Harmonic analysis method for nonlinear evolution equations, I

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ut

8209.9789814360739- tp.indd 1

-Δ

u u+

u

·∇

p +∇

=0

iut + Δu = F(u)

7/1/11 3:37 PM

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ut

u+

-Δ

u+

= ∇p

0

∇ u·

iut + Δu = F(u)

Baoxiang Wang Peking University, China

Zhaohui Huo Chinese Academy of Sciences, China

Chengchun Hao Chinese Academy of Sciences, China

Zihua Guo Peking University, China

World Scientific NEW JERSEY

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8209.9789814360739- tp.indd 2

LONDON

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SINGAPORE

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BEIJING

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SHANGHAI

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HONG KONG

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TA I P E I

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CHENNAI

7/1/11 3:37 PM

Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.

HARMONIC ANALYSIS METHOD FOR NONLINEAR EVOLUTION EQUATIONS, I Copyright © 2011 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Publisher.

For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher. ISBN-13 978-981-4360-73-9 ISBN-10 981-4360-73-2

Printed in Singapore.

ZhangJi - Harmonic analysis Method.pmd

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To our parents

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Preface

The sole purpose of science is the glory of human spirit. J. Jacobi

—— C. G.

During the past thirty years, the nonlinear evolution equations (NLE) have made a great progress by using the harmonic analysis techniques. This book is devoted to introduce the harmonic analysis method for NLE, which is based on lectures given by B. X. Wang at the Peking University during the past seven years. In order to solve an NLE, one needs to choose suitable function spaces as working spaces. It is well known that most function spaces are established by using the Lebesgue integrals as basic tools. However, the Lebesgue integration was not recognized to be of importance in its early stage. As long as Lebesgue attempted to attend a conference, some mathematicians working in analysis said to him: “This is not interested for you, we are discussing the differentiable functions.”, and some experts engaged in geometry told him: “We are talking the surface which has a tangent plane.” One can imagine how Lebesgue was hurt inside. New ideas are usually not so easy to grow up and mathematics brings mathematicians too much sadness and blessedness, which is hard to express by languages. However, the authors believe that, at least from the local history point of view, many important progresses for nonlinear evolution equations have brought us a series of perpetual surprises over the recent thirty years. The harmonic analysis techniques of NLE can be gone back to the pioneering work of R. S. Strichartz in 1977 who discovered the time-space decay of the solutions of the linear wave equation, so called the Strichartz inequality, which is now a fundamental tool in the study of nonlinear dispersive equations. Since 1980s, the NLE, especially the nonlinear dispersive equations have gained a great development by using the harmonic analysis techniques. vii

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Preface

On the other hand, the theory of harmonic analysis is also promoted by the study of NLE. J. Bourgain in 1993 systematically developed the X s,b method which is powerful in handling the derivative nonlinearity in nonlinear dispersive equations, where the X s,b space was previously discovered by J. Rauch, M. Reed and M. Beals for the wave equation. J. Bourgain in 1999 invented the method, so called “separation of the localized energy”, to study the energy scattering of nonlinear Schr¨odinger equations. The “I-method” was introduced by T. Tao’s group to study the global wellposedness of nonlinear dispersive equations for a class of initial data with lower regularities. C. E. Kenig and F. Merle developed the concentration compactness method to study the sharp well posedness and scattering for the focusing nonlinear dispersive equations. Recently, the frequency-uniform decomposition techniques are also applied in the study of NLE. A poem describes the current status of the harmonic analysis techniques of NLE: Resonance to a Book1

—Xi Zhu A square of pool likes an opening mirror Blue sky and white cloud map in it, freely wander Why is the pond crystal and bottomed out It is from the source of alive water More precisely, we will study a class of nonlinear dispersive equations, such as nonlinear Schr¨ odinger equations, nonlinear Klein-Gordon eqautions, KdV equations, as well as the Navier-Stokes equations and the Boltzmann equation. As the first book of this series we will mainly study the local and global wellposedness to the Cauchy problem for those equations. In Chapter 1 we briefly introduce the theories of various function spaces, say Besov and Triebel-Lizorkin spaces. In Chapter 2 we study the NavierStokes equation by using the Littlewood-Paley decomposition to establish some time-space estimates for the linear heat equation. Strichartz type estimates for a class of linear dispersive equations will be systematically set up in Chapter 3. Applying those Strichartz’ inequalities, in Chapter 4 we will consider the wellposedness of the solutions of nonlinear Schr¨odinger and Klein-Gordon equations. In Chapter 5 we introduce the X s,b -method and study the KdV and derivative Schr¨odinger equations. In Chapter 6 we introduce the frequency-uniform decomposition method. It is known that 1A

pool indicates a book.

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Morawetz’ estimates are basic tools for the scattering theory of nonlinear dispersive equations, which will be summarized in Chapter 7 for nonlinear Schr¨ odinger equations. Finally, in Chapter 8 we introduce some fundamental results for the Boltzmann equation. Chapters 1, 2, 3, 4 and 6 are written by B. X. Wang, Chapter 5 is written by Z. H. Guo and Z. H. Huo, Chapter 7 is written by C. C. Hao and Chapter 8 is written by Z. H. Huo. Many results in Chapters 2, 3, 4, 5 and 6 have been reproved or simplified by the authors. The authors would like to genuinely thank Professors Yulin Zhou, Hesheng Sun, Boling Guo, Ling Hsiao, Lizhong Peng and Carlos E. Kenig for their constant supports and they are grateful to Professors Kong Ching Chang, Weiyue Ding and Gang Tian for their helps. B. X. Wang thanks Professors Zhouping Xin and Jiecheng Chen for their invitation to give a series of lectures based on the book at Hong Kong and Hangzhou, respectively. B. X. Wang deeply cherishes the memory of Professor Tingfu Wang for his rudimental guidance in mathematics. Thanks are also due to Ms Ji Zhang for her excellent editorial work.

Beijing, May 1, 2011

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Contents

Preface 1.

s Fourier multiplier, function space Xp,q

1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

2.

vii 1

Schwartz space, tempered distribution, Fourier transform Fourier multiplier on Lp . . . . . . . . . . . . . . . . . . . Dyadic decomposition, Besov and Triebel spaces . . . . . s Embeddings on Xp,q . . . . . . . . . . . . . . . . . . . . . s Differential-difference norm on Xp,q . . . . . . . . . . . . . s Homogeneous space X˙ p,q . . . . . . . . . . . . . . . . . . . Bessel (Riesz) potential spaces Hps (H˙ ps ) . . . . . . . . . . Fractional Gagliardo-Nirenberg inequalities . . . . . . . . s 1.8.1 GN inequality in B˙ p,q . . . . . . . . . . . . . . . . s ˙ 1.8.2 GN inequality in Fp,q . . . . . . . . . . . . . . . .

Navier-Stokes equation 2.1

2.2

2.3 2.4 2.5

1 4 8 13 16 19 22 25 25 29 33

Introduction . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Model, energy structure . . . . . . . . . . . . . 2.1.2 Equivalent form of NS . . . . . . . . . . . . . 2.1.3 Critical spaces . . . . . . . . . . . . . . . . . . Time-space estimates for the heat semi-group . . . . . 2.2.1 Lr → Lp estimate for the heat semi-group . . 2.2.2 Time-space estimates for the heat semi-group Global well-posedness in L2 of NS in 2D . . . . . . . . Well-posedness in Ln of NS in higher dimensions . . . Regularity of solutions for NS . . . . . . . . . . . . . s 2.5.1 Gevrey class and function space E2,1 . . . . . xi

. . . . . . . . . . .

. . . . . . . . . . .

34 34 34 35 35 35 36 39 42 44 44

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Contents

2.5.2 2.5.3 2.5.4 3.

Strichartz estimates for linear dispersive equations

Local and global wellposedness for nonlinear dispersive equations

75

4.1 4.2 4.3

75 78 83 83 84 87 88

Why is the Strichartz estimate useful . . . . . Nonlinear mapping estimates in Besov spaces Critical and subcritical NLS in H s . . . . . . 4.3.1 Critical NLS in H s . . . . . . . . . . 4.3.2 Wellposedness in H s . . . . . . . . . Global wellposedness of NLS in L2 and H 1 . Critical and subcritical NLKG in H s . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

The low regularity theory for the nonlinear dispersive equations 5.1 5.2 5.3 5.4 5.5 5.6 5.7

6.

51

0

52 63 68

4.4 4.5 5.

46 47 48

Lp → Lp estimates for the dispersive semi-group . . . . . Strichartz inequalities: dual estimate techniques . . . . . Strichartz estimates at endpoints . . . . . . . . . . . . . .

3.1 3.2 3.3 4.

s Estimates of heat semi-group in E2,1 . . . . . . . s Bilinear estimates in E2,1 . . . . . . . . . . . . . . Gevrey regularity of NS equation . . . . . . . . .

91

Bourgain space . . . . . . . . . . . . . . . . . . . . . . . Local smoothing effect and maximal function estimates Bilinear estimates for KdV and local well-posedness . . Local well-posedness for KdV in H −3/4 . . . . . . . . . I-method . . . . . . . . . . . . . . . . . . . . . . . . . . Schr¨ odinger equation with derivative . . . . . . . . . . . Some other dispersive equations . . . . . . . . . . . . . .

. . . . . . .

Frequency-uniform decomposition techniques 6.1 6.2 6.3 6.4

Why does the frequency-uniform decomposition work . Frequency-uniform decomposition, modulation spaces 6.2.1 Basic properties on modulation spaces . . . . Inclusions between Besov and modulation spaces . . . NLS and NLKG in modulation spaces . . . . . . . . . 6.4.1 Schr¨ odinger and Klein-Gordon semigroup in modulation spaces . . . . . . . . . . . . . . . . 6.4.2 Strichartz estimates in modulation spaces . . . 6.4.3 Wellposedness for NLS and NLKG . . . . . . .

91 99 104 113 130 145 152 157

. . . . .

. . . . .

157 159 160 164 176

. . 176 . . 179 . . 183

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6.5

7.

Derivative nonlinear Schr¨odinger equations . . . . . 6.5.1 Global linear estimates . . . . . . . . . . . . 6.5.2 Frequency-localized linear estimates . . . . 6.5.3 Proof of global wellposedness for small rough

. . . . . . . . . data

Conservations, Morawetz’ estimates of nonlinear odinger equations Schr¨ 7.1 7.2 7.3 7.4 7.5

8.

xiii

N¨ other’s theorem . . . . . . . . . . . . . Invariance and conservation law . . . . . Virial identity and Morawetz inequality Morawetz’ interaction inequality . . . . Scattering results for NLS . . . . . . . .

. . . . .

187 191 192 197

205 . . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

205 212 214 220 222

Boltzmann equation without angular cutoff

227

8.1

228 228 229 233 235 240 246

8.2 8.3 8.4

Models for collisions in kinetic theory . . . . . . . . . . . 8.1.1 Transport model . . . . . . . . . . . . . . . . . . . 8.1.2 Boltzmann model . . . . . . . . . . . . . . . . . . 8.1.3 Cross section . . . . . . . . . . . . . . . . . . . . . Basic surgery tools for the Boltzmann operator . . . . . . Properties of Boltzmann collision operator without cutoff Regularity of solutions for spatially homogeneous case . .

Appendix A Notations

259

Appendix B Definition of scattering operator

261

Appendix C Some fundamental results

263

C.1 C.2 C.3 C.4 C.5 C.6 C.7 C.8 C.9

Gagliardo-Nirenberg inequality in Sobolev spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . Convexity H¨ older inequality in sequence spaces `sp . . . Inclusion between homogeneous Triebel-Lizorkin spaces Riesz-Thorin interpolation theorem . . . . . . . . . . . . Hardy-Littlewood-Sobolev inequality . . . . . . . . . . . Van der Corput lemma . . . . . . . . . . . . . . . . . . . Littlewood-Paley square function theorem . . . . . . . . Complex interpolation in modulation spaces . . . . . . . Christ-Kiselev lemma . . . . . . . . . . . . . . . . . . .

. . . . . . . . .

263 263 264 265 266 266 266 267 267

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Contents

Bibliography

269

Index

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Chapter 1 s Fourier multiplier, function space Xp,q

The intensive study to the nature is the most plentiful source for the mathematical discovery. —— J. Fourier1

In this chapter we begin by considering the Fourier transform on the Schwartz space S and its dual space S 0 , the material on Fourier transforms is standard and our treatment here will be sketched, see Stein [202], Yosida [256] for instance. On the basis of the Fourier transform, we introduce s s the Fourier multiplier space Mp , Besov spaces Bp,q and Triebel spaces Fp,q and then discuss their elementary properties. The embeddings between these function spaces are very useful in the study of partial differential equations. For the convenience to the readers, we give a self-contained treatment to the theory of function spaces. For some further results on function spaces, one can refer to [13; 224]. Some notations are well-known for readers and not stated in the text, which will be listed in the Appendix A. For convenience to readers, we would like to remind some important relations between Triebel-Lizorkin, Bessel potential spaces Hps (H s = H2s ) and Sobolev spaces Wpm with m ∈ Z+ : s s Fp,2 = Hps , Hpm = Wpm , F2,2 = H s , ∀ s ∈ R, m ∈ Z+ , 1 < p < ∞.

1.1

Schwartz space, tempered distribution, Fourier transform

Let α = (α1 , ..., αn ) be a multi-index, denote Dα = ∂xα11 ...∂xαnn ,

i ∂xαii = ∂ αi /∂xα i , |α| = α1 + ... + αn ,

αn 1 xα = xα 1 ...xn .

1 Joseph Fourier (1768–1830), a French mathematician and physicist, is best known for initiating the investigation of Fourier series and their application to problems of the heat transfer. He is also generally credited with the discovery of the greenhouse effect. He is one of the authority in French analysis school.

1

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s Fourier multiplier, function space Xp,q

2

We denote by C ∞ (Rn ) the set of all infinitely differentiable functions defined on Rn . Denote S = {φ ∈ C ∞ (Rn ) : pk (φ) < ∞}, X pk (φ) = sup (1 + |x|2 )k/2 |Dα φ(x)|. x∈Rn

(1.1)

|α|6k

It is easy to see that pk (·) is a norm on S and so, it is a semi-norm on S . From the theory of the linear topology, S generates a locally convex linear topological space according to the semi-norm system {pk }∞ k=0 , which is said to be the Schwartz space, any function in S is said to be a Schwartz function. The base of zero neighborhoods of S is Bk,ε = {φ ∈ S : pk (φ) < ε}, k ∈ Z+ = {0, 1, 2, ...}, ε > 0.

We denote by S 0 := S 0 (Rn ) the dual space of S and it is a locally convex linear topological space, which is said to be a tempered distribution space. The base of zero neighborhoods of S 0 is ( ) UB,ε =

f ∈ S : sup |f (φ)| < ε , φ∈B

where B is a bounded set in S . Let φ, ψ ∈ S , the Fourier (inverse) transform is defined as follows. Z −n/2 b φ(ξ) = (F φ)(ξ) = (2π) e−ix·ξ φ(x)dx, Rn Z ˇ ψ(x) = (F −1 ψ)(x) = (2π)−n/2 eix·ξ ψ(ξ)dξ, (1.2) Rn

where x · ξ = x1 ξ1 + ... + xn ξn (if there is no confusion, we will write x·ξ = xξ). S plays a crucial role in the theory of the Fourier analysis, which is compatible with the Fourier transform. In fact, we have the following Proposition 1.1. Let φ ∈ S . Then we have α φ(ξ) = i|α| ξ α φ(ξ), d b D

F −1 (F φ) = φ.

α φ(ξ), b = (−i)|α| xd Dα φ(ξ)

Moreover, F (F −1 ) : S → S is a continuously linear bijection, i.e., an isomorphism. Proposition 1.2. Let φ, ψ ∈ S . Then Z Z b φ(x)ψ(x)dx = Rn

Rn

b φ(x)ψ(x)dx.

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We define the convolution φ ∗ ψ on S in the following way Z φ ∗ ψ(x) = φ(x − y)ψ(y)dy. Rn

Proposition 1.3. Let φ, ψ ∈ S . Then b φ[ ∗ ψ = (2π)n/2 φb · ψ,

b φd · ψ = (2π)−n/2 φb ∗ ψ.

We denote by σλ and τh the dilation and the translation operators: σλ φ = φ(λ·),

λ ∈ R \ {0},

τh φ = φ(· − h), h ∈ Rn .

Proposition 1.4. Let φ ∈ S , λ ∈ R \ {0}, h ∈ Rn . Then −ihξ b τd φ, hφ = e ihx φ = τ φ, b e[ h

−n b σd (σ1/λ φ). λ φ = |λ|

For the proofs of the above propositions, except that the proof of F −1 φb = φ needs a special technique, say 2 −|x|2 /2 = e−|ξ| /2 , e\

the other proofs are easily obtained. In the following we consider the Fourier transform of f ∈ S 0 . Inspired by Proposition 1.2, we define fb as follows. b (F f )(φ) = fb(φ) = f (φ), φ ∈ S, −1 ˇ ˇ (F f )(φ) = f (φ) = f (φ), φ ∈ S .

Proposition 1.5. F : S 0 → S 0 is a continuously linear bijection map, i.e., an isomorphism. Now we can define the convolution of f ∈ S 0 and φ ∈ S . Let us recall that for any f, φ, ψ ∈ S , we have Z Z (f ∗ φ)(x)ψ(x)dx = f (x)(φ˜ ∗ ψ)(x)dx, Rn

Rn

where φ˜ = φ(−·). Following this idea, we define (f ∗ φ)(ψ) = f (φ˜ ∗ ψ),

f ∈ S 0 , φ, ψ ∈ S .

Taking notice of the definition of the Fourier transform and the convolution operator on S 0 , we can generalize the other operators to S 0 . We

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first investigate the corresponding operation for Schwartz functions, which can be converted to an integral operation. Then taking into account the abstract form of this integration operation, we can generalize the operation to S 0 . Let f ∈ S 0 , we define Dα f (φ) := f ((−D)α φ), τh f (φ) := f (τ−h φ),

h ∈ Rn ,

σλ f (φ) := f (λ−n σ1/λ φ),

λ > 0.

0

It is known that f ∈ S does not mean that f is a function, the following result indicates that the convolution of f and a Schwartz function must be an infinitely differentiable function. Proposition 1.6. Let f ∈ S 0 and φ ∈ S . Then f ∗ φ is an infinitely differentiable function and ˜ (f ∗ φ)(x) = f (τx φ). Moreover, f ∗ φ grows at most in a polynomial way, i.e., for any α ∈ Zn+ there exists a polynomial Pα , such that |Dα (f ∗ φ)(x)| 6 Pα (x), x ∈ Rn . Proposition 1.7 (Paley-Wiener-Schwartz theorem). (1) φ ∈ S and supp φb ⊂ {ξ : |ξ| 6 b} if and only if, φ(z) (z = x + iy) is an entire analytic function of n complex variables and for any ε > 0, λ > 0, there exists Cε,λ > 0, such that |φ(z)| 6 Cε,λ (1 + |x|)−λ e(b+ε)|y| , x, y ∈ Rn .

(2) f ∈ S 0 and supp fb ⊂ {ξ : |ξ| 6 b} if and only if, f (z) (z = x + iy) is an entire analytic function of n complex variables and for some λ ∈ R and for any ε > 0, there exists Cε > 0 such that |f (z)| 6 Cε (1 + |x|)λ e(b+ε)|y| , x, y ∈ Rn .

Proposition 1.7 is a fundamental result in the framework of the Fourier analysis, which says that if the Fourier transform of a tempered distribution f has a compact support, then it is an analytic function. 1.2

Fourier multiplier on Lp

We denote by Lp := Lp (Rn ) the Lebesgue space, 1 6 p 6 ∞, the norm on Lp is written as k · kp . Let us consider the linear heat equation in Lp , ut − 4u = 0,

u|t=0 = u0 .

(1.3)

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It is easy to see that u = F −1 e−t|ξ| F u0 =: H(t)u0 is the solution of (1.3). For any u0 ∈ Lp , we want to know if H(t)u0 also belongs to Lp . More precisely, does 2

kF −1 e−t|ξ| F u0 kp . ku0 kp ,

16p6∞

(1.4)

hold for all u0 ∈ Lp ? The answer is affirmative (see Chapter 2). One can generalize (1.4) to the following notion. Definition 1.1. Let ρ ∈ S 0 . If there exists C > 0 such that2 kF −1 ρF f kp 6 Ckf kp ,

∀ f ∈ S,

(1.5)

then ρ is said to be a multiplier on Lp . The set of all multipliers on Lp is denoted by Mp , for which the norm is given by kρkMp = sup{kF −1 ρF f kp : f ∈ S , kf kp = 1}.

(1.6)

If ρ ∈ Mp , 1 6 p < ∞, since S is dense in Lp , one can extend F −1 ρF into a bounded operator on Lp and the norm of the extension operator is preserved. The extension operator is still written as F −1 ρF or (F −1 ρ)∗. The multiplier theory is a fundamental tool in the harmonic analysis, and very useful in the definition of various function spaces. In what follows we discuss some basic properties of the multiplier space Mp . Proposition 1.8. Let 1 6 p 6 q 6 2. Then we have (1) (2) (3) (4)

Mp = Mp0 , k · kMp = k · kMp0 ; Mp ⊂ Mq , k · kMq 6 k · kMp ; M2 = L∞ , k · kM2 = k · k∞ ; M1 = {ρ ∈ S 0 : F −1 ρ is a bounded measure}, kρkM1 equals the total variation of F −1 ρ.

Proof. First, we prove (1). Let ρ ∈ Mp . For any f, g ∈ S , kf kp = kgkp0 = 1, we have3 |(F −1 ρF f, g)| 6 kF −1 ρF f kp kgkp0 6 kρkMp .

(1.7)

Noticing that (F −1 ρF f, g) = (F −1 ρ ∗ f ∗ e g )(0), we see that (1.7) implies 0 F −1 ρF g ∈ Lp and ρ ∈ Mp0 with kρkMp0 6 kρkMp . It follows that (1) holds true. 2 Notice

that F −1 ρF f = (F −1 ρ) ∗ f , in view of Proposition 1.6, we see that F −1 ρF f is a smoothRfunction for any f ∈ S . 3 (f, g) = f (x)g(x)dx.

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Secondly, we show the result of (2). Due to 1 6 p 6 q 6 2, we see that p0 > 2. So, one can choose θ ∈ [0, 1] satisfying 1/q = (1 − θ)/p + θ/p0 . If ρ ∈ Mp , in view of (1) we have ρ ∈ Mp0 . Namely, 0

0

F −1 ρF : Lp → Lp ; F −1 ρF : Lp → Lp

(1.8)

are both bounded operators. It follows from Riesz-Thorin’s interpolation theorem (see Appendix) that F −1 ρF : Lq → Lq is bounded and θ kρkMq 6 kρk1−θ Mp kρkMp0 = kρkMp .

(1.9)

Thirdly, we prove (3). If ρ ∈ M2 , by Plancherel’s identity, kF −1 ρF f k2 = kρF f k2 6 kρk∞ kf k2 .

(1.10)

So, kρkM2 6 kρk∞ . Conversely, for any ε > 0, one can take a non-zero measurable closed subset E of Rn , satisfying |ρ(ξ)| > kρk∞ − ε in E. Let f ∈ L2 satisfy supp F f ⊂ E. It follows that kρkM2 > kρk∞ − ε. Finally, we show that (4) holds. By the translation property of F −1 ρF , we easily see that ρ ∈ M∞ if and only if |(F −1 ρ ∗ f )(0)| = (2π)n/2 |(F −1 ρF f )(0)| 6 Ckf k∞ ,

∀ f ∈ S . (1.11)

Noticing that S is dense in C0 (Rn ), from (1.11) and Proposition 1.6 one sees that F −1 ρ is a continuous functional on C0 (Rn )4 . From the construc tion of C0 (Rn )∗ , one has the result. Proposition 1.9. Mp is a Banach algebra.

Proof. Obviously, k · kMp is a norm. By (3) in Proposition 1.8, one sees that Mp ⊂ L∞ . If {ρk } is a Cauchy sequence in Mp , so does in L∞ . We may assume, without loss of generality that it converges to ρ in L∞ . Since L∞ ⊂ S 0 , we conclude that for any f ∈ S , F −1 ρk F f → F −1 ρF f according to the strong topology on S 0 . Recalling that F −1 ρk F f is a Cauchy sequence in Lp ⊂ S 0 , it has a limit ponit g. By the uniqueness of the limit in S 0 one obtains that g = F −1 ρF f . So, kρk − ρkMp → 0 as k → ∞. It follows that Mp is a Banach space. Let ρ1 , ρ2 ∈ Mp . For any f ∈ S , we have kF −1 ρ1 ρ2 F f kp 6 kρ1 kMp kF −1 ρ2 F f kp 6 kρ1 kMp kρ2 kMp kf kp ,

(1.12)

which implies that ρ1 ρ2 ∈ Mp and kρ1 ρ2 kMp 6 kρ1 kMp kρ2 kMp . Hence, Mp is a Banach algebra. 4 We

(1.13)

denote by C0 (Rn ) the space of all continuous functions f (x) that vanish as x → ∞.

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In order to emphasize that Mp depends on Rn , we write Mp = Mp (Rn ). The following proposition indicates that Mp (Rn ) is isometrically invariant under affine transforms5 of Rn . Proposition 1.10. Let a : Rn → Rm (n > m) be a surjective affine transform, and ρ ∈ Mp (Rm ). Then In particular,

kρ(a(·))kMp (Rn ) = kρkMp (Rm ) .

kρ(c ·)kMp (Rn ) = kρkMp (Rn ) ,

kρ(hx, ·i)kMp (Rn ) = kρkMp (R) , Pn where hx, ξi = i=1 xi ξi .

c 6= 0;

x 6= 0,

(1.14) (1.15) (1.16)

Proof. It suffices to consider the case that a : Rn → Rm is a linear transform. Make the coordinate transformation ηi = ai (ξ), 1 6 i 6 m;

ηj = ξj , m + 1 6 j 6 n,

(1.17)

−1

which is written as η = A ξ or ξ = Aη. Let A∗ be the transposed matrix of A. It is easy to see, for any f ∈ S , that F −1 ρ(a(ξ))F f = F −1 ρ(η1 , ..., ηm )F f (A−1 (1.18) ∗ ·) (A∗ ·). It follows from ρ ∈ Mp (Rm ) that for any f ∈ S

kF −1 ρ(a(·))F f kLp (Rn ) = |A|−1 kF −1 ρF f (A−1 ∗ ·)kLp (Rn ) 6 kρkMp (Rm ) kf kLp(Rn ) .

Thus, we have

kρ(a(·))kMp (Rn ) 6 kρkMp (Rm ) .

(1.19) (1.20)

Taking f (A−1 ∗ ·) = f1 (x1 , ..., xm )f2 (xm+1 , ..., xn ), one can conclude that the inverse inequality of (1.20) also holds. In the following we give some criteria how to determine the multiplier in Lp . Proposition 1.11 (Bernstein multiplier theorem). Let L > n/2 be an integer, ∂xαi ρ ∈ L2 , i = 1, ..., n and 0 6 α 6 L. Then we have ρ ∈ Mp , 1 6 p 6 ∞ and !n/2L n X

L 1−n/2L

∂x ρ kρkMp . kρk . (1.21) 2

i

2

i=1

5 An affine transform of Rn is a map F : Rn → Rn of the form F (p) = Ap + q for all p ∈ Rn , where A is a linear transform of Rn .

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8

Proof. ously,

By Proposition 1.8, it suffices to consider the case p = 1. ObvikρkM1 6

For any t > 0,

Denote J(x) = Z

Pn

Z

|x|
Z

Rn

|F −1 ρ(x)|dx.

(1.22)

|F −1 ρ(x)|dx . tn/2 kρk2 .

(1.23)

|xi |L . One has that Z |F −1 ρ(x)|dx = J(x)−1 J(x) F −1 ρ(x) dx i=1

|x|>t

|x|>t

. tn/2−L

Taking t such that kρk2 = t−L the result, as desired.

n X

L

∂x ρ . i 2

(1.24)

i=1

Pn

i=1

k∂xLi ρk2 , from (1.22)–(1.24) we have

Proposition 1.12 (Mihlin multiplier theorem). Let L > n/2 be an integer and ρ ∈ L∞ satisfy |ξ||α| |Dα ρ(ξ)| 6 A,

|α| 6 L, ξ ∈ Rn \ {0}.

(1.25)

Then ρ ∈ Mp for any 1 < p < ∞ and

kρkMp 6 Cp A,

(1.26)

where Cp may depend on p. Proof.

1.3

See [13].

Dyadic decomposition, Besov and Triebel spaces Ingenious ideas are usually beautiful and resonating ideas, the dyadic decomposition is one of them.

We consider the decomposition on Rn , R0 = {ξ : |ξ| < 1},

Rk = {ξ : 2k−1 6 |ξ| < 2k },

k ∈ N.

(1.27)

It is easy to see that {Rk }∞ k=0 is a pairwise disjoint sequence ∞ ∪∞ R , namely, {R } constitutes a decomposition of Rn . k k=0 k=0 k

and Rn = According to this decomposition, we can roughly define the Littlewood-Paley decomposition operators in the following way 4k ∼ F −1 χRk F , k ∈ N ∪ {0},

(1.28)

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where χRk denotes the characteristic function on Rk . The dyadic decomposition operator is a frequency-localized operator, which is one of the most delicate and clever ideas in harmonic analysis. If 4k takes the form as in (1.28), it has an advantage that 4k ’s (k > 0) orthogonalize each other, but the characteristic function in (1.28) has no smoothness and it is hard to use. In order to make multiplier calculations, we need to adopt the smooth version of (1.28). Let ψ : Rn → [0, 1] be a smooth radial cut-off function, say  |ξ| 6 1,  1, ψ(ξ) = smooth, 1 < |ξ| < 2, (1.29)  0, |ξ| > 2. Denote

ϕ(ξ) = ψ(ξ) − ψ(2ξ), and we introduce the function sequence {ϕk }∞ k=0 : ϕk (ξ) = ϕ(2−k ξ), k ∈ N, P∞ ϕ0 (ξ) = 1 − k=1 ϕk (ξ) = ψ(ξ).

(1.30)

(1.31)

Since supp ϕ ⊂ {ξ : 2−1 6 |ξ| 6 2}, we easily see that supp ϕk ⊂ {ξ : 2k−1 6 |ξ| 6 2k+1 }, k ∈ N, supp ϕ0 ⊂ {ξ : |ξ| 6 2}. Define 4k = F −1 ϕk F ,

k ∈ N ∪ {0},

(1.32)

{4k }∞ k=0 is said to be the Littlewood-Paley (or dyadic) decomposition operator. Formally, we find that6 ∞ X k=0

4k = I.

(1.33)

Combining the dyadic decomposition operator with the function spaces7 s `q (Lp ) and Lp (`q ), we can introduce Besov spaces Bp,q and Triebel-Lizorkin s spaces Fp,q , respectively. Let −∞ < s < ∞,

1 6 p, q 6 ∞.

(1.34)

6 Actually, this decomposition works just for any locally integrable function which has some decay at the infinity, and one usually has all the convergence properties of the summation that one needs. In many applications, one can make the a priori assumption that f is Schwartz, in which case the convergence is uniform. However, if the function does not decay, this formula fails. For instance, if f ≡ 1, then all 4k f ’s vanish because 4k 1 = ϕk (0) = ϕ(0) = 0.

P

P q 1/q 7 k(a )k q p := , k(ak )kLp (`q ) := ( k |ak (x)|q )1/q p . k ` (L ) k kak (x)kp

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We define s Bp,q =

s f ∈ S 0 (Rn ) : kf kBp,q <∞ ,

s kf kBp,q :=

X ∞

k=0

2ksq k4k f kqp

1/q

,

(1.35) (1.36)

s Bp,q is said to be the Besov space8 . Assume that

−∞ < s < ∞,

1 6 p < ∞,

1 6 q 6 ∞,

(1.37)

<∞ ,

(1.38)

we define the following s Fp,q

=

0

n

s f ∈ S (R ) : kf kFp,q

X 1/q

∞ ksq

q

, s := 2 |4 f | kf kFp,q k

(1.39)

p

k=0

s Fp,q is said to be the Triebel-Lizorkin space9 . Note that we need replace the `q -norm by the `∞ -norm in the above definition if q = ∞. Besov and Triebel-Lizorkin spaces had been formulated during 1960s– 1980s, which have been widely applied in recent years. Roughly speaking, s is the spatial regularity index; 4k f is the frequency localization of f at the frequency |ξ| ∼ 2k ; the norms of the function sequence {2sk |4k f |} in `q (Lp ) and Lp (`q ) generate Besov and Triebel norms of f , respectively. The dyadic decomposition in the frequency space goes back to the equivalent norm on Lp (Rn ) 0 , kf kp ∼ kf kFp,2

1 < p < ∞.

(1.40)

(1.40) is the well known Littlewood-Paley square function theorem (see Appendix). If there is no explanation, we will assume that conditions (1.34) and s s (1.37) are satisfied for Besov and Triebel spaces Bp,q and Fp,q , respectively. s s s 10 For simplicity, we will use Xp,q to denote Bp,q or Fp,q . s s s s Proposition 1.13. Let Xp,q = Bp,q (Xp,q = Fp,q ). There hold the following inclusions. 8 In the definition of B s , one can also consider the case 0 < p ∧ q < 1, see Triebel p,q [224]. 9F s ∞,q can be found in Triebel [224]. 10 When X appears in the both sides of an inclusion, X should be the same, namely X = B in both sides, or X = F in both sides.

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(1) Let q1 6 q2 . Then s s Xp,q ⊂ Xp,q . 1 2

(1.41)

s+ε s ⊂ Xp,q . Xp,q 1 2

(1.42)

s s s Bp,p∧q ⊂ Fp,q ⊂ Bp,p∨q ,

(1.43)

(2) Let ε > 0 and 1 6 q1 , q2 6 ∞. Then (3) Let p < ∞. Then

where p ∧ q = min(p, q) and p ∨ q = max(p, q). Proof. Since `p ⊂ `p+a , a > 0, we can easily get the result of (1). Let us observe that !1/q2 ∞ X 2skq2 |ak |q2 . sup 2(s+ε)k |ak |. (1.44) k>0

k=0

Taking ak = k4k f kp or ak = |4k f |, we can show that (2) holds with the help of (1). Finally, we prove (3). Denote bk = 2sk 4k f . We divide the proof into the following two cases. Case 1. q 6 p. In view of `q ⊂ `p and Minkowski’s inequality,11 we have kbk k`p (Lp ) 6 kbk kLp (`q ) 6 kbk k`q (Lp ) ,

(1.45)

which implies the result. Case 2. q > p. Analogous to Case 1, we can use Minkowski’s inequality and `p ⊂ `q to get the result, as desired. s s s Proposition 1.14. Let Xp,q be Bp,q or Fp,q . Then s (1) Xp,q is a Banach space; n s (2) S (R ) ⊂ Xp,q ⊂ S 0 (Rn ); s (3) If 1 6 p, q < ∞, then S (Rn ) is dense in Xp,q .

Proof. Since `q (Lp ) and Lp (`q ) are normed spaces, one can conclude that s Xp,q is a normed space. In order to show the result of (1), it suffices to s prove that Xp,q is complete, whose proof will be left to the end. The proof of (2) is separated into the following four steps. s Step 1. We show that S ⊂ Bp,∞ . In fact, for sufficiently large L, M, N ∈ N, s kf kBp,∞ = sup 2sk k4k f kp

k>0

11 Minkowski’s

inequalities P P∞ read i) kP ∞ j=0 fj kp 6 j=0 kfj kp , for any p ∈ [1, ∞]; P ∞ ii) ∞ j=0 kfj kp 6 k j=0 fj kp , for any p ∈ (0, 1) and fj > 0.

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. sup 2sk k(1 + |x|2 )L 4k f k∞ k>0

. sup 2sk k(I − 4)L ϕk F f k1 k>0

. k(1 + | · |2 )M (I − 4)L F f k∞ . pN (F f ).

(1.46)

Since F : S → S is a continuous map, from (1.46) it follows the result, as desired. s s+ε Step 2. We prove that S ⊂ Xp,q . Since S ⊂ Bp,∞ , by Proposition s+ε s s s 1.13, we see that Bp,∞ ⊂ Bp,p∧q ⊂ Bp,q ∩ Fp,q . Thus, we have the desired result. s Step 3. We show that Bp,∞ ⊂ S 0 . For convenience, we denote ϕ−1 ≡ 0. From the construction of ϕk , one has that ϕk ϕk+` ≡ 0 for ` 6= −1, 0, 1. For s any f ∈ Bp,∞ , ψ ∈ S , we can take N ∈ N sufficiently large, |hf, ψi| 6 .

∞ X 1 X

k=0 `=−1 ∞ X 1 X

k=0 `=−1 s . kf kBp,∞

|h4k f, F ϕk+` F −1 ψi| k4k f kp k4k+` ψkp0 ∞ X

k=0

2−sk k4k ψkp0

s . kf kBp,∞ kψkB −s+ε 0 p ,∞

s . kf kBp,∞ pN (F ψ).

(1.47)

For any bounded set B in S , we have pN (F ψ) . 1 for ψ ∈ B. The result follows. s Step 4. Similarly as in Step 2, we can show that Xp,q ⊂ S 0 and the details of the proof are omitted. s Finally, we prove the completeness of Bp,q . In an analogous way we can s get that Fp,q is also complete. Assume that {f` }∞ `=1 is a Cauchy sequence s in Bp,q . By (2) we see that it is also a Cauchy sequence in S 0 . Since S 0 is a complete and locally convex topological linear space, there exists an f ∈ S 0 such that f` → f in the sense of the strongly topology of S 0 . s On the other hand, that {f` }∞ `=1 is a Cauchy sequence in Bp,q implies that p p {4k f` }∞ `=1 is a Cauchy sequence in L . By the completeness of L , there p exists a gk ∈ L such that k4k f` − gk kp → 0,

as ` → ∞.

(1.48)

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Due to 4k f` → 4k f in S 0 as ` → ∞ and Lp ⊂ S 0 , we immediately have gk = 4k f . Hence, (1.48) implies that k4k (f` − f )kp → 0,

as ` → ∞.

s → 0 as ` → ∞. (1.49) and Fatou’s lemma yield kf` − f kBp,q

1.4

(1.49)

s Embeddings on Xp,q

It is known that there is no inclusion between Lp (Rn ) and Lq (Rn ) (p 6= q) and we can not compare k · kp and k · kq . However, if we localize f in a compact subset Ω of Rn in the frequency space, then we have kF −1 χΩ F f kq . kF −1 χΩ F f kp ,

p 6 q.

(1.50)

Combining this fact with the dyadic decomposition operator, we can obtain the inclusions between Besov spaces in a simple way. This is a great advantage of the frequency localized techniques. Let Ω be a compact subset of Rn . Denote SΩ = {f ∈ S : supp F f ⊂ Ω}. Proposition 1.15. Let 1 6 p 6 q 6 ∞. Then we have kf kq . kf kp ,

∀ f ∈ SΩ .

(1.51)

(1.52)

Proof. Let ψ ∈ S satisfy F ψ(ξ) = 1, ∀ ξ ∈ Ω. In view of f ∈ SΩ we see that F f = F f · F ψ. It follows that Z f (x) = C ψ(x − y)f (y)dy. (1.53) Rn

Applied H¨ older’s inequality,

kf k∞ . kψkp0 kf kp . kf kp ,

(1.54)

1−p/q kf kq 6 kf k∞ kf kp/q . kf kp , p

(1.55)

LpΩ = {f ∈ Lp : supp F f ⊂ Ω}.

(1.56)

which implies that for q > p, the result follows.

Denote Using (1.51) and standard approximate techniques, one obtains that Proposition 1.16. Let 1 6 p 6 q 6 ∞. Then kf kq . kf kp ,

∀ f ∈ LpΩ .

(1.57)

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Corollary 1.1. Let 1 6 p 6 q 6 ∞, Bλ = {ξ : |ξ| 6 λ}. Then ∀ f ∈ LpBλ .

kf kq . λn(1/p−1/q) kf kp , Proof.

(1.58)

Due to kf kq = λ−n/q kf (·/λ)kq , n

F (f (·/λ)) = λ (F f )(λ·),

(1.59) (1.60)

we see, for any f ∈ LpBλ , that f (·/λ) ∈ LpB1 . In view of (1.59), (1.60) and Proposition 1.16 we obtain the result, as desired. Let ϕk be defined by (1.31). Since supp ϕk ⊂ B2k+1 , for any 1 6 p1 6 p2 6 ∞, we have, from (1.58), that Bernstein’s inequality holds k4k f kp2 . 2n(1/p1 −1/p2 )k k4k f kp1 .

(1.61)

Assume that s1 and s2 satisfy s1 −

n n = s2 − . p1 p2

(1.62)

By (1.61) and (1.62), we immediately have 2s2 k k4k f kp2 . 2s1 k k4k f kp1 .

(1.63)

Taking the `r -norm in both sides of (1.63), one has that kf kBps2,r . kf kBps1 ,r . 2

1

(1.64)

Thus, we have shown that Theorem 1.1. Let 1 6 p1 6 p2 6 ∞, 1 6 r 6 ∞ satisfy s1 − n/p1 = s2 − n/p2 . Then Bps11 ,r ⊂ Bps22 ,r .

(1.65)

s Considering the inclusions on Fp,q , we have a similar result which is slightly better than that of (1.65).

Theorem 1.2. Let 1 6 p1 < p2 < ∞, 1 6 q, r 6 ∞, −∞ < s2 < s1 < ∞ satisfy s1 − n/p1 = s2 − n/p2 . Then Fps11,q ⊂ Fps22,r . Proof.

(1.66)

By Proposition 1.13, we need to show Fps11,∞ ⊂ Fps22,1 .

(1.67)

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15

We may assume that kf kFps1,∞ = 1. Let us recall an equivalent norm on 1 Lp : Z ∞ kgkpp = p tp−1 |{x : |g(x)| > t}|dt, (1.68) 0

where |E| denotes the measure of the set E. So, ( ) Z A ∞ X p2 p2 −1 ks2 kf kF s2 ∼ t 2 |(4k f )(x)| > t dt x: p2 ,1 0 ( k=0 ) Z ∞ ∞ X ks2 p2 −1 + t 2 |(4k f )(x)| > t dt x: A k=0

=:I + II,

(1.69)

where A 1 is a constant which can be chosen as below. It is easy to see that ∞ X 2ks2 |4k f | . 2K(s2 −s1 ) sup 2ks1 |4k f |. (1.70) k>0

k=K+1

Applying (1.70) we can get the estimate of I (K = −1), Z A p2 −1 ks1 dt I. t x : sup 2 |(4 f )(x)| > ct k k>0

0

.

Z

cA

0

. 1.

p1 −1 ks1 τ x : sup 2 |(4k f )(x)| > τ dτ k>0

(1.71)

Now we estimate II. By Corollary 1.1, k4k f k∞ . 2kn/p1 k4k f kp1 . 2k(n/p1 −s1 ) kf kFps1,∞ . 1

(1.72)

Hence, for K ∈ N ∪ {0}, K X

k=0

2ks2 |4k f | .

K X

2k(s2 −s1 +n/p1 ) . 2Kn/p2 .

(1.73)

k=0

Taking K to be the largest natural number satisfying C2Kn/p2 6 t/2, we have 2K ∼ tp2 /n . It is easy to see that such a K exists if t > A 1. For P ks2 t > A and ∞ |(4k f )(x)| > t, from (1.70) and (1.73) we get k=0 2 C2K(s2 −s1 ) sup 2ks1 |4k f | > k>0

∞ X

k=K+1

2ks2 |4k f | > t/2.

(1.74)

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Combining (1.68) and (1.74), we obtain that Z ∞ p2 −1 ks1 p2 /p1 II . t x : sup 2 |(4k f )(x)| > ct dt k>0 A Z ∞ . τ p1 −1 x : sup 2ks1 |(4k f )(x)| > τ dτ k>0

A0

. 1. The estimates of I and II imply the conclusion.

(1.75)

s Proposition 1.17. Let 1 6 p < ∞, s > n/p and 1 6 q 6 ∞. Let Xp,q be s s Bp,q or Fp,q . Then s 0 Xp,q ⊂ B∞,1 ⊂ L∞ . (1.76)

Proof.

Using the dyadic decomposition and (1.61), we have ∞ ∞ X X k4k uk∞ . 2kn/p k4k ukp kuk∞ 6 .

k=0 ∞ X

k=0

2k(n/p−s)

k=0

!

s s , kukBp,∞ 6 kukXp,q

which is the result, as desired.

1.5

(1.77)

s Differential-difference norm on Xp,q

The earliest version of Besov spaces was defined by an integration with a differential-difference form, not by dyadic decomposition operators. In this section we consider the equivalent differential-difference norm, which is quite convenient when we make nonlinear estimates in the study of nonlinear evolution equations. Denote m X k Mm f (x) = Cm (−1)m−k f (x + kh), (1.78) h k=0

ωpm (t, f )

We have

= sup k Mm h f kp .

(1.79)

|h|6t

Proposition 1.18. Let s > 0, m, N ∈ N and m + N > s, 0 6 N < s; 1 6 p, q 6 ∞. Then n Z ∞ q dt 1/q X N −s m N s . (1.80) kf kBp,q ∼ kf kp + t ωp (t, ∂xj f ) t 0 j=1

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17

Proof. Since ωpm (t, f ) is an increasing function of t > 0, it suffices to show that ! n ∞ q 1/q X X s ∼ kf kp + 2i(s−N ) ωpm (2−i , ∂xNj f ) . (1.81) kf kBp,q j=1

i=−∞

s Let f ∈ Bp,q . We write ρh (ξ) = eihξ − 1. One has that N −1 m k Mm ρh F ∂xNj f kp 6 h ∂xj f kp = kF

In the following we show that12

∞ X

k=0

N kF −1 ρm h ϕk F ∂xj f kp .

m km N k N )2 k4k f kp . kF −1 ρm h ϕk F ∂xj f kp . min(1, |h| 2

(1.82)

(1.83)

Noticing the support set of ϕk (ϕ−1 = 0), N kF −1 ρm h ϕk F ∂xj f kp .

1 X

`=−1

N kF −1 ρm h ϕk+` F 4k ∂xj f kp .

(1.84)

By the definition of ρh we have ρh ∈ Mp . From the fact that Mp is a Banach algebra, it follows that ρm h ∈ Mp . In view of (1.15) in Proposition 1.10, we get ϕk ∈ Mp . Hence, ρm h ϕk ∈ Mp . By (1.84), N N kF −1 ρm h ϕk F ∂xj f kp . k4k ∂xj f kp .

Observing the following identity ρm h ϕk =

ρm h hh, ξim

h ξ , |h| |ξ|

m

(|h||ξ|)m ϕk

(1.85)

(1.86)

and noticing that kρh /hh, ξikMp (Rn ) = k(eiξ − 1)/ξkMp (R) < ∞, it follows that for all k > 1, km kρm |h|m . h ϕk kMp . 2

(1.87)

Obviously, (1.87) also holds for k = 0. Hence, in view of (1.84), we have N km kF −1 ρm |h|m k4k ∂xNj f kp . h ϕk F ∂xj f kp . 2

(1.88)

Applying the technique as in (1.84), we have from (1.15) that k∂xNj 4k f kp . 2N k k4k f kp .

(1.89)

Collecting (1.85), (1.88) and (1.89), we get (1.83). Inserting (1.83) into (1.82), one has that 2i(s−N ) ωpm (2−i , ∂xNj f ) .

∞ X

k=0 12 Notice

2(i−k)(s−N ) (1 ∧ 2(k−i)m )2sk k4k f kp .

that Mh and 4k are different operators.

(1.90)

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18

By (1.90) and Young’s inequality, we obtain n X j=1

∞ X

2

i(s−N )

ωpm (2−i ,

i=−∞

q

∂xNj f )

!1/q

s . . kf kBp,q

(1.91)

s . Thus, the right hand side of (1.80) can be conClearly, kf kp . kf kBp,q trolled by its left hand side. −k Next, we show the remaining part of (1.80). Let ρjk (ξ) = ei2 ξj − 1. It suffices to prove that

k4k f kp . 2−N k

n X j=1

N kF −1 ρm jk F ∂xj f kp ,

k > 1.

(1.92)

In fact, if (1.92) holds, in view of the definition of ωpm (t, f ), we have N m −k kF −1 ρm , ∂xNj f ), jk F ∂xj f kp . ωp (2

k > 1.

(1.93)

It follows from (1.93) and the Besov norm that (ϕ0 ∈ Mp ), s kf kBp,q

n ∞ q X X . kf kp + 2k(s−N ) ωpm (2−k , ∂xNj f ) j=1

k=1

!1/q

,

(1.94)

which implies that the left hand side of (1.80) can be controlled by its right hand side. Finally, we prove (1.92). We need the following lemma Lemma 1.1. There exist smooth χj (j = 1, ..., n) satisfying n X

χj (ξ) = 1, ∀ ξ ∈ {ξ : 1/2 6 |ξ| 6 2};

(1.95)

√ supp χj ⊂ {ξ = (ξ1 , ..., ξn ) : |ξj | > 1/3 n}.

(1.96)

j=1

Proof. We can choose that κ ∈ S (R), ζ ∈ S (Rn−1 ) with supp κ = √ {ξ ∈ R : |ξ| > 1/3 n}, supp ζ = {ξ ∈ Rn−1 : |ξ| 6 3} and with positive values in the interiors of supp κ and supp ζ, respectively. We write ξ j = (ξ1 , ..., ξj−1 , ξj+1 , ..., ξn ) and ( j Pn j Pnκ(ξj )ζ(ξ ) j , j=1 κ(ξj )ζ(ξ ) 6= 0, κ(ξ j )ζ(ξ ) j=1 χj (ξ) = (1.97) Pn j 0, j=1 κ(ξj )ζ(ξ ) = 0. It is easy to see that χj satisfies (1.95) and (1.96).

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s 1.6. Homogeneous space X˙ p,q

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19

Now we can finish the proof of (1.92). By Lemma 1.1, for any k > 1, k4k f kp 6

n X j=1

−k N kF −1 ρ−m ·)ξj−N F (F −1 ρm jk F ∂xj f )kp . jk ϕk χj (2

(1.98)

By Proposition 1.11, (eiξj − 1)−m ϕχj ξj−N ∈ Mp ,

(1.99)

where ϕ is as in (1.30). It follows from Propositions 1.10 and (1.98) that (1.92) holds. As a consequence of Proposition 1.18, we have Corollary 1.2. Let s > 0, s 6∈ N. we denote by [s] the integer part of s, 1 6 p, q 6 ∞. Then ! Z ∞ n q dt 1/q X [s]−s [s] s kf kBp,q ∼ kf kp + t sup k Mh ∂xj f kp , (1.100) t |h|6t 0 j=1

where Mh :=M1h .

It is very convenient to use (1.100) when we make nonlinear mapping estimates. 1.6

˙s Homogeneous space X p,q

Let ϕ be as in (1.30). Analogous to (1.31), we write ϕk (ξ) = ϕ(2−k ξ), It is easy to see that

X k∈Z

ϕk (ξ) = 1,

k ∈ Z.

ξ ∈ Rn \ {0}.

(1.101)

(1.102)

According to (1.32), we introduce the homogeneous dyadic decomposition operators 4k = F −1 ϕk F ,

k ∈ Z.

(1.103)

s Similar to Xp,q , using {4k }k∈Z and function spaces `q (Lp ) and Lp (`q ), one s can define X˙ p,q . Notice that in (1.101)–(1.103), there is no restriction on ξ = 0 and so, one needs to modify the Schwartz space S and it dual S 0 . Denote

S˙ (Rn ) = {f ∈ S (Rn ) : (Dα fb)(0) = 0, ∀α}.

(1.104)

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As a subspace of S , S˙ := S˙ (Rn ) is equipped with the same topology as S . We denote by S˙ 0 := S˙ 0 (Rn ) the dual space of S˙ . We now introduce s the homogeneous space X˙ p,q . Let −∞ < s < ∞, 1 6 p, q 6 ∞. Denote

(1.105)

0 n ˙ f ∈ S (R ) : kf kB˙ s < ∞ ,

s B˙ p,q =

p,q

kf kB˙ s := p,q

X ∞

k=−∞

2ksq k4k f kqp

1/q

,

(1.106) (1.107)

s B˙ p,q is said to be a homogeneous Besov space. Let

−∞ < s < ∞, Define s F˙p,q =

1 6 p < ∞,

1 6 q 6 ∞.

0 n ˙ f ∈ S (R ) : kf kF˙ s < ∞ , p,q

∞ 1/q

X ksq

q

, 2 |4 f | kf kF˙ s := k

p,q k=−∞

(1.108)

(1.109) (1.110)

p

s F˙p,q is said to be a homogeneous Triebel-Lizorkin space. s s s If there is no confusion, we will write X˙ p,q to stand for B˙ p,q or F˙p,q . Using the dilation, one has that

= 2`(n−s/p) kf kX˙ p,q kf (2` ·)kX˙ p,q s , s

(1.111)

s from which one can easily understand why we call X˙ p,q as a homogeneous s s space. X˙ p,q and Xp,q are quite similar. In the following we only state some s results on X˙ p,q , whose proofs follow an analogous way as the nonhomogeneous case. If there is no explanation, we will always assume that conditions s s and F˙p,q , respectively. (1.105) and (1.108) are satisfied for the spaces B˙ p,q s s s s Proposition 1.19. Let X˙ p,q = B˙ p,q (or X˙ p,q = F˙p,q ). Then

(1) (2) (3) (4) (5)

s X˙ p,q is a Banach space; s ˙ S ⊂ X˙ p,q ⊂ S˙ 0 ; s If 1 6 p, q < ∞, then S˙ (Rn ) is dense in X˙ p,q ; s s ˙ ˙ If q1 6 q2 , then Xp,q1 ⊂ Xp,q2 ; s s s Let 1 6 p < ∞. Then B˙ p,p∧q ⊂ F˙p,q ⊂ B˙ p,p∨q .

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Theorem 1.3. 1 6 r, q 6 ∞

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21

Let −∞ < s2 < s1 < ∞, s1 − n/p1 = s2 − n/p2 and B˙ ps11 ,r ⊂ B˙ ps22 ,r ,

F˙ps11,r ⊂ F˙ps22,q .

In the sequel, we will frequently use Theorem 1.3, where the first inclusion can be shown in the same way as in Besov space, the second one is slightly different from the Triebel-Lizorkin space, see Appendix. Theorem 1.4. Let s > 0, m, N ∈ N, N < s, N + m > s and 1 6 p, q 6 ∞. Then we have 1/q n Z ∞ X dt . kf kB˙ s ∼ tq(N −s) ωpm (t, ∂xNj f )q p,q t 0 j=1

In particular, if s > 0 and s ∈ / N, then !1/q Z n ∞ X dt kf kB˙ s ∼ tq([s]−s) sup k Mh ∂x[s]j f kqp . p,q t |h|6t 0 j=1

The following result is a straightforward consequence of Proposition 1.18 and Theorem 1.4, which indicates the relation between homogeneous and nonhomogeneous spaces. Proposition 1.20. Let s > 0 and 1 6 p, q 6 ∞. Then we have B s = Lp ∩ B˙ s . p,q

p,q

The following is an interpolation inequality in Besov spaces, which are very useful in nonlinear estimates, see [79; 89]. older’s inequality). Let 1 6 pi , qi 6 Proposition 1.21 (Convexity H¨ PN PN ∞, 0 6 θi 6 1, σi , σ ∈ R (i = 1, . . . , N ), i=1 θi = 1, σ = i=1 θi σi , PN PN N σi σ ˙ ˙ 1/p = i=1 θi /pi , 1/q = i=1 θi /qi . Then ∩i=1 Bpi ,qi ⊂ Bp,q and for any ˙ σi v ∈ ∩N i=1 Bpi ,qi , kvkB˙ σ 6 p,q

N Y

i=1

kvkθB˙iσi . pi ,qi

σ σ This estimate also holds if one substitutes B˙ p,q by F˙p,q (p, pi 6= ∞).

Proof. Applying H¨ older’s inequality on Lp and `q respectively, one can easily deduces the result: !1/q X kvkB˙ p,q = 2σkq k4k vkqp σ k∈Z

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22

6

X

2

σkq

=

N Y

i=1

k∈Z

N XY

(2

6

N Y

i=1 σ The case F˙p,q is similar.

k4k vkθpiiq

kσi θi

k∈Z i=1

1.7

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!1/q

k4k vkθpii )q

!1/q

kvkθB˙iσi . pi ,qi

˙ s) Bessel (Riesz) potential spaces Hps (H p

Recall that Js = (I − ∆)s/2 and Is = (−∆)s/2 are said to be the Bessel and the Riesz potentials, respectively. Assume that 1 < p < ∞, Denote

−∞ < s < ∞.

o n Hps = f ∈ S 0 : kf kHps := kJs f kp < ∞ ; o n H˙ ps = f ∈ S˙ 0 : kf kH˙ ps := kIs f kp < ∞ .

(1.112)

(1.113) (1.114)

Hps and H˙ ps are said to be Bessel and Riesz potential spaces, respectively. In the following we give the equivalent norms on Hps and H˙ ps : Theorem 1.5 (Littlewood-Paley square function theorem). Let and p satisfy (1.112). Then s Hps = Fp,2 ,

s H˙ ps = F˙p,2

s

(1.115)

with equivalent norms. Theorem 1.5 can be found in Stein [202] and Triebel [224]. Theorem 1.6. Let −∞ < s2 6 s1 < ∞ and 1 < p1 6 p2 < ∞ with s1 − n/p1 = s2 − n/p2 . Then we have Hps11 ⊂ Hps22 ,

Proof.

H˙ ps11 ⊂ H˙ ps22 .

It is a consequence of Theorems 1.5, 1.2 and 1.3.

Proposition 1.22. Let s and p satisfy (1.112). We have

(1.116)

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1.7. Bessel (Riesz) potential spaces Hps (H˙ ps )

(1) (2) (3) (4) (5)

23

Hps is a Banach space; S ⊂ Hps ⊂ S 0 and S (Rn ) is dense in Hps ; s s s s Bp,p ⊂ Hps ⊂ Bp,2 (1 0); Hps ⊂ L∞ (s > n/p).

Remark 1.1. The results in (1)–(3) of Proposition 1.22 also hold for Riesz potential spaces. The proof of Proposition 1.22 follows from Theorem 1.5 s and the corresponding results of Fp,2 . Proposition 1.23. Let s > 0 and 1 < p < ∞. We have Hps = Lp ∩ H˙ ps . Moreover, if s is an integer, then X kf kH˙ s ∼ kDα f kp . p

|α|=s

In the following we establish a modified H¨older inequality, as far as the author have seen, Pecher [190] gave the first result on the modified H¨older inequality. The next two results were obtained in Wang [234] by following Pecher [190]. older inequality I). Let 1 6 p < ∞ Proposition 1.24 (Modified H¨ and 1 < pi < ∞. Assume that αi is a multi-index, |αi | 6 si , ρi > 0, and si − |αi | 1 − , i = 0, 1, ..., N + 1. (1.117) a i = ρi pi n P +1 If ai > 0 and N i=0 ai = 1/p, then we have +1 N +1

NY

Y

|Dαi ui |ρi 6 C kui kρH˙isi .

p

i=0

(1.118)

pi

i=0

Substituted H˙ psii by Hpsii in (1.118), the conclusion also holds. Proof. Let qi = 1/ai . Since 1/p = inequality, that

PN +1 i=0

1/qi , we have, from H¨older’s

+1 N +1 N +1

NY

Y Y

kui kρ˙i|αi | . |Dαi ui |ρi 6 kDαi ui kρρii qi .

i=0

p

i=0

i=0

Hρi qi

(1.119)

Noticing that 1/ρi qi − |αi |/n = 1/pi − si /n, and using Theorem 1.6, we get |α | H˙ psii ⊂ H˙ ρi qii . By (1.119), we have the desired result.

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Proposition 1.25 (Modified H¨ older’s inequality II). Let 1 6 p < ∞, 1 < pi < ∞, |αi | 6 si , ρi > 0, and ai be as in (1.117). We have P (1) If each ai 6= 0, and ai >0 ai = 1/p, then +1 N +1

NY

Y

|Dαi ui |ρi 6 C kui kρHisi . (1.120)

(2) If

P

i=0

ai >0

ai < 1/p 6

p

PN

i=1

pi

i=0

ρi /pi , then (1.120) also holds.

Proof. First, we prove (1). We may assume a0 , ..., aK > 0 and aK+1 , ..., aN +1 < 0. By H¨ older’s inequality, +1 K N +1

NY

Y Y

|Dαi ui |ρi . kDαi ui kρρii qi kDαi ui kρ∞i . (1.121)

i=0

p

i=0

i=K+1

By Proposition 1.24,

K Y

i=0

kDαi ui kρρii qi .

K Y

i=0

kui kρH˙isi .

(1.122)

pi

Since ai < 0 (i = K + 1, ..., N + 1), in view of (5) of Proposition 1.22, we have N +1 N +1 N +1 Y Y Y ρi αi ρi αi kD ui k∞ . kD ui k si −|αi | . kui kρHisi . (1.123) i=K+1

Hpi

i=K+1

i=K+1

pi

Collecting (1.121)–(1.123), we obtain the result. Next, we prove (2). We can assume that a0 , ..., aK > 0, aK+1 = ... = aJ = 0 and aJ+1 , ..., aN +1 < 0. P Case 1. 1/p 6 K i=0 ρi /pi . Take qK+1 , ..., qN 1 (for any i > K and qi > pi /ρi ) satisfying K N +1 K X X 1 X ρi 1 ai + < 6 . (1.124) qi p p i=0 i=0 i i=K+1

Thus, we can choose appropriate qi (i = 0, ..., K) verifying pi /ρi 6 qi 6 1/ai PN +1 and 1/p = i=0 1/qi . By Proposition 1.25, we obtain +1 N +1

NY

Y

αi ρi |D u | 6 C kui kρi|αi | . (1.125)

i i=0

p

i=0

Hρi qi

|α |

Since 1/qi ρi > 1/pi − (s − |αi |)/n and pi 6 qi ρi , we see that Hpsii ⊂ Hρi qii . From (1.125) we have the result. PK PN +1 Case 2. i=0 ρi /pi < 1/p 6 i=0 ρi /pi . Let qi = pi /ρi for i = 0, ..., K. For i = K + 1, ..., N + 1, we can find a qi ∈ [pi /ρi , ∞) satisfying PN +1 1/p = i=0 1/qi . Analogous to the above discussions, we can obtain our result.

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1.8. Fractional Gagliardo-Nirenberg inequalities

1.8

25

Fractional Gagliardo-Nirenberg inequalities

We consider the Gagliardo-Nirenberg (GN) inequality in Besov and Triebel spaces. Using the dyadic decomposition, we obtain a very simple proof for the GN inequality. On the other hand, the GN inequality with fractional order derivatives is of interest for its own sake and has an independent significance. The results in this section are obtained in [99], earlier results in this topic can be found in [23; 162; 183; 233; 242]. 1.8.1

s GN inequality in B˙ p,q

Lemma 1.2. Let 1 6 p, p0 , p1 , q, q0 , q1 6 ∞, s, s0 , s1 ∈ R, 0 6 θ 6 1. Suppose that the following conditions hold: n n n − s = (1 − θ) − s0 + θ − s1 , (1.126) p p0 p1 s 6 (1 − θ)s0 + θs1 , 1 1−θ θ 6 + . q q0 q1

(1.127)

(1.128)

Then the fractional GN inequality of the following type kukB˙ s . kuk1−θ kukθB˙ s1 ˙ s0 B p,q

p0 ,q0

(1.129)

p1 ,q1

holds for all u ∈ B˙ ps00 ,q0 ∩ B˙ ps11 ,q1 . Proof. First, we consider the case 1/q 6 (1 − θ)/q0 + θ/q1 . By (1.127), we have 1 1−θ θ s s0 s1 − − = − (1 − θ) − θ := −η 6 0. p p0 p1 n n n

(1.130)

Take p∗ and s∗ satisfying 1 1 = + η, p∗ p

s∗ = s + nη.

Applying the convexity H¨ older inequality, we have kf kB˙ s∗∗ 6 kf k1−θ kf kθB˙ s1 ˙ s0 B p ,q

p0 ,q0

p1 ,q1

.

∗ s Using the inclusion B˙ ps∗ ,q ⊂ B˙ p,q , we get the conclusion.

(1.131)

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s The result below clarify how the third indices q in B˙ p,q contribute the validity of the GN inequalities.

Theorem 1.7. Let 1 6 q < ∞, 1 6 p, p0 , p1 6 ∞, 0 < θ < 1, s, s0 , s1 ∈ R. Then the fractional GN inequality of the following type θ kukB˙ s . kuk1−θ ˙ s1 ˙ s0 kukB B p,q

p0 ,∞

(1.132)

p1 ,∞

holds if and only if n n n − s = (1 − θ) − s0 + θ − s1 , p p0 p1 n n s0 − 6= s1 − , p0 p1

(1.133) (1.134)

s 6 (1 − θ)s0 + θs1 ,

(1.135)

p0 = p1 if s = (1 − θ)s0 + θs1 .

(1.136)

Proof. (Sufficiency) We can assume that s0 = 0 and the case s0 = 6 0 can be shown by a similar way. Step 1. We consider the case p > p0 ∨ p1 . By definition13 ,  1/q X N sq k4N ukqp  . (1.137) kukB˙ s =  p,q

N dyadic

From (1.135), it follows that n n n n − + s1 − s = (1 − θ) s + − . θ p p1 p0 p Since 0 < θ < 1, (1.134) implies that np − pn1 + s1 − s s + Case 1. We consider the case n n n n s1 − s + − > 0, s + − > 0. p p1 p0 p

(1.138) n p0

−

n p

> 0.

(1.139)

s s Using the inclusion B˙ p,r ⊂ B˙ p,r for any r1 6 r2 , it suffices to consider the 1 2 −1 case q < 1/2, q ∈ N. For brevity, we write K := q −1 . X q2 s kukB˙ p,q 6 (N1s ...NK k4N1 ukp ...k4NK ukp ) s N1 >...>NK

q(1−q)

s × (N1s . . . NK k4N1 ukp . . . k4NK ukp )

13 Here

.

(1.140)

4N is different from the notation as in Section 1.6, which is identical with 4logN

as in (1.103).

2

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27

In view of Bernstein’s inequality, n

k4N ukp 6 N p0

−n p

n

k4N ukp0 , k4N ukp 6 N p1

−n p

k4N ukp1 .

(1.141)

We can choose a ∈ (0, 1], k > 1 satisfying θK = k − 1 + a. Hence, k4N1 ukp ...k4NK ukp

= (k4N1 ukp ...k4Nk−1 ukp k4Nk ukap )(k4Nk uk1−a k4Nk+1 ukp . . . k4NK ukp ) p (1−a)( pn − n p)

. Nk

0

n

× N1p1

−n p

n

−n p

n

p0 Nk+1 n

−n

. . . NKp0 a( pn − n p)

p1 p . . . Nk−1 Nk

1

−n p

k4Nk uk1−a p0 k4Nk+1 ukp0 . . . k4NK ukp0

k4N1 ukp1 . . . k4Nk−1 ukp1 k4Nk ukap1 . (1.142)

Inserting (1.142) into (1.140), we have X 2 s (N1s . . . NK k4N1 ukp . . . k4NK ukp )q kukB˙ s . p,q

N1 >...>NK

q(1−q)θK

× Λ(N1 , ..., NK )kukB˙ s1

p1 ,∞

(1−θ)Kq(1−q)

kukB˙ 0

,

(1.143)

p0 ,∞

where

− n + n −s1 +s n −n a(− n + n −s1 +s) p + p1 −s1 +s Λ(N1 , . . . NK ) = N1 p p1 . . . Nk−1 Nk p p1 n n (1−a)(− n −n − n + n +s q(1−q) p + p0 +s) p + p0 +s × Nk Nk+1 . . . NK p p 0 .

(1.144)

By (1.143), we have . kukB˙ p,q s

X

Λ(N1 , . . . NK )

N1 >...>NK (1−q)θ

× kukB˙ s1

p1 ,∞

K X i=1

(1−θ)(1−q)

kukB˙ 0

(Nis k∆Ni ukp )q

.

(1.145) (1.146)

p0 ,∞

So, it suffices to prove X

N1 >...>NK

Λ(N1 , . . . NK )

K X i=1

(Nis k∆Ni ukp )q . kukqB˙ s .

(1.147)

p,q

In fact, (1.144)–(1.147) imply the result. Finally, we prove (1.147). Applying the condition (1.139), we have X Λ(N1 , . . . NK )(Nks k∆Nk ukp )q N1 >...>NK

.

X (k−1)(s−s1 + pn − np ) (K−k+1−a)(s+ pn − np )+a(s−s1 + pn − np ) q(1−q) 1 0 1 Nk−1 Nk

Nk−1 >Nk

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.

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× Nksq k∆Nk ukqp n n X Nk−1 (k−1)(s−s1 + p1 − p )q(1−q) sq Nk k∆Nk ukqp Nk

Nk−1 >Nk

. kukqB˙ s .

(1.148)

p,q

Case 2. We consider the case n n n n < 0, s + − < 0. (1.149) s1 − s + − p p1 p0 p P P Substituting the summation N1 >...>NK by N1 6...6NK in (1.140) and repeating the procedure as in Case 1, we can get the result, as desired. Up to now, we have shown the results for the following two cases: (i) s = (1 − θ)s0 + θs1 and p0 = p1 ; (ii) s < (1 − θ)s0 + θs1 and p > p0 ∨ p1 . Step 2. We consider the case p < p0 ∨ p1 and s < (1 − θ)s0 + θs1 . Due to θ ∈ (0, 1) and 1/p 6 (1 − θ)/p0 + θ/p1 , we see that p0 6= p1 and p0 ∧ p1 < p < p0 ∨ p1 . Let 0 < ε 1. In view of the result as in Step 1, we see that 1/2

1/2

kf kB˙ s . kf kB˙ s−ε kf kB˙ s+ε . p,q

p,∞

(1.150)

p,∞

Since s0 − n/p0 6= s1 − n/p1 , we can assume that s0 − n/p0 < s1 − n/p1 . It follows that 1/p − s/n ∈ (1/p0 − s0 /n, 1/p1 − s1 /n). Hence, for sufficiently small ε > 0, 1 s0 1 s1 1 s±ε − ∈ − , − . p n p0 n p1 n

It follows that there exist θ± ∈ (0, 1) satisfying 1 s±ε 1 s0 1 s1 − = (1 − θ± ) − + θ± − . p n p0 n p1 n

Due to limε→0 θ± = θ, we see that for sufficiently small ε > 0, s ± ε 6 (1 − θ± )s0 + θ± s1 . Therefore, by Lemma 1.2, we have 1−θ

θ

− s−ε . kf k s0 kf kB˙−s1 kf kB˙ p,∞ ˙ B p0 ,∞

1−θ

p1 ,∞

θ

+ s+ε . kf k s0 kf kB˙ p,∞ kf kB˙+s1 ˙ B p0 ,∞

p1 ,∞

,

(1.151)

.

(1.152)

We easily see that θ = (θ+ + θ− )/2. Inserting (1.151) and (1.152) into (1.150), we have the result, as desired. We omit the proof of the necessity, one can refer to [99] for details.

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29

For the most general case, we have the following Theorem 1.8. Let 0 < p, p0 , p1 , q, q0 , q1 6 ∞, s, s0 , s1 ∈ R, 0 6 θ 6 1. Then the fractional GN inequality of the following type kukB˙ p,q . kuk1−θ kukθB˙ s1 s ˙ s0 B p0 ,q0

p1 ,q1

holds for all u ∈ B˙ ps00 ,q0 ∩ B˙ ps11 ,q1 if and only if n n n − s = (1 − θ) − s0 + θ − s1 , p p0 p1

(1.153)

(1.154)

s 6 (1 − θ)s0 + θs1 , (1.155) 1 1−θ θ 6 + , if p0 6= p1 and s = (1 − θ)s0 + θs1 , (1.156) q q0 q1 1 1−θ θ s0 6= s1 or 6 + , if p0 = p1 and s = (1 − θ)s0 + θs1 , q q0 q1 (1.157) n n 1 1−θ θ s0 − 6= s − or 6 + , if s < (1 − θ)s0 + θs1 . (1.158) p0 p q q0 q1

Proof. 1.8.2

See [99].

s GN inequality in F˙p,q

s In homogeneous Triebel-Lizorkin spaces F˙p,q , we have the following (cf. [99]):

Theorem 1.9. Let 1 6 p, pi , q < ∞, s, s0 , s1 ∈ R, 0 < θ < 1. Then the fractional GN inequality of the following type kukF˙ p,q . kuk1−θ kukθF˙ ps1,∞ s s F˙ 0 p0 ,∞

(1.159)

1

holds if and only if n n n − s = (1 − θ) − s0 + θ − s1 , p p0 p1 s 6 (1 − θ)s0 + θs1 ,

s0 6= s1 if s = (1 − θ)s0 + θs1 .

(1.160) (1.161) (1.162)

Proof. We only prove the sufficiency, which is a reformulation of Oru’s [183] (see also [23]) and the proof of the necessity can be found in [99]. First, we consider the case s < (1 − θ)s0 + θs1 . We can take sufficiently small ε > 0 satisfying s 6 (1 − θ)s∗0 + θs∗1 ,

s∗0 := s0 − ε, s∗1 := s1 − ε.

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s Fourier multiplier, function space Xp,q

30

Since ε 1, we can assume that 1 ε 1 := − > 0, ∗ p0 p0 n Hence,

n n ∗ ∗ − s + θ − s 0 1 , p∗0 p∗1

(1.163)

θ s s∗ s∗ 1 1−θ − ∗ − ∗ = − (1 − θ) 0 − θ 1 := −η 6 0. p p0 p1 n n n

(1.164)

n − s = (1 − θ) p which implies that

1 1 ε := − > 0. ∗ p1 p1 n

Putting 1 1 = + η, p∗ p

s∗ = s + nη,

(1.165)

we see that 1 1−θ θ = + ∗, ∗ ∗ p p0 p1

s∗ = (1 − θ)s∗0 + θs∗1 .

(1.166)

Using H¨ older’s inequality, in an analogous way as in Besov spaces, we have θ kf kF˙ s∗∗ . kf k1−θ . s∗ kf k ˙ s∗ 1 0 F˙ p∗ ,q

p ,q

0

Fp∗ ,q 1

Recalling the inclusions ∗

s F˙ ps00,∞ ⊂ Fp∗0,q , 0

∗

s F˙ps11,∞ ⊂ Fp∗1,q 1

we immediately get the conclusion. Next, we consider the case s = (1 − θ)s0 + θs1 and s0 6= s1 . In this case we easily see that 1/p = (1 − θ)/p0 + θ/p1 . The result follows from Lemma C.1. The following is the GN inequality with fractional derivatives. Corollary 1.3. Let 1 < p, p0 , p1 < ∞, s, s1 ∈ R, 0 6 θ 6 1. Then the fractional GN inequality of the following type θ kukH˙ s . kuk1−θ ˙ s1 Lp0 kukH p

holds if and only if n n − s = (1 − θ) + θ p p0

(1.167)

p1

n − s1 , p1

s 6 θs1 .

