Nonlinear Evolution and Difference Equations of Monotone Type in Hilbert Spaces [1 ed.] 1482228181, 9781482228182

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Nonlinear Evolution and Difference Equations of Monotone Type in Hilbert Spaces [1 ed.]
 1482228181, 9781482228182

Table of contents :
Cover
Title Page
Copyright Page
Table of Contents
Preface
PART I: PRELIMINARIES
1: Preliminaries of Functional Analysis
1.1 Introduction to Hilbert Spaces
1.2 Weak Topology and Weak Convergence
1.3 Reflexive Banach Spaces
1.4 Distributions and Sobolev Spaces
1.4.1 Vector-valued Functions
1.4.2 Lp Spaces
1.4.3 Scalar Distributions and Sobolev Spaces
1.4.4 Vector Distributions and Sobolev Spaces
2: Convex Analysis and Subdifferential Operators
2.1 Introduction
2.2 Convex Sets and Convex Functions
2.3 Continuity of Convex Functions
2.4 Minimization Properties
2.5 Fenchel Subdifferential
2.6 The Fenchel Conjugate
3: Maximal Monotone Operators
3.1 Introduction
3.2 Monotone Operators
3.3 Maximal Monotonicity
3.4 Resolvent and Yosida Approximation
3.5 Canonical Extension
PART II: EVOLUTION EQUATIONS OF MONOTONE TYPE
4: First Order Evolution Equations
4.1 Introduction
4.2 Existence and Uniqueness of Solutions
4.3 Periodic Forcing
4.4 Nonexpansive Semigroup Generated by a Maximal Monotone Operator
4.5 Ergodic Theorems for Nonexpansive Sequences and Curves
4.5.1 Almost Nonexpansive Sequences
4.5.2 Almost Nonexpansive Curves
4.6 Weak Convergence of Solutions and Means
4.7 Almost Orbits
4.8 Sub-differential and Non-expansive Cases
4.9 Strong Ergodic Convergence
4.10 Strong Convergence of Solutions
4.11 Quasi-convex Case
5: Second Order Evolution Equations
5.1 Introduction
5.2 Existence and Uniqueness of Solutions
5.2.1 The Strongly Monotone Case
5.2.2 The Non Strongly Monotone Case
5.3 Two Point Boundary Value Problems
5.4 Existence of Solutions for the Nonhomogeneous Case
5.5 Periodic Forcing
5.6 Square Root of a Maximal Monotone Operator
5.7 Asymptotic Behavior
5.7.1 Ergodic Convergence
5.7.2 Weak Convergence
5.7.3 Strong Convergence
5.7.4 Subdifferential Case
5.8 Asymptotic Behavior for Some Special Nonhomogeneous Cases
5.8.1 Case C ≤ 0
5.8.2 The Case C > 0
6: Heavy Ball with Friction Dynamical System
6.1 Introduction
6.2 Minimization Properties
PART III: DIFFERENCE EQUATIONS OF MONOTONE TYPE
7: First Order Difference Equations and Proximal Point Algorithm
7.1 Introduction
7.2 Boundedness of Solutions
7.3 Periodic Forcing
7.4 Convergence of the Proximal Point Algorithm
7.5 Convergence with Non-summable Errors
7.6 Rate of Convergence
8: Second Order Difference Equations
8.1 Introduction
8.2 Existence and Uniqueness
8.3 Periodic Forcing
8.4 Continuous Dependence on Initial Conditions
8.5 Asymptotic Behavior for the Homogeneous Case
8.5.1 Weak Ergodic Convergence
8.5.2 Strong Ergodic Convergence
8.5.3 Weak Convergence of Solutions
8.5.4 Strong Convergence of Solutions
8.6 Subdifferential Case
8.7 Asymptotic Behavior for the Non-Homogeneous Case
8.7.1 Mean Ergodic Convergence
8.7.2 Weak Convergence of Solutions
8.7.3 Strong Convergence of Solutions
8.8 Applications to Optimization
9: Discrete Nonlinear Oscillator Dynamical System and the Inertial Proximal Algorithm
9.1 Introduction
9.2 Boundedness of the Sequence and an Ergodic Theorem
9.3 Weak Convergence of the Algorithm with Errors
9.4 Subdifferential Case
9.5 Strong Convergence
PART IV: APPLICATIONS
10: Some Applications to Nonlinear Partial Differential Equations and Optimization
10.1 Introduction
10.2 Applications to Convex Minimization and Monotone Operators
10.2.1 Rate of Convergence
10.2.2 Strongly Monotone Case
10.2.3 Using the Second Order Evolution Equation for Approximating a Minimizer
10.3 Application to Variational Problems
10.4 Some Applications to Partial Differential Equations
10.4.1 A Concrete Example of the First Order Equation
10.4.2 An Example of a Second Order Evolution Equation
10.4.3 An Example of a Second Order Difference Equation
Index

Citation preview

Nonlinear Evolution and Difference Equations of Monotone Type in Hilbert Spaces Behzad Djafari-Rouhani Department of Mathematical Sciences University of Texas at El Paso, Texas, USA

Hadi Khatibzadeh Department of Mathematics University of Zanjan, Zanjan, Iran

p, p,

A SCIENCE PUBLISHERS BOOK A SCIENCE PUBLISHERS BOOK

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2019 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper Version Date: 20190114 International Standard Book Number-13: 978-1-4822-2818-2 (Hardback)

This book contains information obtained from authentic and highly regarded sources. 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, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. 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: Djafari-Rouhani, Behzad, author. | Khatibzadeh, Hadi, author. Title: Nonlinear evolution and difference equations of monotone type in Hilbert spaces / Behzad Djafari-Rouhani (Department of Mathematical Sciences, University of Texas at El Paso, Texas, USA), Hadi Khatibzadeh (Department of Mathematics, Zanjan University, Zanjan, Iran). Description: Boca Raton, FL : CRC Press, 2019. | “A science publishers book.” | Includes bibliographical references and index. Identifiers: LCCN 2018060366 | ISBN 9781482228182 (hardback) Subjects: LCSH: Evolution equations, Nonlinear. | Differential equations, Nonlinear. | Differential equations. | Hilbert space. Classification: LCC QA377.3 .D53 2019 | DDC 51/.35--dc23 LC record available at https://lccn.loc.gov/2018060366

Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

Preface This book deals with first and second order evolution equations, as well as difference equations of monotone type. There are already many comprehensive books and papers on the subject. Just to mention a few, the books by V. Barbu, H. Br´ezis, G. Morosanu and N. Apreutesei. We apologize in advance for any contributing authors missing in our references. The novelty of our book is the approach taken of emphasizing the qualitative and long time behavior of the solutions. This approach was first introduced by the first author in his Ph.D. thesis and his subsequent papers, and later continued and extended with the second author in his Ph.D. thesis and subsequent papers to second order evolution equations and difference equations. Namely, we show that in most cases, the existence of solutions is actually equivalent to the zero set of the maximal monotone operator to be nonempty, and give a characterization of the limit as time goes to infinity, of the solution or its averages when either one exists. Most of the results related to this approach appear in book form for the first time. We have tried to make the book self-contained as far as possible, so that in addition to researchers in the field, it can be also fruitfully used by advanced undergraduate and graduate students. The first author would like to dedicate this book to his beloved teacher and mentor, the late Professor Shizuo Kakutani of Yale University. Behzad Djafari-Rouhani Hadi Khatibzadeh

Contents Preface

iii

PART I PRELIMINARIES 1

Preliminaries of Functional Analysis 1.1 1.2 1.3 1.4

2

3

Introduction to Hilbert Spaces Weak Topology and Weak Convergence Reflexive Banach Spaces Distributions and Sobolev Spaces 1.4.1 Vector-valued Functions 1.4.2 L p Spaces 1.4.3 Scalar Distributions and Sobolev Spaces 1.4.4 Vector Distributions and Sobolev Spaces

3 3 6 7 8 8 8 9 11

Convex Analysis and Subdifferential Operators

13

2.1 2.2 2.3 2.4 2.5 2.6

13 13 14 17 18 19

Introduction Convex Sets and Convex Functions Continuity of Convex Functions Minimization Properties Fenchel Subdifferential The Fenchel Conjugate

Maximal Monotone Operators

21

3.1 3.2 3.3 3.4 3.5

21 21 22 23 28

Introduction Monotone Operators Maximal Monotonicity Resolvent and Yosida Approximation Canonical Extension

PART II EVOLUTION EQUATIONS OF MONOTONE TYPE 4

First Order Evolution Equations

33

4.1 Introduction 4.2 Existence and Uniqueness of Solutions

33 33

vi

Nonlinear Evolution and Difference Equations of Monotone Type

5

6

4.3 Periodic Forcing 4.4 Nonexpansive Semigroup Generated by a Maximal Monotone Operator 4.5 Ergodic Theorems for Nonexpansive Sequences and Curves 4.5.1 Almost Nonexpansive Sequences 4.5.2 Almost Nonexpansive Curves 4.6 Weak Convergence of Solutions and Means 4.7 Almost Orbits 4.8 Sub-differential and Non-expansive Cases 4.9 Strong Ergodic Convergence 4.10 Strong Convergence of Solutions 4.11 Quasi-convex Case

39 45

Second Order Evolution Equations

71

45 46 51 53 54 55 57 62 64

5.1 Introduction 5.2 Existence and Uniqueness of Solutions 5.2.1 The Strongly Monotone Case 5.2.2 The Non Strongly Monotone Case 5.3 Two Point Boundary Value Problems 5.4 Existence of Solutions for the Nonhomogeneous Case 5.5 Periodic Forcing 5.6 Square Root of a Maximal Monotone Operator 5.7 Asymptotic Behavior 5.7.1 Ergodic Convergence 5.7.2 Weak Convergence 5.7.3 Strong Convergence 5.7.4 Subdifferential Case 5.8 Asymptotic Behavior for Some Special Nonhomogeneous Cases 5.8.1 Case C ≤ 0 5.8.2 The Case C > 0

71 71 72 79 81 86 87 89 90 91 95 96 99 104 105 113

Heavy Ball with Friction Dynamical System

121

6.1 Introduction 6.2 Minimization Properties

121 121

PART III DIFFERENCE EQUATIONS OF MONOTONE TYPE 7

First Order Difference Equations and Proximal Point Algorithm

129

7.1 7.2 7.3 7.4 7.5 7.6

129 130 131 133 137 142

Introduction Boundedness of Solutions Periodic Forcing Convergence of the Proximal Point Algorithm Convergence with Non-summable Errors Rate of Convergence

vii

Contents

8

9

Second Order Difference Equations

145

8.1 8.2 8.3 8.4 8.5

Introduction Existence and Uniqueness Periodic Forcing Continuous Dependence on Initial Conditions Asymptotic Behavior for the Homogeneous Case 8.5.1 Weak Ergodic Convergence 8.5.2 Strong Ergodic Convergence 8.5.3 Weak Convergence of Solutions 8.5.4 Strong Convergence of Solutions 8.6 Subdifferential Case 8.7 Asymptotic Behavior for the Non-Homogeneous Case 8.7.1 Mean Ergodic Convergence 8.7.2 Weak Convergence of Solutions 8.7.3 Strong Convergence of Solutions 8.8 Applications to Optimization

145 146 160 163 165 165 173 176 180 184 188 188 196 197 203

Discrete Nonlinear Oscillator Dynamical System and the Inertial Proximal Algorithm

207

9.1 9.2 9.3 9.4 9.5

207 208 211 215 217

Introduction Boundedness of the Sequence and an Ergodic Theorem Weak Convergence of the Algorithm with Errors Subdifferential Case Strong Convergence

PART IV APPLICATIONS 10 Some Applications to Nonlinear Partial Differential Equations and Optimization 10.1 Introduction 10.2 Applications to Convex Minimization and Monotone Operators 10.2.1 Rate of Convergence 10.2.2 Strongly Monotone Case 10.2.3 Using the Second Order Evolution Equation for Approximating a Minimizer 10.3 Application to Variational Problems 10.4 Some Applications to Partial Differential Equations 10.4.1 A Concrete Example of the First Order Equation 10.4.2 An Example of a Second Order Evolution Equation 10.4.3 An Example of a Second Order Difference Equation Index

221 221 221 222 227 229 230 231 231 233 233 237

Part I Preliminaries

of 1 Preliminaries Functional Analysis In this chapter, we review the preliminaries and prerequisites that are needed in the sequel. Most proofs are omitted, and the reader is referred to the main references for this chapter that are [ADA, BRE2, DEB-MIK, YOS].

1.1

INTRODUCTION TO HILBERT SPACES

Unless otherwise stated, all vector spaces considered in this book are over the field of real numbers. Definition 1.1.1 Let X be a real vector space. A function k · k : X → R is called a norm on X if the following properties hold: 1) kxk ≥ 0, ∀x ∈ X and kxk = 0 iff x = 0. 2) kαxk = |α|kxk, ∀α ∈ R, ∀x ∈ X. 3) kx + yk ≤ kxk + kyk (triangle inequality). A vector space X with a norm k · k is called a normed linear space. The normed linear space (X, k · k) is said to be a Banach space if it is complete as a metric space, for the metric d(x, y) := kx − yk, induced by the norm k · k. Definition 1.1.2 Let X and Y be Banach spaces. The mapping T : X → Y is called a linear mapping if the following properties hold: 1) T (x + y) = T x + Ty, ∀x, y ∈ X 2) T (λ x) = λ T x, ∀x ∈ Xand λ ∈ R

Proposition 1.1.3 Let X and Y be Banach spaces. For each linear mapping T : X → Y , the following statements are equivalent: (i) T is uniformly continuous (ii) T is continuous at 0 (iii) There is a positive constant M such that kT xk ≤ Mkxk, ∀x ∈ X.

Proof. Obviously (i) implies (ii). Suppose T is continuous at 0, then there exists ry r > 0 such that if kxk ≤ r, then kT xk < 1. Given 0 , y ∈ X, since k kyk k = r, we have ry kT ( kyk )k < 1. Then kTyk ≤ 1r kyk. Hence (ii) implies (iii). Finally, (iii) implies that T is Lipschitz continuous, and therefore (i) is satisfied. 

4

Nonlinear Evolution and Difference Equations of Monotone Type

Definition 1.1.4 For each linear mapping T : X → Y , we define kT k = sup x,0

kT xk . kxk

The linear mapping T is said to be bounded if kT k < +∞. The space of all bounded linear mappings from X to Y with the above norm is denoted by L(X,Y ), and is a normed linear space. L(X,Y ) is a Banach space if Y is a Banach space. From Proposition 1.1.3, we get kT xk ≤ kT kkxk, ∀x ∈ X. In Definition 1.1.4, if Y = R, the space L(X, R) is denoted by X ∗ and is called the dual or conjugate space of X. The elements of the space X ∗ are called bounded linear functionals on X. In this book, we concentrate on some special types of Banach spaces, called Hilbert spaces, that are the immediate generalization of the Euclidean spaces Rn to the infinite dimensional case with very similar properties, and are defined as follows. Definition 1.1.5 Let H be a real vector space. A function (·, ·) : H × H → R is called an inner product on H if the following conditions hold: 1) (αx + y, z) = α(x, z) + (y, z), ∀x, y, z ∈ H, and α ∈ R 2) (x, y) = (y, x), ∀x, y ∈ H 3) (x, x) ≥ 0, ∀x ∈ H, and (x, x) = 0 if and only if x = 0. The vector space H with inner product (·, ·) is called an inner product space. Example 1.1.6 The simplest example of an inner product space is R with multiplication. More generally, Rn with (x, y) = ∑ni=1 xi yi is an inner product space. Example 1.1.7 The space l 2 of all real sequences (x1 , x2 , x3 , . . .) such that 2 ∑+∞ i=1 |xi | < +∞, with inner product defined by +∞

(x, y) := ∑ xi yi , i=1

for each x = (x1 , x2 , . . .) and y = (y1 , y2 , . . .) is an infinite dimensional inner product space. Example 1.1.8 The space L2 (R) of all square integrable functions on R, with inner product defined by Z +∞

( f , g) =

f (x)g(x)dx −∞

is an inner product space. More generally, L2 (Rn ) with inner product defined by Z

( f , g) =

Rn

f (x)g(x)dx

Preliminaries of Functional Analysis

5

is an inner product space. We often use L2 ((a, b)) or L2 (Ω), where Ω is a subset of Rn . Example 1.1.9 The space C([a, b]) of all real-valued continuous functions on [a, b] with inner product defined by Z b

( f , g) =

f (x)g(x)dx a

is an inner product space. We recall below without proof, some of the important identities and inequalities in an inner product space. 1) (Cauchy-Schwarz inequality): Let x, y ∈ H, then p p |(x, y)| ≤ (x, x) (y, y). Moreover, in the above inequality, equality holds if and only if x and y are linearly dependent. 2) kx + yk2 = kxk2 + 2(x, y) + kyk2 . 3) (Parallelogram identity): kx + yk2 + kx − yk2 = 2kxk2 + 2kyk2 . 2 2 4) (Polarization identity): 4(x, y) = pkx + yk − kx − yk . In (H, (·, ·)), defining kxk := (x, x), then k · k is a norm on H that is induced by the inner product (·, ·). Properties 1 and 2 of the norm are easy to show. The triangle inequality follows from Cauchy-Schwarz inequality. The inner product space (H, (·, ·)) is said to be a Hilbert space, if it is complete for the norm induced by its inner product. The space C([a, b]) in Example 1.1.9 is not a Hilbert space, because it is not complete. The spaces in the other Examples 1.1.6–1.1.8 are Hilbert spaces. Other important examples of Hilbert spaces include Sobolev spaces. They will be studied in Section 3 of this chapter. Theorem 1.1.10 (Riesz representation theorem) Let H be a Hilbert space. For every y ∈ H, the mapping f : H → R defined by f (x) = (x, y), ∀x ∈ H, is a bounded linear functional on H with k f k = kyk. Conversely, let f : H → R be a bounded linear functional on H. Then there is a unique element y in H such that f (x) = (x, y), ∀x ∈ H; in addition k f k = kyk. Theorem 1.1.11 (Best approximation theorem) Let H be a Hilbert space and C be a nonempty closed and convex subset of H. Then for each x ∈ H, there is a unique element u ∈ C which is the nearest point to x. That is kx − uk = Min{kx − vk; v ∈ C}. In addition, u is characterized by the following property: ( u∈C (x − u, v − u) ≤ 0, ∀v ∈ C. This unique point u is denoted by PC x, and is called the metric projection of x onto C.

