Nonlinear Second Order Parabolic Equations [1 ed.] 2020042644, 2020042645, 9780367711986, 9780367712846, 9781003150169

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Nonlinear Second Order Parabolic Equations

Nonlinear Second Order Parabolic Equations

Mingxin Wang

First edition published 2021 by CRC Press 6000 Broken Sound Parkway NW, Suite 300, Boca Raton, FL 33487-2742 and by CRC Press 2 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN © 2021 Mingxin Wang CRC Press is an imprint of Taylor & Francis Group, LLC The right of Mingxin Wang to be identified as author of this work has been asserted by him in accordance with sections 77 and 78 of the Copyright, Designs and Patents Act 1988. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, access www.copyright.com or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. For works that are not available on CCC please contact [email protected] Trademark notice: Product or corporate names may be trademarks or registered trademarks and are used only for identification and explanation without intent to infringe. Library of Congress Cataloging-in-Publication Data Names: Wang, Mingxin, 1957- author. Title: Nonlinear second order parabolic equations / Mingxin Wang. Description: First edition. | Boca Raton : CRC Press, 2021. | Includes bibliographical references and index. | Summary: “The parabolic partial differential equations model is one of the most important processes in the process of diffusion in nature. Whether it is the diffusion of energy in space-time, the diffusion of species in ecology, the diffusion of chemicals in biochemical processes, or the diffusion of information in social networks, diffusion processes are ubiquitous and crucial in the physical and natural world as well as our everyday lives. In this graduate textbook, the author starts with classical parabolic problems as well as basic methods and techniques to guide students into the field of second order nonlinear parabolic equations. The author then moves on to discuss key topics such as theory and Schauder theory, the maximum principle and the comparison principle. Topics additionally discussed include periodic boundary value problems, free boundary problems, and semigroup theory. This book is based on tried and tested teaching materials used at the Harbin Institute of Technology over the past 10 years. Special care is taken to make the book suitable for classroom teaching as well as for self-study among graduate students”-- Provided by publisher. Identifiers: LCCN 2020042644 (print) | LCCN 2020042645 (ebook) | ISBN 9780367711986 (hardcover) | ISBN 9780367712846 (paperback) | ISBN 9781003150169 (ebook) Subjects: LCSH: Differential equations, Parabolic. | Differential equations, Partial. Classification: LCC QA377 .W36 2021 (print) | LCC QA377 (ebook) | DDC 515/.3534--dc23 LC record available at https://lccn.loc.gov/2020042644 LC ebook record available at https://lccn.loc.gov/2020042645

ISBN: 978-0-367-71198-6 (hbk) ISBN: 978-0-367-71284-6 (pbk) ISBN: 978-1-003-15016-9 (ebk) Typeset in Computer Modern font by KnowledgeWorks Global Ltd.

Contents Preface Chapter 1.1 1.2

1.3

ix 1  Preliminaries NOTATIONS, AGREEMENTS AND BASIC ASSUMPTIONS p

THE L AND SCHAUDER THEORIES p

1 1 4

1.2.1

L theory

4

1.2.2

Schauder theory

13

1.2.3

Existence and uniqueness of solutions of linear system

16

MAXIMUM PRINCIPLE OF CLASSICAL SOLUTIONS

17

1.3.1

Initial-boundary value problems

17

1.3.2

Cauchy problems

19

1.3.3

Weakly coupled systems

21

1.3.4

Equations with non-classical boundary conditions

23

1.3.5

Equations with non-local terms

24

1.4

MAXIMUM PRINCIPLE OF STRONG SOLUTIONS

28

1.5

WEAK SOLUTION

32

EXERCISES

Chapter 2.1

34

2  Comparison Principle, Regularity and Uniform Estimates

37

COMPARISON PRINCIPLE

37

2.1.1

Classical and strong solutions

38

2.1.2

Weak solutions

39

2.1.3

Quasilinear equations with nonlinear boundary conditions

43

2.2

REGULARITY AND UNIFORM ESTIMATES

44

2.3

UNIFORM BOUNDS FROM BOUNDED Lp NORMS

48

EXERCISES

52

v

vi  Contents

Chapter

3  Semilinear Parabolic Equations

53

3.1

THE UPPER AND LOWER SOLUTIONS METHOD

54

3.2

SOME EXAMPLES

57

3.3

MONOTONICITY AND CONVERGENCE

61

3.4

THE WEAK UPPER AND LOWER SOLUTIONS METHOD

64

EXERCISES

Chapter

69

4  Weakly Coupled Parabolic Systems

73

4.1

LOCAL EXISTENCE AND UNIQUENESS OF SOLUTIONS

73

4.2

THE UPPER AND LOWER SOLUTIONS METHOD

77

4.2.1

Existence and uniqueness of solutions

78

4.2.2

Monotone iterative

80

4.3

APPLICATIONS

84

4.3.1

A competition model

84

4.3.2

A prey-predator model

90

4.3.3

The Belousov-Zhabotinskii reaction model

91

EXERCISES

Chapter 5.1 5.2

94

5  Stability Analysis LOCAL STABILITY DETERMINED BY THE PRINCIPAL EIGENVALUE

97

98

LYAPUNOV FUNCTIONAL METHOD

100

5.2.1

Abstract results

100

5.2.2

A variable-territory prey-predator model

105

5.2.3

A prey-predator model with trophic interactions

109

5.2.4

A system coupled by two PDEs and one ODE

114

5.3

ITERATION METHOD

117

5.4

AVERAGE METHOD

121

5.4.1

Abstract results

121

5.4.2

A special diffusive competition model

123

EXERCISES

128

Contents  vii

Chapter 6.1

6.2

6.3

6  Global Existence and Finite Time Blowup SOME BASIC METHODS

132

6.1.1

Comparison method

132

6.1.2

Kaplan’s first eigenvalue method

135

6.1.3

Energy method

136

6.1.4

Concavity method

139

A SYSTEM WITH DIRICHLET BOUNDARY CONDITIONS

140

6.2.1

Critical exponents

141

6.2.2

Blowup rate estimates

145

A SYSTEM COUPLED IN EQUATION AND BOUNDARY

7.1

147

6.3.1

Properties of the Green function G(x, y, t − s)

148

6.3.2

Critical exponents

150

6.3.3

Blowup on the boundary

155

6.3.4

Blowup rate estimates

157

EXERCISES

Chapter

131

162

7  Time-Periodic Parabolic Boundary Value Problems

165

THE UPPER AND LOWER SOLUTIONS METHOD FOR SCALAR EQUATIONS

165

7.2

TIME-PERIODIC PARABOLIC EIGENVALUE PROBLEMS

169

7.3

THE LOGISTIC EQUATION

173

7.3.1

The case of bounded domain

173

7.3.2

The case of half line

176

7.4

THE UPPER AND LOWER SOLUTIONS METHOD FOR SYSTEMS

181

7.5

A DIFFUSIVE COMPETITION MODEL

184

EXERCISES

Chapter

187

8  Free Boundary Problems from Ecology

189

8.1

DEDUCTION OF FREE BOUNDARY CONDITIONS

190

8.2

EXISTENCE AND UNIQUENESS OF SOLUTIONS

191

8.3

REGULARITY AND UNIFORM ESTIMATES

198

8.4

SOME TECHNICAL LEMMAS

202

8.5

THE LOGISTIC EQUATION

208

8.5.1

208

Longtime behaviour of solution

viii  Contents

8.5.2

Criteria and dichotomy for spreading and vanishing

210

8.6

SPREADING SPEED OF FREE BOUNDARY

215

8.7

A SYSTEM COUPLED BY ONE PDE AND ONE ODE

223

EXERCISES

Chapter 9.1

9.2

235

9  Semigroup Theory and Applications

237

C0 SEMIGROUP THEORY

238

9.1.1

Basic results

238

9.1.2

Solutions of linear problems

241

9.1.3

Mild solutions of semilinear problems

246

ANALYTIC SEMIGROUP THEORY

251

9.2.1

251

Basic results Z

9.2.2

Regularity of function

t

e−(t−s)A f (s)ds

254

0

9.2.3

Real solutions of semilinear problems p

258

9.3

SEMIGROUPS DETERMINED BY −∆ IN L (Ω)

262

9.4

AN EXAMPLE

269

EXERCISES

Appendix

A  Appendix

273

275

Bibliography

277

Index

285

Preface Parabolic partial differential equations model one of the most important processes in the nature world: diffusion. Whether it is the diffusion of energy in space-time, diffusion of species in ecology, diffusion of chemicals in biochemical processes, or the diffusion of information in social networks, diffusion processes are ubiquitous and crucial in the physical and natural world as well as our everyday lives. The aim of this book is to provide graduate students with an introductory and self-contained textbook on second order nonlinear parabolic partial differential equations. We start with classical parabolic problems, basic methods and techniques to guide students into the exciting field of second order nonlinear parabolic equations as quickly and painlessly as possible. The book covers key topics such as Lp theory and Schauder theory, maximum principle, comparison principle, regularity and uniform estimate, initial-boundary value problems of semilinear parabolic scalar equations and weakly coupled parabolic systems, upper and lower solutions method, monotone properties and longtime behaviours of solutions, convergence of solutions and stability of equilibrium solutions, global solutions, and finite time blowup. It also touches on periodic boundary value problems, free boundary problems, and semigroup theory. As a preliminary, Chapter 1 reviews the Lp theory, Schauder theory, Hopf boundary lemma and maximum principle, which are some of the main tools in the study of parabolic partial differential equations. As some of these materials may already be known to the reader from a basic course in linear partial differential equations, we only present the conditions and conclusions and give appropriate explanations and references, without giving the complete proofs for the better-known results. The comparison principle based on the maximum principle will be frequently used in this textbook. The regularity and uniform estimates based on the Lp theory and Schauder theory are important ingredients for the study of convergence of solutions as time goes to infinity and the stability of equilibrium solutions. Chapter 2 is devoted to these topics. Chapters 3 and 4 deal with initial-boundary value problems of semilinear parabolic scalar equations and weakly coupled parabolic systems, respectively. We first introduce the upper and lower solutions method, and then discuss the monotone property and longtime behaviour of solutions. In addition, the use of Lp theory, Schauder theory, and Schauder fixed-point theorem to prove the existence and uniqueness of local solutions is also included. In Chapter 5, we first consider local stability of equilibrium solutions by using the linearized principal eigenvalue and the upper and lower solutions method. We establish that the linear stability implies the local stability. Then we introduce three additional methods: the Lyapunov functional method, the iteration method, and ix

x  Preface

the average method for the investigation of stability of equilibrium solutions, and illustrate them with concrete examples. Chapter 6 involves global solutions and finite time blowup. We first provide four basic methods for dealing with finite time blowup: they are the comparison method, Kaplan’s first eigenvalue method, the energy method, and the concavity method. Then, through some concrete examples, we show how to determine conditions for the global existence and finite time blowup, and discuss the critical exponent, blowup estimates, and blowup on the boundary. In Chapter 7, we focus on time-periodic parabolic boundary value problems. We first introduce the upper and lower solutions method, and then study a diffusive logistic equation (including the unbounded domain case) and a diffusive competition model as applications. For the diffusive logistic equation, the existence and uniqueness of positive solutions are established. Especially in the bounded domain case, we prove that the unique positive solution is stable. For the diffusive competition model, the existence of positive solutions is also established. Meanwhile, some important properties of time-periodic parabolic eigenvalue problems are provided. In Chapter 8, we discuss some free boundary problems coming from ecology. The main content includes existence and uniqueness, regularity and uniform estimates, some technical lemmas, longtime behaviour of solutions, criteria and dichotomy for spreading and vanishing, and the spreading speed of the free boundary. Concrete examples are given as applications. Finally, in Chapter 9, we first give a brief review of the basic results of semigroup theory, and then show how it can be used to prove the existence, uniqueness, extension, and regularity of the mild solution. In the past ten years, materials in this book have been used for the course “Nonlinear Parabolic Equations” for graduate students at Harbin Institute of Technology. Its completion is benefitted by many students who have attended these classes. While working on this book, I was partially supported by NSFC Grants 11071049, 11371113, and 11771110. I am grateful to my students and colleagues for their feedback and helpful comments on the manuscript. The materials presented and the references quoted here are mainly based on the author’s taste and familiarity, which are inevitably biased with many important topics and references not included here. No offence is intended on account of such omissions or oversight. Mingxin Wang, Harbin

CHAPTER

1

Preliminaries

There are many monographs and textbooks on parabolic equations, please refer to [36, 91, 60, 105, 88, 31, 16, 72, 93, 110, 124, 48]. The book [16] received the excellent textbook award from the Ministry of Education of China. This chapter reviews some basic theories and results such as Lp theory, Schauder theory and maximum principle for linear parabolic equations, which are fundamental tools in the study of nonlinear parabolic equations. As some of these materials may be already known to the reader from a basic course in linear partial differential equations, we only present the conditions and conclusions, and give appropriate explanations and references, but not complete proofs for the better known results. The chapter starts with the Lp theory and Schauder theory for the first initial-boundary value problems, second initial-boundary value problems, and mixed initial-boundary value problems. It then moves on to topics such as the Hopf boundary lemma, maximum principle of classical solutions, and strong and weak solutions for scalar equations and systems including problems with non-classical boundary conditions and non-local terms.

1.1

NOTATIONS, AGREEMENTS AND BASIC ASSUMPTIONS

Notations Given two sets A, B ⊂ Rn , we use A to represent the closure of A, and use d(A, B) or dist(A, B) to represent the distance between A and B. Notation A b B means that A is bounded and A ⊂ B. Given a set A ⊂ Rn and a function f : A → Rm , we use f (A) to represent the image of A under f . Let k > 0 be an integer and 0 6 β < 1 be a constant. We say that a domain Ω ⊂ Rn is of class C k+β , or Ω has a C k+β boundary ∂Ω if for each point x0 ∈ ∂Ω there exist a neighbourhood U of x0 and a function Φ ∈ C k+β (U ) such that (i) the inverse function Φ−1 exists and Φ−1 ∈ C k+β (Φ(U )), and (ii) if we set y = (y1 , . . . , yn ) = Φ(x), then Φ(∂Ω ∩ U ) ⊂ {y ∈ Rn : yn = 0} and Φ(Ω ∩ U ) ⊂ {y ∈ Rn : yn > 0}. We say that a domain Ω has interior ball property at point x ∈ ∂Ω, if there exists ¯ ∩ ∂Ω = {x}. If Ω has the interior a ball B with radius r > 0 for which B ⊂ Ω and B 1

2  1. Preliminaries

ball property at each point x ∈ ∂Ω, then we say that Ω has the interior ball property. If Ω is of class C 2 , then it has the interior ball property. However, if Ω is only of class C 1+α with 0 < α < 1, then it may not have the interior ball property. For example, curve x2 = |x1 |1+α does not have the interior ball property at point (0, 0). Let Q ⊂ Rn+1 and operator L be given by L := ∂t − aij (x, t)Dij + bi (x, t)Di + c(x, t),

where symbols aij Dij := ni,j=1 aij Dij , bi Di := ni=1 bi Di . The operator L is called parabolic in Q if matrix (aij )n×n is symmetric and positive definite in Q, i.e., aij = aji in Q and aij (x, t)yi yj > 0 for all (x, t) ∈ Q, 0 6= y ∈ Rn . P

P

A parabolic operator L in Q is called strongly parabolic if there are positive constants λ and Λ such that λ|y|2 6 aij (x, t)yi yj 6 Λ|y|2

for all (x, t) ∈ Q, y ∈ Rn .

For x ∈ Rn and t ∈ R, we write x = (x, t). Given x0 = (x0 , t0 ) and r > 0, we define a lower hemisphere Q(x0 , r) := {x ∈ Rn+1 : |x − x0 | < r, t < t0 }. For a given domain Q ⊂ Rn+1 , we define the parabolic boundary ∂p Q of Q to be a set of all points x0 ∈ ∂Q satisfying that for any r > 0, the lower hemisphere Q(x0 , r) contains points that do not belong to Q, i.e., Q(x0 , r) 6⊂ Q for any r > 0. In Fig. 1.1, the solid line represents the parabolic boundary and the dashed line does not belong to the parabolic boundary. Let Q ⊂ Rn+1 be a bounded domain and x0 ∈ Q. We define S(x0 ) to be a set of all x ∈ Q \ ∂p Q for which we can find a continuous function g : [0, 1] → Q\∂p Q such that g(0) = x0 , g(1) = x and t-component of g is monotonically decreasing. The shaded part in Fig. 1.2 is S(x0 ). Let Ω ⊂ Rn and Q ⊂ Rn+1 be domains, 0 < α < 1 be a constant and k > 0 be an integer. Function spaces C k+α (Ω) and C k+α, (k+α)/2 (Q) are the standard H¨older spaces. Notation u ∈ C k+α (Ω) means that u ∈ C k+α (Ω0 ) for any Ω0 b Ω, and notation u ∈ C k+α, (k+α)/2 (Q) means that u ∈ C k+α, (k+α)/2 (Q0 ) for each Q0 b Q. Let Ω and Q be as above, k be a positive integer and p > 1. Function spaces Wpk (Ω) and Wp2k,k (Q) are the standard Sobolev spaces. We denote the closure of ◦

C0∞ (Ω) in Wpk (Ω) by W kp (Ω), and use Lp (Ω) (Lp (Q)) instead of Lp (Ω) (Lp (Q)). To simplify notations, sometimes we write | · |i+α, Ω = k · kC i+α (Ω) and | · |i+α, Q = k · kC i+α,(i+α)/2 (Q) , i = 0, 1, 2.

1.1. Notations, agreements and basic assumptions  3

In case of no confusions, for simplicity, we write k · kp = k · kp, D = k · kLp (D)

for D = Ω or D = Q

and k · kWp2 = k · kWp2 (Ω) , k · kWp2,1 = k · kWp2,1 (Q) . R

R

Sometimes, we use Ω f (x) instead of Ω f (x)dx to save space. Let X be a Banach space, and f, g ∈ X. We write kf, gkX = kf kX + kgkX . Agreements Throughout this book, unless it is clearly stated, we shall make the following agreements. • All functions are real-valued. • We usually refer to a vector-function (and also a matrix of functions) briefly as a function. • There is a function b(x, t) in the boundary condition ∂n u + bu = g (it is customary to use this notation). However, sometimes in the same problem, this symbol b will also appear in the differential equation. Careful readers are easy to distinguish and will not cause confusion. • Constant 0 < α < 1 may be different in different places. • We use C, C 0 , Ci and ci to represent the generic constants. • Ω ⊂ Rn and Q ⊂ Rn+1 are bounded domains, n and ν are the outward normal vectors of ∂Ω and ∂Q, respectively. For simplicity, we denote the outer normal ∂ ∂ derivatives ∂n and ∂ν by ∂n and ∂ν , respectively. • QT = Ω × (0, T ], ST = ∂Ω × (0, T ] and ∂p QT = ST ∪ (Ω × {0}) with 0 < T < ∞. Obviously, ∂p QT is the parabolic boundary of QT . We call ST the lateral boundary of QT . • Q∞ = Ω × (0, ∞), S∞ = ∂Ω × (0, ∞), R+ = (0, ∞) and R+ = [0, ∞). • Constant 1 < p < ∞. For a given function g defined on ST , we say g ∈ Wp2,1 (QT ) or g ∈ C 2+α,1+α/2 (QT ) if g can be extended to a new function gˆ so that gˆ ∈ Wp2,1 (QT ) or gˆ ∈ C 2+α,1+α/2 (QT ). Basic assumptions Throughout this book, for simplicity, in some places (theorems and lemmas), we will use the following assumptions and abbreviations. (A) The operator L is strongly p parabolic in QT , and aij ∈ C(QT ). Let x = (x, t), y = (y, s) and δ(x, y) = |x − y|2 + |t − s|. Then ω(R) = max i,j

sup

|aij (x) − aij (y)| → 0 as R → 0.

δ(x,y)6R x,y∈QT

Function ω(R) is called modulus of continuity of aij (x, t).

4  1. Preliminaries

(B) Condition (A) holds, and bi , c ∈ L∞ (QT ), f ∈ Lp (QT ). (C) Condition (A) holds, and aij , bi , c, f ∈ C α,α/2 (QT ). (D) ϕ ∈ Wp2 (Ω) and g ∈ Wp2,1 (QT ). (E) ϕ ∈ C 2+α (Ω) and g ∈ C 2+α,1+α/2 (QT ).

1.2

THE Lp AND SCHAUDER THEORIES

The so-called Lp (Schauder) theory includes existence, uniqueness and estimates of Wp2,1 (C 2+α,1+α/2 ) solutions. They are very important in the field of second order parabolic equations. This section is concerned with the following initial-boundary value problems:  L u = f (x, t)      

in QT ,

u=g

on ST ,

u(x, 0) = ϕ(x)

in Ω

(1.1)

and  L u = f (x, t)      

in QT ,

∂n u + bu = g, b > 0

on ST ,

u(x, 0) = ϕ(x)

in Ω,

(1.2)

as well as  L u = f (x, t)       ∂n u + bu = g1 , b > 0  u = g2     

in QT , on Γ 1 × (0, T ], on Γ2 × (0, T ],

u(x, 0) = ϕ(x)

(1.3)

in Ω,

where Γ1 and Γ2 are two disjoint closed subsets of ∂Ω with Γ1 ∪Γ2 = ∂Ω. For problem (1.3), we define B1 u = ∂n u + bu on Γ1 × (0, T ] and B2 u = u on Γ2 × (0, T ]. We call (1.1), (1.2), and (1.3) the first initial-boundary value problem, second initial-boundary value problem, and mixed initial-boundary value problem, respectively. Recall that bi , c ∈ L∞ (QT ), and Ω ⊂ Rn is a bounded domain. 1.2.1 Lp theory

For convenience, we denote A := {n, λ, Λ, T, kbi , ck∞ , ω(R)}. Theorem 1.1 (Lp theory for the first initial-boundary value problem [16, Chapter 5, Theorem 5.2], [60, Chapter 4, Theorem 9.1], [72, Theorem 7.32]) Let conditions (B) and (D) hold, and Ω be of class C 2 . Assume that compatibility condition ϕ(x) = g(x, 0) on ∂Ω

1.2. The Lp and Schauder theories  5

is satisfied. Then problem (1.1) has a unique solution u ∈ Wp2,1 (QT ), and 



kukWp2,1 (QT ) 6 C kf kp,QT + kgkWp2,1 (QT ) + kϕkWp2 (Ω) ,

(1.4)

where constant C depends only on A and Ω. Theorem 1.2 (Lp theory for the second initial-boundary value problem [60, Chapter 4, Section 9]) Let conditions (B) and (D) be fulfilled, Ω be of class C 2 , b ∈ C 1,1/2 (S T ) and b > 0. Assume that compatibility condition ∂n ϕ(x) + b(x, 0)ϕ(x) = g(x, 0) on ∂Ω holds. Then problem (1.2) has a unique solution u ∈ Wp2,1 (QT ), and the estimate (1.4) holds with constant C depending only on A, Ω and |b|1,S T . For the Lp theory with p > 2, under more detailed conditions, in Theorem 1.1 and Theorem 1.2, the requirement ϕ ∈ Wp2 (Ω) in the condition (D) can be weakened to 2−2/p

ϕ ∈ Wp (Ω) and the term kϕkWp2 (Ω) on the right hand side of (1.4) can be replaced by kϕkW 2−2/p (Ω) . The interested reader can refer to [60, Chapter 4, Section 9]. p

In order to study existence and uniqueness of Wp2,1 (QT ) solutions of problem (1.3), we first give the following Lp estimate. Theorem 1.3 (Local Lp estimate) Assume that Ω is of class C 2 . For any given δ > 0, we define Ωδi := {x ∈ Ω : d(x, Γi ) 6 δ}, i = 1, 2. Take 0 < δ  1, so that Ωδ i ∩ ∂Ω = Γi . Denote Diδ = Ωδi × (0, T ]. Assume that conditions (B) and (D) hold with Ω and QT replaced by Ωδi and Diδ , respectively, and b ∈ C 1,1/2 (Γ1 × [0, T ]), b > 0. Let u ∈ Wp2,1 (Diδ ) satisfy  L u = f (x, t)      

in Diδ ,

Bi u = gi

on Γi × (0, T ],

u(x, 0) = ϕ(x)

in Ωδi .

Then for any 0 < σ < δ and 0 6 ε < T , we have the following local estimates: 

kukWp2,1 (Dσ ) 6 C1 kf kp, Diδ + kgi kWp2,1 (Dδ ) + kϕkWp2 (Ωδi ) + kukp, Diδ i



when ε = 0,

i



kukWp2,1 (Ωσ ×(ε,T )) 6 C2 kf kp, Diδ + kgi kWp2,1 (Dδ ) + kukp, Diδ i



when ε > 0,

i

where constants C1 and C2 depend on A and 1/(δ−σ), and also depend on |b|1,Γ1 ×[0,T ] when i = 1, and C2 also depends on 1/ε. We state a lemma before giving the proof of Theorem 1.3.

6  1. Preliminaries

Lemma 1.4 Let b > a > 0 and ϕ(s) be a bounded non-negative function defined in [a, b]. Assume that ϕ satisfies ϕ(t) 6 θϕ(s) +

A +B (s − t)β

for all a 6 t < s 6 b,

(1.5)

where constants θ, A, B, and β are non-negative with θ < 1. Then 

ϕ(ρ) 6 C

A +B (R − ρ)β



for all a 6 ρ < R 6 b,

(1.6)

where C is a constant which depends only on β and θ. Proof. Take t0 = ρ and ti+1 = ti + (1 − τ )τ i (R − ρ), i = 0, 1, . . . , where 0 < τ < 1 is a constant to be determined. Then we have ϕ(ti ) 6 θϕ(ti+1 ) +

A + B, i = 0, 1, . . . [(1 − τ )τ i (R − ρ)]β

by (1.5), as a consequence ϕ(t0 ) 6 θk ϕ(tk ) +

k−1 k−1 X X A i −iβ θ τ + B θi [(1 − τ )(R − ρ)]β i=0 i=0

for all k > 1.

Take τ to satisfy θτ −β < 1 and let k → ∞, then we get inequality (1.6).  Proof of Theorem 1.3. Here we only consider the case i = 1 and ε = 0. For σ 6 ρ < r 6 δ, let η(x) be a smooth function satisfying 0 6 η(x) 6 1, η(x) ≡ 1 in Ωρ1 , η(x) ≡ 0 in Ω \ Ωr1 and |Dk η| 6 C/(r − ρ)k , k = 1, 2. Then function v = ηu satisfies  ∗   L v = ηf − uaij Dij η − 2aij Di uDj η + ubi Di η =: f 

in QT ,

∂n v + bv = g

on ST ,

  

v(x, 0) = η(x)ϕ(x)

in Ω,

where g = g1 on Γ1 × (0, T ] and b = g = 0 on Γ2 × (0, T ]. Compatibility condition ∂n (ηϕ) + b(x, 0)ηϕ = g(x, 0) on ∂Ω is obvious. Denote D1r = Ωr1 × (0, T ] and B = kf kp, D1δ + kg1 kWp2,1 (Dδ ) + kϕkWp2 (Ωδ1 ) . 1 Applying Theorem 1.2 and Theorem A.9 (1), it yields 

kvkWp2,1 (QT ) 6 C kf ∗ kp, QT + kgkWp2,1 (QT ) + kηϕkWp2 (Ω) 

6C 6 Cε



   1 1 r r 1 kDukp, D1r + kuk + kϕk + B p, D1 Wp (Ω1 ) r−ρ (r − ρ)2

2 X i=0

kDi ukp, D1r +

 C(ε)  r + kϕk 1 (Ωr ) kuk + CB. p, D W p 1 1 (r − ρ)2

1.2. The Lp and Schauder theories  7

Noticing η(x) ≡ 1 in Ωρ1 and the above estimate, we have kukWp2,1 (Dρ ) 6 CεkukWp2,1 (Dr ) + 1

1

 C(ε)  kuk + CB. δ + kϕkW 1 (Ωδ ) p, D p 1 1 (r − ρ)2

From Lemma 1.4, it follows that 

kukWp2,1 (Dρ ) 6 C 1

   1 kuk + kϕk + B . δ δ 1 p, D1 Wp (Ω1 ) (r − ρ)2

The desired conclusion is obtained by taking ρ = σ and r = δ. Similar to the proof of Theorem 1.3, we can prove the following theorem.



Theorem 1.5 (Interior Lp estimate in x direction) Assume that u ∈ Wp2,1 (Ω0 × (0, T )) ∩ Lp (QT ) for any Ω0 b Ω, and let u satisfy L u = f (x, t)

in QT .

(1.7)

Then, for any given Ω0 b Ω, we have 



kukWp2,1 (Ω0 ×(0,T )) 6 C kf kp, QT + kukp, QT + ku(·, 0)kWp2 (Ω) , where C is a constant depending only on A and 1/d(Ω0 , ∂Ω). Theorem 1.6 (Global Lp estimate) Let conditions (B) and (D) hold, Ω be of class C 2 and b ∈ C 1,1/2 (Γ1 × [0, T ]). Assume that u ∈ Wp2,1 (QT ) satisfies (1.3). Then we have global estimate 



kukWp2,1 (QT ) 6 C kf kp, QT + kg1 , g2 kWp2,1 (QT ) + kϕkWp2 (Ω) ,

(1.8)

where C is a constant which depends on A and |b|1, Γ1 ×[0,T ] . Proof. Using Theorem 1.3 for i = 1 and i = 2, respectively, and interior Lp estimate in x direction (Theorem 1.5), we can obtain 



kukWp2,1 (QT ) 6 C kf kp, QT + kg1 , g2 kWp2,1 (QT ) + kϕkWp2 (Ω) + kukp, QT . Particularly, 

kut kp,Qs 6 C kf kp, QT + kg1 , g2 kWp2,1 (QT ) + kϕkWp2 (Ω) + kukp, Qs for all 0 6 s 6 T . Denote B = kf kp, QT + kg1 , g2 kWp2,1 (QT ) + kϕkWp2 (Ω) .



(1.9)

8  1. Preliminaries

By using the above estimate and Young’s inequality, Z

p

|u(x, s)| =

Z

Ω

p

|ϕ(x)| +

Z sZ

Ω

0

6 kϕkpp, Ω + p

Ω

Z sZ 0

Z s Z 0

|u|p−1 |ut |

Ω

6 kϕkpp, Ω + (p − 1) 6C

d p |u| dt

Z sZ 0

|u|p +

Z sZ

Ω

0

|ut |p

Ω



|u|p + B p .

Ω

Then the Gronwall inequality gives Z 0

TZ Ω

|u|p 6 C(T )B p .

This, together with (1.9), implies (1.8).



Theorem 1.7 (Lp theory for the mixed initial-boundary value problem) Set g = gi on Γi × (0, T ]. Assume that conditions (B) and (D) hold, Ω is of class C 2 and b ∈ C 1,1/2 (Γ1 × [0, T ]). If compatibility conditions ∂n ϕ(x) + b(x, 0)ϕ(x) = g1 (x, 0) on Γ1 , and ϕ(x) = g2 (x, 0) on Γ2 are satisfied, then problem (1.3) has a unique solution u ∈ Wp2,1 (QT ) and the estimate (1.8) holds. Proof. Global estimate (1.8) implies that (1.3) has at most one solution in Wp2,1 (QT ). Proof of this theorem is similar to those of Theorem 1.1 and Theorem 1.2, hence we omit the details here.  Theorem 1.8 (Interior Lp estimate in t direction) Let Ω be of class C 2 and the condition (B) hold. Assume that, for any 0 < δ < T , u ∈ Wp2,1 (Ω × (δ, T )) ∩ Lp (QT ) satisfies the first three equations of (1.3) for δ 6 t 6 T . Then for any 0 < δ < T we have 



kukWp2,1 (Ω×(δ,T )) 6 C kf kp, QT + kg1 , g2 kWp2,1 (QT ) + kukp, QT , where C is a constant depending on A, |b|1, Γ1 ×[0,T ] and 1/δ. Proof. Take a truncated function η ∈ C ∞ ([0, T ]) such that 0 6 η 6 1, η ≡ 1 in [δ, T ] and η ≡ 0 in [0, δ/2]. Then there exists C 0 > 0 such that |η 0 (t)| 6 C 0 /δ. Function w = uη satisfies  L w = f (x, t)η(t) + uη 0 (t)       ∂n w + bw = g1 η(t)

in QT ,

 w = g2 η(t)     

on Γ2 × (0, T ],

w(x, 0) = 0

on Γ1 × (0, T ], in Ω.

1.2. The Lp and Schauder theories  9

Making use of Theorem 1.7, we have kukWp2,1 (Ω×(δ,T )) 6 kwkWp2,1 (QT ) 

6 C1 kf η(t)kp, QT + kuη 0 (t)kp, QT + kg1 η(t), g2 η(t)kWp2,1 (QT ) 

6 C1 kf kp, QT + C 0 δ −1 kg1 , g2 kWp2,1 (QT ) + C 0 δ −1 kukp, QT 







6 C kf kp, QT + kg1 , g2 kWp2,1 (QT ) + kukp, QT . The proof is complete.



Theorem 1.9 (Interior Lp estimate) Suppose that the condition (B) holds. 2,1 (1) Let u ∈ Wp,loc (QT ) ∩ Lp (QT ) satisfy (1.7). Then, for any Q b QT ,

kukWp2,1 (Q) 6 C (kf kp, QT + kukp, QT ) , where C is a constant which depends only on A and 1/d(Q, ∂p QT ). 2,1 (2) Let p > 2, and u ∈ W22,1 (QT ) satisfy (1.7). Then u ∈ Wp,loc (QT ) and for any Q b QT ,

kukWp2,1 (Q) 6 C (kf kp, QT + kuk2, QT ) ,

(1.10)

where C is the same as above. Proof. (1) For ρ > 0, we define Qρ = {(x, t) ∈ QT : d((x, t), Q) 6 ρ}. Set δ = d(Q, ∂p QT ). For 0 6 ρ < r 6 δ/2, let η be a smooth function with 0 6 η 6 1, η ≡ 1 in Qρ and η ≡ 0 in QT \ Qr . Then |ηt | 6 C/(r − ρ),

|Dk η| 6 C/(r − ρ)k , k = 1, 2,

and the function v = ηu ∈ Wp2,1 (QT ) satisfies  L v = ηf + ηt u − uaij Dij η − 2aij Di uDj η + ubi Di η =: f ∗      

in QT ,

v=0

on ST ,

v(x, 0) = 0

in Ω.

Applying Theorem 1.1 and Theorem A.9 (1), we have kvkWp2,1 (QT ) 6 Ckf ∗ kp, QT 

6C 6 Cε

1 1 kDukp, Qr + kukp, Qr + kf kp, QT r−ρ (r − ρ)2

2 X i=0

kDi ukp, Qr +



C(ε) kukp, Qr + Ckf kp, QT . (r − ρ)2

(1.11)

10  1. Preliminaries

Note that η(x) ≡ 1 in Qρ . From the above estimate it can be deduced that kukWp2,1 (Qρ ) 6 CεkukWp2,1 (Qr ) +

C(ε) kukp, QT + Ckf kp, QT . (r − ρ)2

Then, by the use of Lemma 1.4, 

kukWp2,1 (Qρ ) 6 C

1 kukp, QT + kf kp, QT (r − ρ)2



.

Therefore, 

kukWp2,1 (Q) 6 C

1 kukp, QT + kf kp, QT (r − ρ)2



.

The conclusion can be obtained by taking ρ = 0 and r = δ/2. (2) In this case, taking advantage of Theorem 1.1 and Theorem A.9 (1), we have 

kvkWp2,1 (QT ) 6 C 6 Cε

1 1 kDukp, Qr + kukp, Qr + kf kp, QT r−ρ (r − ρ)2

2 X

kDi ukp, Qr +

C(ε) kukp, Qr + Ckf kp, QT (r − ρ)2

kDi ukp, Qr +

C(ε) kuk2, QT + Ckf kp, QT , (r − ρ)q

i=0

6 Cε

2 X



i=0

where q = 2 + n(p − 2)/(2p). The remaining proof is the same as above.



Theorem 1.10 (Local Lp estimate) Let the condition (B) hold. Assume that Σ ⊂ ST with C 2 boundary ∂Σ, u ∈ Wp2,1 (QT ) satisfies (1.7) and u = 0 on Σ. Then, for any Q ⊂ QT with d(Q, ∂p QT \ Σ) = δ > 0, there is a constant C determined only by A and 1/δ such that kukWp2,1 (Q) 6 C (kf kp, QT + kukp, QT ) . Proof. For 0 6 ρ < r 6 δ/2, we let η be a smooth function with 0 6 η 6 1, η ≡ 1 in Qρ and η ≡ 0 in QT \ Qr as above. Then |ηt | 6 C/(r − ρ)2 ,

|Dk η| 6 C/(r − ρ)k , k = 1, 2,

and function v = ηu ∈ Wp2,1 (QT ) which satisfies (1.11). The remaining proof is the same as that of Theorem 1.9.  Before ending this subsection, we study existence and uniqueness of “weak” solutions. As an example, here we only consider the following special case of (1.1):  u − ∆u = f (x, t)    t

in QT ,

u=0

on ST ,

  

u(x, 0) = 1

in Ω

(1.12)

with corresponding functions g = 0 and ϕ = 1. In this case, the compatibility condition ϕ(x) = g(x, 0) on ∂Ω does not hold.

1.2. The Lp and Schauder theories  11

Theorem 1.11 Assume that Ω is of class C 2 and f ∈ Lp (QT ). Then problem (1.12) has a unique solution u in the sense that u ∈ Wp2,1 (Ω × [τ, T ]) for any 0 < τ < T , and moreover, u satisfies the first two equations of (1.12), and lim ku(·, t) − 1kp, Ω = 0.

(1.13)

t→0 ◦

Proof. Take ϕk ∈ W 1p (Ω)∩Wp2 (Ω) satisfying ϕk → 1 in Lp (Ω) as k → ∞. By Theorem 1.1, the problem  k k   ut − ∆u = f (x, t) 

in QT ,

u =0

on ST ,

  

uk (x, 0) = ϕk (x)

in Ω

k

◦

◦

◦

2,1 ∞ has a unique solution uk ∈ W 2,1 p (QT ), where W p (QT ) is the closure of C (QT ) in Wp2,1 (QT ), and ◦

C ∞ (QT ) := {u ∈ C ∞ (QT ) : u|ST = 0}. Clearly, uk − ul satisfies  k (u − ul )t − ∆(uk − ul ) = 0   

in QT ,

  

in Ω.

uk − ul = 0 k

l

k

l

(u − u )(x, 0) = ϕ (x) − ϕ (x)

on ST ,

Thus, for all 0 < t 6 T , 1d p dt

Z

|uk − ul |p = −(p − 1)

Z

Ω

Ω

|uk − ul |p−2 |∇(uk − ul )|2 6 0,

which implies Z

k

Ω

as well as

Z QT

l p

|u − u | 6

k

l p

Z

|ϕk − ϕl |p → 0

Ω

|u − u | dt 6 T

Z

|ϕk − ϕl |p → 0

Ω

as k, l → ∞. Applying Theorem 1.8, for any 0 < τ < T , we have kuk − ul kWp2,1 (Ω×[τ,T ]) 6 Ckuk − ul kp, QT → 0

(1.14)

12  1. Preliminaries ◦

as k, l → ∞. Therefore, {uk } is a Cauchy sequence in W 2,1 p (Ω × [τ, T ]). Consequently, ◦

◦

uk → uτ in W p2,1 (Ω × [τ, T ]) for some uτ ∈ W 2,1 p (Ω × [τ, T ]). By the uniqueness of limit, uτ = uε in Ω × [τ, T ] when ε < τ . Thus we get a function u having properties: ◦

u = uτ in Ω × [τ, T ], u ∈ W 2,1 p (Ω × [τ, T ]), and lim kuk − ukWp2,1 (Ω×[τ,T ]) = 0

k→∞

for any 0 < τ < T . This implies uk → u in C([τ, T ], Lp (Ω)) (cf. Exercise 1.2). Certainly, uk (·, t) → u(·, t) in Lp (Ω) for any 0 < t 6 T . It is easy to see that u satisfies the first two equations of (1.12). We shall show that u satisfies (1.13). In fact, taking l → ∞ in the inequality of (1.14), we have Z Ω

|uk − u|p 6

Z

|ϕk − 1|p ,

Ω

which implies Z

lim

k→∞ Ω

|uk − u|p = 0 uniformly in t ∈ (0, T ].

As a consequence, there exists k  1 such that Z

|uk − u|p +

Z

Ω

|ϕk − 1|p < ε for all t ∈ (0, T ].

Ω

Since

Z

lim

t→0 Ω

|uk − ϕk |p = 0,

making use of Z Ω

|u − 1|p 6 C

Z

|uk − u|p +

Ω

Z

|ϕk − 1|p +

Ω

Z



|uk − ϕk |p ,

Ω

we can prove (1.13). Existence is proved. We now prove the uniqueness of such solutions. Let u1 and u2 be two such solutions. Then u1 − u2 satisfies (

(u1 − u2 )t − ∆(u1 − u2 ) = 0

in QT ,

u1 − u2 = 0

on ST .

Similar to the above arguments, Z Ω

|u1 (·, t) − u2 (·, t)|p 6

Z Ω

|u1 (·, τ ) − u2 (·, τ )|p , 0 < τ < t 6 T.

As both u1 and u2 satisfy (1.13), letting τ → 0, we have Z Ω

|u1 (·, t) − u2 (·, t)|p = 0, 0 < t 6 T.

Thus u1 = u2 , and the proof is finished.



1.2. The Lp and Schauder theories  13

1.2.2

Schauder theory

For convenience, let us denote B = {n, λ, Λ, T }. Let Ω0 ⊂ Ω00 ⊂ Ω with dist(Ω0 , Ω\Ω00 ) > 0. Denote S1 = ∂Ω∩Ω0 and S2 = ∂Ω∩Ω00 . If S1 = ∅, then Ω00 will be chosen so that S2 = ∅. Let 0 6 t1 < t2 6 T . We take 0 < t0 < t1 if t1 > 0, and t0 = 0 if t1 = 0. Set Q1 = Ω0 × (t1 , t2 ), Q2 = Ω00 × (t0 , t2 ) and S02 = S2 × (t0 , t2 ). We first state the local Schauder estimate. Theorem 1.12 (Local Schauder estimate [60, Chapter 4, Theorem 10.1]) Assume that the condition (A) holds. Let u ∈ C 2+α, 1+α/2 (Q2 ) and L u = f in Q2 . Suppose aij , bi , c, f ∈ C α, α/2 (Q2 ). In addition, if t0 = 0 we assume u|t=0 = ϕ(x) in Ω00 , and if S2 6= ∅ we assume either u=g

on S02 ,

(1.15)

or ∂n u + bu = g on S02 and b ∈ C 1+α, (1+α)/2 (S 02 ).

(1.16)

Then there exists a constant C depending on B, |aij , bi , c|α, Q2 and |b|1+α, S 02 so that 

|u|2+α, Q1 6 C |f |α, Q2 + |g|2+α, S 02 + |ϕ|2+α, Ω00 + kuk∞, Q2



(1.17)



(1.18)

when the boundary condition is (1.15), and 

|u|2+α, Q1 6 C |f |α, Q2 + |g|1+α, S 02 + |ϕ|2+α, Ω00 + kuk∞, Q2

when the boundary condition is (1.16). Moreover, if S2 = ∅ then the right hand sides of (1.17) and (1.18) will not contain norm of g, and the constant C also depends on 1/d(Ω0 , ∂Ω00 ); if t0 > 0 then they will not contain |ϕ|2+α, Ω00 , and the constant C also depends on t0 and t1 . As a consequence of Theorem 1.12, we have the following results. Remark 1.13 Assume that L u = f in QT . (1) Interior Schauder estimate in t direction: Suppose u ∈ C 2+α, 1+α/2 (Ω × (0, T ]). Taking into account that Ω is connected, we can choose Ω0i and Ω00i to have the above properties such that Ω01 ∪ Ω02 = Ω. Take 0 < t0 < t1 < T and t2 = T . Using the estimates (1.17) and (1.18), it follows that 

|u|2+α, Ω×[t1 ,T ] 6 C |f |α, Ω×[t0 ,T ] + |g|2+α, Ω×[t0 ,T ] + kuk∞, Ω×(t0 ,T )



when the boundary condition is u = g, here constant C depends on B, t0 , t1 , |aij , bi , c|α, Ω×[t0 ,T ] ; while 

|u|2+α, Ω×[t1 ,T ] 6 C |f |α, Ω×[t0 ,T ] + |g|1+α, Ω×[t0 ,T ] + kuk∞, Ω×(t0 ,T )



(1.19)

when the boundary condition is ∂n u + bu = g with b ∈ C 1+α, (1+α)/2 (S T ) and b > 0. Here constant C also depends on |b|1+α, ∂Ω×[t0 ,T ] .

14  1. Preliminaries

(2) Interior Schauder estimate in x direction: Suppose u ∈ C 2+α, 1+α/2 (Ω × [0, T ]). Take Ω0 b Ω00 b Ω, t0 = t1 = 0 and t2 = T . Then estimates (1.17) and (1.18) coincide and are reduced to 



|u|2+α, Ω0 ×[0,T ] 6 C |f |α, Ω00 ×[0,T ] + |ϕ|2+α, Ω00 + kuk∞, Ω00 ×(0,T ) , where constant C depends on B, |aij , bi , c|α, Ω00 ×[0,T ] and 1/d(Ω0 , ∂Ω00 ). Theorem 1.14

Assume that Ω is of class C 2+α , b ∈ C 1+α, (1+α)/2 (Γ1 × [0, T ]).

(1) Local Schauder estimate in x direction: Let Ωl ⊂ Ω0l ⊂ Ω with d(Ωl , Ω \ Ω0l ) = ε > 0, and Ωl ∩ ∂Ω = Ω0l ∩ ∂Ω = Γl for l = 1 or l = 2. Denote Dl0 = Ω0l × (0, T ]. Assume that aij , bi , c ∈ C α, α/2 (Dl0 ) and u ∈ C 2+α, 1+α/2 (Dl0 ) satisfies (

L u = f (x, t)

in Dl0 ,

Bl u = gl

on Γl × (0, T ].

Then 



|u|2+α, Ωl ×[0,T ] 6 C |f |α, D0 + |gl |l+α, Γl ×[0,T ] + |u(·, 0)|2+α, Ω0 + kuk∞, Dl0 , l

l

where the constant C depends on B, |aij , bi , c|α, Q0 and 1/ε, and also depends l on |b|1+α, Γ1 ×[0,T ] when l = 1. (2) Interior Schauder estimate in t direction: Let aij , bi , c, f ∈ C α, α/2 (Ω × (0, T ]), u ∈ C 2+α, 1+α/2 (Ω × (0, T ]) satisfy (1.3). Take Ωl and Ω0l as in conclusion (1). Then, for any 0 < δ < τ < T , we have 

|u|2+α, Ωl ×[τ,T ] 6 C |f |α, Ω0 ×[δ,T ] + |gl |l+α, Γl ×[δ,T ] + kuk∞, Ω0l ×[δ,T ]



l

for l = 1, 2. Especially, if we take Ωl and Ω0l such that Ω1 ∪ Ω2 = Ω, then |u|2+α, Ω×[τ,T ] 6 C |f |α, Ω×[δ,T ] +

2 X

!

|gl |l+α, Γl ×[δ,T ] + kuk∞, Ω×(δ,T ) ,

l=1

where the constant C depends only on B, |aij , bi , c|α, Ω×[δ,T ] , 1/(τ − δ) and |b|1+α, Γ1 ×[δ,T ] . Theorem 1.15 (Global Schauder estimate) Define g = gl on Γl × (0, T ]. Let conditions (C) and (E) hold, Ω be of class C 2+α and b ∈ C 1+α, (1+α)/2 (Γ1 × [0, T ]). If u ∈ C 2+α, 1+α/2 (QT ) solves (1.3), then we have the global estimate: |u|2+α, QT 6 C |f |α, QT +

2 X

!

|gl |l+α, Γl ×[0,T ] + |ϕ|2+α, Ω ,

l=1

where the constant C depends on B, |aij , bi , c|α, QT , Ω and |b|1+α, Γ1 ×[0,T ] .

(1.20)

1.2. The Lp and Schauder theories  15

Proof. Taking advantage of Theorem 1.14 (1), we have |u|2+α, Qs 6 C |f |α, QT +

2 X

!

|gl |l+α, Γl ×[0,T ] + |ϕ|2+α, Ω + kuk∞, Qs

(1.21)

l=1

for all 0 6 s 6 T , which implies kut k∞,Qs 6 C |f |α, QT +

2 X

!

|gl |l+α, Γl ×[0,T ] + |ϕ|2+α, Ω + kuk∞, Qs .

(1.22)

l=1

Denote A = |f |α, QT +

2 X

|gl |l+α, Γl ×[0,T ] + |ϕ|2+α, Ω .

l=1

It then follows from (1.22) that, for all x ∈ Ω and 0 6 t 6 τ 6 T , |u(x, t)| = |ϕ(x)| +

t

Z

d |u(x, s)| ds

0

6 kϕk∞, Ω + Z

6 CA + C 6 CA + C

t

Z

|us (x, s)|

0 t

Z0 τ 0

kuk∞, Qs kuk∞, Qs .

Taking the maximum with respect to x ∈ Ω and 0 6 t 6 τ , we have kuk∞,Qτ 6 CA + C

Z 0

τ

kuk∞, Qs

for all 0 < τ 6 T.

Set v(τ ) = kuk∞, Qτ , then the Gronwall inequality gives v(T ) 6 C(T )A. This combines with (1.21) enables us to derive the estimate (1.20).  Theorem 1.16 (Schauder theory for the first initial-boundary value problem [72, Theorem 5.14], [60, Chapter 7, Theorem 10.1]) Let conditions (C) and (E) hold, and Ω be of class C 2+α . Then problem (1.1) has a unique solution u ∈ C(QT ) ∩ C 2,1 (QT ). Moreover, if ϕ and g satisfy compatibility conditions: ϕ(x) = g(x, 0) for x ∈ ∂Ω, gt − aij Dij ϕ + bi Di ϕ + cϕ = f for x ∈ ∂Ω, t = 0, then u ∈ C 2+α, 1+α/2 (QT ) and 



|u|2+α, QT 6 C |f |α, QT + |g|2+α, QT + |ϕ|2+α, Ω , where C is a constant which depends only on B, |aij , bi , c|α, QT and Ω.

16  1. Preliminaries

Theorem 1.17 (Schauder theory for the second initial-boundary value problem [72, Theorem 5.18]) Suppose that conditions (C) and (E) hold, Ω is of class C 2+α and b ∈ C 1+α, (1+α)/2 (S T ). If ϕ and g satisfy the compatibility condition ∂n ϕ + bϕ = g for t = 0, x ∈ ∂Ω, then problem (1.2) has a unique solution u ∈ C 2+α,1+α/2 (QT ), and 

|u|2+α, QT 6 C |f |α, QT + |g|1+α, S T + |ϕ|2+α, Ω



with a constant C relying on B, |aij , bi , c|α, QT , |b|1+α, S T and Ω. Theorem 1.18 (Schauder theory for the mixed initial-boundary value problem) Set g = gl on Γl × (0, T ]. Let conditions (C) and (E) hold, Ω be of class C 2+α and b ∈ C 1+α, (1+α)/2 (Γ1 × [0, T ]). Assume that compatibility conditions ∂n ϕ + bϕ = g1 for t = 0, x ∈ Γ1 , and ϕ = g2 , g2t − aij Dij ϕ + bi Di ϕ + cϕ = f for t = 0, x ∈ Γ2 are satisfied. Then problem (1.3) has a unique solution u ∈ C 2+α, 1+α/2 (QT ), and |u|2+α, QT 6 C |f |α, QT +

2 X

!

|gl |l+α, Γl ×[0,T ] + |ϕ|2+α, Ω ,

l=1

where C is a constant depending on B, |aij , bi , c|α, QT , Ω and |b|1+α, Γ1 ×[0,T ] . Proof. The global estimate (1.20) implies that problem (1.3) has at most one solution in the class of C 2+α, 1+α/2 (QT ). The remaining proof of this theorem is similar to those of Theorems 1.16 and 1.17, and therefore is omitted.  A conclusion similar to Theorem 1.18 is given in [71]. Theorem 1.19 (Interior Schauder estimate) Let the condition (C) hold, and u ∈ W22,1 (QT ) be a solution of (1.7). Then u ∈ C 2+α, 1+α/2 (QT ), and for any Q b QT , 



|u|2+α, Q 6 C |f |α, QT + kuk2, QT ,

(1.23)

where C is a constant which relies on B, |aij , bi , c|α, QT and 1/d(Q, ∂p QT ). This theorem can be proved in the same way as that of Theorem 1.9 (2), and we leave the proof to readers as an exercise. 1.2.3

Existence and uniqueness of solutions of linear system

Give m parabolic operators Lk = ∂t − akij (x, t)Dij + bki (x, t)Di ,

k = 1, . . . , m.

(1.24)

Assume that akij satisfies the condition (A) with the same λ and Λ for all 1 6 k 6 m, and the modulus of continuity ωk (R) of akij are the same as ω(R) defined in (A). Let bki , hkl ∈ L∞ (QT ) with kbki , hkl k∞ 6 Λ, gk and ϕk satisfy the condition (D).

1.3. Maximum principle of classical solutions  17

Theorem 1.20 (Lp theory) Consider the following problem  m X    L u + hkl (x, t)ul = fk (x, t)  k k    l=1 

Bk uk = gk

   uk (x, 0) = ϕk (x)     

in QT , on ST ,

(1.25)

in Ω,

k = 1, . . . , m,

where either Bk uk = uk , or Bk uk = ∂n uk + bk uk with bk > 0. Assume that Ω is of class C 2 , gk and ϕk satisfy requirements of Theorem 1.1 when Bk uk = uk , while bk , gk and ϕk satisfy requirements of Theorem 1.2 when Bk uk = ∂n uk + bk uk . If fk ∈ Lp (QT ), then problem (1.25) has a unique solution u ∈ [Wp2,1 (QT )]m , and kuk kWp2,1 (QT ) 6 C

m  X



kfk kp, QT + kgk kWp2,1 (QT ) + kϕk kWp2 (Ω) ,

k=1

k = 1, . . . , m, where C is a constant depending on n, QT , λ, Λ, max16l6m |bl |1, S T and max16l6m ωl (R). This theorem can refer to Theorem 10.4 of [60, Chapter 7].

1.3 MAXIMUM PRINCIPLE OF CLASSICAL SOLUTIONS An important tool in theory of second order parabolic equations is the maximum principle, which avers that a solution of a given homogeneous linear parabolic equation in a domain must achieve its maximum on parabolic boundary of that domain. The maximum principle will be used to prove uniqueness results for various initialboundary value problems, to obtain L∞ bounds for solutions and their derivatives, and to derive various continuity estimates as well. 1.3.1

Initial-boundary value problems

An available result related to the strong maximum principle is the so-called boundary point lemma, which states that if a non-constant solution takes local strict maximum at some point on boundary, then its outward normal derivative at this point must be positive. Lemma 1.21 (Hopf boundary lemma [91] or [110, Lemmma 2.4.5]) Let aii , bi ∈ L∞ (Q), L be parabolic and c > 0 in Q. Let u ∈ C(Q)∩C 2,1 (Q) satisfy L u 6 0 (> 0) in Q. Suppose u(x0 ) = M = maxQ u (u(x0 ) = m = minQ u) for some x0 = (x0 , t0 ) ∈ ∂Q, and there is a neighbourhood V of x0 such that u < M (u > m) in Q ∩ V , where M > 0 (m 6 0) when c 6≡ 0. We further assume that there exists a ball Br (x∗ ) centred at x∗ = (x∗ , t∗ ) with radius r > 0 and x∗ 6= x0 , such that Br (x∗ ) ⊂ Q and Br (x∗ ) ∩ ∂Q = {x0 }. Then the derivative ∂ν u(x0 ) > 0 (< 0) when it exists, where ν is the outward normal vector of ∂Q at point x0 .

18  1. Preliminaries

Lemma 1.22 (Hopf boundary lemma [72, Lemmma 2.8]) Let R and η be positive constants. We define Γ = Γ(R, η, (x0 , t0 )) = {|x − x0 |2 + η(t0 − t) < R2 , t < t0 }. Assume that aii , bi ∈ L∞ (Γ), L is parabolic in Γ, u ∈ C(Γ) ∩ C 2,1 (Γ) satisfies L u 6 0 (> 0) in Γ. Also suppose that there exists (x∗ , t0 ) with |x∗ −x0 | = R such that u(x∗ , t0 ) > (6) u(x, t), c(x, t)u(x∗ , t0 ) > (6) 0 u(x∗ , t0 ) > ( 0 (< 0) when it exists. Lemma 1.23 (Hopf boundary lemma) Let aii , bi ∈ L∞ (QT ), L be parabolic and c > 0 in QT . Let u ∈ C(QT ) ∩ C 2,1 (QT ) satisfy L u 6 0 (> 0) in QT . Suppose that u(x0 , t0 ) = M = maxQT u (u(x0 , t0 ) = m = minQT u) for some (x0 , t0 ) ∈ ST , and there is a neighbourhood V of (x0 , t0 ) such that u < M (u > m) in QT ∩ V . We further assume that Ω has the interior ball property at x0 and M > 0 (m 6 0) when c 6≡ 0. Then the derivative ∂n u(x0 , t0 ) > 0 (< 0) when it exists. Theorem 1.24 (Strong maximum principle [91] or [72, Theorem 2.7]) Let aii , bi ∈ L∞ (Q), L be parabolic and c > 0 in Q. Assume that u ∈ C(Q) ∩ C 2,1 (Q) satisfies L u 6 0 (> 0) in Q. If u(x0 ) = maxQ u = M (u(x0 ) = minQ u = m) for some x0 ∈ Q \ ∂p Q, and M > 0 (m 6 0) when c 6≡ 0, then u ≡ u(x0 ) in S(x0 ). Corollary 1.25 (Strong maximum principle) Let aii , bi ∈ L∞ (QT ), L be parabolic and c > 0 in QT . Assume that u ∈ C(QT ) ∩ C 2,1 (QT ) satisfies L u 6 0 (> 0) in QT . If u(x0 , t0 ) = maxQT u = M (u(x0 , t0 ) = minQT u = m) for some x0 ∈ Ω, 0 < t0 6 T , and M > 0 (m 6 0) when c 6≡ 0, then u ≡ u(x0 , t0 ) in Qt0 . Lemma 1.26 (Positivity lemma) Let aii , bi ∈ L∞ (QT ), L be parabolic and c be bounded below in QT . Let a, b > 0 with a + b > 0 on ST . For any (x, t) ∈ ST , if a(x, t) 6= 0, then we further assume that Ω has the interior ball property at point x. Suppose u ∈ C(QT ) ∩ C 2,1 (QT ), and when a 6≡ 0 we further assume u ∈ C 1,0 (Ω × (0, T ]). If u satisfies  Lu > 0      

in QT ,

a∂n u + bu > 0

on ST ,

u(x, 0) > 0

in Ω,

(1.26)

then u > 0 in QT . If in addition u(x, 0) 6≡ 0, then u > 0 in QT , and u > 0 in Ω × (0, T ] when a > 0 on ST .

1.3. Maximum principle of classical solutions  19

Proof. Take a constant k large enough so that k + c > 0 in QT , and let w = ue−kt . Then w(x, 0) > 0 in Ω, a∂n w + bw > 0 on ST , and wt − aij Dij w + bi Di w + (k + c)w > 0 in QT . Assume on the contrary that w(x0 , t0 ) = minQT w < 0, then t0 > 0. Moreover, x0 6∈ Ω by Corollary 1.25. Thus x0 ∈ ∂Ω and w > w(x0 , t0 ) in QT . Obviously a(x0 , t0 ) 6= 0. Applying Lemma 1.23 we have ∂n w(x0 , t0 ) < 0, which leads to a∂n w(x0 , t0 ) + bw(x0 , t0 ) < 0. It is a contradiction. In addition, if w(x, 0) 6≡ 0, then w > 0 in QT by Corollary 1.25. Furthermore, if a > 0 on ST , then w > 0 in Ω × (0, T ] by Lemma 1.23.  Here we should remark that, in Lemma 1.26, if it is only assumed that u 6≡ 0 then we cannot conclude u > 0 in QT . For example, take L u = ut − uxx , Ω = (0, 1) and T = 2. Let u(t) = e−1/(t−1) for 1 < t 6 2, u(t) ≡ 0 for 0 6 t 6 1. Then u satisfies (1.26) and u 6≡ 0. However, u > 0 does not hold in QT = (0, 1)×(0, 2]. 1.3.2

Cauchy problems

Let us consider the following Cauchy problem (

Lu = f

in Rn × (0, T ],

u(x, 0) = ϕ(x)

in Rn ,

(1.27)

where the coefficient c of L may be unbounded in Rn × (0, T ]. Lemma 1.27 Assume that aii , bi ∈ L∞ (Rn × (0, T ]), L is parabolic in Rn × (0, T ] and there exists a constant c0 > 0 such that c + c0 > 0 in Rn × (0, T ]. Let u ∈ C 2,1 (Rn × (0, T ]) ∩ C(Rn × [0, T ]) be a solution of (1.27). If f > 0 in Rn × (0, T ], ϕ > 0 in Rn and lim inf u(x, t) > 0 |x|→∞

uniformly in t ∈ [0, T ]. Then u > 0 in Rn × (0, T ]. Proof. Invoking a transform v = ue−c0 t we assume without loss of generality that c > 0. For any given ε > 0, there holds that ϕ(x) + ε > 0 in Rn , and L [u + ε] = L u + cε > 0 in Rn × (0, T ].

By our assumption, there exists R0  1 such that u + ε > 0 on {|x| = R, 0 6 t 6 T } for all R > R0 . Then we can apply Corollary 1.25 to deduce u + ε > 0 in QR T := {|x| < R} × [0, T ]. Hence u > 0 in QR by letting ε → 0. Arbitrariness of R implies T u > 0 in Rn × (0, T ].  Theorem 1.28 (Maximum principle [36, 110]) Let the operator L be parabolic in Rn × (0, T ], and

20  1. Preliminaries

(F)

|aij | 6 M, |bi | 6 M (1 + |x|) and c > −M (1 + |x|2 ).

Assume that u ∈ C 2,1 (Rn × (0, T ]) ∩ C(Rn × [0, T ]) satisfies L u > 0 in Rn × (0, T ], and there exist positive constants B and k such that u > −Bek|x|

2

in Rn × (0, T ].

(1.28)

If u(x, 0) > 0 in Rn , then u > 0 in Rn × (0, T ]. Proof. Take γ, ` > 0 which will be determined later. Define a function (

w = exp

)

1 2k|x|2 + γt , 0 6 t 6 . 1 − `t 2`

A straightforward computation shows that Lw 16k 2 4k 4k 2`k|x|2 =γ− a x x − a + b x + c + . ij i j ii i i w (1 − `t)2 1 − `t 1 − `t (1 − `t)2

Making use of the condition (F),  Lw  > 2k` − (64k 2 n2 + 12kn + 1)M |x|2 + γ − (12kn + 1)M. w

We can take `, γ > 0 so large that Lw > 0. w

(1.29)

For such fixed `, γ, we denote T1 = 1/(2`) and investigate function v = u/w in DT1 . Condition (1.28) implies lim inf |x|→∞ v(x, t) > 0 uniformly in t ∈ [0, T1 ]. A straightforward computation leads to L¯v := vt − aij Dij v + ¯bi Di v + c¯v = f¯,

where ¯bi = bi − 2

X j

aij

Dj w Lw Lu , c¯ = and f¯ = > 0. w w w

According to (1.29), c¯ > 0. By a similar argument in the proof of Lemma 1.27 we have v > 0. As a result, u > 0 in Rn × (0, T1 ]. Since T1 > 0 is a fixed constant, repeating the above procedure we can deduce u > 0 in Rn × (0, T ].  Using Theorem 1.28 we can prove the following theorem, and its proof will be omitted and left to readers. Theorem 1.29 Let L be a parabolic operator in Rn ×(0, T ] and the condition (F) be 2 fulfilled. Then problem (1.27) has at most one solution in the sense that |u| 6 Beβ|x| with some positive constants B and β.

1.3. Maximum principle of classical solutions  21

1.3.3

Weakly coupled systems

Contents of this part are taken from [91, 124]. Let Lk be strongly parabolic operators given by (1.24), and let H(x, t) = (hkl (x, t))m×m be a function matrix of order m. We will establish the maximum principle for the following weakly coupled linear parabolic inequalities Lk uk +

m X

hkl ul 6 0 (> 0),

k = 1, . . . , m.

l=1

It is assumed that akij , bki , hkl are bounded in QT , and functions ak , bk appearing in boundary conditions are non-negative and ak + bk > 0. Denote u = (u1 , . . . , um ). Lemma 1.30 Let Ω be of class C 2 and uk ∈ C(QT ) ∩ C 2,1 (QT ) for all 1 6 k 6 m. And when ak 6≡ 0 for some 1 6 k 6 m, we further assume uk ∈ C 1,0 (Ω × (0, T ]). Suppose that (i) hkl 6 0 for k 6= l, k, l = 1, . . . , m; (ii) Lk uk +

m X

hkl ul < 0 (> 0) in QT , k = 1, . . . , m;

l=1

(iii) u(x, 0) < 0 (> 0) in Ω; (iv) for each 1 6 k 6 m and each point (x, t) ∈ ST , either uk < 0 (> 0), or ak ∂n uk + bk uk 6 0 (> 0) and ak > 0 at (x, t). Then u < 0 (> 0) in QT . Proof. Take a constant α > 0 so that hkk + α > 0 in QT for all k, and let uk = vk eαt . In view of the condition (ii), L k vk +

m X

˜ kl vl < 0 in QT , h

k = 1, . . . , m,

(1.30)

˜ kk = hkk + α > 0, k = 1, . . . , m, h ˜ kl = hkl 6 0, h k= 6 l, k, l = 1, . . . , m.

(1.31)

l=1

where (

Set v = (v1 , . . . , vm ). Thanks to the fact that v(x, 0) < 0 in Ω, there exists δ > 0 such that v(x, t) < 0 for x ∈ Ω and 0 6 t 6 δ. Define 



A = t : t 6 T, v(x, s) < 0 for all x ∈ Ω, 0 6 s 6 t . Then t0 = supA exists and 0 < t0 6 T . Assume, on the contrary, that our conclusion is not valid, then v < 0 in Ω × [0, t0 ) and there exists x0 ∈ Ω such that vk (x0 , t0 ) = 0 for some k. If x0 ∈ Ω, then vk achieves its maximum over Qt0 at (x0 , t0 ), and ∂t vk (x0 , t0 ) > 0, Di vk (x0 , t0 ) = 0, akij Dij vk (x0 , t0 ) 6 0.

22  1. Preliminaries

These imply Lk vk |(x0 ,t0 ) > 0. On the other hand, thanks to vk (x0 , t0 ) = 0 and vl (x0 , t0 ) 6 0 for all l 6= k, it follows from (1.31) that Lk vk (x0 , t0 ) +

m X

˜ kl vl (x0 , t0 ) > 0. h

l=1

This contradicts (1.30). Thus, x0 ∈ ∂Ω and vk (x, t0 ) < 0 in Ω, and subsequently, vk < 0 in Qt0 . Moreover, ak (x0 , t0 ) > 0 by the condition (iv). Due to v 6 0 in Qt0 , we have X ˜ kk vk + ˜ kk vk in Qt . 0 > L k vk + h hkl vl > Lk vk + h 0 l6=k

According to the Hopf boundary lemma, we have ∂n vk (x0 , t0 ) > 0, and so 0 < ak ∂n vk (x0 ,t0 ) = (ak ∂n vk + bk vk ) (x0 ,t0 ) 6 0.



This contradiction shows that v < 0, i.e., u < 0 in QT .



Theorem 1.31 (Maximum principle) Let Ω be of class C 2 and hkl 6 0 in QT for k 6= l, k, l = 1, . . . , m.

(1.32)

Assume uk ∈ C(QT ) ∩ C 2,1 (QT ) for all 1 6 k 6 m. And when ak ≡ 6 0 for some 1 6 k 6 m, we further assume uk ∈ C 1,0 (Ω × (0, T ]). Suppose that the followings hold: (i) Lk uk +

Pm

l=1

hkl ul 6 0 (> 0) in QT for each 1 6 k 6 m;

(ii) u(x, 0) 6 0 (> 0) in Ω; (iii) ak ∂n uk + bk uk 6 0 (> 0) on ST for each 1 6 k 6 m. Then we have (1) u 6 0 (> 0) in QT ; (2) if there exist (x0 , t0 ) ∈ QT and 1 6 k 6 m such that uk (x0 , t0 ) = 0, then uk ≡ 0 in Qt0 . Proof. We first prove (1). Since hkl are bounded, there is β > 0 so that β+

m X

hkl > 0 in QT for all 1 6 k 6 m.

l=1

Then function vk = uk − εeβt satisfies L k vk +

m X l=1

hkl vl = Lk uk +

m X l=1

βt

hkl ul − εe

β+

m X

!

hkl

< 0,

l=1

vk (x, 0) < 0 in Ω, and ak ∂n vk + bk vk = ak ∂n uk + bk uk − εbk e−βt 6 0 on ST .

1.3. Maximum principle of classical solutions  23

By means of Lemma 1.30, vk < 0 in QT . Letting ε → 0 we derive uk 6 0 in QT . Now we prove the conclusion (2). Note that 0 > Lk uk +

m X

hkl ul = Lk uk + hkk uk +

X

hkl ul > Lk uk + hkk uk .

l6=k

l=1

Take γ > 0 to satisfy γ + hkk > 0, and set uk = vk eγt . We have Lk vk + (γ + hkk )vk 6 0.

The desired result can be deduced by the maximum principle.  Similarly, we can prove the Hopf boundary lemma for weakly coupled systems. Theorem 1.32 (Hopf boundary lemma) Let u ∈ [C(QT ) ∩ C 2,1 (QT )]m and u 6 0 (> 0) satisfy the condition (i) of Theorem 1.31. Let H satisfy (1.32). Assume that there exist 1 6 k 6 m and (x0 , t0 ) ∈ ST such that uk (x0 , t0 ) = 0. If Ω has the interior ball property at x0 ∈ ∂Ω, and there is a neighbourhood V of (x0 , t0 ) so that uk < 0 (uk > 0) in QT ∩ V , then ∂n uk (x0 , t0 ) > 0 (< 0) when it exists. We should mention that, in general, the maximum principle does not hold for strongly coupled parabolic systems. In fact, if we take u = −ex+t and v = t − 4(x − 1/2)2 , then it is easily verified that (u, v) satisfies (

ut − uxx 6 0

in (0, 1] × (0, 1],

vt − vxx + 9ux 6 0

in (0, 1] × (0, 1],

and (u, v)|t=0 6 0 for x ∈ [0, 1],

(u, v)|x=0,1 6 0 for t ∈ (0, 1].

However, v|x= 1 = t > 0. 2 In addition, such (u, v) also satisfies ut − uxx 6 0, vt − vxx + 9u 6 0. These discussions show that the maximum principle is not valid for such a system of inequalities. The reason is that h21 = 9 > 0, and the requirement (1.32) is not satisfied. 1.3.4

Equations with non-classical boundary conditions

This subsection is concerned with the maximum principle for initial-boundary value problems of parabolic equations with non-classical boundary conditions. The content of this part is due to [105]. Theorem 1.33 Let Ω be of class C 2 , b ∈ C(S T ), and aij , bi , c ∈ L∞ (QT ). Assume that there exists λ > 0 such that aij (x, t)ξi ξj > λ|ξ|2

for all (x, t) ∈ QT , ξ ∈ Rn .

24  1. Preliminaries

If u ∈ C(QT ) ∩ C 2,1 (QT ) ∩ C 1,0 (Ω × (0, T ]) satisfies  u > aij Dij u + bi Di u + cu    t

in QT ,

∂n u > b(x, t)u

on ST ,

  

u(x, 0) > 0

in Ω,

then u > 0 in QT . Additionally, if u(x, 0) 6≡ 0, then u > 0 in QT . Proof. Take a constant M > 0 so that maxS T |b| 6 M . Let λ be the first eigenvalue of the eigenvalue problem (

−∆φ = λφ

in Ω,

φ=0

on ∂Ω,

(1.33)

and φ be the corresponding positive eigenfunction. Then φ > 0 in Ω and ∂n φ < 0 on ∂Ω. We call (λ, φ) the first eigen-pair of (1.33). There exists k > 0 such that ∂n φ 6 −k on ∂Ω. Let us define h(x) = exp{`(1 + φ(x))},

w(x, t) = u(x, t)h(x),

where ` is a positive constant to be determined. A direct calculation yields  Di h   wt > aij Dij w − 2aij Dj w + bi Di w    h       2Di hDj h − hDij h Di h   + aij − bi +c w 2

h

h

     1   ∂ w + −b − ∂ h w>0  n n   h   

w(x, 0) = u(x, 0)h(x) > 0

As

in QT ,

(1.34)

on ST , in Ω.

1 ∂n h = `∂n φ 6 −`k, h

we have

1 ∂n h > −b + `k > 0 when `k > M. h Applying Lemma 1.26 to (1.34) we get w > 0, and w > 0 in QT when u(x, 0) 6≡ 0.  −b −

1.3.5

Equations with non-local terms

In this subsection we study the maximum principle of initial-boundary value problems with non-local terms. The content in this part is taken from [110]. We first discuss the case of non-local terms appearing in differential equations.

1.3. Maximum principle of classical solutions  25

Theorem 1.34 Let Ω be of class C 2 , a and b be two non-negative continuous functions with a + b > 0 on ST , and let c1 and c2 be two bounded continuous functions with c2 > 0 in QT . If u ∈ C 2,1 (QT ) ∩ C 1,0 (QT ) verifies Z    u − ∆u > c u + c2 (y, t)u(y, t)dy  t 1  Ω

a∂n u + bu > 0    

in QT , on ST ,

u(x, 0) > 0

in Ω,

then u > 0 in QT . In addition, if u(x, 0) 6≡ 0, then u > 0 in QT . Proof. Take c¯i = supQT |ci | for i = 1, 2, and γ > c¯1 + 2¯ c2 |Ω|. Set v = e−γt u, then we have Z    v − ∆v > (c − γ)v + c2 (y, t)v(y, t)dy  t 1  Ω

a∂n v + bv > 0    

in QT , on ST ,

v(x, 0) > 0

in Ω.

Assume that v acquires its negative minimum at some point (x0 , t0 ). In the case of x0 ∈ Ω, we have vt (x0 , t0 ) 6 0 and ∆v(x0 , t0 ) > 0, and hence [γ − c1 (x0 , t0 )]v(x0 , t0 ) > >

Z

c2 (y, t0 )v(y, t0 )dy ZΩ

c2 (y, t0 )v(x0 , t0 )dy Ω

> c¯2 v(x0 , t0 )|Ω|, where we have used 0 6 c2 (y, t0 ) 6 c¯2 and v(x0 , t0 ) < 0. The above inequality implies γ − c1 (x0 , t0 ) 6 c¯2 |Ω|, and thereupon γ 6 c1 (x0 , t0 ) + c¯2 |Ω| 6 c¯1 + c¯2 |Ω|. This contradicts the fact γ > c¯1 + 2¯ c2 |Ω|. In the case of x0 ∈ ∂Ω, we have ∂n v(x0 , t0 ) 6 0. Then the boundary condition implies that a(x0 , t0 ) > 0, b(x0 , t0 ) = 0 and ∂n v(x0 , t0 ) = 0. On the other hand, as v(x0 , t0 ) < 0, there exists a neighbourhood V of (x0 , t0 ) such that v 6 21 v(x0 , t0 ) in QT ∩ V . Noticing c1 − γ < 0, c2 > 0,

and v(x0 , t0 ) < 0,

we see that in QT ∩ V , vt − ∆v > (c1 − γ)v +

Z

c2 (y, t)v(y, t)dy Ω

c1 − γ v(x0 , t0 ) + c2 (y, t)v(x0 , t0 )dy 2 Ω c1 − γ > v(x0 , t0 ) + c¯2 |Ω|v(x0 , t0 ) 2 c1 − γ + 2¯ c2 |Ω| = v(x0 , t0 ) > 0. 2 Z

>

26  1. Preliminaries

By the Hopf boundary lemma, ∂n v(x0 , t0 ) < 0. This contradicts the fact ∂n v(x0 , t0 ) = 0. Thus v > 0, and then u > 0. Since c2 > 0, we have ut − ∆u > c1 u +

Z Ω

c2 (y, t)u(y, t)dy > c1 u in QT .

Therefore, u > 0 in QT so long as u(x, 0) 6≡ 0.  Now we study the case that the boundary conditions have non-local terms. The following theorem can be proved in the same way as that of Theorem 1.33, and therefore it is omitted. Theorem 1.35 Let Ω be of class C 2 , d > 0 be a constant and u ∈ C(QT ) ∩ C 2,1 (QT ) ∩ C 1,0 (Ω × (0, T ]), satisfy    ut − d∆u > 0Z  

∂n u > c1 u +

   

in QT , c2 (y, t)u(y, t)dSy

on ST ,

∂Ω

u(x, 0) > 0

in Ω.

Assume that ci is continuous and bounded for i = 1, 2, and c2 > 0 on ST . Then u > 0 in QT . Furthermore, u > 0 in QT when u(x, 0) 6≡ 0. In the sequel we investigate the case that non-local terms appear in both the differential equation and boundary condition. Lemma 1.36 Let bi , b ∈ L∞ (QT ), c, f > 0 in QT and g > 0 in Ω × ST . Assume that u ∈ C(QT ) ∩ C 2,1 (QT ) satisfies Z    u − ∆u − b D u − bu > f (x, t) c(y, t)udy in QT , t i i    Z Ω

u>     

g(y, x, t)u(y, t)dy

on ST ,

Ω

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

in Ω.

Then u > 0 in QT . Proof. Take ¯b = supQT |b| and γ > ¯b. Then function v = e−γt u satisfies Z    v − ∆v + (γ − b)v − b D v > f (x, t) c(y, t)vdy t i i    Z Ω

v>     

g(y, x, t)v(y, t)dy

in QT , on ST ,

Ω

v(x, 0) = u0 (x) > 0

in Ω.

As v(x, 0) > 0, there exists δ > 0 so that v > 0 in Ω × [0, δ]. Set A := {t 6 T : v > 0 in Ω × [0, t]}, and t0 := sup A,

(1.35)

1.3. Maximum principle of classical solutions  27

then 0 < t0 6 T . If t0 < T , then v > 0 in Ω×(0, t0 ), and v(x0 , t0 ) = 0 for some x0 ∈ Ω. The boundary condition implies v > 0 on ∂Ω × (0, t0 ]. Consequently v achieves its minimum over Qt0 at (x0 , t0 ). It then follows from the first inequality of (1.35) that vt − ∆v + (γ − b)v − bi Di v > 0 in Qt0 . Corollary 1.25 asserts v ≡ 0 in Qt0 . We get a contradiction, and therefore v > 0, i.e., u > 0 in QT .  Lemma Z 1.37 with Ω

Assume that bi , b, f, c ∈ L∞ (QT ) and f, c > 0. Let g > 0 in Ω × ST

g(y, x, t)dy < 1 for all (x, t) ∈ ST . Let u ∈ C(QT ) ∩ C 2,1 (QT ) satisfy Z    u − ∆u − b D u − bu > f (x, t) c(y, t)udy in QT , t i i   Ω  Z

u>     

on ST ,

g(y, x, t)u(y, t)dy Ω

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

in Ω.

Then u > 0 in QT . Proof. Define ¯b = supQT |b| and β = supQT f (x, t) set w = u + εeγt with ε > 0. Then we have

Z

c(y, t)dy. Take γ > ¯b + β, and

Ω

Z    w − ∆w − b D w − bw > f (x, t) c(y, t)wdy t i i    Z Ω

w>     

g(y, x, t)w(y, t)dy

in QT , on ST ,

Ω

w(x, 0) = u0 (x) + ε > 0

in Ω.

As a consequence, w > 0 by Lemma 1.36, i.e., u + εeγt > 0 in QT . The desired result is obtained by sending ε → 0+ .  Theorem 1.38 Assume bi , b, f, c ∈ L∞ (QT ), g ∈ L∞ (Ω × ST ) and f, c, g > 0. If u ∈ C(QT ) ∩ C 2,1 (QT ) satisfies Z    u − ∆u − b D u − bu > f (x, t) c(y, t)udy t i i    Z Ω

u>     

g(y, x, t)u(y, t)dy

in QT , on ST ,

Ω

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

in Ω,

then u > 0 in QT . Proof. As g ∈ L∞ (Ω × ST ), we can take a function z ∈ C 2 (Ω) with properties: z > 0 in Ω, z ≡ 1 on ∂Ω, and Z Ω

z(y)g(y, x, t)dy < 1 for (x, t) ∈ ∂Ω × [0, T ].

28  1. Preliminaries

Let u = z(x)w. Then a straightforward calculation yields 

Di z 2Di z ∆z f (x, t) + bi Di w + b + w+ z z z z(x) 

wt − ∆w > bi +





Z

z(y)c(y, t)wdy Ω

in QT , and Z

w(x, t) > w(x, 0) =

g(y, x, t)z(y)wdy

on ST ,

Ω

u0 (x) >0 z(x)

in Ω.

It follows from our assumptions that bi +

2Di z ∆z Di z f , b+ + bi , , zc ∈ L∞ (QT ), zZ z z z

zg > 0,

g(y, x, t)z(y)dy < 1 for all (x, t) ∈ ST ,

Ω

and f /z > 0, zc > 0. Lemma 1.37 leads to w > 0 in QT , and so u > 0 in QT .



1.4 MAXIMUM PRINCIPLE OF STRONG SOLUTIONS In this section we discuss the maximum principle of strong solutions (solutions in Wp2,1 (QT )) for non-divergence equations. Lemma 1.39 (Hopf boundary lemma for strong solutions, [84, Theorem 3.4]) Let aij , bi ∈ L∞ (QT ), Ω be of class C 2 , L be parabolic and c > 0 in QT . Let u ∈ C(QT )∩ 2,1 Wn+1, loc (QT ) satisfy L u 6 0 (> 0) in QT . Suppose that u(x0 , t0 ) = M = maxQT u (u(x0 , t0 ) = m = minQT u) for some (x0 , t0 ) ∈ ST , and there is a neighborhood V of (x0 , t0 ) such that u < M (u > m) in QT ∩ V , and M > 0 (m 6 0) when c 6≡ 0. Then the derivative ∂n u(x0 , t0 ) > 0 (< 0) when it exists. Theorem 1.40 (1) (Strong maximum principle of strong solutions) Let aij , bi ∈ L∞ (QT ), L be parabolic and c > 0 in QT . Assume that u ∈ C(QT ) ∩ C 1,0 (QT ) 2,1 ∩ Wn+1, loc (QT ) satisfies L u 6 0 (> 0) in QT . If u(x0 , t0 ) = maxQT u = M (u(x0 , t0 ) = minQT u = m) for some x0 ∈ Ω, 0 < t0 6 T , and M > 0 (m 6 0) when c 6≡ 0, then u ≡ u(x0 , t0 ) in Qt0 . (2) (Positivity lemma for strong solutions) Let aij , bi ∈ L∞ (QT ), L be parabolic and c be bounded below in QT . Let a, b ∈ C(ST ) with a, b > 0 and a + b > 0 on ST . For any (x, t) ∈ ST , if a(x, t) 6= 0, then we further assume that ∂Ω is 2,1 of class C 2 near x. Suppose u ∈ C(QT ) ∩ C 1,0 (QT ) ∩ Wn+1, loc (QT ), and when a 6≡ 0 we further assume u ∈ C 1,0 (Ω × (0, T ]). If u satisfies  Lu > 0   

in QT ,

a∂n u + bu > 0

on ST ,

  

u(x, 0) > 0

in Ω,

1.4. Maximum principle of strong solutions  29

then u > 0 in QT . If in addition u(x, 0) 6≡ 0, then u > 0 in QT , and u > 0 in Ω × (0, T ] when a > 0 on ST . Proof. The conclusion (1) can be deduced by Lemma 1.39 directly. The conclusion (2) can be proved by the same way as that of Lemma 1.26 using the conclusion (1) and Lemma 1.39 instead of Corollary 1.25 and Lemma 1.23, respectively. The details are omitted here.  In the following we consider the general case p > 1 + n/2. We first consider the first initial-boundary value problem. Theorem 1.41 Let Ω be of class C 2 , the condition (A) hold and bi , c ∈ C(QT ). Assume that p > 1 + n/2 and u ∈ Wp2,1 (QT ) satisfies  L u = f (x, t) > 0      

in QT ,

u = g(x, t) > 0

on ST ,

u(x, 0) = ϕ(x) > 0

in Ω.