(1.168)

As the end of this chapter, we state some recent results on the generalizations of Besov’s and Triebel-Lizorkin’s spaces.

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Remark 1.2. (1) In [5], the Qα space was introduced for which the norm is given by Z Z 1/2 |f (x) − f (y)|2 dxdy < ∞, kf kQα (Rn ) = sup n+2α I I I |x − y|

where I ranges over all cubes in Rn . Xiao [253] considered the well posedness of the Navier-Stokes equation in Q type spaces. (2) Yang and Yuan [254] gave a unified way to handle the Qα and Triebel-Lizorkin spaces. Indeed, for any p ∈ (1, ∞), q ∈ (1, ∞], s,τ τ, s ∈ R, they introduced the space F˙p,q in the following way   p/q 1/p     Z ∞  1   X sj q s,τ kf kF˙ p,q = sup (2 |4j f (x)|)  dx ,  τ  P dyadic |P |    )  P j=− logl(P  2

where l(P ) denotes the side length of the cube P . They showed α,1/2−α/n s,τ that Qα = F˙2,2 . In a similar way as F˙p,q , Yang and Yuan s,τ [255] introduced Besov type space B˙ p,q . Some recent progress on s,τ s,τ Bp,q and Fp,q spaces can be found in [257]. s (3) Function space like F˙∞,q are not considered in this chapter. It is of importance in the case s = 0 and q = 2, which is equivalent to BMO space introduced by F. John and L. Nirenberg in 1961. An interesting generalization of BMO type on homogeneous-type spaces or on measurable subsets of such spaces was given by X. T. Duong and L. X. Yan [69]. Their spaces are defined by certain maximal functions associated with generalized approximation of the identity, which coincide with the classical BMO spaces of F. John and L. Nirenberg in some special cases.

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Chapter 2

Navier-Stokes equation

To be without some of the things you want is an indispensable part of happiness. —— B. Russell

We consider the initial value problem for a nonlinear evolution equation, which is said to be locally well-posed in Lp if for any initial value u|t=0 ∈ Lp , there exists a unique solution in C(0, T ; Lp) for some T > 0, moreover, the solution map u|t=0 → u is at least a continuous mapping from Lp to C(0, T ; Lp ). Furthermore, if T = ∞, then we say that it is globally wellposed. The Navier-Stokes (NS) equation is a fundamental equation in the theory of fluid mechanics. However, the existence for its global smooth solutions in three spatial dimensions has been open for many years. In two spatial dimensions, the global well-posedness for the NS equation was established by Ladyzhenskaya [154], and Kato [119] gave an alternate proof based on the semi-group method and he also obtained the local well posedness of solutions in Ln for n > 3. We will use harmonic analysis techniques to study the NS equation. First, we consider the Lr → Lp estimate for the heat semi-group H(t) = et∆ and establish the time-space estimates in mixed Lebesgue spaces Lqt Lpx for the solutions of the linear heat equation, those estimates are similar to the Strichartz estimates for the dispersive equations. Next, applying the timespace estimates we show the local well-posedness in Ln for the NS equation in a very simple way. In 2D case, in view of the a priori estimate in L2 , we immediately obtain the global well-posedness of the NS equation in L2 . The method here is also useful for the other nonlinear parabolic equations, such as the semi-linear heat equation, the Ginzburg-Landau equation and so on. 33

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Navier-Stokes equation

Finally, the regularity behavior for the NS equation will be investigated by applying the frequency-uniform decomposition techniques.

2.1 2.1.1

Introduction Model, energy structure

We study the Cauchy problem for the (incompressible) Navier-Stokes (NS) equation ut − ∆u + (u · ∇)u + ∇p = 0, div u = 0, u(0, x) = u0 (x), (2.1) Pn 2 where ∆ = i=1 ∂xi , ∇ = (∂x1 , ..., ∂xn ), div u = ∂x1 u1 + ... + ∂xn un , u = (u1 , ..., un ) and p are real-valued unknown functions of (t, x) ∈ R+ ×Rn , u0 = (u10 , ..., un0 ) denotes the initial value of u at t = 0. In view of its own structure, one can easily deduce that the smooth solution of (2.1) satisfies the following conservation law: Z t 1 1 2 ku(t)k2 + k∇u(s)k22 ds = ku0 k22 , (2.2) 2 2 0 P P n n where kuk22 := i=1 kui k22 , k∇uk22 := i,j=1 k∂xj ui k22 for u = (u1 , ..., un ). Indeed, multiplying u in (2.1) and then integrating by part, we have (2.2). According to (2.2), it is natural to ask what happens if u0 ∈ L2 (Rn ). Using the compactness method, we can obtain the existence of the weak solutions if u0 ∈ L2 (Rn ), however, the uniqueness, the persistence of the regularity of the solution and the continuity of the solution map are hard to obtain, see [159]. 2.1.2

Equivalent form of NS

Let (u, p) be a smooth solution of the NS equation. Taking the divergence of the first equation in (2.1) and noticing that div u = 0, we immediately obtain that ∆p + div[(u · ∇)u] = 0. (2.3) −1 It follows that ∇p = (−∆) ∇div[(u · ∇)u]. For convenience, we write P = I + (−∆)−1 ∇div. (2.4) Solving p from (2.3) and inserting it into (2.1), we get ut − ∆u + P [(u · ∇)u] = 0, u(0, x) = u0 (x). (2.5) From (2.5), we see that the NS equation belongs to a nonlinear parabolic equation.

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2.1.3

35

Critical spaces

If u solves Eq. (2.5), so does uλ = λu(λ2 t, λx) (∀ λ > 0) with initial data uλ (0, x) = λu0 (λx). We say that X := X(Rn ) is a critical space for the NS equation, if the norm of uλ (0, x) in X is invariant for all λ > 0. Taking notice of kuλ (0, ·)kLr (Rn ) = λku0 (λ ·)kLr (Rn ) = λ1−n/r ku0 kLr (Rn ) ,

(2.6) r

we see that r = n is the unique index so that the norm of uλ (0, x) in L (Rn ) is invariant for all λ > 0. So, Ln is a critical space of the NS equation. Using the same way as in the above, we can verify that H˙ n/2−1 (Rn ) is also the unique space so that the norm of uλ (0, x) in H˙ s (Rn ) is invariant. H˙ n/2−1 is another critical space in all H˙ s . On the other hand, noticing that H˙ n/2−1 ⊂ Ln is a sharp inclusion, one can easily understand H˙ n/2−1 to be a critical space. −1 −1 We can calculate that kuλ (0, ·)kB˙ ∞,∞ ∼ ku0 kB˙ ∞,∞ , from this point of −1 ˙ view, we see that B is the largest critical space. ∞,∞

2.2

Time-space estimates for the heat semi-group

We consider the Cauchy problem for the heat equation: ut − ∆u = f,

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

Taking the Fourier transform, we get u bt + |ξ|2 u b = fb,

u b(0) = u b0 .

Solving the ordinary differential equation, we obtain that Z t u(t) = H(t)u0 + H(t − τ )f (τ, ·)dτ,

(2.7)

0

2

where H(t) = et∆ = F −1 e−t|ξ| F . 2.2.1

Lr → Lp estimate for the heat semi-group 2

In H(t), e−t|ξ| as an exponential decay function, which corresponds to the dissipation for H(t), determines that the semi-group H(t) has very good properties. Let us start with an Lr → Lp estimate of H(t), which can be found in [193]. Proposition 2.1. Let 1 6 r 6 p 6 ∞. Then k

n

1

1

k∇k H(t)f kp . t− 2 − 2 ( r − p ) kf kr , k = 0, 1, t > 0.

(2.8)

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Navier-Stokes equation

Proof. First, we show that H(t) : Lr → Lr . Using Young’s inequality, one has that 2

kH(t)f kr 6 kF −1 e−t|ξ| k1 kf kr . kf kr .

(2.9)

Next, we consider kH(t)kLr →L∞ . Again, by Young’s inequality, 2

n

kH(t)f k∞ 6 kF −1 e−t|ξ| kr0 kf kr . t− 2r kf kr .

(2.10)

Now, for any p > r, using H¨older’s inequality, (2.9) and (2.10), we have n

1

1

1−r/p kH(t)f kp 6 kH(t)f k∞ kH(t)f kr/p . t− 2 ( r − p ) kf kr . r

We estimate k∂x1 H(t)f kr . By Young’s inequality, 2

1

k∂x1 H(t)f kr 6 kF −1 (ξ1 e−t|ξ| )k1 kf kr . t− 2 kf kr .

(2.11)

(2.12)

So, 1

k∇H(t)f kr . t− 2 kf kr .

(2.13)

Combining (2.11) with (2.13), we immediately have n

1

1

k∇H(t)f kp = kH(t/2)∇H(t/2)f kp . t− 2 ( r − p ) k∇H(t/2)f kr 1

n

1

1

. t− 2 − 2 ( r − p ) kf kr .

This is the result, as desired.

(2.14)

As a generalization, we can show that for any s > 0 and 1 < r 6 p 6 ∞, s

n

1

1

k(−∆)s/2 H(t)f kp . t− 2 − 2 ( r − p ) kf kr .

2.2.2

(2.15)

Time-space estimates for the heat semi-group

We study the heat semi-group in the mixed space Lγt Lpx and the technique is to extensively use the dyadic decomposition together with the exponential 2 decay of e−t|ξ| . On the heat semi-group, the frequency localization idea goes back to Chemin [29] (see also [31]). The techniques used in this section θ are also adapted to more general semi-groups, say e−t(1+ia)|ξ| with 0 < θ < ∞, see [240]. The following result is due to [240]. Proposition 2.2. Let a > 0, 1 6 r 6 p 6 ∞, 0 < λ 6 ∞ and 2/γ = a + n(1/r − 1/p). Then we have kH(t)f kLγ (R+ ;B˙ 0

p,λ )

6 Ckf kB˙ −a

r, λ∧γ

.

(2.16)

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2

Proof. Noticed that e−t|ξ| is exponentially decaying, in order to extensively use this fact, the frequency localization techniques will be applied. We have 2

k4j H(t)f kr 6 kϕj e−t|ξ| kMr kf kr .

(2.17)

Using the multiplier criteria and noticing that ϕj = ϕ(2 1/2 6 |ξ| 6 2}, one has that 2j

2

kϕj e−t|ξ| kMr = kϕe−t2

|ξ|2

2j

kMr 6 kϕe−t2

|ξ|2

−j

ξ), suppϕ ⊂ {ξ :

kH L ,

(2.18)

where L > n/2. Due to supx>0 xk /ex 6 C, it is easy to see that 2j

kϕe−t2

|ξ|2

So, by (2.17)–(2.19) we get

2j

kH L . e−ct2 .

(2.19)

2j

k4j H(t)f kr . e−ct2 kf kr . P Since 4j = 4j ( `=0,±1 4j+` ), (2.20) implies that

(2.20)

2j

k4j H(t)f kr . e−ct2 k4j f kr .

(2.21)

Taking the `λ -norm in (2.21), we obtain that  1/λ X 2j kH(t)f kB˙ 0 .  e−ct2 k4j f kλr  . r,λ

(2.22)

j

In what follows the proof is separated into two cases. The first case is that γ > λ. Taking the norm in Lγt (R+ ) on (2.22) and using Minkowski’s inequality, we have

∞

1/λ

X −ctλ22k

λ kH(t)f kLγ (R+ ,B˙ 0 ) . e k4 f k k r

γ/λ r,λ Lt

k=−∞

X ∞

−ctλ22k .

e

Lt

k=−∞

Noticing that

−ctλ22k

e

k4k f kλr γ/λ

γ/λ

Lt

. 2−λ2k/γ ,

1/λ

.

(2.23)

(2.24)

from (2.23) and (2.24) we get the consequence in the case p = r and γ > λ. −2/γ+n(1/r−1/p) −2/γ As r < p, applying the inclusion B˙ r,λ ⊂ B˙ p,λ , we can obtain the result. Secondly, we consider the case γ < λ. By (2.22), Z X kH(t)f kγB˙ 0 dt . 2−2j k4j f kγp . (2.25) R+

p,λ

j

We get the result in the case p = r and γ < λ. If r < p, in an analogous way to the first case one can prove the conclusion.

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In fact, the above estimates (2.15) and (2.16) can be developed to the case 0 < r 6 p 6 ∞ and we have no restriction on λ, γ > 0. We have a little bit of surprise about this fact, see [240] for details. Some earlier results related to (2.16) were due to Weissler [250] and Giga [77], where their proof is based on Marcinkiewicz’ interpolation theorem. Corollary 2.1. Let 2/γ(p) = n(1/2 − 1/p). For any 2 6 p < ∞, we have k∇H(t)f kL2 (R+ ;L2 ) . kf kL2 ,

kH(t)f kLγ(p) (R+ ;Lp ) . kf kL2 .

(2.26) (2.27)

0 Proof. Taking λ = r = 2 in Proposition 2.2 and noticing that B˙ p,2 ⊂ Lp , one immediately has the consequence.

For simplicity, we write (A f )(t, x) :=

Z

t

0

H(t − τ )f (τ, x)dτ.

(2.28)

In what follows we consider the estimate of A f . By Proposition 2.1, Z t k n 1 1 k∇k A f kp . (t − τ )− 2 − 2 ( r − p ) kf (τ )kr dτ, k = 0, 1. (2.29) 0

Applying the Hardy-Littlewood-Sobolev inequality, we obtain that

Proposition 2.3. Let 1 6 r 6 p 6 ∞ and 1 < γ, γ1 < ∞ satisfy 1 k n 1 1 k n 1 1 1 = + + − − 1, + − < 1, k = 0, 1. (2.30) γ γ1 2 2 r p 2 2 r p Then we have k∇k A f kLγ (R+ ;Lp ) . kf kLγ1 (R+ ;Lr ) .

(2.31)

Proposition 2.3 can not handle the case γ = ∞, but we have (see [240]) Proposition 2.4. Let 1 6 r 6 ∞, 1 6 q 0 6 λ 6 ∞1 . Then kA f kL∞ (R+ ,B˙ 0

r,λ )

Proof.

0 −2k/q

(2.32)

r,λ

Using (2.21) and Young’s inequality, Z t 2k k4k A f kr . e−c(t−τ )2 k4k f (τ )kr dτ .2

1 p0

. kf kLq0 (R+ ,B˙ −2/q ) .

k4k f kLq0 (R+ ,Lr ) .

stands for the conjugate number of p, i.e. 1/p +

1/p0

= 1.

(2.33)

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39

Taking the `λ -norm in (2.33) and using Minkowski’s inequality, we get ( ∞ ) λ 1/λ X kA f kB˙ 0 . 2−2k/q k4k f kLq0 (R+ ,Lr ) r,λ

k=−∞

. kf kLq0 (R+ ,B˙ −2/q ) ,

(2.34)

r,λ

which implies the result, as desired.

Corollary 2.2. Let 2/γ(p) = n(1/2 − 1/p). For any 2 6 p < ∞, 2/γ(p) < 1, we have k∇A f kL2 (R+ ;L2 ) . kf kLγ(p)0 (R+ ,Lp0 ) ,

(2.35)

kA f kL∞ (R+ ,L2 ) ∩ Lγ(p) (R+ ;Lp ) . kf kLγ(p)0 (R+ ,Lp0 ) .

(2.36)

Clearly, all of the conclusions hold if we substitute the time interval R+ by [0, T ]. 2.3

Global well-posedness in L2 of NS in 2D

For any Banach function space X := X(Rn ) and u = (u1 , ..., un ) ∈ X n , P P we write kuk2X = i kui k2X and k∇uk2X = i,j k∂xi uj k2X . For any vector function u(t) = (u1 (t), ..., un (t)) defined in I and valued in X n , we write Z 1/q kukLq (I,X n ) = ku(t)kqX dt . (2.37) I

If there is no confusion, we will write Lq (I, X) := Lq (I, X n ). We denote by [X]n0 the completion of the set {u ∈ S n : div u = 0} in X n

and by Lq (I, [X]n0 ) the space of all of the vector functions u(t) = (u1 (t), ..., un (t)) in Lq (I, X n ) such that u(t) ∈ [X]n0 for a.e. t ∈ I. It is easy to verify that [Lp ]n0 = {u ∈ [Lp ]n : div u = 0},

Lq (I, [X]n0 ) = {u ∈ Lq (I, [X]n ) : div u(t) = 0, a.e. t ∈ I},

where “div” is in the sense of tempered distributions. Taking n = 2 and p = 4 in Corollaries 2.1 and 2.2, we have k∇H(t)u0 kL2 (0,T ;

L2 )

. ku0 kL2 ,

(2.38)

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Navier-Stokes equation

kH(t)u0 kL4 (0,T ; k∇A f kL2 (0,T ; kA f kL∞ (0,T ;

We write

Lpx,t∈[0,T ]

L4 )

L2 )

. ku0 kL2 ,

. kf kL4/3(0,T ;

L2 )∩L4 (0,T ; L4 )

p

p

(2.39) L4/3 ) ,

(2.40)

. kf kL4/3(0,T ;

L4/3 ) .

(2.41)

= L (0, T ; L ). Define the metric space: n o D = u : kukL4x,t∈[0,T ] + k∇ukL2x,t∈[0,T ] 6 M ,

(2.42)

d(u, v) = ku − vkL4x,t∈[0,T ] + k∇(u − v)kL2x,t∈[0,T ] .

(2.43)

Assume that u0 ∈ [L2 ]20 . We consider the mapping: M : u(t) → H(t)u0 + A P[(u · ∇)u].

(2.44)

Now we fix an M > 0 such that CM = 1/2, where C is the largest constant appeared in the following inequalities. We show that M : (D, d) → (D, d) is a contraction mapping. In fact, for any u ∈ D, kMukL4x,t∈[0,T ] 6 kH(t)u0 kL4x,t∈[0,T ] + CkP[(u · ∇)u]kL4/3

,

(2.45)

x,t∈[0,T ]

k∇MukL2x,t∈[0,T ] 6 k∇H(t)u0 kL2x,t∈[0,T ] + CkP[(u · ∇)u]kL4/3

. (2.46)

x,t∈[0,T ]

In view of Mihlin’s multiplier theorem, P : L4/3 → L4/3 . So, it follows from H¨ older’s inequality that, kP[(u · ∇)u]kL4/3

x,t∈[0,T ]

. kukL4x,t∈[0,T ] k∇ukL2x,t∈[0,T ] 6 M 2 .

(2.47)

By (2.38) and (2.39), there exists a T > 0 such that k∇H(t)u0 kL2 (0,T ;

L2 )

+ kH(t)u0 kL4 (0,T ;

L4 )

6 M/2.

(2.48)

So, kMukL4x,t∈[0,T ] 6 M/2 + CM 2 6 M,

(2.49)

k∇MukL2x,t∈[0,T ] 6 M/2 + CM 2 6 M.

(2.50)

Similarly, 1 d(u, v). (2.51) 2 In view of Banach’s contraction mapping principle, there exists a u ∈ D satisfying d(Mu, Mv) 6

u(t) = H(t)u0 + A P[(u · ∇)u].

(2.52)

In view of the regularity theory, we see that the solution is smooth in the domain (0, T ) × R2 . Since u0 ∈ [L2 ]20 and div P = 0, it follows from the

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integral equation that div u = 0 holds in (0, T ) × Rn . Again, by (2.41), one sees that u ∈ C(0, T ; [L2 ]20 ). The conservation (2.2) can be deduced by a straightforward calculation to the integral equation. It follows that ku(T )k2 6 ku0 k2 . According to the standard semi-group theory, one can extend the solution above step by step and finally finds a maximal time Tm . It suffices to show that Tm = ∞. If not, then Tm < ∞. Tm is maximal implies that kukC([0,Tm);

[L2 ]20 )∩L4x,t∈(0,T

m)

+ k∇ukL2 (0,Tm ;

L2 )

= ∞.

By the energy estimate, 1 ku(t)k22 + 2

Z

t

0

k∇u(s)k22 ds =

1 ku0 k22 , ∀ t < Tm . 2

(2.53)

In view of the Gagliardo-Nirenberg inequality, 1/2

kukL4x,t∈[0,T ] . kukL∞(0,T ;

1/2 L2 ) k∇ukL2 (0,T ; L2 )

. ku0 k2 , ∀ T < Tm . (2.54)

A contradiction. We have shown Theorem 2.1. Let u0 ∈ [L2 ]20 . Then the NS equation (2.5) has a unique solution u satisfying u ∈ C(0, ∞; [L2 ]20 ) ∩ L2 (0, ∞; [H˙ 1 ]20 ),

(2.55)

and 1 ku(t)k22 + 2

Z

0

t

k∇u(s)k22 ds =

1 ku0 k22 , 0 < t < ∞. 2

(2.56)

The contraction mapping argument implies that the solution map is analytic and of course, is continuous and the persistence of the regularity of solutions is also easy to obtain by using the integral equation. So the NS equation is globally well-posed in L2 . Finally, we point out that the method used here can be applied to some other kinds of nonlinear parabolic equations. For examples, ut − ∆u + |∇u|u = 0, u(0, x) = u0 (x) is globally well posed in L2 (R2 ); the Hamilton-Jacobi equation ut − ∆u + |∇u|3/2 = 0, u(0, x) = u0 (x) is locally well-posed in L2 (R2 ).

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2.4

Navier-Stokes equation

Well-posedness in Ln of NS in higher dimensions

For convenience, we still write Lpx,t∈[0,T ] = Lp (0, T ; Lp (Rn )) and use the same notations as in the previous section. By Corollaries 2.1 and 2.2, we have for n > 3, k∇H(t)u0 kL2 (0,T ; L2 ) . ku0 kL2 ,

(2.57)

kH(t)u0 kL2+4/n

. ku0 kL2 ,

(2.58)

k∇A f kL2 (0,T ;

. kf kL(2+4/n)0 ,

(2.59)

x,t∈[0,T ]

kA f kL∞ (0,T ;

L2 )

x,t∈[0,T ]

2+4/n

L2 )∩Lx,t∈[0,T ]

. kf kL(2+4/n)0 .

(2.60)

x,t∈[0,T ]

One may expect that the techniques in the previous section also work very well. Unfortunately, k(u · ∇)ukL(2+4/n)0 6 k∇ukL2x,t∈[0,T ] kukLn+2

x,t∈[0,T ]

x,t∈[0,T ]

.

(2.61)

Ln+2 x,t∈[0,T ] is out of the control of the above linear estimates. So, we need the following Corollary 2.3. We have kH(t)u0 kLn+2

x,t∈[0,T ]

kH(t)u0 kL∞ (0,T ; k∇A f k

Ln+2 x,t∈[0,T ]

k∇A f kL∞ (0,T ; Proof.

. ku0 kn ,

Ln )

(2.62)

. ku0 kn ,

(2.63)

. kf kL(n+2)/2 ,

Ln )

(2.64)

x,t∈[0,T ]

. kf kL(2+n)/2 .

(2.65)

x,t∈[0,T ]

Taking p = r = λ = n and γ = 2 + n in Proposition 2.2, we get kH(t)u0 kL2+n (R+ ;B˙ 2/(2+n) ) . ku0 kB˙ 0

n, n

n,n

. ku0 kn .

(2.66)

2/(2+n) 2/(2+n) 0 Using the inclusions B˙ n,n = F˙n,n ⊂ F˙n+2,2 = Ln+2 , we see that (2.66) implies that (2.62) holds. Obviously, we have (2.63). (2.64) is a straightforward consequence of Proposition 2.3. Taking r = λ = (n + 2)/2 and q 0 = (n + 2)/2 in Proposition 2.4, we get

kA f kL∞ (R+ ; B˙ 2n/(2+n)

(n+2)/2, (n+2)/2

)

. kf kL(n+2)/2 (R+ ; B˙ 0

(n+2)/2, (n+2)/2

).

(2.67)

This implies k∇A f kL∞ (R+ ; B˙ (n−2)/(2+n)

(n+2)/2, (n+2)/2

)

. kf kL(n+2)/2 .

(2.68)

x,t∈R+

(n−2)/(2+n) (n−2)/(2+n) 0 Using the embedding B˙ (n+2)/2, (n+2)/2 = F˙(n+2)/2, (n+2)/2 ⊂ F˙ n,2 = Ln , we see that (2.68) implies that (2.65) holds.

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The above linear estimates can derive the local well-posedness and the global well-posedness with small data in Ln for n > 3. We will use the same notations as in the case n = 2. We have Theorem 2.2. Let u0 ∈ [Ln ]n0 . Then there exists a Tm > 0 such that the NS equation (2.5) has a unique solution u satisfying 2+n n u ∈ C([0, Tm ); [Ln ]n0 ) ∩ L2+n ]0 ). (2.69) loc (0, Tm ; [L If Tm < ∞, then we have kukL2+n(0,Tm ; L2+n ) = ∞. If ku0 kn is sufficiently small, then Tm = ∞. Moreover, if u0 ∈ [L2 ]n0 , then u ∈ C(0, Tm ; [L2 ]n0 ), ∂xi u ∈ L2 (0, Tm ; [L2 ]n0 ), and Z t 1 1 ku(t)k22 + k∇u(s)k22 ds = ku0 k22 , 0 < t < Tm . (2.70) 2 2 0 Put n D = u : kukL2+n

Proof.

x,t∈[0,T ]

d(u, v) = ku − vkL2+n

o 6 δ, kukL∞ ([0,T ];Ln ) 6 2Cku0 kn ,

x,t∈[0,T ]

.

We consider the mapping: M : u(t) → H(t)u0 + A P div (u ⊗ u), by Corollary 2.3, we have . kH(t)u0 kL2+n + ku ⊗ ukL(2+n)/2 kMukL2+n x,t∈[0,T ]

x,t∈[0,T ]

kMukL∞ (0,T ;

Ln )

(2.73)

x,t∈[0,T ]

+

. kH(t)u0 kL2+n

+ δ2,

x,t∈[0,T ]

(2.72)

kuk2L2+n x,t∈[0,T ]

. kH(t)u0 kL2+n

x,t∈[0,T ]

(2.71)

(2.74)

. ku0 kn + ku ⊗ ukL(2+n)/2

x,t∈[0,T ]

. ku0 kn + δ 2 . (2.75) If Cδ 6 1/4, we can show that M is a contraction mapping from D into itself. So, there exists a u satisfying u(t) = H(t)u0 + A P∇ · (u ⊗ u). (2.76) By a standard argument, we see that u is unique in L2+n (0, T ; [L2+n ]n0 ). Moreover, one can extend the solution step by step and find a maximal Tm 2+n n such that u ∈ C([0, Tm ); [Ln ]n0 ) ∩ L2+n ]0 ). loc (0, Tm ; [L 2 n If u0 ∈ [L ]0 , then we can use (2.57)–(2.61) to obtain u ∈ C(0, T ; [L2 ]n0 ) and ∂xi u ∈ L2 (0, T ; [L2 ]n0 ) in a similar way as in the 2D case. According to the regularity theory (see [67] for instance), we see that u is infinitely smooth in the domain (0, T ) × Rn . So, a straightforward computation will lead to (2.2). Moreover, if ku0 kn is small enough, we can take T = ∞ in (2.71) and (2.72).

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The result of Theorem 2.2 has a long history. Weissler [251] gives a detailed Lp -theory in half-space for local solutions. Kato [119] used a different way to obtain the well posedness result in Ln by constructing the resolution norm like supt tθ ku(t)kp . The idea only taking Ln+2 x,t as a resolution space was hidden in some references, see for instance [67]. Here the proof is to n take Ln+2 x,t∈[0,T ] ∩ C([0, T ]; L ) as a resolution space which enables us to get the local well posedness of solutions in Ln , where (2.65) plays a crucial role. Similar bilinear estimate to (2.65) in 3D was shown in [70] in a different way: kA P div (u ⊗ v)kL∞ ([0,T ];L3 ) . kukL5x,t kvkL5x,t .

(2.77)

In 3D case, Kenig and Koch [123] obtained the result of Theorem 2.2 by applying (2.77). 2.5

Regularity of solutions for NS

We will show that, for any t > 0, the solution of the NS equation is infinitely differentiable and in fact, is really analytic. 2.5.1

s Gevrey class and function space E2,1

First, we consider the Gevrey classes. Let s > 0, denote s m! n ∞ n ∃ ∀m ∈ Z+ . Gs (R ) = f ∈ C (R ) : ρ, M > 0 s.t. kf kH˙ m 6 M ρm (2.78) Gs (Rn ) is said to be a Gevrey s-class. Our aim is to show that the solution of the NS equation belongs to the Gevrey 1-class. In order to show this fact, we will use Wiener’s decomposition of Rn . In Chapter 6, we will continuously use Wiener’s decomposition to study the nonlinear dispersive equations. Let Qα be the unit cube with the center at α ∈ Zn , i.e., Qα = α + Q0 , Q0 = {x = (x1 , ..., xn ) : −1/2 6 xi < 1/2}. It is easy to see that Rn = ∪α∈Zn Qα ,

Qα ∩ Qβ = ∅, n

α 6= β.

So, {Qα } constitutes a decomposition of R , which is said to be the uniform decomposition (or Wiener’s decomposition) of Rn . Comparing it with the dyadic decomposition, we see that it is more delicate than the dyadic decomposition of Rn . Putting the uniform decomposition into the frequency

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space and combining it with the `1 (L2 ) space, we have the following ( )

X

s 0 n s|k| −1 b s E2,1 = f ∈ S (R ) : kf kE2,1 = 2

F χQk f < ∞ . (2.79) 2

k∈Zn

s Such a kind of space and its generalized version Ep,q were studied in [247], which is a kind of generalized modulation spaces [86]. It is easy to see s that E2,1 is a Banach space, which has an exponential regularity weight. s From PDE’s point of view, E2,1 has infinite smoothness, which is different from any Besov spaces and any classical modulation spaces. The classical s modulation space M2,1 takes the following norm: ( )

X

s 0 n s −1 M = f ∈ S (R ) : kf kM s = hki F χQ fb < ∞ , (2.80) 2,1

k

2,1

k∈Zn

2

s s where hki = 1 + |k|. M2,1 and its generalizations Mp,q have finite smooths ness, which are similar to Besov spaces (see Chapter 6). E2,1 is of importance in our regularity argument for the parabolic equation, whose infinite regularity can be seen in the following (cf. [247]):

Proposition 2.5. We have G1 =

[

s E2,1 .

(2.81)

s>0 s Proof. First, we show that for any s > 0, E2,1 is a subset of G1 . Indeed, s for any f ∈ E2,1 and 0 < c 1, 0 kf kH˙ m . k∇m f kE2,1 X . kχQk |ξ|m fb k2

k∈Zn

.

X

(|k| +

k∈Zn

.

p n/2)m kχQk fb k2

m! X (cs)m (|k| + C)m s|k| 2 kχQk fb k2 (cs)m m! · 2s|k| n k∈Z

m! s . . kf kE2,1 (2s)m

On the other hand, let f ∈ G1 and L > n/2. Using Taylor’s expansion and H¨ older’s inequality, X s kf kE2,1 = 2s|k| kχQk fb k2 k∈Zn

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Navier-Stokes equation

6 kf k2 +

∞ X (s ln 2)m m! m=0 ∞ X

X

k∈Zn , |k|1

X

m

. kf k2 +

(s ln 2) m! m=0

. kf k2 +

∞ X

k∈Zn , |k|1

m

|k|m kχQk fb k2

m+L f k C m |k|−L kχQk ∇\ 2

(Cs) kf kH˙ L+m m! m=0

∞ X (Cs)m (m + L)! · . m! ρm m=0 P∞ Choosing 0 < s 1, we see that m=0 (Cs/ρ)m (m + L)L is a convergent s series. Thus, f ∈ E2,1 .

. kf k2 +

s Estimates of heat semi-group in E2,1

2.5.2

Using Proposition 2.5, we want to get the solution of the NS equation s u ∈ G1 . It suffices to show that u belongs to E2,1 (s > 0). Hence, one s s . needs to estimate kH(t)u0 kE2,1 and kA f kE2,1 Proposition 2.6. Let H(t) = et∆ . There exists a c > 0 such that for j = 0, 1 and T 6 1, −j/2 ct . t 0 , k∇j H(t)u0 kE2,1 ku0 kE2,1 t 6 T, (2.82)

Z t

j

cτ . H(t − τ )f (τ )dτ . (T + T 1−j/2 ) sup kf (τ )kE2,1 sup

∇

t∈[0,T ]

0

τ ∈[0,T ]

ct E2,1

(2.83)

Proof.

For convenience, we will use the notation k = F −1 χQk F .

By Plancherel’s identity, for |k| > 1,

2

(2.84) 2

kk H(t)u0 k2 = kχQk (ξ)e−t|ξ| F u0 k2 6 e−ct|k| kk u0 k2 .

(2.85)

If k = 0, we see that (2.85) also holds for t 6 1. Taking the summation over all k ∈ Zn , we can get (2.82) hold for j = 0. We now prove (2.82) for j = 1. If k = 0 and t 6 1, 2

k∇k H(t)u0 k2 . kχQk (ξ)e−t|ξ| F u0 k2 6 kk u0 k2 .

For |k| > 1,

2

(2.86)

k∇k H(t)u0 k2 . k|ξ|χQk (ξ)e−t|ξ| F u0 k2 6 t−1/2 e−ct|k| kk u0 k2 . (2.87)

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Combining (2.86) with (2.87), we have k∇k H(t)u0 k2 . t−1/2 e−ct|k| kk u0 k2 .

(2.88)

This proves (2.82) for j = 1. For t ∈ [0, T ] with T 6 1,

Z t Z t

k ∇ H(t − τ )f (τ )dτ . (t − τ )−1/2 e−c(t−τ )|k| kk f (τ )k2 dτ,

0

0

2

(2.89)

which implies that (2.83) holds in the case j = 1. Another case j = 0 is similar to the case j = 1. 2.5.3

s Bilinear estimates in E2,1

By Proposition 2.6, our goal is to show that the solution of the NS equation ct ct . We have belongs to E2,1 , it is necessary to estimate ku ⊗ ukE2,1 λ is a Banach algebra. More precisely, we hace Proposition 2.7. E2,1

kuvkE2,1 λ 6 C2Cλ kukE2,1 λ kvkE λ , 2,1

(2.90)

λ where C is independent of λ > 0 and u, v ∈ E2,1 .

Proof. Noticing that for k = (k1 , ..., kn ) ∈ Zn with |k| = |k1 | + ... + |kn |, we have X kuvkE2,1 λ = 2λ|k| k k (uv)k2 . (2.91) k∈Zn

It is easy to see that

uv =

X

(i u)(j v),

(2.92)

i,j∈Zn

which implies that k (uv) =

X

k (i u j v).

(2.93)

i,j∈Zn

We have F (i u j v) = (χQi u b) ∗ (χQj vb) and

So,

√ supp(χQi u b) ∗ (χQj vb) ⊂ Ω := {ξ : |ξ − i − j| 6 2 n}. √ k (i u j v) = 0, |k − i − j| > 3 n.

(2.94)

(2.95)

It follows that kk (i u j v)k2 = kk (i u j v)k2 χ(|k−i−j|63√n) .

(2.96)

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In view of Plancherel’s identity and H¨older’s inequality, kk (i u j v)k2 6 k i u j vk2 χ(|k−i−j|63√n)

6 k i uk∞ kj vk2 χ(|k−i−j|63√n) . k i uk2 kj vk2 χ(|k−i−j|63√n) .

Hence, kuvkE2,1 λ 6 .

X

k∈Zn

X X

k∈Zn

. 2Cλ

i,j∈Zn

i,j∈Zn

X

i,j∈Zn

The result follows. 2.5.4

X

2λ|k|

(2.97)

kk (i u j v)k2

2λ|k| k i uk2 kj vk2 χ(|k−i−j|63√n)

2λ(|i|+|j|) k i uk2 kj vk2 .

(2.98)

Gevrey regularity of NS equation

ct Once we get that the solution of the NS equation is in E2,1 , then it belongs to the Gevrey 1-class. Let

D = {u :

ct 6 M }, sup ku(t)kE2,1

(2.99)

ct . d(u, v) = sup ku(t) − v(t)kE2,1

(2.100)

06t6t0

06t6t0

Considering the mapping T : u(t) → H(t)u0 −

Z

0

t

H(t − τ )[P ∇ · (u ⊗ u)](τ )dτ,

(2.101)

0 we show that for u0 ∈ E2,1 , there exists a t0 > 0 such that T : (D, d) → (D, d) is a contraction mapping. For convenience, we denote ct . |||u||| = sup ku(t)kE2,1

(2.102)

06t6t0

Let u ∈ D. By Proposition 2.6 and 2.7,

1/2

0 + t0 |||u||| + t0 |||(u ⊗ u)||| |||T u||| . ku0 kE2,1

1/2

0 . ku0 kE2,1 + t0 |||u|||2 .

(2.103) 1/2

0 Take M = 2Cku0 kE2,1 and assume t0 < 1. If Ct0 M 6 1/4, then T u ∈ D. Similarly, for any u, v ∈ D, 1 |||T u − T v||| 6 |||u − v|||. (2.104) 2

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So, there exists a u ∈ D satisfying T u = u. 0 n Theorem 2.3. Let u0 ∈ [E2,1 ]0 . Then there exists a Tm > 0, such that the 0 n ]0 . Moreover, NS equation (2.5) has a unique solution u ∈ C [0, Tm ); [E2,1 there exists a t0 ∈ (0, Tm ) satisfying c(t∧t0 ) n ]0 ,

u(t) ∈ [E2,1

t ∈ [0, Tm ).

(2.105)

0 If Tm < ∞, then sup06t
Remark 2.1. (1) By Theorem 2.3, we can get that the solution of the NS equation belongs to C ∞ ((0, Tm ) × Rn ). (2) Theorem 2.3 implies that the solution of the NS equation belongs to the Gevrey 1-class and so, is really analytic. (3) Theorem 2.3 also describes the disappearing process of the regularity of the solutions to the NS equation. Remark 2.2. The Gevrey regularity for the evolution equations is of importance for its own sake. The Gevrey regularity of the weak solutions for a class of linear and semi-linear Fokker-Planck equations (∂t + v · ∇x − ∆v )u = F (t, x, v, u, ∇v u)

was recently studied by Chen, Li and Xu [35], see also [37] and references therein for a class of the linear model of spatially inhomogeneous Boltzmann equations without an angular cutoff. Remark 2.3. As the end of this chapter, we state some recent progress on NS equation without proofs. (1) Let us mention the result by Koch and Tartaru [151] where the global solutions for NS in 3D are obtained with the small data in the space BM O−1 with the norm: !1/2 Z −3/2 t∆ −1 kukB˙ ∞,∞ + sup R |e u(y)|dydt , x∈R3 , R>0

[0,R2 ]×{y:|x−y|6R}

see also Chemin and Gallagher’s generalizations in [30; 31; 32] for a class of large data.

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ˇ ak [70] showed that any (2) Recently, Escauriaza, Seregin and Sver´ “Leray-Hopf” weak solution in 3D which remains bounded in L3 (R3 ) cannot develop a singularity in finite time. Their proof used a blow-up procedure and reduction to a backwards uniqueness question for the heat equation, and was then completed using Carleman-type inequalities and the theory of unique continuation. Kenig and Koch [123] gave an alternative proof by substituting L3 with H˙ 1/2 . Dong and Du [68] generalized their results in higher spatial dimensions n > 3. −1 (3) Noticing that L3 ⊂ B∞,∞ in 3D is a sharp embedding, for any solution u of the NS equation in C([0, T ∗ ); L3 ), we see that u ∈ −1 C([0, T ∗ ); B∞,∞ ). May [163] prove that if T ∗ < ∞, then there exists a constant c > 0 independent of the solution of NS equation such that −1 lim sup ku(t) − ωkB∞,∞ >c

t→T ∗

for all ω ∈ S . For the Leray-Hopf weak solution, Cheskidov and Shvydkoy [39] obtained similar result. (4) Recently, the Cauchy problem for the 3D anisotropic Navier-Stokes equation ut − (∂x21 + ∂x22 )u + u · ∇u + ∇p = 0, ∇ · u = 0, u|t=0 = u0 is considered in [187; 33] (see also references therein). These equations come from meteorology models. More recent results based on the harmonic analysis method can be found in [155].

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Chapter 3

Strichartz estimates for linear dispersive equations First, let us imagine the large-time-decaying phenomena in a dispersive system... .

We begin to study the dispersive equations. Roughly speaking, the dispersive equation takes the following form ∂t u − iP (D)u = F,

(3.1)

√ (D)u = P (ξ)b u, and where u(t, x) is an unknown function, i = −1, P\ P (ξ) is a real-valued function, which is said to be the dispersion relation of (3.1)1 . If F is a nonlinear function of u, then (3.1) is said to be a nonlinear dispersive equation. In this chapter we study the time-space estimates for the solutions of the linear dispersive equation in mixed Lebesgue spaces Lqt Lpx (or more s general spaces Lq (0, T ; Bp,r )), so called the Strichartz inequalities, which is a starting point to the study of nonlinear dispersive equations. For instant, we consider the Schr¨ odinger equation2 iut + ∆u = f,

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

(3.2)

Pn

where ∆ = i=1 ∂x2i , u(t, x) is a complex valued function of (t, x) ∈ R×Rn , u0 denotes the initial value at t = 0. f is a known complex function of (t, x) ∈ R × Rn . We will use its integral form. Taking the Fourier transform to (3.2), we get

1 We

ib ut − |ξ|2 u b = fb,

u b(0) = u b0 .

emphasize that P (·) must be a real function. Schr¨ odinger (1887-1961) was an Austrian theoretical physicist who achieved fame for his fundamental contributions to quantum mechanics, especially the Schr¨ odinger equation, for which he received the Nobel Prize in 1933. 2 Erwin

51

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Solving the ordinary differential equation, we obtain that Z t u(t) = S(t)u0 − i S(t − τ )f (τ, ·)dτ,

(3.3)

0

2

where S(t) = eit∆ := F −1 e−it|ξ| F . By Plancherel’s identity, we see that kS(t)u0 k2 = ku0 k2 , which means that S(t)u0 is invariant in L2 for any t ∈ R. However, we can show that S(t) satisfies the following Strichartz estimates: kS(t)u0 kLγ(r) (R,Lr ) . ku0 k2 ,

Z t

S(t − τ )f (τ, ·)dτ

0

Lγ(r) (R,Lr )

. kf kLγ(ρ)0 (R,Lρ0 ) ,

where 2 6 r, ρ 6 ∞, 2/γ(·) = n(1/2 − 1/·) ∈ [0, 1), and p0 is the conjugate number of p. Such kinds of estimates are time-decaying versions, which are of importance in the study of nonlinear dispersive equations. In the next chapter, we will further indicate why Strichartz inequalities are useful. 3.1

0

Lp → Lp estimates for the dispersive semi-group 2

Let us consider the Schr¨ odinger semi-group eit∆ := F −1 eit|ξ| F . Since Z |x−y|2 (3.4) eit∆ u0 = ct−n/2 e 4it u0 (y)dy, Rn

1

we immediately have the L → L

∞

decay

keit∆ u0 k∞ . t−n/2 ku0 k1 .

Taken noticing of keit∆u0 k2 = ku0 k2 , an interpolation yields keit∆u0 kp . t−n/2 ku0 kp0 , 0

p > 2, 1/p + 1/p0 = 1.

This is the fundamental Lp → Lp estimate, or the time decay estimate for the Schr¨ odinger semi-group. For the general semi-group, it is not expected to have an analytic expression as in (3.4), we need to look for other ways 0 to get the Lp → Lp estimates. 0 In this section we study the Lp → Lp estimates for a class of dispersive semi-groups U (t) := F −1 exp(itP (ξ))F and use two different ways to 0 consider its Lp → Lp decay. The first method is very effective for homogeneous functions P (·), the second method can deal with nonhomogeneous radial functions P (·). First, we consider the decay estimates for the semi-group Um (t) := F −1 exp(it|ξ|m )F . Some earlier decay estimates on Um (t) were obtained

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53 0

by Pecher [191]. Following Pecher’s proof, we can get an Lp → Lp decay estimate with smooth effects in the case m > 2. Some nontrivial generalized dispersive Lq − Lp estimates can be founded in Cui [52; 53], Sugimoto [207; 208] (see also [195]). We need the following lemma, which is due to Littman [161]. Lemma 3.1. Let v ∈ C0∞ (Rn ), supp v = Ω. P (ξ) : Rn → R is an infinitely smooth function on Ω. For any ξ ∈ Ω, the rank of (∂ 2 P (ξ)/∂ξi ∂ξj )ni,j=1 is at least ρ > 0, Then there exists a K ∈ N such that for any λ ∈ R, X kF −1 eiλP (ξ) vk∞ . (1 + |λ|)−ρ/2 kDα vk∞ . (3.5) |α|6K

Now we derive the decay estimates of Um (t) in homogeneous Besov spaces. Let {4k }k∈Z be as in Sec. 1.6. In view of Young’s inequality (see Appendix), one has that m

m

kF −1 eit|ξ| F 4k f k∞ 6 kF −1 eit|ξ| ϕ(2−k ξ)k∞ kf k1 . Now we estimate kF Lemma 3.1, we have

−1 it|ξ|m

e

ϕ(2

−k

(3.6)

ξ)k∞ . Applying Proposition 1.4 and

m

kF −1 eit|ξ| ϕ(2−k ξ)k∞ km

= 2kn kF −1 eit2

|ξ|m

ϕ(ξ)k∞ . t−ρ/2 2k(n−mρ/2) ,

(3.7)

where ρ denotes the rank of (∂ 2 |ξ|m /∂ξi ∂ξj )ni,j=1 on the support of ϕ (suppϕ ⊂ {ξ : |ξ| ∈ (2−1 , 2)}). It is easy to see that n − 1, m = 1, ρ= (3.8) n, m > 2. So, m

kF −1 eit|ξ| F 4k f k∞ . t−ρ/2 2k(n−mρ/2) kf k1 .

(3.9)

Obviously, by Plancherel’s identity, m

kF −1 eit|ξ| F 4k f k2 6 kf k2 .

(3.10)

In view of Riesz-Thorin’s interpolation theorem, (3.9) and (3.10) imply that for any 2 6 p 6 ∞, 1/p + 1/p0 = 1, 2−k(2n−mρ)(1/2−1/p) kUm (t)4k f kp . t−ρ(1/2−1/p) kf kp0 .

(3.11)

For convenience, we write 2σ(m, p) := (2n − mρ)(1/2 − 1/p).

(3.12)

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Substituting f by

P1

`=−1 4k+` f

in (3.11), we get

2−2σ(m,p)k kUm (t)4k f kp . t−ρ(1/2−1/p)

1 X

k4k+` f kp0 .

(3.13)

kUm (t)f kB˙ −2σ(m,p) . t−ρ(1/2−1/p) kf kB˙ 00 , 2 6 p 6 ∞.

(3.14)

Taking the `q -norm in (3.13), we have p,q

`=−1

p ,q

s s Using the inclusion B˙ p,2 ⊂ F˙ p,2 = H˙ ps and B˙ ps0 ,2 ⊃ F˙ ps0 ,2 = H˙ ps0 , we have

kUm (t)f kH˙ −2σ(m,p) . t−ρ(1/2−1/p) kf kp0 , 2 6 p < ∞. p

(3.15)

Taking m = 1, 2, we immediately have Proposition 3.1. Let n > 2, W (t) = F −1 eit|ξ| F , 2 6 p < ∞, 1 6 q 6 ∞, and 1/p + 1/p0 = 1. Then kW (t)f kB˙ −(n+1)(1/2−1/p) . t−(n−1)(1/2−1/p) kf kB˙ 00 ,

(3.16)

kW (t)f kH˙ −(n+1)(1/2−1/p) . t−(n−1)(1/2−1/p) kf kp0 .

(3.17)

p,q

p ,q

p

2

Proposition 3.2. Let n > 1, S(t) = F −1 eit|ξ| F , 2 6 p < ∞, 1 6 q 6 ∞, and 1/p + 1/p0 = 1. Then we have kS(t)f kB˙ 0 . t−n(1/2−1/p) kf kB˙ 00 ,

(3.18)

kS(t)f kp . t−n(1/2−1/p) kf kp0 .

(3.19)

p,q

p ,q

In the higher order case m > 2, we have Proposition 3.3. Let n > 1, m > 2, 2 6 p < ∞, 1 6 q 6 ∞, and 1/p + 1/p0 = 1. Then we have kUm (t)f kB˙ −2σ(m,p) . t−n(1/2−1/p) kf kB˙ 00 ,

(3.20)

kUm (t)f kH˙ −2σ(m,p) . t−n(1/2−1/p) kf kp0 .

(3.21)

p,q

p ,q

p

Noticed that for m > 2, 2σ(m, p) = n(2 − m)(1/2 − 1/p) < 0, Um (t) gains some regularities in Proposition 3.3. Below, we consider another decay estimate of Um (t) without the smoothness in the case m > 2. We have Proposition 3.4. Let n > 1, m > 2, 2 6 p < ∞, and 1/p + 1/p0 = 1. Then 0

kUm (t)f kp . t−n(1/p −1/p)/m kf kp0 .

(3.22)

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Proof. In view of the dilation property, it suffices to consider the case t = 1. Let {ϕk }∞ k=0 be as in Sec. 1.3. (3.13) implies that for any k > 1, kUm (1)4k f kp . 22σ(m,p)k

1 X

`=−1

k4k+` f kp0 .

(3.23)

By supp ϕ0 ⊂ {ξ : |ξ| 6 2}, we have kUm (1)40 f kp 6 .

1 X

`=0 1 X `=0

kUm (1)40 4` f kp kϕ0 F 4` f kp0 .

1 X `=0

k4` f kp0 .

(3.24)

We write 4−1 = 0. It follows that (3.23) holds for all k > 0. Taking the `2 in (3.23), one has that 0 kUm (1)f kBp,2 . kf kB 00 . p ,2

(3.25)

Similar to (3.15), (3.25) implies that kUm (1)f kp . kf kp0 . In view of the dilation property, we obtain the result, as desired.

(3.26)

Using (3.21), (3.22) and the convexity H¨older inequality (Proposition 1.21), we immediately have Proposition 3.5. Let n > 1, m > 2, 2 6 p < ∞, and 1/p + 1/p0 = 1. Then for any θ ∈ [0, 1], kUm (t)f kH˙ −2σ(m,p)θ . t−(nθ+2n(1−θ)/m)(1/2−1/p) kf kp0 . p

(3.27)

The above method is also valid for some other homogeneous non-radial functions, say P (ξ) = ξ14 ± ... ± ξk4 , k 6 n. Noticing that, if P (ξ) is not a homogeneous function, (3.7) can not be obtained by scaling, we need to look for another way to handle the nonhomogeneous case. Our idea is to simplify P (·) as a radial function P (ξ) := P (|ξ|), which is essentially reduced to one dimensional case. P (|ξ|) can be separated into two parts, lower and higher frequency parts, which correspond to |ξ| . 1 and |ξ| 1, respectively. We further assume that P (ξ) has a different growth as |ξ| . 1 and |ξ| 1, which is sufficient for many semi-groups. In what follows, we always assume that P : (0, ∞) → R is a smooth radial function satisfying

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(H1) there exists an m1 > 0, such that for any α > 2, α ∈ N, |P 0 (r)| ∼ rm1 −1 , |P (α) (r)| . rm1 −α , r > 1;

(H2) there exists an m2 > 0, such that for any α > 3, α ∈ N,

|P 0 (r)| ∼ rm2 −1 , |P (α) (r)| . rm2 −α , 0 < r < 1;

(H3) there exists an α1 such that

|P 00 (r)| ∼ rα1 −2 , r > 1;

(H4) there exists an α2 such that

|P 00 (r)| ∼ rα2 −2 , 0 < r < 1. Remark 3.1. The following are some p examples for which conditions (H1)– (H4) are satisfied: (1) P (ξ) = 1 + |ξ|2 (relevant to the Klein-Gordon 4 2 semi-group); (2) P (ξ) = |ξ| p + |ξ| (relevant to the fourth order Schr¨odinger semi-group); (3) P (ξ) = 1 + |ξ|4 (relevant p to the beam semi-group). Comparing P (|ξ|) = |ξ| with P (|ξ|) = 1 + |ξ|2 , althoughp they have the 2 same growth as t → ∞, we easily see that the rank of (∂ij 1 + |ξ|2 )n×n p 2 is n and 1 + |ξ| is better than |ξ| at ξ = 0. So, we can expect that the 0 Klein-Gordon semi-group has a better Lp → Lp decay. We first consider 1D case, which is easier to find the ideas. The following result is due to [94]. Proposition 3.6. Let n = 1 and U (t) = F −1 eitP (|ξ|) F . We have the following results. (a) Let {4k }k∈Z be defined in Sec. 1.6. Then for any k > 0, kU (t)4k u0 k∞ . 2k ku0 k1 .

Moreover, if P satisfies (H3), then

kU (t)4k u0 k∞ . |t|−θ/2 2k(1−α1 θ/2) ku0 k1 ,

0 6 θ 6 1.

(b) Let {4k }k∈Z be as in Sec. 1.6. Then for any k < 0, one has that kU (t)4k u0 k∞ . 2k ku0 k1 .

Moreover, if P satisfies (H4), then

kU (t)4k u0 k∞ . |t|−θ/2 2k(1−α2 θ/2) ku0 k1 ,

0 6 θ 6 1.

(c) Let 40 be as in Sec. 1.3. Assume that P satisfies (H2) and (H4) with m2 = α2 , then 1 1 −θ kU (t)40 u0 k∞ . (1 + |t|) ku0 k1 , 0 6 θ 6 min , . m2 2

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We need the following Van der Corput lemma, see Appendix. Lemma 3.2. Let ϕ ∈ C0∞ (R), and P ∈ C 2 (R) satisfy |P 00 (ξ)| > λ > 0 for all ξ ∈ supp ϕ. Then Z eiP (ξ) ϕ(ξ)dξ . λ−1/2 (kϕk∞ + kϕ0 k1 ).

Proof. [Proof of Proposition 3.6] First, we prove (a). Using Young’s inequality, one has that kU (t)4k u0 k∞ . kJk k∞ ku0 k1 ,

where Jk (x) = F −1 (eitP (|ξ|) ϕ(2−k |ξ|))(2−k x).

(3.28)

By (3.28), we have kJk k∞ . 2k . k

Denote P1 (ξ) = xξ + tP (2 |ξ|). We have Using Van der Corput lemma, we get

|P100 (ξ)|

(3.29) & |t|2

kJk k∞ . |t|−1/2 2k(1−α1 /2) .

kα1

for ξ ∈ supp ϕ. (3.30)

Making an interpolation between (3.29) and (3.30), we get that for any 0 6 θ 6 1, kJk k∞ . |t|−θ/2 2k(1−θα1 /2) . It follows that (a) holds. The proof of (b) is analogous to (a) and the details will be omitted. We now prove (c). In view of the first conclusion in (b), we have X kU (t)40 u0 k∞ . 2k ku0 k1 . ku0 k1 . (3.31) k<0

We first consider the case m2 < 2. Since min(1/m2 , 1/2) = 1/2, from (b) it follows that X kU (t)40 u0 k∞ . |t|−1/2 2k(1−m2 /2) ku0 k1 . |t|−1/2 ku0 k1 . (3.32) k<0

From (3.31) and (3.32), we can get the conclusions. Next, we discuss the case m2 > 2. It suffices to consider the case m2 > 2 and θ = min(1/m2 , 1/2) = 1/m2 . A straightforward calculation yields d 1 . 2−k(m2 −1) , ξ ∈ suppϕ. (3.33) dξ P 0 (2k |ξ|)

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So, if |x| 6 1, then we have |∂ξm (eixξ ϕ(ξ))| . 1. Integrating by part to Jk (·), we have |Jk (x)| . |t|−1 2k(1−m2 ) . If |x| > 1, let k0 be the minimal integer so that |x| 6 |t|2k0 m2 , then |x| ≈ |t|2k0 m2 . For |k − k0 | > C 1, we have |P10 (ξ)| > c|t|2km2 . Integrating by part to Jk (·), we have |Jk (x)| . |t|−1 2k(1−m2 ) . For |k − k0 | 6 C, noticing that |x| > 1 and m2 > 2, we have (1−m2 /2)/m2 |x| . |t|−1/m2 . |Jk (x)| . |t|−1/2 2k(1−m2 /2) . |t|−1/2 |t|

So, X | Jk (x)| . k60

X

|k−k0 |6C

X

.

|k−k0 |6C

. |t|

−1/m2

|Jk (x)| +

X

|k−k0 |>C

|t|−1/m2 +

|Jk (x)|

X

2k +

2k <|t|−1/m2

X

2k >|t|−1/m2

|t|−1 2k(1−m2 )

,

which finishes the proof of (c).

In higher spatial dimensions, we have Proposition 3.7. Let n > 2. Denote U (t) = F −1 eitP (|ξ|) F . We have the following decay estimates. (a) Let {4k }k∈Z be as in Sec. 1.6, k > 0, and P satisfy (H1). Then kU (t)4k u0 k∞ . |t|−θ 2k(n−m1 θ) ku0 k1 , 0 6 θ 6

n−1 . 2

(3.34)

Moreover, if P satisfies (H3), then kU (t)4k u0 k∞ . |t|−n/2 2k(n−

m1 n α1 −m1 2 − 2

)

ku0 k1 .

(3.35)

(b) Let {4k }k∈Z be as in Sec. 1.6, k < 0, and P satisfy (H2). Then we have n−1 kU (t)4k u0 k∞ . |t|−θ 2k(n−m2 θ) ku0 k1 , 0 6 θ 6 . (3.36) 2 Moreover, if P satisfies (H4), then kU (t)4k u0 k∞ . |t|−n/2 2k(n−

m2 n α2 −m2 2 − 2

)

ku0 k1 .

(3.37)

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(c) Let 40 be as in Sec. 1.3, and P satisfies (H2). Then we have n n−1 −θ , . kU (t)40 u0 k∞ . (1 + |t|) ku0 k1 , 0 6 θ 6 min m2 2 (3.38) Moreover, if P satisfies (H4) with α2 = m2 , then n n , . kU (t)40 u0 k∞ . (1 + |t|)−θ ku0 k1 , 0 6 θ 6 min m2 2 (3.39) Proof. The idea is to use the Bessel function. After making the polar coordinate transform, the estimate is reduced to an oscillating integration in one spatial dimension. Next, using the decay and cycle properties of the Bessel function, analogous to the 1D case, we can get the conclusion. Let Jm (r) be the Bessel function Z 1 (r/2)m Jm (r) = eirt (1 − t2 )m−1/2 dt, m > −1/2. Γ(m + 1/2)π 1/2 −1 Let us state some properties on the Bessel functions, see [82] and [203]. Lemma 3.3. For any 0 < r < ∞, we have (i) Jm (r) 6 Crm , d (r−m Jm (r)) = −r−m Jm+1 (r). (ii) dr (i) is obvious. Integrating by part to Jm , we can get (ii). It is known that, the Fourier transform for a radial f is also radial (see [202]): Z ∞ ˆ f (ξ) = 2π f (r)rn−1 (r|ξ|)−(n−2)/2 J n−2 (r|ξ|)dr. (3.40) 2

0

By Lemma 3.3, for any 0 6 s 6 2 and any k > 0, k ∂ n−1 −(n−2)/2 n−2 (rs)) 6 Ck . (ϕ(r)r (rs) J ∂rk 2

(3.41)

[ ] If m = − n−2 2 , Jm (r) has the following property (see 117, Chapter 1, (1.5) ) r−

n−2 2

J n−2 (r) = cn R(eir h(r)), 2

(3.42)

where h satisfies |∂rk h(r)| 6 ck (1 + r)−

n−1 2 −k

.

(3.43)

So, for any s > 2 and k > 0, n−1 |∂rk ϕ(r)rn−1 h(rs) | 6 ck s− 2 .

(3.44)

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Now we prove (a). By Young’s inequality, kU (t)4k u0 k∞ . kF −1 eitP (|ξ|) ϕ(2−k |ξ|)k∞ ku0 k1 .

Using (3.40), we have

F −1 (eitP (|ξ|) ϕ(2−k |ξ|))(x) k

= 2kn F −1 (eitP (|2 ξ|) ϕ(|ξ|))(2k |x|) Z ∞ k = 2kn eitP (2 r) ϕ(r)rn−1 (r2k s)−(n−2)/2 J n−2 (r2k s)dr 2

0 k

=: Ik (2 s),

where s = |x| and J n−2 (r) denotes the Bessel function. It suffices to show 2

By (i) of Lemma 3.3,

kIk (s)k∞ 6 |t|−θ 2k(n−m1 θ) . kIk (s)k∞ . 2kn .

(3.45)

We consider the following two separated cases. k d Case 1. s 6 2. Denote Dr = ( itP 0 (21k r)2k ) dr . We have Dr (eitP (2 r) ) = k

eitP (2

r)

. By the condition (H1), for any m > 0 and r ∼ 1, dm 1 6 Cm 2−k(m1 −1) . drm P 0 (2k r)

(3.46)

We write ϕ(r) ˜ = ϕ(r)rn−1 . Integrating by part to Ik , for any q ∈ Z+ , we have Z ∞ k − n−2 2 J n−2 (rs)dr Ik (s) = 2kn eitP (2 r) ϕ(r)(rs) ˜ 2 Z0 ∞ k − n−2 2 J n−2 (rs)dr = 2kn Dr (eitP (2 r) )ϕ(r)(rs) ˜ 2 0 kn Z ∞ 2 1 itP (2k r) d − n−2 2 =− k e ϕ(r)(rs) ˜ J n−2 (rs) dr 2 it2 0 dr P 0 (2k r) q X 2kn X = Cq,m k q (it2 ) m=0 q ×

Z

Λqm

0

l1 ,...lq ∈Λm

∞

e

itP (2k r)

q Y

∂rlj

j=1

1 P 0 (2k r)

− n−2 2 J n−2 (rs) ∂rq−m ϕ(r)(rs) ˜ dr, 2

(3.47)

+

where = {l1 , . . . , lq ∈ Z : 0 6 l1 < . . . < lq 6 q, l1 + . . . lq = m}. By (3.41), (3.46) and (3.47), we get that for any q ∈ Z+ , |Ik (s)| . |t|−q 2k(n−m1 q) .

(3.48)

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Interpolating (3.48) with (3.45), we get for any θ > 0, |Ik (s)| . |t|−θ 2k(n−m1 θ) . Case 2. s > 2. By (3.42), Z ∞ k irs eitP (2 r) ϕ(r)(e ˜ h(rs) + e−irs ¯h(rs))dr Ik (s) = cn 2kn Z0 ∞ k kn = cn 2 ei(tP (2 r)+rs) ϕ(r)h(rs)dr ˜ 0 Z ∞ k ¯ + cn 2kn ei(tP (2 r)−rs) ϕ(r) ˜ h(rs)dr 0

=: B1 + B2 .

We can assume, without loss of generality that t > 0 and P 0 (r) > 0. We consider the estimate of B1 . Put P1 (r) = tP (2k r) + rs. Noticing that P10 (r) = t2k P 0 (2k r) + s > ct2km1 , we see that (3.46) also holds if one replaces P by P1 . In view of (3.44), analogous to Case 1, we can get that for any θ > 0, |B1 | . |t|−θ 2k(n−m1 θ) . We now consider the estimate of B2 . Put P2 (r) = tP (2k r) − rs. Notice that if s = t2k P 0 (2k r), then P20 (r) = 0. We divide Case 2 into two subcases. Case 2a. s > 2 supr∈[1/2,2] t2k P 0 (2k r), or s < 21 inf r∈[1/2,2] t2k P 0 (2k r). It is easy to see that |P20 (r)| > ct2km1 if r ∼ 1 and (3.46) still holds if one substitutes P by P2 . Using (3.44), we have for any θ > 0, |B2 | . |t|−θ 2k(n−m1 θ) . Case 2b. ing (3.44),

1 2

inf r∈[1/2,2] t2k P 0 (2k r) 6 s 6 2 supr∈[1/2,2] t2k P 0 (2k r). Us|B2 | . 2kn s−

n−1 2

. t−

n−1 2

2k(n−

(n−1)m1 2

)

.

(3.49)

Making an interpolation between (3.49) and (3.45), we get that for any 0 6 θ 6 n−1 2 , |B2 | 6 t−θ 2k(n−m1 θ) .

(3.50)

If (H3) holds, then |P200 (r)| > t2kα1 . Using Van der Corput lemma, Z ∞ α1 −m1 n d kα1 −1/2 |B2 | . (t2 ) | (ϕ(r)h(rs))|dr ˜ . t−n/2 2k(n− 2 (m1 + n )) . dr 0 (3.51) Thus, we finish the proof of (a).

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The proof of (b) is similar to that of (a) and we omit the details. Finally, n we prove (c). Let 0 6 θ 6 min( mn2 , n−1 2 ). If θ < m2 , then n − m2 θ > 0. In view of (b), we immediately have 2 X kU (t)40 u0 k∞ . |t|−θ 2k(n−m2 θ) k40 u0 k1 k=−∞

. |t|−θ k40 u0 k1 . n n Assume that n−1 2 > m2 and θ = m2 . From the proof of (b), we see that k0 m2 for k0 < 0 and s ∼ t2 > 2, there holds n−1

(n−1)m2

n

|Ik0 (s)| . t− 2 2k0 (n− 2 ) . t− m2 . If |k − k0 | > C 1, then |Ik (s)| . t−α 2k(n−m2 α) , ∀ α > 0. So, choosing α large enough, we have X X |I60 (s)| . |Ik (s)| + |Ik (s)| |k−k0 |6C

.t

− mn

2

+

X

|k−k0 |>C

2kn +

− 1 2k
−

X

t−α 2k(n−m2 α)

− 1 2k >t m2

n

. t m2 , which implies the result, as desired. If (H4) holds and m2 = α2 , the proof is analogous and the details are omitted. Using the above estimates to the Klein-Gordon equation utt + u − ∆u = f (t, x), u(0, x) = u0 (x), ut (0, x) = u1 (x),

which corresponds to the semi-group G(t) = eit(I−∆) lowing

1/2

, we have the fol-

1/2

Proposition 3.8. Let G(t) = eit(I−∆) , θ ∈ [0, 1], 2σ(θ, ·) = (n + 1 + θ)(1/2 − 1/·), and 2/β(θ, ·) = (n − 1 + θ)(1/2 − 1/·). Then we have kG(t)f kB s−2σ(θ,p) . |t|−2/β(θ,p) kf kBps0 ,q . p,q

Remark 3.2. If the wave equation contains a damping term, say utt − ∆u + αut = 0, u(0) = u0 , ut (0) = u1 , where α > 0, the generating semi-group is quite different from the case 2 α = 0 and it is similar to F −1 e−αt/2−it|ξ|−iα t/|ξ| F . In fact, there are many works have been devoted to study the decaying estimates for its solutions; cf. W.K. Wang and Yang [248], W.K. Wang and W.J. Wang [249], Ikehata, Nishihara and Zhao [112], Hayashi, Kaikina and Naumkin [103; 104], for details.

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3.2

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63

Strichartz inequalities: dual estimate techniques Slip of bamboo for writing is eventually replaced by paper.

In this section we study the Strichartz estimates for a class of dispersive semi-groups. Such kinds of estimates go back to the pioneer work of R. S. Strichartz [206] in 1977 for the wave semi-group, afterwards there are a series of generalizations, see Kato [120], Cazenave and Weissler [26] for the Schr¨ odinger equation; Pecher [191], Ginibre and Velo [80] for the wave equation; Brenner [22; 21] for the Klein-Gordon equation; Kenig-PonceVega [126] for KdV and more general dispersive equations. Below, we give a unified approach to the Strichartz inequalities based on the dual estimate method, see [235; 234], which simplifies various Strichartz estimates for the above mentioned semi-groups. As corollaries, the Strichartz estimates for a class of higher order semi-groups are also obtained. Denote Z t −1 itP (ξ) U (t) = F e F, A = U (t − τ )f (τ, ·)dτ, (3.52) 0

n

where P (·) : R → R is a smooth function. In what follows we always assume that X = Lp ,

0 or X = Bp,2 , 2 6 p < ∞.

(3.53)

Assume that U (t) satisfies the following estimate kU (t)f kX α . t−θ kf kX ∗ ,

(3.54)

where α ∈ R, θ ∈ (0, 1), X α := (I − 4)−α/2 X, and X ∗ is the dual space of X. Supposing that (3.54) holds, we can get some interesting estimates for U (·) and A. Using (3.52) and (3.54), we have Z t |t − τ |−θ kf (τ )kX ∗ dτ. (3.55) kAf kX α . 0

Using the Hardy-Littlewood-Sobolev inequality, we immediately obtain that Lemma 3.4. Assume that (3.53) and (3.54) are satisfied. For any T > 0 and s ∈ R, we have kAf kL2/θ (−T,T ;X s+α ) . kf kL(2/θ)0 (−T,T ;(X ∗ )s ) ,

(3.56)

where (2/θ)0 denotes the conjugate number of 2/θ, i.e., θ/2 + 1/(2/θ)0 = 1.

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For convenience, we denote by DT the set of all of the functions which are defined in (−T, T ), valued in S and taken finitely many values, I = (−T, T ). It is easy to see that for 2 6 p < ∞, DT is dense in Lq (I, (X ∗ )s ) (s ∈ R, 1 6 q < ∞). Lemma 3.5. Assume that (3.53) and (3.54) are satisfied. Then kU (t)f kL2/θ (R,X s+α/2 ) . kf kH s . Proof.

In fact,

(3.57)

First, we show that for any T > 0, I = (−T, T ), ϕ ∈ S , ψ ∈ DT , Z T (U (t)ϕ, ψ(t))dt . kϕk2 kψkL(2/θ)0 (I,(X ∗ )−α/2 ) . (3.58) −T

Z

Z

T

T

(U (t)ϕ, ψ(t))dt . kϕk2 U (−t)ψ(t)dt .

−T

−T

(3.59)

2

By Lemma 3.4,

Z

2

T

U (−t)ψ(t)dt

−T

2 Z ! Z T T = ψ(t), U (t − τ )ψ(τ )dτ dt −T −T

Z

T

. kψkL(2/θ)0 (I,(X ∗ )−α/2 ) U (t − τ )ψ(τ )dτ

−T .

L2/θ (I,X α/2 )

kψk2L(2/θ)0 (I,(X ∗ )−α/2 ) .

(3.60)

Combining (3.59) with (3.60), we see that (3.58) holds. Since S is dense 0 in L2 , DT is dense in L(2/θ) (I, (X ∗ )−α/2 ), by (3.58) we immediately have kU (t)ϕkL2/θ (I,X α/2 ) . kϕk2 .

(3.61)

Noticing that the above estimates are independent of T , and letting T → ∞, we get that those estimates also hold if (−T, T ) is substituted by R. Taking ϕ = (I − ∆)s/2 f , we can get the result, as desired. Lemma 3.6. Assume that (3.53) and (3.54) are satisfied. Then for any T > 0, I = (−T, T ), kAf kL∞ (I,H s+α/2 ) . kf kL(2/θ)0 (I,(X ∗ )s ) .

(3.62)

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Proof.

65

Analogous to Lemma 3.5, it suffices to show that for any f ∈ DT , kAf kL∞ (I,H α/2 ) . kf kL(2/θ)0 (I,X ∗ ) . s/2

Let us write Js = (I − ∆)

(3.63)

. We have

kAf k2H α/2 = (AJα/2 f, AJα/2 f )

Z t

. kf kL(2/θ)0 (I,X ∗ ) U (· − τ )f (τ )dτ

0

.

Applying the same techniques as in Lemma 3.4, one has that

Z t

. kf kL(2/θ)0 (I,X ∗ ) . U (· − τ )f (τ )dτ

2/θ

α 0

L

(3.64)

L2/θ (I,X α )

(3.65)

(I,X )

Combining (3.64) and (3.65), we have (3.63).

Lemma 3.7. Assume that (3.53) and (3.54)are satisfied. Then for any T > 0, I = (−T, T ), kAf kL2/θ (I,X s+α/2 ) . kf kL1(I,H s ) . Proof.

(3.66)

Analogous to Lemma 3.5, it suffices to prove that for any f ∈ DT , kAf kL2/θ (I,X α/2 ) . kf kL1 (I,L2 ) .

Let ψ, f ∈ DT , we have Z

Z

T

T

(Af (t), ψ(t))dt . kf kL1 (0,T ;L2 ) U (· − t)ψ(t)dt 0

·

(3.67)

. (3.68) L∞ (I,L2 )

Similar to Lemma 3.6, we have

Z

T

. kψkL(2/θ)0 (I,(X ∗ )−α/2 ) . (3.69) U (· − t)ψ(t)dt

∞

· L (I,L2 ) R0 One can similarly estimate −T (Af (t), ψ(t))dt. So, (3.68) and (3.69) imply that Z T (Af (t), ψ(t))dt . kf kL1(0,T ;L2 ) kψkL(2/θ)0 (I,(X ∗ )−α/2 ) . (3.70) 0 Using (3.70) and the duality, we can directly obtain (3.67).

Remark 3.3. If (3.54) is replaced by

kU (t)f kX˙ α . t−θ kf kX˙ ∗ ,

(3.71) p 0 ˙ ˙ ˙ where X = L or X = Bp,2 , then the results in Lemmas 3.4–3.7 also hold if one replaces the nonhomogeneous spaces by corresponding homogeneous spaces, for instance, the substitution of (3.56) in Lemma 3.5 is that kU (t)f kL2/θ (I,X˙ s+α/2 ) . kf kH˙ s .

We omit the details of those results.

