x ∂y

H = 0 for y < 0,

G|y=0 = 0.

  The surface S = (x, y) ∈ R2 : y = 0 is non-characteristic for the operator ∂ which has constant, thus analytic, coefficients. By the Holmgren theorem every point x0 ∈ S has a neighborhood in which H = 0. It follows that G vanishes in a small ball contained in the set Im z < 0. Since G is holomorphic there, the principle of isolated zeros implies that G = 0. Alternative proof: it follows from Morera’s theorem that if u is a holomorphic function in Im z > 0 and Im z < 0 and is continuous in C then u is holomorphic in C. Here consider the function defined by u(z) = G(z) if Im z ≤ 0 and u(z) = G(−z) for Im z > 0. Then u satisfies all the conditions above (since G = 0 for Im z = 0) therefore u is holomorphic in C. Since u = 0 for Im z = 0 the principle of isolated zeros proves that u = 0. b) The function G(z) = (z2 + μ2 )f (z) + (2z + i)f (z) is holomorphic in Im z < 0. By question 4 G vanishes when Im z = 0. It follows from the previous question that G = 0 in Im z < 0.  f (z) 1 6. a) By the general theory (see the appendix) the integral 2iπ B f (z) dz is equal to n − p where n is the number of zeroes of f and p the number of its poles. Since f is holomorphic in Im z < 0 it has no pole. Therefore, the integral is equal to n ∈ N.

(z) = − z2z+i b) By the question 5b) we have ff (z) 2 +μ2 . Inside B it has a simple pole

(z) = at z = −iμ. The residue at this pole is equal to limz→−iμ (z + iμ) ff (z)

− limz→−iμ  f (z) 1 2iπ

2z+i z−iμ

B f (z) dz 1 2(n+1) .

=

=

1 2μ

− 1. Therefore, by the residue theorem we have

1 2μ

− 1. It follows from question a) that

1 2μ

− 1 = n so

μ= c) It follows from question b) that for all θ ∈ D, θ > 0 satisfying (H − λ)θ = 0 1 we must have λ = − 4(n+1) 2 . The minimum value of λ is then reached for n = 0. This minimum energy is precisely E. Therefore, we must have E = − 14 . d) It follows that

f (z) f (z)

= − 2z+i = −2(z − 2i )−1 . This implies that f (z) = 2 1 z +4

C(z − 2i )−2 , where C ∈ C. 7. a) We have, 

e−irξ ϕ(r) dr = R



+∞

e−r( 2 +iξ ) dr =

1

1

0

d It follows that F (rϕ)(ξ ) = − 1i ( dξ  ϕ )(ξ ) = (ξ − 2i )−2 .

1 2

+ iξ

.

9 Solutions of the Problems

281

  b) The previous question shows that rϕ(r) = F −1 (ξ − 12 )−2 ) . This shows r

r

that w(r) = Crϕ(r) = C1r>0 re− 2 . This implies that v(r) = Cre− 2 so r u(r) = Ce− 2 . Since the set of minimizers is a vector space of one dimension (x) = Ce− 12 |x| with C = we conclude that it is generated by the function ψ  L2 = 1. √1 to have ψ 8π

Solution 43

Part 1 1. We have u ∈ C 2 ([−h, 0], H s (R)) for all s ∈ R, u|y=0 = 0, (∂y u)|y=−h = 0. An integration by parts shows that,

0 = − xy u, u L2 () =

 

 |∂x u(x, y)|2 + |∂y u(x, y)|2 dx dy.



Therefore, u is constant and since u|y=0 = 0 we have u ≡ 0. 2. a) Applying a Fourier transformation in x we see that the equation is equivalent to (∂y2 − ξ 2 ) u(ξ, y) = 0 for all ξ ∈ R. So  u(ξ, y) = A(ξ )ey|ξ | + B(ξ )e−y|ξ | . The boundary conditions show that,  ), A(ξ ) + B(ξ ) = ψ(ξ

|ξ |A(ξ )e−h|ξ | − |ξ |B(ξ )eh|ξ | = 0,

∀ξ ∈ R,

so, A(ξ ) =

eh|ξ |  ), ψ(ξ eh|ξ | + e−h|ξ |

It follows that,  u(ξ, y) =

B(ξ ) =

e−h|ξ |  ), ψ(ξ eh|ξ | + e−h|ξ |

∀ξ ∈ R.

ch((y+h)|ξ |)  ch(h|ξ |) ψ (ξ ).

sh(h|ξ |)  b) We deduce from the above formula that (∂y u)(ξ, 0) = |ξ | ch(h|ξ |) ψ(ξ ) =  ). |ξ |th(h|ξ |)ψ(ξ c) The symbol a(ξ ) = |ξ |th(h|ξ |) is C ∞ in R \ {0} and also near zero since 2k+1 . On the other hand, this symbol is even. It is th(h|ξ |) = +∞ k=0 ck (h|ξ |) therefore sufficient to estimate it for ξ ≥ 0. The symbol ξ belongs to S 1 and th(h|ξ |) is bounded (by 1) and has all its derivatives bounded, so it belongs to S 0 . Therefore, a ∈ S 1 . c) Since |th x| ≤ 1 we have,  2 (ξ )|2 dξ ≤ ψ 2 1 . a(Dx )ψ s− 1 = ξ 2s−1 |ξ |2 (th (h|ξ |))2 |ψ s+ H

2 (R)

H

2 (R)

282

9 Solutions of the Problems

Part 2  η(x) 1. For u ∈ C ∞,0 () and (x, y) ∈ ) we can write u(x, y) = − y ∂y u(x, t) dt so that,  |u(x, y)|2 ≤ |η(x) + h|

η(x) −h

|∂y u(x, t)|2 dt ≤

3h 2



η(x) −h

|∂y u(x, t)|2 dt.

Then,  

η(x)

R −h

|u(x, y)|2 dy dx ≤

3h 2

2  

η(x)

R −h

|∂y u(x, y)|2 dy dx.

Since C ∞,0 () is dense in H 1,0() for the H 1 () norm this inequality extends to H 1 ().  z=0 = ψ, ψ|  z=−h = 0. Now for fixed 2. a) It is obvious, by construction, that ψ| z|ξ | z, by the Parseval equality and the fact that e ≤ 1 for z ≤ 0 we can write,  L2 (R) ≤ (2π)− 2 |χ(z)| ψ  L2 (R) , (·, z) L2 (R) = (2π)− 2 |χ(z)| ez|ξ | ψ ψ 1

1

≤ |χ(z)| ψ L2 (R) .  =  L2 ( Integrating in z we obtain, ψ ) ≤ C ψ L2 (R) . Now, ∂x ψ z|D |

z|D | z|D | x x x  = χ (z)e ∂x ψ and ∂z ψ ψ + χ(z)e |Dx |ψ. Therefore, χ(z)e we have,  22  ≤ C ∂x,z ψ L () Since

0

−h 2|ξ |e

2z|ξ | dz



 )|2 |ξ | |ψ(ξ 2

R

=

 2 2  ≤ C ∂x,z ψ L ()

0

 R

d −h dz



0 −h

e

2z|ξ |

dz

dξ + C ψ 2L2 (R) .

2z|ξ |

e dz ≤ 1 we obtain,

(ξ )|2 dξ + C ψ 2 2 |ξ ||ψ ≤ C

ψ 2 L (R)

1

.

H 2 (R)

(x, y − η(x)). We have ψ|y=η(x) = ψ. On the other hand, b) Set ψ(x, y) = ψ 5h if y ≤ − 6 we have y − η(x) ≤ y + h2 ≤ − h3 , so ψ(x, y) = 0. Setting y − η(x) = z in the integral in y we see that,  L2 (  L2 ( ψ L2 () + ∂y ψ L2 () ≤ C( ψ ) + ∂z ψ ) ).   y − η(x)) − ∂x η(x)(∂z ψ)(x, y − η(x)). It Now, (∂x ψ)(x, y) = (∂x ψ)(x, follows that,  L2 (  L2 ( ∂x ψ L2 () ≤ C ∂x ψ ) + ∂x η L∞ (R) ∂z ψ ) .

9 Solutions of the Problems

283

The desired estimate follows from question a). 3. We have,    |∂x u(x, y)|2 + |∂y u(x, y)|2 dx dy. a(u, v) = 

The coercivity on H 1,0() follows then from question 1. 4. a) By the Lax–Milgram lemma it is sufficient to prove that xy ψ defines a continuous linear form on H 1,0 (). For ϕ ∈ C ∞,0 () set,    xy ψ, ϕ := xy ψ(x, y)ϕ(x, y) dx dy. 

Since ϕ|y=η(x) = 0 and ψ ≡ 0 for y ≤ − 5h 6 , an integration by parts shows that,    xy ψ, ϕ = − ∇xy ψ(x, y) · ∇xy ϕ(x, y) dx dy, 

so that,   | xy ψ, ϕ | ≤ ψ H 1 () ϕ H 1 () ≤ C(1 + η W 1,∞ (R) ) ψ

1

H 2 (R)

ϕ H 1 () .

This shows that xy ψ defines a continuous linear form on C ∞,0 () endowed with the H 1 () norm. By density it can be extended to a continuous linear form on H 1,0(). The Lax–Milgram lemma shows that there exists a unique u ∈ H 1,0() such that,   a(u, v) = xy ψ, v ,

∀v ∈ H 1,0().

  If ϕ ∈ C0∞ () we have a(u, ϕ) = −xy v, ϕ . b) This follows immediately from a). 5. We have ρ(x, 0) = η(x), ρ(x, −h) = −h and ∂z ρ(x, z) = h1 η(x) + 1 ≥ 1 − 1 1 ∞ h η L (R) ≥ 2 . For fixed x, the map z → ρ(x, z) from (−h, 0) to (−h, η(x)) being strictly increasing, is bijective. Therefore, the map, (x, z) → (x, ρ(x, z))  to  and its Jacobian, which is equal to ∂z ρ, is different from is bijective from  zero. Therefore, it is a C ∞ - diffeomorphism. 6. This follows immediately from the chain rule. 7. We have, ∂z U = ∂z 1  φ − ∂x ∂z ρ · 2  φ − ∂x ρ · ∂z 2  φ, = (∂z ρ)21  φ − ∂x ∂z ρ · 2  φ − (∂z ρ)(∂x − 2 )2  φ, φ − ∂x ((∂z ρ)2  = (∂z ρ)(21 + 22 ) φ) = −∂x ((∂z ρ)2  φ).

284

9 Solutions of the Problems

 = {(x, z) : x ∈ R, −h < z < 0} . We have, 8. a) Recall that  U L2 ( φ L2 ( φ L2 ( ) ≤ 1  ) + ∂x ρ L∞ ( ) 2  ) , ≤ 1  φ L2 ( φ L2 ( ) + C ∂x η L∞ (R) 2  ) . It follows from question 4b) that, U L2 ( ) ≤ C(1 + η W 1,∞ (R) ) ψ

H

1 2

.

Now, from question 7 we have, φ L2 () ∂z U L2 ((−h,0),H −1 (R)) ≤ ∂z ρ L∞ (R) 2   ≤ C(1 + η W 1,∞ (R) ) ψ

1

H2

.

b) This follows immediately from Problem 5. 9. By definition we have the equality, (1 − (∂x ρ)2 ) φ(x, z) = (∂x − (∂x ρ)∂y )φ)(x, ρ(x, z)), so that, since ∂x ρ|z=0 = ∂x η, we have, U |z=0 = (∂x − (∂x η)∂y )φ|y=η(x). The surface ! is given by the equation y = η(x). Its unit normal is then the 1 vector n = (1 + (∂x η(x))2 )− 2 (−∂x η(x), 1). The normal derivative is then,

1 ∂ =  ∂y − ∂x η(x)∂x . 2 ∂n 1 + (∂x η(x)) Therefore, we have, U |z=0 =

2

1 + (∂x

η(x))2

∂φ ∂n

|! .

Solution 44

Part 1 1. By the Sobolev embedding we have H 1 (R3 ) ⊂ Lp (R3 ) for 2 ≤ p ≤ 6. Therefore, u3 ∈ L2 (R3 ), which implies that (I − )u ∈ L2 (R3 ) so u ∈ H 2 (R3 ) using the Fourier transform. 2. The function f (u) = u3 is a C ∞ function such that f (0) = 0. Since k ≥ 2 > 32 we can use (3.3) to infer that u3 ∈ H k (R3 ). Then (I − )u = u3 ∈ H k (R3 )

9 Solutions of the Problems

285

which implies, using the Fourier transform, that u ∈ H k+2 (R3 ). Therefore, by induction, u ∈ H m (R3 ) for every m ∈ N. Part 2  Moreover, since E ∈ L1 (R3 ), E  is a continuous 1. Since E is radial so is E. 3   function. Therefore, E(Aξ ) = E(ξ ) for every ξ ∈ R and every orthogonal matrix A. Taking A such that Aξ = (|ξ |, 0, 0) we can write,  e−|x|  dx. e−i|ξ |x1 E(ξ ) = 4π|x| R3 To compute the integral we use the polar coordinates, x1 = r cos θ1 , x2 = r sin θ1 cos θ2 , x3 = r sin θ1 sin θ2 , r > 0, θ1 ∈ (0, π), θ2 ∈ (0, 2π). Then dx = r 2 dr sin θ1 dθ1 dθ2 . It follows that,  π

 1 +∞ −i|ξ |r cos θ1  E(ξ ) = e sin θ1 dθ1 e−r r dr. 2 0 0 Setting s = cos θ1 in the integral in θ1 we obtain,  )= 1 E(ξ 2

+∞  +1

 0

1 = 2i|ξ |

−1



e

−irs|ξ |

ds e−r r dr

+∞ 

 e−r(1−i|ξ |) − e−r(1+i|ξ |) dr =

0

1 . 1 + |ξ |2

 ) = (I  2. We deduce from the previous question that (1 + |ξ |2 )E(ξ − )E = 1 = δ0 . By inverse Fourier transform we have (I − )E = δ0 . Part 3 1. From question 2 in the previous part one can write, u = u δ0 = u (I − )E = (I − )u E = u3 E. Now since E ∈ L1 (R3 ) and u3 ∈ L∞ (R3 ) we deduce that,  u(x) = E(x − y)u(y)3 dy. R3

0 (R3 ) since 2 > 3 , which 2. This follows from the fact that u ∈ H 2 (R3 ) ⊂ C→0 2 means that u tends to zero when |x| → +∞. 3. (i) If x ∈ ωn we have |x| ≥ 2n , so if |y| ≤ 2n−1 we have |y| ≤ 12 |x|; this implies that |x − y| ≥ |x| − |y| ≥ 12 |x| ≥ 2n−1 . Moreover, u3 ∈ L1 (R3 ). Therefore,  1 n−1 −2n−1 I1 ≤ e |u(y)|3 dy ≤ C(u)e−2 . n−1 3 4π2 R

286

9 Solutions of the Problems

(ii) If |x| ≤ 2n+1 and |y| ≥ 2n+2 we have, |x| ≤ 12 |y| so |x − y| ≥ |y| − |x| ≥ 1 n+1 . Therefore, as in (i) we have, 2 |y| ≥ 2 I2 ≤ C(u)e−2

n+1

.

(iii) We have,  I3 ≤

R3

e−|x−y| 1 3 sup |u(y)|3 dy ≤ A 4π|x − y| y∈ωn−1 4π n−1



e−|x| dx ≤ CA3n−1 , |x|

R3

and the same proof holds for I4 and  I5 . Since by question 1 u(x) = 5j =1 Ij we obtain the estimate (8.20) in the statement. 4. Let M ∈ N. By the previous question we have, N +M

An ≤ C

N +M

n=N

+

N +M n=N +1

e −2

n−1

N +M−1 

+ A3N −1 +

n=N

n=N

A3n

A3n +

≤C

+ A3N +M+1

N +M

e

N +M

A3n

n=N

−2n−1

+ A3N −1

+

A3N +M+1

+3

n=N

N +M

! A3n

.

n=N

Now we notice that, since N − n − 1 ≤ −1, e−2

n−1

= e−2

N−2

e−2

n−1 (1−2N−n−1 )

≤ e−2

N−2

e−2

n−2

.

It follows that, +∞ 

e−2

n−1

≤ e−2

N−2

n=N

+∞ 

e−2

n−2

≤ Ce−2

N−2

.

n=N

Now by question 2 we have limn→+∞ An = 0. Therefore, for any ε > 0 such that 3Cε ≤ 12 one can find N0 so large that for n ≥ N ≥ N0 we have A2n ≤ ε. It follows that 3C

N+M  n=N

A3n ≤ 3Cε

N+M  n=N

An ≤

N+M 1  An . 2 n=N

Summing up we obtain, N+M   N−2 1  An ≤ C e−2 + A3N−1 + A3N+M+1 . 2 n=N

9 Solutions of the Problems

287

Now letting M go to +∞ and using the fact that limn→+∞ An = 0, we obtain, +∞ 

 N−2  An ≤ C e−2 + A3N−1 .

n=N

 N−2  5. a) In particular we have AN ≤ C e−2 + A3N−1 ; so taking N ≥ N1 in order to have AN ≤ 1 we obtain,  N−1   N−2 2 + A2N ≤ (1+C ) e−2 + AN = (1+C )BN2 . BN+1 ≤ (1+C ) e−2 This implies, by induction on N ≥ N1 that,

BN ≤ (1 + C )2

N−N1 +1 −1

(BN1 )2

N−N1

=

2N−N1 1 

2 (1 + C ) B . N 1 1 + C

b) Taking moreover N1 so large that, (1 + C )2 BN1 < 1 we deduce from the previous question that, BN ≤

N−N1 ln((1+C )2 B ) 1 N N1 e2 = Ke−δ2 1 + C

with δ > 0.

c) Since AN ≤ BN we have AN ≤ Ke−δ2 for N ≥ N1 . Therefore, for |x| ≥ 2N1 we have x ∈ ωN for some N ≥ N1 which means that 2N ≤ |x| ≤ 2N+1 so, N

|u(x)| ≤ AN ≤ Ke−δ2 ≤ Ke− 2 δ|x| . N

1

Solution 45

Part 1 1. Since the Fourier transform (with respect to the variable x only) of u (t, ·) is compactly supported, it is clear that this function is smooth. Denote by (·, ·)H s the scalar product in H s (Rd ). By multiplying the equation ∂t u − u = 0 by u we find that, (∂t u(t, ·), u (t, ·))H s + (−u(t, ·), u (t, ·))H s = 0.