6

Nonlinear Evolution and Difference Equations of Monotone Type

Proposition 1.1.12 Let C be a nonempty closed and convex subset of a Hilbert space H. Then the mapping PC : H → C has the following properties: 1) (PC x − PC y, x − y) ≥ kPC x − PC yk2 , 2) kPC x − PC yk ≤ kx − yk, for each x, y ∈ H. Corollary 1.1.13 (Best approximation on subspaces) Suppose M ⊂ H is a closed linear subspace of H, and x ∈ H. Then u = PM x is characterized by ( u∈M (x − u, y) = 0, ∀y ∈ M. That is onto M.

PM x

is

the

orthogonal

projection

of

x

Definition 1.1.14 Let a : H × H → R be a bilinear form (that is a linear mapping with respect to each argument separately). a is said to be (i) continuous if there is a constant C such that |a(u, v)| ≤ Ckukkvk, ∀u, v ∈ H (ii) coercive if there is a constant α > 0 such that |a(u, u)| ≥ αkuk2 , ∀u ∈ H (iii) symmetric if a(u, v) = a(v, u), ∀u, v ∈ H Now we recall the following theorem which is very useful for solving linear partial differential equations, and we refer to [BRE2], page 140, for its proof. Theorem 1.1.15 (Lax-Milgram) Assume that a(u, v) is a continuous coercive bilinear form on H. Then, given any φ ∈ H, there exists a unique element u ∈ H such that a(u, v) = (φ , v), ∀v ∈ H. Moreover, if a is symmetric, then u is characterized by the following property: 1 1 u ∈ H and a(u, u) − (φ , u) = min{ a(v, v) − (φ , v)} v∈H 2 2

1.2

WEAK TOPOLOGY AND WEAK CONVERGENCE

Let (X, k · k) be a Banach space with dual X ∗ . For each x ∈ X and f ∈ X ∗ , we denote f (x) by h f , xi. Definition 1.2.1 The weakest topology on the space X such that all elements of X ∗ are continuous is called the weak topology on X.

Preliminaries of Functional Analysis

7

The topology generated by the norm of X is called the strong topology. Obviously every open set in the weak topology is also open in the strong topology and similarly for closed sets, but the converse is not true. The following theorem shows that the converse holds for convex sets. Theorem 1.2.2 (Mazur) Let C be a convex subset of a Banach space X; then C is closed in the strong topology if and only if it is closed in the weak topology. Definition 1.2.3 A sequence {xn } in X is said to be weakly convergent to x ∈ X if h f , xn i → h f , xi, ∀ f ∈ X ∗ as n → +∞. In this text, we denote the weak convergence by *. The following are some important properties of the weak convergence which are also valid in Banach spaces, and we refer the reader to [BRE2] for their proofs. Theorem 1.2.4 1) If xn → x then xn * x. 2) If xn * x, then kxn k is bounded and kxk ≤ lim infn→+∞ kxn k. 3) if xn * x and fn → f in X ∗ , then h fn , xn i → h f , xi. In a Hilbert space H, by Theorem 1.1.10, the weak convergence of a sequence xn to x is equivalent to (xn − x, y) → 0, ∀y ∈ H as n → +∞. The following is from Opial [OPI] and is frequently used as a useful tool to prove weak convergence. Lemma 1.2.5 (Opial) Let {xn } be a sequence in a Hilbert space H and F ⊂ H be nonempty, closed and convex such that the following conditions are satisfied: 1) limn→+∞ kxn − pk exists for each p ∈ F. 2) every weak cluster point of the sequence xn is in F. Then {xn } converges weakly to an element of F. The following criterion is very useful for the strong convergence. Proposition 1.2.6 [BAR1] If H is a Hilbert space and (xn )n ⊂ H is a weakly convergent sequence to some x ∈ H, and if lim sup kxn k ≤ kxk, then xn → x n→∞

strongly in H.

1.3

REFLEXIVE BANACH SPACES

Definition 1.3.1 Let X be a Banach space with dual X ∗ and bidual X ∗∗ (the dual of X ∗ ). Let the mapping J : X → X ∗∗ be defined as follows: For each x ∈ X, hJ(x), y∗ i = hy∗ , xi, ∀y∗ ∈ X ∗ .

8

Nonlinear Evolution and Difference Equations of Monotone Type

Then J is called the canonical embedding of X into X ∗∗ . It is well known that J is an isometry. X is said to be reflexive if J is surjective. Theorem 1.1.10 shows that every Hilbert space is reflexive. The following theorem is useful to prove weak convergence results as shown in future chapters. ˇ Theorem 1.3.2 [YOS] (Eberlein-Smulyan’s theorem) X is a reflexive Banach space if and only if every bounded sequence in X contains a weakly convergent subsequence in X.

1.4

DISTRIBUTIONS AND SOBOLEV SPACES

1.4.1

VECTOR-VALUED FUNCTIONS

Let [0, T ] be a fixed finite interval and X be a real Banach space. The function f : [0, T ] → H is said to be absolutely continuous if for each ε > 0 there exists δ > 0 such that for each partition {0 = t0 < t1 < · · · < tn = T } of the interval [0, T ], ∑ni=1 |ti − ti−1 | < δ implies ∑ni=1 k f (ti ) − f (ti−1 k < ε. It is well-known that if X = R, every absolutely continuous function f (t) on the real interval [0, T ] is almost everywhere differentiable on (0, T ), and can be recovered as the integral of its derivative. Proposition 1.4.1 [BAR1] Let H be a Hilbert space. Then every H-valued absolutely continuous function f (t) on [0, T ] is a.e. differentiable on (0, T ) and f (t) = f (0) +

Z t df 0

ds

(s)ds, 0 ≤ t ≤ T

1.4.2 LP SPACES Let (X, k · k) be a Banach space and U ⊂ Rn be Lebesgue measurable and 1 ≤ p < +∞. The space of all equivalence classes of measurable functions f : U → X with respect to almost everywhere equality such that the function x 7→ k f (x)k p is Lebesgue integrable on U is denoted by L p (U; X). It is a Banach space with the norm Z

k f kL p (U;X) = (

1

k f (x)k p dx) p .

U

L∞ (U; X)

Similarly is the space of all equivalence classes of measurable functions f : U → X such that x 7→ k f (x)k is essentially bounded in U. This is a Banach space with the norm k f kL∞ (U;X) = ess sup k f (x)k. x∈U

R, we denote L p (U; X) by p We denote Lloc (0, +∞; X)

L p (U),

When X = and if U = (a, b), we denote it by L p (a, b; X). for 1 ≤ p ≤ +∞, the space of all equivalence classes with respect to almost everywhere equality of measurable functions f : (0, +∞) → X such that the restriction of f on each finite interval (0, T ) belongs to L p (0, T ; X).

Preliminaries of Functional Analysis

9 R

Theorem 1.4.2 Assume that vol Ω = Ω 1 dx < ∞ and 1 ≤ p ≤ q ≤ ∞. Then the following embedding holds: Lq (Ω) ,→ L p (Ω) . Theorem 1.4.3 Let H be a real Hilbert space with the scalar product (·, ·) , then L2 (Ω; H) is a real Hilbert space with respect to the scalar product hu, vi =

Z

(u (x) , v (x)) dx, ∀u, v ∈ L2 (Ω; H) .



Theorem 1.4.4 (Fatou’s lemma) If Ω ⊆ RN is a measurable set and ( fn )n is a sequence of nonnegative and measurable real functions defined on Ω, then Z

lim inf fn (x) dx ≤ lim inf n→∞

Z

fn (x) dx.

n→∞





Theorem 1.4.5 (Lebesgue’s theorem) If Ω ⊆ RN is a measurable set, { fn }n is a sequence of measurable real functions which is almost everywhere convergent on Ω, and if there exists a nonnegative function g ∈ L1 (Ω) such that | fn (x)| ≤ g (x), ∀n ∈ N, and f or a.e. x ∈ Ω, then Z

fn (x) dx =

lim

n→∞ Ω

1.4.3

Z 

 lim fn (x) dx.

n→∞ Ω

SCALAR DISTRIBUTIONS AND SOBOLEV SPACES

Suppose that U is an open subset of Rn . We denote by C0∞ (U), the space of all infinitely differentiable functions f : U → R whose supp( f ) = {x ∈ U; f (x) , 0} is a compact subset of U. The elements of C0∞ (U) are called test functions. Now suppose that K is a compact subset of U. We denote by DK (U) the set of all f ∈ C0∞ (U) with supp( f ) ⊂ K. The space DK (U) endowed with the family of seminorms pK, j ( f ) =

sup

|Dα f (x)|, j ∈ N

x∈X, |α|≤ j

is a Fr´echet space (i.e. a metrizable locally convex and complete space), where |α| |α| = α1 + · · · + αn and Dα = α1∂ αn for α = (α1 , . . . , αn ) ∈ Nn . If K1 and K2 ∂ x1 ···∂ xn

are two compact subsets of U such that K1 ⊂ K2 , the topology of DK1 (U) coincides with the inductive topology from DK2 (U) over DK1 (U). It is obvious that C0∞ (U) = ∪K⊂U DK (U) where the union is taken over all compact subsets of U. The space C0∞ (U) endowed with the inductive limit topology is denoted by D(U). Definition 1.4.6 A linear and continuous functional on D(U) is called a distribution on U.

10

Nonlinear Evolution and Difference Equations of Monotone Type

The space of all distributions on U (that is the dual space of D(U)) is denoted by D 0 (U). If u : U → R is measurable and Lebesgue integrable on every compact subset of U, then the functional D(U) 3 f 7→

Z

u(x) f (x)dx U

is a distribution on U that we denote again by u. The function u is identified with the corresponding distribution. Definition 1.4.7 The derivative of order α of a distribution u ∈ D 0 (U) is a distribution D α u defined by Dα u( f ) = (−1)|α| u(Dα f ), ∀ f ∈ D(U) Now suppose 1 ≤ p ≤ +∞ and k ∈ N. Definition 1.4.8 The space W k,p (U) = {u; Dα u ∈ L p (U), ∀α ∈ Nn , |α| ≤ k} (where Dα u is the distributional derivative of u) is called the Sobolev space of order k. Theorem 1.4.9 W k,p (U) where 1 ≤ p < +∞ and k ∈ N is a Banach space with the norm 1 kukW k,p (U) = ( ∑ kDα ukLp p (U) ) p |α|≤k

and W k,∞ (U)

for k ∈ N is a Banach space with the norm kukW k,∞ (U) = max|α|≤k kDα ukL∞ (U) 

Proof. See [ADA]

The completion of D(U) with respect to the topology of W k,p (U) is denoted by When p = 2, we denote H k (U) = W k,2 (U) and H0k (U) = W0k,2 (U).

W0k,p (U).

Theorem 1.4.10 Both H k (U) and H0k (U) are Hilbert spaces with the inner product Z

(u, v)H k (U) =



Dα u.Dα vdx.

|α|≤k U

We denote the dual space of H0k (U) by H −k (U). Since D(U) is dense in H0k (U), and the embedding i : D(U) → H0k (U) is continuous, then H −k (U) is embedded in D 0 (U) densely and continuously (by means of the adjoint operator i∗ : H −k (U) → D 0 (U)). If U is a bounded domain in Rn with boundary Γ sufficiently smooth, then H01 (U) = {u ∈ H 1 (U); u|Γ ≡ 0} where u|Γ is the trace of u on Γ.

Preliminaries of Functional Analysis 1.4.4

11

VECTOR DISTRIBUTIONS AND SOBOLEV SPACES

Let U = (a, b) with −∞ ≤ a < b ≤ +∞ and D 0 (a, b; X) be the space of all continuous linear operators from D(a, b) to X. The elements of D 0 (a, b; X) are called vector distributions on (a, b) with values in X. If u : (a, b) → X is Bochner integrable on every bounded interval I ⊂ (a, b), then as in the scalar case, we can define a vector distribution which we can denote again by u as follows Z b

u( f ) = a

f (t)u(t)dt, ∀ f ∈ D(a, b).

Now for u ∈ D 0 (a, b; X), we denote u( j) by the distributional derivative of order j ∈ N i.e. dj f u( j) ( f ) = (−1) j u( j ), ∀ f ∈ D(a, b) dt (0) where by convention u = u. Let k ∈ N and 1 ≤ p ≤ +∞ W k,p (a, b; X) = {u ∈ D 0 (a, b; X); u( j) ∈ L p (a, b; X), j = 0, 1, . . . , k} Theorem 1.4.11 For every k ∈ N and 1 ≤ p < +∞ the space W k,p (a, b; X) with norm k

1

kukW k,p (a,b;X) = ( ∑ ku( j) kLp p (a,b;X) ) p j=0

is a Banach space. Moreover for each k ∈ N, W k,∞ (a, b; X) is a Banach space with norm kukW k,∞ (a,b;X) = max ku( j) kL∞ (a,b;X) 0≤ j≤k

As in the scalar case, for p = 2 we denote H k (a, b; X) for W k,2 (a, b; X). Theorem 1.4.12 Let X be a Hilbert space with inner product (·, ·). Then for every k ∈ N, H k (a, b; X) is also a Hilbert space with the inner product k

(u, v)H k (a,b;X) =



Z b

j=0 a

(u( j) (t), v( j) (t))X dt

k,p We define Wloc (a, b; X) for 1 ≤ p ≤ +∞ and k ∈ N to be the space of all distributions u ∈ D 0 (a, b; X) such that u ∈ W k,p (t1 ,t2 ; X) for each bounded interval (t1 ,t2 ) ⊂ (a, b). For the bounded interval (a, b), 1 ≤ p ≤ +∞ and k ∈ N we denote by Ak,p (a, b; X) the space of all absolutely continuous functions u : (a, b) → X such j that ddt uj exists and is defined almost everywhere for j = 1, 2, . . . , k, and is absolutely

continuous and

dk u dt k

∈ L p (a, b; X).

Theorem 1.4.13 Let v ∈ L p (a, b; X), 1 ≤ p ≤ +∞, then the following conditions are equivalent: (1) v ∈ W k,p (a, b; X) (2) There exists v1 ∈ Ak,p (a, b; X) such that v1 (t) = v(t) almost everywhere on (a, b)

12

Nonlinear Evolution and Difference Equations of Monotone Type

p Definition 1.4.14 If Ω = R+ = [0, ∞) and 1 ≤ p ≤ ∞, then Lloc (0, ∞; X) is the locally convex space of all measurable functions u : [0, ∞) → X with the property that the restriction of u to every interval [0, T ] (T ∈ (0, ∞)) is in L p (0, T ; X) .

We also define the spaces  2,2 Wloc (0, T ; X) = u : [0, T ] → X; u ∈ W 2,2 (η, T − η; X) , ∀ η > 0 and

2,2 2,2 Wloc (0, ∞; X) = {u : [0, ∞) → X; u ∈ Wloc (0, T ; X) , ∀ T > 0}.

Theorem 1.4.15 If X is a real Banach space and u ∈ W 1,p (0, ∞; X) (with 1 ≤ p < ∞), then ku (t) k → 0 as t → ∞.

REFERENCES ADA. R. A. Adams, Sobolev Spaces, Academic Press, New York, 1975. BAR. V. Barbu, Nonlinear Semigroups and Differential Equations in Banach Spaces, Noordhoff, Leyden, 1976. BRE. H. Br´ezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, Universitext. Springer, New York, 2011. DEB-MIK. L. Debnath and P. Mikusinski, Introduction to Hilbert Spaces with Applications, Academic Press, Inc., San Diego, CA, 1999. MOR. G. Morosanu, Nonlinear Evolution Equations and Applications, Editua Academiei (and D. Reidel Publishing Company), Bucuresti, 1988. OPI. Z. Opial, Weak convergence of the sequence of successive approximations for nonexpasive mappings, Bull. Amer. Math. Soc. 73 (1967), 591–597. YOS. K. Yosida, Functional Analysis, third ed., Springer Verlag, Berlin, 1971.

Analysis and 2 Convex Subdifferential Operators 2.1

INTRODUCTION

Convexity has an essential role in optimization, variational inequalities, evolution equations and many other branches of nonlinear analysis. This chapter is a quick review of convex analysis. Our aim is to provide the reader with the essential facts on convex sets and convex functions that will be needed in the subsequent chapters of the book.