Then u > 0 in QT . Proof. Step 1. As u ∈ Wp2,1 (QT ), one has f ∈ Lp (QT ). Moreover, u ∈ C α, α/2 (QT ) as p > 1 + n/2. We assume without loss of generality that c > 0. For any given σ > 0, let v = u + σet . Then v satisfies  t t   L v = (1 + c)σe + L u = (1 + c)σe + f =: fσ > σ 

in QT ,

v = σe + g =: gσ > σ

on ST ,

  

v(x, 0) = σ + ϕ(x) =: ϕσ > σ

in Ω.

t

Step 2. Take τk & 0 satisfying τk+1 < τk , and subdomain Ωk b Ω being of class C 2+α with Ωk b Ωk+1 , Ωk → Ω. Define Qk = Ωk × (τk , T ]. Since v is continuous in QT , there exists k0  1 such that, for all k > k0 , v > σ/2 in QT \ Qk−1 . For such fixed k, take a truncated function η ∈ C ∞ (QT ) so that 0 6 η 6 1, η ≡ 1 in Qk and η ≡ 0 in QT \ Qk+1 . We will see that the following proof is still valid with Qk+3 instead of QT , and hence we may assume that Ω is of class C 2+α . Let w = vη. Then w satisfies (

L w = vηt + ηfσ − vaij Dij η − 2aij Di vDj η + vbi Di η =: f ∗

in QT ,

w=0

on ∂p QT .

Obviously, f ∗ = fσ in Qk and f ∗ = 0 in QT \ Qk+1 . Step 3. Take aεij , bεi , cε , f ε ∈ C α, α/2 (QT ) such that

30  1. Preliminaries

(a) f ε = 0 in QT \ Qk+2 and f ε > σ/2 in Qk ; (b) aεij → aij , bεi → bi , cε → c in C(QT ), and f ε → f ∗ in Lp (QT ) as ε → 0+ . Consider problem (

wtε − aεij Dij wε + bεi Di wε + cε wε = f ε

in QT ,

wε = 0

on ∂p QT .

(1.36)

By Theorem 1.16, this problem has a unique solution wε ∈ C 2+α, 1+α/2 (QT ). We claim that wε → w in Wp2,1 (QT ). In fact, let z ε = w − wε and F ε (x, t) = (aij − aεij )Dij wε + (bεi − bi )Di wε + (cε − c)wε + f ∗ − f ε . Then z ε satisfies (

L zε = F ε ε

z =0

in QT , on ∂p QT .

By use of (1.4), kz ε kWp2,1 (QT ) 6 CkF ε kp, QT . Applying the estimate (1.4) to (1.36), we may assume kwε kWp2,1 (QT ) 6 K for some K and all 0 < ε  1. By the choice of (aεij , bεi , cε , f ε ), it is easy to see that limε→0+ kF ε kp, QT = 0, and this implies limε→0+ kz ε kWp2,1 (QT ) = 0, i.e., limε→0+ wε = w in Wp2,1 (QT ). Step 4. Since Wp2,1 (QT ) ,→ C(QT ), we have limε→0+ wε = w in C(QT ). Recall v > σ/2 in QT \ Qk−1 and η ≡ 1 in Qk . It follows that w > σ/2 in Qk \ Qk−1 . We may assume that wε > σ/3 in Qk \ Qk−1 . Consequently, wε satisfies (

wtε − aεij Dij wε + bεi Di wε + cε wε = f ε > 0

in Qk ,

wε > 0

on ∂p Qk .

k

k

Then wε > 0 in Q by Lemma 1.26, and as a result, v = w > 0 in Q by the fact that limε→0+ wε = w in C(QT ). By the arbitrariness of k we have v > 0 in QT , i.e., u > −σet . It then follows that u > 0 by taking σ → 0+ .  Now we consider the second initial-boundary value problem  L u = f (x, t) > 0   

in QT ,

  

in Ω,

∂n u + bu = g(x, t) > 0 on ST ,

u(x, 0) = ϕ(x) > 0

(1.37)

where b ∈ C 1+α, (1+α)/2 (S T ), b > 0 and g ∈ Wp2,1 (QT ) with p > 1 + n/2. Theorem 1.42 Let Ω be of class C 2+α , the condition (A) hold and bi , c ∈ C(QT ). Assume that either g > 0 or b > 0 on S T , or g ∈ C 2+α, 1+α/2 (QT ). If u ∈ Wp2,1 (QT ) satisfies (1.37), then u > 0 in QT .

1.4. Maximum principle of strong solutions  31

Proof. We first deal with the case that either g > 0 or b > 0 on S T . Similar to the proof of Theorem 1.41 we have f ∈ Lp (QT ) and u ∈ C α, α/2 (QT ), and we may assume c > 0. For any given σ > 0, the function v = u + σet satisfies  L v = (1 + c)σet + L u = (1 + c)σet + f =: fσ > σ    t

  

in QT ,

∂n v + bv = bσe + g =: gσ > 0

on ST ,

v(x, 0) = σ + ϕ(x) =: ϕσ > σ

in Ω.

Since v is continuous in QT , there exists 0 < τ0  1 such that, for all 0 < τ 6 τ0 , v > σ/2 in Ω × [0, 3τ ]. For such fixed τ > 0, take a truncated function η ∈ C ∞ ([0, T ]) so that 0 6 η 6 1, η ≡ 1 in [2τ, T ] and η ≡ 0 in [0, τ ]. Let w = vη. Then w satisfies  L w = vηt + ηfσ =: f ∗   

in QT ,

  

in Ω.

∂n w + bw = ηgσ =: g w(x, 0) = ηϕσ = 0

∗

on ST ,

Obviously, f ∗ = fσ in Ω × [2τ, T ], f ∗ = 0 in Ω × [0, τ ] and g ∗ = gσ on ∂Ω × [2τ, T ], g ∗ = 0 on ∂Ω × [0, τ ]. Take aεij , bεi , cε , f ε ∈ C α, α/2 (QT ) and g ε ∈ C 2+α, 1+α/2 (QT ) such that (a) f ε = 0 in Ω × [0, τ /2], f ε > σ/2 in Ω × [2τ, T ], g ε = 0 on ∂Ω × [0, τ /2], g ε > 21 minS T gσ > 0 on ∂Ω × [2τ, T ]; (b) aεij → aij , bεi → bi , cε → c in C(QT ), f ε → f ∗ in Lp (QT ) and g ε → g ∗ in Wp2,1 (QT ) as ε → 0+ . By Theorem 1.17, the problem  ε w − aεij Dij wε + bεi Di wε + cε wε = f ε    t

in QT ,

  

in Ω

∂n wε + bwε = g ε ε

w (x, 0) = 0

on ST ,

has a unique solution wε ∈ C 2+α, 1+α/2 (QT ). Similar to the proof of Theorem 1.41 we have limε→0+ wε = w in Wp2,1 (QT ), and hence limε→0+ wε = w in C(QT ). Moreover, w > σ/2 in Ω × [2τ, 3τ ]. We may assume that wε > σ/3 in Ω × [2τ, 3τ ]. Consequently, wε satisfies  ε w − aεij Dij wε + bεi Di wε + cε wε = f ε > 0    t

in Ω × [2τ, T ],

  

in Ω.

∂n wε + bwε = g ε > 0 ε

w (x, 2τ ) > σ/3

on ∂Ω × [2τ, T ],

The By Lemma 1.26, wε > 0 in Ω × [2τ, T ]. Same as the proof of Theorem 1.41 we can conclude u > 0.

32  1. Preliminaries

For the case g ∈ C 2+α, 1+α/2 (QT ), we don’t use g ε to approximate g ∗ , but directly use g ∗ . The fact gσ > 0 implies g ∗ > 0 on ST . We can still get wε > 0 in Ω × [2τ, T ] by Lemma 1.26.  2+α, 1+α/2 If the condition g > 0 or b > 0 on S T , or g ∈ C (QT ) in Theorem 1.42 can not be guaranteed, we do not know how to deal with it. However, when aij (x, t) = aij (x) and bi (x, t) = bi (x) are independent of time t, the maximum principle is still valid. Theorem 1.43 Let Ω be of class C 2+α , the condition (A) hold and bi , c ∈ C(QT ). Assume that aij (x, t) = aij (x), bi (x, t) = bi (x) which do not depend on time t, and u ∈ Wp2,1 (QT ) satisfies (1.37). Then u > 0 in QT . Proof. Without loss of generality, we assume c > 1. As Ω is of class C 2 , the following boundary value problem (

−aij (x)Dij h + bi (x)Di h = 0

in Ω,

∂n h = 1

on ∂Ω

has a positive solution h ∈ Wp2 (Ω). Take 0 < ε  1 and let v = u + εh. Then  L v = L u + L (εh) = f + cεh > εh   

in QT ,

∂n v + bv = g + ε + bεh > ε

on ST ,

  

v(x, 0) = ϕ + εh > εh

in Ω.

Similar to the proof of Theorem 1.42, we can show that v > 0, and hence u > 0 by letting ε → 0. 

1.5

WEAK SOLUTION

Consider the following divergence equation ut − Dj (aij Di u + cj u) + bi Di u + cu = f + Di fi

in QT .

Coefficients satisfy the following assumptions: (i) there are positive constants λ and Λ such that λ|y|2 6 aij (x, t)yi yj 6 Λ|y|2 , ∀ (x, t) ∈ QT , y ∈ Rn ; (ii) there exists q > 1 + n/2 such that kc2i + b2i + |c|kq, QT 6 Λ; (iii) kfi k2, QT + kf k 2(n+2) , QT < ∞. n+4

Define Lu := −Dj (aij Di u + cj u) + bi Di u + cu, ◦

1,1 W 1,1 2 (QT ) := {u ∈ W2 (QT ) : u = 0 on ST }.

1.5. Weak solution  33 ◦

For φ ∈ W 1,1 2 (QT ), we denote hLu, φi =

Z

[(aij Di u + cj u)Dj φ + (bi Di u + cu)φ]dx, Ω

and for u, v ∈ L2 (QT ), we denote Z

(u, v) =

u(x, t)v(x, t)dx. Ω

Set V2 (QT ) = L∞ ((0, T ), L2 (Ω)) ∩ L2 ((0, T ), H 1 (Ω)), V21,0 (QT ) = C([0, T ], L2 (Ω)) ∩ L2 ((0, T ), H 1 (Ω)), ◦

1,0 V 1,0 2 (QT ) = {u ∈ V2 (QT ) : u(·, t)|∂Ω = 0 a.e. t ∈ (0, T )}.

Definition 1.44 Assume g ∈ L2 ((0, T ), H 1 (Ω)), u0 ∈ L2 (Ω). Function u ∈ V2 (QT ) is called a weak lower solution (upper solution) of the problem    ut + Lu = f + Di fi  

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

in QT , on ST , in Ω,

(1.38)

if (u − g)+ ∈ L2 ((0, T ), H01 (Ω)) ((g − u)+ ∈ L2 ((0, T ), H01 (Ω))), and t

Z





hLu, φi − (u, φt ) ds 6 (>) (u0 , φ(x, 0)) +

(u, φ) + 0

Z

t





(f, φ) − (fi , Di φ) ds

0

◦

for all φ ∈ W 1,1 2 (QT ) with φ > 0, and for all 0 < t 6 T . If u ∈ V2 (QT ) is both a weak lower solution and a weak upper solution of (1.38), then it is called the weak solution of (1.38). Theorem 1.45 (Existence and uniqueness of weak solutions [16, Chapter 3, Theorem 2.2, Theorem 3.1, Theorem 3.2]) Let (i)–(iii) hold and g = 0, u0 ∈ L2 (Ω). Then ◦

problem (1.38) admits a unique weak solution u ∈ V 1,0 2 (QT ), and


kukV2 (QT ) 6 C kfi k2, QT + kf k 2(n+2) , QT + ku0 k2, Ω . n+4

Theorem 1.46 (Maximum principle of weak solutions [72, Theorem 6.25]) Let the condition (i) hold. Assume f, fi ∈ L2 (QT ), c, bi , ci ∈ L∞ (QT ), g ∈ L2 ((0, T ), H 1 (Ω)), u0 ∈ H 1 (Ω), and Z

T

0

Z 0

(c, φ) + (cj , Dj φ) dt > 0, ∀ φ ∈ C 1 (QT ), φ|ST = 0, φ > 0,

 T



(f, φ) − (fi , Di φ) dt 6 0, ∀ φ ∈ C 1 (QT ), φ|ST = 0, φ > 0,





g 6 0 on ST , u0 6 0 in Ω. If u ∈ V2 (QT ) is a lower solution of (1.38), then u 6 0 in QT .

34  1. Preliminaries

Now let us discuss a special case of (1.38): ut − Dj (aij (x, t)Di u) + bi (x, t)Di u + cu = f (x, t) in QT . It is assumed that aij , bi , c ∈ L∞ (QT ), aij = aji in QT for all 1 6 i, j 6 n, and there exists a constant λ > 0 so that aij (x, t)yi yj > λ|y|2 for all (x, t) ∈ QT , y ∈ Rn . Define

•

C ∞ (QT ) := {u ∈ C ∞ (QT ) : u|∂p QT = 0}, •

•

k,l ∞ and let W k,l p (QT ) be closure of C (QT ) in Wp (QT ).

Theorem 1.47 (Maximum principle of weak solutions) Let u ∈ C([0, T ], L2 (Ω)) ∩ L2 ((0, T ), H 1 (Ω)) ∩ L∞ (Ω × (0, T )). Assume u|∂p QT 6 0 (> 0) and Z

Z

u(x, t)φ(x, t)dx + Ω

Qt





aij Di uDj φ + bi Di uφ + cuφ − uφτ dxdτ 6 (>) 0 •

for all 0 < t 6 T and each test function φ ∈ W 1,1 2 (QT ) with φ > 0 in QT . Then u 6 0 (u > 0) in QT . This theorem is a direct consequence of Theorem 2.7. For further discussions on the existence, uniqueness, regularity, estimation, and maximum principle of weak solutions, readers can refer to monographs [16, 72].

EXERCISES 1.1 Show that if (aij )n×n is a non-negative definite matrix and (bij )n×n is a nonP positive definite matrix, then the number ni,j=1 aij bij is non-positive. 1.2 Let 1 6 p < ∞. Prove Wp2,1 (Ω × [0, T ]) ,→ C([0, T ], Lp (Ω)). 1.3 Prove Theorem 1.3 for the case ε > 0. 1.4 Prove Theorem 1.19. Hint: first, take p > n + 2 and Q1 to satisfy Q b Q1 b QT . Then (1.10) holds for Q = Q1 , and 



kuk∞, Q1 6 CkukWp2,1 (Q1 ) 6 C (kf kp, QT + kuk2, QT ) 6 C |f |α, QT + kuk2, QT . 



Second, prove |u|2+α, Q 6 C |f |α, QT + kuk∞, Q1 . 1.5 Give an example to show that the condition c > 0 is necessary in Theorem 1.24.

Exercises  35

1.6 Let u ∈ C(QT ) ∩ C 2,1 (QT ) satisfy  u − ∆u = −u2 + cu    t   

in QT ,

u=0

on ST ,

u|t=0 = u0 (x)

in Ω,

where c ∈ C(QT ), u0 ∈ C(Ω) and u0 > 0. Prove that o

n

0 6 u 6 max max |c|, max u0 . QT

Ω

1.7 Let u be a classical solution of initial value problem (

ut − ∆u = 0

in Rn × (0, T ],

u(x, 0) = 0

in Rn kl = 0, then l→∞ l 2

and 0 < T < ∞. Set kl = sup|x|6l, 06t6T |u(x, t)|. Prove that if lim u ≡ 0 in Rn × [0, T ]. 1.8 Prove Theorem 1.32 and Theorem 1.35.

1.9 Please give an example to illustrate that the condition c2 > 0 is necessary in both Theorem 1.34 and Theorem 1.35. 1.10 Give the details of the proof of Theorem 1.40.

CHAPTER

2

Comparison Principle, Regularity and Uniform Estimates

In this chapter we mainly study comparison principle, regularity and uniform estimates, and uniform bounds from bounded Lp norms. These techniques help us gain an important handle on solutions of nonlinear parabolic equations. Comparison principle based on the maximum principle plays an important role in estimating upper and/or lower bounds of solutions and judging the sign of solution, as well as establishing the upper and lower solutions method. Regularities and uniform estimates based on the Lp theory and Schauder theory are fundamental problems in the study of parabolic equations. The regularity theory is used to investigate smoothness of solutions (improving regularities of solutions). We will see that for parabolic equations with better coefficients, even if initial value or boundary value is not good enough, their solution will have better interior smoothness. The uniform estimate in time provides a basis for studying the convergence of a solution as time goes to infinity, and its limit is usually a solution of the corresponding equilibrium problem. In order to prove global existence of classical or mild solutions of parabolic equations, some priori bounds in the uniform norm are needed. But usually it is easier to derive the estimate for some Lp -norm. Some interesting examples are given to show that one can use the structure of reaction terms to obtain uniform bounds from bounded Lp norms.

2.1

COMPARISON PRINCIPLE

The comparison principle or comparison method is an important tool in the study of partial differential equations, especially parabolic equations and elliptic equations. It is based on the maximum principle, and it is the basis of the upper and lower solutions method which will be introduced later.

37

38  2. Comparison Principle, Regularity and Uniform Estimates

2.1.1

Classical and strong solutions

In this subsection, L := ∂t − aij (x, t)Dij + bi (x, t)Di .

Theorem 2.1 Let coefficients of L and Ω, a, b satisfy the conditions of Theorem 2,1 1.40 (2). Assume u¯, u ∈ C(QT )∩C 1,0 (QT )∩Wn+1, loc (QT ), and when a 6≡ 0 we further assume u¯, u ∈ C 1,0 (Ω × (0, T ]). Suppose  ¯ − f (x, t, u¯) > L u − f (x, t, u)  Lu 

in QT ,

a∂n u¯ + b¯ u > a∂n u + bu

on ST ,

  

u¯(x, 0) > u(x, 0)

in Ω.

Denote c = minQT min{¯ u, u} and c¯ = maxQT max{¯ u, u}. If f satisfies the Lipschitz condition in u ∈ (c, c¯), i.e., there exists a positive constant M such that |f (x, t, u) − f (x, t, v)| 6 M |u − v| for all (x, t) ∈ QT , u, v ∈ (c, c¯),

(2.1)

then u¯ > u in QT . Moreover, u¯ > u in QT if u¯(x, 0) 6≡ u(x, 0). Proof. For (x, t) ∈ QT , we take cˆ(x, t) = M when u¯(x, t) > u(x, t), and cˆ(x, t) = −M when u¯(x, t) < u(x, t), where M is given by (2.1). Let u = u¯ − u. Then u satisfies  L u + cˆ(x, t)u > 0   

in QT ,

Bu > 0

on ST ,

  

u(x, 0) > 0

in Ω.

By Theorem 1.40 (2) we have u > 0 in QT , and u > 0 in QT if u(x, 0) 6≡ 0. Theorem 2.2



Let Ω be of class C 2 , u¯, u ∈ C(QT ) ∩ Wp2,1 (QT ) satisfy  L u¯ − f (x, t, u¯) > L u − f (x, t, u)   

in QT ,

u¯ > u

on ST ,

  

u¯(x, 0) > u(x, 0)

in Ω.

If p > 1 + n/2, the condition (A) holds, bi , c ∈ C(QT ) and fu (·, u(·)) ∈ C(QT ) for each u ∈ C(QT ), then u¯ > u in QT . Theorem 2.3 Let Ω be of class C 2+α , the condition (A) hold, bi , c ∈ C(QT ), b ∈ C 1+α, (1+α)/2 (S T ) and b > 0. Let p > 1 + n/2, u¯, u ∈ C(QT ) ∩ Wp2,1 (QT ) satisfy  L u¯ − f (x, t, u¯) > L u − f (x, t, u)   

in QT ,

∂n u¯ + b¯ u > ∂n u + bu

on ST ,

  

u¯(x, 0) > u(x, 0)

in Ω.

Assume fu (·, u(·)) ∈ C(QT ) for each u ∈ C(QT ), and either b > 0 on S T , or aij = aij (x) and bi = bi (x). Then u¯ > u in QT .

2.1. Comparison principle  39

2.1.2

Weak solutions

This subsection is concerned with weak solutions of the divergence equation ut − Dj (aij (x, t)Di u) + bi (x, t)Di u = f (x, t, u) in QT ,

(2.2)

where aij , bi ∈ L∞ (QT ), aij = aji in QT for all 1 6 i, j 6 n, and there exists a constant λ > 0 so that aij (x, t)yi yj > λ|y|2 for all (x, t) ∈ QT , y ∈ Rn . •

We define W k,l p (QT ) as in §1.5. Definition 2.4 A weak lower solution (weak upper solution) of equation (2.2) is a measurable function u which satisfies u ∈ L2 ((σ, T ), H 1 (Ω0 )) ∩ L∞ (Ω0 × (σ, T )), and ut , f (·, u(·)) ∈ L2 (Ω0 × (σ, T )) for any 0 < σ < T and any compact subset Ω0 of Ω, and also satisfies Z

Z

u(x, t)φ(x, t) + Qt

Ω

Z

(aij Di uDj φ + bi Di uφ) 6 (>)

(f (x, τ, u)φ + uφτ ) Qt

•

for each 0 < t 6 T and each test function φ ∈ W 1,1 2 (QT ) with φ > 0 in QT . A function u that is both a weak lower solution and a weak upper solution is called a weak solution of (2.2). We should mention that in this definition, the lower (upper) solution may not be defined on ∂p QT . Theorem 2.5 (Comparison principle for weak solutions) Let u¯ and u be a weak upper solution and a weak lower solution of (2.2), respectively. Denote c = inf QT min{¯ u, u} and c¯ = supQT max{¯ u, u}. Assume that f (x, t, u) satisfies one side Lipschitz condition in u ∈ (c, c¯), i.e., there exists a constant k > 0 such that f (x, t, u) − f (x, t, v) 6 k(u − v) for all (x, t) ∈ QT , u, v ∈ (c, c¯), u > v. If lim sup (u − u¯) 6 0,

(2.3)

(x,t)→∂p QT

then u¯ > u in QT . •

Proof. Without loss of generality we assume k = 0. Let u = u − u¯, and φ ∈ W 1,1 2 (QT ) be non-negative. Taking φ as a test function it yields that, for all 0 < t 6 T , Z

Z

(uφ)(x, t) + Ω

Z

[aij Di uDj φ + bi Di uφ] 6

Qt

[f (x, τ, u) − f (x, τ, u¯)]φ +

Qt

Z

uφτ . (2.4) Qt

40  2. Comparison Principle, Regularity and Uniform Estimates

For any given 0 < ε < 1, let v = (u − ε)+ . It follows from the condition (2.3) that lim sup (u − u¯ − ε) = lim sup (u − u¯) − ε 6 −ε. (x,t)→∂p QT

(x,t)→∂p QT

We can choose tε ∈ (0, T ) and Ωε b Ω with tε → 0+ and Ωε → Ω as ε → 0+ , such •

that v = 0 in QT \ Ωε × (2tε , T ). Then v ∈ W 1,1 2 (QT ) is non-negative. As a result, (2.4) is valid with φ replaced by v. That is, Z

Z

Z

uv +

[aij Di uDj v + bi Di uv] 6

Qt

Ω

[f (x, τ, u) − f (x, τ, u¯)]v +

Z

uvτ Qt

Qt

Z

(2.5)

uvτ .

6

Qt

Notice that at the point where v 6= 0, we have u − ε > 0 and hence v = u − ε. As a consequence Z

Z

(u − ε)vτ + ε

uvτ = Qt

Z

Qt

Z

Z

=

vvτ + ε

vτ

Qt

=

1 2

vτ Qt

Qt

Z

v2 + ε

Z

Ω

v. Ω

Substituting this into (2.5) firstly, and letting ε → 0+ secondly, it yields Z Ω

u2+ +

Z Qt

[aij Di u+ Dj u+ + bi Di u+ u+ ] 6

1 2

Z Ω

u2+ .

(2.6)

Thanks to Z

Z

Qt

aij Di u+ Dj u+ > λ

Z

Qt



bi Di u+ u+ 6 λ

|Du+ |2 ,

Qt

Z

|Du+ |2 + C

Qt

Z Qt

u2+ ,

it follows from (2.6) that t 1 u2+ 6 C u2+ . 2 Ω 0 Ω This implies u+ = 0, i.e., u 6 u¯ almost everywhere in QT .  Now we consider the first initial-boundary value problem of divergence equation

Z

Z Z

 u − Dj (aij (x, t)Di u) + bi (x, t)Di u = f (x, t, u)    t   

in QT ,

u=g

on ST ,

u(x, 0) = ϕ(x)

in Ω,

where aij and bi are the same as above, and ϕ ∈ L2 (Ω).

(2.7)

2.1. Comparison principle  41

Definition 2.6 A weak lower solution (weak upper solution) of problem (2.7) is a measurable function u such that u ∈ C([0, T ], L2 (Ω)) ∩ L2 (0, T ; H 1 (Ω)) ∩ L∞ (Ω × (0, T )), f (·, u(·)) ∈ L2 (Ω × (0, T )), u(x, 0) 6 (>) ϕ(x) in Ω,

u 6 (>) g on ST ,

and for all 0 < t 6 T , Z

Z

uφ|τ =t + Qt

Ω

Z

(aij Di uDj φ + bi Di uφ) 6 (>)

(f (x, τ, u)φ + uφτ ) Qt

•

holds for each test function φ ∈ W 1,1 2 (QT ) with φ > 0 in QT . A function u that is both a weak lower solution and a weak upper solution is called a weak solution of (2.7). Theorem 2.7 (Comparison principle for weak solutions) Let u¯, u be weak upper and lower solutions of (2.7), respectively. Denote c = inf QT min{¯ u, u} and c¯ = supQT max{¯ u, u}. Assume that f (x, t, u) satisfies one side Lipschitz condition in u ∈ (c, c¯), i.e., there exists a constant k > 0 such that f (x, t, u) − f (x, t, v) 6 k(u − v) for all (x, t) ∈ QT , u, v ∈ (c, c¯) with u > v. Then u¯ > u in QT . To prove this result, let us do some preparatory work first. For a given function g defined in QT , we make zero extension of g to t < 0 and t > T . We define the Steklov average of g as follows: g h (·, t) =

1 h

Z

t+h

g(·, s)ds, h 6= 0.

t

It is easy to show that (g −h )t = (gs )−h . Moreover, for the given 0 < h < s, if g(x, t) = 0 for t 6 0 and t > s − h, then Z Qs

Lemma 2.8

ug −h dxdt =

Z

uh gdxdt.

(2.8)

Qs

Let 1 6 p < ∞ and 0 < h < δ < T . Then we have

(1) uh ∈ Lp (QT −δ ) and limh→0 uh = u in Lp (QT −δ ) if u ∈ Lp (QT ); (2) Duh ∈ Lp (QT −δ ) and limh→0 Duh = Du in Lp (QT −δ ) if Du ∈ Lp (QT ). Proof of Theorem 2.7. Without loss of generality, we assume k = 0. Let u = u − u¯, •

and φ ∈ W 1,1 2 (QT ) be a non-negative function. Taking φ as a test function, we have that, for all 0 < t 6 T , (2.4) holds.

42  2. Comparison Principle, Regularity and Uniform Estimates •

Let 0 < h  1. For any ζ ∈ W 1,1 2 (QT ) with ζ > 0, and ζ = 0 for t 6 0 and t > T − h, taking t = T and φ = ζ −h in (2.4) and applying (2.8), we have that, due to ζ −h (x, T ) = 0, Z

h

QT

Z

h

[(aij Di u) Dj ζ + (bi Di u) ζ] 6

h

(f (·, u) − f (·, u¯)) ζ +

Z

u(ζ −h )τ . (2.9)

QT

QT

Note that ζ = 0 for t 6 0 and t > T − h. Using (ζ −h )τ = (ζt )−h and (2.8) firstly, and integrating by parts secondly, we obtain Z

u(ζ

−h

Z

h

u ζτ = −

)τ =

QT

Z

QT

(uh )τ ζ.

QT

Therefore, in accordance with (2.9), Z

h

Z

h

[(aij Di u) Dj ζ + (bi Di u) ζ] + QT

Z

h

QT

(u )τ ζ 6

(f (·, u) − f (·, u¯))h ζ.

(2.10)

QT

Clearly, (uh )+ = 0 on ST . For any given 0 < t < T , we will show that (2.10) remains valid if ζ is replaced by (uh )+ ξ(τ ), where ξ(τ ) = 1 for τ < t and ξ(τ ) = 0 for τ > t. For this purpose, we fix ` large enough and choose a continuous non-negative function z` : z` is linear in (0, 1/`) ∪ (t − 1/`, t), z` = 0 in (−∞, 0] ∪ [t, ∞) and z` = 1 •

in (1/`, t − 1/`). Then the function ζ` := (uh )+ z` ∈ W 1,1 2 (QT ), ζ` > 0, and ζ` = 0 for τ 6 0 and τ > T − h provided h < T − t. Therefore (2.10) holds for ζ` . Then we have by letting ` → ∞, Z

h

h

h

Z

h

(uh )τ (uh )+

[(aij Di u) Dj (u )+ + (bi Di u) (u )+ ] + Qt

Z

6

Qt

(f (·, u) − f (·, u¯))h (uh )+ .

(2.11)

Qt

It is easy to see that Z

(uh )τ (uh )+ =

Qt

Z

((uh )+ )τ (uh )+ =

Qt

1 2

Z Ω

(uh )2+ (x, t) −



Z Ω

(uh )2+ (x, 0) ,

and Z

lim

h→0 Ω

(uh )2+ (x, t)

Z

= Ω

u2+ (x, t),

Z

lim

h→0 Ω

(uh )2+ (x, 0)

Z

= Ω

u2+ (x, 0) = 0

because u ∈ C([0, T ], L2 (Ω)). Substituting these conclusions into (2.11), and letting h → 0 and then using Lemma 2.8 we find Z

[aij Di uDj u+ + bi u+ Di u] + Qt

1 2

Z Ω

u2+ (x, t) 6

Z Qt

[f (·, u) − f (·, u¯)] u+ 6 0. (2.12)

Noticing Z

Z

Z

aij Di uDj u+ = Qt

Qt

Z

Qt

Z bi u+ Di u 6 λ

aij Di u+ Dj u+ > λ

Qt

2

|Du+ | + C

Z Qt

Qt

u2+ ,

|Du+ |2 ,

2.1. Comparison principle  43

it can be deduced from (2.12) that 1 2

Z Ω

u2+ (x, t) 6 C

Z Qt

Z tZ

u2+ (x, τ ) = C

0

Ω

u2+ (x, τ ).

This implies u+ = 0, i.e., u 6 u¯ almost everywhere in QT . 2.1.3



Quasilinear equations with nonlinear boundary conditions

In this subsection we consider the following initial-boundary value problem of quasilinear equation with nonlinear boundary conditions  u = ∇(a(u)∇u) + f (u),    t   

x ∈ Ω, t > 0,

∂n u = b(u),

x ∈ ∂Ω, t > 0,

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

x ∈ Ω.

(2.13)

Existence and uniqueness of solutions of (2.13) has been studied by Amann in [7]. Theorem 2.9 ([105]) Assume b ∈ C(R+ ), a, f ∈ C 1 (R+ ) and a > 0 in R+ . Let u, v ∈ C 2,1 (Ω × (0, T )) ∩ C(Ω × [0, T )), u, v > 0 in Ω × [0, T ) and satisfy (

ut − ∇(a(u)∇u) − f (u) > vt − ∇(a(v)∇v) − f (v)

in QT ,

∂n u − b(u) > ∂n v − b(v)

on ST ,

and u(x, 0) > v(x, 0) for x ∈ Ω. If either a0 (s) is increasing in s > 0 or a00 (s) exists and is continuous, then u > v in Ω × [0, T ). Proof. We first assume that a0 (s) is increasing in s > 0. Let w = u − v, then wt − a(u)∆w + [a(v) − a(u)]∆v − a0 (u)(∇v + ∇u) · ∇w +f (v) − f (u) > [a0 (u) − a0 (v)]|∇v|2 . Write g(v) − g(u) = (v − u)

Z

1

g 0 (u + s(v − u))ds,

0

and denote a(x, t) = a(u(x, t)), bi (x, t) = −a0 (u(x, t))(vxi (x, t) + uxi (x, t)), c(x, t) = −∆v

Z 0

1

0

a (u + s(v − u))ds −

Z

1

f 0 (u + s(v − u))ds.

0

Then c is bounded in Ω × [ε, T − ε] for each fixed 0 < ε  1, and w satisfies wt − a∆w + bi wxi + cw > [a0 (u) − a0 (v)]|∇v|2 . If the statement was not valid, then there would exist 0 < t0 < T and x0 ∈ Ω so that u(x0 , t0 ) = v(x0 , t0 ) and u − v > 0 in Ω × [0, t0 ).

44  2. Comparison Principle, Regularity and Uniform Estimates

We claim that x0 6∈ Ω. Otherwise, [a0 (u) − a0 (v)]|∇v|2 > 0 in Ω × [0, t0 ) since u > v in this domain. Take 0 < ε < t0 and k so large that k + c > 0 in Ω × [ε, t0 ]. Then the function z = e−kt w satisfies (

zt − a∆z + bi zxi + (k + c)z > 0

in Ω × [ε, t0 ],

z(x0 , t0 ) = 0, z > 0

in Ω × [ε, t0 ).

Corollary 1.25 avers z ≡ 0 in Ω × [ε, t0 ). This is impossible, and hence x0 6∈ Ω. Therefore, z satisfies  z − a∆z + bi zxi + (k + c)z > 0    t

z = e−kt w > 0

  

in Ω × [ε, t0 ], in Ω × [ε, t0 ),

z(x0 , t0 ) = 0.

The Hopf boundary lemma asserts ∂n z(x0 , t0 ) < 0, i.e., ∂n w(x0 , t0 ) < 0. However,

∂n w(x0 , t0 ) = ∂n u(x0 , t0 ) − ∂n v(x0 , t0 ) > [b(u) − b(v)] (x0 ,t0 ) = 0. This contradiction shows that u > v in Ω × [0, T ). When a00 (s) exists and is continuous, we can write 0

0

2



2

[a (u) − a (v)]|∇v| = |∇v|

Z

1



00

a (v + s(u − v))ds w,

0

and then put it into c(x, t). The remaining proof is the same as above.



2.2 REGULARITY AND UNIFORM ESTIMATES Let Ω be of class C 2+α . We first investigate the following initial-boundary value problem  L u = f (x, t)      

in QT ,

Bu = 0

on ST ,

u(x, 0) = ϕ(x)

in Ω,

(2.14)

where B u = u, or B u = ∂n u + bu with b > 0 on ST and b ∈ C 1+α, (1+α)/2 (S T ), ϕ ∈ Wp2 (Ω) with p > 1 and (i) ϕ(x) = 0 on ∂Ω when B u = u; (ii) ∂n ϕ + b(x, 0)ϕ = 0 on ∂Ω when B u = ∂n u + bu. Theorem 2.10 (Interior regularity and estimate) Under above assumptions, we further assume that the function f and coefficients of L satisfy the requirement (C), which is given in Chapter 1. Then (2.14) has a unique solution u ∈ Wp2,1 (QT ) ∩ C 2+α,1+α/2 (Ω × (0, T ]). Moreover, for any given 0 < σ < T , there is a constant C(σ, T ) such that 



|u|2+α, Ω×[σ,T ] 6 C(σ, T ) |f |α, QT + kϕkWp2 (Ω) .

(2.15)

2.2. Regularity and uniform estimates  45

Proof. We only deal with the case B u = u. Firstly, by Theorem 1.1, problem (2.14) has a unique solution u ∈ Wp2,1 (QT ), and 







kukWp2,1 (QT ) 6 C kf kp, QT + kϕkWp2 (Ω) 6 C |f |α, QT + kϕkWp2 (Ω) . If p > 1 + n/2, then the embedding theorem (Theorem A.7 (1)) implies u ∈ C α, α/2 (QT ) with α = 2 − (n + 2)/p, and 



|u|α, QT 6 CkukWp2,1 (QT ) 6 C |f |α, QT + kϕkWp2 (Ω) . If p = 1+n/2, then the embedding theorem (Theorem A.6 (3)) implies u ∈ Lq (QT ) for any 1 6 q < ∞, and 



kukq, QT 6 CkukWp2,1 (QT ) 6 C |f |α, QT + kϕkWp2 (Ω) . If p < 1 + n/2, then the embedding theorem (Theorem A.6 (2)) implies u ∈ Lp1 (QT ) with p1 = p(n + 2)/(n + 2 − 2p), and 



kukp1 , QT 6 CkukWp2,1 (QT ) 6 C |f |α, QT + kϕkWp2 (Ω) . We only consider the case p < 1 + n/2. The following arguments are the same as those of Theorem 1.8. Take 0 < ε < σ/4 and a truncated function η ∈ C ∞ ([0, T ]) such that 0 6 η 6 1, η ≡ 1 in [ε, T ] and η ≡ 0 in [0, ε/2]. Then there exists C 0 > 0 such that |η 0 (t)| 6 C 0 /ε. As a result, the function w = uη satisfies  0   L w = f (x, t)η(t) + uη (t) 

in QT ,

w=0

on ST ,

  

w(x, 0) = 0

in Ω.

(2.16)

This problem has a unique solution w ∈ Wp2,1 (QT ), and 1 kwkWp2,1 (QT ) 6 Ckf η + uη 0 kp1 , QT 1









6 C |f |α, QT + kukp1 , QT

6 C |f |α, QT + kϕkWp2 (Ω)

according to Theorem 1.1. As a consequence, u ∈ Wp2,1 (Ω × (ε, T )) ,→ Lp2 (Ω × (ε, T )) 1 with p2 = p1 (n + 2)/(n + 2 − 2p1 ). Repeating this process, we can show that u ∈ Wq2,1 (Ω × (σ/4, T )) for some q > 1 + n/2, and 



kukWq2,1 (Ω×(σ/4,T )) 6 C |f |α, QT + kϕkWp2 (Ω) . Hence u ∈ C α, α/2 (Ω × [σ/4, T ]) for some 0 < α < 1, and 



|u|α, Ω×[σ/4,T ] 6 CkukWq2,1 (Ω×(σ/4,T )) 6 C |f |α, QT + kϕkWp2 (Ω) .

46  2. Comparison Principle, Regularity and Uniform Estimates

Take η ∈ C ∞ ([0, T ]) such that 0 6 η 6 1, η ≡ 1 in [σ, T ] and η ≡ 0 in [0, σ/2]. Then there exists C 0 > 0 such that |η 0 (t)| 6 C 0 /σ and |η 0 |C α ([0,T ]) 6 C 0 /σ 1+α . The function w = uη satisfies (2.16). According to Theorem 1.16, this problem has a unique solution w ∈ C 2+α, 1+α/2 (QT ), and 



|w|2+α, QT 6 C|f η + uη 0 |α, QT 6 C |f |α, QT + kϕkWp2 (Ω) .  This implies u ∈ C 2+α, 1+α/2 (Ω × [σ, T ]) and (2.15) holds. In the following, we attempt to give uniform estimates in time t of solution u and its derivatives for problem  u + L[u] = f (x, t, u),    t   

x ∈ Ω,

t > 0,

B[u] = 0,

x ∈ ∂Ω, t > 0,

u(x, 0) = ϕ(x),

x ∈ Ω,

(2.17)

where L[u] = −aij (x)Dij u + bi (x)Di u, and B[u] = u, or B[u] = ∂n u + b(x)u with b ∈ C 1+α (∂Ω) and b > 0. Assume that p > 1+n/2, ϕ ∈ Wp2 (Ω) satisfies the compatibility conditions: ϕ = 0 on ∂Ω when B[u] = u, and ∂n ϕ + b(x)ϕ = 0 on ∂Ω when B[u] = ∂n u + b(x)u. Function u is call a global solution of (2.17) if it is defined for all (x, t) ∈ Q∞ and satisfies (2.17) almost everywhere. Theorem 2.11 (Regularity and uniform estimate) Let Ω be of class C 2+α , C α (Ω) and u be a global solution of (2.17) with c 6 u 6 c¯ for some c, c¯ ∈ R. that f ∈ L∞ (Q∞ × [c, c¯]) satisfies (2.1) with QT replaced by Q∞ . We further f (·, u) ∈ C α,α/2 (Ω × [h, h + 3]) uniformly in u ∈ [c, c¯] and h > 0, i.e., there constant C so that

aij , bi ∈ Assume suppose exists a

|f (x, t, u) − f (y, s, u)| 6 C(|x − y|α + |t − s|α/2 ) for all (x, t), (y, s) ∈ Ω × [h, h + 3], u ∈ [c, c¯] and all h > 0. Then, for any given τ > 0, there is a constant C(τ ) such that kukC 2+α, 1+α/2 (Ω×[τ,∞)) 6 C(τ ).

(2.18)

Proof. We only deal with the case B[u] = ∂n u + b(x)u. For any given 0 < T < ∞, u solves the problem  u + L[u] = f (x, t, u)    t

in QT ,

B[u] = 0

on ST ,

  

u(x, 0) = ϕ(x)

in Ω.

Thanks to c 6 u 6 c¯ and f ∈ L∞ (Q∞ × [c, c¯]), we see that function F (x, t) := f (x, t, u(x, t)) ∈ L∞ (Q∞ ), and then u ∈ Wp2,1 (QT ) by the Lp theory (Theorem 1.2).

2.2. Regularity and uniform estimates  47

Therefore, u ∈ C α,α/2 (QT ) due to p > 1 + n/2, and thereupon f (x, t, u(x, t)) ∈ C α,α/2 (QT ). Thus, u ∈ C 2+α,1+α/2 (Ω × (0, T ]) by Theorem 2.10. Without loss of generality, we assume τ < 1. For integer k > 0, we set uk (x, t) = u(x, k + t), f k (x, t) = f (x, k + t, uk (x, t)) and ϕk (x) = u(x, k). Then c 6 uk , ϕk 6 c¯, f k is bounded uniformly in k, and uk satisfies  k k k   ut + L[u ] = f (x, t), 

x ∈ Ω,

B[u ] = 0,

x ∈ ∂Ω, 0 < t 6 3,

  

uk (x, 0) = ϕk (x),

x ∈ Ω.

k

0 < t 6 3,

Clearly, f k ∈ C α,α/2 (Ω × (0, 3]), uk ∈ C 2+α,1+α/2 (Ω × (0, 3]) for all integer k > 0. Take δ = τ /2 and T = 3 in Theorem 1.8. Then there is a constant C such that 



kuk kWp2,1 (Ω×(τ /2, 3)) 6 C kf k kp, Ω×(0, 3) + kuk kp, Ω×(0, 3) 6 C for all k > 0. Since p > 1 + n/2, the embedding theorem declares |uk |α, Ω×[τ /2, 3] 6 C1 . Assumptions on f show that |f k |α, Ω×[τ /2, 3] 6 C2 . Taking t0 = τ /2, t1 = τ and T = 3 in (1.19), we get 



c| 6 C3 for all k > 0. |uk |2+α, Ω×[τ, 3] 6 C |f k |α, Ω×[τ /2, 3] + |c| + |¯ Thus we have, noticing uk (x, t) = u(x, k + t), kukC 2+α, 1+α/2 (Ω×[k+τ, k+3]) 6 C3 for all k > 0. Estimate (2.18) is obtained immediately. Remark 2.12



An analogous theorem is valid for system version of (2.17).

It can be seen from the proof of Theorem 2.11 that the following theorem holds. Theorem 2.13 (Regularity and uniform estimate) Let Ω be of class C 2 , aij ∈ C(Ω), bi ∈ L∞ (Ω) and u be a global solution of (2.17) with c 6 u 6 c¯ for some c, c¯ ∈ R. Assume that f ∈ L∞ (Q∞ × [c, c¯]) satisfies (2.1) with QT replaced by Q∞ . Then, for any given τ > 0, p > 1 and 0 < α < 1, there exist constants C(τ, p) and C(τ, α) such that kukWp2,1 (Ω×[k+τ, k+3]) 6 C(τ, p) for all k > 0, and kukC 1+α, (1+α)/2 (Ω×[τ,∞)) 6 C(τ, α). Theorem 2.11 and Theorem 2.13 play an important role in understanding longtime behaviours of solutions.

48  2. Comparison Principle, Regularity and Uniform Estimates

2.3

UNIFORM BOUNDS FROM BOUNDED Lp NORMS

In order to get the global existence of solutions of parabolic equations, some priori bounds of the L∞ norm are needed. But for some interesting examples, it is easy to get Lp -bounds for some 1 6 p < ∞. Some special structure of the reaction term can be used to obtain uniform bounds. Here we shall state a result of Rothe ([95]). Suppose that Ω is of class C 2 and 0 < T 6 ∞. Let u satisfy (

ut − d∆u = f (x, t, u),

x ∈ Ω, 0 < t < T,

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

x ∈ Ω,

and either ∂n u = 0, x ∈ ∂Ω, 0 6 t < T, or u = 0, x ∈ ∂Ω, 0 6 t < T. We further assume that either u > 0 and f (x, t, u) 6 c(x, t)(1 + u)m , or |f (x, t, u)| 6 c(x, t)(1 + |u|)m . Define U0 = ku0 k∞, Ω , Cq (T ) = sup kc(·, t)kq, Ω and Up (T ) = sup ku(·, t)kp, Ω . 06t 1 and p > 1/2 satisfy m < 1 + p(2/n − 1/q). Set (m − 1)(1 + 2/n) η = exp , 1 + p(2/n − 1/q) − m 



1 + 2/n (m − 1)(1 + 2/n) exp . 1 + p(2/n − 1/q) − m 1 + 2p(2/n − 1/q) − m 

θ=



Then there exists a constant K = K(n, m, q, p, Ω) so that sup ku(·, t)k∞, Ω 6 K (max{1, Cq (T )})θ (max{1, U0 , Up (T )})η .

06t 0, 6≡ 0 in Ω.

2.3. Uniform bounds from bounded Lp norms  49

Example 2.1 A typical example is a prey-predator model with homogeneous Neumann boundary conditions  ut − d1 ∆u = u(a − u − hv),       vt − d2 ∆v = v(ku − b),

x ∈ Ω,

t > 0,

x ∈ Ω,

t > 0,

 ∂n u = ∂n v = 0,     

x ∈ ∂Ω, t > 0,

u(x, 0) = u0 (x), v(x, 0) = v0 (x),

x ∈ Ω,

where d1 , d2 , a, h, k and b are positive constants, and ∂n u0 = ∂n v0 = 0 on ∂Ω. Let Tmax denote the maximal existence time. Applying the strong maximum principle and comparison arguments, we assert



0 < u 6 max a, max u0 (x) =: B, v > 0 in Ω × (0, Tmax ). Ω

Therefore d dt

Z

(ku + hv) = −b

Ω

6 −b

Z

Z

(ku + hv) + ZΩ

ku(a + b − u)

Ω

(ku + hv) + k(a + b)B|Ω|, Ω

which implies Z Ω

Z

(ku + hv) 6

(ku0 + hv0 ) + Ω

k(a + b)B|Ω| =: C b

for all 0 6 t < Tmax . Certainly sup 06t 0, x ∈ Ω,

where k, h > 0, r, m > 1 and σ, ρ > 0 are constants, and u0 = v0 = 0 on ∂Ω.

50  2. Comparison Principle, Regularity and Uniform Estimates

In the same way as above it can be shown that u, v > 0 and u is bounded in Ω × (0, Tmax ). Here we only deal with the case that σ, ρ > 0. Multiplying two differential equations by uρ and v σ , respectively, and then integrating the result over Ω by parts, we derive 1 d uρ+1 = −d1 ρ |∇u|2 uρ−1 − k ur+ρ v m+σ , ρ + 1 dt Ω Ω Ω Z Z Z 1 d v σ+1 = −d2 σ |∇v|2 v σ−1 + h ur+ρ v m+σ . σ + 1 dt Ω Ω Ω Z

And then

Z

d dt

Z  Ω

Z

k h ρ+1 u + v σ+1 6 0. ρ+1 σ+1 

Therefore Z  Ω

h ρ+1 k u + v σ+1 6 ρ+1 σ+1 

Z  Ω

k h ρ+1 u + v σ+1 , ρ+1 0 σ+1 0 

which implies sup06t 0,

 ∂n u = ∂n v = 0,     

u(x, 0) = u0 (x), v(x, 0) = v0 (x),

(2.19)

x ∈ Ω,

where coefficients are positive constants, u0 , v0 > 0 in Ω and ∂n u0 = ∂n v0 = 0 on ∂Ω. Theorem 2.15 ([95]) Let (u, v) be a solution of (2.19). If 1 6 n 6 3, then there exists a constant C depending on a, b and initial data such that 0 < u(x, t) 6 C and 0 < v(x, t) 6 C for all x ∈ Ω, t > 0. Proof. In the following positive constants Ci depend only on a, b and initial data. By comparison arguments one can deduce u > δ > 0 and v > 0 in Ω × [0, Tmax ). For any given 0 < τ < Tmax , it is not hard to show that max v(x, t) 6 max{max v0 (x), b/δ} =: C1 , Qτ

Ω

and subsequently, v 6 C1 in Ω × [0, Tmax ). It yields a − (b + 1)u + u2 v 6 C2 (1 + u)2 .

(2.20)

2.3. Uniform bounds from bounded Lp norms  51

Adding up the first two equations of (2.19) and integrating the result, we have d dt and hence

Z Ω

Z

(u + v) = a|Ω| −

Ω

Z

u, Ω

u 6 C3 for all t ∈ [0, Tmax ).

Multiplying the equation (u + v)t − ∆(d1 u + d2 v) = a − u by u + v and integrating the result over Ω, we conclude (u + v)2 d + (d1 |∇u|2 + (d1 + d2 )∇u · ∇v + d2 |∇v|2 ) = (a − u)(u + v). dt Ω 2 Ω Ω The quadratic form in ∇u and ∇v is indefinite, however it can be estimated by Z

Z

Z

d1 |∇u|2 + (d1 + d2 )∇u · ∇v + d2 |∇v|2 > −

(d1 − d2 )2 |∇v|2 . 4d1

Consequently (u + v)2 (d1 − d2 )2 + (u + v)2 − |∇v|2 6 (a + v)(u + v) 6 C4 . (2.21) 2 4d1 Ω Ω Ω Ω On the other hand, multiplying the second equation of (2.19) by v and integrating the result, one has d dt

Z

Z

Z

Z

v2 d + d2 |∇v|2 = uv(b − uv) 6 C5 . dt Ω 2 Ω Ω This combined with (2.21) allows us to deduce Z



Z

d 1+ dt

which follows

Z Ω

Z Ω

Z

(d1 − d2 )2 2 v 2(u + v) + 2d1 d2

!

2

6 C6 ,

u2 6 C7 for all t ∈ [0, Tmax ).

When n = 1, 2, 3, Theorem 2.14 can be applied to the equation for u with q = ∞ and p = m = 2, and then estimate sup06t 0,

x ∈ Ω,

t > 0, (2.22)

where f ∈ C 1 satisfies growth condition |f | 6 C(1 + |u|m + |∇u|r ) with m < 1 + 2p/n and r < 1 + p/(n + p), the Lp -bound for solution of (2.22) guarantees the L∞ -bound. The similar conclusion is still valid when we consider (2.22) with nonlinear Neumann boundary condition ∂n u = g(x, t, u) instead of homogeneous Dirichlet condition, provided that g ∈ C 1 satisfies growth condition |g(·, u)| 6 C(1+|u|s ) with s < 1+p/n (see [8, 92, 93]).

52  2. Comparison Principle, Regularity and Uniform Estimates

EXERCISES 2.1 Prove Theorems 2.2 and 2.3. 2.2 Assume that g > 0 in Ω × ST . Let u, v ∈ C(QT ) ∩ C 2,1 (QT ), u > 0 and u, v satisfy Z Z  2 2   ut − ∆u − u udy > vt − ∆v − v vdy    Ω Ω  Z Z

u−      

Ω

g(y, x, t)u(y, t)dy > v −

in QT ,

g(y, x, t)v(y, t)dy

on ST ,

Ω

u(x, 0) > v(x, 0)

in Ω.

Prove u > v in QT . 2.3 Prove (2.8). 2.4 Prove Lemma 2.8. 2.5 Assume that u ∈ C([0, T ], L2 (Ω)). Prove Z

lim

h→0 Ω

(uh )2+ (x, t)

Z

= Ω

u2+ (x, t).

2.6 Prove Theorem 2.10 when B u = ∂n u + bu. 2.7 Use the boundary condition u = 0 instead of ∂n u + b(x)u = 0 and assume f (x, t, 0) = 0. Prove Theorem 2.11. 2.8 For system version of (2.17), state and prove a theorem same as Theorem 2.11. 2.9 Use ut − aij (x, t)Dij u + bi (x, t)Di u instead of ut + L[u] in (2.17) and assume that aij , bi ∈ L∞ (Q∞ ), and aij is uniformly continuous in Q∞ . Prove Theorem 2.13. 2.10 Let Γ1 and Γ2 be given in (1.3). Assume that, in (2.17), B[u] = ∂n u + b(x)u on Γ1 × R+ with b > 0, b ∈ C 1+α (Γ1 ), and B[u] = u on Γ2 × R+ . State and prove the conclusions similar to Theorem 2.11 and Theorem 2.13. 2.11 Let (u, v) be a local strong solution of  m   ut − d1 ∆u = −kuv ,     vt − d2 ∆v = kuv m ,  ∂n u = ∂n v = 0,     

u(x, 0) = u0 (x), v(x, 0) = v0 (x),

x ∈ Ω,

t > 0,

x ∈ Ω,

t > 0,

x ∈ ∂Ω, t > 0, x ∈ Ω,

where k > 0 and m > 1. Prove that (u, v) exists globally and is bounded. 2.12 Prove estimate (2.20).

CHAPTER

3

Semilinear Parabolic Equations

This chapter is devoted to studying initial-boundary value problems of semilinear parabolic equations. Main contents are the upper and lower solutions method, monotone properties and convergence. Some examples will be given as applications. It is shown by applying the fixed point theorem that if a problem admits the upper and lower solutions, then it must have a unique solution located between the upper and lower solutions. Then, by means of the upper and lower solutions, the monotone iterative sequences are constructed to get the unique solution between the upper and lower solutions. The monotone properties and convergence of solutions can also be studied via the upper and lower solutions method. This chapter further introduces the weak upper and lower solutions method, and it is proved that the minimum function of two upper solutions is a weak upper solution and the maximum function of two lower solutions is a weak lower solution. Let us consider the initial-boundary value problem    L u = f (x, t, u) 

in QT ,

Bu = g

on ST ,

  

u(x, 0) = ϕ(x)

in Ω,

(3.1)

where L u = ut − aij Dij u + bi Di u. As a consequence of Theorems 2.1–2.3, we have the following uniqueness result. Theorem 3.1 (Uniqueness) Under conditions of Theorem 2.1, problem (3.1) has at 2,1 most one solution in the class of C(QT ) ∩ C 1,0 (QT ) ∩ Wn+1, loc (QT ) when B u = u, 2,1 1,0 and in the class of C (QT ) ∩ Wn+1, loc (QT ) when B u = ∂n u + bu with b > 0. Under conditions of Theorem 2.2 when B u = u, and conditions of Theorem 2.3 when B u = ∂n u + bu with b > 0, problem (3.1) has at most one solution in the class of C(QT ) ∩ Wp2,1 (QT ).

53

54  3. Semilinear Parabolic Equations

3.1

THE UPPER AND LOWER SOLUTIONS METHOD

The upper and lower solutions method provides a unified and powerful tool for the study of differential equations including ordinary differential equations, partial differential equations, functional differential equations, and combinations thereof. Many monographs and textbooks have an introduction to the upper and lower solutions method ([59, 88, 99, 124]), and there are also many papers devoted to this topic. The notion of upper and lower solutions was used by Keller and Cohen [57] in special cases, and general results were due to [6, 97]. The so-called upper and lower solutions method is, by defining and constructing proper upper and lower solutions, to ensure the existence of solutions. The upper and lower solutions method provides us with estimates of solutions as well. Definition 3.2 Let u ∈ C(QT ) ∩ C 2,1 (QT ) when B u = u, and u ∈ C 1,0 (QT ) ∩ C 2,1 (QT ) when B u = ∂n u + bu. Such a function u is referred to as an upper solution (a lower solution) of (3.1) if  L u > (6) f (x, t, u)   

in QT ,

B u > (6) g

on ST ,

  

u(x, 0) > (6) ϕ(x)

in Ω.

We have the following theorem as a direct consequence of Theorem 2.1. Theorem 3.3 (Ordering of upper and lower solutions) Let aij , bi ∈ L∞ (QT ), L be parabolic in QT , and Ω be of class C 2 when B u = ∂n u + bu. Let u¯ and u be an upper solution and a lower solution of (3.1), respectively. If f satisfies (2.1) with c = minQT min{¯ u, u} and c¯ = maxQT max{¯ u, u}, then u¯ > u in QT . Remark 3.4 The upper and lower solutions defined in Definition 3.2 are called the classical upper and lower solutions. We can also define the strong upper and lower solutions (e.g., the space C 2,1 (QT ) is replaced by Wp2,1 (QT ) or Wp,2,1loc (QT )) and Theorem 3.3 still holds provided that the comparison principle is valid. For example, under conditions of Theorem 2.1, the upper and lower solutions u¯, u ∈ C 1,0 (QT ) ∩ 2,1 C 2,1 (QT ) can be reduced to u¯, u ∈ C 1,0 (QT ) ∩ Wn+1, loc (QT ) when B u = ∂n u + bu, and Theorem 3.3 holds. Moreover, the upper and lower solutions method established later for classical upper and lower solutions is still valid for such strong upper and lower solutions. We state two requirements. (G1) The condition (A) holds, ϕ ∈ Wp2 (Ω) with p > 1 + n/2, bi ∈ L∞ (QT ) and Ω is of class C 2 . Moreover, (i) g ∈ Wp2,1 (QT ) and ϕ(x) = g(x, 0) on ∂Ω when B u = u; (ii) g ∈ Wq2,1 (QT ) with q > n + 2 and q > p, b ∈ C 1,1/2 (S T ) and ∂n ϕ + b(x, 0)ϕ = g(x, 0) on ∂Ω when B u = ∂n u + bu.

3.1. The upper and lower solutions method  55

(G2) Conditions (C) and (E) hold, Ω is of class C 2+α , b ∈ C 1+α, (1+α)/2 (S T ), and (i) and (ii) of (G1) hold. We first use the Schauder fixed point theorem to prove the existence of solutions. Theorem 3.5 (The upper and lower solutions method) Let the condition (G1) hold, and u¯ and u be an upper solution and a lower solution of (3.1), respectively. Define c and c¯ as in Theorem 3.3. If f satisfies (2.1) and f ∈ L∞ (QT × (c, c¯)), then, in the interval hu, ui, problem (3.1) has a unique solution u and u ∈ Wp2,1 (QT ), where n

o

hu, ui = u ∈ C(QT ) : u 6 u 6 u in QT . Proof. By Theorem 3.3, u¯ > u. We only deal with the case B u = u. Step 1. For any given v ∈ hu, ui, the linear problem  L u + M u = f (x, t, v(x, t)) + M v(x, t)   

in QT ,

u=g

on ST ,

  

u(x, 0) = ϕ

in Ω

has a unique solution u ∈ Wp2,1 (QT ), and 

kukWp2,1 (QT ) 6 C kf (·, v(·))kp, QT + M kvkp, QT + kϕkWp2 (Ω) + kgkWp2,1 (QT ) 

6 C1 kf k∞, QT ×(c,¯c) + |c| + |¯ c| + kϕkWp2 (Ω) + kgkWp2,1 (QT )





=: C2 independent of v ∈ hu, ui by the Lp theory (Theorem 1.1). Note that p > 1 + n/2. Using the above estimate and the embedding theorem, we have in addition that |u|α, QT 6 C3 for all v ∈ hu, u¯i. 2,1 Moreover, similar to Theorem 2.10 we have u ∈ C 1+α, (1+α)/2 (QT ) ∩ Wn+1, loc (QT ).

We define an operator F from hu, ui to C α,α/2 (QT ) by F (v) = u. Then |F (v)|α, QT 6 C3 for all v ∈ hu, u¯i. Step 2. We claim that F : hu, u¯i → C(QT ) is a compact operator. Recalling the estimate |F (v)|α, QT 6 C3 for all v ∈ hu, u¯i, it suffices to prove that F : hu, u¯i → C(QT ) is continuous. For vi ∈ hu, u¯i, i = 1, 2, let ui = F (vi ) and w = u1 − u2 . Then w satisfies L w + M w = f (x, t, v1 ) + M v1 − f (x, t, v2 ) − M v2

w=0 w(x, 0) = 0

in QT , on ST , in Ω.

The assumption on f implies |f (x, t, v1 ) + M v1 − f (x, t, v2 ) − M v2 | 6 2M |v1 − v2 |.

56  3. Semilinear Parabolic Equations

Then we have kwkWp2,1 (QT ) 6 2M Ckv1 − v2 kp, QT 6 C1 kv1 − v2 k∞ by the Lp theory (Theorem 1.1). As p > 1 + n/2, the embedding theorem avers |w|α, QT 6 C1 kv1 − v2 k∞

=⇒ kwkC(QT ) 6 C1 kv1 − v2 k∞ ,

where C1 is independent of v1 and v2 . Thus F : hu, u¯i → C(QT ) is continuous. Step 3. Now we prove F : hu, u¯i → hu, u¯i. Suppose v ∈ hu, u¯i and u = F (v). Let w = u¯ − u. Then w satisfies L w + M w > f (x, t, u¯) − f (x, t, v) + M (¯ u − v) > 0

w>0

in QT , on ST ,

w(x, 0) > 0

in Ω.

The maximum principle gives w > 0, i.e., u 6 u¯. Likewise, u 6 u. Consequently F : hu, u¯i → hu, u¯i. Note that hu, u¯i is a bounded and closed convex set of C(QT ). It follows that F has at least one fixed point u ∈ hu, u¯i by the Schauder fixed point theorem (Corollary A.3). Clearly, such a u solves problem (3.1), and u ∈ Wp2,1 (QT ) because u = F (u) ∈ Wp2,1 (QT ). Step 4. The uniqueness of solutions follows from Theorem 3.1.  Now we use the monotone iterative method to prove existence of solutions. This approach is convenient for numerical calculations. Theorem 3.6 (The upper and lower solutions method) Let the condition (G2) hold, and u¯ and u be an upper solution and a lower solution of (3.1), respectively. Define c and c¯ as in Theorem 3.3. If f satisfies (2.1) and f (·, u) ∈ C α, α/2 (QT ) is uniformly ui } and {ui } such that with respect to u ∈ hu, u¯i, then there exist two sequences {¯ u 6 u1 6 . . . 6 ui 6 u¯i 6 . . . 6 u¯1 6 u¯ for all i > 2, and lim ui = lim u¯i =: u

i→∞

i→∞

in C 1+α, (1+α)/2 (QT ) ∩ C 2+β, 1+β/2 (Q) for any Q b QT and any 0 < β < α, here u is the unique solution of (3.1) in the interval hu, u¯i. Furthermore, u ∈ C 2+α, 1+α/2 (QT ) if either B u = ∂n u + bu, or B u = u and the following compatibility condition holds: gt − aij Dij ϕ + bi Di ϕ = f (x, t, ϕ(x)) on ∂Ω × {0}.

(3.2)

Proof. By Theorem 3.3, it holds u¯ > u. Step 1. Note that (G2) implies (G1) for each p > 1. In the proof of Theorem 3.5 we have obtained an operator F : hu, u¯i → Wp2,1 (QT ) satisfying F (u) > u and F (¯ u) 6 u¯, and F (v) is in a bounded set of Wp2,1 (QT ) when v varies in hu, u¯i. Using

3.2. Some examples  57

the same argument as in Step 3 in proving Theorem 3.5, we can prove that F is monotonically increasing, i.e., v, w ∈ hu, ui with v > w implies F (v) > F (w). Step 2. Define two sequences {¯ ui } and {ui } by u¯1 = F (¯ u), u¯i+1 = F (¯ ui ), u1 = F (u), ui+1 = F (ui ). Since F is monotonically increasing, it follows that, by the inductive process, u 6 u1 6 . . . 6 ui 6 . . . u¯i 6 . . . u¯1 6 u¯. Then we have lim u¯i = uˆ and

i→∞

lim ui = u˜ pointwisely in QT .

i→∞

We now show that limi→∞ u¯i = uˆ and limi→∞ ui = u˜ in C 1+α, (1+α)/2 (QT ) ∩ C 2+β, 1+β/2 (Q) for any Q b QT and 0 < β < α. Once this is done, it is not difficult to see that both uˆ and u˜ are solutions of (3.1) by letting i → ∞. We only handle uˆ since u˜ can be handled in a similar way. Owing to 0 < α < 1, we can take a large p such that 0 < α < 1 − (n + 2)/p. Using the uniform estimate of k¯ ui kWp2,1 (QT ) and embedding theorem (Theorem A.7 (3)) we can show that u¯i → uˆ in C 1+α, (1+α)/2 (QT ) for all 0 < α < 1 − (n + 2)/p. Set fi (x, t) = f (x, t, u¯i (x, t)). Recall that f satisfies (2.1) and f (·, u) ∈ C α, α/2 (QT ) is uniformly with respect to u ∈ hu, u¯i. We see that fi ∈ C α, α/2 (QT ) and |fi |α, QT 6 C which is independent of i. Hence, u¯i ∈ C 2+α, 1+α/2 (QT ) and for any Q b QT , 



ui kp, QT 6 C(Q) |¯ ui |2+α, Q 6 C |fi |α, QT + k¯ by Theorem 1.19. Therefore u¯i → uˆ in C 2+β, 1+β/2 (Q) for any 0 < β < α. Step 3. The uniqueness of solutions is guaranteed by Theorem 3.1. Step 4. If B u = ∂n u + bu, then u ∈ C 2+α, 1+α/2 (QT ) by Theorem 1.17. If B u = u and (3.2) holds, then u ∈ C 2+α, 1+α/2 (QT ) by Theorem 1.16. 

3.2

SOME EXAMPLES

We first give a basic fact which will be often used later. Lemma 3.7 Let Ω be of class C 1 , u, v ∈ C 1 (Ω) satisfy u = v = 0 on ∂Ω and v > 0 in Ω. If ∂n v < 0 on ∂Ω, then there exists a constant K > 0 such that u 6 Kv in Ω. Proof. Owing to u, v ∈ C 1 (Ω) and ∂n v < 0 on ∂Ω, there exists K1 > 0 for which ∂n u − K1 ∂n v > 0 on ∂Ω. Noting u − K1 v = 0 on ∂Ω, we can find a Ω-neighbourhood V of ∂Ω such that u − K1 v 6 0 in V . Because of u, v ∈ C 1 (Ω \ V ) and v > 0 in Ω \ V , there is a positive constant K2 such that u 6 K2 v in Ω \ V . Take K = K1 + K2 . Then the desired conclusion holds. See Fig. 3.1. 

58  3. Semilinear Parabolic Equations

Example 3.1 Let u be a lower solution of (3.1). Assume that (G1) holds, and f satisfies (2.1) for c = inf QT u(x, t) and any c¯ > c. If f ∈ L∞ (QT × (c, ∞)), then problem (3.1) has a unique solution u, and u > u in QT .

(3.3)

In fact, let C = supQT ×(c, ∞) f (x, t, u). Then the linear problem  Lu = C   

in Q∞ ,

Bu = g

on S∞ ,

  

u(x, 0) = ϕ(x)

in Ω

admits a unique solution denoted by u¯. Obviously, u¯ is an upper solution of (3.1). Hence (3.1) has a unique solution u and (3.3) holds by the upper and lower solutions method. In the sequel of this section, we always assume that Ω is of class C 2 , the initial datum u0 ∈ Wp2 (Ω) with p > n + 2 and satisfies ∂n u0 + bu0 = 0 on ∂Ω. Example 3.2

Let a > 0 be a constant and u0 > 0, 6≡ 0. Then the problem    ut − ∆u = u(a − u), 

x ∈ Ω,

∂n u = 0,

x ∈ ∂Ω, t > 0,

  

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

x∈Ω

t > 0, (3.4)

has a unique positive global solution u, and lim u(x, t) = a uniformly in Ω.

t→∞

In fact, for any given z0 > 0, the following initial value problem z 0 (t) = z(a − z),

z(0) = z0

has a unique positive solution z = z(t; z0 ) which satisfies limt→∞ z(t; z0 ) = a. Take k = maxΩ u0 (x). Then k > 0 and u¯ = z(t; k) is an upper solution of (3.4). Obviously, u = 0 is a lower solution of (3.4). So, (3.4) has a unique solution u, and 0 6 u(x, t) 6 z(t; k) for all x ∈ Ω, t > 0 by the upper and lower solutions method. Moreover, the maximum principle confirms u(x, t) > 0,

x ∈ Ω, t > 0.

Take δ > 0. Then u(x, δ) > 0 in Ω and the function w(x, t) = u(x, t + δ) satisfies  w − ∆w = w(a − w),    t

x ∈ Ω,

∂n w = 0,

x ∈ ∂Ω, t > 0,

  

w(x, 0) = u(x, δ),

x ∈ Ω.

t > 0,

3.2. Some examples  59

Let us define m = min u(x, δ),

M = max u(x, δ).

Ω

Ω

Then m, M > 0, and comparison argument leads to z(t; m) 6 w(x, t) = u(x, t + δ) 6 z(t; M ) for all t > 0. Notice limt→∞ z(t; m) = limt→∞ z(t; M ) = a. It follows that limt→∞ u(x, t) = a uniformly in Ω. Example 3.3

Let us deal with the following problem  au   ut − ∆u = c(e − 1), 

x ∈ Ω,

∂n u + bu = 0,

x ∈ ∂Ω, t > 0,

  

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

x ∈ Ω,

t > 0, (3.5)

where u0 (x) > 0, a, c and b are constants, meanwhile, a, c > 0 and b > 0. We first discuss the local solution of (3.5). Take B = maxΩ u0 (x), and A > 0 is a constant determined later. Set u¯ = At + B. Then u¯ is an upper solution of (3.5) in QT as long as A > c(ea(At+B) − 1) for 0 < t 6 T. Fix A > c(ea(1+B) − 1). Then the above inequality holds for T = 1/A. Obviously, u = 0 is a lower solution of (3.5). Hence, problem (3.5) has a unique solution u defined in Ω × (0, 1/A] by the upper and lower solutions method. Let λ1 be the first eigenvalue of (

−∆φ = λφ

in Ω,

∂n φ + bφ = 0

on ∂Ω,

(3.6)

and φ1 be the positive eigenfunction corresponding to λ1 . Then φ1 (x) > 0 in Ω. We call (λ1 , φ1 ) the first eigen-pair of (3.6). Theorem 3.8 Assume ac < λ1 . When the initial datum u0 is suitably small, the unique solution u of (3.5) satisfies limt→∞ u(x, t) = 0 uniformly in Ω. Proof. To this aim, we will seek an upper solution of (3.5) with the form: u¯ = ρe−kt φ1 (x), where ρ and k are positive constants to be determined. It suffices to verify that  ¯t − ∆¯ u > c(ea¯u − 1),  u 

x ∈ Ω,

∂n u¯ + b¯ u > 0,

x ∈ ∂Ω, t > 0,

  

u¯(x, 0) > u0 (x),

x ∈ Ω.

t > 0, (3.7)

A straightforward computation shows that (3.7) is satisfied as long as (

−kρe−kt φ1 + λ1 ρe−kt φ1 > c(exp{aρe−kt φ1 } − 1),

x ∈ Ω, t > 0,

ρφ1 (x) > u0 (x),

x ∈ Ω.

(3.8)

60  3. Semilinear Parabolic Equations

A simple analysis indicates that, for any small ε > 0, there exists ρ0 (ε) > 0 such that, when 0 < ρ 6 ρ0 (ε), exp{aρe−kt φ1 (x)} − 1 6 (1 + ε)aρe−kt φ1 (x) for all x ∈ Ω, t > 0. Therefore the first inequality of (3.8) holds provided −kρe−kt φ1 (x) + λ1 ρe−kt φ1 (x) > c(1 + ε)aρe−kt φ1 (x) for all x ∈ Ω, t > 0, which is equivalent to λ1 − k > c(1 + ε)a. As ac < λ1 , we can select 0 < k, ε  1 so that the above inequality holds. For such determined constants k and ε, we take ρ0 (ε) > 0 as above and u¯ = ρe−kt φ1 (x) with 0 < ρ 6 ρ0 (ε). Then u¯ is an upper solution of (3.5) if u0 6 ρφ1 in Ω, which in turn implies 0 6 u(x, t) 6 ρe−kt φ1 (x) → 0 as t → ∞.  Example 3.4

Now we study a problem as follows  u − ∆u = u(1 − u)(u − a),    t   

x ∈ Ω,

t > 0,

∂n u = 0,

x ∈ ∂Ω, t > 0,

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

x ∈ Ω,

(3.9)

where 0 < a < 1 is a constant. Theorem 3.9

Problem (3.9) has a unique non-negative global solution u, and

(1) if maxΩ u0 (x) < a, then limt→∞ u(x, t) = 0 uniformly in Ω; (2) if minΩ u0 (x) > a, then limt→∞ u(x, t) = 1 uniformly in Ω. Proof. Take K > 1 + maxΩ u0 (x). Then any solution u of (3.9) satisfies 0 6 u 6 K by the maximum principle. Obviously, u¯ = K and u = 0 are an upper solution and a lower solution of (3.9), respectively. Therefore (3.9) has a unique solution u which satisfies 0 6 u 6 K by the upper and lower solutions method. Let z0 > 0 and z(t; z0 ) be the unique solution of z 0 = z(1 − z)(z − a), t > 0;

z(0) = z0 .

Then limt→∞ z(t; z0 ) = 0 when z0 < a, while limt→∞ z(t; z0 ) = 1 when z0 > a. Take M = maxΩ u0 (x) and m = minΩ u0 (x). Then z(t; M ) and z(t; m) are an upper solution and a lower solution of (3.9), respectively. So the unique solution u of (3.9) satisfies z(t; m) 6 u(x, t) 6 z(t; M ) by the upper and lower solutions method. If M < a, then m < a and limt→∞ z(t; M ) = limt→∞ z(t; m) = 0. Accordingly, limt→∞ u(x, t) = 0 uniformly in Ω. If m > a, then M > a and limt→∞ z(t; m) = limt→∞ z(t; M ) = 1. Accordingly, limt→∞ u(x, t) = 1 uniformly in Ω. 

3.3. Monotonicity and convergence  61

3.3

MONOTONICITY AND CONVERGENCE

In this section, we discuss monotonicity and convergence of solutions to the following problem  u + L[u] = f (x, u),    t

x ∈ Ω,

B[u] = g(x),

x ∈ ∂Ω, t > 0,

  

u(x, 0) = ϕ(x),

x ∈ Ω,

t > 0, (3.10)

where L[u] = −aij (x)Dij u + bi (x)Di u, and B[u] = u, or B[u] = ∂n u + b(x)u with b ∈ C 1+α (∂Ω) and b > 0, aij satisfies the condition (A), bi ∈ C(Ω). The equilibrium problem of (3.10) is (

L[u] = f (x, u)

in Ω,

B[u] = g(x)

on ∂Ω.

(3.11)

Let us define an upper (a lower) solution of (3.11) as follows. Definition 3.10 Let u ∈ C(Ω) ∩ C 2 (Ω) when B[u] = u, and u ∈ C 1 (Ω) ∩ C 2 (Ω) when B[u] = ∂n u + b(x)u. Such a function u is called an upper solution (a lower solution) of (3.11) if (

L[u] > (6) f (x, u)

in Ω,

B[u] > (6) g(x)

on ∂Ω.

When u is both upper and lower solutions, we say that u is a solution of (3.11). An upper solution (a lower solution) is called a strict upper solution (strict lower solution) if it is not a solution. We first study the monotonicity of solutions in time t. Theorem 3.11 (Monotonicity in time t) Assume that Ω is of class C 2 , and f satisfies (2.1) for any c < c¯. If the initial datum ϕ of (3.10) is a lower (an upper) solution of (3.11), and problem (3.10) has a solution u ∈ C(QT ) ∩ C 2,1 (QT ) for some T > 0. Then u is monotonically increasing (decreasing) in time t, i.e., 0 6 s < t < T implies u(x, s) 6 (>) u(x, t) for all x ∈ Ω. Furthermore, if the initial datum ϕ of (3.10) is a strict lower (upper) solution of (3.11), then u is strictly monotonically increasing (decreasing) in time t, i.e., 0 6 s < t < T implies u(x, s) < (>) u(x, t) for all x ∈ Ω. Proof. Step 1. Obviously, ϕ is a lower solution of (3.10). Because the function f

62  3. Semilinear Parabolic Equations

is locally Lipschitz continuous in u, we have that u(x, t) > ϕ(x) in Ω × [0, T ) by Theorem 3.3. Fix 0 < δ < T and consider the problem  w + L[w] = f (x, w),    t   

x ∈ Ω,

0 < t < T − δ,

B[w] = g(x),

x ∈ ∂Ω, 0 < t < T − δ,

w(x, 0) = u(x, δ),

x ∈ Ω.

(3.12)

Obviously, w(x, t) := u(x, δ + t) is a solution of (3.12). Note u(x, δ) > ϕ(x). We have w > u in Ω × [0, T − δ) by applying the comparison principle to (3.10) and (3.12). That is, u(x, δ + t) > u(x, t) for all x ∈ Ω, 0 6 t < T − δ. Step 2. In addition, if ϕ is a strict lower solution of (3.11), then there is x0 ∈ Ω such that L[ϕ](x0 ) < f (x0 , ϕ(x0 )), or there is x0 ∈ ∂Ω such that B[ϕ](x0 ) < g(x0 ). Let v(x, t) = u(x, t) − ϕ(x). Then    vt + L[v] + M v > 0, 

x ∈ Ω,

B[v] > 0,

x ∈ ∂Ω, 0 < t < T,

  

v(x, 0) = 0,

x ∈ Ω,

0 < t < T,

and for any 0 < t < T , we have vt + L[v] + M v > 0 at (x0 , t) when x0 ∈ Ω, and B[v] > 0 at (x0 , t) when x0 ∈ ∂Ω, where M is the Lipschitz constant of f given by (2.1). The strong maximum principle gives v > 0, i.e., u(x, t) > ϕ(x) for x ∈ Ω and 0 < t < T . Similar to the step 1, it can be shown that u is strictly monotonically increasing in time t.  Now, we study the convergence of solutions as t → ∞. Here, we only consider the case ∂n u + b(x)u = 0. Theorem 3.12 (Convergence of solution) Let conditions of Theorem 2.11 be fulfilled with f = f (x, u). We further assume that the unique solution u of (3.10) is monotonically increasing (decreasing) in time t. Then (3.11) has at least one solution denoted by us (x), and lim ku(·, t) − us (·)kC 2+β (Ω) = 0 for any 0 < β < α.

t→∞

(3.13)

Proof. Clearly, the point-wise limit limt→∞ u(x, t) = us (x) exists since u is bounded and monotone in time t. On the other hand, the set {u(·, t)}t>1 is compact in C 2+β (Ω) owing to the estimate (2.18). So, limt→∞ u(·, t) = us (·) in C 2+β (Ω), which implies that us ∈ C 2+β (Ω) and (3.13) holds. Recalling (2.18), we also have limt→∞ ut (x, t) = 0 for any x ∈ Ω. Thus us (x) solves (3.11) by letting t → ∞ in the first two equations of (3.10).  Theorem 3.13 (Convergence of solution) Let conditions of Theorem 2.11 be fulfilled with f = f (x, u). Suppose that (3.11) admits a unique solution us , and ϕ and ψ are a lower solution and an upper solution of (3.11), respectively, which satisfy ϕ 6 us 6 ψ in Ω. Let uϕ and uψ be solutions of (3.10) respectively with initial data ϕ and ψ. Then, for any 0 < β < α, uϕ (·, t) % us (·) and uψ (·, t) & us (·)

in C 2+β (Ω) as t → ∞.

3.3. Monotonicity and convergence  63

Proof. Based on Theorem 3.11, uϕ and uψ are, respectively, monotonically increasing and decreasing in time t. The comparison principle confirms uϕ 6 uψ as well. Therefore, ϕ 6 uϕ 6 uψ 6 ψ. The desired conclusions can be followed immediately from Theorem 3.12.  Theorem 3.14 Under conditions of Theorem 2.11 with f = f (x, u), we further assume that (3.11) admits a unique solution us (x) ∈ C 2+α (Ω). Let u be the unique solution of (3.10). Then, for any 0 < β < α, the following three kinds of convergence are equivalent: lim u(x, t) = us (x) a.e. in Ω,

(3.14)

lim ku(x, t) − us (x)kC 2+β (Ω) = 0,

(3.15)

t→∞ t→∞

lim ku(x, t) − us (x)kH 1 (Ω) = 0.

t→∞

Proof. It suffices to show that (3.14) implies (3.15). Assume that (3.14) holds. Owing C

to C 2+α (Ω) ,→ C 2+β (Ω) for 0 < β < α, the limit (3.15) can be deduced from the estimate (2.18).  Same as the proof of Theorem 3.14, we have the following conclusion. Theorem 3.15 Under requirements of Theorem 2.13 with f = f (x, u), we further assume that problem (3.11) admits a unique solution us (x) ∈ C 1+α (Ω). Let u be the unique solution of (3.10). Then, for any 0 < β < α, the following three kinds of convergence are equivalent: lim u(x, t) = us (x) a.e. in Ω,

t→∞

lim ku(x, t) − us (x)kC 1+β (Ω) = 0,

t→∞

lim ku(x, t) − us (x)kH 1 (Ω) = 0.

t→∞

Let us study the following problem as an example  u − ∆u = u(a − u),    t   

x ∈ Ω,

t > 0,

u = 0,

x ∈ ∂Ω, t > 0,

u(x, 0) = u0 (x) > 0, 6≡ 0,

x ∈ Ω,

(3.16)

where Ω is of class C 2+α , u0 ∈ C 2 (Ω) and u0 = 0 on ∂Ω. Its equilibrium problem is (

−∆u = u(a − u),

x ∈ Ω,

u = 0,

x ∈ ∂Ω.

(3.17)

In accordance with the upper and lower solutions method, and the regularity theory (Theorem 2.10), it can be shown that (3.16) has a unique non-negative global solution u ∈ C 2+α, 1+α/2 (Ω × R+ ). Moreover, u 6 max{a, maxΩ u0 (x)} by the maximum principle. Let (λ, φ) be the first eigen-pair of (1.33). It is well known that, when a > λ, (3.17) has a unique positive solution us ∈ C 2+α (Ω), and ∂n us < 0 on ∂Ω.

64  3. Semilinear Parabolic Equations

Theorem 3.16 Let u be the unique non-negative global solution of (3.16). If a > λ, then limt→∞ u(x, t) = us (x) in C 2+β (Ω) for any 0 < β < α. Proof. In accordance with the maximum principle and the Hopf boundary lemma, u > 0 in Ω × R+ and ∂n u < 0 on ∂Ω × R+ . Without loss of generality, we may assume that u0 > 0 in Ω and ∂n u0 < 0 on ∂Ω. Note that φ > 0 in Ω and ∂n φ < 0 on ∂Ω. We can find two constants 0 < ε  1 and k > 1 for which εφ(x) 6 u0 (x) 6 kus (x) in Ω. It is easy to verify that εφ and kus are respectively a lower solution and an upper solution of (3.17) as long as 0 < ε  1. Let u1 and u2 be solutions of (3.16) with, respectively, initial data εφ and kus . Then u1 6 u 6 u2 by the comparison principle. Theorem 3.13 asserts that ui (x, t) → us (x) in C 2+β (Ω) as t → ∞.  Theorem 3.17 Let u be the unique non-negative global solution of (3.16). If a 6 λ, then limt→∞ u(x, t) = 0 in C 2+β (Ω) for any 0 < β < α. Z

Proof. Normalize φ by

φ(x) = 1. Multiplying the first equation of (3.16) by φ and Ω

integrating the result over Ω, we conclude Z

Z

Z

uφ = a

ut φ + λ

Ω

Ω

Ω

uφ −

Z Ω

2

u φ6a

Z

2

Z

uφ −

uφ

.

Ω

Ω

Z

Set f (t) =

u(x, t)φ(x). Then Ω

f 0 (t) 6 −f 2 (t) and f (t) > 0 for t > 0; which implies f (t) → 0, i.e.,

Z

f (0) > 0,

u(x, t)φ(x) → 0 as t → ∞. Notice that u is bounded,

Ω

in view of Theorem 2.11 we can find ti with ti → ∞, ˜ ∈ C 2+β (Ω) such that Z and u u(x, ti ) → u˜(x) in C 2+β (Ω). Combining this with u(x, ti )φ(x) → 0, we derive Ω

u˜ ≡ 0. That is, u(x, ti ) → 0 in C 2+β (Ω) as i → ∞. This also implies u(x, t) → 0 in C 2+β (Ω) as t → ∞. 