288

9 Solutions of the Problems

Observe that P is a projector, that is P = P2 . Consequently u = P u and it follows from the Plancherel theorem that, (∂t u(t, ·), u (t, ·))H s = (∂t u(t, ·), P u (t, ·))H s = (∂t P u(t, ·), u (t, ·))H s = (∂t u (t, ·), u (t, ·))H s . Since u is C 1 in time with values in H s (Rd ), we can write, (∂t u (t, ·), u (t, ·))H s =

1 d u (t, ·) 2H s . 2 dt

Similar arguments imply that, (−u(t, ·), u (t, ·))H s = (∇u (t, ·), ∇u (t, ·))H s . This gives the wanted identity, 1 d u (t, ·) 2H s + ∇u (t, ·) 2H s = 0. 2 dt 2. By integrating in time the above identity we find that (u ) is bounded in L∞ ((0, T ); H s (Rd )) and (∇u ) is bounded in L2 ((0, T ); H s (Rd )). This implies that (u ) is bounded in L2 ((0, T ); H s+1(Rd )) which in turn implies that u belongs to L2 ((0, T ); H s+1(Rd )). Part 2 1. This is a consequence of the uniqueness result for solution of the heat equation in the space of tempered distribution. 2. The Bernstein inequality implies that there is C > 0 such that, for any function f ,     j f  ∞ ≤ C2j d2 j f  2 . L L So, it follows from the triangle inequality that, u(t, ·) L∞ ≤ C

 j ≥−1

 d 2j 2 j u(t, ·)L2 .

Now recall that, u(t, ˆ ξ ) = e−t |ξ | g(ξ ˆ ), 2

9 Solutions of the Problems

289

and observe that |ξ | is bounded from below by c2j on the support of the Fourier transform of j g. Therefore, the Plancherel identity implies that,     j u(t, ·) 2 ≤ e−c2 t 22j j g  2 . L L By combining the previous estimates, we conclude that, 

u(t, ·) L∞ ≤ C

d

2j 2 e−c

2 t 22j

j ≥−1

   j g  2 , L

which is the wanted estimate (up to replacing c by c2 ). 3. a) The Cauchy–Schwarz inequality implies that, |hn | ≤ 2



! |fn−k |

k∈Z



! |fn−k | |gk |

2

≤ f 1 (Z)

k∈Z



|fn−k | |gk |2 .

k∈Z

So, summing on n ∈ Z and using the Fubini theorem for positive series, h 22 (Z) ≤ f 21 (Z) g 22 (Z) , which implies the desired result. b) Given j ≥ −1 set, Aj =

+∞  k=−1

ak . 2|k−j |

The Cauchy–Schwarz inequality implies that, ⎛ ⎞1 ⎛ ⎞1 2 2 +∞ ∞  ∞ ∞    aj ak 2⎠ ⎝ 2⎠ ⎝ ≤ aj Aj . 2|k−j |

j =−1 k=−1

j =−1

j =−1

It remains only to use the previous question which gives, ∞  j =−1

A2j ≤ C0

∞ 

ak2

with

k=−1

C0 =



2−|n| < +∞.

n∈Z

c) We start from, u(t, ·) L∞ ≤ C

∞  j =−1

 d 2j  2j 2 e−ct 2 j g L2 ,

290

9 Solutions of the Problems

and take the square of both sides. We then integrate in time, using 

T

e−ct 2 e−ct 2 dt ≤ 2j

2k

0

1 , c(22j + 22k )

to deduce that, u 2L2 ((0,T );L∞ )

≤

∞  ∞  j =−1 k=−1

 kd    1 j d2   2 2 2 k g  2 .  g 2 j L L c(22j + 22k )

Set, 

aj = 2

j

d 2 −1



  j g  2 . L

Then, u 2L2 ((0,T );L∞ ) ≤

∞  ∞  j =−1 k=−1

2j +k aj ak . + 22k )

c(22j

Write, 2j +k 2j +k 2k−j = = c(22j + 22k ) c22j (1 + 22(k−j )) c(1 + 22(k−j )) Now by considering separately the case k ≥ j and j ≥ k we verify that, 2k−j 1 ≤ |k−j | . 2(k−j ) c2 c(1 + 2 ) We thus have proved that, u 2L2 ((0,T );L∞ ) ≤ C

  aj ak . c2|k−j | j

k

The result (8.22) then follows from the previous question.

Solution 46 1. We use the formula  = div ∇ and elementary computations to write,

|∇h|2 |∇h|2 1 1 1 ∇h = (∂t h − h) + 2 = . (∂t − )Log h = ∂t h − div h h h : ;< = h h2 =0

9 Solutions of the Problems

291

Then, (∂t − )(hLog h) = h(∂t − )Log h + (∂t h − h) Log h − 2∇h · ∇Log h : ;< = =0

=h

|∇h|2 h2

−2

|∇h|2 h

=−

|∇h|2 . h

To obtain the third identity, we start again with the Leibniz rule, (∂t − )

  |∇h|2 1 1 = (∂t − ) |∇h|2 + |∇h|2 (∂t − ) − 2∇ |∇h|2 · ∇h. h h h

Since (∂t − )∇h = 0 we get, ⎛ (∂t − ) |∇h|2 = ⎝∂t −

 j

=2

 i

⎞ ∂j2 ⎠



(∂i h)2

i

⎛

∂i h ⎝∂t −



⎞ ∂j2 ⎠ ∂i h −2

j

:

;
0 by assumption, this implies that, for any m ≥ 0, d dt

 [0,2π]d

hm+1 dx ≤ 0.

2. We multiply the Boussinesq equation by ∂t (h2 ) and write the result under the form, 1 2h(∂t h)2 − ∂t (h2 ) div(∇h2 ). 2 Then, we integrate by parts in x we obtain, 1 2



 [0,2π]d

∂t ∇(h2 ) · ∇(h2 ) dx +

[0,2π]d

2h(∂t h)2 dx = 0.

Since h > 0 by assumption this immediately implies that, 1 d 4 dt

 [0,2π]d

2    ∇(h2 ) dx ≤ 0,

which gives the desired inequality. 3. a) Starting from, ∂t (hm |∇h|2 ) = (∂t hm ) |∇h|2 + 2hm ∇h · ∇∂t h, and then using the equation, ∂t hm − mhm−1 div(h∇h) = 0,

9 Solutions of the Problems

299

we deduce that,   d m 2 h |∇h| dx = mhm−1 div(h∇h)|∇h|2 dx dt [0,2π]d [0,2π]d  2hm ∇h · ∇div(h∇h) dx. +

(9.48)

[0,2π]d

b) Directly from the Leibniz rule we have, hm−1 div(h∇h) |∇h|2 = hm−1 |∇h|4 + hm (h) |∇h|2 , and (after some elementary computations),  2 (m − 1)2 hm−1 |∇h|4 . div(hm ∇h) div(h∇h) = div(h(m+1)/2∇h) − 4 So, integrating by parts in the second term of the right-hand side of (9.48) we get, d dt

 1 2 2 |∇h| h dx = (m + (m − 1) ) hm−1 |∇h|4 dx 2 [0,2π]d [0,2π]d   +m hm h|∇h|2 dx − 2 (div(h(m+1)/2∇h))2 dx,



m

[0,2π]d

[0,2π]d

(9.49) which is the desired identity. c) Integrating by parts we have, 

 [0,2π]d

hm |∇h|2 h dx =

[0,2π]d



∇h · hm (h)∇h dx,



=−

[0,2π]d



h · div hm (h)∇h dx,

 =−  −  −

[0,2π]d

[0,2π]d

[0,2π]d

mhm (h) |∇h|2 dx hm+1 ∇h · ∇h dx hm+1 (h)2 dx.

300

9 Solutions of the Problems

Then, we integrate by parts again to compute the contribution of hm+1 ∇h · ∇h. We can write,     hm+1 ∇h · ∇h dx = − div hm+1 ∇h h dx. [0,2π]d

[0,2π]d

Now, we exploit an identity similar to the one used above:    2 (m + 1)2  hm−1 |∇h|4 , − div hm+1 ∇h h = div h(m+1)/2 ∇h 4 which can be verified by computing both sides. This yields, 

 [0,2π]d

hm |∇h|2 h dx = −  +  −  −

[0,2π]d

[0,2π]d

[0,2π]d

[0,2π]d

mhm (h) |∇h|2 dx   2 div h(m+1)/2 ∇h dx (m + 1)2 m−1 h |∇h|4 dx 4 hm+1 (h)2 dx.

Simplifying we deduce that,  [0,2π]d

hm |∇h|2 h dx = −

1 m+1

+

1 m+1

−

m+1 4

 [0,2π]d





[0,2π]d

[0,2π]d

hm+1 (h)2 dx 2   div h(m+1)/2 ∇h dx hm−1 |∇h|4 dx. (9.50)

d) By reporting the previous identity (9.50) in (9.49) we obtain, d dt

 [0,2π]d

hm |∇h|2 dx =

 m2 + 1 2

−

m(m + 1)  4

 [0,2π]d

hm−1 |∇h|4 dx

9 Solutions of the Problems

301

 m hm+1 (h)2 dx m + 1 [0,2π]d

 m − 2− (div(h(m+1)/2∇h))2 dx, m + 1 [0,2π]d

−

which is the desired result. e) We will use the Sobolev inequality proved in the second part with, θ = h(m+3)/2. Then, ⎧   m+3 ⎪ ⎪ div h(m+1)/2∇h , ⎨ θ = 2 ⎪ 1 m + 3 (m−1)/4 ⎪ ⎩ h ∇θ 2 = ∇h(m+3)/4 = ∇h. 4 Consequently, the inequality,    1/2 4 ∇θ  dx ≤ 9 (θ )2 dx, d d 16 [0,2π] [0,2π] reads,

m+3 4

4  [0,2π]d

hm−1 |∇h|4 dx ≤

9 16



 [0,2π]d



m+3 2

2

 2 div h(m+1)/2∇h dx.

This immediately implies the result. f) This follows directly from questions 3d) and 3e). g) This is an elementary computation. 4. The identity (9.47) applied with m = 1 shows that, d dt



 h dx = − 2

[0,2π]d

[0,2π]d

h|∇h|2 dx ≤ 0.

To obtain the second statement, notice that, since 1 ≤ (1 + the previous question yields that, d2 dt 2



d h dx = − dt



2

[0,2π]d

This concludes the proof.

√ 7)/2, the result of

[0,2π]d

h|∇h|2 dx ≥ 0.

302

9 Solutions of the Problems

5. We now study the Boltzmann’s entropy,  H (t) =

[0,2π]d

h(t, x)Log (h(t, x)) dx.

Directly from the above formula and from the equation we can write, d H = dt =

 

[0,2π]d

[0,2π]d

(∂t h + (Log h)∂t h) dx (div(h∇h) + (Log h) div(h∇h)) dx.

  Notice that [0,2π]d div(h∇h) dx = 0; indeed [0,2π] ∂i f (x) dxi = 0 for any C 1 function f : Rd → R which is 2π-periodic with respect to xi . We compute the contribution of the second term by integrating by parts. This gives, d H =− dt

 [0,2π]d

|∇h|2 dx.

d H ≤ 0. Furthermore, one can apply the result of the previous This proves that dt question 3 with m = 0 to obtain,

d d2 H =− 2 dt dt

 [0,2π]d

|∇h|2 dx ≥ 0.

This concludes the proof.

Solution 49 1. Using the Cauchy–Schwarz inequality and the fact that μ(S2 ) = 4π, we can write, |u1 (t, x)|2 ≤

t2 (4π)2

 S2

|g(x − tω)|2 dω,

so that, 

+∞ 0

1 |u1 (·, x)| dt ≤ (4π)2



+∞ 

2

0

S2

|g(x − tω)|2 t 2 dω dt.

9 Solutions of the Problems

303

Setting y = tω we have dy = t 2 dω dt and we obtain, 

+∞ 0

1 |u1 (·, x)| dt ≤ (4π)2



2

R3

|g(x − y)|2 dy =

1 g 2L2 (R3 ) . (4π)2

The estimate of u3 is similar using the fact that |∇x u · ω| ≤ |∇x u|. Let us consider u2 . We have, 

+∞

|u2 (t, x)|2 dt ≤

0

≤

1 (4π)2 1 (4π)2

 

+∞  0

S2

+∞ 

S2

0

|f (x − tω)|2 dω dt, |f (x − tω)|2 2 t dω dt, t2

which, setting y = tω, gives the estimate 

+∞

|u2 (t, x)|2 dt ≤

0

1 (4π)2

 R3

|f (x − y)|2 dy. |y|2

(9.51)

2. For fixed x we can apply the Hardy inequality (see Problem 13) to the function hx (y) = f (x − y). We obtain 

 |f (x − y)|2 dy ≤ 4 |y|2 3

R3



i=1

 2  ∂f     ∂y (x − y) dy. 3 i R

To conclude we have just to set z = x − y and to use (9.51).

Solution 50

Part 1 1. Setting ∂ = ∂t or ∂xj we have ∂v = eu ∂u, ∂ 2 v = eu (∂ 2 u + (∂u)2 ). It follows    that v = eu u + (∂t u)2 − nj=1 (∂xj u)2 = 0. Now, v|t =0 = eu |t =0 = 1 and ∂t v|t =0 = eu ∂t u|t =0 = g. Summing up we have, v = 0,

v|t =0 = 1,

∂t v|t =0 = g.

2. The above problem has a unique solution given by, v(t, x) = 1 + w(t, x),

w(t, x) =

t 4π

 S2

g(x − tω) dω.

304

9 Solutions of the Problems

3. Since g tends to zero at infinity we can write, 

+∞

g(x − tω) = − t

 d [g(x − sω)] ds = ωj (∂xj g)(x − tω) ds. ds 3

j =1

It follows that, |w(t, x)| ≤

 3  1 t  +∞ |(∂xj g)(x − tω)|s 2 ds dω, 2 4π t S2 s j =1

≤

 3  1  +∞ |(∂xj g)(x − tω)|s 2 ds dω, 2 4πt 0 S j =1

3  1  ≤ |(∂xj g)(y)| dy. 4πt R3 j =1

4. a) We deduce from the above inequality that |w(t, x)| 1 4π ∇x g L1 (R3 ) . 1 ∇x g L1 (R3 ) we have, by definition of w, b) If t ≤ 4π |w(t, x) ≤

1 1 ∇x g L1 (R3 ) g L∞ (R3 ) 4π 4π

 S2

dω =




1 ∇x g L1 (R3 ) g L∞ (R3 ) . 4π

5. If ∇x g L1 (R3 ) g L∞ (R3 ) < 4π we have |w(t, x)| < 1 so that v(t, x) > 0 for all (t, x) ∈ R × R3 . Then u(t, x) = ln v(t, x) is the unique solution of problem (8.26). Part 2 1. If |x0 | ≤ ελ and 1 < t0 < (1 − ε)λ we have |x0 − t0 ω| ≤ |x0 | + t0 ≤ λ. So t0 g(x0 − t0 ω) = −1 and w(t0 , x0 ) = − 4π S2 dω = −t0 . xj 2. We have ∂xj g(x) = − |x| ϕ0 (|x|). Taking polar coordinates we obtain, 

 R3

Since

3

|∂xj g(x)| dx =

j =1 |ωj | ≥

 3

2 j =1 ωj

∇x g L1 (R3 ) ≥

1 2

4π δ

 S2

+∞

|ωj | dω 0

ϕ0 (r)r 2 dr.

= 1, according to the form of ϕ0 we have, 

1+δ 1

r 2 dr = 4π(1 + δ +

δ2 ). 3

9 Solutions of the Problems

305

Part 3 1. Let g(x) = −ε, ε > 0. Then w(t, x) = −εt so that v(t, x) = 1 − εt < 0 if t > 1ε .

Solution 51

Part 1 1. a) If x ∈ A we have |x|2 − 4x1 < 0, |x|2 − 2x1 ≥ 0, so 14 |x|2 ≤ x1 ≤ 12 |x|2 . b) In the spherical coordinates this reads 14 ρ ≤ cos θ1 ≤ min(1, 12 ρ). This shows that cos θ1 ≥ 0 thus θ1 ∈ (0, π2 ) and ρ ≤ 4. Since the function arccos :

(−1, 1) → (0, π) is decreasing this implies that arccos min(1, ρ2 ) ≤ θ1 ≤ ρ

arccos 4 .  c) In spherical coordinates one can write, R3 gA (x)2 dx ≤ J1 + J2 , where, 

2  2π

J1 = 0



0 4  2π

J2 =



2

0



arccos( ρ4 ) arccos( ρ2 ) arccos( ρ4 ) 0

ρ 2 sin θ1 dρ dθ1 dθ2 , ρ 4 (1 + |ln ρ|)2α ρ 2 sin θ1 dρ dθ1 dθ2 . ρ 4 (1 + |ln ρ|)2α

We have, 

4

J1 ≤ 2π  ≤π 0

ρ 2 (1 +

0 4

1 arccos( ρ ) [− cos θ1 ]arccos( ρ4 ) dρ 2α 2 |ln ρ|)

1 dρ. ρ(1 + |ln ρ|)2α

 2ln 2 1 Setting s = ln ρ, thus ds = dρ ρ we get, J1 ≤ π −∞ (1+|s|)2α ds < +∞, since 2α > 1. Moreover, it is clear that J2 is a finite integral. This shows that g ∈ L2 (R3 ).  −k+1 < |x| ≤ 2−k+2 = {0 < |x| ≤ 4} and A ∩ A = ∅ if 2. a) Since ∪+∞ k k k=0 2 +∞ k = k we have A = ∪+∞ A and 1 (x) = 1 (x). It follows from the k A A k k=0 k=0 Beppo–Levi theorem that, u(t, x) =

+∞  t  gAk (x − tω) dω ≤ +∞. 4π S2 k=0

b) According to question 1a) the fact that te1 − tω ∈ Ak is equivalent to, 1 1 (i) |te1 −tω|2 < t (1−ω1 ) ≤ |te1 −tω|2 and (ii) 2−k+1 < |te1 −tω| ≤ 2−k+2 . 4 2

306

9 Solutions of the Problems

Since for ω ∈ S2 we have |te1 − tω|2 = 2t 2 (1 − ω1 ) (i) is automatically satisfied if 1 < t < 2 and (ii) is equivalent to 2−2k+1 < t 2 (1−ω1 ) ≤ 2−2k+3 . c) By question b) above we have,  Ik =

dω Bk

  where Bk = ω ∈ S2 : 2−2k+1 < t 2 (1 − ω1 ) ≤ 2−2k+3 . In spherical coordinates we have ω1 = cos θ1 , ω2 = cos θ1 sin θ2 , ω3 = cos θ1 sin θ2 , dω = sin θ1 dθ1 dθ2 , θ1 ∈ (0, π), θ2 ∈ (0, 2π). If ω ∈ Bk we have then, 1 − ε2 ≤ cos θ1 ≤ 1 − ε1 ,

where ε1 = t −2 2−2k+1 ,

ε2 = t −2 2−2k+3 .