2.2

CONVEX SETS AND CONVEX FUNCTIONS

This section is devoted to some definitions and elementary properties of convex sets and functions. Let X be a real Banach space with the dual X ∗ . For each x∗ ∈ X ∗ and x ∈ X, we denote x∗ (x) by hx∗ , xi. Definition 2.2.1 A subset C of a Banach space X is called convex if for all x, y ∈ C and each λ ∈ [0, 1], λ x + (1 − λ )y ∈ C. A ball in a normed linear space, an affine subspace of a linear space and a segment joining two points in a linear space are some examples of convex sets. Also it is clear that the intersection of a family of convex sets is convex. Definition 2.2.2 Let C be a subset of a Banach space X. The “convex hull” of C is the intersection of all convex subsets of X containing C. In other words, it is the smallest convex subset of X containing C. It is denoted by conv(C). The closed convex hull of C is the smallest closed convex subset of X containing C. It is denoted by conv(C) A simplex in a Banach space X is the closed convex hull of a finite number of points in X. Definition 2.2.3 Let C be a convex subset of X. Then f : X → (−∞, +∞] is called: i) convex if f (λ x + (1 − λ )y) ≤ λ f (x) + (1 − λ ) f (y), ∀x, y ∈ C, ∀λ ∈ [0, 1] ii) strictly convex if f (λ x + (1 − λ )y) < λ f (x) + (1 − λ ) f (y), ∀x , y ∈ C, ∀λ ∈ (0, 1) iii) strongly convex if f (λ x + (1 − λ )y) ≤ λ f (x) + (1 − λ ) f (y) − λ (1 − λ )kx − yk2 , ∀x, y ∈ C, ∀λ ∈ (0, 1)

14

Nonlinear Evolution and Difference Equations of Monotone Type Obviously strongly convex ⇒ strictly convex ⇒ convex

The norm k · k of every Hilbert space H is a strictly convex function on H, but it is not strongly convex. However, k · k2 is a strongly convex function on H, and therefore also strictly convex, and hence convex. Proposition 2.2.4 Let f : X → (−∞, +∞] be a convex function. Then dom f := {x ∈ X : f (x) < +∞}, which is called the effective domain of f is a convex set. Definition 2.2.5 The convex function f : X → (−∞, +∞] is called proper if dom f , ∅ Proposition 2.2.6 If f : X → (−∞, +∞] is convex, then the sub-level set Lrf := {x ∈ X : f (x) ≤ r} is a convex set. Definition 2.2.7 Let f : X → (−∞, +∞]. The epi-graph of f is defined as epi f := {(x, r) ∈ X × R : f (x) ≤ r}. Proposition 2.2.8 f : X → (−∞, +∞] is a convex function if and only if epi f is a convex subset of X × R. The following proposition is called Jensen’s inequality. Proposition 2.2.9 Let f : X → (−∞, +∞]. Then f is convex if and only if for all finite families (λi )i∈I in ]0, 1[ such that ∑i∈I λi = 1, and (xi )i∈I in X, we have f (∑ λi xi ) ≤ ∑ λi f (xi ). i∈I

2.3

i∈I

CONTINUITY OF CONVEX FUNCTIONS

This section contains some continuity properties of convex functions. Theorem 2.3.1 Let f : X → (−∞, +∞] be proper and convex and let x0 ∈ dom f . Then the following are equivalent: i) f is locally Lipschitz near x0 . ii) f is continuous at x0 . iii) f is bounded on a neighborhood of x0 . iv) f is bounded above on a neighborhood of x0 . Moreover, if anyone of these conditions holds, then f is continuous and locally Lipschitz on int(dom f ). Proof. (i)⇒(ii)⇒(iii)⇒(iv): Are clear. (iv)⇒(ii): Take ρ > 0 such that η = sup f (B(x0 , ρ)) < +∞. Given ε > 0, choose α ∈ (0, 1) such that α(η − f (x0 )) < ε, and let x ∈ B(x0 , αρ). The convexity of f yields f (x) − f (x0 ) = f (1 − α)x0 +

α(x − (1 − α)x0 )  − f (x0 ) α

Convex Analysis and Subdifferential Operators

15

 (x − x0 ) ) − f (x0 ) α ≤ α η − f (x0 ) . ≤ α f (x0 +

Similarly x α (1 + α)x0 − x + ) − f (x) 1+α 1+α α  α (x0 − x) ≤ f (x0 + ) − f (x) 1+α α  α ≤ (η − f (x0 )) + ( f (x0 ) − f (x)) , 1+α

f (x0 ) − f (x) = f (

which after rearranging implies that f (x0 ) − f (x) ≤ α(η − f (x0 )). Altogether, we showed that | f (x) − f (x0 )| ≤ α(η − f (x0 )) < ε, ∀x ∈ B(x0 , αρ). Therefore f is continuous at x0 . (iii)⇒(i): Choose ρ > 0 such that δ := 2 sup | f (B(x0 , 2ρ))| < +∞. Take distinct points x and y in B(x0 , ρ) and set z = x+(

1 kx − yk kx − yk − 1)(x − y), where α = < . α kx − yk + ρ ρ

Then x = αz + (1 − α)y and kz − x0 k ≤ kz − xk + kx − x0 k = ρ + kx − xo k ≤ 2ρ. Therefore, y and z belong to B(x0 , 2ρ) and hence, by the convexity of f , we have: f (x) = f (αz + (1 − α)y) ≤ f (y) + α( f (z) − f (y)) ≤ f (y) + αδ δ ≤ f (y) + ( )kx − yk. ρ Thus f (x) − f (y) ≤ ( ρδ )kx − yk. Interchanging the roles of x and y, we conclude that | f (x) − f (y)| ≤ ( ρδ )kx − yk, which implies (i). So far, we have shown that the statements (i)–(iv) are equivalent to each other. Now assume that (iv) holds, and say η = sup f (B(x0 , ρ)) < +∞ for some ρ > 0. Then f is continuous and locally Lipschitz near x0 . Take x ∈ int(dom f ) and γ > 0 such that B(x, γ) ⊂ dom f . We may take x , x0 , otherwise there is nothing to prove. Now set y = x0 +

γ 1 (x − x0 ), where α = ∈ (0, 1) 1−α γ + 2kx − x0 k

Then y ∈ B(x, γ). Now take z ∈ B(x, αρ) and set w = x0 + (z−x) α = w ∈ B(x0 , ρ) and z = αw + (1 − α)y. Consequently,

(z−(1−α)y) . α

Then

f (z) ≤ α f (w) + (1 − α) f (y) ≤ αη + (1 − α) f (y) Therefore f is bounded above on B(x, αρ). It follows now from the first part of the proof that f is continuous and locally Lipschitz near x. 

16

Nonlinear Evolution and Difference Equations of Monotone Type

In a finite dimensional Banach space, the local boundedness from above of a convex function follows from Proposition 2.2.9. Therefore, Theorem 2.3.1 implies the following corollary. Corollary 2.3.2 Suppose that X is finite-dimensional and let f : X → R be convex. Then f is continuous on X. Proof. Let x ∈ X. Since X is finite-dimensional, there exist a finite family {y1 , . . . , yn } in X and ρ > 0 such that B(x, ρ) ⊂ conv{y1 , . . . , yn }. Consequently, by Proposition 2.2.9, sup f (B(x, ρ)) ≤ sup f (conv{y1 , . . . , yn }) ≤ Maxi=1,...,n f (yi ) < +∞. The conclusion follows now from Theorem 2.3.1.  Remark 2.3.3 Let f : X → (−∞, +∞], x ∈ X and denote by Ns (resp. Nw ) the family of all neighborhoods of x in the strong (resp. weak) topology. We denote lim inf f (y) = sup

inf

f (y)

lim inf f (y) = sup

inf

f (y)

y→x

W ∈Ns y∈W −{x}

and y*x

W ∈Nw y∈W −{x}

Definition 2.3.4 f : X → (−∞, +∞] is called (weakly) lower semi-continuous at x ∈ X if (lim infy*x f (y) ≥ f (x)) lim infy→x f (y) ≥ f (x). f is called (weakly) lower semi-continuous if it is (weakly) lower semi-continuous at each point of X. Proposition 2.3.5 f : X → (−∞, +∞] is (weakly) lower semi-continuous if Lrf is (weakly) closed for each r ∈ R. The following theorem is an immediate consequence of Mazur’s theorem. Theorem 2.3.6 For any convex function f : X → (−∞, +∞] the following are equivalent: i) f is lower semi-continuous. ii) f is weakly lower semi-continuous. The following theorem provides some additional information on the continuity of convex functions. Theorem 2.3.7 Let f : X → (−∞, +∞] be a proper, convex and lower semicontinuous function. Then f is continuous, and in fact locally Lipschitz, at the points in the interior of dom f . Proof. Assume that int(dom f ) , ∅, and let x ∈ int(dom f ). We have: dom f = f f ∪∞ n=1 Ln . Since f is lower semi-continuous, dom f and Ln are closed subsets of H, it then follows from Baire category theorem, that there is n such that int(Lnf ) , ∅. Suppose that B(x0 , ρ) ⊂ Lnf , for some x0 ∈ Lnf and some ρ > 0. Then we have: f (B(x0 , ρ)) ≤ n, and hence condition (iv) of Theorem 2.3.1 is satisfied. Now the result follows from the last assertion in Theorem 2.3.1. 

Convex Analysis and Subdifferential Operators

2.4

17

MINIMIZATION PROPERTIES

Convex functions play an essential role in optimization theory because of their nice minimization properties. It is important to know whether a local minimum is also a global minimum. Another important question is whether a critical point is a local minimum. The convexity guarantees these properties. In this short section we mention some minimization properties of convex functions. Theorem 2.4.1 Let C be a convex subset of X and f be a convex function on C. Then i) A local minimum point of f is a global minimum. ii) The set of all minimum points of f is convex. iii) The set of all minimum points of f is at most a singleton if f is strictly convex. Proof. (i): Suppose that x0 is a local minimum point of f ; i.e. there exists r > 0 such that f (x0 ) ≤ f (x) for all x ∈ B(x0 , r). Choose y ∈ X arbitrary, and 0 < λ < 1 such that λ ky − x0 k < r. Let x = λ y + (1 − λ )x0 . Then kx − x0 k = λ ky − x0 k < r Therefore x ∈ B(x0 , r). By hypothesis, f (x0 ) ≤ f (x) = f (λ y + (1 − λ )x0 ) ≤ λ f (y) + (1 − λ ) f (x0 ) Hence f (x0 ) ≤ f (y). (ii): Suppose that x0 is a minimum point of f . The set of all minimum points of f is {x ∈ X; f (x) = f (x0 )}. But we have: {x ∈ X; f (x) = f (x0 )} = {x ∈ X; f (x) ≤ f (x0 )} and the last set is convex because f is convex and by using Proposition 2.2.6. (iii): If a strictly convex function f has two distinct minimum points, then the value of f on the segment joining these two points will be strictly less than the minimum value of the function, and this is a contradiction.  We know that for a differentiable function, a minimum point is a critical point, but the converse may not be true. In the following theorem, we show that the converse is also true for convex functions. Theorem 2.4.2 Let C be a convex subset of X and f be a convex differentiable function on C. Then a critical point x0 ∈ C is a global minimum for f . Proof. Let x0 be a critical point of f and let x be an arbitrary point in C. By the convexity of f , we have f (λ x + (1 − λ )x0 ) ≤ λ f (x) + (1 − λ ) f (x0 ), ∀λ ∈ (0, 1) Then

f (λ x + (1 − λ )x0 ) − f (x0 ) ≤ f (x) − f (x0 ) λ

18

Nonlinear Evolution and Difference Equations of Monotone Type

By letting λ → 0, we get 0 = h∇ f (x0 ), x − x0 i ≤ f (x) − f (x0 ) 

which proves the theorem.

The following theorem explores sufficient conditions for the existence of a minimum point for a convex function on a Hilbert space H. It is applied in optimization for instance in Tikhonov regularization methods and the proximal point algorithm for convex functions. Although it holds in the more general setting of reflexive Banach spaces, but for the sake of our future use in the sequel, we recall it in Hilbert spaces. For the proof, see page 71, Corollary 3.23 of [BRE2] or page 34, Theorem 1.10 of [MOR]. Theorem 2.4.3 If f : H → (−∞, +∞] is a proper, convex and lower semi-continuous function on H, satisfying lim f (x) = +∞, kxk→+∞

then there exists an x0 ∈ H, such that f (x0 ) = inf{ f (x) : x ∈ H}.

2.5

FENCHEL SUBDIFFERENTIAL

Proposition 2.5.1 [BAR1, BAR-PRE] If f : X → (−∞, +∞] is a proper, convex and lower semi-continuous function, then f is bounded from below by an affine function, that means there exists y ∈ X ∗ and µ ∈ R such that f (x) ≥ hy, xi + µ, ∀x ∈ X. We now introduce the notion of subdifferential mapping. Definition 2.5.2 Let f : X → (−∞, +∞], and x ∈ dom f , the subdifferential of f at x, ∂ f (x), is defined by ∂ f (x) = {w ∈ X ∗ : f (y) − f (x) ≥ hw, y − xi, ∀y ∈ X}

(2.1)

The elements v ∈ ∂ f (x) are called the subgradients of f at x, while the (possibly multivalued) map x 7→ ∂ f (x) is called the subdifferential of f . The following proposition is straightforward and easy to see. Proposition 2.5.3 f takes a minimum at x, if and only if 0 ∈ ∂ f (x). The next proposition shows that ∂ f is in fact an extension of the usual notion of derivative for nonsmooth functions. For a proof the reader can consult [FER]. Proposition 2.5.4 If x ∈ int(dom f ) and f is differentiable at x, then ∂ f (x) = {∇ f (x)}. Subdifferential operators are defined even for nonconvex functions. The following proposition provides sufficient conditions for their domain to be nonempty. Proposition 2.5.5 [FER] Let f : X → (−∞, +∞] be a proper, convex and lsc function, then ∂ f (x) is nonempty for each x ∈ int(dom f ).

Convex Analysis and Subdifferential Operators

2.6

19

THE FENCHEL CONJUGATE

Definition 2.6.1 Let f : X → (−∞, +∞] be an arbitrary function (not necessarily convex). The Fenchel conjugate of f is the function f ∗ : X ∗ → [−∞, +∞] defined as f ∗ (x∗ ) := sup{hx∗ , xi − f (x)} x∈X

The conjugate of f is always a convex function. We can also define the biconjugate of f , f ∗∗ : X → [−∞, +∞], as follows: f ∗∗ (x) = sup {hx∗ , xi − f ∗ (x∗ )}. x∗ ∈X ∗

Obviously, from the definitions we have f ∗∗ ≤ f . However, equality holds with the conditions of the following theorem. Theorem 2.6.2 (Fenchel-Moreau) [BRE2] Let f : X → (−∞, +∞] be a proper, convex and lsc function, then f ∗∗ = f . In the following propositions, we state the relations between the conjugate of a function f : X → (−∞, +∞] and its subdifferential. Proposition 2.6.3 [LUC] Let f : X → (−∞, +∞]. Then x∗ ∈ ∂ f (x) if and only if f (x) + f ∗ (x∗ ) = hx∗ , xi. Proposition 2.6.4 [LUC] Let f : X → (−∞, +∞]. If ∂ f (x) , ∅, then f (x) = f ∗∗ (x). If f (x) = f ∗∗ (x), then ∂ f (x) = ∂ f ∗∗ (x). Corollary 2.6.5 [LUC] Let f : X → (−∞, +∞]. Then x∗ ∈ ∂ f (x) ⇒ x ∈ ∂ f ∗ (x∗ ) If f (x) = f ∗∗ (x), then x∗ ∈ ∂ f (x) ⇔ x ∈ ∂ f ∗ (x∗ ).

REFERENCES BAR. V. Barbu, Nonlinear Semigroups and Differential Equations in Banach Spaces, Noordhoff, Leyden, 1976. BAR-PRE. V. Barbu and Th. Precupanu, Convexity and Optimization in Banach Spaces, Editors Academiei, Buchrest, 1986 (and D. Reidel Publishing Company). BRE. H. Br´ezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, Universitext. Springer, New York, 2011. FER. J. Ferrera, An Introduction to Nonsmooth Analysis, Elsevier/Academic Press, Amsterdam, 2014. LUC. R. Lucchetti, Convexity and Well-posed Problems. CMS Books in Mathematics/Ouvrages de Mathmatiques de la SMC, 22. Springer, New York, 2006. MOR. G. Morosanu, Nonlinear Evolution Equations and Applications, Editua Academiei (and D. Reidel Publishing Company), Bucuresti, 1988.

Monotone 3 Maximal Operators 3.1

INTRODUCTION

In this chapter we review some essential properties of maximal monotone operators that we will need in this book. These nonlinear and possibly set-valued operators were first introduced by Minty [MIN], and have important applications in partial differential equations, evolution equations, nonlinear semigroups, optimization and variational inequalities. Maximal monotone operators can be defined in Banach spaces, but for the purpose of this book, we will study them only in Hilbert spaces. The main references for this chapter are [BRE1, BRE2, MOR, BAR1].