(3.72)

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Now we apply the above results to a class of linear dispersive equations. We can use (3.14) and (3.15) to get some general results, however, every specific semi-group seems to be important in applications and so, we separately discuss every concrete semi-group, say S(t), W (t) and Um (t). First, we consider the Schr¨ odinger semi-group S(t) = eit∆ and the higher order m/2 semi-group Um (t) = eit(−∆) : Theorem 3.1. Let m > 2,

∞, n 6 m, 2n/(n − m), n > m, n 1 1 1 = − . γ(·) m 2 · ∗

m =

Assume that 2 6 r, p < m∗ , Um (t) = eit(−∆) kUm (t)φkLγ(p) (I,B˙ s

p,2 )

kAUm f kLγ(p) (I,B˙ s

p,2 )

m/2

(3.73) (3.74)

. Then we have

. kφkH˙ s ,

(3.75)

. kf kLγ(r)0 (I,B˙ s0

r ,2

),

(3.76)

Rt where I ⊂ R is an interval, AUm := 0 Um (t − τ ) · dτ . In (3.75) and (3.76), replacing homogeneous Besov spaces by corresponding Besov spaces, Bessel potential spaces and Riesz potential spaces, respectively, the conclusions still hold. It is known that the solution of the wave equation utt − ∆u = f (t, x),

u(0, x) = u0 (x), ut (0, x) = u1 (x)

(3.77)

is relevant to the semi-group W (t), we have 1/2

Theorem 3.2. Let W (t) = eit(−∆) , n > 2, ∞, n = 2, 3, ∗∗ 2 = 2(n − 1)/(n − 3), n > 3,

2 1 1 2σ(·) = = − . n+1 (n − 1)β(·) 2 · ∗∗ Let 2 6 r, p < 2 . Then we have kW (t)φkLβ(p) (I,B˙ s−σ(p) ) . kφkH˙ s , p,2

(3.78) (3.79)

(3.80)

kAW f kLβ(p) (I,B˙ s−σ(p) ) . kf kLβ(r)0 (I,B˙ s+σ(r) ) , (3.81) p,2 r0 ,2 Rt where I ⊂ R is an arbitrary interval, AW := 0 W (t − τ ) · dτ . In (3.80) and (3.81), substituting homogeneous Besov spaces by corresponding Riesz potential spaces, the results also hold.

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67

The solution of the Klein-Gordon equation utt + u − ∆u = f (t, x),

u(0, x) = u0 (x), ut (0, x) = u1 (x)

is relevant to the semi-group G(t) = eit(I−∆)

1/2

(3.82)

, we have the following

1/2

Theorem 3.3. Let G(t) = eit(I−∆) and θ ∈ [0, 1]. For n > 3 − θ, we write 2∗∗ = 2(n − 1 + θ)/(n − 3 + θ) and for n 6 3 − θ, we denote 2∗∗ = ∞. Let 2 6 r, p < 2∗∗ , 2σ(θ, ·) = (n + 1 + θ)(1/2 − 1/·) and 2/β(θ, ·) = (n − 1 + θ)(1/2 − 1/·). Then we have the following estimates: kG(t)φkLβ(θ,p) (I,B s−σ(θ,p) ) . kφkH s ,

(3.83)

p,2

kAG f kLβ(θ,p) (I,B s−σ(θ,p) ) . kf kLβ(θ,r)0 (I,B s+σ(θ,r) ) , r0 ,2

p,2

(3.84)

Rt where I = (−T, T ) ⊂ R is arbitrary, AG := 0 G(t − τ ) · dτ . In (3.83) and (3.84), replacing Besov spaces by corresponding Bessel potential spaces, the results also hold. For the higher order Schr¨odinger semi-group, if we consider the smooth effect, we have m/2

Theorem 3.4. Let Um (t) = eit(−∆) , m > 2 and 2∗ be as in (3.73). We write 1 1 2 =n − , (3.85) γ(·) 2 · 1 1 2σ(m, ·) = n(2 − m) − . (3.86) 2 · Let 2 6 r, p < 2∗ . Then we have

kUm (t)φkLγ(p) (I,B˙ s−σ(m,p) ) . kφkH˙ s ,

(3.87)

p,2

kAUm f kLγ(p) (I,B˙ s−σ(m,p) ) . kf kLγ(r)0 (I,B˙ s+σ(m,r) ) , p,2

r0 ,2

(3.88)

Rt where I ⊂ R is arbitrary, AUm := 0 Um (t−τ )·dτ . In (3.87) and (3.88), replacing homogeneous Besov spaces by corresponding Riesz potential spaces, the conclusions also hold. Proof. The proofs of Theorems 3.1–3.4 are analogous and we only prove Theorem 3.4. By Proposition 3.3, kUm (t)f kB˙ −2σ(m,p) . t−2/γ(p) kf kB˙ 00 . p,2

p ,2

(3.89)

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0 Putting X˙ = B˙ p,2 and α = −2σ(m, p), in view of Lemma 3.5 we can get (3.87). Now we show that (3.88) holds. By Lemmas 3.4, 3.6, 3.7 and Remark 3.3,

kAUm f kLγ(p) (I,B˙ s−σ(m,p) ) . kf kLγ(p)0 (I,B˙ s+σ(m,p) ) ,

(3.90)

kAUm f kL∞ (I,H˙ s ) . kf kLγ(p)0 (I,B˙ s+σ(m,p) ) ,

(3.91)

kAUm f kLγ(p) (I,B˙ s−σ(m,p) ) . kf kL1 (I,H˙ s ) .

(3.92)

p0 ,2

p,2

p0 ,2

p,2

Case I. p ∈ [2, r]. Taking θ ∈ [0, 1] such that 1/p = (1 − θ)/2 + θ/r, we deduce that 1 θ 1−θ = + , σ(m, p) = θσ(m, r) + (1 − θ)σ(m, 2). (3.93) γ(p) γ(r) ∞ In view of the convexity H¨ older inequality, (3.90) and (3.91) imply that θ kAUm f kLγ(p) (I,B˙ s−σ(m,p) ) 6 kAUm f k1−θ s−σ(m,r) s−σ(m,2) kAUm f k γ(r) L (I,B˙ ) L∞ (I,B˙ ) p,2

r,2

2,2

. kf kLγ(r)0 (I,B˙ s+σ(m,r) ) . r0 ,2

(3.94)

Case II. p > r > 2. Take θ ∈ (0, 1) such that 1/r0 = (1 − θ)/2 + θ/p0 . It follows that θ 1−θ 1 = + , σ(m, r) = θσ(m, p) + (1 − θ)σ(m, 2). (3.95) 0 0 γ(r) γ(p) 1 Noticing the complex interpolation, 0 s+σ(m,p) (Lγ(p) (I, B˙ p0 ,2 ),

0 s+σ(m,r) s L1 (I, B˙ 2,2 ))[θ] = Lγ(r) (I, B˙ r0 ,2 ),

(3.96)

(3.90) and (3.92) imply that 0 s+σ(m,r) s−σ(m,p) AUm : Lγ(r) (I, B˙ r0 ,2 ) → Lγ(p) (I, B˙ p,2 )

is a bounded operator. Theorem 3.4 follows. 3.3

(3.97)

Strichartz estimates at endpoints

Let us recall that the starting point in previous section is the estimate (3.54), where we always assume θ ∈ (0, 1). We consider in this section the case θ = 1. Assume that U (t) is defined in (3.52), and for any 2 0 such that kU (t)f kH α(p) . t−θ(p) kf kp0 , p

∀ 2 6 p < ∞.

(3.98)

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Suppose that there exists a p1 > 2 satisfying θ(p1 ) > 1, then there exists a r ∈ (2, p1 ) such that θ(r) = 1 (see (3.118)), i.e., kU (t)f kH α(r) . t−1 kf kr0 .

(3.99)

r

(3.99) is said to be an endpoint estimate. Recall that the condition θ < 1 in previous section aries from the Hardy-Littlewood-Sobolev (HLS) inequality, which is essential for (3.55) and (3.56). This is why we say that θ(r) = 1 is the endpoint case. A natural question is what happens in the endpoint case θ(r) = 1. If fact, if (3.99) occurs, then the Strichartz estimates still hold, i.e., s ) . kφk s−α(r)/2 . kU (t)φkL2 (I,Br,2 H

(3.100)

This estimate was essentially obtained by Keel and Tao [121],3 where they used the techniques of the interpolation on bilinear operators. According to the proof of (3.57), it suffices to show Z T (U (t)ϕ, ψ(t))dt . kϕk2 kψkL2 (I,B −α(r)/2 ) . (3.101) −T r0 ,2 In order to prove (3.101), analogous to (3.59) and (3.60), one needs to show that

Z

2

T

U (−t)ψ(t)dt . kψk2L2 (I,B −α(r)/2 ) . (3.102)

−T

r0 ,2 2

The left hand side of (3.102) can be rewritten as Z T Z T U (−s)ψ(s), U (−t)ψ(t) dsdt. −T

(3.103)

−T

By (3.103), it is natural to introduce the following bilinear operator ZZ L(F, G) := U (−s)F (s), U (−t)G(t) dsdt, (3.104) D

where

D := {(s, t) : s, t ∈ [−T, T ], s 6 t}.

(3.105)

Our goal is to prove that |L(F, G)| . kF kL2 (I,B −α(r)/2 ) kGkL2 (I,B −α(r)/2 ) . r0 ,2

r0 ,2

(3.106)

3 Keel and Tao did not consider the case α(r) 6= 0, here our proof is a modification of Keel and Tao’s proof.

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Noticing that s and t in (3.102) have equal positions, and taking F = G = ψ, we have from (3.106) that (3.100) holds. Now the question reduces to show the bilinear estimate (3.106). Keel and Tao’s idea is to use the dyadic decomposition on t − s to cancel the singularity at t = s. Put Dj := {(s, t) ∈ D : T 2j < t − s 6 T 2j+1 }.

(3.107)

It is easy to see that D = ∪j60 Dj . Denote ZZ Lj (F, G) := U (t − s)F (s), G(t) dsdt.

(3.108)

Dj

So, it suffices to show that X |Lj (F, G)| . kF kL2 (I,B −α(r)/2 ) kGkL2 (I,B −α(r)/2 ) . r0 ,2

j

(3.109)

r0 ,2

In the following we estimate Lj (F, G).

Lemma 3.8. Assume that (3.98) and (3.99) are satisfied, and P = (1/r, 1/r). Let (1/a, 1/b) ∈ B(P, ε) and ε > 0 is sufficiently small. Then we have |Lj (F, G)| . (2j T )−β(a,b) kF kL2 (I,H −α(a)/2 ) kGkL2 (I,H −α(b)/2 ) , a0

b0

(3.110)

where β(a, b) =

1 (θ(a) + θ(b)) − 1. 2

(3.111)

Proof. We can assume that F and G are Schwartz functions which have compact support contained in [−T, T ] on the time variable. We will consider the following three cases: (1) a = b = p ∈ (2, ∞); (2) 2 6 a < r, b = 2; (3) a = 2, 2 6 b < r. First, we consider case (1). Using Young’s and H¨older’s inequalities, Z Z |Lj (F, G)| . (t − s)−θ(p) kF (s)kH −α(p)/2 kG(t)kH −α(p)/2 dsdt I t−s∼2j T j 1−θ(p)

. (2 T )

p0

p0

kF kL2(I, H −α(p)/2 ) kGkL2 (I, H −α(p)/2 ) . p0

p0

It follows that (3.110) holds. Next, we consider the case (2). By H¨older’s inequality,

Z

·−2j T

|Lj (F, G)| . U (· − s)F (s)ds kGkL2 (I, L2 ) .

·−2j+1 T

2 2 L (I,L )

(3.112)

(3.113)

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Applying (3.98) and H¨ older’s inequality, we have

Z

2

·−2j T

U (· − s)F (s)ds

·−2j+1 T

2 2 L (I,L )

.

Z Z I

.

Z Z I

t−2j T

t−2j+1 T 2j T 0

Z

Z

t−2j+1 T

2j T

0

t−2j T

|s − σ|−θ(a) kF (σ)kH −α(a)/2 kF (s)kH −α(a)/2 dσdsdt a0

|s − σ|−θ(a) ×

kF (σ + t − 2j+1 T )kH −α(a)/2 kF (s + t − 2j+1 T )kH −α(a)/2 dσdsdt a0

j

a0

2−θ(a)

. (2 T )

a0

kF k2L2 (I, H −α(a)/2 ) . a0

(3.114)

By (3.113) and (3.114), |Lj (F, G)| . (2j T )1−θ(a)/2 kF kL2 (I, H −α(a)/2 ) kGkL2 (I, L2 ) . a0

(3.115)

Case (3) is similar to case (2). We denote by Σ the open quadrilateral domain with the vertices (1/p, 1/p) (p 1), (1/r, 1/2), (1/2, 1/2) and (1/2, 1/r), which contains B(P, ε). Now we show that (3.110) holds for any (1/a, 1/b) ∈ Σ. In fact, it suffices to consider the case that 1/b > 1/a. We can choose η ∈ (0, 1), (1/p0 , 1/p0 ) and (1/a0 , 1/2) satisfying (1) and (2), such that 1 1 1 1 1 1 , =η , + (1 − η) , . (3.116) a b p0 p0 a0 2 Noticing that U (t) : L2 → L2 and kU (t)ϕk2 = kϕk2 ,

(3.117)

i.e., θ(0) = α(0) = 0. From (3.117) and (3.98) we see that if α(p0 ) 6= 0, then α(a) 6= 0 for any 2 < a < p0 ; if α(p0 ) = 0, then α(a) = 0 for any 2 < a < p0 . We consider the case α(p0 ) 6= 0. It is easy to see that θ(p) α(p) 1/2 − 1/p = = . θ(q) α(q) 1/2 − 1/q

(3.118)

(3.116) and (3.118) imply that (θ(a), θ(b)) = η(θ(p0 ), θ(p0 )) + (1 − η)(θ(a0 ), θ(0)),

(α(a), α(b)) = η(α(p0 ), α(p0 )) + (1 − η)(α(a0 ), α(0)). So, by the complex interpolation, we have the result, as desired.

(3.119) (3.120)

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We need the following (see [13]): Lemma 3.9. Let

  A0 × B0 → C0 , T : A0 × B1 → C1 ,  A1 × B0 → C1

(3.121)

be bounded bilinear operators, where (A0 , A1 ), (B0 , B1 ), (C0 , C1 ) are compatible Banach pairs4 . Let 0 < η, ηi < 1 and 1 6 p, q 6 ∞ satisfy η = η1 +η2 and 1/p + 1/q > 1. Then T : (A0 , A1 )η0 ,p × (B0 , B1 )η1 ,q → (C0 , C1 )η,1

(3.122)

is a bounded operator. Theorem 3.5. Let U (t) and A be as in (3.52) and satisfy (3.98) and (3.99). Assume that 2 6 p, q 6 r. Then s ) . kφkH s−α(p)/2 , kU (t)φkL2/θ(p) (I,Bp,2

(3.123)

kAf kL2/θ(p) (I,B s−α(p)/2 ) . kf kL(2/θ(q))0 (I,B˙ s+α(q) ) . q0 ,2

p,2

(3.124)

Proof. Now we can use Lemmas 3.8 and 3.9 to show (3.100). Take p = q = 2, η = 2/3 and η0 = η1 = 1/3 in Lemma 3.9. Choose a0 = b0 and a1 = b1 satisfying θ(a1 ) = 1 − 2,

θ(a0 ) = 1 + ,

(3.125)

where > 0 is sufficiently small. Put −α(ai )/2

Ai = Bi = L2 (I, Ha0

i

),

0 ,bi ) Ci = `β(a , ∞

i = 0, 1.

By θ(r) = 1, it follows from (3.118) that  1/r = (1 − η0 )/a0 + η0 /a1 ,    α(r) = (1 − η0 )α(a0 ) + η0 α(a1 ),  β(a0 , b1 ) = β(a1 , b0 ),   (1 − η)β(a0 , b0 ) + ηβ(a0 , b1 ) = 0.

(3.126)

(3.127)

In view of α(p0 ) 6= 0, we have α(a0 ) 6= α(a1 ). (3.127) implies that −α(r)/2

(A0 , A1 )η0 ,2 = (B0 , B1 )η1 ,2 = L2 (I, Br0 ,2

),

(C0 , C1 )η,1 = `01 . (3.128)

So, by Lemma 3.9, X |Lj (F, G)| . kF kL2 (I,B −α(r)/2 ) kGkL2 (I,B −α(r)/2 ) . j

r0 ,2

r0 ,2

(3.129)

4 (A , A ) is said to be a compatible Banach pair, if there exists a linear topological 0 1 space A such that A0 , A1 ⊂ A.

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73

We emphasize that in the right hand side of (3.129), the omitted constant is independent of T . As α(p0 ) = 0, the proof is slight different from the above discussions and we omit the details. So, (3.100) is proved and the other cases of (3.123) have been discussed in the previous section. Below we prove (3.124). In fact, (3.129) implies that kAf kL2 (I,B α(r)/2 ) . kf kL2 (I,B −α(r)/2 ) .

(3.130)

r0 ,2

r,2

Using the same way as in the previous section, we have (see Lemmas 3.6, 3.7) kAf kL∞ (I,L2 ) . kGkL2 (I,B −α(r)/2 ) ,

(3.131)

kAf kL2 (I,B α(r)/2 ) . kf kL1(I,L2 ) .

(3.132)

r0 ,2

r,2

Analogous to the proof of Theorem 3.4, we can get (3.124).

Remark 3.4. We emphasize that Theorem 3.5 is independent of T > 0. Thus, by a standard limit argument, we can get that Theorem 3.5 also holds for I = R. Remark 3.5. One needs the condition r < ∞ in Theorem 3.5 and we may further ask what happens if r = ∞, which corresponds to the cases n = 2 and n = 3 for the Schr¨odinger equation and the wave equation, respectively. Generally speaking, Theorem 3.5 does not hold if r = ∞, however, for the radial functions, the conclusion of Theorem 3.5 is still true in the case r = ∞, see Tao [215]. Corollary 3.1. Let n > 3, S(t) = eit∆ , 2∗ = 2n/(n − 2), 2 6 r, p 6 2∗ , and 2/γ(·) = n (1/2 − 1/·). Then we have kS(t)φkLγ(p) (I,B˙ s ) . kφkH˙ s , p,2

Z t

S(t − τ )f (τ )dτ

0

(3.133)

s ) Lγ(p) (I,B˙ p,2

. kf kLγ(r)0 (I,B˙ s0

r ,2

),

(3.134)

where I = (−T, T ) ⊂ R is an arbitrary interval. In (3.133) and (3.92), replacing homogeneous Besov spaces by corresponding Riesz potential spaces H˙ ρs (ρ = r, p), the conclusion still holds. 1/2

Corollary 3.2. Let W (t) = eit(−∆) , n > 3, 2∗∗ = 2(n − 1)/(n − 3), 2 6 r, p 6 2∗∗ , 2σ(·) = (n + 1)(1/2 − 1/·), and 2/β(·) = (n − 1)(1/2 − 1/·). Then we have kW (t)φkLβ(p) (I,B˙ s−σ(p) ) . kφkH˙ s , p,2

(3.135)

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kAW f kLβ(p) (I,B˙ s−σ(p) ) . kf kLβ(r)0 (I,B˙ s+σ(r) ) , r0 ,2

p,2

(3.136)

Rt where I = (−T, T ) ⊂ R is arbitrary, AW := 0 W (t−τ )·dτ . In (3.135) and (3.136), substituting homogeneous Besov spaces by relevant Riesz potential spaces H˙ ρs , the result also holds. 1/2

Corollary 3.3. Let G(t) = eit(I−∆) , θ ∈ [0, 1], n > 3 − θ, 2∗∗ = 2(n − 1 + θ)/(n − 3 + θ), 2 6 r, p 6 2∗∗ , 2σ(θ, ·) = (n + 1 + θ)(1/2 − 1/·), and 2/β(θ, ·) = (n − 1 + θ)(1/2 − 1/·). Then kG(t)φkLβ(θ,p) (I,B s−σ(θ,p) ) . kφkH s ,

(3.137)

p,2

kAG f kLβ(θ,p) (I,B s−σ(θ,p) ) . kf kLβ(θ,r)0 (I,B s+σ(θ,r) ) , p,2

r0 ,2

(3.138)

Rt where I = (−T, T ) ⊂ R is arbitrary, AG := 0 G(t − τ ) · dτ . In (3.137) and (3.138), replacing Besov spaces by corresponding Bessel potential spaces, the conclusion also hold.

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Chapter 4

Local and global wellposedness for nonlinear dispersive equations Stone from another mountain can be harder than jade. the Book of Songs

—— From

In this chapter we will study the local and global wellposedness in H s for a class of nonlinear dispersive equations. In order to solve a nonlinear dispersive equation, one needs to make a delicate balance between the solution and its survival space. The weak solution has too large existing space, which is hard to be unique. The smooth solution exists in a relatively small space and its existence is somehow a problem. So, one of our main task is to look for the most appropriate space to carry out the solution, which strongly associates with the energy and dispersive structures of the equation. To some extent, Strichartz’ inequalities realize the balance between the solution and its survival space, which enable us to obtain the global wellposed results in the energy space for a class of nonlinear dispersive equations, such as nonlinear Schr¨odinger equations, nonlinear Klein-Gordon equations and nonlinear wave equations and so on.

4.1

Why is the Strichartz estimate useful

Let us consider the Cauchy problem for the nonlinear Schr¨odinger equation (NLS) iut + ∆u = f (u),

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

(4.1)

√ Pn where i = −1, f (u) = |u|α u, α > 0, ∆ = i=1 ∂x2i , u(t, x) is a complex valued function of (t, x) ∈ R × Rn , u0 is the initial value of u at t = 0. We easily deduce that the solutions of (4.1) formally satisfy the conservations 75

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of mass and energy1 : ku(t)k22 = ku0 k22 ,

E(u(t)) = E(u0 ),

(4.2)

where E(u(t)) =

2 1 k∇u(t)k22 + ku(t)k2+α 2+α . 2 2+α

(4.3)

According to the conservation laws, it is natural to ask the wellposedness of solutions in L2 (Rn ) and H 1 (Rn ), respectively. For any α > 0, using the compactness or the parabolic regularity method, we can easily prove the global existence of weak solutions of (4.1), however, the uniqueness, persistence and the continuity of the solution map u0 → u are hard to obtain, see [159]. Noticed that S(t) = eit∆ : L2 → L2 , it is natural to solve NLS in k H (k = 0, 1, 2, ...) by using the boundedness of S(t) in L2 . According to the standard semi-group theory, if one can solve NLS in H k , then the uniqueness, persistence and the continuity of the solution map u0 → u can also be obtained by a standard way. However, this method has much more assumptions on nonlinearity and initial data. Let us consider the equivalent integral form of (4.1): Z t u(t) = S(t)u0 − i S(t − τ )f (u(τ ))dτ, (4.4) 0

here we only assume that f (u) is a nonlinear function. Applying kS(t)f k2 = kf k2 , we have Z t ku(t)k2 6 kS(t)u0 k2 + kS(t − τ )f (u(τ ))k2 dτ = ku0 k2 +

Z

0

0

t

kf (u(τ ))k2 dτ.

(4.5)

If we want to get the solution in C(0, T ; L2 ), then one needs |f (u)| 6 C|u|, from which we have kukC(0,T ;L2) . ku0 k + T kukC(0,T ;L2) .

(4.6)

We can easily construct a contraction mapping by choosing T > 0 sufficiently small and show that NLS is well-posed in C(0, T ; L2). 1 Taking the inner product of (4.1) with u and u , and considering its imaginary and t real part, respectively, we can get (4.2).

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|f (u)| 6 C|u| is a rather strong condition on the nonlinearity, which 2 contains f (u) = sin u and f (u) = (e−|u| − 1)u as examples. The power nonlinearity is not included in the above discussions. Similarly, say for n = 1, one can solve NLS in C(0, T ; H 1 ). Assume that f (0) = 0,

|f 0 (u)| 6 C|u|α ,

0 < α < ∞.

Using the embedding H 1 ⊂ L∞ , we have for k = 0, 1, Z t k∇k u(t)k2 6 kS(t)∇u0 k2 + kS(t − τ )∇k f (u(τ ))k2 dτ 0 Z t k . k∇ u0 k2 + k|u|α ∇k uk2 dτ 0 Z t k . k∇k u0 k2 + kukα ∞ k∇ uk2 dτ.

(4.7)

0

So,

k∇k ukC(0,T ;L2 ) . k∇k u0 k2 + T k∇k ukα+1 C(0,T ;L2 ) .

(4.8)

Hence we obtain the local wellposedness in C(0, T ; H 1 ) for NLS. By the conservation of energy, one can extend it to a global one. The above idea can be generalized to any spatial dimensions and we can solve NLS in C(0, T ; H s ) (s > n/2) by assuming that |f (k) (u)| 6 C|u|α+1−k ,

[n/2] < α < ∞,

k = 0, 1, ..., [n/2] + 1,

where s > n/2 is essential to guarantee the inclusion H s ⊂ L∞ . According to the above discussions, for n > 2, we can not solve NLS in H 1 if we only use the L2 estimates. Now, our question is how to show the wellposedness of NLS in the energy spaces in higher spatial dimensions. Using the Strichartz estimates as tools, we can obtain the global wellposedness results of NLS in H 1 for the power nonlinearity |u|α u with α < 2∗ − 2 (2∗ is as in (3.73)), see Kato [120]. If α = 2∗ − 2, the local wellposedness for NLS can be established by resorting to the Strichartz inequalities. On the basis of the local weposedness result, Bourgain [19] in 1999 developed the localized energy separation method and he obtained the global wellposedness of the radial solutions in 3D and 4D. Colliander, Keel, Staffilani, Takaoka and Tao [44] further developed Bourgain’s technique and they removed the radial assumption. In higher spatial dimensions n > 4 and α = 2∗ − 2, the global wellposedness of NLS is solved by Ryckman and Visan [196] and Visan [232]. If α > 2∗ − 2, the global wellposedness of NLS is still open.

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In order to indicate the general idea, now we briefly give an application of the Strichartz inequality to NLS in 2D; cf. Tsutsumi [226], Cazenave and Weissler [25]. We consider the integral equation: Z t u(t) = S(t)u0 − i S(t − τ )|u(τ )|2 u(τ )dτ. (4.9) 0

Recall the Strichartz inequalities:

kS(t)u0 kL4x,t (R2+1 ) . ku0 k2 ,

Z t

S(t − τ )f (τ )dτ

0

So, we have

(4.10)

L4x,t (R2+1 )

. kf kL4/3(R2+1 ) . x,t

(4.11)

kukL4x,t (R2+1 ) . ku0 k2 + k|u|2 ukL4/3(R2+1 ) x,t

. ku0 k2 +

kuk3L4 (R2+1 ) . x,t

(4.12)

So, we can solve (4.9) in the space L4x,t (R2+1 ) at least for small Cauchy data u0 ∈ L2 . To realize (4.12), one needs to make a delicate balance between the nonlinearity |u|2 u and the Strichartz space L4x,t . The above performance contains a very general idea to show the well posedness of NLS. In the next few sections we will mainly use this technique to study NLS and nonlinear Klein-Gordon equations.

4.2

Nonlinear mapping estimates in Besov spaces

In order to solve a nonlinear dispersive equation, after establishing the Strichartz inequalities for the relevant linear dispersive equation, we need 0 to estimate the nonlinear terms in the spaces Lγ(r) (0, T ; Brs0,2 ). The energy structure is not necessary for the local wellposedness and for the global wellposedness with small data. In this section, we consider a nonlinear mapping estimate in Besov spaces. The relevant estimates go back to the works of Pecher [190], Brenner [21], Ginibre and Velo [79]. For NLS, a general nonlinear estimate can be found in Cazenave and Weissler [26]. Wang [236; 234] obtained the nonlinear estimates for the nonlinear KleinGordon equations (NLKG). Now we prove a general result, which covers NLS, NLKG and their higher order versions as special cases, which was obtained in [241]. The nonlinear mapping estimates in Besov spaces rely upon two modified versions of the H¨ older inequality, one is the convexity H¨older’s inequality as in Proposition 1.21, another is the modified H¨older inequality

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79

in Sec. 1.7. We know that |α| X

Dα f (u) =

f (q) (u)

q XY

Dαi u,

(4.13)

Λqα i=1

q=1

where Λqα = α1 + · · · + αq = α, |αq | > · · · > |α1 | > 1 . Assume that f (u) satisfies |f (k) (u)| 6 C|u|p+1−k ,

k = 0, 1, ..., |α|.

(4.14)

Noticing that2 N Y

i=1

ai −

N Y

bi =

i=1

N i−1 X Y

aj

i=1 j=1

we have for any 1 6 r0 < ∞ and |α| > 1,

N Y

j=i+1

bj (ai − bj ),

k Mh Dα f (u)kLr0 .

|α| q

X X n Y

Dαi u

(|uh |p−q + |u|p−q )|uh − u| q=1 Λqα

+

i=1

q q i−1

X Y Y

Dαj uh Dαj uDαi (uh − u)

|uh |p+1−q i=1

=:

Lr 0

j=1

|α| X X q=1 Λqα

kIq kLr0 +

Lr 0

j=i+1

q X i=1

kIIqi kLr0 .

o (4.15)

Lemma 4.1. Let 2 6 r < ∞ and 0 6 δ 6 s0 ∧ s < ∞. Assume that f ∈ C {s−δ} satisfies the following condition: |f (k) (u)| . |u|p+1−k ,

k = 0, 1, ..., {s − δ},

{s − δ} 6 p + 1,

(4.16)

where we assume that {a} = 1 + [a] if a is not an integer; {a} = a if a is an integer. Suppose that 1 s 1 δ 1 1 s0 0 − + − = 0, − > 0. (4.17) p r n r n r r n If s − δ ∈ / N, then we have kf (u)kB˙ s−δ . kukpB˙ s0 kukB˙ s . 0 r ,2

r,2

r,2

(4.18)

If s − δ ∈ N, substituting homogeneous Besov spaces by relevant Riesz potential spaces, (4.18) also holds. 2 We

assume

Q0

j=1

aj =

QN

j=N+1 bj

= 0.

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Proof. If s − δ < 1, the proof is very easy and we omit the details of the proof. It suffices to consider the case s − δ > 1. We assume that s − δ ∈ / N. By the equivalent norm in homogeneous Besov spaces, 1/2 Z ∞ X −2v α 2 dt , (4.19) kf (u)kB˙ s−δ = t sup k M D f (u)k 0 h r L r0 ,2 t |h|6t 0 |α|=[s−δ]

where v = s − δ − [s − δ]. In view of (4.16) we see that k Mh Dα f (u)kLr0 satisfies (4.15). Step 1. we estimate !1/2 Z ∞ −2v 2 dt Aq := t sup kIq kLr0 . (4.20) t |h|6t 0 In (4.16), we consider the estimate of kIq kLr0 . Put 1 s 1 s 0 + β0 − v 0 a0 = (p − q) − , a00 = − , r n r n 1 s0 + βi − |αi | ai = − , i = 1, · · · , q, r n where βi will be chosen below. If s 6 s0 +1, then we can take β0 = β1 = · · · = βq−1 = 0 and βq = s−s0 . Since s− δ > 1, we see that v = s− δ − [s− δ] 6 s− δ − 1 6 s0 , which implies a00 > 0. In view of s − |αq | 6 s0 , we obtain that aq > 1/r − s0 /n > 0. If q > 2, then |αi | 6 s0 for all i = 1, · · · , q − 1. Indeed, in the opposite case one has that |αq | + |αq−1 | > s0 + 1 > s, which is impossible. So, we have P ai > 0, i = 0, 1, ..., q. Notice that a00 + qi=0 ai = 1/r0 . It follows from the modified H¨ older inequality that kIq kLr0 6 Ckukp−1 s kuh − uk 1/a0 kukB ˙s . 0 L B˙ 0 r,2

r,2

(4.21)

Hence, Aq 6 Ckukp−1 s kukB ˙v B˙ 0 r,2

1/a0 ,2 0

kukB˙ s 6 CkukpB˙ s0 kukB˙ s . r,2

r,2

r,2

(4.22)

Below, we consider the case s > s0 + 1. We will choose suitable βi (i = 0, 1, ..., q) satisfying the following four conditions: (a) 0 6 β0 6 v, 0 6 βi 6 |αi |, i = 1, · · · , q, (b) s0 + β0 > v, s0 + βi > |αi |, i = 1, · · · , q, Pq (c) i=0 βi = s − s0 , (d) s0 + βi 6 s, i = 0, · · · , q. Conditions (a) and (c) imply condition (d), So, it suffices to show that there exist β0 , · · · , βq satisfying (a)–(c).

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If q = 1, then we can take β0 = v, β1 = s − s0 − v. Due to δ 6 s0 , we have β1 6 [s − δ] and s0 + β1 > [s − δ]. So, conditions (a)–(c) are satisfied. We consider the case q > 2 which is divided into the following three subcases. First, we assume that |αq | > s − s0 − v. Let β0 = v, β1 = · · · = βq−1 = 0 and βq = s − s0 − v. Obviously, conditions (a) and (c) hold. On the other hand, it is easy to see that s0 +βq > |αq | and s0 +β0 > v. If |αq−1 | > s0 , then |αq |+ |αq−1 | > s− v > [s− δ], a contraction. So we have s0 + βq−1 > |αq−1 |, which implies the condition (b) holds. Secondly, we consider the case s0 6 |αq | < s − s0 − v. Put β0 = v, βi = |αi | for i = 1, · · · , q − 1 and βq = |αq | + δ − s0 . A straightforward Pq calculation yields i=0 βi = s − s0 . In view of δ 6 s0 and |αq | > s0 , we see that 0 6 βq 6 |αq |, which leads the condition (a) holds. Noticing that s0 + βq = |αq | + δ, we easily see that condition (b) holds. Thirdly, we consider the case |αq | 6 (s − s0 − v) ∧ s0 . In the current Pq case, we easily see that i=1 |αi | = [s − δ] = s − δ − v > s − s0 − v. Hence, Pq we can choose βi ∈ [0, |αi |] (i = 1, · · · , q) satisfying i=1 βi = s − s0 − v. Put β0 = v. Obviously, conditions (a) and (c) hold. From |αi | 6 |αq | 6 s0 it follows that the condition (b) holds. Therefore, we have chosen β0 , · · · , βq satisfying conditions (a)–(d). We have, q 1 s 1 s + β − v + Pq (β − |α |) X 0 0 i 0 i=1 i − + − a00 + ai = p r n r n i=0 1 s 1 δ 1 0 =p − + − = 0. r n r n r Applying the modified H¨ older inequality, kIq kLr0 6

Ckukp−q s0 kuh B˙ r,2

q Y

− ukL1/a00

i=1

kukB˙ s0 +βi .

(4.23)

r,2

By (4.23), we have Aq 6 Ckukp−q s kukB ˙v B˙ 0

1/a00 ,2

r,2

6 Ckukp−q s B˙ 0

r,2

q Y

i=0

q Y

i=1

kukB˙ s0 +βi

kukB˙ s0+βi . r,2

r,2

(4.24)

One can choose θi (i = 0, 1, . . . , q) satisfying s0 + βi = θi s0 + (1 − θi )s. It Pq Pq is easy to see that 0 6 θi 6 1, i=0 θi = q and i=0 (1 − θi ) = 1. Using

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the convexity H¨ older inequality, we have q Y

i=1

kukB˙ s0+βi 6 CkukqB˙ s0 kukB˙ s . r,2

(4.25)

r,2

r,2

So, we obtain the estimate of Aq . Step 2. We estimate !1/2 Z ∞ q q X X dt . Bq := Bqi := t−2v sup kIIqi k2Lr0 t |h|6t 0 i=1 i=1

(4.26)

The estimates of Bq are similar to those of Aq and we only sketch the proof. (I) We consider the case s 6 s0 + 1. If q = 1, the estimate of Bq is very easy and we omit the details. We now consider the estimates of Bqi , i 6= q, q > 2. Put 1 s 0 a0 = (p + 1 − q) − , r n 1 s0 − |αj | aj = − , j 6= i, j = 1, · · · , q − 1, r n 1 s0 − v − |αi | 1 s − |αq | ai = − , aq = − . r n r n It is easy to see that s0 > v + |αj |. If not, then s − δ > |αq | + |αj | + v > s0 + 1 > s, which is impossible. So, aj > 0, j = 1, · · · , q − 1. By s 6 s0 + 1 we see that aq > 0. Using the modified H¨older inequality, we can get the estimate of kIIqi kLr0 . Thus, we obtain the estimates of Bqi for i 6= q. Put 1 s 1 s0 − |αj | 0 − , aj = − , j = 1, · · · , q − 1, a0 = (p + 1 − q) r n r n 1 s0 − v − |αq | aq = − , r n then the estimate of Bqq follows. (II) We consider the case s > s0 + 1. Put 1 s 1 s0 + βj − |αj | 0 a0 = (p + 1 − q) − , aj = − , j = 1, · · · , q, r n r n where β1 , · · · , βq can be chosen as in Step 1. So, (a) 0 6 βi 6 |αi |;

(b) s0 + βi > |αi |;

(c)

q X j=1

βj = s − s0 − v.

Applying the modified H¨ older inequality, Y i Bq 6 Ckukp+1−q kuk s0 +βj kukB˙ s0+βi +v . s0 ˙ ˙ B r,2

j6=i

Br,2

r,2

(4.27)

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83

Since s0 + βj + v 6 s, 1 6 j 6 q, we claim that there exist θi (i = 1, . . . , q) satisfying s0 + βj = θj s0 + (1 − θj )s, j 6= i, j = 1, ..., q, s0 + βi + v = θi s0 + (1 − θi )s. Pq Pq Since i=1 θi = q −1 and i=1 (1−θi ) = 1, in view of the convexity H¨older inequality we obtain the estimates of Bqi for i = 1, · · · , q. Remark 4.1. (i) If δ = s0 = 0 and s > 0, then the result of (4.18) can be slightly improved by kf (u)kB˙ s0 6 CkukpLr kukB˙ s . r,2

r ,2

(4.28)

(ii) Lemma 4.1 covers the nonlinear estimates for NLS and NLKG as in [19; 26; 41; 173; 191; 236; 237]), where Cazenave and Weissler [26] considered the case δ = 0, s0 = s; Wang [236; 237] discussed the nonlinear estimates for NLKG. 2 (iii) If the nonlinearity has an exponential growth, say sinh u, (e|u| − 1)u, the nonlinear mapping estimates can not be covered by Lemma 4.1, one can refer to Nakamura and Ozawa [173], and Wang [239].

4.3 4.3.1

Critical and subcritical NLS in H s Critical NLS in H s

We consider the initial value problem for NLS, iut + ∆u = f (u),

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

(4.29)

σ

where f (u) = c|u| u. If u is a solution of (4.29), then uλ (t, x) = λ2/σ u(λ2 t, λx) also solves (4.29) with initial data λ2/σ u0 (λ·) at t = 0. Let us observe that kuλ kH˙ s (Rn ) = λ2/σ+s−n/2 kukH˙ s (Rn ) .

This implies that σ = 4/(n−2s) is the unique index such that the norm of uλ in H˙ s is invariant for all λ > 0. From this point of view, σ = 4/(n − 2s) > 0 is said to be the critical power in H˙ s (or in H s ) for NLS. For s = n/2 − 2/σ, H˙ s (H s ) is said to be the critical space for NLS. When s < n/2, corresponding to the critical case, σ < 4/(n − 2s) is said to be a subcritical power in H s . If s > n/2, σ < ∞ is said to be a subcritical power in H s . σ > 4/(n − 2s) is said to be a supercritical power in H s for NLS.

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Wellposedness in H s

We consider the following equivalent integral equation with respect to (4.29), Z t S(t − τ )f (u(τ ))dτ, (4.30) u(t) = S(t)u0 − i 0

where S(t) = eit∆ . According to the Strichartz inequalities, if u0 ∈ H s , then s (3.75) and (3.76) indicate that the solution should belong to Lγ(p) (I, Bp,2 ). Assume that |f (k) (u)| 6 C|u|σ+1−k , k = 0, 1, ..., [s] + 1, (4.31) where 0 6 s < n/2, [s] denotes the integer part of s, 0 < σ 6 4/(n − 2s). We have (see [26]) Theorem 4.1. Let 0 6 s < n/2, 0 < σ < 4/(n − 2s). Assume that f ∈ C [s]+1 satisfies (4.31), [s] 6 σ. If u0 ∈ H s , then there exists a T ∗ := T ∗ (ku0 kH s ) > 0 such that (4.30) has a unique solution γ(r) s u ∈ Lloc ([0, T ∗ ); Br,2 ), (4.32) where n(2 + σ) r= . (4.33) n + sσ ∗ Moreover, for any 2 6 p 6 2 (p 6= ∞), we have γ(p) s u ∈ Lloc ([0, T ∗ ); Bp,2 ), (4.34) ∗ and if T < ∞, then s ) = ∞, kukLγ(r) ([0,T ∗ );Br,2 (4.35) ku(t)kH s & (T ∗ − t)1−σ(n−2s)/4 , 0 < t < T ∗ .

(4.36)

In Theorem 4.1, we obtain the local wellposedness in H s , where σ is a subcritical power in H s . If σ is a critical power in H s , we can get the global wellposedness for NLS with small data. Theorem 4.2. Let 0 6 s < n/2 and σ = 4/(n − 2s). Assume that f ∈ C [s]+1 satisfies (4.31), [s] 6 σ. If u0 ∈ H s , then there exists a T ∗ := T ∗ (u0 ) > 0 such that (4.30) has a unique solution u verifying (4.32), where r = n(2 + σ)/(n + sσ) and γ(r) = 2 + σ. Moreover, for any 2 6 p 6 2∗ (p 6= ∞), the solution satisfies (4.34). If T ∗ < ∞, then (4.35) holds. If ku0 kH˙ s is sufficiently small, then the above solution is a global one, i.e., T ∗ = ∞ and kuk∩26p62∗, p6=∞ Lγ(p) (0,∞;B˙ s ) . Cku0 kH˙ s , (4.37) p,2

s ) . Cku0 kH s . kuk∩26p62∗, p6=∞ Lγ(p) (0,∞;Bp,2

(4.38)

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Remark 4.2. If T ∗ < ∞ in Theorem 4.1, then T ∗ has a lower bound which depends on ku0 kH s , i.e., T ∗ > δ(ku0 kH s ) > 0. However, if T ∗ < ∞ in Theorem 4.2, then we can not get the lower bound of T ∗ which only depends on ku0 kH s , namely, T ∗ may depend on not only ku0 kH s , but also the choice of u0 in H s , which is easily seen from the scaling u(t, x) → uλ (t, x) := λ2/σ u(λ2 t, λx). Proof. Now we prove Theorems 4.1 and 4.2. Let T > 0, M > 0 and δ > 0 which will be chosen later. Put n o s s ) 6 M D = u ∈ Lγ(r) (0, T ; Br,2 ) : kukLγ(r) (0,T ;B˙ s ) 6 δ, kukLγ(r) (0,T ;Br,2 , r,2

(4.39)

which is equipped with the metric d(u, v) = ku − vkLγ(r) (0,T ;Lr ) .

(4.40)

Considering the mapping T : u(t) → S(t)u0 − i

Z

0

t

S(t − τ )f (u(τ ))dτ,

(4.41)

we show that T : (D, d) → (D, d) is a contraction mapping for some T, δ, M > 0. Since 1 s 1 1 − + = 0, (4.42) σ r n r r by Lemma 3.1 we have σ+1 kf (u)kB˙ s0 . kukB ˙s ,

(4.43)

s . kf (u)kBrs0 ,2 . kukσB˙ s kukBr,2

(4.44)

r ,2

r,2

r,2

So, kf (u)kLγ(r)0 (0,T ;B˙ s0

r ,2

)

. T 1−σ(n−2s)/4 kukσ+1 . Lγ(r) (0,T ;B˙ s )

(4.45)

r,2

In view of Theorem 3.1, one has that kT ukLγ(r) (0,T ;B˙ s

r,2 )

. ku0 kH˙ s + T 1−σ(n−2s)/4 kukσ+1 . Lγ(r) (0,T ;B˙ s )

(4.46)

r,2

Similarly, we have kT ukLγ(r) (0,T ;Br,2 s )

s ) . ku0 kH s + T 1−σ(n−2s)/4 kukσLγ(r) (0,T ;B˙ s ) kukLγ(r) (0,T ;Br,2 r,2

kT u − T vkLγ(r) (0,T ;Lr ) ,

(4.47)

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. T 1−σ(n−2s)/4 (kukσ + kvkσ )Lγ(r) (0,T ;B˙ s ) ku − vkLγ(r) (0,T ;Lr ) . r,2

(4.48)

Case 1. 0 < σ < 4/(n − 2s). Put δ = 2Cku0 kH˙ s , M = 2Cku0 kH s .

(4.49)

We take T > 0 satisfying 2CT 1−σ(n−2s)/4 M σ 6 1/2.

(4.50)

(4.46)–(4.48) imply that kT ukLγ(r) (0,T ;B˙ s

. δ,

(4.51)

kT ukLγ(r) (0,T ;B˙ s

. M,

(4.52)

r,2 ) r,2 )

1 kT u − T vkLγ(r) (0,T ;Lr ) ku − vkLγ(r) (0,T ;Lr ) . (4.53) 2 So, T : (D, d) → (D, d) is a contraction mapping. So, there exists a u ∈ D satisfying (4.30). Again, in view of the Strichartz inequalities (3.75) and (3.76), we have kukLγ(p) (0,T ;B˙ s

p,2 )

. ku0 kH˙ s + T 1−σ(n−2s)/4 kukσ+1 , Lγ(r) (0,T ;B˙ s )

(4.54)

r,2

s from which we obtain that u ∈ Lγ(p) (0, T ; Bp,2 ). By a standard argument, we can extend the solution above. Considering the mapping Z t T : u(t) → S(t − T )u(T ) − i S(t − τ )f (u(τ ))dτ, (4.55) T

and noticing that u(T ) ∈ H s , we can use the same way as in the above to solve (4.55). Repeating this argument step by step, we find a maximal T ∗ > 0 satisfying (4.33)–(4.36). The uniqueness of the solution can be shown by following the same way as that of (4.48) and we omit the details. Case 2. σ = 4/(n − 2s). By Strichartz estimate (3.75), we see that s ) → 0, as T → 0. kS(t)u0 kLγ(r) (0,T ;Br,2

Put

n o s s ) 6 M D0 = u ∈ Lγ(r) (0, T ; Br,2 ) : kukLγ(r) (0,T ;Br,2 ,

(4.56)

(4.57)

which is equipped with the metric as in (4.40). Take M > 0 satisfying 2CM σ 6 1/2

(4.58)

s ) 6 M/2. kS(t)u0 kLγ(r) (0,T ;Br,2

(4.59)

and T > 0 such that

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87

So, analogous to the above, one can show the local wellposedness of Theorem 4.2. For the global wellposedness with small data, it suffices to take T = ∞ in (D, d) as in Case 1. In fact, analogous to (4.46)–(4.48), we have kT ukLγ(r) (0,∞;B˙ s

r,2 )

. ku0 kH˙ s + kukσ+1 , Lγ(r) (0,∞;B˙ s )

(4.60)

r,2

s ) kT ukLγ(r) (0,∞;Br,2

s ), . ku0 kH s + kukσLγ(r) (0,∞;B˙ s ) kukLγ(r) (0,∞;Br,2 r,2

(4.61)

kT u − T vkLγ(r) (0,∞;Lr ) ,

. (kukσ + kvkσ )Lγ(r) (0,∞;B˙ s ) ku − vkLγ(r) (0,∞;Lr ) .

(4.62)

δ = 2Cku0 kH˙ s 1, M = 2Cku0 kH s .

(4.63)

r,2

Put

Using the contraction mapping argument, we can get the result. 4.4

Global wellposedness of NLS in L2 and H 1

We consider the initial value problem for NLS: iut + ∆u = λ|u|σ u,

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

(4.64)

where λ ∈ R. λ > 0 is the defocusing case and λ < 0 is the focusing case. Recall that the solution of NLS formally satisfies ku(t)k22 = ku0 k22 ,

E(u(t)) = E(u0 ),

(4.65)

2λ 1 k∇u(t)k22 + ku(t)k2+α (4.66) 2+α . 2 2+α From the conservation of energy we see that for the focusing NLS, the 2λ kinetic energy 21 k∇uk22 and the potential energy 2+λ kuk2+σ 2+σ have opposite signs and for the defocusing NLS, the kinetic energy and the potential energy are both positive. So, the defocusing case is better than the focusing case. Roughly speaking, the defocusing NLS has global solutions and the focusing NLS has blowup phenomena. E(u(t)) =

Theorem 4.3. Let 0 < σ < 4/n. If u0 ∈ L2 , then (4.64) has a unique solution γ(r) u ∈ C([0, ∞); L2 ) ∩ ∩2
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Theorem 4.4. Let 2∗ be as in (3.73), λ > 0 and 0 < σ < 2∗ − 2. If u0 ∈ H 1 , then (4.64) has a unique solution γ(r) u ∈ C([0, ∞); H 1 ) ∩ ∩2
Theorems 4.3 and 4.4 are corollaries of Theorem 4.1 and the conservation laws in (4.65). Since we have gotten the local wellposedness in Theorem 4.1, using the standard argument, one can get that the local solution of NLS (4.64) satisfies (4.65) at any local time. By (4.36), we have T ∗ = ∞. For the energy critical case σ = 4/(n − 2), n > 3, Theorem 4.4 also holds, see Chapter 7. 4.5

Critical and subcritical NLKG in H s

We study the Cauchy problem for the nonlinear Klein-Gordon equation (NLKG): utt + (m2 − ∆)u = f (u),

u(0, x) = u0 (x), ut (0, x) = u1 (x),

(4.69)

where m2 > 0. The main results in this section are the following Theorem 4.5. Let n > 2, 1/2 6 s < n/2 and 0 < σ < 4/(n − 2s). Assume that f ∈ C [s+1/2] satisfies |f (k) (u)| 6 C|u|σ+1−k ,

k = 0, 1, ..., [s + 1/2],

(4.70)

[s + 1/2] 6 σ + 1. If (u0 , u1 ) ∈ H s × H s−1 , then there exists a T ∗ := T ∗ (ku0 kH s , ku1 kH s−1 ) > 0 such that (4.69) has a unique solution γ(r) s−β(r) u ∈ C([0, T ∗ ); H s ) ∩ ∩r∈(2,2∗ ) Lloc ([0, T ∗ ); Br,2 ) , (4.71)

where 2β(r) = (n + 1)(1/2 − 1/r), 2/γ(r) = (n − 1)(1/2 − 1/r), and 2(n − 1)/(n − 3), n > 3, ∗ 2 = ∞, n = 2, 3. Moreover, if T ∗ < ∞, then ku(t)kH s & (T ∗ − t)−δ/p , 0 < t < T ∗ ,

(4.72)

where δ = 1 for 0 < σ < 2/(n − 2s), δ = (4 − σ(n − 2s))/2 for 2/(n − 2s) 6 σ < 4/(n − 2s).

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In Theorem 4.5 we get the local wellposedness in H s , where σ is a subcritical power in H s . If σ is a critical power in H s , we have the following global wellposedness result with small data: Theorem 4.6. Let n > 2, 1/2 6 s < n/2 and σ = 4/(n− 2s). Assume that f ∈ C [s+1/2] satisfies (4.70), [s + 1/2] 6 σ + 1. If (u0 , u1 ) ∈ H s × H s−1 , then there exists a T ∗ := T ∗ (u0 , u1 ) > 0 such that (4.69) has a unique solution u verifying (4.71). Moreover, if k(u0 , u1 )kH s ×H s−1 is sufficiently small, then T ∗ = ∞. If the initial data belong to energy spaces, we have Theorem 4.7. Let n > 2, 0 < σ < 4/(n − 2) and f (u) = |u|σ u. If (u0 , u1 ) ∈ H 1 × L2 , then (4.69) has a unique solution γ(r) s−β(r) u ∈ C([0, ∞); H 1 ) ∩ ∩r∈(2,2∗ ) Lloc ([0, ∞); Br,2 ) , (4.73) and

E(u, ut ) :=

m2 1 2 1 k∇uk22 + kuk22 + kut k22 + kuk2+σ 2+σ = E(u0 , u1 ). 2 2 2 2+σ (4.74)

Theorems 4.5 and 4.6 can be developed to the nonlinear wave equation a (NLW), i.e., m2 = 0 in (4.69). Indeed, substituting H s , Bp,2 by relevant s ˙a ˙ H , Bp,2 , we see that Theorems 4.5 and 4.6 also hold for the following NLW utt − ∆u = f (u),

u(0, x) = u0 (x), ut (0, x) = u1 (x).

(4.75)

We have similar results to Theorems 4.5–4.7 for NLKG in one spatial dimension, whose proofs are slightly different from the case n > 2. For NLW, one can use the L2 estimates to get the local wellposedness in one spatial dimension. However, the Strichartz estimates fail in 1D and Theorem 4.6 can not be generalized to one spatial dimension. If σ = 4/(n − 2), the global wellposedness result will be discussed in Chapter 7. Proof. [Sketch Proof of Theorem 4.5] We only consider the case 2/(n − 2s) 6 σ 6 4/(n − 2s). Put ρ=

2(n − 1)(2 + σ) . (n − 1)(2 + σ) + 4 − 2σ(n − 2s)

ρ satisfies the following identity, 1 s − β(ρ) 1 1 − 2β(ρ) 1 − + − = 0. σ ρ n ρ n ρ

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By Lemma 4.1, we have kf (u)kB˙ s−1+β(ρ) . kukσ+1 . ˙ s−β(ρ)

(4.76)

Bρ,2

ρ0 ,2

Consider the equivalent integral equation of (4.69), Z t K(t − τ )f (u(τ ))dτ, u(t) = K 0 (t)u0 + K(t)u1 +

(4.77)

0

where K(t) = (m2 − ∆)−1/2 sin(m2 − ∆)1/2 and K 0 (t) = cos(m2 − ∆)1/2 . Put s−β(ρ)

D = {u ∈ Lγ(ρ) (0, T ; Bρ,2

) : kukLγ(ρ) (0,T ;B s−β(ρ) ) 6 M }, ρ,2

and d(u, v) = ku − vkLγ(ρ) (0,T ;Lρ ) . Using the Strichartz inequalities together with (4.76), we can show that Z t 0 T : u(t) → K (t)u0 + K(t)u1 + K(t − τ )f (u(τ ))dτ (4.78) 0

is a contraction mapping from (D, d) into itself. For the details, see Wang [236; 237].

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Chapter 5

The low regularity theory for the nonlinear dispersive equations From PDE point of view, one of the main reasons that we like Fourier transform is that it makes the differential operation into fundamental algebraic multiplication.

This chapter is devoted to introduce some methods in studying the low regularity theory for the Cauchy problems of the nonlinear dispersive equations: X s,b method and I-method. By the low regularity theory we mean to study the Cauchy problems assuming rough initial data. For instance, whether the Cauchy problems are locally well-posed in H s and how low can s be? How about global well-posedness? The low regularity theory is important for several reasons: First, for Sobolev H s , if s is smaller, then the space H s is larger; thus the scope of initial data is wide, for example, it can include Dirac measure–δ function which belongs to H −1/2− , > 0. Secondly, for some equations, conservation laws of L2 and H 1 can be easily obtained, so the task of extending a local existence result to a global existence result can be easier if one is working at low regularities than high regularities. In this chapter, we take the Korteweg de-Vries equation and the Schr¨ odinger equation with derivative as examples to study the low regularity solution for the nonlinear dispersive equations.

5.1

Bourgain space

In this section, we introduce Bourgain space and its properties. The Cauchy problem for a general dispersive equation usually has the following form: ( ∂t u − iφ(D)u = f (u, ∂ α u), (x, t) ∈ Rn × R (5.1) u(x, 0) = u0 (x) ∈ H s (Rn ). 91

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where u(x, t) : Rn → C is a unknown function, u0 (x) is a given data, ∂ α = (∂xα11 , ..., ∂xαnn ), and φ(D) is a Fourier multiplier: Z φ(D)u = (2π)−n/2 eixξ φ(ξ)b u(ξ)dξ. Rn

Here φ(ξ) is a real-valued continuous function1 , which we refer as dispersion relation of the equation or phase function. Nonlinear term f is a function of multi-variables, for example f is a polynomials or usually of the form |u|p u. Many dispersive equations can be reduced to the form (5.1), for inodinger equation (with dispersion relation φ(ξ) = |ξ|2 ), Kostance, Schr¨ rteweg de-Vries equation (with dispersion relation φ(ξ) = ξ 3 ), and so on. In the previous several chapters we have discussed some space-time structures, such as Strichartz-type space Lqt Lrx , smoothing effect-type Lqx Lrt . In this chapter we will apply a new class of space-time structure X s,b . To briefly introduce the motivation, we view (5.1) as a perturbation of a linear equation as before, hence consider first the corresponding linear equation: ∂t u − iφ(D)u = 0.

(5.2)

Denote S(t)f = Fx−1 eitφ(ξ) Fx f . Taking Fourier transform with respect to both space and time on both side of (5.2), then it becomes u(ξ, τ ) = 0. (τ − φ(ξ))b

(5.3)

Unless particularly specified, u b or F u will always denote the Fourier transform of the function u(x, t) with respect to x and t, Fx u denotes the one only on x.2 From (5.3) we see that u b is supported on the surface {(ξ, τ ) : τ = φ(ξ)}. One can imagine that this surface is intimately related to the equation (5.1). By viewing x and t equally, we then introduce the Sobolev-type space for (5.1) as Sobolev space for the Laplacian equation. Definition 5.1. Assume φ : Rd → R is a continuous function. Let s, b ∈ R. d s,b The space Xτs,b (Rd ×R) or X s,b , is then =φ(ξ) (R ×R), simply denoted as X defined to be the closure of the Schwartz functions S(Rd+1 ) under the norm s

b

kukX s,b = khξi hτ − φ(ξ)i F ukL2ξ L2τ . 1 We

(5.4)

emphasize that φ must be real-valued. there is no confusion, we also denote by u b the Fourier transform of u for function u : Rn → C. 2 If

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These modern forms of the spaces were first used to systematically study nonlinear dispersive wave problems by Bourgain [15; 14]. Klainerman and Machedon [146] used similar ideas in their study of the nonlinear wave equation. The spaces appeared earlier in a dierent setting in the works [194], [7] of Rauch, Reed, and M. Beals. Then the bilinear estimates in these space were deeply studied by Kenig, Ponce and Vega [132], and by Tao [216] in an abstract setting. Bourgain spaces X s,b can exploit deeply dispersive effect of the equation, and analyze the frequency interactions of waves evolving from the equation. Thus it is usually a powerful tool in the well-posedness study, see [132; 218]. We take the nonlinear term uux as an example, then Bourgain space methods reduce to the following two crucial estimates:

Z t

ψ(t)

S(t − s)f (s)ds 6Ckf kX s,b−1 , (5.5)

0

X s,b

k∂x (uv)kX s,b−1 6CkukX s,b kvkX s,b ,

(5.6)

where ψ(t) is a smooth cut-off function. Intuitively, we can view these two inequalities in this way: s is the degree of regularity for the space variable, and b is that R t for the operator ∂t − iφ(D). From (5.5) we see the nonlinear evolution 0 S(t − s) · ds gains one derivative with respect to ∂t − iφ(D), and (5.6) shows this gain can compensate the loss of derivative on the space variable. How much regularity can be obtained through this process depends on the strength of dispersive effect of the equation. More precisely, we will show (5.5) always hold for any φ when 1/2 < b < 1, hence the main task is to show the second inequality which is the main topics of the next section. This section devotes to prove some basic properties of Bourgain X s,b . Assume ψ ∈ C0∞ (R), suppψ ⊂ [−2, 2], and equals to 1 on [−1, 1]. Let ψδ (·) = ψ(·/δ) for any δ > 0. We will use the following lemma, which is proved in Chapter 3. Lemma 5.1. If 0 < α < 1, 1 < p < ∞, then k(−∆)α/2 (f g)kLp . kf kL∞ k(−∆)α/2 gkLp + kgkL∞ k(−∆)α/2 f kLp . (5.7) From the definition of X s,b , it is easy to see that Lemma 5.2. Let s, b ∈ R. Then

kf kX s,b = hξis ke−itφ(ξ) (Fx f )(ξ, t)kHtb L2 . ξ

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Proposition 5.1. Assume s ∈ R, 1/2 < b < b0 < 1, 0 < δ < 1. Then 0

kψδ (t)f kX s,b .δ 1/2−b kf kX s,b ,

kψδ (t)f kX s,b−1 .δ b −b kf kX s,b0 −1 .

Proof. We may assume s = 0. It follows from Lemma 5.2 and Lemma 5.1 that

kψδ (t)f kX 0,b = ke−itφ(ξ) ψδ (t)Fx (f )kHtb L2 ξ

2 . kψδ (t)kHtb ke−itφ(ξ) Fx (f )(ξ, t)kL∞ t Lξ

−itφ(ξ) + kψδ (t)kL∞ ke Fx (f )(ξ, t)kHtb L2 . t ξ

By simple calculations one easily get for any λ > 0

kf (λt)kH˙ b = λb−1/2 kf (t)kH˙ b , kf (λt)kH b .(λ1/2 + λb−1/2 )kf (t)kH b . (5.8)

The first inequality follows from (5.8) and Sobolev’s inequality H b ,→ L∞ . Next we show the second inequality. For simplicity, let c = 1 − b and d = 1 − b0 , and then 0 6 d < c < 1/2. From Lemma 5.2 it suffices to prove kψδ (t)hkH −c .δ c−d khkH −d . t

t

By duality it suffices to prove kψδ (t)gkHtd .δ c−d kgkHtc , ∀ g ∈ Htc .

(5.9)

It follows from H¨ older’s inequality and Sobolev’s inequality that kψδ (t)gkL2t .δ c kgkHtc .

(5.10)

kψδ (t)gkH˙ d .δ c−d kgkHtc .

(5.11)

Then it suffices to prove t

From (5.10) and Gagliardo-Nirenberg inequality we get c−d θ θ kψδ (t)gkH˙ d .kψδ (t)gk1−θ kψδ (t)gk1−θ ˙ c kψδ (t)gkL2 .δ ˙ c kgkHtc , H H t

t

t

t

where θ = (c − d)/c, thus it suffices to show kψδ (t)gkH˙ c .kgkHtc . t

(5.12)

Indeed, kψδ (t)gkH˙ c .(kψδ kL∞ kgkHtc + k((−∆)c/2 ψδ )gkL2t ) t t

.(kgkHtc + k((−∆)c/2 ψδ )gkL2t ).

By H¨ older’s inequality and Sobolev’s inequality one get k((−∆)c/2 ψδ )gkL2t .k(−∆)c/2 ψδ kL1/c kgkHtc .kgkHtc . In conclusion, the proof of the proposition is completed.

(5.13)

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Proposition 5.2. (a) Let s ∈ R, u0 ∈ H s (Rn ), 1/2 < b < 1, 0 < δ < 1. Then kψδ (t)S(t)u0 kX s,b .δ 1/2−b ku0 kH s .

(5.14)

s,b−1

(b) Let s ∈ R, f ∈ X and 1/2 < b < 1. Then

Z t

0 0

ψδ (t) S(t − t )f (t ) .δ 1/2−b kf kX s,b−1 .

0

Proof.

(5.15)

X s,b

First we show (a). It follows from Lemma 5.2 that kψδ (t)S(t)u0 kX s,b = kψδ (t)kHtb ku0 kHxs .

Using (5.8), we have kψδ (t)kHtb 6 (δ 1/2−b + δ 1/2 )kψ(t)kHtb .δ 1/2−b ,

(5.16)

then we get (5.14). For (b), It follows from Lemma 5.2 that the left-hand side of (5.15) equals to

Z t

s

−it0 φ(ξ) 0 0

hξi ψδ (t) e Fx (f )(ξ, t )dt .

2 b 0

Lξ Ht

Htb−1

Thus to prove (b) it suffices to prove for any g ∈

Z t

0 0 1/2−b

ψδ (t) kgkH b−1 . g(t )dt

6 Cδ

Let h(t) = ψδ (t)

Z

t

0

h(t) = ψδ (t)

t

Htb

0

g(t0 )dt0 . Simple calculation implies that Z tZ 0

R

0

eit τ b g(τ )dτ dt0 = ψδ (t)

Z

R

eitτ − 1 gb(τ )dτ. iτ

To control kh(t)kHtb , we divide h into two parts h(t) = h1 (t) + h2 (t), where Z Z eitτ − 1 eitτ − 1 h1 (t) = ψδ (t) gb(τ )dτ, h2 (t) = ψδ (t) gb(τ )dτ. iτ iτ |τ |61 |τ |>1

Since ktn ψδ (t)kH b = δ n k(t/δ)n ψ(t/δ)kH b 6 δ n δ 1/2−b 4n , then from Taylor’s expansion we get Z

X (itτ )n

kh1 kHtb = ψδ (t) g(τ )dτ b b n!(iτ ) Ht |τ |61 n>1 Z X ktn ψδ (t)kH b (iτ )n b g(τ ) . dτ .kgkH b−1 . (5.17) t n! (iτ ) |τ |61 n>1

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Next we need to control kh2 kHtb . Divide it into two parts

Z Z

eitτ gb(τ ) g(τ )dτ b

dτ kh2 kHtb . ψδ (t)

+ ψδ (t)

iτ iτ |τ |>1 |τ |>1 b Ht

:= I + II. Htb

For the term I we have I.kψδ kH b kgkH b−1 t

Z

2(1−b)

|τ |>1

hτ i |τ |2

dτ

1/2

.δ 1/2−b kgkH b−1 . t

(5.18)

For the term II we have

ˆ(τ ) −1 χ|τ |>1 g

II = ψδ (t)Ft (t)

b iτ Ht

−1 χ|τ |>1 gˆ(τ )

−1 χ|τ |>1 gˆ(τ )

∞ .kψδ kH b Ft

∞ + kψδ kL Ft

iτ iτ

Htb

L

.δ 1/2−b kgkH b−1 .

(5.19)

t

Combining (5.17), (5.18)(5.19), then we obtain (5.15) if 1/2 < b < 1.

Lemma 5.3. Assume Y is a space-time Banach space such that for all u0 ∈ L2 (Rn ) and τ0 ∈ R keitτ0 S(t)u0 kY .ku0 kL2 (Rn ) . Then when 1/2 < b < 1, for any u ∈ X 0,b kukY .kukX 0,b . Proof. get

From the inverse Fourier transform and change of variables we

u(x, t) = (2π)−(n+1)/2

Z

u b(ξ, τ )eixξ+itτ dξdτ Z Z −(n+1)/2 itτ = (2π) e u b(ξ, τ + φ(ξ))eixξ eitφ(ξ) dξdτ. Rn+1

R

Rn

Thus from Minkowski’s inequality and the hypothesis on Y we get Z Z kukY . kb u(ξ, τ + φ(ξ))kL2ξ dτ . hτ i−b khτ ib u b(ξ, τ + φ(ξ))kL2ξ dτ. R

R

The lemma follows from the condition b > 1/2 and Cauchy-Schwartz inequality.

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If b = 1/2, there are counter-examples to show both Proposition 5.2 (b) and Lemma 5.3 fail. We leave this to the readers as an exercise. When b = 1/2, one has a good substitute for X s,b : Besov-type Bourgain space F s . We will use this space combined with a special low frequency structure to solve the H −3/4 GWP problem for the KdV equation in Section 5.4. Now we introduce the space F s . For convenience, for k ∈ N ∪ {0}, denote I0 = [−2, 2];

Ik = [2k−1 , 2k+1 ], k > 1.

(5.20)

Assume {ϕk }∞ k=0 is the sequence of functions for the non-homogeneous dyadic decomposition which are constructed in Section 1.3, and {4k }∞ k=0 are the corresponding Littlewood-Paley projector operators. For k ∈ N∪{0} we define the dyadic X s,b -type space Xk (R2 ): ( ) f ∈ L2 (R2 ) : f (ξ, τ ) is supported in Ik × R and P∞ Xk = . (5.21) kf kXk = j=0 2j/2 kϕj (τ − φ(ξ)) · f (ξ, τ )kL2ξ,τ < ∞ Then we resemble these space in a Littlewood-Paley manner X kuk2F s = 22sk kϕk (ξ)F uk2Xk ,

(5.22)

k>0

kuk2N s =

X

k>0

22sk khτ − φ(ξ)i−1 ϕk (ξ)F uk2Xk .

(5.23)

These Besov-type X s,b space was first introduced by Tataru [221], and then widely used by Ionescu and Kenig [113; 115], Tao [219] and the authors [96; 90; 91; 92]. F s is a good substitute of X s,1/2 for the following two reasons: Firstly, it is easy to see X s,1/2+ ⊂ F s ⊂ X s,1/2 , and moreover, F s space can control many space-time norm such as Strichartz-type space Lpt Lqx (while X s,1/2 fails logarithmicly), secondly, F s has the same scale in time as X s,1/2 . This is very similar as the comparison between Besov space n/2 B2,1 and Sobolev space H n/2 , H n/2+ . First we prove the linear estimates in the space F s . Proposition 5.3. (a) Let s ∈ R and u0 ∈ H s . Then kψ(t)S(t)u0 kF s .ku0 kH s . (b) Assume s ∈ R, k ∈ N ∪ {0} and hτ − φ(ξ)i−1 F u ∈ Xk . Then there exists C > 0 independent of k such that

Z t

F ψ(t)

6 Ckhτ − φ(ξ)i−1 F ukX . S(t − s)u(s)ds k

0

Xk

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Proof. To prove (a), from the definition of F s it suffices to show that for any k ∈ N ∪ {0} kϕk (ξ)F (ψ(t)S(t)u0 )kXk .kϕk (ξ)b u0 (ξ)kL2 . By the definition of Xk we get kϕk (ξ)F (ψ(t)S(t)u0 )kXk =

X j>0

b )c 2j/2 kϕk (ξ)ϕj (τ )ψ(τ u0 (ξ)kL2

.kψkB 1/2 kϕk (ξ)b u0 (ξ)kL2 , 2,1

which implies (a). Next we show (b). It follows from simple calculation that Z t F ψ(t) S(t − s)u(s)ds (ξ, τ ) =C

Z

0

F u(ξ, τ 0 )

R

0

b − τ 0 ) − ψ(τ b − φ(ξ)) ψ(τ dτ 0 . 0 τ − φ(ξ)

Let f (ξ, τ ) = F u(ξ, τ 0 )hτ 0 − φ(ξ)i−1 . For fk ∈ Xk we define the operator T: Z b − τ 0 ) − ψ(τ b − φ(ξ)) ψ(τ T (fk )(ξ, τ ) = fk (ξ, τ 0 ) hτ 0 − φ(ξ)idτ 0 . 0 τ − φ(ξ) R Then it suffices to prove that

kT fk kXk 6 Ckfk kXk hold uniformly for k > 0. Denote fk] (ξ, τ ) = fk (ξ, τ + φ(ξ)), (T fk )] (ξ, τ ) = (T fk )(ξ, τ + φ(ξ)). By change of variables we get Z b − τ 0 ) − ψ(τ b ) ψ(τ hτ 0 idτ 0 . (T fk )] (ξ, τ ) = fk] (ξ, τ 0 ) 0 τ R

It is easy to see that ψ(τ b ) b − τ 0 ) − ψ(τ 0 hτ i 6 C[(1 + |τ |)−4 + (1 + |τ − τ 0 |)−4 ]. 0 τ

] For j ∈ N∪{0}, define fk,j = fk (ξ, τ )ϕj (τ −φ(ξ)) and fk,j = fk,j (ξ, τ +φ(ξ)). Then we have "Z #1/2 ] ] −4 j/2 0 2 0 (T fk,j ) (ξ, τ )|.(1 + |τ |) 2 |fk,j (ξ, τ )| dτ Ij

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+

2 X

ϕj+l (τ )

l=−2

Z

Ij

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99

] |fk,j (ξ, τ 0 )|(1 + |τ − τ 0 |)−4 dτ 0

:=I + II. Hence kT fk kXk . .

X

2j

0

/2

j 0 >0

X

j 0 ,j>0

kϕj 0 (τ )(T fk )] (ξ, τ )kL2

0

2j /2 kϕj 0 (τ )(T fk,j )] (ξ, τ )kL2 .

For the term I, it is obvious that X 0 2j /2 kϕj 0 (τ )IkL2 .kfk kXk . j 0 ,j>0

For the term II, it follows from Young’s inequality that X 0 2j /2 kϕj 0 (τ )IIkL2 .kfk kXk . |j 0 −j|.1

Therefore, the proposition is proved.

We give an analogue of Lemma 5.3 in the case b = 1/2 in the following lemma which can be proved similarly. Lemma 5.4. Assume Y is space-time Banach space such that for all u0 ∈ L2 (Rd ) and τ0 ∈ R keitτ0 S(t)u0 kY 6 Cku0 kL2 (Rn ) .

Then for any k ∈ N ∪ {0} and u ∈ F 0

\ k4k (u)kY .k4 k (u)kXk .

5.2

Local smoothing effect and maximal function estimates

In the last Section, we already see that the elements of Bourgain space X s,b are very close to the linear solutions u = S(t)u0 with initial data u0 ∈ H s . Thus to use X s,b we need to get the estimates for the linear solutions. In studying the dispersive equations with derivatives in the nonlinear terms, there are two important kinds of estimates known as Local smoothing effect and maximal function estimates. In this section, we introduce these two

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estimates using S α (t) = Fx−1 eit|ξ| ξ Fx , 1 6 α 6 2 as examples. The results were due to Kenig-Ponce-Vega, for example see [127; 126]. Theorem 5.1. Assume 1 6 α 6 2. Then 1/2 Z 1/2 Z |c u0 (ξ)|2 α 2 dξ , sup |S (t)u0 | dt . |ξ|α x R R Z 1/4 Z 1/2 sup |S α (t)u0 |4 dx . |c u0 (ξ)|2 |ξ|1/2 dξ . R t∈R

(5.24) (5.25)

R

Proof. Let ω(ξ) = |ξ|α ξ. Then ω is invertible and the inverse is denoted by ω −1 . By change of variable η = ω(ξ), we have Z Z −1 u b0 (ω −1 (η)) α ixξ itω(ξ) S (t)u0 = C e e u ˆ0 (ξ)dξ = C eixω (η) eitη 0 −1 dη. ω (ω (η)) R R

Using Plancherel’s equality and making change of varible η = ω(ξ) (noting ω 0 (ξ) = (α + 1)|ξ|α ), we get Z Z |c u0 (ω −1 (η))|2 |c u0 (ξ)|2 α 2 kS (t)u0 kL2 . dη. dξ. 0 −1 (η))|2 t |ξ|α R |ω (ω R

Therefore, (5.24) is proved. Next we prove (5.25). The main idea is to permute the position of x, t, namely, view t as space variable and x is time variable, then L4x L∞ t becomes Strichartz-type estimate, and (4, ∞) is usually admissible pair. From the proof of the first inequality we see S α (t)u0 can be viewed as the linear solution to the dispersive equation with dispersive relation ω −1 (η), −1 c (η)) 0 (ω and with the initial data whose Fourier transform is uω 0 (ω −1 (η)) . It is easy to check that ω −1 (ξ) satisfies (H1), (H2), (H3), (H4) in Section 2.1, and m1 = m2 = α1 = α2 = 1/(α + 1). Then we get the Strichartz estimate kFx−1 eitω

−1

(ξ)

Fx f kL4t L∞ .kf k ˙ x

1−

H2

1 4(α+1)

.