If k is large cos θ1 is close to one and θ1 to zero so this is equivalent to, arccos(1 − ε1 ) ≤ ϕ ≤ arccos(1 − ε2 ). Therefore, we have, 

2π

Ik = 0



arccos(1−ε2 ) arccos(1−ε1 )

arccos(1−ε )

sin θ1 dθ1 dθ2 = 2π [− cos θ1 ]arccos(1−ε21 ) = 2π(ε2 − ε1 ).

We have just to notice that ε2 − ε1 = 6t −2 2−2k . 1 ≥ d) On Ak for large k we have, |x|2 (1+|ln |x||)α According to question 2c) we have for large k, mk ≥ c0 t −2 2−2k c

22k−4 (1+|−k+2|ln 2)α

2k

2 ≥ c (1+k) α.

22k c1 t −2 ≥ . (1 + k)α (1 + k)α

3. We  deduce from the previous questions that u(t, te1 ) is infinite since u(t, te1 ) ≥ k≥k0 mk and α ≤ 1. It follows that (8.28) cannot hold. Indeed by question 1c) the left-hand side is finite while the left-hand side is infinite since, u L2 ((0,T ),L∞ (R3 )) ≥ u(·, te1 ) L2 (0,T ) = +∞. Part 2 1. a) By the Cauchy–Schwarz inequality we have, t2 μ(S2 ) |un (t, x) − u(t, x)| ≤ 16π 2



2

S2

|(gn − g)(x − tω)|2 dω.

9 Solutions of the Problems

307

Integrating in t between 0 and +∞ we obtain,  un (·, x) − u(·, x) 2L2 (R+ ) ≤ C t

 ≤C

+∞  S2

0

R3

(gn − g)(x − tω)|2 t 2 dt dω,

|(gn − g)(x − y)|2 dy = C gn − g 2L2 (R3 ) .

The right-hand side converges to zero when n goes to +∞, therefore (un ) 3 2 + converges to u in L∞ x (R , Lt (R )). 2 + ∞ 3 b) Since L (R , L (R )) is the dual of the Banach space L2 (R+ , L1 (R3 )) the Banach–Alaoglu theorem asserts that the closed ball B(0, M) in L2 (R+ , L∞ (R3 )) in weak-star compact. Therefore, there exists a subsequence (uσ (n) ) which converges to v ∈ L2 (R+ , L∞ (R3 )) in the weakstar topology which means that,   uσ (n) , ϕ → v, ϕ ,

∀ϕ ∈ L2 (R+ , L1 (R3 )).

(9.52)

3 2 + c) By question 1a) we know that (un ) converges to u in L∞ x (R , Lt (R )). This

+ 3 implies that (un ) converges to u in D (R × R ). Moreover, taking ϕ in C0∞ (R+ × R3 ) in (9.52) we see that (uσ (n) ) converges to v in D (R+ × R3 ). Therefore, u = v ∈ L2 (R+ , L∞ (R3 )). This is impossible by Part 1 since we have proved there that the L2 (R+ , L∞ (R3 )) norm of u was infinite. Thus the sequence (un ) is not bounded in L2 (R+ , L∞ (R3 )). d) This follows immediately from the last sentence above. 2. a) To prove the first claim we have just to set x = λy thus dx = λ3 dy, when we compute the L2 (R+ ) norm of gnλk . Now setting s = λt we can write,

uλnk (t, x)

t −3 λ 2 = 4π



 S2

gnk

x−tω λ

dω = λ

 x 1 1 = λ− 2 unk s, = λ− 2 unk λ



1

+∞ 0

uλnk (t, ·) 2L∞ (R3 ) dt

−1

 S2

gnk

x λ

 − s ω dω,

t x , . λ λ

It follows that uλnk (t, ·) L∞ (R3 ) = λ− 2 unk 

s 4π

− 12



t

+∞

=λ

0



λ,·

L∞ (R3 ) and,

unk

t , · 2L∞ (R3 ) dt λ

= unk 2L2 (R+ ,L∞ (R3 )) . λ 2 , To prove the last claim we notice that g, nk (λξ ). Therefore, nk (ξ ) = λ g 3



 gnλk 2H s (R3 ) = λ3

R3

(1 + |ξ |2 )s | gnk (λξ )|2 dξ =

 R3

η s 1 + | |2 | gnk (η)|2 dη. λ

308

9 Solutions of the Problems

It follows that,  gnλk 2H s (R3 )

− (2π)

3

gnk 2L2 (R3 )

=

. R3

/ η s 2 1 + | |2 − 1 |g, nk (η)| dη. λ

Setting for x ≥ 0, f (x) = (1 + x)s we have f (0) = 1 and by the Taylor formula we get, 

1

f (x) − 1 = sx

(1 + μx)s−1 dμ.

0

Setting x = | ηλ |2 and taking λ ≥ 1 we obtain,     2 s  2  1  η 2   s    1 +  η   ≤ s  η  − 1 (1+μ   )s−1 dμ ≤ 2 |η|2 max (1, (1+|η|2 )s−1 ).   λ λ λ λ 0

It follows that, gnλk 2H s (R3 ) − (2π)3 gnk 2L2 (R3 )  s 2 ≤ 2 |η|2 max (1, (1 + |η|2 )s−1 )|g, nk (η)| dη, λ R3 and the right-hand side tends to zero when λ goes to +∞ since gnk ∈ S(R3 ). b) If (8.31) was true we would have with a fixed C and for all λ ≥ 1, uλnk L2 (R+ ,L∞ (R3 )) ≤ C gnλk H s (R3 ) . By the previous question this would imply, unk L2 (R+ ,L∞ (R3 )) ≤ C gnλk H s (R3 ) . Again by question a) above the right-hand side converges to (2π)3 C gnk L2 (R3 ) when λ goes to +∞. Therefore, we would have, unk L2 (R+ ,L∞ (R3 )) ≤ C gnk L2 (R3 ) . But this is impossible since the right-hand side is bounded since (gnk ) converges to g in L2 (R3 ) while the left-hand side tends to +∞ by question 1d).

9 Solutions of the Problems

309

Solution 52 1. We know (see the course reminder) that for t = 0, S(t)u0 L2 (Rd ) = u0 L2 (Rd ) , 1

S(t)u0 L∞ (Rd ) ≤

d

|4πt| 2

∀u0 ∈ L2 (Rd ),

u0 L1 (Rd ) ,

∀u0 ∈ L1 (Rd ).

So S(t) is continuous from L2 (Rd ) to L2 (Rd ) with norm ≤ (2π)− 2 and from 1 L1 (Rd ) to L∞ (Rd ) with norm ≤ d . By interpolation S(t) is continuous d

|4πt | 2

from Lp (Rd ) to Lq (Rd ) where, for 0 < θ < 1, Then,

1 p

+

1 1−θ q = 2 d − 2 1−θ

+

1−θ 2 +θ − d2 θ [|4πt| ] =

1 p

=

1−θ 2

+

θ 1 ∞, q

=

1−θ 2

+ θ1 .

= 1. So q = p . Moreover, its norm is bounded −d( 1 − 1 )

by C[(2π) ] C |t| 2 p since θ2 = 12 − p1 . 2. Recall that the Sobolev embedding ensures that H 1 (Rd ) can be continuously 2d embedded in Lp (Rd ) for 12 ≥ p1 ≥ 12 − d1 = d−2 2d . Therefore, for 2 ≤ p < d−2 there exists C0 > 0 such that, v Lp (Rd ) ≤ C0 v H 1 (Rd ) ,

∀v ∈ H 1 (Rd ).

(9.53)

Let u0 ∈ H 1 (Rd ) and ε > 0. There exists ϕ ∈ C0∞ (Rd ) such that u0 − ϕ H 1 (Rd ) ≤ 2Cε 0 . Then, S(t)u0 − S(t)ϕ H 1 (Rd ) = u0 − ϕ H 1 (Rd ) ≤ From (9.53) for 2 ≤ p
0 we have limt →+∞ C|t| T > 0 such that for t ≥ T we have,

−d( 21 − p1 )

−d( 21 − p1 )

S(t)ϕ Lp (Rd ) ≤

ϕ Lp (Rd ) .

ϕ Lp (Rd ) = 0. So there exists

ε . 2

(9.55)

It follows from (9.54) and (9.55) that for t ≥ T we have S(t)u0 Lp (Rd ) ≤ ε.

310

9 Solutions of the Problems

Solution 53 In all what follows we set L2 = L2 (Rd ). 1. Set F (t) = (f (t), g(t))L2 . Let t0 ∈ R. We have, F (t0 + h) − F (t0 ) − (∂t f (t0 ), g(t0 ))L2 − (f (t0 ), ∂t g(t0 ))L2 = I + I I + I I I, h

f (t0 + h − f (t0 ) − ∂t f (t0 ), g(t0 + h) , I = h L2 I I = (∂t f (t0 ), g(t0 + h) − g(t0 ))L2 ,

g(t0 + h − g(t0 ) − ∂t g(t0 ) . I I I = f (t0 ), h L2 By the Cauchy–Schwarz inequality, since f ∈ C 1 (R, L2 ), we have for |h| ≤ 1,     f (t0 + h) − f (t0 )  − ∂t f (t0 ) |I | ≤   2 sup g(t) L2 → 0 when h → 0. h L |t −t0 |≤1 Similarly I I I → 0. The term I I tends to zero since g ∈ C 0 (R, L2 ). 2. By the previous question we have, ∂t (Bu(t), u(t))L2 = (B∂t u(t), u(t))L2 + (Bu(t), ∂t u(t))L2 , = (Biu(t), u(t))L2 + (Bu(t), iu(t))L2 , = i (Bu(t), u(t))L2 − i (Bu(t), u(t))L2 = (i [B, ]u(t), u(t))L2 . 3. Integrating the previous inequality between 0 and T we get, 

T 0

(i [B, ]u(t), u(t))L2 dt = (Bu(T ), u(T ))L2 − (Bu(0), u(0))L2 .

Since B is of order zero, therefore continuous from L2 to L2 and u(t) L2 = u0 L2 for all t ∈ R, there exists C > 0 such that,    

T 0

   ≤ C u0 2 2 . dt [B, ]u(t), u(t)) (i L2  L

4. a) This follows from the fact that ξ ∈ S 1 ,

x

dy 0 y2σ

∈ S 0 , ξ1 ∈ S −1 .

9 Solutions of the Problems

311

b) We have b ∈ S 0 , −ξ 2 ∈ S 2 . By the symbolic calculus the bracket i [B, ] is of order 0+2−1 = 1 and, modulo  a zeroorder term, its symbol is equal to the Poisson bracket i · 1i b, −ξ 2 = ξ 2 , b = 2ξ ∂x b. Since ∂x b = 12 x12σ ξξ    2 ξ 1 + R, R ∈ Op(S 0 ). Setting  = we eventually have i [B, ] = Op ξ 2σ    x Op (ξ ) we have −1 = Op ξ1 so that i [B, ] = −−1 ∂x2 x−2σ + R, R ∈ Op(S 0 ). 1 c) Set Q = − 2 ∂x . We have Q∗ = −Q and by Fourier transform, Q2 = −1 2  ∂x ∈ Op(S 1 ). Then,   (i [B, ]u(t), u(t))L2 = − Q2 x−2σ u(t), u(t)

L2

+ (Ru(t), u(t))L2 . (9.56)

We have − Q2 x−2σ u(t), u(t) L2 = (1) + (2) where,   (1) = − [Q2 , x−σ ] x−σ u(t), u(t) 2 , L

(2) = Q x−σ u(t), Q x−σ u(t) L2 . Since Q2 ∈ Op(S 1 ) and x−σ ∈ Op(S 0 ) the operator [Q2 , x−σ ] is of order zero. Therefore, |(1)| ≤ C u(t) 2L2 . Now, (2) = Q x−σ u(t) 2L2 . Using (9.56) and the fact that R ∈ Op(S 0 ) we see that there exists C > 0 such that, Q x−σ u(t) 2L2 ≤ | (i [B, ]u(t), u(t))L2 | + C u(t) 2L2 . Integrating this inequality between 0 and T and using question 3 we obtain, 

T 0

Q x−σ u(t) 2L2 dt ≤ C(1 + T ) u0 2L2 .

(9.57)

Now set v = x−σ u. We have for fixed t,  2 ξ  | v (t, ξ )|2 dξ v(t) 1 = H

2

 =

R



|ξ |≤1

√  ≤ 2

ξ  | v (t, ξ )|2 dξ +

|ξ |≥1

 | v (t, ξ )| dξ + 2

|ξ |≤1

ξ  | v (t, ξ )|2 dξ,

|ξ |≥1

ξ  | v (t, ξ )|2 dξ = (1) + (2).

312

9 Solutions of the Problems

We have, (1) ≤ C v(t) 2L2 = C x−σ u(t) 2L2 ≤ C u(t) 2L2 = C u0 2L2 . 2

For |ξ | ≥ 1 we have ξ 2 ≤ 2ξ 2 so that ξ  ≤ 2 ξξ  . It follows that,  (2) ≤ 2 R

ξ2 | v (t, ξ )|2 dξ ≤ C Q x−σ u(t) 2L2 . ξ 

Integrating between 0 and T and using (9.57) we obtain eventually, 

T

x−σ u(t) 2

H

0

1 2

dt ≤ C(1 + T ) u0 L2 .

5. a) Applying the inequality obtained in question 4 to uj − uk we see that (x−σ uk ) is a Cauchy sequence in L2 ((0, T ), H 2 ). Since this space is 1 complete we deduce that (x−σ uk ) converges to v in L2 ((0, T ), H 2 ), therefore in D ((0, T ) × Rd ). b) We have, 1



T

 uk (t)

0

− u(t) 2L2

T

dt = 0

F −1 (e−it ξ ϕk ) − F −1 (e−it ξ  ϕ ) 2L2 dt, 2

= (2π)

−1



T 0

2

e−it ξ ϕk − e−it ξ  ϕ 2L2 dt, 2

2

= T ϕk − ϕ 2L2 → 0. Now the convergence in L2 ((0, T ) × Rd ) implies that in D ((0, T ) × Rd ). c) Since the function x−σ is C ∞ , it follows from question b) that the sequence (x−σ uk ) converges to (x−σ u) in D ((0, T ) × Rd ). From question a) we 1 have x−σ u = v ∈ L2 ((0, T ), H 2 ). We have just to pass to the limit in the estimate 

T 0

x−σ uk (t) 2

H

1 2

dt ≤ C(1 + T ) ϕk 2L2 .

Solution 54

Part 1 1. First of all we have, S(t)u0 H s = u0 H s =

1 R. 2

9 Solutions of the Problems

313

Now, using the estimate (ii) we obtain for u ∈ B and t ∈ (−T , T ),  t     S(t − σ )[f (σ )], dσ    0

 Hs

t

≤ 

t

≤ 

S(t − σ )[f (u(σ ))] H s dσ,

0

f (u(σ )) H s dσ,

0 t

≤

F ( u(σ ) H s ) u(σ ) H s dσ ≤ F (R)RT .

0

Taking T such that F (R)T ≤ 2. For u1 , u2 ∈ B we have,

1 2

we obtain A(u(t) H s ≤ R for all t ∈ (−T , T ). 

A(u1 ) − A(u2 ) L∞ (I,H s ) ≤

T −T

f (u1 (σ )) − f (u2 (σ )) H s dσ.

Since s > d2 , H s (Rd ) is an algebra and there exists C > 0 such that uv H s ≤ C u H s v H s . Using the Taylor formula we deduce that,  A(u1 ) − A(u2 ) L∞ (I,H s ) ≤ C



T −T

1

df (λu1 (σ ) + (1 − λ)u2 (σ )) H s

0

× u1 (σ ) − u2 (σ )) H s dλ dσ, where df is the differential of f . Since u1 , u2 ∈ B we have, λu1 (σ ) + (1 − λ)u2 (σ ) H s ≤ λ u1 (σ ) H s + (1 − λ) u2 (σ ) H s ≤ R. It follows that there exists F : R+ → R+ increasing such that, A(u1 ) − A(u2 ) L∞ (I,H s ) ≤ CF (R)2T u1 − u2 L∞ (I,H s ) . We have just to take T > 0 such that 2CF (R)T < 1. 3. The space L∞ (I, H s ) ∩ C 0 (I, H s ) endowed with the L∞ (I, H s ) norm is a Banach space. The set B is closed in this space. Therefore, it is a complete metric space. If T is small enough the map A sends B into B and it is contractive. By the fixed point theorem there exists a unique u ∈ B such that u = A(u). 4. We have for t ∈ (T , T ), 

t

u(t) = S(t)u0 − i 0

S(t − σ )[f (u(σ )] dσ.