3.2

MONOTONE OPERATORS

Definition 3.2.1 A nonlinear possibly set-valued mapping A : H → 2H is said to be (i) monotone if (x∗ − y∗ , x − y) ≥ 0, (ii) α-strongly monotone for α > 0, if (x∗ − y∗ , x − y) ≥ αkx − yk2 , for each x, y ∈ H and x∗ ∈ Ax and y∗ ∈ Ay. A is said to be (iii) strictly monotone if (x∗ − y∗ , x − y) > 0, for each x, y ∈ H such that x , y. The domain of A is defined as D(A) := {x ∈ H : A(x) , ∅}. From now on, we denote a monotone operator by A : D(A) ⊂ H → H, which assigns to each x ∈ D(A), a subset A(x) of H. The range of A is defined as R(A) := {x∗ ∈ H : ∃x ∈ D(A) such that x∗ ∈ A(x)} The graph of a monotone operator A is defined as the following subset of H × H: G(A) := {[x, y] : x ∈ D(A) and y ∈ A(x)}. Each monotone operator is usually identified by its graph. For each monotone operator A, its inverse A−1 whose graph is defined as G(A−1 ) := {[y, x] : [x, y] ∈ A} is also a monotone operator.

22

3.3

Nonlinear Evolution and Difference Equations of Monotone Type

MAXIMAL MONOTONICITY

Definition 3.3.1 A monotone operator A : D(A) ⊂ H → H is said to be maximal monotone if G(A) (the graph of A) is not properly contained in the graph of any other monotone operator in H. Theorem 3.3.2 Let A : D(A) ⊂ H → H be a monotone operator. Then A is maximal monotone if and only if for every λ > 0 (equivalently, for some λ > 0), R(I + λ A) = H. Proof. We first prove the necessity. It suffices to prove this for λ = 1. Given z0 ∈ H, we should find x0 ∈ H such that (y − (z0 − x0 ), x − x0 ) ≥ 0 for all [x, y] ∈ A. Then the maximality of A implies that z0 − x0 ∈ Ax0 . For [x, y] ∈ A, define the weakly compact set Cx,y by Cx,y = {x0 ∈ H : (y + x0 − z0 , x − x0 ) ≥ 0} It suffices to show that the family {Cx,y }[x,y]∈A has the finite intersection property. To this end take [xi , yi ] ∈ A for i = 1, 2, · · · , n. Let ∆ = {(λ1 , · · · , λn ) : λi ≥ 0; ∑ni=1 λi = 1} denote the n-dimensional simplex and consider the function f : ∆ × ∆ → R defined by n

f (λ , µ) = ∑ µi (yi + x(λ ) − z0 , x(λ ) − xi ) i=1

with x(λ ) = ∑ni=1 λi xi . Clearly f is convex and continuous with respect to the first argument and linear with respect to the second one. The Von Neumann Minimax Theorem (see Theorem 1.1 of [BRE1]) implies the existence of λ0 ∈ ∆ such that max f (λ0 , µ) = max min f (λ , µ) ≤ max f (µ, µ). µ∈∆ λ ∈∆

µ∈∆

µ∈∆

Now the monotonicity of A implies n

f (µ, µ) = ∑ µi (yi , x(µ) − xi ) + (x(µ) − z0 , x(µ) − x(µ)) i=1 n

=



µi µ j (yi , x j − xi )

i, j=1 n

=

1 µi µ j (yi − y j , x j − xi ) ≤ 0 2 i,∑ j=1

which shows that f (λ0 , µ) ≤ 0 for all µ ∈ ∆. Choosing for µ the extreme points of ∆, we get (yi +x(λ0 )−z0 , x(λ0 )−xi ) ≤ 0 for all i, which implies that x(λ0 ) ∈ ∩ni=1Cxi ,yi . Conversely, take [u, u∗ ] ∈ H × H such that (u∗ − v∗ , u − v) ≥ 0 for all [v, v∗ ] ∈ A. We shall prove that [u, u∗ ] ∈ A. Since I + A is surjective, there is [w, w∗ ] ∈ A such that w + w∗ = u + u∗ . Then (u∗ − w∗ , u − w) = −ku − wk2 ≥ 0 which implies that u = w and u∗ = w∗ , and therefore [u, u∗ ] ∈ A. 

Maximal Monotone Operators

23

Remark 3.3.3 If A is maximal monotone, then for each x ∈ D(A) A(x) = {y ∈ H; (y − v, x − u) ≥ 0, ∀[u, v] ∈ A} Therefore for each x ∈ D(A), the set A(x) is closed and convex in H. By Theorem 1.1.11, it has an element with minimum norm that we denote by A0 (x). Therefore, for each x ∈ D(A), A0 (x) is identified by the following two properties: (1) A0 (x) ∈ A(x) (2) kA0 (x)k = inf{kyk : y ∈ A(x)}. A0 : D(A) ⊂ H → H is a single-valued monotone operator which is called the “minimal section” of A. Proposition 3.3.4 Assume that A : D(A) ⊂ H → H is maximal monotone. Then A is demiclosed, i.e. if xn → x ∈ H, yn * y ∈ H and [xn , yn ] ∈ A, then [x, y] ∈ A. Proof. Let [xn , yn ] ∈ A, then we have (xn − u, yn − v) ≥ 0,

∀[u, v] ∈ A.

Passing to the limit in this inequality, we get: (x − u, y − v) ≥ 0,

∀[u, v] ∈ A.

Then the maximality of A implies that [x, y] ∈ A. This completes the proof.



If A and B are two monotone operators, then A + B with the domain D(A) ∩ D(B) is defined by (A + B)(x) = {x∗ + y∗ ; x∗ ∈ A(x) and y∗ ∈ B(x)}. It is clear that A + B is also a monotone operator, but it is not necessarily maximal monotone. The following theorem gives a sufficient condition for the maximal monotonicity of A + B. For the proof see [MOR]. Theorem 3.3.5 Let A : D(A) ⊂ H → H and B : D(B) ⊂ H → H be two maximal monotone operators such that 0 ∈ int(D(A) − D(B)), then A + B is also maximal monotone.

3.4

RESOLVENT AND YOSIDA APPROXIMATION

Let A be a maximal monotone operator and λ > 0. We define Jλ := (I +λ A)−1 which is called the resolvent of A, and Aλ := λ1 (I − Jλ ) which is called the Yosida approximation of A. By Theorem 3.3.2, D(Jλ ) = D(Aλ ) = H, ∀λ > 0. In the following theorem, we collect some properties of the resolvent and the Yosida approximation of maximal monotone operators that we will need in this book.

24

Nonlinear Evolution and Difference Equations of Monotone Type

Theorem 3.4.1 Let A : D(A) ⊂ H → H be a maximal monotone operator. Then, for every λ > 0, (1) Jλ is nonexpansive (i.e. Lipschitz with constant 1). (2) Aλ (x) ∈ A(Jλ (x)), ∀x ∈ H. (3) Aλ is monotone and Lipschitz with constant λ1 . (4) kAλ (x)k ≤ kA0 (x)k, ∀x ∈ D(A). (5) limλ →0 Aλ (x) = A0 (x), ∀x ∈ D(A). (6) D(A) is convex. (7) limλ →0 Jλ (x) = ProjD(A) x, ∀x ∈ H. Proof. (1) From the definition of resolvent, we have Jλ (x) − Jλ (y) + λ [A(Jλ (x)) − A(Jλ (y))] 3 x − y Multiplying the above inclusion by Jλ (x) − Jλ (y), and using the monotonicity of A, we get: kJλ (x) − Jλ (y)k2 ≤ (x − y, Jλ (x) − Jλ (y)). This implies (1). (2) Let y = Aλ (x). By the definition of the Yosida approximation, we have: x − λ y = Jλ (x). Now the definition of Jλ implies that y ∈ A(x − λ y) which yields the result. (3) Since Jλ is nonexpansive, then I − Jλ is monotone and therefore Aλ is monotone too. By the definition of Jλ and Aλ we have: x − y = Jλ (x) − Jλ (y) + λ [Aλ (x) − Aλ (y)]. Multiplying both sides by Aλ (x) − Aλ (y), and using the monotonicity of A, we get: λ kAλ (x) − Aλ (y)k2 ≤ (x − y, Aλ (x) − Aλ (y)) which gives the desired result. (4) By the definitions of Jλ and Aλ , we have Aλ (x) = λ −1 (Jλ (x + λ A0 (x)) − Jλ (x)), ∀x ∈ D(A), λ > 0. Now (1) yields the result. (5) Let x ∈ D(A). By Part (4), and Theorem 1.3.2, there exists a sequence λn → 0 such that Aλn (x) converges weakly to a point p. The monotonicity of A implies that (Aλn (x) − v, Jλn (x) − u) ≥ 0, ∀[u, v] ∈ A. By (4), Jλn (x) → x, therefore the above inequality implies that: (p − v, x − u) ≥ 0, ∀[u, v] ∈ A. Since A is maximal, p ∈ A(x). Again (4) implies that kpk ≤ lim sup kAλ (x)k ≤ kA0 (x)k. λ →0

Maximal Monotone Operators

25

Therefore p = A0 (x). Now Proposition 1.2.6 of Chapter 1 implies that Aλ (x) → A0 (x) as λ → 0. (6) and (7) Let C = conv(D(A)) and x ∈ H. The monotonicity of A implies that: (Aλ (x) − v, Jλ (x) − u) ≥ 0, ∀[u, v] ∈ A. Now by the definition Aλ we get: (x − Jλ (x) − λ v, Jλ (x) − u) ≥ 0, ∀[u, v] ∈ A This yields kJλ (x)k2 ≤ (x − λ v, Jλ (x) − u) + (Jλ (x), u), ∀[u, v] ∈ A

(3.1)

Now, let us choose a sequence λn → 0 such that Jλn (x) converges weakly to q. Taking the limit in the above inequality we get: kqk2 ≤ lim sup kJλn (x)k2 ≤ (x, q − u) + (q, u), ∀u ∈ D(A) n→+∞

i.e. (x − q, u − q) ≤ 0, ∀u ∈ D(A). This inequality obviously holds for all u ∈ C. Theorem 1.2.2 shows that C is weakly closed and since Jλ (x) ∈ C, we deduce that q ∈ C. Therefore by Theorem 1.1.11, the inequality (x − q, u − q) ≤ 0, ∀u ∈ C implies that q = ProjC x. Since the sequence λn is arbitrary, we see that Jλ (x) * ProjC x as λ → 0. Now (3.1) yields lim sup kJλ (x)k2 ≤ (x, q − u) + (q, u), ∀u ∈ C. λ →0

By taking u = q, we get lim sup kJλ (x)k ≤ kqk. λ →0

Now Theorem 1.2.6 implies that Jλ (x) → ProjC x as λ → 0. Finally, we prove that C = D(A). Since Jλ (x) ∈ D(A), ∀x ∈ H, and Jλ (z) → z for all z ∈ C, we see that C ⊂ D(A). Therefore D(A) = C.  An important class of maximal monotone operators consists of the subdifferential of proper, convex and lower semicontinuous functions. This important result was proved by Rockafellar in [ROC]. Theorem 3.4.2 Let ϕ : H → (−∞, +∞] be a proper, convex and lsc function. Then (1) ∂ ϕ is a maximal monotone operator. (2) D(ϕ) = D(∂ ϕ).

26

Nonlinear Evolution and Difference Equations of Monotone Type

Proof. (1) Monotonicity is easy. We prove the maximality. By Theorem 3.3.2, it suffices to show that R(I + ∂ ϕ) = H. Let x∗ ∈ H and consider the convex function 1 ˜ ϕ(x) = kxk2 + ϕ(x) − (x, x∗ ) 2 which is also proper and lsc from H to (−∞, +∞]. By Theorem 2.5.1, ˜ lim ϕ(x) = +∞.

kxk→∞

˜ 0 ) ≤ ϕ(x), ˜ Hence by Theorem 2.4.3, there exists x0 ∈ D(ϕ) such that ϕ(x ∀x ∈ H. Then 1 1 ϕ(x0 ) − ϕ(x) ≤ (x∗ , x0 − x) + kxk2 − kx0 k2 2 2 ≤ (x∗ − x, x0 − x), ∀x ∈ H. Now taking x = tx0 + (1 − t)v, 0 < t < 1, v ∈ H in this inequality, and using the convexity of ϕ, we obtain ϕ(x0 ) − ϕ(v) ≤ (x∗ − x0 , x0 − v) + (1 − t)kx0 − vk2 . Taking the limit as t → 1, in this inequality, we get ϕ(x0 ) − ϕ(v) ≤ (x∗ − x0 , x0 − v), ∀v ∈ H, which means that x∗ ∈ x0 + ∂ ϕ(x0 ), and proves the surjectivity of I + ∂ ϕ, completing the proof. (2) Let x ∈ D(ϕ). We know that Jλ x ∈ D(∂ ϕ). On the other hand, by Part (7) of Theorem 3.4.1, Jλ (x) → x as λ → 0. Therefore x ∈ D(∂ ϕ) which shows that D(ϕ) ⊂ D(∂ ϕ). The reverse inclusion is trivial because D(∂ ϕ) ⊂ D(ϕ).  Now we define the resolvent and Yosida approximation of a proper, convex and lsc function ϕ. Definition 3.4.3 Let ϕ : H → (−∞, +∞] be a proper, convex and lsc function. The Moreau-Yosida approximation of the function ϕ is defined by: ϕλ (x) = inf{

1 kx − yk2 + ϕ(y); y ∈ H}, x ∈ H, λ > 0. 2λ

Theorem 3.4.4 Let ϕ : H → (−∞, +∞] be a proper, convex and lsc function, and A = ∂ ϕ. Then for each λ > 0, ϕλ is convex and 1 (1) ϕλ (x) = 2λ kx − Jλ (x)k2 + ϕ(Jλ (x)), ∀x ∈ H, ∀λ > 0, where Jλ is the resolvent of A. (2) ϕ(Jλ (x)) ≤ ϕλ (x) ≤ ϕ(x), ∀x ∈ H, ∀λ > 0. (3) limλ →0 ϕλ (x) = ϕ(x), ∀x ∈ H. (4) Aλ = ∂ ϕλ , ∀λ > 0, where Aλ is the Yosida approximation of A. (5) ϕλ is differentiable on H.

Maximal Monotone Operators

27

Proof. Let λ > 0 and x ∈ H. Consider the function ψ(y) =

1 kx − yk2 + ϕ(y), y ∈ H. 2λ

Since by Theorem 2.5.1, ϕ has an affine minorant, we have lim ψ(y) = +∞

kyk→∞

Then ψ achieves its minimum and therefore ϕλ is finite at each point of H. Convexity of ϕλ is easy to check. (1) We have 1 (y − x) + ∂ ϕ(y) ⊂ ∂ ψ(y), ∀y ∈ D(∂ ψ) λ On the other hand y = Jλ (x) is the unique solution of 0∈

1 (y − x) + ∂ ϕ(y). λ

Therefore 0 ∈ ∂ ψ(Jλ (x)). Then by Proposition 2.5.3, Jλ (x) is the minimizer of ψ. (2) is a consequence of (1) and the definition of ϕλ . (3) We first prove this for x ∈ D(ϕ). By Part (7) of Theorem 3.4.1, for each x ∈ D(ϕ) limλ →0 Jλ x = x. Now the result follows from the lower semicontinuity of ϕ, and by taking the limit as λ → 0 in (2). To complete the proof, we should show that for every x < D(ϕ), limλ →0 ϕλ (x) = +∞. Suppose to the contrary there is a sequence λn → 0 such that ϕλn (x) ≤ C < +∞. (3.2) Then by Part (2), ϕ(Jλn (x)) ≤ C.

(3.3)

On the other hand, since by Theorem 2.5.1, ϕ has an affine minorant and {Jλn (x)} is a bounded sequence, then {ϕ(Jλn (x))} is also a bounded sequence. This fact, together with (3.2), (3.3) and Part (1) imply that Jλn (x) → x. Letting n → +∞ in (3.3), the lower semicontinuity of ϕ implies that x ∈ D(ϕ), which is a contradiction. (4) By Part (1) 1 1 kx − Jλ (x)k2 − ky − Jλ (y)k2 + ϕ(Jλ (x)) − ϕ(Jλ (y)) 2λ 2λ 1 1 ≤ kx − Jλ (x)k2 − ky − Jλ (y)k2 2λ 2λ 1 + (x − Jλ (x), Jλ (x) − Jλ (y)) λ 1 = (x − Jλ (x), x − y) − kx − yk2 λ

ϕλ (x) − ϕλ (y) =

28

Nonlinear Evolution and Difference Equations of Monotone Type + 2(x − y, Jλ (x) − Jλ (y)) − kJλ (x) − Jλ (y)k2 1 ≤ (x − Jλ (x), x − y). λ

Therefore we showed that: ϕλ (x) − ϕλ (y) ≤

1 (x − Jλ x, x − y), ∀y ∈ H λ

which is the desired result by the definition of Aλ . (5) By (4), ϕλ (y) − ϕλ (x) ≤ (Aλ (y), y − x), ∀x, y ∈ H, λ > 0 By Part (3) of Theorem 3.4.1, we have: 0 ≤ ϕλ (y) − ϕλ (x) − (Aλ (x), y − x) ≤ (Aλ (y) − Aλ (x), y − x) 1 ≤ kx − yk2 , ∀x, y ∈ H, λ > 0. λ Consequently, for every λ > 0, ϕλ is Fr´echet differentiable on H and ∂ ϕλ (x) = ∇ϕλ (x) = Aλ (x), ∀x ∈ H.  Remark 3.4.5 By Part 1 of Theorem 3.4.4, Jλ (x) = Argmin{

1 kx − yk2 + ϕ(y); y ∈ H}, 2λ

and it is called the resolvent of the convex function ϕ. The following result is important in the theory of nonlinear operators [ROC], and its proof can be found in [BAR1]. Proposition 3.4.6 Let A = ∂ f , where f : H → (−∞, +∞] is a proper, convex and lower semi-continuous function. Then the following assertions are equivalent: (a) lim f (x) /kxk = +∞; kxk→∞ x∈D( f )

(b) R (A) = H and A−1 is bounded.