Thus we have 2 !1/2 Z 1− 1 u ˆ0 (ω −1 (η) 2 4(α+1) kS (t)u0 kL4x L∞ . dη . |η| t ω 0 (ω −1 (η)) α

R

By change of variable η = ω(ξ) we get

kS α (t)u0 kL4x L∞ .ku0 kH˙ 1/4 , t which is (5.25) as desired.

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101

The proof we gave here is a simplified argument for the concrete example. In [126], Kenig, Ponce and Vega proved a more precise estimate for a general class of dispersive equation 1/2 !1/2 0 Z 41 Z ω (ξ) 2 4 dξ . . |c u0 (ξ)| sup |S(t)u0 (x)| dx ω 00 (ξ) R R t∈R Next we give another maximal function estimate for S α (t), which was proved in [127]. s Theorem 5.2. Assume s > α+1 4 , α > 1 and u0 ∈ H . Then Z ∞ 1/2 sup |S α (t)u0 |2 dx 6 Cku0 kH s ,

(5.26)

−∞ t∈[−T,T ]

where the constant C > 0 depends on T . To prove Theorem 5.2, we need the following lemma. Lemma 5.5. Assume ϕ is a C ∞ function supported in [2k−1 , 2k+1 ] and k ∈ N. For α > 1, define the function Hkα (·)  k  2 , |x| 6 1, Hkα (x) = 2k/2 |x|−1/2 , 1 6 |x| 6 C2αk ,  1/(1 + x2 ), |x| > C2αk .

Then, if |t| 6 2 there exists C > 0 independent of x, t and k such that Z ei(t|ξ|α ξ+xξ) ϕ(ξ)dξ 6 CHkα (x). (5.27) R

Moreover, if k = 0 and ϕ is smooth function supported in [−2, 2], then the same conclusion holds. R α Proof. Let u(x, t) = R ei(t|ξ| ξ+xξ) ϕ(ξ)dξ. First we have the trivial estimate for any x, t ∈ R |u(x, t)| 6 C2k . Thus it remains to consider the case |x| > 1. Assume k = 0. It follows from integration by parts and |t| 6 2 that Z 2 i(t|ξ|α ξ) −2 ixξ d e ϕ(ξ) dξ .|x|−2 . |u(x, t)|. x e dξ 2 R

Thus it remains to consider the case k > 1.

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Define the set Ω = {ξ ∈ supp ϕ : |(α + 1)t|ξ|α + x| 6 |x|/2}. Choose a cut-off function ζ ∈ C ∞ supported in Ω such that ζ = 1 when |(α + 1)t|ξ|α + x| 6 |x|/3, for example we may set (α + 1)t|ξ|α + x , ζ =η |x| where η is a fixed smooth cut-off function. Since |t| 6 2, if ξ ∈ Ω, then |x| ∼ αt|ξ|α 6 C2αk , hence h(ξ, x) = tξ|ξ|α + xξ satifies |h00 (ξ)| = Ct|ξ|α−1 > C2−k |x|. It follows from Van der Corput’s lemma that Z eih(ξ,x) ϕ(ξ)ζdξ 6 C2k/2 |x|−1/2 .

If ξ ∈ supp (1 − ζ), then |h0 (ξ)| = |(α + 1)t|ξ|α + x| > |x|/3. Integrating by parts, we get for |x| > 1 Z eih(ξ,x) (1 − ζ)ϕ(ξ)dξ 6 C . 1 + x2

In view of the estimates above, the lemma is proved.

Using the lemma above, we can prove the following lemma which implies Theorem 5.2 immediately. Lemma 5.6. If s > ∞ X

1+α 4 ,

sup

j=−∞ |t|61

α > 1, then sup

j<x<j+1

|S α (t)u0 |2

1/2

6 Cku0 kH s .

(5.28)

α

Proof. Define Skα (t)u0 = Fx−1 eitξ|ξ| ϕk (ξ)Fx u0 , where {ϕk }∞ k=0 is nonhomogeneous dyadic functions. It suffices to prove ∞ X

sup

sup

j=−∞ |t|61 j<x<j+1

|Skα (t)u0 |2

1/2

By duality we need to prove

Z 1

∞ Z h X

(1+α)k α

.2 4 S (t)g(·, t)dt k

−1

L2

j=−∞

1

−1

.2

Z

j

(1+α)k 4

ku0 kL2 .

j+1

|g(x, t)|dxdt

2 i1/2

.

(5.29)

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103

By duality (see [223]), (5.29) follows from the following: ∞ X

sup

j=−∞ |t|61

.2

Z

sup j<x<j+1

−1

∞ Z h X (1+α)k

Z

1

2

−1

j=−∞

1

2 1/2 Skα (t − t0 )g(·, t0 )dt0

j+1

|g(x, t)|dxdt

j

2 i1/2

.

(5.30)

To prove (5.30), we get from Lemma 5.5 that Z 1 Z 1 Z α 0 0 0 α Sk (t − t )g(x, t )dt . Hk (y) |g(x − y, t0 )|dt0 dy −1

.

−1

∞ X

Hkα (l)

l=−∞

Z

l+1

l

Z

1

−1

|g(x − y, t0 )|dt0 dy.

Thus the left-hand side of (5.30) is bounded by ∞ h X j=−∞

sup j6x<j+1

∞ X

Z

Hkα (l)

Z

l+1

l

l=−∞

1

|g(x − y, t0 )|dt0 dy

−1

2 i1/2

(5.31)

By change of variables we get (5.31).

∞ ∞ X h X j=−∞

Hkα (l)

l=−∞

Z

Z

l−j+2

l−j−1

−1

It follows from Minkowski’s inequality that ∞ Z l−j+2 Z ∞ h X X α Hk (l) (5.31). .

l=−∞ ∞ X l=−∞

.2

l−j−1

j=−∞

∞ Z h X Hkα (l) j=−∞

∞ Z h X (α+1)k

1

2

j=−∞

−1

j+1

j

Z

Z

|g(z, t0 )|dt0 dz

2 i1/2

1 −1

1

−1

j+1

j

1

|g(z, t0 )|dt0 dz

.

2 i1/2

2 i1/2 |g(x, t0 )|dt0 dx

|g(x, t0 )|dxdt0

2 i1/2

,

where in the last inequality we use the following simple fact: [2αk ]

X l=1

l

−1/2

.

[2αk ] Z l+1 X l=1

x−1/2 dx.2αk/2 .

l

Thus we complete the proof of the lemma.

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The maximal function associated to dispersive equation was first proposed by Carleson [24]. We take Schr¨odinger equation as an example, ( iut + ∆u = 0, u(x, 0) = u0 (x).

From Plancherel’s equality we know as t → 0 the solution u = eit∆ u0 → u0 in H s . Carleson asked the following question: under what condition one has u(x, t) → u0 (x) almost everywhere? To answer the question, he asked further: what is least regularity on u0 ∈ H s so that sup|t|61 |eit∆ u0 (x)| is locally integrable? This problem was completely solved in one dimension, but remains unclear in high dimension. A general conjecture states as following: in any dimension u(x, t) → u0 (x) almost everywhere if and only if s > 1/4. In one dimension we already show if s > 1/2 keit∂xx u0 kL2x L∞ .ku0 kH s . |t|61 A conjecture of Kenig and Vega states that it still hold when s = 1/2. Actually we show the inequality holds when s = 1/2 and kukH s is replaced s by B2,1 . For these problems we refer the readers to [126; 127]. 5.3

Bilinear estimates for KdV and local well-posedness

In this section we consider Korteweg-de Vries (KdV) equation, and prove the local well-posedness by showing the bilinear estimates in X s,b . KdV equation is given by the following form: ( ∂t u + ∂x3 u + 6u∂x u = 0, (x, t) ∈ R2 , (5.32) u(x, 0) = u0 (x) ∈ H s (R). Then the dispersion relation for KdV equation is φ(ξ) = ξ 3 . KdV equation is the fundamental equation in shallow water theory, which was founded in 1895 by Korteweg and de Vries. From the discussion in Section 5.1 we know the main task is to prove the bilinear estimate. First we prove Theorem 5.3. If −3/4 < s < −1/2, then there exists b ∈ (1/2, 1) such that for any b0 ∈ (1/2, b] with b − b0 6 min{−s − 1/2, 1/4 + s/3} we have k∂x (u1 u2 )kX s,b−1 6 Cku1 kX s,b0 ku2 kX s,b0 .

(5.33)

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105

We will prove Theorem 5.3 by adopting the [k; Z]-multiplier ideas. The method is quite standard now. From the definition of X s,b , (5.33) is equal to 0

0

khξis hτ −ξ 3 ib−1 ξ(b u ∗b v )(ξ, τ )kL2ξ,τ .khξis hτ −ξ 3 ib u bkL2ξ,τ khξis hτ −ξ 3 ib vbkL2ξ,τ . For the convenience, we denote

Λs,b (ξ, τ ) = hξis hτ − ξ 3 ib .

Then (5.33) is equal to

−1

ξΛs,b−1 (ξ, τ )[(Λ−1 s,b u) ∗ (Λs,b v)](ξ, τ )

L2ξ,τ

.kukL2(R2 ) kvkL2 (R2 ) .

(5.34)

∞ Assume {ϕk }∞ k=0 ({χk }k=−∞ ) is the nonhomogeneous (homogeneous) dyadic functions, the intervals {Ik }∞ k=0 are given by (5.20). Decompose ξi , τi − ξi3 into dyadic pieces, then by duality we get (5.34) is equivalent to X h2k3 is 2j3 (b−1) 2k3 ki ∈Z,ji ∈Z+ ,

×

Z

h2k1 is 2j1 b0 h2k2 is 2j2 b0

[1Dk3 ,j3 (uk1 ,j1 ∗ vk2 ,j2 )f ](ξ, τ )dξdτ .kukL2 kvkL2 kf kL2 ,

(5.35)

where L2 = L2 (R2 ), for k ∈ Z, j ∈ Z+ we denote

Dk,j = {(ξ, τ ) : ξ ∈ [2k−1 , 2k+1 ], τ − ξ 3 ∈ Ij },

and uk1 ,j1 = χk1 (ξ)ϕj1 (τ − ξ 3 )u. Thus to prove Theorem 5.3, we need to prove first Z . (ξ, τ )(u ∗ v ) · f dξdτ 1 k1 ,j1 k2 ,j2 2 Dk3 ,j3 R

So far, the issues reduce to some elementary calculus. We rewrite it in a symmetric form, then the bilinear estimates will follow from the trilinear estimates below Z 3 Y f1 (ξ1 , τ1 )f2 (ξ2 , τ2 )f3 (ξ3 , τ3 ).C(ki , ji ) kfi kL2 , (5.36) Γ3

i=1

where fi are non-negative functions supported in Dki ,ji , and Γ3 = {ξ1 + ξ2 + ξ3 = 0, τ1 + τ2 + τ3 = 0},

endowed with induced measure 3 . The estimates of the form (5.36) were first studied by Bourgain [14; 15] using the relations between X s,b and 3

R

Γ3

Πi=1,2,3 fi (ξi , τi ) =

R

R4

f1 (ξ1 , τ1 )f2 (ξ2 , τ2 )f3 (−ξ1 − ξ2 , −τ1 − τ2 )dξ1 dξ2 dτ1 dτ2 .

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other space-time norm, then by Kenig, Ponce, Vega [132] using elementary calculus. Later, Tao [216] systematically studied that in an abstract setting by using dyadic decomposition. We will study the estimate of (5.36) in the next lemma, while for a general class of dispersion relation φ(ξ) = |ξ|α ξ we refer the readers to [92]. Our proofs are simplified version of the [k; Z]-multiplier, quite elementary but a little complicated. We suggest the readers to see the crucial ideas in the calculation. For convenience, if a1 , a2 , a3 ∈ R, we use amin 6 amed 6 amax to denote the maximum, medium, minimum among a1 , a2 , a3 . We also denote (f1 ∗ f2 ) · f3 = f1 ∗ f2 · f3 . Lemma 5.7. Assume ki ∈ Z, ji ∈ Z+ , and fki ,ji ∈ L2 (R2 ) are noni negative function supported in ∪jl=0 Dki ,l and kfki ,ji k2 6 1, i = 1, 2, 3. Then (a) for any k1 , k2 , k3 ∈ Z and j1 , j2 , j3 ∈ Z+ Z fk1 ,j1 ∗ fk2 ,j2 · fk3 ,j3 dξdτ 6 C2jmin /2 2kmin /2 ; (5.37) R2

Z

(b) if kmin 6 kmax − 10, then for i = 1, 2, 3

R2

fk1 ,j1 ∗ fk2 ,j2 · fk3 ,j3 dξdτ 6 C2(j1 +j2 +j3 )/2 2−kmax /2 2−(ji +ki )/2 ; (5.38)

(c) if kmin > kmax − 10 > 10, then Z fk1 ,j1 ∗ fk2 ,j2 · fk3 ,j3 dξdτ 6 C2jmin /2 2jmed /4 2−kmax /4 .

(5.39)

R2

Proof. For f, g, h ∈ L2 (R2 ), let J(f, g, h) = of variables we get

R

R2

f ∗ g · hdξdτ . By change

|J(f, g, h)| = |J(g, f, h)| = |J(fe, h, g)| = |J(f, e g , h)|,

where fe(ξ, µ) = f (−ξ, −µ). Note that φ(ξ) = ξ 3 is an odd function4 , Thus feki ,ji and fki ,ji have the same support. By Cauchy-Schwartz inequality and support properties of functions fki ,ji we get J(fk1 ,j1 , fk2 ,j2 , fk3 ,j3 ) Z jmin /2 .2 kfk1 ,j1 (ξ1 , ·)kL2τ kfk2 ,j2 (ξ2 , ·)kL2τ kfk3 ,j3 (ξ1 + ξ2 , ·)kL2τ dξ1 dξ2 R2

4 If

not odd, there are less symmetries, but the methods still work.

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.2kmin /2 2jmin /2

3 Y

i=1

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kfki ,ji kL2 ,

which is (a). For (b), from the support properties we know J(fk1 ,j1 , fk2 ,j2 , fk3 ,j3 ) = 0 unless |kmax − kmed | 6 5.

(5.40)

From symmetry we may assume k1 6 k2 6 k3 , hence |k2 − k3 | 6 5. Then we divide it into three cases: j1 = jmax , j2 = jmax , j3 = jmax . If j3 = jmax , J(fk1 ,j1 , fk2 ,j2 , fk3 ,j3 ) can be rewritten into the following form: Z J(fk1 ,j1 , fk2 ,j2 , fk3 ,j3 ) = fk]1 ,j1 (ξ1 , τ1 )fk]2 ,j2 (ξ2 , τ2 ) R4

× fk]3 ,j3 (ξ1 + ξ2 , τ1 + τ2 + Ω(ξ1 , ξ2 ))dξ1 dξ2 dτ1 dτ2 ,

where fk]i ,ji (ξ, τ ) = fki ,ji (ξ, τ + ξ 3 ), i = 1, 2, 3, and Ω(ξ1 , ξ2 ) = ξ13 + ξ23 − (ξ1 + ξ2 )3 = −3ξ1 ξ2 (ξ1 + ξ2 ). By change of variable ξ20 = ξ1 + ξ2 and H¨older’s inequality we get J(fk1 ,j1 , fk2 ,j2 , fk3 ,j3 ) Z = fk]1 ,j1 (ξ1 , τ1 )fk]2 ,j2 (ξ2 − ξ1 , τ2 ) R4

.

Z

× fk]3 ,j3 (ξ2 , τ1 + τ2 + Ω(ξ1 , ξ2 − ξ1 ))dξ1 dξ2 dτ1 dτ2

R2

kfk]1 ,j1 (·, τ1 )kL2 (R) kfk]2 ,j2 (·, τ2 )kL2 (R)

× kfk]3 ,j3 (ξ2 , τ1 + τ2 + Ω(ξ1 , ξ2 − ξ1 ))kL2 (|ξi |∼Ni , i=1,2) dτ1 dτ2 . (5.41)

By change of variable µ2 = τ1 + τ2 + Ω(ξ1 , ξ2 − ξ1 ) and noting ∂ξ1 [Ω(ξ1 , ξ2 − ξ1 )] ∼ N22 , we get

kfk]3 ,j3 (ξ2 , τ1 + τ2 + Ω(ξ1 , ξ2 − ξ1 ))kL2 (|ξi |∼Ni , i=1,2) . N2−1 kfk]3 ,j3 kL2 (R2 ) . (5.42)

It follows from (5.41), (5.42) and H¨older’s inequality that J(fk1 ,j1 , fk2 ,j2 , fk3 ,j3 ) . 2−k3

3 Y

i=1

kfki ,ji kL2 (R2 ) .

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If j2 = jmax , from symmetry it is identical to the case j3 = jmax . If j1 = jmax , it follows from symmetry that J(fk1 ,j1 , fk2 ,j2 , fk3 ,j3 ) Z = fk]1 ,j1 (ξ1 , τ2 + τ3 + Ω(ξ2 , ξ1 − ξ2 ))fk]2 ,j2 (ξ2 , τ2 )fk]3 ,j3 (ξ1 − ξ2 , τ3 ). 4 R Note that ∂ξ2 [Ω(ξ2 , ξ1 − ξ2 )] ∼ N2 N1 , then repeating the argument we get 3 Y J(fk1 ,j1 , fk2 ,j2 , fk3 ,j3 ) . 2−(k1 +k3 )/2 kfki ,ji kL2 (R2 ) , i=1

hence (b) is proved. Now we prove (c). For the simplicity of notations, we denote fi = fk]i ,ji , i = 1, 2, 3. We may assume j1 6 j2 6 j3 and rewrite J(f1 , f2 , f3 ) into the following form Z f1 (ξ1 , µ1 )f2 (ξ2 − ξ1 , µ2 − µ1 − ξ13 − (ξ2 − ξ1 )3 ) R4

× f3 (ξ2 , µ2 − ξ23 )dξ1 dµ1 dξ2 dµ2 .

It follows from Cauchy-Schwartz inequality that J(f1 , f2 , f3 ) is bounded by

Z

3 3

· kf3 kL2 .

2 f1 (ξ1 , µ1 )f2 (ξ2 − ξ1 , µ2 − µ1 − ξ1 − (ξ2 − ξ1 ) )dξ1 dµ1 2 R

Lξ

The integral area in

L2ξ2 ,µ2

2 ,µ2

is

E = {(ξ1 , µ1 ) : |µ1 |.2j1 , ξ13 + (ξ2 − ξ1 )3 = µ2 − µ1 + O(2j2 )},

where |ξ2 | ∼ 2k2 . Since ξ13 + (ξ2 − ξ1 )3 = 3ξ2 (ξ1 − ξ2 /2)2 + ξ23 /4, then 3ξ2 (ξ1 − ξ2 /2)2 + ξ23 /4 − µ2 + µ1 = O(2j2 ),

moreover, if ξ23 /4 − µ2 + µ1 > 0 then

|ξ1 − ξ2 /2|.2(j2 −k2 )/2 ,

or, if ξ23 /4 − µ2 + µ1 > 0, for some θ > 0,

(ξ1 − ξ2 /2 + θ)(ξ1 − ξ2 /2 − θ) = O(2j2 −k2 ),

and then |ξ1 − ξ2 /2 + θ| . 2(j2 −k2 )/2 or |ξ1 − ξ2 /2 − θ| . 2(j2 −k2 )/2 . In any case we always have |E|.2j1 2(j2 −k2 )/2 . Thus from Cauchy-Schwartz inequality we have J(f1 , f2 , f3 ) .2j1 /2 2(j2 −k2 )/4 kf1 (ξ1 , µ1 )f2 (ξ2 − ξ1 , µ2 − µ1 − ξ13 − (ξ2 − ξ1 )3 )kL2ξ

2 ,µ2 ,ξ1 ,µ1

.2

j1 /2 (j2 −k2 )/4

2

,

which is (c), and therefore, the lemma is proved.

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109

Now we prove Theorem 5.3 using the lemma. In order to better understand how the conditions are imposed, we divide the proof into many cases. For each case one has a condition, then intersection of all conditions would ensure the theorem. Proof. [Proof of Theorem 5.3] From the previous discussion it suffices to prove (5.35). We may assume kuk2 = kvk2 = 1. Then we need to show X

sup kmax >1

kmin 6kmax ,ji ∈Z+ ,

h2k3 is 2j3 (b−1) 2k3

1D

· u k 0 0 1 ,j1 ∗ vk2 ,j2 L2 k ,j k s j b k s j b 3 3 h2 1 i 2 1 h2 2 i 2 2 ξ,τ

(5.43)

is bounded. Actually, from the support properties of uk1 ,j1 , vk2 ,j2 we see 1Dk3 ,j3 · uk1 ,j1 ∗ vk2 ,j2 = 0 unless |kmax − kmed | 6 5, 2jmax ∼ max(2jmed , 2k1 +k2 +k3 ).

(5.44)

Thus one may assume (5.44) in (5.43). Now we prove (5.35) using (5.43). Divide the left-hand side of (5.35) into several parts, we may assume (k2 , k3 ) = (kmed , kmax ), and k2 , k3 > 1, then from (5.43) we get Z X h2k3 is 2j3 (b−1) 2k3 [1Dk3 ,j3 (uk1 ,j1 ∗ vk2 ,j2 )f ](ξ, τ )dξdτ h2k1 is 2j1 b0 h2k2 is 2j2 b0 |k2 −k3 |65,ji ∈Z+ , X . kukL2 k4k2 vkL2 k4k3 f kL2 .kukL2 kvkL2 kf kL2 , |k2 −k3 |65

which is (5.35). Next we study (5.43) according to the frequency interacP P tions. Fix kmax , then the summation below ki ∈Z means kmin ∈Z . Case 1 (low-low interactions): kmax 6 100. It follows from Lemma 5.7 (a) that (5.43).

X

ki ∈Z,ji ∈Z+

h2k3 is 2j3 (b−1) 2k3 2jmin /2 2kmin /2 .1. h2k1 is 2j1 b0 h2k2 is 2j2 b0

Case 2 (high-low interaction): k2 > 100, |k3 − k2 | 6 5, k1 6 k2 − 10 (or k1 , k2 exchange). If j3 = jmax , then we get from Lemma 5.7 (b) that (5.43).

X

ki ∈Z,ji ∈Z+

2j3 (b−1) 2k3 min(2(j1 +j2 )/2 2−k2 , 2j1 /2 2k1 /2 ). h2k1 is 2j1 b0 2j2 b0

Divide the summation into three parts, the first part is k1 6 −2k2 , denoted by I, the second part is k1 > −2k2 and 2k1 /2 6 2j2 /2−k2 , denoted by II, the

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third part is k1 > −2k2 and 2k1 /2 > 2j2 /2−k2 , denoted by III. It is easy to see that I.1. For II, summing on j1 , j2 , j3 we get X

II.

−2k2 6k1 6k2

2(k1 +2k2 )(b−1) . h2k1 is

Then it is easy to get that II.1 if 1 + s 1 and k1 6 1. The term III can be similarly handled and we get III.

X

−2k2 6k1 6k2

2(k1 +2k2 )(b−1) . h2k1 is

Thus if 1 + s < b 6 1 + s/3 then III.1. If j2 = jmax , then from a similar argument and using Lemma 5.7 (b) we can get if 0 6 b − b0 6 1/2 + s/3, X

(5.43).

ki ∈Z,ji ∈Z+

2j3 (b−1) 2k3 min(2(j1 +j3 )/2 2−k2 , 2j1 /2 2k1 /2 ).1. h2k1 is 2j1 b0 2j2 b0

If j1 = jmax , it follows from Lemma 5.7 (b) that if 0 6 b − b0 6 1/4, (5.43).

X

ki ∈Z,ji ∈Z+

2j3 (b−1) 2k3 min(2(j2 +j3 )/2 2−(k2 +k1 )/2 , 2j2 /2 2k1 /2 ).1. h2k1 is 2j1 b0 2j2 b0

Case 3 (high-low interaction II): k2 > 100, |k3 − k2 | 6 10, k1 > k2 − 10. Similarly, it follows from Lemma 5.7 (c) that if s > −3/4, 1/2 1

Case 4 (high-high interactions): k2 > 100, |k1 − k2 | 6 5, k3 6 k2 − 10. If j3 = jmax , from Lemma 5.7 (c) we get 1/2 < b < 5/4 + s, (5.43) .

X

ki ∈Z,ji ∈Z+

.

h2k3 is 2j3 (b−1) 2k3 2(j1 +j2 )/2 2−k1 /2 2−k3 /2 h2k1 is 2j1 b0 h2k2 is 2j2 b0

X h2k3 is max(1, 2k3 +2k1 )b−1 2k3 /2 .1. 22sk1 2k1 /2

ki ∈Z

If j1 = jmax , from Lemma 5.7 (c) we get if b − b0 6 −1/2 − s, (5.43) .

X

ki ∈Z,ji ∈Z+

h2k3 is 2j3 (b−1) 2k3 2(j2 +j3 )/2 2−k1 h2k1 is 2j1 b0 h2k2 is 2j2 b0

(5.45)

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.

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111

X h2k3 is max(1, 2k3 +2k1 )b−b0 −1/2 2k3 −k1 .1. 22sk1

(5.46)

ki ∈Z

The case j2 = jmax is identical to the case j1 = jmax . Taking all the conditions, we proved that if s ∈ (−3/4, −1/2), 1/2 < b < 3/4 + s/3, 1/2 < b0 6 b with b − b0 6 −1/2 − s, then (5.33) holds. Therefore, Theorem 5.3 is proved. If s is large, one has similar estimates. Actually, the same methods above also shows the following theorem. For a ∈ R, a+ denotes a + for a fixed 0 < 1. Theorem 5.4. If −3/4 < s 6 0, then there exists b ∈ (1/2, 1) such that k∂x (u1 u2 )kX s,b−1 6 Cku1 kX s,b ku2 kX s,b

(5.47)

and k∂x (u1 u2 )kX s,b−1 .ku1 kX −3/4+,b ku2 kX s,b + ku1 kX s,b ku2 kX −3/4+,b . (5.48) Remark 5.1. The condition s > −3/4 in Theorem 5.4 is necessary. When s 6 −3/4, (5.47) fails for any b ∈ R. The case s < −3/4 was due to Kenig, Ponce and Vega [132], and the case s = −3/4 was due to Nakanishi, Takaoka and Tsutsumi [180]. Now we prove the local well-posedness for KdV equation using the bilinear estimates. First assume u0 ∈ H s , −3/4 < s < −1/2, and define the operator and set: Z t Φu0 (u) = ψ1 (t)S(t)u0 − ψ1 (t) S(t − t0 )ψT (t0 )∂x (u2 )(t0 )dt0 , 0

B = {u ∈ X s,b : kukX s,b 6 2Cku0 kH s }. We will show if T is sufficiently small, the map Φu0 is a contraction mapping in B. From Proposition 5.2, Proposition 5.1 and Theorem 5.3, we have for 1/2 < b < b0 < 1 0

kΦu0 (u)kX s,b 6Cku0 kH s + CT b −b kuk2X s,b 0

6Cku0 kH s + 4C 2 T b −b ku0 k2H s .

Thus choose T such that 0

4CT b −b ku0 kH s 6 1/2,

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then we have Φu0 (B) ⊂ B. Assume (u1 , u2 ) ∈ B, and we have kΦu0 (u1 ) − Φu0 (u2 )kX s,b 6 1/2ku1 − u2 kX s,b . Thus, Φu0 is a contraction mapping in B and there exists a unique u ∈ B such that Φu0 (u) = u: Z t u(t) = ψ1 (t)S(t)u0 − ψ1 (t) S(t − t0 )ψT (t0 )∂x (u2 )(t0 )dt0 . 0

It is easy to see when t ∈ [−T, T ] Z t u(t) = S(t)u0 − S(t − t0 )∂x (u2 )(t0 )dt0 , 0

hence u is a solution to KdV equation (5.32) in [−T, T ] and u ∈ XTs,b , where XTs,b is defined as following: kukX s,b = inf{ke ukX s,b : u e(t) = u(t)t ∈ [−T, T ]}. T

Next we will show u is the unique solution in XTs,b by using the ideas in [165]. Assume u1 , u2 ∈ XTs,b are two solutions to KdV equation with the same initial data u0 , we will prove u1 (t) = u2 (t), t ∈ [−T, T ]. From symmetry it suffices to prove u1 (t) = u2 (t), t ∈ [0, T ]. For δ > 0 which will be determined later, i = 1, 2, define uei    ui (t), t ∈ [0, δ], uei = ui (2δ − t), t ∈ [δ, 2δ], (5.49)   u , otherwise. 0

Thus t → uei (t) is continuous, and ψ(t)e ui (t) ∈ X s,b , u e1 (t) − u e2 (t) = 0 if t ∈ R \ [0, 2δ]. Since u1 , u2 are solutions to KdV equation, then when t ∈ [0, δ] Z t u1 (t) − u2 (t) = −ψ1 (t) S(t − t0 )ψδ (t0 )∂x [(e u1 − u e2 )(u1 + u2 )](t0 )dt0 . 0

For T > 0 define

kukX s,b = inf{ke ukX s,b : u e(t) = u(t) if t ∈ [0, T ]}. T+

Thus we get from Proposition 5.2, Proposition 5.1 and Theorem 5.3 that 0

u1 − u e2 kX s,b . ku1 − u2 kX s,b 6 δ b −b (kukX s,b + kvkX s,b )ke δ+

T

T

From the constructing we know

ke u1 − u e2 kX s,b 6 2ku1 − u2 kX s,b , δ+

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113

then we have 0

ku1 − u2 kX s,b 6 Cδ b −b (kukX s,b + kvkX s,b )ku1 − u2 kX s,b . δ+

T

T

δ+

b0 −b

Choose δ such that Cδ (kukX s,b + kvkX s,b ) < 1/2, then we get u1 (t) = T T u2 (t) if t ∈ [0, δ]. Repeating this procedure, we obtained the uniqueness in [0, T ]. If s is large, then we use the scaling invariance. It is easy to see that KdV equation has the following scaling invariance: for any λ > 0 u(x, t) → λ2 u(λx, λ3 t), u0 (x) → λ2 u0 (λx).

(5.50)

Then H˙ −3/2 is critical in the sense that: kλ2 u0 (λ·)kH˙ −3/2 = ku0 kH˙ −3/2 . From the fact 3

kλ2 u0 (λx)kH −3/4+ .λ 2 + ku0 kH −3/4+ + λ3/4+ ku0 kH −3/4+ , thus choosing λ sufficiently small, we may assume kφkH −3/4+ 6 0 1.

(5.51)

The rest argument is similar to the case −3/4 < s < −1/2, and we leave it to the readers. Therefore, we prove Theorem 5.5. Assume s ∈ (−3/4, 0], u0 ∈ H s , Then there exists T = T (ku0 kH −3/4+ ) > 0 and b > 1/2 such that KdV equation (5.32) has a unique solution u(x, t) ∈ XTs,b ⊂ C([−T, T ]; H s ). Moreover, ∀ R > 0, the map u0 → u(t) is Lipschitz continuous from {φ ∈ H s , kφkH s 6 R} to C([−T, T ]; H s ). 5.4

Local well-posedness for KdV in H −3/4

In Section 5.3, we consider the local well-posedness of Korteweg-de Vries equation in H s with s > −3/4. It is known that the bilinear estimate (5.47) is invalid if s 6 −3/4, one can refer to [132]. Moreover, KdV equation is not locally well-posed in H s for s < −3/4, the solution operator fails to be uniformly continuous with respect to the H s norm [47; 134]. Thus there exists a natural question: whether is the Cauchy problem of KdV equation locally well-posed in H −3/4 ? In this section, we give the answer, one can refer to [90]. From the results in Section 5.3, it follows that the space X s,b with b > 1/2 is unsuitable to be chosen as the work space for the local wellpopsedness in H −3/4 . We try to choose a new space instead of X s,b with

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b > 1/2. From the definition of Bourgain’s spaces X s,b , it is expected to 1/2 obtain more elaborate estimates if Htb with b > 1/2 is replaced by B2,1 in spaces X s,b . In Section 5.1, we have introduced this space F s , where φ(ξ) = ξ 3 . The key approach is prove whether the following bilinear in spaces F s holds k∂x (uv)kN s 6 C(kukF s kvkF s + kvkF s kukF s ).

(5.52)

By the definitions of F s and N s , in order to obtain the bilinear estimate (5.52), it suffices to show that the following dyadic bilinear estimate [ [ [ [ khτ − ξ 3 i−1 ϕk3 (ξ)ξ(4 k1 u ∗ 4k2 v)kXk3 6C(k1 , k2 , k3 )k4k1 ukXk1 k4k2 vkXk2 . [ [ By examining the supports of the functions, ϕk3 (ξ)(4 k1 u∗ 4k2 v) ≡ 0 unless | max(k1 , k2 , k3 ) − med(k1 , k2 , k3 )| 6 5. Thus we have the following several cases depending on the relative sizes k1 , k2 , k3 : high × low → high, low × high → high, high × high → low, high × high → high, low × low → low. For each case, we have the corresponding estimtaes. Using Lemma 5.4, Theorem 5.1 and the proof of Lemma 5.6, we have the following proposition. Proposition 5.4. If k ∈ Z+ , j ∈ N, I ⊂ R and |I|.1, then 2 .kF [4k (u)]kX , k4k (u)kL∞ k t Lx

.23k/4 kF [4k (u)]kXk , k4k (u)kL2x L∞ t∈I k4k (u)kL4x L∞ .2k/4 kF [4k (u)]kXk , t −j 2 .2 kF [4j (u)]kXj . k4j (u)kL∞ x Lt

Now we prove dyadic bilinear estimates according to frequency interactions. Proposition 5.5. (a) If k > 10, |k − k2 | 6 5, then d d [ [ khτ − ξ 3 i−1 ϕk (ξ)ξ(4 0 u ∗ 4k2 v)kXk .k40 vkX0 k4k2 vkXk2 .

(5.53)

(b) If k > 10, |k − k2 | 6 51 6 k1 6 k − 9, then

3 −k/2−k1 [ [ [ [ khτ − ξ 3 i−1 ϕk (ξ)ξ(4 k4k1 ukXk1 k4 k1 u ∗ 4k2 v)kXk .k 2 k2 vkXk2 . (5.54)

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115

For the simplicity of notations, we assume k = k2 . We always

Proof. denote

uk,j = ϕk (ξ)ϕj (τ − ξ 3 )b u. First we show (a). From the definition of Xk we get X k d d khτ − ξ 3 i−1 ϕk (ξ)ξ(4 2−j/2 k1Dk,j (u0,j1 ∗ vk,j2 )k2 . 0 u ∗ 4k v)kXk .2 j,j1 ,j2 >0

Similar to the proof of Lemma 5.7 (b), we get

k1Dk,j (u0,j1 ∗ vk,j2 )k2 .2−k 2(j1 +j2 )/2 ku0,j1 k2 kvk,j2 k2 . Thus k d d khτ − ξ 3 i−1 ϕk (ξ)ξ(4 0 u ∗ 4k v)kXk .2

X

j,j1 ,j2 >0

2−j/2 k1Dk,j (u0,j1 ∗ vk,j2 )k2

d [ .k4 0 vkX0 k4k2 vkXk2 .

Then (a) is proved. For (b), from the definition of Xk we have X k d [ 2−j3 /2 k1Dk,j3 (uk1 ,j1 ∗ vk,j2 )k2 . khτ − ξ 3 i−1 ϕk (ξ)ξ(4 k1 u ∗ 4k v)kXk .2 ji >0

(5.55)

From the support properties we may assume jmax > 2k + k1 − 10 in the right-hand side of (5.55). We may also assume j1 , j2 , j3 6 10k, otherwise (b) follows from Lemma 5.7 (a) immediately. It follows from Lemma 5.7 (b) that X 2k 2−j3 /2 k1Dk,j (uk1 ,j1 ∗ vk,j2 )k2 j3 ,j1 ,j2 >0

X

. 2k

j3 ,j1 ,j2 >0

.2

k

X

2−j/2 2jmin /2 2−k/2 2−k1 /2 2jmed /2 kuk1 ,j1 k2 kvk,j2 k2

jmax >2k+k1 −10 3 −k/2 −k1

.k 2

2

d [ k 3 2−k/2 2−k1 /2 2−jmax /2 k4 k1 ukXk1 k4k vkXk

d [ k4 k1 ukXk1 k4k vkXk .

Therefore, the proposition is proved.

Proposition 5.6. If k > 10, |k − k2 | 6 5 and k − 9 6 k1 6 k + 10, then

−3k/4 d d [ [ khτ − ξ 3 i−1 ϕk1 (ξ)ξ(4 k4k ukXk k4 k u ∗ 4k2 v)kXk1 . 2 k2 vkXk2 . (5.56)

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Proof.

We assume k = k2 . From the definition of Xk1 we get d d khτ − ξ 3 i−1 ϕk1 (ξ)ξ(4 k u ∗ 4k v)kXk1 X . 2k1 2−j1 /2 k1Dk1 ,j1 (uk,j2 ∗ vk,j3 )k2 .

(5.57)

ji >0

As before, we may assume jmax > 3k − 20 and j1 , j2 , j3 6 10k. It follows from Lemma 5.7 (c) that X 2−j1 /2 k1Dk1 ,j1 (uk,j2 ∗ vk,j3 )k2 2k1 j1 ,j2 ,j3 >0

.

X

+

j1 =jmax

X

+

j2 =jmax

:= I + II + III.

X

j3 =jmax

−j1 /2 3k/4 jmin /2 j /4 2 2 2 2 med kuk,j2 k2 kvk,j3 k2

The term I is easy to control, we omit the details. By symmetry we only need to control II, and dividing it into two parts we get X X II. + 2−j1 /2 23k/4 2jmin /2 2jmed /4 kuk,j2 k2 kvk,j3 k2 j2 =jmax ,j1 6j3 j2 =jmax ,j1 >j3

:=II1 + II2 .

For II1 , by summing on j1 we get X II1 . 2−j1 /2 23k/4 2j1 /2 2j3 /4 kuk,j2 k2 kvk,j3 k2 j2 =jmax ,j1 6j3

.

X

23k/4 2j3 /2 kuk,j2 k2 kvk,j3 k2

X

2−j1 /2 23k/4 2j3 /2 2j1 /4 kuk,j2 k2 kvk,j3 k2

j2 >3k−20,j3 >0

d [ . 2−3k/4 k4 k ukXk k4k2 vkXk2 .

For II2 we have II2 .

j2 =jmax ,j1 >j3

d [ . 2−3k/4 k4 k ukXk k4k2 vkXk2 .

Therefore, the proposition is proved.

Proposition 5.7. If 0 6 k1 , k2 , k3 6 100, then [ [ [ [ khτ − ξ 3 i−1 ϕk1 (ξ)ξ(4 k2 u ∗ 4k3 v)kXk1 .k4k2 ukXk2 k4k3 vkXk3 .

(5.58)

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Proof.

By the definition we get

3 −1

khτ −ξ i

117

k1 [ [ ϕk1 (ξ)ξ(4 k2 u∗4k3 v)kXk1 . 2

X 2−j1 /2 k1Dk1 ,j1 (uk2 ,j2 ∗vk3 ,j3 )k2 .

ji >0

From the support properties we have 1Dk1 ,j (uk2 ,j1 ∗vk3 ,j2 ) = 0 unless |jmax − jmed | 6 10 or jmax 6 1000. It follows from Lemma 5.7 (a) that [ [ [ [ khτ − ξ 3 i−1 ϕk1 (ξ)ξ(4 k2 u ∗ 4k3 v)kXk1 .k4k2 ukXk2 k4k3 vkXk3 .

Therefore, the proposition is proved.

Actually, for the low-low interactions, we can prove a stronger result. Proposition 5.8. If 0 6 k1 , k2 , k3 6 100, then [ 2 k4k vkL∞ L2 . khτ − ξ 3 i−1 ϕk1 (ξ)ξ F [ψ(t)4k2 u] ∗ 4 k3 v kXk1 .k4k2 ukL∞ 3 x t t Lx

Proof. It follows from the definition of Xk1 , Plancherel’s equality and Bernstein’s inequality that [ khτ − ξ 3 i−1 ϕk1 (ξ)ξ(F [ψ(t)4k2 u] ∗ 4 k3 v)kXk1 X 2 k4k vkL∞ L2 , . 2k1 2−j3 /2 kψ(t)4k2 u · 4k3 vkL2t L2x .k4k2 ukL∞ 3 x t Lx t j3 >0

which implies the proposition.

The last one is the high-high to low interactions. Proposition 5.9. (a) If k > 10|k − k2 | 6 5, then

−3k/2 d d [ [ k4k ukXk k4 khτ − ξ 3 i−1 ϕ0 (ξ)ξ(4 k u ∗ 4k2 v)kX0 . k2 k2 vkXk2 .

(b) If k > 10, |k − k2 | 6 5, 1 6 k1 6 k − 9, then

d [ khτ − ξ 3 i−1 ϕk1 (ξ)ξ(4 k u ∗ 4k2 v)kXk1

d [ .(2−3k/2 + k2−2k+k1 /2 )k4 k ukXk k4k2 vkXk2 .

Proof. First we show (a), assuming k = k2 . The left-hand side of the inequality in (a) is bounded by 0 X

k3 =−∞

2k3

X

j1 ,j2 ,j3 >0

2−j3 /2 k1Dk3 ,j3 · (uk,j1 ∗ vk,j2 )k2 .

(5.59)

We may assume k3 > −10k and j1 , j2 , j3 6 10k. We only consider the worst case |j3 − 2k − k3 | 6 10. It follows from Lemma 5.7 (b) that d d khτ − ξ 3 i−1 ϕ0 (ξ)ξ(4 k u ∗ 4k v)kX0

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.

0 X

X

k3 =−10k j1 ,j2 >0

2−k 2−k3 /2 2k3 2−k/2 2−k3 /2 2j1 /2 2j2 /2 kuk,j1 k2 kvk,j2 k2

d d .k2−3k/2 k4 k ukXk k4k vkXk .

(5.60)

Thus, (a) is proved. Now we prove (b), assuming k = k2 . From definition we have X k1 d d khτ − ξ 3 i−1 ϕk1 (ξ)ξ(4 2−j1 /2 k1Dk1 ,j1 (uk,j2 ∗ vk,j3 )k2 . k u ∗ 4k v)kXk1 .2 ji >0

(5.61)

We may assume jmax > 2k + k1 − 10 and j1 , j2 , j3 6 10k. We will control the right-hand side of (5.61) case by case. If j1 = jmax , it follows from Lemma 5.7 (b) that X 2k1 2−j1 /2 k1Dk1 ,j1 · (uk,j2 ∗ vk,j3 )k2 j1 ,j2 ,j3 >0

X

. 2k1

X

j1 >2k+k1 −10 j2 ,j3 >0

.2

−3k/2

2−j1 /2 2−k/2 2−k1 /2 2(j2 +j3 )/2 kuk,j2 k2 kvk,j3 k2

d [ k4 k ukXk k4k2 vkXk2 .

If j2 = jmax , we have better estimate for the characterization multiplier. It follows from Lemma 5.7 (b) that X 2−j1 /2 k1Dk1 ,j1 · (uk,j2 ∗ vk,j3 )k2 2k1 j1 ,j2 ,j3 >0

X

. 2k1

X

j2 >2k+k1 −10 j1 ,j3 >0

.k2

−2k k1 /2

2

2−j1 /2 2−k 2(j1 +j3 )/2 kuk,j2 k2 kvk,j3 k2

d [ k4 k ukXk k4k2 vkXk2 ,

where in the last inequality we use j1 6 10k. By symmetry, the case j3 = jmax is identical to the case j2 = jmax . Thus the proposition is proved. Next we prove the bilinear estimates in F s . Theorem 5.6. Fix s ∈ (−3/4, 0]. Then ∀ s 6 σ 6 0, there exists C > 0 such that ∀ u, v ∈ F σ k∂x (uv)kN σ 6 C(kukF s kvkF σ + kvkF s kukF σ ).

(5.62)

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Proof.

119

From definition we have X k∂x (uv)k2N σ = 22σk3 khτ − ξ 3 i−1 ϕk3 (ξ)ξ(b u ∗ vb)k2Xk . 3

k3 ∈Z+

Applying Littlewood-Paley decomposition to u, v we get

khτ − ξ 3 i−1 ϕk3 (ξ)ξ(b u ∗ vb)kXk3 X 3 −1 [ [ . khτ − ξ i ϕk3 (ξ)ξ(4 k1 u ∗ 4k2 v)kXk3 .

(5.63)

k1 ,k2 ∈Z+

We may assume |kmax − kmed | 6 5. By symmetry we may also assume k1 6 k2 . Then the right-hand side of (5.63) is bounded 4 X

X

j=1 k1 ,k2 ∈Aj

[ [ khτ − ξ 3 i−1 ϕk3 (ξ)ξ(4 k1 u ∗ 4k2 v)kXk3 ,

(5.64)

where Aj , j = 1, 2, 3, 4, is defined as following A1 = {k2 > 10, |k2 − k3 | 6 5, k1 6 k2 − 10};

A2 = {k2 > 10, |k2 − k3 | 6 5, k2 − 9 6 k1 6 k2 + 10}; A3 = {k2 > 10, |k2 − k1 | 6 5, k3 6 k1 − 10}; A4 = {k1 , k2 , k3 6 100}.

Thus, (5.62) follows from Propositions 5.5-5.9, the condition −3/4 < s 6 0 and discrete Young’s inequality. With this bilinear estimates and the scaling (5.50), we can also obtain the LWP of KdV equation in H s for s > −3/4. From the proof of Theorem 5.6 we see the condition s > −3/4 was only needed in Proposition 5.9 (a). Thus to study the endpoint case s = −3/4, one naturally asks whether Theorem 5.6 holds at s = −3/4, more precisely, whether the bound in Proposition 5.9 (a) can be improved to 2−3k/2 ? The answer is negative by the following counter-example which was given by N. Kishimoto [141] using the one in [180]. Proposition 5.10. If k > 200 and |k − k2 | 6 5, then there exists u, v ∈ F −3/4 such that − 3k d d [ [ khτ − ξ 3 i−1 ϕ0 (ξ)ξ(4 k u ∗ 4k2 v)kX0 & log(k)2 2 k4k ukXk k4k2 vkXk2 . (5.65)

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Proof. Let N = 2k , choose m ∈ N sufficiently large such that 2m N 1/2 . For j = 0, 1, · · · , m, define Rj0 ⊂ R2 to be the parallelogram with vortex (τ, ξ) = (0, 0), (1, 0), (3 · 2−j N 3/2 + 1, 2−j N −1/2 ), (3 · 2−j N 3/2 , 2−j N −1/2 ). Then let Rj := ((N + 2j N −1/2 )3 , N + 2j N −1/2 ) + Rj0 . Choose u b=b v=N

m X j=0

2j/2 aj 1Rj ∪(−Rj ) ,

3 where aj > 0. It is easy to see that ∪m j=0 Rj ⊂ {(τ, ξ) : |τ − ξ | 6 10}, thus we get 2 d [ N −3/2 k4 k ukXk k4k2 vkXk2 ∼ N

m X j=0

[N −3/4 2j/2 aj |Rj |1/2 ]2 ∼

m X

a2j .

j=0

On the other hand, if 1 6 j 6 m

1Rj ∗ 1−R0 &|Rj |1(τ (j) ,ξ(j) )−1/2R00 ∼ 2−j N −1/2 1(τ (j) ,ξ(j) )−1/2R00 ,

(5.66)

where (τ (j) , ξ (j) ) = ((N + 2j N −1/2 )3 , N + 2j N −1/2 ) − ((N + N −1/2 )3 , N + N −1/2 ). d [ Thus khτ − ξ 3 i−1 ϕ0 (ξ)ξ 4 k u ∗ 4k2 vkX0 is larger than ∞ X

2

−j 0 /2

j 0 =0

3

kϕ (τ − ξ )ϕ0 (ξ)ξ(N j0

m X j=1

2j/2 aj 1Rj ) ∗ (N a0 1−R0 )kL2 .

(5.67)

From (5.66) we get (5.67) is larger than N 3/2 a0

∞ X

j 0 =0

0

2−j /2 kϕj 0 (τ − ξ 3 )ϕ0 (ξ)ξ

m X j=1

2−j/2 aj 1(τ (j) ,ξ(j) )−1/2R00 kL2 . (5.68)

It is easy to see that if (τ, ξ) ∈ (τ (j) , ξ (j) ) − 1/2R0 (j > 1), then |ξ| ∼ 2j N −1/2 |τ | ∼ 2j N 3/2 , hence (5.68) is equivalent to X 0 N 3/2 a0 2−j /2 kϕj 0 (τ − ξ 3 )ϕ0 (ξ)ξ2−j/2 aj 1(τ (j) ,ξ(j) )−1/2R00 kL2 ∼

j 0 >0,2j N 3/2 ∼2j0 m X N 3/2 a0 (2j N 3/2 )−1/2 2j N −1/2 2−j/2 aj |R00 |1/2 j=1

∼ a0

m X

aj .

j=1

Taking aj = 1/(j + 1), we complete the proof of the proposition.

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121

The proposition above shows that if s = −3/4, one can not obtain local well-posedness by using F s or X s,b . From the discussion we see the only problem is the low frequency structure P60 u. Fortunately, by using a new low frequency structure, we can overcome the logarithmic divergence in Proposition 5.9 (a). We will use the following structure: kukX¯ 0 = kukL2xL∞ . t

From Proposition 5.4 we have

kϕ0 (t)40 ukX¯ 0 .k40 ukX0 .

(5.69)

k40 ukLq|t|6T Lrx ∩Lrx Lq|t|6T .T k40 ukL2x L∞ . |t|6T

(5.70)

Thus it is weaker than X0 , but on the other hand, we have for any 1 6 q 6 ∞ and 2 6 r 6 ∞ For −3/4 6 s 6 0, define

F¯ s = {u ∈ S 0 (R2 ) : kuk2F¯ s =

X

k>1

22sk kϕk (ξ)F uk2Xk + k40 uk2X¯ 0 < ∞}.

Assume T > 0, define the local space F¯ s (T ): kukF¯ s (T ) = inf {k40 ukL2xL∞ +k(I−40 )wkF¯ s , w(t) = u(t), t ∈ [−T, T ]}. |t|6T w∈F¯ s

Now we will show local well-posedness for KdV equation at s = −3/4. Theorem 5.7. Assume s > −3/4 and φ ∈ H s . Then (a) Existence. There exists T = T (ku0kH −3/4 ) > 0 and a solution u to the Cauchy problem (5.32) such that u ∈ F¯ s (T ) ⊂ C([−T, T ] : H s ).

(b) Uniqueness. The solution map ST : u0 → u is the unique continuous extension of the smooth solution map H ∞ → C([−T, T ] : H ∞ ). (c) Lipschitz continuity. For any R > 0, the map u0 → u is Lipschitz continuous from {u0 ∈ H s : ku0 kH s < R} to C([−T, T ] : H s ). (d) Persistence. If u0 ∈ H s for σ > s, then u ∈ H σ . Now we indicate our ideas in constructing F¯s . The starting point is the bilinear estimates in F s : k∂x (uv)kN s 6 CkukF s kvkF s .

This estimate fails at s = −3/4. On the other hand, we expect contraction principle still work at s = −3/4. Thus we need a new space F¯ −3/4 , which of course satisfies the following F¯ −3/4 ⊂ C(R : H −3/4 ).

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From the equivalent integral equation of (5.32): Z t u = S(t)u0 + C S(t − s)∂x (u2 )(s)ds. 0

Localizing in time, we get

u = Tu0 (u) = ψ(t)S(t)u0 + Cψ(t)

Z

t 0

S(t − s)∂x (u2 )(s)ds,

(5.71)

where ψ(t) = ϕ0 (t). By an iteration, u(0) = 0; · · · ; u(n+1) = Tu0 (u(n) ); · · ·

we obtain a sequence {u(n) }. If contraction principle works in F¯ −3/4 and u0 satisfies ku0 kH −3/4 1, then {u(n) } is a Cauchy sequence in C(R : H −3/4 ). Then a basic requirement is u(n) ∈ C(R : H −3/4 ),

∀ n ∈ N.

(5.72)

Next we check (5.72) for {u(n) }. For n = 0, 1, it is obvious that u , u(1) ∈ C(R; H −3/4 ). The case n = 2 is nontrivial. From the definition we have Z t u(2) = ψ(t)S(t)u0 + Cψ(t) S(t − s)∂x (S(s)u0 · S(s)u0 )ds. (0)

0

(2)

It suffices to prove (I − 40 )(u ) ∈ C(R; H −3/4 )40 (u(2) ) ∈ C(R; L2 ). For the high frequency part Z t ψ(t) S(t − s)∂x (S(s)u0 · S(s)u0 )ds 0 Z t = ψ(t) S(t − s)∂x [ψ(s/2)S(s)u0 · ψ(s/2)S(s)u0 ]ds. 0

From the linear estimate

kψ(t/2)S(t)u0 kF s .ku0 kH s , then by dyadic bilinear estimates and Proposition 5.3, we get for k ∈ N, 4k (u(2) ) ∈ Xk ⊂ C(R; H −3/4 ). For the low frequency part, we can not obtain 40 (u(2) ) ∈ X0 , due to the logarithmic divergence. By calculation we get Z t Fx ψ(t) S(t − s)40 ∂x [4k1 u(s)4k2 v(s)]ds (ξ) 0 Z t Z 3 3 isξ 3 \ \ = ψ(t)η0 (ξ)iξ ei(t−s)ξ eisξ1 4 k1 u0 (ξ1 )e 2 4k2 v0 (ξ2 )ds 0

ξ=ξ1 +ξ2

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3

= ψ(t)η0 (ξ)eitξ ξ

Z

123 3

ξ=ξ1 +ξ2

:= Fx (I) + Fx (II).

3

3

1 − e−it(ξ −ξ1 −ξ2 ) \ \ 4k1 u0 (ξ1 )4 k2 v0 (ξ2 ) ξ 3 − ξ13 − ξ23

Since in the hyperplane ξ = ξ1 + ξ2 one has ξ 3 − ξ13 − ξ23 = 3ξξ1 ξ2 , then we get Z \ \ 4 k1 u0 (ξ1 )4k2 v0 (ξ2 ) itξ 3 . Fx (I) = ψ(t)ϕ0 (ξ)e 3ξ1 ξ2 ξ=ξ1 +ξ2 Applying Theorem 5.2 we get

Z

\ \ 4 u (ξ ) 4 v (ξ )

k1 0 1 k2 0 2 kIkL2x L∞ 6 C

t

ξ=ξ1 +ξ2

3ξ1 ξ2

L2ξ

6 C2−3k1 /2 ku0 kL2 kv0 kL2 .

For the term II we have

Fx (II) = ψ(t)ϕ0 (ξ)

Z

3

ξ=ξ1 +ξ2

3

−eit(ξ1 +ξ2 ) \ \ 4k1 u0 (ξ1 )4 k2 v0 (ξ2 ). 3ξ1 ξ2

Applying Theorem 5.2 we get 3

3

kIIkL2xL∞ 6Cket∂x ∂x−1 4k1 u0 · et∂x ∂x−1 4k2 v0 kL2x L∞ t t 3

3

6Cket∂x ∂x−1 4k1 u0 kL4x L∞ ket∂x ∂x−1 4k2 v0 kL4x L∞ t t 6C2−3k1 /2 ku0 kL2 kv0 kL2 .

Thus, we proved 40 (u(2) ) ∈ L2x L∞ t . Next we consider the case n > 3. If n > 3, do we have (I − 40 )(u(n) ) ∈ F s and 40 (u(n) ) ∈ L2x L∞ t ? To answer this, we have the following proposition: ¯ 0 estimate). Assume |k1 − k2 | 6 5 and k1 > 10. Proposition 5.11 (X 0 ¯ Then u, v ∈ F

Z t

ψ(t)

S(t − s)4 ∂ [4 u(s)4 v(s)]ds 0 x k1 k2

L2x L∞ t

0

.2

3k − 21

[ [ k4 k1 ukXk1 k4k2 ukXk2 .

Rt Proof. Denote Q(u, v) = ψ(t) 0 S(t − s)40 ∂x [4k1 u(s)4k2 v(s)]ds. Direct computations show that Z b b − ξ3 ) ψ(τ − τ 0 ) − ψ(τ F [Q(u, v)] (ξ, τ ) =c ϕ0 (ξ)iξ τ 0 − ξ3 R

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× dτ 0

Z

ξ=ξ1 +ξ2 ,τ 0 =τ1 +τ2

[ [ 4 k1 u(ξ1 , τ1 )4k2 v(ξ2 , τ2 ).

Fix ξ ∈ R, divide Γ := {ξ = ξ1 + ξ2 , τ 0 = τ1 + τ2 } as following Γ1 ={|ξ|.2−2k1 } ∩ Γ;

Γ2 ={|ξ| 2−2k1 , |τi − ξi3 | 3 · 22k1 |ξ|, i = 1, 2} ∩ Γ; Γ3 ={|ξ| 2−2k1 , |τ1 − ξ13 |&3 · 22k1 |ξ|} ∩ Γ;

Γ4 ={|ξ| 2−2k1 , |τ2 − ξ23 |&3 · 22k1 |ξ|} ∩ Γ.

Then we get Z t F ψ(t) · S(t − s)40 ∂x [4k1 u(s)4k2 v(s)]ds (ξ, τ ) = A1 + ... + A4 , 0

where

Ai =C

Z

R

Z b − τ 0 ) − ψ(τ b − ξ3 ) ψ(τ 0 [ [ 40 (ξ)iξ 4 k1 u(ξ1 , τ1 )4k2 v(ξ2 , τ2 )dτ . τ 0 − ξ3 Γi

First we consider the estimate of A1 . It follows from Proposition 5.4 and Proposition 5.3 (b) that

Z

0

hτ − ξ 3 i−1 ϕ0 (ξ)ξ

. [ [ kF −1 (A1 )kL2x L∞ . 4 u(ξ , τ ) 4 v(ξ , τ ) k1 1 1 k2 2 2

t A1

X0

−2k1

Since in Γ1 one has |ξ|.2 , then we get

Z

0

hτ − ξ 3 i−1 ϕ0 (ξ)iξ

[ [ u(ξ , τ ) P v(ξ , τ ) 4 2 2 k1 1 1 k2

Γ1 X0 X X X −j3 /2 k3 . 2 2 k1Dk3 ,j3 · (uk1 ,j1 ∗ vk2 ,j2 )kL2 . k3 6−2k1 +10 j3 >0

Using (5.37) we get kF −1 (A1 )kL2x L∞ . t

X

j1 >0,j2 >0

X

2−j3 /2 2k3 2jmin /2 2k3 /2 kuk1 ,j1 kL2 kvk2 ,j2 kL2

k3 6−2k1 +10 ji >0

[ [ .2−3k1 k4 k1 ukXk1 k4k2 ukXk2 , which suffices for the bound of A1 . Next we consider A3 . As for A1 , it follows from Proposition 5.4 and Proposition 5.3 (b) that

Z

0

−1 3 −1

[ [ kF (A3 )kL2x L∞ . hτ − ξ i ϕ0 (ξ)ξ 4k1 u(ξ1 , τ1 )4k2 v(ξ2 , τ2 )

t A3 ∪A4 X0 X X X −j3 /2 k3 2 . 2 2 k1Dk3 ,j3 · (uk1 ,j1 ∗ vk2 ,j2 )kL . k3 60 j3 >0

j1 ,j2 >0

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125

We may assume j3 6 10k1 and |τ1 − ξ13 |&3|ξξ1 ξ2 |. It follows from Lemma 5.7 (b) that X X 2k3 2j2 /2 2−k1 kuk1 ,j1 kL2 kvk2 ,j2 kL2 kF −1 (A3 )kL2x L∞ . t k3 60 j1 >k3 +2k1 −10,j2 ,j3 >0

[ [ .k1 2−2k1 k4 k1 ukXk1 k4k2 ukXk2 ,

which gives the bound for A3 . Similarly, we can control A4 . Now we consider A2 , which is the main term. Simple calculations show that Z t Z 3 Ft−1 (A2 ) =ψ(t) ei(t−s)ξ ϕ0 (ξ)iξ eis(τ1 +τ2 ) 0 R2 Z × uk1 (ξ1 , τ1 )vk2 (ξ2 , τ2 ) dτ1 dτ2 ds ξ=ξ1 +ξ2

where uk1 (ξ1 , τ1 ) =ϕk1 (ξ1 )1{|τ1 −ξ13 |3·22k1 |ξ|} u b(ξ1 , τ1 ), vk2 (ξ2 , τ2 ) =ϕk2 (ξ2 )1{|τ2 −ξ23 |3·22k1 |ξ|} vb(ξ2 , τ2 ).

By change of variable τ10 = τ1 − ξ13 , τ20 = τ2 − ξ23 , we get Z t Z −1 itξ 3 −isξ 3 eis(τ1 +τ2 ) Ft (A2 ) =ψ(t)e ϕ0 (ξ)iξ e R2 0 Z 3 3 × eisξ1 uk1 (ξ1 , τ1 + ξ13 )eisξ2 vk2 (ξ2 , τ2 + ξ23 ) dτ1 dτ2 ds. ξ=ξ1 +ξ2

By interchange of the integral we get it is equal to Z Z 3 3 3 eit(ξ1 +ξ2 −ξ ) − e−it(τ1 +τ2 ) itξ 3 it(τ1 +τ2 ) ψ(t)e ϕ0 (ξ)ξ e τ1 + τ2 − ξ 3 + ξ13 + ξ23 R2 ξ=ξ1 +ξ2

× uk1 (ξ1 , τ1 + ξ13 )vk2 (ξ2 , τ2 + ξ23 ) dτ1 dτ2 := Ft−1 (II1 ) − Ft−1 (II2 ).

For the term II2 , Ft−1 (II2 ) Z Z itξ 3 = ψ(t)e ϕ0 (ξ)ξ R2

ξ=ξ1 +ξ2

uk1 (ξ1 , τ1 + ξ13 )vk2 (ξ2 , τ2 + ξ23 ) dτ1 dτ2 . τ1 + τ2 − ξ 3 + ξ13 + ξ23

Since in the integral area |τ1 +τ2 −ξ 3 +ξ13 +ξ23 | ∼ |ξξ1 ξ2 |, then from Theorem 5.2 we get kF −1 (II2 )kL2x L∞ t

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.

Z

R

.2

uk1 (ξ1 , τ1 + ξ13 )vk2 (ξ2 , τ2 + ξ23 )

dτ1 dτ2 ξ

τ1 + τ2 − ξ 3 + ξ13 + ξ23 ξ=ξ1 +ξ2 L2

Z

2

−

3k1 2

ξ

[ [ k4 k1 ukXk1 k4k2 ukXk2 .

To prove 5.11, it suffices to prove [ [ kF −1 (II1 )kL2x L∞ .2−3k1 /2 kP k1 ukXk1 kPk2 ukXk2 . t Compare II1 with II10 defined as following: Z Z −1 0 itξ 3 it(τ1 +τ2 ) Ft (II1 ) =ψ(t)e ϕ0 (ξ)ξ e × uk1 (ξ1 , τ1 +

For II10 we have Ft−1 (II10 ) =

Z

R2

×

R2 ξ13 )vk2 (ξ2 , τ2

3

3

3

eit(ξ1 +ξ2 −ξ ) −ξ 3 + ξ13 + ξ23

ξ=ξ1 +ξ2 + ξ23 ) dτ1 dτ2 .

ψ(t)ϕ0 (ξ)eit(τ1 +τ2 ) 1{|ξ||τ1 |2−2k1 } 1{|ξ||τ2 |2−2k1 }

Z

3

ξ=ξ1 +ξ2

3

eit(ξ1 +ξ2 ) F (fτ1 )(ξ1 )F (gτ2 )(ξ2 ) dτ1 dτ2 , −3ξ1 ξ2

where for τ1 , τ2 ∈ R we denote

3 3 [ [ F (fτ1 )(ξ) = 4 k1 u(ξ, τ1 + ξ ), F (gτ2 )(ξ) = 4k2 v(ξ, τ2 + ξ ).

Since (We may need a smooth cut-off 1{|ξ|λ} ) ∀ λ > 0 , kFx−1 1{|ξ|λ} Fx ukL2x L∞ .kukL2xL∞ t t From Theorem 5.2 we get Z . kS(t)∂x−1 fτ1 S(t)∂x−1 fτ2 kL2x L∞ dτ1 dτ2 kF −1 (II10 )kL2x L∞ t t 2 R Z . kS(t)∂x−1 fτ1 )kL4x L∞ kS(t)∂x−1 fτ2 kL4x L∞ dτ1 dτ2 t t R2

.2

−

3k1 2

[ [ k4 k1 ukXk1 k4k2 ukXk2 ,

which gives the bound of II10 . To finish the proof of Proposition 5.11, it remains to prove [ [ kF −1 (II1 − II10 )kL2x L∞ .2−3k1 /2 kP k1 ukXk1 kPk2 ukXk2 . t Since in the integral area |τi | 22k1 |ξ|, i = 1, 2, thus in the hyperplane n ∞ X 1 1 1 τ1 + τ2 − = . τ1 + τ2 − ξ 3 + ξ13 + ξ23 −ξ 3 + ξ13 + ξ23 3ξξ1 ξ2 3ξξ1 ξ2 n=1

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then we get Ft−1 (II1 − II10 ) Z Z = ψ(t)ϕ0 (ξ)ξ eit(τ1 +τ2 )

1 3ξξ 1 ξ2 ξ=ξ1 +ξ2 n=1

R2

× e

it(ξ13 +ξ23 )

∞ X

τ1 + τ2 3ξξ1 ξ2

n

uk1 (ξ1 , τ1 + ξ13 )vk2 (ξ2 , τ2 + ξ23 ) dτ1 dτ2 .

Decomposing dyadically on the low frequency, {χk (ξ)}∞ k=−∞ denote homogeneous decomposition, we get Ft−1 (II1 − II10 ) ∞ Z X = eit(τ1 +τ2 ) n=1

×

R2

Z

ξ=ξ1 +ξ2

X

3

3

ψ(t)eit(ξ1 +ξ2 ) χk3 (ξ)

2k3 2−2k1 max(|τ1 |,|τ2 |) n τ1 + τ2 uk1 (ξ1 , τ1 + ξ13 )vk2 (ξ2 , τ2

3ξξ1 ξ2

+ ξ23 )

3ξ1 ξ2

dτ1 dτ2 .

We rewrite it into the following form Ft−1 (II1 − II10 ) ∞ Z X = eit(τ1 +τ2 ) n=1

R2

× 2−nk3

Z

X

3

2k3 2−2k1 max(|τ1 |,|τ2 |) n uk1 (ξ1 , τ1 τ1 + τ2

ξ=ξ1 +ξ2

3

ψ(t)eit(ξ1 +ξ2 ) χk3 (ξ)(ξ/2k3 )−n

3ξ1 ξ2

+ ξ13 )vk2 (ξ2 , τ2 + ξ23 ) dτ1 dτ2 . 3ξ1 ξ2

0 Since χk3 (ξ)(ξ/2k3 )−n is the multiplier for the L2x L∞ t , then as for II1 we get

kF −1 (II1 − II10 )kL2x L∞ t ∞ Z X X . n=1

R2

2k3 2−2k1 max(|τ1 |,|τ2 |)

Z

it(ξ13 +ξ23 )

× ψ(t)e

ξ=ξ1 +ξ2

C n |τ1 + τ2 |n 2−nk3

F (fτ1 )(ξ1 )F (gτ2 )(ξ2 )

dτ1 dτ2 .

3ξ1n+1 ξ2n+1 L2 L∞ x

t

By Theorem 5.2 and summing on k3 we get for some M 1 kF −1 (II1 − II10 )kL2x L∞ t ∞ Z X X . n=1

×2

R2

2k3 2−2k1

−3k1 /2

max(|τ1 |,|τ2 |)

C n |τ1 + τ2 |n 2−nk3 2−2nk1

kF (fτ1 )kL2 kF (gτ2 )kL2 dτ1 dτ2

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.

∞ X

(C/M )n

Z

R2

n=1

2−3k1 /2 kF (fτ1 )kL2 kF (gτ2 )kL2 dτ1 dτ2

[ [ .2−3k1 /2 kP k1 ukXk1 kPk2 ukXk2 . Therefore, we complete the proof of the proposition.

Inspired by the proposition, then we may use the structure L2x L∞ t for the low frequency part. It remains to control the high-low interaction. Proposition 5.12. If k > 10, |k − k2 | 6 5, then for any u, v ∈ F¯ 0 d [ [ khτ − ξ 3 i−1 ϕk (ξ)ξ(4 k4 0 u ∗ 4k2 v)kXk .k40 ukL2x L∞ k2 vkXk2 . t

Proof.

(5.73)

We may assume k = k2 . From the definition of Xk we get d d khτ − ξ 3 i−1 ϕk (ξ)ξ(4 0 u ∗ 4k v)kXk X k −j/2 d [ .2 2 k 40 u ∗ 4 k2 vkL2ξ,τ .

(5.74)

j>0

It follows from Plancherel’s equality and Proposition 5.4 that k d d [ 2 .k40 ukL2 L∞ k4 k4k ukL∞ 2k k4 k vkXk , 0 u ∗ 4k2 vkL2ξ,τ .2 k40 ukL2x L∞ x t t x Lt

thus, the proposition is proved.

Next we prove a crucial bilinear estimate which is key to prove Theorem 5.7. For u, v ∈ F¯ s we define bilinear operator Z t B(u, v) = ψ(t/4) S(t − τ )∂x ψ 2 (τ )u(τ ) · v(τ ) dτ. (5.75) 0

Consider the following integral equation Z t u = ψ(t)S(t)u0 + ψ(t/4) S(t − τ )∂x ψ 2 (τ )u(τ ) · u(τ ) dτ.

(5.76)

0

In order to apply the contraction principle, the rest task is to show the boundedness B : F¯ s × F¯ s → F¯ s .

Proposition 5.13. Assume −3/4 6 s 6 0. Then there exists C > 0 such that kB(u, v)kF¯ s 6 C(kukF¯ s kvkF¯ −3/4 + kukF¯ −3/4 kvkF¯ s ) hold for all u, v ∈ F¯ s .

(5.77)

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From the definition of F s we have X kB(u, v)k2F¯ s = k40 B(u, v)k2X¯ 0 + 22k1 s kϕk1 (ξ)F [B(u, v)]k2Xk1 . (5.78)

Proof.

k1 >1

First we consider the second term on the right-hand side of (5.78). By applying Littlewood-Paley decomposition to u, v we get X kϕk1 (ξ)F [B(u, v)]kXk1 . kϕk1 (ξ)F [B(4k2 (u), 4k3 (v))]kXk1 . k2 ,k3 >0

(5.79)

It follows from Proposition 5.3 (b) that the right-hand side of (5.79) is bounded by X \ \ khτ − ξ 3 i−1 ϕk1 (ξ)ξ ψ(t)4 (5.80) k2 u ∗ ψ(t)4k3 v)kXk1 . k2 ,k3 >0

From symmetry we may assume k2 6 k3 in (5.80). It suffices to prove X X 2 1/2 \ \ 22k1 s khτ − ξ 3 i−1 ϕk1 (ξ)ξ ψ(t)4 k2 u ∗ ψ(t)4k3 v)kXk1 k1 >1

k2 ,k3 >0

.kukF¯ −3/4 kvkF¯ s .

(5.81)

If kmax 6 20, then from Proposition 5.8 we get (5.80) is bounded by X 2 k4k vkL∞ L2 , k4k2 ukL∞ 3 t Lx t x kmax 620

2 .k4k ukX which is sufficient to give (5.81), since in this case k4k ukL∞ k t Lx 2 for k > 1, and k4k ukL∞ .k4 uk for k = 0. Now we assume kmax > 20 ¯ k Xk t Lx in (5.81). There are three cases. If |k1 − k3 | 6 5, k2 6 k1 − 10, then we use Proposition 5.5 (a) for k2 = 0, and (b) for k2 > 1; if |k1 − k3 | 6 5, k1 − 9 6 k2 6 k3 , then we use Proposition 5.6; if |k2 − k3 | 6 5, 1 6 k1 6 k2 − 5, then we use Proposition 5.9 (b). Therefore, we get (5.81). It remains to prove

kB(u, v)kX¯ 0 6 C(kukF¯ s kvkF¯ −3/4 + kukF¯ −3/4 kvkF¯ s ).

(5.82)

By applying Littlewood-Paley decomposition to u, v we get X kB(u, v)kX¯ 0 6 kB(4k2 u, 4k3 v)kX¯ 0 . k2 ,k3 >0

If max(k2 , k3 ) 6 10, then from Proposition 5.14 and Proposition 5.8 we get 2 k4k vkL∞ L2 , kB(4k2 u, 4k3 v)kX¯ 0 .k4k2 ukL∞ 3 x t Lx t

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which is sufficient to give (5.82). If max(k2 , k3 ) > 10, then |k2 − k3 | 6 5. It follows from Proposition 5.11 that X kB(u, v)kX¯ 0 6 2−3k2 /2 kF (4k2 u)kXk2 kF (4k3 v)kXk3 |k2 −k3 |65, k2 ,k3 >10

.kukF¯ −3/4 kvkF¯ −3/4 which gives (5.82). Therefore, the proposition is proved.

Similarly, as the proof of 5.3 (a), it is easy to prove the following proposition by Lemma 5.4. Proposition 5.14. Assume s ∈ R and φ ∈ H s . Then there exists C > 0 such that kψ(t)S(t)φkF¯ s 6 CkφkH s . To prove Theorem 5.7, we need the bilinear form contraction principle which is easy to prove. The proof is left to the readers as an exercise. Lemma 5.8. Assume (X, k · k) is a Banach space with the norm k · k. Let B : X × X→X be a bilinear operator satisfying kB(x1 , x2 )k 6 ηkx1 kkx2 k,

∀ x1 , x2 ∈ X.

Then for all y ∈ X with 4ηkyk < 1, the equation x = y + B(x, x) has unique 1 . solution x ∈ X such that kxk < 2η Using Lemma 5.8, Proposition 5.13 and Proposition 5.14, we can prove Theorem 5.7. 5.5

I-method

In the last two sections, we studied the local well-posedness for the KdV equation. In this section, we study the global well-posedness, namely extend the local solution to a global one. KdV equation (5.32) is completely integrable, and hence has infinite conservation laws. With these conservation laws, one can derive some a priori estimates of the solution. For example, if u is a smooth solution to the KdV equation (5.32), then for k ∈ N ∪ {0} and T > 0 ku(t)kH k 6 C(T, ku0 kH k ), ∀ t ∈ [−T, T ].