314

9 Solutions of the Problems

Set v(t) = S(t)u0 . We have v ∈ C 0 (R, H s ) ∩ C 1 (R, H s−2 ) and, (i∂t + )v(t) = 0,

v|t =0 = S(0)u0 = u0 .

t Now set w(t) = −i 0 S(t − σ )[f (u(σ )] dσ. By the linear theory we have w ∈ C 0 (I, H s ) ∩ C 1 (I, H s−2 ) and, (i∂t + )w = f (u(t)),

w|t =0 = 0.

Therefore, u = w + w is the unique solution of the problem, (i∂t + )u(t) = f (u(t)),

u|t =0 = u0 .

Part 2 1. We have 0 ∈ F since u1 |t =0 = u2 |t =0 = u0 . Moreover, F is closed since u1 , u2 ∈ C 0 (I, H s ). 2. a) Let t0 ∈ F. We have, 

t+t0

uj (t + t0 ) = S(t + t0 )u0 − i

S(t + t0 − σ )[f (uj (σ ))] dσ = S(t + t0 )u0

0



t0

−i



t+t0

S(t + t0 − σ )[f (uj (σ ))] dσ − i

0

S(t + t0 − σ )[f (uj (σ ))] dσ.

t0

Using the Hint we can write,  S(t + t0 )u0 − i

t0

S(t + t0 − σ )[f (uj (σ ))] dσ

0

%  = S(t) S(t0 )u0 − i

$ S(t0 − σ )[f (uj (σ ))] dσ = S(t)uj (t0 ).

t0

0

On the other hand, setting σ = σ − t0 in the integral we obtain,  −i

t+t0



t

S(t + t0 − σ )[f (uj (σ ))] dσ = −i

t0



S(t − σ )[f (uj (σ + t0 ))] dσ ,

0 t

= −i

S(t − σ )[f (vj (σ ))] dσ .

0

Therefore, we have,  vj (t) = S(t)uj (t0 ) − i 0

t

S(t − σ )[f (vj (σ ))] dσ .

9 Solutions of the Problems

315

b) By hypothesis we have t0 ∈ F so u1 (t0 ) = u2 (t0 ). Using the fact that S(t − σ ) is an isometry on H s we have for t ∈ (−ε, ε), 

t

v1 (t) − v2 (t) H s ≤

S(t − σ )[f (v1 (σ )) − f (v2 (σ ))] H s dσ,

0



t

≤

f (v1 (σ )) − f (v2 (σ )) H s dσ,

0

 ≤

ε

−ε

f (v1 (σ )) − f (v2 (σ )) H s dσ.

c) Using the Taylor formula and the fact that H s is an algebra we deduce that,  v1 (t) − v2 (t)

Hs

≤C

ε

−ε



1

df (λv1 (σ ) + (1 − λ)v2 (σ )) H s

0

· v1 (σ ) − v2 (σ ) H s dλ dσ. Therefore, we obtain, 

1

v1 − v2 L∞ ((−ε,ε),H s ) ≤ 2Cε

df (λv1 + (1 − λ)v2 ) L∞ ((−ε,ε),H s ) dλ

0

· v1 − v2 L∞ ((−ε,ε),H s ) . 1 If we choose ε > 0 such that 2Cε 0 df (λv1 +(1−λ)v2 ) L∞ ((−ε,ε),H s ) < 1, we obtain that v1 (t) = v2 (t) for t ∈ (−ε, ε) which, coming back to the definition of vj shows that there exists ε > 0 such that u1 (t) = u2 (t) for t ∈ (t0 − ε, t0 + ε), which proves that F is open. By connectedness we have F = (−T , T ) that is u1 = u2 on (−T , T ). Part 3 1. By construction we have u L∞ (I,H s ) ≤ R, v L∞ (I,H s ) ≤ R. Now we write, 

t

u(t) − v(t) = S(t)(u0 − v0 ) − i

S(t − σ )[f (u(σ )) − f (v(σ ))] dσ.

0

Using the fact that S(t) is an isometry on H s we deduce,  u(t) − v(t) H s ≤ u0 − v0 H s +

T −T

f (u(σ )) − f (v(σ )) H s dσ.

Using again the Taylor formula and the fact that u L∞ (I,H s ) ≤ R, v L∞ (I,H s ) ≤ R we get, u − v L∞ (I,H s ) ≤ u0 − v0 H s + C(R)T u − v L∞ (I,H s ) .

316

9 Solutions of the Problems

Taking T0 such that C(R)T0 < 1 and T ≤ T0 we absorb the second term in the right-hand side by the left-hand side and we obtain, u − v L∞ (I,H s ) ≤ C (R, T0 ) u0 − v0 H s .

Solution 55 1. By definition [A, B] = AB − BA. Therefore > [P , Lj ] = −2 ∂xj +

d 

? ∂x2k

, xj

= −2 ∂xj + 2 ∂xj = 0.

k=1

It follows that [i∂t + x , Lα ] = 0, for all α ∈ Nd . 2. If g ∈ L2 the problem i∂t u + x u = 0 with u|t =0 = g has a unique solution u ∈ C 0 (R, L2 (Rd )). Consider v = Lα u ∈ C 0 (R, S ). It is easy to see that, v0 = x α u0 = x α g ∈ L2 (Rd ). On the other hand, using question 1, we have (i∂t + x ) v = [i∂t + x , Lα ] u + Lα (i∂t + x ) u = 0. It follows that Lα u ∈ C 0 (R, L2 (Rd )) for all α ∈ Nd . Now, (x + 2it ∂x )α u =



aαβ (t, x) ∂xβ u + (2it)|α| ∂xα u,

(9.58)

|β|≤|α|−1

where aαβ is a polynomial. Since the C ∞ (Rd ) functions act on the spaces s (Rd ), an induction on |α|, starting from |α| = 0 and from the fact that u ∈ Hloc C 0 (R, L2 (Rd )), shows that, for t = 0, we have ∂xα u(t, ·) ∈ L2loc (Rd ). Therefore, s (Rd ) for all s ∈ N and t = 0. Since ∩ H s (Rd ) ⊂ C ∞ (Rd ) we u(t, ·) ∈ Hloc loc s≥0

obtain the result.

Solution 56

1. We have, f(x) =

 R

=

e−ixξ e−|ξ | dξ =



0 −∞



+∞

eξ(1−ix) dξ +

1 1 2 + = . 1 − ix 1 + ix 1 + x2

0

e−ξ(1+ix) dξ,

9 Solutions of the Problems

317

2. We have  g = g1 . On the other hand, by the Lebesgue differentiation of integrals theorem we can write,   d gλ (x) = e−ixξ |ξ |e−λ|ξ | dξ = − e−ixξ e−λ|ξ | dξ dλ      d 1 d 2λ −i xλ η −|η| =− e dξ = − e dλ λ dλ λ2 + x 2 =−

2 4λ2 + · λ2 + x 2 (λ2 + x 2 )2

Taking λ = 1 we deduce that,   F |ξ |e−|ξ | = −

2 4 + . 1 + x 2 (1 + x 2 )2     2 2 1 3. a) Using question 1 we can write, e−|ξ | = F −1 1+x = so, F 2 2π 1+x 2  ) = πe−|ξ | , Q(ξ s  belongs to L2 (R) for all s ∈ R. which shows that (1 + |ξ |2 ) 2 Q  b) For all t ∈ R we have, |D u(t, ξ ). On the other hand, x |u(t, ξ ) = |ξ |   y −ixξ  u(t, ξ ) = 4c e Q(c(x + ct)) dx = 4 e−i( c −ct )ξ Q(y) dy,

 = 4eict ξ

ξ

 e−iy c Q(y) dy = 4eict ξ Q

ξ ξ = 4πeict ξ e−| c | , c

ξ

ict ξ e −| c | .  so |D x |u(t, ξ ) = 4π|ξ |e Taking the inverse Fourier transform and using question 2 we can write,

  1 2 ixξ ictξ −| ξc | 2 eiη(c t+cx) |η|e−|η| dη, |Dx |u(t, x) = dξ = 2c · 4π e |ξ |e e 2π

4 2 2 = 2c − + , 1 + (c2 t + cx)2 (1 + (c2 t + cx)2 )2 1 = −cu(t, x) + u(t, x)2 . 2

c) We deduce from the previous question that, 1 ∂x |Dx |u = −c∂x u + ∂x (u2 ). 2 To conclude we have just to notice that ∂t u = c∂x u.

318

9 Solutions of the Problems

Solution 57

Part 1 1. Since u ∈ L2 (Rd × ), for almost all x ∈ Rd the function ω → u(x, ω) belongs 1 to L2 () which is contained in μ() < +∞.  L (), since 2 2. a) We have |U (x)| ≤ μ()  |u(x, ω)|2 dμ(ω) ∈ L1 (Rd ).    

, ϕ = U,  b) We have for ϕ ∈ S(Rd ), U ϕ  = Rd  u(x, ω) dμ(ω)  ϕ (x) dx. 1 d Since the function (x, ω) →  ϕ (x)u(x, ω) belongs to L (R × ) we deduce from the Fubini theorem that,

    , ϕ = u(x, ω) ϕ (x) dx dμ(ω). U 

Rd

Since for almost all ω ∈  the function x → u(x, ω) belongs to L2 (Rd ) the Parseval theorem implies that,

      u(ξ, ω)ϕ(ξ ) dξ dμ(ω). U, ϕ = 

Rd

Since for almost all ω the function ξ →  u(ξ, ω)ϕ(ξ ) belongs to L1 (Rd ) the Fubini theorem implies that,

0 1       u(ξ, ω) dμ(ω) ϕ(ξ )dξ =  u(ξ, ω) dμ(ω), ϕ . U, ϕ = Rd







(ξ ) = So U u(ξ, ω) dμ(ω); the estimate is a consequence of the Cauchy–  Schwarz inequality and the fact that μ() < +∞. 3. Applying the Cauchy–Schwarz inequality we obtain,  |I1 (ξ )| ≤ μ {ω ∈  : |pω (ξ ) ≤ 1}

| u(ξ, ω)|2 dμ(ω).

2



Since pω is homogeneous of degree one in ξ we have,      ξ  1 1 ≤ c0 2δ , μ {ω ∈  : |pω (ξ ) ≤ 1} = μ ω ∈  : pω ≤ |ξ |  |ξ | |ξ | using ( ) with a = 1/|ξ |. 4. We have just to write,  I2 (ξ ) =

|pω (ξ )|≥1

1 |pω (ξ )| u(ξ, ω) dμ(ω), |pω (ξ )|

and then use the Cauchy–Schwarz inequality and the homogeneity of pω .

9 Solutions of the Problems

319

5. a) By the change of variables theorem and condition ( ) we have, 

 

1{ω:|pω (θ)|≤a} (ω)dμ(ω) =

R+

1{0≤t ≤a}(t) dν(t) ≤ c0 a 2δ .

So ν {0 ≤ t ≤ a} ≤ c0 a 2δ . b) The first equality is an immediate consequence of the change of variables theorem, and the second one follows from the Lebesgue dominated convergence theorem. 6. We have   N−1 k+1 1  |ξ |2 k JN (ξ ) ≤ ≤t ≤ , ν |ξ | |ξ | |ξ |2 k2 k=1

≤

N−1  k=1

 N−1    1  1 k+1 k − . ν 0 ≤ t ≤ ν 0 ≤ t < k2 |ξ | k2 |ξ | k=1

In the first sum of the right-hand side set k = k + 1; we isolate the term k = N and in the second sum we isolate the term k = 1. We obtain 

  N−1   1 N 1 1 k ν 0≤t ≤ − 2 ν 0≤t≤ JN (ξ ) ≤ + (N − 1)2 |ξ | (k − 1)2 k |ξ | k=2   1 −ν 0≤t ≤ ≤ A + B − C. |ξ | Using ( ) we obtain, since 2δ ≤ 2, A≤

N 2δ c1 1 c ≤ . 0 (N − 1)2 |ξ |2δ |ξ |2δ

Now, B≤

N−1  k=2

since δ < 1. Eventually, C ≤

c2 2k − 1 2δ c0 k ≤ 2δ , 2 2δ − 1) |ξ | |ξ |

k 2 (k c0 . |ξ |2δ

7. We deduce from the previous questions that |I2 (ξ )|2 ≤ |ξc|2δ and from question 2 that, 

 (ξ )|2 ≤ C |ξ |2δ |U | u(ξ, ω)|2 dμ(ω) + |pω (ξ )|2 | u(ξ, ω)|2 dμ(ω) . 



320

9 Solutions of the Problems

 2  )|2 ≤ C | Since |U(ξ  u(ξ, ω)| dμ(ω), integrating these inequalities with respect to ξ and using Parseval equality we obtain,   U 2H δ (Rd ) ≤ C u 2L2 (Rd ×) + Op(pω )u 2L2 (Rd ×) . Part 2 1. Assume that there exists ξ0 ∈ Rd such that g(ξ0 ) > 0. Then there exists ε > 0  ∈ C ∞ (Rd ) such that g(ξ ) > 0 for |ξ − ξ0 | ≤ ε. Let ψ ∈ S(Rd ) be such that ψ 0 1   with ψ (ξ ) = 1 if |ξ − ξ0 | ≤ 2 ε, ψ (ξ ) = 0 if |ξ − ξ0 | ≥ ε. Then,    2 2   |ψ (ξ )| g(ξ ) dξ = |ψ (ξ )| g(ξ ) dξ ≥ g(ξ ) dξ > 0, |ξ −ξ0 |≤ 12 ε

|ξ −ξ0 |≤ε

Rd

which contradicts our hypothesis. 2. Apply (

) to u(x, ω) = ψ(x)ϕ(ω) where ψ belongs to S(Rd ) and ϕ ∈ L2 (). Using the fact that, 2  Op(pω )u 2L2 (Rd ×) ≤ C Op(p ω )u L2 (Rd ×) ,   (ξ )|2 |ϕ(ω)|2 dξ dω, |pω (ξ )|2 |ψ ≤C  Rd

we deduce that the quantity, ! 2      2 2δ 2 2  )| |ξ |  ϕ(ω) dμ(ω) − C |ψ(ξ |ϕ(ω)| (1 + |pω (ξ )| ) dμ(ω) dξ   Rd





is negative. Since this holds for any function ψ ∈ S(Rd ) and since the function inside the parentheses in the integral is continuous we deduce from the previous question that we have,  2    |ϕ(ω)|2 (1 + |pω (ξ )|2 ) dμ(ω) ≤ 0, ∀ξ ∈ Rd |ξ |2δ  ϕ(ω) dμ(ω) − C 



(9.59) for all ϕ ∈ L2 (Rd ). Take a > 0, ξ = ηa where η ∈ Sd−1 and ϕ(ω) = 1{|pω (η)|≤a} . We have  2   1 |ξ |2δ  ϕ(ω) dμ(ω) = 2δ (μ {ω ∈  : |pω (η| ≤ a})2 , a    1 |ϕ(ω)|2(1 + |pω (ξ )|2 ) dμ(ω) = (1 + 2 |pω (η|2 ) dμ(ω), a {|pω (η|≤a}  ≤ 2μ {ω ∈  : |pω (η| ≤ a} . Applying the inequality (9.59) we obtain ( ).

9 Solutions of the Problems

321

Solution 58

Part 1 1. We begin by a necessary condition which will give the uniqueness. Assume that f is a solution of (8.35). Consider the differential system, t˙(s) = 1, t (0) = 0,

x(s) ˙ = v, x(0) = x0 ∈ Rd .

We have obviously t (s) = s and x(s) = x0 + sv. Then,

d ∂f + v · ∇x f (s, x0 + sv, v) = 0. [f (s, x0 + sv, v)] = ds ∂t Therefore, f (s, x0 + sv, v) = f (0, x0 , v) = f 0 (x0 , v) which shows that for any x ∈ Rd we have f (s, x, v) = f 0 (x − sv, v). On the other hand, if f 0 ∈ C 1 (Rdx × Rdv ), setting f (t, x, v) = f 0 (x − tv, v) we obtain easily a solution of problem (8.35). 2. According to question 1 we have,   p f (t) Lp (Rd ×Rd ) = |f 0 (x − tv, v)|p dx dv. x

v

Rdx

Rdv

We have just to perform the change of variables y = x − tv in the integral in x to conclude. 3. According to the expression of the solution f we can write,   |ρ(t, x)| ≤ |f 0 (x − tv, v)| dv ≤ sup |f 0 (x − tv, w)| dv. Rd

Rd w∈Rd

Setting v = x − tv in the above integral we obtain dv = |ρ(t, x)| ≤ 4. a) Set A :=

  R Rd

Rd

sup |f 0 (v , w)| dv . w∈Rd



R Rdx



 =



|ρ(t, x)||(t, x)| dx dt. We have,  

A≤

1 td

dv so,

1 td

Rdy

Rdv

Rdv

|f 0 (x − tv, v)||(t, x)| dv dx dt 

|f 0 (y, v)|

|(t, y + tv)| dt

dv dy,

R

     ≤ f La (Rdy ×Rdv )  |(t, y + tv)| dt   0

R

La (Rdy ×Rdv )

.

322

9 Solutions of the Problems

2r b) Recall that a = r+1 . Then a =

a = 2r . It follows that,

2r r−1 .