3.4

THE WEAK UPPER AND LOWER SOLUTIONS METHOD

For the sake of simplicity, we only focus on (2.7), and rewrite it here:  u − Dj (aij Di u) + bi Di u = f (x, t, u)    t   

in QT ,

u=g

on ST ,

u(x, 0) = ϕ(x)

in Ω,

(3.18)

where aij , bi ∈ L∞ (QT ), aij = aji in QT for all 1 6 i, j 6 n, and there exists a constant λ > 0 so that aij (x, t)yi yj > λ|y|2 for all (x, t) ∈ QT , y ∈ Rn .

3.4. The weak upper and lower solutions method  65

Also, weak lower solution, weak upper solution and weak solution of (3.18) are described in Definition 2.6. We first focus on the weak upper and lower solutions method. For convenience and to facilitate understanding, we give the conclusion and proof under relatively strong conditions. Theorem 3.18 (Weak upper and lower solutions method) Assume that the condition (G1) holds with p > 2, and f ∈ L∞ (QT × (c, c¯)) satisfies (2.1) for any constants c < c¯. Let u¯ and u be respectively the bounded weak upper solution and weak lower solution of (3.18). Then, in the interval hu, ui, problem (3.18) has a unique solution u ∈ Wp2,1 (QT ), where hu, ui = {u ∈ L∞ (QT ) : u 6 u 6 u in QT } . Proof. Firstly, u 6 u¯ by Theorem 2.7. Let us define L u = ut − Dj (aij Di u) + bi Di u.

For any given v ∈ hu, ui, the linear problem  L u + M u = f (x, t, v(x, t)) + M v(x, t)   

in QT ,

u=g

on ST ,

  

u(x, 0) = ϕ(x)

in Ω

(3.19)

has a unique solution u ∈ Wp2,1 (QT ) by the Lp theory (Theorem 1.1). As in the proof of Theorem 3.5 we conclude that kukWp2,1 (QT ) 6 C and |u|α, QT 6 C for all v ∈ hu, u¯i, and the operator F : hu, u¯i → C(QT ) defined by F (v) = u is compact. Next we prove F : hu, u¯i → hu, u¯i. Suppose v ∈ hu, u¯i and u = F (v). Clearly, u is a weak solution of (3.19). Note that u¯ and u are respectively the weak upper solution and weak lower solution of (3.18), and the function f (·, s) + M s is increasing in s and u 6 v 6 u¯. It follows that u¯ and u are respectively the weak upper solution and weak lower solution of (3.19), and hence u 6 u 6 u¯ in QT by Theorem 2.7. Consequently F : hu, u¯i → hu, u¯i. Because hu, u¯i is a bounded and closed convex set of L∞ (QT ), the operator F has at least one fixed point u ∈ hu, u¯i by the Schauder fixed point theorem (Corollary A.3), which in turn solves problem (3.18). Clearly u ∈ Wp2,1 (QT ) due to u = F (u) ∈ Wp2,1 (QT ). The uniqueness follows from Theorem 2.7.  In applications, we usually encounter such a situation that two better upper solutions (lower solutions) u1 and u2 have been constructed, however they do not meet our requirements, and min{u1 , u2 } (max{u1 , u2 }) might be suitable for our problems. Naturally, we want to know whether min{u1 , u2 } (max{u1 , u2 }) is a weak upper (lower) solution. We will answer this question in the following.

66  3. Semilinear Parabolic Equations

Theorem 3.19 Assume that p > 2 and f ∈ L∞ (QT ×(c, c¯)) for any constants c < c¯. If u¯1 , u¯2 ∈ Wp2,1 (QT )∩C(QT ) are two upper solutions of (3.18), then u¯ = min{¯ u1 , u¯2 } 2,1 is a weak upper solution of (3.18). If u1 , u2 ∈ Wp (QT ) ∩ C(QT ) are two lower solutions of (3.18), then u = max{u1 , u2 } is a weak lower solution of (3.18). Proof. We only prove the conclusion for weak lower solutions, and the assertion for weak upper solutions is analogous. Step 1. Let u1 , u2 ∈ Wp2,1 (QT )∩C(QT ) be two lower solutions of (3.18). Then they are weak lower solutions of (3.18). To simplify notations, we denote ui by ui (i = 1, 2), and u by u in the sequel. It will be illustrated that u is a weak lower solution of (3.18). Write u = u1 + (u2 − u1 )+ . We find that, for any 0 < t 6 T ,

Du =

  Du1  

in At1 := {(x, s) ∈ Qt : u1 (x, s) > u2 (x, s)},

Du1 = Du2

in At0 := {(x, s) ∈ Qt : u1 (x, s) = u2 (x, s)},

   

Du2

in At2 := {(x, s) ∈ Qt : u2 (x, s) > u1 (x, s)}.

As p > 2, it is easy to see that u ∈ C([0, T ], L2 (Ω)) ∩ L2 ((0, T ), H 1 (Ω)) ∩ L∞ (Ω × (0, T )), f (·, u(·)) ∈ L2 (Ω × (0, T )), u(x, 0) 6 ϕ(x) in Ω, u 6 g on ST . It remains to show that, for all 0 < t 6 T , Z Qt

Z

(aij Di uDj φ + bi Di uφ) 6

(f (x, s, u)φ + uφs ) −

Qt

Z

uφ|s=t

(3.20)

Ω

•

for each test function φ ∈ W 1,1 2 (QT ) with φ > 0 in QT . Step 2. Let v = u2 − u1 ∈ W22,1 (QT ). Thanks to classical density results for there exists a sequence {vl } ⊂ C 2 (QT ) such that vl → v in W22,1 (QT ) as l → ∞. By passing to a sub-sequence, we may assume that W22,1 (QT ),

kvl − vkW 2,1 (QT ) 6 l−3 for all l > 1.

(3.21)

2

Select a function θ : R → R satisfying (a) θ ∈ C ∞ (R); (b) θ is nondecreasing in R; (c) 0 6 θ(τ ) 6 1; (d) θ(τ ) = 0 for τ 6 0, θ(τ ) = 1 for τ > 1. Set θl (τ ) = θ(lτ ). Obviously, θl meets with (a)–(c) above, and θl (τ ) = 0 for τ 6 0,

θl (τ ) = 1 for τ > 1/l.

Let M = max[0,1] {θ0 + |θ00 |}. Then we have 0 6 θl0 (τ ) = lθ0 (lτ ) 6 M l,

|θl00 (τ )| 6 M l2 .

(3.22)

3.4. The weak upper and lower solutions method  67 •

Now for any φ ∈ C ∞ (QT ) with φ > 0, we define φ1 = φl1 = (1 − θl (vl ))φ,

φ2 = φl2 = θl (vl )φ.

•

It is easy to see that φ1 , φ2 ∈ W 1,1 2 (QT ) are non-negative. Since uk (k = 1, 2) are weak lower solutions of (3.18), by the definition we see that for k = 1, 2, Z

Z

Qt

(aij Di uk Dj φk + bi Di uk φk ) 6





f (·, uk )φk + uk φk,s −

Qt

Z

uk φk |s=t .

Ω

That is, Z



Qt

Z

6

aij Di u1 (1 − θl (vl ))Dj φ − φθl0 (vl )Dj vl + bi Di u1 (1 − θl (vl ))φ 



[f (·, u1 )φ1 + u1 φ1,s ] −

Qt

Z

u1 φ1 |s=t



(3.23)

Ω

and Z



Qt

Z

6

aij Di u2 θl (vl )Dj φ + φθl0 (vl )Dj vl + bi Di u2 θl (vl )φ 



Z

[f (·, u2 )φ2 + u2 φ2,s ] −

Qt



u2 φ2 |s=t .

(3.24)

Ω

It then follows from (3.23) and (3.24) that Z

Z

aij Di v[θl (vl )Dj φ + φθl0 (vl )Dj vl ]

aij Di u1 Dj φ + Qt

Qt

Z

[bi Di u1 φ + bi Di vθl (vl )φ] −

+

Z

Qt

Qt

Z

6

Z







f (·, u2 ) − f (·, u1 ) θl (vl )φ −

f (·, u1 )φ + Qt

u1 φs + vθl (vl )φs + vθl0 (vl )vl,s φ



Qt

Z

(u1 φ + vθl (vl )φ)|s=t . (3.25) Ω

Step 3. Since f (x, t, u2 ) = f (x, t, u1 ) for a.e. (x, t) ∈ At0 , we conclude Z





f (·, u2 ) − f (·, u1 ) θl (vl )φ =

Qt

Z At1 ∪At2





f (·, u2 ) − f (·, u1 ) θl (vl )φ.

Notice v > 0 in At2 . It follows that, by passing to a sub-sequence, θl (vl ) → 1 a.e. in At2 as l → ∞. Similarly, θl (vl ) → 0 a.e. in At1 as l → ∞. Due to the dominated convergence theorem, we find that Z

lim

l→∞ At 2





f (·, u2 ) − f (·, u1 ) θl (vl )φ =

and

Z

lim

l→∞

At1



Z



At2





f (·, u2 ) − f (·, u1 ) φ

f (·, u2 ) − f (·, u1 ) θl (vl )φ = 0.

68  3. Semilinear Parabolic Equations

Therefore Z

lim



l→∞ Qt



f (·, u2 ) − f (·, u1 ) θl (vl )φ =

Z





f (·, u2 ) − f (·, u1 ) φ.

At2

(3.26)

Analogously, we have Z Z    lim aij Di vθl (vl )Dj φ = aij Di vDj φ,    At2  l→∞ Qt   Z Z      bi Di vθl (vl )φ = bi Di vφ, lim  l→∞  t Qt

A2

Z

Z

   lim vθl (vl )φs = vφs ,    l→∞ Qt At2    Z Z     lim  vθl (vl )φ|s=t =  t l→∞ Ω

(3.27)

vφ.

A2 ∩{s=t}

Next, we use θl0 (vl ) > 0 and φ > 0 to obtain Z Qt

aij Di vφθl0 (vl )Dj vl

Z

= Qt

Z

>

Qt

[aij Di vθl0 (vl )Dj vφ + aij Di vθl0 (vl )Dj (vl − v)φ] aij Di vθl0 (vl )Dj (vl − v)φ.

(3.28)

By (3.21) and (3.22) it yields lim kDj (vl − v)θl0 (vl )φk2,Qt = 0,

l→∞

and hence Z

lim

l→∞ Qt

aij Di vθl0 (vl )Dj (vl − v)φ = 0.

Consequently, from (3.28) we derive Z

lim inf l→∞

Qt

aij Di vφθl0 (vl )Dj vl > 0.

(3.29)

Analogously, we have Z

lim

l→∞ Qt

vθl0 (vl )vl,s φ = lim

Z

l→∞ At 2

θl0 (vl )vvs φ.

(3.30)

We claim Z

lim

l→∞ At 2

θl0 (vl )vvs φ = 0.

In fact, write θl0 (vl ) = θl0 (vl ) − θl0 (v) + θl0 (v) = θl00 (v ∗ )(vl − v) + θl0 (v).

(3.31)

Exercises  69

Noticing (3.21), (3.22), and v, φ ∈ L∞ (QT ), we have Z

lim

l→∞ At 2

θl00 (v ∗ )(vl − v)vvs φ = 0

as above. Owing to |θl0 (v)v| 6 M in At2 and liml→∞ θl0 (v(x, t))v(x, t) = 0 for each (x, t) ∈ At2 , it can be obtained by the dominated convergence theorem that Z

lim

l→∞ At 2

θl0 (v)vvs φ = 0.

Therefore (3.31) holds. Step 4. We now take l → ∞ in (3.25) and make use of (3.26), (3.27), (3.29), (3.30) and (3.31) to infer that Z

Z

aij Di u1 Dj φ +

At2

Qt

Z

6

Z

f (·, u1 )φ + Qt

Z

bi Di u1 φ + Qt



At2

Z

aij Di vDj φ + 

f (·, u2 ) − f (·, u1 ) φ −

Z

At2

bi Di vφ −

u1 φ|s=t −

Ω

Z

u1 φs −

Qt

Z At2

vφs

Z

vφ,

(3.32)

At2 ∩{s=t}

which implies (3.20) (cf. Exercise 3.12). •

•

∞ For the given φ ∈ W 1,1 2 (QT ) with φ > 0, we can choose φ` ∈ C (QT ) with φ` > 0, 1,1 such that φ` → φ in W2 (QT ). Then (3.20) holds for all such φ` . Letting ` → ∞ we conclude that (3.20) holds for such φ. 

Remark 3.20

Assume that conditions of Theorems 3.18 and 3.19 hold.

(1) We have max{u1 , u2 } 6 min{¯ u1 , u¯2 }. (2) For ui , u¯i , using the upper and lower solutions method we know that problem (3.18) has a unique solution u ∈ Wp2,1 (QT ) satisfying ui 6 u 6 u¯i , i = 1, 2. u1 , u¯2 }. Hence, max{u1 , u2 } 6 u 6 min{¯

EXERCISES 3.1 Prove that, for the upper and lower solutions defined in Remark 3.4, Theorem 3.3 still holds. Hint: apply Theorems 1.41, 1.42 and 1.43. 3.2 Let u0 ∈ C 2 (Ω) and u0 = 0 on ∂Ω. Using the upper and lower solutions method to prove that there is T > 0 such that problem  u − ∆u = u2    t

u=0

  

has a unique solution.

in QT on ST

u(x, 0) = u0 (x) in Ω

70  3. Semilinear Parabolic Equations

3.3 Let (λ, φ) be the first eigen-pair of (1.33). Assume a > λ and 0 < ε < a − λ. Prove that the solution u of  u − ∆u = au − u2 ,    t

x ∈ Ω, t > 0,

u = 0,

x ∈ ∂Ω, t > 0,

  

u(x, 0) = εφ(x),

x∈Ω

is strictly increasing in time t. 3.4 Consider an initial-boundary value problem  u − ∆u = u(a − u),    t   

x ∈ Ω,

t > 0,

B u = 0,

x ∈ ∂Ω, t > 0,

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

x ∈ Ω,

where B u = ∂n u or B u = u, a is a constant, u0 ∈ C 1 (Ω) ∩ Wp2 (Ω) with p > 1 and B u0 = 0 on ∂Ω. (1) Prove that this problem has a unique global solution u and u satisfies 0 6 u 6 max{a, maxΩ u0 }. Especially, please explain requirements of the Hopf boundary lemma when we use it. (2) Let λ be the first eigenvalue of −∆φ = λφ in Ω,

B φ = 0 on ∂Ω.

Prove that a < λ implies lim u(x, t) = 0 uniformly in Ω.

t→∞

3.5 Consider problem  u − ∆u = au − uk ,    t   

x ∈ Ω,

t > 0,

u = 0,

x ∈ ∂Ω, t > 0,

u(x, 0) = u0 (x) > 0, 6≡ 0,

x ∈ Ω,

(P3.1)

where constant k > 1, Ω is of class C 2+α , u0 ∈ C(Ω) ∩ Wp2 (Ω) with p > 1 and u0 = 0 on ∂Ω. (1) Use the upper and lower solutions method and regularity theory to show that problem (P3.1) has a unique global solution u ∈ C 2+α,1+α/2 (Ω × R+ ). (2) Let λ be the first eigenvalue of (1.33) and a > λ. Let us be the unique positive solution of (

−∆u = au − uk in Ω u=0

on ∂Ω.

Prove that the unique solution u of (P3.1) satisfies limt→∞ u(x, t) = us (x) in C 2 (Ω).

Exercises  71

3.6 Apply the regularity and uniform estimates given in §2.2 to prove that the unique solution of (3.4) satisfies limt→∞ u(x, t) = a in C 2 (Ω). 3.7 Apply the regularity and uniform estimates given in §2.2 to prove Theorem 3.9. 3.8 Consider the following linear problem  u − ∆u = au,    t   

x ∈ Ω,

t > 0,

u = 0,

x ∈ ∂Ω, t > 0,

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

x ∈ Ω,

where a > 0 is a constant and u0 ∈ Wp2 (Ω) with p > 1 and u0 > 0, 6≡ 0. Let (λ, φ) be the first eigen-pair of (1.33). Prove the following conclusions. (1) Solution u of this problem exists uniquely and globally in time t and u ∈ C 2+α,1+α/2 (Ω × R+ ). Moreover, u > 0 in Ω × R+ and ∂n u < 0 on ∂Ω × R+ . (2) If a < λ, then limt→∞ u(x, t) = 0 uniformly in Ω. (3) If a = λ, then for any τ > 0, there exist positive constants ε < K such that εφ(x) 6 u(x, t) 6 Kφ(x) in Ω × [τ, ∞). (4) If a > λ, then limt→∞ u(x, t) = ∞ for x ∈ Ω. 3.9 Let µ1 be the principal eigenvalue of (

L[ϕ] = µϕ

in Ω,

B[ϕ] = 0

on ∂Ω

and ϕ1 be the corresponding positive eigenfunction, where the operators L and B are as in (3.10). Let us (x) be a solution of (3.11). Assume that there exist δ > 0 and 0 < θ < µ1 such that |fu (x, u)| 6 µ1 − θ

for x ∈ Ω, us (x) − δ 6 u 6 us (x) + δ.

Prove that there exists ρ > 0 such that when |ϕ(x) − us (x)| 6 ρϕ1 (x), problem (3.10) has a unique solution uϕ , and |uϕ (x, t) − us (x)| 6 ρe−(µ1 −θ)t ϕ1 (x) for all (x, t) ∈ Ω × R+ . 3.10 Let q ∈ L∞ (Ω) and λ1 (q) be the principal eigenvalue of the operator −∆ + q in Ω with the homogeneous Dirichlet boundary condition. Assume that f > 0, g > 0 and u0 > 0, 6≡ 0. Prove that if λ1 (q) < 0 then solution u of    ut − ∆u + q(x)u = f (x, t) 

in Q∞

u = g(x, t)

on S∞

  

u(x, 0) = u0 (x)

in Ω

satisfies limt→∞ u(x, t) = ∞ uniformly in any compact subset of Ω.

72  3. Semilinear Parabolic Equations

3.11 Let λ be the first eigenvalue of (1.33) and f ∈ C 1 (R) satisfy lim sup |u|→∞

f (u) < λ. u

Prove that the following problem  u − ∆u = f (u),    t   

x ∈ Ω,

t > 0,

u = 0,

x ∈ ∂Ω,

t > 0,

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

x∈Ω

has a unique global solution u and u is bounded. 3.12 Prove that (3.32) implies (3.20). 3.13 Let Γ1 and Γ2 be given in (1.3). Set B u = ∂n u + bu, b > 0 on Γ1 × (0, T ],

and B u = u on Γ2 × (0, T ],

and consider initial-boundary value problem    L u = f (x, t, u) 

in QT ,

Bu = g

on ST ,

  

u(x, 0) = ϕ(x)

in Ω.

Establish the upper and lower solutions method (definitions of upper and lower solutions, existence and uniqueness of solutions derived from the upper and lower solutions, monotonicity and convergence of solution).

CHAPTER

4

Weakly Coupled Parabolic Systems

As a continuation of the previous chapter, this chapter deals with the initial-boundary value problems of weakly coupled semilinear parabolic systems. First, the Lp theory, Schauder theory and Schauder fixed point theorem are used to investigate the local existence and uniqueness of solutions to initial-boundary value problems of weakly coupled nonlinear systems. Second, by applying the Schauder fixed point theorem it is proved that if a problem admits a pair of coupled upper and lower solutions, then it must have a unique solution located between upper and lower solutions. Third, by means of the coupled upper and lower solutions, we construct the monotone iterative sequences to obtain the unique solution. Examples are included to illustrate this method.

4.1

LOCAL EXISTENCE AND UNIQUENESS OF SOLUTIONS

In this section we intend to use the Lp theory, Schauder theory and Schauder fixed point theorem to study the local existence and uniqueness of solutions to initialboundary value problem of weakly coupled parabolic systems. Let us consider the following initial-boundary value problem  Lk uk = fk (x, t, u)       Bk uk = gk  uk (x, 0) = ϕk (x)     

in QT , on ST , in Ω,

(4.1)

k = 1, . . . , m,

where Ω is of class C 2 , u = (u1 , . . . , um ), and Lk = ∂t − akij (x, t)Dij + bki (x, t)Di , Bk = ak ∂n + bk (x, t), 1 6 k 6 m.

Assume that either (i) ak = 0, bk = 1; or (ii) ak = 1, bk > 0 and bk ∈ C 1, 1/2 (S T0 ) for some T0 > 0. Define f = (f1 , . . . , fm ), g = (g1 , . . . , gm ), ϕ = (ϕ1 , . . . , ϕm ). 73

74  4. Weakly Coupled Parabolic Systems

Let us discuss the strong solution (solution in [Wp2,1 (QT )]m ) first. We further assume that coefficients of Lk are bounded and satisfy the condition (A) which is given in §1.1, g ∈ [Wp2,1 (QT0 )]m and ϕ ∈ [Wp2 (Ω)]m with p > 1. Define n

o

UT0 (ϕ) = u ∈ [Lp (QT0 )]m : ku − ϕk[Lp (QT0 )]m 6 1 . Theorem 4.1 (Local existence and uniqueness) Assume that, for each 1 6 k 6 m, fk : QT0 × UT0 (ϕ) → Lp (QT0 ) is continuous with respect to u ∈ UT0 (ϕ), i.e., kfk (·, ui ) − fk (·, u)kp, QT0 → 0 when ui , u ∈ UT0 (ϕ) and kui − uk[Lp (QT0 )]m → 0. If compatibility conditions Bk ϕk = gk for x ∈ ∂Ω, t = 0 and all 1 6 k 6 m

hold, then there exists 0 < T 6 T0 such that problem (4.1) has at least one solution u ∈ [Wp2,1 (QT )]m . Furthermore, if f satisfies the Lipschitz condition with respect to u ∈ UT0 (ϕ), i.e., there is a constant M > 0 such that, for each 1 6 k 6 m, kfk (·, u) − fk (·, v)kp,QT0 6 M ku − vk[Lp (QT0 )]m for all u, v ∈ UT0 (ϕ), then the solution is unique. Proof. By our assumption, there is a constant C1 = C1 (T0 ) > 0 such that kfk (·, u)kp, QT0 6 C1 for all u ∈ UT0 (ϕ). For the given u ∈ UT (ϕ), we set u = ϕ in Ω × [T, T0 ]. Then u ∈ UT0 (ϕ), and the linear problem  L v = fk (x, t, u)    k k   

in QT0 ,

Bk vk = gk

on ST0 ,

vk (x, 0) = ϕk (x)

in Ω

has a unique solution vk ∈ Wp2,1 (QT0 ), and kvk kWp2,1 (QT

0



)

6 C kfk (·, u)kp, QT0 + kgk kWp2,1 (QT ) + kϕk kWp2 (Ω)



0



6 C C1 + kgk kWp2,1 (QT ) + kϕk kWp2 (Ω)



(4.2)

0

by Theorem 1.7. We set v = (v1 , . . . , vm ), and define an operator F : UT (ϕ) → [Lp (QT0 )]m ⊂ [Lp (QT )]m by letting F (u) = v. Let ui , u ∈ UT (ϕ) and kui − uk[Lp (QT )]m → 0 as i → ∞. Set ui = u = ϕ in Ω × [T, T0 ]. Then kui − uk[Lp (QT0 )]m = kui − uk[Lp (QT )]m → 0 as i → ∞, and fk (·, ui ) = fk (·, u) in Ω × [T, T0 ]. Thus kfk (·, ui ) − fk (·, u)kp, QT = kfk (·, ui ) − fk (·, u)kp, QT0 → 0

(4.3)

4.1. Local existence and uniqueness of solutions  75

as i → ∞ by our assumption. Let v i = F (ui ) and v = F (u). Then v i − v satisfies, for each 1 6 k 6 m,  i i   Lk (vk − vk ) = fk (x, t, u ) − fk (x, t, u)      

in QT ,

Bk (vki − vk ) = 0

on ST ,

vki (x, 0) − vk (x, 0) = 0

in Ω.

By using the Lp theory (Theorem 1.7) and (4.3) in turn, it can be obtained that kvki − vk kWp2,1 (QT ) 6 Ckfk (·, ui ) − fk (·, u)kp, QT → 0 as i → ∞, which implies kF (ui ) − F (u)k[Wp2,1 (QT )]m → 0 as i → ∞. Therefore, F is continuous. This, together with the estimate (4.2), concludes that C

F : UT (ϕ) → [Lp (QT )]m is compact since [Wp2,1 (QT )]m ,→ [Lp (QT )]m . Now, we prove F : UT (ϕ) → UT (ϕ) provided 0 < T  1. In fact, the estimate (4.2) implies that, for v = F (u) and each 1 6 k 6 m,

kvk − ϕk kWp2,1 (QT

0

)

6 C for all u ∈ UT (ϕ).

We can find q > p such that Wp2,1 (QT0 ) ,→ Lq (QT0 ). Then kvk − ϕk kq, QT0 6 C, and kvk − ϕk kp, QT 6 |QT |(q−p)/(pq) kvk − ϕk kq, QT 6 CT (q−p)/(pq) kvk − ϕk kq, QT0 for all u ∈ UT (ϕ). This shows that F (u) ∈ UT (ϕ) if 0 < T  1. Therefore, F has at least one fixed point u ∈ UT (ϕ) by the Schauder fixed point theorem (Corollary A.3), and then problem (4.1) has at least one solution u ∈ UT (ϕ). Clearly, u ∈ [Wp2,1 (QT )]m on account that u = F (u). In the following we prove the uniqueness under assumption that f satisfies the Lipschitz condition with respect to u ∈ UT0 (ϕ). Let 0 < T  1 and u, w ∈ [Wp2,1 (QT )]m ∩ UT (ϕ) be two solutions of (4.1). We extend u and w by defining u = w = ϕ in Ω × [T, T0 ]. Obviously, u, w ∈ UT0 (ϕ). Thus, we can define u˜ = F (u) ∈ [Lp (QT0 )]m , w˜ = F (w) ∈ [Lp (QT0 )]m . It is easy to see that u, u˜ are solutions of linear initial-boundary value problem  Lk Uk = fk (x, t, u(x, t))       Bk Uk = gk  Uk (x, 0) = ϕk (x)     

k = 1, . . . , m,

in QT , on ST , in Ω,

76  4. Weakly Coupled Parabolic Systems

and w, w˜ are solutions of linear initial-boundary value problem    Lk Wk = fk (x, t, w(x, t))     Bk Wk = gk  Wk (x, 0) = ϕk (x)     

in QT , on ST , in Ω,

k = 1, . . . , m.

The uniqueness of solutions to linear initial-boundary value problem implies u˜ = u, w˜ = w in QT . For each 1 6 k 6 m, k˜ uk − w˜k kWp2,1 (QT

0

)

6 Ckfk (·, u) − fk (·, w)kp, QT0 6 Cku − wk[Lp (QT0 )]m = Cku − wk[Lp (QT )]m .

Similar to the above, k˜ uk − w˜k kq, QT0 6 Ck˜ uk − w˜k kWp2,1 (QT

0

)

for some q > p. Then

kuk − wk kp, QT = k˜ uk − w˜k kp, QT 6 CT (q−p)/(pq) k˜ uk − w˜k kq, QT 6 CT (q−p)/(pq) k˜ uk − w˜k kq, QT0 6 CT (q−p)/(pq) ku − wk[Lp (QT )]m for each 1 6 k 6 m. This implies u = w if 0 < T  1.  2+α, 1+α/2 m Next, we deal with the classical solution (solution in [C (QT )] ). Suppose that, for QT0 , coefficients of Lk satisfy the condition (C) which is given in §1.1, gk ∈ C 2+α, 1+α/2 (QT0 ) and ϕk ∈ C 2+α (Ω). Define VT0 (ϕ) := {u ∈ [C(QT0 )]m : ku − ϕk[L∞ (QT0 )]m 6 1}. Theorem 4.2 Let Ω be of class C 2+α and b ∈ [C 1+α, (1+α)/2 (S T0 )]m . Assume that f is H¨older continuous with respect to (x, t) ∈ QT0 and u ∈ VT0 (ϕ). If compatibility conditions ∂n ϕk + bk ϕk = gk for x ∈ ∂Ω, t = 0 if ak = 1, ϕk = gk , gkt − akij Dij ϕk + bki Di ϕk = fk (·, ϕk ) for x ∈ ∂Ω, t = 0 if ak = 0 hold, then there exists 0 < T 6 T0 such that problem (4.1) has at least one solution u ∈ [C 2+α, 1+α/2 (QT )]m . Furthermore, if f satisfies the Lipschitz condition with respect to u ∈ VT0 (ϕ), i.e., there is a constant M > 0 such that, for each 1 6 k 6 m, |fk (x, t, u) − fk (x, t, v)| 6 M |u − v| for all (x, t) ∈ QT0 , u, v ∈ VT0 (ϕ), then the solution is unique. The proof of Theorem 4.2 is left to readers as an exercise.

4.2. The upper and lower solutions method  77

4.2

THE UPPER AND LOWER SOLUTIONS METHOD

In §3.1 we have mentioned that the upper and lower solutions method is a powerful tool for the study of differential equations. Moreover, the upper and lower solutions method not only provides existence but also gives estimates of solutions. For the scalar case, we have seen that the upper and lower solutions method is completely based on comparison principle. Regarding systems, in §1.3.3 we specifically emphasized that the maximum principle holds only for certain special circumstances. For this reason, when investigating systems, we cannot use the same manner as in the scalar case to define upper and lower solutions, we need to define them according to the structure of systems. Define [u]k = (u1 , . . . , uk−1 , uk+1 , . . . , um ). Then (4.1) can be written as  Lk uk = fk (x, t, uk , [u]k )       Bk uk = gk  uk (x, 0) = ϕk (x)     

in QT , on ST ,

(4.4)

in Ω,

1 6 k 6 m.

It is assumed that coefficients of Lk are bounded in QT and satisfy the condition (A) which is given in §1.1. In this section we always assume u¯k , uk ∈ C(QT ) ∩ C 2,1 (QT )

if ak = 0,

u¯k , uk ∈ C 1,0 (QT ) ∩ C 2,1 (QT )

if ak = 1.

For two given functions w, z, satisfying w 6 z, i.e., wk 6 zk for all 1 6 k 6 m, let us define an order interval hw, zi = u ∈ [C(QT )]m : wk 6 uk 6 zk , 1 6 k 6 m . 

Definition 4.3



For u¯ = (¯ u1 , . . . , u¯m ) and u = (u1 , . . . , um ), we define uk , uk }, ψ¯k = max{¯ ψ k = min{¯ uk , uk }, 1 6 k 6 m, ψ = (ψ 1 , . . . , ψ m ), ψ¯ = (ψ¯1 , . . . , ψ¯m ).

We say that (¯ u, u) is a pair of coupled upper and lower solutions of (4.4) if the following holds:  ¯k ¯k > fk (x, t, u¯k , [u]k ) for all [ψ]k 6 [u]k 6 [ψ]   Lk u     ¯k  L u 6 fk (x, t, uk , [u]k ) for all [ψ]k 6 [u]k 6 [ψ]   k k

B u¯ > g > B u

k k k k k     u¯k (x, 0) > ϕk (x) > uk (x, 0)     

in QT , in QT , on ST ,

(4.5)

in Ω,

1 6 k 6 m.

Sometimes we call that u¯ and u are the coupled upper and lower solutions of (4.4).

78  4. Weakly Coupled Parabolic Systems

Theorem 4.4 (Ordering of upper and lower solutions) Let (¯ u, u) be a pair of coupled upper and lower solutions of (4.4). We define ck = minQT min{¯ uk , uk }, uk , uk } for 1 6 k 6 m, and c¯k = maxQT max{¯ c = (c1 , . . . , cm ) and c¯ = (¯ c1 , . . . , c¯m ). Suppose that there exists a constant M > 0 such that |fk (x, t, u) − fk (x, t, v)| 6 M |u − v|

(4.6)

for all (x, t) ∈ QT , c 6 u, v 6 c¯ and 1 6 k 6 m. Then u¯ > u in QT . Proof. Set wk = u¯k − uk . Then Bk wk > 0 on ST and wk (x, 0) > 0 in Ω. Taking [u]k = [¯ u]k in the first two inequalities of (4.5), and u = u¯, v = (uk , [¯ u]k ) in (4.6), we have Lk wk > fk (x, t, u¯k , [¯ u]k ) − fk (x, t, uk , [¯ u]k ) > −M |wk | in QT . By means of Lemma 1.26, it turns out that wk > 0, i.e., u¯k > uk in QT . 4.2.1



Existence and uniqueness of solutions

Theorem 4.5 (The upper and lower solutions method) Assume that, for each 1 6 k 6 m, functions gk , ϕk , bk and coefficients of Lk satisfy the condition (G1) given in §3.1. Let (¯ u, u) be a pair of coupled upper and lower solutions of (4.4), c and c¯ be defined in Theorem 4.4 and fk ∈ L∞ (QT × hc, c¯i) satisfy (4.6). Then, in the order interval hu, u¯i, (4.4) has a unique solution u ∈ [Wp2,1 (QT )]m . If we further assume that, for each 1 6 k 6 m, functions gk , ϕk , bk and coefficients of Lk satisfy the condition (G2) given in §3.1, and fk (·, u) ∈ C α, α/2 (QT ) is uniform in u ∈ hc, c¯i. Then such a solution u ∈ [C 1+α, (1+α)/2 (QT ) ∩ C 2+α, 1+α/2 (QT )]m . In addition, we have uk ∈ C 2+α, 1+α/2 (QT ) if either ak = 1, or ak = 0 and gkt − akij Dij ϕk + bki Di ϕk = fk (x, t, ϕk ) on ∂Ω × {0}. Proof. By Theorem 4.4, u 6 u¯. The following proof will be divided into several steps. Step 1. For any given v ∈ hu, u¯i, function fk (·, v) + M vk ∈ L∞ (QT ). Then, by the Lp theory, the linear problem  L u + M uk = fk (x, t, v(x, t)) + M vk (x, t)    k k   

in QT ,

Bk uk = gk

on ST ,

uk (x, 0) = ϕk (x)

in Ω

has a unique solution uk ∈ Wp2,1 (QT ), and 

kuk kWp2,1 (QT ) 6 C kfk (·, v)kp, QT + M kvk kp, QT + kgk kWp2,1 (QT ) + kϕk kWp2 (Ω) 

6 C1 kfk k∞, QT ×hc, c¯i + |c| + |¯ c| + kgk kWp2,1 (QT ) + kϕk kWp2 (Ω) 6 C2 for all v ∈ hu, u¯i.





4.2. The upper and lower solutions method  79

As p > 1 + n/2, the embedding theorem causes |uk |α, QT 6 C3 for all v ∈ hu, u¯i, 1 6 k 6 m.

(4.7)

2,1 Moreover, similar to Theorem 2.10 we have uk ∈ C 1+α, (1+α)/2 (QT ) ∩ Wn+1, loc (QT ) 2,1 k k 1+α, (1+α)/2 when a = 0, and uk ∈ C (Ω × (0, T ]) ∩ Wq, loc (QT ) when a = 1. Step 2. Define an operator F by

F (v) = (u1 , . . . , um ) =: u.

We shall show that F : hu, u¯i → [C(QT )]m is compact. Recall the estimate (4.7). It suffices to prove that F : hu, u¯i → [C(QT )]m is continuous. For v i ∈ hu, u¯i, we put ui = F (v i ), i = 1, 2, and wk = u1k − u2k . Then wk satisfies Lk wk + M wk = fk x, t, v 1 + M vk1 − fk x, t, v 2 − M vk2 in QT ,





Bk wk = 0 on ST and wk (x, 0) = 0 in Ω.

The assumption on fk demonstrates |fk x, t, v 1 + M vk1 − fk x, t, v 2 − M vk2 | 6 2M |v 1 − v 2 |.





Hence, by the Lp estimate and embedding theorem, we have |wk |α, QT 6 Ckwk kWp2,1 (QT ) 6 2M Ckv 1 − v 2 kp, QT 6 C 0 kv 1 − v 2 k∞ . This illustrates that F : hu, u¯i → [C(QT )]m is continuous. Step 3. Now we prove F : hu, u¯i → hu, u¯i. Suppose v ∈ hu, u¯i and F (v) = u. Thanks to u 6 v 6 u¯, we see that  L u¯ + M u¯k > fk (x, t, u¯k , [v]k ) + M u¯k    k k

in QT ,

Bk u¯k > gk

on ST ,

  

u¯k (x, 0) > ϕk (x)

in Ω.

Therefore, function wk = u¯k − uk satisfies Lk wk + M wk > fk (x, t, u¯k , [v]k ) − fk (x, t, vk , [v]k ) + M (¯ uk − vk ) > 0 in QT , Bk wk > 0 on ST , and wk (x, 0) > 0 in Ω. Theorem 1.40 asserts wk > 0, i.e., uk 6 u¯k . Likewise, uk 6 uk . This accomplishes that F : hu, u¯i → hu, u¯i.

Step 4. Note that hu, u¯i is a bounded and closed convex set of [C(QT )]m , by the Schauder fixed point theorem (Corollary A.3), F has at least one fixed point u ∈ hu, u¯i, which in turn implies that u solves (4.4) and u ∈ [Wp2,1 (QT )]m owing to u = F (u) ∈ [Wp2,1 (QT )]m . Step 5. The proof of uniqueness of solutions is analogous to that of Theorem 4.4. Step 6. We have known the solution u ∈ [Wp2,1 (QT )]m ,→ [C α, α/2 (QT )]m due to p > 1 + n/2. Set Fk (x, t) = fk (x, t, u(x, t)). Then, under our further assumptions, it can be seen that Fk ∈ C α, α/2 (QT ). Hence, uk ∈ C 2+α, 1+α/2 (QT ) by Theorem 1.19. Furthermore, if ak = 1 then uk ∈ C 2+α, 1+α/2 (QT ) by Theorem 1.17, if ak = 0 and the compatibility condition holds then wk ∈ C 2+α, 1+α/2 (QT ) by Theorem 1.16. 

80  4. Weakly Coupled Parabolic Systems

4.2.2

Monotone iterative

In Definition 4.3, we acknowledge that the requirements for upper and lower solutions are relatively strong. However, for some special cases, we can simplify requirements on upper and lower solutions and obtain better results. Suppose that [u]k is an union of two disjoint subsets [u]ik and [u]dk , which consist of ik and dk components, respectively, with ik +dk = m−1. Under this decomposition, we can rewrite (4.4) in the form    Lk uk = fk (x, t, uk , [u]ik , [u]dk )     Bk uk = gk  uk (x, 0) = ϕk (x)     

in QT , on ST , in Ω,

(4.8)

1 6 k 6 m.

Definition 4.6 Let w 6 z. The system f = (f1 , . . . , fm ) is said to be mixed quasimonotonous in the order interval hw, zi if there exists a decomposition as above such that for each 1 6 k 6 m and any fixed (x, t) ∈ QT , the function fk (x, t, uk , [u]ik , [u]dk ) is monotonically increasing in [u]ik and monotonically decreasing in [u]dk for all u ∈ hw, zi. If ik = 0 for all k, we say that f is quasi-monotonically decreasing. If dk = 0 for all k, we say that f is quasi-monotonically increasing. Definition 4.7 Give u and u¯, we define ψ and ψ¯ as in Definition 4.3. Assume that ¯ We say that (¯ the system f = (f1 , . . . , fm ) is mixed quasi-monotonous in hψ, ψi. u, u) is a pair of coupled upper and lower solutions of (4.8) if the following holds:  Lk u¯k > fk (x, t, u¯k , [¯ u]ik , [u]dk )       u]dk ) L u 6 fk (x, t, uk , [u]ik , [¯    k k

B u¯ > g > B u

k k k k k     u¯k (x, 0) > ϕk (x) > uk (x, 0)     

in QT , in QT , on ST , in Ω,

1 6 k 6 m.

Sometimes we call that u¯ and u are coupled upper and lower solutions of (4.8). Let us emphasize that, in fact, the coupled upper and lower solutions are defined together with the mixed quasi-monotonicity of the system. Besides, it is easy to verify that Definition 4.3 and Definition 4.7 are equivalent if f is mixed quasi-monotonous ¯ and satisfies the Lipschitz condition (4.6). in hψ, ψi Theorem 4.8 (Ordering of upper and lower solutions) Under conditions of Definition 4.7, if f satisfies the Lipschitz condition (4.6), then u¯ > u in QT . Proof. First, derivatives

∂fk ∂fk exist almost everywhere and are bounded, and >0 ∂uj ∂uj

4.2. The upper and lower solutions method  81

for uj ∈ [u]ik ,

∂fk 6 0 for uj ∈ [u]dk by assumptions. Set wk = u¯k − uk . Then wk ∂uj

satisfies u]dk ) Lk wk > fk (·, u¯k , [¯ u]ik , [u]dk ) − fk (·, uk , [u]ik , [¯ =

X ∂fk (·) X ∂fk (·) wk + wj − ∂uk ∂uj u ∈[u] u ∈[u] j

ik

j

dk

∂fk (·) wj . ∂uj

Define hkk = −

∂fk ∂fk ∂fk , hkj = − if uj ∈ [u]ik , and hkj = if uj ∈ [u]dk . ∂uk ∂uj ∂uj

Then hkj ∈ L∞ (QT ), hkj 6 0 when k 6= j, and Lk wk +

m X

hkj wj > 0 in QT .

j=1

Obviously, Bk wk > 0 on ST and wk (x, 0) > 0 in Ω. Theorem 1.31 declares wk > 0, i.e., u¯k > uk in QT .  Theorem 4.9 (The upper and lower solutions method) Suppose that, for each 1 6 k 6 m, functions gk , ϕk , bk and coefficients of Lk satisfy the condition (G2) given in §3.1. Let (¯ u, u) be a pair of coupled upper and lower solutions of (4.8), c and c¯ be defined in Theorem 4.4 and fk (·, u) ∈ C α, α/2 (QT ) be uniform in u ∈ hc, c¯i. Assume that f has mixed quasi-monotonicity property in hu, u¯i, and f satisfies (4.6). Then  i i there exist two monotone sequences u and u¯ satisfying u 6 ui 6 ui+1 6 u¯i+1 6 u¯i 6 u¯ for all i > 1. In addition, lim ui = lim u¯i = u ∈ [Wp2,1 (QT ) ∩ C 2+α,1+α/2 (QT )]m ,

i→∞

i→∞

and u is the unique solution of (4.8) in the order interval hu, u¯i. Furthermore, uk ∈ C 2+α,1+α/2 (QT ) if either ak = 1, or ak = 0 and gkt − akij Dij ϕk + bki Di ϕk = fk (x, t, ϕk (x)), x ∈ ∂Ω, t = 0. Proof. First, u 6 u¯ by Theorem 4.8. The proof will be divided into five steps. Step 1. Constructions of ui and u¯i . For any given v, w ∈ hu, u¯i and each 1 6 k 6 m, the linear initial-boundary value problem  L u + M uk = fk (x, t, vk , [v]ik , [w]dk ) + M vk    k k

in QT ,

Bk uk = gk

on ST ,

  

uk (x, 0) = ϕk (x)

in Ω

82  4. Weakly Coupled Parabolic Systems

admits a unique solution uk ∈ Wp2,1 (QT ) by the Lp theory. Define an operator F by F (v, w) = u, and set u1 = F (u, u¯), u¯1 = F (¯ u, u), ui+1 = F (ui , u¯i ), and u¯i+1 = F (¯ ui , ui ). We claim that sequences ui and u¯i are monotonic and convergent. In fact, since u¯ satisfies 



 Lk u¯k + M u¯k > fk (x, t, u¯k , [¯ u]ik , [u]dk ) + M u¯k       Bk u ¯k > gk  uk (x, 0) > ϕk (x)     

in QT , on ST , in Ω,

1 6 k 6 m,

it is easy to see that function vk = u¯k − u¯1k satisfies  L v + M vk > 0    k k

in QT ,

Bk vk > 0

on ST ,

  

vk (x, 0) > 0

in Ω.

The maximum principle gives vk > 0, i.e., u¯k > u¯1k . In the same way, uk 6 u1k . Let wk = u¯1k − u1k . Noting u 6 u¯, the monotonicity of fk shows that wk satisfies Lk wk + M wk = fk (·, u¯k , [¯ u]ik , [u]dk ) − fk (·, uk , [u]ik , [¯ u]dk ) + M (¯ uk − uk )

> fk (·, u¯k , [u]ik , [u]dk ) − fk (·, uk , [u]ik , [u]dk ) + M (¯ uk − uk ) >0

in QT ,

Bk wk = 0

on ST ,

wk (x, 0) = 0

in Ω.

By the maximum principle, wk > 0, i.e., u¯1k > u1k . Set zk = u¯1k − u¯2k . Then zk satisfies Lk zk + M zk = fk ·, u¯k , [¯ u]ik , [u]dk − fk ·, u¯1k , [¯ u1 ]ik , [u1 ]dk + M u¯k − u¯1k




u1 ]ik , [u]dk + M u¯k − u¯1k > fk ·, u¯k , [¯ u1 ]ik , [u]dk − fk ·, u¯1k , [¯













>0

in QT ,

Bk zk = 0

on ST ,

zk (x, 0) = 0




in Ω.

Thus zk > 0, i.e., u¯1k > u¯2k by the maximum principle. Inductively u 6 ui 6 u¯i 6 u¯ for all i. As a result, point-wise limits limi→∞ ui = u˜ and limi→∞ u¯i = uˆ exist. Evidently, u˜ 6 uˆ.

4.2. The upper and lower solutions method  83

Step 2. For any given Q b QT , we shall show that ui → u˜ and u¯i → uˆ in [C 2 (Q)]m . Simplify fki (x, t) := fk x, t, uik (x, t), [ui ]ik (x, t), [¯ ui ]dk (x, t) ,


∗ fki (x, t) := fk x, t, u¯ik (x, t), [¯ ui ]ik (x, t), [ui ]dk (x, t) .




∗ As u 6 ui , u¯i 6 u¯, we conclude kfki k∞ , kfki k∞ 6 C, and then by the Lp estimate,

kuik kWp2,1 (QT ) 6 C1 (p),

k¯ uik kWp2,1 (QT ) 6 C1 (p).

(4.9)

C

Take p > n + 2. Noticing Wp2,1 (QT ) ,→ C 1+α, (1+α)/2 (QT ) for each 0 < α < 1 − (n + 2)/p, it is deduced that ui → u˜ and u¯i → uˆ in C 1+α, (1+α)/2 (QT ). Thereby both u˜ and uˆ satisfy initial and boundary conditions of (4.8). ∗ Besides, fki , fki ∈ C α, α/2 (QT ), and |fki |α, QT 6 C2 ,

∗ |fki |α, QT 6 C2 .

Applying Theorem 1.19 and (4.9) we have that uik , u¯ik ∈ C 2+α, 1+α/2 (QT ), and for each Q b QT , we can find a constant C3 > 0 such that |uik |2+α, Q 6 C3 ,

|¯ uik |2+α, Q 6 C3 for all i.

C

Since C 2+α, 1+α/2 (Q) ,→ C 2,1 (Q), there exist sub-sequences of u i and u¯i converg ing to u˜ and uˆ in [C 2,1 (Q)]m , respectively. Because ui and u¯i are monotonic and bounded, we have ui → u˜ and u¯i → uˆ in [C 2,1 (Q)]m by the uniqueness of limits. Step 3. Letting i → ∞ in 



Lk ui+1 + M ui+1 = fk x, t, uik , [ui ]ik , [¯ ui ]dk + M uik k k

in QT ,

Lk u¯i+1 + M u¯i+1 = fk x, t, u¯ik , [¯ ui ]ik , [ui ]dk + M u¯ik k k

in QT ,





we conclude that u˜ and uˆ satisfy Lk u˜k = fk x, t, u˜k , [˜ u]ik , [ˆ u]dk




in QT ,

Lk uˆk = fk x, t, uˆk , [ˆ u]ik , [˜ u]dk




in QT .

Step 4. Now, we prove u˜ = uˆ, which implies that u˜ is a solution of (4.8). Let w = u˜ − uˆ. Recall u˜ 6 uˆ and (4.6). It deduces that, for 1 6 k 6 m, Lk wk = fk x, t, u˜k , [˜ u]ik , [ˆ u]dk − fk x, t, uˆk , [ˆ u]ik , [˜ u]dk






> −M |˜ uk − uˆk | + |[˜ u]ik − [ˆ u]ik | + |[ˆ u]dk − [˜ u]dk | = M wk + M

X

wj

in QT ,

j6=k

Bk wk = 0 on ST , wk (x, 0) = 0 in Ω.

Theorem 1.31 confirms w > 0, i.e., u˜ > uˆ. And then u˜ = uˆ.

84  4. Weakly Coupled Parabolic Systems

Step 5. At last, we prove the uniqueness. Suppose that w ∈ hu, u¯i is a solution of problem (4.8). Then w = F (w, w). Similar to arguments of monotonic properties  i  i of u and u¯ , we can show ui 6 w 6 u¯i for all i. This causes u˜ 6 w 6 uˆ, and thereby u˜ = w = uˆ.  Before ending this section, it is worthy to point out that, similar to Remark 3.4, we can define strong upper and lower solutions for weakly coupled parabolic system, and the upper and lower solutions method is still effective. Moreover, for divergence weakly coupled parabolic system, we can also define weak upper and lower solutions, and conclusions similar to Theorems 3.18 and 3.19 hold.

4.3

APPLICATIONS

In this section, we shall give some examples as applications of the upper and lower solutions method. 4.3.1

A competition model

We first consider a competition model  ut − ∆u = u(a − u − hv),       vt − ∆v = v(b − v − ku),

x ∈ Ω,

t > 0,

x ∈ Ω,

t > 0,

 B u = B v = 0,     

x ∈ ∂Ω, t > 0,

u = u0 > 0, v = v0 > 0,

x ∈ Ω,

(4.10)

t = 0,

where a, b, h and k are positive constants, Ω is of class C 2 , B w = w or B w = ∂n w, initial data u0 , v0 ∈ Wp2 (Ω) with p > 1 + n/2, and u0 |∂Ω = v0 |∂Ω = 0 when B w = w, u0 , v0 ∈ C 1 (Ω), ∂n u0 |∂Ω = ∂n v0 |∂Ω = 0 when B w = ∂n w. Let us denote f1 (u, v) = u(a − u − hv) and f2 (u, v) = v(b − v − ku). Then (f1 , f2 ) is quasi-monotonically decreasing for u, v > 0, and f1 and f2 are Lipschitz continuous in the bounded region of u, v. Let u¯, u, v¯, v > 0. Then ((¯ u, v¯), (u, v)) is a pair of coupled upper and lower solutions of (4.10) if   u¯t − ∆¯ u > u¯(a − u¯ − hv),      v − ∆v 6 v(b − v − k¯ u),

x ∈ Ω, t > 0,

 ut − ∆u 6 u(a − u − h¯ v ),     

x ∈ Ω, t > 0,

t

v¯t − ∆¯ v > v¯(b − v¯ − ku),

x ∈ Ω, t > 0, x ∈ Ω, t > 0

and B u¯ > 0 > B u, B v¯ > 0 > B v,

x ∈ ∂Ω, t > 0,

u¯ > u0 > u, v¯ > v0 > v,

x ∈ Ω,

t = 0.

(4.11)

4.3. Applications  85

In (4.11), u¯ and v are coupled together, and u and v¯ are coupled together. These four inequalities can be decomposed into two groups of independent inequalities. Take M1 = max{a, maxΩ u0 } and M2 = max{b, maxΩ v0 }. Then any solution (u, v) of (4.10) in the class of [C 2,1 (Ω × (0, Tmax ))]2 satisfies 0 6 u 6 M1 and 0 6 v 6 M2 in Ω × (0, Tmax ) by the maximum principle, where Tmax is the maximal existence time of (u, v). Moreover, if u0 6≡ 0 and v0 6≡ 0, then u, v > 0 in Ω × (0, Tmax ) by the positivity lemma. On the other hand, it is easy to check that (¯ u, v¯) = (M1 , M2 ) and (u, v) = (0, 0) are coupled upper lower solutions of (4.10). Therefore, (4.10) has a unique global solution (u, v) with 0 6 u 6 M1 , 0 6 v 6 M2 and u, v ∈ C 2+α,1+α/2 (Ω× R+ ) by the upper and lower solutions method and the interior regularity (cf. Theorem 2.10). Case 1: B w = w. Let (λ, φ) be the first eigen-pair of (1.33) with kφk∞ = 1. Then we have the following results. Theorem 4.10 If a 6 λ, then limt→∞ u(x, t) = 0 uniformly in Ω. Analogously, if b 6 λ, then limt→∞ v(x, t) = 0 uniformly in Ω. The proof of this theorem is similar to that of Exercise 4.7. Theorem 4.11

Let θa (x) be the unique positive solution of (3.17) with a > λ.

(1) If a > λ, b 6 λ and u0 6≡ 0, then limt→∞ u(x, t) = θa (x) and limt→∞ v(x, t) = 0 uniformly in Ω. (2) If a 6 λ, b > λ and v0 6≡ 0, then limt→∞ u(x, t) = 0 and limt→∞ v(x, t) = θb (x) uniformly in Ω. Proof. We only prove the conclusion (1), as the proof of conclusion (2) is the same. Assume a > λ and b 6 λ. Then limt→∞ v(x, t) = 0 uniformly in Ω by Theorem 4.10. The comparison principle together with Theorem 3.16 confirm that lim sup u(x, t) 6 θa (x) uniformly in Ω.

(4.12)

t→∞

Take 0 < ε  1 so that a − hε > λ. Then there exists Tε  1 such that v 6 ε in Ω × (Tε , ∞). Thereby (

ut − ∆u > u(a − hε − u),

x ∈ Ω,

u = 0,

x ∈ ∂Ω, t > Tε .

t > Tε ,

The conditions u0 (x) > 0, 6≡ 0 imply u(x, Tε ) > 0 in Ω. Let uε and θaε be, respectively, the unique positive solutions of  ε u − ∆uε = uε (a − hε − uε ),    t

uε = 0,

  

ε

u (x, Tε ) = u(x, Tε ),

x ∈ Ω,

t > Tε ,

x ∈ ∂Ω, t > Tε , x∈Ω

86  4. Weakly Coupled Parabolic Systems

and (

−∆θaε = θaε (a − hε − θaε ),

x ∈ Ω,

θaε = 0,

x ∈ ∂Ω.

The comparison principle asserts that u > uε in Ω×(Tε , ∞). On account of u(x, Tε ) > 0 in Ω, we conclude limt→∞ uε (x, t) = θaε (x) uniformly in Ω by Theorem 3.16, which implies lim inf u(x, t) > θaε (x) uniformly in Ω. t→∞

(4.13)

By continuity and uniqueness of solutions we can prove limε→0 θaε (x) = θa (x) uniformly in Ω. Thus, by (4.13), lim inf t→∞ u(x, t) > θa (x) uniformly in Ω. Recalling  (4.12), we find limt→∞ u(x, t) = θa (x) uniformly in Ω. In the following, let us investigate the case a, b > λ. We first handle a special case a = b and h, k < 1. Take ε, σ > 0 satisfying h(1 + ε) < 1 − σ and k(1 + ε) < 1 − σ. Let u¯s = (1 + ε)θa , v¯s = (1 + ε)θa , us = σθa and v s = σθa . After careful calculations, we have  −∆¯ us > u¯s (a − u¯s − hv s )   

in Ω,

−∆v s 6 v s (a − v s − k¯ us )

in Ω,

  

u¯s = v s = 0

on ∂Ω

(4.14)

and  −∆us 6 us (a − us − h¯ vs )      

in Ω,

−∆¯ vs > v¯s (a − v¯s − kus )

in Ω,

us = v¯s = 0

on ∂Ω.

(4.15)

Let (¯ u, v) be the unique solution of (4.10) with (u0 , v0 ) = (¯ us , v s ), and set w = u¯ − u¯s and z = v s − v. In view of (4.14), it then follows that  wt − ∆w 6 A(x, t)w + h¯ us z,       zt − ∆z 6 B(x, t)z + kv w, s

 w = z = 0,     

w(x, 0) = z(x, 0) = 0,

x ∈ Ω,

t > 0,

x ∈ Ω,

t > 0,

x ∈ ∂Ω, t > 0, x ∈ Ω,

where A(x, t) = a − (¯ u + u¯s ) − hv and B(x, t) = a − (v + v s ) − k¯ u. Theorem 1.31 gives w 6 0 and z 6 0, i.e., u¯ 6 u¯s and v > v s . Using arguments in proving Theorem 3.11, we can show that u¯ and v are, respectively, monotonically decreasing and monotonically increasing in time t.

4.3. Applications  87

Denote (u, v¯) as the unique solution of (4.10) with (u0 , v0 ) = (us , v¯s ). In the same way, it can be deduced that u > us , v¯ 6 v¯s , and u and v¯ are, before and after respectively, monotonically increasing and monotonically decreasing in time t. Furthermore, since u¯s > us and v¯s > v s , similarly to the above it can be shown that us 6 u¯, v 6 v¯s , u 6 u¯s and v s 6 v¯. From the above discussions, we see that point-wise limits lim u¯(x, t) = u∗ (x),

t→∞

lim v¯(x, t) = v ∗ (x),

t→∞

lim u(x, t) = u∗ (x),

t→∞

lim v(x, t) = v∗ (x)

t→∞

exist. Clearly, us (x) 6 u∗ (x), u∗ (x) 6 u¯s (x) and v s (x) 6 v∗ (x), v ∗ (x) 6 v¯s (x). Following the proof of Theorem 2.11 step by step, we can show that each one of u¯, u, v¯ and v satisfies the estimate (2.18). Repeating arguments of Theorem 3.12, it can be derived that, as t → ∞, u¯(x, t) → u∗ (x), u(x, t) → u∗ (x), v¯(x, t) → v ∗ (x) and v(x, t) → v∗ (x) (4.16) in C 2 (Ω). Also, as t → ∞, u¯t (x, t) → 0, ut (x, t) → 0, v¯t (x, t) → 0 and v t (x, t) → 0 in C(Ω). Therefore, both (u∗ , v∗ ) and (u∗ , v ∗ ) are solutions of  −∆u = u(a − u − hv)   

in Ω,

−∆v = v(a − v − ku)

in Ω,

  

u=v=0

on ∂Ω.

(4.17)

Hence, we have the following theorem. Theorem 4.12 Assume that a = b > λ and h, k < 1. Then problem (4.17) has at least one positive solution. Moreover, if u0 ≡ 6 0 and v0 6≡ 0, then the unique solution (u, v) of (4.10) satisfies   inf u(x, t) 6 lim sup u(x, t) 6 u∗ (x),  u∗ (x) 6 lim t→∞ t→∞

∗   v∗ (x) 6 lim inf v(x, t) 6 lim sup v(x, t) 6 v (x). t→∞

(4.18)

t→∞

Proof. It is sufficient to prove (4.18). Take 0 < ε  1 so that h(1 + ε) < 1 and k(1 + ε) < 1. In consideration of (4.12), there exists T  1 such that u(x, T ) 6 (1 + ε)θa (x) and v(x, T ) 6 (1 + ε)θa (x) for all x ∈ Ω. Besides, we can take 0 < σ  1 verifying that h(1 + ε), k(1 + ε) 6 1 − σ, and u(x, T ), v(x, T ) > σθa (x) in Ω. Thus, we have that u 6 u 6 u¯ and v 6 v 6 v¯ for t > T and x ∈ Ω by the upper and lower solutions method. Recalling (4.16), the conclusion (4.18) is followed.  For a general case we have the following result.

88  4. Weakly Coupled Parabolic Systems

Theorem 4.13 Assume a, b > λ. If a − λ > hb and b − λ > ka, then problem (4.17) has at least one positive solution. Moreover, if u0 6≡ 0 and v0 6≡ 0, then the unique solution (u, v) of (4.10) satisfies (4.18). Proof. Under our assumptions, there exists ε > 0 for which a − λ > hb(1 + ε) and b − λ > ka(1 + ε). In consideration of (4.12), there exists T  1 such that u(x, T ) 6 (1 + ε)θa (x) and v(x, T ) 6 (1 + ε)θb (x) for all x ∈ Ω. Take 0 < δ < min{a − λ − hb(1 + ε), b − λ − ka(1 + ε)} satisfying δφ(x) 6 min{u(x, T ), v(x, T )} for all x ∈ Ω. Define u¯s = (1 + ε)θa , us = δφ, v¯s = (1 + ε)θb and v s = δφ. us , v s ) and (us , v¯s ) Noting θa (x) < a and θb (x) < b in Ω, it can be verified that (¯ satisfy (4.14) and (4.15), respectively. Following the proof of Theorem 4.12 step by step, the desired result can be obtained.  Facts (4.14) and (4.15) show that ((¯ us , v¯s ), (us , v s )) is a pair of coupled ordered upper and lower solutions of (4.17). Same as the scalar case, solutions of (4.10) with initial data (¯ us , v s ) and (us , v¯s ) are monotone in time t. A system with this property is usually called a monotonic system. Case 2: B = ∂n . If parameters satisfy h < a/b < 1/k, then (4.10) has a unique positive constant equilibrium solution (A, B).

(4.19) 

a−bh b−ak 1−hk , 1−hk



=:

Theorem 4.14 Assume B = ∂n and (4.19) holds. Then (A, B) is globally asymptotically stable, i.e., the unique solution (u, v) of (4.10) satisfies lim u(x, t) = A and

t→∞

lim v(x, t) = B

t→∞

uniformly in Ω provided u0 6≡ 0 and v0 6≡ 0. Proof. Firstly, the condition (4.19) implies a > hb and b > ka. Take ε > 0 satisfying a > h(b + ε) and b > k(a + ε). Note that u and v satisfy ut − ∆u 6 u(a − u) and vt − ∆v 6 v(b − v), x ∈ Ω, t > 0,

4.3. Applications  89

respectively. The comparison principle and conclusions of Example 3.2 in §3.2 assert lim supt→∞ u(x, t) 6 a and lim supt→∞ v(x, t) 6 b uniformly in Ω. Then there exists t0  1 such that u(x, t0 ) 6 a + ε and v(x, t0 ) 6 b + ε for all x ∈ Ω.

(4.20)

Clearly, u(x, t0 ) > 0 and v(x, t0 ) > 0 in Ω by the positivity lemma. Take σ > 0 small such that 



σ < min A, B, a − h(b + ε), b − k(a + ε), min u(x, t0 ), min v(x, t0 ) . Ω

Ω

We choose 0 < q  1 satisfying q(A − σ) < σ[a − σ − h(b + ε)] and q(B − σ) < σ[b − σ − k(a + ε)], and define u¯ = A + (a + ε − A)e−q(t−t0 ) , u = A − (A − σ)e−q(t−t0 ) , t > t0 , v¯ = B + (b + ε − B)e−q(t−t0 ) , v = B − (B − σ)e−q(t−t0 ) , t > t0 . Careful computations show that u¯, v¯, u and v satisfy (4.11) for t > t0 , and (

u¯(x, t0 ) = a + ε > u(x, t0 ), u(x, t0 ) = σ 6 u(x, t0 ), x ∈ Ω, v¯(x, t0 ) = b + ε > v(x, t0 ), v(x, t0 ) = σ 6 v(x, t0 ), x ∈ Ω.

(4.21)

Obviously, ∂n u¯ = ∂n u = ∂n v¯ = ∂n v = 0 on ∂Ω for all t > t0 . Subsequently the unique solution (u, v) of (4.10) satisfies A − (A − σ)e−q(t−t0 ) 6 u(x, t) 6 A + (a + ε − A)e−q(t−t0 ) , B − (B − σ)e−q(t−t0 ) 6 v(x, t) 6 B + (b + ε − B)e−q(t−t0 ) for all x ∈ Ω and t > t0 by the upper and lower solutions method. Since q > 0, the desired conclusions are followed.  If parameters satisfy a > hb and ka > b,

(4.22)

then the possible non-negative and non-trivial constant equilibrium solution of (4.10) is (a, 0). Theorem 4.15 Assume B = ∂n . If (4.22) holds, then (a, 0) is globally asymptotically stable, i.e., the unique solution (u, v) of (4.10) satisfies lim u(x, t) = a and

t→∞

uniformly in Ω provided u0 6≡ 0.

lim v(x, t) = 0

t→∞

90  4. Weakly Coupled Parabolic Systems

Proof. If v0 ≡ 0, then v ≡ 0, and the conclusion is obvious. Now, we assume that v0 6≡ 0. Same as the proof of Theorem 4.14, we can find 0 < ε, σ  1, and t0  1 to guarantee (4.20) and a − σ > h(b + ε), k(a − σ) > b + ε,



σ < min minΩ u(x, t0 ), minΩ v(x, t0 ) . Take 0 < β  1 so that β 6 kσ + ε and (a − σ)β 6 σ[a − σ − h(b + ε)]. Set q = ka − b + σ + kε > 0 and u¯ = a + εe−a(t−t0 ) , u = a − (a − σ)e−β(t−t0 ) , t > t0 , v¯ = (b + ε)e−β(t−t0 ) , v = σe−q(t−t0 ) , t > t0 . After careful calculations, we can verify that u¯, v, u, and v¯ satisfy (4.11) for t > t0 and (4.21) holds. Clearly, ∂n u¯ = ∂n u = ∂n v¯ = ∂n v = 0 on ∂Ω for t > t0 . Therefore, the unique solution (u, v) of (4.10) satisfies a − (a − σ)e−β(t−t0 ) 6 u(x, t) 6 a + εe−a(t−t0 ) , σe−q(t−t0 ) 6 v(x, t) 6 (b + ε)e−β(t−t0 ) in Ω × [t0 , ∞) by the upper and lower solutions method. The desired conclusion is followed.  4.3.2

A prey-predator model

In this subsection we consider a prey-predator model  ut − ∆u = u(a − u − hv),       vt − ∆v = v(ku − b),

x ∈ Ω,

t > 0,

x ∈ Ω,

t > 0,

 u = v = 0,     

x ∈ ∂Ω, t > 0,

u = u0 > 0, v = v0 > 0,

x ∈ Ω,

(4.23)

t = 0,

where a, b, h and k are positive constants, Ω is of class C 2 , and initial data u0 , v0 ∈ ◦

W 1p (Ω) ∩ Wp2 (Ω) with p > 1 + n/2. This is a mixed quasi-monotonic system for u, v > 0. u, v¯), (u, v)) is a pair of coupled upper and lower soluLet u¯, u, v¯, v > 0. Then ((¯ tions of (4.23) if   ¯t − ∆¯ u > u¯(a − u¯ − hv), u     v − ∆v 6 v(ku − b), t

 ut − ∆u 6 u(a − u − h¯ v ),     

v¯t − ∆¯ v > v¯(k¯ u − b),

x ∈ Ω, t > 0, x ∈ Ω, t > 0, x ∈ Ω, t > 0, x ∈ Ω, t > 0

(4.24)

4.3. Applications  91

and (

u¯ > 0 > u, v¯ > 0 > v,

x ∈ ∂Ω, t > 0,

u¯ > u0 > u, v¯ > v0 > v,

x ∈ Ω,

t = 0.

In (4.24), u¯, u, v¯ and v are coupled together. We cannot decompose (4.24) into two groups of independent inequalities like the competition model. Let u = v = 0, u¯ = M := max{a, maxΩ u0 (x)}, and v¯ be a solution of  v = v¯(kM − b),   v¯t − ∆¯ 

x ∈ Ω,

v¯ = 0,

x ∈ ∂Ω, t > 0,

  

v¯(x, 0) = v0 (x),

x ∈ Ω,

t > 0,

which exists uniquely and globally in time t. Evidently, ((¯ u, v¯), (u, v)) is a pair of coupled upper and lower solutions of (4.23), and then (4.23) has a unique global solution by the upper and lower solutions method. Theorem 4.16

Let (λ, φ) be the first eigen-pair of (1.33).

(1) If a 6 λ, then limt→∞ u(x, t) = limt→∞ v(x, t) = 0 uniformly in Ω. (2) If a > λ and ka < b, then limt→∞ u(x, t) = us (x) and limt→∞ v(x, t) = 0 uniformly in Ω provided u0 6≡ 0, where us (x) is the unique positive solution of (3.17). The proof of Theorem 4.16 is left to readers as an exercise. 4.3.3

The Belousov-Zhabotinskii reaction model

In this subsection, we study the following Belousov-Zhabotinskii reaction model  ut − d1 ∆u = u(1 − u − cv),        vt − d2 ∆v = bw − v − kuv,  

w − d ∆w = h(u − w),

t 3      u = v = w = 0,    

(u, v, w) = (u0 , v0 , w0 ),

x ∈ Ω,

t > 0,

x ∈ Ω,

t > 0,

x ∈ Ω,

t > 0,

(4.25)

x ∈ ∂Ω, t > 0, x ∈ Ω,

t = 0,

where coefficients are all positive constants, Ω is of class C 2 , and initial data u0 , v0 , w0 satisfy u0 , v0 , w0 ∈ C01 (Ω) ∩ Wp2 (Ω) with p > 1 + n/2 and u0 , v0 , w0 > 0, 6≡ 0 in Ω. This is a mixed quasi-monotonic system for u, v, w > 0. ¯ w > 0. Then ((¯ u, v¯, w), ¯ (u, v, w)) is a pair of coupled upper and Let u¯, u, v¯, v, w, lower solutions of (4.25) if

92  4. Weakly Coupled Parabolic Systems

  u¯t − d1 ∆¯ u > u¯(1 − u¯ − cv),      uv, v t − d2 ∆v 6 bw − v − k¯       w − d3 ∆w 6 h(u − w),

x ∈ Ω, t > 0,

 v ), ut − d1 ∆u 6 u(1 − u − c¯       v, v¯t − d2 ∆¯ v > bw¯ − v¯ − ku¯     

x ∈ Ω, t > 0,

x ∈ Ω, t > 0, x ∈ Ω, t > 0,

t

(4.26)

x ∈ Ω, t > 0,

w¯t − d3 ∆w¯ > h(¯ u − w), ¯

x ∈ Ω, t > 0

and (

u¯ > 0 > u, v¯ > 0 > v, w¯ > 0 > w,

x ∈ ∂Ω, t > 0,

u¯ > u0 > u, v¯ > v0 > v, w¯ > w0 > w,

x ∈ Ω,

t = 0.

In (4.26), u¯, u, v¯, v, w¯ and w are coupled together, and so (4.26) cannot be decomposed into a simple system of inequalities. Definition 4.17 Problem (4.25) is referred to as permanent, if initial data u0 , v0 , w0 > 0, 6≡ 0 in Ω, then for any Ω0 b Ω we have lim inf min u(x, t) > 0, lim inf min v(x, t) > 0 and lim inf min w(x, t) > 0. t→∞

t→∞

Ω0

t→∞

Ω0

Ω0

Let (λ, φ) be the first eigen-pair of (1.33). Normalize φ by kφk∞ = 1. Theorem 4.18 ([105]) If d1 λ < 1 − bc, then (4.25) is permanent. Proof. The proof will be divided into three steps. Step 1. Owing to d1 λ < 1 − bc < 1, the elliptic problem (

−d1 ∆u = u(1 − u),

x ∈ Ω,

u = 0,

x ∈ ∂Ω

has a unique positive solution uˆs (x). Because of h > 0, the linear problem (

−d3 ∆w + hw = hˆ us (x),

x ∈ Ω,

w = 0,

x ∈ ∂Ω

has a unique positive solution wˆs (x). Analogously, the linear problem (

−d2 ∆v + v = bwˆs (x),

x ∈ Ω,

v = 0,

x ∈ ∂Ω

admits a unique positive solution vˆs (x). That is, (ˆ us , vˆs , wˆs ) is the unique positive solution of  −d1 ∆u = u(1 − u),       −d2 ∆v = bw − v,  −d3 ∆w = h(u − w),     

u = v = w = 0,

x ∈ Ω, x ∈ Ω, x ∈ Ω, x ∈ ∂Ω.

(4.27)

4.3. Applications  93

Step 2. Since initial data u0 , v0 , w0 ∈ C 1 (Ω), invoking Lemma 3.7, we can find a constant K  1 so that (u0 (x), v0 (x), w0 (x)) 6 K(ˆ us (x), vˆs (x), wˆs (x)) in Ω. Evidently, (K uˆs , K vˆs , K wˆs ) is an upper solution of (4.27). Let (ˆ u, vˆ, w) ˆ be the unique solution of  uˆt − d1 ∆ˆ u = uˆ(1 − uˆ),       vˆ − d2 ∆ˆ v = bwˆ − vˆ,    t

wˆ − d ∆wˆ = h(ˆ u − w), ˆ

t 3     u = v = w = 0,     

(ˆ u, vˆ, w) ˆ = (K uˆs , K vˆs , K wˆs ),

x ∈ Ω,

t > 0,

x ∈ Ω,

t > 0,

x ∈ Ω,

t > 0,

x ∈ ∂Ω, t > 0, x ∈ Ω,

t = 0.

Then ((ˆ u, vˆ, w), ˆ (0, 0, 0)) is a pair of coupled upper and lower solutions of (4.25), and therefore 0 6 u 6 uˆ, 0 6 v 6 vˆ, 0 6 w 6 wˆ by the upper and lower solutions method. As u0 , v0 , w0 > 0, 6≡ 0 in Ω, we have u, v, w > 0 in Ω × R+ by the positivity lemma. Using the same discussions as in the proof of Theorem 3.11 we can prove that (ˆ u, vˆ, w) ˆ is monotonically decreasing in time t, and lim (ˆ u(x, t), vˆ(x, t), w(x, ˆ t)) = (ˆ us (x), vˆs (x), wˆs (x)) t→∞

uniformly in Ω. On the other hand, 0 < uˆs < 1, 0 < wˆs < 1 and 0 < vˆs < b in Ω by the maximum principle. Then there exists t0  1 such that 0 < u < 1, 0 < v < b, 0 < w < 1

in Ω × [t0 , ∞).

Furthermore, in consideration of Lemma 3.7, there exists ε > 0 so that u(x, t0 ) > εφ(x), v(x, t0 ) > εφ(x), and w(x, t0 ) > εφ(x) for all x ∈ Ω. Step 3. Take 0 < θi < ε (i = 1, 2, 3) satisfying θ1 6 1 − bc − d1 λ, θ3 6 hθ1 /(h + d3 λ) and θ2 6 bθ3 /(1 + k + d2 λ). We can show that ((1, b, 1), (θ1 φ, θ2 φ, θ3 φ)) is a pair of coupled upper and lower solutions of (4.25) with initial time t0 and initial datum (u(·, t0 ), v(·, t0 ), w(·, t0 )). Consequently, any solution (u, v, w) of (4.25) satisfies (θ1 φ(x), θ2 φ(x), θ3 φ(x)) 6 (u(x, t), v(x, t), w(x, t)) 6 (1, b, 1) in Ω × [t0 , ∞) by the upper and lower solutions method. This indicates that (4.25) is permanent. 

94  4. Weakly Coupled Parabolic Systems

EXERCISES 4.1 Prove Theorem 4.2. 4.2 Verify that Definition 4.3 and Definition 4.7 are equivalent if f is mixed quasi¯ and satisfies the Lipschitz condition (4.6). monotonous in hψ, ψi 4.3 In the proofs of Theorems 4.14 and 4.15, verify that functions u¯, v, u and v¯ satisfy (4.11). 4.4 Prove Theorem 4.16. ◦

4.5 Let u0 , v0 ∈ W 1p (Ω)∩Wp2 (Ω) with p > 1+n/2. Use the upper and lower solutions method to prove that there is T > 0 such that the problem  ut − d1 ∆u = u(a + uv)       vt − d2 ∆v = v(b + kuv)

in QT ,

 u=v=0     

on ST ,

in QT ,

u(x, 0) = u0 (x), v(x, 0) = v0 (x)

in Ω

has a unique solution, where d1 , d2 , a, b and k are positive constants. ◦

4.6 Let u0 , v0 ∈ W 1p (Ω) ∩ Wp2 (Ω) with p > 1 + n/2, and u0 , v0 > 0. Use the upper and lower solutions method to prove that the problem  ut − d1 ∆u = u(a − v),      v − d ∆v = v(b + ku), t

2

 u = v = 0,    

u = u 0 , v = v0 ,

x ∈ Ω, x ∈ Ω, x ∈ ∂Ω, x ∈ Ω,

t > 0, t > 0, t > 0, t=0

has a unique global solution, where d1 , d2 , a, b and k are as above. 4.7 Consider the following initial-boundary value problem    uit − di ∆ui = ui [ai + fi (u1 , u2 )] , in Q∞ ,  

ui = 0 u(x, 0) = ϕi (x), i = 1, 2

on S∞ , in Ω,

(P4.1)

◦

where ai , di are positive constants, ϕi ∈ W 1p (Ω) ∩ Wp2 (Ω) with p > 1 + n/2 and ϕi > 0, i = 1, 2. Assume that fi : [0, ∞) × [0, ∞) → R satisfies fi (0, 0) = 0,

∂fi (u1 , u2 ) < 0, i, j = 1, 2 ∂uj

and a1 + f1 (c1 , 0) 6 0,

a2 + f2 (0, c2 ) 6 0

for some positive constants c1 and c2 . Let λ be the first eigenvalue of (1.33). Prove that if ai < di λ then the solution of (P4.1) satisfies limt→∞ ui (x, t) = 0 uniformly in Ω, i = 1, 2.

Exercises  95

4.8 Verify that ((1, b, 1), (θ1 φ, θ2 φ, θ3 φ)) is a pair of coupled upper and lower solutions of (4.25) with initial time t0 and initial datum (u(·, t0 ), v(·, t0 ), w(·, t0 )) in proving Theorem 4.18. 4.9 Use the upper and lower solutions method to investigate existence, uniqueness and longtime behaviours of global solutions of the following problem    u   − c1 u , x ∈ Ω, ut − d1 ∆u = r1 u 1 −    a1 + b1 v      v

vt − d2 ∆v = r2 v 1 −

    u = v = 0,    

a2 + b 2 u

− c2 v , x ∈ Ω,

t > 0, t > 0,

x ∈ ∂Ω, t > 0, x ∈ Ω,

u = u0 > 0, v = v0 > 0,

t = 0,

◦

where di , ri , ai , bi and ci are positive constants, u0 , v0 ∈ W 1p (Ω) ∩ Wp2 (Ω) with p > 1 + n/2. 4.10 Let Γ1 and Γ2 be given in (1.3). Consider the problem  Lk uk = fk (x, t, u)       Bk uk = gk  uk (x, 0) = ϕk (x)     

in QT , on ST , in Ω,

1 6 k 6 m,

where Lk are as in (4.1), and for each 1 6 k 6 m, Bk uk = ∂n uk + bk uk on Γ1 × (0, T ], Bk uk = uk on Γ2 × (0, T ],

or Bk uk = ∂n uk + bk uk on Γ2 × (0, T ], Bk uk = uk on Γ1 × (0, T ].

As Exercise 3.13, discuss the upper and lower solutions method for such a problem. 4.11 Let Σ ⊂ R2 be a domain, and the system {f, g} is quasi-monotonous in Σ (quasi-monotonically increasing or quasi-monotonically decreasing). Consider the following problem  ut − ∆u = f (u, v)       vt − ∆v = g(u, v)

in QT ,

 u=v=0     

on ST ,

u(x, 0) = u0 (x), v(x, 0) = v0 (x)

◦

in QT ,

(P4.2)

in Ω,

where u0 , v0 ∈ W 1p (Ω) ∩ Wp2 (Ω) with p > 1 + n/2, and (u0 (x), v0 (x)) ∈ Σ for all x ∈ Ω. Discuss the following issues.

96  4. Weakly Coupled Parabolic Systems

(1) Define the weak solution and coupled weak upper and lower solutions of (P4.2) in Σ † . (2) Attach appropriate conditions to the system {f, g} and prove the ordering of weak upper and lower solutions. (3) Let ((¯ u, v¯), (u, v)) be a pair of coupled weak upper and lower solutions of problem (P4.2). Prove that (P4.2) has a unique weak solution (u, v) between (u, v) and (¯ u, v¯). (4) Let ((¯ u1 , v¯1 ), (u1 , v 1 )) and ((¯ u2 , v¯2 ), (u2 , v 2 )) be two pairs of coupled classical (strong) upper and lower solutions of (P4.2) in Σ. Define u¯ = min{¯ u1 , u¯2 }, v¯ = min{¯ v1 , v¯2 }, u = max{u1 , u2 }, v = max{v 1 , v 2 }. Explore some conditions on the system {f, g} and u¯i , v¯i , ui , v i (i = 1, 2) to prove that ((¯ u, v¯), (u, v)) is a pair of coupled weak upper and lower solutions of (P4.2) in Σ. (5) Give a statement similar to Remark 3.20 for problem (P4.2).

†

A solution (u, v) of (P4.2) in Σ means that (u, v) is a solution of (P4.2) and the range of (u, v) is in Σ.

CHAPTER

5

Stability Analysis

A basic question in the study of nonlinear parabolic equations is whether a timedependent solution, as time increases, remains in a neighbourhood of an equilibrium solution and whether it converges to this equilibrium solution as t → ∞. From a practical point of view, it is important to know that, for a given equilibrium solution us , whether there is a set of initial data such that their corresponding time-dependent solutions converge to us as t → ∞. This leads to a question of stability and asymptotic stability of an equilibrium solution and its stability region. The stability of an equilibrium solution is not only an important part in the study of partial differential equations, but also has an obvious physical meaning. When an equilibrium solution is globally asymptotically stable, it means that the corresponding physical phenomenon that the equation models is stable and has a unique stable state. In Chapter 3 and Chapter 4, it has been introduced how to deal with the stability of an equilibrium solution by using the upper and lower solutions method. In this chapter, we first include the local stability of equilibrium solutions by using the linearized principal eigenvalue and upper and lower solutions method. The result that linear stability implies local stability is established. Then we introduce three other ways to establish the global stability of equilibrium solutions, namely, Lyapunov functional method, iteration method and average method. Throughout this chapter, we assume that Ω is of class C 2+α and initial data satisfy the corresponding compatibility conditions. Consider the following problem of weakly coupled parabolic system (

ut + L[u] = f (x, u)

in Q∞ ,

B[u] = g(x)

on S∞ ,

(5.1)

where L[u] = (L1 [u1 ], . . . , Lm [um ]), B[u] = (B1 [u1 ], . . . , Bm [um ]) and Lk [uk ] = −akij (x)Dij uk + bki (x)Di uk , Bk [uk ] = uk or Bk [uk ] = ∂n uk +bk (x)uk . We assume that ∂t +Lk is strongly parabolic in QT and bki ∈ L∞ (Ω) for all 1 6 k 6 m and 1 6 i 6 n. The equilibrium problem of (5.1) is (

L[u] = f (x, u)

in Ω,

B[u] = g(x)

on ∂Ω.

(5.2) 97

98  5. Stability Analysis

Let u(x, t; u0 ) be a solution of (5.1) with initial datum u0 . Definition 5.1 Let X be a Banach space and Y be a subset of X. A solution us ∈ Y of (5.2) is said to be locally stable in Y , if for any given ε > 0, there exists a constant δ > 0 such that when u0 ∈ Bδ (us ; Y ), the solution u(x, t; u0 ) satisfies {u(x, t; u0 )}t>0 ⊂ Y and ku(x, t; u0 ) − us (x)kX 6 ε for all t > 0, where Bδ (us ; Y ) is a neighbourhood of us in Y . Otherwise, we call that us is unstable in Y . If us is locally stable in Y and lim ku(x, t; u0 ) − us (x)kX = 0,

t→∞

(5.3)

we call that us is locally asymptotically stable in Y . If this limit is uniform with respect to u0 ∈ Bδ (us ; Y ), we say that us is locally uniformly asymptotically stable in Y . If {u(x, t; u0 )}t>0 ⊂ Y and (5.3) holds for all u0 ∈ Y , we say that us is globally asymptotically stable in Y . Let us be a solution of (5.2). The eigenvalue problem of the linearization of (5.2) at us is (

−L[φ] + fu (x, us )φ = λφ

in Ω,

B[φ] = 0

on ∂Ω.

(5.4)

Definition 5.2 We say that us is linearly stable if each eigenvalue λ of (5.4) has a negative real part. If problem (5.4) has an eigenvalue whose real part is positive, then we say that us is not linearly stable. We say that us is degenerate if λ = 0 is an eigenvalue of (5.4). Otherwise, us is non-degenerate.

5.1

LOCAL STABILITY DETERMINED BY THE PRINCIPAL EIGENVALUE

For scalar case, i.e., m = 1, we intend to use the upper and lower solutions method to prove that the local stability of us can be determined by the sign of the principal eigenvalue of (5.4), namely, the linear stability implies the local stability. Let λ1 (us ) be the principal eigenvalue of (5.4) and φ be the corresponding positive eigenfunction with kφk∞ = 1. It is well-known that λ1 (us ) is real. Theorem 5.3 Let f ∈ C α,1 (Ω × R) and fu (x, ·) ∈ C(R) be uniformly with respect to x ∈ Ω. Let u be the unique solution of (5.1) with initial datum u0 . (1) If λ1 (us ) < 0, then there exist constants 0 < ρ, k  1 such that |u(x, t) − us (x)| 6 ρe−kt φ(x) for all t > 0, x ∈ Ω provided that it holds at t = 0.