(5.83)

With these a priori estimates and local well-posedness, one can easily obtain the following: KdV equation is globally well-posed in H k , where k ∈ N ∪

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{0}. In view of the local well-posedness results, it is a natural question: what about global well-posedness in H s for s ∈ [−3/4, 0)? For the regularity below L2 , there is no conservation law, and hence the global well-posedness do not hold automatically. I-method is an effective method to study this kind of problem. It was introduced by J. Colliander, M. Keel, G. Staffilani, H. Takaoka, and T. Tao [42], inspired by Bourgain’s frequency decomposition techniques [17]. We explain briefly the basic ideas of I-method. We recall the proof of (5.83) in the case k = 0, 1. Multiplying u on both sides of equation (5.32), and then integrating on x, we get d ku(t)k2L2 = 0, dt which is (5.83) at k = 0. For k = 1, it is easy to see that d ku(t)k2H 1 = 0 dt fails, the reason is the quantity ku(t)kH 1 does not exploit the structure of the equation. A natural idea is to find some other quantity Q1 (u(t)) such that d [Q1 (u(t))] = 0. Q1 (u(t)) ∼ ku(t)k2H 1 , dt If such Q1 (u(t)) exists, then we get ∀ t ∈ R ku(t)kH 1 ∼ kQ1 (u(t))k = kQ1 (u0 )k.ku0 kH 1 , which is (5.83) at k = 1. Now we know such quantity exists, for instance, we can take Z 1 Q1 (u(t)) = u2x − u3 + Cu2 dx. 3 R

d Actually, obviously dt [Q(u(t))] = 0, it suffices to show Q1 (u(t)) ∼ ku(t)k2H 1 . By Gagliardo-Nirenberg’s inequality we have 5/2

1/2

kuk3L3 .kukL2 kux kL2 . Then using ab 6

ap bq + , 1 < p 6 q < ∞, 1/p + 1/q = 1, p q

and taking q = 4, we get 10/3

kuk3L3 6

3kukL2 4

4/3

+

3ku0 kL2 kuk2L2 kux k2L2 kux k2L2 6 + . 4 4 4

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Finally take C = ku0 kL2 . From this, for general H s norm, we also need to find quantity Qs (u(t)), they have similar properties as Q1 (u(t)). However, since such quantity does not exist for s < 0, then we relax the condition to Qs (u(t)) increase much slower than the norm kukH s during the evolution. The issues that how to construct Qs (u(t)) and control the increasing rate are systematically studied in ’I-method’. Given m : Rk → C, m is said to be symmetric if m(ξ1 , · · · , ξk ) = m(σ(ξ1 , · · · , ξk )) for all σ ∈ Sk , where Sk is the permutation group for k elements. The symmetrization of m is defined by 1 X m(σ(ξ1 , ξ2 , · · · , ξk )). [m]sym (ξ1 , ξ2 , · · · , ξk ) = k! σ∈Sk

For each m, we define a k-linear functional acting on k functions u1 , · · · , uk (m is said to be k-multiplier) as following Z Λk (m; u1 , · · · , uk ) = m(ξ1 , · · · , ξk )c u1 (ξ1 ) · · · u ck (ξk ). ξ1 +···+ξk =0

Usually we apply Λk on k functions which are all u. For convenience, Λk (m; u, · · · , u) is simply denoted by Λk (m). From symmetries we see Λk (m) = Λk ([m]sym ). By using the KdV equation (5.32) we can get the following proposition.

Proposition 5.15. Assume u satisfies (5.32) and m is a symmetric function. Then dΛk (m) ik = Λk (mhk ) − Λk+1 m(ξ1 , · · · , ξk−1 , ξk + ξk+1 )(ξk + ξk+1 ) , dt 2 where

hk = i(ξ13 + ξ23 + · · · + ξk3 ). Next we introduce the I-operator. Assume m : R → R is a real-valued even function, then we define Fourier multiplier operator Iu as following c Iu(ξ) = m(ξ)b u.

Define the modified energy EI2 (t)

EI2 (t) = kIuk2L2 .

(5.84)

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Since m is real-valued even function and u is real-valued, then by Plancherel’s equality we get EI2 (t) = Λ2 (m(ξ1 )m(ξ2 )). It follows from Proposition 5.15 that d 2 E (t) = Λ2 (m(ξ1 )m(ξ2 )h2 ) − iΛ3 (m(ξ1 )m(ξ2 + ξ3 )(ξ2 + ξ3 )). dt I Since ξ13 +ξ23 = 0 for ξ1 +ξ2 = 0, then the first term vanishes. Symmetrizing the second term we get d 2 E (t) = Λ3 (−i[m(ξ1 )m(ξ2 + ξ3 )(ξ2 + ξ3 )]sym ). dt I Denote M3 (ξ1 , ξ2 , ξ3 ) = −i[m(ξ1 )m(ξ2 + ξ3 )(ξ2 + ξ3 )]sym . Define a new modified energy EI3 (t) = EI2 (t) + Λ3 (σ3 ), where symmetric function σ3 will be set momentarily. The role of σ3 is to make a cancelation. It follows from Proposition 5.15 that 3 d 3 EI (t) = Λ3 (M3 ) + Λ3 (σ3 h3 ) − iΛ4 (σ3 (ξ1 , ξ2 , ξ3 + ξ4 )(ξ3 + ξ4 )). dt 2 (5.85) Take σ3 = −M3 /h3 such that the two trilinear terms in (5.85) cancels. Denote 3 M4 (ξ1 , ξ2 , ξ3 , ξ4 ) = −i [σ3 (ξ1 , ξ2 , ξ3 + ξ4 )(ξ3 + ξ4 )]sym , 2 then we have d 3 E (t) = Λ4 (M4 ). dt I Similarly we define a new modified energy EI4 (t) = EI3 (t) + Λ4 (σ4 ), where σ4 = −

M4 . h4

It is easy to get d 4 E (t) = Λ5 (M5 ) dt I

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where M5 (ξ1 , . . . , ξ5 ) = −2i[σ4 (ξ1 , ξ2 , ξ3 , ξ4 + ξ5 )(ξ4 + ξ5 )]sym . This procedure can be made for a finite times, but we stop here since it suffices for our purposes. In order to control the norm kukH s , it seems to convenient to take m(ξ) = hξis . However, it is difficult to control its increase. Basing on the L2 conservation, we take m as following: m is a smooth even function which has the form 1, |ξ| < N, m(ξ) = (5.86) N −s |ξ|s , |ξ| > 2N. It is easy to see that if N → ∞ then Iu → u, and hence the increase rate of the modified energy EIk (t) tends to 0 as N → ∞. The basic question is how fast it tends to 0. If m is of the form (5.86), then it is easy to see m2 satisfies  2 2 0 0   m (ξ) ∼ m (ξ ) |ξ| ∼ |ξ |, 2

(m2 )0 (ξ) = O( m|ξ|(ξ) ),

(5.87)

  (m2 )00 (ξ) = O( m2 (ξ) ). |ξ|2

Next we need to estimate the multipliers M3 , M4 , M5 . We will use two mean value formulas: if |η|, |λ| |ξ| |a(ξ + η) − a(ξ)|.|η| sup |a0 (ξ 0 )|,

(5.88)

|ξ 0 |∼|ξ|

and |a(ξ + η + λ) − a(ξ + η) − a(ξ + λ) + a(ξ)|.|η||λ| sup |a00 (ξ 0 )|.

(5.89)

|ξ 0 |∼|ξ|

First we give the estimate of M3 . Proposition 5.16. Let m be given by (5.86). In the set {ξ1 + ξ2 + ξ3 = 0, |ξi | ∼ Ni } where Ni is dyadic, we have |M3 (ξ1 , ξ2 , ξ3 )|. max(m2 (ξ1 ), m2 (ξ2 ), m2 (ξ3 )) min(N1 , N2 , N3 ).

Proof. From symmetry we may assume N1 = N2 > N3 . If N3 &N1 , then it follows directly from the definition of m. If N3 N1 , from the mean-value formula we get m2 (ξ1 )ξ1 − m2 (ξ1 + ξ3 )(ξ1 + ξ3 ) + m2 (ξ3 )ξ3 6 max(m2 (ξ1 ), m2 (ξ2 ))N3 . Thus the proposition is proved.

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To estimate M4 , first we extend the multiplier σ3 from the hyperplane to the entire space. Proposition 5.17. If m is of the form (5.86), then for any dyadic numbers λ 6 µ there exists an extension of σ3 from {ξ1 + ξ2 + ξ3 = 0, |ξ1 | ∼ λ, |ξ2 |, |ξ3 | ∼ µ} to the entire space {(ξ1 , ξ2 , ξ3 ) ∈ R3 , |ξ1 | ∼ λ, |ξ2 |, |ξ3 | ∼ µ} such that |∂1β1 ∂2β2 ∂3β3 σ3 (ξ1 , ξ2 , ξ3 )| 6 Cm2 (λ)µ−2 λ−β1 µ−β2 −β3 ,

(5.90)

where the constant C is independent of λ, µ. Proof.

Since on the hyperplane {(ξ1 , ξ2 , ξ3 ) : ξ1 + ξ2 + ξ3 = 0} we have h3 = i(ξ13 + ξ23 + ξ33 ) = 3iξ1 ξ2 ξ3 ,

then |h| ∼ λµ2 . From M3 (ξ1 , ξ2 , ξ3 ) = − i[m(ξ1 )m(ξ2 + ξ3 )(ξ2 + ξ3 )]sym

=i(m2 (ξ1 )ξ1 + m2 (ξ2 )ξ2 + m2 (ξ3 )ξ3 ),

we will discuss case by case. If λ ∼ µ, then we extend σ3 as following σ3 (ξ1 , ξ2 , ξ3 ) = −

i(m2 (ξ1 )ξ1 + m2 (ξ2 )ξ2 + m2 (ξ3 )ξ3 ) , 3iξ1 ξ2 ξ3

which suffices for the purpose. If λ µ, then we extend σ3 as following σ3 (ξ1 , ξ2 , ξ3 ) = −

i(m2 (ξ1 )ξ1 + m2 (ξ2 )ξ2 − m2 (ξ1 + ξ2 )(ξ1 + ξ2 )) . 3iξ1 ξ2 ξ3

Then (5.90) follows from (5.88) and (5.87).

Next we give the estimate of σ4 . The proof here was due to [96]. Compared to the proof given in [42], the proof here is more flexible to a general class of dispersive equations. But the ideas are the same. Proposition 5.18. Let m be of the form (5.86). Then on the area |ξi | ∼ Ni , |ξj + ξk | ∼ Njk , where Ni , Njk are dyadic numbers, we have m2 (min(Ni , Njk )) |M4 (ξ1 , ξ2 , ξ3 , ξ4 )| . . |h4 | (N + N1 )(N + N2 )(N + N3 )(N + N4 )

(5.91)

Proof. From symmetry we may assume N1 > N2 > N3 > N4 . Since ξ1 + ξ2 +ξ3 +ξ4 = 0, then N1 ∼ N2 . We may also assume N1 ∼ N2 &N , otherwise M4 vanishes, since m2 (ξ) = 1 if |ξ| 6 N , and if max(N12 , N13 , N14 ) N1 ,

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then ξ3 ≈ −ξ1 , ξ4 ≈ −ξ1 , which contradicts to ξ1 + ξ2 + ξ3 + ξ4 = 0. The right-hand side of (5.91) is actually m2 (min(Ni , Njk )) . N1 2 (N + N3 )(N + N4 )

(5.92)

Since on the hyperplane ξ1 + ξ2 + ξ3 + ξ4 = 0 h4 = ξ13 + ξ23 + ξ33 + ξ43 = 3(ξ1 + ξ2 )(ξ1 + ξ3 )(ξ1 + ξ4 ), thus we get CM4 (ξ1 , ξ2 , ξ3 , ξ4 ) = [σ3 (ξ1 , ξ2 , ξ3 + ξ4 )(ξ3 + ξ4 )]sym = σ3 (ξ1 , ξ2 , ξ3 + ξ4 )(ξ3 + ξ4 ) + σ3 (ξ1 , ξ3 , ξ2 + ξ4 )(ξ2 + ξ4 ) + σ3 (ξ1 , ξ4 , ξ2 + ξ3 )(ξ2 + ξ3 ) + σ3 (ξ2 , ξ3 , ξ1 + ξ4 )(ξ1 + ξ4 ) + σ3 (ξ2 , ξ4 , ξ1 + ξ3 )(ξ1 + ξ3 ) + σ3 (ξ3 , ξ4 , ξ1 + ξ2 )(ξ1 + ξ2 ) = [σ3 (ξ1 , ξ2 , ξ3 + ξ4 ) − σ3 (−ξ3 , −ξ4 , ξ3 + ξ4 )](ξ3 + ξ4 )

+ [σ3 (ξ1 , ξ3 , ξ2 + ξ4 ) − σ3 (−ξ2 , −ξ4 , ξ2 + ξ4 )](ξ2 + ξ4 )

+ [σ3 (ξ1 , ξ4 , ξ2 + ξ3 ) − σ3 (−ξ2 , −ξ3 , ξ2 + ξ3 )](ξ2 + ξ3 )

:= I + II + III.

(5.93)

We will get (5.91) by studying different cases. Case 1. |N4 |& N2 . Case 1a. N12 , N13 , N14 &N1 . For this case, it follows directly from (5.90) that |M4 (ξ1 , ξ2 , ξ3 , ξ4 )| m2 (N4 ) . . |h4 | N1 N2 N3 N4

(5.94)

|I| m2 (min(N4 , N12 )) . . |h4 | N1 N2 N3 N4

(5.95)

II m2 (N4 ) . . h4 N 1 N 1 N 1 N 3

(5.96)

Case 1b. N12 N1 , N13 &N1 , N14 &N1 . First we consider the contribution of I. From (5.90) we get

For the contribution of II, if N12 &N3 , then we use (5.88) and (5.90), else if N12 N3 , then we use (5.88)(5.90). Thus we get

From symmetry, the contribution of III is identical to that of II. Case 1c. N12 N1 , N13 N1 , N14 &N1 . Since N12 N1 , N13 N1 , then N1 ∼ N2 ∼ N3 ∼ N4 .

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First we consider the contribution of I. Since I =[σ3 (ξ1 , ξ2 , ξ3 + ξ4 ) − σ3 (−ξ3 , ξ2 , ξ3 + ξ4 )](ξ3 + ξ4 )

+ [σ3 (−ξ3 , ξ2 , ξ3 + ξ4 ) − σ3 (−ξ3 , −ξ4 , ξ3 + ξ4 )](ξ3 + ξ4 )

=I1 + I2 .

Using (5.90) and (5.88) we get I1 I2 m2 (N12 ) I . + . . |h4 | |h4 | |h4 | N14

The contribution of II is identical to that of I. Finally we consider III. Since

III =1/2[σ3 (ξ1 , ξ4 , ξ2 + ξ3 ) − σ3 (−ξ2 , −ξ3 , ξ2 + ξ3 )

− σ3 (−ξ3 , −ξ2 , ξ2 + ξ3 ) + σ3 (ξ4 , ξ1 , ξ2 + ξ3 )](ξ2 + ξ3 ),

then using (5.89) four times we get III m2 (N1 ) . . |h4 | N14

Case 1d. N12 N1 , N13 &N1 , N14 N1 . This case is identical to Case 1c. Case 2. N4 N/2. It is obvious that in this case m2 (min(Ni , Njk )) = 1, and N13 ∼ |ξ1 + ξ3 | = |ξ2 + ξ4 | ∼ N1 . Case 2a. N1 /4 > N12 &N/2. Since N4 N/2 and |ξ3 + ξ4 | = |ξ1 + ξ2 |&N/2, then N3 &N/2. From |h4 | ∼ N12 N12 , we control the six terms in (5.93) respectively and get 1 |M4 | . . |h4 | N12 N3 N

(5.97)

|I| |σ3 (ξ1 , ξ2 , ξ3 + ξ4 )| 1 . . 4. |h4 | N12 N1

(5.98)

Case 2b. N12 N/2. Since N12 = N34 N/2 and N4 N/2, then N3 N/2, and N13 ∼ N14 ∼ N1 . For the contribution of I, since N3 , N4 , N34 N/2, then from the definition of m we have σ3 (−ξ3 , −ξ4 , ξ3 + ξ4 ) = 0. Using (5.90) we get

Next we consider the contribution of II and III. We can not deal with the two terms separately, but need to exploit a cancelation between the two terms. Rewrite II + III as following II + III =[σ3 (ξ1 , ξ3 , ξ2 + ξ4 ) − σ3 (−ξ2 , −ξ4 , ξ2 + ξ4 )](ξ2 + ξ4 )

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+ [σ3 (ξ1 , ξ4 , ξ2 + ξ3 ) − σ3 (−ξ2 , −ξ3 , ξ2 + ξ3 )](ξ2 + ξ3 )

=[σ3 (ξ1 , ξ3 , ξ2 + ξ4 ) − σ3 (−ξ2 , −ξ4 , ξ2 + ξ4 )]ξ4

+ [σ3 (ξ1 , ξ4 , ξ2 + ξ3 ) − σ3 (−ξ2 , −ξ3 , ξ2 + ξ3 )]ξ3 + [σ3 (ξ1 , ξ3 , ξ2 + ξ4 ) − σ3 (−ξ2 , −ξ4 , ξ2 + ξ4 )

+ σ3 (ξ1 , ξ4 , ξ2 + ξ3 ) − σ3 (−ξ2 , −ξ3 , ξ2 + ξ3 )]ξ2

=J1 + J2 + J3 .

(5.99)

For J1 , since |[σ3 (ξ1 , ξ3 , ξ2 + ξ4 ) − σ3 (−ξ2 , −ξ4 , ξ2 + ξ4 )]ξ4 | |J1 | 6 |h4 | |h4 |

(5.100)

if N12 N3 (in this case N3 ∼ N4 ), then use (5.88) twice, otherwise use (5.88) and (5.90), thus we get |J1 | 1 . . |h4 | N14

J2 is identical to J1 . Now consider J3 . First we assume N12 &N3 . From the symmetry of σ3 we get J3 =[σ3 (ξ1 , ξ3 , ξ2 + ξ4 ) − σ3 (−ξ2 − ξ3 , ξ3 , ξ2 )

+ σ3 (ξ1 , ξ4 , ξ2 + ξ3 ) − σ3 (−ξ2 − ξ4 , ξ4 , ξ2 )]ξ2 .

From (5.88) and that N12 &N3 we get |J3 | 1 . . |h4 | N14

If N12 N3 , then N3 ∼ N4 . Rewrite J3 as following J3 =[σ3 (−ξ2 , ξ3 , ξ2 + ξ4 ) − σ3 (−ξ2 , −ξ4 , ξ2 + ξ4 )

+ σ3 (ξ1 , ξ4 , ξ2 + ξ3 ) − σ3 (ξ1 , −ξ3 , ξ2 + ξ3 )]ξ2 + [σ3 (ξ1 , ξ3 , ξ2 + ξ4 ) − σ3 (−ξ2 , ξ3 , ξ2 + ξ4 )

+ σ3 (ξ1 , −ξ3 , ξ2 + ξ3 ) − σ3 (−ξ2 , −ξ3 , ξ2 + ξ3 )]ξ2

=J31 + J32 .

It follows from (5.88) that |J32 | 1 . . |h4 | N14

(5.101)

It suffices to control J31 . Since in this case m2 (ξ3 ) = m2 (ξ4 ) = 1, then we get J31 =[σ3 (−ξ2 , ξ3 , ξ2 + ξ4 ) − σ3 (ξ1 , −ξ3 , ξ2 + ξ3 )

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− σ3 (−ξ2 , −ξ4 , ξ2 + ξ4 ) + σ3 (ξ1 , ξ4 , ξ2 + ξ3 )]ξ2

=

−m2 (ξ2 )ξ2 + ξ3 + m2 (ξ2 + ξ4 )(ξ2 + ξ4 ) ξ2 −ξ2 ξ3 (ξ2 + ξ4 ) −m2 (ξ2 )ξ2 − ξ4 + m2 (ξ2 + ξ4 )(ξ2 + ξ4 ) ξ2 − ξ2 ξ4 (ξ2 + ξ4 ) m2 (ξ1 )ξ1 + ξ4 + m2 (ξ2 + ξ3 )(ξ2 + ξ3 ) + ξ2 ξ1 ξ4 (ξ2 + ξ3 ) m2 (ξ1 )ξ1 − ξ3 + m2 (ξ2 + ξ3 )(ξ2 + ξ3 ) − ξ2 . −ξ1 ξ3 (ξ2 + ξ3 )

Noting that there is a cancelation, then we get ξ3 + ξ4 −m2 (ξ2 )ξ2 + m2 (ξ2 + ξ4 )(ξ2 + ξ4 ) ξ2 ξ3 ξ4 ξ2 (ξ2 + ξ4 ) ξ3 + ξ4 m2 (ξ1 )ξ1 + m2 (ξ2 + ξ3 )(ξ2 + ξ3 ) + ξ2 . ξ3 ξ4 ξ1 (ξ2 + ξ3 )

J31 = −

(5.102)

Rewrite (5.102) as following −m2 (ξ2 )ξ2 + m2 (ξ2 + ξ4 )(ξ2 + ξ4 ) + m2 (ξ1 )ξ1 + m2 (ξ2 + ξ3 )(ξ2 + ξ3 ) ξ2 ξ2 (ξ2 + ξ4 ) ξ3 + ξ4 2 ξ3 + ξ4 + [m (ξ1 )ξ1 + m2 (ξ2 + ξ3 )(ξ2 + ξ3 )] × ξ3 ξ4 ξ3 ξ4 1 1 × + ξ2 . ξ1 (ξ2 + ξ3 ) ξ2 (ξ2 + ξ4 ) −

Thus, for the first term we use (5.89), and (5.88) for the second term, finally we get |J31 | 1 . , |h4 | N14 Therefore, the proposition is proved. The estimate of M5 follows immediately from the estimate of σ4 . Proposition 5.19. If m is of the form (5.86), then m2 (N∗45 )N45 |M5 (ξ1 , . . . , ξ5 )|. , (N + N1 )(N + N2 )(N + N3 )(N + N45 ) sym where N∗45 = min(N1 , N2 , N3 , N45 , N12 , N13 , N23 ).

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Next we will prove two properties. One is to show the modified energy EI4 (t) increase slowly, the other is to prove it is close to EI2 (t). To control the increase rate of EI4 (t), it suffices to control its derivative d 4 E (t) = Λ5 (M5 ). dt I Proposition 5.20. Assume I ⊂ R and |I|.1. Let 0 6 k1 6 . . . 6 k5 and k4 > 10. Then Z Z 5 5 Y Y 5(k1 +k2 +k3 ) −k4 −k5 12 .2 2 k4\ P (w )(x, t)dxdt ki i kj (wj )kXkj , I j=1

i=1

(5.103)

¯0. where on the right-hand side if kj = 0 then replace Xkj by X

Proof. It follows from H¨ older inequality that the left-hand side of (5.103) is bounded by 3 Y

i=1

2 · k4k (w5 )kL∞ L2 . k4ki (wi )kL3x L∞ · k4k4 (w4 )kL∞ 5 t∈I x Lt x t

2 and k4k (w5 )kL∞ L2 , we use Proposition 5.4. For the term k4k4 (w4 )kL∞ 5 x t x Lt For the term k4ki (wi )kL3x L∞ , we use interpolation between t∈I k4ki (wi )kL2x L∞ and k4ki (wi )kL4x L∞ , then use Proposition 5.4. t∈I t∈I

Proposition 5.21. Assume δ.1. Let m be given by (5.86) with s = −3/4. Then Z 5 δ Y Λ5 (M5 ; u1 , · · · , u5 )dt .N −15/4 kI(uj )kF¯ 0 (δ) . (5.104) 0 j=1

Proof. It follows from Proposition 5.19 that the left-hand side of (5.104) is bounded by X Z δ N45 m2 (N∗45 ) Λ5 ; Q (N + N1 )(N + N2 )(N + N3 )(N + N45 ) 5i=1 m(Ni ) 0 ki >0 5 Y 15 Pk1 u1 , · · · , Pk5 u5 dt .N − 4 kuj kF¯ 0 (δ) , i=1

ki

N∗45 (N +N45 )

where Ni = 2 . Cancel 6 1 and consider only the worst case m2 (N∗45 ) = 1. it suffices to prove R Q P δ 3 1 1 1 k1 ,··· ,k5 >0 0 Λ5 i=1 (N +Ni )m(Ni ) m(N4 ) m(N5 ) ; Pk1 u1 , · · · , Pk5 u5 dt

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141 15

.N − 4

Q5

i=1

kuj kF¯ 0 (δ) .

From symmetry we may assume N1 > N2 > N3 N4 > N5 and then at least two of Ni satisfy Ni &N . Fix the extension u ei of ui such that ke ui kF¯ 0 .2kui kF¯ 0 (δ) , still denoted by ui . From (5.86) and s = −3/4, it is easy to see (N +Ni1)m(Ni ) .N −3/4 hNi i−1/4 and 1 3/4 3/4 .N −3/2 N4 N5 . m(N4 )m(N5 ) Hence we need to control ! 3 XZ δ Y 15 3/4 3/4 Λ5 hNi i−1/4 N4 N5 ; u1 , · · · , u5 dt. N− 4 ki

0

(5.105)

i=1

If N2 ∼ N1 &N , N4 .N2 , then consider the worst case N1 > N2 > N4 > N5 > N3 . We get from (5.103) that 15

(5.105).N − 4

X Ni

15

.N − 4

5 Y

j=1

7/6

7/6

hN1 i−5/4 hN2 i−5/4 hN3 i1/6 N4 N5

5 Y

i=1

[ kP ki ukXki

kIuj kF¯ 0 (δ) .

(5.106)

The rest cases N4 ∼ N5 &N , N1 .N5 or N1 ∼ N4 &N can be handled similarly. The following proposition shows that EI4 (t) is very close to EI2 (t). Proposition 5.22. Assume I the Fourier multiplier with symbol m given by (5.86) and s = −3/4. Then |EI4 (t) − EI2 (t)|.kIu(t)k3L2 + kIu(t)k4L2 . Proof.

Since EI4 (t) = EI2 + Λ3 (σ3 ) + Λ4 (σ4 ), it suffices to prove |Λ3 (σ3 ; u1 , u2 , u3 )|.

3 Y

j=1

kIuj (t)kL2 ,

(5.107)

and |Λ4 (σ4 ; u1 , u2 , u3 , u4 )|.

4 Y

j=1

kIuj (t)kL2 .

(5.108)

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We may assume ubj is non-negative. To prove (5.107), it suffices to prove 2 Y 3 2 2 Λ3 m (ξ1 )ξ1 + m (ξ2 )ξ2 + m (ξ3 )ξ3 ; u1 , u2 , u3 . kuj k2 . (5.109) ξ1 ξ2 ξ3 m(ξ1 )m(ξ2 )m(ξ3 ) i=1 Making dyadic frequency decomposition, we get the left-hand side of (5.109) is bounded by X m2 (ξ1 )ξ1 + m2 (ξ2 )ξ2 + m2 (ξ3 )ξ3 Λ3 ; 4k1 u1 , 4k2 u2 , 4k3 u3 . ξ1 ξ2 ξ3 m(ξ1 )m(ξ2 )m(ξ3 ) ki >0

(5.110)

Denote Ni = 2ki . From symmetry we may assume N1 > N2 > N3 , and hence N1 ∼ N2 &N . Case 1. N3 N . Now m3 (N3 ) = 1, hence we have X N s N s Λ3 (5.112). ; 4 u , 4 u , 4 u k 1 k 2 k 3 1 2 3 1+s 1+s N1 N1 ki >0 X −1/4 −1/4 . N2 ; 4k1 u1 , 4k2 u2 , 4k3 u3 . Λ3 N1 ki >0

It suffices to prove XZ ki >0

−1/2

ξ1 +ξ2 +ξ3 =0,|ξi |∼Ni

N1

3 Y

i=1

Define vi (x), whose Fourier transform is −1/6

vbi (ξ) = Ni

ηki (ξi )ubi (ξi ).

3 Y

i=1

kui kL2 .

ubi (ξ)χ{|ξ|∼Ni } (ξ).

From Sobolev’s embedding theorem we get kvi kL3 .kui kL2 , thus use H¨ older’s inequality we get 3 XZ Y −1/2 N1 ηki (ξi )ubi (ξi ) ki >0

.

X

ξ1 +ξ2 +ξ3 =0,|ξi |∼Ni −1/6

N1

3 Y 1/6

N3

i=1

ki >0

kvi kL3 .

i=1

3 Y

i=1

kui kL2 .

Case 2. N3 &N . It is easy to see that ! 3 −3/4 −3/4 Y X N3 N ; Pk1 u1 , Pk2 u2 , Pk3 u3 . kui kL2 . (5.112). Λ3 1/2 N ki >0

1

i=1

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Thus (5.107) is proved. Next show (5.108). It suffices to prove Y 4 σ4 . Λ4 ; u , u , u , u kuj k2 . (5.111) 1 2 3 4 m(ξ1 )m(ξ2 )m(ξ3 )m(ξ4 ) i=1 By dyadic frequency decomposition, we get the left-hand side of (5.111) is bounded by X σ4 Λ4 ; 4k1 u1 , 4k2 u2 , 4k3 u3 , 4k4 u4 . m(ξ1 )m(ξ2 )m(ξ3 )m(ξ4 ) ki >0

(5.112)

Denote Ni = 2ki . From symmetry we may assume N1 > N2 > N3 > N4 , and hence N1 ∼ N2 &N . In the integration area σ4 N −3 1 .Q . . Q 4 m(ξ1 )m(ξ2 )m(ξ3 )m(ξ4 ) 1/4 4 N i=1 (N + Ni )m(Ni ) i=1

i

We get from H¨ older’s inequality that X N −3 (5.112). k4k1 u1 kL2 k4k2 u2 kL2 k4k3 u3 kL∞ k4k4 u4 kL∞ Q4 1/4 N ki >0 i=1 i .

4 Y

i=1

kuj k2 .

Thus we proved the proposition.

Now we are ready to extend the local solution to a global one. First we need a variant local well-posedness which can be proved similarly. Proposition 5.23. Let −3/4 6 s 6 0. Assume φ satisfies kIφkL2 (R) 6 20 1. Then there exists a unique solution to (5.32) on [−1, 1] such that kIukF¯ 0 (1) 6 C0 ,

(5.113)

where C is independent of N . For any given u0 ∈ H −3/4 and time T > 0, our purpose is to construct a solution on [0, T ]. If u is a solution to KdV equation with initial data u0 , then for any λ > 0, uλ (x, t) = λ−2 u(x/λ, t/λ3 ) is also a solution to KdV equation with initial data u0,λ = λ−2 u0 (x/λ). Simple calculations show that 3

kIu0,λ kL2 .λ− 2 −s N −s ku0 kH s .

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For fixed N (N will be determined later), we take λ ∼ N − 3+2s such that 3

λ− 2 −s N −s kφkH s = 0 < 1.

For simplicity of notations, we still denote uλ by u, u0,λ by u0 , and assume kIu0 kL2 6 0 . The purpose is then to construct solutions on [0, λ3 T ]. By Proposition 5.23 we get a solution on [0, 1], and we need to extend the solution. It suffices to control the modified energy EI2 (t) = kIuk2L2 . First we control EI2 (t) for t ∈ [0, 1], we will prove EI2 (t) < 420 . By standard bootstrap, we may assume EI2 (t) < 520 . Then from Proposition 5.22 we get EI4 (0) = EI2 (0) + O(30 ) and EI4 (t) = EI2 (t) + O(30 ). From Proposition 5.21 we get for all t ∈ [0, 1]

EI4 (t) 6 EI4 (0) + C50 N −15/4 .

Thus kIu(1)k2L2 = EI4 (1) + O(30 ) 6EI4 (0) + C50 N −15/4 + O(30 )

=20 + C50 N −15/4 + O(30 ) < 420 .

Thus the solution u can be extended to t ∈ [0, 2]. Iterating this procedure M steps, then we get for t ∈ [0, M + 1] EI4 (t) 6 EI4 (0) + CM 50 N −15/4

as long as M N −15/4 .1. Thus we get EI2 (M ) = EI4 (t) + O(30 ) = 20 + O(30 ) + CM 50 N −15/4 < 420 , hence the solution can be extended to t ∈ [0, N 15/4 ]. Take N (T ) sufficiently large such that N 15/4 > λ3 T ∼ N 3 T. Therefore, the u is extended to [0, λ3 T ]. In the end of this section, we prove some properties of the global solution. By the scaling we get sup ku(t)kH s ∼λ3/2+s

t∈[0,T ]

sup t∈[0,λ3 T ]

kuλ (t)kH s 6 λ3/2+s

sup t∈[0,λ3 T ]

kIuλ (t)kL2 ,

kIφλ kL2 .N −s kφλ kH s ∼ N −s λ−3/2−s kφkH s .

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From the proof we have sup t∈[0,λ3 T ]

kIuλ (t)kL2 .kIφλ kL2 ,

and hence sup ku(t)kH s .N −s kφkH s .

t∈[0,T ]

Take λ such that kIφλ kL2 ∼ 0 1, thus we get 2 2s kφkH s 3+2s − 3+2s N . λ = λ(N, 0 , kφkH s ) ∼ 0 15

6s

Take N such that N 4 > λ3 T ∼ckφkH s ,0 N − 3+2s T , then N ∼ T 4/3 . Thus the global solution u(x, t) satisfies ku(t)kH −3/4 . (1 + |t|)kφkH −3/4 . For the case s ∈ (−3/4, 0], we left the proof to the readers. 5.6

Schr¨ odinger equation with derivative

In above several sections, we use the Bourgain’s spaces to prove some bilinear estimates with respect to KdV equation. However, in some cases, for example: local well-posedness of the KdV equation in H −3/4 , it is not enough to get the local result directly using the Bourgain’s spaces. In this section, we consider the following Cauchy problem of the Schr¨odinger equation with derivative such that the reader can understand more the method of the Bourgain’s spaces ( iut + uxx = iλ(|u|2 u)x , (x, t) ∈ R2 (5.114) u(x, 0) = u0 (x). From Section 5.1 in this chapter, it follows that dispersion relation of (5.114) φ(ξ) = −ξ 2 . Thus we can obtain the following the definition of the norm of the Bourgain’s spaces X s,b with respect to Schr¨odinger equation with derivative: kukX s,b = khξi2 hτ + ξ 2 ib u b(ξ, τ )kL2 .

In order to prove the well-posedness of the Cauchy problem (5.114), from the standard method, it suffices to show k∂x (u¯ v w)kX s,b−1 .kukX s,b kvkX s,b kwkX s,b .

(5.115)

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But the above inequality is invalid for any s, b ∈ R, one can refer to [91]. In this section, we will use gauge transform to improve the derivative nonlinearity, one can refer to [105; 184; 213]. The result in this section was first given by Takaoka[213]. Via the gauge transformation which was first used by Hayashi [102] (see also [106]) v(x, t) = Gλ (u)(x, t) = e−iλ

Rx

−∞

|u(y,t)|2 dy

u(x, t),

(5.116)

(5.114) is formally rewritten as the Cauchy problem ( 2 i∂t v + ∂x2 v = −iλv 2 ∂x v¯ − |λ|2 |v|4 v,

(5.117)

v(x, 0) = v0 (x),

where v0 (x) = e−iλ transformation,

Rx

−∞

|u0 (y)|2 dy

u(x, t) = eiλ

Rx

−∞

u0 (x). The corresponding inverse gauge

|v(y,t)|2 dy

v(x, t) = G−λ (v).

Since the map Gλ is a bicontinuous map from H s to H s , we can show that the global well-posedness of (5.114) in H s is equivalent to that of (5.117). in the following, we only consider the Cauchy problem (5.117). Notice that the nonlinear term of the original equation iλ(|u|2 u)x = 2iλ|u|2 ux + iλu2 u ¯x , via the gauge transformation, the derivative in the nonlinearity |u|2 ux has been replaced by the quintic nonlinearity |v|4 v. The Strichartz estimate can control the nonlinearity |v|4 v easy. But, at first sight, there still exists a derivative in the nonlinearity u2 u ¯x in equation (5.117). However a derivative of the complex conjugate of the solution u can be handled while a derivative of u cannot since ω(ξ) = −ξ 2 is a even function. From the definition of Bourgain’s spaces, it follows that for complex conjugate of u s

b

s

b

k¯ ukX s,b = khξi hτ − ξ 2 i F ukL2ξ L2τ , kukX s,b = khξi hτ + ξ 2 i F ukL2ξ L2τ . For the nonlinearityu2 u ¯x ,we can prove that following estimate holds for s > 1/2 and b > 1/2 kuv w ¯x kXs,b−1 .kukXs,b kvkXs,b kwkXs,b .

(5.118)

But for |u|2 ux , the above inequality is invalid. Now we mainly estimate the nonlinear term v 2 ∂x v¯. Similarly with the KdV equation, we also need some space-time estimates (Strichartz estimate, local smoothing effect and maximal function estimate) of Schr¨odinger

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eqution, the proofs of these estimates are similar with one of the BenjaminOno equation (the dispersion relation ω(ξ) = −|ξ|ξ) in Section 5.2. 2

Lemma 5.9. The group of Schr¨ odinger equation S(t) = Fx−1 eitξ Fx satisfies : kS(t)v0 kL6x L6t .kv0 kL2 ,

kD

1/2

2 .kv0 kL2 , S(t)v0 kL∞ x Lt

kD−1/4 S(t)v0 kL4x L∞ .kv0 kL2 . t Using Lemma 5.3 and Lemma 5.9, we can easily obtain the following lemma: Lemma 5.10. Let u ∈ X 0,b with 1/2 1/2, 1/2 1/2. Then kv1 v2 ∂x v¯3 kXs,b−1 6 Ckv1 kXs1 ,b0 kv2 kXs2 ,b0 kv3 kXs3 ,b0 . Proof.

(5.119)

By duality and the Plancherel identity, it suffices to show that Q 4 Y hξ4 is hτ4 − ξ42 ib−1 ξ3 4i=1 fi (ξi , τi ) . kfi kL2 , (5.120) Q 3 2 0 2 0 2 0 s Γ4 hτ1 + ξ1 ib hτ2 + ξ2 ib hτ3 − ξ3 ib j=1 hξi i i=1

Z

for all fi ∈ L2 (R2 ) > 0, i = 1, 2, 3, 4; where the hyperplane Γ4 is defined by Γ4 = {(ξ, τ ) ∈ R4 × R4 : ξ1 + ξ2 + ξ3 + ξ4 = 0, τ1 + τ2 + τ3 + τ4 = 0}

which we endow with the standard measure: θi = (ξi , τi ) Z Z h(θ1 , θ2 , θ3 , θ4 ) = h(θ1 , θ2 , θ3 , −θ1 − θ2 − θ3 )dθ1 dθ2 dθ3 . R6

Γ4

From Plancherel identity, it follows that Z Y Z Y 4 4 fi (ξi , τi ) = F −1 (fi )(x, t)dxdt. Γ4 i=1

(5.121)

R2 i=1

By symmetry it suffices to estimate the integral in the domain |ξ1 | 6 |ξ2 |. We define |ξ|max , |ξ|sub , |ξ|thd , |ξ|min be the maximum, second largest,

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third largest and minimum of |ξ1 |, |ξ2 |, |ξ3 |, |ξ4 |; and define σj± = τj ± ξj2 , j = 1, 2, 3, 4. Let fk fi −1 , i = 1, 2; F = F , k = 3, 4. Fi = F −1 k hτ + ξ 2 ib0 hτ − ξ 2 ib0 K(ξ1 , ξ2 , ξ3 , ξ4 ) =

|ξ3 |hξ4 is . hξ1 is hξ2 is hξ3 is

In order to obtain the boundedness of the integral in the left side of (5.120), we split the domain of integration in several pieces as in [91]. Case 1. If max(|ξ1 |, |ξ2 |, |ξ3 |, |ξ4 |) 6 10, then K(ξ1 , ξ2 , ξ3 , ξ4 ) 6 C. By Lemma 5.10 and (5.121), the left side of (5.120) restricted to this domain is bounded by Q4 Z 3 4 Y Y i=1 fi (ξi , τi ) 6. .kf k kF k kfi kL2 . 4 2 i L 0 0 0 2 b 2 b 2 b Γ4 hτ1 + ξ1 i hτ2 + ξ2 i hτ3 − ξ3 i i=1 i=1

Case 2. If |ξ|max 1 and |ξ|thd |ξ|sub , then from ξ1 + ξ2 + ξ3 + ξ4 = 0 it follows that |ξ|max ∼ |ξ|sub . We split this case into the following several subcases. Subcase 2a. If |ξ2 | |ξ3 | ∼ |ξ4 |, then recall the following identity on the hyperplane Γ4 h(ξ1 , ξ2 , ξ3 , ξ4 ) = σ1+ + σ2+ + σ3− + σ4− = 2(ξ1 + ξ3 )(ξ2 + ξ3 ). (5.122) This implies that max(|σ1+ |, |σ2+ |, |σ3− |, |σ4− |)&|ξ3 |2 . By symmetry, we only consider |σ4− | = max(|σ1+ |, |σ2+ |, |σ3− |, |σ4− |) and |σ2+ | = max(|σ1+ |, |σ2+ |, |σ3− |, |σ4− |). If |σ4− | = max(|σ1+ |, |σ2+ |, |σ3− |, |σ4− |), then for b 6 3/4 and s > 1/4, we have |ξ3 ||ξ3 |2(b−1) |ξ3 |1/2 K(ξ1 , ξ2 , ξ3 , ξ4 ) . . . − 1−b s s hξ1 i hξ2 i hξ1 is hξ2 is |σ4 |

By Lemma 5.10, (5.121) and H¨older inequality, the left side of (5.120) restricted to this domain is bounded by Q4 Z |ξ3 ||ξ3 |1/2 i=1 fi (ξi , τi ) Q2 2 b0 2 b0 2 b0 s Γ4 i=1 hξi i hτ1 + ξ1 i hτ2 + ξ2 i hτ3 − ξ3 i

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149

2. .kJ −s F1 kL4x L∞ kJ −s F2 kL4x L∞ kf4 kL2x,t kΛ1/2 F3 kL∞ t t x Lt

4 Y

kfi kL2 . i=1 0 hσ4− i1−b hσ2+ ib

If |σ2+ | = − b0 hσ4 i hσ2+ i1−b ,

max(|σ1+ |, |σ2+ |, |σ3− |, |σ4− |), then from & similarly with above, for b 6 3/4 and s > 1/4, the left side of (5.120) restricted to this domain is bounded by Q Z |ξ3 ||ξ3 |2(b−1) 4i=1 fi (ξi , τi ) Q2 2 b0 2 b0 2 b0 s Γ4 i=1 hξi i hτ1 + ξ1 i hτ3 − ξ3 i hτ4 − ξ4 i 2. .kJ −s F1 kL4x L∞ kF4 kL4x L∞ kf2 kL2x,t kΛ1/2 F3 kL∞ t t x Lt

4 Y

i=1

kfi kL2 ,

which follows from Lemma 5.10, (5.121) and H¨older inequality. Subcase 2b. Assume |ξ2 | ∼ |ξ3 | ∼ |ξ|max or |ξ2 | ∼ |ξ4 | ∼ |ξ|max . By symmetry, we can assume |ξ2 | ∼ |ξ3 | ∼ |ξ|max . If |ξ2 + ξ3 | 6 1 or |ξ4 | . |ξ1 |, then hξ1 i ∼ hξ4 i. For s > 1/2, we have K(ξ1 , ξ2 , ξ3 , ξ4 ) 6 C.

Similarly with Case 1, the left side of (5.120) restricted to this domain is bounded by Q4 Z 4 Y i=1 fi (ξi , τi ) kfi kL2 . . 0 0 0 2 b 2 b 2 b Γ4 hτ1 + ξ1 i hτ2 + ξ2 i hτ3 − ξ3 i i=1 If |ξ2 + ξ3 | > 1 and |ξ4 | |ξ1 |, then

max(|σ1+ |, |σ2+ |, |σ3− |, |σ4− |)&|ξ1 + ξ3 ||ξ2 + ξ3 | ∼ hξ3 ihξ4 i.

By symmetry, we only consider |σ4− | = max(|σ1+ |, |σ2+ |, |σ3− |, |σ4− |). Then for s > 1/2 and b 6 3/4 |ξ3 |1/2 |ξ3 |1/2 K(ξ1 , ξ2 , ξ3 , ξ4 ) . . . − 1−b s 1−b 1−b hξ1 i hξ3 i hξ4 i hξ1 i1/4 hξ2 i1/4 |σ4 | Similarly with Subcase 2a, we can obtain the result. Subcase 2c. If |ξ1 | ∼ |ξ3 | ∼ |ξ|max or |ξ1 | ∼ |ξ4 | ∼ |ξ|max , then similarly with Subcase 2b, we can obtain the result. Subcase 2d. If |ξ1 | ∼ |ξ2 | ∼ |ξ|max , then max(|σ1+ |, |σ2+ |, |σ3− |, |σ4− |) &|ξ3 |2 , similarly with Subcase 2a, we can obtain the result. Case 3. If |ξ|min |ξ|thd ∼ |ξ|sub ∼ |ξ|max , then we have max(|σ1+ |, |σ2+ |, |σ3− |, |σ4− |)&|ξ3 |2 .

Similarly with Subcase 2a, we can obtain the result. Case 4. If |ξ|min ∼ |ξ|thd ∼ |ξ|sub ∼ |ξ|max , similarly with Case 1, we can obtain the result. This completes the proof of theorem.

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Next we estimate the nonlinearity |u|4 u. From the above standard argument, it suffices to show that ku1 u2 u3 u ¯4 u ¯5 kX s,b−1 .

5 Y

i=1

kui kX s,b .

(5.123)

Notice that there exists the condition s > 1/2 when we estimate the nonlinearity u2 u ¯x , so we do not need to obtain the sharp index of s for (5.123), the index s > 1/2 is enough for the local well-posedness. But the author believe that the sharp index of s for (5.123) is s > 0. Lemma 5.11. Let s ∈ R, 1/2 < b 6 1. Then kψ(t/T )ukX s,b−1 .kukL2/(3−2b) H s . x

T

Proof. Without loss of generality, we assume s = 0. Using the defi2/(3−2b) nition of X s,b , imbedding inequality Lt ,→ Htb−1 and Minkowski’s inequality, we have

kψ(t/T )ukX 0,b−1 = ke−itω(ξ) ψ(t/T )(Fx u)(ξ, t)kH b−1 L2 t ξ

−itω(ξ)

. ke ψ(t/T )(Fx u)(ξ, t)kL2/(3−2b) L2 t

ξ

.kukL2/(3−2b) L2 . x

T

This completes the proof of lemma.

Theorem 5.9. Let s > 1/2 and 1/2 0 such that kψ(t/T )u1u2 u3 u ¯4 u¯5 kX s,b−1 .T θ Proof.

5 Y

i=1

kui kX s,b .

(5.124)

Assume u1 = · · · = u5 = u, it suffices to show that kψ(t/T )|u|4ukX s,b−1 .T θ kuk5X s,b .

(5.125)

Using Lemma 5.11, Lemma 5.10 and the Leibniz rule for fractional derivatives, we have 2 kuk 2/(3−2b) s kψ(t/T )|u|4 ukX s,b−1 .k|u|4 ukL2/(3−2b) H s .ku4 kL∞ L H T Lx T

x

θ 5 .T θ kuk4L∞ L8x kukL4T H∞ s .T kuk s,b . X

T

∞

T

This completes the proof.

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151

Using the proof of Theorem 5.5 in Section 5.3, we have Proposition 5.24. Let 1/2 6 s < 1. For v0 ∈ H s , then there exists b > 1/2 and T = T (kv0 kH 1/2 ) > 0 such that the Cauchy problem (5.117) admits a unique local solution v ∈ XTs,b . For the original equation (5.114), we have Theorem 5.10. Let λ ∈ R and s > 1/2. For u0 ∈ H s (R), then there exists b > 1/2 and T = T (ku0 kH 1/2 ) > 0 such that the Cauchy problem (5.114) admits a unique local solution Gλ (u) ∈ XTs,b .

u ∈ C([−T, T ] : H s (R)),

Moreover, given t ∈ (0, T ), the map u0 → u(t) is Lipschitz continuous from H s to C(0, T ; H s ). Proof.

For u0 ∈ H s and 1/2 6 s < 1, we define v0 ∈ H s v0 (x) = e−iλ/2

Rx

−∞

|u0 (y)|2 dy

(n) v0

u0 (x). (n)

Assume is a sequence in H ∞ satisfying v0 → v0 in H s . Let vn be a solution of the Cauchy problem (5.117) obtained in Proposition 5.24. Define un (x, t) =G−λ (vn )(x, t), (n)

u0 (x) =eiλ/2

Rx

−∞

(n)

|v0

(y)|2 dy (n) v0 (x).

From the definition of un , it follows that 2 = kvn kL∞ L2 . kun kL∞ x T Lx T

Using the Leibniz rule for fractional derivatives and Sovolev inequality, for 1/2 6 s < 1, we have

(n) x

s iλ/2 R−∞

|v0 (y)|2 dy s 2 .kD vn kL∞ L2 + D kDs un kL∞ e vn ∞ 2 L x x T T LT Lx

Rx (n)

iλ/2 −∞ |v0 (y)|2 dy s s q 2 + D .kD vn kL∞ e

∞ p kvn kL∞ T Lx T Lx LT Lx

s

kvn k2L∞ L2p 1 x T

.(1 +

kvn k2L∞ s )kvn kL∞ H s , x T T Hx

2 + .kD vn kL∞ T Lx

kvn kL∞ H 1/2 T

x

where 2 < p < ∞, 1/p + 1/q = 1/2 and 1/p1 = 1/p + 1 − s. Similarly with above, we have 2 s .(1 + kvn kL∞ H s + kvm kL∞ H s ) kvn − vm kL∞ H s . kun − um kL∞ x x x T Hx T T T

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It suffices to show that the map Gλ is Lipschitz continuous from H s to Lp for any 2 6 p < ∞. In fact, we have

Rx x 2

iλ/2 R−∞

|f (y)|2 dy f (x) − eiλ/2 −∞ |g(y)| dy g(x)

e Lp

Rx x 2 2

iλ/2 R−∞

|f (y)| dy .kf − gkH s + e − eiλ/2 −∞ |g(y)| dy ∞ (kf kH s + kgkH s ). L

0

Using the fact that |f (x) − f (y)| 6 kf kL1 , we have

Rx x 2

iλ/2 R−∞

|f (y)|2 dy − eiλ/2 −∞ |g(y)| dy

e ∞

L x

iλ/2 R−∞

(|f (y)|2 −|g(y)|2 )dy . e − 1 ∞ L

.k|f |2 − |g|2 kL1 .kf − gkH s (kf kH s + kgkH s ).

Let n → ∞, we can show that u is a solution of the Cauchy problem (5.114). This completes the proof of theorem. 5.7

Some other dispersive equations

Except for KdV equation and Schr¨odinger equation, there exist some other important dispersive equations which have been considered by lots of mathematicer recently. In this section, the authors give some methods for some other dispersive equations such that the readers can understand more the method of Bourgain’s spaces. From the bilinear estimates of KdV equation and trilinear estimates of Schr¨ odinger equation with derivative, it follows that we can use not only the method of Tao’s [k; Z] multiplier [216], but also the method of the local smoothing effects and maximal function estimates [213] to consider the well-posedness of dispersive equations. We take the bilinear estimates of KdV equation for example to show the difference of the two methods. First, we can use the method of Tao’s [k; Z] multiplier to prove Theorem 5.3. That is k∂x (u1 u2 )kX s,b−1 6 Cku1 kX s,b0 ku2 kX s,b0 , s > −3/4, b, b0 > 1/2. (5.126) However, (5.126) holds with the condition s > −5/8 using only Strichartz estimate, the local smoothing effects and the maximal function estimates of KdV equation, one can refer to [128]. If −5/8 > s > −3/4, Kenig, Ponce and Vega [132] used the following inequalities to obtain (5.126). Z ∞ dx C 1 6 , b> (5.127) 2b (1 + |x − β|)2b 2b (1 + |x − α|) (1 + |α − β|) 2 −∞

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Z

∞

−∞ ∞

Z

−∞

Z

153

C 1 dx √ 6 (5.128) 1 , b > 2 (1 + |x|)2b a − x 2 (1 + |a|) C 1 dx 6 , b> 2 (1 + |x − α|)2(1−b) (1 + |x − β|)2b (1 + |α − β|)2(1−b) (5.129)

|x|6β

dx (1 +

|x|)2(1−b)

1

√ 6 a−x

C(1 + β)2(b− 2 ) (1 + |a|)

1 2

, b>

1 . 2

(5.130)

As in Section 5.3 of this chapter, we first define ξ = ξ1 + ξ2 , τ = τ1 + τ2 σ = τ − ξ 3 , σ1 = τ1 − ξ13 , σ2 = τ2 − ξ23 . We split the the hyperplane {(ξ, ξ1 , ξ2 ) × (τ, τ1 , τ2 ) ∈ R3 × R3 : ξ = ξ1 + ξ2 , τ = τ1 + τ2 } into suitable several parts. For the case: |ξ1 | ∼ |ξ2 | |ξ| and |σ| > |σ1 |, |σ2 |, we only use the inequality (5.127); for other cases, similarly with the paper [128], we use the Strichartz estimates, the local smoothing effects and the maximal function estimates, thus we can also obtain (5.126). This can simplify the proof of the paper [132], the readers can try it. From this, it follows that the interaction: high × high → low is worst when one consider low regularity solution in Sobolev spaces of negative indice. Moreover, corresponding to the inequality (5.127), there exists the following inequality, which is also important when one consider the local well-posedness of the dispersive equations Z C 1 1 dx 6 ,
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ν with λ1 = − 3µ 4 , λ2 = −2µ+ 2 , λ3 = −4µ−ν, λ5 = −2µ+ν, where µ is also a real constant. The equation (5.132) describes the three-dimensional motion of an isolated vortex filament, which is embedded in inviscid incompressible fluid fulfilled in an infinite region. Note that the dispersion relation of the equation (5.132) φ(ξ) = νξ 4 − ξ 2 has non-zero singular points. we can not directly use the local smoothing effects and the maximal function estimates of Section 5.2 in this chapter to consider it. So we need use the following Fourier restriction operators Z Z PNf = eixξ fˆ(ξ)dξ, PN f = eixξ fˆ(ξ)dξ, ∀N > 0, (5.133) |ξ|>N

|ξ|6N

to eliminate the singularity of the phase function. Using the Fourier restriction operators and the results of Section 5.2, we have 3

Z

2 6 CkϕkL2 , kDx2 P 2a S(t)ϕkL∞ x Lt

(5.134)

kDx P a S(t)ϕkL4x L∞ 6 CkϕkL2 , t 1/2 sup |P a S(t)u0 |2 dx 6 CT,s ku0 kH s , s > 1,

(5.135) (5.136)

∂t u + H∂xx u = µ∂x (uk ),

(5.137)

− 41

∞

−∞ [−T,T ]

where a depends on ν. Then using (5.134), (5.135), (5.136) and the Strichartz estimates, and the Bourgain’s spaces, we can obtain the local well-posedness of the Cauchy problem (5.132). One can refer to [109; 110]. For some other dispersive equation, in order to obtain the wellposedness, we can not directly use the Bourgain’s spaces X s,b to construct the contraction mapping. So we need new method to consider them. In this book, we do not introduce them in details, one can refer to Guo’s thesis [93]. For example: the generalized Benjamin-Ono (BO) equation: u(x, 0) = u0 (x),

whereR k ∈ Z, µ ∈ R, and H is the Hilbert transform H(f )(x) = f (y) 1 d b π p.v. R x−y dy. From Hf = −isgn(ξ)f (ξ), it follows that the dispersion relation of BO equation ω(ξ) = −|ξ|ξ. This model was first introduced by Benjamin [10] and Ono [182], and it describe one-dimensional internal waves in deep water. Unlike Schr¨odinger equation, if initial data u0 is a real number, then the solution u of (5.137) is also a real number. The real and complex BO equation are different. Compared with KdV equation, BO equation has weaker dispersive effect. If k = 2, the Cauchy problem of BO equation (5.137) is badly behaved with respect to Picard iterative methods in standard Sobolev spaces. But

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for the definition of some weaker well-posedness (for example: the solution mapping is only continuous), some local results of the real BO equation were obtained. For now, the best local result was obtained by Kenig and Ionescu[113], who obtained local well-posedness in L2 . Their method based on the gauge transform and the Besov-type Bourgain’s spaces X s,b . If k = 3, the best result of the well-posedness was obtained by Kenig and Takaoka [136], who used the dyadic gauge transform to prove that the Cauchy problem is globally well-posed in H 1/2 . Moreover,they showed that the index s = 1/2 is sharp in the sense: the solution map u0 → u as mapping from H s to C([−T, T ]; H s ) is no longer uniformly continuous when s < 1/2. Recently, in the paper [91], the author followed the idea in [113], used the contraction mapping to obtain the global well-posedness in H 1/2 with small data in L2 norm, here the author did not use the gauge transform. So this method can apply to complex BO equation. KdV equation and BO equation can be viewed as the special cases of the following dispersion generalized BO equation: ∂t u + |∂x |1+α ∂x u = µ∂x (uk ),

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

(5.138)

2

where 0 6 α 6 1, u : R →R and |∂x | is the Fourier multiplier operator with symbol |ξ|. This model is very interesting from mathematical view, it let us understand the relation between the dispersion effect and the wellposedness. From the intensity of dispersion effect, it lies between KdV equation and BO equation, for example: if 0 < α < 1, then it is similar with BO equation, H s assumption on the initial data is insufficient for the local well-posedness via Picard iteration by showing the solution mapping fails to be C 2 smooth from H s to C([−T, T ]; H s ) at the origin for any s, one can refer to [167]. For the methods to obtained the well-posedness for (5.138), the method of refined energy was used in [127], the method of short time X s,b in [115] was used in [92], and a para-differential renormalization technique was used in [108]. Notice that the spatial dimension of the dispersive equations considered above in this chapter is one. How about the application of the Bourgain’s spaces to dispersive equations in higher dimension? For example: the Bourgain’s spaces can be applied to the higher dimensional Schr¨odinger equation. Compared with one dimension case, cases in higher dimension are more delicate. For instance, a quadratic non-linear Schr¨odinger equation(n = 1, 2) iut + ∆u = N (u, u ¯), u(x, 0) = u0 (x), n

where (x, t) ∈ R × R, N (u, u ¯) = c1 |u|2 + c2 u2 + c3 u¯2 . Using only the Strichartz estimates, one can obtain the well-posedness in H s with s > 0.

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However, by the Bourgain’s spaces, it was showed that it is well-posed in H s with s > −3/4(c1 = 0) and H s with s > −1/4(c1 6= 0) respectively, one can refer to [131; 41]. These results can be improved, one can refer to [9; 142]. For other nonlinearity |u|p u to non-linear Schr¨odinger equation, one can not obtain the better result of low regularity solution using Bourgain’s spaces, but one can obtain the better result of global well-posedness using I-method, one can refer to [45]. Finally, we introduce KP equation, which can be viewed as two dimensional KdV equation. The KP equation is also an important shallow water model. The Cauchy problem of KP-I equation as follows: ∂t u + ∂x3 u − ∂x−1 ∂y2 u + ∂x (u2 /2) = 0; u(x, y, 0) = φ(x, y), 3

(5.139)

where u(x, y, t) : R → R. The KP-I equation and The KP-II equation, in which the sign of the term ∂x−1 ∂y2 u in (5.139) is + instead of −, arise in physical contexts as models for the propagation of dispersive long waves with weak transverse effects. The KP-II equation is well understood from the point of view of well-posedness, one can refer to [16; 214; 212; 97; 98]. For the KP-I equation, Molinet, Saut and Tzvetkov [169] showed that it is badly behaved with respect to Picard iterative methods in standard Sobolev spaces H s1 ,s2 with any s1 , s2 ∈ R. On the positive side, it is known that the KP-I initial value problem is globally well-posed in the second energy spaces [122; 168]; these global well-posedness results rely on refined energy methods. Recently, Ionescu, Kenig and Tataru [115] obtained the global well-posedness in the natural energy space E1 = {φ ∈ L2 (R2 ), ∂x φ ∈ L2 (R2 ), ∂x−1 ∂y φ ∈ L2 (R2 )}, they introduce a new method, which can be looked as the blend of Bourgain’s spaces method and energy method. Recently, the authors in [95] remove the condition ∂x−1 ∂y φ ∈ L2 (R2 ). Compared with the result of KP-II equation, the global wellposedness of KP-I equation in L2 is a open problem.

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Chapter 6

Frequency-uniform decomposition techniques W. Orlicz had a small apartment and he once applied to the city administration for a bigger one. The answer of an employee was: “Your apartment is really small but we cannot accept your claim since we know that you have your own spaces!” —— Lech Maligranda2

In this chapter we study the NLS and the NLKG by using the frequencyuniform decomposition techniques, see [243; 245; 246; 247]. Let Qk be the unit cube with the center at k, {Qk }k∈Zn constitutes a decomposition of Rn . Such a kind of decomposition goes back to the work of N. Wiener [252], and we say that it is the Wiener decomposition of Rn . We can roughly write k ∼ F −1 χQk F ,

k ∈ Zn ,

(6.1)

where χE denotes the characteristic function on the set E. Since Qk is a translation of Q0 , k (k ∈ Zn ) have the same localized structures in the frequency space, which are said to be the frequency-uniform decomposition operators. Similar to Besov spaces, one can use {k }k∈Zn and `q (Lp ) to generate a class of function spaces, so called modulation spaces for which the norm is defined by !1/q X sq q s kf kMp,q = hki k k f kp . k∈Zn

6.1

Why does the frequency-uniform decomposition work

Comparing with the dyadic decomposition, the frequency-uniform decomposition has at least two advantages for the Schr¨odinger semi-group: 2 W. Orlicz (1903-1990), Polish mathematician and he is known for his Orlicz spaces, L. Maligranda is one of his students, see http://www.sm.luth.se/˜lech.

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(1) k eit∆ : Lp → Lp satisfies a uniform truncated decay; (2) k eit∆ is uniformly bounded in Lp . It is known that S(t) = eit∆ : Lp → Lp if and only if p = 2. This is one of the main reasons that we can not solve NLS in Lp (p 6= 2). However, if we consider the frequency-uniform decomposition, there holds kk S(t)f kp . (1 + |t|)n|1/2−1/p| kk f kp .

(6.2) n

Taking the summation to the above inequality over all k ∈ Z , we can get the relevant estimate in modulation spaces, which enable us to solve NLS 0 in modulation spaces Mp,1 , 1 6 p 6 ∞. Now we give a comparison between frequency-uniform and dyadic decompositions. Recalling that 4k ∼ F −1 χ{|ξ|∼2k } F

and noticing that |{ξ : |ξ| ∼ 2k }| = O(2nk ) and |Qk | = 1, we see that their Bernstein’s estimates are quite different: k4k f kq . 2n(1/p−1/q)k k4k f kp ,

kk f kq . kk f kp ,

p 6 q.

We find that there is no regularity increasement for the Bernstein estimate of k , which leads to the Schr¨odinger semi-group S(t) = eit∆ satisfies the following truncated decay estimate, kk S(t)f kp . (1 + |t|)−n(1/2−1/p) kk f kp0 ,

p > 2, 1/p + 1/p0 = 1. (6.3)

0

Comparing it with the classical Lp → Lp estimate

k4k S(t)f kp . |t|−n(1/2−1/p) k4k f kp0 ,

(6.4)

we see that the singularity at t = 0 disappears in (6.3). Recalling that in the classical Strichartz inequalities, in order to handle the singularity of |t|−n(1/2−1/p) at t = 0, we need condition n(1/2−1/p) 6 1, which arises from the Hardy-Littlewood-Sobolev inequality and it is an essential condition. To solve NLS with the nonlinearity |u|σ u, one need to improve the spatial regularity at least at the H˙ n/2−2/σ level for σ > 4/n. Considering the Strichartz inequalities with the frequency-uniform decomposition, due to the truncated decay, we can remove the condition n(1/2 − 1/p) 6 1. As a result, in solving NLS with the nonlinearity |u|σ u, it is not necessary to improve the spatial derivative regularity if σ is large. The frequency-uniform decomposition is more delicate than the dyadic decomposition and it is easier to handle the derivatives in the nonlinearity, say for k = (k1 , ..., kn ) with k1 1 and j 1, k∂x1 k f kp ∼ k1 kk f kp ,

k∂x1 4j f kp . 2j k4j f kp .

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159

Using the frequency-uniform decomposition techniques, the initial data s s should belong to modulation spaces Mp,q . The modulation space Mp,q s coincides with the Sobolev space H in the case p = q = 2. Recently, this technique was also developed to the Navier-Stokes equation and the dissipative nonlinear electrohydrodynamic system [247; 116; 56]. 6.2

Frequency-uniform decomposition, modulation spaces

Roughly speaking, dyadic decomposition operators combined with function spaces `q (Lp ) generate Besov spaces, frequency-uniform decomposition operators joint with function spaces `q (Lp ) produce modulation spaces. The modulation space was introduced by Feichtinger [73] in 1983, from the history point of view, it was defined by the short-time Fourier transform3 . During the past twenty years, the frequency-uniform decomposition had not been attached the importance to applications and it is even not mentioned in Gr¨ ochenig’s book [86]. However, from PDE point of view, the combination of frequency-uniform decomposition operators and Banach function spaces `q (X(Rn ))4 seems to be important in making nonlinear estimates, which contains an automatic decomposition on high-low frequencies. We now give an exact definition on frequency-uniform decomposition operators. Since χQk can not make differential operations, one needs to replace χQk in (6.1) by a smooth cut-off function. Let ρ ∈ S (Rn ), ρ : Rn → [0, 1] be a smooth function verifying ρ(ξ) = 1 for |ξ|∞ 6 1/2 and ρ(ξ) = 0 for |ξ|∞ > 15 . Let ρk be a translation of ρ, k ∈ Zn .

ρk (ξ) = ρ(ξ − k), 3 Let

(6.6)

g ∈ S (Rn ), Vg f (x, ω) =

Z

e−itω g(t − x)f (t)dt. Rn

Vg f is said to be the short-time Fourier transform of f . The norm on modulation spaces is given by kf koM s p,q

Z

=

Rn

One can prove that k · koM s

p,q

Z

Rn

p

|Vg f (x, ω)| dx

q/p

sq

hωi dω

!1/q

.

(6.5)

s and k · kMp,q are equivalent norms; cf. [73] for a proof on

modulation spaces defined in an Abel group and [246] for a straightforward proof. 4 X is a Banach function space defined in Rn . 5 For ξ = (ξ , ..., ξ ), |ξ| n ∞ := maxi=1,...,n |ξi |. 1

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P n We see that ρk (ξ) = 1 in Qk and so, k∈Zn ρk (ξ) > 1 for all ξ ∈ R . Denote !−1 X σk (ξ) = ρk (ξ) ρk (ξ) , k ∈ Zn . (6.7) k∈Zn

Then we have  |σk (ξ)| > c, ∀ ξ ∈ Qk ,    supp σk ⊂ {ξ : |ξ − k|∞ 6 1}, P  ∀ ξ ∈ Rn , n σk (ξ) ≡ 1,   k∈Z α |D σk (ξ)| 6 C|α| , ∀ ξ ∈ Rn , α ∈ (N ∪ {0})n .

(6.8)

Hence, the set

Υn = {{σk }k∈Zn : {σk }k∈Zn satisfies (6.8)}

(6.9)

is non-void. If there is no confusion, in the sequel we will write Υ = Υn . Let {σk }k∈Zn ∈ Υ. Denote k := F −1 σk F ,

k ∈ Zn .

(6.10)

{k }k∈Zn are said to be frequency-uniform decomposition operators. For k ∈ Zn , we denote |k| = |k1 | + ... + |kn |, hki = 1 + |k|. Let s ∈ R, 0 < p, q 6 ∞,   !1/q   X s s Mp,q (Rn ) = f ∈ S 0 (Rn ) : kf kMp,q = hkisq k k f kqp <∞ .   n k∈Z

(6.11)

0 s For simplicity, we write Mp,q = Mp,q . Mp,q is said to be the modulation space.

6.2.1

Basic properties on modulation spaces

As indicated in Proposition 1.16, if we consider a function f with a compact support set in the frequency space, then we can compare kf kp with kf kq , which is one of the advantages of the frequency localizations. The following is a refinement of Proposition 1.16. Lemma 6.1. Let Ω be a compact subset of Rn , diam Ω < 2R, 0 0 which only depends on p, q and R such that

where

LpΩ

kf kq 6 Ckf kp ,

= {f ∈ L : suppfˆ ⊂ Ω}. p

∀ f ∈ LpΩ ,

(6.12)

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Proof. We emphasize that the constant in (6.12) is independent of the b B(0,R) = 1. position of Ω. Let ψ ∈ S (Rn ) satisfy supp ψb ⊂ B(0, 2R) and ψ| Due to diam Ω < 2R, we can take ξ0 satisfting Ω ⊂ B(ξ0 , R). For any b − ξ0 ), which implies that f ∈ S (Rn ) ∩ LpΩ , we have fb = fb · ψ(· Z f (x) = c f (x − y)eiξ0 y ψ(y)dy. Rn

If 1 6 p 6 ∞, using Young’s inequality, we can get the conclusion. If 0 < p < 1, Z 1−p kf k∞ 6 Cψ sup |f (x − y)| |f (x − y)|p dy, y

Rn

we obtain that for q = ∞ and 0 < p < 1, the result holds. For general 0 < p 6 q < ∞, by H¨ older’s inequality, Lq norm can be controlled by Lp ∞ and L norms, from which we get the result, as desired. Lemma 6.2 (LpΩ -multiplier). Let Ω ⊂ Rn be a compact subset, 0 < r 6 ∞ and σr = n(1/(r ∧ 1) − 1/2). If s > σr , then there exists a C > 0 such that kF −1 ϕF f kr 6 CkϕkH s kf kr

(6.13)

holds for all f ∈ LrΩ and ϕ ∈ H s . Proof. If r > 1, in view of Bernstein’s multiplier estimates, we have the result, see Proposition 1.11. If r < 1, the proof is similar to the case r > 1, cf. [224]. Proposition 6.1 (Completeness). Let 0 < p, q 6 ∞ and s ∈ R. s (1) Mp,q is a (quasi-) Banach space. Moreover, if 1 6 p, q 6 ∞, then s Mp,q is a Banach space. s (2) S (Rn ) ⊂ Mp,q ⊂ S 0 (Rn ). s (3) Let 0 < p, q < ∞, then S (Rn ) is dense in Mp,q .

Proof. Analogous to Besov spaces, we can prove the consequence and the details are omitted. Proposition 6.2 (Equivalent norm). Let {σk }k∈Zn , {ϕk }k∈Zn ∈ Υ. s Then {σk }k∈Zn and {ϕk }k∈Zn generate equivalent norms on Mp,q .

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We have the translation identity, (F −1 mF f )(x) = eixk F −1 m(· + k)F (e−iky f (y)) (x).

Proof.

For convenience, we denote

σk := F −1 σk F ,

−1 ϕ ϕk F . k := F

Noticing the almost orthogonality of σk X σk = σk ϕ k+` ,

(6.14)

|`|∞ 61

we have kσk f kp 6

X

|`|∞ 61

σ kϕ k+` (k f )kp .

Using the multiplier estimate ϕ σ kϕ k+` (k f )kp . kσk kH s kk+` f kp ,

s > n(1/(1 ∧ p) − 1/2),

and kσk kH s 6 C, we immediately have X kσk f kp . kϕ k+` f kp . |`|∞ 61

{σ }

{ϕ }

k k 6 kf kMp,q So, kf kMp,q s . s

Proposition 6.2 indicates that one can choose {σk }k∈Zn ∈ Υn according to our requirement. In applications of PDE, it is convenient to us the following {σk }k∈Zn ∈ Υn . Let {ηk }k∈Z ∈ Υ1 , we denote σk (ξ) := ηk1 (ξ1 )...ηkn (ξn ),

(6.15)

then we have {σk }k∈Zn ∈ Υn . the above σk (ξ) realizes the separation of different variables. Proposition 6.3 (Embedding). Let s1 , s2 ∈ R and 0 < p1 , p2 , q1 , q2 6 ∞. (1) If s2 6 s1 , p1 6 p2 and q1 6 q2 , then Mps11,q1 ⊂ Mps22,q2 . s1 s2 (2) If q2 < q1 and s1 − s2 > n/q2 − n/q1 , then Mp,q ⊂ Mp,q . 1 2 Proof.

Recall that kk f kp2 6

X

|`|∞ 61

kF −1 σk F (k+` f )kp2 .

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163

Using Lemma 6.1, we Xhave X kk f kp2 . kF −1 σk F (k+` f )kp1 . kk+` f kp1 . q1

|`|∞ 61 q2

|`|∞ 61

In view of ` ⊂ ` , we can get the conclusion of (1). Now we prove (2). Using H¨older’s inequality, !1/q2 X s q q 2 2 2 s2 kf kMp,q = hki k k f kp 2

k∈Zn

s1 6 kf kMp,q

1

X

k∈Zn

(s2 −s1 )q1 q2 /(q1 −q2 )

hki

!(q1 −q2 )/q1 q2

Noticing that ∞ X X hki(s2 −s1 )q1 q2 /(q1 −q2 ) . hiin−1+(s2 −s1 )q1 q2 /(q1 −q2 ) , k∈Zn

.

(6.16)

(6.17)

i=0

s1 − s2 > n/q2 − n/q1 implies that the right hand side of (6.17) is a con vergent series. It follows that (2) holds.

Proposition 6.4 (Dual space). Let s ∈ R and 0 < p, q < ∞. If p > 1, we denote 1/p + 1/p0 = 1; If 0 1, Proposition 6.4 is similar to that of Besov spaces, however, if 0 < p < 1, the result is quite different from that of Besov spaces. The details of the proof of Proposition 6.4 can be found in [246] by following the proof of the relevant result in Besov spaces. Remark 6.1. If p, q ∈ [1, ∞], Propositions 6.1 and 6.4 were obtained by Feichtinger [73]. In [246; 247], the cases 0 < p < 1 and 0 < q < 1 were considered. The proofs of the results in this section are due to [246; 247]. Soon after the work [247], Kobayashi [148] independently defined Mp,q for all 0 < p, q 6 ∞ and obtained Proposition 6.1. Almost at the same time as [246], Kobayashi [149] discussed the dual space of Mp,q and obtained partial results of Proposition 6.4: if 0 < p < 1 or 1 < q < ∞, he obtained Mp0 ,q0 ⊂ (Mp,q )∗ ⊂ M∞,∞ . For the other cases, he showed (Mp,q )∗ = Mp0 ,q0 . Recently, by using the molecular decomposition techniques of modulation spaces, Kobayashi and Sawano [150] reconsidered the s dual space of Mp,q and they also obtained the result of Proposition 6.4. It is worth to mention that Triebel [225] introduced a class of generalized modulation spaces for all indices 0 < p, q 6 ∞, however, those spaces have no complete norms, which seems harder to use in the study of PDEs.