Since

1 r

= 1−

=

1 r

r−1 r

we have

 

2      = |(t, x + tv)| dt  .  r d d  R L (Rdx ×Rdv ) L (Rx ×Rv )

 2    I = |(t, x + tv)| dt   a R

c) We have,       I = |(t, x + tv)||(s, x + sv)| ds dt  

,

R R

Lr (Rdx ×Rdv )

R R

Lr (Rdx ×Rdv )

      = |(t, x + (t − s)v)||(s, x)| ds dt     ≤ R R

,

(t, x + (t − s)v)(s, x) Lr (Rd ×Rd ) ds dt. x

v

Setting x + (t − s)v = y in the integral in v we obtain,   I≤ R R

  ≤

R R



 Rdx

Rdy

1 d

|t − s| r

1

|(t, y)|r |(s, x)|r dx dy |t − s|d

! 1 r

ds dt,

(t, ·) Lr (Rd ) (s, ·) Lr (Rd ) ds dt.

d) This follows from the Hölder inequality. d e) Since 1 < r < d−1 we have, 0 < r1 = 1 − we have,

1 r

< d1 . Thus

d r

< 1. Moreover,

1 d 1 1 2 1 1 1 − + 1 = − d(1 − ) + 1 = − + 1 = − + 1 = . q r q r q q q q Thus b = q . We can apply the Hardy–Littlewood–Sobolev inequality (see the inequality (3.21) in Chap. 2) to get the result. f) This follows immediately from questions a) and d) above. 5. We have q < +∞, r < +∞. It follows that C0∞ (Rt × Rdx ) is dense in



Lq (Rt , Lr (Rdx )) and Lq (Rt , Lr (Rdx )) is the dual of Lq (Rt , Lr (Rdx )). Then, ρ Lq (R+ ,Lr (Rd )) = t

x

sup

∈C0∞ (Rt ×Rdx )

 

 ρ(t, x)(t, x) dx dt  ,  Lq (Rt ,Lr (Rd ))



R+ Rd

≤ C(d) f La (Rdx ×Rdv ) 0

by the previous question.

x

9 Solutions of the Problems

323

Part 2 a ) if and only if w = Ov, v ∈ E a . Then |w| = |v| and 1. We have w ∈ O(Eτ,ξ τ,ξ   −1 |τ + O w, ξ | = |τ + w, Oξ  | ≤ a. Moreover, setting v = O −1 w in the integral we get,  a ) μ(Eτ,ξ

=

 a Eτ,ξ

χ(v) dv =

χ(O −1 w) dw. a ) O(Eτ,ξ

2. For every ξ ∈ Rd \ {0} one can find an orthogonal matrix O such that Oξ = (0, . . . , 0, |ξ |) and we apply question 1. a 3. We have obviously Eτ,ξ ⊂ B(0, M) so that, a μ(Eτ,ξ ) ≤ μ(B(0, M)) ≤ 2μ(B(0, M))a,

since a > 12 . 4. a) If |ξ | ≤ ε0 < 1 we have τ 2 = 1 − |ξ |2 ≥ 1 − ε02 and, |τ + v · ξ | ≥

2

1 − ε02 − |v||ξ | ≥

2 1 − ε02 − Mε0 .

√ The function g(ε) = 1 − ε2 − Mε is such that g(0) = 1, so there exists ε0 ∈ (0, 1) depending only on M such that g(ε0 ) > 12 > a. b) If |ξ | > ε0 we can write,          τ    τ a a     {w : |τ + |ξ |wd | ≤ a} = w :  + wd  ≤ ⊂ w: + wd  ≤ |ξ | |ξ | |ξ | ε0 Using question 2 we have therefore,   τ  + wd  ≤ :  |ξ |     τ

 ≤ μ1 |w | ≤ M,  + wd  ≤ |ξ |

 a ) ≤ μ1 |w| ≤ M μ(Eτ,ξ

a ε0 a ε0

  ≤ C(d, M)

a . ε0

Here we have set w = (w , wd ), w ∈ Rd−1 . 5. This is an immediate application of the result proved in Part 1 of Problem 57 with δ = 12 .

324

9 Solutions of the Problems

Solution 59

Part 1 1. This follows from the development of each determinant with respect to the i t h line. 2. If cij = cof(bij ) and adj(B) = (dij ) we have dij = cj i = cof(bj i ) so that,

dB (t) adj(B(t)) dt

= ij

n  dbik k=1

dt

(t) dkj (t) =

n  dbik k=1

dt

(t) cof(bj k )(t).

Then, Tr

 n  n dB dbik d adj(B(t)) = (t) cof(bik (t)) = (t). dt dt dt i=1 k=1

3. Since that,

dB dt

= A(t)B(t)and B(t) adj(B(t) = (det B(t)) Id = (t)Id we deduce d (t) = Tr (A(t)B(t) adj(B(t))) = (t) Tr(A(t)). dt

Part 2   1. a) Assume that X(t0 , y) = X(t0 , y ). Set E = t ∈ [0, T ] : X(t, y) = X(t, y ) . By hypothesis t0 ∈ E. On the other hand, by continuity E is closed. Let us show that it is open in [0, T ]. Let t ∈ E and consider the differential systems (setting f˙ = df dt ) Y˙ (t) = v(t, Y (t)),

Y (0) = X(t, y),

(9.60)

˙ Z(t) = v(t, Z(t)),

Z(0) = X(t, y ).

(9.61)

Since t ∈ E we have X(t, y) = X(t, y ). The uniqueness of the solution of this system implies that there exists ε > 0 such that Y (t) = Z(t) for |t −t| ≤ ε. But X(t, y) is the unique solution of (9.60) and X(t, y ) the unique solution of (9.61). Therefore, we have X(t, y) = X(t, y ) for |t −t| ≤ ε which shows that B(t, ε) ⊂ E. It follows that E = [0, T ] that is 0 ∈ E, so y = y . b) Using the continuity of the solution with respect to the initial data we see that the map y → X(t, y) is C 1 and that we have, d dt



∂Xj ∂yk

(t, y) =

d  ∂vj =1

∂x

(t, X(t))

∂X (t, y). ∂yk

9 Solutions of the Problems

325



The matrix B(t) =

∂Xj ∂yk

form (8.36) with A(t) = deduce that,



 is therefore solution of an equation of the  and B(0) = Id. Using Part 1 we

∂vj ∂x (t, X(t))

 t

∂Xj Tr (A(s)) ds , (t, y) = det (t, y) = (t0 ) exp ∂yk t0  t

= exp divx v(s, X(s, y)) ds .

0

c) The map y → t (y) : 0 → t = t (0 ) is then bijective and its determinant is nonzero. The inverse function theorem implies that it is a C 1 diffeomorphism. 2. a) Set in the integral x = t (y), so dx = |(t, y)| dy = (t, y) dy since (t, y) > 0. We obtain, 



V (t) =

dx =

(t, y) dy.

t

0

b) Now 0 is bounded and t → (t, y) is C 1 . By the Lebesgue differentiation theorem we can differentiate inside the integral. Using Part 1 we obtain, V (t) =

 0

d (t, y) dy = dt

 (t, y)(divx v)(t, X(t, y)) dy. 0

c) It follows from question b), setting x = t (y), and from the hypothesis that,



V (t) =

(div v)(t, x) dx = 0,

∀t ∈ [0, T ).

t

Therefore, V (t) = V (0) for all t ∈ [0, T ). Part 3 1. We have seen in question 2b) of Part 1 that, V (t0 ) =

 (divx v)(t0 , X(t0 , y))(t0 , y) dy. 0

By the choice of 0 and the fact that (t0 , y) > 0 we see that the function inside the integral is strictly positive. So V (t0 ) > 0. The Taylor formula shows then that for t > t0 , |t − t0 | small enough, we have V (t) > V (t0 ).

326

9 Solutions of the Problems

Solution 60

1. We have just to differentiate the j t h equation of (8.38) with respect to xi then the i t h equation with respect to xj and make the difference. 2. We have, ∂i vk ∂k vj − ∂j vk ∂k vi = ∂i vk (∂k vj − ∂j vk ) + ∂j vk (∂i vk − ∂k vi ) = ∂i vk ωkj + ∂j vk ωki . 3. It is the translation in terms of matrices of the result in question 1. 4. a) We have ∂s C(s) = [∂s A + (v · ∇x )A](s, X(s, ·)) so that this equality follows from the previous question. b) Integrating the above equation between 0 and s and taking the matrix norms of both members we obtain,  s B(σ, X(σ, ·) C(σ ) dσ. C(s) ≤ C(0) + 2 0

5. The Gronwall inequality implies that,  C(s) ≤ C(0) exp 2

s

B(σ, X(σ, ·) dσ .

0

We have C(0) = A(0, y) = (curl v)(0, y) = (curl v0 )(y), so if curl v0 ≡ 0, the above inequality shows that A(s, X(s, y)) = 0 for all (s, y) ∈ [0, T ) × Rd so (curl v)(s, x) = 0 for all (s, x) ∈ [0, T ) × Rd .

Solution 61 1. Set for 1 ≤ j ≤ 3, wj (ξ ) =

ξj  f (ξ ). i|ξ |2



 ξ  |wj (ξ )| dξ ≤ 2s

R3

2

{|ξ |≤1}

For all s ∈ R we have,

 ξ 2s |f(ξ )|2 ξ 2s |f(ξ )|2 dξ, dξ + |ξ |2 {|ξ |>1}

and the right-hand side is finite since

1 |ξ |2

∈ L1loc (R3 ) and f ∈ S(R3 ). Set vj =

F −1 wj . Then vj ∈ H +∞ (R3 ). Let us show that v = (vj ) solves the question.  3 We have, 3j =1 ∂ iξj wj (ξ ) = f(ξ ). So div v = f. j vj (ξ ) = j =1  2 2 −|x| 2. a) Choose f = e . Then R3 (div v)(x) dx = R3 e−|x| dx = 0.

9 Solutions of the Problems

327

b) Consider the function v described in question a). If we had ∂j vj ∈ L1 (R3 ) for    j = 1, 2, 3, we would have R3 (div v)(x) dx = 3j =1 R3 ∂j vj (x) dx = 0 since, because vj ∈ H +∞ (R3 ), we have lim|x|→+∞ vj (x) = 0. 3. Set A = ξ σ +1 |ξ1|  u 2L2 (R3 ) . Write,  A=

|ξ |≤1

ξ 2σ +2 | u(ξ )|2 dξ + |ξ |2

 |ξ |>1

ξ 2σ +2 | u(ξ )|2 dξ = A1 + A2 . |ξ |2

We have,  A1 ≤

C  u 2L∞ (R3 )

|ξ |≤1

dξ ≤ C u 2L1 (R3 ) . |ξ |2

 A2 ≤ C

R3

ξ 2σ | u(ξ )|2 dξ = C u 2H σ (R3 ) .

Solution 62 1. If f (x) = C ∈ C we have f(k) = 0 if k = 0 and f(0) = C(2π)2 so that f H s (T2 ) = |C|(2π)2 . If f (x) = sin nx1 = 2i1 (einx1 − e−inx1 ) we have, 1 f(k) = 2i





2π

e 0

i(n−k1 )x1

2π

dx1 −

e

−i(n+k1 )x1

0



2π

dx1

e−ik2 x2 dx2 .

0

The right-hand side vanishes for k2 = 0. If k2 = 0 it vanishes also if k1 = n and 2 2 k1 = −n. Eventually f(−n, 0) = − 2πi , f(n, 0) = 2πi . It follows that, √ f H s (T2 ) = 2 2π 2 ns . 2. a) First of all, since u1 does not depend on x1 and u2 on x2 , we have A = div Un,ω = 0. Now,

∂t Un,ω = ω n−s sin(nx2 − ωt), ω n−s sin(nx1 − ωt)  

Un,ω · ∇x Un,ω = −(ωn−1 + n−s cos(nx1 − ωt))n−s+1 sin(nx2 − ωt ,   − ωn−1 + n−s cos(nx2 − ωt))n−s+1 sin(nx1 − ωt) .

328

9 Solutions of the Problems

It follows that

B = ∂t Un,ω + Un,ω · ∇x Un,ω  = − n−2s+1 cos(nx1 − ωt) sin(nx2 − ωt),  − n−2s+1 cos(nx2 − ωt) sin(nx1 − ωt) .

(9.62)

Then we have,

C = div Un,ω · ∇x Un,ω = 2n−2s+2 sin(nx1 −ωt) sin(nx2 −ωt).

(9.63)

b) We have,  (sin(nx1 − ωt) sin(nx2 − ωt)) = −2n2 sin(nx1 − ωt) sin(nx2 − ωt)   so  2n1 2 wn = −wn . c) This follows from b).  1 ik,x , we have 2 ik,x =  d) Since R(x) = (2π) 2 k∈Z2 R(k)e k∈Z2 −|k| R(k)e  = 0 if k = 0 so R(x) = 1 2 R(0).  0. This implies that R(k) (2π)

e) We have,

 ∇x −1 div Un,ω · ∇x Un,ω = − n−2s+1 cos(nx1 − ωt) sin(nx2 − ωt),  − n−2s+1 sin(nx1 − ωt) cos(nx2 − ωt) . (9.64) f) This follows from (9.62) and (9.64). 3. 0 Un0 = Un,1 ,

0 Vn0 = Un,−1 .

Then Un0 − Vn0 = ( n2 , n2 ) so that Un0 − Vn0 (H s (T2 )

2

)

=

lim Un0 − Vn0 H s (T2 ) 2 = 0. ) (

n→+∞

√ 8 2π 2 n .

Therefore, (9.65)

4. Denote by Un and Vn the solutions of the Euler system corresponding to these data. By (8.43) we have,   Un (t, x) = n−1 + n−s cos(nx2 − t), n−1 + n−s cos(nx1 − t) ,   Vn (t, x) = −n−1 + n−s cos(nx2 + t), −n−1 + n−s cos(nx1 + t) .

9 Solutions of the Problems

329

Then, Un (t, x) − Vn (t, x) =

2 + n−s ( cos(nx2 − t) − cos(nx2 + t)), n

2 −s + n (cos(nx1 − t) − cos(nx1 + t)) . n

q−p Using the formula cos p − cos q = 2 sin( q+p 2 ) sin( 2 ) we can write,

Un (t, x) − Vn (t, x) =

2 + 2n−s sin(nx2 ) sint(t), n

2 + 2n−s sin(nx1 ) sin(t) . n

If f (x) = sin(nx2 ) or f (x) = sin(nx1 ) we have, from question 1 f H s (T2 ) = C ns . Therefore, the (H s (T2 ))2 norm of each component of Un −Vn is bounded below by C1 | sin(t)| − Cn2 . It follows that, Un (t, ·) − Vn (t, ·) (H s (T2 ))2 ≥ C3 | sin(t)| −

C4 . n

Let t > 0 be arbitrary small. This construction contradicts the uniform continuity of the map U0 → U (t) from H s (T2 ) to H s (T2 ). Indeed denying the uniform continuity reads, ∃ ε0 > 0 : ∀δ > 0 ∃ U0 , V0 ∈ (H s (T2 ))2 : U0 − V0 (H s (T2 ))2 ≤ δ and U (t) − V (t) (H s (T2 ))2 ≥ ε0 . We have just to take ε0 =

C3 2 |sint|, δ

≤

√ 8 2π 2 n ,

n large enough.

Solution 63

Part 1  1. a) It follows from (8.45) that, |∂t ρ(t, x0 )| ≤ M 3j =1 |vj (t, x0 )|+3M|ρ(t, x0)|. We deduce from (8.46) if ρ(t, x0 ) = 0, (dividing by ρ(t, x0 )) and from (C2 ) that, |∂t vk (t, x0 )| ≤ M

3 

|vj (t, x0 )| + 3M|ρ(t, x0)|,

j =1

|∂t vk (t, x0 )| ≤ M

3  j =1

|vj (t, x0 )|

if ρ(t, x0 ) = 0.

if ρ(t, x0 ) = 0,

330

9 Solutions of the Problems

b) We have, ⎛ |∂t u(t, x0 )| ≤ 2 ⎝|∂t ρ(t, x0 )||ρ(t, x0 )| +

3  j =1

⎛ ≤ 2 ⎝|∂t ρ(t, x0 )| +

3 

⎞ |∂t vj (t, x0 )||vj (t, x0 )|⎠ . ⎞⎛

|∂t vj (t, x0 )|⎠ ⎝|ρ(t, x0 )| +

j =1

3 

⎞ |vj (t, x0 )|⎠ ,

j =1

We have then just to use question a) and the Cauchy Schwarz inequality to conclude. t 2. For t ∈ [0, T0 ] we deduce from question 1b) that, u(t) ≤ u(0) + C 0 u(s) ds. Since |x0| > R and The Gronwall inequality implies that u(t) ≤ u(0)eCT0 . supp ρ0 ∪ supp v0 ⊂ B(0, R), we have u(0) = ρ0 (x0 )2 + 3j =1 |vj,0 (x0 )|2 = 0. It follows that u(t) = 0 so ρ(t, x0 ) = vj (t, x0 ) = 0 for all t ∈ [0, T0 ] and 1 ≤ j ≤ 3. In other words ρ(t, ·) and v(t, ·) vanish identically when |x| > R and t ∈ [0, T ); this proves our claim. Part 2 1. The first identity follows immediately from (8.45). To obtain the second one we have just to multiply (8.45) by vk and to add it to (8.46). Denote by A the lefthand side of the third identity. We have, A = |v|2 ∂t ρ + 2ρ

3 

vk ∂t vk + 3ρ 2 ∂t ρ + ρ|v|2 (div v) + 3ρ 3 (div v)

k=1

+ |v|2 (v · ∇)ρ + 2ρ

3  3 

vj vk ∂j vk + 9ρ 2 (v · ∇)ρ =

j =1 k=1

8 

A .

=1

From (8.45) we get A1  + A4 = −|v|2 (v · ∇)ρ. Now multiplying (8.46) by vk and summing we obtain, ρ k=1 vk ∂t vk + ρ 3k=1 vk (v · ∇)vk + 3ρ 2 (v · ∇)ρ = 0. It follows that A2 + A7 = −6ρ 2 (v · ∇)ρ. Using (8.45) we obtain A3 = −3ρ 2 (v · ∇)ρ − 3ρ 3 (div v). Now we have A5 = 3ρ 3 (div v), A6 = |v|2 (v · ∇)ρ, A8 = 9ρ 2 (v · ∇)ρ. Therefore, 8 

A = −|v|2 (v · ∇)ρ−6ρ 2 (v · ∇)ρ − 3ρ 2 (v · ∇)ρ − 3ρ 3 (div v)

=1

+ 3ρ 3 (div v) + |v|2 (v · ∇)ρ + 9ρ 2 (v · ∇)ρ = 0.