5.1. Local stability determined by the principal eigenvalue  99

(2) If λ1 (us ) > 0, then there exists ρ > 0 such that for any 0 < σ < 1 we can find 0 < k  1 for which 



u(x, t) > us (x) + ρ 1 − σe−kt φ(x) for all t > 0, x ∈ Ω provided that it holds at t = 0. (3) In the case λ1 (us ) = 0, we assume that fuu is continuous. If fuu (x, us (x)) > 0 for all x ∈ Ω, then us is not locally asymptotically stable. Proof. We first prove the conclusion (1). Let u¯ = us (x) + ρe−kt φ(x). Then we have u¯t + L[¯ u] − f (x, u¯) = L[us ] + ρe−kt L[φ] − kρe−kt φ − f (x, u¯) = f (x, us ) − f (x, u¯) + [fu (x, us ) − λ1 (us ) − k] ρe−kt φ. Using λ1 (us ) < 0 we see that when ρ, k > 0 are small enough, f (x, us ) − f (x, u¯) + [fu (x, us ) − λ1 (us ) − k] ρe−kt φ = [fu (x, us ) − fu (x, us + θφ) − λ1 (us ) − k] ρe−kt φ > 0, 0 6 θ 6 ρ, as fu (x, us ) − fu (x, us + θφ) → 0 when ρ → 0. Therefore, u¯ is an upper solution provided u0 (x) 6 us (x) + ρφ(x). Analogously, u = us (x) − ρe−kt φ(x) is a lower solution provided that 0 < ρ, k  1 and u0 (x) > us (x) − ρφ(x). By the upper and lower solutions method, the unique solution u of (5.1) with initial datum u0 satisfies us (x) − ρe−kt φ(x) 6 u(x, t) 6 us (x) + ρe−kt φ(x) for all x ∈ Ω, t > 0, if us (x) − ρφ(x) 6 u0 (x) 6 us (x) + ρφ(x). The proof of (2) is left to readers as an exercise. Now we prove the conclusion (3). Assume λ1 (us ) = 0 and fuu (x, us (x)) > 0 for all x ∈ Ω. Let ε > 0 and u¯ = us + εφ. Then we have, by using Taylor’s formula, L[¯ u] − f (x, u¯) = L[us ] + εL[φ] − f (x, us + εφ) = f (x, us ) + εfu (x, us )φ − f (x, us + εφ) 1 = − ε2 φ2 fuu (x, us + τ φ) with 0 6 τ 6 ε 2 us (x)+εφ(x) in Ω × R+ . Which implies that us is not locally asymptotically stable.  Example 5.1

Consider the problem (

−∆u = (a − u)2 (1 − u) =: f (u)

in Ω,

∂n u = 0

on ∂Ω,

where 0 < a < 1 is a constant.

100  5. Stability Analysis

It has two positive constant equilibria a and 1. From relations f 0 (u) = (u − a)(2 + a − 3u) and f 00 (u) = 2 + 4a − 6u, we see that f 0 (a) = 0, f 0 (1) < 0 and f 00 (a) > 0. Thus, λ1 (a) = 0 and λ1 (1) = f 0 (1) < 0. It follows from Theorem 5.3 that us = a is not locally asymptotically stable and us = 1 is locally stable.

5.2

LYAPUNOV FUNCTIONAL METHOD

The Lyapunov function (functional) method is a powerful tool in the study of stability of equilibrium point (solution) of differential equations, especially parabolic partial differential equations. In this section we shall introduce the Lyapunov functional method to study global stability of the unique equilibrium solution of reactiondiffusion systems including the system coupled by reaction-diffusion equations and ordinary differential equations. 5.2.1

Abstract results

Let τ1 , . . . , τm be non-negative constants, and u1 (x, t), . . . , um (x, t) be functions of (x, t). We define τ = (τ1 , . . . , τm ), u(x, t) = (u1 (x, t), . . . , um (x, t)) and uτ (x, t) = (u1 (x, t − τ1 ), . . . , um (x, t − τm )). Let l, k and m be positive integers with 1 6 l < k < m. Consider the following initial-boundary value problem  τ   uit − di ∆ui = fi (x, u, u ), x ∈ Ω,     u = fi (x, u, uτ ), x ∈ Ω,    it

u = 0,

i     ∂n ui = 0,     

ui (x, t) = ui0 (x, t),

t > 0,

1 6 i 6 k,

t > 0,

k + 1 6 i 6 m,

x ∈ ∂Ω, t > 0,

1 6 i 6 l,

x ∈ ∂Ω, t > 0,

l + 1 6 i 6 k,

x ∈ Ω,

(5.5)

−τi 6 t 6 0, 1 6 i 6 m,

where Ω is of class C 2 , di > 0 are constants (1 6 i 6 k), and ui0 (·, t) ∈ C α (Ω) uniformly with respect to t ∈ [−τi , 0], i = 1, . . . , m. This is an initial-boundary value problem of system coupled by reaction-diffusion equations and ordinary differential equations. Define di = 0 for k + 1 6 i 6 m. We make the following assumptions. (H1) The corresponding equilibrium problem of (5.5)  −di ∆ui = fi (x, u, u)   

in Ω,

ui = 0

on ∂Ω, 1 6 i 6 l,

  

∂n ui = 0

on ∂Ω, l + 1 6 i 6 k

has a unique solution u˜(x) = (˜ u1 (x), . . . , u˜m (x)).

1 6 i 6 m,

5.2. Lyapunov functional method  101

(H2) Problem (5.5) has a unique global solution u. Moreover, there exist positive constants α and ρ with 0 < α < 1, such that, for all 1 6 i 6 m, ui − u˜i , di ∇(ui − u˜i ) ∈ C α ([ρ, ∞), L2 (Ω)) and kui − u˜i kC α ([ρ,∞), L2 (Ω)) + di k∇(ui − u˜i )kC α ([ρ,∞), L2 (Ω)) 6 C, where C is a constant depending only on α and ρ. Here, function u ∈ C α ([ρ, ∞), L2 (Ω)) means that ku(·, t) − u(·, s)k2, Ω < ∞. |t − s|α t,s>ρ, t6=s sup

We mention that, in general, the regularity and estimate in (H2) can be obtained by Theorems 2.11 and 2.13. Theorem 5.4 (Lyapunov functional method [114]) Let conditions (H1) and (H2) hold. Assume that there exists a function E(u, u˜) such that, along the orbit u of (5.5), the functional Z F (t) =

E(u(x, t), u˜(x)) Ω

is non-negative and satisfies F 0 (t) 6 −A

k Z X i=1

m Z X

|∇ui (x, t) − ∇˜ ui (x)|2 − B

Ω

|ui (x, t) − u˜i (x)|2

i=l+1 Ω

for some positive constants A, B. Then lim kui (·, t) − u˜i (·)kH 1 (Ω) = 0

t→∞

lim kui (·, t) − u˜i (·)k2,Ω = 0

t→∞

for 1 6 i 6 k,

for k + 1 6 i 6 m.

Theorem 5.4 can be deduced from the following lemma directly, and the proof is omitted here. Lemma 5.5

Let a and b be positive constants, ϕ ∈ C 1 ([a, ∞)) and ϕ be bounded Z ∞

from below. Assume that ψ > 0,

h(t)dt < ∞ and

a

ϕ0 (t) 6 −bψ(t) + h(t).

(5.6)

If either ψ ∈ C 1 ([a, ∞)) and ψ 0 (t) 6 k in [a, ∞) for some constant k > 0, or ψ is uniform continuous in [a, ∞), then limt→∞ ψ(t) = 0. Proof. We only deal with the case that ψ ∈ C 1 ([a, ∞)) and ψ 0 (t) 6 k in [a, ∞). Assume that the conclusion were not valid. Then there exist ε > 0 and ti % ∞ such

102  5. Stability Analysis

that ψ(ti ) > ε and ti − ti−1 > ε/(2k) for all i. Take 0 < σ < ε/(2k). The condition ψ 0 (t) 6 k guarantees ψ(ti ) − ψ(t) 6 k(ti − t) for all t ∈ [ti − σ, ti ], and then ψ(t) > ψ(ti ) − k(ti − t) > ε − kσ > ε/2 for all t ∈ [ti − σ, ti ]. Integrating (5.6) from a to t > tN with large N and using the condition ψ > 0, we obtain immediately ϕ(t) 6 ϕ(a) − b 6 ϕ(a) − b

Z

t

Z

t

ψ(s)ds + a N X Z ti i=1

h(s)ds a

Z

ψ(s)ds +

ti −σ

t

h(s)ds a

6 ϕ(a) − N bεσ/2 +

t

Z a

h(s)ds for all t > tN .

It follows that ϕ(t) → −∞ as t → ∞ due to the fact that

Z

∞

h(t)dt < ∞. A

a

contradiction is attained, and the proof is complete.  τ τ If fi (x, u, u ) = fi (u, u ) does not depend on x, i = 1, . . . , m, then the ordinary differential system corresponding to (5.5) is  y = fi (y, y τ ),    it

t > 0, −τi 6 t 6 0,

yi (t) = yi0 (t),

  

(5.7)

i = 1, . . . , m.

When we find a Lyapunov function of (5.7), a natural question is that, does it have to be a Lyapunov functional of (5.5)? In the following we will answer this question in some sense. For simplicity, we only consider the case τi = 0 and write fi (u, uτ ) = fi (u), 1 6 i 6 m. Let Σ ⊂ Rm be an invariant set with respect to both (5.7) and (5.5), i.e., y(0) ∈ Σ implies y(t) ∈ Σ for all t > 0, and u(x, 0) ∈ Σ for all x ∈ Ω implies u(x, t) ∈ Σ for all x ∈ Ω and t > 0. Let y ∗ ∈ Σ be the unique equilibrium point of (5.7). Lemma 5.6 Suppose τi = 0 and fi = fi (u), 1 6 i 6 m. Assume that y(0) ∈ Σ, u(x, 0) ∈ Σ for all x ∈ Ω, and (5.7) has a Lyapunov function V (y), i.e., m X

Vyi (y)fi (y) 6 0 for all y ∈ Σ.

i=1

Denote D = diag(d1 , . . . , dm ). Suppose that along the orbit of (5.5), m X i,j=1

∇ui di Vui uj (u) · ∇uj > 0,

Z

l X

∂Ω i=1



di Vui (u) ui =0 ∂n ui dS 6 0.

5.2. Lyapunov functional method  103

Then functional Z

F (t) =

V (u(x, t)) Ω

satisfies F 0 (t) 6 0 for all t > 0. Proof. A straightforward computation shows that along the orbit of (5.5), 0

F (t) = =

Z X m Ω i=1 Z X m Ω i=1

Z

=

Vui (u)uit di Vui (u)∆ui +

l X

∂Ω i=1

+

Z X m Ω i=1

Vui (u)fi (u)



di Vui (u) ui =0 ∂n ui dS −

Z X m Ω i=1

Z

m X

Ω i,j=1

∇ui di Vui uj (u) · ∇uj

Vui (u)fi (u)

6 0 for all t > 0. The proof is complete.



Example 5.2 We consider an initial-boundary value problem with homogeneous Dirichlet boundary condition of scalar equation    ut − ∆u = kf (u),    

x ∈ Ω,

t > 0,

u = 0,

x ∈ ∂Ω, t > 0,

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

x ∈ Ω,

(5.8)

where Ω is of class C 2+α , k > 0 is a constant, u0 ∈ Wp2 (Ω) with p > 1, and f ∈ C 1 (R) satisfies f (0) = 0, f 0 (0) > 0 and f 0 (v) 6 f 0 (0) for all v ∈ R. Using the interior estimate in t direction we can show that ∇ut ∈ L2 (Ω) for t > 0. It is clear that us = 0 is a equilibrium solution of (5.8). The eigenvalue problem of linearization of (5.8) at us = 0 is (

∆φ + kf 0 (0)φ = µφ

in Ω,

φ=0

on ∂Ω.

(5.9)

Let λ be the first eigenvalue of (1.33). Then the first eigenvalue of (5.9) is µ1 = kf 0 (0) − λ. If kf 0 (0) > λ, then µ1 > 0, and so us = 0 is not linearly stable. If kf 0 (0) < λ, then µ1 < 0, and so every eigenvalue µ of (5.9) is negative. Consequently, us = 0 is linearly stable and is also locally asymptotically stable. Fortunately, the following theorem shows that us = 0 is also globally asymptotically stable.

104  5. Stability Analysis

Theorem 5.7 If kf 0 (0) < λ, then us = 0 is globally asymptotically stable with respect to problem (5.8). Proof. Define Z

W (t) =

|∇u(x, t)|2 dx.

Ω

Recall that k∇vk22 6

1 k∆vk22 , ∀ v ∈ H01 (Ω) ∩ H 2 (Ω). λ

Then, along the orbit of (5.8), we have W 0 (t) = 2

Z

∇u · ∇ut dx

Ω

= −2 = −2

Z

∆uut dx ZΩ

|∆u|2 dx − 2

Ω

Z

= 2k

Z

kf (u)∆udx Ω

f 0 (u)|∇u|2 dx − 2

Z

Ω

Z

6 2k

|∆u|2 dx

Ω 0

2

f (u)|∇u| dx − 2λ

Z

Ω

|∇u|2 dx.

Ω

By the assumption that f 0 (v) 6 f 0 (0) for all v ∈ R, it follows that W 0 (t) 6 2(kf 0 (0) − λ)

Z

|∇u|2 dx = −2(λ − kf 0 (0))W (t).

Ω

0

Thanks to λ−kf (0) > 0 and W (t) > 0, we have limt→∞ W (t) = 0, i.e., limt→∞ u = 0 in H01 (Ω). By Theorem 2.11, kukC 2+α, 1+α/2 (Ω×[1,∞)) 6 C. We can use this fact and the compactness arguments to deduce that limt→∞ u = 0 in C 2 (Ω).  In the following three subsections we will use the Lyapunov functional to study the global asymptotic stability of the constant equilibrium solution in conjunction with specific examples. For the convenience, we shall use the notation z(x, y) = x − y − y ln

x y

with x, y > 0.

Recently, Ni et al. [86] made an extension of Lyapunov functional method to study the global asymptotic stability of non-constant equilibrium solution in heterogeneous environment. The models discussed in [86] include the scalar parabolic equations  u = d(x)∆u + uf (x, u),    t   

x ∈ Ω, t > 0,

∂n u = 0,

x ∈ ∂Ω, t > 0,

u(x, 0) = ϕ(x) > 0, 6≡ 0,

x∈Ω

5.2. Lyapunov functional method  105

and  u = d(x)div[a(x)∇Φ(x, u)] + f (x, u),    t

x ∈ Ω, t > 0,

∂n Φ(x, u) = g(x, u),

x ∈ ∂Ω, t > 0,

  

u(x, 0) = ϕ(x) > 0, 6≡ 0,

x ∈ Ω,

as well as the Lotka-Volterra competition model with k species    k  X    u = d (x)∆u + u m (x) − a (x)u i i i i ij j ,   it

x ∈ Ω, t > 0, 1 6 i 6 k,

 ∂n ui = 0,     u (x, 0) = ϕ (x) > 0, 6≡ 0, i i

x ∈ ∂Ω, t > 0, 1 6 i 6 k, x ∈ Ω, 1 6 i 6 k,

j=1

where for each i, either di (x) > 0 or di (x) ≡ 0 in Ω. 5.2.2

A variable-territory prey-predator model

A so-called variable territory model posits that the self-limitation of predator depends inversely on the availability of the prey. Under suitable re-scaling, a variable-territory prey-predator model can be written as ([117])  ut − d1 ∆u = u(β − au − bv),       vt − d2 ∆v = v(u − 1 − v/u),

x ∈ Ω,

t > 0,

x ∈ Ω,

t > 0,

 ∂n u = ∂n v = 0,     

x ∈ ∂Ω, t > 0,

u(x, 0) = u0 (x), v(x, 0) = v0 (x),

(5.10)

x ∈ Ω,

where a, b, β > 0 are constants, u0 , v0 ∈ C 2 (Ω) and u0 , v0 > 0 in Ω. Consider the initial value problem of ODE system   u¯0 (t) = u¯(β − a¯ u − bv),     0   u (t) = u(β − au − b¯ v ),     0  v¯ (t) = v¯(¯ u − 1 − v¯/¯ u),       0

v (t) = v(u − 1 − v/u),

 u¯(0) = maxΩ u0 (x),       u(0) = minΩ u0 (x),       v¯(0) = maxΩ v0 (x),     

v(0) = minΩ v0 (x).

Clearly, ((¯ u, v¯), (u, v)) is a pair of coupled upper and lower solutions of (5.10). In order to show the global existence of (¯ u, u, v¯, v), we set w = v¯/¯ u and z = v/u. Then

106  5. Stability Analysis

(¯ u, u, v¯, v, w, z) satisfies   u¯0 (t) = u¯(β − a¯ u − bv),      u0 (t) = u(β − au − b¯ v ),       v¯0 (t) = v¯(¯ u − 1 − w),  v 0 (t) = v(u − 1 − z),       w0 (t) = w[(a + 1)¯ u + bv − (1 + β) − w],      0

v − (1 + β) − z] z (t) = z[(a + 1)u + b¯

with positive initial data u¯(0) = max u0 (x), u(0) = min u0 (x), v¯(0) = max v0 (x), Ω

Ω

Ω

v(0) = min v0 (x), w(0) = v¯(0)/¯ u(0), z(0) = v(0)/u(0). Ω

Clearly, this problem has a unique global positive solution (¯ u, u, v¯, v, w, z). Therefore, (¯ u, u, v¯, v) exists globally. By the upper and lower solutions method, (5.10) has a unique global solution (u, v) which satisfies u(t) 6 u(x, t) 6 u¯(t) and v(t) 6 v(x, t) 6 v¯(t). Moreover, it is clear that u¯(t) 6 max{β/a, max u0 (x)} =: A and v¯(t) 6 max{A2 , max v0 (x)} =: B, Ω

Ω

and thereby, the solution (u, v) of (5.10) satisfies 0 < u 6 A and 0 < v 6 B

in Ω × [0, ∞).

(5.11)

The system (5.10) has a positive constant equilibrium solution if and only if β > a, and in that case, the positive constant equilibrium solution (˜ u, v˜) is uniquely given by b − a + [(a − b)2 + 4bβ]1/2 β − a˜ u u˜ = and v˜ = . 2b b Obviously, u˜ > 1. We point out that contents of this part take from [117], and the choice of k(s) in proving Theorem 5.8 is inspired by [24]. Theorem 5.8 (5.10).

If β > a, then (˜ u, v˜) is globally asymptotically stable for problem

Proof. Step 1. We define f1 (u, v) = u(β − au − bv), f2 (u, v) = v(u − 1 − v/u)

5.2. Lyapunov functional method  107

and 2a + b(s + u˜ − 1) , b2 Z u s − u˜ V (u, v) = k(s) 2 ds + z(v, v˜). s u ˜ k(s) =

(5.12)

After careful computations we have Vu (u, v)f1 (u, v) + Vv (u, v)f2 (u, v) = k(u)

u2 − u − v u − u˜ (β − au − bv) + (v − v˜) u u

=−

v − v˜ k(u) [a(u − u˜)2 + b(u − u˜)(v − v˜)] + [(u + u˜ − 1)(u − u˜) − (v − v˜)] u u

=−

ak(u)(u − u˜)2 − [u + u˜ − 1 − bk(u)](u − u˜)(v − v˜) + (v − v˜)2 . u

(5.13)

It is well known that if (u + u˜ − 1 − bk(u))2 < 4ak(u),

(5.14)

then the quadratic form ak(u)ξ 2 − [u + u˜ − 1 − bk(u)]ξη + η 2 is positive unless ξ = η = 0. The inequality (5.14) can be rewritten as b2 k 2 (u) − k(u)[4a + 2b(u + u˜ − 1)] + (u + u˜ − 1)2 < 0.

(5.15)

Obviously, the equation b2 k 2 − k[4a + 2b(u + u˜ − 1)] + (u + u˜ − 1)2 = 0 has two positive roots: q 1 k1 = k1 (u) = 2 2a + b(u + u˜ − 1) − 2 a2 + ab(u + u˜ − 1) , b   q 1 k2 = k2 (u) = 2 2a + b(u + u˜ − 1) + 2 a2 + ab(u + u˜ − 1) , b 



and (5.15) holds for all k1 (u) < k(u) < k2 (u). It is easy to see that the function k(u) given by (5.12) satisfies k1 (u) < k(u) < k2 (u). Accordingly, the inequality (5.14) holds. Furthermore, as 0 6 u 6 A, we declare ak(u)ξ 2 − [u + u˜ − 1 − bk(u)]ξη + η 2 > ε(|ξ|2 + |η|2 ) for some ε > 0. This, together with (5.13), derives 



Vu (u, v)f1 (u, v) + Vv (u, v)f2 (u, v) 6 −C |u − u˜|2 + |v − v˜|2 .

(5.16)

108  5. Stability Analysis

Step 2. Let (u, v) denote the solution of (5.10), and define Z

W (t) =

V (u(x, t), v(x, t)), f (u) = k(u) Ω

u − u˜ . u2

Then we have v˜ −d1 f (u)|∇u| − d2 2 |∇v|2 v

Z 

0

W (t) = Ω

0

2



Z

+

[Vu (u, v)f1 (u, v) + Vv (u, v)f2 (u, v)] .

(5.17)

Ω

Now we calculate f 0 (u). Firstly, rewrite f (u) as 2a + b(˜ u − 1) f (u) = 2 b



1 u˜ − 2 u u



+

u − u˜ . bu

In view of u˜(˜ u − 1) = v˜ and β = a˜ u + b˜ v , it is easy to calculate 2a + b(˜ u − 1) 2˜ u−u u˜ + 2 2 3 b u bu 1 = 2 3 (4a˜ u − 2au + 2b˜ v + bu) b u 2 > 2 3 (2a˜ u + b˜ v − au) b u   2a β = 2 3 u˜ + − u . b u a

f 0 (u) =

(5.18)

On the other hand, the first equation of (5.10) implies ut − d1 ∆u 6 u(β − au), and so lim sup max u(·, t) 6 β/a by Lemma 5.14. Then one can take T  1 such that t→∞

Ω

u 6 u˜/2 + β/a in Ω × [T, ∞). This estimate combined with (5.18) allows us to produce f 0 (u) >

a˜ u a˜ u > 2 3 b2 u3 b A

in Ω × [T, ∞).

(5.19)

Step 3. Since v is bounded, it follows from (5.16), (5.17) and (5.19) that 0

W (t) 6 −C

Z 

2

2

|∇u| + |u − u˜|



−C

Z 

|∇v|2 + |v − v˜|2



Ω

Ω

for t > T . Certainly, 0

W (t) 6 −C

Z Ω

|u − u˜|2 =: −ψ(t) for all t > T.

(5.20)

5.2. Lyapunov functional method  109

Using the equation of u and the estimate (5.11), we have ψ 0 (t) = 2C

Z ZΩ

= 2C

ut (u − u˜) (d1 ∆u + βu − au2 − buv)(u − u˜)

Ω

= −2Cd1

Z

|∇u|2 + 2C

Z

Ω

Z

6 2C

(βu − au2 − buv)(u − u˜)

Ω 2

(βu − au − buv)(u − u˜)

Ω

6k for some constant k > 0. Applying Lemma 5.5 to (5.20) we conclude ψ(t) → 0, and then u → u˜ in L2 (Ω) as t → ∞. Recalling (5.11) and Theorem 2.13, it achieves ku(·, t)kC 1+α (Ω) 6 M for all t > T + 1, and then lim ku(·, t) − u˜kC 1 (Ω) = 0.

t→∞

As a result, u > u˜/2 in Ω × [T1 , ∞) for some T1 > T . Hence u − 1 − v/u is bounded in Ω × [T1 , ∞). Similar to discussions about u, it can be deduced that lim kv(·, t) − v˜kC 1 (Ω) = 0.

t→∞

Step 4. Above arguments show that u˜/2 6 u 6 A and 0 < v 6 B in Ω × [T0 , ∞) with T0  1. Invoking Theorem 2.11 we can get further that ku(·, t)kC 2+α (Ω) 6 C, kv(·, t)kC 2+α (Ω) 6 C for all t > T0 , which implies lim ku(·, t) − u˜kC 2 (Ω) = lim kv(·, t) − v˜kC 2 (Ω) = 0.

t→∞

t→∞

The proof is complete. 5.2.3



A prey-predator model with trophic interactions

In this part we treat the following prey-predator model with trophic interactions of three levels ([68]):   ut − d1 ∆u = uf1 (u, v),       vt − d2 ∆v = vf2 (u, v, w),  

x ∈ Ω, t > 0, x ∈ Ω, t > 0,

wt − d3 ∆w = wf3 (v, w),

x ∈ Ω, t > 0,

     ∂n u = ∂n v = ∂n w = 0,     (u, v, w) = (u , v , w ) > 0, 0 0 0

x ∈ ∂Ω, t > 0, x ∈ Ω, t = 0,

(5.21)

110  5. Stability Analysis

where f1 (u, v) = 1 − u −

av , v+ρ

cw bu −γ− , v+ρ w+σ mv f3 (v, w) = − β, w+σ

f2 (u, v, w) =

u, v and w are population densities of resource (level 0 species), consumer (level 1 species) and top predator (level 2 species) respectively. We assume u0 , v0 , w0 ∈ Wp2 (Ω). Then problem (5.21) has a unique global solution (u, v, w). The content of this part is due to [68]. We first analyze positive solutions of the algebraic system f1 (u, v) = 0, f2 (u, v, w) = 0, f3 (v, w) = 0.

(5.22)

Let (˜ u, v˜, w) ˜ be a positive solution of (5.22). Then (˜ u, v˜, w) ˜ satisfies  ρ + (1 − a)˜ v    u˜ = ,   v˜ + ρ      m˜ v − βσ w˜ = ,  β        b˜ v ρ + (1 − a)˜ v cβσ    = (γ + c)˜ v− .  2

(˜ v + ρ)

(5.23)

m

Let us define v = βσ/m,

f (v) = (γ + c)v(v + ρ)2 − cv(v + ρ)2 − bρv − b(1 − a)v 2 .

Then (˜ u, v˜, w) ˜ is a positive solution of (5.23) if and only if f (v) = 0 has a positive root v˜ and (1) v˜ > v when a 6 1; (2) v < v˜ < ρ/(a − 1) =: v¯ when a > 1. Notice f (±∞) = ±∞,

f (−ρ) = abρ2 > 0 and f (0) = −cρ2 v < 0.

There exist three numbers v1 , v2 , v˜ satisfying −∞ < v1 < v2 < 0 < v˜ such that f (v1 ) = f (v2 ) = f (˜ v ) = 0, f 0 (v1 ) > 0, f 0 (v2 ) < 0, f 0 (˜ v ) > 0, f (v) < 0 in (−∞, v1 ) ∪ (v2 , v˜), f (v) > 0 in (v1 , v2 ) ∪ (˜ v , ∞). This indicates that the system (5.23) has at most one positive solution, and (1) when a 6 1, system (5.23) has a positive solution if and only if f (v) < 0; (2) when a > 1, system (5.23) has a positive solution if and only if f (v) < 0 < f (¯ v ).

5.2. Lyapunov functional method  111

A direct calculation gives f (v) = (γ + c)v(v + ρ)2 − cv(v + ρ)2 − bρv − b(1 − a)v 2 = v γ(v + ρ)2 − bρ − b(1 − a)v . 



It follows that, when a 6 1, the system (5.23) has a positive solution if and only if γ(v + ρ)2 − bρ − b(1 − a)v < 0.

(5.24)

When a > 1 and βσ/m = v < v¯ = ρ/(a − 1), a direct computation gives v + ρ)2 − bρ¯ v − b(1 − a)¯ v2 f (¯ v ) = (γ + c)¯ v (¯ v + ρ)2 − cv(¯ = (¯ v + ρ)2 (γ + c)¯ v − cv > 0. 



As a result, when a > 1, (5.23) has a positive solution if and only if (5.24) holds and βσ/m = v < v¯ = ρ/(a − 1).

(5.25)

So far, we get Theorem 5.9 When a 6 1, the system (5.22) has a positive solution if and only if (5.24) holds. When a > 1, (5.22) has a positive solution if and only if (5.24) and (5.25) hold. Furthermore, in any case, (5.22) has at most one positive solution. Theorem 5.10 If (5.22) has a positive solution (˜ u, v˜, w), ˜ then, as the equilibrium solution of (5.21), it is globally asymptotically stable. Proof. Define h= and

b(˜ v + ρ) , aρ

k=

cσ m(w˜ + σ)

Z

W (t) = Ω

[hz(u, u˜) + z(v, v˜) + kz(w, w)] ˜ .

After careful calculations, we get h(u − u˜)f1 (u, v) + (v − v˜)f2 (u, v, w) + k(w − w)f ˜ 3 (v, w) !

b˜ u(v − v˜)2 mk˜ v (w − w) ˜ 2 = − h(u − u˜) + + , (˜ v + ρ)(v + ρ) (w˜ + σ)(w + σ) 2

and then W 0 (t) =

Z  Ω

=−

Z Ω

−

Z Ω

v − v˜ k(w − w) ˜ h(u − u˜) ut + vt + wt u v w



|∇u|2 |∇v|2 |∇w|2 d1 h˜ u 2 + d2 v˜ 2 + d3 k w˜ u v w2

!

!

b˜ u(v − v˜)2 mk˜ v (w − w) ˜ 2 h(u − u˜) + + . (˜ v + ρ)(v + ρ) (w˜ + σ)(w + σ) 2

112  5. Stability Analysis

Repeating the last part in proving Theorem 5.8, we can show that lim ku − u˜kC 2 (Ω) = lim kv − v˜kC 2 (Ω) = lim kw − wk ˜ C 2 (Ω) = 0.

t→∞

t→∞

t→∞

The proof is complete.  Now, suppose that (5.22) has no positive solution. Then, possible non-negative solutions of (5.22) are (0, 0, 0), (1, 0, 0) and (u∗ , v ∗ , 0), where (u∗ , v ∗ ) is a possible positive solution of u∗ = 1 −

bu∗ = γ. v∗ + ρ

av ∗ , v∗ + ρ

(5.26)

Hence, (u∗ , v ∗ ) exists if and only if ρ < b/γ. And in this case, v ∗ is uniquely given by 1 v∗ = 2

b(1 − a) − 2ρ + γ

s

!

b2 (a − 1)2 4ρab . + γ2 γ

Theorem 5.11 Suppose that (5.22) has no positive solution and (5.26) has a positive solution (u∗ , v ∗ ). Then the non-negative equilibrium solution (u∗ , v ∗ , 0) of (5.21) is globally asymptotically stable. Proof. Step 1. We first verify that v ∗ satisfies mv ∗ 6 βσ.

(5.27)

When a 6 1, owing to that (5.22) has no positive solution, we must have γ(v + ρ)2 − bρ − b(1 − a)v > 0 by Theorem 5.9. Since v ∗ satisfies γ(v ∗ + ρ)2 − bρ − b(1 − a)v ∗ = 0, it is deduced that v ∗ 6 v, i.e., (5.27) holds. When a > 1, we infer that at least one of (5.24) and (5.25) does not hold by Theorem 5.9. If (5.24) is not valid, same as the case a 6 1 we can get (5.27). If (5.25) is not valid, then βσ/m > ρ/(a − 1). Now, if (5.27) does not hold, then v ∗ > βσ/m > ρ/(a − 1) and accordingly s

b2 (a − 1)2 4ρab 2ρ b(a − 1) 2aρ b(a − 1) + > + 2ρ + = + , 2 γ γ a−1 γ a−1 γ

which implies 2aρ/(a − 1) < 0. This is a contradiction, and then (5.27) holds. Step 2. Similar to the proof of Theorem 5.10, we take p=

b(v ∗ + ρ) , aρ

q=

c m

5.2. Lyapunov functional method  113

and

Z

U (t) = Ω

[pz(u, u∗ ) + z(v, v ∗ ) + qw] .

Notice (5.27) and u, v, w > 0. Careful calculations lead to v − v∗ p(u − u∗ ) ut + vt + qwt u v Ω Ω Ω Z Z 2 2 |∇v| |∇u| = −d1 pu∗ − d2 v ∗ 2 2 Ω v Ω u   Z  av av ∗ − + p(u − u∗ ) u∗ − u + ∗ v +ρ v+ρ Ω     ∗ bu bu cw mv ∗ +(v − v ) − − −β + qw v + ρ v∗ + ρ w + σ w+σ

U 0 (t) =

Z

Z

= −d1 pu

∗

Z Ω

−

Z " Ω

|∇u|2 − d2 v ∗ u2

Z

Z Ω

|∇v|2 −p v2

Z

(u − u∗ )2

Ω

#

βqw2 w bu∗ (v − v ∗ )2 + + q(βσ − mv ∗ ) . ∗ (v + ρ)(v + ρ) w + σ w+σ

Repeating the last part in proving Theorem 5.8, we can show that lim ku − u∗ kC 2 (Ω) = lim kv − v ∗ kC 2 (Ω) = lim kwkC 2 (Ω) = 0.

t→∞

t→∞

The proof is complete. Similarly, we can prove the following theorem.

t→∞



Theorem 5.12 If both (5.22) and (5.26) do not have any positive solution, then the equilibrium solution (1, 0, 0) of (5.21) is globally asymptotically stable. Above results provide a complete qualitative description of dynamical behaviour of (5.21). In particular, we note that in consideration of asymptotic behaviour, either populations of three species tend to a (constant) positive equilibrium, or one or more of predator species must become extinct. Further, it elaborates that, if conditions are favorable for the positive equilibrium (˜ u, v˜, w) ˜ to exist, then its existence dominates the entire dynamical behaviour in the sense that all positive solutions are globally attracted to this equilibrium. If this positive equilibrium does not exist, but the semipositive equilibrium (u∗ , v ∗ , 0) exists, then this semi-positive equilibrium dominates the dynamical behaviour in the above sense. In the absence of these equilibria, we see the extinction of two predator species. Accordingly, one can discern a hierarchy of dynamical behaviour that is characterized by the preservation of the maximum number of species in the system.

114  5. Stability Analysis

5.2.4

A system coupled by two PDEs and one ODE

A two-predators-one-prey system with strategy may be written as an ODE system ([87]) with u = (u1 , u2 , u3 ) that satisfies    u2 u3 du1   = u1 − 1 =: f1 (u),    dt u1 + u2       du bu1 u3 2 = u2 − a =: f2 (u),  u1 + u2  dt       du3 (1 + b)u1 u2    = u3 r − u3 − =: f3 (u),

(5.28)

u1 + u2

dt

where d1 , d2 , a, b, r are positive constants, u1 , u2 and u3 are population densities of two predator species and a prey species, respectively. It is easy to see that (5.28) has a positive constant equilibrium solution if and only if rb > a + b. In this case, the positive constant equilibrium solution is uniquely given by u˜1 = (a + b)

rb − (a + b) , b2 (1 + b)

u˜2 = (a + b)

rb − (a + b) , ab(1 + b)

u˜3 =

a+b . b

To take into account the in-homogeneous distribution of predators and prey in different spatial locations within a fixed bounded domain Ω ⊂ Rn at any given time, and the natural tendency of each species to diffuse to areas of smaller population concentration, we have to pay attention to the role of diffusion. Let us consider that two predators are birds and one prey is worm. It is well known that diffusions of birds are very fast, while that of worm is so slow that we can omit its diffusion. Based on these reasons, we study the following system coupled by two reaction-diffusion equations and one ordinary differential equation  u1t − d1 ∆u1 = f1 (u),         u2t − d2 ∆u2 = f2 (u), 

u = f (u),

3t 3       ∂n u1 = ∂n u2 = 0,   

x ∈ Ω,

t > 0,

x ∈ Ω,

t > 0,

x ∈ Ω,

t > 0,

(5.29)

x ∈ ∂Ω, t > 0,

ui (x, 0) = ui0 (x), i = 1, 2, 3, x ∈ Ω,

where f1 (u), f2 (u) and f3 (u) are given in (5.28), ui0 ∈ Wp2 (Ω) for some p > 1 + n/2 and i = 1, 2, u30 ∈ C γ (Ω) for some 0 < γ < 1 and u10 (x) > 0, u20 (x) > 0 and u30 (x) > 0 in Ω. Same as Theorem 4.1, we can show that problem (5.29) has a unique local solution (u1 , u2 , u3 ) satisfying u1 , u2 ∈ Wp2,1 (QT ) and u3 ∈ C α,1 (QT ) for some 0 < T < ∞ and 0 < α < min{γ, 2 − (n + 2)/p}. We leave details to readers as an exercise (Exercise 5.9).

5.2. Lyapunov functional method  115

Let Tmax be the maximal existence time of (5.29). It is easy to see from the third equation of (5.29) that u3 6 max{r, maxΩ u30 (x)}. Integrating equations of (5.29) over Ω and adding the result, we have d dt

Z Ω

(u1 + u2 + u3 ) 6 −

Z

(u1 + au2 + u3 ) + C. Ω

It follows that kui (·, t)k1, Ω is bounded in [0, Tmax ), i = 1, 2. Using Theorem 2.14, we can obtain that kui (·, t)k∞, Ω is also bounded in [0, Tmax ), i = 1, 2. Therefore, Tmax = ∞ and (u1 , u2 , u3 ) is bounded. Theorem 5.13 ([114]) The positive constant equilibrium solution (˜ u1 , u˜2 , u˜3 ) of (5.29) is globally asymptotically stable in the sense that lim kui (·, t) − u˜i kC 1 (Ω) = 0, i = 1, 2;

t→∞

lim ku3 (·, t) − u˜3 k2, Ω = 0.

t→∞

Proof. Applying Theorem 2.13 to equations of u1 and u2 , we have, for some 0 < α < 1, k∇ui kC α/2 ([1,∞), C(Ω)) + kui kC (1+α)/2 ([1,∞), C(Ω)) 6 C,

(5.30)

i = 1, 2. It then follows from the differential equation of u3 that ku3 kC 1 ([1,∞), C(Ω)) 6 C. We set p =

(5.31)

b2 b(a + b) ,q= , and define a a(1 + b) Z

V (t) = Ω

[pz(u1 , u˜1 ) + z(u2 , u˜2 ) + qz(u3 , u˜3 )].

Then V (t) > 0 for all t > 0. Using (5.29) and integrating by parts, we obtain pd1 u˜1 d2 u˜2 2 2 2 |∇u1 | + u2 |∇u2 | u Ω 1 2  Z  f2 (u) f3 (u) f1 (u) + (u2 − u˜2 ) + q(u3 − u˜3 ) . + p(u1 − u˜1 ) u1 u2 u3 Ω

V 0 (t) = −

Z 



Through careful calculations we can get f1 (u) f2 (u) f3 (u) + (u2 − u˜2 ) + q(u3 − u˜3 ) u1 u2 u3 2 (bu1 − au2 ) a+b =− − (a + b − bu3 )2 . a(u1 + u2 ) ab(1 + b) p(u1 − u˜1 )

Therefore, 0

V (t) = −

Z  Ω

−

Z Ω

pd1 u˜1 d2 u˜2 |∇u1 |2 + 2 |∇u2 |2 2 u1 u2

 !

(bu1 − au2 )2 a+b + (a + b − bu3 )2 . a(u1 + u2 ) ab(1 + b)

116  5. Stability Analysis

As u2i 6 C for i = 1, 2, 3, it follows that C0 V (t) 6 − C 0

Z

2

2

(|∇u1 | + |∇u2 | ) −

Ω

Z Ω

a+b (bu1 − au2 )2 + (a + b − bu3 )2 a(u1 + u2 ) ab(1 + b)

=: −ψ1 (t) − ψ2 (t).

!

(5.32)

According to (5.30) and (5.31), we have kψ1 kC α/2 ([1,∞)) 6 C and kψ2 kC 1 ([1,∞)) 6 C. Applying Lemma 5.5 to (5.32) we conclude that ψ1 (t), ψ2 (t) → 0 as t → ∞. Therefore, Z

lim

t→∞ Ω

Z

lim

t→∞ Ω

(|∇u1 |2 + |∇u2 |2 ) = 0,

(bu1 − au2 )2 = lim t→∞ (u1 + u2 )

Z

(5.33)

(a + b − bu3 )2 = 0.

Ω

The latter demonstrates that Z

lim

t→∞ Ω

|au2 − bu1 | = lim

Z

t→∞ Ω

(u3 − u˜3 )2 = 0.

(5.34)

From (5.33) and the Poincar´e inequality, we deduce Z

lim

t→∞ Ω

[(u1 − u¯1 )2 + (u2 − u¯2 )2 ] = 0,

(5.35)

1 f for f ∈ L1 (Ω). The second limit of (5.34) implies u¯3 (t) → u˜3 . where f¯ = |Ω| Ω Noticing that u¯1 (t) and u¯2 (t) are bounded, and using the first limit of (5.34), we infer that there exist a sequence {tk } with tk → ∞ and a non-negative constant uˆ1 such that Z

u¯03 (tk ) → 0, u¯1 (tk ) → uˆ1 and u¯2 (tk ) → uˆ1 b/a.

(5.36)

At t = tk , we write |Ω|¯ u03 =

u1 u2 u1 + u2 Ω  Z  Z b(1 + b) 1+b u1 (bu1 − au2 ) +˜ u3 r − u˜3 − u1 + . a+b b u1 + u2 Ω Ω Z





(u3 − u˜3 ) r − u3 − u˜3 − (1 + b)

(5.37)

Noticing (5.34), (5.36) and r > u˜3 , it follows from (5.37) that uˆ1 6= 0 and r − u˜3 − uˆ1 b(1 + b)/(a + b) = 0, and hence uˆ1 = u˜1 . Consequently, lim u¯i (tk ) = u˜i ,

k→∞

i = 1, 2.

(5.38)

Since kui (·, tk )kC 1+α (Ω) 6 C, there exist a sub-sequence of {tk }, still denoted by it self, and non-negative functions wi ∈ C 1 (Ω), such that lim kui (·, tk ) − wi (·)kC 1 (Ω) = 0,

k→∞

i = 1, 2.

This combined with (5.35) and (5.38) allows us to derive wi ≡ u˜i . Therefore, lim kui (·, t) − u˜i kC 1 (Ω) = 0,

t→∞

Theorem 5.13 is thus proved.

i = 1, 2. 

5.3. Iteration method  117

5.3

ITERATION METHOD

The idea of iteration method is to construct iterative sequences and to prove these sequences converge to the unique equilibrium solution. We only take the following ratio-dependent prey-predator model ([116]) as an example to introduce this method:    v    , ut − d1 ∆u = u 1 − ku −   cu + v        u

vt − d2 ∆v = v −a − bv +

cu + v      ∂n u = ∂n v = 0,      u(x, 0) = u (x), v(x, 0) = v (x), 0 0

,

x ∈ Ω,

t > 0,

x ∈ Ω,

t > 0,

(5.39)

x ∈ ∂Ω, t > 0, x ∈ Ω,

where d1 , d2 , c, k, a and b are positive constants, u0 , v0 ∈ Wp2 (Ω) and u0 , v0 > 0 in Ω. By the upper and lower solutions method, (5.39) has a unique global solution (u, v) and 0 < u, v 6 C in Ω × [0, ∞) for some constant C. Problem (5.39) has a positive constant equilibrium solution (˜ u, v˜) if and only if ac < 1, and in this case it is uniquely given by u˜ =

c(a + b˜ v) , k

v˜ =

c(1 − ac) . k + bc2

We first state a lemma which plays an important role in the construction of iterative sequences and can be proved by the comparison principle and known results of ODEs. Details will be omitted. Lemma 5.14 Let f ∈ C 1 ([0, ∞)) with f > 0, and d, β, γ be constants with d > 0 and β > 0. Assume that T ∈ [0, ∞) and w ∈ C 2,1 (Ω × (T, ∞)) ∩ C 1,0 (Ω × [T, ∞)) is a positive function. (1) If γ > 0 and w satisfies (

wt − d∆w 6 (>) w1+β f (w)(γ − w)

in Ω × (T, ∞),

∂n w = 0

on ∂Ω × [T, ∞), 

then lim sup max w(·, t) 6 γ t→∞

Ω



lim inf min w(·, t) > γ . t→∞

Ω

(2) If γ 6 0 and w satisfies (

wt − d∆w 6 w1+β f (w)(γ − w)

in Ω × (T, ∞),

∂n w = 0

on ∂Ω × [T, ∞),

then lim sup max w(·, t) 6 0. t→∞

Ω

Theorem 5.15 ([116]) If parameters appearing in (5.39) satisfy ac < 1 and k < 2bc2 + 2ack, then (˜ u, v˜) is globally asymptotically stable for problem (5.39).

(5.40)

118  5. Stability Analysis

Proof. Our intention is to use Lemma 5.14 to construct iterative sequences and to prove that these sequences are monotonic and converge to (˜ u, v˜). For s > 0, we define h(v; s) = −bv 2 − (a + bcs)v + (1 − ac)s. Then h(v; s) = 0 has two roots: p

(a + bcs)2 + 4b(1 − ac)s > 0, 2b p −(a + bcs) − (a + bcs)2 + 4b(1 − ac)s < 0. ψ− (s) = 2b ψ(s) =

−(a + bcs) +

Therefore, h(v; s) = b(v − ψ− (s))(ψ(s) − v). Step 1. It follows from the first equation of (5.39) that ut − d1 ∆u 6 u(1 − ku). Then by Lemma 5.14, lim sup max u(·, t) 6 1/k =: u¯1 . t→∞

Ω

Step 2. For any given ε > 0, we can find T1ε  1 so that u 6 u¯1 + ε in Ω × [T1ε , ∞). Then we have that, for x ∈ Ω and t > T1ε , vt − d2 ∆v 6 −av − bv 2 +

(¯ u1 + ε)v c(¯ u1 + ε) + v

v h(v; u¯1 + ε) c(¯ u1 + ε) + v bv = (v − ψ− (¯ u1 + ε))(ψ(¯ u1 + ε) − v) c(¯ u1 + ε) + v

=

u1 + ε) by Lemma by the second equation of (5.39). Hence lim sup max v(·, t) 6 ψ(¯ t→∞

Ω

5.14, and the arbitrariness of ε gives lim sup max v(·, t) 6 ψ(¯ u1 ) =: v¯1 . t→∞

Ω

Step 3. A straightforward computation shows that k¯ v1 < c if and only if k < 2bc2 + 2ack. Thanks to the condition (5.40), one can choose ε0 > 0 so small that c > k(¯ v1 + ε) for all 0 < ε 6 ε0 . Take T2ε  1 such that v 6 v¯1 + ε in Ω × [T2ε , ∞). It then follows from the equation of u that 2

ut − d1 ∆u > u



c c − k(¯ v1 + ε) − cku − k = u2 in Ω × [T2ε , ∞). cu + v¯1 + ε cu + v¯1 + ε 

5.3. Iteration method  119

Using Lemma 5.14, we have lim inf min u(·, t) > [c − k(¯ v1 + ε)]/(ck), and then the t→∞

Ω

arbitrariness of ε informs lim inf min u(·, t) > (c − k¯ v1 )/(ck) =: u1 > 0. t→∞

Ω

Step 4. For any given 0 < ε < u1 , we choose T3ε  1 such that u > u1 − ε in Ω × [T3ε , ∞). Making use of the second equation of (5.39) and careful calculations, we obtain that, for x ∈ Ω and t > T3ε , vt − d2 ∆v > −av − bv 2 +

(u1 − ε)v c(u1 − ε) + v

v h(v; u1 − ε) c(u1 − ε) + v bv (v − ψ− (u1 − ε))(ψ(u1 − ε) − v). = c(u1 − ε) + v =

Lemma 5.14 gives lim inf min v(·, t) > ψ(u1 − ε), and hence, by the arbitrariness of ε, t→∞

Ω

lim inf min v(·, t) > ψ(u1 ) =: v 1 . t→∞

Ω

Step 5. For any given 0 < ε < v 1 , we take T4ε  1 so that v > v 1 −ε in Ω×[T4ε , ∞). Consequently, for x ∈ Ω and t > T4ε , ut − d1 ∆u 6 u2



c c − k(v 1 − ε) − cku − k = u2 . cu + v 1 − ε cu + v 1 − ε 

Same as above, lim sup max u(·, t) 6 (c − kv 1 )/(ck) =: u¯2 . t→∞

Ω

Step 6. Define ϕ(s) = (c − ks)/(ck). Then ϕ0 (s) < 0, ψ 0 (s) > 0, and constants u¯1 , v¯1 , u1 , v 1 and u¯2 constructed above satisfy v 1 = ψ(u1 ) < ψ(¯ u1 ) = v¯1 , u1 = ϕ(¯ v1 ) < ϕ(v 1 ) = u¯2 6 u¯1 ,

(5.41)

u1 6 lim inf min u(·, t) 6 lim sup max u(·, t) 6 u¯2 , t→∞

Ω

t→∞

Ω

v 1 6 lim inf min v(·, t) 6 lim sup max v(·, t) 6 v¯1 . t→∞

Ω

t→∞

Ω

Step 7. Applying inductive method, we can construct four sequences {ui }, {v i }, {¯ ui } and {¯ vi } by vi ), v i = ψ(ui ), u¯i+1 = ϕ(v i ), v¯i = ψ(¯ ui ), ui = ϕ(¯

(5.42)

and ui , v i , u¯i , v¯i satisfy   inf min u(·, t) 6 lim sup max u(·, t) 6 u¯i ,  ui 6 lim t→∞ Ω

t→∞

Ω

  v i 6 lim inf min v(·, t) 6 lim sup max v(·, t) 6 v¯i . t→∞

Ω

t→∞

Ω

(5.43)

120  5. Stability Analysis

On account that ϕ0 (s) < 0 and ψ 0 (s) > 0 for s > 0, it follows from (5.41) and (5.42) that v i−1 < v i = ψ(ui ) < ψ(¯ ui ) = v¯i < v¯i−1 , ui−1 < ui = ϕ(¯ vi ) < ϕ(v i ) = u¯i+1 < u¯i inductively. Therefore, sequences {ui }, {v i }, {¯ ui } and {¯ vi } have point-wise limits: lim ui = u,

i→∞

lim v i = v,

i→∞

lim u¯i = u¯ and

i→∞

lim v¯i = v¯.

i→∞

(5.44)

Obviously, 0 < u 6 u¯ and 0 < v 6 v¯, and u, v, u¯, v¯ satisfy u = ϕ(¯ v ), u¯ = ϕ(v), v = ψ(u) and v¯ = ψ(¯ u).

(5.45)

By a direct calculation we see that (5.45) is equivalent to v )/(ck), u¯ = (c − kv)/(ck), u = (c − k¯

(5.46)

v − u¯/(c¯ u + v¯) = 0. a + bv − u/(cu + v) = 0, a + b¯

(5.47)

It follows from (5.46) that cu + v¯ = c/k = c¯ u + v, which implies c(¯ u − u) = v¯ − v.

(5.48)

Step 8. Now, we prove v¯ = v, and then get u¯ = u by (5.48). Substituting the second equation of (5.46) into that of (5.47), we produce a + b¯ v−

c − kv = 0, c(c − kv) + ck¯ v

that is, v + bc(c − kv)¯ v + bck¯ v 2 = c − kv. ac(c − kv) + ack¯

(5.49)

Similarly, by the first equation of (5.46) and that of (5.47), we have ac(c − k¯ v ) + ackv + bc(c − k¯ v )v + bckv 2 = c − k¯ v.

(5.50)

Subtracting (5.50) into (5.49) results in v − v) + 2ack(¯ v − v) = k(¯ v − v). bck(¯ v 2 − v 2 ) + bc2 (¯

(5.51)

The proof will be accomplished by a contradiction argument. Suppose otherwise that v¯ 6= v. Then v¯ > v and k = bck(¯ v + v) + bc2 + 2ack

(5.52)

in accordance with (5.51). The condition (5.40) asserts k < 2bc2 +2ack. This together with (5.52) alleges bc2 > bck(¯ v + v), i.e., v¯ + v < c/k. Hence, by (5.48), cu + v¯ = c/k > v¯ + v

=⇒ v < cu.

(5.53)

5.4. Average method  121

Owing to (5.47) we have as well b(¯ v − v) =

u u¯ (¯ u − u)v − u(¯ v − v) − . = c¯ u + v¯ cu + v (c¯ u + v¯)(cu + v)

It follows from (5.48) immediately that b(¯ v − v) =

1 v− c (¯

v)v − u(¯ v − v) (¯ v − v)(v − cu) = , (c¯ u + v¯)(cu + v) c(c¯ u + v¯)(cu + v)

which implies v > cu since v¯ − v > 0. This contradicts (5.53). And subsequently, v¯ = v and u¯ = u, which in turn yield u¯ = u = u˜ and v¯ = v = v˜. Using (5.43) and (5.44), we finally get lim (u(x, t), v(x, t)) = (˜ u, v˜) t→∞

uniformly in Ω.

5.4



AVERAGE METHOD

Let u denote the unique solution of the following initial-boundary value problem    uit − di ∆ui = fi (x, t, u),     ∂n ui = 0,

x ∈ Ω,

 ui (x, 0) = ui0 (x),     

x ∈ Ω,

t > 0,

x ∈ ∂Ω, t > 0,

(5.54)

i = 1, . . . , m,

where di > 0. Define 1 u¯i (t) = |Ω|

Z

ui (x, t),

u¯(t) = (¯ u1 (t), . . . , u¯m (t)).

Ω

Then u¯i satisfies u¯0i (t) =

1 |Ω|

Z

fi (x, t, u(x, t)), t > 0; Ω

u¯i (0) =

1 |Ω|

Z

ui0 (x). Ω

This is an initial value problem of the first order ODEs. In some cases, it is possible to obtain limit of u¯i (t) as t → ∞. The average method, just as what its name suggests, is to determine the limit of ui (x, t) by that of u¯i (t). 5.4.1

Abstract results

Let u be the unique global solution of (5.54) and u(x, t) ∈ Σ for some bounded domain Σ ⊂ Rm and all (x, t) ∈ Q∞ . We define K = sup{|fu (x, t, u)| : (x, t) ∈ Q∞ , u ∈ Σ}, F (t) = sup{|∇x f (x, t, u)| : x ∈ Ω, u ∈ Σ}. Let η > 0 be the second eigenvalue of −∆ in Ω with the homogeneous Neumann boundary condition and d = min{d1 , . . . , dm }.

122  5. Stability Analysis

Theorem 5.16

Let f be bounded in Q∞ × Σ. If σ = dη − K > 0, then

|Ω| t σs 2 e F (s)ds , t > 0, σ 0   Z 1 −σt |Ω| t σs 2 2 2 ku(·, t) − u¯(t)k2 6 e e F (s)ds , t > 0. k∇u0 k2 + η σ 0 



Z

k∇u(·, t)k22 6 e−σt k∇u0 k22 +

(5.55) (5.56)

If we further assume that limt→∞ F (t) = 0, then, for any 0 < α < 1, lim ku(·, t) − u¯(t)kC α (Ω) = 0.

(5.57)

t→∞

Proof. Firstly, by the Poincar´e inequality, k∇wk22 > ηkw − wk ¯ 22 ,

k∆wk22 > ηk∇wk22

(5.58)

for all w ∈ H 2 (Ω) with ∂n w = 0 on ∂Ω. Denote v(t) = k∇u(·, t)k22 ,

D = diag(d1 , . . . , dm ).

Then, by the second inequality of (5.58) and the Young inequality, it derives v 0 (t) = 2

Z

h∇u, ∇ut i

Ω

Z

h∇u, ∇(D∆u + f )i

=2 Ω

= −2

Z

h∆u, D∆ui + 2

Z

Ω

6 −2dη

h∇u, fu · ∇u + ∇x f i

Ω

Z

|∇u|2 + 2K

Ω

Z

|∇u|2 + σ

Ω

Z

|∇u|2 +

Ω

|Ω| 2 F (t) σ

|Ω| 2 = −σv(t) + F (t), σ which implies |Ω| −σt t σs 2 e e F (s)ds. σ 0 This is exactly (5.55). Using this fact and the first inequality of (5.58) we get (5.56) immediately. If in addition limt→∞ F (t) = 0, it then follows from (5.55) and (5.56) that v(t) 6 v(0)e−σt +

Z

lim k∇u(·, t)k2 = lim ku(·, t) − u¯(t)k2 = 0.

t→∞

(5.59)

t→∞

Since f is bounded in Q∞ ×Σ and u(x, t) ∈ Σ for all (x, t) ∈ Q∞ , we have that |∇u| is bounded in Ω × [1, ∞) by Theorem 2.13. Take p > max{n, 2} satisfying α < 1 − n/p. Then we have 2/p

ku(·, t) − u¯(t)kC α (Ω) 6 Cku(·, t) − u¯(t)kWp1 (Ω) 6 C 0 ku(·, t) − u¯(t)kH 1 (Ω) , t > 1. This, together with (5.59), leads to (5.57).



5.4. Average method  123

5.4.2

A special diffusive competition model

To understand effects of migration and spatial heterogeneity, Dockery et al. ([21]) and Lou ([80]) studied the following initial-boundary value problem of a diffusive competition model  ut − d1 ∆u = u(m(x) − u − v),       vt − d2 ∆v = v(m(x) − u − v),

x ∈ Ω, t > 0,

 ∂n u = ∂n v = 0,     

x ∈ ∂Ω, t > 0,

u(x, 0) = u0 (x), v(x, 0) = v0 (x),

x ∈ Ω, t > 0, x ∈ Ω,

and established various interesting results about equilibrium solutions and their stabilities. Here, we deal with a special case m(x) ≡ 1, i.e.,  ut − d1 ∆u = u(1 − u − v),       vt − d2 ∆v = v(1 − u − v),  ∂n u = ∂n v = 0,     

u(x, 0) = u0 (x), v(x, 0) = v0 (x),

x ∈ Ω,

t > 0,

x ∈ Ω,

t > 0,

x ∈ ∂Ω, t > 0,

(5.60)

x ∈ Ω.

Obviously, for each s ∈ (0, 1), (s, 1 − s) is a positive equilibrium solution of (5.60). On the other hand, it is easily shown that any positive equilibrium solution of (5.60) must be constant and has form (s, 1 − s) for some s ∈ (0, 1). That is, segment {(a, b) : a, b ∈ (0, 1), a + b = 1} is the set of positive equilibrium solutions of (5.60). Our main aim is to investigate longtime behaviour of solutions of (5.60). Contents of this part are due to [85]. In this subsection, we assume u0 , v0 ∈ Wp2 (Ω) ∩ C(Ω) and u0 , v0 > 0, 6≡ 0. Theorem 5.17 (1) Problem (5.60) has a unique bounded non-negative global solution (u, v). Moreover, u¯0 u0 6 lim inf min u(x, t) 6 lim sup max u(x, t) 6 , t→∞ x∈Ω u0 + v¯0 u¯0 + v 0 t→∞ x∈Ω v0 v¯0 6 lim inf min v(x, t) 6 lim sup max v(x, t) 6 , t→∞ u¯0 + v 0 u0 + v¯0 t→∞ x∈Ω x∈Ω where u¯0 = maxΩ u0 (x), u0 = minΩ u0 (x), v¯0 = maxΩ v0 (x) and v 0 = minΩ v0 (x). (2) There exists a constant C > 0 such that ku(·, t)kC 2+α (Ω) , kv(·, t)kC 2+α (Ω) 6 C for all t > 1.

(5.61)

Proof. Since problem (5.60) is a quasi-monotonically decreasing system for u, v > 0, the conclusion (1) can be derived by the upper and lower solutions method directly. The conclusion (2) is a consequence of Remark 2.12. 

124  5. Stability Analysis

Lemma 5.18

Let (u, v) be the unique solution of (5.60) and w = u + v. Then lim sup w(t) ¯ 6 1.

(5.62)

w(T ¯ ) 6 1 for some T > 0, which implies w(t) ¯ 6 1 for all t > T,

(5.63)

w(t) ¯ > 1 for all t > 0.

(5.64)

t→∞

Moreover, we have that either

or

Proof. It is easy to see that 1 |Ω|

w¯ 0 (t) =

Z

w(1 − w).

Ω

By using the H¨older inequality we have w¯ 0 (t) 6 w(1 ¯ − w). ¯ The desired conclusions can be derived from this inequality directly.  Recalling Theorem 5.17 (1), we may assume u, v 6 1 in the following discussions. Set Z Z E(t) = (u − u¯)2 + (v − v¯)2 . Ω

Ω

Similar to the discussion in proving Theorem 5.16, we can find a constant k > 0 such that 0

E (t) 6 −2 Lemma 5.19

Z

(d1 |∇u|2 + d2 |∇v|2 ) + kE(t).

(5.65)

Ω

If d1 , d2 > 2/η, then Z

lim

t→∞ Ω

2

(u − u¯) = lim

Z

t→∞ Ω

(v − v¯)2 = 0.

(5.66)

Proof. Set f (u, v) = u(1 − u − v) and g(u, v) = v(1 − u − v). Then |fu , fv , gu , gv | 6 2. Invoking notations in §5.4.1, we have K 6 2 and F (t) ≡ 0. If d1 , d2 > 2/η, then the limit (5.66) can be followed by Theorem 5.16.  k+2 , then there is a constant 0 6 s 6 1 such that 2η limt→∞ u(x, t) = s and limt→∞ v(x, t) = 1 − s uniformly in Ω.

Theorem 5.20

If d1 , d2 >

Proof. Let ε be a constant to be determined, and set Z

Q(t) =

u + εE(t). Ω

Using (5.65), we have E 0 (t) 6 −2η

k+2 E(t) + kE(t) 6 −2E(t). 2η

(5.67)

5.4. Average method  125

When (5.63) holds, we have, by taking ε = −1 and using (5.67), that Q0 (t) >

Z



u(1 − w) + 2[(u − u¯)2 + (v − v¯)2 ]



2[(u − u¯)2 + (v − v¯)2 ] − (u − u¯)2 − (u − u¯)(v − v¯)



Ω

Z

>



Ω

1 > 2

Z

(u − u¯)2 + 3(v − v¯)2 > 0, t > T.



Ω



This implies that Q(t) is convergent as t → ∞ since Q(t) is bounded. By (5.66), there exists a constant 0 6 s1 6 1 such that u¯ converges to s1 as t → ∞. Using (5.66) again we conclude limt→∞ ku(·, t) − s1 k2, Ω = 0. When (5.64) holds, taking ε = 1/2 and using (5.67) one has Q0 (t) 6

Z



u(1 − w) − [(u − u¯)2 + (v − v¯)2 ]



Ω

6−

Z



2(u − u¯)2 + (u − u¯)(v − v¯) + (v − v¯)2



Ω

3 6− 2

1 (u − u¯) − 2 Ω

Z

2

Z Ω

(v − v¯)2 6 0.

Then, arguing as above, we likewise deduce limt→∞ ku(·, t) − s1 k2, Ω = 0. Similarly, there exists a constant 0 6 s2 6 1 such that limt→∞ kv(·, t)−s2 k2, Ω = 0. Recalling (5.61), it derives that lim ku(·, t) − s1 kC(Ω) = lim kv(·, t) − s2 kC(Ω) = 0.

t→∞

t→∞

As a result, limt→∞ kw(·, t)−s1 −s2 kC(Ω) = 0. In addition, limt→∞ w(t) ¯ = s1 +s2 6 1 by (5.62). We claim that s1 + s2 = 1. If s1 + s2 < 1, then there is T  1 such that w < δ := (1 + s1 + s2 )/2 < 1 in Ω × [T, ∞). As a consequence, d dt

Z

Z

u= Ω

Ω

u(1 − w) > (1 − δ)

Z Ω

u for t > T,

from which it yields that limt→∞ u¯(t) = ∞ since u¯(T ) > 0. It is a contradiction.  In the following, we consider a special case d1 = d2 =: d. Then w = u + v satisfies  w − d∆w = w(1 − w),    t   

x ∈ Ω,

t > 0,

∂n w = 0,

x ∈ ∂Ω, t > 0,

w(x, 0) = w0 (x),

x∈Ω

with w0 (x) = u0 (x) + v0 (x). Clearly, lim w(x, t) = 1 uniformly in Ω.

t→∞

(5.68)

Without loss of generality, we assume w0 > 0 in Ω. Take m = minΩ w0 (x) and M = maxΩ w0 (x). Then m > 0 and met M et 6 w(x, t) 6 . 1 − m + met 1 − M + M et

126  5. Stability Analysis

Problem (5.60) reduces to    ut − d∆u = u(1 − w(x, t)),     vt − d∆v = v(1 − w(x, t)),

x ∈ Ω,

t > 0,

x ∈ Ω,

t > 0,

x ∈ ∂Ω, t > 0,

 ∂n u = ∂n v = 0,     

x ∈ Ω.

u(x, 0) = u0 (x), v(x, 0) = v0 (x),

It will be shown that lim u(x, t) = s,

t→∞

lim v(x, t) = 1 − s uniformly in Ω

t→∞

for some s ∈ [0, 1]. For this motivation, let us first study a new problem  ϕ − d∆ϕ = ϕ(1 − w(x, t)),    t

x ∈ Ω,

∂n ϕ = 0,

x ∈ ∂Ω, t > 0,

  

ϕ(x, 0) = ϕ0 (x) ∈ C(Ω),

x ∈ Ω.

t > 0, (5.69)

In the sequel, we will prove that the unique solution ϕ of (5.69) converges to a constant uniformly in Ω as t → ∞. Let ψ denote the unique solution of  ψ − d∆ψ = 0,    t   

x ∈ Ω,

t > 0,

∂n ψ = 0,

x ∈ ∂Ω, t > 0,

ψ(x, 0) = ϕ0 (x),

x ∈ Ω.

Then limt→∞ ψ(x, t) = ϕ¯0 uniformly in Ω. Moreover, the upper and lower solutions method gives ψ(x, t)

et et 6 ϕ(x, t) 6 ψ(x, t) , 1 − M + M et 1 − m + met

(5.70)

which implies 1 1 ϕ¯0 6 lim inf ϕ(x, t) 6 lim sup ϕ(x, t) 6 ϕ¯0 . t→∞ M m t→∞

(5.71)

Theorem 5.21 Let d1 = d2 = d, and let (u, v) be the unique solution of (5.60) and ϕ be the unique solution of (5.69) with w(x, t) = u(x, t) + v(x, t). If ϕ0 > 0, then ϕ converges to a constant uniformly in Ω as t → ∞. Proof. For any given δ > 0, let us denote mδ = min w(x, δ), Mδ = max w(x, δ), ϕδ (x) = ϕ(x, δ). Ω

Ω

It follows from (5.68) that lim mδ = lim Mδ = 1.

δ→∞

δ→∞

(5.72)

5.4. Average method  127

Similar to derivations of (5.70) and (5.71), we have 1 ϕ¯δ 6 lim inf ϕ(x, t + δ) = lim inf ϕ(x, t) t→∞ t→∞ Mδ 6 lim sup ϕ(x, t) = lim sup ϕ(x, t + δ) 6 t→∞

t→∞

1 ϕ¯δ , mδ

(5.73)

1 ϕδ (x). In accordance with (5.72) and the boundedness of ϕ¯δ in δ, |Ω| Ω it then follows that Z

where ϕ¯δ =



0 6 lim sup ϕ(x, t) − lim inf ϕ(x, t) 6 t→∞

t→∞

1 1 − mδ Mδ



ϕ¯δ → 0

as δ → ∞. This implies that ϕ is uniformly convergent in Ω as t → ∞, and then ϕ¯δ converges to a constant as δ → ∞. Recalling (5.73), we assert that ϕ converges to a  constant uniformly in Ω as t → ∞. Let ϕ(x, t; φ) be the unique solution of (5.69) with initial function ϕ0 (x) = φ(x). Then, for any k1 , k2 ∈ R, there holds ϕ(x, t; k1 φ1 + k2 φ2 ) = k1 ϕ(x, t; φ1 ) + k2 ϕ(x, t; φ2 ).

(5.74)

Noticing 0 6 u0 6 w0 and 0 6 v0 6 w0 in Ω, the following theorem can be derived from Theorem 5.21 and the limit (5.68). Theorem 5.22 uniformly in Ω.

There is 0 6 s 6 1 for which limt→∞ (u(x, t), v(x, t)) = (s, 1 − s)

Especially, we can get more accurate results under some requirements on initial densities. (1) If w0 ≡ C for some constant C > 0, then

Theorem 5.23

lim u(x, t) = u¯0 /C and

t→∞

lim v(x, t) = v¯0 /C uniformly in Ω.

t→∞

In particular, for the corresponding ODE problem of (5.60), if u0 + v0 > 0 then lim u(t) = u0 /(u0 + v0 ) and

t→∞

lim v(t) = v0 /(u0 + v0 ).

t→∞

(2) If u0 ≡ kv0 for some constant k > 0, then lim u(x, t) = k/(1 + k) and

t→∞

lim v(x, t) = 1/(1 + k) uniformly in Ω.

t→∞

Proof. Conclusions can be derived from (5.68), (5.71) and (5.74).



Remark 5.24 For the case w0 (x) ≡ C > 0, or for the corresponding ODE problem of (5.60), we can get the exact limits of u and v. Unfortunately, for the general case, we can only prove that u and v have limits as t → ∞ and cannot give the exact values of them.

128  5. Stability Analysis

EXERCISES 5.1 Prove conclusion (2) of Theorem 5.3. 5.2 Let L and B be as in (3.10) and λ1 be the first eigenvalue of (

L[φ] = λφ

in Ω,

B[φ] = 0

on ∂Ω.

Let f ∈ C 1 (Ω × R) and us (x) be a solution of (

L[u] = f (x, u)

in Ω,

B[u] = 0

on ∂Ω.

By using the upper and lower solutions method, prove the following conclusions. (1) If fu (x, us (x)) < λ1 for all x ∈ Ω, then us (x) is locally asymptotically stable. (2) If fu (x, us (x)) > λ1 for all x ∈ Ω, then us (x) is unstable. 5.3 Let λ be the first eigenvalue of (1.33). Prove that 1 k∇uk22 for all u ∈ H01 (Ω), λ 1 k∇uk22 6 k∆uk22 for all u ∈ H01 (Ω) ∩ H 2 (Ω), λ kuk22 6

and 1/λ is the best constant. 5.4 Consider problem (

−∆u = u(a − u)2 (1 − u)

in Ω,

∂n u = 0

on ∂Ω,

where 0 < a < 1 is a constant. Discuss local stabilities of 0, a and 1. 5.5 Assume that

∂f1 ∂f2 > 0 and > 0. Let (u1s , u2s ) be a solution of ∂u2 ∂u1  −∆u1 = f1 (u1 , u2 )      

in Ω,

−∆u2 = f2 (u1 , u2 )

in Ω,

u1 = u2 = 0

on ∂Ω.

Denote aij =

∂fi (u1s , u2s ), i, j = 1, 2. ∂uj

Let λ0 be the principle eigenvalue of  ∆φ1 + a11 φ1 + a12 φ2 = λφ1   

in Ω,

∆φ2 + a21 φ1 + a22 φ2 = λφ2

in Ω,

  

φ1 = φ2 = 0

on ∂Ω,

Exercises  129

and φ = (φ1 , φ2 ) be the corresponding eigenfunction with φ1 , φ2 > 0 in Ω. (1) Prove that (u1s , u2s ) is locally stable if λ0 < 0, while it is unstable if λ0 > 0. ∂f1 ∂f2 (2) When 6 0 and 6 0, what kind of conclusion can we draw? ∂u2 ∂u1 5.6 Prove Lemma 5.5 under the condition that ψ is uniformly continuous in [a, ∞). 5.7 Let h : [0, ∞) → R be uniformly continuous, and lim

Z

t→∞ 0

t

h(s)ds exist. Prove

lim h(t) = 0.

t→∞

5.8 Prove Theorem 5.12. 5.9 Assume ui0 ∈ Wp2 (Ω) for some p > 1 + n/2 and i = 1, 2, u30 ∈ C γ (Ω) for some 0 < γ < 1, and u10 , u20 > 0, u30 > 0 in Ω. Using the space C(QT0 ) instead of Lp (QT0 ), and imitating Theorem 4.1 to prove that problem (5.29) has a unique local solution (u1 , u2 , u3 ) satisfying u1 , u2 ∈ Wp2,1 (QT ) and u3 ∈ C α, 1 (QT ) for some 0 < T < ∞ and 0 < α < min{γ, 2 − (n + 2)/p}. 5.10 Continue to study problem (5.39) and assume ac < 1. Prove that if k < bc2 + 2ack then the positive constant equilibrium solution (˜ u, v˜) =   c(a + b˜ v ) c(1 − ac) , is globally asymptotically stable by using the Lyapunov k k + bc2 functional method. 5.11 Give the proof of Lemma 5.14. 5.12 Consider a prey-predator model with a transmissible disease in the prey population    ut − d1 ∆u = au(1 − u − v) − uv − muw,     v − d2 ∆v = uv − bvw − cv,    t

w − d ∆w = θmuw − θbvw − kw,

t 3     ∂n u = ∂n v = ∂n w = 0,     

u = u0 (x), v = v0 (x), w = w0 (x),

x ∈ Ω, t > 0, x ∈ Ω, t > 0, x ∈ Ω, t > 0,

(P5.1)

x ∈ ∂Ω, t > 0, x ∈ Ω, t = 0,

where coefficients are positive constants, u0 , v0 , w0 ∈ Wp2 (Ω) and u0 , v0 , w0 > 0, 6≡ 0. (1) Prove that if c < 1 then (P5.1) has a non-negative constant equilibrium  a(1 − c) solution E ∗ = c, , 0 which is called the predator-free equilibrium a+1 solution. (2) Assume c < 1 and mc + ab(1 − c) < k/θ. Take advantage of the Lyapunov functional method to prove that the predator-free equilibrium solution E ∗ is globally asymptotically stable.

130  5. Stability Analysis

5.13 Give details of the proof of Theorem 5.23. 5.14 Consider a Lotka-Volterra model of a two-predator and one-prey system    vw   , u − d ∆u = u −1 + 1  t  u+v        buw   vt − d2 ∆v = v −a + ,   u+v   (1 + b)uv   w − d ∆w = w r − w − ,  t 3   u+v     ∂n u = ∂n v = ∂n w = 0,    

u = u0 (x), v = v0 (x), w = w0 (x),

x ∈ Ω,

t > 0,

x ∈ Ω,

t > 0,

x ∈ Ω,

t > 0,

(P5.2) x ∈ ∂Ω, t > 0, x ∈ Ω,

t = 0,

where coefficients are positive constants, u0 , v0 , w0 ∈ Wp2 (Ω) and u0 , v0 , w0 > 0 in Ω. (1) Utilize Theorem 2.14 to prove that the solution (u, v, w) of (P5.2) exists globally and is uniformly bounded. (2) Prove that (P5.2) has a positive constant equilibrium solution if and only if rb > a + b, and in this case the positive constant equilibrium solution is uniquely given by u˜ = (a + b)

rb − (a + b) rb − (a + b) a+b , v˜ = (a + b) , w˜ = . b2 (1 + b) ab(1 + b) b

(3) Use the Lyapunov functional method to prove that (˜ u, v˜, w) ˜ is globally asymptotically stable. 5.15 Consider a predator-prey model with hyperbolic mortality and nonlinear prey harvesting  suv bu   ut − d1 ∆u = u(1 − u) − − ,    a+u ρ+u     uv rv 2

vt − d2 ∆v =

a+u      ∂ u = ∂ v = 0, n n    

−

1 + rv

,

u(x, 0) = u0 (x), v(x, 0) = v0 (x),

x ∈ Ω,

t > 0,

x ∈ Ω,

t > 0,

(P5.3)

x ∈ ∂Ω, t > 0, x ∈ Ω,

where s, a, b, ρ and r are positive constants, u0 , v0 ∈ Wp2 (Ω) and u0 , v0 > 0, 6≡ 0. (1) Prove that if b < ρ < a then (P5.3) has a unique positive constant equilibrium solution (u∗ , v ∗ ). (2) Assume b < ρ < a and s < ra2 , b < ρ(1 − s/(ra2 )), ρ > 1, a(1 − ρ) + ρ < b. Use the iteration method to prove that (u∗ , v ∗ ) is globally asymptotically stable.

CHAPTER

6

Global Existence and Finite Time Blowup

Blowup phenomena occur in various types of nonlinear evolution equations such as the Schr¨odinger equations, hyperbolic equations as well as parabolic equations. The phenomenon of finite time blowup has important applications in many engineering fields. For global existence and finite time blowup of solutions, there are some natural questions: • will the blowup occur in finite time? • how to determine conditions for the global existence and finite time blowup of solutions? • is there a critical exponent and how to determine it? • when the blowup occurs in finite time, where are blowup points, and can the blowup occur in a region? • when the blowup occurs in finite time, how large is the blowup time? • what is the profile of solution near blowup time? The global existence and finite time blowup of solutions for nonlinear parabolic equations and systems have been studied in numerous papers. Survey papers [63, 33, 11, 18] are the good starting point to get a sense of current development of global existence and finite time blowup theory. There are several books (e.g., [88, 96, 37, 93, 48]) with very detailed and extensive study on global existence and finite time blowup theory. These are very nice reference books for experts and researchers who want to quote results in this field. In this chapter, we shall introduce some basic concepts and methods to deal with global existence and finite time blowup of solutions to parabolic differential equations. We first present simple comparison methods. Some early papers also appeared in the 1970s, e.g., [101, 45, 102, 10, 58, 9]. Then we introduce a simple eigenvalue method from [55, 32]. We also cover the energy method from [15] and concavity method from [64, 65, 62]. Finally, illustrated by specific examples, we systematically present critical exponents, boundary blowup and estimates of blowup rates. 131

132  6. Global Existence and Finite Time Blowup

Throughout this chapter, we always assume that Ω is of class C 2 , and initial data u0 , v0 ∈ Wp2 (Ω) ∩ C(Ω) with p > 1 satisfy the corresponding compatibility conditions.

6.1

SOME BASIC METHODS

Consider the problem  u − ∆u = f (u),    t

x ∈ Ω,

a∂n u + bu = 0,

x ∈ ∂Ω, t > 0,

  

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

x ∈ Ω,

t > 0, (6.1)

where a, b > 0 are constants with a + b > 0, f ∈ C 1 and u0 > 0 in Ω. We take n

o

n

o

n

o

σ < min 0, inf u0 (x) , A > max 0, sup u0 (x) , Ω

n

β < min 0,

min

2σ6u62A

Ω

f (u) , B > max 0,

o

max f (u) .

2σ6u62A

Then u¯ = Bt + A and u = βt + σ are, respectively, an upper solution and a lower solution of (6.1) in [0, T ] so long as T 6 min{|σ/β|, A/B}. Thus problem (6.1) has a unique solution in Ω × [0, T ] by the upper and lower solutions method. Repeating this process we can prove that there exists 0 < Tmax 6 ∞ for which (6.1) has a unique solution u defined in Ω × [0, Tmax ), and Tmax < ∞ implies − lim supt→Tmax maxΩ |u(·, t)| = ∞. Such a Tmax is called the maximal existence time. When Tmax = ∞, we call u a global solution of (6.1), or we say that solution u of (6.1) exists globally; when Tmax < ∞, we say that u blows up in finite time and such a finite time Tmax is called the blowup time. This phenomenon is also called blowup in finite time. In this section we shall introduce four methods dealing with the finite time blowup. 6.1.1

Comparison method

Let us start with a comparison method. We first study problem (3.5). For convenience, we rewrite it here  u − ∆u = c(eau − 1),    t   

x ∈ Ω,

t > 0,

∂n u + bu = 0,

x ∈ ∂Ω, t > 0,

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

x ∈ Ω,

where u0 > 0 in Ω, a, c and b are constants with a, c > 0 and b > 0. Theorem 6.1 ([124]) Let (λ1 , φ1 ) be the first eigen-pair of problem (3.6). If ac > λ1 , u0 (x) > δφ1 (x) in Ω

(6.2)

6.1. Some basic methods  133

for some constant δ > 0, then there exists 0 < Tmax < ∞ for which (6.2) has a unique solution u defined in Ω × [0, Tmax ), and Tmax 6

1 , u(x, t) > δ(1 − γt)−1 φ1 (x), and γ

where

lim max u(·, t) = ∞,

− t→Tmax

Ω

1 γ = cδa2 min φ1 (x). 2 Ω

This indicates that u blows up in finite time. Proof. We first look for a lower solution which blows up in finite time. Take u = p(t)φ1 (x) with p(t) > 0. To our purpose, we require (∂t − ∆)p(t)φ1 (x) = [p0 (t) + λ1 p(t)]φ1 (x) 6 c(eap(t)φ1 (x) − 1), p(0) 6 δ. Since 1 1 c(eapφ1 − 1) > capφ1 + ca2 p2 φ21 > [cap + ca2 p2 min φ1 (x)]φ1 , 2 2 Ω it is enough to demand that p(t) > 0 satisfies 1 p0 (t) + (λ1 − ca)p 6 ca2 p2 min φ1 (x), p(0) 6 δ. 2 Ω As λ1 − ca 6 0, it suffices to take p(t) as the unique solution of 1 p0 = ca2 p2 min φ1 (x), p(0) = δ, ¯ 2 Ω that is, δ , 1 − γt

p(t) =

t ∈ [0, 1/γ).

Therefore, u = p(t)φ1 (x) is a lower solution of (6.2) which blows up in finite time, and so does the solution of (6.2) by the comparison principle.  In the following, we study an initial-boundary value problem of semilinear equation with nonlinear boundary condition  u − ∆u = up ,    t q

  

x ∈ Ω, t > 0,

∂n u = u ,

x ∈ ∂Ω, t > 0,

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

x ∈ Ω,

(6.3)

where p, q are positive constants, and u0 > 0 in Ω. Theorem 6.2 ([105]) Assume max{p, q} 6 1. Then every solution of (6.3) exists globally.

134  6. Global Existence and Finite Time Blowup

Proof. We seek a global upper solution u¯ of (6.3). Let h(x) be a solution of ∆h = k =

|∂Ω| |Ω|

in Ω,

∂n h = 1 on ∂Ω.

Without loss of generality, we assume h(x) > 0 in Ω. Define ` = max |∇h(x)|2 , r = 1 + max u0 (x), σ = r(k + r` + 1), Ω

Ω

and set u¯ = reσt+rh(x) . We then find that u¯ satisfies  u¯ > ∆¯ u + u¯p ,    t q

  

x ∈ Ω, t > 0,

∂n u¯ > u¯ ,

x ∈ ∂Ω, t > 0,

u¯(x, 0) > u0 (x),

x ∈ Ω,

and certainly u¯ is a desired upper solution which exists globally in time t.



Theorem 6.3 ([105]) If max{p, q} > 1, then the solution u of (6.3) blows up in finite time. Proof. We first consider the case p > 1. Take a > 0 so large that a−1/(p−1) 6 minΩ u0 (x), and set u = [a − (p − 1)t]−1/(p−1) . Then u satisfies  u = ∆u + up ,    t

x ∈ Ω, t > 0,

∂n u 6 uq ,

x ∈ ∂Ω, t > 0,

  

u(x, 0) 6 u0 (x),

x ∈ Ω.

Therefore, u > u and u blows up in finite time. Now, we handle the case q > 1. Because Ω is bounded, we can take d > 0 suitable large such that |x1 | < d for all x ∈ Ω. Set min u0 (x) = 2ε > 0, g(s) = [ε1−q − (q − 1)s]−1/(q−1) . Ω

Then g 0 (s) = g q (s) and g(0) = ε. Choose s0 > 0 satisfying g(s0 ) = 2ε. Denote c = min{1, s0 /(2d)}, σ = c2 qεq−1 and u(x, t) = g(σt + c(x1 + d)). Then u blows up in finite time, and u(x, 0) = g(c(x1 + d)) 6 g(2cd) 6 g(s0 ) = 2ε 6 u0 (x), x ∈ Ω.

6.1. Some basic methods  135

A straightforward computation yields ut = σg 0 , ux1 = cg 0 , ux1 x1 = c2 g 00 = c2 qg q−1 g 0 and uxj = uxj xj = 0 when j > 1. Thus we have ∂n u 6 cg 0 = cg q = cuq 6 uq , x ∈ ∂Ω, t > 0, ut − ∆u = σg 0 − c2 qg q−1 g 0 6 g 0 (σ − c2 qεq−1 ) = 0, x ∈ Ω, t > 0. Above arguments show that u is a lower solution of (6.3), and the proof is complete.  6.1.2

Kaplan’s first eigenvalue method

Here we are going to introduce the first eigenvalue method which was introduced in 1963 by Kaplan [55]. As one will see from the proof, it is a very simple method and yet applied to a large class of equations. We first study problem (6.2). Let (λ1 , φ1 ) be the first eigen-pair of eigenvalue problem (3.6). From Theorem 6.1, we have known that if ac > λ1 and u0 (x) > 0, 6≡ 0, then the unique solution of (6.2) blows up in finite time. In this part, we will prove that even if ac < λ1 , its solution may blow up in finite time. Theorem 6.4 Assume ac < λ1 . Then solution u of (6.2) will blow up in finite time for large initial datum u0 (x). Z

Proof. Normalize φ1 by

φ1 (x) = 1. Multiplying the first equation of (6.2) by φ1 Ω

and integrating the result over Ω, we get Z

Z

Z

ut φ1 + λ1

uφ1 = c

Ω

Ω

au

(e Ω

− 1)φ1 > c

Z Ω

a2 u2 au + 2

!