3.5

CANONICAL EXTENSION

Let Ω be a Lebesgue measurable set in Rn , n ≥ 1 and A : D(A) ⊂ H → H be a ¯ ⊂ L2 (Ω; H) → L2 (Ω; H) by maximal monotone operator. We define A¯ : D(A) ¯ = {u ∈ L2 (Ω; H); ∃v ∈ L2 (Ω; H) such that v(ζ ) ∈ A(u(ζ )), a.e. ζ ∈ Ω}, D(A)

Maximal Monotone Operators

29

¯ ¯ A(u) = {v ∈ L2 (Ω; H); v(ζ ) ∈ A(u(ζ )), a.e. ζ ∈ Ω}, ∀u ∈ D(A). If either Ω has finite measure or 0 ∈ A(0), then A¯ is maximal monotone. Monotonicity is easy to see, and for maximality, we note that under either assumption above, ¯ for each f ∈ L2 (Ω; H), we have u(t) = (I + A)−1 ( f (t)) ∈ L2 (Ω; H) and u + A(u) 3 f. 2 ¯ The operator A is called the canonical extension of A to L (Ω; H) or the realization of A on L2 (Ω; H). The canonical extensions of (I + λ A)−1 and Aλ to L2 (Ω; H) are ¯ −1 and (A) ¯ λ. (I + λ A) Theorem 3.5.1 Let U ⊂ Rn with µ(U) < +∞ and let ϕ : H → (−∞, +∞] be a proper, convex and lsc function. Let the function Φ : L2 (U; H) → (−∞, +∞] be defined by (R ϕ(u(s))ds, ϕ(u) ∈ L1 (U) Φ(u) = U +∞, otherwise R

(Since ϕ is bounded from below by an affine function, it follows that U ϕ(u(s))ds ¯ where cannot assume the value −∞). Then Φ is proper, convex and lsc and ∂ Φ = A, A¯ denotes the canonical extension of A = ∂ ϕ to L2 (U; H). Moreover, we have Z

Φλ (u) = and

L2 (U;H)

D(Φ)

U

ϕλ (u(s))ds, ∀λ > 0, ∀u ∈ L2 (U; H) H

= {u ∈ L2 (U; H); u(s) ∈ D(ϕ) , a.e. s ∈ U}

(3.4)

(3.5)

Proof. Obviously Φ is convex, and since D(Φ) contains at least the constant function u ≡ h, with h ∈ D(ϕ), Φ is proper. To show that Φ is lsc on L2 (U; H), we prove that for every r ∈ R, the sub-level set {u ∈ L2 (U; H) : Φ(u) ≤ r} is closed in L2 (U; H). To this end, let {um } be a sequence such that Φ(um ) ≤ r and um → u in L2 (U; H). We may assume that um (t) → u(t), a.e. t ∈ U. Now take [x0 , y0 ] ∈ ∂ ϕ and define the function ˜ ϕ(x) = ϕ(x) − ϕ(x0 ) − (y0 , x − x0 ). ˜ Then ϕ˜ is convex, lsc and ϕ˜ ≥ 0 on H. Moreover for each u ∈ L2 (U; H), ϕ(u) is measurable. Hence, by making use of Fatou’s lemma, we have Z

lim inf

m→+∞ U

˜ m (t))dt ≥ ϕ(u

Z

˜ ϕ(u(t))dt.

U

This obviously implies that Φ(u) ≤ r and therefore Φ is lsc. To show that A¯ = ∂ Φ, it ¯ Then v(t) ∈ A(u(t)), a.e. t ∈ U and so suffices to show that A¯ ⊂ ∂ Φ. Let [u, v] ∈ A. ϕ(u(t)) − ϕ(w(t)) ≤ (v(t), u(t) − w(t)), a.e. t ∈ U, ∀w ∈ D(Φ) This shows that ϕ(u) ∈ L1 (U) and Φ(u) − Φ(w) ≤

Z U

(v(t), u(t) − w(t))dt, ∀w ∈ D(Φ)

30

Nonlinear Evolution and Difference Equations of Monotone Type

and hence v ∈ ∂ Φ(u). Since [u, v] is arbitrary, we have A¯ ⊂ ∂ Φ. Now we prove (3.4). 1 ¯ −1 (u)k2 2 ¯ −1 ku − (I + λ A) L (U;H) + Φ((I + λ A) (u)) 2λ Z Z 1 ku(t) − (I + λ A)−1 (u(t))k2 dt + ϕ((I + λ A)−1 (u(t)))dt = 2λ U U

Φλ (u) =

Z

= U

ϕλ (u(t))dt, ∀u ∈ L2 (U; H).

Finally, we prove (3.5). We know that: D(Φ) ⊂ {u ∈ L2 (U; H); u(t) ∈ D(ϕ), a.e. t ∈ U}. Therefore it suffices to show that for any u ∈ L2 (U; H) with u(t) ∈ D(ϕ), a.e. t ∈ U, we have u ∈ D(Φ) (here the closure is with respect to L2 (U; H)). Let u be such a function and set uλ (t) = (I + λ A)−1 (u(t)), λ > 0, a.e. t ∈ U Then, by Theorems 3.4.1 and 3.4.2, we get lim kuλ (t) − u(t)k = 0, a.e. t ∈ U

λ →0

(3.6)

On the other hand, since (I + λ A)−1 is nonexpansive, kuλ (t)k ≤ ku(t)k +C, a.e. t ∈ U

(3.7) L2 (U; H).

where C is a constant. By Theorem 1.4.5 and (3.7) and (3.6), uλ → u in ¯ −1 (u) ∈ D(A), ¯ this implies that u ∈ D(Φ) in L2 (U; H), which Since uλ = (I + λ A) shows that (3.5) holds. 

REFERENCES BAR. V. Barbu, Nonlinear semigroups and Differential Equations in Banach Spaces, Noordhoff, Leiden, 1976. BAR-PRE. V. Barbu and Th. Precupanu, Convexity and Optimization in Banach Spaces, D. Reidel Publishing Co., Dordrecht; Editura Academiei Bucharest, 1986. BRE1. H. Br´ezis, Op´erateurs maximaux monotones et semi-groupes de contractions dans les espaces de Hilbert (French) North-Holland Mathematics Studies, No. 5. Notas de Matem´atica (50). North-Holland Publishing Co., Amsterdam-London; American Elsevier Publishing Co., Inc., New York, 1973. BRE2. H. Br´ezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, Universitext. Springer, New York, 2011. MOR. G. Morosanu, Nonlinear Evolution Equations and Applications, D. Reidel Publishing Co., Dordrecht; Editura Academiei Bucharest, 1988. MIN. G. J. Minty, Monotone (nonlinear) operators in Hilbert space, Duke Math. J. 29 (1962) 341–346. ROC. R. T. Rockafellar, On the maximal monotonicity of subdifferential mappings, Pacific J. Math. 33 (1970), 209–216.

Part II Evolution Equations of Monotone Type

Order Evolution 4 First Equations 4.1

INTRODUCTION

In this chapter we consider the nonhomogeneous Cauchy problem for a maximal monotone operator of the form ( −u0 (t) ∈ A(u(t)) + f (t), a.e. on (0, +∞) u(0) = u0 ∈ D(A) and study the existence, uniqueness, periodicity and the asymptotic behavior of the solutions. In the next section we concentrate on the existence and uniqueness of solutions. After which study the existence of periodic solutions when f (t) is a periodic function, and A = ∂ ϕ, where ϕ is a proper, convex and lsc function on H. Here, we also study the asymptotic convergence of bounded solutions to periodic solutions. Solutions to the above Cauchy problem in the homogeneous case form a semigroup of nonexpansive mappings whose infinitesimal generator is the maximal monotone operator A. In the nonhomogeneous case, the trajectory of the solutions are almost nonexpansive curves (as will be defined later). Therefore, in order to study the asymptotic behavior of the solutions, we study the asymptotic behavior of nonexpansive and almost nonexpansive curves. The material in this chapter is generally adapted from [MOR, DJA1, DJA2, DJA3, DJA4, KHA-MOH].

4.2

EXISTENCE AND UNIQUENESS OF SOLUTIONS

Consider the following Cauchy problem: du + A(u) 3 f (t), 0 < t < T dt

(4.1)

u(0) = u0

(4.2)

where A : D(A) ⊂ H → H is a (possibly multi-valued) operator, u0 ∈ H and f is a given function. First we introduce two concepts of solutions. Definition 4.2.1 Let f ∈ L1 (0, T ; H). The function u ∈ C([0, T ]; H) is called a strong solution (or breifly, solution) of (4.1) and (4.2) if: (1) u is absolutely continuous on each compact subinterval of (0, T ). (2) u(t) ∈ D(A) for almost every t ∈ (0, T ). (3) u(0) = u0 and u satisfies (4.1) for a.e. t ∈ (0, T ).

34

Nonlinear Evolution and Difference Equations of Monotone Type

Definition 4.2.2 A function u ∈ C([0, T ]; H) is a weak solution for (4.1) and (4.2) if there exist sequences {un } ⊂ W 1,∞ (0, T ; H) and { fn } ⊂ L1 (0, T ; H) such that: n (1) du dt (t) + A(un (t)) 3 f n (t), a.e. t ∈ [0, T ), n = 1, 2, · · · (2) un → u in C([0, T ]; H) (3) u(0) = u0 , fn → f in L1 (0, T ; H) Now we state a Gronwall type Lemma that we will need in the sequel. Lemma 4.2.3 Suppose that ψ ∈ L1 (a, b) (−∞ < a < b < +∞) with ψ(t) ≥ 0 for a.e. t ∈ (a, b), and let C be a real constant. If h ∈ C[a, b] satisfies the following inequality 1 2 1 h (t) ≤ C2 + 2 2 then |h(t)| ≤ |C| +

Z t

ψ(s)h(s)ds, ∀t ∈ [a, b]

a

Z t

ψ(s)ds, ∀t ∈ [a, b].

a

Proof. Set g(t) = 12 C2 +

Rt

a ψ(s)h(s)ds.

Then obviously

g0 (t) = ψ(t)h(t) ≤ ψ(t)|h(t)|, for a.e. t ∈ (a, b). p From the assumption we have |h(t)| ≤ 2g(t), therefore: p g0 (t) ≤ ψ(t) 2g(t) p Dividing both sides of the above inequality by 2g(t) and then integrating from a to t, we get: Z t Z t g0 (s) p ds ≤ ψ(s)ds a a 2g(s) This shows that: 1

1

(2g(t)) 2 ≤ (2g(a)) 2 +

Z t

ψ(s)ds = |C| +

a

Since by assumption |h(t)| ≤

Z t

ψ(s)ds a

p 2g(t), we get: |h(t)| ≤ |C| +

Z t

ψ(s)ds, a

which is the desired result.



Theorem 4.2.4 Suppose that A : D(A) ⊂ H → H is maximal monotone, u0 ∈ D(A) and f ∈ W 1,1 (0, T ; H), then the Problem (4.1) and (4.2) has a unique strong solution u ∈ W 1,∞ (0, T ; H). Moreover, u is differentiable from the right at each point of [0, T ) and d+u (t) = ( f (t) − A(u(t)))o , ∀t ∈ [0, T ) (4.3) dt

First Order Evolution Equations

35

t df d+u k (t)k ≤ k( f (0) − A(u0 ))o k + k (s)kds, ∀t ∈ [0, T ) (4.4) dt ds 0 where ( f (t)−A(u(t)))o is the least norm element of the set f (t)−A(u(t)). Moreover, if u, v are solutions corresponding to (u0 , f ) and (v0 , g) in D(A) × W 1,1 (0, T ; H), then: Z t ku(t) − v(t)k ≤ ku0 − v0 k + k f (s) − g(s)kds, 0 ≤ t ≤ T (4.5)

Z

0

Proof. First we prove (4.5) which implies the uniqueness of the solution. It follows from the monotonicity of A that: 1 d ku(t) − v(t)k2 ≤ k f (t) − g(t)kku(t) − v(t)k, a.e. t ∈ (0, T ) 2 dt By integrating this inequality on [0,t] we obtain: 1 1 ku(t) − v(t)k2 ≤ ku0 − v0 k2 + 2 2

Z t

k f (s) − g(s)kku(s) − v(s)kds, ∀t ∈ [0, T ].

0

Now an application of Lemma 4.2.3 implies (4.5). Next we show the existence of the solution to problem (4.1) and (4.2). A classical result from the theory of ordinary differential equations shows the existence of a unique solution uλ ∈ C1 ([0, T ]; H) for the Cauchy problem duλ (t) + Aλ (uλ (t)) = f (t), 0 ≤ t ≤ T dt

(4.6)

uλ (0) = u0

(4.7)

for each λ > 0, where Aλ is the Yosida approximation of A, because Aλ is Lipschitz continuous on H. Now since Aλ is monotone we get: 1 d ku (t + h) − uλ (t)k2 ≤ k f (t + h) − f (t)kkuλ (t + h) − uλ (t)k, ∀t,t + h ∈ [0, T ] 2 dt λ and therefore integrating this inequality from 0 to t, we obtain: 1 1 ku (t +h)−uλ (t)k2 ≤ kuλ (h)−u0 k2 + 2 λ 2

Z t 0

k f (s+h)− f (s)kkuλ (s+h)−uλ (s)kds,

for all t,t + h ∈ [0, T ]. By Lemma 4.2.3, we get kuλ (t + h) − uλ (t)k ≤ kuλ (h) − u0 k +

Z t

k f (s + h) − f (s)kds,

(4.8)

0

for all t,t + h ∈ [0, T ]. Now if we divide (4.8) by h > 0 and take the limit as h → 0, we get: k

duλ (t)k ≤ k f (0)−Aλ (u0 )k+ dt

Z t df

k

0

ds

(s)kds ≤ k f (0)k+kAo (u0 )k+

Z T df

k

0

ds

(s)kds, (4.9)

36

Nonlinear Evolution and Difference Equations of Monotone Type

for 0 ≤ t ≤ T . Therefore by (4.6) kAλ (uλ (t))k ≤ k f (t)k + k f (0)k + kAo (u0 )k +

Z T df

k

0

ds

(s)k ≤ Const, 0 ≤ t ≤ T

(4.10) Next we prove the convergence of uλ in C([0, T ]; H) as λ → 0. To this aim, multiplying the equation duµ duλ (t) − + Aλ (uλ (t)) − Aµ (uµ (t)) = 0, λ , µ > 0, t ∈ [0, T ] dt dt by uλ (t) − uµ (t), we get: 1 d ku (t) − uµ (t)k2 = 2 dt λ  − Aλ (uλ (t)) − Aµ (uµ (t)), λ Aλ (uλ (t)) − µAµ (uµ (t))  − Aλ (uλ (t)) − Aµ (uµ (t)), Jλ (uλ (t)) − Jµ (uµ (t)) , 0 ≤ t ≤ T

(4.11)

Now it follows easily from (4.10) and (4.11) and the fact that  Aλ (uλ (t)) ∈ A Jλ (uλ (t)) , that

d ku (t) − uµ (t)k2 ≤ C(λ + µ), ∀λ , µ > 0 0 ≤ t ≤ T dt λ for some constant C. Therefore 1

1

kuλ (t) − uµ (t)k ≤ ct 2 (λ + µ) 2 , 0 ≤ t ≤ T

(4.12)

1

where c = C 2 . But (4.12) states that the net {uλ } is Cauchy and therefore there is u ∈ C([0, T ]; H) such that uλ → u as λ → 0 in C([0, T ]; H)

(4.13)

It follows from (4.9) that u ∈ W 1,∞ (0, T ; H) and duλ du → dt dt

(4.14)

in the weak star topology of L∞ (0, T ; H) as λ → 0. But then ∗

Aλ (uλ ) * f −

du dt

(4.15)

in L∞ (0, T ; H) as λ → 0. On the other hand from (4.10) we can see that Jλ (uλ ) → u in C([0, T ]; H)

(4.16)

First Order Evolution Equations

37

Denote the extension of A in L2 (0, T ; H) by A¯ which is a maximal monotone operator; then A¯ is demiclosed. This fact together with (4.15) and (4.16) implies that f−

du ¯ ∈ A(u) dt

i.e. u satisfies (4.1) almost everywhere. Moreover, (4.7) and (4.13) show that u(0) = u0 , and (4.16) implies that u(t) ∈ D(A), ∀t ∈ (0, T ] because A is demiclosed. Now we prove that u has right derivatives and satisfies (4.3). Suppose that t0 is an arbitrary point in [0, T ). Multiplying both sides of d (u(t0 + h) − u(t0 )) ∈ f (t0 + h) − A(u(t0 + h)) dh for almost every h > 0 with t0 + h < T , by u(t0 + h) − u(t0 ), then by the monotonicity of A, we get the following inequality: 1 d ku(t0 + h) − u(t0 )k2 ≤ {k( f (t0 ) − A(u(t0 )))o k + k f (t0 + h) − f (t0 )k}ku(t0 + h) − u(t0 )k 2 dh