9 Solutions of the Problems

331

2. By hypothesis there exists R > 0 such that supp(ρ0 , v0 ) ⊂ B(0, R). From Part 1 we have supp(ρ(t, ·), v(t, ·)) ⊂ B(0, R). So, 

  ρ(t, x)|v(t, x)|2 + ρ(t, x)3 dx.

E(t) = B(0,R)

The functions inside the integral being C 1 in t we can, by the Lebesgue differentiation theorem, differentiate inside the integral. Using the previous question we obtain, 

/ ∂ . ρ(t, x)|v(t, x)|2 + ρ(t, x)3 dx ∂t 3  /  ∂ . ρ(t, x)|v(t, x)|2 + 3ρ(t, x)3 dx = 0, =− ∂xj

∂E (t) = ∂t

0 ≤ t < T,

j =1

since the functions inside the bracket have compact support in x.

Therefore, we have E = ρ0 (x)|v0 (x)|2 + ρ0 (x)3 dx > 0 since ρ0 is ≥ 0 and ρ0 ≡ 0. 3. As above we can differentiate inside the integral. Using the first relation in (8.48) we obtain, ∂ ∂t



 ρ(t, x) dx =

 ∂ρ (t, x) dx = − ∂t 3

j =1



∂ (ρ(t, x)vj (t, x)) dx = 0, ∂xj

since the functions ρ and v have compact support x. 4. We have,   H (t) = |x|2ρ(t, x) dx ≤ R 2 ρ(t, x) dx B(0,R)



= R2

ρ0 (x) dx = M < +∞.

5. The fact that these integrals are C 1 has been considered in question 2. So we have, using the first relation in (8.48), H (t) =



 ∂ρ (t, x) dx = − ∂t 3

|x|2

j =1

 |x|2

∂ (ρ(t, x)vj (t, x)) dx. ∂xj

332

9 Solutions of the Problems

Integrating by parts and using the fact that ρ, v have compact support in x we get, H (t) = 2

3  

ρ(t, x)xj vj (t, x) dx.

j =1

6. a) By the same argument as in 2. we see that the right-hand side of H (t) is a C 1 function in t and that we have,

H (t) =

3  

xk ∂t (ρvk ) (t, x) dx.

k=1

We use the second equality in (8.48). We obtain,

H (t) = −

3  3   j =1 k=1

 ∂ (ρvj vk )(t, x) dx − xk ∂xj 3

k=1

 xk

/ ∂ . ρ(t, x)3 dx. ∂xk

Integrating by parts we see that the first integral vanishes if j = k so,

H (t) =



3  

ρ(t, x)vk (t, x) dx + 2

ρ(t, x)3 dx.

k=1

b) This follows from the fact that ρ ≥ 0. 7. By the Taylor formula with integral reminder and the previous question we have, 

H (t) = H (0) + tH (0) + t

1

2 0

1 (1 − λ)H

(λt) dλ ≥ H (0) + tH (0) + Et 2 . 2

8. From question 4 we have H (t) ≤ M. So, H (0) + tH (0) + 12 Et 2 ≤ M for all t < T . Since from question 2 we have E > 0 this inequality shows that T must be bounded.

Solution 64

1. Recall that, " W

1+α,∞

(R , R ) = u ∈ W 3

3

1,∞

# |∇x u(x) − ∇x u(y)| (R , R ) : sup < +∞ . |x − y|α x=y 3

3

9 Solutions of the Problems

333

First of all we have, u(t, ·) L∞ ≤ C( f L∞ + h L∞ ). Then ∂x1 u(t, ·) L∞ ≤ h L∞ and ∂x2 u(t, ·) L∞ ≤ C( f L∞ + T h L∞ f L∞ ). Now, ∂x1 u(t, x) − ∂x1 u(t, y) = (0, 0, h (x1 − tf (x2 )) − h (y1 − tf (y2 )). Since f, h ∈ W 1+α,∞ (R) we have h ∈ W α,∞ , f ∈ L∞ . Therefore, we have, |∂x1 u(t, x) − ∂x1 u(t, y)| ≤ h W α,∞ |x1 − y1 − t (f (x2 ) − f (y2 )|α ,

α ≤ h W α,∞ |x1 − y1 | + T f L∞ |x2 − y2 | ≤ C(f, h, T )|x − y|α . Eventually we have, ∂x2 u(t, x) − ∂x2 u(t, y) = (f (x2 ) − f (y2 ), 0, −tf (x2 )h (x1 − tf (x2 )) + tf (y2 )h (y1 − tf (y2 )). Set (1) := f (y2 )h (y1 − tf (y2 )) − f (x2 )h (x1 − tf (x2 )). Then (1) = A + B where, A = (f (y2 ) − f (x2 ))h (y1 − f (y2 )), B = f (x2 )(h (y1 − tf (y2 )) − h (x1 − tf (x2 )). We obtain, |(1)| ≤ h L∞ f W α,∞ |x2 − y2 |α + f L∞ h W α,∞ |y1 − x1 + t (f (x2 ) − f (y2 ))|α ≤ C(f, h, T )|x − y|α . It follows that, |∂x2 u(t, x) − ∂x2 u(t, y)| ≤ C(f, h, T )|x − y|α . This shows that u ∈ W 1+α,∞ . The computation is exactly the same for v. 2. We have, u0 = (f (x2 ), 0, h(x1 )). Then div u0 = ∂x1 u01 + ∂x2 u02 + ∂x3 u03 = 0. On the other hand, since u1 (t, x) = f (x2 ), u2 (t, x) = 0, u3 (t, x) = h(x1 − tf (x2 )) we have, u · ∇x = f (x2 )∂x1 + h(x1 − tf (x2 ))∂x3 . It follows that, ∂t u1 + (u · ∇x )u1 = 0,

∂t u2 + (u · ∇x )u2 = 0,

∂t u3 + (u · ∇x )u3 = −f (x2 )h (x1 − tf (x2 )) + f (x2 )h (x1 − tf (x2 )) = 0. Therefore, u satisfies the system (8.49) with P = 1.

334

9 Solutions of the Problems

3. We have u0 − v0 = (f (x2 ) − g(x2 ), 0, 0). Let g ∈ W 1+α,∞ (R). Let δ > 0. Set f (x) = g(x) + δ. Then f (x) = g(x) for all x ∈ R and, u0 − v0 W 1+α,∞ = f − g W 1+α,∞ = δ 1 W 1+α,∞ = δ. 4. Keeping the notation in the statement we have, A(t) ≥ ∂x1 (u(t, ·) − v(t, ·)) W α,∞ = h (· − tf (·)) − h (· − tg(·)) W α,∞ .   5. Since (x, y) : x1 = y1 , x, y ∈ [−a, a]2 ⊂ {(x, y) : x = y} we have, U W α,∞ ≥

sup

x1 =y1 x,y∈[−a,a]2

|U (x) − U (y)| . |x1 − x2 |α

Now if |x1 | ≤ a we have |x1 − tf (x2 )| ≤ a + T f L∞ ≤ 2a. Therefore, by our choice of h we have h (x1 − tf (x2 )) = |x1 − tf (x2 )|α . This gives the desired lower bound for A(t). 6. When x2 = y2 = c we obtain from the previous question that A(t) is greater than, sup

x1 =y1 x1 ,y1 ∈[−a,a]

||x1 − tf (c)|α − |x1 − tg(c)|α − |y1 − tf (c)|α + |y1 − tg(c)|α | . |x1 − y1 |α

Then for t ∈ (0, T ] one can take x1 = tg(c), y1 = tf (c) since for t > 0, x1 = tg(c) = tf (c) = y1 , and |x1 | ≤ T g L∞ ≤ a, |y1 | ≤ T f L∞ ≤ a. We obtain, A(t) ≥ 2

(t|g(c) − f (c)|)α = 2. (t|g(c) − f (c)|)α

7. Fix v0 = (g(x2 ), 0, 0). We have proved that there exists c0 (= 2) such that, for all δ > 0 there exists u0 ∈ W 1+α,∞ such that, u0 − v0 W 1+α,∞ ≤ δ and sup u(t, ·) − v(t, ·) W 1+α,∞ ≥ c0 . t ∈[0,T ]

This contradicts the continuity of the map u0 → u at the point v0 from W 1+α,∞ to L∞ ([0, T ], W 1+α,∞ ).

9 Solutions of the Problems

335

Solution 65

Part 1 1. By using the equation, we have d [u(s, x(s))] = (∂s u + x(s)∂ ˙ x u)(s, x(s)) = (∂s u + u ∂x u)(s, x(s)) = 0. ds Therefore, u(s, x(s)) = u(0, x(0)) = u0 (y). Coming back to the equation of the characteristic we find that x(s) ˙ = u0 (y) so x(s) = y + su0 (y) for all s ∈ [0, +∞). 2. Since u(s, x(s)) = u0 (y) we deduce from question 1 that, u(s, y + u0 (y)) = u0 (y). Part 2 ∂F

1. We have, ∂F ∂y (t, y) = 1+tu0 (y).If infR u0 ≥ 0 we have ∂y (t, y) ≥ 1 for all t ≥ 0 and if infR u 0 = −α, with 0 < α < +∞, we have 1 + tu 0 (y) ≥ 1 − tα > 0 if t < α1 . 2. a) For t ∈ [0, T ∗ ) the map y → F (t, y) from R → R is then strictly increasing. It is therefore invertible and there exists a map x → k(t, x) such that F (t, y) = x ⇐⇒ y = κ(t, x). The fact that the map κ is C 1 follows from the inverse function theorem. b) Obvious. 3. According to question 2 Part 1 set u(t, x) = u0 (κ(t, x)). Then u is a C 1 function on [0, T ∗ ) × R. Differentiating the equality κ(t, y + tu0 (y)) = y with respect to t and setting y + tu0 (y) = x we find, ∂t κ(t, x) + u0 (κ(t, x))∂x κ(t, x) = 0.

(9.66)

It follows from the form of u and (9.66) that we have, ∂t u(t, x) = u 0 (κ(t, x))∂t κ(t, x) = −u 0 (κ(t, x))u0(κ(t, x))∂x κ(t, x), = −u(t, x)u 0(κ(t, x))∂x κ(t, x), ∂x u(t, x) = u 0 (κ(t, x))∂x κ(t, x). Therefore, ∂t u + u ∂x u = 0 in [0, T ∗ ) × R. Moreover, since κ(0, x) = x we have u(0, x) = u0 (x). This proves the existence of a solution. The uniqueness follows from the necessary condition proved in question 2 Part 1. 4. Let suppu0 ⊂ [a, b]. We show by contradiction that there exists x0 ∈ R such that u 0 (x0 ) < 0. Indeed assume that u 0 (x) ≥ 0 for every x ∈ R. Then u0 is increasing so u0 (x) ≥ u0 (a) = 0 for x ≥ a and since u0 ≡ 0 there exists x1 > a

336

9 Solutions of the Problems

such that u0 (x1 ) > 0. Then for x ≥ x1 we have, 

1

u0 (x) − u0 (x1 ) = 0

u 0 (tx + (1 − t)x1 )(x − x1 ) dt ≥ 0.

Therefore, u0 (x) ≥ u0 (x1 ) > 0 which contradicts the fact that u0 = 0 for x ≥ b. 5. We have by definition κ(t, y + tu0 (y)) = y. Differentiating this equality with respect to y and setting y + tu0 (y) = x that is y = κ(t, x) we obtain the expression of ∂x κ. Differentiating the above equality with respect to t we obtain the expression of ∂t κ. Part 3 1. a) Since q(t) = (∂x u)(t, x0 + tu0 (x0 )) we have, q (t) = ∂t (∂x u)(t, x0 + tu0 (x0 )) + u0 (x0 )(∂x2 u)(t, x0 + tu0 (x0 )).

(9.67)

Differentiating the equation (8.50) with respect to t and setting x = x0 + tu0 (x0 ) we obtain, ∂t (∂x u)(t, x0 + tu0 (x0 )) + u(t, x0 + tu0 (x0 ))(∂x2 u)(t, x0 + tu0 (x0 )) + (∂x u)2 (t, x0 + tu0 (x0 )) = 0. Since u(t, x0 + tu0 (x0 )) = u0 (x0 ) we deduce from the above equation that, ∂t (∂x u)(t, y) + u0 (x0 )(∂x2 u)(t, y) = −(∂x u)2 (t, y) where y = x0 + tu0 (x0 ). Using (9.67) we obtain q (t) = −q(t)2 . Moreover, q(0) = (∂x u)(0, x0 ) =

u0 (x0 ) < 0 by the choice of x0 . It follows that q is decreasing so, since q(0) < 0 we have q(t) < 0 for 0 ≤ t < Tx∗0 . Therefore, the equation on q is equivalent to, d q (t) =− dt q(t)2



1 q(t)

Integrating between 0 and t > 0 we obtain

= −1. 1 q(t )

= t+

1 . u 0 (x0 )

Since q < 0

1 . Moreover, in this this equality shows that q exists only for 0 ≤ t < − u (x 0 0) interval we have,

q(t) =

1 t+

1 u 0 (x0 )

=

1 . t − Tx∗0

9 Solutions of the Problems

337

b) This equality shows in particular that limt →Tx∗ = −∞. Now if x0 changes we 0

see that our solution can exist as a C 1 function for t smaller than the smallest Tx∗0 that is for, t < inf Tx∗0 = inf x0 ∈R

x0 ∈R

−

1 u 0 (x0 )



= − sup

x0 ∈R

1 u 0 (x0 )

=

1 = T ∗. − infx0 ∈R u 0 (x0 )

Part 4 1. This follows from the fact that if u0 ∈ H 2 then u 0 ∈ H 1 ⊂ C 0 ∩ L∞ and that we have proved in question 2b) Part 2 that if T u 0 L∞ < 1 then T < T ∗ . 2. Setting κ(t, x) = y ⇔ x = y + tu0 (y) in the integral we obtain,  u(t, ·) 2L2 =

 |u0 (κ(t, x))|2 dx =

|u0 (y)|2 |1 + tu 0 (y)| dy,

≤ (1 + T u 0 L∞ (R) ) u0 2L2 . Now we have, ∂x u(t, x) = u 0 (κ(t, x))∂x κ(t, x). Using question 6 Part 2 we have, |∂x κ(t, x)| ≤ 1−T u1 ∞ . Therefore, the same computation as above gives, 0 L

∂x u(t, ·) 2L2 ≤

1 + T u 0 L∞ u 2 2 . (1 − T u 0 L∞ )2 0 L

3. We have ∂x2 u(t, x) = (1) + (2) where, (1) = ((∂x κ)(t, x))2 u

0 (κ(t, x),

(2) = (∂x2 κ)(t, x)u 0 (κ(t, x)).

The term (1) can be estimated by the same method as that used in the previous question. We find, 1

(1) L2

(1 + T u 0 L∞ ) 2

≤ u 2 . (1 − T u 0 L∞ )2 0 L

Let us estimate (2). Differentiating the equality in question 6 Part 2 with respect to x we find, (∂x2 κ)(t, x) = −

tu

0 (κ(t, x))∂x κ(t, x) tu

0 (κ(t, x)) = − . (1 + tu 0 (κ(t, x)))2 (1 + tu 0 (κ(t, x)))3

It follows that, |(2)| ≤

T u 0 L∞ (R) |u

(κ(t, x))|. (1 − T u 0 L∞ (R) )3 0

(9.68)

338

9 Solutions of the Problems

We may therefore apply the same method as above to prove that 1

(2) L2 ≤

T u 0 L∞ (1 + T u 0 L∞ ) 2

u0 L2 . (1 − T u 0 L∞ )3

4. a) Let |x| ≥ R. Set yn = κ(tn , x) ⇔ x = yn + tn u0 (yn ). and y0 = κ(t0 , x) ⇔ x = y0 + t0 u0 (y0 ). We have, |κ(tn , x)| = |yn | ≥ |x| − (t0 + 1) u0 L∞ (R) ≥ R0 , |κ(t0 , x)| = |y0 | ≥ |x| − t0 u0 L∞ (R) ≥ R0 , so ϕ(κ(tn , x)) = ϕ(κ(t0 , x)) = 0. b) For n ≥ n0 , using question 6 Part 2 we can write, In ≤ |BR | ϕ L∞ |κ(tn , x) − κ(t0 , x)| ≤ |BR | ϕ L∞ sup | t∈[0,T ]

≤ |BR | ϕ L∞

∂κ (t, x)||tn − t0 | ∂t

u0 L∞ (R) |tn − t0 |. 1 − T u 0 L∞

Therefore, limn→+∞ In = 0. 5. Let v ∈ L2 . There exists (ϕk ) ⊂ C0∞ such that ϕk → v in L2 . We write, Jn ≤ C v(κ(tn , ·))−ϕk (κ(tn , ·)) 2L2 + C ϕk (κ(tn , ·)) − ϕk (κ(t0 , ·)) 2L2 +C ϕk (κ(t0 , ·)) − v(κ(t0 , ·)) 2L2 = (1) + (2) + (3). Setting successively y = κ(tn , x) then y = κ(t0 , x) we see that,  (1) + (3) ≤ 2C R

|v(y) − ϕk (y)|2 dy (1 + (t0 + 1)) u 0 L∞ ).