φ1 .

Z

uφ1 . Then, in view of the H¨older inequality, it can be inferred that h(t)

Set h(t) = satisfies

Ω

h0 (t) + λ1 h(t) > ach(t) + Z

h(0) =

ca2 2 h (t), t > 0, 2

u0 (x)φ1 > 0. Ω

It is not hard to show that h(t) blows up in finite time provided that h(0) is suitably large, and so does u.  Next, we consider a special case of (6.1) in the form of  u − ∆u = f (u),    t

x ∈ Ω,

u = 0,

x ∈ ∂Ω, t > 0,

  

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

x ∈ Ω,

t > 0, (6.4)

where f ∈ C 1 and uZ0 > 0 in Ω. Let (λ, φ) be the first eigen-pair of the eigenvalue problem (1.33) with

φ(x) = 1. Then λ > 0 and φ > 0 in Ω. Ω

136  6. Global Existence and Finite Time Blowup

Theorem 6.5 Assume that there exist positive constants a, b and p > 1 such that f (s) > asp − bs for all s > 0. Let u be the unique solution of (6.4). If 

Z

u0 (x)φ(x) > Ω

λ+b a

1/(p−1)

,

then u blows up in finite time. Proof. Firstly, u > 0. The function Z

g(t) = Ω

u(x, t)φ(x) > 0

and satisfies Z

0

Z

g (t) =

ut φ =

[∆u + f (u)]φ

ZΩ

=

ZΩ

u∆φ +

f (u)φ

Ω

Ω

Z

= −λg(t) +

f (u)φ ZΩ

> −λg(t) +

[aup − bu]φ

Ω

= −(λ + b)g(t) + a

Z

φup .

Ω

It follows that Z

g(t) 6

φup

1/p Z

(p−1)/p

Z

=

φ

Ω

Ω

φup

1/p

Ω

according to the H¨older inequality, and hence g 0 (t) > g(t)[ag p−1 (t) − (λ + b)]. − Since ag p−1 (0) > λ + b, the above inequality implies limt→Tmax g(t) = ∞ for some 0 < Tmax < ∞. Consequently, u blows up in finite time. 

6.1.3

Energy method Z

In this subsection, we continue to study problem (6.4). Set F (u) =

Lemma 6.6

1 2

Z Ω

|∇u(x, t)|2 −

f (s)ds, and 0

define an energy functional with the form E(u(t)) =

u

Z

F (u(x, t)). Ω

Let u be the unique solution of (6.4) and T > 0. If

u ∈ C((0, T ), C(Ω)) ∩ C((0, T ), H01 (Ω) ∩ H 2 (Ω)) ∩ C 1 ((0, T ), L2 (Ω)), then

Z tZ

(∆u + f (u))uτ + E(u(t)) = E(u(s)) for all 0 < s < t < T. s

Ω

6.1. Some basic methods  137

Proof. Based on the denseness, it can be assumed that u ∈ C 1 ([s, t], H01 (Ω) ∩ H 2 (Ω)). Accordingly, we have Z

Z

(−∇u · ∇ut + f (u)ut ) = −

(∆u + f (u))ut = Ω

Ω

d E(u(t)). dt

Integration of the above equation from s to t derives the desired result. Lemma 6.7



Let u be the unique solution of (6.4) defined in (0, T ). Then d dt Z tZ s

Z

u2 + 2

Ω

Z

|∇u|2 = 2

Z

Ω

uf (u),

(6.5)

Ω

u2τ + E(u(t)) = E(u(s)) for all 0 < s < t < T.

Ω

(6.6)

Proof. Multiplying the first equation of (6.4) by u and integrating the result over Ω, we obtain (6.5) immediately. The equation (6.6) can be deduced from Lemma 6.6 and (6.4).  Theorem 6.8 ([15]) Suppose that there are positive constants K and ε such that uf (u) > (2 + ε)F (u) for all |u| > K. Set η = min|u|6K uf (u) and γ = max|u|6K F (u). If u0 ∈ H01 (Ω) and (2 + ε)E(u0 ) < |Ω|[η − (2 + ε)γ],

(6.7)

then the unique solution u of (6.4) blows up in finite time. Z

u2 (x, t)dx. Then by (6.5),

Proof. Write g(t) = Ω

g 0 (t) = −2

Z

|∇u|2 + 2

Z {|u|6K}

Ω

> −2

Z

Z

uf (u) + 2

2

|∇u| + 2

uf (u) {|u|>K}

Z

Z

uf (u) + 2(2 + ε) {|u|6K}

Ω

F (u).

(6.8)

{|u|>K}

As u0 ∈ H01 (Ω), we can take s = 0 in (6.6) and get Z

F (u) = −

{|u|>K}

Z

F (u) + {|u|6K}

1 2

Z Ω

|∇u|2 − E(u0 ) +

Z tZ 0

Ω

u2t .

(6.9)

As η 6 0 = uf (u)|u=0 = F (0) 6 γ, by the definitions of η and γ one has Z

2 {|u|6K}

uf (u) − 2(2 + ε)

Z {|u|6K}

F (u) > 2|Ω|[η − (2 + ε)γ].

(6.10)

138  6. Global Existence and Finite Time Blowup

It follows from (6.8)–(6.10) that g 0 (t) > 2(2 + ε)

Z tZ 0

Ω

u2t + ε

Z

|∇u|2 + 2 |Ω|[η − (2 + ε)γ] − (2 + ε)E(u0 ) . 



Ω

This, together with the assumption (6.7), deduces Z tZ

0

u2t

g (t) > 2(2 + ε) 0

Z

Set h(t) =

Ω

Z

Z tZ

2

+ε Ω

|∇u| > 2(2 + ε)

0

Ω

u2t .

(6.11)

t

g(s)ds. By using the Cauchy-Schwarz inequality we have 0 0

t

Z

0

h (t) − h (0) =

Z tZ

0

g (s) = 2 0

uut 0

Z t Z

62

0

u2

Ω

1/2 Z t Z

Ω

= 2h1/2 (t)

0

Z t Z 0

Ω

u2t

Ω

u2t

1/2

1/2

.

This combined with (6.11) allows us to yield Z tZ

00

h(t)h (t) > 2(2 + ε)h(t)

0

Ω

u2t > (1 + ε/2)(h0 (t) − h0 (0))2 .

(6.12)

On the contrary, we assume that u exists globally. It follows from (6.11) that h00 (t) = g 0 (t) > 0. Consequently, h0 (t) > h0 (t1 ) > h0 (0) for every t > t1 > 0, and then h0 (0) < (1 − σ)h0 (t) for all t > t1 and h(t) → ∞ as t → ∞, where 0 < σ < 1. In view of (6.12), it declares that h(t)h00 (t) > (1 + ε/2)σ 2 (h0 (t))2 for all t > t1 , i.e., 0 ε h00 (t) 2 h (t) > 1 + σ , t > t1 . h0 (t) 2 h(t)





It yields that h0 (t) → ∞ since h(t) → ∞. Hence, there exists t2 > t1 so that 

ε ε [h0 (t) − h0 (0)]2 > 1 + (h0 (t))2 for t > t2 . 2 4 



1+



Combining this with (6.12), we get ε (h0 (t))2 , t > t2 . h(t)h (t) > 1 + 4 00





Let η(t) = h−ε/4 (t), then η 00 (t) 6 0 for all t > t2 .

(6.13)

Obviously, limt→∞ η(t) = 0 since limt→∞ h(t) = ∞. Then there exists t0 > t2 for which η 0 (t0 ) < 0. According to (6.13), we have η 0 (t) 6 η 0 (t0 ) < 0 for all t > t0 , and furthermore η(t) → −∞ as t → ∞. This is a contradiction. Therefore, u must blow up in finite time. 

6.1. Some basic methods  139

6.1.4

Concavity method

We first state a lemma which is a basic conclusion of convex functions and its proof will be omitted. Let J ∈ C 2 ([0, T )) ∩ C([0, T ]), J(0) > 0, J 0 (0) < 0, and J 00 (t) 6 0 in

Lemma 6.9 [0, T ). If

T > −J(0)/J 0 (0), then there exists T∗ : 0 < T∗ 6 −J(0)/J 0 (0) such that J(T∗ ) = 0. Theorem 6.10 ([105]) Suppose f ∈ C 1 (R) and f (u) > 0 for u > 0. Let u0 ∈ H01 (Ω) ∩ H 2 (Ω), and u0 > 0, 6≡ 0 and satisfy ∆u0 + f (u0 ) > 0, 6≡ 0 in Ω.

(6.14)

We further assume that there exists ψ ∈ C 1 ([0, ∞)) ∩ C 3 (R+ ) such that (i) ψ(0) = ψ 0 (0) = 0, ψ(u) > 0, 6≡ 0, ψ 00 (u) > 0, ψ 000 (u) > 0 and ψ 0 (u) > 0 for all u > 0; (ii) ψ 0 (u)f 0 (u) − ψ 00 (u)f (u) > 0 for all u > 0; (iii) there exists θ > 0 so that ψ(u)ψ 00 (u) >

θ+1 0 2 2 (ψ (u))

for u > 0.

If u is a solution of (6.4) and satisfies u ∈ C 1 ((0, Tmax ), H01 (Ω) ∩ H 2 (Ω)) ∩ C([0, Tmax ), C(Ω)), where Tmax 6 ∞ is the maximal existence time of u, then Tmax < ∞. Proof. Firstly, by (6.14) and the maximum principle we see that u is strictly monotone increasing in t, and ut (·, t) > 0, 6≡ 0 for t ∈ (0, Tmax ). Thus u > 0, and u > 0 in Ω × (0, Tmax ). Set Z I(t) =

ψ(u(x, t)). Ω

Then I(t) > 0 for all t ∈ [0, Tmax ), and Z

0

ψ 0 (u(x, t))ut (x, t) > 0 for all 0 < t < Tmax .

I (t) = Ω

Without loss of generality, we assume I 0 (0) > 0. As u = 0 on ∂Ω, a direct calculation causes I 0 (t) =

Z

ψ 0 (u)ut =

Z

Ω

=−

ψ 0 (u)(∆u + f (u))

Ω

Z

00

Z

2

ψ (u)|∇u| + Ω

Z

Z

0

=2

ψ (u)ut + Ω

00

Z

2ψ

Ω

Z

>2

Ω

00

2

ψ (u)|∇u| −

Z

Ω



I (t) =

ψ 0 (u)f (u)

Ω

00

(u)u2t

Ω 0

0

+ [ψ (u)f (u) − ψ (u)f (u)]ut + ψ 000 (u)|∇u|2 ut

ψ 00 (u)u2t , 0 6 t < Tmax .

00

ψ 0 (u)f (u), 0 6 t < Tmax ,

140  6. Global Existence and Finite Time Blowup

Define J(t) = I −θ (t) for t ∈ [0, Tmax ), where θ is given by the condition (iii). Clearly, J(0) = I −θ (0) > 0, J 0 (0) = −θI −(θ+1) (0)I 0 (0) < 0, and J 00 (t) = θI −(θ+2) (t)[(θ + 1)(I 0 (t))2 − I(t)I 00 (t)], 0 6 t < Tmax . Applying the Schwarz inequality and the condition (iii), we obtain Z

00

I(t)I (t) > >

Z

ψ(u) Ω

Ω

Z q

2ψ 00 (u)u2t

2ψ(u)ψ 00 (u)u

2 t

Ω

> (θ + 1)(I 0 (t))2 , 0 6 t < Tmax . Hence, J 00 (t) 6 0 for all 0 6 t < Tmax . By means of Lemma 6.9, if Tmax > −J(0)/J 0 (0) = I(0)/(θI 0 (0)),

(6.15)

then there exists T∗ 6 I(0)/(θI 0 (0)) such that J(T∗ ) = 0, i.e., Z

lim−

t→T∗

ψ(u(x, t)) = ∞,

Ω

which implies limt→T∗− ku(·, t)k∞ = ∞. This contradicts (6.15), and so Tmax 6 I(0)/(θI 0 (0)) < ∞. 

6.2

A SYSTEM WITH DIRICHLET BOUNDARY CONDITIONS

By Theorem 6.5 we see that the unique solution of  u − ∆u = up ,    t   

x ∈ Ω, t > 0,

u = 0,

x ∈ ∂Ω, t > 0,

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

x∈Ω

(6.16)

blows up in finite time when u0 (x) is suitably large. However, the solution u of (6.16) exists globally and limt→∞ u(x, t) = 0 uniformly in Ω when u0 (x) is suitably small. In fact, let (λ, φ) be the first eigen-pair of problem (1.33). Take 0 < ε  1 satisfying εp−1 maxΩ φp−1 (x) 6 λ/2. Then u¯ = εe−λt/2 φ(x) is an upper solution of (6.16) when u0 (x) 6 εφ(x), and as a consequence, solution u of (6.16) exists globally and limt→∞ u(x, t) = 0 uniformly in Ω. In this section we discuss the following problem  ut − ∆u = uρ v p ,       vt − ∆v = uq v β ,

x ∈ Ω,

t > 0,

x ∈ Ω,

t > 0,

 u = v = 0,     

x ∈ ∂Ω, t > 0,

u = u0 (x), v = v0 (x),

x ∈ Ω,

(6.17)

t = 0,

where p, q, ρ, β are non-negative constants, u0 , v0 > 0 in Ω. It will be shown that (6.17) exhibits very different properties from (6.16).

6.2. A system with Dirichlet boundary condition  141

6.2.1

Critical exponents

The content of this part refers to [106]. Theorem 6.11 (Critical exponents) Let (λ, φ) be the first eigen-pair of (1.33). (1) Assume ρ, β 6 1

and pq 6 (1 − ρ)(1 − β).

(6.18)

Then every solution of (6.17) exists globally for any initial data. If (6.18) does not hold, then all solutions of (6.17) exist globally for small initial data, and blow up in finite time for large initial data. (2) Suppose that either ρ > 1, p > 0, q = 0, β = 1, λ < 1, ρ 6 1 + p(1 − λ)/λ,

(6.19)

β > 1, q > 0, p = 0, ρ = 1, λ < 1, β 6 1 + q(1 − λ)/λ.

(6.20)

or

Furthermore, when ρ = 1 + p(1 − λ)/λ in (6.19), or β = 1 + q(1 − λ)/λ in (6.20), we also assume λ < 2/3. Then, for any initial data u0 , v0 > 0, 6≡ 0, the corresponding solution of (6.17) blows up in finite time. (3) If one of (6.19) and (6.20) is not valid, then every solution of (6.17) exists globally for small initial data. Proof. Step 1. Assume that (6.18) holds. When pq = 0, the conclusion is obvious. 1−ρ1−β > 1. There exist 0 < m, l < 1 such that When pq > 0, we have p q (1 − ρ)/p > l/m, (1 − β)/q > m/l.

(6.21)

Denote k = 1/m + 1/l, and let w be the unique global solution of the linear problem  wt − ∆w = kw,   

x ∈ Ω,

x ∈ ∂Ω, t > 0,

w = 1,

  

1/m

w = 1 + u0

t > 0,

1/l

(x) + v0 (x),

x ∈ Ω,

(6.22)

t = 0.

As is well known, w exists globally and w > 1. Define u¯ = wm and v¯ = wl . In view of (6.21) and (6.22), we can verify u¯t > ∆¯ u + u¯ρ v¯p ,

v¯t > ∆¯ v + u¯q v¯β for all x ∈ Ω, t > 0.

Obviously, u¯, v¯ > 0 on ∂Ω × [0, ∞) and u¯(x, 0) > u0 (x), v¯(x, 0) > v0 (x) in Ω. The comparison principle gives (u, v) 6 (¯ u, v¯), and so (u, v) exists globally. Here, (u, v) 6 (¯ u, v¯) means that u 6 u¯ and v 6 v¯.

142  6. Global Existence and Finite Time Blowup

Step 2. Assume that (6.19) holds and u0 , v0 > 0, 6≡ 0. Then v satisfies    vt − ∆v = v, 

x ∈ Ω,

v = 0,

x ∈ ∂Ω, t > 0,

  

v(x, 0) = v0 (x),

x ∈ Ω.

t > 0,

According to the maximum principle and Lemma 3.7, we may think that v0 > εφ in Ω for some ε > 0. Therefore, v > εe(1−λ)t φ, and of course  u − ∆u > εp ep(1−λ)t φp uρ ,    t   

x ∈ Ω,

t > 0,

u = 0,

x ∈ ∂Ω, t > 0,

u(x, 0) = u0 (x) > 0, 6≡ 0,

x ∈ Ω.

(6.23)

When ρ < 1+p(1−λ)/λ, we choose Ω0 b Ω so that the first eigenvalue λ0 of −∆ in Ω0 with the homogeneous Dirichlet boundary condition satisfies ρ < 1 + p(1 − λ0 )/λ0 . Denote a = minΩ0 φ(x). Then a > 0, and ut − ∆u > εp ap eλ0 (ρ−1)t uρ , x ∈ Ω0 , t > 0.

(6.24)

Z

Let φ0 be the eigenfunction corresponding to λ0 and satisfy

φ0 (x) = 1. MultiΩ0

plying (6.24) by φ0 and integrating the result over Ω0 firstly, and using the H¨older inequality secondly, we have d dt

Z Ω0

uφ0 > −λ0

Z

p p λ0 (ρ−1)t

uφ0 + ε a e

uφ0

Ω0





Z Ω0

,

Ω0

which implies d eλ0 t dt

ρ

Z



uφ0 > εp ap eλ0 t

ρ

Z

uφ0

.

Ω0

Z

Z

Due to the fact that

u(x, 0)φ0 (x) > 0 and ρ > 1, it derives that Ω0

u(x, t)φ0 (x) Ω0

blows up in finite time, and so does u. If ρ = 1 + p(1 − λ)/λ and λ < 2/3, then (ρ − 1 − p)/(ρ − 1) > −1. Since ∂n φ < 0 on ∂Ω, it is clear that Z φ(ρ−1−p)/(ρ−1) (x) < ∞.

b := Ω

The H¨older inequality gives Z

Z Ω

uφ 6

ρ 1+p

1/ρ Z

u φ

φ

Ω (ρ−1)/ρ

(ρ−1−p)/(ρ−1)

(ρ−1)/ρ

Ω

Z

ρ 1+p

=b

1/ρ

u φ

.

Ω

Multiplying (6.23) by φ and integrating the result over Ω, from above we have d dt

Z Ω

uφ > −λ

Z

p 1−ρ λ(ρ−1)t

uφ + ε b Ω

Z

e

ρ

uφ Ω

.

6.2. A system with Dirichlet boundary condition  143

Similar to the above, it can be deduced that u blows up in finite time. When (6.20) holds, the proof is similar. Step 3. If one of (6.18), (6.19), and (6.20) does not hold, we first prove that each solution (u, v) of (6.17) exists globally for small initial data. To this end, let ψ be the unique solution of −∆ψ = 1 in Ω, ψ = 0 on ∂Ω. Then 0 6 ψ 6 C for some constant C > 0. Case 1: ρ > 1 and q > 0. We can find positive constants a, b satisfying a > aρ bp (1 + C)ρ+p and b > aq bβ (1 + C)q+β .

(6.25)

Then the pair (¯ u, v¯) = (a(1 + ψ), b(1 + ψ)) is an upper solution of (6.17) as long as u0 6 a(1 + ψ), v0 6 b(1 + ψ). Consequently, (u, v) exists globally. When β > 1 and p > 0, the proof is similar. Case 2: q = 0 and β 6= 1. Similarly to the case 1, there exists b > 0 such that v 6 b(1 + ψ) whenever v0 6 b(1 + ψ), and then ut − ∆u 6 bp (1 + ψ)p uρ . If ρ 6 1, then u exists globally. If ρ > 1, then u¯ = a(1 + ψ(x)) satisfies u¯t − ∆¯ u > bp (1 + ψ)p u¯ρ provided that 0 < aρ−1 6 b−p (1 + C)−(p+ρ) . Of course u 6 u¯ when u0 6 a(1 + ψ), and thereby (u, v) exists globally when u0 6 a(1 + ψ) and v0 6 b(1 + ψ). Case 3: q = 0, β = 1 and λ > 1. In such a case, we have v 6 φ provided v0 (x) 6 φ(x). Similar to the case 2, it can be proved that u exists globally if u0 is properly small. Case 4: ρ > 1, p > 0, q = 0, β = 1, λ < 1 and ρ > 1 + p(1 − λ)/λ. Similar to arguments in Step 2, it can be shown that v 6 φe(1−λ)t if v0 6 φ. Therefore ut 6 ∆u + φp ep(1−λ)t uρ 6 ∆u + Cep(1−λ)t uρ . Since (ρ − 1)λ > p(1 − λ), we can find 0 < θ < λ for which (ρ − 1)θ > p(1 − λ). Choosing a > 0 small enough, by a direct computation, we know that u¯ = ae−θt φ(x) satisfies u¯t − ∆¯ u > Cep(1−λ)t u¯ρ . The comparison principle yields u 6 u¯ so long as u0 6 aφ, and hence (u, v) exists globally when u0 6 aφ and v0 6 φ. Case 5: ρ 6 1, β 6 1 and pq > (1 − ρ)(1 − β). In such a case, we can find two positive constants a and b satisfying (6.25). Similar to the proof of Case 1, (u, v) exists globally provided u0 6 a(1 + ψ) and v0 6 b(1 + ψ). For other cases, their proofs can be argued similarly. Step 4. It will be shown that if (6.18) does not hold, then each solution (u, v) of (6.17) blows up in finite time for large initial data.

144  6. Global Existence and Finite Time Blowup

Case 1: ρ > 1. It is well known that v > e−λt φ whenever v0 > φ. Accordingly, u satisfies ut − ∆u > e−λpt φp uρ .

(6.26)

If p > ρ − 1, then θ := p/(ρ − 1) > 1. Multiplying (6.26) by φθ and integrating the result over Ω one has d dt

Z

θ

Ω

uφ > −λθ > −λθ

Z

Z

θ

uφ + θ(θ − 1)

uφ

Ω

Z

θ−2

2

|∇φ| + e

Z

−λpt

Ω 1−ρ −λpt

θ

uφ + |Ω|

uρ φp+θ

Ω

Z

e

θ

ρ

uφ

Ω

,

Ω

and consequently, d e−λθt dt 

Z

θ



−λθt

> −2λθe

uφ Ω

Z

θ

uφ + |Ω|

1−ρ



−λθt

θ

e

Ω

Z

Z

ρ

uφ

.

Ω

u0 (x)φθ (x) is sufficiently large.

Thereby, u blows up in finite time when Ω

If p < ρ − 1, multiplying (6.26) by φ and integrating the result over Ω, we have d dt

Z Ω

uφ > −λ

Z

Z

uφ + Ω

φ(ρ−1−p)/(ρ−1)

1−ρ

e−λpt

Ω

Z

ρ

uφ

.

Ω

Z

Thus, u blows up in finite time so long as

u0 (x)φ(x) is sufficiently large. Ω

For the case when β > 1, the proof is similar to the case ρ > 1. Case 2: ρ, β 6 1 and pq > (1 − ρ)(1 − β). We choose m, l > 1 satisfying (p + 1 − β)/(q +1−ρ) = m/l. Then (1−ρ)/p < l/m, (1−β)/q < m/l and σ := m(ρ−1)+lp = l(β − 1) + mq > 0. Take k = min{1/m, 1/l}, and write w as the unique non-negative solution of  w − ∆w = kw1+σ ,    t   

x ∈ Ω0 ,

t > 0,

w = 0,

x ∈ ∂Ω0 , t > 0,

w(x, 0) = w0 (x) > 0,

x ∈ Ω0 ,

where Ω0 b Ω. Then w blows up in finite time Tmax provided that initial datum w0 is suitable large. Set u = wm and v = wl . After careful calculations we obtain ∆u + uρ v p = mwm−1 ∆w + m(m − 1)wm−2 |∇w|2 + wmρ+pl 1 > mwm−1 (∆w + w1+σ ) m > mwm−1 wt = ut ,

x ∈ Ω0 , 0 < t < Tmax ,

∆v + u v > v t ,

x ∈ Ω0 , 0 < t < Tmax .

q β

Define w0 = 0 in Ω\Ω0 . If u0 > φ+w0m and v0 > φ+w0l , then the maximum principle yields u, v > 0 in Ω0 × (0, Tmax ), and then the comparison principle causes u > u and v > v in Ω0 × (0, Tmax ). Therefore, (u, v) blows up in finite time. 

6.2. A system with Dirichlet boundary condition  145

6.2.2

Blowup rate estimates

We consider a special case of (6.17), namely a problem as follows:  ut − ∆u = uρ v p ,       vt − ∆v = uq v β ,

x ∈ BR , t > 0,

 u = v = 0,     

|x| = R, t > 0,

x ∈ BR , t > 0,

u = u0 (x), v = v0 (x),

(6.27)

x ∈ BR , t = 0,

where BR = BR (0) is a ball in Rn centred at the origin with radius R > 0. Initial data u0 and v0 are continuous and non-negative functions which vanish on the boundary |x| = R. The content of this part is due to [107]. Theorem 6.12 (Blowup rate estimates) Let (u, v) be a smooth solution of (6.27), and let the following conditions hold: (i) ρ = 0 or ρ > 1, β = 0 or β > 1, and p, q > 1 with p + 1 > β and q + 1 > ρ; (ii) u0 , v0 are radially symmetric and decreasing; (iii) u and v blow up simultaneously in finite time Tmax < ∞, and ut , vt > 0. Then there are positive constants c and C such that  −θ −θ   c(Tmax − t) 6 max u(x, t) = u(0, t) 6 C(Tmax − t) ,

0 6 t < Tmax ,

−σ −σ   c(Tmax − t) 6 max v(x, t) = v(0, t) 6 C(Tmax − t) ,

0 6 t < Tmax ,

06|x|6R

06|x|6R

where θ=

p+1−β > 0, pq − (ρ − 1)(β − 1)

σ=

(6.28)

q+1−ρ > 0. pq − (ρ − 1)(β − 1)

Proof. Firstly, as u0 and v0 are radially symmetric and decreasing, so do u and v, i.e., u = u(r, t) and v = v(r, t) with r = |x| and ur 6 0, vr 6 0. Step 1. Since u and v blow up simultaneously in finite time Tmax , and ut > 0 and vt > 0, it follows that, for any 0 < t0 < Tmax , either ut (x, t0 ) 6≡ 0 or vt (x, t0 ) 6≡ 0 in BR . Otherwise, (u(x, t0 ), v(x, t0 )) is a positive equilibrium solution of (6.27), and thereby (u, v) cannot blow up in finite time. Let ϕ = ut and ψ = vt . Then we have  ϕt = ∆ϕ + ρuρ−1 v p ϕ + puρ v p−1 ψ, x ∈ BR , t0 6 t < Tmax ,       ψt = ∆ψ + quq−1 v β ϕ + βuq v β−1 ψ, x ∈ BR , t0 6 t < Tmax ,  ϕ = ψ = 0,     

ϕ(x, t0 ) > 0, ψ(x, t0 ) > 0,

|x| = R, t0 6 t < Tmax , ¯R , x∈B

¯R . The maximum principle shows that and ϕ(x, t0 ) 6≡ 0 or ψ(x, t0 ) 6≡ 0 in B ϕ > 0, ψ > 0 in BR × (t0 , Tmax ), ∂n ϕ < 0, ∂n ψ < 0 on ∂BR × (t0 , Tmax ).

146  6. Global Existence and Finite Time Blowup

Using Lemma 3.7 we can show that, for any t0 < t1 < Tmax , there exists ε > 0 such that ϕ > εuρ v p and ψ > εuq v β for t = t1 and x ∈ BR . That is, ∆u + uρ v p > εuρ v p and ∆v + uq v β > εuq v β for t = t1 , |x| 6 R. Without loss of generality, we may assume that t1 = 0, i.e., ∆u0 + uρ0 v0p > εuρ0 v0p and ∆v0 + uq0 v0β > εuq0 v0β ,

|x| 6 R.

(6.29)

Set w = ut − εuρ v p and z = vt − εuq v β . In view of that ∇u · ∇v = ur vr > 0, from the condition (i) and (6.29), we have wt − ∆w = (ut − ∆u)t − ερuρ−1 v p (ut − ∆u) − εpuρ v p−1 (vt − ∆v) +εuρ−2 v p−2 ρ(ρ − 1)|v∇u|2 + 2ρpuv∇u · ∇v + p(p − 1)|u∇v|2 



> (ut − ∆u)t − ερuρ−1 v p (ut − ∆u) − εpuρ v p−1 (vt − ∆v) = ρuρ−1 v p w + puρ v p−1 z, q−1 β

zt − ∆z > qu

q β−1

v w + βu v

z,

|x| < R, 0 < t < Tmax , |x| < R, 0 < t < Tmax

and w(x, 0) > 0, z(x, 0) > 0,

|x| < R,

w(x, t) = z(x, t) = 0, |x| = R, 0 < t < Tmax . The maximum principle asserts that w > 0 and z > 0, i.e., ut > εuρ v p and vt > εuq v β . In particular, ut (0, t) > εuρ (0, t)v p (0, t) and vt (0, t) > εuq (0, t)v β (0, t), 0 < t < Tmax .

(6.30)

On the other hand, the condition (ii) implies ∆u(0, t) 6 0 and ∆v(0, t) 6 0, i.e., ut (0, t) 6 uρ (0, t)v p (0, t) and vt (0, t) 6 uq (0, t)v β (0, t), 0 < t < Tmax .

(6.31)

Step 2. It follows from (6.30) and (6.31) that 1 εuρ−q v p−β vt (0, t) 6 ut (0, t) 6 uρ−q v p−β vt (0, t). ε

(6.32)

Integrating the right-hand side of (6.32) from 0 to t, we obtain t t 1 1 uq+1−ρ (0, s) 0 6 v p+1−β (0, s) 0 . q+1−ρ ε(p + 1 − β)

Since v(0, t) → ∞ as t → Tmax and p + 1 > β, it confirms uq+1−ρ (0, t) 6 C1 v p+1−β (0, t) whenever 0 < Tmax − t  1

(6.33)

6.3. A system coupled in equation and boundary  147

for some C1 > 0. Analogously, by the left-hand side of (6.32), we have v p+1−β (0, t) 6 C2 uq+1−ρ (0, t) whenever 0 < Tmax − t  1

(6.34)

for some C2 > 0. It then follows from (6.33), (6.34) and (6.30) that ut (0, t) > ε1 u(1+θ)/θ (0, t) and vt (0, t) > ε2 v (1+σ)/σ (0, t). Thereby, for some positive constant C, u(0, t) 6 C(Tmax − t)−θ and v(0, t) 6 C(Tmax − t)−σ for 0 < Tmax − t  1. Similarly, combining (6.33), (6.34) and (6.31) we get, for positive constant c, u(0, t) > c(Tmax − t)−θ and v(0, t) > c(Tmax − t)−σ for 0 < Tmax − t  1. Estimates (6.28) are achieved, and hence the proof is complete.



6.3 A SYSTEM COUPLED IN EQUATION AND BOUNDARY In this section, we study the following parabolic system coupled in an equation and a boundary condition  u − ∆u = v p , vt − ∆v = 0, x ∈ Ω,    t

t > 0,

  

t = 0,

q

∂n u = 0,

x ∈ ∂Ω, t > 0,

∂n v = u ,

x ∈ Ω,

u = u0 (x), v = v0 (x),

(6.35)

where p, q are positive constants; initial data u0 , v0 ∈ C 2 (Ω), and u0 , v0 > 0, 6≡ 0 in Ω which satisfy ∂n u0 = 0 and ∂n v0 = uq0 on ∂Ω. Let G(x, y, t − s) be the Green function for heat equation with the homogeneous Neumann boundary condition, i.e., for each fixed x ∈ Ω, G(x, y, t) satisfies (

Gt − ∆y G = 0,

y ∈ Ω,

∂ny G = 0,

y ∈ ∂Ω, t > 0

t > 0,

and

lim[G(x, y, t) − Γ(x − y, t)] y∈Ω = 0 in the sense of measure,

t→0

where Γ(x − y, t) =

   

1 |x − y|2 exp − 4t (4πt)n/2

  

0,

!

, t > 0,

t 6 0, (x, t) 6= (y, 0).

Then we have the following representation formulae Z tZ

Z

u(x, t) =

G(x, y, t)u0 (y)dy + Ω

Z

v(x, t) =

G(x, y, t − s)v p (y, s)dyds,

0 Ω Z tZ

G(x, y, t)v0 (y)dy + Ω

0

G(x, y, t − s)uq (y, s)dSy ds.

(6.36) (6.37)

∂Ω

Invoking above representation formulae and the contraction mapping principle, we can establish local existence for solutions of (6.35). Contents of this section refer to [19, 49, 50, 108].

148  6. Global Existence and Finite Time Blowup

6.3.1

Properties of the Green function G(x, y, t − s)

It is clear that G(x, y, t − s) = G(y, x, t − s). We fix ε > 0. Thanks to the maximum principle, G(x, y, ε) > 0 for x, y ∈ Ω, and thereby min G(x, y, ε) = c0 (ε) > 0. x,y∈Ω

By using of the maximum principle again, it deduces G(x, y, t − s) > c0 (ε), x, y ∈ Ω, t − s > ε.

(6.38)

Let w be the unique solution of  w − ∆w = 0,    t

x ∈ Ω,

∂n w = 0,

x ∈ ∂Ω, t > 0,

  

w(x, 0) = 1,

x ∈ Ω.

t > 0,

Then w = 1, and by the representation formula we have Z

1 = w(x, t) =

x ∈ Ω, t > 0.

G(x, y, t)dy, Ω

(6.39)

Now, we are going to demonstrate that there exists a constant c0 > 0 such that Z ∂Ω

G(x, y, t − s)dSy > c0 ,

x ∈ Ω, t > s > 0.

(6.40)

To this aim, we first prove that, for some small ε0 > 0, c (ε ) √0 0 6 t−s

C0 G(x, y, t − s)dSy 6 √ , t −s ∂Ω

Z

x ∈ ∂Ω, 0 < t − s 6 ε0 .

(6.41)

It is well-known that (see [38]) there exist constants C > c > 0 such that C |x − y|2 G(x, y, t − s) 6 exp −c t−s (t − s)n/2

!

, x, y ∈ Ω, 0 < t − s 6 1.

(6.42)

The upper bound in (6.41) can be deduced from (6.42). In the following, we will derive the lower bound in (6.41). It causes from the construction of G(x, y, t − s) that G(x, y, t − s) = Γ(x − y, t − s) + h(x, y, t − s), where h(x, y, τ ) solves  h = ∆y h,    τ   

y ∈ Ω,

τ > 0,

∂ny h = −∂ny Γ,

y ∈ ∂Ω, τ > 0,

h(x, y, 0) = 0,

y ∈ Ω.

(6.43)

6.3. A system coupled in equation and boundary  149

Obviously, 2(x − y) · ny |x − y|2 −∂ny Γ(x − y, τ ) = − n/2 exp − 4τ π (4τ )1+n/2

!

.

The representation formula and the estimate (6.42) lead to, for x ∈ ∂Ω, Z

|h(x, y, τ )|dSy

∂Ω

Z

6

∂Ω

Z

τ

Z

τ

Z ∂Ω

0

Z

Z

= 0

∂Ω

∂Ω

|x − z|2 |y − z|2 C|(x − z) · nz | exp − − c 4s τ −s (τ − s)n/2 s1+n/2

!

C|(x − z) · nz | |x − z|2 |y − z|2 exp − − c 4s τ −s (τ − s)n/2 s1+n/2

!

dSz dsdSy dSy dSz ds. (6.44)

Clearly, |y − z|2 exp −c τ −s ∂Ω

Z

!

dSy 6 C 0 (τ − s)(n−1)/2 .

(6.45)

As Ω is of class C 2 , we can find 0 < s0  1 and C1 > 0 such that for all x ∈ ∂Ω |(x − z) · nz |dSz 6 C1 |x0 − z 0 |2 dz 0

for z ∈ ∂Ω ∩ {|z − x| < s0 },

(6.46)

where z 0 = (z1 , . . . , zn−1 ), x0 = (x1 , . . . , xn−1 ). Consequently, Z ∂Ω∩{|z−x| √ for 0 < τ 6 1, x ∈ ∂Ω τ ∂Ω

Z

for some positive constant c. We choose ε0 > 0 small enough so that c c √ − C4 > √ for 0 < τ 6 ε0 . τ 2 τ

(6.49)

(6.50)

The lower bound in (6.41) can be deduced by (6.43) and (6.48)–(6.50). Combining (6.38) and (6.41), we conclude (6.40). 6.3.2

Critical exponents

In this subsection, we will show that every solution of (6.35) is global if and only if pq 6 1. Theorem 6.13

Assume pq 6 1. Then every solution of (6.35) exists globally.

Proof. We hope to seek a global upper solution (¯ u, v¯) of problem (6.35). Let h(x) be a solution of    ∆h = k = |∂Ω| |Ω|  

∂n h = 1

in Ω, on ∂Ω.

Without loss of generality, we may assume h(x) > 0 in Ω. Let ` = maxΩ {h(x) + |∇h(x)|}, m = max{ku0 k∞ , kv0 k∞ }, γ = mq−1 , σ = max{mp−1 epγ` , γ(k + γ`2 )/q}, and define u¯ = meσt ,

v¯ = meqσt+γh(x) .

By a series of computations we find that (¯ u, v¯) satisfies  u¯ > ∆¯ u + v¯p , v¯t > ∆¯ v,    t

x ∈ Ω,

∂n u¯ > 0, ∂n v¯ > u¯q ,

x ∈ ∂Ω, t > 0,

  

u¯(x, 0) > u0 (x), v¯(x, 0) > v0 (x),

x ∈ Ω.

t > 0,

This indicates that (¯ u, v¯) is a desired upper solution.  To establish blowup results, we need the following relationship between solution components u and v.

6.3. A system coupled in equation and boundary  151

Lemma 6.14 Let (u, v) be a non-negative solution of (6.35) with the maximal existence time Tmax 6 ∞. Then there exists a positive constant c = c(p, q, Ω) such that (1) when p, q > 1, it holds Z

Z t Z

q

∂Ω

u (x, t)dSx > c

q

p

v (y, s)dyds 0

for all 0 < t < Tmax ;

(6.51)

Ω

(2) when p > 1 > q and pq > 1, we have Z

q

∂Ω

q(1−p)

pq

Z t Z

u (x, t)dSx > ct

for all 0 < t < Tmax ; (6.52)

v(y, s)dyds 0

Ω

(3) when q > 1 > p and pq > 1, if Tmax > 1 then Z

p

Ω

p(1−q)

pq

Z t Z

for all 1 6 t < Tmax .

u(y, s)dSy ds

v (x, t) > ct

∂Ω

0

(6.53)

Proof. By using the maximum principle and Hopf boundary lemma, we know that u, v > 0 in Ω × (0, Tmax ). (1) Proof of (6.51). By means of (6.36), (6.40) and the H¨older inequality, it can be deduced that Z

uq (x, t)dSx >

∂Ω

Z t Z

Z ∂Ω

0 1−q

> |∂Ω|

= |∂Ω|1−q

G(x, y, t − s)v p (y, s)dyds

dSx

Ω

Z

Z tZ

∂Ω 0 Z t Z 0

>

q

cq0 |∂Ω|1−q

G(x, y, t − s)v (y, s)dydsdSx

Ω

v p (y, s)

q

Z

G(x, y, t − s)dSx dyds

∂Ω

Ω

Z t Z

q

p

for all 0 < t < Tmax .

v (y, s)dyds 0

q

p

Ω

(2) Proof of (6.52). Making use of the H¨older inequality and (6.39), it concludes p

Z t Z Z tZ 0

Ω

G(x, y, t − s)v p (y, s)dyds >

0

G(x, y, t − s)v(y, s)dyds

Ω

Z t Z 0

= t1−p

p−1

G(x, y, t − s)dyds

Ω tZ

Z

0

p

G(x, y, t − s)v(y, s)dyds

.

Ω

This together with (6.36) and (6.40) derives Z ∂Ω

q

q(1−p)

u (x, t)dSx > c1 t

> c2 tq(1−p) >

pq

Z t Z

Z ∂Ω

0

G(x, y, t − s)v(y, s)dyds

Z t Z

q(1−p) c2 cpq 0 t

Z

v(y, s) 0

pq

G(x, y, t − s)dSx dyds

∂Ω

Ω

pq

Z t Z

v(y, s)dyds 0

dSx

Ω

Ω

for all 0 < t < Tmax .

152  6. Global Existence and Finite Time Blowup

(3) Proof of (6.53). Note that G(x, y, t − s) is bounded when t − s > ε0 , and (6.41) holds when t − s 6 ε0 . It follows that Z tZ 0

∂Ω

G(x, y, t − s)dSy ds 6 C1 t, x ∈ Ω, 1 6 t < Tmax .

This together with (6.37), the H¨older inequality and (6.39) derives that Z Ω

Z Z t Z

p

v (x, t)dx >

pq

R R t

Z

>

G(x, y, t − s)u (y, s)dSy ds

∂Ω

0

Ω

∂Ω

0

Ω

p

q

G(x, y, t − s)u(y, s)dSy ds

R R t

G(x, y, t − s)dSy ds

∂Ω

0

> (C1 t)p(1−q) p(1−q)

Z t Z

pq

Z

G(x, y, t − s)dxdSy ds

u(y, s) ∂Ω

0

p(q−1)

Ω

pq

Z t Z

= (C1 t)

u(y, s)dSy ds ∂Ω

0

for all 1 6 t < Tmax .

Proofs are complete.



Theorem 6.15 Suppose pq > 1. Then all non-negative and non-trivial solutions of (6.35) blow up in finite time. Proof. By using the maximum principle and the Hopf boundary lemma, we see that u and v are positive. In view of the preceding lemma, we treat three cases respectively. Case 1: p, q > 1. Introduce a function Z tZ

F (t) =

v(x, s)dxds. 0

Ω

Utilizing (6.37) and (6.39), we see that F (t) satisfies Z tZ Z

G(x, y, s)v0 (y)dydxds

F (t) >

0

Ω

Ω

Z tZ

=

Z

v0 (y) 0

G(x, y, s)dxdyds

Ω

Ω

Z tZ

=

v0 (y)dyds =: c3 t. 0

(6.54)

Ω

Moreover, integrating by parts and recalling (6.51) we conclude F 00 (t) =

Z

Z

vt (x, t)dx = Z t Z

>c

uq (x, t)dSx

∂Ω

Ω

0

Ω

q

v p (y, s)dyds

> c4 tq(1−p) F pq (t).

Since F 0 (t) > 0, we may multiply the above inequality by F 0 (t) and integrate by parts, and keep p > 1 in mind, to obtain (F 0 (t))2 > c25 tq(1−p) F pq+1 (t),

6.3. A system coupled in equation and boundary  153

or, equivalently, F 0 (t) > c5 tq(1−p)/2 F (pq+1)/2 (t).

(6.55)

Choose a constant δ satisfying 1 min{pq + 1, q + 3}. 2

1 c6 t(q+1)/2−δ F δ (t), i.e., F 0 (t)F −δ (t) > c6 t(q+1)/2−δ . It follows that 



F 1−δ (t) 6 F 1−δ (η) + c7 η (q+3)/2−δ − t(q+3)/2−δ → −∞ as t → ∞. This indicates that F (t) cannot be global, so (u, v) cannot be global either. Case 2: p > 1 > q. Define F (t) as above. Recalling (6.52), the proof is essentially the same as that in Case 1, and it is omitted here. Case 3: q > 1 > p. Assume conversely that the solution (u, v) of (6.35) exists globally. Thanks to (6.40), Z

Z Z

Z

G(x, y, t)u0 (y)dSx dy

G(x, y, t)u0 (y)dydSx = ∂Ω

Z ∂Ω

Ω

Ω

u0 (y)dy, t > 0.

> c0

(6.56)

Ω

On the other hand, applying (6.40) and (6.53) successively we derive that, for t > 1, Z

Z tZ

∂Ω 0

Ω

Z tZ

p

G(x, y, t − s)v (y, s)dydsdSx > c0

0

Z

v p (y, s)dyds

Ω t

p(1−q)

pq

Z s Z

u(y, τ )dSy dτ

s

> c1

0

0

∂Ω

which, combined with (6.36) and (6.56), leads to Z ∂Ω

Z

u(x, t)dSx > a + b

t

s

p(1−q)

Z

s

pq

Z

u(y, τ )dSy dτ

0

0

∂Ω

for some positive constants a and b. Set Z tZ

H(t) = ∂Ω

0

u(y, s)dSy ds, t > 1.

Then H 0 (t) > a + b

Z 0

t

sp(1−q) H pq (s)ds, t > 1.

ds, t > 1

ds,

154  6. Global Existence and Finite Time Blowup

Integrating above inequality from 1 to t, then we assert that, with a1 = a/2 Z tZ

H(t) > a1 t + b

1

Z

= a1 t + b

τ

sp(1−q) H pq (s)dsdτ

1 t

(t − s)sp(1−q) H pq (s)ds

1 p(1−q)

Z

> a1 t + bt

1

t

(t − s)H pq (s)ds, t > 2.

Choosing τ > 2, we have H(t) > a1 τ + b1 τ p(1−q)

t

Z τ

(t − s)H pq (s)ds, τ 6 t 6 2τ

with b1 = 2p(1−q) b. Thus, by the comparison principle, H(t) > h(t) in [τ, 2τ ], where h(t) is a solution of h(t) = a1 τ + b1 τ

p(1−q)

t

Z τ

(t − s)hpq (s)ds, τ 6 t 6 2τ.

Obviously, h(t) satisfies h00 (t) = b1 τ p(1−q) hpq (t), τ 6 t 6 2τ, h(τ ) = a1 τ, h0 (τ ) = 0. Multiplying the differential equation by h0 (t) and integrating the result from τ to t we ascertain h0 (t) = b2 τ p(1−q)/2 [hpq+1 (t) − hpq+1 (τ )]1/2 with b2 = b2 τ

p

2b1 /(pq + 1), and thereby,

(p−pq+2)/2

Z

h(2τ )

= h(τ )

dz p = h(1−pq)/2 (τ ) pq+1 z − hpq+1 (τ )

Z

h(2τ ) h(τ )

dy p

1

y pq+1

−1

.

It is easy to see that Z 1

h(2τ ) h(τ )

dy p

y pq+1

Z

−1

6

1

2

dy √ + y−1

Z 2

∞

dy p

y pq+1

−1

√ Z 62+ 2

2

∞

dy y

pq+1 2

=: b3 .

Thus, we get, recalling h(τ ) = a1 τ , b2 τ (p−pq+2)/2 6 b3 h(1−pq)/2 (τ ) = b3 (a1 τ )(1−pq)/2 =: b4 τ (1−pq)/2 , which is equivalent to τ (p+1)/2 6 b4 /b2 . This is impossible for sufficiently large τ , and the proof is complete. 

6.3. A system coupled in equation and boundary  155

6.3.3

Blowup on the boundary

In this subsection, it is always assumed that (u, v) is a solution of (6.35) which blows up in finite time Tmax < ∞. We shall derive boundary blowup results. Theorem 6.16

When p, q > 1, blowup can occur only on the boundary.

Proof. The proof will be divided into three steps. Step 1. We first establish another relationship between the solution components u and v. It follows from (6.37) and (6.39) that Z Ω

v p (x, t)dx >

Z Z t Z Ω

> |Ω|

Z t Z

q

Z

u (y, s) 0

= |Ω|1−p

p

∂Ω

0 1−p

G(x, y, t − s)uq (y, s)dSy ds

∂Ω

Z t Z 0

p

G(x, y, t − s)dxdSy ds

Ω

uq (y, s)dSy ds

p

.

(6.57)

∂Ω

Step 2. Set A(t) = ku(·, t)kq, ∂Ω and B(t) = kv(·, t)kp, Ω . By virtue of (6.51) and (6.57), we have, with c = |Ω|1−p , t

A(t) > c1/q

Z

B(t) > c1/p

Z

0

0

t

B p (s)ds =: c1/q f (t), 0 6 t < Tmax , Aq (s)ds =: c1/p g(t), 0 6 t < Tmax .

As a consequence, f 0 (t) > cg p (t), g 0 (t) > cf q (t),

0 6 t < Tmax ,

(6.58)

(f g)0 (t) > cg p+1 (t) + cf q+1 (t), 0 6 t < Tmax .

(6.59)

which implies

Define µ = (p + 1)/(pq − 1) and θ = (q + 1)/(pq − 1). We shall use the idea of [74] to prove f (t) 6 C(Tmax − t)−µ ,

g(t) 6 C(Tmax − t)−θ , 0 6 t < Tmax .

(6.60)

In fact, set k = (1 + p + q + pq)/(2 + p + q), then 1/(k − 1) = µ + θ,

k/(p + 1) + k/(q + 1) = 1,

and k > 1 since pq > 1. Application of the Young inequality in (6.59) confirms (f g)0 (t) > C(c, p, q)(f g)k (t), 0 6 t < Tmax . Integrating the above inequality from t to Tmax , we obtain f (t)g(t) 6 C(Tmax − t)−1/(k−1) = C(Tmax − t)−µ−θ , 0 6 t < Tmax .

(6.61)

156  6. Global Existence and Finite Time Blowup

Now, we prove (6.60) by contradiction, and assume that there exist sequences {ti } − and {Ci } with ti → Tmax and Ci → ∞ as i → ∞ such that f (ti ) > Ci (Tmax − ti )−µ .

(6.62)

Integrating two inequalities in (6.58) from ti to t, and using the monotonicity of f (t) and (6.62), we conclude that, for ti 6 t < Tmax , Z

g(t) > g(ti ) +

t

ti Z t

f (t) > f (ti ) +

cf q (s)ds > cf q (ti )(t − ti ) > cCiq (Tmax − ti )−µq (t − ti ), cg p (s)ds > cp+1 Cipq (Tmax − ti )−µpq

ti −1 p+1

= (p + 1) c

Z

t

(s − ti )p ds

ti

Cipq (Tmax − ti )−µpq (t − ti )p+1 .

It then follows that f (t)g(t) > (p + 1)−1 cp+2 Cipq+q (Tmax − ti )−µq(p+1) (t − ti )p+2 . Taking t = t∗ = (Tmax + ti )/2 in the above inequality, it yields f (t∗ )g(t∗ ) > (p + 1)−1 cp+2 Cipq+q (Tmax − ti )−µq(p+1) (1/2)p+2 (Tmax − ti )p+2 = (p + 1)−1 (c/2)p+2 Cipq+q (Tmax − ti )−µ−θ = (p + 1)−1 (c/2)p+2 Cipq+q 2−µ−θ (Tmax − t∗ )−µ−θ . This contradicts (6.61) since Ci → ∞. As a result, the first estimate of (6.60) is valid. The second estimate of (6.60) can be achieved in a similar way. Step 3. For any open subset Ω1 ⊂ Ω with ∂Ω1 ∈ C 2 and dist(Ω1 , ∂Ω) = ε > 0, we can take Ω2 b Ω so that Ω1 b Ω2 , dist(Ω1 , ∂Ω2 ) > ε/3 and dist(Ω2 , ∂Ω) > ε/3. It is well known that for any ε > 0, 0 6 G(x, y, t − s) 6 Cε

for x, y ∈ Ω, |x − y| > ε/3, 0 6 s < t < Tmax .

This together with (6.37) and the second estimate of (6.60) deduces Z

max v(·, t) 6 C0 + Cε Ω2

t

Aq (s)ds

0

= C0 + Cε g(t) 6 C1 (Tmax − t)−θ , 0 6 t < Tmax .

(6.63)

We shall use the idea of [49] to construct a control function in order to show that v cannot blow up in the interior of Ω1 . Define d(x) = d(x, ∂Ω1 ) and φ(x) = d2 (x), x ∈ Nσ (∂Ω1 ) := {x ∈ Ω1 : d(x) < σ}. Owing to ∂Ω1 ∈ C 2 , the function φ ∈ C 2 (Nσ (∂Ω1 )) if σ is small enough. By the direct calculation we see that there exists a constant C > 0 such that ∆φ −

(θ + 1)|∇φ|2 > −C φ

in Nσ (∂Ω1 )

6.3. A system coupled in equation and boundary  157

provided that σ is small enough. We next extend φ to a function in Ω1 , still denoted by itself, such that φ ∈ C 2 (Ω1 ) and φ > c0 > 0 in Ω1 \ Nσ (∂Ω1 ). Naturally, ∆φ −

(θ + 1)|∇φ|2 > −ρ in Ω1 φ

for some ρ > 0. Set w=

C , x ∈ Ω1 , 0 < t < Tmax . [φ(x) + ρ(Tmax − t)]θ

Careful computations yield wt − ∆w =

Cθ [φ + ρ(Tmax − t)]1+θ

(θ + 1)|∇φ|2 ρ + ∆φ − φ + ρ(Tmax − t)

!

> 0.

Take C to be large enough so that w(x, 0) > v(x, 0) in Ω1 and Cρ−θ > C1 . Then w ∂Ω1 = Cρ−θ (Tmax − t)−θ > v ∂Ω1



in accordance with (6.63). The comparison principle gives v(x, t) 6 w(x, t) =

C , [φ(x) + ρ(Tmax − t)]θ

x ∈ Ω1 , 0 6 t < Tmax .

This shows that v cannot blow up in the interior of Ω1 , and so v cannot blow up in the interior of Ω by the arbitrariness of Ω1 . Analogously, u cannot blow up in the interior of Ω.  6.3.4

Blowup rate estimates

In this part, we study a special case of (6.35) of the form  p   ut − ∆u = v , 

∂n u = 0,

  

vt − ∆v = 0, q

∂n v = u ,

x ∈ BR ,

t > 0,

x ∈ ∂BR , t > 0,

(6.64)

u(x, 0) = u0 (x), v(x, 0) = v0 (x), x ∈ BR ,

here BR is a ball in Rn centred at the origin with radius R > 0, initial data u0 (x) = u0 (r), v0 (x) = v0 (r) with r = |x| are positive, radially symmetric C 2 functions and satisfy v00 (r) > 0 for 0 6 r < R, u00 (R) = 0, v00 (R) = uq0 (R). Under above conditions, it is easy to show that any solution (u, v) of (6.64) is positive and radially symmetric, and satisfies vr (r, t) > 0. By results of §6.3.2, all solutions of (6.64) exist globally if and only if pq 6 1. It is assumed that pq > 1, and components u and v of the solution (u, v) of (6.64) blow up simultaneously in the same time Tmax < ∞. For any t > 0, we denote h(t) := max u(R, τ ), f (t) := max max u(·, τ ), g(t) := max max v(·, τ ). 06τ 6t

06τ 6t B R

06τ 6t B R

158  6. Global Existence and Finite Time Blowup

Then f (t) > h(t), and h(t), f (t) and g(t) are monotonically increasing in t and f (t), g(t) → ∞ as t → Tmax . Moreover, the maximum principle gives g(t) = max06τ 6t v(R, τ ). Denote γ = maxB R u00 (r), and let w(r, t) = ur (r, t) + γ. Then we have n−1 n−1 n−1 wr − w+ γ + pv p−1 vr r r2 r2 n−1 n−1 > wrr + wr − w, 0 < r < R, 0 < t < Tmax . r r2

wt = wrr +

Since w(0, t) = w(R, t) = γ > 0, and w(r, 0) > 0 for 0 6 r 6 R, the maximum principle implies w > 0, i.e., ur > −γ. For any t > 0, we choose r(t) ∈ [0, R] so that u(r(t), t) = maxB R u(·, t). Integrating the inequality ur (r, t) > −γ from r(t) to R with respect to r we conclude u(R, t) > u(r(t), t) − γ(R − r(t)) > u(r(t), t) − γR. Consequently, when t closes to Tmax , 1 h(t) = max u(R, τ ) > max u(r(τ ), τ ) − γR = f (t) − γR > f (t). 06τ 6t 06τ 6t 2

(6.65)

Define ρ=

2+p 1 + 2q and σ = . 2(pq − 1) 2(pq − 1)

(6.66)

We shall use the idea of [17] (the proof of (2.2) there) to prove the following lemma. Lemma 6.17

There exists a positive constant ε such that

εg 1/σ (t) 6 f 1/ρ (t), εf 1/ρ (t) 6 g 1/σ (t) for all t ∈ (Tmax /2, Tmax ).

(6.67)

Proof. On the contrary we assume that the first inequality of (6.67) was not valid. Then there would exist ti → Tmax such that g −1/σ (ti )f 1/ρ (ti ) → 0.

(6.68)

For each ti , we can choose (ˆ xi , tˆi ) ∈ ∂BR × (0, ti ] for which v(ˆ xi , tˆi ) = g(ti ). On ˆ account that g(ti ) → ∞, it can be inferred that ti → Tmax as i → ∞. Define λi = λ(ti ) = g −1/(2σ) (ti ) and (

ˆi , λ2i s + tˆi ), ϕi (y, s) = λ2ρ i u(λi Ri y + x ψi (y, s) = λ2σ ˆi , λ2i s + tˆi ), i v(λi Ri y + x

(y, s) ∈ Ωi × Ii (Tmax ), (y, s) ∈ Ωi × Ii (Tmax ),

(6.69)

6.3. A system coupled in equation and boundary  159

where ˆ −2 ˆ Ωi = {y : λi Ri y + xˆi ∈ BR }, Ii (Tmax ) = (−λ−2 i ti , λi (Tmax − ti )), and Ri is a rotation operator such that (−1, 0, . . . , 0) is the exterior normal vector of ∂Ωi at y = 0. By use of expressions of ρ and σ, direct calculations tell us (

ϕis − ∆ϕi = ψip , ψis − ∆ψi = 0, y ∈ Ωi , s ∈ Ii (Tmax ), ∂n ϕi = 0,

∂n ψi = ϕqi ,

y ∈ ∂Ωi , s ∈ Ii (Tmax )

(6.70)

and (

ψi (0, 0) = 1, 0 6 ψi (y, s) 6 1, 0 6 ϕi (y, s) 6 f (ti )g −ρ/σ (ti ),

ˆ y ∈ Ωi , s ∈ (−λ−2 i ti , 0], ˆ y ∈ Ωi , s ∈ (−λ−2 i ti , 0].

(6.71)

For any given k > 0, we can find i0 (k) > 0 such that ˆ Σk := {|y − y k | < k} ⊂ Ωi , [−k, 0] ⊂ (−λ−2 i ti , 0] for all i > i0 (k), where y k = (k, 0, . . . , 0). Recalling (6.70) and (6.71), and using the interior Lp estimate and the interior Schauder estimate successively, we obtain k(ϕi , ψi )kC 2+µ,1+µ/2 (Σk ×[−k,0]) 6 Ck for all i > i0 (k). Using the Cantor-Hilbert diagonal method, we can prove that there exist a subsequence of {(ϕi , ψi )}, which is still denoted by itself, and non-negative functions ϕ and ψ such that (ϕi , ψi ) → (ϕ, ψ) in C 2,1 (Σk × [−k, 0]) for any k > 0, and (ϕ, ψ) satisfies ϕs − ∆ϕ = ψ p , y ∈ Rn+ := {y ∈ Rn , y1 > 0}, s ∈ (−∞, 0].

(6.72)

Obviously, ϕ and ψ are continuous at (0, 0). Moreover, ϕ ≡ 0 and ψ(0, 0) = 1 by (6.68) and (6.71). This contradicts (6.72). Remembering (6.65), the second inequality of (6.67) can be proved similarly.  Lemma 6.18 ([34]) Consider the following problem in half space  ϕ − ∆ϕ = ψ p , ψs − ∆ψ = 0,    s

−∂y1 ϕ = 0, −∂y1 ψ = ϕq ,

  

y ∈ Rn+ , s > 0, y1 = 0,

s > 0,

(6.73)

ϕ(y, s) > 0, ψ(y, s) > 0.

Let ρ, σ be given by (6.66). If either max{ρ, σ} > n/2, or max{ρ, σ} = n/2 and p, q > 1, then all non-trivial solutions are not global. Considering that the proof of Lemma 6.18 is too long, we omit details here.

160  6. Global Existence and Finite Time Blowup

Theorem 6.19 (Blowup rate estimates) (1) There exists a constant c > 0 such that f (t) > c(Tmax − t)−ρ , g(t) > c(Tmax − t)−σ , 0 < t < Tmax .

(6.74)

(2) Assume that one of the followings holds: (2a) ut , vt > 0; (2b) max{ρ, σ} > n/2; (2c) max{ρ, σ} = n/2 and p, q > 1. Then there exists a constant C > 0 such that f (t) 6 C(Tmax − t)−ρ , g(t) 6 C(Tmax − t)−σ , 0 < t < Tmax .

(6.75)

Proof. Let G(x, y, t−s) be the Green function for heat equation with the homogeneous Neumann boundary condition. Then we conclude that, for 0 < s < t < Tmax , Z

u(x, t) =

G(x, y, t − s)u(y, s)dy +

BR

Z

v(x, t) =

Z tZ s

G(x, y, t − s)v(y, s)dy +

Z tZ s

BR

G(x, y, t − τ )v p (y, τ )dydτ,

(6.76)

BR

G(x, y, t − τ )uq (R, τ )dSy dτ. (6.77)

∂BR

(1) Taking advantage of (6.67), it follows from (6.76) that t

Z

f (t) 6 f (s) +

s

g p (t)dτ 6 f (s) + C(t − s)f pσ/ρ (t), 0 < s < t < Tmax .

Owing to our assumption, f (t) → ∞ as t → Tmax . For any given s ∈ (0, Tmax ), we can select a t ∈ (s, Tmax ) such that f (t) = 2f (s). Accordingly, we have 2f (s) 6 f (s) + C(t − s)f pσ/ρ (s) 6 f (s) + C(Tmax − s)f pσ/ρ (s), 0 < s < Tmax , which implies f (s) > c(Tmax − s)−ρ for some constant c > 0. The first inequality of (6.74) is obtained, and then the second inequality of that is derived immediately by utilizing (6.67). (2) When (2a) holds we have h(t) = u(R, t), f (t) = max u(·, t), g(t) = v(R, t). BR

Recalling (6.65) and (6.67), it follows from (6.77) that Z

g(t) > c

s

t

f q (τ ) √ dτ > c t−τ

Z s

t

g qρ/σ (τ ) √ dτ, 0 < s < t < Tmax . t−τ

(6.78)

We shall use the idea of [49] to prove the second inequality in (6.75). Once this is done, using the second inequality of (6.67), we can get the first inequality in (6.75) immediately. Denote η = qρ/σ and Z t g η (τ ) √ J(t) = dτ. Tmax − τ s

6.3. A system coupled in equation and boundary  161

Then g(t) > cJ(t) by (6.78), and so J 0 (t) = √

cη J η (t) g η (t) >√ , Tmax − t Tmax − t

which implies Z

J(Tmax )

J(t)

dJ > cη Jη

Z

Tmax

√

t

dt Tmax − t

p

= 2cη Tmax − t.

Note that η > 1 due to pq > 1. The above inequality leads to p p J 1−η (t) J 1−η (Tmax ) > 2cη Tmax − t + > 2cη Tmax − t, η−1 η−1

and thereafter, J(t) 6 C(Tmax − t)−1/[2(η−1)] .

(6.79)

On the other hand, for t∗ = 2t − Tmax (we assume that t is close to Tmax here), Z

J(t) >

t

t∗

√

g η (τ ) dτ > g η (t∗ ) Tmax − τ

Z

t

t∗

p dτ √ = C 0 g η (t∗ ) Tmax − t. Tmax − τ

Combining this inequality with (6.79), we get g(t∗ ) 6 C1 (Tmax − t)−1/[2(η−1)] = C2 (Tmax − t∗ )−1/[2(η−1)] , and it is exactly the second inequality of (6.75) due to 1/[2(η − 1)] = σ. When (2b) or (2c) holds, we first use ideas of [35, 47] to prove the second inequality in (6.75), and then get the first inequality in (6.75) by using the second inequality of (6.67). For any t0 ∈ (Tmax /2, Tmax ), define t+ 0 := max{t ∈ (t0 , Tmax ) : g(t) = 2g(t0 )} and λ(t0 ) = g −1/(2σ) (t0 ). We claim λ−2 (t0 )(t+ 0 − t0 ) 6 M,

Tmax /2 < t0 < Tmax

(6.80)

for some constant M > 0 which is independent of t0 . If (6.80) was not valid, then − there would exist ti → Tmax satisfying λ−2 (ti )(t+ i − ti ) → ∞. Same as the proof of Lemma 6.17, we can choose (ˆ xi , tˆi ) ∈ ∂BR × (0, ti ] for which v(ˆ xi , tˆi ) = g(ti ). Define ϕi (y, s) and ψi (y, s) as in (6.69). Then (6.70) holds, and (6.71) becomes + ψi (0, 0) = 1, 0 6 ψi (y, s) 6 λ2σ (ti )g(t+ i ) = 2, y ∈ Ωi , s ∈ Ii (ti ), −ρ ρ/σ 0 6 ϕi (y, s) 6 λ2ρ (ti )f (t+ , y ∈ Ωi , s ∈ Ii (t+ i )6ε 2 i ),

where (6.67) was used in the last inequality. Same as the proof of Lemma 6.17, we can find two C 2,1 functions ϕ(y, s) and ψ(y, s) to solve (6.73) which satisfy ψ(0, 0) = 1,

162  6. Global Existence and Finite Time Blowup

ψ > 0 and ϕ > 0. Since the assumption (2b) or (2c) holds, by Lemma 6.18, (ϕ, ψ) blows up in finite time. It is a contradiction, and hence (6.80) holds. Now we use an idea from [47, Lemma 3.1] to deduce the second inequality of + + (6.75). Fix t0 ∈ (Tmax /2, Tmax ), and denote t1 = t+ 0 , t2 = t1 , . . . , ti+1 = ti , . . . . Then g(ti+1 ) = 2g(ti ), and ti+1 − ti 6 M g −1/σ (ti ) by (6.80). Clearly, ti → Tmax and Tmax − t0 =

∞ X

(ti+1 − ti ) 6 M

i=0

∞ X

g −1/σ (ti )

i=0

= M g −1/σ (t0 )

∞ X

2−i/σ = Cg −1/σ (t0 ).

i=0

This implies the second inequality of (6.75).  In the study of blowup rates, blowup sets, blowup profiles and blowup boundary layers of blowup solutions for parabolic systems, a basic prerequisite is that all components of the solution must blow up simultaneously. This phenomenon is called simultaneous blowup. This book does not cover this aspect. Interested readers can refer to [100, 94, 67, 76] and the references therein.

EXERCISES 6.1 For problem (6.27), prove that if initial data u0 , v0 ∈ C 2 (Ω) satisfy −∆u0 6 uρ0 v0p , −∆v0 6 uq0 v0σ , x ∈ BR (0), then ut , vt > 0. 6.2 Prove (6.46). 6.3 Let f (s)Zbe a continuous function in s > 0 and f (s) > 0. Prove that if the ∞ ds < ∞ then any solution u of integral f (s) 0  u − ∆u = f (u),    t

x ∈ Ω, t > 0, x ∈ ∂Ω, t > 0,

∂n u = 0,

  

u(x, 0) = u0 (x) > 0, 6≡ 0, x ∈ Ω

must blow up in finite time. 6.4 (Kalantarov-Ladyzhenskaya [53]) Suppose that f 00 (t)f (t) − (1 + ρ)(f 0 (t))2 > af (t)f 0 (t) − bf 2 (t), t > 0 with constants ρ, a, b > 0. Prove that if 0

f (0) > 0, f (0) + then f (t) blows up in finite time.

a−

p

a2 + 4ρb f (0) > 0, 2ρ

Exercises  163

6.5 Consider the following problem  p−1 u,   ut − ∆u = |u| 

x ∈ Ω, t > 0,

u = 0,

x ∈ ∂Ω, t > 0,

  

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

x ∈ Ω,

where p > 1. Investigate finite time blowup of solution by use of the eigenfunction method, the energy method and the concavity method, respectively, and compare their advantages and disadvantages. 6.6 Let p > q > 1 and b > 0. Consider following problem  u − ∆u = up − buq , x ∈ Ω,    t

u = 0,

  

t > 0,

x ∈ ∂Ω, t > 0,

u(x, 0) = u0 (x) > 0, x ∈ Ω,

where u0 ∈ C 2 (Ω) and u0 = 0 on ∂Ω. Please study global existence and finite time blowup of solution (using the eigenfunction method, the energy method and the concavity method, respectively). 6.7 Let p, q > 1 and pq > 1. Investigate global existence and finite time blowup of solution of  ut − ∆u = v p − u,       vt − ∆v = uq − v,  u = v = 0,     

x ∈ Ω, t > 0, x ∈ Ω, t > 0, x ∈ ∂Ω, t > 0,

u(x, 0) > 0, v(x, 0) > 0, x ∈ Ω.

CHAPTER

7

Time-Periodic Parabolic Boundary Value Problems

“It seems at first that this fact [the existence of periodic solutions] could not be of any practical interest whatsoever [however] what renders these periodic solutions so precious is that they are, so to speak, the only breach through which we may try to penetrate a stronghold previously reputed to be impregnable.” —–Henri Poincar´e The importance of periodic solutions in understanding behaviours of ordinary differential equations is well known. In recent decades, time-periodic solutions of partial differential equations have gained prominence, and have been studied by employing a variety of methods. This chapter focuses on the time-periodic parabolic boundary value problems. The upper and lower solutions method for time-periodic parabolic boundary value problems, and the principal eigenvalue of periodic parabolic eigenvalue problems form the bulk of the chapter. As applications, a diffusive logistic equation (including the unbounded domain case) and a diffusive competition model are systematically studied. For the diffusive logistic equation, the existence and uniqueness of positive solutions are established; in the bounded domain case, it is proved that the unique positive solution is globally asymptotically stable in the periodic sense. For the diffusive competition model, the existence of positive solutions is also established. Throughout this chapter we always assume that coefficients of L are timeperiodic with period T and satisfy the condition (C) given in §1.1, and Ω is a bounded domain of class C 2+α .

7.1

THE UPPER AND LOWER SOLUTIONS METHOD FOR SCALAR EQUATIONS

We first study the following time-periodic parabolic boundary value problem:  L u = f (x, t, u)   

in QT ,

Bu = 0

on ST ,

  

u(x, 0) = u(x, T )

in Ω,

(7.1)

165

166  7. Time-Periodic Parabolic Boundary Value Problems

where B u = u or B u = ∂n u + bu with b > 0 and b ∈ C 1+α, (1+α)/2 (S T ), f and b are time-periodic with period T . Nonlinear periodic diffusion equations of the form (7.1) arise naturally, such as in problems stemming from population ecology, where data depends periodically on time (seasonal or daily variations, for example). The basis of comparison principle is the maximum principle. We have realized the importance and roles of comparison principle in the study of parabolic problems. For time-periodic parabolic boundary value problems, there is also the maximum principle which takes the following version. Theorem 7.1 (Maximum principle) Assume c > 0, and c 6≡ 0 when B = ∂n . Let u ∈ C(QT ) ∩ C 2,1 (QT ) when B u = u and u ∈ C 1,0 (QT ) ∩ C 2,1 (QT ) when B u = ∂n u + bu. If u satisfies   Lu > 0 

in QT ,

Bu > 0

on ST ,

  

u(x, 0) > u(x, T )

in Ω,

then u > 0 in QT . Moreover, if one of the above three inequalities is not identical, then u > 0 in QT . Proof. Suppose, to the contrary, that u(x0 , t0 ) = minQT u < 0. We may assume t0 > 0 on account of u(x0 , T ) 6 u(x0 , 0). Using the Hopf boundary lemma, we can conclude x0 ∈ Ω. Then the strong maximum principle leads to u ≡ u(x0 , t0 ) in Qt0 . Certainly, u(x, 0) ≡ u(x0 , t0 ) in Ω. By the assumption, u(x, T ) 6 u(x, 0) ≡ u(x0 , t0 ) in Ω. As a consequence, t0 = T and u ≡ u(x0 , t0 ) < 0 in QT . Note that c > 0, and c 6≡ 0 when B = ∂n . This is impossible. Hence u > 0 in QT . If there is (x0 , t0 ) ∈ QT such that u(x0 , t0 ) = 0 = minQT u, then u ≡ 0 in Qt0 by Corollary 1.25. Hence u(x, T ) ≡ u(x, 0) ≡ 0 in Ω due to u(x, 0) > u(x, T ) > 0 in Ω, and in turn u ≡ 0 in QT . This indicates that L u ≡ 0 in QT and B u ≡ 0 on ST .  Existence and uniqueness of solutions of initial-boundary value problem for linear equation and the comparison principle are foundations of establishing the upper and lower solutions method of initial-boundary value problem for semilinear equations. However, existence and uniqueness of solutions are not completely clear for timeperiodic parabolic boundary value problem of linear equations. To establish the upper and lower solutions method of (7.1), we must adopt a different approach from initialboundary value problems. Based on the maximum principle (Theorem 7.1), we define upper and lower solutions of (7.1) as follows. Definition 7.2 Let u ∈ C(QT ) ∩ C 2,1 (QT ) when B u = u and u ∈ C 1,0 (QT ) ∩ C 2,1 (QT ) when B u = ∂n u + bu. Such a function u is referred to as an upper solution (a lower solution) of (7.1) if  L u > (6) f (x, t, u)   

in QT ,

B u > (6) 0

on ST ,

  

u(x, 0) > (6) u(x, T )

in Ω.

7.1. The upper and lower solutions method for scalar equations  167

In order to set up the upper and lower solutions method, it is generally sufficient for initial-boundary value problems to have upper and lower solutions, and for boundary value problems of elliptic equations to have ordered upper and lower solutions. However, these are not enough for time-periodic parabolic boundary value problem (7.1), and a new hypothesis is needed, that is, there must be a better function between u(x, 0) and u¯(x, 0). We set ◦ Wp,2 B (Ω) = W 1p (Ω) ∩ Wp2 (Ω) when B u = u, and Wp,2 B (Ω) = u ∈ Wp2 (Ω) : ∂n u + b(x, 0)u 







∂Ω

=0

when B u = ∂n u + bu. Theorem 7.3 (The upper and lower solutions method) Suppose that u¯ > u are an upper solution and a lower solution of (7.1), respectively. Let f (·, u) ∈ C α, α/2 (QT ) be uniform with respect to u ∈ [c, c¯] and fu ∈ C(QT × [c, c¯]), where c = minQT u and c¯ = maxQT u¯. When B u = u, we further assume f (x, 0, 0) = 0 on ∂Ω. Fix p > 1+n/2. If there exists at least one u0 ∈ Wp,2 B (Ω) satisfying u(x, 0) 6 u0 (x) 6 u¯(x, 0), then problem (7.1) has at least one solution u ∈ C 2+α, 1+α/2 (Ω × [0, T ]) which satisfies u 6 u 6 u¯

in QT .

Proof. The idea here is inspired by [46, Section 21]. We only deal with the case B u = u. Step 1. Define a set n

o

V = u0 ∈ Wp,2 B (Ω) : u(x, 0) 6 u0 (x) 6 u(x, 0) in Ω, ku0 kWp2 (Ω) 6 C(T, c, c¯) , where C(T, c, c¯) is a positive constant which will be given in the following (7.4). Then V is a bounded and closed convex set of Wp2 (Ω) and V 6= ∅. For any given u0 ∈ V , let us investigate the problem  L u = f (x, t, u)   

in QT ,

u=0

on ST ,

  

u(x, 0) = u0 (x)

in Ω.

(7.2)

Obviously, u¯ and u are an upper solution and a lower solution of (7.2), respectively. Thus (7.2) has a unique solution u ∈ Wp2,1 (QT ) which satisfies u 6 u 6 u. It follows that u(x, T ) > u(x, T ) > u(x, 0) and u(x, T ) 6 u¯(x, T ) 6 u¯(x, 0). Set F (x, t) = f (x, t, u(x, t)). Then F ∈ C α, α/2 (QT ) as u ∈ Wp2,1 (QT ) ,→ C α, α/2 (QT ). Using Theorem 2.10, we conclude u ∈ Wp2,1 (QT ) ∩ C 2+α, 1+α/2 (Ω × (0, T ]).

168  7. Time-Periodic Parabolic Boundary Value Problems ◦

This leads to u(x, T ) ∈ W 1p (Ω) ∩ Wp2 (Ω) = Wp,2 B (Ω). Define an operator F : V → Wp,2 B (Ω) ∩ hu(x, 0), u¯(x, 0)i by F (u0 ) = u(x, T ).

In the following, we will prove that F : V → V is compact. Step 2. Thanks to u 6 u 6 u, it follows |F (x, t)| 6 C(c, c¯) := max |f (x, t, s)| for all u0 ∈ V. QT ×[c,¯ c]

In view of Theorem 1.8, for any 0 < δ < T , there exists C(δ) > 0 so that kukWp2,1 (Ω×(δ,T )) 6 C(δ) (kF kp, QT + kukp, QT ) 6 C(δ, c, c¯) for all u0 ∈ V. The embedding theorem affirms |u|α, Ω×[δ,T ] 6 C(δ, c, c¯) for all u0 ∈ V, which implies |F |α, Ω×[δ,T ] 6 C(δ, c, c¯) for all u0 ∈ V. For any 0 < δ < τ < T , taking advantage of Theorem 2.10 and Theorem 1.14 (2), we can find a constant C(δ, τ ) such that 

|u|2+α, Ω×[τ,T ] 6 C(δ, τ ) |F |α, Ω×[δ,T ] + kuk∞, Ω×(δ,T )



6 C(δ, τ, c, c¯) for all u0 ∈ V. Certainly, kF (u0 )kC 2+α (Ω) = ku(x, T )kC 2+α (Ω) 6 C(δ, τ, c, c¯) for all u0 ∈ V.

(7.3)

Take δ = T /3 and τ = T /2. Then we have ku(x, T )kWp2 (Ω) 6 C 0 ku(x, T )kC 2+α (Ω) 6 C(T, c, c¯) for all u0 ∈ V.

(7.4)

This shows that F : V → V . Step 3. Now, we show that F : V → V is compact. Recalling the above estimate (7.3), it is sufficient to prove that F : V → V is continuous. For ui0 ∈ V with ui0 → u0 in Wp2 (Ω) as i → ∞, let ui be the unique solution of (7.2) with ui (x, 0) = ui0 (x) and wi = ui − u. Write f (x, t, ui ) − f (x, t, u) = wi

Z

1

fu (x, t, u + swi )ds =: −ci wi .

0

Then ci ∈ L∞ (QT ) and wi satisfies  L wi + ci wi = 0   

in QT ,

wi = 0

on ST ,

  

wi (x, 0) = ui0 (x) − u0 (x)

in Ω,

7.2. Time-periodic parabolic eigenvalue problem  169

and in accordance with Theorem 1.1, kwi kWp2,1 (QT ) 6 Ckui0 − u0 kWp2 (Ω) , which implies |ui − u|α, QT 6 Ckui0 − u0 kWp2 (Ω) → 0 as i → ∞. n

It follows that |ui |α, QT

o

n

o

is bounded, and so is |f (·, ui (·))|α, QT . We have known

that f (·, ui (·)) → f (·, u(·)) in C(QT ), and as a result of compactness, f (·, ui (·)) → f (·, u(·)) in C γ, γ/2 (QT ) for any 0 < γ < α. The application of Schauder theory (Theorem 1.14 (2)) deduces 



|ui − u|2+γ, Ω×[T /3,T ] 6 C |f (x, t, ui ) − f (x, t, u)|γ, QT + kui − uk∞, QT → 0. Certainly, |ui (x, T ) − u(x, T )|2+γ, Ω → 0, and then F : V → V is continuous. Step 4. Applying the Schauder fixed point theorem (Corollary A.3), F has at least one fixed point u0 ∈ V , i.e., solution u of (7.2) with such u0 as its initial datum satisfies u(x, T ) = u0 (x) = u(x, 0). Therefore, u is a solution of (7.1). Moreover, since u = F (u0 ) ∈ C 2+α, 1+α/2 (Ω × (0, T ]), we see that, with g = 0, u(x, T ) = g(x, T ) for x ∈ ∂Ω, gt − aij Dij u + bi Di u + cu = f (x, t, u) for x ∈ ∂Ω, t = T. These imply the following compatibility conditions with ϕ(x) = u(x, 0): ϕ(x) = g(x, 0) for x ∈ ∂Ω, gt − aij Dij ϕ + bi Di ϕ + cϕ = f (x, t, u) for x ∈ ∂Ω, t = 0 because u(x, 0) = u(x, T ) and f is time-periodic with period T . Thus, we have u ∈ C 2+α, 1+α/2 (QT ) by the Schauder theory.  We should mention that the uniqueness in Theorem 7.3 is not valid. This is the most difference from initial-boundary value problems of parabolic equations. For time-periodic parabolic boundary value problems, some good properties of initialboundary value problems may not hold, and however, some properties of elliptic boundary value problems are inherited. From Step 2 in the proof of the following Theorem 7.10 we shall see that if a lower solution u (an upper solution u¯) satisfies the compatibility condition at t = 0 and x ∈ ∂Ω, then we can take u(x, 0) (¯ u(x, 0)) as initial datum to construct an iteration sequence z(x, t+iT ) so that z(x, t+iT ) → u(x, t) as i → ∞ and u(x, t) is the minimal (maximal) solution of (7.1) located between u and u¯.

7.2

TIME-PERIODIC PARABOLIC EIGENVALUE PROBLEMS

In this section, we will briefly discuss the time-periodic parabolic eigenvalue problems  L φ = λφ      

in QT ,

Bφ = 0

on ST ,

φ(x, 0) = φ(x, T )

in Ω,

(7.5)

170  7. Time-Periodic Parabolic Boundary Value Problems

where the operator B is defined as at the first paragraph of §7.1, and b = b(x) is independent of t. Some contents of this part refer to [46]. The notion of principal eigenvalue for time-periodic parabolic boundary value problems was introduced by Lazer [61] (see also [14]). Same as elliptic operators, a number λ (real or complex) is called an eigenvalue of (7.5) if problem (7.5) has a non-trivial solution for such λ, and the corresponding non-trivial solution is called the corresponding eigenfunction to λ. A real eigenvalue λ is called principal eigenvalue of (7.5) if its corresponding eigenfunction is positive. Theorem 7.4 ([46, Proposition 14.4]) The principal eigenvalue λ1 of (7.5) exists uniquely. Furthermore, if c > 0, and c 6≡ 0 when B = ∂n , then λ1 > 0. In the case c = 0 and B = ∂n , we have λ1 = 0. Define two spaces X0 , X2 : n

o

X0 = w ∈ C α, α/2 (QT ) : w is T -periodic in time t , n

o

X2 = w ∈ C 2+α, 1+α/2 (QT ) : B w = 0 on ST , and w is T -periodic in time t , as well as an operator L: D(L) = X2 , L = L : D(L) → X0 . The eigenvalue problem (7.5) can be formulated as an abstract eigenvalue problem Lφ = λφ in X0 .

(7.6)

We next study variation of the principal eigenvalue with respect to the coefficient c, the boundary condition and the domain Ω. The following theorem will play an important role in the study of variation of principal eigenvalue. Theorem 7.5 ([46, Theorem 16.6]) For the inhomogeneous problem Lφ − λφ = h,

h ∈ X0 , h > 0, 6≡ 0,

(7.7)

let λ1 be the principal eigenvalue of (7.6) and assume λ1 > 0. (1) If 0 < λ < λ1 , then problem (7.7) has a unique solution φ ∈ X0 and φ > 0. (2) If λ > λ1 , then (7.7) has no positive solution, and no solution at all for λ = λ1 . We first explore the dependence of the principal eigenvalue on coefficient c. Theorem 7.6 (Monotonicity of principal eigenvalue) Let λ1 = λ1 (c) be the principal eigenvalue of (7.6). Then λ1 (c) is strictly increasing in c.

7.2. Time-periodic parabolic eigenvalue problem  171

Proof. Suppose c1 6 c2 and c1 6≡ c2 . Let φi be the corresponding positive eigenfunction to λ1 (ci ), i = 1, 2. We may think that λ1 (c1 ) > 0 and λ1 (c2 ) > 0. Assume conversely that λ1 (c1 ) > λ1 (c2 ). Write L = L (c) := L0 + c. Then we conclude L (c2 )(φ1 − φ2 ) = L0 φ1 + c2 φ1 − λ1 (c2 )φ2

>, 6≡ L0 φ1 + c1 φ1 − λ1 (c2 )φ2 = λ1 (c1 )φ1 − λ1 (c2 )φ2 > λ1 (c2 )(φ1 − φ2 ), namely L (c2 )(φ1 − φ2 ) − λ1 (c2 )(φ1 − φ2 ) =: h > 0, 6≡ 0.

This is a contradiction with Theorem 7.5 (2).  In order to emphasize the dependence of principal eigenvalue on boundary conditions, when B u = u we write L = LD and λ1 = λD 1 , when B u = ∂n u we write N L = LN and λ1 = λ1 , and when B u = ∂n u + b(x)u with b ∈ C 1+α (∂Ω) and b > 0, 6≡ 0 we write L = LR and λ1 = λR 1. R D Theorem 7.7 (Monotonicity of principal eigenvalue) Relations λN 1 < λ1 < λ1 hold.