Now intergrating this inequality on [0, h] and applying Lemma 4.2.3, we get: ku(t0 + h) − u(t0 )k ≤ hk( f (t0 ) − A(u(t0 )))o k +

Z h 0

k f (t0 + s) − f (t0 )kds, ∀h > 0 (4.17)

with t0 + h < T . This gives 1 lim sup ku(t0 + h) − u(t0 )k ≤ k( f (t0 ) − A(u(t0 )))o k + h→ 0 h

(4.18)

On the other hand since u is a strong solution of (4.1) and (4.2), for each [x, y] ∈ A, we have 1 1 ku(t) − xk2 ≤ ku(s) − xk2 + 2 2

Z t

( f (s) − y, u(s) − x)ds 0 ≤ s ≤ t ≤ T

0

This implies that 1 1 (u(t0 + h) − u(t0 ), u(t0 ) − x) ≤ ku(t0 + h) − xk2 − ku(t0 ) − xk2 2 2 ≤

Z t0 +h

( f (s) − y, u(s) − x)ds,

(4.19) (4.20)

t0

for all [x, y] ∈ A and for all h > 0 such that t0 + h < T . Now taking into acount (4.18), we can choose a subsequence hn → 0 such that u(t0 + hn ) − u(t0 ) * p, in H. hn

38

Nonlinear Evolution and Difference Equations of Monotone Type

Now taking h = hn in (4.20) and dividing the resulting inequality by hn and taking the limit as n → +∞, we conclude that (p − f (t0 ) + y, x − u(t0 )) ≥ 0, ∀[x, y] ∈ A Therefore by using (4.18) and the maximality of A we see that p = ( f (t0 ) − A(u(t0 )))o

(4.21)

Therefore p does not depend on the choice of the sequence hn and so u(t0 + h) − u(t0 ) * ( f (t0 ) − A(u(t0 )))o as h → 0. h But this fact together with (4.18) shows that u is right differentiable at t0 and d+u (t0 ) = ( f (t0 ) − A(u(t0 )))o dt i.e. u satisfies (4.3). A similar argument gives ku(t + h) − u(t)k ≤ ku(h) − u0 k + ku(h) − u0 k ≤

Z h 0

Z t

k f (s + h) − f (s)kds, ∀t,t + h ∈ [0, T ] (4.22)

0

k( f (s) − A(u0 ))o kds, ∀h ∈ [0, T ]

(4.23)

Now from estimates (4.22) and (4.23) we conclude that: ku(t + h) − u(t)k ≤

Z h 0

k( f (s) − A(u0 ))o kds +

Z t

k f (s + h) − f (s)kds, ∀t,t + h ∈ [0, T ]

0

and this obviously implies (4.4), which completes the proof of the theorem.



Theorem 4.2.5 Suppose that A : D(A) ⊂ H → H is maximal monotone and u0 ∈ D(A) and f ∈ L1 (0, T ; H). Then there exists a unique weak solution of (4.1) and (4.2) satisfying 1 1 ku(t) − xk2 ≤ ku(s) − xk2 + 2 2

Z t

( f (s) − y, u(s) − x)ds,

(4.24)

s

for all 0 ≤ s ≤ t ≤ T and [x, y] ∈ A. Moreover, if u and v are weak solutions corresponding to (u0 , f ), (v0 , g) ∈ D(A) × L1 (0, T ; H), then the following inequality holds 1 1 ku(t) − v(t)k2 ≤ ku(s) − v(s)k2 + 2 2

Z t s

( f (r) − g(r), u(r) − v(r))dr, 0 ≤ s ≤ t ≤ T (4.25)

First Order Evolution Equations

39

Proof. Suppose that u0 ∈ D(A) and f ∈ L1 (0, T ; H). Then there exist sequences {un0 } ⊂ D(A) and { fn } ⊂ W 1,1 (0, T ; H) such that un0 → u0 in H and fn → f in L1 (0, T ; H). Now from Theorem 4.2.4, we know that for each n ∈ N, un ∈ W 1,∞ (0, T ; H) exists such that ( dun dt (t) + A(un (t)) 3 f n (t), a.e. t ∈ (0, T ) (4.26) un (0) = un0 Moreover estimate (4.5) implies that: kun (t) − um (t)k ≤ kun0 − um 0 k+

Z T 0

k fn (s) − fm (s)kds

Therefore u ∈ C([0, T ]; H) exists such that un → u in C([0, T ]; H) and in particular u(0) = u0 . Obviously the function u is only a weak solution of (4.1) and (4.2). It remains to prove (4.24) and (4.25). Obviously these hold for strong solutions, and therefore passing to the limit, they also hold for weak solutions. Note that uniqueness is an immediate consequence of (4.25), and the proof is now complete. 

4.3

PERIODIC FORCING

Lemma 4.3.1 Suppose that ϕ : H → (−∞, +∞] is a proper, convex and lsc function and u ∈ W 1,2 (t0 , T ; H) such that u(t) ∈ D(∂ ϕ), a.e. t ∈ (t0 , T ) and g ∈ L2 (t0 , T ; H) exists such that g(t) ∈ ∂ ϕ(u(t)), a.e. t ∈ (0, T ) where −∞ < t0 < T < +∞. Then the function t 7→ ϕ(u(t)) is absolutely continuous on [t0 , T ] and for almost every t ∈ (t0 , T ) we have d du ϕ(u(t)) = (h, (t)), ∀h ∈ ∂ ϕ(u(t)) dt dt

(4.27)

Proof. Obviously, for each λ > 0 the function t 7→ ϕλ (u(t)) is differentiable a.e. on 1 (t0 , T ) (remember from Chapter 2, ϕλ (x) := infy∈H {ϕ(y) + 2λ kx − yk2 }, which is called the Yosida regularization of ϕ) and d du ϕλ (u(t)) = (∇ϕλ (u(t)), (t)) dt dt du = ((∂ ϕ)λ (u(t)), (t)), a.e. t ∈ (t0 , T ) dt Therefore ϕλ (u(t)) − ϕλ (u(s)) =

Z t s

((∂ ϕ)λ (u(r)),

du (r))dr, s,t ∈ [t0 , T ] dr

Note that for almost all t ∈ (t0 , T ) lim (∂ ϕ)λ (u(t)) = (∂ ϕ)o (u(t))

λ →0

(4.28)

40

Nonlinear Evolution and Difference Equations of Monotone Type

and k(∂ ϕ)λ (u(t))k ≤ k(∂ ϕ)o (u(t))k ≤ kg(t)k consequently ∂ ϕλ (u) → (∂ ϕ)o (u) in L2 (t0 , T ; H), therefore by taking the limit in (4.28) as λ → 0, we get ϕ(u(t)) − ϕ(u(s)) =

Z t

((∂ ϕ)o (u(r)),

s

du (r))dr, s,t ∈ [t0 , T ] dr

This implies that the function t 7→ ϕ(u(t)) is absolutely continuous on [t0 , T ]. Now choose t1 ∈ [t0 , T ] such that both u(t) and ϕ(u(t)) are differentiable at t = t1 and u(t1 ) ∈ D(∂ ϕ), then for each h ∈ ∂ ϕ(u(t1 )), we have ϕ(u(t1 )) − ϕ(v) ≤ (h, u(t1 ) − v), ∀v ∈ H. Now taking v = u(t1 ± ε) in the above inequality where ε > 0, then dividing the resulting inequality by ε and taking the limit as ε → 0, we get d du ϕ(u(t)) |t=t1 = (h, (t1 )), dt dt 

which completes the proof.

Lemma 4.3.2 Suppose that u is a weak solution of (4.1) with A = ∂ ϕ and f ∈ L2 (0, T ; H), then Z T

tk 0

du 2 k dt ≤ dt

Z T 0

tk f (t)k2 dt + 2(ku0 − x0 k +

Z T

k f (t)kdt)2 ,

0

where x0 ∈ (∂ ϕ)−1 (0) ˜ Proof. Take [x0 , y0 ] ∈ ∂ ϕ. We define ϕ(x) = ϕ(x) − ϕ(x0 ) − (y0 , x − x0 ). Then problem (4.1) is equivalent to: du ˜ + ∂ ϕ(u) 3 f (t) − y0 , 0 < t < T dt Hence without loss of generality we may assume that min ϕ(x) = ϕ(x0 ) = 0 x∈H

and so ϕ(x) ≥ 0. Multiplying both sides of (4.1) by t du dt and using Lemma 4.3.1, we get du d du tk (t)k2 + t ϕ(u(t)) = t( f (t), (t)), a.e. t ∈ (0, T ) dt dt dt Then after integration from 0 to T , we obtain Z T

tk 0

du 2 k dt + T ϕ(u(T )) = dt

Z T

t( f (t), 0

du (t))dt + dt

Z T

ϕ(u(t))dt 0

First Order Evolution Equations

41

Since ϕ ≥ 0, a straightforward computation shows that Z T

tk 0

du (t)k2 dt ≤ dt

Z T

tk f (t)k2 dt + 2

Z T

0

ϕ(u(t))dt

(4.29)

0

On the other hand by (4.1), we have ϕ(u(t)) ≤ ( f (t) −

du (t), u(t) − x0 ), a.e. t ∈ (0, T ) dt

Therefore Z T 0

1 ϕ(u(t))dt ≤ ku0 − x0 k2 + 2

Z T 0

k f (t)kku(t) − x0 kdt

(4.30)

Now multiplying (4.1) by u(t) − x0 and integrating on [0,t], by using Lemma 4.2.3, we get ku(t) − x0 k ≤ ku0 − x0 k +

Z T

k f (t)kdt, 0 ≤ t ≤ T

(4.31)

0

From (4.30) and (4.31), we get Z T 0

ϕ(u(t))dt ≤ (ku0 − x0 k +

Z T

k f (t) |dt)2

(4.32)

0

Now (4.29) and (4.32) imply that Z T 0

du tk k2 dt ≤ dt

Z T 0

2

tk f (t)k dt + 2(ku0 − x0 k +

Z T

k f (t)kdt)2 ,

(4.33)

0

which is the desired inequality.



Theorem 4.3.3 Suppose that A = ∂ ϕ where ϕ : H → (−∞, +∞] is a proper, convex 2 (R+ ; H) is a T −periodic function (T > 0). Then the and lsc function and f ∈ Lloc Equation (4.1) has a bounded solution if and only if it has at least a periodic solution with period T > 0. In this case, all solutions of (4.1) are bounded on R+ , and for each solution z(t), t ≥ 0 there is a T -periodic solution p of (4.1) such that z(t) − p(t) * 0, as t → +∞.

(4.34)

Moreover, any two periodic solutions differ by a constant and dzn dp → , as n → +∞ dt dt

(4.35)

in L2 (0, T ; H), where zn (t) = z(t + nT ), n = 1, 2, · · · . Proof. First note that if z(t), t ≥ 0 is a bounded solution of (4.1) on R+ , then every other solution v(t), t ≥ 0 is also bounded because kz(t) − v(t)k ≤ kz(0) − v(0)k.

42

Nonlinear Evolution and Difference Equations of Monotone Type

Now we intend to prove that the boundedness of solutions on R+ implies the existence of a periodic solution. To this aim, define Q : D(A) → D(A) as Q(x) = z(T ; x) where z(t; x), t ≥ 0 is the solution of (4.1) with initial condition x ∈ D(A). It is easy to see that Q is nonexpansive. For each x ∈ D(A) the sequence {Qn (x)} is bounded because Qn (x) = z(nT ; x). By Browder-Petryshyn fixed point theorem, Q has at least a fixed point. But this means that (4.1) has at least a T -periodic solution. To prove the relations (4.34) and (4.35) we use Opial’s lemma. Suppose that z(t), t ≥ 0 is a bounded solution of (4.1). Denote the set of all T -periodic solutions of (4.1) by F. Define zn : [0, 2T ] → H by zn (t) = z(t + nT ), n = 1, 2, · · · . Extending each p ∈ F by periodicity on the interval [0, 2T ], we deduce that F is a subset of L2 (0, 2T ; H). We have: kzn (t) − p(t)k ≤ kzn (0) − p(0)k = kzn−1 (T ) − p(T )k ≤ kzn−1 (t) − p(t)k ≤ kzn−1 (0) − p(0)k,

0 ≤ t ≤ T, p ∈ F

(4.36)

Therefore lim kzn (t) − p(t)k = l(p)

(4.37)

n→+∞

exists uniformly on [0, T ], and consequently 1

kzn − pkH → ρ(p) := (2T ) 2 l(p), ∀p ∈ F

(4.38)

where H = L2 (0, 2T ; H). Therefore the first condition in Opial’s lemma is satisfied. Now without loss of generality suppose that Minϕ = ϕ(x0 ) = 0 where x0 ∈ D(A). By Lemma 4.3.2, we have Z 2T

tk 0

dzn (t)k2 dt ≤ dt

Z 2T 0

tk f (t)k2 dt + 2(kzn (0) − x0 k +

Z 2T 0

k f (t)kdt)2 ≤ C (4.39)

This implies that Z 2T

tk T

i.e.

dzn (t)k2 dt ≤ C dt

dzn } is bounded in H (4.40) dt Suppose that q ∈ H is a weak cluster point of {zn } i.e. znk * q ∈ H . By (4.40), dq dt ∈ H and dznk dq * in H (4.41) dt dt Let A¯ be the canonical extension of A to H . By Theorem 3.5.1, A¯ = ∂ Φ where Φ : H → (−∞, +∞] is defined by (R 2T ϕ(v)dt if ϕ(v) ∈ L1 (0, 2T ) Φ(v) = 0 +∞, otherwise {

First Order Evolution Equations

43

It is obvious that Φ(zn ) − Φ(p) ≤ ( f − Φ(p) − Φ(v) ≤ ( f −

dzn , zn − p)H , ∀p ∈ F dt

dp , p − v)H , ∀v ∈ D(Φ), ∀p ∈ F dt

Consequently, dzn d p dp − , zn − p)H + ( f − , zn − v)H , ∀v ∈ D(Φ), ∀p ∈ F dt dt dt (4.42) On the other hand from (4.37) we have Φ(zn ) − Φ(v) ≤ −(

dzn d p 1 − , zn − p)H = ( )(kzn (2T ) − p(2T )k2 − kzn (0) − p(0)k2 ) → 0, ∀p ∈ F dt dt 2 (4.43) Now by using (4.42) and (4.43) we can see that (

Φ(q) − Φ(v) ≤ ( f −

dp , q − v)H , ∀v ∈ D(Φ) dt

which implies that dp ¯ ∈ A(q), ∀p ∈ F (4.44) dt By Corollary 2.6.5, since ∂ Φ∗ = A¯ −1 , where Φ∗ is the conjugate function of Φ, we have Φ∗ ( f − z0n ) − Φ∗ ( f − p0 ) ≤ (zn , −z0n + p0 )H f−

and Φ∗ ( f − p0 ) − Φ∗ ( f − v) ≤ (p, −p0 + v)H , v ∈ H , where z0n =

dzn dt

and p0 =

dp dt .

Summing up the two inequalities above, we get:

Φ∗ ( f − z0n ) − Φ∗ ( f − v) ≤ −(zn − p, z0n − p0 )H − (p, z0n − v)H 1 = (kzn (0) − p(0)k2 − kzn (2T ) − p(2T )k2 ) 2 − (p, z0n − v)H Now letting n → +∞, by (4.37), and since Φ∗ is lsc, we obtain: Φ∗ ( f − q0 ) − Φ∗ ( f − v) ≤ −(p, q0 − v)H Again by Corollary 2.6.5, this shows that f−

dq ¯ ∈ A(p), ∀p ∈ F dt

(4.45)

Therefore by Lemma 4.3.1 we have d d p dq Φ(q) = ( f − , ) dt dt dt

(4.46)

44

Nonlinear Evolution and Difference Equations of Monotone Type

and

d dq d p dp dp Φ(p) = ( f − , ) = ( f − , ) dt dt dt dt dt for almost every t ∈ (0, 2T ) and p ∈ F. Therefore 2T Φ(q(2T )) −

Z 2T

Z 2T

Φ(q)dt = 0

0

Z 2T

= 0

d p dq , )tdt dt dt dq dp dq d p  ( f , ) + k k2 − 2( , ) tdt dt dt dt dt

(4.47)

(f −

(4.48)

On the other hand, by taking v = q in (4.42) we obtain lim sup Φ(znk ) ≤ Φ(q) k→+∞

and therefore since Φ is lsc, we get: lim Φ(znk ) = Φ(q)

(4.49)

k→+∞

Similarly, we also have: 2T Φ(zn (2T )) −

Z 2T 0

Z 2T

Φ(zn )dt =

0

(f −

dzn dzn , )tdt dt dt

(4.50)

Now in (4.50), replacing n by nk , substracting it from (4.48) and taking the limit as k → +∞, we get Z 2T

lim sup k→+∞

tk 0

 dznk d p 2 − k dt = lim sup 2T Φ(q(2T )) − Φ(znk (2T )) dt dt k→+∞

(4.51)

On the other hand, since znk * q in H , then by (4.41) we get znk (t) * q(t), as k → +∞, ∀t ∈ [0, 2T ]

(4.52)

Now it follows from (4.51), (4.52) and the fact that Φ is lsc, that Z 2T

lim

k→+∞ 0

and therefore

tk

dznk d p 2 − k dt = 0, ∀p ∈ F dt dt

(4.53)

dq d p = , ∀p ∈ F dt dt Therefore by (4.44), q ∈ F. Moreover this shows that every two periodic solutions differ by a constant. Thus all conditions in Opial’s lemma are satisfied, therefore p ∈ F exists such that zn * p in H . In fact this means that q = p. Everything that was proved for znk is valid for zn . Then (4.53) with zn instead of znk implies (4.35) because Z T Z 2T dzn d p 2 dzn d p T k − k dt ≤ tk − kdt dt dt dt dt 0 T Finally (4.34) follows from the fact that zn (t) * p(t) as n → +∞ uniformly with respect to t ∈ [0, T ]. The proof is now complete. 