∞ Let ε > 0. We fix k so large that (1) + (3) ≤ 2ε 3 . Since ϕk ∈ C0 (R) we may use the previous question to infer that for n large enough we have (2) ≤ ε3 . Therefore, Jn ≤ ε. 6. We write,

w(tn , x) − w(t0 , x) = f (tn , x) (v(κ(tn , x)) − v(κ(t0 , x))) + (f (tn , x) − f (t0 , x)) v(κ(t0 , x)). The first term tends to zero in L2 using question 5 above and the fact that f is uniformly bounded. The second term tends to zero in L2 by

9 Solutions of the Problems

339

the dominated convergence theorem. Indeed, for fixed x the quantity | (f (tn , x) − f (t0 , x)) v(κ(t0 , x))|2 tends to zero and it is bounded by, 4 f 2L∞ ([0,T )×R) |v(κ(t0 , x))|2 , which is integrable. 7. Recall that u(t, x) = u0 (κ(t, x)) where u0 ∈ H 2 . The fact that u ∈ C 0 ([0, T ), L2 ) follows immediately from question 5 taking v = u. Now we have, ∂x u(t, x) = ∂x κ(t, x)u 0 (κ(t, x)).

(9.69)

Since ∂x κ(t, x) is continuous, bounded, and u 0 ∈ L2 it follows from question 6 above that ∂x u ∈ C 0 ([0, T ), L2 ). We are left with the second derivative. We have, ∂x2 u(t, x) = (∂x κ(t, x))2u

0 (κ(t, x)) + ∂x2 κ(t, x)u 0 (κ(t, x)).

(9.70)

Therefore, by (9.68) we have, ∂x2 u(t, x) = (∂x κ(t, x))2 u

0 (κ(t, x)) − t (∂x κ(t, x))3 u

0 (κ(t, x))u 0(κ(t, x)). We use question 6 above. The first term in the right-hand side belongs to C 0 ([0, T ), L2 ) since (∂x κ(t, x))2 is continuous, bounded, and u

0 ∈ L2 . The same is true for the second term since t (∂x κ(t, x))3 is continuous, bounded, and u 0 u

0 ∈ L2 since u 0 ∈ H 1 ⊂ L∞ and u

0 ∈ L2 . Part 5 1. The uniform continuity of this map reads: for all ε > 0 there exists δ > 0 such that,   u0 , v0 ∈ H 2 and u0 − v0 H 2 ≤ δ ⇒ sup u(t, ·) − v(t, ·) H 2 ≤ ε. t ∈[0,T ]

Therefore, the nonuniform continuity will be proved if: there exists ε0 > 0 such that, ∀n ∈ N ∃u0n , vn0 ∈ H 2 : u0n −vn0 H 2 ≤

1 but sup un (t, ·)−vn (t, ·) H 2 > ε0 . n t ∈[0,T ]

This proves our claim. −1

2. a) This follows from the equalities, (u0n ) (x) = λn 2 χ (λn x) and (vn0 ) (x) = (u0n ) (x) = εn χ (x). b) This follows from the discussion made in Part 2.

340

9 Solutions of the Problems

3. We have 1 − T (u0n ) L∞ ≤ |1 + t (u0n ) (κ(t, x))| ≤ 1 + T (u0n ) L∞ . Since by hypothesis T (u0n ) L∞ ≤ 12 we deduce that, 12 ≤ |1 + t (u0 ) (κ(t, x))| ≤ 32 . The same holds for vn0 . The first estimate follows from question 6 Part 2. Now again by question 6 Part 2 we have, t (u0 )

(κ(t, x))∂x κ1 (t, x) ∂ 2 κ1 (t, x) = − n . 2 ∂x (1 + t (u0n ) (κ1 (t, x)))2 1

Now by (8.52) we have (u0n )

(x) = λ 2 χ

(λn x). By the above estimates we deduce that,  2  1  ∂ κ1    ≤ Ctλn2 |χ

(λn κ1 (t, x))|. (t, x)  ∂x 2  Since vn0 (x) = u0n (x) + εn χ(x) we have in the same way,  2  1  ∂ κ2 

2    ∂x 2 (t, x) ≤ Ct (λn |χ (λn κ2 (t, x))| + εn ). 4. Setting κ1 (t, x) = y ⇔ x = y + tu0n (y) in the integral we obtain, (u0n )

(κ1 (t, ·)) 2L2 = λn



|χ(λn y)|2 |1 + t (u0n ) (y)| dy.

By question 2a) we have fixed T such that T (u0n ) L∞ ≤ 12 . It follows that, (u0n )

(κ1 (t, ·)) 2L2 ≥

1 λn 2

 |χ(λn y)|2 dy =

1 2

 |χ(y)|2 dy.

j

5. Set Ij (t) = (∂x A)(t, ·) L2 , j = 0, 1, 2. Setting κ2 (t, x) = y in the integral and using the fact that T (vn0 ) L∞ ≤ 12 we obtain,  I0 (t) ≤ εn

|χ(y)|2|1 + t (vn0 ) (y)| dy

1 2

≤ Cεn .

   2  The estimate of I1 (t) is analogue since by (8.54) we have  ∂κ (t, x)  ≤ 2. ∂x Eventually,   I2 (t) ≤ εn (∂x κ2 (t, ·))2 χ (κ2 (t, ·)) L2 + ∂x2 κ2 (t, ·)χ

(κ2 (t, ·)) L2 . The first term in the right-hand side is estimated as I1 . For the second term, we 1

use question 3. Since, |∂x2 κ2 (t, x)| ≤ Ct (λn2 |χ

(λn κ2 (t, x))| + 1), its square can

9 Solutions of the Problems

341

be bounded by, Cεn2 (1 + T (vn0 ) L∞ (R) )T 2



(λn |χ

(λn y)|2 + 1)|χ

(y)|2 dy ≤ C(T )εn2 .

Therefore, we have proved that An (t, ·) H 2 ≤ C(T )εn which gives the conclusion. 6. Setting as before κj (t, x) = y in the integral we obtain,   0 2 −3 −3 un (κj (t, ·)) L2 = λ |χ(λn κj (t, x)) dx ≤ Cλn |χ(λn y)|2 dy → 0 when n → +∞. Using the same method and question 3 we can write,  ∂x [u0n (κj (t, ·))] 2L2 ≤ Cλ−1 |χ (λn y)|2 dy → 0, n → +∞. n 7. a) Using question 3 and setting as usual y = κj (t, x) in the integral we can write,  2    ∂ 2 κj  0  −1 |χ (λn y)|2 t 2 (λn |χ

(λn y)|2 +1) dy. (un ) (κj (t, ·)) 2 (t, ·) ≤ Cλn   2 ∂x L

Setting z = λn y we see that the right-hand side tends to zero when n → +∞. b) This follows from (8.56), questions 5, 6, 7a) and the fact that, ∂x2 vn (t, x) = (u0n )

(κ2 (t, x))



∂κ2 (t, x) ∂x

2

+ (u0n ) (κ2 (t, x))



∂ 2 κ2 (t, x) ∂x 2

2 .

c) First of all we have, κ2 (t, y + tvn0 (y)) = y. Moreover, y + tu0n (y) = x ⇔ y = κ1 (t, x). Thus if |λn κ1 (t, x)| ≤ 1 we have λn |y| ≤ 1 so χ(y) = 1. Since ∗ 2 u0n (y)−vn0 (y) = −εn χ(y) we can write κ2 (t, y +tu0n (y)) = y −tεn ∂κ ∂x (t, x ). 0 Setting y + tun (y) = x we deduce from question 3 that,    ∂κ2  ∗   |λn κ2 (t, x)| ≥ λn εn t  (t, x ) − λn |y| ≥ Cλn εn t − 1 ∂x Therefore, if n is large enough we obtain, |λn κ2 (t, x)| ≥ 2 since λn εn → +∞ and t > 0. d) This follows from the equality (α − β)2 = α 2 + β 2 − 2αβ and the fact that αβ = 0 since they have disjoint supports. We may apply the previous question to α = f, β = g. e) This follows from the previous question and questions 3 and 4. where we have proved that |∂x κ1 (t, x ∗ )| is bounded from below and (u0n )

(κ1 (t, ·)) L2 ≥ c > 0.

Chapter 10

Classical Results

In this chapter we recall some results which are used in the problems.

10.1 Some Classical Formulas 10.1.1 The Leibniz Formula • Let N be an integer N ≥ 1,  be an open subset of Rd , and u, v be two C N functions on . Then for α ∈ Nd , |α| ≤ N we have,  α ∂ (uv) = ∂ β u ∂ α−β v. β α

β≤α

Here β ≤ α means βj ≤ αj , 1 ≤ j ≤ d;

α

β

=

α! β!(α−β)! ,

α! = α1 ! . . . αd !.

10.1.2 The Taylor Formula with Integral Reminder • Let  be an open subset of Rd , N be an integer, N ≥ 1, and u ∈ C N (). Let x, y be two points in  and assume that the segment that joins them is contained in . Then we have,  ∂ α u(y)(x − y)α u(x) = |α|≤N−1

 (x − y)α  1 +N (1 − t)N−1 ∂ α u(tx + (1 − t)y) dt. α! 0 |α|=N

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 T. Alazard, C. Zuily, Tools and Problems in Partial Differential Equations, Universitext, https://doi.org/10.1007/978-3-030-50284-3_10

343

344

10 Classical Results

10.1.3 The Faa–di-Bruno Formula • Let  be an open subset of Rdx , k, N ∈ N, N ≥ 1, k ≥ 1, U = (u1 , . . . , uk ) ∈

k N C (, R) and F ∈ C N (Rky ). Then for every α ∈ Nd , 1 ≤ |α| ≤ N, we have, 

∂xα (F (U (x))) =

cα,β (∂yβ F )(U (x))

s  kj 9  ∂x j u(x) , j =1

1≤|β|≤|α|

where cα,β are constants and in the sum we have, s ≥ 1,

|kj | ≥ 1,

|j | ≥ 1,

s 

kj = β,

j =1

s 

|kj |j = α.

j =1

• In dimension d = 1 we can give the expression of the real coefficients which occur in the above sum. Indeed we have the following formula. d Set D = dx . Let n ∈ N∗ and let F, g be two C n functions on R. Then we have, D [F (g(x))] = n



n

n! Dg(x) k1 D g(x) kn k (D F )(g(x)) ··· , k1 ! · · · kn ! 1! n!

where, in the sum, we have k1 + 2k2 + · · · + nkn = n and k = k1 + · · · + kn .

10.2 Elements of Integration In this section we recall several important results in integration theory.

10.2.1 Convergence Theorems Let (X, T , μ) be a measured space. In what follows a.e. will stand for almost everywhere with respect to μ. • The Beppo–Levi theorem. Let (fn )N∈N be an increasing sequence of real integrable functions. Then 

 lim

n→+∞ X

fn dμ =

( lim fn ) dμ ≤ +∞.

X n→+∞

Notice that we have the same result for a decreasing sequence.

10.2 Elements of Integration

345

• The Fatou lemma. Let (fn )n∈N be a sequence of positive measurable functions on X. Then 

 lim inf fn dμ ≤ lim inf

X n→+∞

n→+∞ X

fn dμ ≤ +∞.

• The Lebesgue dominated convergence theorem. Let (fn )n∈N be a sequence of integrable functions such that, (i) the sequence (fn (x))n∈N converges a.e. in C, (ii) there exists g positive integrable such that |fn (x)| ≤ g(x), a.e.∀n ∈ N. Then there exists f integrable such that limn→+∞ fn (x) = f (x) a.e. and, 

 lim

n→+∞ X

fn dμ =

f (x) dμ. X

10.2.2 Change of Variables in Rd Let  ⊂ Rdx and O ⊂ Rdξ be two open subsets such that there exists a C 1 diffeomorphism ϕ : O → . Recall that for ξ ∈ O the Jacobian Jϕ (ξ ) of ϕ is the determinant of the differential of ϕ at ξ. • A function f :  → C is integrable on  if and only if the function (f ◦ ϕ)|Jϕ | : O → C is integrable on O and we have, 

 f (x) dx = 

O

f (ϕ(ξ ))|Jϕ (ξ )| dξ.

10.2.3 Polar Coordinates in Rd • They consist to use the coordinates (r, ω) where r ∈ (0, +∞) and ω ∈ Sd−1 where Sd−1 is the unit sphere in Rd . If we call dω the Lebesgue measure on Sd−1 then we have, 

 Rd

+∞ 

f (x) dx = 0

f (rω)r d−1 dω dr. Sd−1

346

10 Classical Results

In particular if f is a radial function, this means that f (x) = F (|x|), we have, 

 Rd

f (x) dx = |Sd−1 |

+∞

F (r)r d−1 dr,

0

where |Sd−1 | is the Lebesgue measure of the unit sphere. The diffeomorphism related to these coordinates can be written explicitly. Set, x1 = r cos θ1 , x2 = r sin θ1 cos θ2 , . . . , xd−1 = r sin θ1 sin θ2 · · · sin θd−2 cos θd−1 xd = r sin θ1 sin θ2 · · · sin θd−2 sin θd−1. Then we obtain a diffeomorphism from the open set 

 r ∈ (0, +∞), θj ∈ (0, π), 1 ≤ j ≤ d − 2, θd−1 ∈ (0, 2π) ,

to Rd \ A where A has Lebesgue measure zero. Then the Jacobian of this diffeomorphism is, r d−1 (sin θ1 )d−2 (sin θ2 )d−3 · · · (sin θd−2 ). • Notice that in the polar coordinates (r, ω) in Rd the Laplacian takes the form, =

1 d −1 ∂ ∂2 + 2 ω , + 2 ∂r r ∂r r

where ω is a second order operator in ω ∈ Sd−1 called the “Laplace–Beltrami” operator.

10.2.4 The Gauss–Green Formula This is a formula of integration by parts in more than one dimension. It is sometimes called the divergence theorem. Let  be an open subset of Rd and ∂ be its boundary. Assume that one has,    = x ∈ Rd : ρ(x) < 0 ,

  ∂ = x ∈ Rd : ρ(x) = 0 ,

(10.1)

where ρ is a real C 1 function on Rd such that grad ρ(x) =0 for all x ∈ ∂.  For instance the unit ball B in Rd can be written as B = x ∈ Rd : |x|2 − 1 < 0 and ρ(x) = |x|2 − 1 satisfies the above condition. Its boundary is Sd−1 . Of course the exterior of the ball B can also be written as (10.1) by taking ρ(x) = 1 − |x|2.

10.2 Elements of Integration

347

grad ρ(x) Let x ∈ ∂. Then the vector n(x) = grad ρ(x) is called the unit exterior normal to the boundary at the point x. Recall that for k ∈ N, C k () is the space of restrictions to  of elements of C k (Rd ). If  is not bounded we shall set C0k () = C k () ∩ C0k (Rd ) which means that we consider C k functions on  having compact support in Rd . However, C0k () = C k () if  is bounded. Then we have the following result.

• There exists a positive measure dσ on the boundary ∂ such that for every functions ϕ ∈ C01 () and f1 , . . . , fd ∈ C 1 () we have, 





divF (x) ϕ(x) dx = − 

F (x) · gradϕ(x) dx + 

ϕ(x) n(x) · F (x) dσ, ∂

(10.2) d

∂fj j =1 ∂xj

where F (x) = (f1 (x), . . . , fd (x)), divF =

and X · Y =

Moreover, the same formula is true if ϕ ∈ and f1 , . . . , fd ∈ 1 In particular if ϕ ≡ 1 and F ∈ C0 () we have, C 1 ()



d

j =1 Xj Yj . 1 C0 ().

 divF (x) dx =



n(x) · F (x) dσ. ∂

Taking F = grad u where u ∈ C02 () and ϕ ∈ C 2 () we deduce from (10.2) the so-called Green formula for the Laplacian, 

 

where  =

u(x)ϕ(x) dx + 

d

∂2 j =1 ∂x 2 j





u(x)ϕ(x) dx =

∂

is the Laplacian and

∂ ∂n

∂ϕ ∂u (x)ϕ(x) − u(x) (x) ∂n ∂n

10.2.5 Integration on a Graph • Let O be an open subset of Rd−1 and let S be a graph over O given by,   S = x = (x , xd ) : x ∈ O : xd = ψ(x ) . Then the Lebesgue measure dσ on S is given for ϕ ∈ C 0 (S) by,  ϕ(x) dσ = S

O

dσ,

is the exterior normal derivative

to the boundary.



2 ϕ(x , ψ(x )) 1 + |∇x ψ(x )|2 dx .

348

10 Classical Results

10.3 Elements of Differential Calculus • The inverse function theorem. Let E, F be two Banach spaces. Let  be an open subset of E and x0 ∈ E. Let f be a C 1 map from  to F . Assume that df (x0 ) (the differential of f at x0 ) is an isomorphism from E to F . Then there exists V neighborhood of x0 and W neighborhood of f (x0 ) such that f is a C 1 diffeomorphism from V to W. • The implicit function theorem. Let E, F, G be three Banach spaces and let  be a neighborhood of (x0 , y0 ) ∈ E × F. Let f :  → G be a C 1 map. Assume that there exists a linear and continuous map L : G → F such that fy (x0 , y0 ) ◦ L = I dG . Then there exists a C 1 map g from a neighborhood of x0 to F such that f (x, g(x)) = f (x0 , y0 ). Moreover, if fy (x0 , y0 ) is bijective then g is unique and f (x, y) = f (x0 , y0 ) is equivalent to y = g(x) for (x, y) close to (x0 , y0 ). • The Hadamard theorem. Recall first that a map f : Rd → Rd is said to be proper if the inverse image of a compact set is compact. This is equivalent to say that lim|x|→+∞ |f (x)| = +∞. Let f : Rd → Rd be a C 1 map. Then the two following conditions are equivalent. (i)

f is a diffeomorphism from Rd to Rd .

(ii) f is proper and its Jacobian is different from zero everywhere.

10.4 Elements of Differential Equations We recall in this section some basic facts about systems of differential equations.

10.4.1 The Precise Cauchy–Lipschitz Theorem • Let a, b be two strictly positive real numbers and (t0 , x0 ) ∈ R × Rd . Set   Q = (t, x) ∈ R × Rd : |t − t0 | ≤ a, |x − x0 | ≤ b . Let f : Q → Rd be continuous and let M > 0 be such that, |f (t, x)| ≤ M,

∀(t, x) ∈ Q.