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Inclusions between Besov and modulation spaces

From the definitions, we see that Besov spaces and modulation spaces are rather similar, both of them are the combinations of frequency decomposition operators and function spaces `q (Lp ). In fact, we have the following inclusion results. Theorem 6.1 (Embedding). Let 0 < p, q 6 ∞ and s1 , s2 ∈ R. We have the following results. s1 s2 (1) Bp,q ⊂ Mp,q if and only if s1 > s2 + τ (p, q), where 1 1 1 1 − , n + −1 ; τ (p, q) = max 0, n q p q p s1 s2 (2) Mp,q ⊂ Bp,q if and only if s1 > s2 + σ(p, q), where 1 1 1 1 σ(p, q) = max 0, n − , n 1− − . p q p q

1 q

p=2

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S1 S3 ( 12 , 12 )

S2 (0, 0)

Fig. 6.1 τ (p, q) =

(1, 0)

The distribution of τ (p, q) R2+ : τ (p, q) = n( q1 − n( p1

+

1 q

− 1) in S3 .

1 p

1 ) p

in S1 ; τ (p, q) = 0 in S2 ;

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165

1 q

1

=

q

1

R1 1

(0, 1)

p

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p 1

+

q

= 1

R2

( 12 , 12 ) R3

(0, 0)

Fig. 6.2

(1, 0)

1 p

The distribution of σ(p, q): σ(p, q) = 0 in R1 ; σ(p, q) = n( p1 −

σ(p, q) = n(1 −

1 p

−

1 ) q

1 ) q

in R2 ;

in R3 .

The inclusions between Besov and modulation spaces in the cases (1/p, 1/q) ∈ [0, 1]2 were first discussed by Gr¨obner [85] and he has never published his results. When (1/p, 1/q) is in the vertices of the square [0, 1]2 , Gr¨ obner’s results are optimal. Afterwards, Toft [222] obtained the sufficiency of Theorem 6.1 in the cases (1/p, 1/q) ∈ [0, 1]2 . Sugimoto and Tomita [210] showed the necessity of the first inclusion of Theorem 6.1 in the cases (1/p, 1/q) ∈ [0, 1]2 , and by duality they obtained the second inclusion is also sharp if (1/p, 1/q) ∈ [0, 1]2 and p, q 6= ∞. Sugimoto and Tomita’s idea is to use Feichtinger’s norm and the dilation property of modulation spaces. In [245; 246; 247] the authors proved the conclusions of Theorem 6.1 by using frequency-uniform decomposition techniques. Corollary 6.1. We have the following inclusions. s+n/2

B2,1

s s ⊂ M2,1 ⊂ B2,1 ,

s+n s s B∞,1 ⊂ M∞,1 ⊂ B∞,1 .

The above embedding theorem is of importance for the study of nonlinear PDEs and we give the details of the proof. We can first prove Theorem 6.1 for some special p, q and then by the interpolation to carry out the other cases. The following proof is a collection of [85; 247; 246;

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245], which is different from those of [210; 222]. Lemma 6.3. Let 0 < p, q 6 ∞. We have

0 Mp,q ⊂ Bp,q , ∀ q 6 p ∧ 1,

n(1/(p∧1)−1/q) 0 Mp,q ⊂ Bp,q , ∀ q > 1 ∧ p.

Proof.

First, we show the first inclusion ∞ X k4k f kqp . kf kqB 0 =

(6.19)

p,q

k=0

√ Case 1. q < 1 and p > 1. Let ak = max(0, 2k−1 − n) and bk = √ 2k+1 + n. Noticing that for |i| ∈ 6 [ak , bk ], 4k i f = 0, we obtain that  q X X k4k (i f )kp  . ki f kqp . k4k f kqp 6  i∈Zn , |i|∈[ak ,bk ]

i∈Zn , |i|∈[ak ,bk ]

(6.20)

From (6.20) we deduce the result. Case 2. q < 1 and p < 1. From p < 1 and q/p 6 1 it follows that  q/p Z X k4k f kqp 6  |4k (i f )(x)|p dx i∈Zn , |i|∈[ak ,bk ]

6

i∈Zn ,

X

|i|∈[ak ,bk ]

k4k (i f )kqp .

(6.21)

In view of the multiplier estimate on LpΩ , 4k i : LpB(i,√n) → LpB(i,√n) . By (6.21) we get the conclusion. Next, we show the second inclusion. We consider the following two cases. Case 1. p > 1. Denote √ Λ0 = {k ∈ Zn : B(k, n) ∩ {ξ : |ξ| ∈ [0, 2)} 6= ∅}, (6.22) √ n j−1 j+1 Λj = {k ∈ Z : B(k, n) ∩ {ξ : |ξ| ∈ [2 , 2 )} 6= ∅}, j > 1. (6.23) In view of the multiplier estimate on LpΩ , X X X k4j f kp . k4j k+` k f kp . kk f kp . k∈Λj |`|∞ 61

(6.24)

k∈Λj

Since q > p ∧ 1, we have

(a1 + ... + am )q 6 mq−1 (aq1 + ... + aqm ).

(6.25)

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167

For k ∈ Λj , one has that |k| ∼ 2j and Λj overlaps at most O(2nj ) cubes, which follows that  q ∞ X X  kk f kp  kf kqB 0 . p,q

j=0

.

∞ X

k∈Λj

j=0

.

X

2jn(q−1)

k∈Λj

∞ X

X

j=0 k∈Λj

. kf kq

kk f kqp

hkin(q−1) kk f kqp

n(1−1/q)

Mp,q

.

(6.26)

Case 2. p < 1. In view of q/p > 1, we have q/p

(a1 + ... + am )q/p 6 mq/p−1 (a1

+ ... + aq/p m ).

(6.27)

LpΩ ,

By the multiplier estimate on  X XZ q  k4j f kp .

Rn

k∈Λj `∈Λ

.2

jn(q/p−1)

X

k∈Λj

q/p

|4j k+` k f |p dx kk f kqp .

Similar to Case 1, we have the conclusion, as desired.

(6.28)

Lemma 6.4. We have n(1/q−1/2)

B2,q Proof.

⊂ M2,q , ∀ 0 < q 6 2.

By Plancherel’s identity, kf kqM2,q ∼

X

k∈Zn

kχQk fbkq2 .

(6.29)

Denote Λj = {k ∈ Zn : |k| ∈ [2j−1 , 2j )}. Let L 1. One has that kf kqM2,q .

X

|k|62L

kχQk fbkq2 +

∞ X X

j=L k∈Λj

kχQk fbkq2 .

It is easy to see that Λj has at most O(2nj ) elements. So,

q

X

∞ X

q q jn(1−q/2) b kf kM2,q . kf k2 + 2 χQk f

k∈Λj

j=L 2

(6.30)

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. kf kq2 + .

∞ X

j=L

2jn(1−q/2) kχ|·|∈(2j−2 ,2j+1 ) fbkq2

kf kq n(1/q−1/2) , B2,q

(6.31)

which is the result, as desired.

Lemma 6.5. Let 1 6 p, q 6 ∞. We have σ(p,q) 0 Mp,q ⊂ Bp,q , σ(p, q) = max 0, n Proof.

1 1 − p ∧ p0 q

.

(6.32)

By the dual versions of Lemmas 6.3 and 6.4, we have n 0 Mp,∞ ⊂ Bp,∞ , 1 6 p 6 ∞, n(1/2−1/q) M2,q

0 ⊂ B2,q , 2 6 q 6 ∞.

(6.33) (6.34)

Taking p = 1, ∞ and q = ∞, we have n/2

n 0 0 n 0 M1,∞ ⊂ B1,∞ , M2,∞ ⊂ B2,∞ , M∞,∞ ⊂ B∞,∞ .

(6.35)

By the complex interpolation on (6.35) (see Appendix), 0

max(n/p, n/p ) 0 Mp,∞ ⊂ Bp,∞ , 1 6 p 6 ∞.

(6.36)

Lemma 6.3 also implies that 0 Mp,1 ⊂ Bp,1 , 1 6 p 6 ∞.

(6.37)

Recall that 0 M2,2 = B2,2 .

(6.38)

Making a complex interpolation on (6.36)–(6.38), we obtain the result, as desirted. When 0 < p < 1, we need the following multiplier estimate (cf. Peetre [192] and Triebel [224]). Proposition 6.5. Let Ω ⊂ Rn be a compact set, and 0 0 and 0 0.

LpB(0,b)

and M ∈

n(1/p−1/2) B2,p ,

where C > 0 is independent

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Proof.

169

If f ∈ LpB(0,b) , then f (b−1 ·) ∈ LpB(0,1) . In view of (F −1 M F f )(x) = [F −1 M (b ·)f\ (b−1 ·)](bx),

it follows from Proposition 6.5 that Corollary 6.2 holds.

Lemma 6.6. Let 0 < p 6 ∞. Then we have n(1/(p∧1)−1) Bp,∞ ⊂ Mp,∞ .

(6.39)

Proof. First, we consider the case 0 < p < 1. By Corollary 6.2, for any |k| 1, |k| ∈ [2j−1 , 2j ),

4

X

−1 kk f kp = F σk ϕj+` F f

`=−4

p

. kσk (2j+5 ·)kB n(1/p−1/2) 2,p

4 X

`=−4

k4j+` f kp .

(6.40)

Using the scaling in Besov spaces (cf. [224]) s , s 6 λs−n/p kgkBp,q kg(λ ·)kBp,q

λ & 1,

we obtain that6 kσk (2j+5 ·)kB n(1/p−1/2) . 2jn(1/p−1) kσk kB n(1/p−1/2) . 2jn(1/p−1) . 2,p

2,p

(6.41)

Inserting (6.41) into (6.40), we immediately have kk f kp . 2

jn(1/p−1)

4 X

`=−4

k4j+` f kp .

(6.42)

By (6.42), we get (6.39). Now we consider the case p > 1. By Young’s inequality, for |k| 1,

4

X

−1 kk f kp = F σk ϕj+` F f

`=−4

. kF −1 σk k1

4 X

`=−4

p

k4j+` f kp .

4 X

`=−4

k4j+` f kp .

This implies the result. 6 We

can assume that σk = σ0 (· − k) in the definition of

(6.43)

s . Mp,q

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Proof. [Proof of Theorem 6.1] (Sufficiency) First, we show that s+σ(p,q) s Mp,q ⊂ Bp,q . By Lemma 6.5, we see that the conclusion holds if 1 6 p, q 6 ∞. By Lemma 6.3 we have the result if 0 0} and (see Fig. 6.1), S1 = {(1/p, 1/q) ∈ R2+ : 1/q > 1/p, 1/p 6 1/2};

S2 = {(1/p, 1/q) ∈ R2+ : 1/q 6 1/p, 1/p + 1/q 6 1}; S3 = R2+ \ (S1 ∪ S2 ).

We first consider the case (1/p, 1/q) ∈ S3 . We have τ (p, q) = n(1/p + 1/q − 1). Take (p0 , q0 ) and (p1 , q1 ) satisfying 1 1 = + p0 p 1 1 = , p1 2

Let θ = q1 ( p1 +

1 q

1 1 , = 0; q q0 1 1 1 1 = + − . q1 p q 2

− 12 )−1 , we have

1 1−θ θ 1 1−θ θ = + , = + ; p p0 p1 q q0 q1 1 1 1 1 1 + − 1 = (1 − θ) −1 + − θ. p q p0 q1 2 By Lemmas 6.4 and 6.6, n(1/q1 −1/2)

B2,q1

⊂ M2,q1 ,

0 −1) Bpn(1/p ⊂ Mp0 ,∞ . 0 ,∞

A complex interpolation yields n(1/p+1/q−1) Bp,q ⊂ Mp,q .

Secondly, we consider the case (1/p, 1/q) ∈ S1 . If (1/p, 1/q) ∈ S˙ 1 (S˙ 1 denotes the set of all inner points of S1 ), then (1/p, 1/q) can be lying in the segment by connecting (1/∞, 1/∞) and a point (1/p1 , 1/q1 ) in the line {(1/p, 1/q) : p = 2, q < 2}. By a complex interpolation, we have n(1/q−1/p) Bp,q ⊂ Mp,q .

(6.44)

Thirdly, if (1/p, 1/q) ∈ S2 , the dual version of the first inclusion implies the result, as desired. (Necessity) We need to show that for any 0 < η 1, τ (p,q)−η Bp,q 6⊂ Mp,q .

(6.45)

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Case 1.1. (1/p, 1/q) ∈ S3 . Let f = F −1 ϕj , j 1. We have kf kq τ (p,q)−η = Bp,q

1 X

`=−1

2q(τ (p,q)−η)(j+`) kF −1 ϕj+` ϕj kqp

. 2q(n/q−η)j .

(6.46)

Assume without loss of generality that 3 j+1 5 j−1 ·2 6 |ξ| 6 · 2 . ϕj (ξ) = 1, if ξ ∈ Dj := ξ : 4 4

Noticing that

√ Λj = k ∈ Zn : B(k, n) ⊂ Dj

contains at least O(2jn ) elements, we have X X kf kqMp,q = kk F −1 ϕj kqp > kF −1 σk ϕj kqp & 2nj . k∈Zn

(6.47)

k∈Λj

By (6.46) and (6.47), kf kMp,q & 2ηj kf kB τ (p,q)−η , p,q

which implies that (6.45) holds. Case 1.2. (1/p, 1/q) ∈ S2 . We consider the case q = ∞. Taking k(j) = (2j , 0, ..., 0) and f = F −1 σk(j) , we see that −η . kf kMp,∞ & 1 & 2ηj kf kBp,∞

If q < ∞, we show that

ε ε . Mp,q 6⊂ Bp,r ∪ B∞,∞

(6.48)

k(j) = (2N j , 0, ..., 0) ∈ Zn , ∞ X b F (ξ) = 2−εN j fb(ξ − k(j)).

(6.49)

Let f ∈ S (Rn ) be a Schwartz function satisfying suppfb ⊂ {ξ : |ξi | < 1/2, i = 1, ..., n}. Let N 1, 0 < ε 1,

(6.50)

j=1

Noticing that suppFb ⊂ ∪∞ j=1 Qk(j) , we have

k F = 0 if k 6= k(j) + `,

Since N 1, we see that ∞ X X kF kqMp,q 6 k k(j)+` F kqp j=1 |`|∞ 61

|`|∞ 6 1.

(6.51)

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6

∞ X X

j=1 |`|∞ 61

=

∞ X X

j=1 |`|∞ 61

2−εN qj kF −1 σk(j)+` fb(· − k(j))kqp 2−εN qj k` f kqp . 1,

(6.52)

which leads to F ∈ Mp,q . On the other hand, letting s > ε, we have kF krBp,r > s &

&

∞ X j=1

2srj k 4j F krp

∞ X

2srN j k 4N j F krp

j=1

2(s−ε)rN j k F −1 ϕN j χQk(j) fb(· − k(j))krp ,

j=1 ∞ X

(6.53)

where ϕj = ϕ(2−j ·). We can assume that ϕ(ξ) = 1 if |ξ| ∈ [3/4, 5/4]. Hence, (6.54) F −1 ϕN j χQk(j) fb(· − k(j)) = F −1 χQk(j) fb(· − k(j)). By (6.53) and (6.54),

& kF krBp,r s

∞ X j=1

2(s−ε)rN j = ∞.

(6.55)

s The above discussion also implies that F 6∈ B∞,∞ . ˜ k := Case 1.3. (1/p, 1/q) ∈ S1 . We can assume that σk (ξ) = 0 if ξ ∈ 6 Q {ξ : |ξi − ki | 6 5/8, 1 6 i 6 n}, and the dyadic decomposition function sequence satisfying ϕj (ξ) = 1 if ξ ∈ Dj (= {ξ : 45 · 2j−1 6 |ξ| 6 43 · 2j+1 }). Let ˜ k ⊂ Dj }, j 1. Aj = {k ∈ Zn : Q (6.56)

It is easy to see that Aj has at most O(2nj ) elements. Let f ∈ S (Rn ) be a radial Schwartz function satisfying suppfˆ ⊂ B(0, 1/8), X g(x) = eixk (τk f )(x), τk f = f (· − k). (6.57) k∈Aj

d Taking notice of supp τd k f ⊂ B(0, 1/8), we see that supp τk (τk f )∩supp σ` = ∅, k 6= `. So,  1/q X kgkMp,q >  kF −1 σk F gkqp  k∈Aj

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

=

X

k∈Aj

173

1/q

q kF −1 σ0 (τd k f )kp 

& 2jn/q .

On the other hand, supp gˆ ⊂ {ξ : 2j−1 6 |ξ| 6 2j+1 }. Hence, !1/q 1 X nj(1−q/p)−ηjq −1 q kgkB n(1/q−1/p)−η 6 2 kF ϕj+` F gkp . p,q

(6.58)

(6.59)

`=−1

Using the multiplier estimates and H¨older’s inequality, we have 2/p

(6.60)

. 2nj/2 .

(6.61)

kgk2 . kF −1 ϕj+` F gkp . kgkp 6 kgk1−2/p ∞ By Plancherel’s identity,  Z  kgk2 = kˆ g k2 =

X

Rn k∈A j

1/2

|τk (e−ikξ fˆ(ξ))|2 dξ 

We can further assume that f (x) = f (|x|) is a decreasing function on |x|. In view of f ∈ S (Rn ), we have |f (x − k)| . (1 + |x − k|)−N ,

N 1.

(6.62)

Denote B0 = {k ∈ Aj : |x − k| 6 2},

Bi = {k ∈ Aj : 2i < |x − k| 6 2i+1 }.

Bi contains at most O(2n i ) elements. It follows from (6.62) that X X XX 2ni |f (2i )| . 2(n−N )i . 1. |g(x)| 6 |f (x − k)| . f (0) + i>1

i>0 k∈Bi

i>0

(6.63)

Collecting (6.60), (6.61) and (6.63), we have kF −1 ϕj+` F gkp . 2nj/p .

(6.64)

Inserting (6.64) into (6.59) and using (6.58), we immediately have kgkB n(1/q−1/p)−η . 2nj/q−ηj . 2−ηj kgkMp,q . p,q

(6.65)

This implies the conclusion. Now we show the necessity of the second inclusion. It suffices to show σ(p,q)−η 0 Mp,q 6⊂ Bp,q ,

∀η > 0.

Notice that (see Fig. 6.2) R1 = {(1/p, 1/q) ∈ R2+ : 1/q > 1/p, 1/p + 1/q > 1},

(6.66)

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R2 = {(1/p, 1/q) ∈ R2+ : 1/q 6 1/p, 1/p > 1/2}, R3 = R2+ \ (R1 ∪ R2 ).

We consider the following three cases. Case 2.1. (1/p, 1/q) ∈ R1 . We have discussed it in Case 1.1. Case 2.2. (1/p, 1/q) ∈ R3 . Assume that there exists an η > 0 such that n(1−1/p−1/q)−η 0 Mp,q ⊂ Bp,q , we will get a contradiction. If 1 6 p, q < ∞, by −n(1/p0 +1/q0 −1)+η

duality, Bp00 ,q0 ⊂ Mp0 ,q0 , which contradicts the first inclusion. If p = ∞ or q = ∞, one can use the same way as in Case 2.1 to get the result. In fact, letting f = F −1 ϕj , we have 0 kf kBp,q > kF −1 ϕj ϕj kp & 2nj(1−1/p) .

(6.67)

On the other hand, kf kM n/p0 6 suphkin(1−1/p) kF −1 σk ϕj kp . 2nj(1−1/p) , p,∞

k



kf kM n/q0 6  ∞,q

X

|k|∈[2j−1 ,2j+1 ]

1/q

2nj(q−1) kF −1 σk ϕj kq∞ 

(6.68) . 2nj .

(6.69)

From (6.67)–(6.69) it follows that (6.66) holds in the cases p = ∞ or q = ∞. Case 2.3. (1/p, 1/q) ∈ R2 . Take f ∈ S (Rn ) satisfying f (0) = 1, supp fˆ ⊂ Q0 . Choose 0 < a 1 (which will be fixed in (6.75) below). Denote fa (x) = f (x/a). We easily see that supp fˆa ⊂ Q0,a := {ξ : |ξi | 6 1/2a, 1 6 i 6 n}. Recall that Dj = {ξ : 45 · 2j−1 6 |ξ| 6 43 · 2j+1 } contains at most O(an 2jn ) pairwise disjoint cubes Qk(i),a := k(i) + Q0,a , i = 1, ..., O(an 2jn ). We write Aj = {k(i) : i = 1, ..., O(an 2jn )}, X g(x) = eixk (τk fa )(x). (6.70) k∈Aj

For any N 1, |f (x)| 6 CN (1 + |x|)−N ,

(6.71)

|fa (x)| 6 CN aN |x|−N .

(6.72)

it follows that

By the continuity of f (x) and f (0) = 1, we deduce that there exists a % > 0 such that7

7 In

|fa (x)| > 1/2,

x ∈ B(0, %).

fact, % can be chosen as % = a%0 , %0 > 0 is independent of a.

(6.73)

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In view of (6.70) and (6.73), we get, for any Xx ∈ B(k(i), %), that |g(x)| > |fa (x − k(i))| − |fa (x − k)| >

1 − 2

k∈Aj \{k(i)}

X

k∈Aj \{k(i)} `

|fa (x − k)|.

(6.74)

Denote Aj,` := {k ∈ Aj : 2 6 |k − k(i)| < 2`+1 }. We can further assume that f (x) is a decreasing function on |x|. Since Aj,` has at most O(an 2`n ) elements, weX have for any x ∈ B(k(i), %), X X |fa (x − k)| 6 |fa (x − k)| `>1 k∈Aj,`

k∈Aj \{k(i)}

6C

X `>1

.

X

an 2n` |fa (2` − %)|

CN an+N 2(n−N )` 6 1/4,

(6.75)

`>1

where N > n + 1 and CCN an+N 6 1/4. Hence, it follows from (6.74) and (6.75) that |g(x)| > 1/4, x ∈ B(k(i), %). (6.76) By (6.76), we have

n nj

O(aX 2 )

1 n/p nj/p

χ 2 , (6.77) kg(x)kp > B(k(i),%) & (a%)

4

i=1 p

where % and a are independent of j 1. We can assume that ϕj (ξ) = 1 for ξ ∈ Dj . Since supp gˆ ⊂ Dj , we have F −1 ϕj F g = g. Hence, (6.77) implies that 0 kgkBp,q > kF −1 ϕj F gkp & (a%)n/p 2nj/p . (6.78) On the other hand,  1/q X kgk n(1/p−1/q) =  2nj(q/p−1) kF −1 σk F gkqp  Mp,q

|k|∈[2j−1 ,2j+1 ]

62

nj/p

sup

|k|∈[2j−1 ,2j+1 ]

kF −1 σk F gkp .

(6.79)

Since supp σk overlaps at most finite many supp τ` (τd ` fa ), using multiplier estimates, we have kF −1 σk F gkp . kf kp . (6.80) Hence, (6.79) and (6.80) imply that kgkM n(1/p−1/q) . 2nj/p . (6.81) p,q By (6.78) and (6.81) we immediately have (6.66). We finish the proof of Theorem 6.1.

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6.4

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NLS and NLKG in modulation spaces

As indicated in §6.1, the dispersive semi-group combined with the frequency-uniform decomposition operator has some advantages and we discuss them in this section. The results of this section can be found in [11; 12; 50; 247; 246]. 6.4.1

Schr¨ odinger and Klein-Gordon semigroup in modulation spaces

Let S(t) = eit4 denote the Schr¨odinger semi-group. In [247], Wang, Zhao and Guo obtained the uniform boundedness for the Ginzburg-Landau semigroup L(t) = e(a+i)t4 (a > 0) in modulation spaces and their proof is also adapted to the Schr¨ odinger semi-group (a = 0 in L(t)). First, we show that k S(t) : Lp → Lp is uniformly bounded on k ∈ Zn by following the proof in [247]. X 2 kk S(t)f kp 6 kF −1 σk+` eit|ξ| σk fˆkp |`|∞ 61

6

X

|`|∞ 61

2

kF −1 (σk+` eit|ξ| )k1 kk f kp .

(6.82)

2

Hence, it suffices to estimate kF −1 (σk eit|ξ| )k1 . Using the multiplier estimate, we have 2

2

kF −1 (σk eit|ξ| )k1 = kF −1 (σ0 eit|ξ| )k1 X 2 1−n/2L n/2L . kσ0 k2 kDα (σ0 eit|ξ| )k2 |α|=L

. (1 + |t|

n/2

).

(6.83)

Noticing that kk S(t)f k2 = kk f k2 ,

(6.84)

by a complex interpolation, we immediately have kk S(t)f kp . (1 + |t|)n|1/2−1/p| kk f kp .

(6.85)

s Proposition 6.6 (Uniform boundedness of S(t) in Mp,q ). Let s ∈ R, 1 6 p 6 ∞ and 0 < q < ∞. Then we have s . s . (1 + |t|)n|1/2−1/p| kf kMp,q kS(t)f kMp,q

(6.86)

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177

Shortly after the work [247], Proposition 6.6 is independently obtained by B´enyi, Gr¨ ochenig, Okoudjou and Rogers in [12] and their result contains α more general semi-group eit(−∆) with α 6 1, whose proof is based on the short-time frequency analysis technique. Miyachi, Nicola, Riveti, Taracco and Tomita [164] were able to consider the case α > 1, Chen, Fan and Sun α [38] obtained some refined estimates for eit(−∆) with any α > 0 by using the oscillatory integral estimates in higher spatial dimensions. Now we consider the truncated decay of S(t); cf. [246]. Let us recall 0 the Lp − Lp estimate of S(t), kS(t)f kp . |t|−n(1/2−1/p) kf kp0 , where 1/p + 1/p0 = 1. We also have X k k S(t)f kp . k k+` f kp0 , |`|∞ 61

2 6 p 6 ∞, 2 6 p 6 ∞.

(6.87)

(6.88)

Indeed, by Young’s and H¨ older’s inequalities, we have X kk S(t)f kp 6 kF −1 σk σk+` exp(−it|ξ|2 )F f kp . `∈Λ

6

X `∈Λ

kσk+` exp(−it|ξ|2 )σk F f kp0 . kk f kp .

Combing (6.87) with (6.88), one has that kk S(t)f kp . (1 + |t|)−n(1/2−1/p) kk f kp0 ,

2 6 p 6 ∞.

(6.89)

Multiplying both sides of (6.89) by hkis and then taking `q norm, we have s kS(t)f kMp,q . (1 + |t|)−n(1/2−1/p) kf kMps0 ,q .

Proposition 6.7. Let s ∈ R, 2 6 p < ∞, 1/p + 1/p0 = 1 and 0 < q < ∞. Then we have s kS(t)f kMp,q . (1 + |t|)−n(1/2−1/p) kf kMps0 ,q .

(6.90)

Propositions 6.6 and 6.7 are optimal in the sense that the powers of time variable are sharp, cf. [51]. Now we consider the truncated decay estimate 1/2 for the Klein-Gordon semi-group G(t) = eitω where ω = I − ∆. X kk G(t)f kp 6 kF −1 σk+` eithξi σk fˆkp |`|∞ 61

6

X

|`|∞ 61

kF −1 (σk+` eithξi )k1 kk f kp .

(6.91)

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So, it suffices to estimate kF −1 (σk eithξi )k1 . Using the Fourier multiplier estimate, we have kF −1 (σk eithξi )k1 = kF −1 (σ0 eithξ+ki )k1 X 1−n/2L n/2L . kσ0 k2 kDα (σ0 eithξ+ki )k2 |α|=L

. (1 + |t|

n/2

).

(6.92)

Noticing that kk G(t)f k2 = kk f k2 ,

(6.93)

by a complex interpolation, we immediately have kk G(t)f kp . (1 + |t|)n|1/2−1/p| kk f kp .

(6.94)

Proposition 6.8. Let s ∈ R, 1 6 p 6 ∞ and 0 < q < ∞. Then we have s s . kG(t)f kMp,q . (1 + |t|)n|1/2−1/p| kf kMp,q

(6.95)

0

It is known that G(t) satisfies the following Lp − Lp estimate kG(t)f kH −2σ(p) . |t|−n(1/2−1/p) kf kp0 ,

(6.96)

p

where 2 6 p < ∞,

2σ(p) = (n + 2)

1 1 − 2 p

.

(6.97)

From (6.96) it follows that k k G(t)f kH −2σ(p) . |t|−n(1/2−1/p) k k f kp0 . p

(6.98)

Applying the multiplier estimate, k k (I − 4)δ/2 gkp . hkiδ kgkp .

(6.99)

In view of (6.98) and (6.99), we have X k k G(t)f kp . hki2σ(p) k k+` G(t)f kH −2σ(p) p

`∈Λ

. hki2σ(p) |t|−n(1/2−1/p)

X `∈Λ

k k+` f kp0 .

On the other hand, by H¨ older’s and Young’s inequality, 2 1/2 k k G(t)f kp . kσk eit(1+|ξ| ) fbkp0 X 2 1/2 . kσk eit(1+|ξ| ) F k+` f kp0

`∈Λ

(6.100)

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.

X `∈Λ

.

X `∈Λ

179

kF k+` f kp k k+` f kp0 .

(6.101)

So, for any θ ∈ [0, 1], it follows from (6.100) and (6.101) that X k k+` f kp0 . k k G(t)f kp . hki2σ(p)θ |t|−nθ(1/2−1/p)

(6.102)

`∈Λ

Noticing that σ(p) > 0, by (6.101) we have X k k G(t)f kp . hki2σ(p)θ k k+` f kp0 .

(6.103)

`∈Λ

Combining (6.102) and (6.103), we get

k k G(t)f kp . hki2σ(p)θ (1 + |t|)−nθ(1/2−1/p)

X `∈Λ

k k+` f kp0 .

(6.104)

Multiplying both sides of (6.104) by hkis and then taking `q norm, we immediately obtain that Proposition 6.9. Let s ∈ R, 2 6 p < ∞, 1/p + 1/p0 = 1, 0 < q < ∞, θ ∈ [0, 1] and σ(p) is as in (6.97). Then we have s kG(t)f kMp,q . (1 + |t|)−nθ(1/2−1/p) kf kM s+2σ(p)θ .

(6.105)

p0 ,q

6.4.2

Strichartz estimates in modulation spaces

For convenience, we write kf k`s,q (Lγ (I,Lp )) =

X

k∈Zn

sq

hki

kk f kqLγ (I,Lp )

!1/q

,

(6.106)

q p q γ p p p `q (Lγ (I, Lp )) := `0,q (L (I, L )), ` (Lx,t∈I ) := ` (L (I, L )). Recall that the truncated decay can be generalized to the following estimate α kU (t)f kMp,q . (1 + |t|)−δ kf kMp0 ,q ,

(6.107)

U (t) = F −1 eitP (ξ) F ,

(6.108)

where 2 6 p < ∞, 1 6 q < ∞, α = α(p) ∈ R, δ = δ(p) > 0, α and δ are independent of t ∈ R, U (t) is a dispersive semi-group, and P (·) : Rn → R is a real valued function. In the sequel we will assume that U (t) satisfies conditions (6.107) and (6.108), from which we can get some Strichartz inequalities for U (t) in modulation spaces.

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Generally speaking, all of the Strichartz estimates in Sec. 3.2 hold if we α replace X ∗ and X α by Mp0 ,2 and Mp,2 in Sec. 3.2, respectively. However, do not forget that the estimate in (6.107) contains no singularity at t = 0, which leads that we can remove the restriction condition δ 6 1 in the Strichartz estimates on modulation spaces and the results are much better than those in Sec. 3.2. Proposition 6.10 (Strichartz inequalities). Let U (t) satisfy (6.107) and (6.108). For any γ > 2 ∨ (2/δ), we have kU (t)f k`α/2,q (Lγ (R,Lp )) . kf kM2,q .

(6.109)

In addition, if γ > q, then we have kU (t)f kLγ (R,M α/2 ) . kf kM2,q .

(6.110)

p,q

Proof. The proof is similar to that as in Sec. 3.2. However, we need to carefully handle the indices α, p, q and γ. First, we consider the case 1 < q < ∞ to show that Z (U (t)f, ψ(t))dt . kf kM2,q kψk`−α/2,q0 (Lγ 0 (R,Lp0 )) (6.111)

R

n

S (R ) and ψ ∈ C0∞ (R, S (Rn )). Noticing that S (Rn ) and 0 0 −α/2,q0 are dense in M2,q and ` (Lγ (R, Lp )), respectively,

holds for all f ∈ C0∞ (R, S (Rn )) (6.111) implies (6.109). By duality,

Z

Z

U (−t)ψ(t)dt (U (t)f, ψ(t))dt . kf kM2,q

R

R

For any k ∈ Zn ,

2 Z

k U (−t)ψ(t)dt

R

2 Z

. k k ψkLγ 0 (R,Lp0 ) U (t − s)ψ(s)ds

k

R

.

(6.112)

.

(6.113)

M2,q0

Lγ (R,Lp )

Recall that {k }k∈Zn are almost orthogonal. By (6.107), the definition of k · kMp0 ,q together with the multiplier estimate, we have X k k U (t)f kp . hti−δ hki−α k k k+` f kMp0 ,q −δ

. hti

−α

hki

|`|∞ 61

k k f kp0 .

(6.114)

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If δ 6= 1, applying (6.114), Young’s and Hardy-Littlewood-Sobolev’s inequalities, we obtain that8

Z

k . hki−α k k ψkLγ 0 (R,Lp0 ) . (6.115) U (t − s)ψ(s)ds

γ

p R

L (R,L )

If δ = 1 and γ > 2, we can use Young’s inequality to get that (6.115) holds. If γ = 2 and δ = 1, applying the same way as in Sec. 3.3, we can obtain (6.115). Hence, in view of (6.113) and (6.115), we have

Z

−α/2

k U (−t)ψ(t)dt k k ψkLγ 0 (R,Lp0 ) . (6.116)

. hki R

2

0

Both sides in (6.116) are taken the norm in `q ,

Z

U (−s)ψ(s)ds . kψk`−α/2,q0 (Lγ 0 (R,Lp0 )) .

R

(6.117)

M2,q0

(6.112) and (6.117) imply (6.111). If γ > q, by Minkowski’s inequality, one sees that the left hand side of (6.110) can be bounded by the left hand side of (6.109). Now we consider the case q = 1. It suffices to show that Z (U (t)f, ψ(t))dt . kf kM2,q kψkc−α/2(Lγ 0 (R,Lp0 )) (6.118)

R

n

holds for all f ∈ S (R ) and ψ ∈ procedure, we can prove our result. Denote (U f )(t) =

Z

0

C0∞ (R, S (Rn )).

Repeating the above

t

U (t − s)f (s, ·)ds.

(6.119)

Proposition 6.11. Let U (t) satisfy (6.107) and (6.108). For any γ > 2 ∨ (2/δ), we have kU f k`q (L∞ (R,L2 )) . kf k`−α/2,q (Lγ 0 (R,Lp0 )) .

(6.120)

In addition, if γ 0 6 q, then kU f kL∞ (R,M2,q ) . kf kLγ 0 (R,M −α/2 ) . p0 ,q

(6.121)

Proof. We only sketch the proof. Applying the same way as in (6.113), (6.115) and (6.116), we see that kk U f k22 . hki−α k k f k2Lγ 0 (R,Lp0 ) .

(6.122)

(6.122) implies (6.120). By Minkowski’ s inequality, it follows from (6.120) that (6.121) holds. 8 Due to ht − si−δ < |t − s|−δ , if δ < 1, γ = 2/δ, one can apply Hardy-LittlewoodSobolev’s inequality.

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Proposition 6.12. Let U (t) satisfy (6.107) and (6.108). For any γ > 2 ∨ (2/δ), we have kU f k`α/2,q (Lγ (R,Lp )) . kf k`−α/2,q (Lγ 0 (R,Lp0 )) .

(6.123)

In addition, we assume that q 6 2 if δ = 1 and γ = 2, then we have for any γ > 2 ∨ (2/δ), kU f kLγ (R,M α/2 ) . kf kLγ 0 (R,M −α/2 ) . p,q

Proof. have

(6.124)

p0 ,q

We only give an outline of the proof of (6.124). By (6.107), we kU f kM α/2 . p,q

Z

t

0

ht − si−δ kf (s)kM −α/2 ds.

(6.125)

p0 ,q

If δ 6= 1, or δ = 1 and γ > 2, then we can use the same way as in (6.115) to get the conclusion. If δ = 1 and γ = 2, it follows from (6.115) and Minkowski’s inequality that (6.124) holds. Proposition 6.13. Assume that U (t) satisfies (6.107) and (6.108), γ > max(2/δ, 2). Then we have kU f k`α/2,q (Lγ (R,Lp )) . kf k`q (L1 (R,L2 )) .

(6.126)

In addition, if γ > q, then kU f kLγ (R,M α/2 ) . kf kL1 (R,M2,q ) .

(6.127)

p,q

Proof. have

Assume that f, ψ ∈ C0∞ (R, S (Rn )). By Proposition 6.11, we Z Z t U (t − τ )f (τ )dτ, ψ(t) dt R+ 0

Z ∞

. kf kL1 (R,M2,q ) U (· − t)ψ(t)dt

·

L∞ (R,M2,q0 )

. kf kL1 (R,M2,q) kψk`−α/2,q0 (Lγ 0 (R,Lp0 )) .

(6.128)

−α/2,q

0

0

0

Since ψ ∈ C0∞ (R, S (Rn )) is dense in ` (Lγ (R, Lp )) and −α/2 γ0 p0 c (L (R, L )), by duality, we get the result, as desired. The Schr¨ odinger semi-group corresponds to the cases α = 0, δ = n(1/2 − 1/p) and 2 6 p < ∞. Taking q = 1 in Propositions 6.10–6.12, we immediately have Corollary 6.3. Let 2 6 p < ∞, γ > 2 ∨ γ(p), and 1 1 2 =n − . γ(p) 2 p

(6.129)

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183

Rt Let S(t) = eit∆ , A = 0 S(t − s) · ds. Then kS(t)ϕk`1 (Lγ (R,Lp )) . kϕkM2,1 ,

kA f k`1 (Lγ (R,Lp ))∩`1 (L∞ (R,L2 )) . kf k`1 (Lγ 0 (R,Lp0 )) .

(6.130) (6.131)

Similar to Corollary 6.3, we have Corollary 6.4. Let 2 6 p < ∞, θ ∈ (0, 1], 1 6 q < ∞, 1 1 1 1 2 = nθ − , 2σ = (n + 2)θ − . (6.132) γθ (p) 2 p 2 p Rt Let G(t) be as in (6.96), G = 0 G(t − s) · ds. Then for any γ > 2 ∨ γθ (p), we have kG(t)ϕk`−σ,q (Lγ (R,Lp )) . kϕkM2,q , (6.133)

kG f k`−σ,q (Lγ (R,Lp ))∩`q (L∞ (R,L2 )) . kf k`σ,q (Lγ 0 (R,Lp0 )) .

(6.134)

Related Strichartz estimates in Wiener amalgam spaces for the Schr¨odinger equation were obtained by Cordero and Nicola [49]. 6.4.3

Wellposedness for NLS and NLKG

We consider the Cauchy problem for NLS, iut + ∆u = f (u), u(0, x) = u0 (x). (6.135) n 0 ∞ Noticing that B∞,1 ⊂ M∞,1 ⊂ B∞,1 ⊂ L are sharp embeddings, up to 0 now, we can not get the wellposedness of NLS in L∞ or in B∞,1 . However, we can obtain the local wellposedness of NLS in M∞,1 . We have (see [11; 50]) Theorem 6.2. Let n > 1, f (u) = λ|u|κ u, κ ∈ 2N, λ ∈ R, u0 ∈ Mp,1 and 1 6 p 6 ∞. Then there exists a T > 0 such that (6.135) has a unique solution u ∈ C([0, T ), Mp,1 ). Moreover, if T < ∞, then lim supt%T ku(t)kMp,1 = ∞. 2

If the nonlinearity has an exponential growth, say f (u) = λ(e|u| − 1)u, n/2 0 the result in Theorem 6.2 also holds. Noticing that B2,1 ⊂ M2,1 ⊂ B2,1 ∩ n C(R ) are sharp embeddings, we can get that NLS is global wellposed in M2,1 if initial data are sufficiently small. Theorem 6.3. Let n > 1, f (u) = λ|u|κ u, κ ∈ 2N, λ ∈ R, κ > 4/n, u0 ∈ M2,1 and there exists a sufficiently small δ > 0 such that ku0 kM2,1 6 δ. Then (6.135) has a unique solution u ∈ C(R, M2,1 ) ∩ `1 (Lpx,t∈R ), (6.136)

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where p ∈ [2 + 4/n, 2 + κ] ∩ N, `1 (Lpx,t∈R ) is as in (6.106). 2

Theorem 6.4. Let n > 2, f (u) = λ(e%|u| − 1)u, λ ∈ C and % > 0. Assume that u0 ∈ M2,1 and there exists a sufficiently small δ > 0 such that ku0 kM2,1 6 δ. Then (6.135) has a unique solution u ∈ C(R, M2,1 ) ∩ `1 (L4x,t∈R ).

(6.137)

The proof of Theorem 6.2 relies upon the algebra structure of Mp,1 . Lemma 6.7. Let 1 6 p 6 ∞, then Mp,1 is a Banach algebra. The proof of Lemma 6.7 can be shown by following the same technique as s and we omit the details. that of E2,1 [Proof of Theorem 6.2] Denote Z t A f (t, x) = S(t − τ )f (τ, x)dτ.

Proof.

0

Consider the mapping T : u(t) → S(t)u0 − iA f (u).

(6.138)

We write D = {u : kukC([0,T ]:Mp,1) 6 M },

d(u, v) = ku − vkC([0,T ]:Mp,1 ) ,

where M = 2Cku0 kMp,1 . If u ∈ D, by Proposition 6.6 and Lemma 6.7, kT ukC([0,T ]:Mp,1) . (1 + T )n/2 ku0 kMp,1 + T kukκ+1 C([0,T ]:Mp,1) . (6.139)

We can choose a sufficiently small 0 < T < 1 such that CT M κ 6 1/2. It follows that T : (D, d) → (D, d) is a contraction mapping. The left part of the proof is standard and we omit the details of the proof. Lemma 6.8. Let 1 6 p, pi , γ, γi 6 ∞ satisfy 1 1 1 = + ... + , p p1 pN

1 1 1 = + ... + . γ γ1 γN

(6.140)

Then we have ku1 u2 ...uN k`1 (Lγ (R,Lp )) 6 C N

N Y

i=1

kui k`1 (Lγi (R,Lpi )) .

(6.141)

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Proof.

185

It suffices to consider the case N = 2. We have X k k (u1 u2 )kp 6 kk (i u1 j u2 )kp .

(6.142)

i,j∈Zn

Noticing that k (i u1 j u2 ) = 0 if |k−i−j| > k0 (n), where k0 (n) depends only on n, we have from (6.142) that X kk (i u1 j u2 )kp χ|k−i−j|6k0 (n) . (6.143) k k (u1 u2 )kp 6 i,j∈Zn

Applying Bernstein’s estimate and H¨older’s inequality, by (6.143) we have X k i u1 kp1 k j u2 kp2 χ|k−i−j|6k0 (n) . (6.144) k k (u1 u2 )kp 6 i,j∈Zn

So, by (6.144), H¨ older’s and Minkowski’s inequalities, we get that ku1 u2 k`1 (Lγ (R,Lp ))  1/γ γ X Z X  k i u1 (t)kp1 k j u2 (t)kp2 χ|k−i−j|6k0 (n) dt . k∈Z

.

R

X X

k∈Z i,j∈Zn

i,j∈Zn

k i u1 (t)kLγ1 (R,Lp1 ) k j u2 (t)kLγ2 (R,Lp2 ) χ|k−i−j|6k0 (n) .

(6.145)

By Young’s inequality, it follows from (6.145) that ku1 u2 k`1 (Lγ (R,Lp )) . ku1 k`1 (Lγ1 (R,Lp1 )) ku2 k`1 (Lγ2 (R,Lp2 )) .

By the induction and (6.146), we can get the result, as desired.

(6.146)

Taking p ∈ N, p ∈ [2 + 4/n, 2 + κ] and X = `1 (L∞ (R, L2 )) ∩ `1 (Lp (R, Lp )), we easily see that 2 6n p

1 1 − 2 p

=

n(p − 2) 2p

(6.147)

(6.148)

and “ = ” in (6.148) holds if and only if p = 2 + 4/n. Using Corollary 6.3, we have kS(t)u0 kX . ku0 kM2,1 ,

(6.149)

kA f kX . kf (u)k`1 (Lp0

x,t∈R )

.

(6.150)

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Proof. [Proof of Theorem 6.3] Let X be as in (6.147) and T be as in (6.138). By (6.149) and (6.150), kT ukX . ku0 kM2,1 + kf (u)k`1 (Lp0

x,t∈R )

.

(6.151)

Since p ∈ [2 + 4/n, 2 + κ], we have

p−1 κ+2−p 1 = + . p0 p ∞

By Lemma 6.8,

π(u1+κ ) 1 p0 ` (L

x,t∈R )

. kukp−1 `1 (Lp

x,t∈R )

(6.152)

kuk2+κ−p `1 (L∞

x,t∈R )

.

(6.153)

Since

k i uk∞ . k i uk2 , i ∈ Zn , in view of (6.153) and (6.154), we have

π(u1+κ ) 1 p0

(6.154)

. kuk1+κ X .

(6.155)

1+κ kT ukX . ku0 kM2,1 + kukX .

(6.156)

` (Lx,t∈R )

It follows from (6.151) and (6.155) that

Putting D = {u : kukX 6 M },

d(u, v) = ku − vkX ,

(6.157)

we easily see that, if M > 0 is sufficiently small, ku0 kM2,1 . M/2, then T : (D, d) → (D, d) is a contraction mapping, which leads that (6.135) has a solution u ∈ X. The left part of the proof is standard and we omit the details. The proof of Theorem 6.4 is similar to that of Theorem 6.3. Let Y = `1 (L∞ (R, L2 )) ∩ `1 (L4x,t∈R ).

(6.158)

Then we have kT ukY . ku0 kM2,1 +

∞ X %k

k=1

k!

Using Lemma 6.8, one has that

1+2k

u

1 4/3 . C 2k+1 kuk31 4 ` (L ` (L )

x,t∈R

k|u|2k uk`1 (L4/3

x,t∈R )

kuk2k−2 `1 (L∞

x,t∈R )

x,t∈R )

.

(6.159)

. C 2k+1 kuk2k+1 . Y (6.160)

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Hence, kT ukY . ku0 kM2,1 +

∞ X C 2k+1 k=1

k!

kuk2k+1 . Y

(6.161)

Applying (6.161) and combining the proof of Theorem 6.3, we can get the conclusion of Theorem 6.4. We now consider the initial value problem for NLKG, utt + (I − ∆)u + f (u) = 0,

u(0) = u0 , ut (0) = u1 .

(6.162)

Analogous to NLS, we have Theorem 6.5. Let n > 1, f (u) = λ|u|κ u, κ ∈ 2N, λ ∈ R, (u0 , u1 ) ∈ −1 Mp,1 × Mp,1 and 1 6 p 6 ∞. Then there exists a T > 0 such that (6.135) −1 has a unique solution (u, ut ) ∈ C([0, T ), Mp,1 ) × C([0, T ), Mp,1 ). Moreover, if T < ∞, then lim supt%T (ku(t)kMp,1 + kut (t)kM −1 ) = ∞. p,1

If the nonlinearity has an exponential growth, the corresponding results as in Theorem 6.5 also hold. Theorem 6.6. Let n > 1, f (u) = λu1+κ , κ ∈ N and κ > 4/n. Put σ=

n+2 . n(2 + κ)

(6.163)

σ−1 σ Assume that (u0 , u1 ) ∈ M2,1 × M2,1 and there exists a sufficiently small σ δ > 0 such that ku0 kM2,1 + ku1 kM σ−1 6 δ. Then (6.162) has a unique 2,1 solution σ u ∈ C(R, M2,1 ) ∩ `1 (L2+κ x,t∈R ).

(6.164)

Theorem 6.7. Let n > 2, f (u) = sinh u − u and σ = (n + 2)/4n. Assume σ−1 σ that (u0 , u1 ) ∈ M2,1 × M2,1 and there exists a sufficiently small δ > 0 such σ + ku1 k σ−1 6 δ. Then (6.162) has a unique solution that ku0 kM2,1 M 2,1

σ u ∈ C(R, M2,1 ) ∩ `1 (L4x,t∈R ).

6.5

(6.165)

Derivative nonlinear Schr¨ odinger equations

We study the initial value problem for the derivative nonlinear Schr¨odinger equation (gDNLS) iut + ∆± u = F (u, u ¯, ∇u, ∇¯ u),

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

(6.166)

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where u is a complex valued function of (t, x) ∈ R × Rn , ∆± u =

n X

εi ∂x2i ,

εi ∈ {1, −1},

i = 1, ..., n,

(6.167)

∇ = (∂x1 , ..., ∂xn ), F : C2n+2 → C is a series of z ∈ C2n+2 , X F (z) = F (z1 , ..., z2n+2 ) = cβ z β , cβ ∈ C,

(6.168)

i=1

m+16|β|<∞

2 6 m < ∞, m ∈ N, supβ |cβ | < ∞9 . The typical nonlinear term is ¯, ∇u, ∇¯ u) = |u|2~λ · ∇u + u2 ~µ · ∇¯ u + |u|2 u, F (u, u see [48; 63; 227]. Another model is F (u, u ¯, ∇u, ∇¯ u) = (1 + |u|2 )−1 (∇u)2 u¯ =

∞ X

k=0

(−1)k |u|2k (∇u)2 u ¯,

|u| < 1,

which is an equivalent version of the Schr¨odinger flow [61; 88; 114; 261]. The non-elliptic gDNLS arises in the strongly interacting manybody systems near the criticality, where anisotropic interactions are manifested by the presence of the non-elliptic case, as well as additional residual terms which involve cross derivatives of the independent variables [48; 63; 227]. Some water wave and completely integrable system models in higher spatial dimensions are also non-elliptic, cf. [1; 152; 258; 259]. A large amount of work has been devoted to the study of gDNLS, see [100; 101; 114; 125; 126; 130; 133; 135; 143; 147; 107; 186; 209; 199]. Since the nonlinearity in gDNLS contains derivative terms and the Strichartz inequalities can not absorb any derivatives, gDNLS can not be solved if we use only the Strichartz estimate. One needs to look for some other ways to handle the derivative terms in the nonlinearity. Up to now, three kinds of methods seem to be very useful for gDNLS. One is to use the energy estimate to deal with the derivatives in the nonlinearity, the second way is to use Bourgain’s space X s,b and the third technique is Kato’s smooth effect estimates. Of course, there are some connections between these methods. We will use the smooth effect estimates together with the frequencyuniform decomposition techniques to study gDNLS and we show that it is globally wellposed and scattering in a class of modulation spaces. 9 In

fact, we only need the condition |cβ | 6 C |β| .

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189

For convenience to the readers, we sketch the ideas. Denote Z t P εj ξj2 it∆± −1 it n j=1 S(t) = e =F e F , A f (t, x) = S(t − τ )f (τ, x)dτ. 0

For the sake of brevity, we simplify the nonlinearity as F (u, u ¯, ∇u, ∇¯ u) = ~λ·∇(|u|2 u). According to the equivalent integral form of gDNLS, one needs to solve u(t) = S(t)u0 − iA ~λ · ∇(|u|2 u). On the basis of the smooth effect of the Schr¨odinger semi-group in 1D, we easily get its smooth effect estimates in all dimensions,

1/2

. ku0 k2 , (6.169)

Dxi S(t)u0 ∞ 2 2 1+n Lx L(x i

k∂xi A f kL∞ L2 xi

where kf kLpx1 Lp2 i

(xj )j6=i

p Lt 2

(xj )j6=i

j )j6=i

Lt (R

. kf kL1x

L2t (R1+n )

= kf kLpx2 ,...,x 1

)

L2(x

i

j )j6=i

L2t ,

p L 2 (R×Rn−1 ) i−1 ,xi+1 ,...,xn t

(6.170)

p

Lx1i (R)

. (6.171)

(6.169) and (6.170) are scaling-invariant and so, they are optimal estimates. 2 2 In view of (6.170), one should choose L∞ xi L(xj )j6=i Lt as a framework to handle the partial derivative ∂xi in the nonlinearity. According to the integral equation, we have 2 k∂x1 ukL∞ x L(x 1

L2 j )j6=1 t

. kDx1/2 u0 k2 + 1

n X i=1

k∂xi (|u|2 u)kL1x

1

L2(x

j )j6=1

L2t .

(6.172)

So, one needs to make the following two kinds of nonlinear estimates, I = k∂x1 (|u|2 u)kL1x

1

L2(x

j )j6=1

L2t (R1+n ) ,

II = k∂x2 (|u|2 u)kL1x

1

L2(x

j )j6=1

L2t .

Let us consider the estimates of I. Using H¨older’s inequality, we have 2 I 6 k∂x1 ukL∞ x L(x 1

j )j6=1

Hence, we need to estimate kukL2x kukL2x

1

L∞ (x

j )j6=1

L∞ t

6 kS(t)u0 kL2x

1

1

2 L2t kukL2x L∞ 1 (x

L∞ (x

j )j6=1

j )j6=1

L∞ (x

j )j6=1

. L∞ t

L∞ t

. L∞ t

(6.173)

By the integral equation,

+ k∇A (|u|2 u)kL2x

1

L∞ (x

j )j6=1

. L∞ t

Unfortunately, we can not get a straightforward global estimates. So, we consider the frequency-localized version of kS(t)u0 kL2x L∞ : L∞ t (x ) 1

kk S(t)u0 kL2x

1

L∞ (x

j )j6=1

L∞ t

1/2

. hk1 i

j j6=1

kk u0 k2 .

(6.174)

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Since (6.174) is only a frequency-local version, one needs to localize (6.172) and introduce the following X 2 kuk`1,s (L∞ L2 hk1 is kk ukL∞ (6.175) L2t , L2 ) := x L(x ) x1

hi

kuk`1 (L2x

1

t

(xj )j6=1

L∞ (x

j )j6=1

1

k∈Zn , |k1 |>4

X

:=

L∞ t )

k∈Zn

kk ukL2x

1

L∞ (x

j )j6=1

j j6=1

. L∞ t

(6.176)

Noticing that the smooth effect estimate in the lower frequency part is worse than that of the Strichartz estimate, we have thrown away the lower frequency on the ξ1 orientation in (6.175). If we do not consider the sharp estimate, (6.174) implies that X ∞ ) . kuk hk1 i1/2 kk ∇(|u|2 u)kL1t L2x . kuk`1 (L2x L∞ 1/2 + L M t (x )

1

2,1

j j6=1

k∈Zn

Making the nonlinear estimate, we have kuk`1 (L2x

1

L∞ (x

j )j6=1

L∞ t )

. kukM 1/2 + 2,1

X

k∈Zn

3/2

hki

kk ukL3t L6x

!3

,

which means that we need the following norm X hki3/2 kk ukL3t L6x . kuk`1,3/2 (L3 L6 ) := t

x

(6.177)

(6.178)

k∈Zn

The local version of (6.172) is k∂x1 uk`1,2

hi

2 (L∞ x1 L(x

. ku0 kM 5/2 + 2,1

j )j6=1

n X i=1

L2t )

hk1 i2 kk ∂xi (|u|2 u)kL1x

1

L2(x

j )j6=1

L2t .

(6.179)

After making nonlinear estimates, we find that the right hand side of (6.179) can be estimated by (6.175), (6.176) and (6.178). The estimate of II is more complicated and one needs to consider the interaction between the partial ∞ 2 2 derivative ∂x2 and the space `1,2 hi (Lx1 L(xj )j6=1 Lt ), see below for details. We now state the global wellposedness and scattering results for gDNLS. Let us recall the anisotropic Lebesgue space Lpx1i Lp(x2j )j6=i Lpt 2 is defined by (6.171). For k = (k1 , ..., kn ), we write kukXαs =

n X

X

hki is−1/2 ∂xα` k u L∞ L2 xi

i, `=1 k∈Zn , |ki |>4

+

n X

X

∂xα k u m ∞ ` L L

i, `=1 k∈Zn

xi

(xj )j6=i

L∞ t

,

(xj )j6=i

L2t

(6.180)

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kukSαs = kukX s =

n X X

`=1 k∈Zn

X

α=0,1

191

hkis−1 ∂xα` k u L∞ L2 T L3 L6 , t

kukXαs ,

kukS s =

x

X

α=0,1

t

(6.181)

x

kukSαs .

(6.182)

5/2

Theorem 6.8. Let n > 3, m = 2, u0 ∈ M2,1 and there exists a suitably small δ > 0 such that ku0 kM 5/2 6 δ. Then (6.166) has a unique solution 2,1

5/2

u ∈ C(R, M2,1 ) ∩ X 5/2 ∩ S 5/2 , kukX 5/2 ∩S 5/2 . δ. Moreover, the scattering 5/2

operator S of (6.166) carries a whole zero neighborhood in C(R, M2,1 ) into 5/2

C(R, M2,1 ). 6.5.1

Global linear estimates

Proposition 6.14 (Smooth effect). For any i = 1, ..., n, we have the following estimates

1/2

. ku0 k2 , (6.183)

Dxi S(t)u0 ∞ 2 2 Lx L(x i

k∂xi A f kL∞ 2 x L i

(xj )j6=i

j )j6=i

L2t

Lt

. kf kL1x

k∂xi A f kL∞ L2x . kDx1/2 f kL1x i t

i

i

L2(x

L2(x

j )j6=i

j )j6=i

L2t ,

L2t .

(6.184) (6.185)

Proof. By standard dual estimates, (6.183) implies (6.185). So, it suffices ¯ = (x2 , ..., xn ). By Plancherel’s to show the first two inequalities. Denote x identity and Minkowski’s inequality,

−1 itε1 ξ12

kS(t)u0 kL∞ 6 e F (F u ) (6.186)

F 2 2 x1 x ¯ 0 2 ∞ 2. ξ1 ¯ Lt x Lx Lξ¯Lx1 Lt

1

Recall the smooth effect of S(t) in one spatial dimension,

−1 itξ2

Fξ e Fx u0 ∞ 2 1+1 . kDx−1/2 u0 kL2 (R) . Lx Lt (R

)

(6.187)

So, (6.186), (6.187) together with Plancherel’s equality yield (6.183). Denote ξ1 −1 Ft,x f. (6.188) u = cFt,x 2 |ξ|± − τ ¯ 2 . By Plancherel’s We can assume that |ξ|2 = ξ 2 +ε2 ξ 2 +...+εn ξ n := ξ 2 +|ξ| identity,

±

1

2

n

1

±

−1

ξ1

kukL∞ L2x¯ L2 6 F F f . t,x ξ

2 2 1 ¯ x1 t ξ1 + |ξ|± − τ L2 L∞ L2 ¯ ξ

x1

τ

(6.189)

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¯ 2 , the right hand side of Making the change of the variable τ → µ + |ξ| ± (6.189) becomes

−1 ξ1 ¯2 −it|ξ|

F ±F F (e f ) (6.190) x2 ,...,xn

2 ∞ 2.

ξ1 ξ 2 − µ t,x1 1

Lξ¯Lx1 Lµ

Recalling the smooth effect in

one spatial dimension

−1 ξ

F

τ,ξ ξ 2 − τ Ft,x f ∞ 2 1+1 . kf kL1xL2t (R1+1 ) , Lx Lt (R

(6.191)

)

in view of (6.189), (6.190) and (6.191) we get that

¯2

−it|ξ|

±F kukL∞ . f

e 2 2 x ,...,x 2 n ¯ Lt x Lx

L2ξ¯L1x1 L2t

1

.

(6.192)

Using Minkowski’s inequality and Plancherel’s identity, we immediately have (6.193) kukL∞ 2 L2 . kf kL1 L2 L2 . x ¯ t ¯ t x x1 L x R∞ 1 Noticing that ∂x1 A f = u − ∂x1 S(t) −∞ S(s)sgn(s)f (s)ds, we easily see that in (6.193), substituting u by ∂x1 A f , the result also holds. 6.5.2

Frequency-localized linear estimates

We consider the frequency-localized versions of the smooth effect estimate, the maximal function estimate and their relations to the Strichartz estimate for the Schr¨ odinger semi-group. Let {ηk }k∈Z ∈ Υ1 and σk (ξ) := ηk1 (ξ1 )...ηkn (ξn ).

(6.194)

We have {σk }k∈Zn ∈ Υn . In the sequel we will always use the expression of σk in (6.194). Recall that k = F −1 σk F . For convenience, we denote P ek = |`|∞ 61 k+` . Lemma 6.9. Let Dxσi = (−∂x2i )σ/2 . For any σ ∈ R, k = (k1 , ..., kn ) ∈ Zn with |ki | > 4, we have kk Dxσi ukLpx1 Lpx2 ,...,xn Lpt 2 . hki iσ kk ukLpx1 Lpx2 ,...,xn Lpt 2 . 1

2

1

2

If σ ∈ N, replacing Dxσi by ∂xσi , the above estimate holds for all k ∈ Zn . Using (6.194), we have 1 Z X σ σ Fξ−1 (η (ξ )|ξ | ) (yi )(k u)(xi − yi )dyi . k Dxi u = k +` i i i i

Proof.

`=−1

R

By Young’s inequality and

kFξ−1 (ηki +` (ξi )|ξi |σ )kL1 (R) . hki iσ , i

we immediately have the result, as desired.

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Proposition 6.15 (Maximal function estimate). Let 4/n < q 6 ∞ and q > 2. Then we have kk S(t)u0 kLqxi L∞ (x

j )j6=i

. hki i1/q kk u0 kL2 (Rn ) .

L∞ t

(6.195)

Proof. For convenience, we denote x ¯ = (x2 , ..., xn ). By duality, it suffices to show that 2 ¯ q/2 ∞ n . hk1 i2/q . kF −1 eit|ξ|± ηk1 (ξ1 )η¯ (ξ)k k

Lx1 Lx¯,t (R )

k S(t) satisfies the following decay ¯2 ¯ L∞ (Rn−1 ) . (1 + |t|)−(n−1)/2 , kFξ¯−1 eit|ξ|± ηk¯ (ξ)k x ¯ 2

kFξ−1 eitξ1 ηk1 (ξ1 )kL∞ . (1 + |t|)−1/2 . x (R) 1 1

On the other hand, integrating by parts, we can get that for |x1 | > 4|t|hk1 i, 2

So, for any |x1 | > 1,

|Fξ−1 eitξ1 ηk1 (ξ1 )| . |x1 |−2 . 1

2

¯ . (1 + |x1 |)−2 + hk1 in/2 (hk1 i + |x1 |)−n/2 . |F −1 eit|ξ|± ηk1 (ξ1 )ηk¯ (ξ)|

This implies that

2 ¯ q/2 ∞ n kF −1 eit|ξ|± ηk1 (ξ1 )ηk¯ (ξ)k L L (R ) x1

n/2

. 1 + hk1 i 2/q

. hk1 i

x ¯,t

−n/2

k(hk1 i + |x1 |)

kLq/2 (R) x1

.

It follows that (6.195) holds.

By Proposition 6.14, we have Proposition 6.16 (Frequency-localized smooth effect). For k = (k1 , ..., kn ) ∈ Zn , we have kk A ∂xi f kL∞ L2 xi

(xj )j6=i

L2t

. kk f kL1x

kk A ∂xi f kL∞ L2 . hki i1/2 kk f kL1x t

x

i

i

L2(x

L2(x

L2t ,

(6.196)

L2t .

(6.197)

j )j6=i

j )j6=i

any

Proof. By Proposition 6.14, we have (6.196). In view of Proposition 6.14 and Lemma 6.9, we obtain that (6.197) holds for all |ki | > 3. If |ki | 6 2, from Proposition 6.14 it follows that

kk A ∂xi f kL∞ L2 . Dx−1/2 A ∂ f

∞ 2 k x i i t x Lt Lx

. kk f kL1x

which implies the result, as desired.

L2 L2 i (xj )j6=i t

,

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Proposition 6.17. (Relations of Strichartz, smooth effect and maximal function estimates). Let 2 6 r < ∞, 2/γ(r) = n(1/2 − 1/r) and γ > γ(r) ∨ 2. We have kk S(t)u0 kLγt Lrx . kk u0 kL2 (Rn ) ,

(6.198)

kk A f kL∞ L2 ∩ Lγ Lr . kk f kLγ 0 Lr0 , t

x

t

x

t

kk A ∂xi f kLγ Lrx . hki i1/2 kk f kL1x t

(6.199)

x

i

L2(x

j )j6=i

L2t ,

(6.200)

1/2 kk A ∂xi f kL∞ L2 kk f kLγ 0 Lr0 , 2 . hki i xi (xj )j6=i Lt x t

α+1/2

k A ∂ α f 2 ∞ . hki i kk f kL1t L2x . xi L L L∞ xi

(xj )j6=i

t

(6.201) (6.202)

Proof. (6.198) and (6.199) are corollaries of the Strichartz estimate. By (6.195) we can get (6.202). For convenience, we write Z Z k S(t − τ )f (τ )dτ, ψ(t) dt . (6.203) Lk (f, ψ) := R

R

Now we prove (6.200). Applying the Strichartz inequality, Lemma 6.9 and Proposition 6.16, we have

˜ γ 0 r0 Lk (∂x1 f, ψ) . hki i1/2 kk f kL1 L2 L2 k ψ x1

1/2

. hki i

kk f kL1x

1

x2 ,...,xn

Lt Lx

t

L2x2 ,...,xn L2t

kψkLγ 0 Lr0 . t

x

(6.204)

By the duality, (6.210) and Christ-Kiselev’s lemma (see Appendix), we can get (6.200). Changing the role of f and ψ, we immediately get that (6.201) holds for r > 2. If r = 2, (6.201) is a straightforward consequence of the smooth effect estimate of S(t). Corollary 6.5. Let 2 6 q < ∞, q > 4/n and 4/n 6 p < ∞. We have the following results.

1/2

. kk u0 kL2 (Rn ) , (6.205)

Dx1 k S(t)u0 2 2 L∞ x1 Lx2 ,...,xn Lt

kk S(t)u0 kLqx

1

∞ L∞ x2 ,...,xn Lt

. hki i1/q kk u0 kL2 (Rn ) ,

(6.206)

kk S(t)u0 kL2+p ∩ L∞ L2 . kk u0 kL2 (Rn ) ,

(6.207)

kk A ∂x1 f kL∞ L2

(6.208)

t,x

x1

t

x

2 x2 ,...,xn Lt

. kk f kL1x 1/2

kk A f kL∞ L2 ∩ L2+p . hk1 i t

x

t,x

kk A ∂x1 f kL∞ L2 x1

kk A ∂x1 f kLqx

1

2 x2 ,...,xn Lt

∞ L∞ x2 ,...,xn Lt

1

L2x2 ,...,xn L2t ,

kk f kL1x

1

L2x2 ,...,xn L2t ,

. hk1 i1/2 kk f kL(2+p)/(1+p) , t,x

. hk1 i1+1/q kk f kL1t L2 ,

(6.209) (6.210) (6.211)

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195

kk A f kL∞ L2 ∩ L2+p . kk f kL(2+p)/(1+p) . t

x

t,x

(6.212)

t,x

3 6 Moreover, if L2+p x,t is replaced by Lt Lx in (6.207), (6.209) and (6.212), then corresponding results also hold.

By (6.208) in Corollary 6.5, the operator k ∂x1 A : L1x1 L2x2 ,...,xn L2t → and the partial derivative ∂x1 is successfully absorbed. How2 2 ever, A in the space L∞ x1 Lx2 ,...,xn Lt can not handle the partial derivative ∂x2 . So, one needs to use another way to deal with ∂x2 A when it appears 2 2 in the space L∞ x1 Lx2 ,...,xn Lt . 2 2 L∞ x1 Lx2 ,...,xn Lt

Proposition 6.18. Let i = 2, ..., n, 2 6 q 6 ∞, q > 4/n, 2 6 r < ∞, 2/γ(r) = n(1/2 − 1/r), γ > γ(r), γ > 2. Then we have kk ∂xi A f kL∞ L2 x1

2 x2 ,...,xn Lt

kk ∂xi A f kL∞ L2 x1

kk ∂xi A f kLqx Proof.

1

2 x2 ,...,xn Lt

∞ L∞ x2 ,...,xn Lt

. k∂xi ∂x−1 k f kL1x 1

1

L2x2 ,...,xn L2t ,

(6.213)

k f kLγ 0 Lrx0 , k∂xi Dx−1/2 1

(6.214)

. hki ihk1 i1/q kk f kL1t L2x .

(6.215)

.

(6.213) is a corollary of Proposition 6.14. We have Z Z L(∂x2 f, ψ) := S(t − τ )∂x2 f (τ )dτ, ψ(t) dt R

Z R

−1/2

6 S(−τ )∂x2 Dx1 f (τ )dτ

R

Z

1/2

× Dx1 S(−t)ψ(t)dt

R

L2 (Rn )

.

(6.216)

L2 (Rn )

In view of the Strichartz inequality and Proposition 6.14, f kLγ 0 Lr0 kψkL1x L(∂x2 f, ψ) . k∂x2 Dx−1/2 1 t

x

1

L2x2 ,...,xn L2t .

(6.217)

If r > 2, by the duality, Christ-Kiselev’s lemma and (6.217), we have (6.214). If r = 2, in view of the smooth effect estimate of S(t), we see that (6.214) holds. By Proposition 6.15, we have (6.215). Lemma 6.10. Let ψ : [0, ∞) → [0, 1] be a smooth cut-off function satisfying ψ(x) = 1 for |x| 6 1; and ψ(x) = 0 for |x| > 2. Let ψ1 (ξ) = ψ(ξ2 /2ξ1 ), ψ2 (ξ) = 1 − ψ(ξ2 /2ξ1 ), ξ ∈ Rn . For any σ > 0, we have

X

∂ A f hk1 iσ Fξ−1 ψ F

∞ 2 1 x ,x k x 1 2 2 1 ,ξ2 2 k∈Zn , |k1 |>4

.

Lx1 Lx2 ,...,xn Lt

X

hk1 iσ kk f kL1

k∈Zn , |k1 |>4

2 2 x1 Lx2 ,...,xn Lt

.

(6.218)

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For σ > 1,

X

hk1 iσ Fξ−1 ψ2 Fx1 ,x2 k ∂x2 A f 1 ,ξ2

2 2 L∞ x1 Lx2 ,...,xn Lt

k∈Zn , |k1 |>4

.

X

hk2 iσ kk f kL1x

k∈Zn , |k2 |>4

Proof.

1

L2x2 ,...,xn L2t

.

(6.219)

For simplicity, we denote x¯ = (x2 , ..., xn ) and

I = Fξ−1 ψ1 Fx1 ,x2 k ∂x2 A f ∞ 2 2 , 1 ,ξ2 Lx1 Lx¯ Lt

II = Fξ−1 ψ2 Fx1 ,x2 k ∂x2 A f 1 ,ξ2

2 L2 L∞ ¯ t x1 L x

.

Let ηk be as in Lemma (6.194). For k ∈ Zn , |k1 | > 4, using the almost orthogonality of k , we have

X Y

F −1 ψ ξ2 ξ2

I. η (ξ )F ∂ A f . ki +`i i x1 ,x2 k x1

ξ1 ,ξ2

2ξ1 ξ1 i=1,2

|`1 |,|`2 |61 ∞ 2 2 Lx1 Lx¯ Lt

(6.220)

Denote

(f ~12 g)(x) =

Z

R2

f (t, x1 − y1 , x2 − y2 , x3 , ..., xn )g(t, y1 , y2 )dy1 dy2 . (6.221)

For any Banach function space X defined in R1+n , we have kf ~12 gkX 6 kgkL1y

1 ,y2

(R2 )

sup kf (·, · − y1 , · − y2 , ·, ..., ·)kX .

(6.222)

y1 ,y2

Hence, in view of (6.220) and (6.222),

Y X

ξ ξ 2 2 −1

ηki +`i (ξi ) I.

Fξ1 ,ξ2 ψ 2ξ1 ξ1

i=1,2 |`1 |,|`2 |61

kk ∂x1 A f kL∞ 2 L2 . ¯ x Lx 1

t

L1 (R2 )

(6.223)

Using the multiplier estimates, for any |k1 | > 4, we have

Y

−1

ξ ξ 2 2

F

ψ η (ξ ) k +` i i i

ξ1 ,ξ2

2ξ1 ξ1 i=1,2

1 2 L (R )

 

X Y

ξ ξ 2 2 α

 . ηki +`i (ξi ) . 1.

D ψ 2ξ1 ξ1

i=1,2 |α|62

L2 (R2 )

(6.224)

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By Proposition 6.16, (6.223) and (6.224), one has that I . kk f kL1

2 L2 ¯ t x1 L x

, |k1 | > 4.

(6.225)

Now we consider the estimate of II. Applying Proposition 6.18, we have

(ξ /ξ )ψ F f II . Fξ−1

1 2 2 2 1 2 x ,x k 1 2 1 ,ξ2 Lx1 Lx¯ Lt

Y X

ξ2 ξ2 −1

ηki +`i (ξi ) .

Fξ1 ,ξ2 1 − ψ 2ξ1 ξ1 i=1,2

|`1 |,|`2 |61

L1 (R2 )

× kk f kL1

2 L2 ¯ t x1 L x

.

(6.226)

Notice that suppψ2 ⊂ {ξ : |ξ2 | > 2|ξ1 |}. If |k1 | > 4, then |k2 | > 6 and in the summation of the left hand side of (6.219), one has that |k2 | > |k1 |. So, X X hk1 iσ II 6 hk2 iσ−1 hk1 iII. k∈Zn , |k1 |>4

k∈Zn , |k1 |>4

We have

Y

−1 ξ2 ξ2

F 1 − ψ η (ξ ) ki +`i i

ξ1 ,ξ2 2ξ1 ξ1 i=1,2

1 2

L (R )

 

Y X

ξ ξ 2 2 α  −1

 ηki +`i (ξi ) .

D Fξ1 ,ξ2 1 − ψ 2ξ1 ξ1 i=1,2

|α|62

L2 (R2 )

−1

. hk2 ihk1 i

.

(6.227)

Combining (6.226) and (6.227), we can get the estimate of II. 6.5.3

Proof of global wellposedness for small rough data

Denote (i)

%1 (u) =

X

k∈Zn , (i) %2 (u)

=

X

k∈Zn (i)

%3 (u) =

X

k∈Zn

Let X :=

  

2 hki i2 kk ukL∞ x L(x i

|ki |>4

∞ kk ukLm x L(x i

j )j6=i

j )j6=i

L2t ,

, L∞ t

hki i3/2 kk ukL3t L6x ∩ L∞ 2. t Lx

u ∈ S 0 : kukX :=

3 X X n X

`=1 α=0,1 i,j=1

(i)

%` (∂xαj u) 6 δ

  

.