First Order Evolution Equations

4.4

45

NONEXPANSIVE SEMIGROUP GENERATED BY A MAXIMAL MONOTONE OPERATOR

Definition 4.4.1 Let C be a nonempty closed subset of H. A continuous semigroup of contractions (or breifly, a semigroup of contractions) on C is a family of operators {S(t) : C → C; t ≥ 0} such that: (1) S(t + s)x = S(t)S(s)x, ∀x ∈ C, ∀t ≥ 0 (2) S(0)x = x, ∀x ∈ C (3) for each x ∈ C the mapping t → S(t)x is continuous on [0, +∞) (4) kS(t)x − S(t)yk ≤ kx − yk, ∀x, y ∈ C, ∀t ≥ 0 If besides 4 all of the other above conditions are satisfied, then S(t) is called a continuous semigroup or briefly, a semigroup. The infinitesimal generator of a semigroup {S(t) : C → C; t ≥ 0}, say G, is defined by S(h)x − x G(x) = lim (4.54) + h h→0 where D(G) is the set of all x ∈ C for which the strong limit in (4.54) exists in H. Suppose that A : D(A) ⊂ H → H is a maximal monotone operator. Consider the following Cauchy problem du + A(u) 3 0, t ≥ 0 dt u(0) = u0

(4.55) (4.56)

From Theorem 4.2.4, we know that for every u0 ∈ D(A), there exists a strong solution u(t), t ≥ 0 of (4.55) and (4.56). Set S(t)u0 = u(t), t ≥ 0 It is easy to check that for each t ≥ 0, S(t) is nonexpansive on D(A) and therefore it can be uniquely extended to D(A). Moreover it is obvious that S(t) : D(A) → D(A); t ≥ 0 is a continuous semigroup of nonexpansive mappings as was defined in Definition 4.4.1. Taking into account (4.3) we can also see that the infinitesimal generator of this semigroup is −Ao , where Ao is the minimal section of A. The given semigroup is called the semigroup generated by −A. On the other hand, by a result of Crandall and Pazy (see [BRE] p. 114), we know that if C is a closed convex subset of H, and {S(t) : C → C; t ≥ 0} is a semigroup of nonexpansive mappings, then there exists a unique maximal monotone operator A with D(A) = C, such that (S(t))t≥0 coincides with the semigroup generated by −A.

4.5

ERGODIC THEOREMS FOR NONEXPANSIVE SEQUENCES AND CURVES

Definition 4.5.1 A mapping T : D ⊂ H → H is called nonexpansive if kT x − Tyk ≤ kx − yk, ∀x, y ∈ D,

46

Nonlinear Evolution and Difference Equations of Monotone Type

where D is a nonempty subset of H. The set of all fixed points of T is denoted by F(T ), therefore F(T ) = {x ∈ D; T x = x}. The first fixed point theorem for nonexpansive self-mappings was proved in 1965, simultaneously by Browder [BRO], G¨ohde [GOH] and Kirk [KIR], extending Banach’s contraction principle. Theorem 4.5.2 Suppose that T : D → D is a nonexpansive mapping, where D is a nonempty, closed and convex subset of a Hilbert space H. Then T has a fixed point, and F(T ) is a closed and convex subset of H. Although in the Banach fixed point theorem, all orbits converge to the unique fixed point of T , this fact does not hold for a nonexpansive mapping, and orbits may not converge at all. As a simple example, consider T : [−1, 1] → [−1, 1] defined by T x = −x. Clearly, for each non-zero point in the interval [−1, 1], the Picard iterates do not converge. Baillon [BAI] proved that the Cesaro means of the Picard iterates of any nonexpansive mapping T , always converge weakly to a fixed point of T , provided that F(T ) , ∅. In the following, we state Baillon’s mean ergodic theorem. Theorem 4.5.3 Let C be a closed convex subset of a Hilbert space H, and T be a nonexpansive mapping from C into itself. If the set F(T ) of fixed points of T is nonempty, then for each x ∈ C, the Cesaro means Sn (x) :=

1 n−1 k ∑T x n k=0

converge weakly to some y ∈ F(T ). If we define P : C → C by Px = y, then P is a nonexpansive retraction of C onto F(T ) such that PT = T P = P and Px is contained in the closed convex hull of {T n x : n = 1, 2, ldots} for each x ∈ C. This retraction is called an “ergodic retraction”. If C is not convex, then F(T ) may be empty, and then Baillon’s proof is not applicable. To avoid the convexity assumption on C, we are going to introduce the notion of nonexpansive sequences. 4.5.1

ALMOST NONEXPANSIVE SEQUENCES

Definition 4.5.4 A sequence {xi } in a Hilbert space H is called a nonexpansive sequence if kxi+1 − x j+1 k ≤ kxi − x j k, for all i, j ≥ 0. Similarly a nonexpansive curve is defined as follows: Definition 4.5.5 A curve u : R+ → H is called nonexpansive if ku(t + h) − u(s + h)k ≤ ku(t) − u(s)k, ∀t, s ≥ 0, ∀h ≥ 0.

First Order Evolution Equations

47

Let {xn } be a sequence in H; let sn = 1n ∑n−1 i=0 xi . Definition 4.5.6 The sequence {xn } is said to be almost nonexpansive (in short ANES) if kxi+k − x j+k k2 ≤ kxi − x j k2 + ε(i, j), ∀i, j, k ≥ 0, where lim ε(i, j) = 0.

i, j→+∞

Obviously every nonexpansive sequence is an ANES. Remark 4.5.7 A bounded sequence {xn } in H that satisfies kxi+k − x j+k k ≤ kxi − x j k + ε1 (i, j), ∀i, j, k ≥ 0 where lim ε1 (i, j) = 0

i, j→+∞

is an ANES. Definition 4.5.8 Given a bounded sequence {xn } in H, the asymptotic center c of {xn } is defined as follows (see [EDE]): for every q ∈ H, let ϕ(q) = limn→+∞ sup kxn − qk2 . Then ϕ is a continuous strictly convex function on H, satisfying ϕ(q) → +∞ as kqk → +∞. Thus ϕ achieves its minimum on H at a unique point c, called the asymptotic center of the sequence xn . Remark 4.5.9 a) ωw (xn ) denotes the weak ω-limit set of xn . b) F(xn ) (in short F, when there is no confusion) denotes the following (possibly empty) subset of H: F = {q ∈ H :

lim kxn − qkexists}.

n→+∞

Note that if F , ∅, then the sequence {xn } is bounded. Lemma 4.5.10 F is a closed convex (possibly empty) subset of H. Proof. The closedness follows from the inequality kxn − qk − kxn+k − qk = kxn − qk − kxn − qm k + kxn − qm k − kxn+k − qm k + kxn+k − qm k − kxn+k − qk ≤ kxn − qk − kxn − qm k + kxn − qm k − kxn+k − qm k + kxn+k − qm k − kxn+k − qk ≤ 2kqm − qk + kxn − qm k − kxn+k − qm k . The convexity follows from the equalities kxn − (λ q1 + (1 − λ )q2 )k2 = kλ (xn − q1 ) + (1 − λ )(xn − q2 )k2

48

Nonlinear Evolution and Difference Equations of Monotone Type = λ 2 kxn − q1 k2 + (1 − λ )2 kxn − q2 k2 + 2λ (1 − λ )(xn − q1 , xn − q2 )

and 2(xn − q1 , xn − q2 ) = kxn − q1 k2 + kxn − q2 k2 − kq1 − q2 k2 .  Proposition 4.5.11 For an ANES {xn } in H, the weak limit p of any weakly convergent subsequence {sml } of sn (if any) belongs to F(i.e. limn→+∞ kxn − pk exists). Proof. For each k, i ≥ 0 and each m ≥ 1, by polarization identity, we have: 2(xk − xk+m , xi − p) = 2(xk − p, xi − p) − 2(xk+m − p, xi − p) = kxk − pk2 + kxi − pk2 − kxk − xi k2 − kxk+m − pk2 − kxi − pk2 + kxk+m − xi k2 = kxk − pk2 − kxk+m − pk2 + kxk+m − xi k2 − kxk − xi k2 Now summing both sides from i = 0 to i = n − 1 and dividing by n, we get: 2(xk − xk+m , sn − p) = kxk − pk2 − kxk+m − pk2 + −

1 n−1 ∑ kxk − xi k2 n i=0

= kxk − pk2 − kxk+m − pk2 + +

1 n−1 ∑ kxk+m − xi k2 n i=0

1 m−1 ∑ kxk+m − xi k2 n i=0

1 n−1 1 n−1 kxk+m − xi k2 − ∑ kxk − xi k2 ∑ n i=m n i=0

= kxk − pk2 − kxk+m − pk2 +

1 m−1 ∑ kxk+m − xi k2 n i=0

+

1 n−m−1 1 n−m−1 kxk+m − xi+m k2 − ∑ ∑ kxk − xi k2 n i=0 n i=0



1 n−1 ∑ kxk − xi k2 n i=n−m

≤ kxk − pk2 − kxk+m − pk2 + +

1 n−1 ∑ ε(i, k). n i=0

1 m−1 ∑ kxk+m − xi k2 n i=0

First Order Evolution Equations

49

Let ε > 0 be given, and choose N0 so that ε(i, j) ≤ ε, ∀i, j ≥ N0 . Then taking k ≥ N0 , and replacing n by ml and letting l → +∞, we get kxk+m − pk2 ≤ kxk − pk2 + ε, ∀k ≥ N0 , ∀m ≥ 1. Hence lim sup kxn − pk2 ≤ kxk − pk2 + ε, ∀k ≥ N0 . n→+∞

Thus lim sup kxn − pk2 ≤ lim inf kxn − pk2 + ε, ∀ε > 0, n→+∞

n→+∞

which implies that limn→+∞ kxn − pk exists, and therefore p ∈ F.



Lemma 4.5.12 If p1 , p2 ∈ F and q1 , q2 ∈ ωw ({xn }), then (p1 − p2 , q1 − q2 ) = 0. In particular, F ∩ convωw ({xn })contains at most one point. Proof. Assume xnk * q1 as k → +∞ and xml * q2 as l → +∞, we have: lim kxn − pi k2 = ϕ(pi ), for i = 1, 2,

n→+∞

and kxn − p2 k2 = kxn − p1 k2 + kp1 − p2 k2 + 2(xn − p1 , p1 − p2 ) thus ϕ(p2 ) = ϕ(p1 ) + kp1 − p2 k2 + 2(q1 − p1 , p1 − p2 ) ϕ(p2 ) = ϕ(p1 ) + kp1 − p2 k2 + 2(q2 − p1 , p1 − p2 ). Hence by subtraction, we get (q1 − q2 , p1 − p2 ) = 0.



Theorem 4.5.13 Let {xn } be an ANES in H. Then the following are equivalent: i) F , ∅. ii) lim infn→+∞ ksn k < +∞. iii) sn converges weakly to p ∈ H. Moreover under these conditions, we have: a) convωw ({xn }) ∩ F = {p}. b) p is the asymptotic center of the sequence {xn }. Proof. (i)⇒(ii): If F , ∅, then {xn } is bounded hence ksn k is also bounded. (ii)⇒(iii): Assume lim infn→+∞ ksn k < +∞. Then by the reflexivity of H, sn has a weakly convergent subsequence {snk } to some p ∈ H. Let us prove that sn * p as n → +∞. In fact if there was another subsequence sml * q ∈ H as l → +∞, then by Proposition 4.5.11, we have p ∈ F and q ∈ F. (Thus kxn k is bounded.) Hence the sequence αk = kxk − pk2 − kxk − qk2 = kpk2 − kqk2 + 2(xk , q − p)

50

Nonlinear Evolution and Difference Equations of Monotone Type

has a limit as k → +∞. Thus limk→+∞ (xk , q − p) exists and therefore (p, q − p) = (q, q − p), which implies kq − pk2 = 0 and hence q = p. Hence, every weakly convergent subsequence of sn converges weakly to p, therefore (since ksn k is bounded), sn converges weakly to p. (iii)⇒(i): This follows from Proposition 4.5.11 since p ∈ F. Let us now prove (a). We Obviously have p ∈ convωw ({xn }) ∩ F, and the conclusion follows from Lemma 4.5.12. Finally, let us prove (b). For every u ∈ H, we have: kxn − pk2 = kxn − uk2 + ku − pk2 + 2(xn − u, u − p), hence 1 n−1 1 n−1 2 kx − pk = i ∑ ∑ kxi − uk2 + ku − pk2 + 2(sn − u, u − p). n i=0 n i=0 Since p ∈ F, we have lim kxn − pk2 = ϕ(p),

n→+∞

therefore letting n → +∞, we get that ϕ(p) ≤ lim sup kxn − uk2 − kp − uk2 = ϕ(u) − kp − uk2 , ∀u ∈ H. n→+∞

This implies that p is the asymptotic center of {xn }.



Proposition 4.5.14 Assume {xn } is an ANES in H which is weakly asymptotically regular (i.e. xn+1 − xn * 0). Then ωw ({xn }) ⊂ F (in particular ωw ({xn }) , ∅ ⇒ kxn k is bounded). Proof. By a similar proof as in Proposition 4.5.11, we get 2(xk − xk+m ,

1 n−1 ∑ xn j +i − p) ≤ kxk − pk2 − kxk+m − pk2 n i=0 +

1 m−1 1 n−1 kx − x k + n+i k+m ∑ ∑ ε(k, n j + i − m) n i=0 n i=m

≤ kxk − pk2 − kxk+m − pk2 + +

M(m, k) n

1 n−1 ∑ ε(k, n j + i − m), n i=m

where M(m, k) is a constant depending only on m and k. Let ε > 0 be given, and choose N0 so that ε(i, j) < ε, ∀i ≥ N0 , ∀ j ≥ N0 .

First Order Evolution Equations

51

Then taking k ≥ N0 , m ≥ 1, n ≥ 1 fixed, and letting j → +∞, we get 0 ≤ kxk − pk2 − kxk+m − pk2 +

M(m, k) + ε, ∀k ≥ N0 , ∀m, n ≥ 1. n

Now letting n → +∞, we get kxk+m − pk2 ≤ kxk − pk2 + ε, ∀k ≥ N0 , ∀m ≥ 1. Therefore by the same argument as in Proposition 4.5.11, we conclude that limn→+∞ kxn − pk exists and thus p ∈ F.  Theorem 4.5.15 Let {xn } be a weakly asymptotically regular ANES. Then the following are equivalent: i) F , ∅. ii) lim infn→+∞ kxn k < +∞. iii) xn converges weakly to p ∈ H. Proof. (i) ⇒ (ii): If F , ∅, then kxn k is bounded. (ii) ⇒ (iii): Assume lim infn→+∞ kxn k < +∞. Then since a Hilbert space is reflexive, xn has a weakly convergent subsequence {xnk } to some p ∈ H. Now if xnk * p ∈ H as k → +∞ and xml * q ∈ H as l → +∞, then by Proposition 4.5.11, we have p ∈ F and q ∈ F. Hence we must have q = p. This follows by the same argument as in Theorem 4.5.15 since limn→+∞ (xn , q − p) exists. Thus every weakly convergence subsequence of xn converges weakly to p, therefore (since kxn k is bounded), xn converges weakly to p. (iii) ⇒ (i): This follows from Proposition 4.5.14 since p ∈ F.  4.5.2

ALMOST NONEXPANSIVE CURVES

Let u ∈ RC([0, +∞[; H); in the sequel we refer to such u as a curve in H. Let σT := T1 0T u(t)dt. In this section, we prove some theorems for almost non-expansive curves in H, similar to those for sequences proved in the previous subsection, by giving appropriate definitions for curves. Since the proofs are similar to those of the previous subsection, we will simply state the theorems and omit their proofs. Definition 4.5.16 The curve {u(t)} is almost non-expansive (shortly, ANEC), if ku(r + h) − u(s + h)k2 ≤ ku(r) − u(s)k2 + ε(r, s), where lim ε(r, s) = 0.

r,s→+∞

Remark 4.5.17 A bounded sequence {u(t)} in H that satisfies ku(r + h) − u(s + h)k ≤ ||u(r) − u(s)| + ε1 (r, s), ∀r, s, h ≥ 0, where lim ε1 (r, s) = 0

r,s→+∞

is an ANEC.