(10.3)

10.4 Elements of Differential Equations

349

Assume moreover that, ∃C > 0 : |f (t, x) − f (t, y)| ≤ C|x − y|,

∀(t, x), (t, y) ∈ Q.

(10.4)

Then the Cauchy problem, dx (t) = f (t, x(t)), dt

x(t0 ) = x0 ,

(10.5)

has a unique solution x(t) defined for t ∈ J = [t0 − T , t0 + T ], where T = b min(a, M ), such that (s, x(s)) ∈ Q for all s ∈ J. • Notice that, in this version, the time T of existence of the solution does not depend on the constant C in (10.4). • Differentiability with respect to the initial conditions. Assume conditions (10.3) and (10.4) satisfied. Then there exists a neighborhood V of (t0 , x0 ) such that for every (s, y) ∈ V the problem, dx (t) = f (t, x(t)), dt

x(s) = y,

(10.6)

b has a unique solution defined for t ∈ J = [t0 − T , t0 + T ] where T = min(a, M ) and this solution denoted by x(t; s, y) is continuous with respect to (s, y) in V . Moreover, if f ∈ C k (Q) for k ∈ N, k ≥ 1 then the solution is C k with respect to (s, y) and C k+1 with respect to t.

10.4.2 The Cauchy–Arzela–Péano Theorem • We keep the notations above. Let f : Q → Rd be a continuous function satisfying (10.3). Then the problem (10.5) has a solution defined on J . • Notice that in this case we lose the uniqueness of the solution. Moreover, we have the same differentiability results with respect to the initial conditions.

10.4.3 Global Theory • Let x1 (resp. x2 ) be a solution of (10.5) on an interval J1 (resp. J2 .) We say that x2 extends x1 if J1 ⊂ J2 and x1 = x2 on J1 . A maximal solution of (10.5) is a solution which has no nontrivial extension. • Let I ⊂ R be an interval,  ⊂ Rd be an open set, (t0 , x0 ) ∈ I × Rd and let f : I ×  → Rd be a continuous function.

350

10 Classical Results

Then the problem (10.5) has a maximal solution defined on an open interval J = (T∗ , T ∗ ) ⊂ I. Moreover, if f is locally Lipschitz then this solution is unique. • Set I = (a, b) where −∞ ≤ a < b ≤ +∞ and assume  = Rd . Let x be a maximal solution of (10.5) defined on (T∗ , T ∗ ) ⊂ (a, b). Then, either

T∗ = b

or T ∗ < b

and lim∗ |x(t)| = +∞,

(ii) either

T∗ = a

or T∗ > a

and lim |x(t)| = +∞.

(i)

t →T

t →T∗

• Notice that in the above results we have a true limit and not only lim.

10.4.4 The Gronwall Inequality Here is a very useful lemma. • Let ϕ, k be two nonnegative and continuous functions on an interval [a, b] ⊂ R. Assume that there exist two constants A ≥ 0, B ≥ 0 such that, 

t

ϕ(t) ≤ A + B

k(s)ϕ(s) ds,

∀t ∈ [a, b].

 t k(s) ds , ϕ(t) ≤ A exp B

∀t ∈ [a, b].

a

Then,

a

10.5 Elements of Holomorphic Functions In this section we recall some basic facts of the theory of holomorphic functions in one variable. In C, the variable is denoted by z = x + iy with x, y ∈ R. Moreover, in all what follows  will denote an open subset of C. • The Cauchy–Riemann operator. It is defined by 1 ∂= 2



∂ ∂ +i ∂x ∂y

.

10.5 Elements of Holomorphic Functions

351

• Holomorphic functions. A function f :  → C is said to be holomorphic if, (i)

f is a C 1 function with respect to x, y,

(ii) ∂f = 0 in .

The vector space of holomorphic functions on  is denoted by H(). • The maximum modulus principle. Assume  is connected and let f ∈ H(). If there exists z0 ∈  such that |f (z0 )| ≥ |f (z)| for all z ∈ , then f is constant on . • The principle of isolated zeroes. Assume  is connected and let f, g ∈ H(). Let S be a subset of  containing an accumulation point. If f (z) = g(z) for z ∈ S then f = g. • The index. Let γ be a closed path in  and z0 ∈ / γ . The index of z0 with respect to γ is defined by, Ind(γ , z0 ) =

1 2iπ

 γ

dζ . ζ − z0

It is an integer. For instance if γ is the boundary of the disc D = {z : |z| < R} then Ind(γ , z0 ) = 1 if z0 is inside D and Ind(γ , z0 ) = 0 if it is outside. • Simply connected domain. In C this is a connected open set  such that c (its complementary) has no bounded connected component. Star shaped domains are simply connected. An annulus is not simply connected. • The Cauchy integral formula. Assume  is simply connected and let f ∈ H(). Let γ be a closed path. Then,  f (z) dz = 0,

(i) γ

(ii) ∀z0 ∈ , z0 ∈ / γ,

1 Ind(γ , z0 )f (z0 ) = 2iπ

 γ

f (z) dz. z − z0

• Characterization of the holomorphy. The two following conditions are equivalent: (i)

f ∈ H(),

(ii) ∀z0 ∈ , ∃r > 0 : f (z) =

+∞ 

an (z − z0 )n , ∀z ∈ {z ∈ C : |z − z0 | < r} ,

n=0

where the series is absolutely convergent for |z − z0 | < r. • Convergence of sequences of holomorphic functions. Let (fn )n∈N ⊂ H(). Assume that for each compact subset K of  the sequence converges uniformly on K to a function f . Then f belongs to H().

352

10 Classical Results

•  Laurent series. Residues. Poles. A Laurent series is a series of the form f (z) = +∞ n n=−∞ an (z − z0 ) . It is said  to be absolutely (resp.  uniformly) convergent in a set A ⊂ C if the series n≥0 an (z − z0 )n and n 0, r > 0. The residue of f at z0 is defined by, Res (f, z0 ) = a−1 . If f is holomorphic in  \ {z0 } we shall say that f has a pole of order m ≥ 1 if the function (z − z0 )m f (z) is holomorphic in  and does not vanish at z0 . If f has a pole of order m ≥ 1 at z0 then, 1 Res (f, z0 ) = lim (m − 1)! z→z0



d dz

m−1



(z − z0 )m f (z) .

• The Residue formula. Let  be a simply connected domain and A = {z1 , . . . , zn } be a subset of . Let γ be a closed path in  such that zj ∈ / γ, j = 1 . . . n. Let f ∈ H( \ A). Then,   f (z) dz = 2iπ Res (f, zj ) Ind(γ , zj ). γ

a∈A

• Meromorphic functions. Let  be an open subset of C. A function f is said to be meromorphic in  if there exists A ⊂  with no accumulation point such that, (i) f ∈ H( \ A),

(ii)

every point of A is a pole of f.

Meromorphic functions are quotient of two holomorphic functions. • Interior of a path. Let  be an open subset of C and let γ be a closed path in . We say that γ has an interior if Ind (γ , a) = 0 or 1 for every a ∈ / γ . Then the set {a ∈  : Ind (γ , a) = 1} will be called the interior of γ . This notion of interior coincides obviously with the usual one if for instance γ is the boundary of a disc. Assume  simply connected and let γ ⊂  be a closed path. Let f be a meromorphic function in  with only a finite number of zeros and poles / γ , j = 1, . . . , n. Then, z1 , . . . , zn such that zj ∈ 1 2iπ

 γ

f (z) dz = N − P , f (z)

where N (resp. P ) is the number of zeros (resp. poles) of f (counted with their multiplicity) in the interior of γ .

References

1. Alazard, T.: Analyse et équations aux dérivées partielles. Lectures Notes. Available at http:// talazard.perso.math.cnrs.fr (2020) 2. Alinhac, S.: Hyperbolic Partial Differential Equations. Universitext. Springer, Dordrecht (2009) 3. Alinhac, S., Gérard, P.: Pseudo-Differential Operators and the Nash-Moser Theorem. Graduate Studies in Mathematics, vol. 82. American Mathematical Society, Providence, RI (2007) (Translated from the 1991 French original by Stephen S. Wilson) 4. Ambrosio, L., Caffarelli, L.A., Brenier, Y., Buttazzo, G.M., Villani, C.: In: Caffarelli, L.A., Salsa, S. (eds.) Optimal Transportation and Applications. Lectures from the C.I.M.E. Summer School held in Martina Franca, September 2–8, 2001. Lecture Notes in Mathematics, vol. 1813. Springer/Centro Internazionale Matematico Estivo (C.I.M.E.), Berlin/Florence (2003) 5. Bahouri, H., Chemin, J.-Y., Danchin, R.: Fourier Analysis and Nonlinear Partial Differential Equations. Grundlehren der Mathematischen Wissenschaften (Fundamental Principles of Mathematical Sciences), vol. 343. Springer, Heidelberg (2011) 6. Bergh, J., Löfström, J.: Interpolation Spaces. An Introduction. Grundlehren der Mathematischen Wissenschaften, vol. 223. Springer, Berlin (1976) 7. Brezis, H.: Functional Analysis, Sobolev Spaces and Partial Differential Equations. Universitext. Springer, New York (2011) 8. Carleman, T.: Sur un problème d’unicité pur les systèmes d’équations aux dérivées partielles à deux variables indépendantes. Ark. Mat., Astr. Fys. 26(17), 9 (1939) 9. Evans, L.C.: Partial Differential Equations. Graduate Studies in Mathematics, vol. 19, 2nd edn. American Mathematical Society, Providence, RI (2010) 10. Gérard, P.: Microlocal defect measures. Commun. Partial Differ. Equ. 16(11), 1761–1794 (1991) 11. Gilbarg, D., Trudinger N.S.: Elliptic Partial Differential Equations of Second Order. Grundlehren der Mathematischen Wissenschaften (Fundamental Principles of Mathematical Sciences), vol. 224, 2nd edn. Springer, Berlin (1983) 12. Himonas, A.A., Misiołek, G.: Non-uniform dependence on initial data of solutions to the Euler equations of hydrodynamics. Commun. Math. Phys. 296(1), 285–301 (2010) 13. Hörmander, L.: Hypoelliptic second order differential equations. Acta Math. 119, 147–171 (1967) 14. Hörmander, L.: The Analysis of Linear Partial Differential Operators. I. Distribution Theory and Fourier Analysis. Classics in Mathematics. Springer, Berlin (2003) (Reprint of the second (1990) edition [Springer, Berlin; MR1065993 (91m:35001a)].)

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 T. Alazard, C. Zuily, Tools and Problems in Partial Differential Equations, Universitext, https://doi.org/10.1007/978-3-030-50284-3

353

354

References

15. Hörmander, L.: The Analysis of Linear Partial Differential Operators. IV. Fourier Integral Operators. Classics in Mathematics. Springer, Berlin (2009) (Reprint of the 1994 edition) 16. John, F.: Partial Differential Equations. Applied Mathematical Sciences, vol. 1, 4th edn. Springer, New York (1982) 17. Kato, T.: The Cauchy problem for quasi-linear symmetric hyperbolic systems. Arch. Ration. Mech. Anal. 58(3), 181–205 (1975) 18. Lerner, N.: Carleman Inequalities. An Introduction and More. Grundlehren der Mathematischen Wissenschaften (Fundamental Principles of Mathematical Sciences), vol. 353. Springer, Cham (2019) 19. Lieb, E.H., Loss, M.: Analysis. Graduate Studies in Mathematics, vol. 14, 2nd edn. American Mathematical Society, Providence, RI (2001) 20. Makino, T., Ukai, S., Kawashima, S.: Sur la solution à support compact de l’équations d’Euler compressible. Jpn. J. Appl. Math. 3(2), 249–257 (1986) 21. Métivier, G.: Para-Differential Calculus and Applications to the Cauchy Problem for Nonlinear Systems. Centro di Ricerca Matematica Ennio De Giorgi (CRM) Series, vol. 5. Edizioni della Normale, Pisa (2008) 22. Misiołek, G., Yoneda, T.: Continuity of the solution map of the Euler equations in Hölder spaces and weak norm inflation in Besov spaces. Trans. Am. Math. Soc. 370(7), 4709–4730 (2018) 23. Nishida, T.: A note on a theorem of Nirenberg. J. Differ. Geom. 12(4), 629–633 (1978), 1977. 24. Rudin, W.: Functional Analysis. International Series in Pure and Applied Mathematics, 2nd edn. McGraw-Hill, Inc., New York (1991) 25. Smoller, J.: Shock Waves and Reaction-Diffusion Equations. Grundlehren der Mathematischen Wissenschaften (Fundamental Principles of Mathematical Sciences), vol. 258, 2nd edn. Springer, New York (1994) 26. Sogge, C.D.: Lectures on Non-Linear Wave Equations, 2nd edn. International Press, Boston, MA (2008) 27. Sogge, C.D.: Hangzhou Lectures on Eigenfunctions of the Laplacian. Annals of Mathematics Studies, vol. 188. Princeton University Press, Princeton, NJ (2014) 28. Stein, E.M.: Harmonic Analysis: Real-Variable Methods, Orthogonality, and Oscillatory Integrals. Princeton Mathematical Series, vol. 43. Princeton University Press, Princeton, NJ (1993) (With the assistance of Timothy S. Murphy, Monographs in Harmonic Analysis, III.) 29. Tao, T.: Nonlinear Dispersive Equations. Local and Global Analysis. CBMS Regional Conference Series in Mathematics, vol. 106. American Mathematical Society, Providence, RI (2006) (Published for the Conference Board of the Mathematical Sciences, Washington, DC) 30. Taylor, M.E.: Partial Differential Equations II. Qualitative Studies of Linear Equations. Applied Mathematical Sciences, vol. 116, 2nd edn. Springer, New York (2011) 31. Taylor, M.E.: Partial Differential Equations III. Nonlinear Equations. Applied Mathematical Sciences, vol. 117, 2nd edn. Springer, New York (2011) 32. Zuily, C.: Problems in Distributions and Partial Differential Equations. North-Holland Mathematics Studies, vol. 143. North-Holland Publishing Co./Hermann, Amsterdam/Paris (1988) (Translated from the French.) 33. Zuily, C.: Distributions et équations aux dérivées partielles. Dunod, Paris (2002) 34. Zworski, M.: Semiclassical Analysis. Graduate Studies in Mathematics, vol. 138. American Mathematical Society, Providence, RI (2012)

Index

A Adjoint operator, 9 self-, 9 Approximation of the identity, 11 Ascoli, 7

B Banach -Alaoglu, 6 fixed point theorem, 5 Hahn-, 7 isomorphism theorem, 5 -Steinhaus, 6 Benjamin–Ono equation, 159 Bernstein inequalities, 16, 23 Bicharacteristics, 67 BMO space, 52 Boussinesq equation, 147 Burgers equation, 116, 174

C Carleman inequality, 86 Cauchy -Arzela–Peano theorem, 349 -Kovalewski theorem, 102 -Lipschitz theorem, 348 Closed graph theorem, 5

D Dirichlet–Neuman operator, 137 Distribution, 12

with compact support, 13 convergence of a sequence of, 13 derivation of a, 13 support of a, 12

E Euler equations compressible, 121, 170 incompressible, 117 vorticity, 119

F Faa–di-Bruno formula, 344 Fefferman–Phong inequality, 66 Fourier transform, 15 and convolution, 16 of distributions with compact support, 15 in Lebesgue spaces, 15 in S , 15 in S , 15 Fréchet space, 4 Fundamental solution, 13

G Garding inequalities, 65 sharp-, 65 weak-, 65 Gauss–Green formula, 346 Gronwall inequality, 350

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2020 T. Alazard, C. Zuily, Tools and Problems in Partial Differential Equations, Universitext, https://doi.org/10.1007/978-3-030-50284-3

355

356 H Hadamard theorem, 348 Hardy inequality, 51 -Littlewood–Sobolev inequality, 39 Harmonic oscillator, 130 Heat equation, 108 entropy for, 143 maximum principle for, 108 Hilbert space, 7 transform, 77 Hölder inequality, 10 spaces of fractional order, 32 spaces of integer order, 31 Holmgren theorem, 103 Holomorphic functions, 350 Hydrogen atom, 133

I Incompressibility, 118 Interpolation, 37

K Kinetic equation, 162

L Laplace equation, 104 Dirichlet problem for, 106 estimates of eigenfunctions for, 107 Harnack inequality for, 105 Hopf lemma for, 105 hypoellipticity of, 104 maximum principle for, 105 mean value for-, 104 spectral theory for, 106 Lax–Milgram lemma, 8 Leibniz formula, 343 Lichtenstein, 120 Locally convex space, 3

M Mean curvature equation, 145 Microlocal analysis, 61 defect measure, 71 Monge–Ampère equation, 125

Index N Navier–Stokes equations, 122

P Paradifferential operators, 69 Paraproducts, 35 Poincaré inequality, 30 Poisson bracket, 64 Polar coordinates, 345 Propagation of singularities, 68 Pseudo-differential operators, 62 action on Sobolev spaces, 65 adjoint of a, 64 composition of, 64 image of a, 63 kernel of a, 63

R Residue theorem, 352 Riesz, 8 -Thorin, 38

S Schauder fixed point theorem, 5 Schrödinger equation, 113 infinite speed propagation for, 114 Strichartz estimate for, 114 Sobolev spaces, 25 compactness in, 28 duality on, 27 embeddings of, 26 local, 28 on an open set, 29 operation on, 26 on Rd , 25 on the torus, 31 traces in, 28 uniformly local, 44 Spectral theory, 8, 9 Spectrum, 9 Stationary phase method, 16 Symbols classical, 62 of a DO, 61 of a paradifferential operator, 68

T Taylor formula, 343

Index W Wave equation, 109 conservation of energy, 111 end point Strichartz estimate, 153 finite speed propagation for, 110 Huygens principle for , 110 influence domain for , 111 Strichartz estimates for, 111 Wave front set, 66 and pdo, 67 Weak convergence, 7

357 Weak Lebesgue spaces, 52 Weyl law, 107 Y Young inequality, 10 Yudovitch, 121

Z Zygmund spaces, 35