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Considering the mapping T : u(t) → S(t)u0 − iA F (u, u ¯, ∇u, ∇¯ u), we show that T : X → X is a contraction mapping. Since kukX = k¯ ukX , we can assume that X F (u, u ¯, ∇u, ∇¯ u) = F (u, ∇u) := cκν uκ (∇u)ν , m+16κ+|ν|<∞

where (∇u)ν = uνx11 ...uνxnn . For convenience, we write v1 = ... = vκ = u, vκ+1 = ... = vκ+ν1 = ux1 , ..., vκ+|ν|−νn +1 = ... = vκ+|ν| = uxn . By (6.183), for α = 0, 1, we have X (i) %1 (∂xαj S(t)u0 ) . hki i1/2 hkj i2 kk u0 kL2 (Rn ) 6 ku0 kM 5/2 . 2,1

k∈Zn , |ki |>4

In view of (6.206) and (6.207), for α = 0, 1, we have (i)

(i)

%2 (∂xαj S(t)u0 ) + %3 (∂xαj S(t)u0 ) . ku0 kM 5/2 . 2,1

So, kS(t)u0 kX . ku0 kM 5/2 . 2,1

(i) In order to estimate %1 (A ∂xαj (v1 ...vκ+|ν| )), i, j = 1, ..., n, it suffices to con(1) (1) sider the estimates of %1 (A ∂xα1 (v1 ...vκ+|ν| )) and %1 (A ∂xα2 (v1 ...vκ+|ν| )).

Applying the frequency-uniform decomposition, we have X k k(1) v1 ...k(κ+|ν|) vκ+|ν| k (v1 ...vκ+|ν| ) = (i)

S1

+

X (i) S2

k k(1) v1 ...k(κ+|ν|) vκ+|ν| ,

(6.228)

where (i)

(1)

(κ+|ν|)

S1 := {(k (1) , ..., k (κ+|ν|) ) : |ki | ∨ ... ∨ |ki (i) S2

:= {(k

(1)

, ..., k

(κ+|ν|)

):

(1) |ki |

∨ ... ∨

| > 4},

(κ+|ν|) |ki |

We will frequently use the almost orthogonality of k , k k(1) v1 ...k(κ+|ν|) vκ+|ν| = 0,

6 4}.

if |k − k (1) − ... − k (κ+|ν|) |∞ > κ + |ν| + 1.

(6.229)

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By (6.196) and (6.201), (1)

%1 (A ∂xα1 (v1 ...vκ+|ν| )) X X . hk1 i2 kk k(1) v1 ...k(κ+|ν|) vκ+|ν| kL1x k∈Zn , |k1 |>4

+

X

1

L2x¯ L2t

(1)

S1

hk1 i5/2

k∈Zn , |k1 |>4

X (1)

S2

kk k(1) v1 ...k(κ+|ν|) vκ+|ν| k

κ+|ν|+1 κ+|ν|

Lt,x

=: I + II.

(6.230)

Applying the almost orthogonality of k in (6.229), we have X (1) 2 L2 I . C κ+|ν| hk1 i2 kk(1) v1 kL∞ ¯ t x Lx 1

(1)

k(1) ∈Zn , |k1 |>2

×

κ+|ν|

X

k(2) ,...,k(κ+|ν|) ∈Zn

Y

i=2

kk(i) vi kLκ+|ν|−1 L∞ L∞ . x1

x ¯

(6.231)

t

By H¨ older’s inequality and Lemma 6.1, kk(i) vi kLκ+|ν|−1 L∞ L∞ x1

6 kk(i) vi k

x ¯ t m κ+|ν|−1 ∞ L∞ Lm ¯ x1 L x t

1−

m

1−

m

κ+|ν|−1 kk(i) vi kL∞ x,t

m

κ+|ν|−1 . . kk(i) vi kLκ+|ν|−1 m L∞ L∞ kk(i) vi k ∞ 2 L L x ¯ x t t

1

(6.232)

x

Hence, in view of vi = u or vi = uxj , it follows from (6.231) and (6.232) that I . (CkukX )κ+|ν| .

(6.233)

(1)

By the definition of S2 , one has that |k1 | 6 C(κ + |ν|) in the summation of II. Again, by H¨ older’s inequality and Lemma 6.1, κ+|ν|

kk(1) v1 ...k(κ+|ν|) vκ+|ν| k

κ+|ν|+1 κ+|ν| Lx,t

6

Y

i=1

kk(i) vi kLκ+|ν|+1 x,t

κ+|ν|

.

Y

i=1

kk(i) vi kL2+m x,t

T

2, L∞ t Lx

(6.234)

which implies that II . (CkukX )κ+|ν| .

(6.235)

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Frequency-uniform decomposition techniques (1)

Now we estimate %1 (A ∂xα2 (v1 ...vκ+|ν| )). The case α = 0 has been discussed in the above, it suffices to consider the case α = 1. Let ψi (i = 1, 2) be as in Lemma 6.10 and Pi = F −1 ψi F . We have (1)

%1 (A ∂x2 (v1 ...vκ+|ν| )) X 2 L2 6 hk1 i2 kP1 k (A ∂x2 (v1 ...vκ+|ν| ))kL∞ ¯ t x Lx 1

k∈Zn , |k1 |>4

X

+

k∈Zn ,

2 L2 hk1 i2 kP2 k (A ∂x2 (v1 ...vκ+|ν| ))kL∞ ¯ t x Lx 1

|k1 |>4

=: III + IV.

(6.236)

Using the decomposition in (6.228), we have X X 2 L2 III 6 hk1 i2 kP1 k (A ∂x2 (k(1) v1 ...k(κ+|ν|) vκ+|ν| ))kL∞ ¯ t x Lx 1

k∈Zn , |k1 |>4

+

X

(1)

S1

hk1 i2

k∈Zn , |k1 |>4

X

2 L2 kP1 k (A ∂x2 (k(1) v1 ...k(κ+|ν|) vκ+|ν| ))kL∞ ¯ t x Lx 1

(1)

S2

=: III1 + III2 . By Lemma 6.10, X III1 .

(6.237)

X

n (1) S1 k∈Z ,

hk1 i2 kk (k(1) v1 ...k(κ+|ν|) vκ+|ν| )kL1x

1

L2x¯ L2t .

|k1 |>4

(6.238) (1)

(1)

(κ+|ν|)

By the symmetry, we can assume that |k1 | = max(|k1 |, ..., |k1 (1) S1 . So,

|) in

III1 . C κ+|ν| (1) S1 ,

X

κ+|ν| (1)

2 L2 hk1 i2 kk(1) v1 kL∞ ¯ t x Lx 1

(1) |k1 |>4

Y

i=2

kk(i) vi kLκ+|ν|−1 L∞ L∞ x1

x ¯

t

κ+|ν| (1)

. C κ+|ν| %1 (v1 )

Y

(1)

(1)

(%2 (vi ) + %3 (vi )) . (CkukX )κ+|ν| .

(6.239)

i=2

Applying (6.214) and noticing that |k1 | 6 C in III2 , we obtain that III2 .

X

hk1 i5/2

k∈Zn , |k1 |>4, |k2 |.|k1 |

X (1)

S2

kk (k(1) v1 ...k(κ+|ν|) vκ+|ν| )kL(2+m)/(1+m) t,x

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201

κ+|ν|

. C κ+|ν|

Y

(1)

%3 (vi ) 6 (CkukX )κ+|ν| .

(6.240)

i=1

We have shown that III . (CkukX )κ+|ν| .

(6.241)

Now we estimate IV . Using (6.228), we have X X 2 L2 hk1 i2 kP2 k (A ∂x2 (k(1) v1 ...k(κ+|ν|) vκ+|ν| ))kL∞ IV 6 ¯ t x Lx 1

k∈Zn , |k1 |>4

+

X

(2)

S1

hk1 i2

k∈Zn , |k1 |>4

X (2)

2 L2 kP2 k (A ∂x2 (k(1) v1 ...k(κ+|ν|) vκ+|ν| ))kL∞ ¯ t x Lx 1

S2

=: IV1 + IV2 .

(6.242)

By Lemma 6.10, X X IV1 . (2) S1

hk2 i2 kk (k(1) v1 ...k(κ+|ν|) vκ+|ν| )kL1x

k∈Zn , |k2 |>4

1

L2x¯ L2t .

(6.243)

(κ+|ν|)

We can choose some k (i) , say k (κ+|ν|) such that |k2 | does not attain (i) or does not uniquely attain max16i6κ+|ν| |k2 |, then we can take another (1)

(i)

k (i) , say k (1) such that k2 = max16i6κ+|ν| |k2 |. By H¨older’s inequality, kk(1) v1 ...k(κ+|ν|) vκ+|ν| kL1x

1

L2x¯ L2t

6 kk(1) v1 ...k(κ+|ν|−1) vκ+|ν|−1 kL2x,t kk(κ+|ν|) vκ+|ν| kL2x

1

∞ L∞ x ¯ Lt

κ+|ν|

Y

2 L2 6 kk(1) v1 kL∞ ¯ t x Lx 1

i=2

2 2 kk(i) vi k(L∞ t Lx )∩(Lx

1

. L∞ x ¯,t )

(6.244)

Combining (6.243) with (6.244), we get IV1 . (CkukX )κ+|ν| .

(6.245)

In order to estimate IV2 , we apply (6.214) to obtain that X X hk2 i5/2 kP2 k (k(1) v1 ...k(κ+|ν|) vκ+|ν| )kL(2+m)/(1+m) IV2 . k∈Zn , |k1 |>4

. C κ+|ν|

X (2) S2

t,x

(2)

S2

κ+|ν|

kk(1) v1 ...k(κ+|ν|) vκ+|ν| kL(2+m)/(1+m) . kukX t,x

.

(6.246) Hence, in view of (6.245) and (6.246), we have IV . (CkukX )κ+|ν| .

(6.247)

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Collecting (6.233), (6.235), (6.241) and (6.247), we have shown that n X X

(i)

%1 (A ∂xαj (uκ (∇u)ν )) . (CkukX )κ+|ν| .

(6.248)

α=0,1 i,j=1 (i)

Now we estimate %3 (A (uκ (∇u)ν )). By (6.212), n X

X

(i)

%3 (A (uκ (∇u)ν )) .

k∈Zn

i=1

hki3/2 kk (uκ (∇u)ν )k

2+m 1+m Lt,x

.

(6.249)

Using Lemma 6.8, one can control the right hand side of (6.249), X hki3/2 kk (v1 ...vκ+|ν| )k 2+m 1+m Lt,x

k∈Zn

.C

κ+|ν|

. C κ+|ν|

. C κ+|ν|

m+1 Y

X

i=1

k∈Zn

m+1 Y

X

i=1

k∈Zn

κ+|ν|

n X

Y

i=1

3/2

hki

kk vi kL2+m t,x

hki3/2 kk vi kL2+m t,x

!

(1)

%3 (vi )

i=1

!

!

κ+|ν|

Y

X

i=m+2

k∈Zn

κ+|ν|

X

Y

k∈Zn

i=m+2

3/2

hki

kk vi kL∞ t,x

!

2 hki3/2 kk vi kL∞ t Lx

6 (CkukX )κ+|ν| .

!

(6.250)

(1)

Now we estimate %2 (A ∂xα1 (uκ (∇u)ν )). X (1) %2 (A ∂xα1 (v1 ...vκ+|ν| )) . hki3/2 kk v1 ...vκ+|ν| kL1t L2x .

(6.251)

k∈Zn

Analogous to (6.250), using Lemma 6.8, we can control (6.251), κ+|ν| (1) %2 (A

∂xα1 (v1 ...vκ+|ν| ))

.C

κ+|ν|

Y

i=1

X

k∈Zn

3/2

hki

3 6 kk vi kL∞ 2 t Lx ∩Lt Lx

. (CkukX )κ+|ν| .

(6.252)

Now we estimate n X (i) ρ3 (A ∂x1 (v1 ...vκ+|ν| )) i=1

.

X

|k|64



+

kk (v1 ...vκ+|ν| )kL1t L2x X

k∈Zn , |k1 |=kmax >4

+... +

!

X



 hki3/2

k∈Zn , |kn |=kmax >4

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203

3 6 × kk A ∂x1 v1 ...vκ+|ν| kL∞ 2 t Lx ∩ Lt Lx

=: Υ0 (u) + Υ1 (u) + ... + Υn (u).

(6.253)

Υ0 (u) has been estimated as above. It suffices to estimate Υi (u), i = 1, 2. Using (6.228) and Corollary 6.5, X X kk k(1) v1 ...k(κ+|ν|) vκ+|ν| kL1x L2x¯ L2t Υ1 (u) . hk1 i2 1

k∈Zn , |k1 |>4

X

+

(1)

S1

hk1 i5/2

k∈Zn , |k1 |>4

X (1)

S2

kk k(1) v1 ...k(κ+|ν|) vκ+|ν| k

κ+|ν|+1 κ+|ν|

,

Lt,x

(6.254)

which reduces to (6.230). Hence, Υ1 (u) . (CkukX )κ+|ν| .

(6.255)

We now estimate Υ2 (u). Taking notice of |k1 | 6 |k2 | in the summation of Υ2 (u) and using (6.209) and (6.212), we have X X Υ2 (u) . hk2 i2 kk k(1) v1 ...k(κ+|ν|) vκ+|ν| kL1x L2x¯ L2t 1

k∈Zn , |k2 |>4

+

X

(2)

S1

hk2 i5/2

k∈Zn , |k2 |>4

X (2)

S2

kk k(1) v1 ...k(κ+|ν|) vκ+|ν| k

κ+|ν|+1 κ+|ν|

.

Lt,x

(6.256)

The first term in the right hand side has been estimated in (6.244). The second term is the same as in II of (6.230). Summarizing the above estimates, we get that10 X kT ukX 6 Cku0 kM 3/2 + `2n+2 C ` kuk`X . (6.257) m+16`<∞

We can use the contraction mapping argument to finish the proof and omit the details. Remark 6.2. (1) There are some recent works which have been devoted to the study of the derivative NLKG [54; 55; 76; 144; 185; 200]: utt + u − ∆u = F (u, ut , uxx , uxt ),

where the methods are quite different from those of derivative NLS. We do not know whether the frequncy-uniform decomposition techniques are applicable to the derivative NLKG. 10 Notice

that |cβ | 6 C |β| .

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(2) We now state some results on the Zakharov system i∂t u + ∆u = nu, ∂t2 n − ∆n = ∆|u|2 . In 2D, Bourgain and Colliander [20] used the Fourier restriction norm to prove the local well-posedness and global well-posedness for initial data in the spaces H 1 × L2 × H −1 . Fang, Pecher and Zhong [71] applied the I-method to obtain the global well-posedness in H s × L2 × H −1 with 3/4 < s < 1. Recently, Bejenaru, Herr, Holmer, and Tataru [8] showed the local well-posedness for the Schr¨ odinger data in L2 and the wave data in H −1/2 × H −3/2 . In higher spatial dimensions, the local well posed result can be found in Ginibre, Tsutsumi and Velo [81].

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Chapter 7

Conservations, Morawetz’ estimates of nonlinear Schr¨ odinger equations This result is too beautiful to be false; it is more important to have beauty in one’s equations than to have them fit experiment. —— Paul Dirac

On mentioning evolution partial differential equations, we will naturally think whether they satisfy some physical conservation laws, such as the mass conservation, the energy conservation and the momentum conservation, and so on. For example, as indicated in (4.2), we see that NLS (4.1) satisfies the conservations of mass and energy. In this chapter, we will derive some conservation laws by using N¨other’s theorem and Morawetz’ inequalities by the virial identity. It is known that Morawetz’ estimates are fundamental tools in the study of the scattering theory for NLS, see for instance, [27; 40; 43; 218]. 7.1

N¨ other’s theorem

The energy conservation law means that it does not vary with time, in other word, it has a certain symmetry with respect to the time. This fundamental connection between conservation laws and symmetries was first discovered in 1915 (published in 1918) by Emmy N¨other1 and was other’s theorem”. In short, N¨other’s theorem states informally called the “N¨ that any differentiable symmetry of the action of a physical system has a corresponding conservation law. N¨other’s theorem is important, both because of the insight it gives into conservation laws, and also as a practical calculational tool. It allows us to determine the conserved quantities from 1 Amalie Emmy N¨ other (1882-1935) was a German-born mathematician known for her ground-breaking contributions to abstract algebra and theoretical physics. She revolutionized the theories of rings, fields, and algebras. In physics, N¨ other’s theorem explains the fundamental connection between symmetry and conservation laws.

205

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the observed symmetries of a physical system. The latter turns out to be easier to find than the former. Application of N¨ other’s theorem allows physicists to gain powerful insights into any general theories in physics, by just analyzing the various transformations that would make the form of the laws involved invariant. For example, the invariance of physical systems with respect to the spatial translation (in other words, the laws of physics do not vary with locations in space) gives the conservation law of the linear momentum; the invariance with respect to the rotation gives the conservation law of the angular momentum; the invariance with respect to the time translation gives the well-known conservation law of energy. But how to derive the associated conserved quantities of its if we know that an equation has a symmetry group? Let us begin by recalling some concepts of the calculus of variations. Suppose the interval I ⊂ R, the generalized coordinate q is dependent on the time t. Assume that L : Rnq × Rnq˙ × I 7→ R is smooth and convex with respect to the first slot q˙ (The dot indicates the derivative with respect to the time.), then we call the function L to be a Lagrangian 2 of the system. For some q satisfying the related boundary conditions, we define the action functional Z t1 ˙ I[q] = L(q(t), q(t), t)dt. t0

Hamilton’s principle states that the true evolution of a system described by generalized coordinates q between two specified states q(t0 ) and q(t1 ) at two specified times t0 and t1 is an extremum (i.e., a stationary point, a minimum, maximum or saddle point) of the action functional I[q], in other word, the first variation of the action functional I[q] is zero. Let q(t) represent the true evolution of the system between two specified states q(t0 ) and q(t1 ) at two specified times t0 and t1 . And let ε(t) be a small perturbation that is zero at the endpoints of the trajectory def

ε(t0 ) = ε(t1 ) == 0. (7.1) To the first order in the perturbation ε(t), the change in the action functional δIZ would be Z t1 t1 ∂L ∂L ˙ − L(q, q)) ˙ dt = + ε˙ · dt, δI = (L(q + ε, q˙ + ε) ε· ∂q ∂ q˙ t0 t0

2 It is named after Joseph Louis Lagrange (1736-1813), who was an Italian-born mathematician and astronomer, making significant contributions to all fields of analysis, to number theory, and to classical and celestial mechanics. He is one of the founders of the calculus of variations. Roughly speaking, the Lagrangian of a dynamical system is a function that summarizes the dynamics of the system. In classical mechanics, the Lagrangian is defined as the kinetic energy of the system minus its potential energy.

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where we have expanded the Lagrangian L to the first order in the perturbation ε(t). Applying the integration by parts to the last term results in t Z t1 ∂L d ∂L ∂L 1 + ε· −ε· dt. δI = ε · ∂ q˙ t0 ∂q dt ∂ q˙ t0

By the boundary condition (7.1), the first term vanishes, and then Z t1 ∂L d ∂L δI = ε· − dt = 0. ∂q dt ∂ q˙ t0 Noticing that we do not make any assumptions to the generalized coordinates q except that all of the generalized coordinates are independent of each other. Thus, we can apply the fundamental theorem of variational calculus to obtain an Euler-Lagrange equation d ∂L ∂L − = 0, (7.2) dt ∂ q˙ ∂q which is a system involving n second-order equations with respect to q(t). The Legendre transformation 3 of L is ˙ H(q(t), p(t), t) = max[p(t) · q˙ − L(q(t), q(t), t)]. q˙

˙ = p(t) · q˙ − L(q(t), q(t), ˙ Let M (p, q) t). In order to reach the maximum ˙ we have to require that the first-order partial of M with respect to q, ˙ with respect to q˙ is zero and that its second partial derivative of M (p, q) derivative is less than zero (this can be guaranteed by the convexity of L, 2 ˙ ˙ i.e. ∂∂ q˙L2 > 0). Thus, if p(t) = ∂L ∂ q˙ (q(t), q(t), t), then M (p, q) reaches the maximum, i.e.   p(t) = ∂L (q(t), q(t), ˙ t), ∂ q˙  H(q(t), p(t), t) = p(t)q˙ − L(q(t), q(t), ˙ t),

where H is called the Hamiltonian. The second equation yields ∂H ∂L ∂H ˙ =− , = q. ∂q ∂q ∂p Thus, the Euler-Lagrange equation (7.2) can be rewritten as  ∂H    p˙ = − ∂q ,  ∂H   q˙ = , ∂p 3 Legendre

(7.3)

transformation is an operation that transforms one real-valued function of a real variable into another. Specifically, the Legendre transformation of a function f is the function f ? defined by f ? (p) = maxx px − f (x) .

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which are called Hamilton’s equation. This is a system of 2n first-order equations in (p(t), q(t)). Denote u(t) = (p(t), q(t)) ∈ R2n , Hu = (Hp , Hq ) and define the matrix 0 −I J= , I 0 where I denotes the unit matrix, then Hamilton’s equation (7.3) can be written as4 2

u˙ = J Hu .

(7.4)

Notice that J = −I and Hamilton’s equations are defined on an even dimensional space. It suggests that there may be a nice connection with complex numbers. For (x, y) ∈ R2 , define z = x + iy, z¯ = x − iy, i2 = −1. Define ∂z = 21 (∂x − i∂y ) and ∂z¯ = 12 (∂x + i∂y ). It is clear that ∂z z = ∂z¯z¯ = 1 and ∂z z¯ = ∂z¯z = 0, that is, z¯ and z are independent of each other. Let z = q + ip, then (7.3) yields z˙ = q˙ + ip˙ = −i(Hq + iHp ) = −2iH¯z .

We give an example. Let I be a time interval, u : Rn × I 7→ C, u and its derivatives are smooth and vanish at the infinity. Assume that the Hamiltonian is defined by Z Z 1 1 |∇u|2 dx = ∇u∇¯ udx. H = H[u, u ¯] = 2 R 2 R The usual calculus of variations argument shows Z H[u, u ¯ + τ v¯] − H[u, u ¯] 1 1 lim = (−∆u)¯ v dx = h−∆u, vi. τ →0 τ 2 R 2

Therefore, Hu¯ [u, u ¯] = − 21 ∆u and the associated Hamilton’s equation is u˙ = −2iHu¯ = i∆u,

which is precisely the linear Schr¨odinger equation. Now, we consider the n-dimensional nonlinear Schr¨odinger equation iut + ∆u = F 0 (|u|2 )u.

(7.5)

Assume that u and its derivatives are smooth and vanish as |x| → +∞. The nonlinearity F 0 (·) is a smooth real-valued function of its argument and define Z λ F (λ) = F 0 (s)ds. 0

4 It

is convenient to write u and Hu as column vectors in the computation.

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Define the Lagrangian i (¯ uut − u¯ ut ) − [|∇u|2 + F (|u|2 )], 2 and the associated action functional Z t1 Z I[u] = L dxdt, ∀u ∈ A, L = L(u, u ¯, ut , u ¯t , ∇u, ∇¯ u) =

t0

(7.6)

(7.7)

Rn

where A is some appropriate class of admissible functions. The usual calculus of variations argument shows that if u is a smooth critical point of I(·), u satisfies the associated Euler-Lagrange equation n

X ∂ ∂L ∂ ∂L ∂L − − = 0. ∂u ∂t ∂(∂t u) j=1 ∂xj ∂(∂xj u)

(7.8)

Substituting (7.6) into the above equation and taking the complex conjugate, we can obtain the nonlinear Schr¨odinger equation (7.5). Now, we give another statement for N¨other’s theorem: Theorem 7.1 (N¨ other’s theorem). If the action functional is invariant under a family of transformations, then the solutions of the associated Euler-Lagrange equation satisfy a conservation law. We mainly consider the case of the one parameter family of transformations and apply N¨ other’s theorem to the nonlinear Schr¨odinger equation (7.5). For convenience, denote ξ = (t, x) = (ξ0 , ξ1 , · · · , ξn ), ∂0 = ∂t , ∂ = (∂t , ∇x ) = (∂0 , ∂1 , · · · , ∂n ), and u = (u1 , u2 ) = (u, u ¯), and denote the integral domain D := [t0 , t1 ] × Rn . A one parameter group of transformations T ε is defined by ˜ u, ε), ξ 7→ ξ(ξ,

˜ (ξ, u, ε), u 7→ u

(7.9)

˜ are differentiable with respect to ε, and it degenwhere we assume ξ˜ and u erates into an identical transformation for the case ε = 0. For infinitesimal ε, denote ξ˜ = ξ + δξ,

˜ = u + δu, u

(7.10)

where δξ and δu are functions of (ξ, u, ε), both δξ = O(ε) and δu = ∂(ξ˜0 ,··· ,ξ˜n ) O(ε) tend to zero as ε → 0, and Jacobi’s determinant ∂(ξ = 0 ,··· ,ξn ) Pn ∂(δξ)j ˜ the domain D ˜ (ξ), 1 + j=0 ∂ξj + o(ε). Applied T ε , u(ξ) changes into u ˜ and the action functional becomes into D, Z I[u] = L(u, ∂u)dξ (7.11) D

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turns into ˜ u] = I[˜ =

Z

˜ D

Z

D

˜ )dξ˜ = L(˜ u, ∂˜u ˜ )dξ + L(˜ u, ∂˜u

Z

Z

D

D



˜ ) 1 + L(˜ u, ∂˜u L(u, ∂u)

n X ∂(δξ)j j=0

n X ∂(δξ)j j=0

∂ξj

∂ξj



+ o(ε) dξ

dξ + o(ε),

(7.12)

˜ respectively. where ∂ and ∂˜ denote the differential with respect to ξ and ξ, The last two terms can be obtained by expanding with the help of (7.6) and subsuming the higher order terms of ε in o(ε) in the expansion. We only consider the transformations under which I is invariant. Denote ˜ u] − I[u] δI :=I[˜ Z Z n X ∂(δξ)j ˜ ) − L(u, ∂u) dξ + = L(˜ u, ∂˜u L(u, ∂u) dξ + o(ε), ∂ξj D D j=0

(7.13)

where the first integrand can be written as ˜ ) − L(u, ∂u) L(˜ u, ∂˜u   n 2 X X ∂L ∂L ˜ − uk (ξ)) + ˜ − ∂j uk (ξ)  . (7.14)  (˜ uk (ξ) ∂˜j u ˜k (ξ) = ∂uk ∂(∂ u ) j k j=0 k=1

Denote

˜ − uk (ξ) = δ˜ uk ≡ u ˜k (ξ) Since

n X

∂j uk (δξ)j + δuk (ξ).

n ˜ X ∂(δξ)l ˜ ˜ ∂ ξl = ˜ ∂˜l u˜k (ξ) (δjl + )∂l u ˜k (ξ) ∂ξj ∂ξj l=0 l=0 ! n X ∂(δξ)l ˜ ˜ ˜ = ∂j + ∂l u ˜k (ξ), ∂ξj

˜ = ∂j u˜k (ξ)

(7.15)

j=0

n X

(7.16)

l=0

we have

˜ − ∂j uk (ξ) =(∂˜j − ∂j )˜ ˜ + ∂j (˜ ˜ − uk (ξ)) ∂˜j u ˜k (ξ) uk (ξ) uk (ξ) =−

n X ∂(δξ)l l=0

∂ξj

˜ + ∂j ∂˜l u ˜k (ξ)

n X l=0

∂l uk (δξ)l + δuk

!

.

(7.17)

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211

Hence,

  n 2 X X ∂L  ˜ ) − L(u, ∂u) = ∂j uk (δξ)j + δuk  L(˜ u, ∂˜u ∂uk j=0 k=1

+

2 X n X

k=1 j=0

∂L ∂(∂j uk )

−

n X ∂(δξ)l

∂ξj

l=0

n X

˜ + ∂j ∂˜l u ˜k (ξ)

∂l uk (δξ)l + δuk

l=0

2

k=1

and ∂L ∂ ∂j δuk = ∂(∂j uk ) ∂ξj

n

X ∂L ∂L ∂j uk + ∂ 2 uk ∂uk ∂(∂l uk ) lj l=0

∂L δuk ∂(∂j uk )

∂ − ∂ξj

∂L ∂(∂j uk )

.

(7.18)

Noticed that ∂ ∂(δξ)j X (L(δξ)j ) = L + ∂ξj ∂ξj

!!

!

δuk ,

(δξ)j , (7.19)

(7.20)

it is easily to obtain ˜ ) − L(u, ∂u) L(˜ u, ∂˜u

n 2 n X X ∂(δξ)j X ∂L ∂ L (L(δξ)j ) − + δuk ∂ξj ∂ξj ∂uk j=0 j=0

=

k=1

2 X n X ∂L ∂ ∂L ∂ δuk − δuk + ∂ξj ∂(∂j uk ) ∂ξj ∂(∂j uk ) j=0 k=1

−

2 X n X k=1 j=0

n

∂L X ∂(δξ)l ˜ ˜ − ∂l uk ). (∂l u ˜k (ξ) ∂(∂j uk ) ∂ξj

(7.21)

l=0

Therefore, it yields   Z X 2 n X ∂L ∂ ∂L   δuk dξ δI = − ∂uk j=0 ∂ξj ∂(∂j uk ) D k=1 ! Z X n 2 X ∂ ∂L + L(δξ)j + δuk dξ + o(ε). ∂(∂j uk ) D j=0 ∂ξj

(7.22)

k=1

(7.8) implies n

X ∂ ∂L − ∂uk j=0 ∂ξj

∂L ∂(∂j uk )

= 0,

k = 1, 2.

(7.23)

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Since D is arbitrary, we have the following result in order to guarantee that I remains invariant under the infinitesimal transformation T ε . Theorem 7.2. Let ξ = (t, x1 , · · · , xn ), u = (u1 , u2 ). If the action functional (7.11) is invariant under the infinitesimal transformation T ε : ˜ u, ε), u 7→ u ˜ (ξ, u, ε), then the following conservation law holds: ξ 7→ ξ(ξ, ! n 2 X ∂ X ∂L L(δξ)j + δuk = 0, (7.24) ∂ξj ∂(∂j uk ) j=0 k=1

˜ − uk (ξ) − ∇ξ uk · δξ. where δuk (ξ) = u˜k (ξ)

Integrating with respect to the spatial variables yields the following result. Theorem 7.3. If the action functional (7.11) is invariant under the infinitesimal transformation t 7→ t˜ = t + δt(t, x, u),

x 7→ x ˜ = x + δx(t, x, u), u(t, x) 7→ u ˜(t˜, x ˜) = u(t, x) + δu(t, x),

(7.25) (7.26) (7.27)

then Z

∂L ∂L (ut δt + ∇u · δx − δu) + (¯ ut δt + ∇¯ u · δx − δ¯ u) − Lδt dx ∂ut ∂u ¯t Rn is a conserved quantity. In particular, in the case of nonlinear Schr¨ odinger i ∂L i ∂L ¯, ∂ u¯t = − 2 u, it implies that equations, since ∂ut = 2 u Z i i u ¯(ut δt + ∇u · δx − δu) − u(¯ ut δt + ∇¯ u · δx − δ¯ u) − Lδt dx = const. 2 Rn 2 (7.28) 7.2

Invariance and conservation law

In this subsection, we apply the formalism (7.28) to identify some invariant quantities of the action functional for nonlinear Schr¨odinger equations and thereby infer some conservation laws for nonlinear Schr¨odinger equations. (i) Invariance by the phase shift : u ˜ = eiε u. For the infinitesimal quantity ε, let δu = iεu, δt = 0, and δx = 0. From (7.28), we get the mass conservation law (or the charge conservation law, or the L2 -norm conservation law) Z N (t) := |u(t, x)|2 dx = const. (7.29) Rn

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(ii) Invariance by the time translation: t 7→ t+δt, δx = 0 and δu = δ¯ u= 0, where δt is an infinitesimal quantity independent of (t, x, u, u ¯). (7.28) and (7.6) yield the energy conservation law Z H(t) := |∇u(t, x)|2 + F (|u(t, x)|2 )dx = const. (7.30) Rn

(iii) Invariance by the spatial translation: x 7→ x + δx and δt = δu = δ¯ u = 0, where δx is an infinitesimal quantity independent of (t, x, u, u ¯). (7.28) implies the momentum conservation law Z P~ (t) := i (u(t, x)∇¯ u(t, x) − u¯(t, x)∇u(t, x))dx = constant vector. Rn

(7.31)

(iv) Invariance by the spatial rotation: δx = δθ(~a × x) and δt = δu = δ¯ u = 0, which mean the spatial variables rotate anticlockwise for a angle of δθ along the axis ~a where δθ is an infinitesimal quantity independent of (t, x, u, u ¯). From (7.28), we have the angular momentum conservation law Z i x × (¯ u∇u − u∇¯ u)dx = constant vector. (7.32) Rn

(v) Invariance by the Galilean transformation:   x 7→ x ˜ = x − ~ct,   t 7→ t˜ = t,   2˜ 1 1  u 7→ u ˜(t˜, x ˜) = e−i[ 2 ~c·˜x+ 4 |~c| t] u(t˜, x˜ + ~ct˜),

(7.33)

in other words, for infinitesimal velocity ~c, let

i δu = − ~c · xu(t, x). 2 From (7.28), it is easily to gain the renormalized centroid conservation law Z x|u(t, x)|2 dx − tP~ (t) = constant vector. (7.34) δt = 0,

δx = −~ct,

Rn

(vi) Invariance by the pseudo-conformal transformation:  x  x 7→ x ˜= ,   `(t)    Z t  1 t 7→ t˜ = dτ, 2 (τ )  ` 0     `t |x|2   u 7→ u ˜(t˜, x ˜) = `n/2 u(t, x) exp(−i ). ` 4

(7.35)

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Assume `tt = 0 and F 0 (|˜ u|2 ) = `2 F 0 (|u|2 ), then the equation is invariant under this pseudo-conformal transformation. Let ε ∈ (0, 1/|t|) be an infinitesimal quantity and `(t) = 1 − εt, then δx = εtx, δt = εt2 and δu = ε(− n2 t + 4i |x|2 )u. From (7.28), we have the pseudo-conformal conservation law Z |xu + 2it∇u|2 + 4t2 F (|u|2 ) dx = const. (7.36) Rn

Remark 7.1. 1) In fact, x + 2it∇ is a Galilean operator, which is com|x|2

mutable with i∂t + ∆. Let J(t) = x + 2it∇ and M (t) = e 4it , then we have the following relations between J(t) and the Schr¨odinger semigroup S(t) = eit∆ or M (t): J(t) = S(−t)xS(t),

J(t)u = 2itM (−t)∇(M (t)u).

2) For the general nonlinearity F 0 (|u|2 )u, we can differentiate (7.36) with respect to the time t, Z d |xu + 2it∇u|2 + 4t2 F (|u|2 ) dx dt Rn Z =4t (n + 2)F (|u|2 ) − nF 0 (|u|2 )|u|2 dx, (7.37) Rn

to obtain, for any time t in the time interval where the solution exists, the following pseudo-conformal conservation law Z |xu + 2it∇u|2 + 4t2 F (|u|2 ) dx Rn Z Z t Z 2 = |xu(0, x)| dx + 4 τ (n + 2)F (|u|2 ) − nF 0 (|u|2 )|u|2 dxdτ. Rn

0

Rn

(7.38)

7.3

Virial identity and Morawetz inequality

In the previous subsection, it involves |u|2 , x|u|2 , |x|2 |u|2 , · · · , in the conservation laws. Have they some kind of rules to follow? What will happen if we replace the coefficient in front of |u|2 by a general function? Can we obtain some other conservation laws or estimates?

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Now, let a(t, x) be an arbitrary real-valued function on R1+n , and define the virial potential Va (t) and the Morawetz5 action Ma (t) associated to a as Z Va (t) := a(t, x)|u(t, x)|2 dx, Ma (t) := ∂t Va (t). Rn

For convenience, denote G(|u|2 ) = F 0 (|u|2 )|u|2 − F (|u|2 ). By (7.5) and the integration by parts, we have Z Ma (t) = ∂t Va (t) = at |u|2 + 2Im(¯ u∇a · ∇u) dx, (7.39) n ZR ∂t Ma (t) = ∂tt Va (t) = (att − ∆2 a)|u|2 dx Rn Z +4 Im(¯ u∇at · ∇u)dx Rn

+4

Z

n X

ajk Re(uk u ¯j )dx

Rn j, k=1

+2

Z

∆aG(|u|2 )dx.

(7.40)

Rn

In the time independent case a(t, x) = a(x), we have Z Z n X ∂tt Va (t) = − ∆2 a|u|2 dx + 4 ajk Re(uk u¯j )dx Rn

+2

Z

Rn j, k=1

∆aG(|u|2 )dx.

(7.41)

Rn

On the other hand, in the time dependent case a(t, x) 6= a(x), we get, for any positive function Q = Q(t, x), that 4Im(¯ u∇at · ∇u) = |Q−1 (∇at )u − 2iQ∇u|2 − Q−2 |∇at |2 |u|2 − 4Q2 |∇u|2 . Thus, ∂tt Va (t) =

Z

(att − ∆2 a − Q−2 |∇at |2 )|u|2 dx Z + |Q−1 (∇at )u − 2iQ∇u|2dx Rn

Rn

5 Cathleen

Synge Morawetz (1923- ) is a mathematician born in Toronto, Canada. Morawetz’ research was mainly in the study of the partial differential equations governing fluid flow, particularly those of mixed type occurring in transonic flow. She is a professor emerita at the Courant Institute of Mathematical Sciences at the New York University, where she had also served as the director from 1984 to 1988.

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+4

Z

n X

(ajk − Q2 δjk )Re(uk u ¯j )dx

Rn j, k=1

+2

Z

∆aG(|u|2 )dx.

(7.42)

Rn

The problem is then to choose Q and a such that as many as possible of these terms are non-negative up to some manageable errors. Next, we show some examples. Example 7.1 (Virial identity). Let a(t, x) = |x|2 , then ajk = 2δjk and ∆2 a = 0, thus from (7.41), it yields the virial identity Z Z Z ∂tt |x|2 |u|2 dx = 8n |∇u|2 dx + 4n G(|u|2 )dx. Rn

Rn

Rn

2 For the defocusing case F (|u|2 ) = p+1 |u|p+1 where p > 1, i.e. G(|u|2 ) = R p−1 p+1 , it implies that the convexity of Rn |x|2 |u|2 dx can be controlled p+1 |u| 2 |u|p+1 with p > 1 + n4 , by the energy. For the focusing case F (|u|2 ) = − p+1 R p+1 i.e. G(|u|2 ) = − p−1 , the quantity ∂tt Rn |x|2 |u|2 dx can be bounded p+1 |u| from above by a multiple of the energy, thus if the energy is negative and R 2 |x| |u|2 dx is finite, the solution of the equation necessarily has a blowup Rn in finite time.

Example 7.2 (Morawetz’ inequality). Let a(t, x) = |x|, n > 1, then δjk xj xk x , ajk = |x| − |x| and ∆a = n−1 ∇a = |x| 3 |x| . From (7.39) and (7.41), we have Z x Ma (t) =2 Im(¯ u · ∇u)dx, |x| Rn Z Z 1 |∇ /0 u|2 ∂t Ma (t) = − (n − 1) (∆ )|u|2 dx + 4 dx |x| |x| Rn Rn Z G(|u|2 ) + 2(n − 1) dx, |x| Rn where |∇ /y u|2 := |∇u|2 − |

x−y · ∇u|2 . |x − y|

1 ) = −4πδ, and so In the three-dimensional case n = 3, we have ∆( |x| Z |∇ /0 u|2 + G(|u|2 ) ∂t Ma (t) =8π|u(t, 0)|2 + 4 dx. |x| R3

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Integrating over [T∗ , T ∗ ] with respect to t, we get the Morawetz identity Z T∗ Z T∗ Z |∇ /0 u|2 + G(|u|2 ) dxdt + 4π |u(t, 0)|2 dt 2 |x| T∗ T∗ R3 Z Z x x · ∇u(T ∗ , x))dx − Im(¯ u(T∗ , x) · ∇u(T∗ , x))dx. = Im(¯ u(T ∗ , x) |x| |x| R3 R3

In the three-dimensional defocusing case, it implies the following Morawetz inequality Z T∗ Z Z T∗ |∇ /0 u|2 + G(|u|2 ) dxdt + 2π |u(t, 0)|2 dt |x| T∗ R3 T∗ .

sup

T∗ 6t6T ∗

ku(t)k2H˙ 1/2 .

(7.43)

This inequality can be used to demonstrate scattering results for defocusing Schr¨ odinger equations. 1 ) = In higher dimensional defocusing case (n > 4), we have −∆( |x| (n−3) |x|3

> 0 and the Morawetz inequality Z T∗ Z Z T∗ Z |u|2 |∇ /0 u|2 (n − 1)(n − 3) dxdt + 4 dxdt 3 |x| T∗ Rn |x| T∗ Rn Z T∗ Z G(|u|2 ) dxdt + 2(n − 1) |x| T∗ Rn .

sup

T∗ 6t6T ∗

ku(t)k2H˙ 1/2 .

(7.44)

1 But in lower dimensions n = 1, 2, the distribution −∆( |x| ) is quite nasty so that we can not obtain a simple result as in higher dimensional cases.

Examplep7.3 (Nakanishi-Morawetz inequality). Let a(t, x) = λ := |(t, x)| = t2 + |x|2 , then we can compute t xj txj |x|2 at = , aj = , atj = − 3 , att = 3 , λ λ λ λ δjk t2 δjk |x|2 − xj xk n |x|2 n−1 t2 ajk = 3 + , ∆a = − = + . λ λ3 λ λ3 λ λ3 Substituting these into (7.39) and (7.42), we obtain Z t 2 x Ma (t) = |u| + 2Im(¯ u · ∇u) dx, λ λ Rn Z 2 2 |x| t |x|2 ∂t Ma (t) = ( 3 − ∆2 a − 6 2 )|u|2 dx λ Q Rn λ

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Z

tx u − 2iQ∇u|2 dx Qλ3 Z |∇ /0 u|2 |x|2 t2 dx +4 ( 3 − Q2 )|∇u|2 + λ3 Rn λ Z +2 ∆aG(|u|2 )dx.

+

Rn

|−

Rn

2

Taking Q := as ∂t Ma (t) = −

t2 λ3 ,

we can eliminate some terms to reduce the above identity

Z

∆2 a|u|2 dx +

Rn

Z

Z

|xu + 2it∇u|2 dx + 4 λ3

Rn

Z

Rn

|∇ /0 u|2 |x|2 dx λ3

n−1 t2 +2 ( + 3 )G(|u|2 )dx. λ λ Rn Noticing ∆2 a = O(λ−3 ), we have, for G(|u|2 ) > 0, that Z Z |xu + 2it∇u|2 |∇ /0 u|2 |x|2 dx + 4 dx λ3 λ3 Rn Rn Z t2 G(|u|2 ) +2 dx λ3 Rn Z . ∂t Ma (t) + O(λ−3 )|u|2 dx. Rn

Integrating over (1, T ] with respect to t (we can deal with [−T, −1) in a similar way), we obtain Z TZ Z TZ |∇ /0 u|2 |x|2 |xu + 2it∇u|2 dxdt + 4 dxdt λ3 λ3 1 Rn 1 Rn Z TZ t2 G(|u|2 ) +2 dxdt λ3 1 Rn Z TZ . Ma (T ) − Ma (1) + O(λ−3 )|u|2 dxdt . C(E(0), N (0)) +

Z

1 T

1

Z

Rn

O(λ−3 )|u|2 dxdt,

Rn

where H(t) and N (t) are the energy and the L2 -norm defined before, respectively. For |t| > 1, we have Z Z Z Z |u|2 |u|2 dxdt . dxdt . N (0). 3 3 |t|>1 Rn λ |t|>1 Rn |t|

Taking T → ∞, we get Z Z Z Z |xu + 2it∇u|2 |∇ /0 u|2 |x|2 dxdt + 4 dxdt 3 λ λ3 |t|>1 Rn |t|>1 Rn

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+2

Z

|t|>1

Z

Rn

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t2 G(|u|2 ) dxdt λ3

.C(E(0), N (0)). 2 In particular, let F (|u|2 ) = p+1 |u|p+1 . Assume that p ∈ (1, ∞) if n = p−1 4 ) if n > 3, then F 0 (|u|2 )|u|2 − F (|u|2 ) = p+1 |u|p+1 . 1, 2, and p ∈ (1, 1 + n−2 From the previous estimates, it yields Z Z t2 |u|p+1 dxdt . C(E(0), N (0)). (7.45) 3 |t|>1 Rn |(t, x)|

For the case |t| 6 1, we split it into two cases. In the case |x| > 1, we have ZZ ZZ t2 |u|p+1 dxdt . |u|p+1 dxdt . C(E(0), N (0)). (7.46) |t|61 |t|61 |(t, x)|3 |x|>1 |x|>1 In the case |x| 6 1, we get ZZ ZZ t2 |u|p+1 |u|p+1 dxdt . dxdt |t|61 |(t, x)|3 |t|61 |(t, x)| |x|61 |x|61 ZZ |u|p+1 . dxdt |t|61 |t|1−(p+1)ε |x|(p+1)ε |x|61 Z k|x|−ε ukp+1 Lp+1 (|x|61) dt . 1−(p+1)ε |t| |t|61 Z 1 1 kukp+1 . H 1 (|x|61) dt 1−(p+1)ε −1 |t| . C(E(0), N (0)),

(7.47)

6

where we have used the following Hardy’s inequality k|x|−ε ukLq (|x|61) 6 CkukH 1 , ∀2 6 q < 2∗ . (7.48) 2n/(n − 2), n > 3, n/q − n/2∗ , n 6= 2, Here 2∗ := 0<ε< In fact, by ∞, n 6 2, n/2q, n = 2. H¨ older’s7 inequality, we have, for n 6= 2, that k|x|−ε ukLq (|x|61) 6k|x|−ε kLq2∗ /(2∗ −q) (|x|61) kukL2∗ (|x|61))

6 Godfrey Harold Hardy (1877-1947) was a prominent English mathematician, known for his achievements in number theory and mathematical analysis. 7 Otto Ludwig H¨ older (1859-1937) was a German mathematician born in Stuttgart. He is famous for many things including: H¨ older’s inequality, the Jordan-H¨ older theorem, H¨ older’s theorem and the H¨ older condition which is used e.g. in the theory of partial differential equations and function spaces.

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6C

Z

1

r

∗

−ε 2q2 ∗ −q +n−1

0

6CkukH 1 .

dr

∗ −q 2q2 ∗

kukL2∗ (|x|61))

For the case n = 2, from H¨ older’s inequality, we get k|x|−ε ukLq (|x|61) 6k|x|−ε kL2q (|x|61) kukL2q (|x|61)) Z 1 1/2q 6C r−2qε+n−1 dr kukL2q (|x|61)) 0

6CkukH 1 .

From (7.45), (7.45) and (7.45), we obtain the Nakanishi-Morawetz inequality (cf. [176]) Z t2 |u|p+1 dxdt . C(E(0), N (0)), ∀n ∈ N. (7.49) 3 R1+n |(t, x)| 7.4

Morawetz’ interaction inequality

Let a(x) be an arbitrary real-valued function on Rn , and define the virial interaction potential V a (t) and Morawetz’ interaction potential M a (t) associated to a as Z Z V a (t) := a(x − y)|u(t, x)|2 |u(t, y)|2 dxdy, M a (t) := ∂t V a (t). Rn

Rn

From (7.5) and the integration by parts, we have Z Z h ∇a(x − y) · Im(∇u(t, x)¯ u(t, x))|u(t, y)|2 Ma (t) = 2 n n R R i − Im(∇u(t, y)¯ u(t, y))|u(t, x)|2 dxdy,

(7.50)

and

∂t Ma (t) Z Z =−2 ∆2 a(x − y)|u(t, x)|2 |u(t, y)|2 dxdy Rn Rn Z Z +2 ∆a(x − y)[G(|u(x)|2 )|u(y)|2 + G(|u(y)|2 )|u(x)|2 ]dxdy Rn

+4

Z

Rn

Rn

Z

n X

Rn j, k=1

h ajk (x − y) Re(uk (t, x)¯ uj (t, x))|u(t, y)|2 + Re(uk (t, y)¯ uj (t, y))|u(t, x)|2

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− Im(uk (t, y)¯ u(t, y))Im(uj (t, x)¯ u(t, x)) i − Im(uk (t, x)¯ u(t, x))Im(uj (t, y)¯ u(t, y)) dxdy. (7.51)

Denote

Aj (t, x, y) := u(t, x)¯ uj (t, y) + uj (t, x)¯ u(t, y), Bj (t, x, y) := u(t, x)uj (t, y) + uj (t, x)u(t, y), then, the last integral can be reduced to Z Z n X 2 ajk (x − y)[Aj (t, x, y)Ak (t, x, y) + Bj (t, x, y)Bk (t, x, y)]dxdy, Rn

Rn j, k=1

where we can exchange k and j since ajk = akj . For convenience, we denote the above quantity by I. If a is a radial 0 0 (|x|) (|x|) xk xj function, i.e. a(x) = a(|x|), then ajk (x) = a |x| δjk +(a00 (|x|)− a |x| ) |x| |x| . From the Cauchy-Schwartz inequality, we obtain Z Z n a0 (|x − y|) X I =2 [|Aj (t, x, y)|2 + |Bj (t, x, y)|2 ]dxdy |x − y| j=1 Rn Rn Z Z n 2 a0 (|x − y|) h X xj − yj +2 (a00 (|x − y|) − ) Aj (t, x, y) |x − y| |x − y| n n R R j=1

>2

Z

Rn

Z

n 2 i X xj − yj Bj (t, x, y) dxdy + |x − y| j=1

n 2 h X xj − yj Aj (t, x, y) a00 (|x − y|) |x − y| Rn j=1

n X 2 i xj − yj + Bj (t, x, y) dxdy. |x − y| j=1

It is clear that I > 0 if a00 (λ) > 0 for λ > 0. Hence, (7.51) and (7.50) imply Theorem 7.4 (Morawetz’ interaction inequality). Let a(x) be a realvalued radial convex function, and u(t, x) be a solution of (7.5), then it holds Z Z Rn

+

Z

(−∆2 a(x − y))|u(t, x)|2 |u(t, y)|2 dxdy Z ∆a(x − y)[G(|u(t, x)|2 )|u(t, y)|2 + G(|u(t, y)|2 )|u(t, x)|2 ]dxdy

Rn

Rn

Rn

1 6 ∂t Ma (t), 2 where Ma (t) is defined in (7.50).

(7.52)

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Next, we consider an example of three dimensional nonlinear Schr¨ odinger equations. Example 7.4 (Morawetz’ interaction inequality of 3D NLS). 2 x , ∆a = |x| > 0, and −∆2 a = 8πδ(x). Let n = 3, a(x) = |x|, then ∇a = |x| From (7.52), we get Z T∗ Z 8π |u(t, x)|4 dxdt T∗

+2

Z

R3

T∗

T∗

Z

R3

Z

R3

G(|u(t, x)|2 )|u(t, y)|2 + G(|u(t, y)|2 )|u(t, x)|2 dxdydt |x − y|

1 6 (Ma (T ∗ ) − Ma (T∗ )) Z2 Z h x−y = Im( · ∇u(T ∗ , x)¯ u(T ∗ , x))|u(T ∗ , y)|2 |x − y| Rn Rn i x−y − Im( · ∇u(T ∗ , y)¯ u(T ∗ , y))|u(T ∗ , x)|2 dxdy |x − y| Z Z h x−y − Im( · ∇u(T∗ , x)¯ u(T∗ , x))|u(T∗ , y)|2 |x − y| Rn Rn i x−y − Im( · ∇u(T∗ , y)¯ u(T∗ , y))|u(T∗ , x)|2 dxdy |x − y| . N 2 (0)

sup

T∗ 6t6T ∗

7.5

ku(t)k2˙ 1 . H2

(7.53)

Scattering results for NLS

The Morawetz estimates are time-decaying estimates, which are very useful for us to study the existence of the scattering operators. In this section, we only state some results on the existence of the scattering operators for NLS iut + ∆u = |u|p u, u(0, x) = u0 (x). (7.54) The existence of the scattering operators in the H 1 -subcritical but L2 supercritical cases was shown by Nakanishi [176] in 1 and 2 spatial dimensions and by Ginibre and Velo [79] in higher spatial dimensions n > 3: Theorem 7.5. Let p ∈ (4/n, 4/(n − 2)) for n > 3, and p ∈ (4/n, ∞) for n = 1, 2. For any initial datum u0 ∈ H 1 with E(u0 ) < ∞, (7.54) has a p(n+2)/2 unique global solution u ∈ Ct0 (H˙ x1 ) ∩ Lt,x satisfying Z ∞Z |u(t, x)|p(n+2)/2 dxdt 6 C(E(u0 )), −∞

Rn

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where E(u(t)) =

Z

Rn

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2 1 |∇u(t, x)|2 + |u(t, x)|2+p dx 2 2+p

is the energy conserved quantity. Moreover, there exist finite-energy solutions u± of the linear Schr¨ odinger equation (i∂t + ∆)u± = 0 such that ku± (t) − u(t)kH˙ 1 → 0, as t → ±∞, x

and the mapping u0 7→ u± (0) is a homeomorphism from H˙ 1 (Rn ) to itself. Next, we introduce briefly the related results for the scattering theory of H 1 -critical defocusing NLS 4

iut + ∆u = |u| n−2 u, u(0, x) = u0 (x).

(7.55)

In 1999, J. Bourgain [19] proved the global wellposedness of the solution with radial initial data in 3 and 4 dimensions. Then M. Grillakis [84] gave a different proof for Bourgain’s results in 2000. In 2005, these results were generalized to any spherically symmetric initial data in any dimensions n > 3 by T. Tao [217]. At the same time (the paper was published in 2008), J. Colliander, M. Keel, G. Staffilani, H. Takaoka, and T. Tao [44] obtained the global well-posedness and the scattering theory for any general initial data in three dimensions based on the induction arguments on the energy of Bourgain’s. They applied the induction argument not only on the frequency space but also on the physical space, and replaced the original Morawetz’ inequality (7.43) by Morawetz’ interaction inequality (7.53) in order to deal with the non-radial cases. In addition, the Morawetz’ interaction inequality, together with the almost conserved quantity of the momentum which controlled the mass in the frequency space, described the possibilities of the energy concentration. For the higher dimensional case, it is generalized by E. Ryckman and M. Visan [196] for n = 4 and by M. Visan [232] for n > 5. Thus, the scattering in the energy class is now known for (7.55) with any general initial data in all dimensions n > 3, namely, as stated in the following results [44; 196; 232]. Theorem 7.6. Let n > 3. For any initial datum u0 ∈ H 1 with E(u0 ) < ∞, 2(n+2) then (7.55) yields a unique global solution u ∈ Ct0 (H˙ x1 ) ∩ Lt,xn−2 such that Z ∞ Z 2(n+2) |u(t, x)| n−2 dxdt 6 C(E(u0 )), −∞

Rn

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where E(u(t)) =

Z

Rn

2n n−2 1 |∇u(t, x)|2 + |u(t, x)| n−2 dx 2 2n

is the energy conserved quantity. The above theorem implies the scattering results [44; 232; 196]: Corollary 7.1. Let n > 3. Assume that u0 has finite energies, u is the 2(n+2) unique global solution of (7.55) on u ∈ Ct0 (H˙ x1 ) ∩ Lt,xn−2 . Then, there exist

finite-energy solutions u± of the linear Schr¨ odinger equation (i∂t + ∆)u± = 0 such that ku± (t) − u(t)kH˙ 1 → 0, t → ±∞, x

and the mapping u0 7→ u± (0) is a homeomorphism from H˙ 1 (Rn ) to itself. Along with the perfect solution of the scattering theory of the energycritical defocusing NLS, most researchers transfer their interests into another important problem – the global well-posedness and scattering theories of the mass-critical defocusing NLS 4

iut + ∆u = |u| n u, u(0, x) = u0 (x).

(7.56)

At present, there are some results in this aspect: • n > 1, for sufficiently small initial datum u0 ∈ L2 (Rn ), the solution of (7.56) is global well-posed and scatters. • n > 2, for radial initial data u0 ∈ L2 (Rn ), the solution of (7.56) is global well-posed and scatters, cf. [138; 140]. • n > 1, for any initial data u0 ∈ L2 (Rn ), the solution of (7.56) is global well-posed and scatters, cf. [64; 65; 66]. For other nonlinearities, there are also some interesting results. K. Nakanishi established the scattering theory for the mixed power nonlinearity |u|p1 u+|u|p2 u for 4/n < p1 6 p2 < 4/(n−2) and 4/n 6 p1 < p2 6 2∗ −2 in the energy space H 1 , cf. [176; 179; 220]. For the nonlinearity (V ∗ |u|2 )u, i.e. the Hartree equation, he also proved the existence of the scattering operator in H 1 for V ∈ Lp1 + Lp2 , p1 , p2 > 1 and n/4 < p2 6 p1 < n/2, cf. [175]. For the nonlinear Schr¨odinger equation with the following nonlinearity X λ2k f (u) = µ |u|2k u, µ, λ > 0, n = 1, 2, (7.57) (2k)! k>1+2/n

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the scattering operator also exists under some assumptions, cf. [244]. In addition, there is a great process for focusing nonlinear Schr¨odinger equations iut + ∆u = −|u|p u, (7.58) u(0, x) = u0 (x). For H˙ 1 -critical nonlinearities in n = 3, 4, 5 dimensions, i.e. p = 4/(n − 2), the solution of (7.58) is global well-posed and scatters for any radial initial datum u0 ∈ H˙ 1 (Rn ) whose energy and kinetic energy are less than those of the ground state, cf. [124]; For the n > 5 dimensional case, the condition “radial” can be eliminated and it holds that the solution of (7.58) is global well-posed and scatters for any initial datum u0 ∈ H˙ 1 (Rn ) whose energy and kinetic energy are less than those of the ground state, cf. [139]. Otherwise, in the n > 4 dimensions, for a global solution u to the focusing mass-critical nonlinear Schr¨odinger equation (7.58) (where p = 4/n) with spherically symmetric H 1 initial data and mass equal to that of the ground state Q, u does not scatter then, up to the phase rotation and the scaling, u is the solitary wave eitQ . And the only spherically symmetric minimalmass non-scattering solutions are, up to the phase rotation and the scaling, the pseudo-conformal ground state and the ground state solitary wave, cf. [137]. Recently, scattering for the focusing energy-subcritical NLS has been studied by Fang, Xie and Cazenave [72].

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Chapter 8

Boltzmann equation without angular cutoff

The kinetic theory of the gas is a theory devoted to the study of evolutionary behaviors of the gas in the one-particle phase space of position and velocity. Recently, the kinetic theory is getting more and more recognized to be significant both in mathematics and practical applications as a key theory connecting the microscopic and macroscopic theory of gases and fluids. In this sense, the kinetic theory is in-between or mesoscopic. In the macroscopic scales where the gas and fluid are regarded as a continuum, their motion is described by the macroscopic quantities such as macroscopic mass density, bulk velocity, temperature, pressure, stresses, heat flux and so on. The Euler and Navier-Stokes equations, compressible or incompressible, are the most famous equations among governing equations proposed so far in fluid dynamics. The extreme contrary is the microscopic scale where the gas, fluid, and hence any matter, are looked at as a many-body system of microscopic particles (atom/molecule). Thus, the motion of the system is governed by the coupled Newton equations, within the framework of the classical mechanics. The number of the involved equations is 6N if the total number of the microscopic particles is N . Although the Newton equation is the first principle of the classical mechanics, it is not of practical use because the number of the equations is so enormous (N ∼ the Avogadro number 6 × 1023 ) that it is hopeless to specify all the initial data, and we must appeal to statistics. On the other hand, the macroscopic (fluid dynamical) quantities mentioned above are related to statistical average of quantities depending on the microscopic state. Thus, the kinetic theory that gives the mesoscopic descriptions of the gas and fluid is noticed to be a key theory that links the microscopic and

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macroscopic scales. The Boltzmann equation, which is the subject of these notes, is the most classical but fundamental equation in the mesoscopic kinetic theory. The goal of this chapter is to introduce, and make a preliminary discussion of the Boltzmann equation. We mainly use harmonic analysis to consider the Boltzmann equation without angular cutoff. We will show that the Boltzmann collision operator without angular cutoff behaves like a singular integral operator or pseudo-differential operator whose leading term is characterized by the operator (−∆)ν/2 . This was first pointed out by Pao [188], see also [228], and was formulated explicitly by Lions [160]. The optimal Sobolev exponent ν/2 is due to Villani [231]. In the middle of 1990s, Desvillettes managed to prove the regularity of solutions to some simplified models for the spatially homogeneous problems, cf. [57; 58]. Around 2000s, the regularity induced by the grazing collision was analyzed in terms of the entropy production integral, cf. the work [2] and others. In particular, [2] establishes several elegant formulations associated with the collision operator which have been essentially used to the study of the spatially homogeneous problem.

8.1

Models for collisions in kinetic theory

In this section, we will first give the derivation of the Boltzmann equation, in details, one can refer to [28; 59; 229].

8.1.1

Transport model

The object of kinetic theory is the modelling of a gas (or plasma, or any system made up of a large number of particles) by a distribution function in the particle phase space. This phase space includes macroscopic variables, i.e. the position in physical space, but also microscopic variables, which describe the state of the particles. In the present survey, we shall restrict ourselves, most of the time, to systems made of a single species of particles (no mixtures), and which obey the laws of classical mechanics (non-relativistic, non-quantum). Assume that the gas is contained in a (bounded or unbounded) domain Ω ⊂ Rn (n = 3 in applications) and observed on a time interval [0; T ], or [0; +∞). Then, under the above simplifying assumptions, the corresponding

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kinetic model is a nonnegative function f (t; x; v), defined on [0; T ]× Ω× Rn. Here the space Rn = Rnv is the space of possible velocities, and should be thought of as the tangent space to Ω. f (t; x; v) is the phase space density of particles which at time t and point x move with velocity v, where x = (x1 , x2 , ...xn ) and v = (v1 , v2 , ...vn ). When there is no interaction between the particles and their surrounding environment (including themselves), they will move at a constant velocity and along straight lines. In other words, for all times t and s, point x and velocity v, a particle which at time t sits at point x and move with velocity v will sit at time t + s at point x + vs and will keep its velocity v. This entails that f (t + s, x + vs, v) = f (t, x, v), for any s,

(8.1)

or, after differentiation with respect to s ∂t f + v · ∇x f = 0.

(8.2)

(8.2) is free transport model. v·∇x is the transport operator, where v·∇x = Pn ∂ k=1 vk ∂xk . (8.2) implies that the change rate of density function is zero if there is no collisions of the particles. When a given force F (t, x) acts on the particles (such a force can also depend on v in specific situations, for example when the particles are charged and feel the action of a magnetic field, or when the force is the drag force due to a surrounding gas), the particles will follow the trajectories of the following system of differential equations: x(t) ˙ = v(t)

(8.3)

v(t) ˙ = F (t, x(t))

(8.4)

and the corresponding partial differential equation satisfied by f (that is, the PDE whose characteristic curves are exactly the solutions of (8.3) and (8.4)), is the Vlasov equation ∂t f + v · ∇x f + F (t, x) · ∇v f = 0, Pn where F (t, x) · ∇v = k=1 Fk ∂v∂k . 8.1.2

(8.5)

Boltzmann model

When the forces acting on the particles are mainly due to the collisions of the particles between themselves, one is led to write down the Boltzmann equation. (Very often, particles will be assumed to interact via a given

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interaction potential φ(r), r distance between particles). We first shall make several postulates: (1). We assume that particles interact via binary collisions : this is a vague term describing the process in which two particles happen to come very close to each other, so that their respective trajectories are strongly deviated in a very short time. Underlying this hypothesis is an implicit assumption that the gas is dilute enough that the effect of interactions involving more than two particles can be neglected. Typically, if we deal with a three-dimensional gas of n hard spheres of radius r, this would mean N r3 1, N r2 ' 1. (8.6) (2). Moreover, we assume that these collisions are localized both in space and time, i.e. they are brief events which occur at a given position x and a given time t. This means that the typical duration of a collision is very small compared to the typical time scale of the description, and also quantities such as the impact parameter (see below) are negligible in front of the typical space scale (say, a space scale on which variations due to the transport operator are of order unity). (3). Next, we further assume these collisions to be elastic : momentum and kinetic energy are preserved in a collision process. (4). We also assume collisions to be microreversible. This word can be understood in a purely deterministic way : microscopic dynamics are time-reversible; or in a probabilistic way : if let (v, v∗ ) stand for the velocities before collision, and (v 0 , v∗0 ) stand for the velocities after collision; the probability that velocities (v, v∗ )are changed into (v 0 , v∗0 ) in a collision process, is the same as the probability that (v 0 , v∗0 ) are changed into (v, v∗ ). (5). Finally, we make the Boltzmann chaos assumption : the velocities of two particles which are about to collide are uncorrelated. Roughly speaking, this means that if we randomly pick up two particles at position x, which have not collided yet, then the joint distribution of their velocities will be given by a tensor product (in velocity space) of f with itself. Note that this assumption implies an asymmetry between past and future: indeed, in general if the pre-collisional velocities are uncorrelated, then post-collisional velocities have to be correlated! Thus we only take into account interactions between particles of rarefied gas, equation (8.2) can be written by a generalized form (8.7) ∂t f + v · ∇x f = Q(f )(v). (8.7) means that change rate of density function of particles depends on itself, is a function of f , and is not zero when there exist collision between particles. Next we will derive the form of Q(f ).

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We denote by f2 (v1 , v2 ) the joint density of two particles with respective velocities v1 and v2 . It follows (due to the assumption that the collisions are binary) that we must take into account only two distinct phenomena which modify the number density of particles with velocity v. First, because of a possible collision with a particle of velocity v∗ , a particle which had v for velocity will end up with a velocity v 0 (its partner in the collision will end up with velocity v∗0 ). Secondly, some particle with a velocity w will encounter a particle with velocity w∗ and will end up with a velocity v after the collision (its partner in the collision will end up with velocity w∗0 ). We now denote by p(v1 , v2 → v3 , v4 ) the (density of) probability that for two particles sitting at the same point x at a given time t, a collision occurs and transforms the ingoing velocities v1 and v2 in the outgoing velocities v3 and v4 (we shall see that in the so-called non cutoff case, this quantity is in fact far from being a probability density, since it is not integrable). We see that Q(f ) is the sum of two terms −Q− (f ) and Q+ (f ) which respectively correspond to the two phenomena described above. According to their definition Q− (f ) and Q+ (f ) write Z Z Z f2 (v, v∗ )p(v, v∗ → v 0 , v∗0 )dv∗0 dv 0 dv∗ , Q− (f )(v) = (8.8) +

Q (f )(v) =

v∗

v0

Z Z

Z

w

w∗

v∗0

w∗0

f2 (w, w∗ )p(w, w∗ → v, w∗0 )dw∗0 dw∗ dw.

(8.9)

From the assumption (4), it follows that ∀ v1 , v2 , v3 , v4 , p(v1 , v2 → v3 , v4 ) = p(v3 , v4 → v1 , v2 ). Moreover, using the assumption (5), since v and v∗ are irrelative variables, w and w∗ are irrelative variables; we have that f2 (v, v∗ ) = f (v)f (v∗ ) and f2 (w, w∗ ) = f (w)f (w∗ ). Let Q(f ) = Q(f, f ) = Q+ (f, f ) − Q− (f, f ), where Z Z Z − f (v)f (v∗ )p(v, v∗ → v 0 , v∗0 )dv∗0 dv 0 dv∗ , Q (f, f )(v) = (8.10) Q+ (f, f )(v) = We have

Z Q(f, f )(v) =

v∗

v∗

v0

v∗

Z Z

Z

v0

v∗0

v∗0

f (v 0 )f (v∗0 )p(v 0 , v∗0 → v, v∗ )dv∗0 dv 0 dv∗ .

(8.11)

Z Z f (v 0 )f (v∗0 ) − f (v)f (v∗ ) p(v, v∗ → v 0 , v∗0 )dv∗0 dv 0 dv∗ . v0

v∗0

(8.12)

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Let us explain (8.12) a little bit. This operator can formally be split, in a self-evident way, into a gain and a loss term, Q(f ; f ) = Q+ (f ; f )−Q−(f ; f ): The loss term counts all collisions in which a given particle of velocity v will encounter another particle, of velocity v∗ . As a result of such a collision, this particle will in general change its velocity, and this will make less particles with velocity v. On the other hand, each time particles collide with respective velocities v 0 and v∗0 , then the v 0 particle may acquire v as new velocity after the collision, and this will make more particles with velocity v : this is the meaning of the gain term. Next, we only need to give the expression of p(v, v∗ → v 0 , v∗0 ). Using the conservation of momentum and kinetic energy in a collision, we have v + v∗ = v 0 + v∗0 , (8.13) |v|2 + |v∗ |2 = |v 0 |2 + |v∗0 |2 . Since this is a system of n + 1 scalar equations for 2n scalar unknowns, it is natural to expect that its solutions can be defined in terms of n − 1 parameters. Here is a convenient representation of all these solutions, which we shall sometimes call the σ-representation: v0 =

|v − v∗ | v + v∗ |v − v∗ | v + v∗ + σ, v∗0 = − σ. 2 2 2 2

(8.14)

Here the parameter σ ∈ S n−1 varies in the n − 1 unit sphere. One can see the following figure

v0 σ θ/2

v∗

θ k

v∗0

Fig. 8.1

The geometry of collisions

v

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The Galilean invariance which holds in the context of binary collisions entails that the measure p(v, v∗ → v 0 , v∗0 ) can only depend on |v − v∗ | and v−v∗ 0 0 |v−v∗ | · σ, p(v, v∗ → v , v∗ ) can be written by B(·, ·). We now can write down the final form of Boltzmann’s collision operator : Z Z v − v∗ Q(f, f ) = B(|v − v∗ |, · σ){f (v∗0 )f (v 0 ) − f (v∗ )f (v)}dσdv∗ , |v − v∗ | Rn S n−1 (8.15) where function B is called the Boltzmann collision cross section (or the collision kernel). For convenience, we shall also use the bilinear form Q(g, f ) related to the quadratic form Q(f, f ) and defined by Z Z Q(g, f ) = B(|v − v∗ |, σ){g(v∗0 )f (v 0 ) − g(v∗ )f (v)}dσdv∗ . (8.16) Rn

S n−1

For simplicity, we sometimes use standard abbreviations: g(v∗0 )f (v 0 ) = g∗0 f 0 and g(v∗ )f (v) = g∗ f . Finally, we write down the standard form of the Boltzmann equation ∂t f + v · ∇x f = Q(f, f )(v),

(8.17)

where Q(f, f ) is given by (8.15). The spatially homogeneous Boltzmann equation describes the behavior of a dilute gas, in which the velocity distribution of particles is assumed to be independent of the position; it reads ∂t f = Q(f, f )(v),

(8.18)

where the unknown f = f (t, v) is assumed to be nonnegative, and stands for the density of particles at time t with velocity v. These equations are called homogeneous since f is assumed to be independent of the position. This model is simple, in this chapter we only consider (8.18). 8.1.3

Cross section

In most cases, the collision kernel B cannot be expressed explicitly. However, to capture its main properties, it is usually assumed to have the form: v − v∗ π , σi, 0 6 θ 6 , |v − v∗ | 2 (8.19) where the deviation angle θ is the angle between pre- and post-collisional velocities. Using n-dimensional spherical coordinates, we have Z Z π2 n−2 B(|v − v∗ |, cos θ) sinn−2 θdθ, B(|v − v∗ |, σ)dσ = |S | B(|v − v∗ |, σ) = Φ(|v − v∗ |)b(cos θ), cos θ = h

S n−1

0

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where |S n−2 | denotes the surface area of unit sphere in n dimension. The collision kernel should be computed in terms of the interaction potential φ. Here, we consider the important potential: inverse-power law potentials. It is possible to (almost) explicitly compute the cross section if potential φ(r) is inverse-power law potentials, that is, the inter particle force is proportional to r−s (with r denoting the interparticle distance and s > 2). In such a case (and in dimension 3) B writes s−5

B(|v − v∗ |, σ) = |v − v∗ | s−1 b(cos θ)

(8.20)

with b > 0 a smooth function except at point 1 and satisfying: Z π2 K sinn θb(cos θ)dθ < ∞ and sinn−2 θb(cos θ) ≈ s+1 (8.21) 0 θ s−1 when θ → 0, K > 0. Since s+1 s−1 > 1, b is not integrable at θ = 0. This means Z Z π2 n−2 b(k · σ)dσ = |S | sinn−2 θb(cos θ)dθ = ∞. S n−1

0

Because of the difficulties entailed by this singularity, Grad has proposed to introduce an angular cutoff near θ = 0. It means that we replace B in (8.20) and (8.21) by a new cross section cross

s−5 ˜b(cos θ), ˜ − v∗ |, σ) = |v − v∗ | s−1 B(|v (8.22) where ˜b is smooth, or at least such that sinn−2 θ˜b(cos θ) is integrable near θ = 0. In our model case, this means Z Z π2 ˜b(k · σ)dσ = |S n−2 | sinn−2 θ˜b(cos θ)dθ < ∞.

S n−1

0

In the sequel,we shall speak of cutoff cross sections (cutoff potentials or Grad’s angular cut-off) when B is locally integrable, and of non cutoff cross sections (or non cutoff potentials) when B has a singularity like in (8.21). In this chapter, we mainly consider the Boltzmann equation without cutoff cross section. Note that the separation Q(f, f ) = Q+ (f, f )−Q− (f, f ) make sense if B is cutoff cross section, where Z Z v − v∗ Q+ (f, f ) = B(|v − v∗ |, · σ)f (v∗0 )f (v 0 )dσdv∗ , (8.23) |v − v∗ | Rn S n−1 Z Z v − v∗ Q− (f, f ) = B(|v − v∗ |, · σ)f (v∗ )f (v)dσdv∗ . (8.24) |v − v∗ | n n−1 R S If B is non cutoff cross section, for convenience, we sometimes need to calculate the gain and loss terms Q± in the collision operator separately. This

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separation is not allowed for the non-cutoff cross section but is legitimate as an intermediate step in the calculation. In this chapter, we mainly consider the inverse-power law potentials, the cross section B have the form: B(|v − v∗ |, σ) = |v − v∗ |γ b(cos θ).