Proof. Adding a multiple of φ on both sides of (7.6), we may assume c > 0 in QT , R D R and then λN 1 , λ1 , λ1 > 0. Let φR be the corresponding positive eigenfunction to λ1 . Then R LR φ R = λ R 1 φR and λ1 φR > 0 in QT , D R and φR > 0 on ST . Take 0 < ε < min λR 1 , λ1 , minQT c . Then (λ1 − ε)φR > 0. Applying Theorem 7.5 (1), we see that inhomogeneous problem





LD v − εv = (λR 1 − ε)φR has a unique positive solution v. Obviously v = 0 on ST . Thus, the function z = φR −v satisfies  L z − εz = 0   

in QT ,

z>0

on ST ,

  

z(x, 0) = z(x, T )

in Ω.

As c(x, t) − ε > 0, by Theorem 7.1, z > 0, i.e., φR > v in QT . Therefore, LD v − εv = R (λR 1 − ε)φR > (λ1 − ε)v, which implies that v solves LD v − λR 1 v =: h > 0, v > 0. D By Theorem 7.5 (2), it holds λR 1 < λ1 . R The proof of λN 1 < λ1 is left to readers as an exercise. Finally, we discuss the dependence of the principal eigenvalue on domain Ω.



172  7. Time-Periodic Parabolic Boundary Value Problems

Theorem 7.8 (Monotonicity of principal eigenvalue) Suppose that Ω1 ⊂ Ω2 and Ω1 6= Ω2 . Let LD (Ωi ) be the corresponding operator in Ωi with the homogeneous Dirichlet boundary condition (B u = u), and λD 1 (Ωi ) be the principal eigenvalue of D LD (Ωi ), i = 1, 2. Then λD (Ω ) > λ (Ω ). 1 2 1 1 Proof. Let φi be the corresponding positive eigenfunction to λD 1 (Ωi ), i = 1, 2. Similar to the proof of Theorem 7.7, we may assume that c > 0 in Ω2 ×[0, T ] and λD 1 (Ωi ) > 0, i = 1, 2. Then φ2 > 0, 6≡ 0 on ∂Ω1 × [0, T ], and LD (Ω2 )φ2 = λD 1 (Ω2 )φ2

and λD 1 (Ω2 )φ2 > 0 in Ω2 × [0, T ].

D D As above, we take 0 < ε < min λD 1 (Ω1 ), λ1 (Ω2 ), minΩ2 ×[0,T ] c . Then (λ1 (Ω2 ) − ε)φ2 > 0. In view of Theorem 7.5 (1), the inhomogeneous problem





LD (Ω1 )v − εv = (λD 1 (Ω2 ) − ε)φ2 has a unique positive solution v. Certainly, v = 0 on ∂Ω1 × (0, T ]. As a result, the function z = φ2 − v satisfies    L z − εz = 0 

in Ω1 × (0, T ],

z > 0, 6≡ 0

on ∂Ω1 × (0, T ],

  

z(x, 0) = z(x, T )

in Ω1 .

As c(x, t) − ε > 0, Theorem 7.1 gives z > 0, i.e., φ2 > v in Ω1 × (0, T ]. Therefore, D LD (Ω1 )v − εv = (λD 1 (Ω2 ) − ε)φ2 > (λ1 (Ω2 ) − ε)v. This shows that v is a positive solution of LD (Ω1 )v − λD 1 (Ω2 )v =: h > 0. D According to Theorem 7.5 (2), λD 1 (Ω2 ) < λ1 (Ω1 ). In the following we consider a special case

   φt − dφxx − a(x, t)φ = λφ, 

0 < x < `, 0 < t 6 T,

B φ(0, t) = φ(`, t) = 0,

0 < t 6 T,

  

φ(x, 0) = φ(x, T ),

0 < x < `,



(7.8)

where ` > 0, B u = βu − (1 − β)ux and β is a constant satisfying 0 6 β 6 1. The function a ∈ C α,α/2 (QT ) is time-periodic with period T . Theorem 7.9 The principal eigenvalue of (7.8) exists uniquely, denoted by λ(`; d, a). And the following statements hold: (1) λ(`; d, a) is strictly monotone decreasing in a and `; (2) when β = 1 we denote λ(`; d, a) by λD (`; d, a). Then λ(`; d, a) 6 λD (`; d, a) for all 0 6 β 6 1; (3) limd→∞ λ(`; d, a) = lim`→0+ λ(`; d, a) = ∞.

7.3. The logistic equation  173

Proof. The proof of existence and uniqueness of λ(`; d, a) is similar to Theorem 7.4. Theorem 7.5 holds for problem (7.8). Conclusions (1) and (2) can be proved by the same ways as those of Theorems 7.6–7.8. In the following we prove the conclusion (3). Define a ˆ = max[0,`]×[0,T ] a(x, t). Then λ(`; d, a) > λ(`; d, a ˆ) since λ(`; d, a) is monotone decreasing in a. Because a ˆ is a constant, we know that λ(`; d, a ˆ) is the principal eigenvalue of (

−dφ00 − a ˆφ = λφ, 0 < x < `, B φ(0) = φ(`) = 0,

and limd→∞ λ(`; d, a ˆ) = ∞. Thereafter, limd→∞ λ(`; d, a) = ∞. Define a ˜ = max[0,1]×[0,T ] a(x, t). Similarly to above, λ(`; d, a) > λ(`; d, a ˜) for all 0 < ` 6 1, and λ(`; d, a ˜) is the principal eigenvalue of elliptic problem (

−dφ00 − a ˜φ = λφ, 0 < x < `, B φ(0) = φ(`) = 0.

As lim`→0+ λ(`; d, a ˜) = ∞, we have lim`→0+ λ(`; d, a) = ∞.



7.3 THE LOGISTIC EQUATION In this section, we will use the principle eigenvalue and the upper and lower solutions method of time-periodic parabolic boundary value problem, together with the comparison principle of initial-boundary value problem to study the time-periodic problem of logistic equation. Some contents of this part take from [113]. 7.3.1

The case of bounded domain

In this subsection we study the time-periodic parabolic boundary value problem  u − d∆u = a(x, t)u − b(x, t)u2    t

in QT ,

Bu = 0

on ST ,

  

u(x, 0) = u(x, T )

in Ω,

(7.9)

where B u = βu + (1 − β)∂n u with constant 0 6 β 6 1. We assume that functions a, b ∈ C α, α/2 (QT ) are time-periodic with period T , b1 6 b 6 b2 in QT for some positive constants b1 and b2 . Let λ(a) be the principal eigenvalue of the following time-periodic parabolic eigenvalue problem

Theorem 7.10

 φ − d∆φ − a(x, t)φ = λφ    t

in QT ,

Bφ = 0

on ST ,

  

φ(x, 0) = φ(x, T )

in Ω.

(7.10)

If λ(a) < 0, then problem (7.9) has a unique positive solution u.

174  7. Time-Periodic Parabolic Boundary Value Problems

Moreover, for any v0 ∈ C 1 (Ω) ∩ Wp2 (Ω) satisfying B v0 = 0 on ∂Ω and v0 > 0 in Ω, the unique solution v of the initial-boundary value problem  v − d∆v = a(x, t)v − b(x, t)v 2    t

in Q∞ ,

Bv = 0

on S∞ ,

  

v(x, 0) = v0 (x)

in Ω

(7.11)

satisfies lim v(x, t + iT ) = u(x, t) uniformly in QT .

i→∞

(7.12)

This indicates that the unique positive solution u of (7.9) is globally asymptotically stable in the time-periodic sense. Proof. The proof will be divided into three steps: Step 1. The existence. Let φ be a positive eigenfunction corresponding to λ(a). Take ε > 0, and set u¯ = M and u = εφ. Then u¯ and u are ordered upper and lower solutions of (7.9) so long as ε  1 and M  1. Obviously, εφ(x, 0) ∈ Wp2 (Ω), B (εφ(x, 0)) = 0 on ∂Ω and u(x, 0) 6 εφ(x, 0) 6 u¯(x, 0) in Ω. According to Theorem 7.3, problem (7.9) has at least one solution u which satisfies εφ 6 u 6 M . Step 2. We now prove uniqueness. The strategy of this proof is to construct a monotone sequence to get a minimal positive solution and then use a standard comparison method. Let u be a positive solution of (7.9). Invoking Lemma 3.7, there exists 0 < ε  1 such that u(x, 0) > εφ(x, 0) in Ω. Let z be the unique solution of (7.11) with initial datum εφ(x, 0). Note that the function ξ := εφ satisfies ξt − d∆ξ < a(x, t)ξ − b(x, t)ξ 2

in QT .

We have εφ = ξ 6 z 6 u by the comparison principle. For any integer i > 0, let us define z i (x, t) = z(x, t + iT ) for (x, t) ∈ QT . Since a and b are time-periodic with period T , we see that z i satisfies  i z − d∆z i = a(x, t)z i − b(x, t)(z i )2    t

in QT ,

  

in Ω.

Bzi = 0 i

z (x, 0) = z(x, iT )

on ST ,

Note that z(x, 0) = εφ(x, 0) = εφ(x, T ) 6 z(x, T ) = z 1 (x, 0) 6 u(x, T ) = u(x, 0). We can adopt the comparison principle to produce εφ 6 z 6 z 1 6 u in QT which in turn asserts z 1 (x, 0) = z(x, T ) 6 z 1 (x, T ) = z 2 (x, 0) 6 u(x, T ) = u(x, 0). As above, z 1 6 z 2 6 u in QT . Applying the inductive method, we can prove that z i

7.3. The logistic equation  175

is monotonically increasing in i and εφ 6 z i 6 u in QT for all i. Consequently, there exists a positive function uε such that z i → uε pointwisely in QT as i → ∞. Clearly, uε (x, 0) = uε (x, T ) since z i+1 (x, 0) = z i (x, T ). Based on the regularity theory and compactness argument, it can be shown that z i → uε in C 2,1 (QT ), εφ 6 uε 6 u in QT and uε satisfies the first two equations of (7.9). This shows that uε is a positive solution of (7.9). To prove uniqueness, it suffices to illustrate that uε = u. Firstly, we can find a constant δ > 0 so that uε > δu in QT (refer to the proof of Lemma 3.7). Then the infimum uε (x, t) σ = x∈Ω inf u(x, t) 06t6T exists and is positive. Clearly, δ 6 σ 6 1 and uε > σu in QT as we have known uε 6 u. If we can show σ = 1, then uε = u. Assume on the contrary that σ < 1. Define ϕ = uε − σu. Then ϕ > 0 and B ϕ = 0 on ST ,

ϕ(x, 0) = ϕ(x, T ) in Ω.

A straightforward calculation alleges ϕt − d∆ϕ = a(x, t)ϕ − b(x, t) u2ε − σu2




> a(x, t)ϕ − b(x, t) u2ε − σ 2 u2




> a(x, t)ϕ − 2b(x, t)uε ϕ. By the maximum principle (Theorem 7.1), ϕ > 0 in Ω × [0, T ]. As above, there exists ρ > 0 for which ϕ > ρu, and thus uε > (σ + ρ)u in QT . This is a contradiction with the definition of σ, and thereupon uε = u. Step 3. Proof of (7.12). Let v be the unique positive solution of (7.11). Using v(x, τ ) instead of v0 (x) for the fixed τ > 0, we may assume that ∂n v0 < 0 on ∂Ω when β = 1, and v0 (x) > 0 in Ω when 0 6 β < 1. Using Lemma 3.7, we conclude that there exist 0 < ε  1 and k  1 such that εφ(x, 0) 6 v0 (x) 6 ku(x, 0) for all x ∈ Ω. Clearly, ku is an upper solution of (7.9) as k > 1 and b > 0. Let z and Z be solutions of (7.11) with initial data εφ(x, 0) and ku(x, 0), respectively. Then z 6 v 6 Z in Q∞ by the comparison principle. On the other hand, from the discussion of Step 2, we see that lim z(x, t + iT ) = u(x, t) uniformly in QT .

i→∞

Analogously, lim Z(x, t + iT ) = u(x, t) uniformly in QT .

i→∞

Recalling z 6 v 6 Z, we get (7.12), and the proof is complete.



176  7. Time-Periodic Parabolic Boundary Value Problems

7.3.2

The case of half line

In this subsection, we study the existence and uniqueness of positive solutions to the following time-periodic parabolic boundary value problem in half line  u − duxx = a(x, t)u − b(x, t)u2 ,    t

0 < x < ∞, 0 < t 6 T,

B u(0, t) = 0,

0 < t 6 T,

  

u(x, 0) = u(x, T ),

0 < x < ∞,

(7.13)

where B u = βu − (1 − β)ux and 0 6 β 6 1 is a constant. Define Π` = [0, `] × [0, T ] and Π∞ = [0, ∞) × [0, T ] for simplicity. We assume that functions a and b satisfy (I)

a, b ∈ C α, α/2 (Π∞ ) ∩ L∞ (Π∞ ) are time-periodic with period T , a is positive somewhere, and b1 6 b 6 b2 for some positive constants b1 and b2 .

The strategy in the following arguments is to approximate (7.13) with a timeperiodic parabolic boundary value problem in a bounded interval (0, `). We first consider the following problem with ` > 0  u − duxx = a(x, t)u − b(x, t)u2 ,    t   

0 < x < `, 0 < t 6 T,

B u(0, t) = u(`, t) = 0,

0 < t 6 T,

u(x, 0) = u(x, T ),

0 < x < `.

(7.14)

Let λ(`; d, a) be the principal eigenvalue of (7.8). Same as Theorem 7.10, we have Theorem 7.11 If λ(`; d, a) < 0, then (7.14) has a unique positive solution u` . Moreover, for any given v0 ∈ C 1 ([0, `]) ∩ Wp2 ((0, `)) satisfying βv0 (0) − (1 − β)v00 (0) = v0 (`) = 0 and v0 > 0 in (0, `), the unique solution v of the initial-boundary value problem  v − dvxx = a(x, t)v − b(x, t)v 2 ,    t

0 < x < `, t > 0,

B v(0, t) = v(`, t) = 0,

t > 0,

  

v(x, 0) = v0 (x),

0 1 and xi satisfying xi → ∞ as i → ∞, so that a(x, t) > ςxρ in [xi , kxi ] × [0, T ]. Then λ(∞; d, a) < 0 for any d > 0.

7.3. The logistic equation  177

Proof. Let λD (a) and γ(a), respectively, be the principal eigenvalues of    φt − dφxx − a(x, t)φ = λφ, 

0 < x < kxi , 0 < t 6 T,

φ(0, t) = φ(kxi , t) = 0,

0 < t 6 T,

  

φ(x, 0) = φ(x, T ),

0 < x < kxi

and  ψ − dψxx − a(x, t)ψ = γψ,    t   

xi < x < kxi , 0 < t 6 T,

ψ(xi , t) = ψ(kxi , t) = 0,

0 < t 6 T,

ψ(x, 0) = ψ(x, T ),

xi < x < kxi .

Applying Theorem 7.9, we conclude λ(kxi ; d, a) 6 λD (a) < γ(a). Owing to ρ 6 0 and k > 1, one has a(x, t) > ςxρ > ςk ρ xρi in [xi , kxi ] × [0, T ]. Since γ(a) is monotonically decreasing in a, it follows γ(a) 6 γ(ςk ρ xρi ), and then λ(kxi ; d, a) < γ(ςk ρ xρi ) for all i > 1.

(7.15)

Let ψ(x) be the positive eigenfunction corresponding to γ(ςk ρ xρi ), and set y = x/xi and Ψ(y) = ψ(x). Then Ψ satisfies (

ρ ρ ρ ρ −dx−2 i Ψyy − ςk xi Ψ = γ(ςk xi )Ψ,

1 < y < k,

Ψ(1) = Ψ(k) = 0. Denote λ∗ as the principal eigenvalue of (

−u00 = λu, 1 < y < k, u(1) = u(k) = 0.

Then we conclude −2 ρ ρ ∗ ρ 2+ρ γ(ςk ρ xρi ) = dλ∗ x−2 ) < 0 when i  1, i − ςk xi = xi (dλ − ςk xi

since 2 + ρ > 0 and xi → ∞ as i → ∞. This combined with (7.15) derives that λ(∞; d, a) < 0.  Theorem 7.13 Assume that there exist constants a0 , b0 , a0 , b0 > 0 and −2 < ρ 6 0 such that  max[0,T ] a(x, t) min[0,T ] a(x, t)    a0 = lim inf , a0 = lim sup , ρ ρ x→∞

x

   b0 = lim inf min b(x, t), x→∞ [0,T ]

x b = lim sup max b(x, t). x→∞

0

x→∞

[0,T ]

(7.16)

178  7. Time-Periodic Parabolic Boundary Value Problems

Then (7.13) has a unique positive solution u ∈ C 2+α, 1+α/2 (Π∞ ). Moreover, a0 u(x, t) u(x, t) a0 6 lim inf 6 lim sup 6 x→∞ b0 xρ xρ b0 x→∞

(7.17)

uniformly in [0, T ]. Proof. Here we only handle the case 0 < β < 1. This proof will be divided into three steps. In the first one, we construct the minimal positive solution of (7.13). The estimate (7.17) will be given in the second step. Finally, we show the uniqueness of positive solutions. It is worth mentioning that (7.16) implies the assumption (I1). Step 1. The existence. In this step, we shall construct a positive solution u and prove that it is the minimal one. Let ` > 0 and λ(`; d, a) be the principal eigenvalue of (7.8). Since the assumption (I1) holds, by use of Proposition 7.12, there exists `0  1 so that λ(`; d, a) < 0 when ` > `0 . For each fixed ` > `0 , utilizing Theorem 7.11, problem (7.14) has a unique positive solution u` . Obviously, u` 6 kak∞ /b1 by the maximum principle, where b1 is given by (I). When `∗ > `, it is easy to see that u`∗ is an upper solution of (7.14). Let φ be the positive eigenfunction of (7.8) corresponding to λ(`; d, a) and ε > 0 be a constant. Then εφ is a positive lower solution of (7.14), εφ 6 u`∗ provided ε  1, and εφ(x, 0) ∈ Wp2 ((0, `)), B (εφ(0, 0)) = εφ(`, 0) = 0. Thus u`∗ > u` since u` is the unique positive solution of (7.14). This shows that u` is monotonically increasing in `. Making use of the regularity theory and compactness argument, it can be proved that there exists a positive function u ∈ C 2,1 (Π∞ ) such that lim`→∞ u` = u in C 2,1 (ΠL ) for any L > 0 and u solves (7.13). Let u be a positive solution of (7.13). Then u 6 kak∞ /b1 by the maximum principle. Obviously, u is an upper solution of (7.14) for any given ` > 0. As above, u` 6 u in Π` for all ` > `0 , and hence u 6 u in Π∞ . This shows that u is the minimal positive solution of (7.13). Step 2. Proof of (7.17). For any positive solution u of (7.13), we have known that u 6 u 6 kak∞ /b1 in Π∞ . Set a(x) = min a(x, t), a ¯(x) = max a(x, t), [0,T ]

[0,T ]

b(x) = min b(x, t), ¯b(x) = max b(x, t). [0,T ]

[0,T ]

Remembering λ(∞; d, a ¯) 6 λ(∞; d, a) < 0 and invoking [22, Theorem 7.12], we have that the problem (

−dw¯ 00 = a ¯(x)w¯ − b(x)w¯ 2 ,

0 < x < ∞,

w¯ 0 (0) = 0 has a unique positive solution w, ¯ and lim sup x→∞

w(x) ¯ a0 6 . xρ b0

(7.18)

7.3. The logistic equation  179

For ` > `0 and θ = kak∞ /b1 , where b1 is given by (I), the problem (

−dw¯ 00 = a ¯(x)w¯ − b(x)w¯ 2 ,

0 < x < `,

w¯ 0 (0) = 0, w(`) ¯ =θ

(7.19)

has a unique positive solution w¯` , and w¯` is globally asymptotically stable. Therefore, the solution v of  v − dvxx = a ¯(x)v − b(x)v 2 ,    t   

0 < x < `, t > 0,

vx (0, t) = 0, v(`, t) = θ,

t > 0,

v(x, 0) = θ,

0 0, where w is a positive solution of (7.21) and w0 (0) > 0. Obviously, w` is a lower solution of (7.14). Hence w` 6 u` since u` is the unique

180  7. Time-Periodic Parabolic Boundary Value Problems

positive solution of (7.14). This implies w 6 u 6 u in Π∞ on account of w` → w and u` → u as ` → ∞. If w0 > 0 in [0, ∞), then w(x) → w∗ as x → ∞ for some positive constant w∗ . As ρ 6 0, it is immediate to get lim inf x→∞

u(x, t) w(x) > lim inf ρ > w∗ ρ x→∞ x x

(7.22)

uniformly in [0, T ]. In this case we claim that ρ = 0 and lim inf a(x) w∗ >

x→∞

lim sup ¯b(x)

=

a0 . b0

(7.23)

x→∞

In fact, if ρ < 0, then limx→∞ a(x, t) = 0 uniformly in [0, T ] by (7.16), and consequently limx→∞ a(x) = 0. This is impossible by noting that function ¯b has a positive lower bound, w0 > 0 in [0, ∞) and w is bounded from above. Therefore, ρ = 0. Now, we are going to prove (7.23). If it was not valid, then there would exist ε > 0 and x∗  1 so that w(x) < a0 /b0 − ε for all x > x∗ . For such an ε > 0, we can find δ > 0 and x0 > x∗ such that a0 − δ a0 − 0 + ε =: σ > 0, b0 + δ b and a(x) > a0 − δ and ¯b(x) < b0 + δ for all x > x0 . It then follows   −dw00 = w a(x) − ¯b(x)w !

a(x) a0 > w¯b(x) − 0 +ε ¯b(x) b > σ¯b(x)w for all x > x0 ,

which is impossible since σ > 0 and ¯b(x) has a positive lower bound. Hence, (7.23) holds. If w0 (x0 ) = 0 for some x0 > 0, we set a∗ (x) = a(x + x0 ), ¯b∗ (x) = ¯b(x + x0 ), v(x) = w(x + x0 ) for x > 0, a∗ (x) = a(x0 − x), ¯b∗ (x) = ¯b(x0 − x), v(x) = w(x0 − x) for x < 0. Then the function v(x) satisfies −dv 00 = a∗ (x)v − ¯b∗ (x)v 2

in R.

(7.24)

Applying [22, Theorem 7.12], we conclude that v(x) is the unique positive solution of (7.24) and satisfies a∗ (x) a(x) lim inf ρ v(x) ρ x→∞ x→∞ x x = a0 . lim inf ρ > = ∗ ¯ ¯ x→∞ x b0 lim sup b (x) lim sup b(x) lim inf x→∞

x→∞

7.4. The upper and lower solutions method for systems  181

Accordingly, lim inf x→∞

w(x) v(x) u(x, t) a0 > lim inf ρ = lim inf ρ > 0 x→∞ x→∞ xρ x x b

(7.25)

uniformly in [0, T ]. It follows from (7.20), (7.22), (7.23) and (7.25) that u satisfies (7.17). Step 3. The uniqueness. Let u be a positive solution of (7.13). Then u > u, here u is the minimal positive solution of (7.13) obtained in Step 1. In order to prove uniqueness, it suffices to show that u ≡ u. If not, then u >, 6≡ u. It follows from (7.17) that there exists k > 1 such that u 6 ku in Π∞ . To arrive at a contradiction, we turn to a technique introduced by Marcus and V´eron [81]. Define z = u − (2k)−1 (u − u). Then k+1 2k 1 u >, 6≡ z > u and z+ u = u. (7.26) 2k 2k + 1 2k + 1 Noticing that function y 2 is convex for y ∈ R+ , we conclude u2 6

2k 2 1 z + u2 2k + 1 2k + 1

from the second formula of (7.26). Since b is positive, a direct computation gives zt − dzxx > a(x, t)z − b(x, t)z 2 . Obviously, for the large `, B z(0, t) = 0, z(`, t) > 0 in [0, T ] and z(x, 0) = z(x, T ) in [0, `].

Hence, z is an upper solution of (7.14). Let u` be the unique positive solution of (7.14). As above, u` 6 z in Π` . As a consequence, u 6 z in Π∞ due to the fact that liml→∞ u` = u in C 2,1 (ΠL ) for any L > 0. This contradicts the first inequality of (7.26). Therefore, u ≡ u, and the uniqueness is derived. 

7.4

THE UPPER AND LOWER SOLUTIONS METHOD FOR SYSTEMS

In this section, we study the upper and lower solutions method for the time-periodic parabolic boundary value problems of weakly coupled parabolic systems. Let ik , dk and [u]ik , [u]dk be defined as in §4.2.2. Now, we discuss the following time-periodic parabolic boundary value problem of weakly coupled parabolic system  L u = fk (x, t, uk , [u]ik , [u]dk )    k k    Bk uk = 0  uk (x, 0) = uk (x, T )     

in QT , on ST , in Ω,

(7.27)

1 6 k 6 m,

where Lk is strongly parabolic operator, coefficients of Lk are time-periodic with period T and satisfy the condition (C) which is given in §1.1, Bk = ak ∂n + bk , either ak = 0 and bk = 1, or ak = 1 and bk > 0 with bk ∈ C 1+α, (1+α)/2 (S T ), fk and bk are as well time-periodic with period T .

182  7. Time-Periodic Parabolic Boundary Value Problems

Definition 7.14 Given u¯ and u with u¯k , uk ∈ C(QT ) ∩ C 2,1 (QT ) when ak = 0, and u¯k , uk ∈ C 1,0 (QT ) ∩ C 2,1 (QT ) when ak = 1. Assume that u 6 u¯ and f is mixed quasi-monotonous in hu, u¯i. We say that (¯ u, u) is a pair of coupled ordered upper and lower solutions of (7.27) if  Lk u¯k > fk (x, t, u¯k , [¯ u]ik , [u]dk )       u]dk )  Lk uk 6 fk (x, t, uk , [u]ik , [¯  

in QT , in QT ,

B u¯ > 0 > B u

k k k k     u¯k (x, 0) > u¯k (x, T ), uk (x, 0) 6 uk (x, T )     

on ST , in Ω,

1 6 k 6 m.

Sometimes we call that u¯ and u are coupled ordered upper and lower solutions of (7.27). Theorem 7.15 (The upper and lower solutions method) Let u¯ and u be coupled ordered upper and lower solutions of (7.27) and f be mixed quasi-monotonous in ∂fk hu, u¯i. Assume that fk (·, u) ∈ C α, α/2 (QT ) are uniform in u ∈ hc, c¯i and ∂u ∈ C(QT × j hc, c¯i) for all 1 6 k, j 6 m, where c and c¯ are defined as in §4.2 (Theorem 4.4). Fix a number p > 1 + n/2. If there exists a function u0 ∈ Wp,2 B (Ω) satisfying u(x, 0) 6 u0 (x) 6 u(x, 0) in Ω, then (7.27) has at least one solution u ∈ [C 2+α, 1+α/2 (Ω × [0, T ])]m , and u 6 u 6 u¯, where Wp,2 B (Ω) = Wp,2 B1 (Ω) × · · · × Wp,2 Bm (Ω), and Wp,2 Bk (Ω) is defined as in §7.1 for 1 6 k 6 m. Proof. This proof is a combination of proofs of Theorem 7.3 and Theorem 4.9. Step 1. Define a set (

V =

u0 ∈

Wp,2 B (Ω)

:

u(x, 0) 6 u0 (x) 6 u(x, 0) in Ω, kuk0 kWp2 (Ω) 6 C(T, c, c¯), 1 6 k 6 m

)

,

where C(T, c, c¯) is a positive constant which will be given in (7.29) below. Then V is a bounded and closed convex set of [Wp2 (Ω)]m and V 6= ∅. For any given u0 ∈ V , let us consider initial-boundary value problem of a weakly coupled system  Lk uk = fk (x, t, uk , [u]ik , [u]dk )       Bk uk = 0  uk (x, 0) = uk0 (x)     

in QT , on ST , in Ω,

(7.28)

1 6 k 6 m.

It is easily checked that (¯ u, u) is a pair of coupled upper and lower solutions of (7.28). Hence, problem (7.28) has a unique solution u in hu, u¯i by Theorem 4.9. Certainly u(x, 0) 6 u(x, T ) 6 u(x, T ) 6 u¯(x, T ) 6 u¯(x, 0).

7.4. The upper and lower solutions method for systems  183

The remaining proof is similar to Theorem 7.3, and we shall give details for the completeness. Step 2. Set Fk (x, t) = fk (x, t, u(x, t)) and α0 = min{1, 2 − (n + 2)/p}. Since ul ∈ Wp2,1 (QT ) ,→ C α, α/2 (QT ) for all 1 6 l 6 m, 0 < α < α0 , it asserts Fk ∈ C α, α/2 (QT ), and hence uk ∈ Wp2,1 (QT ) ∩ C 2+α, 1+α/2 (Ω × (0, T ]) by Theorem 2.10. Certainly, uk (x, T ) ∈ Wp,2 Bk (Ω). Define an operator F : V → Wp,2 B (Ω) ∩ hu(x, 0), u¯(x, 0)i by F (u0 ) = u(x, T ).

As u 6 u 6 u, it deduces |Fk (x, t)| 6 C(c, c¯) := max |fk (x, t, u)| for all u0 ∈ V. QT ×[c, c¯]

In view of Theorem 1.8, for any 0 < δ < T , there exists a positive constant Ck (δ) such that kuk kWp2,1 (Ω×(δ,T )) 6 Ck (δ) (kFk kp, QT + kuk kp, QT ) 6 Ck (δ, c, c¯) for all u0 ∈ V. Then, by the embedding theorem (Theorem A.7), |uk |α, Ω×[δ,T ] 6 Ck (δ, c, c¯) for all u0 ∈ V, which implies |Fk |α, Ω×[δ,T ] 6 Ck (δ, c, c¯) for all u0 ∈ V. Set δ = T /3 and τ = T /2. Taking advantage of Theorem 1.14 (2), there is a constant Ck for which (2.15) holds, i.e., 

|uk |2+α, Ω×[T /2, T ] 6 Ck |Fk |α, Ω×[T /3, T ] + kuk k∞, Ω×(T /3, T )



6 Ck (T, c, c¯) for all u0 ∈ V. In particular, kuk (·, T )kC 2+α (Ω) 6 Ck (T, c, c¯) for all u0 ∈ V, 1 6 k 6 m, and consequently kF (u0 )kC 2+α (Ω) = ku(x, T )kC 2+α (Ω) 6 C(T, c, c¯) for all u0 ∈ V.

(7.29)

It follows that F : V → V . Step 3. We shall show that F : V → V is a compact operator. By the above estimate, it suffices to prove that F : V → V is continuous. Let ui0 ∈ V with ui0 → u0 in [Wp2 (Ω)]m as i → ∞. Denote ui := ui (x, t) as the unique solution of (7.28) with ui (x, 0) = ui0 (x). Set wi = ui − u and write fk (x, t, ui ) − fk (x, t, u) =

m X ∂fk

∂uj j=1

(x, t, v i (x, t))wji =: −

m X j=1

cikj (x, t)wji ,

184  7. Time-Periodic Parabolic Boundary Value Problems

where v i (x, t) is between ui (x, t) and u(x, t). Clearly, cikj ∈ L∞ (QT ). It then follows that  m X  i   L w + cikj (x, t)wji = 0 k  k 

in QT ,

j=1

 Bk wki = 0     i

wk (x, 0) =

on ST , uik0 (x)

− uk0 (x)

in Ω,

and Theorem 1.20 gives kwki kWp2,1 (QT ) 6 Ckuik0 − uk0 kWp2 (Ω) . This implies |ui − u|α, QT = |wi |α, QT 6 Ckui0 − u0 kWp2 (Ω) → 0, and thereafter fk (·, ui (·)) → fk (·, u(·)) in C(QT ) as i → ∞ for all 1 6 k 6 m. In almost exactly the same way as the proof of Theorem 7.3, we can verify that fk (·, ui (·)) → fk (·, u(·)) in C γ, γ/2 (QT ) and |uik − uk |2+γ, Ω×[T /3,T ] → 0 as i → ∞ for any 0 < γ < α, which illustrates that kuik (x, T ) − uk (x, T )kC 2+γ (Ω) → 0 for all 1 6 k 6 m. Thus F : V → V is continuous. Step 4. Repeating the last part in proving Theorem 7.3, we can show that (7.27) has at least one solution u and u possesses the desired properties. 

7.5 A DIFFUSIVE COMPETITION MODEL As an application we study the following time-periodic parabolic boundary value problem of competition model  2   ut − d1 ∆u = a(x, t)u − u − kuv     vt − d2 ∆v = b(x, t)v − v 2 − huv

in QT ,

 B1 u = B2 v = 0     

on ST ,

u(x, 0) = u(x, T ), v(x, 0) = v(x, T )

in QT ,

(7.30)

in Ω,

where di , k and h are positive constants, and Bi w = βi w +(1−βi )∂n w with constants 0 6 βi 6 1, i = 1, 2. We assume that functions a, b ∈ C α, α/2 (QT ) are time-periodic with period T . The content of this part refers to [118]. Let λi (a) denote the principal eigenvalue of (7.10) with d = di and B = Bi , i = 1, 2. Define a0 = maxQT a(x, t) and b0 = maxQT b(x, t). From discussions of §7.3 we see that the strategy (key point) is to make sure that the principal eigenvalue is negative. The same is reasonable for the model (7.30). Theorem 7.16 Assume a0 , b0 > 0 and λ1 (a − kb0 ) < 0, λ2 (b − ha0 ) < 0. Then there exist four positive time-periodic functions u∗ , u∗ , v ∗ , v∗ ∈ C 2+α, 1+α/2 (QT ), for which both (u∗ , v∗ ) and (u∗ , v ∗ ) are positive solutions of (7.30). Moreover, any positive solution (u, v) of (7.30) satisfies u∗ 6 u 6 u∗ and v∗ 6 v 6 v ∗ in QT .

(7.31)

7.5. A diffusive competition model  185

Proof. The proof will be divided into four steps. Step 1. Construction of a pair of coupled ordered upper and lower solutions ((¯ u, v¯), (u, v)) of (7.30). First of all, λ1 (a) < λ1 (a − kb0 ) < 0 since b0 > 0. Based on Theorem 7.10, the following problem  u − d1 ∆u = a(x, t)u − u2    t

in QT ,

B1 u = 0

on ST ,

  

u(x, 0) = u(x, T )

in Ω

(7.32)

has a unique positive solution, denoted by u¯. Moreover, the maximum principle gives u¯ 6 a0 in QT . Noticing λ2 (b − ha0 ) < 0, the following problem  v − d2 ∆v = [b(x, t) − ha0 ]v − v 2    t   

in QT ,

B2 v = 0

on ST ,

v(x, 0) = v(x, T )

in Ω

(7.33)

admits a unique positive solution, denoted by v. Similarly, the following problems  2   vt − d2 ∆v = b(x, t)v − v    

in QT ,

B2 v = 0

on ST ,

v(x, 0) = v(x, T )

in Ω

(7.34)

and  2   ut − d1 ∆u = [a(x, t) − kb0 ]u − u    

in QT ,

B1 u = 0

on ST ,

u(x, 0) = u(x, T )

in Ω

(7.35)

have unique positive solutions v¯ and u, respectively, and v¯ 6 b0 . Using the upper and lower solutions method and uniqueness of positive solutions for the logistic equation, we can prove u 6 u¯ and v 6 v¯. Taking into account of (7.32)–(7.35), it is easy to verify that ((¯ u, v¯), (u, v)) is a pair of coupled ordered upper and lower solutions of (7.30). Theorem 7.15 avers that problem (7.30) has at least one positive solution (u, v) satisfying u 6 u 6 u¯ and v 6 v 6 v¯. Step 2. Construction of (u∗ , v∗ ). Our intention is to use functions u¯ and v obtained above to construct a monotone sequence {(wi , z i )} and then determine its limit. Let (w, z) be the unique solution of the initial-boundary value problem  wt − d1 ∆w = w (a(x, t) − w − kz)       zt − d2 ∆z = z (b(x, t) − z − hw)

in Q∞ ,

 B1 w = B2 z = 0     

on S∞ ,

w(x, 0) = u¯(x, 0), z(x, 0) = v(x, 0)

in Q∞ , in Ω.

Then w and z are positive, and the comparison principle gives u 6 w 6 u¯ and v 6 z 6 v¯ in Q∞ .

(7.36)

186  7. Time-Periodic Parabolic Boundary Value Problems

For any chosen non-negative integer i, we define wi (x, t) = w(x, t + iT ) and z i (x, t) = z(x, t + iT ). Noting that a and b are time-periodic functions with period T , it is easy to see that (wi , z i ) satisfies  i
wt − d1 ∆wi = wi a(x, t) − wi − kz i    
  z i − d2 ∆z i = z i b(x, t) − z i − hwi t

 B1 wi = B2 z i = 0      i

in QT , in QT , on ST ,

w (x, 0) = w(x, iT ), z i (x, 0) = z(x, iT )

in Ω.

Remembering w1 (x, 0) = w(x, T ) 6 u¯(x, T ) = u¯(x, 0) = w(x, 0), z 1 (x, 0) = z(x, T ) > v(x, T ) = v(x, 0) = z(x, 0), the comparison principle declares w1 6 w and z 1 > z in QT . And then w2 (x, 0) = w1 (x, T ) 6 w(x, T ) = w1 (x, 0), z 2 (x, 0) = z 1 (x, T ) > z(x, T ) = z 1 (x, 0). As above, w2 6 w1 and z 2 > z 1 in QT . Utilizing the inductive method, we can show that wi and z i are, respectively, monotonically decreasing and monotonically increasing in i. As a consequence, limits limi→∞ wi (x, t) = u∗ (x, t) and limi→∞ z i (x, t) = v∗ (x, t) exist. In addition, u∗ , v∗ > 0 in QT . Obviously, u∗ (x, 0) = u∗ (x, T ) and v∗ (x, 0) = v∗ (x, T ) for x ∈ Ω, since wi+1 (x, 0) = wi (x, T ) and z i+1 (x, 0) = z i (x, T ). By use of the regularity theory and compactness argument, it can be proved that wi → u∗ and z i → v∗ in C 2,1 (QT ) as i → ∞, and (u∗ , v∗ ) satisfies the first three equations of (7.30). Therefore, (u∗ , v∗ ) is a solution of (7.30). Step 3. Construction of (u∗ , v ∗ ). The idea is similar to Step 2. Let (ϕ, ψ) be the unique solution of the initial-boundary value problem    ϕt − d1 ∆ϕ = ϕ (a(x, t) − ϕ − kψ)     ψt − d2 ∆ψ = ψ (b(x, t) − ψ − hϕ)

in Q∞ ,

 B1 ϕ = B2 ψ = 0,     

on S∞ ,

ϕ(x, 0) = u(x, 0), ψ(x, 0) = v¯(x, 0)

in Q∞ , in Ω.

Then ϕ and ψ are positive. Similar to above, lim ϕ(x, t + iT ) = u∗ (x, t) and

i→∞

lim ψ(x, t + iT ) = v ∗ (x, t) in C 2,1 (QT ),

i→∞

u∗ , v ∗ > 0 in QT , and (u∗ , v ∗ ) solves (7.30).

Exercises  187

Obviously, u∗ > u∗ and v ∗ > v∗ . Step 4. Proof of (7.31). Let (u, v) be a positive solution of (7.30). We only prove u 6 u∗ and v > v∗ , as the proofs of u > u∗ and v 6 v ∗ are similar. Recall that u satisfies  u − d1 ∆u < a(x, t)u − u2    t

in QT ,

B1 u = 0

on ST ,

  

u(x, 0) = u(x, T )

in Ω,

and u¯ is the unique positive solution of (7.32). We can deduce u 6 u¯ by the upper and lower solutions method and the uniqueness of u¯. Moreover, u 6 a0 in QT by the maximum principle. Consequently, v satisfies  v − d2 ∆v > [b(x, t) − ha0 ]v − v 2    t   

in QT ,

B2 v = 0

on ST ,

v(x, 0) = v(x, T )

in Ω.

It follows that v > v since v is the unique positive solution of (7.33). Noting that (w, z) is the unique positive solution of (7.36), w(x, 0) = u¯(x, 0) > u(x, 0) and z(x, 0) = v(x, 0) 6 v(x, 0), we conclude u 6 w and v > z by the comparison principle. Certainly u(x, t) = u(x, t + iT ) 6 w(x, t + iT ) → u∗ (x, t), v(x, t) = v(x, t + iT ) > z(x, t + iT ) → v∗ (x, t). The proof is finished.



EXERCISES 7.1 Consider the time-periodic parabolic boundary value problem  L u = f (x, t, u)   

in QT ,

B u = g(x, t)

on ST ,

  

u(x, 0) = u(x, T )

in Ω,

where L , B and f are as in (7.1), and g is time-periodic with period T and g ∈ Wp2,1 (QT ). Establish the upper and lower solutions method. R 7.2 Prove λN 1 < λ1 in Theorem 7.7.

7.3 Consider the time-periodic parabolic boundary value problem  u − d∆u = a(x, t)u − b(x, t)uk    t

in QT ,

u=0

on ST ,

  

u(x, 0) = u(x, T )

in Ω,

(P7.1)

where k > 2 is an integer, functions a, b ∈ C α,α/2 (QT ) are time-periodic with period T , and b1 6 b 6 b2 in QT for some positive constants b1 , b2 . Let λ(a) be the principal eigenvalue of (7.10) with boundary condition φ = 0 on ST . Prove the following conclusions.

188  7. Time-Periodic Parabolic Boundary Value Problems

(a) Suppose that k > 3 is odd. If λ(a) < 0, then problem (P7.1) admits a unique positive solution and a unique negative solution. If λ(a) > 0, then problem (P7.1) admits only the trivial solution u = 0. (b) Suppose that k > 2 is even. If λ(a) < 0, then problem (P7.1) has a unique positive solution, and does not have any negative solution. If λ(a) > 0, then problem (P7.1) has no positive solution. (c) Let λ(a) < 0 and u be the unique positive solution of problem (P7.1). Discuss stability of u in the time-periodic sense. 7.4 Prove that problem (7.19) has a unique positive solution w¯` and w ¯` is globally asymptotically stable. Hint: use the upper and lower solutions method to prove that (7.19) has a minimal positive solution, and then use the integration method or the method of Step 3 in proving Theorem 7.13 to prove the uniqueness. 7.5 Complete the proof of Step 4 of Theorem 7.15. 7.6 Use the upper and lower solutions method to discuss the existence of positive solutions to a time-periodic parabolic boundary value problem    v   u − ∆u = u a(x, t) − u − b(x, t)  t  v+ρ       u w    vt − ∆v = v c(x, t) − f (x, t) −   v+ρ w+σ   v   − h(x, t) wt − ∆w = w g(x, t)   w+σ      u=v=w=0    

(u, v, w)|t=0 = (u, v, w)|t=T

in QT , in QT , in QT , on ST , in Ω,

where functions a, b, c, f, g, h ∈ C α, α/2 (QT ) are time-periodic with period T , ρ and σ are positive constants.

CHAPTER

8

Free Boundary Problems from Ecology

Free boundary problems deal with systems of partial differential equations, where a part of the domain boundary is unknown. They arise in many contexts in the real world. Some examples include heat transfer problems with phase-changes such as from liquid to solid, oxidation or damage of surfaces of materials, infiltration of groundwater, distribution of oil in the process of oil exploitation, wound healing, tumor growth, option pricing, spreads of (invasive) species and epidemic. The earliest free boundary model was put forward by Joseph Stefan when he studied the interface between ice and water in the process of icing. Sometimes, this kind of problem is called the Stefan problem, and the corresponding free boundary condition is called the Stefan condition. This chapter only focuses on free boundary problems from ecological models. The main content includes the existence and uniqueness, regularity and uniform estimates of solutions, longtime behaviour of solutions, criteria and dichotomy for spreading and vanishing, and the asymptotic speed of the free boundary. Examples of ecological models are studied in detail. Throughout this chapter, for convenience, we use function u(t, x) and the Banach spaces Wp1,2 (Q), C (i+α)/2, i+α (Q) (i = 0, 1, 2) instead of u(x, t) and Wp2,1 (Q), C i+α, (i+α)/2 (Q), respectively. Most theoretical approaches are based on or started with single-species models. In consideration of the heterogeneous environment, the following problem  u − d∆u = a(t, x)u − b(t, x)u2 ,    t

t > 0, x ∈ Ω,

B u = 0,

t > 0, x ∈ ∂Ω,

  

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

x∈Ω

is a typical model to describe spread, persistence and extinction of new or invasive species and has received an astonishing amount of attention. In this model, u(t, x) represents population density, constant d > 0 denotes diffusion (dispersal) rate, a(t, x) and b(t, x) respectively represent intrinsic growth rate and self-limitation coefficient

189

190  8. Free Boundary Problems from Ecology

of the species, Ω is a bounded domain of Rn , boundary operator B = β + (1 − β)∂n with constant 0 6 β 6 1, n is the outward unit normal vector of boundary ∂Ω. In most spreading processes in the natural world, a spreading front can be observed. When a new or invasive species initially occupies a region Ω0 with density u0 (x), it is natural to expect that as time t increases Ω0 will evolve into an expanding region Ω(t) with an expanding front ∂Ω(t), and initial function u0 (x) will evolve into a positive function u(t, x) vanishing on moving boundary ∂Ω(t).

8.1 DEDUCTION OF FREE BOUNDARY CONDITIONS The content of this section refers to the reference [12]. In the process of population range expansion, population density near the propagating front is assumed to be close to zero. According to Fick’s first law, diffusion flux of u is J = −d∇u. Outflow per unit area per unit time through boundary ∂Ω(t) is −d∇u · n, where n is the outward unit normal vector of ∂Ω(t). For a small increment ∆t, during the period from t to t + ∆t, individual numbers entering region Ω(t + ∆t) \ Ω(t) through boundary ∂Ω(t) are approximate to −d∇u · n × ∆t (the contribution of reaction term can be ignored as it is close to zero near the boundary). As the front enters a new unpopulated environment, pioneering members at the front, with very low population density, are particularly vulnerable (note that since only one species is considered here, some existing interacting species are regarded as part of environment). Thus, it is plausible to assume that as expanding front propagates the population suffers a loss of k units per unit volume at the front. For simplicity, we assume that k is a constant for a given species in a given homogeneous environment. In order to simplify mathematics, here we only consider one dimensional case, i.e., n = 1. Let x = g(t) and x = h(t) be left and right free boundaries, respectively. Set Ω(t) = (g(t), h(t)), then ∂Ω(t) = {g(t), h(t)} and Ω(t + ∆t) \ Ω(t) = (g(t + ∆t), g(t)) ∪ (h(t), h(t + ∆t)). Average densities of u in regions (g(t + ∆t), g(t)) and (h(t), h(t + ∆t)) are dux (t, g(t))∆t g(t) − g(t + ∆t)

and

−dux (t, h(t))∆t , h(t + ∆t) − h(t)

respectively. Evidently, they tend to −dux (t, g(t))/g 0 (t) and − dux (t, h(t))/h0 (t) as ∆t → 0, respectively. These limits are known as diffusion pressures on free boundaries, which are average population densities there. If we assume that such densities near fronts are kept at the “preferred density level”, then the above two quantities should be equal to k. We thus have the following free boundary conditions g 0 (t) = −µux (t, g(t)) and h0 (t) = −µux (t, h(t)) with µ = dk −1 . If environments on the left and right are different, then the above free boundary conditions can be written as g 0 (t) = −µ1 ux (t, g(t)) and h0 (t) = −µ2 ux (t, h(t)) with µi = dki−1 .

8.2. Existence and uniqueness of solutions  191

If the preferred density level k is not a constant, then the above free boundary conditions become g 0 (t) = −µ1 (t, g(t))ux (t, g(t)) and h0 (t) = −µ2 (t, h(t))ux (t, h(t)).

8.2 EXISTENCE AND UNIQUENESS OF SOLUTIONS In this section, we will establish a general result on existence and uniqueness of solutions. Most materials of this section are taken from [115]. Another method for the existence and uniqueness was given in [25]. We consider the following free boundary problem  ut − duxx = f (t, x, u),       B u(t, 0) = u(t, h(t)) = 0, 0

 h (t) = −µux (t, h(t)),     

t > 0, 0 < x < h(t), t > 0,

(8.1)

t > 0,

h(0) = h0 , u(0, x) = u0 (x), 0 < x < h0 ,

where h0 denotes the size of initial habitat, B u = βu − (1 − β)ux with constant 0 6 β 6 1, and initial function u0 (x) satisfies u0 ∈ Wp2 ((0, h0 )) with p > 3, u0 > 0, 6≡ 0 in (0, h0 ), and B u0 (0) = u0 (h0 ) = 0. Function f satisfies (J) f (t, x, 0) > 0. For any given τ , l, k > 0, f ∈ L∞ ((0, τ ) × (0, l) × (0, k)) and there exists a constant L1 (τ, l, k) such that, for all t ∈ [0, τ ], x ∈ [0, l] and u, v ∈ [0, k], |f (t, x, u) − f (t, x, v)| 6 L1 (τ, l, k)|u − v|. Denote h∗ = −µu00 (h0 ) and n

o

K = h0 , h∗ , ku0 kWp2 ([0,h0 ]) . In order to save space, for any given interval I ⊂ R+ = [0, ∞), we write simply I × [0, h(t)] =

[

{t} × [0, h(t)], I × (0, h(t)) =

t∈I

[

{t} × (0, h(t)).

t∈I

For any given 0 < T < ∞, we set DT = (0, T ] × (0, h(t)). And for any given bounded set Q ⊂ R2 and constant 0 < α < 1, we write | · |i+α, Q := k · kC (i+α)/2, i+α (Q) , i = 0, 1, 2. Theorem 8.1

Assume that the condition (J) holds and fix 0 < α < 1 − 3/p.

192  8. Free Boundary Problems from Ecology

(1) There is a constant 0 < T  1 depending only on K such that (8.1) has at least one solution (u, h) ∈ Wp1,2 (DT ) × C 1+α/2 ([0, T ]) and u > 0 in DT , h0 (t) > 0 in (0, T ]. Moreover, there exists a constant C(K) > 0 such that kukWp1,2 (DT ) + |u|1+α, DT + khkC 1+α/2 ([0,T ]) 6 C(K).

(8.2)

(2) If, in addition, f is also locally Lipschitz continuous in x, i.e., for any given τ , l, k > 0, there exists a constant L2 (τ, l, k) such that, for all t ∈ [0, τ ], x, y ∈ [0, l] and u ∈ [0, k], |f (t, x, u) − f (t, y, u)| 6 L2 (τ, l, k)|x − y|,

(8.3)

then the solution (u, h) of (8.1) is unique. (3) Under above assumptions, if we further assume that, for any given τ > 0, f (·, x, u) ∈ C α/2 ([0, τ ]) uniformly in (x, u) when (x, u) varies in a bounded set of R+ × R+ , i.e., for any given τ , l, k > 0, there exists a constant L3 (τ, l, k) such that kf (·, x, u)kC α/2 ([0,τ ]) 6 L3 (τ, l, k) for all x ∈ [0, l], u ∈ [0, k],

(8.4)

then the unique solution u ∈ C 1+α/2, 2+α ((0, T ] × [0, h(t)]). What we need to emphasize here is that condition (8.3) is not needed for the uniqueness of solutions of the initial value problem and initial-boundary value problem with known boundary. Condition (8.3) reveals the important difference between free boundary problem and initial value problem and initial-boundary value problem with known boundary. Proof of Theorem 8.1. (1) We first straighten the free boundary. Let y = x/h(t), v(t, y) = u (t, h(t)y) . Then (v, h) satisfies    vt − ρ(t)vyy − ξ(t)yvy = f (t, h(t)y, v), 

0 < t 6 T, 0 < y < 1,

  

0 0 and f (t, h(t)y, v) is Lipschitz continuous in v, we conclude v > 0 in (0, T ] × (0, 1) and vy (t, 1) < 0 in (0, T ] by Theorem 1.40 (2) and Lemma 1.39. On the other hand, by continuous dependence on the given data, v continuously depends on h ∈ ΘT in the space C (1+α)/2, 1+α (∆T ). For such a definite function v, ˜ = h(t; ˜ h). Then the initial value problem (8.6) has a unique solution, denoted by h(t) ˜h(0) = h0 , h ˜ 0 (0) = h∗ and ˜ 0 (t) > 0, h

˜ 0 ∈ C α/2 ([0, T ]), h

˜ 0k α kh 6 C2 (K) for all h ∈ ΘT . C 2 ([0,T ])

(8.8)

˜ continuously depends on v ∈ C (1+α)/2, 1+α (∆T ) in the space C 1 ([0, T ]), and Clearly, h so does on h ∈ ΘT . Now we define an operator F : ΘT → C 1 ([0, T ]) by ˜ F(h) = h. Obviously, F is continuous in ΘT , and h ∈ ΘT is a fixed point of F if and only if (v, h) solves (8.5).

194  8. Free Boundary Problems from Ecology

According to (8.8), we know that F is compact, and ˜ 0 − h∗ kC([0,T ]) 6 kh ˜ 0k α kh T α/2 6 C2 (K)T α/2 . C 2 ([0,T ]) As a result, F maps ΘT into itself if h0 −2/α , C2 (K) . T 6 min 1, 2(1 + h∗ ) 



Consequently, F has at least one fixed point h ∈ ΘT by the Schauder fixed point theorem (Corollary A.3), and then problems (8.5) and (8.6) have at least one solution (v, h) defined in [0, T ]. Clearly, h ∈ C 1+α/2 ([0, T ]), h0 (t) > 0 for t > 0, and function u(t, x) = v(t, h−1 (t)x) satisfies u ∈ Wp1,2 (DT ) ∩ C (1+α)/2,1+α (DT ), u > 0 in DT . Thus (u, h) solves (8.1). From the above arguments we see that the estimate (8.2) holds. (2) Let (ui , hi ) ∈ Wp1,2 (DT ) × C 1+α/2 ([0, T ]) (i = 1, 2) be two solutions of (8.1) and 0 < T  1. Then ui > 0 in DT , h0i (t) > 0 in (0, T ], and (ui , hi ) satisfies the estimate (8.2). We may assume that h0 6 hi 6 h0 + 1 in [0, T ], ui 6 ku0 k∞ + 1 in [0, T ] × [0, hi (t)], i = 1, 2. Define vi (t, y) = ui (t, hi (t)y) ,

0 6 t 6 T, 0 6 y 6 1

for i = 1, 2, and set v = v1 − v2 , h = h1 − h2 . Then  vt − ρ1 (t)vyy − ξ1 (t)yvy − a(t, y)v = [ρ1 (t) − ρ2 (t)] v2yy       + [ξ1 (t) − ξ2 (t)] yv2y + b(t, y)yh, 0 < t 6 T, 0 < y < 1,  B1 v(t, 0) = −βhv2 (t, 0), v(t, 1) = 0, 0 < t 6 T,     

v(0, y) = 0,

(8.9)

0 0, and (u, h) ∈ Wp1,2 (DT ) × C 1+α/2 ([0, T )) be the unique solution of (8.1) verifying u 6 M for some M > 0. If f is bounded in R+ × R+ × (0, M ), then there exists a positive constant K, which depends on M but not on T , such that h0 (t) 6 2µM K =: C(M ) in [0, T ).

196  8. Free Boundary Problems from Ecology

Proof. The idea of this proof comes from [25, Lemma 2.2]. Set A=

sup

f (t, x, u),

R+ ×R+ ×(0,M )

and define a comparison function by h

w(t, x) = M 2K(h(t) − x) − K 2 (h(t) − x)2

i

for some appropriate positive constant K > 1/h0 over the region Λ = {(t, x) : 0 < t < T, h(t) − K −1 < x < h(t)}. First of all, one can easily compute that, for any (t, x) ∈ Λ, wt = 2M K[1 − K(h(t) − x)]h0 (t) > 0, −wxx = 2M K 2 . It follows that, when K 2 >

A 2dM ,

wt − dwxx > 2dM K 2 > f (t, x, u) = ut − duxx

in Λ.

It is clear that w(t, h(t) − K −1 ) = M > u(t, h(t) − K −1 ), w(t, h(t)) = 0 = u(t, h(t)) for all 0 < t < T . Taking advantage of u0 (x) = −

Z

h0

x

u00 (y)dy 6 − min u00 (x)(h0 − x), x ∈ [h0 − K −1 , h0 ], [0,h0 ]

w(0, x) > M K(h0 − x), x ∈ [h0 − K −1 , h0 ], we see that if M K > − min[0,h0 ] u00 (x), then u0 (x) 6 w(0, x) in [h0 − K −1 , h0 ]. Applying the comparison principle we deduce w > u in Λ. It then leads to ux (t, h(t)) > wx (t, h(t)) = −2M K, and hence h0 (t) 6 2µM K.  Theorem 8.3 Assume that conditions (J) and (8.3) hold and f is bounded in R+ × R+ × (0, k) for any given 0 < k < ∞. Let Tmax be given by (8.11). Then Tmax < ∞ implies that either lim sup max u(t, ·) = ∞, − [0, h(t)] t→Tmax

or lim ku(t, ·)kWp2 ((0, h(t))) = ∞.

− t→Tmax

8.2. Existence and uniqueness of solutions  197

Proof. Assume that Tmax < ∞ and there exists a positive constant M such that u6M

in [0, Tmax ) × [0, h(t)].

By Lemma 8.2, h0 (t) 6 C in (0, Tmax ), and then h0 6 h(t) 6 h0 + CTmax . Let v(t, y) = u(t, h(t)y). For any given T < Tmax , applying the Lp theory (Theorem 1.7) to (8.5) we can get kvkWp1,2 (∆T ) 6 C1 (Tmax ) which is independent of T . It follows that v ∈ Wp1,2 (∆Tmax ) and kvkWp1,2 (∆Tmax ) 6 C1 (Tmax ). Therefore, h ∈ C 1+α/2 ([0, Tmax ]) and khkC 1+α/2 ([0,Tmax ]) 6 C2 (Tmax ) by (8.6). By the Fubini theorem, v(t, ·) ∈ Wp2 ((0, 1)) for almost t ∈ [0, Tmax ]. Thus, u(t, ·) ∈ for almost t ∈ [0, Tmax ]. We claim that

Wp2 ((0, h(t)))

lim ku(t, ·)kWp2 ((0,h(t))) = ∞.

− t→Tmax

If this was not valid, then there would exist ti ∈ (0, Tmax ) with ti % Tmax and a positive constant C such that ku(ti , ·)kWp2 ((0,h(ti ))) 6 C for all i. Regarding ti and (u(ti , x), h(ti )) as an initial time and an initial datum, similar to the proof of Theorem 8.1, we can find a constant 0 < T  1, which depends only on h(ti ), h0 (ti ), ku(ti , ·)kWp2 ((0,h(ti ))) and kf k∞, R+ ×R+ ×(0, M +1) , such that (8.1) has a unique solution (ui , hi ) in [ti , ti + T ]. By the uniqueness, (u, h) = (ui , hi ) for ti 6 t 6 min{ti + T, Tmax }. This shows that the solution (u, h) of (8.1) can be extended uniquely to [0, ti + T ]. Noting that h(ti ), h0 (ti ) 6 C2 (Tmax ),

ku(ti , ·)kWp2 ((0,h(ti ))) 6 C,

and C2 (Tmax ), C do not depend on i, we can choose T independent of i. Clearly, ti + T > Tmax when i is large. This is a contradiction with the definition of Tmax .  Theorem 8.4 Assume that conditions (J), (8.3) and (8.4) hold, and f is bounded in R+ × R+ × (0, k) for any given k > 0. Let Tmax be given by (8.11). If Tmax < ∞, then lim sup max u(t, ·) = ∞. − [0, h(t)] t→Tmax

Proof. Assume, on the contrary, that Tmax < ∞ and there exists M such that u6M

in [0, Tmax ) × [0, h(t)].

Let v(t, y) = u(t, h(t)y). Similar to the above, we have v ∈ Wp1,2 (∆Tmax ), h ∈ C 1+α/2 ([0, Tmax ]) and kvkWp1,2 (∆T

max )

6 C1 (Tmax ), khkC 1+α/2 ([0,Tmax ]) 6 C1 (Tmax ).

Moreover, we can apply Theorem 2.10 to (8.5) to get v ∈ C 1+α/2, 2+α ((0, Tmax ] ×

198  8. Free Boundary Problems from Ecology

[0, 1]), and for any given 0 < ε < Tmax , there holds |v|2+α, [ε,Tmax ]×[0,1] 6 C2 (ε, Tmax ). Therefore, u ∈ C 1+α/2, 2+α ((0, Tmax ] × [0, h(t)]) and |u|2+α, [ε,Tmax ]×[0,h(t)] 6 C3 (ε, Tmax ). This shows that solution (u, h) exists in [0, Tmax ]. Choose ti ∈ (0, Tmax ) with ti % Tmax , and regard ti and (u(ti , x), h(ti )) as an initial time and an initial datum. Similar to the second half of the proof of Theorem 8.3 we can get a contradiction with the definition of Tmax , and the proof is complete.  2 Now, we consider a special case f (t, x, u) = a(t, x)u − b(t, x)u . Theorem 8.5 Let f (t, x, u) = a(t, x)u−b(t, x)u2 , and a, b ∈ (C α/2, 1 ∩L∞ )(R+ ×R+ ) satisfy b(t, x) > b1 > 0. Then the solution (u, h) of (8.1) exists globally and uniquely. Moreover, u ∈ C 1+α/2, 2+α ((0, T ] × [0, h(t)]) and   0 < u 6 max {ku0 k∞ , kak∞ /b1 } =: M

in D∞ ,

 0 < h0 (t) 6 2µM K

in R+ ,

(8.12)

where D∞ = R+ × (0, h(t)) and   s 0  1 min[0,h0 ] u0 kak∞ K = max , , − .  h0  2d M

Proof. Let Tmax be the maximal existence time of (u, h). Using the maximum principle directly we have 0 < u 6 M in DTmax , and hence Tmax = ∞ by Theorem 8.4. Thanks to Lemma 8.2, h0 (t) 6 2µM K in [0, ∞).  Assume that problem (8.1) has a unique global solution (u, h). Then the limit limt→∞ h(t) := h∞ 6 ∞ exists since h0 (t) > 0. When h∞ = ∞, we call that u spreads successfully or that spreading happens. When h∞ < ∞ and lim max u(t, ·) = 0, we t→∞ [0, h(t)]

call that u vanishes or that vanishing happens.

8.3 REGULARITY AND UNIFORM ESTIMATES Regularity and uniform estimates are important elements in the study of parabolic partial differential equations. They will be used to study longtime behaviour of solutions and to determine dichotomy for spreading and vanishing. Materials of this section are taken from references [113, 120]. Theorem 8.6 (Regularity and uniform estimate) Take f (t, x, u) = a(t, x)u −
2 0,1 ∞ b(t, x)u . Assume a, b ∈ C ∩ L (R+ × R+ ) and b > b1 > 0 for some constant b1 . Then the unique global solution (u, h) of (8.1) satisfies |u|1+α, R+ ×[0,h(t)] 6 C

(8.13)

for some constant C > 0, which implies kh0 kC α/2 (R+ ) 6 C.

(8.14)

8.3. Regularity and uniform estimates  199

If, in addition, a, b ∈ (C α/2,1 ∩ L∞ )(R+ × R+ ), then we have u ∈ C 1+α/2, 2+α (R+ × [0, h(t)]), h ∈ C 1+(1+α)/2 (R+ ),

(8.15)

and there exists a constant C > 0 such that |u|2+α, [1,∞)×[0, h(t)] 6 C.

(8.16)

Proof. The idea of this proof is the same as that in Theorem 2.11, and we will divide it into several steps. Step 1. Noticing Theorem 8.5, from the proof of Theorem 8.1 we can see that u ∈ C (1+α)/2, 1+α (R+ × [0, h(t)]) and h ∈ C 1+α/2 (R+ ). Step 2. Proof of (8.15). Same as the proof of Theorem 8.1, we define y = x/h(t) and v(t, y) = u(t, h(t)y). Then (v, h) satisfies (8.5) for all T > 0. It is easily checked that function F (t, y) = f (t, h(t)y, v(t, y)) is in C α/2, α (R+ × R+ ). Recall h ∈ C 1+α/2 (R+ ). For any given τ > 0 and 0 < ε  1, applying Theorem 2.10 to problem (8.5) in [ε, τ ] × [0, 1] we have v ∈ C 1+α/2, 2+α ([ε, τ ] × [0, 1]). Thus u ∈ C 1+α/2, 2+α ([ε, τ ] × [0, h(t)]). Due to the arbitrariness of ε, one achieves u ∈ C 1+α/2, 2+α (Dτ0 ) =⇒ ux ∈ C (1+α)/2, 1+α (Dτ0 ), where Dτ0 = (0, τ ] × [0, h(t)]. Hence, by condition that h0 (t) = −µux (t, h(t)), we have h0 ∈ C (1+α)/2 ((0, τ ]), and thus (8.15) is proved. Step 3. Proof of (8.13). Applying the Lp estimate and embedding theorem to (8.5) directly, we can easily get an estimate |v|1+α, [0,2]×[0,1] 6 C.

(8.17)

In order to introduce more methods to readers, for two cases that h∞ < ∞ and h∞ = ∞, we shall use different processes. Case 1: h∞ < ∞. For any chosen integer i > 0, we define v i (t, y) = v(t + i, y) and i h (t) = h(t + i). Then v i satisfies  i i v − ρi vyy − ξ i yvyi = f i (t, y),    t

0 < t 6 3, 0 < y < 1,

  

0 < y < 1,

i i

[βh v − (1 − i

β)vyi ](t, 0)

v (0, y) = u(i, h(i)y),

i

= v (t, 1) = 0,

0 < t 6 3,

(8.18)

where ρi = ρ(t + i), ξ i = ξ(t + i) and f i (t, y) = f (t + i, hi (t)y, v i (t, y)). Making use of the estimate (8.12) we know that functions v i , ρi , ξ i and f i are uniformly bounded, and 4dµ ω i (r) := max |ρi (s) − ρi (t)| 6 3 M Kr → 0 as r → 0 06s,t63, |s−t|6r h0 uniformly with respect to i. On the other hand, ρi (t) > dh−2 ∞ for all i > 0 according to h(t) 6 h∞ < ∞.

200  8. Free Boundary Problems from Ecology

Remember boundary conditions of v i (t, y) at y = 0, 1 and 0 < (hi (t))0 6 2µM K. Choosing p > 3, we can apply Theorem 1.8 to problem (8.18), and then use the embedding theorem to get |v i |1+α, [1,3]×[0,1] 6 C. This implies |v|1+α, Ai 6 C, where Ai = [i + 1, i + 3] × [0, 1]. As these Ai overlap each other and C is independent of i, we have |v|1+α, [1,∞)×[0,1] 6 C. This together with (8.17) yields |v|1+α, [0,∞)×[0,1] 6 C. In view of ux (t, x) = h−1 (t)vy (t, x/h(t)),

0 6 h0 (t) 6 2µM K,

the estimate (8.13) is obtained. Case 2: h∞ = ∞. The process is to find two integers i0 and k, and make estimates in [0, i0 ] × [0, h(t)], [i0 , ∞) × [0, h(t) − k] and [i0 , ∞) × [h(t) − k, h(t)], respectively. (a) The estimate in [0, i0 ] × [0, h(t)]. For any fixed i0 < ∞, in the same way as Case 1, we can show |u|1+α, [0,i0 ]×[0,h(t)] 6 C(i0 ).

(8.19)

(b) The estimate in [i0 , ∞) × [0, h(t) − k]. Since ρi (t) = ρ(t + i) → 0 as i → ∞, it is useless to study problem (8.18) as above in order to get an estimate of |u|1+α, [i0 , ∞)×[0, h(t)−k] . We shall handle u directly. Same as above, for integer i > 0, we let ui (t, x) = u(t + i, x) and hi (t) = h(t + i). Then ui satisfies  i u − duixx = g i (t, x),    t

0 < t 6 3, 0 < x < hi (t),

  

0 < x < h(i),

B ui (t, 0) = ui (t, hi (t)) = 0, i

u (0, x) = u(i, x),

0 < t 6 3,

(8.20)

where g i (t, x) = f (t + i, x, ui (t, x)). According to Theorem 8.5, we know that ui and g i are bounded uniformly on i and hi (t) 6 h(i) + 2µM Kt 6 h(i) + 6µM K for all 0 6 t 6 3. Denote 6

k = 6µM K + 2.

hi (t)

h(i) h(i) − 2 hi (t) − k -t 3

Then h(i) − 2 > hi (t) − k for all i > 0 and 0 6 t 6 3. As h∞ = ∞, there exists an i0 > 0 so that h(i0 ) > k + 1 (see Fig. 8.1). In view of (8.19) we only need deal with the ui for i > i0 . Evidently,

Fig. 8.1

hi (t) − 3 > h(i) − 3 > k − 2 = 6µM K for all 0 6 t 6 3, i > i0 .

8.3. Regularity and uniform estimates  201

Choose p > 3. For any integer 0 6 l 6 h(i) − 3, we can adopt the local Lp estimate (Theorems 1.3 and 1.9) to problem (8.20) and derive that there exists a positive constant C, independent of l and i, such that kui kWp1,2 ([1,3]×[l, l+2]) 6 C for all i > i0 . And then the embedding theorem avers |ui |1+α, [1,3]×[l, l+2] 6 C for all i > i0 , 0 6 l 6 h(i) − 3. Since these intervals [l, l + 2] overlap each other and C is independent of l, we declare |ui |1+α, [1,3]×[0, [h(i)]−1] 6 C for all i > i0 , where [h(i)] is an integral part of h(i). The above estimate asserts |ui |1+α, [1,3]×[0, h(i)−2] 6 C for all i > i0 . Note that h(i) − 2 > hi (t) − k for i > i0 and 0 6 t 6 3. The above estimate implies |ui |1+α, [1,3]×[0, hi (t)−k] 6 C for all i > i0 , which implies |u|1+α, Bi 6 C for all i > i0 , where Bi = [i + 1, i + 3] × [0, h(t) − k]. As these rectangles Bi overlap each other and C is independent of i, it follows from the above estimates that |u|1+α, [i0 , ∞)×[0, h(t)−k] 6 C,

(8.21)

where i0 = i0 + 1. (c) The estimate in [i0 , ∞) × [h(t) − k, h(t)]. We shall show |u|1+α, [i0 , ∞)×[h(t)−k, h(t)] 6 C.

(8.22)

Once this is done, joining this with (8.19) and (8.21) we can derive (8.13). In order to attain the estimate (8.22), we introduce a variable transformation: y = h(t) − x, w(t, y) = u(t, h(t) − y). Then w satisfies  w − dwyy + h0 (t)wy = F (t, y), t > 0, 0 < y < h(t),    t   

w(t, 0) = B ∗ w(t, h(t)) = 0,

t > 0,

w(0, y) = u(0, h0 − y),

0 < y < h0 ,

where F (t, y) = f (t, h(t) − y, w(t, y)), B ∗ w = βw + (1 − β)wy . As above, for every integer i > i0 , let wi (t, y) = w(t + i, y) and hi (t) = h(t + i). Then wi satisfies  i i w − dwyy + (hi (t))0 wyi = F i (t, y), 0 < t 6 3, 0 < y < hi (t),    t

wi (t, 0) = B ∗ wi (t, hi (t)) = 0,

  

i

w (0, y) = w(i, y),

0 < t 6 3,

0 < y < h(i),

(8.23)

202  8. Free Boundary Problems from Ecology

where F i (t, y) = F (t + i, y). According to Theorem 8.5, we know that wi , (hi (t))0 and F i are bounded uniformly with respect to i. Set Γi = [i + 1, i + 3] × [0, k]. Remember hi (t) − k > h(i0 ) − k > 1 for all i > i0 , 0 6 t 6 3. Applying the interior Lp estimate to (8.23) firstly and the embedding theorem secondly, we conclude |wi |1+α, Γ0 6 C. Accordingly, |w|1+α, Γi 6 C. Since these rectangles Γi overlap each other and C is independent of i, it follows that |w|1+α, [i0 ,∞)×[0,k] 6 C, i0 = i0 + 1. Notice that 0 6 y 6 k is equivalent to that h(t) − k 6 x 6 h(t). By means of 0 < h0 (t) 6 2µM K and ux (t, x) = −wy (t, y), the estimate (8.22) is followed. Step 4. The estimate (8.16) can be proved by the similar way. 

8.4

SOME TECHNICAL LEMMAS

In this section, we will give four basic conclusions. The first one will be used to study asymptotic behaviour of u when h∞ < ∞, which indicates that if species cannot spread into [0, ∞), then it will die out in the long run. The second one concerns with the comparison principle, which, as well known, is an important tool. The last two are very useful in determining the criteria for spreading and vanishing. Lemma 8.7 Let d, C, µ and g0 be positive constants, and m(t, x) be a bounded function. Let v ∈ Wp1,2 ((0, T )×(0, g(t))) for each 0 < T < ∞, v, vx ∈ C(R+ ×[0, g(t)]) and g ∈ C 1 (R+ ). Assume that (v, g) satisfies    vt − dvxx + m(t, x)vx > −Cv, t > 0, 0 < x < g(t),     v > 0, t > 0, x = 0,   

v = 0, g 0 (t) > −µv ,

x     v(0, x) = v0 (x) > 0, 6≡ 0,     

t > 0, x = g(t), x ∈ (0, g0 ),

g(0) = g0

and lim g 0 (t) = 0,

(8.24)

kv(t, ·)kC 1 ([0, g(t)]) 6 M for all t > 1

(8.25)

lim g(t) = g∞ < ∞,

t→∞

t→∞

for some constant M > 0. Then lim

max v(t, x) = 0.

t→∞ 06x6g(t)

(8.26)

Proof. Firstly, the maximum principle gives v(t, x) > 0 for t > 0 and 0 < x < g(t), hence vx (t, g(t)) 6 0. Furthermore, applying the Hopf boundary lemma to problem of w(t, y) = v(t, g(t)y) we can get wy (t, 1) < 0. It demonstrates vx (t, g(t)) < 0, and then g 0 (t) > 0 for t > 0.

8.4. Some technical lemmas  203

The idea of the following arguments comes from [112, 121, 122]. On the contrary, we assume that there exist σ > 0 and {(ti , xi )}∞ i=1 , with 0 6 xi < g(ti ) and ti → ∞ as i → ∞, so that v(ti , xi ) > 4σ, i = 1, 2, . . . .

(8.27)

Then there exist a sub-sequence of {xi }, noted by itself, and x0 ∈ [0, g∞ ] such that xi → x0 as i → ∞. We claim x0 6= g∞ . Otherwise, xi − g(ti ) → 0 as i → ∞. By use of the inequality (8.27) firstly and the inequality (8.25) secondly, it derives that 4σ 6 |v(ti , xi ) − v(ti , g(ti ))| = |vx (ti , x¯i )||xi − g(ti )| 6 M |xi − g(ti )|, where x¯i ∈ (xi , g(ti )). This is a contradiction since xi − g(ti ) → 0. The strategy in the following arguments is to construct a series of lower solutions and use the comparison principle. In view of (8.25) and (8.27), there exists δ > 0 so that x0 + δ < g∞ and v(ti , x) > 2σ for all x ∈ [x0 , x0 + δ] for all large i. Since g(ti ) → g∞ as i → ∞, without loss of generality, we may think that g(ti ) > x0 + δ for all i. Let

t τi

6

ri (t) = x0 + δ + t − ti .

g(t)

Σi

Then ri (ti ) = x0 + δ < g(ti ). Since g 0 (t) > 0 and g∞ < ∞ = ri (∞), there exists a unique τi > ti such that g(τi ) = ri (τi ). See Fig. 8.2. Now, we define

ri (t)

ti •p x0

•p

g(ti )

x0 + δ

• -p x g∞

Fig. 8.2

Σi = {(t, x) : ti < t < τi , x0 < x < ri (t)}, vi (t, x) = σe−k(t−ti ) (cos yi (t, x) + cos θ) for (t, x) ∈ Σi with yi (t, x) = (π − θ)

2(x − x0 ) − (δ + t − ti ) , δ + t − ti

where θ (0 < θ < π/8) and k are positive constants to be chosen later. It is obvious that |yi (t, x)| 6 π − θ for (t, x) ∈ Σi , which implies vi (t, x) > 0 in Σi . We want to compare v(t, x) and vi (t, x) in Σi . Firstly, it is clear that v(t, x0 ) > 0 = vi (t, x0 ), v(t, ri (t)) > 0 = vi (t, ri (t)) for all t ∈ [ti , τi ] and v(ti , x) > 2σ > vi (ti , x) for all x ∈ [x0 , x0 + δ]. Thus, if positive constants θ and k can be chosen independent of i such that vit − dvixx + m(t, x)vix 6 −Cvi

in Σi ,

(8.28)

204  8. Free Boundary Problems from Ecology

then we can deduce v > vi in Σi by applying the comparison principle to v and vi over Σi . Since v(τi , g(τi )) = 0 = vi (τi , ri (τi )) and g(τi ) = ri (τi ), it follows that vx (τi , g(τi )) 6 vix (τi , ri (τi )). Due to the fact that θ < π/8 and δ + τi − ti < g∞ , it concludes vix (τi , ri (τi )) = −

2σ(π − θ) −k(τi −ti ) 7σπ −kg∞ e sin(π − θ) 6 − e sin θ. δ + τi − ti 4g∞

The free boundary condition −µvx (τi , g(τi )) 6 g 0 (τi ) implies g 0 (τi ) > −µvix (τi , ri (τi )) >

7µσπ −kg∞ e sin θ. 4g∞

This contradicts (8.24) since τi → ∞, and subsequently, (8.26) holds. In the following, we prove that if θ and k satisfy 3dπ 2 g∞ kmk∞ sin θ < 2 , + 2 δ δ g∞  2 2π(g∞ + δkmk∞ ) 2π + 2 , k >C +d δ δ (cos θ − cos 2θ)

θ
− cos 2θ when (t, x) ∈ Σ2i . Recall that θ < π/8, vi (t, x) > 0 and x − x0 6 g∞ in Σi . Taking advantage of (8.29) and (8.30), we conclude 3dπ 2 g∞ kmk∞ + sin 2θ − cos θ 6 0 I 6 2π 2 δ2 δ g∞ 



8.4. Some technical lemmas  205

when (t, x) ∈ Σ1i , and I 6 (C + d(2π/δ)2 − k)(cos θ − cos 2θ) + 2πδ −2 (g∞ + δkmk∞ ) < 0 when (t, x) ∈ Σ2i . The proof is complete. Now, we give a comparison principle.



Lemma 8.8 (Comparison principle) Assume that f satisfies conditions (J) and ¯ ∈ C 1 ([0, T ]), h ¯ > 0 in [0, T ], and u¯ ∈ C 0,1 ([0, T ] × [0, h(t)]) ¯ (8.3). Let h ∩ C 1,2 ([0, T ] × ¯ (0, h(t))) satisfy  u¯ − d¯ uxx > f (t, x, u¯),    t

¯ 0 < t 6 T, 0 < x < h(t),

¯ B u¯(t, 0) > 0, u¯(t, h(t)) = 0, 0 < t 6 T, ¯ h (t) > −¯ µu¯x (t, h(t)), 0 < t 6 T.

   ¯0

¯ Let (u, h) be the unique solution of (8.1) defined in [0, T ]. If µ 6 µ ¯, h0 6 h(0), ¯ ¯ u¯(0, x) > 0 in [0, h(0)], and u¯(0, x) > u0 (x) > 0 in (0, h0 ). Then h 6 h in [0, T ] and u 6 u¯ in [0, T ] × [0, h(t)]. Proof. The idea of this proof comes from [25], and the strategy is to use the classical maximum principle and comparison principle. For small ε > 0, let (uε , hε ) be the unique solution of (8.1) with h0 , µ, u0 (x) replaced by hε0 = (1 − ε)h0 , µε = (1 − ε)µ, uε0 (x) ∈ Wp2 ([0, hε0 ]), respectively, where uε0 (x) satisfies B uε0 (0) = uε0 (hε0 ) = 0 and 0 < uε0 (x) 6 u0 (x) in (0, hε0 ], lim uε0 ((1 − ε)x) = u0 (x) in Wp2 ([0, h0 ]).

ε→0

Then hε ∈ C 1+α/2 ([0, T ]). ¯ in (0, T ]. Obviously, this is reasonable for small t > 0. We shall prove that hε < h ¯ Arguing indirectly, it is assumed that we can find a first 0 < τ 6 T such that hε < h ¯ in (0, τ ) and hε (τ ) = h(τ ). Then ¯ 0 (τ ). h0ε (τ ) > h

(8.31)

¯ ¯ As u¯(0, x) > 0 in [0, h(0)], and u¯(0, x) > 0 in (0, h0 ), we have u¯ > 0 in (0, τ ] × (0, h(t)) ¯ by the maximum principle, and u¯x (t, h(t)) < 0 by the Hopf boundary lemma. And then the comparison principle asserts u¯ > uε in (0, τ ] × (0, hε (t)) since u¯|x=hε (t) > ¯ ), and 0 = uε |x=hε (t) . Obviously, u¯(τ, hε (τ )) = uε (τ, hε (τ )) = 0 since hε (τ ) = h(τ then u¯x (τ, hε (τ )) 6 uεx (τ, hε (τ )). Notice u¯x (τ, hε (τ )) < 0 and µε < µ 6 µ ¯. It follows 0 0 ¯ ¯ that hε (τ ) < h (τ ). This contradicts (8.31). As a result, hε < h in (0, T ]. Apply the comparison principle to uε and u¯ over (0, T ] × (0, hε (t)) to deduce uε < u¯ in (0, T ] × (0, hε (t)). As the unique solution of (8.1) depends continuously on parameters in (8.1), and (uε , hε ) converges to (u, h), the unique solution of (8.1), the desired result then follows ¯ by letting ε → 0 in inequalities: uε < u¯ and hε < h. 

206  8. Free Boundary Problems from Ecology

¯ in Lemma 8.8 is usually called an upper solution of (8.1). We can The pair (¯ u, h) define a lower solution by reversing all inequalities in the obvious places. Moreover, one can easily obtain an analogue of Lemma 8.8 for lower solutions. Applying Lemma 8.8, we can prove that the solution (u, h) of (8.1) is strictly monotone increasing in µ > 0. Details are left to readers as an exercise. The next two lemmas are very useful in determining the criteria for spreading and vanishing. ¯ 0 , H > 0, Lemma 8.9 Let C > 0 and p > 3 be constants. For any given constants h 2 ¯ 0 ]) satisfying B u¯0 (0) = u¯0 (h ¯ 0 ) = 0 and u¯0 > 0 in and any function u¯0 ∈ Wp ([0, h 0 0 ¯ ¯ (0, h0 ), there exists µ > 0 so that when µ > µ and (¯ u, h) satisfies  u¯t − d¯ uxx > −C u¯,      ¯  Bu ¯(t, 0) = u¯(t, h(t)) = 0,

¯ t > 0, 0 < x < h(t),

 ¯ ¯ 0 (t) > −µ¯ h ux (t, h(t)),      ¯

t > 0,

t > 0,

¯ 0, 0 < x < h ¯ 0, u¯(0, x) = u¯0 (x), h(0) = h

¯ > H. we must have limt→∞ h(t) Proof. This proof is inspired by [90], and the strategy is to construct a lower solution and use the comparison principle. By results of §8.2 we see that the linear problem  vt − dvxx = −Cv,       B v(t, 0) = v(t, r(t)) = 0, 0

 r (t) = −µvx (t, r(t)),     

t > 0, 0 < x < r(t), t > 0, t > 0,

(8.32)

¯ 0, 0 < x < h ¯0 v(0, x) = u¯0 (x), r(0) = h

admits a unique global solution (v, r), and v ∈ C 1+α/2, 2+α (R+ × [0, h(t)]), r0 (t) > 0 for t > 0. By Lemma 8.8, u¯(t, x) > v(t, x),

¯ > r(t) for all t > 0, x ∈ [0, r(t)]. h(t)

(8.33)

In the following, we are going to prove that for all large µ, (8.34)

r(2) > H.

¯ 0 /2, r0 (t) > 0 and To this end, we first choose a smooth function r(t) with r(0) = h r(2) = H. We then discuss the following initial-boundary value problem  v − dv xx = −Cv,    t   

t > 0, 0 < x < r(t),

B v(t, 0) = v(t, r(t)) = 0,

t > 0,

v(0, x) = v 0 (x),

¯ 0 /2. 0 µ0 , (8.37) r0 (t) 6 −µv x (t, r(t)) for all 0 6 t 6 2. ¯ 0 /2 < r(0), and using (8.32) and (8.35)–(8.37), On the other hand, noting r(0) = h it is deduced by the comparison principle that v(t, x) > v(t, x), r(t) > r(t) for all t ∈ [0, 2], x ∈ [0, r(t)]. It particularly illustrates that r(2) > r(2) = H, and then (8.34) holds. Making use of (8.33) and (8.34) we obtain ¯ > lim r(t) > r(2) > H. lim h(t) t→∞

t→∞

The proof is complete.  We are next to give a condition to judge vanishing. For any given T0 > 0, consider problem    ut − duxx 6 uf (t, x), t > T0 , 0 < x < h(t),     ux (t, 0) > 0,

t > T0 ,

 u = 0, h0 (t) = −µux , t > T0 , x = h(t),     

u(T0 , x) > 0,

(8.38)

0 6 x < h(T0 ),

where f (t, x) and h(t) are given continuous function and moving boundary, respectively. Lemma 8.10 Assume that (u, h) satisfies (8.38) and u is always positive for t > T0 . If f satisfies lim sup sup f (t, x) < 0, t→∞

06x6h(t)

then h∞ := limt→∞ h(t) < ∞. Proof. This proof comes from [51]. A direct calculation gives d dt

Z

h(t)

Z

h(t)

u(t, x) = 0

ut (t, x) + h0 (t)u(t, h(t))

0

Z

6

h(t)

Z

h(t)

duxx (t, x) + 0

d 6 − h0 (t) + µ

u(t, x)f (t, x) 0

Z

h(t)

u(t, x)f (t, x). 0

By the assumption, there exists τ > T0 such that f (t, x) 6 0 for all t > τ and 0 6 x 6 h(t). Integration of the above inequality from τ to t yields Z

06

h(t)

u(t, x) 0 h(τ )

d (h(τ ) − h(t)) + µ 0 Z h(τ ) d 6 u(τ, x) + (h(τ ) − h(t)) , µ 0 Z

6

Z tZ

h(s)

u(s, x)f (s, x)ds

u(τ, x) +

τ

0

208  8. Free Boundary Problems from Ecology

which implies µ h(τ ) h(t) 6 h(τ ) + u(τ, x), d 0 < ∞, and the proof is concluded. Z

Accordingly, h∞

8.5

t > τ. 

THE LOGISTIC EQUATION

Here, as an application of abstract conclusions obtained in above sections, we investigate a diffusive logistic equation with a free boundary and sign-changing intrinsic growth rate in heterogeneous time-periodic environment  ut − duxx = a(t, x)u − b(t, x)u2 ,       B u(t, 0) = 0, u(t, h(t)) = 0,

t > 0, 0 < x < h(t),

 h0 (t) = −µux (t, h(t)),     

t > 0,

h(0) = h0 , u(0, x) = u0 (x),

t > 0,

(8.39)

0 < x < h0 ,

where B u = βu − (1 − β)ux with constant 0 6 β 6 1, and u0 (x) satisfies u0 ∈ C 2 ([0, h0 ]), u0 > 0 in (0, h0 ),

B u0 (0) = u0 (h0 ) = 0.

Functions a and b satisfy (K) a, b ∈ (C α/2,1 ∩ L∞ )([0, T ] × R+ ) are time-periodic with period T , a is positive somewhere in [0, T ] × R+ , and b1 6 b(t, x) 6 b2 in [0, T ] × R+ for some positive constants b1 and b2 . By Theorem 8.5, problem (8.39) has a unique global solution (u, h). In this section, we will analyze longtime behaviours of (u, h), criteria for spreading and vanishing, and spreading-vanishing dichotomy. Contents of this part are due to [113]. 8.5.1

Longtime behaviour of solution

In this subsection, we investigate longtime behaviour of solution. It will be shown that if h∞ < ∞, i.e., the species u cannot spread over [0, ∞), then it must die out in the long run; if h∞ = ∞, i.e., the species u can spread over [0, ∞), then it must tend to the unique positive solution U of (7.13) locally uniformly in [0, T ] × [0, ∞) in the time-periodic sense. The estimate (8.14) implies limt→∞ h0 (t) = 0 when h∞ < ∞. Applying the estimate (8.13) and Lemma 8.7, we obtain the following theorem immediately. Theorem 8.11 (Vanishing) Let (u, h) be the unique solution of (8.39). When h∞ < ∞, we must have lim max u(t, x) = 0. t→∞ 06x6h(t)

Theorem 8.12 (Spreading) Assume that the condition (7.16) holds. Let U be the unique positive solution of (7.13) and (u, h) be the unique solution of (8.39). If h∞ = ∞, then limi→∞ u(t + iT, x) = U (t, x) uniformly in [0, T ] × [0, L] for any L > 0.

8.5. The logistic equation  209

Proof. In current circumstances, the condition (I1) is fulfilled. Let λ(`; d, a) be the principal eigenvalue of (7.8). In view of Proposition 7.12, there is an `0  1 such that λ(`, d, a) < 0 for all ` > `0 . We divide this proof into two parts, i.e., lim sup u(t + iT, x) 6 U (t, x) uniformly in [0, T ] × [0, L]

(8.40)

lim inf u(t + iT, x) > U (t, x) uniformly in [0, T ] × [0, L].

(8.41)

i→∞

and i→∞

Step 1. Take θ = ku0 k∞ + b11 kak∞ , where b1 is given by the condition (K). Then u < θ by the maximum principle. For any fixed ` > `0 , there exists an integer m  1 such that h(t) > ` for all t > mT. Since λ(`; d, kak∞ ) 6 λ(`; d, a) < 0, by the upper and lower solutions method for elliptic equations, we can prove that the boundary value problem (

−dVxx = kak∞ V − b1 V 2 ,

0 < x < `,

B V (0) = 0, V (`) = θ

has a unique positive solution V (x), and V (x) 6 θ in [0, `]. In consideration of regularities of u(mT, x) and V (x) with respect to x, we can find a constant k > 1 such that u(mT, x) 6 kV (x) for all 0 6 x 6 `. Noticing u(t, `) < kθ = kV (`), we can apply the comparison principle to u and kV , and then derive u(t, x) 6 kV (x) for all t > mT and 0 6 x 6 `. Since k > 1, it is easy to see that function v := kV satisfies −dvxx > kak∞ v − b1 v 2 > a(t, x)v − b(t, x)v 2 . Let w` be the unique solution of  w − dwxx = a(t, x)w − b(t, x)w2 ,    t

t > mT, 0 < x < `,

  

0 < x < `.

B w(t, 0) = 0, w(t, `) = kV (`) = kθ, t > mT,

w(mT, x) = kV (x),

The comparison principle gives u 6 w` 6 kV for t > mT and 0 6 x 6 `. For any integer i > m, we define w`i (t, x) = w` (t + iT, x) for (t, x) ∈ [0, T ] × [0, `]. Similar to the proof of Theorem 7.10, it can be shown that w`i → W` in C 1,2 ([0, T ] × [0, `]) as i → ∞, where W` is a positive solution of  2   Wt − dWxx = a(t, x)W − b(t, x)W , 

0 < t 6 T, 0 < x < `,

B W (t, 0) = 0, W (t, `) = kθ,

0 < t 6 T,

  

W (0, x) = W (T, x),

0 < x < `.

On account of u(t + iT, x) 6 w` (t + iT, x) = w`i (t, x) in [0, T ] × [0, `], it yields lim sup u(t + iT, x) 6 W` (t, x) uniformly in [0, T ] × [0, `]. i→∞

(8.42)

210  8. Free Boundary Problems from Ecology

Since w` 6 kθ in [0, `]×[mT, ∞), we see that w` is monotonically decreasing in ` by the comparison principle, and hence W` is monotonically decreasing in `. Remembering that U is the unique positive solution of (7.13), and arguing as Step 1 in proving Theorem 7.13, we can verify that W` → U in C 1,2 ([0, T ] × [0, L]) as ` → ∞. This fact combined with (8.42) allows us to derive (8.40). Step 2. Let ` > `0 and v` be the unique solution of  v − dvxx = a(t, x)v − b(t, x)v 2 ,    t

t > mT, 0 < x < `,

B v(t, 0) = v(t, `) = 0,

t > mT,

  

v(mT, x) = u(mT, x),

0 < x < `.

Then u > v` in [mT, ∞) × [0, `]. Since λ(`; d, a) < 0, problem  2   Vt − dVxx = a(t, x)V − b(t, x)V , 

0 < t 6 T, 0 < x < `,

B V (t, 0) = V (t, `) = 0,

0 < t 6 T,

  

V (0, x) = V (T, x),

0 v` in [mT, ∞) × [0, `] for all ` > `0 .  8.5.2

Criteria and dichotomy for spreading and vanishing

We first give a necessary condition for vanishing. Let λ(`; d, a) be the principal eigenvalue of (7.8). Lemma 8.13 h∞ = ∞.

If h∞ < ∞, then λ(h∞ ; d, a) > 0. And hence, λ(h0 ; d, a) 6 0 implies

Proof. We assume λ(h∞ ; d, a) < 0 to get a contradiction. By continuity of λ(`; d, a) in ` and h(t) → h∞ , there exists τ  1 for which λ(h(τ ); d, a) < 0. Let z be the unique solution of  z − dzxx = a(t, x)z − b(t, x)z 2 ,    t

t > τ, 0 < x < h(τ ),

B z(t, 0) = z(t, h(τ )) = 0,

t > τ,

  

z(τ, x) = u(τ, x),

0 < x < h(τ ).

Then u > z in [τ, ∞) × [0, h(τ )] by the comparison principle. Since λ(h(τ ); d, a) < 0, it infers from Theorem 7.11 that z(t+iT, x) → Z(t, x) uniformly in [0, T ]×[0, h(τ )] as i → ∞, where Z is the unique positive solution of time-periodic parabolic boundary value problem  Z − dZxx = a(t, x)Z − b(t, x)Z 2 ,    t

0 < t 6 T, 0 < x < h(τ ),

B Z(t, 0) = Z(t, h(τ )) = 0,

0 < t 6 T,

  

Z(0, x) = Z(T, x),

0 < x < h(τ ).

8.5. The logistic equation  211

Since u > z in [τ, ∞) × [0, h(τ )], it is deduced immediately that lim inf u(t + iT, x) > Z(t, x) for all (t, x) ∈ [0, T ] × [0, h(τ )]. i→∞

This is a contradiction with Theorem 8.11, and the proof is complete.



Lemma 8.14 If λ(h0 ; d, a) > 0, then there exists µ0 > 0 such that h∞ < ∞ for µ 6 µ0 . As a consequence, λ(h∞ ; d, a) > 0 for µ 6 µ0 by Lemma 8.13. Proof. Let φ(t, x) be the positive eigenfunction of (7.8) with ` = h0 corresponding to λ(h0 ; d, a) =: λ1 . Note that φx (t, h0 ) < 0, φ(t, 0) > 0 in [0, T ] when β < 1, while φ(t, 0) = 0 and φx (t, 0) > 0 in [0, T ] when β = 1. In accordance with regularities of φ, there exists a constant C > 0 such that xφx (t, x) 6 Cφ(t, x) for all (t, x) ∈ [0, T ] × [0, h0 ],

(8.43)

please refer to the proof of Lemma 3.7. Let 0 < δ, σ < 1 and k > 0 be constants, which will be determined later. Set −σt

s(t) = 1 + 2δ − δe

Z

t

, τ (t) = 0

s−2 (ρ)dρ, t > 0

and v(t, x) = ke−σt φ (τ (t), y) , y = y(t, x) =

x , 0 6 x 6 h0 s(t). s(t)

First of all, for any given 0 < ε  1, since a is uniformly continuous in [0, T ]×[0, 3h0 ] and time-periodic with period T , we can find 0 < δ0 (ε)  1 such that, for all 0 < δ 6 δ0 (ε) and 0 < σ < 1, −2 s (t)a (τ (t), y(t, x)) − a(t, x) 6 ε for all t > 0, 0 6 x 6 h0 s(t).

(8.44)

Remembering (8.43), (8.44) and λ1 > 0, a direct calculation alleges a(τ, y) yφy (τ, y) σδ −σt λ1 vt − dvxx − a(t, x)v = v − σ − a(t, x) − e + 2 s2 (t) φ(τ, y) s(t) s (t) 



> (λ1 /4 − σ − ε − Cσ)v > 0 for all t > 0, 0 < x < h0 s(t)

(8.45)

provided 0 < σ, ε  1. Evidently, v(t, h0 s(t)) = ke−σt φ(τ (t), h0 ) = 0. If either β = 0 or β = 1, then B v(t, 0) = 0. If 0 < β < 1, then βφ(τ (t), 0) = (1 − β)φy (τ (t), 0), and φy (τ (t), 0) > 0. Therefore B v(t, 0) = (1 − β)ke−σt φy (τ (t), 0)[1 − 1/s(t)] > 0

due to s(t) > 1. In conclusion, B v(t, 0) > 0 and v(t, h0 s(t)) = 0 for all t > 0.

(8.46)

212  8. Free Boundary Problems from Ecology

Fix 0 < σ, ε  1 and 0 < δ 6 δ0 (ε). Based on regularities of u0 (x) and φ(0, x), we can choose a constant k  1 so that u0 (x) 6 kφ (0, x/(1 + δ)) = v(0, x) for all 0 6 x 6 h0 . As h0 s0 (t) = h0 σδe−σt and vx (t, h0 s(t)) = such that, for all 0 < µ 6 µ0 ,

1 −σt φy (τ (t), h0 ), s(t) ke

(8.47)

there exists µ0 > 0

−µvx (t, h0 s(t)) 6 h0 s0 (t) for all t > 0.

(8.48)

Notice (8.45)–(8.48). Using Lemma 8.8 we conclude h(t) 6 h0 s(t) for all t > 0 and 0 < µ 6 µ0 . Therefore, h∞ 6 h0 (1 + 2δ), and the proof is complete.  In order to study the spreading phenomenon and establish sharp criteria, we shall introduce some sets and analyze their properties. For any given d > 0, we define 



Σd = ` > 0 : λ(`; d, a) = 0 . By the monotonicity of λ(`; d, a) in `, we see that Σd contains at most one element. For any given ` > 0, we define + Σ− ` = d > 0 : λ(`; d, a) 6 0 , Σ` = d > 0 : λ(`; d, a) > 0 .









Then Σ+ ` 6= ∅ by Theorem 7.9 (3). Here we should remark that because λ(`; d, a) is not monotone in d (cf. Theorem 2.2 of [52]), it is useless to define Σ` as the manner treating Σd . Remark 8.15

For any fixed d > 0, since

lim λ(`; d, a) = ∞,

`→0+

lim λ(`; d, a) = λ(∞; d, a) exists,

`→∞

we know that Σd 6= ∅ is equivalent to λ(∞; d, a) < 0 by the monotonicity of λ(`; d, a) in `. Proposition 8.16 Z

(I2)

Assume that

T

a(t, xˆ)dt > 0 for some xˆ > 0. 0

Then the set Σ− ˆ. ` is non-empty for all ` > x T

Z

Proof. Since

a(t, xˆ)dt > 0, by the continuity of a(t, x) we can find 0 < ε < 0

min{ˆ x, ` − xˆ} for which

Z

T

0

aε (t)dt > 0, where aε (t) = minx∈I¯ε a(t, x) and Iε =

(ˆ x − ε, xˆ + ε). ˆ be the principal eigenvalue of the time-periodic parabolic eigenvalue problem Let λ  ˆ ˆ ˆ ˆˆ   φt − dφxx − aε (t)φ = λφ, 

ˆ x) = 0, φ(t,

  ˆ ˆ x), φ(0, x) = φ(T,

x ∈ Iε , 0 < t 6 T, x = xˆ ± ε, 0 < t 6 T, x ∈ I¯ε .

(8.49)

8.5. The logistic equation  213

ˆ by the monotonicity of principal eigenvalue. Then λ(`; d, a) 6 λ ˆ by decomposing φ(t, ˆ x) = ξ(t)ϕ(x), where ξ(t) is a function to be We calculate λ determined, and ϕ(x) > 0 is the principal eigenfunction of (

−ϕ00 = γϕ, x ∈ Iε , ϕ(ˆ x ± ε) = 0

with principal eigenvalue γ > 0. It follows from (8.49) that ˆ − dγ]ξ(t), ξ 0 (t) = [aε (t) + λ and consequently ˆ

ξ(t) = ξ0 e(λ−dγ)t exp

t

Z



aε (s)ds . 0

ˆ x) is time-periodic with period T , we must have ξ(0) = ξ(T ), i.e., In order that φ(t, ˆ − dγ)T + (λ

Z

T

aε (s)ds = 0, 0

ˆ = dγ − 1 hence λ T

Z

T

aε (s)ds. Accordingly 0

1 λ(`; d, a) 6 dγ − T ˆ Owing to as λ(`; d, a) 6 λ.

Z

Z

T

aε (s)ds 0

T

aε (s)ds > 0, we can find d0 > 0 so that λ(`; d, a) < 0 0

for all 0 < d 6 d0 . This implies (0, d0 ] ⊂ Σ−  ` . We will study sharp criteria and dichotomy for spreading and vanishing in two cases. Case 1: we fix d and consider h0 and µ as varying parameters. Firstly, recalling the estimate (8.12), as consequences of Lemmas 8.9, 8.13 and 8.14, we have Corollary 8.17 Assume Σd 6= ∅, and let h∗0 = h∗0 (d) ∈ Σd , i.e., λ(h∗0 ; d, a) = 0. Then the following conclusions hold: (1) if h∞ < ∞, then h∞ 6 h∗0 . Of course, h0 > h∗0 implies h∞ = ∞ for all µ > 0; (2) when h0 < h∗0 , there exist µ0 > µ0 > 0 such that h∞ 6 h∗0 for µ 6 µ0 , while h∞ = ∞ for µ > µ0 . With regard to sharp criteria for spreading and vanishing we have the following theorem. Theorem 8.18

Assume Σd 6= ∅ and let h∗0 = h∗0 (d) ∈ Σd . Then we conclude

(1) h0 > h∗0 implies h∞ = ∞ for all µ > 0;

214  8. Free Boundary Problems from Ecology

(2) if h0 < h∗0 , then there exists µ∗ > 0 so that h∞ = ∞ for µ > µ∗ , while h∞ 6 h∗0 for µ 6 µ∗ . Proof. The verdict (1) is exactly Corollary 8.17 (1). Now we prove the conclusion (2). Assume h0 < h∗0 . Write (uµ , hµ ) in place of (u, h) to clarify the dependence of the solution of (8.39) on µ. Define Σ∗ = {µ > 0 : hµ∞ 6 h∗0 } . According to Corollary 8.17 (2), (0, µ0 ] ⊂ Σ∗ and Σ∗ ∩ [µ0 , ∞) = ∅. Therefore, µ∗ := sup Σ∗ ∈ [µ0 , µ0 ]. In accordance with this definition and the monotonicity of (uµ , hµ ) in µ, we find that hµ∞ = ∞ when µ > µ∗ , and so Σ∗ ⊂ (0, µ∗ ]. ∗ Now, let us show that µ∗ ∈ Σ∗ . Otherwise, hµ∞ = ∞. There exists T > 0 so that ∗ hµ (T ) > h∗0 . Utilizing the continuous dependence of (uµ , hµ ) on µ, one can find an ε > 0 such that hµ (T ) > h∗0 for µ ∈ (µ∗ − ε, µ∗ + ε). It follows that, for all such µ, lim hµ (t) > hµ (T ) > h∗0 .

t→∞

Therefore, [µ∗ − ε, µ∗ + ε] ∩ Σ∗ = ∅, and then sup Σ∗ 6 µ∗ − ε. This contradicts the definition of µ∗ .  Noting that (7.16) implies (I1), and combining Theorems 8.11, 8.12 and 8.18, we immediately obtain the following spreading-vanishing dichotomy and sharp criteria for spreading and vanishing. Theorem 8.19 Assume (7.16) holds. Let h∗0 be the unique positive root of λ(`; d, a) = 0 and (u, h) the unique solution of (8.39). Then either (1) spreading: h∞ = ∞ and limi→∞ u(t+iT, x) = U (t, x) uniformly in [0, T ]×[0, L] for any L > 0, where U (t, x) is the unique positive solution of (7.13); or (2) vanishing: h∞ 6 h∗0 and lim

max u(t, x) = 0.

t→∞ 06x6h(t)

Moreover, (3) h0 > h∗0 implies h∞ = ∞ for all µ > 0; (4) if h0 < h∗0 , then there exists µ∗ > 0 so that h∞ = ∞ for µ > µ∗ , while h∞ 6 h∗0 for µ 6 µ∗ . Case 2: we fix h0 , and regard d and µ as variable parameters. Theorem 8.20

(1) When d ∈ Σ− h0 , we have h∞ = ∞ for all µ > 0.

(2) For any fixed d ∈ Σ+ h0 , there exists µ0 = µ0 (d) > 0 such that h∞ < ∞ as long as 0 < µ 6 µ0 . If, in addition, Σd 6= ∅ for such d, then there exists µ∗ > 0, such that h∞ = ∞ when µ > µ∗ , and h∞ < ∞ when µ 6 µ∗ .

8.6. Spreading speed of free boundary  215

Proof. (1) When d ∈ Σ− h0 , there holds λ(h0 ; d, a) 6 0, which implies Σd 6= ∅ and h0 > h∗0 (d). Theorem 8.18 (1) confirms h∞ = ∞ for all µ > 0. (2) For the fixed d ∈ Σ+ h0 , we have λ(h0 ; d, a) > 0. By Lemma 8.14, there exists µ0 > 0 such that h∞ < ∞ for µ 6 µ0 . If, in addition, Σd 6= ∅, then there exists H  1 for which λ(H; d, a) < 0. Taking advantage of Lemma 8.9, there exists µ0 > 0 such that h∞ > H provided µ > µ0 , which implies λ(h∞ ; d, a) < λ(H; d, a) < 0. As a result, h∞ = ∞ for µ > µ0 by Lemma 8.13. The remaining proof is the same as Theorem 8.18.  Remark 8.21 (1) When the condition (I1) holds, we have Σd 6= ∅ for all d > 0 by Theorem 7.9 (3) and Proposition 7.12. (2) When the condition (I2) holds, from the proof of Proposition 8.16, we see that if h0 > xˆ, then Σ− 6 ∅ and there exists d0 > 0 so that d ∈ Σ− h0 = h0 for 0 < d 6 d0 . Note that (7.16) implies (I1). Invoking Theorems 8.11, 8.12 and 8.20, and Remark 8.21, we have the following spreading-vanishing dichotomy and sharp criteria. Theorem 8.22 Assume that conditions (7.16) and (I2) are satisfied and h0 > xˆ. Let (u, h) be the unique solution of (8.39). Then either (1) spreading: h∞ = ∞ and limi→∞ u(t+iT, x) = U (t, x) uniformly in [0, T ]×[0, L] for any L > 0, where U (t, x) is the unique positive solution of (7.13); or (2) vanishing: h∞ < ∞ and lim

max u(t, x) = 0.

t→∞ 06x6h(t)

Moreover, (3) d ∈ Σ− h0 implies h∞ = ∞ for all µ > 0; ∗ ∗ (4) if d ∈ Σ+ h0 , then there exists µ > 0 such that h∞ = ∞ when µ > µ , while ∗ h∞ < ∞ when µ 6 µ .

8.6

SPREADING SPEED OF FREE BOUNDARY

This section concerns with spreading speed of free boundary. As an example, here we only treat the following problem.  ut − duxx = au − bu2 ,       B u(t, 0) = 0, u(t, h(t)) = 0,  h0 (t) = −µux (t, h(t)),     

h(0) = h0 , u(0, x) = u0 (x),

t > 0, 0 < x < h(t), t > 0, t > 0,

(8.50)

0 < x < h0 ,

where a and b are positive constants. This problem has a unique global solution (u, h) by the conclusions of §8.2. It will be shown that when spreading occurs, the expanding front x = h(t) moves at a constant speed for large time, namely h(t) = (k0 + o(1)) t as t → ∞.

216  8. Free Boundary Problems from Ecology

This constant k0 will be called the asymptotic spreading speed, and it is determined in Proposition 8.23 below. The fact limt→∞ h(t)/t = k0 will be proved by using modifications of solution of the following elliptic problem (8.53). Contents of this part are taken from [12, 25]. We first remark that, since a and b are positive constants, the unique positive solution U of (7.13) does not depend on t, and lim U (x) = a/b.

x→∞

(8.51)

Moreover, the conclusion of Theorem 8.12 becomes lim u(t, x) = U (x) uniformly in [0, L] for any L > 0.

t→∞

(8.52)

Assume that (u, h) is the unique solution of (8.50) and h(t) → ∞ as t → ∞. Set v(t, x) = u(t, h(t) − x). Then  v − dvxx + h0 (t)vx = v(a − bv),    t    v(t, 0) = 0, h0 (t) = µvx (t, 0),

t > 0, 0 < x < h(t),

 βv(t, h(t)) = (β − 1)vx (t, h(t)),     

t > 0,

t > 0,

h(0) = h0 , v(0, x) = u0 (h0 − x), 0 < x < h0 .

Note that limt→∞ h(t) = ∞. If h0 (t) approaches a constant k, and v(t, x) approaches a positive function Uk (x) as t → ∞, then Uk (x) must be a positive solution of (

−dU 00 + kU 0 = aU − bU 2 , x > 0, U (0) = 0

(8.53)

and satisfies Uk (0) = k/µ. To study asymptotic behaviours of h(t) we should investigate problem (8.53). √ Proposition 8.23 For any given 0 6 k < 2 ad, problem (8.53) has a unique positive solution, √ denoted by Uk (x). Moreover, for each µ > 0, there exists a unique k0 = k0 (µ) ∈ (0, 2 ad) such that µUk0 0 (0) = k0 . √ Proof. Since 0 6 k < 2 ad, it is well known that for any given constant θ > 0 and all large ` > 0, the problem (

−dU 00 + kU 0 = aU − bU 2 , 0 < x < `, U (0) = 0, U (`) = θ

(8.54)

has a unique positive solution, denoted by U`θ . Similar to the proof of the existence in Theorem 7.13, we can show that problem (8.53) has at least one positive solution U (x). The proof is left to the reader as an exercise. By the strong maximum principle

8.6. Spreading speed of free boundary  217

for elliptic equations, we see that any non-trivial and non-negative solution of (8.53) satisfies 0 < U (x) < a/b for x > 0. Next, we claim that U (x) is monotonically increasing and limx→∞ U (x) = a/b. Indeed, by use of (8.53) we conclude k

k

(de− d x U 0 )0 = −e− d x (a − bU )U < 0 in R+ .

(8.55)

Since U (x) is bounded, we can find a sequence xi → ∞ such that U 0 (xi ) → 0 as i → ∞. Hence, making use of (8.55), k

k

e− d x U 0 (x) > lim e− d xi U 0 (xi ) = 0 for x > 0. i→∞

Of course U 0 (x) > 0, and then σ = limx→∞ U (x) exists. In the light of (8.55) we easily find σ = a/b. Let us now prove uniqueness. Suppose that both U1 and U2 are positive solutions of (8.53). Then for any ε > 0, function wi = (1 + ε)Ui satisfies −dwi00 + kwi0 > awi − bwi2 in R+ , i = 1, 2. On the basis of limx→∞ wi (x) = (1 + ε)a/b and limx→∞ Ui (x) = a/b, we can find `0 > 0 large enough so that w1 (`) > U2 (`), w2 (`) > U1 (`) for all ` > `0 . By means of the upper and lower solutions method and the uniqueness of positive solutions of (8.54), it can be proved that (1 + ε)U1 (x) > U2 (x), (1 + ε)U2 (x) > U1 (x) for all 0 6 x 6 ` and ` > `0 , which implies (1+ε)U1 > U2 and (1+ε)U2 > U1 in [0, ∞). Letting ε → 0 we conclude U1 = U2 , and the uniqueness is obtained. Finally, if 0 6 k1 < k2 , then, due to Uk0 1 (x) > 0, one has −dUk001 + k2 Uk0 1 > −dUk001 + k1 Uk0 1 = aUk1 − bUk21 , x > 0. Hence, for any ε > 0, function w = (1 + ε)Uk1 satisfies −dw00 + k2 w0 > aw − bw2 , x > 0. As above, w > Uk2 , i.e., (1 + ε)Uk1 > Uk2 , and then Uk1 > Uk2 in [0, ∞) by letting ε → 0. Besides, applying the strong maximum principle for elliptic equations, we can still get Uk1 > Uk2 in R+ . Notice Uk1 (0) = Uk2 (0) = 0. Adopting the Hopf boundary lemma to differential equations satisfied by Uk1 , Uk2 and Uk1 −Uk2 , respectively, it can be derived that Uk0 i (0) > 0 and Uk0 1 (0) > Uk0 2 (0). Hence, both sequences {Uk (x)} and {Uk0 (0)} are monotonically decreasing in k > 0. Thus, for any fixed µ > 0, function σ(k) = k − µUk0 (0)

218  8. Free Boundary Problems from Ecology

is strictly monotonically increasing in k > 0, and σ(0) = −µU00 (0) < 0. We claim that √ lim √ σ(k) = 2 ad.

(8.56)

k%2 ad

√ Once this is done, there exists a unique k0 = k0 (µ) ∈ (0, 2 ad) such that σ(k0 ) = 0. √ To prove (8.56), we choose an arbitrary increasing sequence {ki } such that ki → 2 ad as i → ∞ and consider the corresponding sequence of solutions {Uki } of (8.53). Since 0 < Uki (x) < a/b for x > 0, and Uki (x) is decreasing in i, the limit U∗ (x) := lim Uki (x) i→∞

exists and satisfies 0 6 U∗ (x) < a/b for x > 0. Moreover, applying standard Lp estimates to the equation for Uki and then the embedding theorem (Theorem A.4 (3)), 1 we easily see that Uki → U∗ in Cloc (R+ ). Hence U∗ satisfies in the weak sense (and also classical sense) √ ( −dU∗00 + 2 adU∗0 = aU∗ − bU∗2 , 0 < x < ∞, U∗ (0) = 0. Define W∗ (x) =

 b −x q e U∗ d/a x . a

Then one readily checks that W∗00 = ex W∗2 , 0 6 W∗ < 1 in (0, ∞), and W∗ (0) = 0. We claim that U∗ ≡ 0. Otherwise, by the strong maximum principle, we have U∗ (x) > 0 for x > 0 and U∗0 (0) > 0. It follows that W∗00 (x) = ex W∗2 (x) > 0 for x > 0, and W∗0 (0) > 0, which implies W∗ (x) → ∞ as x → ∞. This is a contradiction to the fact that W∗ (x) < 1 for x > 0, and thus U∗ ≡ 0. √ Since {ki } is an arbitrary increasing to 2 ad, the above discussions √ sequence √ 1 show that Uk → 0 as k → 2 ad in Cloc (R+ ). In particular, Uk0 (0) → 0 as k → 2 ad. Therefore √ lim √ σ(k) = 2 ad, k%2 ad

as we wanted.



Theorem 8.24 (Asymptotic spreading speed) Let (u, h) be the unique solution of h(t) (8.50). If h∞ = ∞, then limt→∞ = k0 , where k0 is uniquely determined in t Proposition 8.23.

8.6. Spreading speed of free boundary  219

Proof. We first prove lim sup t→∞

h(t) 6 k0 . t

(8.57)

Clearly, lim supt→∞ u(t, x) 6 a/b uniformly for x > 0. Thus, for any given ε > 0 small, there exists T = Tε > 0 such that u(t, x) 6 a(1 + ε)/b for all t > T, x > 0. Let Uk0 (x) be the unique positive solution of (8.53) with k = k0 and k0 = µUk0 0 (0). Since Uk0 (x) → a/b as x → ∞, there exists x0  1 so that a Uk0 (x) > for all x > x0 . (1 + ε)b We now define ξ(t) = (1 + ε)2 k0 t + x0 + h(T ), 2

v(t, x) = (1 + ε) Uk0 (ξ(t) − x),

t > 0, t > 0, 0 6 x 6 ξ(t).

Then v(t, ξ(t)) = (1 + ε)2 Uk0 (0) = 0, ξ 0 (t) = (1 + ε)2 k0 , −µvx (t, ξ(t)) = µ(1 + ε)2 Uk0 0 (0) = (1 + ε)2 k0 , and so ξ 0 (t) = −µvx (t, ξ(t)). Clearly, B v(t, 0) > 0

by noting that v(t, 0) = (1 + ε)2 Uk0 (ξ(t)) > 0, vx (t, 0) = −(1 + ε)2 Uk0 0 (ξ(t)) < 0. Moreover, for 0 6 x 6 h(T ), v(0, x) = (1 + ε)2 Uk0 (ξ(0) − x) > (1 + ε)2 Uk0 (x0 ) > a(1 + ε)/b > u(T, x), and v(0, x) > 0 for h(T ) < x < ξ(0). A straightforward computation shows that vt − dvxx = (1 + ε)2 (Uk0 0 ξ 0 − dUk000 ) = (1 + ε)2 ((1 + ε)2 k0 Uk0 0 − dUk000 ) > (1 + ε)2 (k0 Uk0 0 − dUk000 ) (due to Uk0 0 > 0) = (1 + ε)2 (aUk0 − bUk20 ) > av − bv 2

for t > 0, 0 < x < ξ(t).

220  8. Free Boundary Problems from Ecology

The comparison principle (Lemma 8.8) concludes h(t + T ) 6 ξ(t) for t > 0, and consequently h(t) ξ(t − T ) lim sup 6 lim = k0 (1 + ε)2 . t→∞ t t t→∞ Taking ε → 0 we get (8.57). In the following we will prove lim inf t→∞

h(t) > k0 t

by constructing a suitable lower solution. Consider (

−dV 00 + k0 V 0 = aV − bV 2 , 0 < x < `, V (0) = V (`) = 0.

Similar to above, for all large `, this problem has a unique positive solution, denoted by V` , and V` (x) < Uk0 (x) < a/b for 0 < x 6 `. Moreover, V` (x) is monotonically increasing in `, and as ` → ∞, V` (x) → V∞ (x) which solves (8.53) with k = k0 . Then V∞ = Uk0 by the uniqueness of positive solutions of (8.53). Furthermore, a simple regularity and compactness consideration show that 1 lim V` = Uk0 in Cloc (R+ ). `→∞

Accordingly, for any given 0 < ε  1, we can find `0 = `0 (ε)  1 such that V`00 (0) > (1 − ε)Uk0 0 (0) = (1 − ε)k0 /µ. Define (

V0 (x) =

V`0 (x),

0 6 x 6 x0 ,

V`0 (x0 ), x > x0 ,

where x0 ∈ (0, `0 ) satisfies V`0 (x0 ) = max[0,`0 ] V`0 (x). From the equation for V`0 we see k0

that e− d x V`00 (x) is monotonically decreasing in (0, `0 ), see (8.55). Certainly, V`00 (x) changes sign exactly once in this interval. It follows that such x0 is unique, V`00 (x) > 0 for x ∈ (0, x0 ), and V`00 (x) < 0 for x ∈ (x0 , `0 ). As a consequence, V0 (x) 6 V`0 (x0 ) < Uk0 (x0 ) < a/b for x > 0, V00 (x) = 0 for x > x0 ,

V00 (x) > 0 for 0 6 x < x0 .

On the other hand, it is easy to check that V0 satisfies (in the weak sense) (

−dV000 + k0 V00 6 aV0 − bV02 ,

x > 0,

V0 (0) = 0,

x > 0.

V0 (x) < a/b,

8.6. Spreading speed of free boundary  221

According to (8.51), there exists L > 0 such that U (x) > (1 − ε/2)a/b for all x > L. Due to h∞ = ∞, using (8.52) we can choose T = T (ε, x0 , L) so large that h(t) > L + x0 ,

u(t, x) > (1 − ε)a/b for t > T, x ∈ [L, L + x0 ].

Define η(t) = (1 − ε)2 k0 t + x0 ,

t > 0,

w(t, x) = (1 − ε)V0 (η(t) − x), t > 0, 0 6 x 6 η(t), and set u¯(t, x) = u(t + T, x + L),

¯ = h(t + T ) − L. h(t)

Obviously, u¯t − d¯ uxx = a¯ u − b¯ u2

¯ for t > 0, 0 < x < h(t),

and ¯ η(0) = x0 < h(T ) − L = h(0), w(0, x) = (1 − ε)V0 (x0 − x) < (1 − ε)a/b 6 u(T, x + L) = u¯(0, x), 0 6 x 6 x0 , w(t, 0) = (1 − ε)V0 (η(t)) < (1 − ε)a/b 6 u(t + T, L) = u¯(t, 0), t > 0, ¯ w(t, η(t)) = 0 = u¯(t, h(t)), t > 0, η 0 (t) = (1 − ε)2 k0 < µ(1 − ε)V`00 (0) = −µwx (t, η(t)), t > 0, ¯ 0 (t) = −µ¯ ¯ h ux (t, h(t)), t > 0. Moreover, due to V00 > 0, we have wt − dwxx = (1 − ε)[(1 − ε)2 k0 V00 − dV000 ] 6 (1 − ε)(k0 V00 − dV000 ) 6 (1 − ε)(aV0 − bV02 ) 6 aw − bw2 ,

t > 0, 0 < x < η(t).

Hence, by the comparison principle, u(t + T, x + L) > w(t, x) and h(t + T ) − L > η(t) for t > 0, 0 6 x 6 η(t), which leads to

η(t − T ) h(t) > lim = (1 − ε)2 k0 . t→∞ t t The arbitrariness of ε implies lim inf t→∞

lim inf t→∞

and the proof is now complete.

h(t) > k0 , t 

222  8. Free Boundary Problems from Ecology

Next, we analyze the dependence of k0 on parameters a, b, µ and d. From the proof of Proposition 8.23 we know that k0 is the unique solution of k − µUk0 (0) = 0. Clearly, when √ all other parameters are fixed, k0 increases with µ, k0 → 0 as µ → 0, and k0 → 2 ad as µ → ∞. On the other hand, one easily sees by a comparison argument that for fixed k, Uk (·) increases with a and decreases with b, and it follows that Uk0 (0) increases with a and decreases with b. This implies that k0 increases with a and decreases with b. Combining these analysis, we find that for fixed d, if k0 is viewed as a function of (µ, a, b), namely k0 = k0 (µ, a, b), then µ1 > µ2 , a1 > a2 and b1 6 b2 imply k0 (µ1 , a1 , b1 ) > k0 (µ2 , a2 , b2 ), and the strict inequality holds when (µ1 , a1 , b1 ) 6= (µ2 , a2 , b2 ). Proposition 8.25 Let k0 be the spreading speed determined by Proposition 8.23. Then k0 lim √ = 2 ad

aµ →∞ bd

k0 bd 1 lim √ =√ . 3 ad aµ

and

aµ →0 bd

Proof. By Proposition 8.23 above, for any λ ∈ [0, 2), the problem −W 00 + λW 0 = W − W 2 in (0, ∞), W (0) = 0 has a unique positive solution denoted by Wλ , and for each α > 0, there exists a unique λ0 (α) ∈ (0, 2) such that λ0 (α) = αWλ0 0 (α) (0) and W00 (0) > 0,

lim Wλ0 (0) = 0.

λ→2−

Clearly, the pair (λ0 (α), λ0 (α)/α) satisfies lim (λ0 (α), λ0 (α)/α) = (0, W00 (0)) and

α→0

lim (λ0 (α), λ0 (α)/α) = (2, 0).

α→∞

A simple calculation confirms that for each k > 0, q  b W √k (x) = Uk d/a x . ad a

And consequently, b W √k (0) = a ad 0

and µUk0 0 (0) = k0 ⇐⇒

s

d 0 U (0) a k

aµ 0 k0 W √k0 (0) = √ . bd ad ad

(8.58)

8.7. A system coupled by one PDE and one ODE  223

It follows that aµ k √ 0 = λ0 . bd ad 



This combined with (8.58) allows us to derive k0 k0 bd = W00 (0). lim √ = 2 and aµlim √ aµ →0 ad ad bd √ We show next that W00 (0) = 1/ 3. Indeed, W0 satisfies aµ →∞ bd

−W000 = W0 − W02 , W0 > 0 in (0, ∞), W0 (0) = 0 and W0 (∞) = 1. It follows that ∞ ∞ 1 1 (W0 − W02 )W00 dx = , W000 W00 dx = (W00 (0))2 = − 2 6 0 0 √ which implies W00 (0) = 1/ 3, and the proof is complete.

Z

8.7

Z



A SYSTEM COUPLED BY ONE PDE AND ONE ODE

In this section we introduce a free boundary problem of a system coupled by one PDE and one ODE   ut = f1 (t, x, u, v),       vt = dvxx + f2 (t, x, u, v),      u(t, x) = v(t, x) = 0,   0

g (t) = −µv (t, g(t)),

x    0   h (t) = −βvx (t, h(t)),      u(0, x) = u0 (x), v(0, x) = v0 (x),     

t > 0, g(t) < x < h(t), t > 0, g(t) < x < h(t), t > 0, x = g(t), h(t), t > 0,

(8.59)

t > 0, −h0 6 x 6 h0 ,

h(0) = −g(0) = h0 .

This model describes the relationship between two species, such as viruses and healthy cells, in which the viruses have the ability of diffusion and the healthy cells do not. Contents of this section are taken from [79]. Let us define a couple of notations for the sake of arguments. Let a < b and T > 0 be constants, and g, h ∈ C([0, T ]) satisfy g(t) < h(t) for all 0 6 t 6 T . We define • C 1− ([a, b]) to be the Lipschitz continuous function space in [a, b]; T • Dg,h = {0 < t 6 T, g(t) < x < h(t)}; T

• function u ∈ C 1,1− (Dg,h ) to mean that u is continuously differentiable in t ∈ [0, T ] and is Lipschitz continuous in x ∈ [g(t), h(t)] for all t ∈ [0, T ]; • function u ∈ C 1−,1− ([0, T ]×[g(T ), h(T )]) to mean that u is Lipschitz continuous in t ∈ [0, T ] and in x ∈ [g(T ), h(T )].

224  8. Free Boundary Problems from Ecology

We assume that initial functions u0 and v0 satisfy • (u0 , v0 ) ∈ C 1− ([−h0 , h0 ]) × Wp2 ((−h0 , h0 )) with p > 3, u0 (±h0 ) = v0 (±h0 ) = 0, u0 , v0 > 0 in (−h0 , h0 ), and v00 (h0 ) < 0, v00 (−h0 ) > 0, and denote by L0 the Lipschitz constant of u0 in x. Let α = 1 − 3/p. It is assumed that (f1 , f2 ) satisfies (L) f1 (t, x, 0, v) > 0 for all v > 0 and f2 (t, x, u, 0) > 0 for all u > 0. For any given τ, l, k1 , k2 > 0, fi (·, 0, 0) ∈ L∞ ((0, τ ) × (−l, l)), and there exists a constant L(τ, l, k1 , k2 ) > 0 such that |fi (t, x, u1 , v1 ) − fi (t, y, u2 , v2 )| 6 L(τ, l, k1 , k2 )(|x − y| + |u1 − u2 | + |v1 − v2 |) for all t ∈ [0, τ ], x, y ∈ [−l, l], u1 , u2 ∈ [0, k1 ], v1 , v2 ∈ [0, k2 ] and i = 1, 2. It is easy to see that the condition (L) implies fi ∈ L∞ ((0, τ ) × (−l, l) × (0, k1 ) × (0, k2 )) for any given τ , l, k1 , k2 > 0. Conditions that u(t, g(t)) = u(t, h(t)) = 0 in (8.59) look like boundary conditions of u, but they do actually play roles of initial conditions of u at points x = g(t) and x = h(t), respectively. It can be seen from the following proof of Theorem 8.26 that the method and process of proving existence and uniqueness of solutions are quite different from problem (8.1). Theorem 8.26 Under above assumption (L), there exists a T > 0 such that problem (8.59) has a unique solution (u, v, g, h) defined in [0, T ]. Moreover, g, h ∈ C 1+α/2 ([0, T ]), g 0 (t) < 0, h0 (t) > 0 in [0, T ], T

T

T T ) ∩ C (1+α)/2, 1+α (Dg,h ), u, v > 0 in Dg,h . u ∈ C 1,1− (Dg,h ), v ∈ Wp1,2 (Dg,h

Proof. The proof process is very long, and we will divide it into several steps. As parameters d, µ and β are fixed, in the following arguments we do not emphasize the dependence of estimations on them. Step 1. We denote  1 min − βv00 (h0 ), µv00 (−h0 ) > 0, 2 A = max u0 + 1, B = max v0 + 1, L1 = L(1, 2h0 , A, B),

σ=

[−h0 ,h0 ]

[−h0 ,h0 ]

C = {h0 , A, B, kv0 kWp2 ((−h0 ,h0 )) , v00 (±h0 ), kf2 k∞, Π1 ×(0,A)×(0,B) , L1 },

and for any given 0 < T < ∞, we define ΠT = [0, T ] × [−2h0 , 2h0 ], ∆∗T = [0, T ] × [−1, 1], XTu0 = {φ ∈ C(ΠT ) : φ(0, x) = u0 (x), 0 6 φ 6 A} .

8.7. A system coupled by one PDE and one ODE  225

We say u ∈ Cx1− (ΠT ) if there is a constant L(u, T ) such that |u(t, x1 ) − u(t, x2 )| 6 L(u, T )|x1 − x2 | for all x1 , x2 ∈ [−2h0 , 2h0 ] and t ∈ [0, T ]. For the given u ∈ X1u0 ∩ Cx1− (Π1 ), we consider the following problem  vt = dvxx + f2 (t, x, u(t, x), v), 0 < t 6 1, g(t) < x < h(t),        v(t, g(t)) = v(t, h(t)) = 0, 0 6 t 6 1,      g 0 (t) = −µvx (t, g(t)), 0 6 t 6 1,   h0 (t) = −βvx (t, h(t)),       v(0, x) = v0 (x),     

0 6 t 6 1,

(8.60)

|x| 6 h0 ,

h(0) = −g(0) = h0 .

Due to properties of f2 and u, using a similar argument in proving Theorem 8.1 we can show that there exists 0 < T0 6 1, independent of L(u, 1), the Lipschitz constant of u in x, such that (8.60) has a unique solution (v, g, h)‡ which satisfies g, h ∈ C 1+α/2 ([0, T0 ]),

T0

T0 v ∈ Wp1,2 (Dg,h ) ∩ C (1+α)/2, 1+α (Dg,h ),

and |g(t)|, h(t) 6 2h0 in [0, T0 ] and T0 kg, hkC 1+α/2 ([0,T0 ]) , |v|1+α, DT0 6 K(T0−1 ), 0 < v 6 B in Dg,h ,

(8.61)

g,h

where the constant K(T0−1 ) is independent of L(u, 1). Moreover, we can choose T0 to be independent of u ∈ X1u0 ∩ Cx1− (Π1 ). In fact, let ∗ f2 (t, x, v) = max06u6A f2 (t, x, u, v) and (v ∗ , g ∗ , h∗ ) be the unique solution of (8.60) with f2 (t, x, u(t, x), v) replaced by f2∗ (t, x, v). We denote T ∗ as the maximum existence time of (v ∗ , g ∗ , h∗ ). Then the comparison principle shows that v 6 v ∗ , g > g ∗ , h 6 h∗ and T0 > T ∗ . 
2/α Take T0∗ = min T0 , σ/K(T0−1 ) . Then T0∗ does not depend on u ∈ X1u0 ∩ Cx1− (Π1 ) and L(u, 1), and (8.61) holds with T0 replaced by T0∗ . For convenience, we also write T0∗ as T0 . Noting −g 0 (0) > 2σ and h0 (0) > 2σ. Using the C α/2 ([0, T0 ]) estimates of g 0 (t), h0 (t) given in (8.61), we can find a 0 < T∗ 6 T0 , which depends only on K(T0−1 ), such that −g 0 (t) > σ, h0 (t) > σ f or all 0 6 t 6 T∗ . Clearly, T∗ does not depend on u ∈ X1u0 ∩ Cx1− (Π1 ) and L(u, 1), and (8.61) holds with T0 replaced by T∗ . We also write T∗ as T0 . Define u˜0 (x) = u0 (x) when |x| 6 h0 , and u˜0 (x) = 0 when |x| > h0 . Then u˜0 ∈ C 1− ([g(T0 ), h(T0 )]) since u0 ∈ C 1− ([−h0 , h0 ]). For functions g(t) and h(t) obtained ‡ Using the Schauder fixed point theorem to determine T0 and get the existence of solutions first, we can see that T0 is independent of L(u, 1). Then prove the uniqueness of solutions.

226  8. Free Boundary Problems from Ecology

above, it is easy to see that their inverse functions g −1 (x) and h−1 (x) exist for x ∈ [g(T0 ), h(T0 )]. We set (see Fig. 8.3) t 6 T0

x g(t)

x

tx

•

tx • −h0

• h0

• h(t)

tx =

-x

 −1 g (x)   

if x ∈ [g(T0 ), −h0 ),

  

if x ∈ (h0 , h(T0 )],

0

−1

h (x)

if |x| 6 h0 ,

(8.62)

Fig. 8.3

which is Lipschitz continuous in x. For the function v(t, x) obtained above, and for every g(T0 ) < x < h(T0 ), we consider the following initial value problem  u ˜t = f1 (t, x, u˜, v(t, x)),

t x < t 6 T0 ,

(8.63)

u ˜(tx , x) = u˜0 (x).

Note properties of f1 and v. By the standard theory for ODE, we can prove that there exists 0 < T < T0 , which depends on L1 , A, B and K, such that for each g(T ) 6 x 6 h(T ), the solution u˜(t, x) of (8.63) exists uniquely in [tx , T ] as a function T of t. Consequently, we get a function u˜ of (t, x), which is defined in Dg,h . Moreover, T

u˜ ∈ C 1,1− (Dg,h ), of which the Lipschitz constant will be calculated in the next step, and u˜(t, g(t)) = u˜(t, h(t)) = 0 for 0 6 t 6 T . Make zero extension of u˜ to [0, tx ] for every g(T ) 6 x 6 h(T ). Then u˜ ∈ C 1−,1− ([0, T ] × [g(T ), h(T )]). Step 2. Now, we intend to get an estimate of the Lipschitz constant of u˜ in x. Since −g 0 (t), h0 (t) > σ > 0 for all 0 6 t 6 T0 , we have k(g −1 )0 kC([g(T0 ),−h0 ]) 6 σ −1 and k(h−1 )0 kC([h0 ,h(T0 )]) 6 σ −1 .

(8.64)

Set F1 (s, x) = f1 (s, x, u˜(s, x), v(s, x)). It follows from the first equation of (8.63) that, for tx < τ 6 T , Z τ

u˜(τ, x) = u˜(tx , x) +

F1 (s, x)ds. tx

For any given (t, x1 ), (t, x2 ) ∈ ΠT , we divide arguments into several cases. T

Case 1: (t, x1 ), (t, x2 ) ∈ Dg,h with −h0 6 x2 < x1 . Then t > tx1 > tx2 > 0, and consequently, for any tx1 6 τ 6 t, |˜ u(τ, x1 ) − u˜(τ, x2 )| 6 |˜ u(tx1 , x1 ) − u˜(tx2 , x2 )| + Z

Z

tx1

|F1 (s, x2 )|ds

tx2

τ

+

|F1 (s, x1 ) − F1 (s, x2 )|ds.

tx1

In view of the condition (L), it is easy to derive that |F1 (s, x1 ) − F1 (s, x2 )| 6 L1 (|x1 − x2 | + |˜ u(s, x1 ) − u˜(s, x2 )| + |v(s, x1 ) − v(s, x2 )|),

8.7. A system coupled by one PDE and one ODE  227

where L1 = L(1, 2h0 , A, B). It yields |˜ u(τ, x1 ) − u˜(τ, x2 )| 6 T L1 k˜ u(·, x1 ) − u˜(·, x2 )kC([tx1 ,t]) + |˜ u(tx1 , x1 ) − u˜(tx2 , x2 )| +L1 |x1 − x2 | + C1 |tx1 − tx2 | Z

τ

+L1

|v(s, x1 ) − v(s, x2 )|ds

(8.65)

tx1

as τ 6 T 6 1, where C1 = kf1 k∞, Π1 ×(0,A)×(0,B) . By use of (8.61), Z

τ

tx1

T |x1 − x2 | 6 C2 |x1 − x2 | |v(s, x1 ) − v(s, x2 )|ds 6 T kvx k∞, Dg,h

as τ 6 T 6 1. If tx1 > 0 and tx2 > 0, then u˜(tx1 , x1 ) = u˜(tx2 , x2 ) = 0, and |tx1 − tx2 | = |h−1 (x1 ) − h−1 (x2 )| 6 k(h−1 )0 kC([h0 , h(T )]) |x1 − x2 | 6 σ −1 |x1 − x2 | by (8.64). If tx1 > 0 and tx2 = 0, then x2 ∈ [−h0 , h0 ], x1 > h0 and u˜(tx1 , x1 ) = 0. Let L0 be the Lipschitz constant of u0 in x. It then follows that |tx1 − tx2 | = |h−1 (x1 ) − h−1 (h0 )| 6 σ −1 |x1 − h0 | 6 σ −1 |x1 − x2 |, |˜ u(tx1 , x1 ) − u˜(tx2 , x2 )| = |u0 (h0 ) − u0 (x2 )| 6 L0 |h0 − x2 | 6 L0 |x1 − x2 |. If tx1 = tx2 = 0, i.e., x1 , x2 ∈ [−h0 , h0 ], then |˜ u(tx1 , x1 ) − u˜(tx2 , x2 )| = |u0 (x1 ) − u0 (x2 )| 6 L0 |x2 − x1 |. Substituting these estimates into (8.65), we conclude |˜ u(τ, x1 ) − u˜(τ, x2 )| 6 T L1 k˜ u(·, x1 ) − u˜(·, x2 )kC([tx1 ,t]) + M |x1 − x2 |, where M = L0 + L1 + C1 σ −1 + C2 L1 . Take the maximum of |˜ u(τ, x1 ) − u˜(τ, x2 )| in [tx1 , t] to yield k˜ u(·, x1 ) − u˜(·, x2 )kC([tx1 ,t]) 6 T L1 k˜ u(·, x1 ) − u˜(·, x2 )kC([tx1 ,t]) + M |x1 − x2 |, which implies |˜ u(t, x1 ) − u˜(t, x2 )| 6 k˜ u(·, x1 ) − u˜(·, x2 )kC([tx1 ,t]) 6 2M |x1 − x2 |

(8.66)

provided 0 < T 6 min{1, 2L1 1 }. T

Case 2: (t, x1 ), (t, x2 ) ∈ Dg,h with x2 < x1 6 h0 . Similar to above, (8.66) holds. T

Case 3: (t, x1 ), (t, x2 ) ∈ Dg,h with x2 < −h0 < h0 < x1 . Then |˜ u(t, x1 ) − u˜(t, x2 )| 6 |˜ u(t, x1 ) − u˜(t, h0 )| + |˜ u(t, h0 ) − u˜(t, x2 )| 6 4M |x1 − x2 |.

(8.67)

228  8. Free Boundary Problems from Ecology T

Case 4: (t, x1 ), (t, x2 ) 6∈ Dg,h . Then u˜(t, x1 ) = u˜(t, x2 ) = 0. T

T

Case 5: (t, x1 ) 6∈ Dg,h and (t, x2 ) ∈ Dg,h . We may assume x1 > h(t). It is clear T

that (t, h(t)), (t, x2 ) ∈ Dg,h and u˜(t, x1 ) = u˜(t, h(t)) = 0. Hence, we have |˜ u(t, x1 ) − u˜(t, x2 )| 6 |˜ u(t, h(t)) − u˜(t, x2 )| 6 4M |h(t) − x2 | 6 4M |x1 − x2 |. In conclusion, the estimate (8.67) always holds provided 0 < T 6 min{1, 2L1 1 }. Define YTu0 = {φ ∈ XTu0 : |φ(t, x) − φ(t, y)| 6 4M |x − y| for − 2h0 6 x, y 6 2h0 }. Obviously, YTu0 is complete with metric d(φ1 , φ2 ) = maxΠT |φ1 − φ2 |. For any given u ∈ YTu0 , we extend u to [T, 1] × [−2h0 , 2h0 ] by setting u(t, x) = u(T, x). Then u ∈ X1u0 ∩ Cx1− (Π1 ). Define a mapping Γ by Γ(u) = u˜, where u˜ is given in Step 1. Above discussions show that Γ maps YTu0 into itself. Step 3. We hope to show that Γ is a contraction mapping in YTu0 for 0 < T  1. Let (vi , gi , hi ) be the unique local solution of (8.60) with u = ui ∈ YTu0 , i = 1, 2, and define tix by the manner (8.62) with (g, h) = (gi , hi ). Let u˜i be the unique solution of (8.63) with tx = tix , v = vi and T0 = T . Then u˜i (t, x) = u˜i (tix , x) +

Z

t

tix

f1 (s, x, u˜i , vi )ds for x ∈ [gi (T ), hi (T )].

Make zero extensions of u˜i and vi in ([0, T ] × R) \ DgTi ,hi , and set e =u U = u1 − u2 , U ˜1 − u˜2 , h = h1 − h2 , g = g1 − g2 , ΩT = DgT1 ,h1 ∪ DgT2 ,h2 . e (t, x)| in all possible cases. Fix (t, x) ∈ ΩT , we now estimate |U Case 1: x ∈ (g1 (t), h1 (t)) \ (g2 (t), h2 (t)). In such case, either g1 (t) < x 6 g2 (t) or h2 (t) 6 x < h1 (t), so we have u˜1 (t1x , x) = 0, u˜2 (t, x) = 0. Thus Z t e |U (t, x)| = |˜ u1 (t, x)| = f1 (s, x, u˜1 , v1 )ds 6 C1 |t − t1x |, t1x

where C1 = kf1 k∞, Π1 ×(0,A)×(0,B) . When h2 (t) 6 x < h1 (t), we have 0 < t1x < t and h1 (t) > h1 (t1x ) = x > h2 (t). Therefore, e (t, x)| 6 C1 |t − t1 | = C1 |h−1 (h1 (t)) − h−1 (h1 (t1 ))| |U x x 1 1 0 1 6 C1 k(h−1 1 ) kC([0,T ]) |h1 (t) − h1 (tx )|

6 C1 σ −1 |h1 (t) − h1 (t1x )| 6 C1 σ −1 |h1 (t) − h2 (t)| 6 C1 σ −1 khkC([0,T ]) ,

8.7. A system coupled by one PDE and one ODE  229

where σ > 0 is given in (8.64). When g1 (t) < x 6 g2 (t), we can analogously obtain e (t, x)| = |˜ |U u1 (t, x)| 6 C1 σ −1 kgkC([0,T ]) .

Case 2: x ∈ (g2 (t), h2 (t)) \ (g1 (t), h1 (t)). Similar to Case 1, we can get e (t, x)| = |˜ |U u2 (t, x)| 6 C1 σ −1 kg, hkC([0,T ]) .

Case 3: x ∈ (g1 (t), h1 (t)) ∩ (g2 (t), h2 (t)). When x ∈ [−h0 , h0 ], we have t1x = t2x = 0 and u˜1 (t1x , x) = u˜2 (t2x , x) = u˜0 (x). Hence e (t, x)| 6 |U

Z

t

|f1 (s, x, u˜1 , v1 ) − f1 (s, x, u˜2 , v2 )|ds

0





ek 6 T L1 kU C(ΩT ) + kv1 − v2 kC(ΩT ) .

When x ∈ (g1 (t), h1 (t)) ∩ (g2 (t), h2 (t)) \ [−h0 , h0 ], we have t1x > 0, t2x > 0 and u˜1 (t1x , x) = u˜2 (t2x , x) = 0. Without loss of generality, we assume x > h0 and t2x > t1x > 0. Then h1 (t2x ) > h1 (t1x ) = x = h2 (t2x ), x ∈ (g1 (s), h1 (s)) ∩ (g2 (s), h2 (s)) for all t2x < s 6 t and x ∈ (g1 (t2x ), h1 (t2x )) \ (g2 (t2x ), h2 (t2x )). Consequently e (t2 , x)| = |˜ |U u1 (t2x , x)| 6 C1 σ −1 kg, hkC([0,T ]) x

in line with the conclusion of Case 1. Integrating the differential equation of u˜i from t2x to t, we obtain u˜1 (t, x) = u˜1 (t2x , x) +

Z

t

t2x

Z

f1 (s, x, u˜1 , v1 )ds, u˜2 (t, x) =

t

t2x

f1 (s, x, u˜2 , v2 )ds.

It follows that e (t, x)| 6 |˜ |U u1 (t2x , x)| + e (t2 , x)| + 6 |U x

Z

t

t2x Z t

|f1 (s, x, u˜1 , v1 ) − f1 (s, x, u˜2 , v2 )|ds |f1 (s, x, u˜1 , v1 ) − f1 (s, x, u˜2 , v2 )|ds

0





ek 6 C1 σ −1 kg, hkC([0,T ]) + T L1 kU C(ΩT ) + kv1 − v2 kC(ΩT ) .

In conclusion, 



e (t, x)| 6 C1 σ −1 kg, hk e |U C([0,T ]) + T L1 kU kC(ΩT ) + kv1 − v2 kC(ΩT ) .

(8.68)

We will show in the following Step 4 that if 0 < T  1 then there exists a positive constant C such that kg, hkC 1 ([0,T ]) 6 CkU kC(ΠT ) and kv1 − v2 kC(ΩT ) 6 CkU kC(ΠT ) .

(8.69)

Once this is done, noticing g(0) = h(0) = 0, the first inequality of (8.69) implies kg, hkC([0,T ]) 6 T kg, hkC 1 ([0,T ]) 6 T CkU kC(ΠT ) .

230  8. Free Boundary Problems from Ecology

Then combining with (8.68), we have 1 ek kU C(ΠT ) 6 kU kC(ΠT ) if 0 < T  1. 3 This demonstrates that Γ is a contraction mapping in YTu0 . Hence, Γ has a unique fixed point u in YTu0 by the contraction mapping principle. Let (v, g, h) be the unique solution of (8.60) with such u. Then (u, v, g, h) is a solution of (8.59), and it is T the unique one provided u ∈ YTu0 . Moreover, we can see that u ∈ C 1,1− (Dg,h ) and T

T v ∈ C 1,2 (Dg,h ) ∩ C (1+α)/2, 1+α (Dg,h ).

Step 4. Now we show the estimate (8.69). Before our statement, some preparations are needed. Let 1 xi (t, y) = [(hi (t) − gi (t))y + hi (t) + gi (t)], 2 2 h0 (t) + gi0 (t) h0i (t) − gi0 (t) ξi (t) = , ζi (t, y) = i + y, hi (t) − gi (t) hi (t) − gi (t) hi (t) − gi (t) wi (t, y) = ui (t, xi (t, y)), zi (t, y) = vi (t, xi (t, y)) and f2i (t, y, w, z) = f2 (t, xi (t, y), w, z) for i = 1, 2. Then  2 i   zi,t = dξi zi,yy + ζi zi,y + f2 (t, y, wi (t, y), zi (t, y)),

z (t, ±1) = 0,

i   z (0, y) = v (h y) =: z (y), i 0 0 i0

0 < t 6 T, |y| < 1, 0 6 t 6 T, |y| 6 1.

Recalling (8.61), we have 1 1 6 ξi (t) 6 , t ∈ [0, T ], 2h0 h0 2[hi (t) − hi (s) − (gi (t) − gi (s))] sup |ξi (t) − ξi (s)| = sup [hi (t) − gi (t)][hi (s) − gi (s)] |t−s| 0. If the initial value problem  0 0  φ (t) = f1 (φ, ψ), ψ (t) = f2 (φ, ψ), t > 0,  φ(0) = max u0 > 0, ψ(0) = max v0 > 0 [−h0 ,h0 ]

[−h0 ,h0 ]

234  8. Free Boundary Problems from Ecology

has a global solution (φ, ψ), then the unique solution (u, v, g, h) of (8.59) also exists globally. Proof. Clearly, φ(t) > 0, ψ(t) > 0 for all t > 0. Let Tmax be the maximal existence time of (u, v, g, h). For any fixed 0 < T < Tmax , applying the comparison principle in T region Dg,h , we conclude: T

   u(t, x) 6 φ(t) 6 max φ(t) =: M (T )

in Dg,h ,

  v(t, x) 6 ψ(t) 6 max ψ(t) =: N (T )

in Dg,h .

[0, T +1]

[0, T +1]

T

(8.83)

We have known in proving Theorem 8.26 that h0 (t) > 0. Set A=

sup

f2 (u, v).

[0,M (T )]×[0,N (T )]

As in the proof of Lemma 8.2,   s 2  AN (T ) h0 (t) 6 2β max , , − min v00 (x) for all 0 6 t 6 T,  h0  [0,h0 ] 2d   s 2  AN (T ) g 0 (t) > −2µ max , , max v00 (x) for all 0 6 t 6 T.  h0  [−h0 ,0] 2d

Recalling the estimate (8.83) and above estimates, and using a similar method to the proof of Theorem 8.5 we conclude Tmax = ∞.  Theorem 8.28 Let conditions (L) and (L1) hold. Assume also that there exists k0 > 0 such that f1 (u, v) < 0 for all u > k0 and v > 0, and suppose that for any given η > 0, there exists Θ(η) > 0 such that f2 (u, v) < 0 for 0 6 u 6 η, v > Θ(η), then the unique solution (u, v, g, h) of (8.59) exists globally. Proof. It is easy to see that u(t, x) 6 max u0 + k0 =: η, v(t, x) 6 max v0 + Θ(η). [−h0 ,h0 ]

[−h0 ,h0 ]

The remaining proof is the same as that of Theorem 8.27.  In the past decade, research on free boundary models originating from species invasion and epidemic spread has developed rapidly, and a large number of research results have appeared. This book only focuses on the reaction-diffusion equation and one system coupled by one PDE and one ODE. In these two respects, in addition to the work mentioned above, see also [23, 28, 27, 41, 3, 56, 66, 83, 20, 30, 54, 78, 69] and references therein. With regard to the work of systems, we just refer the reader to [26, 121, 43, 44, 123, 29, 122, 119, 77] for the competition models, [112, 120] for the predator models, and references therein. In particular, papers [44, 119, 77, 120] focus on the

Exercises  235

case where two components have different free boundaries, and the paper [29] deals with the propagation speed of free boundary of a competitive model. There have also been many studies on epidemic free boundary models, readers can refer to [51, 39, 75] and references therein. At present, there are also many advances in the study of free boundary problems of non-local diffusion equations, and systems with non-local vs. local diffusions. The processing method is different from free boundary problems of general reactiondiffusion equations. Interested readers can refer to [13, 70, 103, 104, 125] for non-local diffusion problems and references therein.

EXERCISES 8.1 Prove the estimate (8.16). 8.2 Prove that solution (u, h) of (8.1) is strictly monotonically increasing in µ > 0. 8.3 Study the following free boundary problem with double free boundaries and sign-changing intrinsic growth rate in heterogeneous time-periodic environment   ut − duxx = a(t, x)u − b(t, x)u2 ,      u(t, g(t)) = u(t, h(t)) = 0,       g 0 (t) = −ρux (t, g(t)), 0

 h (t) = −µux (t, h(t)),       u(0, x) = u0 (x),     

t > 0, g(t) < x < h(t), t > 0, t > 0, t > 0, −h0 6 x 6 h0 ,

g(0) = −h0 , h(0) = h0 ,

where a, b ∈ C α/2,1 ∩ L∞ ([0, T ] × R) with α ∈ (0, 1), and are time-periodic with period T , a is positive somewhere in [0, T ]×R, and b1 6 b 6 b2 in [0, T ]×R for some positive constants b1 and b2 . Investigate the following problems:


(1) existence and uniqueness of global solution; (2) regularity and uniform estimates; (3) longtime behaviour of solution; (4) criteria and dichotomy for spreading and vanishing. 8.4 Prove that problem (8.53) has at least one positive solution. 8.5 Prove that there exists 0 < T0  1 such that (8.60) has a unique solution (v, g, h) which satisfies g, h ∈ C 1+α/2 ([0, T0 ]),

T0 v ∈ Wp1,2 (Dg,h )∩C

1+α , 1+α 2

T0

(Dg,h ).

236  8. Free Boundary Problems from Ecology

8.6 Consider the following free boundary problem of predator-prey model with double free boundaries   ut − uxx = u(1 − u + av),       u(t, x) ≡ 0,      vt − dvxx = v(b − v − cu),       u = 0, g 0 (t) = −µux ,  u = 0, h0 (t) = −µux ,       g(0) = −h0 , h(0) = h0 ,       u(0, x) = u0 (x),     

v(0, x) = v0 (x),

t > 0, g(t) < x < h(t), t > 0, x 6∈ (g(t), h(t)), t > 0, x ∈ R, t > 0, x = g(t), t > 0, x = h(t), x ∈ [−h0 , h0 ], x ∈ R,

where a, b, c, d, h0 and µ are given positive constants. The initial functions u0 (x), v0 (x) satisfy: u0 ∈ C 2 ([−h0 , h0 ]), u0 (±h0 ) = 0, u0 > 0 in (−h0 , h0 ), v0 ∈ Cb (R) and v0 > 0 in R, here Cb (R) is the space of continuous and bounded functions in R. Same as Exercise 8.3, investigate the following problems: (1) existence and uniqueness of global solution; (2) regularity and uniform estimates; (3) longtime behaviour of solution; (4) criteria and dichotomy for spreading and vanishing.

CHAPTER

9

Semigroup Theory and Applications

Semigroup theory provides a unified and powerful tool for the study of differential equations on Banach spaces covering systems described by ordinary differential equations, partial differential equations, functional differential equations, and their combinations. The basic idea of the semigroup method is to rewrite an evolutionary partial differential equation as an abstract ordinary differential equation, and to deal with the evolutionary partial differential equation by imitating the initial value problem of the ordinary differential equation. Considering an initial value problem of an ordinary differential system (

y 0 (t) + Ay(t) = f (t, y), t > 0, y(0) = y0 ∈ Rm ,

where A is a matrix of order m. Based on the variation-of-constants formula, its solution can be represented as y(t) = e

−tA

Z

y0 +

t

e−(t−s)A f (s, y(s))ds.

0

Given an initial-boundary value problem of parabolic equation (system), for example,    ut − ∆u = f (x, t, u), 

x ∈ Ω,

u = 0,

x ∈ ∂Ω, t > 0,

  

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

x ∈ Ω,

t > 0,

we can rewrite it as an initial value problem of abstract ordinary differential equation (

u0 (t) + Au(t) = f (t, u), t > 0, u(0) = u0 ,

where A is an operator from a subspace X0 (of a Banach space X) to X. Similarly 237

238  9. Semigroup Theory and Applications

to the ordinary differential equation, its solution can be written formally as u(t) = e−tA u0 +

Z

t

e−(t−s)A f (s, u(s))ds.

0

The family of operators {e−tA }t>0 is called a semigroup, and such a function u is called a mild solution. Then, by studying the properties of the semigroup, we can determine whether this mild solution satisfies the abstract ordinary differential equation. This chapter is devoted to a brief review of basic results of semigroup theory and to establishing the existence, uniqueness, extension and regularity of mild solutions. An initial-boundary value problem of a system coupling parabolic equations and ordinary differential equations is systematically investigated. For abstract integrals (Bochner integrals) and abstract derivatives (Fr´echet derivatives and Gˆateaux derivatives) of abstract functions (functions from a measurable space to a Banach space), readers can refer to textbooks on nonlinear functional analysis and evolution equations such as [15, 42]. Throughout this chapter, X is a Banach space equipped with the norm k · k.

C0 SEMIGROUP THEORY

9.1 9.1.1

Basic results

In this subsection, we review some basic results of C0 semigroup theory. Their proofs will be omitted, and we refer to [89, 15, 82, 31, 109, 124] for details. Let Ω be a smooth bounded domain and ϕ ∈ C 2 (Ω), and let u(x, t) be the unique solution of    ut − ∆u = 0,  

u = 0, u(x, 0) = ϕ(x),

x ∈ Ω, t > 0, x ∈ ∂Ω, t > 0, x ∈ Ω.

Set u(x, t) = Q(t)ϕ(x). Then Q(t) has the following properties: (a) Q(0) = I, the identity operator; (b) Q(t)Q(s) = Q(t + s) for all t, s > 0; (c) limt→0+ kQ(t)ϕ − ϕkC(Ω) = 0 for all ϕ ∈ C 2 (Ω); Q(t)ϕ − ϕ u(x, t) − u(x, 0) = lim = ut (x, 0) = ∆u(x, 0) = ∆ϕ(x). t&0 t&0 t t

(d) lim

Definition 9.1 For t > 0, let Q(t) : X → X be continuously linear operators having above properties (a)–(c) with ϕ ∈ C 2 (Ω) replaced by ϕ ∈ X in (c). The family of operators {Q(t)}t>0 is called a strongly continuous semigroup or a C0 semigroup (of operators) of X. For simplicity, we denote Q(t) := {Q(t)}t>0 henceforth.

9.1. C0 semigroup theory  239

Definition 9.2

Let Q(t) be a C0 semigroup of X. We put 

D(A) = v ∈ X : lim+ t→0

Q(t)v − v exists t



and call the linear operator −A : D(A) → X defined by −Av = lim+ t→0

Q(t)v − v t

the infinitesimal generator of the semigroup Q(t). For a given operator A, the resolvent set and spectrum of A are denoted by ρ(A) and σ(A), respectively. Theorem 9.3 Let Q(t) be a C0 semigroup of X and −A the infinitesimal generator of Q(t). Then the following conclusions are valid. (1) There exist constants M > 0 and a ∈ R so that kQ(t)k 6 M e−at for all t > 0. (2) For each v ∈ X, we have Q(t)v ∈ C([0, ∞), X) and Z

t

Q(s)vds ∈ D(A),

Q(t)v − v = −A

Z

0

t

Q(s)vds. 0

(3) (Existence of solution) For v ∈ X, we define u(t) = Q(t)v, t > 0. When v ∈ D(A), the derivative u0 (t) exists and u0 (t) = −Au(t) = −Q(t)Av, i.e., A commutes with Q(t). In particular, if v ∈ D(A) then so is Q(t)v. In that sense, there is no loss of regularity of Q(t)v when compared with v. We can also write Q0 (t) = −Q(t)A in D(A). (4) (λ − A)−1 Q(t) = Q(t)(λ − A)−1 for all λ ∈ ρ(A) and t > 0, where ρ(A) is the resolvent set of A. (5) A is closed and D(A) is dense in X. (6) (Uniqueness of solution) Let τ ∈ (0, ∞] and u : [0, τ ) → X be a continuous function. If for 0 < t < τ , u(t) ∈ D(A), u0 (t) exists and u0 (t) = −Au(t), then u(t) = Q(t)u(0) in [0, τ ). (7) (Uniqueness of semigroup) If T (t) is a C0 semigroup of X, and its infinitesimal generator is still −A, then T (t) = Q(t) for all t > 0. The following two theorems give the necessary and sufficient conditions for a linear operator to be an infinitesimal generator of a C0 semigroup. Theorem 9.4 (Hille-Yosida Theorem) Let A : D(A) → X be a closed and denselydefined linear operator. Then −A is the infinitesimal generator of a C0 semigroup Q(t) satisfying kQ(t)k 6 1 if and only if (−∞, 0) ⊂ ρ(A) and k(A − λ)−1 k 6

1 for all λ < 0. |λ|

(9.1)

240  9. Semigroup Theory and Applications

A C0 semigroup Q(t) satisfying kQ(t)k 6 1 is called a contraction semigroup. Theorem 9.5 (Hille-Yosida Theorem) Let A : D(A) → X be a linear operator. Then −A is the infinitesimal generator of a C0 semigroup Q(t) satisfying kQ(t)k 6 M e−at for some constants M and a, if and only if (1) D(A) is dense in X; (2) (−∞, a) ⊂ ρ(A) and k(A − λ)−k k 6 M (a − λ)−k for all λ < a, k = 1, 2, . . . . Remark 9.6

(9.2)

(1) Inequality (9.2) indicates that A is closed.

(2) Under conditions of Theorem 9.5 we have  −k

t

lim sup Q(t)v − 1 + A v = 0 for 0 < T < ∞, v ∈ X.

k

k→∞ t∈[0,T ]

Based on this limit, it is reasonable to write Q(t) = e−tA . Here we consider an example. Example 9.1 Let Ω be of class C 2 and X = {u ∈ C(Ω) : u|∂Ω = 0}. Define an operator A by D(A) = {u ∈ H 2 (Ω) ∩ X : ∆u ∈ X}, Au = −∆u

for u ∈ D(A).

Theorem 9.7 The operator −A is the infinitesimal generator of a contraction semigroup in X. Proof. Clearly, A is densely-defined because C0∞ (Ω) is dense in X. Now we prove that A is closed. Let uk ∈ D(A) and uk → u, Auk → v in X. Then uk ∈ H 2 (Ω) ∩ X, Auk , u, v ∈ L2 (Ω), and uk → u, Auk → v in L2 (Ω). Thus, we have kuk − ul kH 2 (Ω) 6 CkA(ul − uk )k2 → 0 as k, l → ∞, which implies uk → u in H 2 (Ω) ∩ X, and consequently Auk + ∆u = −∆(uk − u) → 0 in L2 (Ω). Since Auk → v in X, we have −∆u = v ∈ X. This implies u ∈ D(A) and Au = v. So, the operator A is closed. In the following we shall prove that (9.1) holds. Given f ∈ L∞ (Ω) ⊂ L2 (Ω) and λ < 0. The boundary value problem (

−∆u − λu = −λf u=0

in Ω, on ∂Ω

9.1. C0 semigroup theory  241

has a unique solution u ∈ H 2 (Ω) ∩ H01 (Ω). Let M = kf k∞ . Then, −∆(u − M ) − λ(u − M ) = −λ(f − M ). Note that v = (u − M )+ ∈ H01 (Ω) and ∇v = χ{u>M } ∇u. Multiplying the above equation by v and integrating the result over Ω, we conclude −λ

Z

v 2 dx +

Z

|∇v|2 dx = −λ

{u>M }

Ω

Z Ω

(f − M )vdx 6 0.

This combined with the fact λ < 0 yields v = 0, and so u 6 M . Similarly, u > −M . Hence, u ∈ L∞ (Ω), and kuk∞ 6 kf k∞ . Using −∆u = λ(u − f ) we also have −∆u ∈ L∞ (Ω), which implies u ∈ C(Ω) by the imbedding theorem. Since u ∈ H01 (Ω), it yields −∆u = λ(u − f ) ∈ X. Therefore, u ∈ D(A). This indicates that λ ∈ ρ(A). On the other hand, the equation −∆u − λu = −λf is equivalent to u = (−∆ − λI)−1 (−λf ) = −λ(−∆ − λI)−1 f. Hence we have |λ|k(−∆ − λI)−1 f k∞ = kuk∞ 6 kf k∞ , which implies k(−∆ − λI)−1 k 6

1 . |λ|

Consequently, (9.1) holds, and the conclusion is followed from Theorem 9.4. 9.1.2



Solutions of linear problems

This subsection is concerned with the linear problem (

u0 (t) + Au(t) = f (t), 0 < t < T, u(0) = u0 ∈ X.

(9.3)

It is assumed throughout this subsection that (i) −A is the infinitesimal generator of a C0 semigroup Q(t) in X; (ii) 0 < T < ∞ and f ∈ L1 ((0, T ), X). Definition 9.8 followings hold:

A function u ∈ C([0, T ), X) is called a real solution of (9.3) if the

(1) for any fixed 0 < t < T , u0 (t) exists, u(t) ∈ D(A) and satisfies u0 (t) + Au(t) = f (t); (2) u(0) = u0 , i.e., limt→0+ u(t) = u0 in X. Theorem 9.9

If u is a real solution of (9.3), then Z

u(t) = Q(t)u0 + 0

t

Q(t − s)f (s)ds for all t ∈ [0, T ).

(9.4)

242  9. Semigroup Theory and Applications

Proof. Take t ∈ (0, T ) and denote v(s) = Q(t − s)u(s),

0 < s 6 t.

For s, s + h ∈ (0, T ] and h > 0, applying Theorem 9.3 (3) we conclude v(s) − v(s − h) h

I − Q(h) u(s) − u(s − h) Q(t − s)u(s) + Q(t − s + h) h h

= h→0+

−→ AQ(t − s)u(s) + Q(t − s)u0 (s) Q(t − s)f (s),

=

i.e., v 0 (s) = Q(t − s)f (s). Integrating this differential equation from 0 to t and using lims→0+ v(s) = Q(t)u0 we get (9.4).  t

Z

Lemma 9.10

Q(t − s)f (s)ds. Then F ∈ C([0, T ), X).

Let F (t) = 0 1

Proof. Since f ∈ L ((0, T ), X), function F (t) =

Z

Z

Q(t − s)f (s)ds is well defined.

0

For t ∈ [0, T ) and 0 < l  1, we have F (t + l) − F (t) =

t

t+l

Q(t + l − s)f (s)ds −

0

Z

t

Q(t − s)f (s)ds

0

= (Q(l) − I)

t

Z

Q(t − s)f (s)ds +

Z

t+l

Q(t + l − s)f (s)ds,

t

0

which implies lim [F (t + l) − F (t)] = 0.

l→0+

For t ∈ (0, T ), l < 0 and |l|  1, as F (t + l) − F (t) = (I − Q(−l))

Z

t+l

Q(t + l − s)f (s)ds −

Z

t

Q(t − s)f (s)ds,

t+l

0

it follows that lim [F (t + l) − F (t)] = 0.

l→0−

Accordingly, F ∈ C([0, T ), X). Based on Lemma 9.10, the function u defined by (9.4) is in C([0, T ), X).



Definition 9.11 The function u defined by (9.4) is called a mild solution of (9.3). A mild solution u of (9.3) is called a strong solution if u ∈ L1 ((0, T ), D(A)) ∩ W11 ((0, T ), X) satisfies (

u0 (t) + Au(t) = f (t) u(0) = u0 .

a.e. in (0, T ),

(9.5)

A mild solution u of (9.3) is called a classical solution of (9.3) if u ∈ C([0, T ], D(A)) ∩ C 1 ([0, T ], X) and satisfies (9.3).

9.1. C0 semigroup theory  243

By this definition, a classical solution must be a real solution. The following theorem gives an equivalent form of the mild solution. Theorem 9.12 Function u is a mild solution of (9.3) if and only if u ∈ C([0, T ), X) Z t

u(s)ds ∈ D(A) and

satisfies 0

u(t) − u0 + A

t

Z

t

Z

f (s)ds for all t ∈ [0, T ).

u(s)ds = 0

(9.6)

0

Proof. Let u be given by (9.4). For t ∈ [0, T ), one has t

Z

t

Z

u(r)dr =

t

Z

Q(r)u0 dr +

0

0 t

=

Z

ds

0

t

Z

Z

Q(r − s)f (s)dr t−s

Q(r)f (s)dr.

ds

Q(r)u0 dr +

0

0

0

t

s

0 t

=

Q(r − s)f (s)ds

0 t

Z

Q(r)u0 dr + Z

r

dr

0

Z

Z

Making use of Theorem 9.3 (2) we conclude t

Z

Q(r)u0 dr ∈ D(A) and A

0

Q(r)f (s)dr ∈ D(A) and A

Z

Z

t−s

Q(r)f (s)dr = f (s) − Q(t − s)f (s).

0

0

Thus

Q(r)u0 dr = u0 − Q(t)u0 ,

0

t−s

Z

t

Z

t

u(r)dr ∈ D(A), and

0

Z

A

t

u(r)dr = u0 − Q(t)u0 +

0

Z

t

[f (s) − Q(t − s)f (s)]ds

0

= u0 − u(t) +

t

Z

f (s)ds. 0

This shows that every mild solution of (9.3) satisfies (9.6). Conversely, let v ∈ C([0, T ), X) be a solution of (9.6) and set w(t) =

Z

t

(v − u).

0

Then w0 + Aw = 0 and w(0) = 0. Theorem 9.3 (6) asserts w ≡ 0, i.e., v ≡ u.



Corollary 9.13 Suppose u0 ∈ X and f ∈ L1 ((0, T ), X). If u ∈ L1 ((0, T ), D(A)) or u ∈ W11 ((0, T ), X), then u is a mild solution of problem (9.3) if and only if u is a strong solution of (9.3). Proof. We first assume that u is a mild solution of (9.3), i.e., u is given by (9.4). Then u satisfies (9.6) by Theorem 9.12. Z t Z t 1 If u ∈ L ((0, T ), D(A)), then A u(s)ds = Au(s)ds. Thus we have in the 0

0

light of (9.6), u(t) − u0 +

Z

t

Z

Au(s)ds = 0

0

t

f (s)ds for all t ∈ [0, T ).