(8.25)

If γ > 0, it is called hard potentials; in general, we consider the case of 0 < γ 6 2. If γ = 0, it is called Maxwellian potentials. If −n < γ < 0, it is called soft potentials; for −2 < γ < 0, it is called moderately soft potentials, for γ 6 −2 it is called very soft potentials. Note that if s = 2 then φ(r) becomes Coulomb potential, which leads to Fokker-Planck-Landau equation, we do not consider this model in this book. 8.2

Basic surgery tools for the Boltzmann operator

In this section, we give some basic tools which one often needs for a fine study of the Boltzmann operator. In details, one can refer to [59; 229]. 1.Symmetrization of the collision kernel. In view of formulas (8.14), the quantity f∗0 f 0 − f∗ f is invariant under the change of variables σ → −σ. From the physical point of view this reflects the undiscernability of particles. Thus one can replace (from the very beginning, if necessary) B by its symmetrized version ¯ σ) = {B(z, σ) + B(z, −σ)}Iz·σ>0 . B(z,

This is the reason why we assume that angle θ varies from 0 to π2 in (8.19). In fact, we get rid of collisions with deflexion angle larger than π2 by this ¯ in the above equality. symmetrization trick. B in (8.19) is indeed B 2. Pre-postcollisional change of variables. A universal tool in the Boltzmann theory is the involutive change of variables with unit Jacobian v−v∗ (v, v∗ , σ) → (v 0 , v∗0 , k), where k = |v−v is the unit vector along v − v∗ . ∗| Indeed, denote the mapping Φ : (v, v∗ , σ) → (v 0 , v∗0 , k), then it’s easy to verify that Φ2 = I. Thus the Jacobian of Φ is 1. Then for any function f we have Z Z Z f (v, v∗ , v 0 , v∗0 , σ)dσdv∗ dv Rn Rn S n−1 Z Z Z (8.26) v − v∗ = f (v 0 , v∗0 , v, v∗ , )dσdv∗ dv. |v − v∗ | Rn Rn S n−1

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3. The change of variables (v, v∗ , σ) → (v∗ , v, σ) is clearly involutive and has unit Jacobian. For function f = f (v, v∗ , v 0 , v∗0 , σ), one has Z Z Z f (v, v∗ , v 0 , v∗0 , σ)dσdv∗ dv Rn Rn S n−1 Z Z Z (8.27) = f (v∗ , v, v 0 , v∗0 , σ)dσdv∗ dv. Rn

Rn

S n−1

Next, we give some applications of the two change of variables above. We first give the conservation laws of the Boltzmann equation. As a consequence, if ϕ(v) is an arbitrary continuous function of the velocity v, then we get the following various weak formulations for Boltzmann’s kernel Q Z Q(g, f )ϕ(v)dv (8.28) Rn Z Z v − v∗ · σ){g(v∗0 )f (v 0 ) − g(v∗ )f (v)}ϕ(v)dσdv∗ dv = B(|v − v∗ |, |v − v∗ | 2n n−1 R S Z Z v − v∗ = B(|v − v∗ |, · σ)g(v∗ )f (v)(ϕ(v 0 ) − ϕ(v))dσdv∗ dv. |v − v | 2n n−1 ∗ R S In the special case g = f , we have that Z Q(f, f )ϕ(v)dv Rn Z Z 1 Bf (v∗ )f (v)(ϕ(v∗0 ) + ϕ(v 0 ) − ϕ(v∗ ) − ϕ(v))dσdv∗ dv. = 2 Rn ×Rn S n−1

From the mathematical point of view, it is interesting because expressions like (8.28) may be well-defined in situations where Q(f, f ) is not. From the physical point of view, it expresses the change in the integral R Q(f, f )(v)dv which is due to the action of collisions. Let f be a soluRn tion of the Boltzmann equation (8.17), set in the whole space Rn to simplify. By the conservative properties of the transport operator v · ∇x , Z Z d f (t, x, v)ϕ(v)dxdv = Q(f, f )ϕ(v)dxdv (8.29) dt Rn Rn

and the right-hand side is just the x-integral of any one of the expressions in formulas (8.28). As an immediate consequence, whenever ϕ satisfies the functional equation ∀(v, v∗ , σ) ∈ Rn × Rn × S n−1 , ϕ(v 0 ) + ϕ(v∗0 ) = ϕ(v) + ϕ(v∗ ), then, at least formally, d dt

Z

Rn

f (t, x, v)ϕ(v)dxdv = 0

(8.30)

(8.31)

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along solutions of the Boltzmann equation. The words at least formally of course mean that the preceding equations must be rigorously justified with the help of some integrability estimates on the solutions to the Boltzmann equation. Pluging ϕ(v) = 1, v1 , ..., vn , |v|2 /2, in formula (8.28) and (8.31), by (8.30), we get the conservation of mass, momentum and energy at the level of the Boltzmann operator: Z Z Z |v|2 dv = 0, Q(f, f )(v)dv = 0, Q(f, f )(v)vj dv = 0, Q(f, f )(v) 2 Rn Rn Rn (8.32) or, d dt

Z

d f (t, x, v)dxdv = dt Rn

Z

d f (t, x, v)vj dv = dt Rn

Z

Rn

f (t, x, v))

|v|2 dv = 0. 2 (8.33)

Next, we give the Boltzmann’s H theorem. Without caring about integrability issues, we plug ϕ(v) = log f (v) into (8.28), Z Q(f, f )(v) log f (v)dv = −D(f ), Rn

where D is the entropy dissipation functional. Using the fact that (F − G)(log F − log G) > 0, we have Z Z v − v∗ 1 B(|v − v∗ |, · σ) f (v∗0 )f (v 0 ) − f (v∗ )f (v) D(f ) = 4 Rn ×Rn S n−1 |v − v∗ | 0 (f (v∗ )f (v 0 ) × log dσdv∗ dv > 0. (8.34) f (v∗ )f (v) Now, we introduce Boltzmann’s H functional, Z H(f ) = f log f dxdv. n Rn x ×Rv

Of course, the transport operator −v ·∇x does not contribute in any change of the H functional in time. As a consequence, if f = f (t, x, v) is a solution of the Boltzmann equation, then H(f ) will evolve in time because of the effects of the collision operator: Z d H(f (t, ·, ·)) = − D(f (t, ·, ·))dx 6 0. (8.35) dt Rn x This is the first part of Boltzmann’s H-theorem. Next we assume that B(|v − v∗ |, σ) > 0 for almost all (v, v∗ , σ), which is always the case in

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applications of interest. Then equality in Boltzmann’s H theorem occurs if and only if for almost all x, v, v∗ , σ f (v∗0 )f (v 0 ) = f (v∗ )f (v).

(8.36)

Then, it is possible to prove (under suitable, but rather weak assumptions on B and f ) that D(f ) = 0 ⇐⇒ Q(f, f ) = 0 ⇐⇒ f = M (v), M (v) = Mρ,u,T (v) =

2 ρ − |v−u| 2T , e (2πT )n/2

(8.37) (8.38)

where M is called the Maxwellian and is known to describe the velocity distribution of a gas in an equilibrium state with the mass density ρ(x), bulk velocity u(x) and temperature T . Here, (ρ, u, T ) are taken to be parameters. If (ρ, u, T ) are constants,then M is called a global ( absolute) Maxwellian ; if they are functions of (x, t), then it is called a local Maxwellian. Evidently, the global Maxwellian is a stationary solution Boltzmann equation (8.17) without force F . This is the second part of Boltzmann’s H-theorem. 4. Bobylev’s identities . We now turn to more intricate tools introduced by Bobylev, who first made Fourier transform an extremely powerful tool in the study of the Boltzmann operator with Maxwellian collision kernel. Now we give the Fourier transform of Q(g, f ) in (8.19). We first perform the calculation of the Fourier transform of the gain term in a general Boltzmann collision operator Q+ (g, f ). For any test function ϕ(v), using the pre-postcollisional change of variables in (8.26), we have Z Q+ (g, f )ϕ(v)dv n R Z Z (8.39) v − v∗ = B(|v − v∗ |, · σ)g(v∗ )f (v)ϕ(v 0 )dσdv∗ dv. |v − v∗ | Rn ×Rn S n−1 Plugging ϕ(v) = e−ivξ in (8.39), we have

F (Q+ (g, f ))(ξ) Z Z |v−v∗ | v+v∗ v − v∗ = B(|v − v∗ |, · σ)g(v∗ )f (v)e−i 2 ·ξ e−i 2 σ·ξ dσdv∗ dv. |v − v∗ | R2n S n−1 (8.40) Using the following general equality: Z Z F (k · σ, l · σ)dσ = F (l · σ, k · σ)dσ, |l| = |k| = 1, S n−1

S n−1

(8.41)

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(due to the existence of an isometry on S n−1 exchanging l and k) we have Z |v−v∗ | v − v∗ · σ)e−i 2 σ·ξ dσ B(|v − v∗ |, |v − v | n−1 ∗ S Z (8.42) |ξ| ξ = B(|v − v∗ |, · σ)e−i 2 σ·(v−v∗ ) dσ. |ξ| S n−1 Then, using (8.42) and the Fourier inversion formula, we have

F (Q+ (g, f ))(ξ) Z Z |ξ| v+v∗ ξ · σ)g(v∗ )f (v)e−i 2 ·ξ e−i 2 σ·(v−v∗ ) dσdv∗ dv = B(|v − v∗ |, |ξ| 2n n−1 ZR ZS + − ξ = B(|v − v∗ |, · σ)g(v∗ )f (v)e−ivξ e−iv∗ ξ dσdv∗ dv |ξ| 2n n−1 ZR ZS ξ B(|v − v∗ |, = · σ) |ξ| R2n + − × gˆ(η∗ )fˆ(η)eivη eiv∗ η∗ e−ivξ e−iv∗ ξ dη∗ dη dσdv∗ dv Z = gˆ(η∗ )fˆ(η) R2n ×S n−1 Z + − ξ × B(|v − v∗ |, · σ)eiv(η−ξ ) eiv∗ (η∗ −ξ ) dv∗ dv dσdη∗ dη, |ξ| (8.43) where ξ − |ξ|σ ξ + |ξ|σ , ξ− = . (8.44) 2 2 R ˆ Let δ be the Dirac measure, B(|ξ|, cos θ) = Rn B(|q|, cos θ)e−iq·ξ denotes the Fourier transform of B in the relative velocity variable. By the change of variables q = v − v∗ , Z + − ξ B(|v − v∗ |, · σ)eiv(η−ξ ) eiv∗ (η∗ −ξ ) dv∗ dv |ξ| R2n Z − ξ = B(|q|, · σ)eiv(η∗ +η−ξ) e−iq(η∗ −ξ ) dqdv (8.45) |ξ| 2n R ˆ ∗ − ξ − |, ξ · σ)δ(η = ξ − η∗ ). = B(|η |ξ| ξ+ =

By (8.45), we give the Fourier transform of Q+ (g, f ) Z ˆ ∗ − ξ − |, ξ · σ))dσdη∗ . F (Q+ (g, f ))(ξ) = gˆ(η∗ )fˆ(ξ − η∗ )B(|η |ξ| Rn ×S n−1 (8.46)

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Let ξ∗ = η∗ − ξ − , weZ find F (Q+ (g, f ))(ξ) =

Rn ×S n−1

ˆ ∗ |, gˆ(ξ − + ξ∗ )fˆ(ξ + − ξ∗ )B(|ξ

ξ · σ))dσdξ∗ . |ξ| (8.47) case, that is,

In Maxwellian potentials (Maxwellian molecules) B(|z|, cos θ) = b(cos θ), we have ˆ ∗ |, cos θ) = δ(ξ∗ = 0)b(cos θ). B(|ξ (8.48) and as a consequence Z ξ · σ)dσ. F (Q+ (g, f ))(ξ) = gˆ(ξ − )fˆ(ξ + )b( (8.49) |ξ| n−1 S − For the Fourier transform of Q (g, f ), if B(|z|, cos θ) = b(cos θ), then we easily have Z ξ − F (Q (g, f ))(ξ) = gˆ(0)fˆ(ξ)b( · σ)dσ, (8.50) |ξ| n−1 R S which follows from the fact that S n−1 b(k · σ) does not depend on k. 8.3

Properties of Boltzmann collision operator without cutoff

In this section, we will consider the properties of the Boltzmann collision operator Q(g, f ) without cutoff. We prove that for a given distribution function g ∈ L1 , the Boltzmann operator Q(g, f ) behaves essentially as a fractional power of the Laplacian: Q(g, f ) = −Cg (−∆)ν/2 f + more regular terms. In details, one can refer to [2]. In the following discussion, we also adopt the notations for the weighted function spaces, kf kLpr = kf (v)hvir kLZp , 1 6 p 6 ∞, r ∈ R, kf kLlogL =

f log(1 + f )dv.

Rn

n

2 +1 (Rn ), we have Wefirst give the weak formulation of Q(g, f ). For f ∈ H Q(g, f ), f L2 Z v − v∗ = B(|v − v∗ |, · σ)g(v∗ )f (v)(f (v 0 ) − f (v))dσdv∗ dv |v − v | 2n n−1 ∗ R ×S Z 1 v − v∗ = B(|v − v∗ |, · σ)g(v∗ )(f 2 (v 0 ) − f 2 (v))dσdv∗ dv 2 R2n ×S n−1 |v − v∗ | Z 1 v − v∗ − B(|v − v∗ |, · σ)g(v∗ )(f (v 0 ) − f (v))2 dσdv∗ dv. 2 R2n ×S n−1 |v − v∗ | (8.51)

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For the first term in (8.51), we have the following lemma. Lemma 8.1. (Cancellation) For a.e. v∗ ∈ Rn , Z B(|v − v∗ |, k · σ)(f 2 (v 0 ) − f 2 (v))dσdv = (f 2 ∗ A)(v∗ ),

(8.52)

Rn ×S n−1

where

k=

A(z) = S

n−2

Z

π 2

sinn−2 θ

0

v − v∗ , |v − v∗ |

(8.53)

1 |z| B( , cos θ) − B(|z|, cos θ) dθ. cosn 2θ cos θ2 (8.54)

Proof. We do the calculation as if B were integrable and apply a limiting procedure to conclude in the general case, that is, we choose a sequence ˜j in S n−1 , such that limj→∞ B ˜j = B. In fact the integrable function B right-hand side of (8.52) should be taken as a definition of the left-hand side. Recall that v + v∗ |v − v∗ | |v − v∗ | v0 = + σ = v∗ + (k + |k|σ). 2 2 2 For each σ and with v∗ still fixed, we perform the change of variables v → v 0 . This change of variables is well-defined on the set cos θ > 0, and it follows either by a direct calculation or by using the cylindrical symmetry of this transformation that its Jacobian determinant is dv 0 1 1 1 (k 0 · σ) (8.55) = I + k ⊗ σ = n (1 + k · σ) = n−1 , dv 2 2 2 2

where k 0 =

v 0 −v∗ |v 0 −v∗ | .

Then

θ 1 > √ . 2 2 0 Define the inverse transformation ψσ : v → ψσ (v 0 ) = v, from the figure 0 ∗| 0 8.1, it follows that |v∗ − ψσ (v 0 )| = |vk−v 0 ·σ . Then changing the name v for v, k 0 · σ = cos

|v − v∗ | . (8.56) k·σ Applying this change of variable to the first part in the left-hand side of (8.52), then changing the name v 0 for v, Z B(|v − v∗ |, k · σ)g(v∗ )f 2 (v 0 )dσdv |v∗ − ψσ (v)| =

Rn ×S n−1

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= = =

Z

Z

Z

Rn ×S n−1

dv |dσdv 0 dv 0

B(|ψσ (v 0 ) − v∗ |, 2(k 0 · σ)2 − 1)g(v∗ )f 2 (v 0 )

k0 ·σ> √12

k·σ> √12

B(|v − v∗ |, 2(k 0 · σ)2 − 1)g(v∗ )f 2 (v 0 )|

B(|ψσ (v) − v∗ |, 2(k · σ)2 − 1)g(v∗ )f 2 (v)

2n−1 dσdv 0 (k 0 · σ)2

2n−1 dσdv. (k · σ)2

(8.57)

Using the fact that 2n−2 sinn−2 2θ cosn−2 2θ = sinn−2 θ and n-dimensional spherical coordinates, we have Z 2n−1 B(|ψσ (v) − v∗ |, 2(k · σ)2 − 1) dσ (k · σ)2 k·σ> √12 Z |v − v∗ | 2n−1 = B( , 2(k · σ)2 − 1) dσ k·σ (k · σ)2 k·σ> √12 Z π4 |v − v∗ | 2n−1 n−2 sinn−2 θB( = |S | , cos 2θ) 2 dθ cos θ cos θ 0 Z π2 |v − v | θ 2n−2 ∗ = |S n−2 | sinn−2 B( , cos θ) 2 θ dθ θ 2 cos 2 cos 2 0 Z π2 n−2 sin θ |v − v∗ | B( , cos θ)dθ. (8.58) = |S n−2 | θ n cos 2 cos θ2 0 Define A(v − v∗ ) Z n−2 =S

π 2

0

sinn−2 θ

|v − v∗ | 1 B( , cos θ) − B(|v − v |, cos θ) dθ. ∗ θ θ cosn 2 cos 2 (8.59)

This completes the proof.

Remark 8.1. If B(|v − v∗ |, k · σ) = |v − v∗ |γ b(k · σ), then A(v − v∗ ) Z n−2 =S

π 2

0

=S

n−2

Z

0

π 2

sinn−2 θ

1 |v − v∗ | γ γ ( ) b(cos θ) − |v − v | b(cos θ) dθ ∗ cosn θ2 cos θ2

sinn−2 θ|v − v∗ |γ b(cos θ)(

1

cosγ+n 2θ

− 1)dθ.

(8.60)

Note that if γ > −n, then (8.60) is nonnegative; if γ = −n, it is Coulomb potentials, they require a more careful analysis. Using Lemma 8.1, we find

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that if 0 6 γ 6 2 Z R2n ×S n−1

B(|v − v∗ |, k · σ)g(v∗ )(f 2 (v 0 ) − f 2 (v))dσdvdv∗

243

(8.61)

6 CkgkL12 kf kL21 , θ 2

which follows from the fact that 1 − cosn

6 n(1 − cos 2θ ) = 2n sin 4θ .

For the second term in (8.51), for simplicity, we only consider Maxwellian molecules case, that is, cross section B(|v − v∗ |, k · σ) = b(k · σ). Lemma 8.2. Assume g ∈ L1 (Rn ) and f ∈ L2 (Rn ), the following Plancherel-type identity holds Z v − v∗ b( · σ)g(v∗ )(f (v 0 ) − f (v))2 dσdv∗ dv |v − v∗ | R2n ×S n−1 Z ξ (8.62) = b( · σ) gˆ(0)|fˆ(ξ)|2 + gˆ(0)|fˆ(ξ + )|2 |ξ| Rn ×S n−1 − gˆ(ξ − )fˆ(ξ + )fˆ(ξ) − gˆ(ξ − )fˆ(ξ + )fˆ(ξ) dξdσ.

Proof. We shall do the proof only in the case when b is integrable and the result will follow by monotonicity. First, we have (f (v 0 ) − f (v))2 = f 2 (v 0 ) − 2f (v 0 )f (v) + f 2 (v).

(8.63)

0

We begin with the middle term 2f (v )f (v) in (8.63). By the prepostcollisional change of variables and Parseval’s identity and Bobylev’s identity, we have Z Z b(k · σ)g(v∗ )f (v 0 )f (v)dσdv∗ dv = Q+ (g, f )f dv

1 1 + (Q (g, f ), f )L2 + (f, Q+ (g, f ))L2 2 2 (8.64) 1 1 = (F Q+ (g, f ), F f )L2 + (F f, F Q+ (g, f ))L2 2Z 2 1 ξ − ˆ + ˆ = b( · σ) gˆ(ξ )f (ξ )f (ξ) + gˆ(ξ − )fˆ(ξ + )fˆ(ξ) dξdσ. 2 |ξ| R For the third term f 2 (v) in (8.63), using the fact that S n−1 b(k · σ)dσ does not depend on the unit vector k, we have Z Z Z Z b(k · σ)dσ g(v∗ )dv∗ f 2 (v)dv b(k · σ)g(v∗ )f 2 (v)dσdv∗ dv = S n−1 Rn Rn Z ξ = b( · σ)ˆ g (0)|fˆ(ξ)|2 dξdσ. (8.65) |ξ| =

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For the first term f 2 (v 0 ) in (8.63), we first make the change of variables (v − v∗ , v∗ ) → (v1 , v∗ ), change the name v1 for v, and then use the change of variables v → v 0 as in Lemma 8.1 to obtain Z v + v∗ |v − v∗ | v − v∗ · σ)g(v∗ )f 2 ( + σ)dσdv∗ dv b( |v − v∗ | 2 2 Z v1 v1 + |v1 |σ 2 · σ)g(v∗ )|τ−v∗ f ( )| dσdv∗ dv1 = b( |v1 | 2 Z (8.66) v v + |v|σ 2 · σ)g(v∗ )|τ−v∗ f ( )| dσdv∗ dv = b( |v| 2 Z n−1 2 = b(ψ(v 0 , σ)g(v∗ ) v0 |τ−v∗ f (v 0 )|2 dσdv∗ dv 0 , ( |v0 | · σ)2 where v0 ψ(v 0 , σ) = 2( 0 · σ)2 − 1, τ−v∗ f = f (v∗ +). |v | R Using the fact that S n−1 b(k · σ) does not depend on k and |F (τh f )| = |F (f )|, and reversing the change of variables ξ → ξ + as in Lemma 8.1, we have Z Z 2n−1 (8.66) = g(v∗ ) b(ψ(ξ, σ) ξ |f (ξ)|2 dσdξ dv∗ ( |ξ| · σ)2 (8.67) Z ξ + 2 · σ)|f (ξ )| dσdξ. = gˆ(0) b( |ξ|

Collecting (8.63), (8.64), (8.65) and (8.67), we can obtain (8.62).

Corollary 8.1. Assume g ∈ L1 (Rn ), f ∈ L2 (Rn ) and f > 0. Then Z v − v∗ b( · σ)g(v∗ )(f (v 0 ) − f (v))2 dσdv∗ dv |v − v∗ | 2n n−1 R ×S Z Z (8.68) ξ − 2 ˆ > |f (ξ)| b( · σ)(ˆ g (0) − |ˆ g(ξ )|)dσdξ. |ξ| Rn S n−1 Proof.

Using Lemma 8.2 and the following inequality, |fˆ(ξ + )|2 + |fˆ(ξ)|2 > |fˆ(ξ)|2 ,

we can obtain Corollary 8.1.

Lemma 8.3. Assume that b satisfies (8.21), Then there exists a positive constant Cg depending on n, kgkL11 , kgkL log L and b, such that for |ξ| > 1 Z ξ b( · σ)(ˆ g (0) − |ˆ g (ξ − )|)dσ > Cg |ξ|ν . (8.69) |ξ| S n−1

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Lemma 8.3 is a consequence of the two lemmas below, next we will use Cg to denote a positive constant depending on n, kgkL11 , kgkL log L and b. Lemma 8.4. There exists a positive constant C˜g depending on n, kgkL11 and kgkL log L , such that for all ξ ∈ R gˆ(0) − gˆ(ξ) > C˜g (|ξ|2 ∧ 1).

Proof.

(8.70)

For some θ ∈ R, we have

gˆ(0) − gˆ(ξ) Z = g(v)(1 − cos(v · ξ + θ))dv Rn Z v·ξ+θ )dv =2 g(v) sin2 ( 2 Rn Z > 2 sin2 ε

(8.71)

g(v)dv

|v|6r,∀p∈Z,|v·ξ+θ−2pπ|>2ε

Z n o kgkL11 > 2 sin2 ε kgkL1 − − g(v)dv r |v|6r,∀p∈Z,|v·ξ+θ−2pπ|62ε Z n o 1 kgk L 1 > 2 sin2 ε kgkL1 − − sup g(v)dv . r |A|6 4ε (2r)n−1 (1+ r|ξ| ) A π

|ξ|

If |ξ| > 1, Lemma 8.4 holds with the following constant Z o n kgkL11 − sup g(v)dv , C˜g = 2 sin2 ε kgkL1 − r |A|64ε(2r)n−1 + 2ε (2r)n A

(8.72)

π

ε > 0 and r > 0 being chosen in such a way that this quantity is positive. ε If |ξ| 6 1, let δ = |ξ| in (8.71), Lemma 8.4 holds with the following constant Z o kgkL11 sin2 (δ|ξ|) n 2 1 − Cg = 2δ inf kgk − sup g(v)dv , L r δ 2 |ξ|2 r |ξ|61 |A|64ε(2r)n−1 (1+ π ) A

(8.73) where δ > 0 and r > 0 are chosen in such a way that this quantity is positive. Lemma 8.5. If b satisfies sinn−2 θb(cos θ) ≈ then for |ξ| > 1,

Z

S n−1

b(

K(ν) as θ → 0, K(ν) > 0; θ1+ν

ξ · σ)(|ξ − |2 ∧ 1)dσ > K(ν)|ξ|ν . |ξ|

(8.74)

(8.75)

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Proof.

Recall that |ξ − |2 =

ξ |ξ|2 (1 − · σ). 2 |ξ|

Passing to n-dimensional spherical coordinates, we find for some θ0 > 0, Z ξ b( · σ)(|ξ − |2 ∧ 1)dσ |ξ| S n−1 Z π2 |ξ|2 = S n−2 sinn−2 θb(cos θ) (1 − cos θ) ∧ 1 dθ (8.76) 2 0 Z θ0 K |ξ|2 θ2 dθ ( ∧ 1) 1+ν . > |S n−2 | 2 2 θ 0

By the change of variables θ → |ξ|θ, the integral in (8.76) is also Z θ0 2 θ dθ |ξ|ν ( ∧ 1) 1+ν 2 θ 0 so that when |ξ| > 1, Lemma 8.5 holds with Z θ0 2 K θ dθ K(v) = |S n−2 | ( ∧ 1) 1+ν . 2 2 θ 0

(8.77)

By Lemma 8.2, Corollary8.1 and Lemma 8.3, for the second term in (8.51), we have Z v − v∗ · σ)g(v∗ )(f (v 0 ) − f (v))2 dσdv∗ dv > Cg kf k2H ν/2 . b( |v − v∗ | R2n ×S n−1 (8.78) 8.4

Regularity of solutions for spatially homogeneous case

In this section, we will give the application of the results of Section 3 in this chapter; and prove the regularity of solutions for the spatially homogeneous Boltzmann equation without angular cutoff ∂t f (t, v) = Q(f, f )(t, v), t > 0, v ∈ Rn , (8.79) f (0, v) = f0 (v), where cross section B B(|v − v∗ |,

v − v∗ v − v∗ · σ) = |v − v∗ |γ b( · σ), |v − v∗ | |v − v∗ |

sinn−2 θb(cos θ) ≈

K(ν) when θ → 0, K(ν) > 0. θ1+ν

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From (8.33), it follows that the solutions f (t, v) of the Cauchy problem (8.79) have the following conservation of mass, momentum and energy, Z Z f (t, v)dv = f0 (v)dv, (8.80) Rn

Z

Rn

f (t, v)vj dv =

Rn

Z

f (t, v)

Rn

Z

f0 (v)vj dv, j = 1, 2...n,

(8.81)

Rn

|v|2 dv = 2

Z

f0 (v)

Rn

|v|2 dv. 2

(8.82)

For the existence of weak solutions for Cauchy problem (8.79), in 1998 Villani [230] showed that if the initial data have the finite mass, energy and entropy, Z f0 (v)[1 + |v|2 + log(1 + f0 (v))]dv < +∞, (8.83) Rn

then he constructed a weak solution f (t, v) of the Cauchy problem (8.79), which satisfies the following: f (t, v) > 0, f (t, v) ∈ C(R+ , S 0 ); t > 0, f (t, v) ∈ L12 ∩ LlogL,

Z

f (t, v)ψ(v)dv =

Rn

f (t, v) ∈ L1 ([0, T ]; L12+γ ),

(8.85)

f (0, v) = f0 (v),

(8.86)

Z

f0 (v)ψ(v)dv, ψ(v) = 1, v1 , ..., vn ,

Rn

Z

Rn

Z

(8.84)

f (t, v)logf (t, v)dv 6

Z

f0 (v)logf0 (v)dv,

|v|2 , 2

(8.87) (8.88)

Rn

Z Z t Z f (t, v)ϕ(t, v)dv − f0 (v)ϕ(0, v)dv − f (τ, v)∂τ ϕ(τ, v)dv dτ Rn Rn 0 Rn Z t Z dτ Q(f, f )(τ, v)ϕ(τ, v)dv, ∀ϕ(t, v) ∈ C 1 (R+ ; C0∞ (Rn )), (8.89) = 0

Rn

where the last integral in the right-hand side being defined by the following formulae Z Q(f, f )(τ, v)ϕ(τ, v)dv Rn Z Z 1 = Bf (v∗ )f (v)(ϕ(v∗0 ) + ϕ(v 0 ) − ϕ(v∗ ) − ϕ(v))dvdv∗ dσ. 2 R2n S n−1

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These make sense provided that f satisfies (8.84) and test function ϕ ∈ L∞ ([0, T ]; W 2,∞ ). In this section, we assume that a weak solution to the Boltzmann equation (8.79) has already been constructed and that it satisfies the usual entropic estimates, to say, Z f (t, v)[1 + |v|2 + log(1 + f (t, v))]dv < +∞, (8.90) Rn

we are interested in regularity issues associated to such solutions. That is, is this weak solution more regular than the initial datum, and if so, can we have estimates on this regularity? In this section, we will show that the weak solutions above constructed f (t, v) of the Cauchy problem (8.79) with Maxwellian molecules case belongs to Hv+∞ . For hard potentials case, there exist some results only in modified hard potentials cases now, in details, one can refer to [4; 60; 111]. For soft potentials cases, there seems no any results now. In this section, we will introduce two methods of multiplier to consider it. That is, we first choose a team of suitable multiplier, then mainly give the sharp estimates of the commutators of the collision operator Q(f, f ) and pseudo-differential operators composed by multipliers. First, we give the method of the Littlewood-Paley decomposition, in details, one can refer to [3]. The Littlewood-Paley decomposition is defined as in Section 3 in Chapter 1: Assume that supp ϕk ⊂ {ξ : 2k−1 6 |ξ| 6 P∞ 2k+1 }, k ∈ N, supp ϕ0 ⊂ {ξ : |ξ| 6 2}, k=0 ϕk = 1; we can assume ϕk is a radial function. Recall that for every multi-index α, there exists a positive number Cα such that 2k|α| |Dα ϕk (ξ)| 6 Cα , k = 0, 1, 2...; ξ ∈ Rn .

(8.91)

Then the Littlewood-Paley projection operator is defined by n d ˆ 4 k f (ξ) = ϕk (ξ)f (ξ), k = 0, 1, 2...; ξ ∈ R .

Lemma 8.6. Assume that the initial data f0 (v) satisfies (8.83), cross section B(|v − v∗ |, k · σ) = b(k · σ) with b satisfying the following K sinn−2 θb(cos θ) ≈ 1+ν when θ → 0, K > 0. (8.92) θ Let f (t, v) be any weak non negative solution of the Cauchy problem (8.79), satisfying (8.80)-(8.82) and (8.90). Then (4k Q(f, f ), 4k f )L2 − (Q(f, 4k f ), 4k f )L2 6 Ckf0 kL1 (

k+1 X

j=k−2

k4k f kL2 ), (8.93)

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(4k f, 4k Q(f, f ))L2 − (4k f, Q(f, 4k f ))L2 6 Ckf0 kL1 (

k+1 X

j=k−2

k4k f kL2 ). (8.94)

Remark 8.2. Let f (t, v) be a weak solution of the Boltzmann equation. If we choose 4k f as test function (4k f ∈ L∞ ([0, T ]; W 2,∞ )), then by the definition of weak solutions, the inner products in the left side of (8.93) and (8.94) make sense. Unless f ∈ H ∞ (Rn ), f (t, v) cannot be chosen as a test function in the definition of weak solutions. This is also one of the reasons why we use the method of multiplier. Proof. We only prove (8.93), the proof of (8.94) is similar with one of (8.93). Applying Fourier transform on Boltzmann equation (8.79), using (8.49) and (8.50) (Bobylev identity), we have Z ξ \ ∂t fˆ(ξ) = Q(f, f )(ξ) = b( · σ) fˆ(ξ − )fˆ(ξ + ) − fˆ(0)fˆ(ξ) dσ. (8.95) |ξ| S n−1 Multiplying (8.95) by ϕk (ξ), we have \ d ∂t 4 k f (ξ) = 4k Q(f, f )(ξ) Z ξ · σ) fˆ(ξ − )fˆ(ξ + )ϕk (ξ) − fˆ(0)fˆ(ξ)ϕk (ξ) dσ. = b( |ξ| S n−1 (8.96) Then, d d ∂t (4 k f , 4k f )L2 =

Z

Rn

d d (∂t 4 k f )4k f dξ +

Z

Rn

d d 4 k f ∂t 4k f dξ

\ d d = (4k\ Q(f, f ), 4 k f )L2 + (4k f , 4k Q(f, f ))L2 Z Z ξ = b( · σ) fˆ(ξ − )fˆ(ξ + )ϕ2k (ξ)fˆ(ξ) − fˆ(0)fˆ(ξ)ϕ2k (ξ)fˆ(ξ) dσdξ |ξ| n n−1 ZR ZS ξ + b( · σ) fˆ(ξ − )fˆ(ξ + )ϕ2k (ξ)fˆ(ξ) − fˆ(0)fˆ(ξ)ϕ2k (ξ)fˆ(ξ) dσdξ. |ξ| Rn S n−1 (8.97)

Moreover, we have Z \ Q(f, 4k f )(ξ) =

S n−1

b(

ξ · σ) fˆ(ξ − )ϕk (ξ + )fˆ(ξ + ) − fˆ(0)ϕk (ξ)fˆ(ξ) dσ, |ξ| (8.98)

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\ \ d d (Q(f, 4k f ), 4 k f )L2 + (4k f , Q(f, 4k f ))L2 ZZ ξ = b( · σ) fˆ(ξ − )ϕk (ξ + )fˆ(ξ + )ϕk (ξ)fˆ(ξ) − fˆ(0)fˆ(ξ)ϕ2k (ξ)fˆ(ξ) dσdξ |ξ| ZZ ξ + b( · σ) fˆ(ξ − )ϕk (ξ + )fˆ(ξ + )ϕk (ξ)fˆ(ξ) − fˆ(0)fˆ(ξ)ϕ2k (ξ)fˆ(ξ) dσdξ. |ξ| (8.99) Thus, from (8.97) and (8.99) it follows that \ d d (4k\ Q(f, f ), 4 k f )L2 − (Q(f, 4k f ), 4k f )L2 Z Z ξ = b( · σ)fˆ(ξ − )fˆ(ξ + )(ϕk (ξ) − ϕk (ξ + ))ϕk (ξ)fˆ(ξ)dσdξ. |ξ| n n−1 R S (8.100) From the definitions of ξ + and ξ − , it follows that ξ + |ξ|σ θ ξ π ξ+ = , |ξ + | = |ξ| cos , · σ = cos θ, 0 6 θ 6 , 2 2 |ξ| 2 |ξ|2 θ 6 |ξ + |2 6 |ξ|2 , |ξ|2 − |ξ + |2 = |ξ − |2 = |ξ|2 sin2 . 2 2 From the integral in (8.100), we note that 2k−1 6 |ξ| 6 2k+1 . From the definition of ξ + , it follows that 2k−2 6 |ξ + | 6 2k+1 . Thus + d + \ + \ \ + fˆ(ξ + ) = 4 k−2 f (ξ ) + 4k−1 f (ξ ) + 4k f (ξ ) + 4k+1 f (ξ ).

(8.101)

Plugging this expression of f (ξ + ) in (8.100), and by the fact that |fˆ(ξ − )| 6 kf kL1 6 kf0 kL1 , (8.100) is bounded by Z Z ξ + k d d kf0 kL1 b( · σ)|4 k f (ξ )||Aξ ||4k f (ξ)|dσdξ |ξ| k−1 k+1 n−1 2 6|ξ|62 S Z Z ξ + k d \ 1 + kf0 kL b( · σ)|4 k−2 f (ξ )||Aξ ||4k f (ξ)|dσdξ |ξ| k−1 k+1 n−1 2 6|ξ|62 S Z Z ξ + k d \ + kf0 kL1 b( · σ)|4 k−1 f (ξ )||Aξ ||4k f (ξ)|dσdξ |ξ| k−1 k+1 n−1 2 6|ξ|62 S Z Z ξ + k d \ + kf0 kL1 b( · σ)|4 k+1 f (ξ )||Aξ ||4k f (ξ)|dσdξ, |ξ| k−1 k+1 n−1 2 6|ξ|62 S (8.102) where Akξ = ϕk (ξ) − ϕk (ξ + ). By (8.91), we have

|Akξ | = |ϕk (ξ) − ϕk (ξ + )| 6 |ϕ˜k (|ξ|) − ϕ˜k (|ξ + |)| .

1 1 |ξ|2 − |ξ + |2 θ + (|ξ| − |ξ |) . . sin2 . 2k 2k |ξ| + |ξ + | 2

(8.103)

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We only estimate the first term in (8.102), the estimates of the other terms in (8.102) can be obtained similarly. By (8.103) and H¨older inequality, (8.102) is bounded by Z Z ξ + k d d b( · σ)|4 kf0 kL1 k f (ξ )||Aξ ||4k f (ξ)|dσdξ |ξ| 2k−1 6|ξ|62k+1 S n−1 Z Z ξ θ d + d · σ) sin2 |4 . kf0 kL1 |4 b( k f (ξ)| k f (ξ )|dσdξ |ξ| 2 Rn S n−1 Z 1/2 nZ Z 2 o1/2 ξ θ d + 2 d . kf0 kL1 |4 b( · σ) sin2 |4 dξ k f (ξ)| dξ k f (ξ )|dσ |ξ| 2 . kf0 kL1 k4k f k2L2 .

(8.104)

which follows from the change of variables ξ → ξ

+

as in Lemma 8.1.

Theorem 8.1. Under the hypothesis of Lemma 8.6, Let f (t, v) be any weak non negative solution of the Cauchy problem (8.79), satisfying (8.90). Then for any s ∈ R+ and t > 0, one has f (t, v) ∈ H s (Rn ). Proof.

First, we have

\ d d ∂t k4k f kL2 = (4k\ Q(f, f ), 4 k f )L2 + (4k f , 4k Q(f, f ))L2 .

(8.105)

From (8.51) it follows that Q(f, 4k f ), 4k f 2 L Z v − v∗ · σ)f (v∗ )4k f (v)(4k f (v 0 ) − 4k f (v))dσdv∗ dv = b( |v − v∗ | R2n ×S n−1 Z 1 v − v∗ = b( · σ)f (v∗ )((4k f )2 (v 0 ) − (4k f )2 (v))dσdv∗ dv 2 R2n ×S n−1 |v − v∗ | Z 1 v − v∗ − b( · σ)f (v∗ )(4k f (v 0 ) − 4k f (v))2 dσdv∗ dv 2 R2n ×S n−1 |v − v∗ | = I1 − I2 .

(8.106)

For the first term in (8.106), using Lemma 8.1 in this chapter, we get Z I1 = b(k · σ)((4k f )2 (v 0 ) − (4k f )2 (v))dσdvdv∗ 2n n−1 R ×S Z (8.107) =A (4k f )2 (v∗ )dv∗ , Rn

where

A = S n−2

Z

π 2

0

sinn−2 θ

1 cosn

θ 2

− 1 b(cos θ)dθ.

(8.108)

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From 1 − cosn

θ 2

6 n(1 − cos 2θ ) = 2n sin2

θ 4

it follows that

I1 . k4k f k2L2 .

(8.109)

For the second term in (8.106), using Corollary 8.1 and Lemma 8.3 in this chapter, we have Z v − v∗ 1 b( · σ)f (v∗ )(4k f (v 0 ) − 4k f (v))2 dσdv∗ dv I2 = 2 R2n ×S n−1 |v − v∗ | Z > Cf0 (1 + |ξ|ν )ϕk (ξ)fˆ(ξ)dξ > Cf0 k4k f k2 ν2 . H

Rn

(8.110)

Thus collecting (8.105), (8.106), (8.109), (8.110) and Lemma 8.6, we get ∂t k4k f kL2 + Cf0 k4k f k2H ν2 6 Ckf0 kL1 (

k+1 X

j=k−2

k4j f k2L2 ).

(8.111)

Dividing(8.111) by 2kn , we have k4k f k2L2 ∂t k4k f kL2 + (Cf0 2kν − Ckf0 kL1 ) kn 2 2kn k4 2 2 k4k−1 f kL2 k4k+1 f k2L2 k−2 f kL2 6 Ckf0 kL1 + + . 2(k−2)n 2(k−1)n 2(k+1)n

(8.112)

k4k f k2

L2 , Ck = Cf0 2kν − Ckf0 kL1 and β = Ckf0 kL1 . Then Let Uk (t) = 2kn (8.112) can be rewritten by:

∂t Uk (t) + Ck Uk (t) 6 β(Uk−2 (t) + Uk−1 (t) + Uk+1 (t)).

(8.113)

From Bernstein’s inequality (Polyya Plancherel Nikoolski inequality), it follows that k4k f kL2 6 C(ϕ)2nk k4k f kL1 6 C(ϕ)2nk kf kL1 6 C(ϕ)2nk kf0 kL1 . For k > 0, t > 0, Uk (t) 6 M, with M = C(ϕ)kf0 kL1 .

(8.114)

By the definition of Ck , we show that Ck is nondecreasing, Ck > 0 with large enough k (without loss of generality, we can assume that Ck > 0 with k > k0 > 3). From (8.113) and (8.114), it follows that Uk (t) satisfy the conditions of the following Lemma 8.7. Then for all integer p > 1, there exist constants Ap and Dp such that Uk (t) 6 M Ap e−Ck−2(p−1)t + M Dp

1 ,t > 0 (Ck−2(p−1) )p

(8.115)

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for all k > 2(p − 1) + k0 + 2. Thus, for fixed s > 0, t > 0 and an integer p > 1 which we shall choose just below; using the definition of Sobolev spaces, we have ∞ X kf k2H s ∼ 22ks k4k f k2L2 k=0

2(p−1)+k0 +1

=

X

22ks k4k f k2L2 +

k=0 2(p−1)+k0 +1

=

X

k=0

22ks k4k f k2L2 +

∞ X

k=2(p−1)+k0 +2 ∞ X

22ks k4k f k2L2 (8.116) 2k(2s+n) Uk .

k=2(p−1)+k0 +2

In order to prove f ∈ H s , it remains to show that the last series appearing above is a convergent one. From (8.115) it follows that ∞ X 2k(2s+n) Uk k=2(p−1)+k0 +2

6

∞ X

2k(2s+n) M Ap e−Ck−2(p−1)t

(8.117)

k=2(p−1)+k0 +2

+

∞ X

2k(2s+n) M Dp

k=2(p−1)+k0 +2

1 . (Ck−2(p−1) )p

Using the definition of Ck , we have Ck−2(p−1) ∼ C(f0 , p)2kνp , if k is large enough.

(8.118)

Choosing p such that νp > 2s + n + 1 for instance, yields that the two series on the right hand side of the last inequality are convergent. Lemma 8.7. (Iteration) Let β, M be two non negative numbers. Given positive integer k0 , assume that {Ck }k>k0 is a sequence of positive numbers, and is non-decreasing in k, and satisfies: there exists a positive α such that Ck+1 − Ck > α for all k > k0 . {Uk }k>k0 = {Uk (t)}k>k0 is another sequence with t ∈ R+ satisfying: ∂t Uk (t) + Ck Uk (t) 6 β(Uk−2 (t) + Uk−1 (t) + Uk+1 (t)), 0 6 Uk (t) 6 M, ∀k > k0 + 2, ∀t > 0.

Then for any integer p > 1, there exist constants Ap , Dp such that, for all k > 2(p − 1) + k0 + 2, one has 1 , t > 0. Uk (t) 6 M Ap e−Ck−2(p−1)t + M Dp (Ck−2(p−1) )p

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Proof. It is done by iteration. First start from the fact that Uk 6 M , and using this in the right hand side of the differential inequality, leads to the above conclusion, for p = 1. Then, start from the above first iteration, and replace again in the rhs of the above inequality, and just repeat the process. Next, we will introduce another method of multiplier, in details, one can refer to [171]. For convenience, we assume n = 3. Let f (t, v) be a weak solution of the Cauchy problem (8.79). For any fixed T0 > 0, it follows that f (t, v) ∈ L1 (R3 ) ⊂ H −2 (R3 ), t ∈ [0, T0 ]. For t ∈ [0, T0 ], N > 0 and 0 < δ < 1, define multiplier: N t−4 N T0 + 4 Mδ (t, ξ) = (1 + |ξ|2 ) 2 (1 + δ|ξ|2 )−N0 , N0 = 2 and the corresponding pseudo-differential operator Mδ (t, Dv ) is defined by Mδ (t, Dv ) = F −1 Mδ (t, ξ)F . Then for any δ ∈ (0, 1),

Mδ (t, Dv )2 f ∈ L∞ ([0, T0 ]; W 2,∞ (R3 )), Mδ (t, Dv )f ∈ C([0, T0 ]; L2 (R3 )).

Lemma 8.8. Under the hypothesis of Lemma 8.6, Let f (t, v) be any weak non negative solution of the Cauchy problem (8.79), satisfying (8.90). Then (Q(f, f ), Mδ2 f )L2 − (Q(f, Mδ f ), Mδ f )L2 6 Cf kf0 kL1 kMδ2 f kL2 ,

(8.119)

where the constant Cf is independent of 0 < δ < 1. Proof.

Similarly with the proof of Lemma 8.6, we have

(Q(f, f ), Mδ2 f )L2 − (Q(f, Mδ f ), Mδ f )L2

2 [ \ \ [ = (Q(f, f ), M δ f )L2 − (Q(f, Mδ f ), Mδ f )L2 Z Z ξ = b( · σ)fˆ(ξ − )fˆ(ξ + )(Mδ (t, ξ) − Mδ (t, ξ + ))Mδ (t, ξ)fˆ(ξ)dσdξ. |ξ| n n−1 R S (8.120)

It suffices to prove Mδ (t, ξ) − Mδ (t, ξ + ) 6 N0 2

N T0 +4 2

θ Mδ (t, ξ + ) sin2 . 2

Define ˜ δ (t, s) = (1 + s) N t−4 2 M (1 + δs)−N0 , s = |ξ|2 , s+ = |ξ + |2 , so that ˜ δ (t, |ξ|2 ). Mδ (t, ξ) = M

(8.121)

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From the mean value theorem, it follows that there exists s+ < s˜ < s such that ˜ ˜ δ (t, s) − M ˜ δ (t, s+ ) = ∂ Mδ (t, s˜)(s − s+ ). M ∂s ˜ δ, By the definition of M Nt − 4 ˜δ ∂M δN0 ˜ (t, s) = − Mδ (t, s). ∂s 2(1 + s) 1 + δs

By the following

we have

M N T0 +4 s θ δs ˜ δ (t, s˜) , 6 1; 6 2 2 , s − s+ = s sin2 , + ˜ 1 + s 1 + δs 2 Mδ (t, s ) ˜ δ (t, s) − M ˜ δ (t, s+ ) 6 N0 2 M

Then Z Z

N T0 +4 2

˜ δ (t, s+ ) sin2 θ . M 2

ξ · σ)fˆ(ξ − )fˆ(ξ + )(Mδ (t, ξ) − Mδ (t, ξ + ))Mδ (t, ξ)fˆ(ξ)dσdξ |ξ| Z Z ξ θ 6 b( · σ) sin2 |fˆ(ξ − )||fˆ(ξ + )|Mδ (t, ξ + )Mδ (t, ξ)|fˆ(ξ)|dσdξ |ξ| 2 n n−1 R S Rn

b(

S n−1

6 Ckf0 kL1 kMδ f k2L2 .

Thus using Lemma 8.8, we give another proof of Theorem 8.1. We note that any weak solution f has the following properties: Mδ2 f ∈ L∞ ([0, T0 ]; W 2,∞ (R3 )) and Mδ f ∈ C([0, T0 ]; L2 (R3 )). From the definition of weak solution, we choose ψ(t, v) = Mδ2 f (t, v) as test function. For any t ∈ (0, T0 ), Z Z f (t, v)Mδ2 f (t, v)dv − f0 (v)Mδ2 f (0, v)dv R3 R3 Z t Z − dτ f (τ, v)∂τ (Mδ2 f (τ, v))dv (8.122) 0 R3 Z t Z = dτ Q(f, f )(τ, v)Mδ2 f (τ, v)dv. R3

0

Moreover, we have Z t Z dτ f (τ, v)∂τ (Mδ2 f (τ, v))dv R3 t

0

= lim

h→0

Z

0

dτ

Z

R3

(f (τ, v) + f (τ + h, v))

Mδ2 f (τ + h, v) − Mδ2 f (τ, v) dv 2h

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= lim

h→0

Z

t

dτ

0

n

Z

R3

M f (τ + h, v) δ Mδ (τ + h)f (τ, v) + Mδ (τ + h)f (τ + h, v) 2h M f (τ, v) o δ − Mδ (τ )f (τ, v) + Mδ (τ )f (τ + h, v) dv 2h Z t Z (Mδ f )2 (τ + h, v) − (Mδ f )2 (τ, v) dv = lim dτ h→0 0 2h R3 Z t Z 1 n + lim dτ Mδ (τ + h)(f (τ, v))Mδ (τ + h)(f (τ + h, v)) h→0 0 R3 2h o − Mδ (τ )(f (τ, v))Mδ (τ )(f (τ + h, v)) dv

= J1 + J2 . For J1 , we have

(8.123)

Z

t

Z

(Mδ f )2 (τ + h, v) − (Mδ f )2 (τ, v) dv h→0 0 2h R3 Z Z h oZ 1 n t+h = lim dτ − dτ (Mδ f )2 (τ, v)dv h→0 2h 0 0 R3 Z Z 1 1 = (Mδ f )2 (t, v)dv − (Mδ f )2 (0, v)dv. (8.124) 2 R3 2 R3 For J2 , we have Z t Z 1 n Mδ (τ + h)(f (τ, v))Mδ (τ + h, v)(f (τ + h, v)) J2 = lim dτ h→0 0 R3 2h o − Mδ (τ )(f (τ, v))Mδ (τ )(f (τ + h, v)) dv Z t Z 1 n = lim dτ f (τ, v)Mδ2 (τ + h, v)(f (τ + h, v)) h→0 0 R3 2h o − f (τ, v)Mδ2 (τ )(f (τ + h, v)) dv Z t Z n o 1 = lim dτ f (τ, v) Mδ2 (τ + h) − Mδ2 (τ ) (f (τ + h, v)) dv h→0 0 R3 2h Z t Z 1 = dτ f (τ, v)(∂τ Mδ2 (τ ))(f (τ, v))dv. (8.125) 2 0 3 R By Z(8.123), (8.124) and (8.125), Z (8.122) can be rewritten by 1 1 2 f (t, v)Mδ f (t, v)dv − f0 (v)Mδ2 f (0, v)dv 2 R3 2 R3 Z t Z Z t Z = dτ f (τ, v)(∂τ Mδ2 )(f (τ, v))dv + dτ (Q(f, f ), Mδ2 f )L2 (τ, v)dv. J1 = lim

0

R3

dτ

0

R3

(8.126)

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Similarly with the first proof of Theorem 8.1, or by the content of Section 3 in this chapter, we have n o (8.127) kMδ f k2H ν2 6 Cf (−Q(f, Mδ f ), Mδ f )L2 + kMδ f k2L2 . From Lemma 8.8, it follows that n o ef (−Q(f, f ), M 2 f )L2 + kMδ f k2 2 . kMδ f k2H ν2 6 C δ L

(8.128)

Since

∂t Mδ (t, ξ) = N Mδ (t, ξ) loghξi, we obtain Z t Z Z t 1 2 k(log Λ) 2 Mδ f (τ )k2L2 dτ, dτ f (τ, v)(∂τ Mδ (τ ))(f (τ, v))dv 6 2N 0

R3

0

(8.129)

−1

−1

where log Λ = F loghξiF and Λ = F hξiF . This together with (8.126) and (8.128), imply Z t ν 1 kMδ f (t)k2L2 + kΛ 2 Mδ f (τ )k2L2 dτ 2Cf 0 Z t Z t 1 2 2 2 6 kMδ f (0)kL2 + 2N k(log Λ) Mδ f (τ )kL2 dτ + kMδ f (τ )k2L2 dτ. 0

0

(8.130)

Using loghξi 6 hξi and Gagliardo-Nirenberg interpolation inequality, for any ε > 0, we have Z t 1 ν 2 −ε kΛ 2 Mδ f (τ )k2L2 dτ kMδ f (t)kL2 + 2Cf 0 (8.131) Z t 2 2 6 kMδ f (0)kL2 + Cε,N kMδ f (τ )kL2 dτ. 0

1 4Cf

By choosing ε = > 0, there exists a constant Cf,N depending only on Cf , N , T0 and being independent of δ ∈ (0, 1), such that for any t ∈ (0, T0 ), Z t (8.132) kMδ f (t)k2L2 6 kMδ f (0)k2L2 + Cf,N kMδ f (τ )k2L2 dτ. 0

Then Gronwall inequality yields

kMδ f (t)k2L2 6 eCf,N,t kMδ (0)f0 )k2L2 . Since kMδ f (t)kL2 = k(1 − δ∆)−N0 f (t)kH N t−4 ,

(8.133)

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Boltzmann equation without angular cutoff

and kMδ (0)f0 kL2 (R3 ) = k(1 − δ∆)−N0 f0 kH −4 (R3 ) 6 kf0 kH −4 (R3 ) 6 Ckf0 kL1 (R3 ) , we obtain ˜ Cf,N,t kf0 kL1 (R3 ) , k(1 − δ∆)−N0 f (t)kH N t−4 6 Ce

where the constant C˜ is independent of δ ∈ (0, 1). Finally, for any given t > 0 since N can be arbitrarily large, by letting δ → 0, we have f (t) ∈ H +∞ (R3 ). Remark 8.3. Recently, the existence of the classical global solutions of the Boltzmann equation without angular cut-off was obtained by Gressman and Strain [83]. There are some recent works by Chen, Li and Xu [34; 36] which have been devoted to the study of the regularity for the Landau equation, which can be regarded as a limit of the Boltzmann equation when the collisions become grazing. Mouhot and Villani [172] for the first time establish the Landau damping in a nonlinear context.

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Appendix A

Notations

We list some notations used in this book, most of them are familiar for the PDE readers. (1) R = the set of real, C = the set of complex number, N = the set of natural number, Z = the set of integers, Z+ = N ∪ {0}, R+ = [0, ∞). (2) a ∨ b = max(a, b); a ∧ b = min(a, b). (3) p0 denotes the duality number of p, i.e., 1/p+1/p0 = 1, ∀ p ∈ [1, ∞]. (4) C > 1 and 0 < c < 1 express universal constants that can be different at different places. (5) A . B means A 6 CB; A ∼ B means A . B and B . A. (6) Lp := Lp (Rn ) is the Lebesgue space, Z 1/p kf kp := kf kLp(Rn ) = |f (x)|p dx Rn

(7) Let f = (f1 , ..., fn ) be a vector function, kf kp = (kf1 k2p + ... + kfn k2p )1/2 . (8) For x = (x1 , ..., xn ) ∈ Rn , we write |x| = (|x1 |2 + ... + |xn |2 )1/2 , and sometimes we also denote |x| = |x1 | + ... + |xn |. (9) Let A ⊂ Rn , we denote by |A| the measure of A. (10) Let T and S be operators, we use the notation T ∼ S to express that T can be roughly regarded as S (when S is easily understood).

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Appendix B

Definition of scattering operator

We give an exact definition of the scattering operator by the taking NLS (7.54) as an example. Definition B.1. Let X be a Banach space, S(t) = eit∆ be the evolution semigroup associated to (7.54), and u(t) be the global solution of (7.54) with the initial datum u0 ∈ X. If the limit u+ = lim S(−t)u(t) t→∞

exists in X, we say that u+ is the asymptotic state of u0 at +∞. Also, if the limit u− = lim S(−t)u(t) t→−∞

exists in X, we say that u− is the asymptotic state of u0 at −∞. In other words, u(t) behaves as t → ±∞ like the solutions S(t)u± of the linear Schr¨ odinger equation (or the free Schr¨odinger equation, i.e. iut + ∆u = 0). −1 The inverse operators Ω± = U± of the operators U± : u0 7→ u± are called the forward/backward wave operators. Note that the uniqueness aspect of the H 1 -wellposedness theory ensures that the wave operators are injective. If they are also surjective, in other word, if every H 1 -wellposed solution is global and scatters in H 1 as t → +∞, we say that we also have the asymptotic completeness. If the forward wave operator and the backward wave operator exist simultaneously, the mapping S = Ω−1 + ◦ Ω− : u− 7→ u+ is called the scattering operator. Remark B.1. The scattering theory involves two essential factors: the existence of the forward/backward wave operators and the asymptotic completeness. Generally speaking, the existence of wave operators is relatively easy to establish as long as the power p is suitable, and especially if a smallness condition is assumed, both in focusing and defocusing cases. However, 261

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262

Definition of scattering operator

the asymptotic completeness is a bit harder and requires some decay estimates even in the defocusing case. Remark B.2. To reduce the number of cases slightly, we shall only consider scattering from t = 0 to t = +∞ or vice versa. One can certainly consider scattering back and forth between t = 0 to t = −∞, or between t = −∞ and t = +∞, but the theory is more or less the same in each of these cases. In general, for the well-posedness and the scattering theory of nonlinear Schr¨ odinger equations, we usually use the integral version of the equation, that is, the Duhamel formula Z t u(t) = S(t − t0 )u(t0 ) − i S(t − τ )(|u(τ )|p−1 u(τ ))dτ. (B.1) t0

For the scattering theory of the equation (7.54) in H 1 , we need to show the global solution of (7.54) with the initial datum u(0) = u0 ∈ H 1 scatters to a solution S(t)u+ of the associated linear equation as t → +∞ in H 1 , that is, or equivalently,

ku(t) − S(t)u+ kH 1 → 0, kS(−t)u(t) − u+ kH 1 → 0,

as t → +∞, as t → +∞.

In other words, we require that the function S(−t)u(t) converges in H 1 as t → +∞. From the Duhamel formula (B.1), we know Z t S(−τ )(|u(τ )|p−1 u(τ ))dτ. S(−t)u(t) = u0 − i 0

Thus, u scatters in H 1 as t → +∞ if and only if the improper integral Z ∞ S(−τ )(|u(τ )|p−1 u(τ ))dτ (B.2) 0

is conditionally convergent in H 1 , in which case the asymptotic state u+ is given by the formula Z ∞ u+ = u0 − i S(−τ )(|u(τ )|p−1 u(τ ))dτ. (B.3) 0

We can regard the asymptotic state u+ as a nonlinear perturbation of the initial state u0 . Comparing (B.1) with (B.3), we eliminate u0 to obtain the identity Z ∞ u(t) = S(t)u+ + i S(t − τ )(|u(τ )|p−1 u(τ ))dτ, (B.4) t

which can be viewed as the limiting case t0 = +∞ of (B.1).

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Appendix C

Some fundamental results

C.1

Gagliardo-Nirenberg inequality in Sobolev spaces

The Gagliardo-Nirenberg inequality is a fundamental tool in the study of partial differential equations, some special cases of which were discovered by Gagliardo [75], Ladyzhenskaya [153] and Nirenberg [181]. The general version can be stated as follows. Theorem C.1. Let 1 6 p, p0 , p1 6 ∞, `, m ∈ N∪{0}, ` < m, `/m 6 θ 6 1, ` 1−θ 1 m 1 = + +θ − . (C.1) p n p0 p1 n If m − ` − n/p0 is an integer, we further assume that `/m 6 θ < 1. Then for any u ∈ C0∞ (Rn ), X X kDα ukLp (Rn ) . kuk1−θ kDα ukθLp1 (Rn ) . (C.2) p n 0 L (R ) |α|=`

|α|=m

The proof of the Gagliardo-Nirenberg inequality is based on the globalderivative analysis in Lp spaces, which is rather complicated, see [74; 87], for instance.

C.2

Convexity H¨ older inequality in sequence spaces `sp

Analogous to the convexity H¨older inequality in Triebel-Lizorkin spaces, we have the following convexity H¨older inequality in sequence spaces `sp . Lemma C.1. Let 0 < q 6 ∞, −∞ < s1 , s0 < ∞ with s0 6= s1 , 0 < θ < 1, s = (1 − θ)s0 + θs1 . We have s1 j k2sj aj k`q . k2s0 j aj k1−θ aj kθ`∞ . (C.3) `∞ k2 263

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Proof. We can assume that {aj } = {aj }j>0 , aj > 0 and s1 < s0 . Put Ci = supj>0 2si j aj , i = 0, 1. It suffices to consider the case C1 > 0. We have C1 6 C0 . Take j0 such that C0 /2s0 j , j > j0 , min(C0 /2s0 j , C1 /2s1 j ) = C1 /2s1 j , j 6 j0 . It is easy to see that C0 ∼ C1 2(s0 −s1 )j0 . So,

s1 j k2s0 j aj k1−θ aj kθ`∞ ∼ C1 2(s0 −s1 )j0 (1−θ) . `∞ k2

(C.4)

On the other hand, aj 6 C0 /2s0 j , j > j0 ;

aj 6 C1 /2s1 j , 0 6 j 6 j0 .

A simple calculation yields k2sj aj k`q . C1 2(s0 −s1 )j0 (1−θ) , which implies the result, as desired. C.3

Inclusion spaces

between

(C.5)

homogeneous

Triebel-Lizorkin

The inclusion among homogeneous Triebel-Lizorkin spaces is very useful but Triebel [224] only claimed that it is probably true. Here we give the details of the proof, which is analogous to that of Theorem 1.2. Theorem C.2. Let 1 6 p1 < p2 < ∞, 1 6 r, q 6 ∞ and −∞ < s2 < s1 < ∞ satisfy s1 − n/p1 = s2 − n/p2 . Then we have F˙ps11,q ⊂ F˙ps22,r .

Proof.

(C.6)

By `r ⊂ `q for q > r, it suffices to show that F˙ ps11,∞ ⊂ F˙ps22,1 .

(C.7)

We can assume that kf kF˙ps1,∞ = 1. Recalling the equivalent norm on Lp , 1 we have ( ) Z ∞ X p2 2ks2 |(4k f )(x)| > t dt, kf kF˙ s2 ∼ tp2 −1 x : (C.8) p2 ,1 0 k∈Z

where |{· · · }| denotes the measure of the set {· · · }. It is easy to see that ∞ X

k=K+1

2ks2 |4k f | . 2K(s2 −s1 ) sup 2ks1 |4k f |. k

(C.9)

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C.4. Riesz-Thorin interpolation theorem

265

By Corollary 1.1, k4k f k∞ . 2kn/p1 k4k f kp1 . 2k(n/p1 −s1 ) kf kFps1,∞ . 1

(C.10)

Hence, for any K ∈ Z, K X

k=−∞

2ks2 |4k f | .

K X

2k(s2 −s1 +n/p1 ) . 2Kn/p2 .

(C.11)

k=−∞

Choosing K ∈ Z such that C2Kn/p2 ∼ t/2, we have 2K ∼ tp2 /n . If P ks2 |(4k f )(x)| > t, then it follows from (C.9) and (C.11) that k∈Z 2 ∞ X

C2K(s2 −s1 ) sup 2ks1 |4k f | > k∈Z

k=K+1

2ks2 |4k f | > t/2.

Collecting (C.8) and (C.12), we have Z ∞ kf kpF˙2s2 . tp2 −1 x : sup 2ks1 |(4k f )(x)| > ctp2 /p1 dt p2 ,1 k 0 Z ∞ p1 −1 ks1 dτ . τ x : sup 2 |(4 f )(x)| > τ k k

0

. 1,

which implies the result, as desired.

C.4

(C.12)

Riesz-Thorin interpolation theorem

Theorem C.3. Let 1 6 pi , qi 6 ∞, p0 6= p1 , q0 6= q1 satisfy T : Lpi → Lqi ,

i = 0, 1.

Assume that θ ∈ (0, 1) satisifies 1−θ θ 1 = + , p p0 p1

(C.13)

1 1−θ θ = + . q q0 q1

Then we have T : Lp → Lq and θ kT kLp→Lq 6 kT k1−θ Lp0 →Lq0 kT kLp1 →Lq1 .

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Hardy-Littlewood-Sobolev inequality

Up to now, the singular integration is still a difficult problem in the theory of harmonic analysis. Hardy-Littlewood-Sobolev inequality is a fundamental tool in this subject, cf. [202]. Let 0 < α < n, Z f (y) Iα f (x) = dy. |x − y|n−α n R Proposition C.1. Let 1 < p, q < ∞ and 0 < α < n satisfy 1/p = 1/q + α/n. Then we have kIα f kq . kf kp . C.6

(C.14)

Van der Corput lemma

Lemma C.2. Let ϕ ∈ C0∞ (R). Assume that P ∈ C 2 (R) satisfies that for any ξ ∈ supp ϕ, |P (k) (ξ)| > 1. Moreover, we assume that the following alternative condition holds (i) k > 2; (ii) k = 1 and P 0 (x) is a monotone function. Then we have Z eiλP (ξ) ϕ(ξ)dξ . λ−1/k (kϕk∞ + kϕ0 k1 ). C.7

Littlewood-Paley square function theorem

The Littlewood-Paley square function theorem is one of the most important results in the early stage of harmonic analysis. It seems that TriebelLizorkin spaces are also inspired by this theorem. Proposition C.2. Let 1 < p < ∞, s ∈ R. Then kukH˙ s ∼ kukF˙ s ,

s . kukHps ∼ kukFp,2

(C.15)

kukLp ∼ kukF˙ 0 ,

0 . kukLp ∼ kukFp,2

(C.16)

p

p,2

In particular, p,2

The proof of the Littlewood-Paley square function theorem can be found in [202].

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C.8. Complex interpolation in modulation spaces

C.8

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Complex interpolation in modulation spaces

We have the following Theorem C.4. Let 0 < p, q, pi , qi 6 ∞, s, si ∈ R with i = 0, 1 and s = (1 − θ)s0 + θs1 ,

1−θ θ 1 = + , p p0 p1

1 1−θ θ = + . q q0 q1

(C.17)

Then we have s (Mps00,q0 , Mps11,q1 )θ = Mp,q .

Recall that the result of Theorem C.4 is quite similar to Besov spaces, indeed, if (C.17) holds, then we have s (Bps00 ,q0 , Bps11 ,q1 )θ = Bp,q .

(C.18)

See [245]. C.9

Christ-Kiselev lemma

The Christ-Kiselev lemma (cf. [46]) is very useful for the study of nonlinear dispersive equations. There are a series of generalizations to the ChristKiselev lemma in recent years, see Molinet-Ribaud [166] and Smith-Sogge [201]. The following result is due to Wang-Han-Huang [243]. Denote Z ∞ Z t 0 0 0 T f (t) = K(t, t )f (t )dt , Tre f (t) = K(t, t0 )f (t0 )dt0 . (C.19) −∞

0

If T : Y1 → X1 implies that Tre : Y1 → X1 , then T : Y1 → X1 is said to be a well restricted operator. Proposition C.3. Let T be as in (C.19). We have the following conclusions. (1) If ∧3i=1 pi > (∨3i=1 qi ) ∨ (q1 q3 /q2 ), then T : Lqx11 Lqx22 Lqt 3 (R3 ) → Lpx11 Lpx22 Lpt 3 (R3 ) is a well restricted operator. (2) If q1 < ∧3i=1 pi , then T : Lqt 1 Lqx21 Lqx32 (R3 ) → Lpx11 Lpx22 Lpt 3 (R3 ) is a well restricted operator. (3) If p1 > (∨3i=1 qi ) ∨ (q1 q3 /q2 ), then T : Lqx11 Lqx22 Lqt 3 (R3 ) → Lpt 1 Lpx21 Lpx32 (R3 ) is a well restricted operator. (4) If ∧3i=1 pi > (∨3i=1 qi ) ∨ (q1 q3 /q2 ), then T : Lqx11 Lqx22 Lqt 3 (R3 ) → Lpx22 Lpx11 Lpt 3 (R3 ) is a well restricted operator.

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In higher spatial dimensions, we have similar results. Proposition C.4. Let T be as in (C.19). We have the following results. (1) If min(p1 , p2 , p3 ) > max(q1 , q2 , q3 , q1 q3 /q2 ), then T : Lqx11 Lqx22 ,...,xn Lqt 3 (Rn+1 ) → Lpx11 Lpx22 ,...,xn Lpt 3 (Rn+1 ) is a well restriction operator. (2) If p0 > (∨3i=1 qi ) ∨ (q1 q3 /q2 ), then

T : Lqx11 Lqx22 ,...,xn Lqt 3 (Rn+1 ) → Lpt 0 Lpx11 ...Lpxnn (Rn+1 )

is a well restriction operator. (3) If q0 < min (p1 , p2 , p3 ), then T : Lqt 0 Lqx11 ...Lqxnn (Rn+1 ) → Lpx11 Lpx22 ,...,xn Lpt 3 (Rn+1 ) is a well restriction operator. (4) If min(p1 , p2 , p3 ) > max(q1 , q2 , q3 , q1 q3 /q2 ), then T : Lqx12 Lqx21 ,x3 ,...,xn Lqt 3 (Rn+1 ) → Lpx11 Lpx22 ,...,xn Lpt 3 (Rn+1 ) is a well restriction operator.

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Index

s Bp,q ,9 ∞ C (Rn ), 2 s E2,1 , 45 s Fp,q , 9 Lp (`q ), 281 Lpx1i Lp(x2j )j6=i Lpt 2 , 189 s Mp,q , 160 s Xp,q , 10 k , 160 S, 2 S 0, 2 s , 20 B˙ p,q s F˙p,q , 20 s X˙ p,q , 20 S˙ , 20 S˙ 0 , 20 `q (Lp ), 9 γ p `s,q (L (R, L )), 179 φ ∗ ψ, 3 ˜ 3 φ, 4k , 9 e k , 192 {σk }k∈Zn , 160 p ∧ q, p ∨ q, 11

Boltzmann equation Global (absolute) Maxwellian, 238 Bobylev’s identities , 238 Boltzmann collision operator, 233 Boltzmann equation, 233 Cross section, 233 Cutoff cross section, 234 Hard potentials, 235 Homogeneous Boltzmann equation, 233 Local Maxwellian, 238 Maxwellian potentials, 235 Non cutoff cross section, 234 Soft potentials, 235 Christ-Kiselev lemma, 267 Complex interpolation in modulation spaces, 267 Convexity H¨ older’s inequality, 21 Critical power in H˙ s for NLS, 83 Critical space H˙ s for NLS, 83 Critical space for NS equation, 35 Dilation operator, 3 Dispersion, 51 Dispersion relation, 92 Dyadic decomposition operator, 9

Action functional, 206 Affine transform, 7 Almost orthogonality of k , 198 Asymptotic completeness, 261 Asymptotic state, 261

Embedding theorem Embedding between Besov and modulation spaces, 164 Embedding in homogeneous

Bernstein’s inequality, 14 281

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282

Triebel-Lizorkin spaces, 264 Embeddings on Besov spaces, 14 Embeddings on homogeneous Besov space, 21 Embeddings on homogeneous Triebel space, 21 Embeddings on Triebel spaces, 14 Fourier (inverse) transform F (F −1 ), 2 Fourier multiplier, 5 Fourier multiplier space Mp , 5 Free transport, 229 Frequency-uniform decomposition operator, 160 Function space BM O−1 , 49 Anisotropic Lebesgue space, 189 s Besov space Bp,q ,9 s Homogeneous Besov spaceB˙ p,q , 20 Homogeneous Triebel-Lizorkin s space F˙p,q , 20 s Modulation space Mp,q , 160 Schwartz spaceS := S (Rn ), 2 Space of tempered distributions S 0 = S 0 (Rn ), 2 s Triebel space Fp,q ,9 Gagliardo-Nirenberg inequality Fractional derivative cases, 25 Sobolev space cases, 263 Galilean operator, 214 Galilean transformation, 213 Gauge transformation, 146 Gevrey class, 44 Hamilton’s equation, 208 Hamilton’s principle, 206 Hamiltonian, 207 Hardy inequality, 219 Hardy-Littlewood-Sobolev inequality, 266 Heat semi-group Lr → Lp estimate, 36 Time-space mixed estimate, 36

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Index

Isomorphism, 2 Klein-Gordon semi-group Strichartz estimates in modulation spaces, 183 truncated decay in modulation spaces, 178, 179 Lagrangian, 206 Legendre transformation, 207 Linear PDE Klein-Gordon equation, 62, 67 Linear heat equation, 35 Schr¨ odinger equation, 51 Wave equation, 66 Littlewood-Paley decomposition operator, 9 Littlewood-Paley square function theorem, 22, 266 Minkowski’s inequality, 11 Morawetz action, 215 Morawetz inequality, 216 Multi-index α, 1 Multiplier theorem Bernstein multiplier theorem, 7 Mihlin multiplier theorem, 8 N¨ other’s theorem, 205 N¨ other’s theorem, 209 Nakanishi-Morawetz inequality, 217, 220 NLKG, 88 NLS, 75 NLW, 89 Nonlinear PDE Derivative nonlinear Klein-Gordon equation, 203 Hamilton–Jacobi equation, 41 Navier-Stokes equation, 34 Nonlinear dispersive equation, 75 Nonlinear Klein–Gordon equation, 88 Nonlinear Schr¨ odinger equation, 75 Nonlinear wave equation, 89 Semi-linear parabolic equation, 41

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Index

Zakharov system, 204 NS, 34 Paley-Wiener-Schwartz theorem, 4 Pseudo-conformal conservation law, 214 Pseudo-conformal transformation, 213 Riesz-Thorin interpolation theorem, 265 Scattering operator, 222, 261 Schr¨ odinger semi-group Local smoothing effect, 99 Maximal function estimate, 193 Maximal function estimates, 99 Smooth effect, 191 Strichartz estimate in modulation spaces, 182 truncated decay in modulation spaces, 177 Uniform boundedness in modulation spaces, 176 Semi-group Heat semi-group, 35, 36 Klein-Gordon semi-group, 62, 67, 74, 178

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283

Schr¨ odinger semi-group, 54, 67, 73, 176 Schr¨ odinger semi-group with higher order, 54, 67 Wave semi-group, 54, 66, 73 Short time Fourier transform, 159 Strichartz inequality, 51, 63, 68 Tempered distribution Composition of distribution, 4 Dilation of distribution, 4 Generalized derivative of distribution, 4 Translation of distribution, 4 Translation operator, 3 Van der Corput lemma, 266 Virial identity, 216 Virial potential, 215 Vlasov equation, 229 Wave operators, 261 Wiener decomposition, 157