244  9. Semigroup Theory and Applications

Owing to the fact that f, Au ∈ L1 ((0, T ), X), we see that u(t) is differentiable almost everywhere, and the first equation of (9.5) holds. Accordingly, u ∈ W11 ((0, T ), X). If u ∈ W11 ((0, T ), X), then, by use of (9.6), t

Z

F (t) = A 0

u(s)ds ∈ W11 ((0, T ), X). 1 h

Since A is a closed operator, we have that 1 Au(t) ← A h

!

t+h

Z

u(s)ds = t

t+h

Z

u(s)ds → u(t), and

t

F (t + h) − F (t) → F 0 (t) a.e. in [0, T ] h

as h → 0. It follows that u(t) ∈ D(A) and F 0 (t) = Au(t) almost everywhere in [0, T ]. This fact and (9.6) affirm that u satisfies the first equation of (9.5). Moreover, u ∈ L1 ((0, T ), D(A)) since u0 , f ∈ L1 ((0, T ), X). Conversely, when u is a strong solution of (9.3), i.e., u ∈ L1 ((0, T ), D(A)) ∩ W11 ((0, T ), X) satisfies (9.5), similar to the proof of Theorem 9.9, we can show that u satisfies (9.4).  1 Corollary 9.13 indicates that when the mild solution u ∈ L ((0, T ), D(A)) or u ∈ W11 ((0, T ), X), it is exactly a strong solution. Theorem 9.14 Let u0 ∈ D(A), f ∈ C([0, T ], X) and u be given by (9.4). If either f ∈ L1 ((0, T ), D(A)) or f ∈ W11 ((0, T ), X), then u ∈ C([0, T ], D(A)) ∩ C 1 ([0, T ], X) and satisfies (9.3). This indicates that u is a classical solution. Proof. Step 1. Define Z

t

v(t) =

Q(t − s)f (s)ds =

Z

t

Q(s)f (t − s)ds, t ∈ [0, T ].

0

0

We shall show v ∈ C 1 ([0, T ), X). In fact, when f ∈ L1 ((0, T ), D(A)), for t ∈ [0, T ) and h ∈ (0, T − t], we have v(t + h) − v(t) = h

Z

t

0

Z

=

Z

Q(h) − I 1 f (s)ds + h h

Z

t

Z

t

Q(t − s)

0

=

t+h

Q(h) − I 1 Q(t − s) f (s)ds + h h

Q(h) − I h

Z

1 h

Q(t − s)f (s)ds +

0

Q(t + h − s)f (s)ds

t h

Q(s)f (t + h − s)ds

0 h

Q(s)f (t + h − s)ds.

0

Keeping f (s) ∈ D(A) in mind, it is easy to see that lim+

h→0

Q(h) − I h

Z

t

Q(t − s)f (s)ds = −A

t

Z

0

Q(t − s)f (s)ds

0

=−

Z 0

t

Q(t − s)Af (s)ds.

9.1. C0 semigroup theory  245

Consequently, in accordance with f ∈ C([0, T ], X), d+ v(t) =− dt

Z

t

Q(t − s)Af (s)ds + f (t) for all t ∈ [0, T ).

0

Now we consider the case f ∈ W11 ((0, T ), X). For t ∈ [0, T ) and h ∈ (0, T − t], we have v(t + h) − v(t) = h

t

Z 0

Q(h) f (t + h − s) − f (t − s) ds + Q(s) h h

Notice that lim+

h→0

Z

h

Q(t − s)f (s)ds.

0

f (t + h − ·) − f (t − ·) = f 0 (t − ·). h

Similar to above, d+ v(t) = dt

Z

t

Q(s)f 0 (t − s)ds + Q(t)f (0) for all t ∈ [0, T ).

0 +

In conclusion, the right derivative d dtv(t) ∈ C([0, T ), X) and hence v ∈ C 1 ([0, T ), X). − v(T ) − v(T ) By the same way as above, the left derivative d dt exists and d dt = 0 1 limt→T − v (t). Hence v ∈ C ([0, T ], X). Step 2. Let t ∈ [0, T ) and h ∈ [0, T − t]. Then Q(h) − I 1 v(t) = h h =

Z

t

Q(t + h − s)f (s)ds −

0

v(t + h) − v(t) 1 − h h

Z

1 h

Z

t

Q(t − s)f (s)ds

0

t+h

Q(t + h − s)f (s)ds.

t

Taking h → 0+ , it then yields v(t) ∈ D(A) and −Av(t) = v 0 (t) − f (t) for t ∈ [0, T ). Since A is a closed operator, equation −Av(T ) = v 0 (T ) − f (T ) is valid. As a consequence, v ∈ C([0, T ], D(A)). Applying Theorem 9.3 (3) one has u(t) = Q(t)u0 + v(t) ∈ C([0, T ], D(A)) ∩ C 1 ([0, T ], X), u(0) = u0 and u0 (t) = −AQ(t)u0 − Av(t) + f (t) = −Au(t) + f (t) for all t ∈ [0, T ]. The proof is finished.



246  9. Semigroup Theory and Applications

9.1.3

Mild solutions of semilinear problems

Let −A be the infinitesimal generator of a C0 semigroup Q(t) in X. Now, we consider the following semilinear problem (

u0 (t) + Au(t) = F (t, u(t)), t ∈ [0, T ),

(9.7)

u(0) = u0 ∈ X. Assume that (i) there exist M > 1 and a ∈ R such that kQ(t)k 6 M e−at for all t > 0;

(ii) T ∈ (0, ∞], O ⊂ X is an open set, u0 ∈ O and F : [0, T ) × O → X is continuous. For any given τ ∈ (0, T ) and z ∈ O, there exist δ = δ(τ, z) > 0 and L = L(τ, z) < ∞ such that kF (t, u) − F (t, v)k 6 Lku − vk for all u, v ∈ Bδ (z), t ∈ [0, τ ].

(9.8)

Let r ∈ (0, T ]. A function u ∈ C([0, r), O) is referred to as a mild solution of (9.7) in [0, r) if u solves the following integral equation t

Z

Q(t − s)F (s, u(s))ds for all t ∈ [0, r).

u(t) = Q(t)u0 +

(9.9)

0

Clearly, the function f (t) = F (t, u(t)) satisfies f ∈ L1 ((0, T ), X). Thanks Zto Theorem t

u(s)ds ∈

9.12, u is a mild solution of (9.7) in [0, r) if and only if u ∈ C([0, r), O),

0

D(A) and u(t) − u0 + A

t

Z

Z

u(s)ds = 0

t

F (s, u(s))ds in [0, r). 0

Let Sm (r) be a set of mild solutions of (9.7) in [0, r). Sometimes, we write Sm (r, u0 ) instead of Sm (r) to clarify dependence of Sm (r) on u0 . Firstly, we give the local existence of mild solutions. Theorem 9.15 (Local existence) For any given u0 ∈ O, there exists r ∈ (0, T ) such that problem (9.7) has a mild solution u ∈ Sm (r). Moreover, lim sup kuk (t) − u(t)k = 0,

k→∞ 06t6r

where uk ∈ C([0, r], O) are defined by u0 (t) ≡ u0 and Z

uk (t) = Q(t)u0 +

t

Q(t − s)F (s, uk−1 (s))ds, k = 1, 2, . . . .

(9.10)

0

Proof. This proof is exactly the Picard iteration. We pick τ fixed such that 0 < τ < T . Since O is open, we can choose 0 < δ  1 so that Bδ (u0 ) ⊂ O and (9.8) is satisfied. Then there exists M > 0 for which kQ(t)k 6 M, kF (t, u)k 6 M for all 0 6 t 6 τ and u ∈ Bδ (u0 ).

(9.11)

9.1. C0 semigroup theory  247

According to the continuity of Q(t), there exists 0 < t0 6 τ such that kQ(t)u0 − u0 k 6 δ/2 for all 0 6 t 6 t0 . Then using (9.11), we derive that, when 0 6 t 6 r = min{t0 , δM −2 /3}, t

Z

ku1 (t) − u0 k 6 kQ(t)u0 − u0 k +

kQ(t − s)F (s, u0 )kds

0

6 δ/2 + M 2 t < δ, which implies u1 ∈ Bδ (u0 ). Analogously, ku2 (t) − u0 k 6 kQ(t)u0 − u0 k + 6 δ/2 + M 2 t < δ

Z

t

kQ(t − s)F (s, u1 (s))kds

0

for 0 6 t 6 r.

Inductively, it can be deduced that uk (t) ∈ Bδ (u0 ) for all k > 1, 0 6 t 6 r. Using (9.8) and (9.11) we conclude that, for 0 6 t 6 r,

Z t

ku2 (t) − u1 (t)k = Q(t − s)[F (s, u1 (s)) − F (s, u0 )]ds

0 Z t

ku1 (s) − u0 kds 6 M Lδt,

6 ML

Z t 0

ku3 (t) − u2 (t)k = Q(t − s)[F (s, u2 (s)) − F (s, u1 (s))]ds

0 Z t

6 ML

ku2 (s) − u1 (s)kds

0

6 (M L)2 δ

Z 0

t

1 sds = (M L)2 δt2 . 2

Taking advantage of the inductive method, we can prove that, for all k > 1, kuk+1 (t) − uk (t)k 6

1 δ (M L)k δtk 6 (M Lr)k , 0 6 t 6 r. k! k!

Hence for any given integer N > 1 and all 0 6 t 6 r, one has kuk+N (t) − uk (t)k 6

N −1 X i=0

kuk+i+1 (t) − uk+i (t)k 6 δ

N −1 X i=0

1 (M Lr)k+i . (k + i)!

According to Lemma 9.10, uk ∈ C([0, r], X). This and the above inequality assert that {uk (t)} is a Cauchy sequence in C([0, r], X). Since uk ∈ C([0, r], Bδ (u0 )), there is a u ∈ C([0, r], Bδ (u0 )) such that uk (t) → u(t) in C([0, r], Bδ (u0 )). It then follows that u is a mild solution of (9.7) in [0, r] by letting k → ∞ in (9.10).  Secondly, we give the uniqueness of mild solution. Theorem 9.16 (Uniqueness) Let 0 < µ 6 r 6 T , w ∈ Sm (µ), u ∈ Sm (r). Then w(t) = u(t) for all 0 6 t < µ.

248  9. Semigroup Theory and Applications

Proof. Noting that w(t) and u(t) are continuous and w(0) = u(0) = u0 , there exists 0 < t1 6 µ so that w(t), u(t) ∈ Bδ (u0 ) for all 0 6 t 6 t1 , where δ = δ(t1 , u0 ) is given by the condition (ii). Let M = max06t6t1 kQ(t)k. In view of (9.8) we have Z

ku(t) − w(t)k 6

t

kQ(t − s)[F (s, u(s)) − F (s, w(s))]kds

0 t

Z

6 ML

0

ku(s) − w(s)kds, 0 6 t 6 t1 .

This leads to w(t) = u(t) in [0, t1 ] because of u(0) = w(0). Define τ = sup {s : 0 < s < µ, u(t) = w(t) in [0, s]} . Clearly, τ > t1 . If τ < µ, then u(t) = w(t) in [0, τ ] by continuities of u(t) and w(t). Regard τ and u(τ ) as an initial time and an initial datum, respectively. Similarly to above it can be proved that w(t) = u(t) in [0, τ + ε] for some ε > 0. This contradicts the definition of τ . Hence, τ = µ and the proof is finished.  Thirdly, we show the extension of mild solution. Theorem 9.17 (Extension) Let u0 ∈ O. Then there exist Tmax := Tmax (u0 ) ∈ (0, T ] and a mild solution u ∈ Sm (Tmax ) with property: for any given µ ∈ (0, T ], the set Sm (µ) 6= ∅ implies Tmax > µ. This means that Tmax is the maximal existence time. Z

Tmax

Furthermore, if Tmax < T , then either

kF (t, u(t))kdt = ∞, or there exists

0

w ∈ ∂O so that limt→Tmax u(t) = w. Such a solution u is called the maximal defined solution. Proof. The first conclusion is followed from Theorems 9.15 and 9.16. Details are omitted here. Now, we prove the second conclusion. Suppose Tmax < T and Z

Tmax

kF (t, u(t))kdt < ∞.

0

Denote f (s) = F (s, u(s)). The above inequality indicates f ∈ L1 ((0, Tmax ), X), and then in line with Lemma 9.10, Z

t

Q(t − s)f (s)ds ∈ C([0, Tmax ], X).

0

Equation (9.9) implies u ∈ C([0, Tmax ], X). Accordingly, limit limt→Tmax u(t) = v ¯ We use a contradiction argument to show that v 6∈ O. exists. Obviously, v ∈ O. Otherwise, as O is an open set, there exist t0 > 0 and δ = δ(Tmax + t0 ) > 0 such that Tmax + t0 < T and Bδ (v) ⊂ O. Let K=

max

06t6Tmax +t0

kQ(t)k +

max

06t6Tmax +t0 , ku−vk6δ

kF (t, u)k, δ0 =

δ , 2(K + 2)

and choose ti % Tmax such that ku(ti ) − vk < δ0 . In line with the continuity of Q(t)v, there exists ε > 0 so small that kQ(t)v − vk < δ0 for all 0 6 t 6 ε.

9.1. C0 semigroup theory  249

As a result, kQ(t)u(ti ) − u(ti )k 6 kQ(t)(u(ti ) − v)k + kQ(t)v − vk + ku(ti ) − vk 6 kQ(t)kku(ti ) − vk + kQ(t)v − vk + δ0 , 6 δ/2 for all 0 6 t 6 ε, i > 1. From the proof of Theorem 9.15 we know that there exist a positive constant r, depending only on ε, δ and K, and a function ui ∈ C([0, r], O), such that Z

ui (t) = Q(t)u(ti ) +

t

Q(t − s)F (ti + s, ui (s))ds, t ∈ [0, r].

0

Define

(

wi (t) =

u(t)

for 0 6 t 6 ti ,

ui (t − ti )

for ti 6 t 6 ti + r.

It is obvious that wi (t) is a mild solution of (9.7) in [0, ti + r] and wi (t) ∈ Sm (Tmax + r/2) when i  1. According to Theorem 9.16, we have u(t) = wi (t) in [0, Tmax ) for large i. This contradicts the first conclusion, and the proof is complete.  Finally, we study the continuous dependence of mild solution on the initial data. Theorem 9.18 (Continuous dependence) Let 0 < τ < µ 6 T and u ∈ Sm (µ). Then there exist ε, C > 0 such that for each v0 ∈ Bε (u0 ) we can find a mild solution v ∈ Sm (τ, v0 ) satisfying kv(t) − u(t)k 6 Ckv0 − u0 k for all t ∈ [0, τ ].

(9.12)

Proof. Firstly, for any fixed t ∈ [0, τ ], according to the condition (ii) (think of τ = t, u(t) = z), there exist δ(t) = δ(τ, u(t)) > 0 and L(t) = L(τ, u(t)) < ∞ so that Bδ(t) (u(t)) ⊂ O and kF (s, x) − F (s, y)k 6 L(t)kx − yk for all x, y ∈ Bδ(t) (u(t)), s ∈ [0, τ ]. Take µ(t) > 0 such that ku(s) − u(t)k < δ(t)/2 for −µ(t) + t < s < t + µ(t). Secondly, noting that [0, τ ] is compact, there exists ti ∈ [0, τ ] such that [0, τ ] ⊂

N [

[ti − µ(ti ), ti + µ(ti )].

i=1

Let us define L = max L(ti ) and δ = min 16i6N

16i6N

δ(ti ) . 4

When t ∈ [0, τ ] and x, y ∈ B2δ (u(t)), we have x, y ∈ O, and subsequently kF (t, x) − F (t, y)k 6 Lkx − yk for all t ∈ [0, τ ], x, y ∈ B2δ (u(t)). Take K = max kQ(t)k and 0 < ε < 06t6τ

δ −KLτ e . K

(9.13)

250  9. Semigroup Theory and Applications

For each v0 ∈ Bε (u0 ), in the light of Theorem 9.17, there exist Tmax := Tmax (v0 ) ∈ (0, T ] and the maximal defined solution v ∈ Sm (Tmax , v0 ). Thirdly, we define 



σ = sup s : 0 < s 6 min{τ, Tmax }, kv(t) − u(t)k < 2δ in [0, s] . Then σ > 0, and for any 0 < t < σ, we have kv(t) − u(t)k 6 kQ(t)(v0 − u0 )k + Z

6 Kε + KL

Z

t

kQ(t − s)kkF (s, v(s)) − F (s, u(s))kds

0

t

kv(s) − u(s)kds.

0

Thus, by the Gronwall inequality, kv(t) − u(t)k 6 εKeKLt 6 εKeKLτ < δ for all 0 < t < σ. As a result, this inequality holds when t = σ, and v(t) ∈ Bδ (u(t)) for all 0 6 t 6 σ. Applying (9.13) and the definition of Tmax we have that Tmax > τ and σ = τ . Hence, for any 0 < t 6 τ , kv(t) − u(t)k 6 kQ(t)(v0 − u0 )k + 6 Kkv0 − u0 k + KL

Z

t

kQ(t − s)kkF (s, v(s)) − F (s, u(s))kds

0

Z

t

kv(s) − u(s)kds.

0

As above, we can get (9.12). Here is an example of applications of the above theorems.



Example 9.2 Let Ω be of class C 2 and 0 < T < ∞. Consider the following initialboundary value problem    ut − ∆u = f (x, t) 

in QT ,

u=0

on ST ,

  

u(x, 0) = u0 (x)

in Ω.

(9.14)

Let X = {v ∈ C(Ω) : v|∂Ω = 0} and define the operator A as in Example 9.1. Then −A is the infinitesimal generator of a contraction semigroup Q(t) by Theorem 9.7. Assume that f ∈ L1 ((0, T ), X). We have the following conclusions. (1) If u0 ∈ X, then problem (9.14) has a unique mild solution u ∈ C([0, T ), X) by Lemma 9.10. (2) If u0 ∈ D(A) := {v ∈ H 2 (Ω) ∩ X : ∆v ∈ X}, and either f ∈ L1 ((0, T ), D(A)) or f ∈ W11 ((0, T ), X). Then problem (9.14) has a unique classical solution u ∈ C([0, T ], D(A)) ∩ C 1 ([0, T ], X) by Theorem 9.14.

9.2. Analytic semigroup theory  251

(3) Consider the nonlinear version of (9.14), i.e.,  u    ut − ∆u = 1 + u2 

u=0    

in QT , on ST ,

u(x, 0) = u0 (x)

(9.15)

in Ω.

If u0 ∈ X, then problem (9.15) has a unique mild solution u ∈ C([0, T ], X), i.e., the integral equation Z

u(t) = Q(t)u0 +

t

Q(t − s)

0

u(s) ds for all t ∈ [0, T ] 1 + u2 (s)

has a unique solution u ∈ C([0, T ], X) by Theorems 9.15–9.17.

9.2

ANALYTIC SEMIGROUP THEORY

Let Q(t) be a C0 semigroup in X and −A its infinitesimal generator. For any given v ∈ D(A), we see from Theorem 9.3 (3) and (6) that problem (

u0 (t) + Au(t) = 0, t > 0, u(0) = v

(9.16)

has a unique real solution u(t) = Q(t)v (it is actually a classical solution). However, when v ∈ X \ D(A), we do not know whether (9.16) has a real solution. To solve this problem, we introduce sectorial operator and analytic semigroup. 9.2.1

Basic results

In this subsection, we state some basic results of analytic semigroup theory. Their proofs will be omitted, and interested readers can refer to [89, 82, 124] for details. Definition 9.19 Let Q(t) be a C0 semigroup of X. We call Q(t) a differentiable semigroup if for any given x ∈ X, function Q(t)x is differentiable in t > 0. A C0 semigroup Q(t) of X is referred to as a real analytic semigroup if for any given x ∈ X and ` ∈ X ∗ , function `(Q(t)x) is analytic for t > 0, where X ∗ is the dual space of X. We make conventions: arg 0 = 0, arg z ∈ (−π, π] when z ∈ C \ {0}, and denote z = |z|ei arg z , where C is complex plane. Definition 9.20 Let ϕ1 < 0 < ϕ2 , 4 := {z : ϕ1 < argz < ϕ2 } and Q(z) be a bounded linear operator for each z ∈ 4. The family {Q(z)}z∈4 is called an analytic semigroup if the followings hold: (1) for any given x ∈ X and ` ∈ X ∗ , function `(Q(t)z) is analytic in 4; (2) Q(0) = I and lim43z→0 Q(z)v = v for every v ∈ X;

252  9. Semigroup Theory and Applications

(3) Q(z1 + z2 ) = Q(z1 )Q(z1 ) for z1 , z2 ∈ 4. Theorem 9.21 Let Q(t) be a differentiable semigroup in X and −A its infinitesimal generator. Then, for any v ∈ X, (9.16) has a unique solution u(t) = Q(t)v. Proof. Because Q(t) is differentiable in t > 0, we know that, for any given v ∈ X, the limit 1 1 lim [Q(h)Q(t)v − Q(t)v] = lim+ [Q(t + h)v − Q(t)v] h→0+ h h→0 h exists. Therefore u(t) = Q(t)v ∈ D(A) and

d+ u(t) = −Au(t). dt

−

u(t) Analogously, d dt = −Au(t). Hence u0 (t) = −Au(t). The uniqueness is followed by Theorem 9.3 (6). 

Corollary 9.22 Assume that −A is the infinitesimal generator of an analytic semigroup. Then, for each v ∈ X, problem (9.16) has a unique solution. Definition 9.23 Suppose that A : X ⊃ D(A) → X is a linear operator. We call that A is a sectorial operator if the followings hold: (1) A is closed and D(A) is dense in X; (2) there exist a ∈ R and θ ∈ (0, π/2) such that sector Sa,θ := {λ ∈ C : |arg(λ − a)| > θ} ⊂ ρ(A); (3) there exists M > 1 so that k(A − λ)−1 k 6 M/|λ − a| for λ ∈ Sa,θ . Definition 9.24 Assume that A is a sectorial operator. For any complex number z ∈ C with | arg z| < π2 − θ, we define linear operators e−zA in X as follows: e−zA = I when z = 0, while −zA

e

1 = 2πi

Z

(φ(t) − A)−1 e−φ(t)z φ0 (t)dt

R

when z 6= 0, where φ(t) = b + |t| cos ϕ − it sin ϕ, t ∈ R with b < a, ϕ ∈ (θ, π/2 − | arg z|). Theorem 9.25 If A is a sectorial operator, then e−zA is an analytic semigroup. Conversely, if −A is the infinitesimal generator of an analytic semigroup Q(z), then A is a sectorial operator and Q(z) = e−zA . Theorem 9.26

Let A be a sectorial operator.

(1) For all integer k > 0 and t > 0, we have 1◦ e−tA u ∈ D(Ak ) for every u ∈ X;

9.2. Analytic semigroup theory  253 

d 2◦ Ak e−tA = − dt

k

e−tA ;

3◦ Ak e−tA u = e−tA Ak u for every u ∈ D(Ak ). (2) For any given τ ∈ R+ , there exists a constant C = C(τ ) > 0 such that kAe−tA k 6 Ct−1 , 0 < t 6 τ,

(9.17)

ke−tA − e−sA k 6 Cs−1 (t − s), 0 < s 6 t 6 τ,

(9.18)

kAe−tA − Ae−sA k 6 Ct−1 s−1 (t − s), 0 < s 6 t 6 τ.

(9.19)

Definition 9.27 Let −A be the infinitesimal generator of a C0 semigroup Q(t) in X. If there exist positive constants a and M such that kQ(t)k 6 M e−at for all t > 0, then we call that A ∈ B (X). For every chosen A ∈ B (X) and α < 0, we define a linear operator Aα in X by 1 A v= Γ(−α) α

Z

∞

t−α−1 Q(t)vdt, v ∈ X,

0

where Γ(−α) = a−α

Z

∞

t−α−1 e−at dt.

0

For any given A ∈ B (X) and α > 0, we define Aα = I when α = 0, and Aα the inverse operator of A−α when α > 0. Theorem 9.28

Suppose A ∈ B (X). Then the followings hold:

(1) Aα is closed and D(Aα ) is dense in X. Moreover, 0 ∈ ρ(Aα ) and (Aα )−1 = A−α for all α > 0; (2) Aα Q(t)v = Q(t)Aα v for all α ∈ R, t > 0 and v ∈ D(Aα ); (3) D(Aα ) ⊂ D(Aβ ) when α > β; (4) if α, β ∈ R and v ∈ D(Aβ ) ∩ D(Aα+β ), then Aα+β v = Aα Aβ v; (5) kAα vk 6 2(M + 1)kAvkα kvk1−α for all α ∈ [0, 1] and v ∈ D(A); (6) there exists a constant C > 0 such that λt

λt 1−α

kv − e Q(t)vk 6 C(1 + e )

eλt − 1 (1 + |λ|) λ

for all λ ∈ R, α ∈ [0, 1], v ∈ D(Aα ) and t > 0.

!α

kAα vk

254  9. Semigroup Theory and Applications

Theorem 9.29 Let A be a sectorial operator, a and b be positive constants with a 6 b. If Reλ > b for all λ ∈ σ(A), then A − a ∈ B (X), and for any given k > 1 we can find a positive constant C(k) for which k(A − a)α e−tA k 6 C(k)t−α e−bt for all t > 0, α ∈ [0, k]. Definition 9.30 Assume that A is a sectorial operator and a is a constant so that Reλ > a for all λ ∈ σ(A). For α > 0, we define X α = D((A − a)α ) and kvkα = k(A − a)α vk, v ∈ X α . Space X α is referred to as a fractional power space of X. Given a1 > a2 . If Reλ > a1 for all λ ∈ σ(A), then D((A − a1 )α ) = D((A − a2 )α ) and k(A − a1 )α vk = k(A − a2 )α vk. These demonstrate that the fractional power space X α and the norm k · kα do not depend on the choice of a. Theorem 9.31 Let A be a sectorial operator and integer k > 1. Then there exists a positive constant C such that 



ke−(t+h)A v − e−tA vkα 6 C e−a(t+h) + e−at hδ tβ−α−δ kvkβ for all h, t > 0, δ ∈ [0, 1], α ∈ [0, k], β ∈ [0, α + δ] and v ∈ X β . Theorem 9.32 Suppose that A is a sectorial operator in X, A ∈ B (X) and 0 < α 6 1. If v ∈ D(Aα ) then k(e−tA − I)vk 6

C α α t kA vk. α

Theorem 9.33 Let A ∈ B (X) and B : X → Y be a closed linear operator with D(B) ⊃ D(A). If there exist β ∈ [0, 1) and C > 0 so that kBxkY 6 CkAxkβ kxk1−β for all x ∈ D(A), then, when α > β, we have D(B) ⊃ D(Aα ) and BA−α ∈ L(X, Y ). Z

9.2.2

Regularity of function

t

e−(t−s)A f (s)ds

0

Let X be a Banach space, A be a sectorial operator in X and Reλ > a for all λ ∈ σ(A). Assume that T ∈ (0, ∞) and f : [0, T ] → X is bounded. Set Z

v(t) :=

t

e−(t−s)A f (s)ds.

0

Main aim of this part is to study properties of v(t). Lemma 9.34

v ∈ C 1−α ([0, T ], X α ) ∩ C α ([0, T ], X) for all α ∈ (0, 1).

9.2. Analytic semigroup theory  255

Proof. Rewrite v(t + h) − v(t) =

Z

t

t+h

Z

g1 (s, t)ds +

g2 (s, t)ds, 0 6 t 6 t + h 6 T,

t

0

where 



g1 (s, t) = e−(t+h−s)A − e−(t−s)A f (s) and g2 (s, t) = e−(t+h−s)A f (s). Let A1 = A − a. Then we have t+h−s

Z

g1 (s, t) = t−s

=−

Z

d −τ A e f (s)dτ dτ

t+h−s

Ae−τ A f (s)dτ

t−s Z t+h−s

e−τ A f (s)dτ,

= −A

t−s

Aα1 g1 (s, t) = −AA−1 1

Z

t+h−s

t−s

A1+α e−τ A f (s)dτ. 1

As f is bounded, applying Theorem 9.29 we declare kAα1 g1 (s, t)k Z 0

t

Z

6 C1

t+h−s

τ −(1+α) dτ =

t−s

C1 [(t − s)−α − (t + h − s)−α ], α

C1 kAα1 g1 (s, t)kds 6 h1−α . α(1 − α)

Similarly, t+h

Z t

kAα1 g2 (s, t)kds 6 C2

t+h

Z

(t + h − s)−α ds =

t

C2 1−α h . 1−α

Thus we have kv(t + h) − v(t)kα = kAα1 [v(t + h) − v(t)]k Z

6

0

t

kAα1 g1 (s, t)kds

t+h

Z

kAα1 g2 (s, t)kds

+ t

6 Ch1−α , which implies v ∈ C 1−α ([0, T ], X α ) for all 0 < α < 1. Thanks to kv(t + h) − v(t)k 6 kAα−1 k × kv(t + h) − v(t)k1−α , 1 it yields v ∈ C α ([0, T ], X).



If we define t

Z

G(t) =

e−(t−s)A [f (s) − f (t)]ds, t ∈ [0, T ],

0

then Z

v(t) = G(t) +

t

−(t−s)A

e 0

Z

f (t)ds = G(t) + 0

t

e−sA f (t)ds.

(9.20)

256  9. Semigroup Theory and Applications

Lemma 9.35 Let 0 < µ < 1, 0 6 α < µ and f ∈ C µ ([0, T ], X). Then AG(t) ∈ C µ−α ([0, T ], X α ). Proof. Take 0 6 t < t + h 6 T . Then we have G(t + h) − G(t) =

Z

t

Z

g1 (s, t)ds + 0

t

Z

t+h

g2 (s, t)ds +

g3 (s, t)ds,

(9.21)

t

0

where 



g1 (s, t) = e−(t+h−s)A − e−(t−s)A (f (s) − f (t)), g2 (s, t) = e−(t+h−s)A (f (t) − f (t + h)), g3 (s, t) = e−(t+h−s)A (f (s) − f (t + h)). Write 







A e−(t+h−s)A − e−(t−s)A = e−(t−s)A/2 A e−(h+(t−s)/2)A − e−(t−s)A/2 , then it follows from Theorem 9.29 and (9.19) that kAg1 (s, t)kα 6 C2 h(t − s + 2h)−1 (t − s)µ−α−1 . This in turn gives Z

t

µ−α

kAg1 (s, t)kα ds 6 C2 h

0

Z 0

t/h

(2 + r)−1 rµ−α−1 dr 6 C3 hµ−α .

(9.22)

Noticing t

Z

Z

g2 (s, t)ds = 0

t

e−(τ +h)A (f (t) − f (t + h))dτ,

0

and taking advantage of Theorem 9.3 (2), we have Z

A

t

g2 (s, t)ds = e−hA (1 − e−tA )(f (t) − f (t + h)),

0

and thereafter, by taking δ = β = 0 in Theorem 9.31, we have

Z t

g2 (s, t)ds 6 Ch−α kf (t) − f (t + h)k 6 C4 hµ−α .

A 0

(9.23)

α

On the other hand, thanks to Theorem 9.29 and (9.17), it deduces that Z t

t+h

kAg3 (s, t)kα ds 6 C5

Z

t+h

(t + h − s)µ−α−1 ds = C5 (µ − α)−1 hµ−α .

t

We combine this with (9.21)–(9.23) to derive G(t + h) − G(t) ∈ D(A) and AG ∈ C µ−α ([0, T ], X α ). This completes the proof.



9.2. Analytic semigroup theory  257

Theorem 9.36

Let µ ∈ (0, 1), α ∈ [0, µ), ε ∈ (0, T ) and f ∈ C µ ([0, T ], X). Then Z t

the function v(t) =

e−(t−s)A f (s)ds has the followings properties.

0

(1) v 0 (t) + Av(t) = f (t) in [0, T ] and v 0 ∈ C([0, T ], X); (2) v ∈ C([0, T ], X 1 ) and Av ∈ C µ ([ε, T ], X); (3) v : (0, T ) → X α is differentiable and v 0 ∈ C µ−α ([ε, T ], X α ). Proof. Invoking Lemma 9.35, Theorem 9.3 (2) and (9.20), we conclude that v(t) ∈ D(A) and Av(t) = AG(t) + f (t) − e−tA f (t) for all t ∈ [0, T ]. As a result, Av ∈ C([0, T ], X) and A

Z

t

(9.24)

t

Z

v(s)ds =

Av(s)ds. According to the

0

0

definition of v(t), it is not difficult to see that v(t) solves (

v 0 (t) + Av(t) = f (t), 0 < t 6 T, v(0) = 0.

Therefore, in line with Theorem 9.12, Z

t

t

Z

Av(s)ds = v(t) + A

v(t) +

Z

v(s)ds =

0

0

t

f (s)ds, 0

hence v 0 (t) + Av(t) = f (t) in [0, T ] and v 0 ∈ C([0, T ], X). Conclusion (1) is accomplished. As f ∈ C µ ([0, T ], X), inequality (9.18) implies e−tA f (t) ∈ C µ ([ε, T ], X). Taking α = 0 in Lemma 9.35, we have AG(t) ∈ C µ ([0, T ], X) and then, by (9.24), Av ∈ C µ ([ε, T ], X). It has been shown that Av ∈ C([0, T ], X). Write A1 v = Av − av, then A1 v ∈ C([0, T ], X), i.e., v ∈ C([0, T ], X 1 ). Conclusion (2) is proved. The conclusion (1) and (9.24) lead to v 0 (t) + AG(t) = e−tA f (t).

(9.25)

It follows from Lemma 9.35 that AG(t) ∈ C µ−α ([0, T ], X α ). We claim e−tA f (t) ∈ C µ−α ([ε, T ], X α ). Once this is done, then v 0 ∈ C µ−α ([ε, T ], X α ) by (9.25). Let A1 = A − a as above. We have Aα1 v(t

+ h) −

Aα1 v(t)

=

Aα1 [v(t

+ h) − v(t)] =

Aα1

Z

t+h

0

Z

v (s)ds = t

t

t+h

Aα1 v 0 (s)ds,

and thus 1 Aα v(t + h) − Aα1 v(t) = lim lim 1 h→0 h h→0 h

Z t

t+h

Aα1 v 0 (s)ds = Aα1 v 0 (t),

258  9. Semigroup Theory and Applications

i.e., v : (0, T ) → X α is differentiable. The conclusion (3) is obtained. Now we prove e−tA f (t) ∈ C µ−α ([ε, T ], X α ). For h > 0, we have ke−(t+h)A f (t + h) − e−tA f (t)kα 6 ke−(t+h)A [f (t + h) − f (t)]kα + k(e−(t+h)A − e−tA )f (t)kα . In view of Theorem 9.29, it follows that ke−(t+h)A [f (t + h) − f (t)]kα 6 C(t + h)−α kf (t + h) − f (t)k 6 Chµ−α . Choose δ = µ − α and β = 0 in Theorem 9.31, then, for ε 6 t 6 T , k(e−(t+h)A − e−tA )f (t)kα 6 Chµ−α t−µ kf (t)k 6 Cε−µ max kf (t)khµ−α 6 Chµ−α . 06t6T

Therefore, ke−(t+h)A f (t + h) − e−tA f (t)kα 6 Chµ−α . The proof is complete. 9.2.3



Real solutions of semilinear problems

Let X be a Banach space and A be a sectorial operator in X. We continue to discuss the semilinear problem (9.7). Assume that (i) α ∈ [0, 1) and O ⊂ X α is a nonempty open set; (ii) κ ∈ (0, 1], T ∈ (0, ∞], F : [0, T ) × O → X and satisfies: for any given t ∈ [0, T ) and u ∈ O, there exist positive constants L = L(t, u) and δ = δ(t, u) such that kF (t2 , u2 ) − F (t1 , u1 )k 6 L(|t2 − t1 |κ + ku2 − u1 kα )

(9.26)

for all ti ∈ [0, T ) and ui ∈ O satisfying |ti − t| < δ and kui − ukα < δ. For r ∈ (0, T ], let RS(r) be the set of all functions u ∈ C([0, r), O) solving (9.7) in the sense of Definition 9.8. When u ∈ RS(r), we call that u is a real solution of (9.7) defined in [0, r). Theorem 9.37 satisfies

Let r ∈ (0, T ]. Then u ∈ RS(r) if and only if u ∈ C([0, r), O) and

u(t) = e−tA u(0) +

Z

t

e−(t−s)A F (s, u(s))ds for all t ∈ [0, r).

(9.27)

0

Moreover, if u ∈ RS(r), then the following assertions are valid. (1) Functions u0 (t), Au(t), F (t, u(t)) : (0, r) → X are locally H¨older continuous. (2) If the constant κ given in (9.26) satisfies κ > 1 − α, then for any 0 6 η < 1 − α, function u : (0, r) → X η is differentiable and its derivative u0 : (0, r) → X η is locally H¨older continuous. (3) If γ ∈ [α, 1] and u(0) ∈ X γ , then u ∈ C([0, r), X γ ).

9.2. Analytic semigroup theory  259

Proof. Step 1. Obviously, u ∈ C([0, r), O) implies F (t, u(t)) =: f (t) ∈ C([0, r), X). Applying Theorem 9.9 we see that (9.27) holds when u ∈ RS(r). Conversely, we assume that u ∈ C([0, r), O) and satisfies (9.27). Rewrite (9.27) as u(t) = e−tA u(0) + g(t), where Z

g(t) =

t

e−(t−s)A f (s)ds,

(9.28)

f (t) = F (t, u(t)).

0

For any given 0 < a < b < r, since f ∈ C([0, r), X), taking advantage of Lemma 9.34, we produce g ∈ C 1−β ([0, b], X β ) ∩ C β ([0, b], X) for all β ∈ (0, 1).

(9.29)

On the other hand, for all a 6 s 6 t 6 b and β ∈ (0, 1), Theorem 9.31 alleges ke−tA u(0) − e−sA u(0)kβ 6 Ctα−1 (t − s)1−β ku(0)kα 6 Caα−1 (t − s)1−β . Combining this with (9.28) and (9.29), we get u ∈ C 1−β ([a, b], X β ), which in turn asserts that f is H¨older continuous in [a, b], i.e., f ∈ C τ ([a, b], X) for some τ ∈ (0, 1). Then g(t) satisfies conclusions (1)–(3) of Theorem 9.36 with [0, T ], µ and α replaced respectively by [a, b], τ and 0 < α0 < τ . As j(t) := e−tA u(0) satisfies j 0 (t) + Aj(t) = 0 by Theorem 9.26, we have u0 (t) + Au(t) = F (t, u(t)) for t ∈ [a, b]. The arbitrariness of a, b implies u ∈ RS(r). Step 2. Assume u ∈ RS(r). Then, for a 6 s 6 t 6 b, we have kAe−tA u(0) − Ae−sA u(0)k 6 Ct−1 s−1 (t − s)ku(0)k 6 Ca−2 ku(0)k(t − s) by (9.19). This combines with Theorem 9.36 (2) shows that Au ∈ C τ ([a, b], X). We have known F (t, u(t)) =: f (t) ∈ C τ ([a, b], X). The arbitrariness of a, b implies that the conclusion (1) holds. If the number κ given in (9.26) satisfies κ > 1 − α, then for any τ ∈ (0, 1) with τ 6 1 − α, there holds f ∈ C τ ([a, b], X). Let us define w(t) = e−tA u(a), Z

v(t) = w(t) +

t

e−(t−s)A f (a + s)ds, t ∈ [0, b − a].

0

Then in accordance with Theorem 9.26, we have that w0 (t) exists, w(t) ∈ D(A) and satisfies w0 (t) + Aw(t) = 0. According to Theorem 9.36, v 0 (t) exists, v(t) ∈ D(A), v 0 (t) + Av(t) = f (a + t) in [0, b − a],

260  9. Semigroup Theory and Applications

and Aw ∈ C τ ([ε, b − a], X) ∩ C τ −η ([ε, b − a], X η ), Av ∈ C τ ([ε, b − a], X), v 0 ∈ C τ −η ([ε, b − a], X η ) for any 0 < ε < b − a, 0 6 η < τ . Remembering (9.27), it is easy to see that v(t) = u(t + a). The arbitrariness of a and b implies the conclusion (2). When γ ∈ (α, 1), in view of (9.28), (9.29) and e−tA u(0) ∈ C((0, r), X γ ), we see that the conclusion (3) holds. When γ = 1, taking β = 1 and δ = 1 − α in Theorem 9.31, we deduce e−tA u(0) ∈ C 1−α ([0, b], X α ). Making use of (9.28) and (9.29), u ∈ C β ([0, b], X α ) for some β ∈ (0, 1). Thus f ∈ C τ ([0, b], X) for some τ ∈ (0, 1). Theorem 9.36 avers Au ∈ C([0, b], X). Based on u ∈ RS(r) and the arbitrariness of b we conclude u ∈ C([0, r), X). The conclusion (3) holds for γ = 1.  Now we give the local existence and uniqueness of real solutions. Theorem 9.38 (Local existence and uniqueness of real solutions) If u0 ∈ O, then there exists r ∈ (0, T ) such that problem (9.7) has a unique real solution u defined in [0, r), i.e., u ∈ RS(r). Furthermore, lim max kuk (t) − u(t)kα = 0,

k→∞ 06t6r

where uk ∈ C([0, r), O) is defined as follows: u1 (t) = u0 , uk+1 (t) = e

−tA

Z

t

u0 +

e−(t−s)A F (s, uk (s))ds, t ∈ [0, r), k = 1, 2, . . . .

0

Proof. Noticing that O is open, there exists δ > 0 so that Bδ (u0 ) ⊂ O. We define Y := {v : v ∈ C([0, r], X α )}, kvkY := max kv(t)kα 06t6r

and 



V := v : v ∈ Y, kv(t) − u0 kY 6 δ . For any given v ∈ V , let us define G(v)(t) = e−tA u0 +

Z

t

e−(t−s)A F (s, v(s))ds.

0

Obviously, v ∈ X α implies G(v) ∈ X α . Since u0 ∈ X α , it yields e−tA u0 ∈ C([0, r], X α ). In view of Theorem 9.29 and the condition (9.26), we can find a 0 < r  1 such that for any v ∈ V and 0 6 t 6 r, −tA

kG(v)(t) − u0 kα 6 ke

u0 − u0 kα +

Z

t

k(A − a)α e−(t−s)A kkF (s, v(s))kds

0

t δ 6 + C max kF (s, u)k (t − s)−α ds 2 0 [0,r]×Bδ (u0 ) δ C = + max kF (s, u)kt1−α 2 1 − α [0,r]×Bδ (u0 ) 6 δ,

Z

9.2. Analytic semigroup theory  261

and for any u, v ∈ V , kG(u)(t) − G(v)(t)kα 6

t

Z

k(A − a)α e−(t−s)A kkF (s, u(s)) − F (s, v(s))kds

0 t

Z

6 CL

(t − s)−α ku(s) − v(s)kα ds

0

6 CLku(s) − v(s)kY

Z

t

(t − s)−α ds

0

1 ku − vkY . 2

6

This shows G : V → V , and G is a contraction map. Hence, by the contraction mapping principle, there exists a unique u ∈ V such that G(u) = u. And so, u ∈ RS(r) by Theorem 9.37. Above arguments also indicate that u1 := G(u0 ) ∈ V , and then uk+1 := G(uk ) ∈ V for all k > 0. Moreover, as u satisfies (9.27), it is deduced that kG(uk )(t) − u(t)kα 6

Z

t

k(A − a)α e−(t−s)A k × kF (s, uk (s)) − F (s, u(s))kds

0

Z

6 CL

t

(t − s)−α kuk (s) − u(s)kα ds

0

6 CL max kuk (s) − u(s)kα 06s6r

Z

t

(t − s)−α ds

0

1 6 max kuk (s) − u(s)kα for all 0 6 t 6 r 2 06s6r by use of Theorem 9.29 and the condition (9.26). Hence max kuk+1 (s) − u(s)kα 6

06s6r

1 max kuk (s) − u(s)kα for all k > 0. 2 06s6r

This leads to max kuk (s) − u(s)kα 6 2−k max ku0 (s) − u(s)kα → 0

06s6r

06s6r

as k → ∞. The proof is complete.  Same as the mild solution, we have the following extension result for the real solution. Theorem 9.39 (Extension) Let u0 ∈ O. Then there exist Tmax := Tmax (u0 ) ∈ (0, T ] and a unique u ∈ RS(Tmax ) with the property: for any given µ ∈ (0, T ], if there exists a v ∈ RS(µ) satisfying v(0) = u0 , then we must have Tmax > µ. Furthermore, Z

Tmax < T implies that either

Tmax

kF (t, u(t))kα dt = ∞, or there exists w ∈ ∂O such

0

that limt→Tmax ku(t)−wkα = 0. Such a time Tmax is called the maximal existence time, and such a solution u is called the maximal defined solution. The proof of this theorem is the same as Theorem 9.17, and details are left to readers as an exercise.

262  9. Semigroup Theory and Applications

9.3

SEMIGROUPS DETERMINED BY −∆ IN Lp (Ω)

As the application, here we consider an example. Let Ω be of class C 2+γ with γ ∈ (0, 1), and 1 < p < ∞. Define X = Lp (Ω) and ◦

D(Ap ) = u ∈ W 1p (Ω) ∩ Wp2 (Ω) , 



Ap u = −∆u for u ∈ D(Ap ) for the homogenous Dirichlet boundary condition, and n

o

D(Ap ) = u ∈ Wp2 (Ω) : ∂n u = 0 on ∂Ω , Ap u = (−∆ + 1)u for u ∈ D(Ap ) for the homogenous Neumann boundary condition. It will be proved that Ap is a sectorial operator. To this aim, we first give a theorem to judge whether an operator is a sectorial operator. Theorem 9.40 Let B be a closed linear operator and D(B) dense in X. We denote the adjoint operator of B by B ∗ . Assume that there exist R0 , C > 0 and θ ∈ (0, π/2) such that for each λ ∈ Sθ := {λ ∈ C : |λ| > R0 , | arg λ| > θ}, there holds kuk 6

C k(λ − B)uk for all u ∈ D(B), |λ|

(9.30)

¯ − B ∗ )v = 0 only has zero solution, where λ ¯ is the conjugate number and equation (λ of λ. Then B is a sectorial operator. Before proving Theorem 9.40, we first present a lemma. Lemma 9.41 Let A be a closed linear operator and D(A) dense in X. If the equation A∗ v = 0 only has zero solution, then range R(A) of A is dense in X. Proof. Firstly, R(A) is a linear space of X. If R(A) is not dense in X, then there exists u0 ∈ X so that d = d(u0 , R(A)) > 0. According to the Hahn-Banach theorem, there exists f ∈ X ∗ satisfying hf, u0 i = d, kf k = 1 and hf, ui = 0 for all u ∈ R(A). This implies hA∗ f, ui = hf, Aui = 0 for all u ∈ D(A). Because D(A) = X and f is continuous, we have hA∗ f, ui = 0 for any u ∈ X. This leads to A∗ f = 0, and then f = 0 by our assumption. It is a contradiction, and the proof is finished.  Proof of Theorem 9.40. Given λ ∈ Sθ . We first show R(λ − B) = X. Since λ − B is ¯ − B ∗ )v = 0 only has zero solution, Lemma 9.41 closed and equation (λ − B)∗ v = (λ

9.3. Semigroups determined by −∆ in Lp (Ω)  263

indicates that R(λ − B) is dense in X. So, R(λ − B) = X once we prove R(λ − B) is closed. Assume vi ∈ R(λ − B) and vi → v. Take ui ∈ D(B) satisfying (λ − B)ui = vi . According to (9.30), one has kui − uj k 6

C kvi − vj k → 0 |λ|

as i, j → ∞. Thus, {ui } is a Cauchy sequence, and ui → u ∈ X. Remembering that λ − B is closed, it yields u ∈ D(B) and v = (λ − B)u, i.e., v ∈ R(λ − B). Therefore, R(λ − B) is closed, and then R(λ − B) = X. Using (9.30) again, we know that λ ∈ Sθ implies inverse (λ − B)−1 exists and k(λ − B)−1 uk 6

C kuk for all u ∈ D((λ − B)−1 ) = R(λ − B) = X, |λ|

which implies k(λ − B)−1 k 6 C/|λ|. Obviously, there exists a ∈ R such that Sa,θ := {λ ∈ C : | arg(λ − a)| > θ} ⊂ Sθ . Thus, for λ ∈ Sa,θ , k(λ − B)−1 k 6

C C |λ − a| M = 6 . |λ| |λ − a| |λ| |λ − a|

Hence, Theorem 9.40 is proved. In order to apply Theorem 9.40, we state two estimates.



Lemma 9.42 (Lp estimate for elliptic equations) Let 1 < p < ∞. Then there exists a constant C > 0 so that kukWp2 (Ω) 6 CkAp ukp for all u ∈ D(Ap ).

(9.31)

This lemma is exactly Lemma 9.17 of [40] for the homogeneous Dirichlet boundary condition, and Theorem 6.29 of [73] for the homogeneous Neumann boundary condition. Lemma 9.43 ([2, Theorem 4.4]) Let 1 < p < ∞. Then there exist positive constants C, R and 0 < θ < π/2 such that, when |λ| > R and | arg λ| > θ, there holds kukp 6

C k(λ − Ap )ukp for all u ∈ D(Ap ). |λ|

Now we prove that Ap is closed and Aq is adjoint operator of Ap for q = p/(p − 1). Lemma 9.44

If 1 < p < ∞, then Ap is closed and D(Ap ) is dense in Lp (Ω).

264  9. Semigroup Theory and Applications

Proof. Since C0∞ (Ω) ⊂ D(Ap ) and C0∞ (Ω) = Lp (Ω), it follows that D(Ap ) = Lp (Ω), i.e., D(Ap ) is dense. Next, we prove that Ap is closed. Let ul ∈ D(Ap ) and ul → u, Ap ul → v in Lp (Ω). Thanks to Lemma 9.42, kul − uk kWp2 (Ω) 6 CkAp (ul − uk )kp → 0, which implies ul → u in Wp2 (Ω), and hence u ∈ D(Ap ). As kAp (ul − u)kp 6 kul − ukWp2 (Ω) → 0, it follows that Ap u = v, and then Ap is closed. Lemma 9.45



If 1 < p < ∞, then Aq is adjoint operator of Ap for q = p/(p − 1).

Proof. Integrating by parts, hAp u, vi = hu, Aq vi for all u ∈ D(Ap ), v ∈ D(Aq ).

(9.32)

So, D(Aq ) ⊂ D((Ap )∗ ) and Aq v = (Ap )∗ v for v ∈ D(Aq ). Let v ∈ D((Ap )∗ ) and w = (Ap )∗ v. Then v ∈ Lq (Ω) and hAp u, vi = hu, wi for all u ∈ D(Ap ).

(9.33)

Noticing D(Aq ) = Lq (Ω), we can find a sequence {vk } ⊂ D(Aq ) so that kvk −vkq → 0 as k → ∞, which implies hAp u, vk i → hAp u, vi. This combined with (9.32) and (9.33) allows us to derive hu, Aq vk i = hAp u, vk i → hAp u, vi = hu, wi for all u ∈ D(Ap ).

(9.34)

By virtue of D(Ap ) = Lp (Ω), it follows from (9.34) that Aq vk * w. Therefore v ∈ D(Aq ), i.e., D((Ap )∗ ) ⊂ D(Aq ) since Aq is closed. Thus (Ap )∗ = Aq , and the proof is finished.  We shall prove that Ap is a sectorial operator and −Ap generates an analytic semigroup. Theorem 9.46 Assume 1 < p < ∞. Then Ap is a sectorial operator in Lp (Ω) and −Ap is the infinitesimal generator of an analytic semigroup Qp (t) = e−tAp . Proof. Utilizing Lemmas 9.44 and 9.45, we know that Ap is closed, D(Ap ) is dense in Lp (Ω) and (Ap )∗ = Aq . Applying Lemma 9.43 to Ap and Aq , respectively, there exist θ ∈ (0, π/2) and positive constants R, C such that when |λ| > R and | arg λ| > θ, there hold kukp 6 C|λ|−1 k(λ − Ap )ukp for all u ∈ D(Ap ), ¯ − Aq )vkq for all v ∈ D(Aq ). kvkq 6 C|λ|−1 k(λ ¯ − (Ap )∗ )v = 0 only has zero solution. The second inequality implies that equation (λ It follows from Theorem 9.40 that Ap is a sectorial operator. 

9.3. Semigroups determined by −∆ in Lp (Ω)  265

In the following we shall investigate properties of the semigroup Qp (t). Making use of Lr (Ω) ,→ Lp (Ω) for r > p, we see that Ap is an extension of Ar for r > p and Ar u = Ap u

for u ∈ D(Ar ),

Qr (t)u = Qp (t)u

for u ∈ Lr (Ω).

p Taking advantage of Theorem 9.26 (1), Qp (t)u ∈ D(Am p ) for all u ∈ L (Ω), m > 1 and t > 0. Exploiting the semigroup property and iteration method, we can deduce Qp (t)u ∈ C 2+α (Ω) for all u ∈ Lp (Ω) and t > 0. This leads to that for each 1 < q 6 ∞,

Qp (t)u ∈ D(Aq ), Qp (t)u = Qq (t − ε)Qp (ε)u for all u ∈ Lp (Ω), t > ε > 0. Let λ∗ be the smallest positive eigenvalue of A2 . Then Reλ > λ∗ for all λ ∈ σ(Ap ). Accordingly, we can define the fractional power space: X α = D(Aαp ), 0 6 α 6 1. Now, we prove the following embedding results. Theorem 9.47

Let Ω be of class C 1 and 0 6 α 6 1. Then we have

1 n 1 1 + − and p 6 q; 2 2 p q   n 1 1 (2) X α ,→ Lq (Ω) when α > − and p 6 q; 2 p q (3) X α ,→ C µ (Ω) when 0 6 µ < 2α − n/p. (1) X α ,→ Wq1 (Ω) when α >





Moreover, all above embedding are continuous. Proof. We first prove the conclusion (1). Taking m = 2 and k = 1 in Theorem A.9 (1) and noticing (9.31), we have that if 1/2 6 θ 6 1 and q > p, 1 − n/q = θ(2 − n/p) − n(1 − θ)/p = 2θ − n/p,

(9.35)

then kBukq = kukWq1 (Ω) 6 CkukθWp2 (Ω) kukp1−θ 6 CkAp ukθp kukp1−θ for all u ∈ D(Ap ), where B=

X

Dβ , D(B) = Wq1 (Ω) ⊂ X = Lp (Ω).

|β|61

Clearly, B is a closed linear operator and D(B) ⊃ D(Ap ). According to the hypothesis we can choose 1/2 6 θ < α such that (9.35) is fulfilled. It then follows by Theorem 9.33 that p q D(B) ⊃ D(Aαp ), BA−α p ∈ L(L (Ω), L (Ω)).

266  9. Semigroup Theory and Applications

Thanks to −α p q X α = D(Aαp ) = R(A−α p ), BAp ∈ L(L (Ω), L (Ω)),

we have X α ,→ D(B) = Wq1 (Ω). Notice α α α kBukq = kBA−α p Ap ukq 6 CkAp ukp = Ckukα for all u ∈ X .

It is clear that the embedding X α ,→ Wq1 (Ω) is continuous. Similarly, we can prove conclusions (2) and (3). Lemma 9.48



For each u0 ∈ L2 (Ω), there holds kQ2 (t)u0 k2 6 ku0 k2 e−λ

∗t

for all t > 0.

Proof. Denote ∗

u(t) = Q2 (t)u0 and f (t) = e2λ t ku(t)k22 , t > 0. We first handle the homogeneous Dirichlet boundary condition and assume u0 ∈ H01 (Ω) ∩ H 2 (Ω). Utilizing the Poincer´e inequality and straightforward calculations we arrive at ∗

e−2λ t f 0 (t) = 2λ∗

Z

u2 (t) + 2

Ω

= 2λ

∗

= 2λ∗

Z

Z

u(t)u0 (t)

Ω

Z

2

u (t) + 2 ZΩ

u2 (t) − 2

Ω

u(t)∆u(t) ZΩ

|∇u(t)|2

Ω

6 0, which implies kQ2 (t)u0 k2 = ku(t)k2 6 ku0 k2 e−λ

∗t

for all u0 ∈ H01 (Ω) ∩ H 2 (Ω), t > 0.

This estimate still holds for all u0 ∈ L2 (Ω) based on denseness. For the homogeneous Neumann boundary condition, we first assume u0 ∈ H 2 (Ω) and ∂n u0 = 0 on ∂Ω. As above, e−2t f 0 (t) = 2

Z

u2 (t) + 2

ZΩ

=2

2

Z ZΩ

u (t) + 2 Ω

= −2

u(t)u0 (t) u(t)[∆u(t) − u(t)]

Ω

Z Ω

|∇u(t)|2 6 0,

and then kQ2 (t)u0 k2 6 ku0 k2 e−t for all u0 ∈ L2 (Ω), t > 0. The proof is complete.



9.3. Semigroups determined by −∆ in Lp (Ω)  267

Lemma 9.49 ([95]) Semigroups Qp (t) have the following properties. (1) Qp (t) is positivity preserving, i.e., u > 0 in Ω implies Qp (t)u > 0 in Ω for all t > 0. (2) Assume 1 < p 6 q 6 ∞. Then for any given δ > 0, there exists a constant C = C(n, p, q, δ, Ω) such that 



∗

kQp (t)ukq 6 C 1 + t−δ−n(1/p−1/q)/2 e−λ t kukp for all u ∈ Lp (Ω). (9.36) For heat equations with homogeneous Dirichlet boundary conditions, one may choose δ = 0. (3) For any given µ ∈ (0, 2) and δ > 0, there exists a constant C = C(n, µ, δ, Ω) such that 



∗

kQp (t)ukC µ (Ω) 6 C 1 + t−δ−µ/2 e−λ t kuk∞ for all u ∈ L∞ (Ω). Proof. (1) This is the weak maximum principle. (2) In the following, constants C and Ci depend on exponents involved and Ω. If 1 < p 6 q 6 ∞ satisfy n(1/p − 1/q)/2 < 1, then by Theorem 9.47 (2), kwkq 6 C1 kAαp wkp for all w ∈ D(Aαp )

(9.37)

provided n(1/p − 1/q)/2 < α 6 1. Taking a = b = 0 in Theorem 9.29 one has kAαp Qp (t)ukp 6 C2 t−α kukp for all t > 0.

(9.38)

Because of δ > 0, we can choose α such that n(1/p − 1/q)/2 < α 6 1 and α < δ + n(1/p − 1/q)/2. Then combining (9.37) and (9.38), we discover kQp (t)ukq 6 C3 t−δ−n(1/p−1/q)/2 kukp for all 0 < t 6 3.

(9.39)

The restriction n(1/p − 1/q)/2 < 1 can be removed by exploiting the semigroup property and iteration. We left its proof to readers as an exercise. Hence (9.39) holds for all 1 < p 6 q 6 ∞ and δ > 0. Certainly, 



∗

kQp (t)ukq 6 C 1 + t−δ−n(1/p−1/q)/2 e−λ t kukp for all 0 < t 6 3.

(9.40)

Now we treat the case t > 3. If p > 2, then Qp (t)v = Q2 (t)v for v ∈ Lp (Ω) and t > 0. Notice q > p > 2. In view of (9.39) and Lemma 9.48 it avers kQp (t)ukq = kQ2 (t)ukq = kQ2 (1)Q2 (t − 1)ukq 6 C4 kQ2 (t − 1)uk2 6 C5 e−λ ∗

6 Ce−λ t kukp .

∗ (t−1)

kuk2

268  9. Semigroup Theory and Applications

When 1 < p 6 2, we have Qp (1)u ∈ L2 (Ω). As above, kQp (t)ukq = kQp (1)Qp (t − 2)Qp (1)ukq 6 C6 kQp (t − 2)Qp (1)ukp = C6 kQ2 (t − 2)Qp (1)ukp 6 C7 kQ2 (t − 2)Qp (1)uk2 6 C8 e−λ

∗ (t−2)

∗

kQp (1)uk2 6 Ce−λ t kukp .

Anyhow, for each p > 1, there holds ∗

kQp (t)ukq 6 Ce−λ t kukp for all t > 3.

(9.41)

Combining (9.41) and (9.40), we obtain (9.36). For the case of the homogeneous Dirichlet boundary condition, we define (

v0 (x) =

|u(x)|,

x ∈ Ω,

0,

x ∈ Rn \ Ω,

and let v(x, t) be the unique solution of (

vt − ∆v = 0,

x ∈ Rn , t > 0,

v(x, 0) = v0 (x),

x ∈ Rn .

Then Z

v(x, t) = Rn

1 |x − y|2 exp − 4t (4πt)n/2

!

v0 (y)dy,

and an explicit calculation shows kv(·, t)kq 6 Ct−n(1/p−1/q)/2 kv0 kp , 1 6 p 6 q 6 ∞. Taking advantage of a comparison argument, we can get |Q(t)u| 6 ve−λ

∗t

for all x ∈ Ω, t > 0,

and hence (9.36) is valid for δ = 0. (3) When µ < 2α − n/p and w ∈ D(Aαp ), we have kwkC µ (Ω) 6 C9 kAαp wkp by Theorem 9.47. This leads to kQp (t)ukC µ (Ω) 6 C9 kAαp Q(t)ukp . Because of 0 < µ < 2 and δ > 0, we can take α < 1 and p  1 such that µ < 2α − n/p and α < δ + µ/2. Then, by use of (9.38), kQp (t)ukC µ (Ω) 6 C9 kAαp Qp (t)ukp 6 C10 t−α kukp .

(9.42)

And so, for 0 < t 6 4, ∗

kQp (t)ukC µ (Ω) 6 C10 t−(µ/2+δ) kukp 6 Ct−(µ/2+δ) e−λ t kuk∞ . When t > 4, it follows from (9.42) and (9.41) that kQp (t)ukC µ (Ω) = kQp (1)Qp (t − 1)ukC µ (Ω) 6 C11 kQp (t − 1)ukp ∗ (t−1)

6 C12 e−λ The conclusion (3) is obtained.

∗

kukp 6 Ce−λ t kuk∞ . 

9.4. An example  269

9.4

AN EXAMPLE

As an application of the semigroup theory, in this section, we study the following initial-boundary value problem ([95])  uit − di ∆ui = fi (x, t, u), x ∈ Ω,       ujt = fj (x, t, u), x ∈ Ω,  ui = 0 or ∂n ui = 0,     

ul (x, 0) = ul0 (x),

t > 0, 1 6 i 6 k, t > 0, k + 1 6 j 6 m,

x ∈ ∂Ω, t > 0, 1 6 i 6 k, x ∈ Ω,

(9.43)

1 6 l 6 m,

where, for each 1 6 i 6 k, di > 0 is a constant, and for each 1 6 l 6 m, ul0 ∈ C γ (Ω) for some 0 < γ < 1, fl : Ω × [0, ∞) × Rm → R is measurable and satisfies (M) For any bounded set B ⊂ Ω × [0, ∞) × Rm , fl is bounded in B and |fl (x, t, u) − fl (x, t, v)| 6 L(B)|u − v| for all (x, t, u), (x, t, v) ∈ B. Let Qp (t) = e−tAp , where Ap is defined at the beginning of §9.3. Define Γi (t) = Qp (di t) L∞ (Ω) for 1 6 i 6 k, and Γi (t) = I for k + 1 6 i 6 m. Then operators Γ (t) = (Γ1 (t), . . . , Γm (t)) : [L∞ (Ω)]m → [L∞ (Ω)]m satisfy kΓ (t)k 6 1 for all t > 0. For 1 6 i 6 k, we set f˜i (x, t, u) = fi (x, t, u)+di ui when the boundary is ∂n ui |∂Ω = 0, while f˜i (x, t, u) = fi (x, t, u) when the boundary is ui |∂Ω = 0. For k + 1 6 i 6 m, we set f˜i (x, t, u) = fi (x, t, u). Let 0 < T 6 ∞. A mild solution of the initial-boundary value problem (9.43) is a measurable function u : Ω × [0, T ) → Rm satisfying u(·, t) ∈ [L∞ (Ω)]m and Z

u(·, t) = Γ (t)u0 +

t

Γ (t − s)f˜(·, s, u(·, s))ds, t ∈ [0, T ),

0

or precisely, for t ∈ [0, T ), Z t    Γi (t − s)f˜i (·, s, u(·, s))ds, 1 6 i 6 k, u (·, t) = Γ (t)u + i i0  i 0 Z t    ui (·, t) = ui0 + fi (·, s, u(·, s))ds, k + 1 6 i 6 m.

(9.44)

0

Theorem 9.50 ([95]) Under the condition (M), we have the following conclusions. (1) For each initial function u0 ∈ [L∞ (Ω)]m , there exists Tmax > 0 such that (9.43) has a unique mild solution in [0, Tmax ). (2) The “existence time Tmax ” can be chosen maximal: either Tmax = ∞, or Tmax < ∞ and lim ku(·, t)k∞ = ∞.

t→Tmax

(9.45)

(3) Consider the existence time Tmax as a functional of the initial data u0 . Then inf{Tmax (u0 ) : ku0 k 6 C} > 0 for any C > 0.

270  9. Semigroup Theory and Applications

Proof. The existence part of (1) and the conclusion (3) can be proved by the standard Picard iteration, and we just refer the reader to the proof of Theorem 9.15. The uniqueness part of (1) can be proved by a similar way as that of Theorem 9.16. They are left to the reader as an exercise. Now, we prove the conclusion (2). Assume that the maximal existence time Tmax is finite. First we show that assumption sup 06t µ + n/p. Using Theorem 9.47 (3), we know that the new norm kuk∗ := kuk∞ +

m X

di kui kC µ (Ω) 6 kuk∞ + C

i=1

m X

di kAαp ui kp .

i=1

We have known that sup σ6t tj . It follows from the second equation of (9.44) that kui (tl ) − ui (tj )k∞ 6 |tl − tj | sup kfi (s, u)k∞ ΣC

for k + 1 6 i 6 m,

which implies, as l, j → ∞, kui (tl ) − ui (tj )k∞ → 0 for k + 1 6 i 6 m.

(9.48)

9.4. An example  271

For 1 6 i 6 k. Using the first equation of (9.44), we achieve ui (tl ) − ui (tj ) = (Γi (tl − tj ) − 1)ui (tj ) Z

tl −tj

+

Γi (tl − tj − s)f˜i (tj + s, u(tj + s))ds,

0

and then α+β kAαp (ui (tl ) − ui (tj ))kp 6 kA−β ui (tj )kp p (Γi (tl − tj ) − 1)kp,p kAp

+ sup kf˜i (s, u)k∞

Z

tl −tj

0

ΣC

kAαp Γi (s)kp,p ds,

(9.49)

where k · kp,p = k · kL(Lp ,Lp ) . Based on Theorem 9.28 (2) and Theorem 9.32, −β kA−β p (Γi (t) − 1)vkp = k(Γi (t) − 1)Ap vkp 6

C β t kvkp for all v ∈ Lp (Ω), β

β which implies kA−β p (Γi (t) − 1)kp,p 6 Kt , and consequently β kA−β p (Γi (tl − tj ) − 1)kp,p 6 K(tl − tj ) → 0 as l, j → ∞.

(9.50)

Taking advantage of Theorem 9.29, we conclude kAαp Γi (s)kp,p 6 Cs−α , and then Z

tl −tj

0

kAαp Γi (s)kp,p ds 6 C(tl − tj )1−α → 0 as l, j → ∞

(9.51)

because of 0 < α < 1. Remembering (9.47), it follows from (9.49)–(9.51) that kAαp (ui (tl ) − ui (tj ))kp → 0 for 1 6 i 6 k as l, j → ∞. Hence, by use of Theorem 9.47 (3), it derives kui (tl ) − ui (tj )kC µ (Ω) 6 CkAαp (ui (tl ) − ui (tj ))kp → 0 for 1 6 i 6 k.

(9.52)

Recalling the definition of kuk∗ , it follows from (9.48) and (9.52) that ku(tl ) − u(tj )k∗ → 0 as l, j → ∞. Thus, limt→Tmax u(t) exists in the norm k · k∗ . Therefore, the mild solution can be prolonged to an interval [0, Tmax + ε), which contradicts the maximality of Tmax . This indicates that (9.46) cannot hold, and thereby lim sup ku(t)k∞ = ∞.

(9.53)

t→Tmax

Let us assume lim inf ku(t)k∞ < r < ∞, t→Tmax

(9.54)

and we are going to derive a contradiction. According to (9.53) and (9.54), there exist two sequences {tl } and {sl } with properties: tl < sl , liml→∞ tl = liml→∞ sl = Tmax and ku(tl )k∞ = r, ku(sl )k∞ = r + 1, ku(t)k∞ 6 r + 1 in [tl , sl ].

272  9. Semigroup Theory and Applications

From the integral equation u(sl ) = Γ (sl − tl )u(tl ) +

Z

sl −tl

Γ (sl − tl − τ )f˜(tl + s, u(tl + s))ds,

0

one gets ku(sl )k∞ 6 ku(tl )k∞ + |sl − tl | sup kf˜(τ, u(τ ))k∞ , Σr+1

where Σr+1 = {τ : 0 6 τ < Tmax , ku(τ )k∞ 6 r + 1}, and hence r + 1 6 r due to the fact that sl − tl → 0, which is a contradiction. Thus (9.54) cannot hold, and (9.45) is obtained.  Note: Before closing this chapter, we point out that conclusions and solution spaces in this chapter are different from those obtained by the Lp theory and Schauder theory. For example, consider the initial-boundary value problem (9.15). • The mild solution u of (9.15) is only the continuous solution in time t of integral equation u(t) = et∆ u0 +

Z 0

t

e(t−s)∆

u(s) ds for all t ∈ [0, T ), 1 + u2 (s)

and its weak derivatives ut and D2 u may not exist. For the strong solution of (9.15) here, although weak derivatives ut and D2 u exist, they may not belong to Lp (QT ). That is, the strong solution u of (9.15) may not belong to Wp2,1 (QT ). For the classical solution of (9.15) here, although weak derivatives ut and D2 u exist and are continuous from [0, T ] to X = Lp (Ω), they may not belong to C(QT ), i.e., the classical solution u may not belong to C 2,1 (QT ). • When we use the Lp theory or Schauder theory to discuss the existence and ◦

uniqueness of solutions of (9.15), we should assume that u0 ∈ Wp2 (Ω) ∩ W 1p (Ω) for the Lp theory and u0 ∈ C 2+α (Ω) for the Schauder theory. The corresponding solution spaces are Wp2,1 (QT ) and C 2+α, 1+α/2 (QT ), respectively. When we use the semigroup theory to discuss the existence and uniqueness of ◦

solutions of (9.15), we can take X = Lp (Ω) and D(−∆) = Wp2 (Ω) ∩ W 1p (Ω). Then −∆ is a sectorial operator in Lp (Ω) and ∆ is the infinitesimal generator of an analytic semigroup Qp (t) = et∆ by Theorem 9.46. For any 0 6 α < 1 and u0 ∈ X α , problem (9.15) has a unique solution u ∈ C([0, T ], X α ) by Theorem 9.38. Because there is no necessary embedding relationship between spaces Wp2 (Ω) ∩ ◦

W 1p (Ω) and (Lp (Ω))α , the Lp theory (Schauder theory) and semigroup method can not replace each other. In order to establish the existence and uniqueness of solutions, generally speaking, the semigroup method has a lower requirement for initial values, for example, u0 ∈ Lp (Ω), while Lp theory and Schauder theory have a higher requirement for initial values.

Exercises  273

• Generally speaking, solutions obtained by the semigroup method are not necessarily weak solutions in the sense of distribution. Therefore, we cannot use standard interior regularity methods to improve the interior smoothness of this solution. • Sometimes some people call the solution obtained by semigroup method as a semigroup solution.

EXERCISES 9.1 Take X = L2 (Ω) and D(A) = H 2 (Ω) ∩ H01 (Ω) in Example 9.1. Prove that the operator A is the infinitesimal generator of a contraction semigroup in L2 (Ω). 9.2 Let Ω be of class C 2 and p > max{1, n/2}. Define an operator C by 

D(C) =

u∈

Wp2 (Ω)

∩

◦

W 1p (Ω)



∩ C(Ω) : ∆u ∈ C(Ω) ,

Cu = −∆u for u ∈ D(A). Prove that the operator −C is the infinitesimal generator of a contraction semigroup in C(Ω). 9.3 Consider the initial-boundary value problem (9.15). Let D(C) be as above and assume u0 ∈ D(C). Prove that problem (9.15) has a unique global classical solution u ∈ C([0, ∞), D(C)) ∩ C 1 ([0, ∞), C(Ω)). 9.4 Prove Theorem 9.39. 9.5 Prove that (9.39) holds for all 1 < p 6 q 6 ∞ and δ > 0 without the restriction n(1/p − 1/q)/2 < 1. 9.6 Let Ω = (0, 1) and X = L2 (Ω). Define a linear operator A: Au = −u00

for u ∈ D(A) := H 2 (Ω) ∩ H01 (Ω).

Prove (1) D(A1/2 ) = H01 (Ω) and kA1/2 uk = ku0 k for all u ∈ H01 (Ω); (2) if α > 1/4, then X α ,→ L∞ (Ω), i.e., there exists constant C > 0 such that kuk∞ 6 CkAα uk for all u ∈ D(Aα ). 9.7 Let A be a sectorial operator and Re(σ(A)) > 0. Prove that the followings are equivalent: (1) for any given α > 0, operator A−α is compact; (2) for any given t > 0, operator e−tA is compact. 9.8 Prove conclusions (2) and (3) of Theorem 9.47.

274  9. Semigroup Theory and Applications

9.9 Prove conclusions (1) and (3) of Theorem 9.50. 9.10 Let Ω be of class C 2+α . Consider the problem  u − ∆u = u|∇u|2 ,    t

x ∈ Ω,

u(x, t) = 0,

x ∈ ∂Ω, t > 0,

  

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

x ∈ Ω.

t > 0,

Give an appropriate condition regarding initial datum u0 (x) and study the local existence, uniqueness and regularity of solutions.

APPENDIX

A

Appendix

In this appendix, we review the fixed point theorem, embedding theorems and interpolation inequalities. Theorem A.1 (Contraction mapping principle) Let X be a nonempty complete metric space, and let F : X → X be a strict contraction mapping, i.e., d(F(u), F(v)) 6 kd(u, v) for all u, v ∈ X with k < 1. Then F has a unique fixed point, F(u) = u. A continuous operator between two Banach spaces is called a compact operator (or a completely continuous operator) if images of bounded sets are precompact (that is, their closures are compact). Theorem A.2 (Schauder fixed point theorem [98]) Let V be a compact convex set in a Banach space X, and let F be a continuous operator of V into itself. Then F has at least one fixed point in V , that is, F(u) = u for some u ∈ V . Corollary A.3 (Schauder fixed point theorem) Let V be a closed convex set in a Banach space X, and let F be a continuous operator of V into itself such that the image F(V ) is precompact. Then F has at least one fixed point in V . Now we review the Sobolev embedding theorems and interpolation inequalities. We C use the symbol X ,→ Y to indicate that the Banach space X is embedded compactly in the Banach space Y . We always assume that Ω ⊂ Rn is a bounded open set. Theorems of this part can be found in [1, 16, 111]. Theorem A.4

Let Ω be of class C 1 , k be a positive integer, and 1 6 p 6 ∞. C

(1) If kp < n, then Wpk (Ω) ,→ Lq (Ω) for all 1 6 q < np/(n−kp), and the embedding constant depends only on n, k, p, q and Ω. C

(2) If kp = n, then Wpk (Ω) ,→ Lq (Ω) for all 1 6 q < ∞, and the embedding constant depends only on n, k, q, and Ω. 275

276  Nonlinear Second Order Parabolic Equations C

n

(3) If kp > n, then Wpk (Ω) ,→ C k−[ p ]−1+α (Ω) for all 0 < α < α0 , where α0 =

h i  n +1− p

n p

 any α∗ ∈ (0, 1)

when n/p is not an integer , when n/p is an integer,

and the embedding constant depends only on n, k, p, α and Ω. ◦

Remark A.5 When we use W kp (Ω) instead of Wpk (Ω), conclusions of Theorem A.4 hold without assuming that Ω is a C 1 domain. In the following theorems, the embedding constants depend only on n, p, d−1 , T −1 and ∂Ω, where d = diam(Ω) is the diameter of Ω. Theorem A.6 Let p > 1, n > 2 and Ω be of class C 2 . (1) Assume p < n + 2. Then for q = p(n + 2)/(n + 2 − p), we have kDukq, QT 6 CkukWp2,1 (QT ) for all u ∈ Wp2,1 (QT ). (2) If p < 1 + n/2, then Wp2,1 (QT ) ,→ Lq (QT ) with q = p(n + 2)/(n + 2 − 2p). (3) If p = 1 + n/2, then Wp2,1 (QT ) ,→ Lq (QT ) for all 1 6 q < ∞. Theorem A.7 Let Ω be of class C 2 . (1) If 1 + n/2 < p < n + 2, then Wp2,1 (QT ) ,→ C α, α/2 (QT ) with α = 2 − (n + 2)/p. (2) If p = n + 2, then Wp2,1 (QT ) ,→ C α, α/2 (QT ) for any 0 < α < 1. (3) If p > n + 2, then Wp2,1 (QT ) ,→ C 1+α, (1+α)/2 (QT ) with α = 1 − (n + 2)/p. C

Theorem A.8 Let k, l be constants and 0 6 l < k. Then C k,k/2 (QT ) ,→ C l,l/2 (QT ). Theorem A.9 (Interpolation inequality [111, Theorems 2.11.4 and 2.11.5]) Let k, m be integers, Ω be of class C m with m > 1, 1 6 p, r 6 ∞ and u ∈ Wpm (Ω) ∩ Lr (Ω). (1) Let 0 6 k < m and q > nr/(n + rk) such that Wpm (Ω) ,→ Wqk (Ω). If k/m 6 θ 6 1 and k − n/q = θ(m − n/p) − n(1 − θ)/r, then kukWqk (Ω) 6 CkukθWpm (Ω) kuk1−θ r . (2) Assume 0 < θ 6 1. If µ satisfies 0 < µ 6 θ (m − n/p) − n(1 − θ)/r, and this inequality is strict when µ is an integer, then kukC µ (Ω) 6 CkukθWpm (Ω) kuk1−θ r .

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Index A b B, 1 Aα , 253 Ap , D(Ap ), 262 C (i+α)/2, i+α (Q), 189 T C 1,1− (Dg,h ), 223 C 1−,1− ([0, T ] × [g(T ), h(T )]), 223 C 1− ([a, b]), 223 Cx1− (ΠT ), 225 C k+α, (k+α)/2 (Q), C k+α, (k+α)/2 (Q), 2 C k+α (Ω), C k+α (Ω), 2 D(A), 239 T Dg,h , 223 DT , 191 G(x, y, t − s), 147 I × [0, h(t)], I × (0, h(t)), 191 Q, ∂p Q, 2 Q(t), {Q(t)}t>0 , 238 QT , Q∞ , ST , S∞ , ∂p QT , 3 S(x0 ), 2 Tmax , maximal existence time, 49, 132, 195, 248, 261 2 Wp,Bk (Ω), 182 ◦

Wpk (Ω), Wp2k,k (Q), W kp (Ω), 2 Wp1,2 (Q), 189 X α , k·kα , 254 [ · ]C α/2, α (∆T ) , 195 [u]k , 77 [u]ik , [u]dk , 80, 181 Π` , Π∞ , 176 n, ν, ∂n , ∂ν , 3 z(x, y), 104 R Ω f (x), 3 k·kWp2 , k·kWp2,1 , k·kp,D , k·kp , 3 hw, zi, 77 hu, ui, 55 |·|i+α,Ω , |·|i+α,Q , 2 R+ , R+ , 3 YTu0 , 228

L, 2 ω(R), δ(x, y), 3 C

,→, 275 •

•

C ∞ (QT ), W k,l p (QT ), 34 ◦

W 1,1 2 (QT ), 33 ρ(A), σ(A), 239 c, c¯, 78, 182 ck , c¯k , 78 d(A, B), 1 g ∈ Wp2,1 (QT ), g ∈ C 2+α,1+α/2 (QT ), 3 h∗ , 191 h∞ , 198 k0 , k0 (µ), 216 u(t, x), 189 A, 4 B, 13 K, 191 R(A), 262 Γ1 , Γ2 , 4 x, Q(x0 , r), 2 B (X), 253 B1 , B2 , 4 dist(A, B), 1 e−zA , 252 ik , dk , 80, 181 (A), 3 (B), (C), (D), (E), 4 (F), 20 (G1), (G2), 55 (H1), 100 (H2), 101 (I), 176 (I1), 176 (I2), 212 (J), 191 (K), 208 (L1), 233 Lp estimate, 5 285

286  Index

global ∼, 7 interior ∼, 9 interior ∼ in x direction, 7

interior ∼ in t direction, 8 local ∼, 5, 10 Lp theory, 4, 17 ∼ for the first initial-boundary value problem, 4 ∼ for the mixed initial-boundary value problem, 8 ∼ for the second initial-boundary value problem, 5

fractional power space, 254 free boundary condition, 189, 190 free boundary problems, 191 Green function, 147 Hille-Yosida Theorem, 239, 240 Hopf boundary lemma, 17, 18, 23 ∼ for strong solutions, 28

infinitesimal generator, 239 initial habitat, 191 initial-boundary value problem, 4 first ∼, 4 asymptotic spreading speed, 216, 218 mixed ∼, 4 average method, 121 second ∼, 4 interior ball property, 1, 2 blows up in finite time, 132 interior regularity and estimate, 44 blowup, 132 iteration method, 117 ∼ in finite time, 132 iterative sequences, 117, 118 ∼ on the boundary, 155 ∼ rate estimates, 145, 157, 160 Kaplan’s first eigenvalue method, 135 ∼ time, 132 lateral boundary, 3 compact operator, 275 Lipschitz condition, 38, 74, 76 comparison method, 37, 132 one side ∼, 39, 41 comparison principle, 37, 205 Lipschitz continuous, 192 ∼ for weak solutions, 39, 41 local existence, 246 concavity method, 139 local existence and uniqueness, 74, continuous dependence, 249 260 contraction mapping principle, 275 longtime behaviour, 208 convergence of solution, 62 lower solution, 54, 61, 166 critical exponents, 141, 150 weak ∼, 33, 39, 41, 65, 66 Lyapunov function, 102 degenerate, 98 Lyapunov functional method, 100, divergence equation, 32, 39, 40 101 eigenfunction, 170 maximal defined solution, 248, 261 eigenvalue, 170 maximum principle, 19, 22, 166 embedding theorem, 275 ∼ of strong solutions, 28 energy method, 136 ∼ of weak solutions, 33, 34 extension, 248, 261 strong ∼, 18 first eigen-pair, 24, 59 strong ∼ of strong solutions, 28 first eigenvalue, 59 modulus of continuity, 3 first eigenvalue method, 135 monotonicity in time t, 61

Index  287

analytic ∼, 251 contraction ∼, 240 differentiable ∼, 251 non-classical boundary conditions, 23 real analytic ∼, 251 non-degenerate, 98 strongly continuous ∼, 238 non-local terms, 24 semilinear problem, 246 nonlinear boundary conditions, 43 sharp criteria, 213–215 solution, 4 parabolic, 2 classical ∼, 242, 244 ∼ boundary, 2, 3 global ∼, 46, 132 strongly ∼, 2 mild ∼, 238, 242, 243, 246 permanent, 92 real ∼, 241, 258 positive eigenfunction, 59 semigroup ∼, 273 positivity lemma, 18 strong ∼, 242 ∼ for strong solutions, 28 weak ∼, 33, 39, 41, 65 positivity preserving, 267 spread successfully, 198 principal eigenvalue, 170 spreading, 198, 208, 214, 215 spreading-vanishing dichotomy, 214, quasi-, 80 215 mixed ∼monotonic system, 90 stable, 98 mixed ∼monotonicity, 80 globally asymptotically ∼, 88, 98, mixed ∼monotonous, 80 174 ∼monotonically decreasing, 80 linearly ∼, 98 ∼monotonically increasing, 80 locally asymptotically ∼, 98 regularity, 46, 47, 198 locally uniformly asymptotically ∼, 98 Schauder estimate, 13 locally ∼, 98 global ∼, 14 not linearly ∼, 98 interior ∼, 16 Stefan condition, 189 interior ∼ in t direction, 13, 14 Stefan problem, 189 interior ∼ in x direction, 14 strict lower solution, 61 local ∼, 13 strict upper solution, 61 local ∼ in x direction, 14 Schauder fixed point theorem, 275 time-periodic, 166 Schauder theory, 13 ∼ parabolic boundary value ∼ for the first initial-boundary problem, 165, 173, 181 value problem, 15 ∼ parabolic eigenvalue problem, ∼ for the mixed initial-boundary 169, 173 value problem, 16 ∼ sense, 174 ∼ for the second initial-boundary uniform bound from bounded Lp value problem, 16 norm, 48 sectorial operator, 252 uniform estimate, 46, 47, 198 semigroup, 238 uniqueness, 53, 247 C0 ∼, 238 monotonicity of principal eigenvalue, 170–172

288  Index

unstable, 98 upper and lower solutions ∼ method, 54–56, 78, 81, 167, 182 a pair of coupled ordered ∼, 182 a pair of coupled ∼, 77, 80 coupled ordered ∼, 182 coupled ∼, 77, 80 ordering of ∼, 54, 78, 80 weak ∼, 41 weak ∼ method, 65 upper solution, 54, 61, 166 weak ∼, 33, 39, 41, 65, 66 vanishes, 198 vanishing, 198, 208, 214, 215 weakly coupled parabolic system, 73, 97, 181