52

Nonlinear Evolution and Difference Equations of Monotone Type

Definition 4.5.18 Given a bounded curve {u(t)} in H, the asymptotic center c of {u(t)} is defined as follows (see [EDE]): for every q ∈ H, let ϕ(q) = limt→+∞ sup ku(t) − qk2 . Then ϕ is a continuous strictly convex function on H, satisfying ϕ(q) → +∞ as kqk → +∞. Thus ϕ achieves its minimum on H at a unique point c, called the asymptotic center of the curve u(t). Remark 4.5.19 a) ωw ({u(t)}) denotes the weak ω-limit set of {u(t)}. b) F({u(t)}) (in short F, when there is no confusion) denotes the following (possibly empty) subset of H: F = {q ∈ H :

lim ku(t) − qkexists}.

t→+∞

Note that if F , ∅, then the curve {u(t)} is bounded. Lemma 4.5.20 F is a closed convex (possibly empty) subset of H. Proof. Similar to Lemma 4.5.10.



Proposition 4.5.21 For an ANEC {u(t)} in H, the weak limit p of any weakly convergent subnet {σTl } of σT (if any) belongs to F(i.e. limt→+∞ ku(t) − pk exists). Proof. Similar to Proposition 4.5.11.



Lemma 4.5.22 If p1 , p2 ∈ F and q1 , q2 ∈ ωw ({u(t)}), then (p1 − p2 , q1 − q2 ) = 0. In particular, F ∩ convωw ({u(t)}) contains at most one point. Proof. Similar to Lemma 4.5.12.



Theorem 4.5.23 Let {u(t)} be an ANEC in H. Then the following are equivalent: i) F , ∅. ii) lim inft→+∞ kσT k < +∞. iii) σT converges weakly to p ∈ H. Moreover under these conditions we have: a) convωw ({u(t)}) ∩ F = {p}. b) p is the asymptotic center of the curve {u(t)} Proof. Similar to Theorem 4.5.13.



Proposition 4.5.24 Assume {u(t)} is an ANEC in H which is weakly asymptotically regular (i.e. u(t + h) − u(t) * 0). Then ω({u(t)}) ⊂ F (in particular ωw ({u(t)}) , ∅ ⇒ ku(t)k is bounded). Proof. Similar to Proposition 4.5.14.



Theorem 4.5.25 Let {u(t)} be a weakly asymptotically regular ANEC. Then the following are equivalent: i) F , ∅. ii) lim inft→+∞ ku(t)k < +∞. iii) u(t) converges weakly to p ∈ H. Proof. Similar to Theorem 4.5.15.



First Order Evolution Equations

4.6

53

WEAK CONVERGENCE OF SOLUTIONS AND MEANS

In this section, by using the results of the previous sections on almost nonexpansive sequences and curves, we study the ergodic convergence of solutions to the following quasi-autonomous dissipative system ( du dt + Au 3 f (4.57) u(0) = u0  1 (0, +∞); H . We prove where A is a monotone operator in H, u0 ∈ H, and f ∈ Lloc the weak ergodic  convergence of the weak solution u(t) of this equation when f ∈ L1 (0, +∞); H , or more generally f − f∞ ∈ L1 (0, +∞); H , for some f∞ ∈ H. First we recall the following lemmas. See Br´ezis [BRE].  Lemma 4.6.1 Let A be a monotone operator in H, and f , g ∈ L1 (0, T ); H ; then if u and v are respectively weak solutions of the equations du dv + Au 3 f and + Av 3 g on [0, T ] dt dt we have ku(t) − v(t)k ≤ ku(s) − v(s)k +

Z t

k f (θ ) − g(θ )kdθ , ∀0 ≤ s ≤ t ≤ T

s

 1 (0, +∞); H . Then Corollary 4.6.2 Let A be a monotone operator in H and f ∈ Lloc if u is a weak solution of the equation du dt + Au 3 f , we have ku(r + h) − u(s + h)k ≤ ku(r) − u(s)k +

Z s+h

k f (θ + (r − s)) − f (θ )kdθ , ∀h ≥ 0, ∀r ≥ s ≥ 0

s

Proof. It is enough to apply Lemma 4.6.1 with g(t) = f (t + (r − s)) and v(t) = u(t + (r − s)).  Proposition 4.6.3 If for each T > 0, u is a weak solution of the system (4.57) on [0, T ], and if supt≥0 ku(t)k < +∞ and Z +∞

lim

s,r→+∞r>s s

k f (θ + (r − s)) − f (θ )kdθ = 0,

then the curve u(t) is ANEC in H. Proof. This follows from Corollary 4.6.2 and Remark 4.5.17 by taking (R +∞ k f (θ + (r − s)) − f (θ )kdθ if r ≥ s ε1 (r, s) = Rs+∞ k f (θ + (s − r)) − f (θ )kdθ if s ≥ r r 

54

Nonlinear Evolution and Difference Equations of Monotone Type

The following theorem which is the main result of this section, follows from Theorem 4.5.23 and Proposition 4.6.3. Theorem 4.6.4 If u is a weak solution of the system (4.57) on every  interval [0, T ], and satisfies supt>0 ku(t)k < +∞, and if f − f∞ ∈ L1 (0, +∞); H for some f∞ ∈ H, R then σT = T1 0T u(t)dt converges weakly to the asymptotic center of the curve u(t). Finally, taking into account Proposition 4.6.3 and Theorem 4.5.25, we have the following theorem for the weak convergence of solutions to (4.57). Theorem 4.6.5 If u is a weak solution of the system (4.57) on every interval [0, T ], and satisfies supt>0 ku(t)k < +∞, and for each h ≥ 0, u(t + h) − u(t) * 0, and if f −  f∞ ∈ L1 (0, +∞); H for some f∞ ∈ H, then u(t) converges weakly to the asymptotic center of the curve u(t).

4.7

ALMOST ORBITS

First we define the concept of an almost-orbit of a semi-group, first introduced by Miyadera [MIY]. Let C be a nonempty closed and convex subset of H, and {T (t) : C → C; t ≥ 0} be a nonlinear semi-group (defined in Section 4). An almost orbit for T (t) is a function u : R+ → C such that lim [sup ku(t + h) − T (h)u(t)k] = 0.

t→+∞ h≥0

In the following, we show that every almost-orbit of a semi-group of nonexpansive mappings is an ANEC. Lemma 4.7.1 Let C be a nonempty closed and convex subset of H, and u(·) be a bounded almost orbit of {T (t) : C → C; t ≥ 0}. Then u(·) is an ANEC in H. Proof. Let φ (t) = suph≥0 ku(t + h) − T (h)u(t)k. Then limt→+∞ φ (t) = 0. Now we have: ku(t + h) − u(s + h)k ≤ φ (t) + φ (s) + ku(t) − u(s)k, ∀h,t, s ≥ 0 The result follows therefore by letting ε(t, s) = φ (t) + φ (s).



Theorem 4.7.2 Let {T (t) : C → C; t ≥ 0} be a non-expansive semi-group. If every orbit of T (t) converges strongly (resp. weakly) as t goes to ∞, so does every almost orbit of T (t). Proof. Let τ be the corresponding topology (weak or strong topology), and suppose that τ − limt→+∞ T (t)x exists for all x. Let u be an almost-orbit of T (t). Let s ≥ 0 and set ζ (s) = τ − limt→+∞ T (t − s)u(s). We have ζ (s + h) − ζ (s) = τ − lim {T (t − s − h)u(s + h) − T (t − s)u(s)} t→+∞

First Order Evolution Equations

55

But for all t ≥ s + h the quantity kT (t − s − h)u(s + h) − T (t − s)u(s)k is bounded above by ku(s + h) − T (h)u(s)k. By the τ−lower semi-continuity of the norm, we get kζ (s + h) − ζ (s)k ≤ ku(s + h) − T (h)u(s)k. Since u is an almost-orbit of T (t), the right hand side tends to 0 as s → +∞, uniformly with respect to h ≥ 0. Therefore ζ (s) is a Cauchy net and converges strongly to a limit ζ∞ . Now u(s + h) − ζ∞ = [u(s + h) − T (h)u(s)] + [T (h)u(s) − ζ (s)] + [ζ (s) − ζ∞ ]. Given ε > 0, we can choose s large enough so that the first and the third terms on the right hand side are less than ε in norm, uniformly in h for the first term. Next, for such a fixed s, we let h → ∞ so that the second term τ− converges to zero. Then u(t) is τ− convergent to ζ∞ as t → +∞.  For a maximal monotone operator A : D(A) ⊂ H → H, by Part (6) of Theorem 3.4.1, D(A) is closed and convex. Suppose {T (t) : D(A) → D(A);t ≥ 0} is a nonlinear semi-group generated by ( −u0 (t) ∈ A(u(t)) (4.58) u(0) = x ∈ D(A)  Proposition 4.7.3 If f ∈ L1 (0, +∞); H , then every solution to (4.57) is an almost orbit of the semi-group T (t) generated by (4.58). Proof. From (4.57) and (4.58), we have 1 d ku(t + τ) − T (τ)u(t)k2 ≤ k f (t + τ)kku(t + τ) − T (τ)u(t)k 2 dτ Therefore

d ku(t + τ) − T (τ)u(t)k ≤ k f (t + τ)k dτ Integrating the above inequality from τ = 0 to τ = s, we get ku(t + s) − T (s)u(t)k ≤

Z s 0

k f (t + τ)kdτ =

Z t+s t

k f (τ)kdτ ≤

Z ∞

k f (τ)kdτ.

t

The result follows now by taking the supremum over s ≥ 0, and then letting t → +∞ in the above inequality. 

4.8

SUB-DIFFERENTIAL AND NON-EXPANSIVE CASES

In this section, we study the weak convergence of solutions to (4.57) for two special cases. First, for a monotone operator that is the sub-differential of a proper convex and lower semi-continuous function ϕ, and then for operators of the form I −T where I is the identity operator, and T is a non-expansive mapping on H. In both cases, we prove that if Argminϕ , ∅ (resp. F(T ) , ∅), then the trajectory of the solution to (4.57) converges weakly to a minimum point of ϕ (resp. to a fixed point of T ).

56

Nonlinear Evolution and Difference Equations of Monotone Type

Theorem 4.8.1 Suppose that u(t) is a solution to (4.57), then we have: (i) If A = ∂ ϕ, where ϕ is a proper, convex and lower semi-continuous function with Argminϕ , ∅, then u(t) converges weakly to a minimum point of ϕ. (ii) If A = I − T , where I is the identity operator, and T is a non-expansive mapping on the Hilbert space H, with F(T ) , ∅, then u(t) converges weakly to a fixed point of T . Proof. By Theorem 4.7.2 and Proposition 4.7.3, it is enough to consider the homogeneous first order evolution Equation (4.58), and by Theorem 4.5.25 we only need to show the asymptotic regularity of the solutions to (4.58). For any monotone operator, since the trajectory of the solutions to (4.58) is non-expansive, i.e. the function t 7→ ku(t + h) − u(t)k is non-increasing, then ku0 (t)k is also non-increasing. (i) First, let’s consider the case A = ∂ ϕ. By Lemma 4.3.1, we have: d ϕ(u(t)) = (−u0 (t), u0 (t)) = −ku0 (t)k2 dt Therefore Z T

ku0 (t)k2 dt ≤ ϕ(u(0)) − ϕ(u(T )) ≤ ϕ(u(0)) − ϕ(p)

0

where p ∈ Argminϕ. By letting T → +∞, we get Z +∞

ku0 (t)k2 dt < +∞

0

Since ku0 (t)k is non-increasing, it follows that ku0 (t)k goes to zero as t → +∞. Now ku(t + h) − u(t)k = k

Z t+h

u0 (s)dsk ≤

t

Z t+h

1

Z t+h

ku0 (s)kds ≤ h 2 (

t

1

ku0 (s)k2 ds) 2 → 0

t

as t → +∞. This shows that u(t) is asymptotically regular, and completes the proof of (i). (ii) Now we consider the case A = I − T , where T is a non-expansive mapping on H. By the non-expansiveness of T kTu(t) − pk ≤ ku(t) − pk, ∀p ∈ F(T ) Now by (4.58), we have ku0 (t) + u(t) − pk2 ≤ ku(t) − pk2 Hence ku0 (t)k2 ≤ −

d ku(t) − pk2 dt

Integrating from t = 0 to T , we get Z T 0

ku0 (t)k2 dt ≤ ku(0) − pk2 − ku(T ) − pk2

First Order Evolution Equations

57

Therefore Z ∞

ku0 (t)k2 dt < +∞

0

Since we know that ku0 (t)k is non-increasing, it follows that ku0 (t)k → 0 as t → +∞, and therefore with a similar proof as in case (i), u(t) is asymptotically regular.  Remark 4.8.2 The weak convergence of solutions to (4.57) may be shown also with a more general condition on the monotone operator and the semi-group, which is the demi-positivity of the semi-group generated by (4.58). We refer the reader to [BRU] (see also [MOR]) for the definition and the result.

4.9

STRONG ERGODIC CONVERGENCE

This section is devoted to the strong convergence of the sequence of means for solutions to (4.57), to the asymptotic center of the trajectory of solutions, as well as to a zero of the monotone operator A if A is maximal monotone. First we prove some strong ergodic convergence results for sequences. Recall from the introduction in Section 5, that the first weak ergodic convergence theorem for iterates of a non-expansive mapping was proved by Baillon [BAI]. He also proved that if the non-expansive mapping is odd, then the convergence of the means is strong. This result was extended by Br´ezis and Browder [BRE-BRO] to a more general summation method called strongly regular summation method. Set yn = ∑∞j=0 an, j x j , where an, j ≥ 0, ∑∞j=0 an, j = 1, an, j → 0 as n → +∞ for fixed j, and σn = ∑∞j=0 |an, j+1 − an, j | → 0 as n → +∞. In this section, we study the strong convergence of the sequence yn , where xn is a sequence satisfying: ( (x j , x j+l ) ≤ (xk , xk+l ) + ε(k, l, j − k), ∀k, l ≥ 0 and j ≥ k, (4.59) with ε bounded and limk,l,m→+∞ ε(k, l, m) = 0 This condition which was introduced by Djafari Rouhani [DJA4] is more general than the conditions assumed by Br´ezis and Browder [BRE-BRO] and Wittmann [WIT]. Let C = {q ∈ H; limn→+∞ (xn , q) exists}, and C1 = {q ∈ H; (xn , q) ≥ (xn+1 , q), ∀n ≥ 0}. The proof of the following theorem is similar to that of Theorem 4.5.13. Theorem 4.9.1 Let xn be a bounded sequence in H such that the weak limit of every weakly convergent subsequence of yn belongs to C. Then yn converges weakly to the unique element p ∈ C ∩ convωw ({xn }). Moreover, if limn→+∞ kxn k exists, then p is the asymptotic center of the sequence xn . Proposition 4.9.2 If in Theorem 4.9.1 we make the stronger assumptions that kxn k is non-increasing and replacing C by C1 , then in addition, the sequence zn = PC1 (−xn ) converges strongly to some z ∈ C1 with kzk ≤ kpk and kp + zk2 ≤ kpk2 − kzk2 .

58

Nonlinear Evolution and Difference Equations of Monotone Type

Proof. It is clear for the assumption that for any q ∈ C1 the sequence kxn + qk is non-increasing. We have kxn+1 + zn+1 k2 ≤ kxn+1 + zn k2 ≤ kxn + zn k2 where the first inequality holds since zn+1 = PC1 (−xn+1 ) and the second holds because zn ∈ C1 . Hence the sequence kxn + zn k2 is non-increasing. Now for all q ∈ C1 , we have: 1 (−xn − zn , q − zn ) = (kxn + zn k2 + kq − zn k2 − kxn + qk2 ) ≤ 0 2 Replacing n by n + k and q by zn ∈ C1 , we get: kzn − zn+k k2 ≤ kxn+k − zn k2 − kxn+k − zn+k k2 ≤ kxn + zn k2 − kxn+k − zn+k k2 → 0, uniformly in k ≥ 0. Therefore zn is a Cauchy sequence in H, thus strongly convergent to some z ∈ C1 . Since the projection map PC1 satisfies (PC1 x − PC1 y, x − y) ≥ kPC1 − PC1 yk2 , ∀x, y ∈ H, putting y = 0 ∈ C1 and x = −xn we get: (zn , xn ) ≤ −kzn k2 , hence ∞





j=0

j=0

j=0

∑ an, j (z j , x j ) = ∑ an, j (z j − z, x j ) + (z, ∑ an, j x j ) ≤ −kzn k2

Letting n → +∞, we get: kzk2 ≤ −(z, p) ≤ kzkkpk from which it follows that kzk ≤ kpk and kp + zk2 = kpk2 + kzk2 + 2(z, p) ≤ kpk2 − kzk2 completing the proof of the proposition.  Example 4.9.3 By taking xn = 1+ n1 in the real numbers, we have zn = 0 for all n ≥ 0 whereas xn → 1, showing therefore that in Proposition 4.9.2 we may have kzk < kpk Corollary 4.9.4 Let xn be a sequence in H satisfying (4.59). Then the sequence yn = ∑∞j=0 an, j x j converges weakly to the unique element p ∈ C ∩ conv(ωw {xn }) and (xn , p) → kpk2 . Moreover if limn→+∞ kxn k exists, then p is the asymptotic center of the sequence xn . Proof. We note that (4.59) implies that the sequence xn is bounded. Hence yn has a weakly convergent subsequence. Assume ynr * q. Let ε > 0 given; choose n0 so that (x j , x j+l ) ≤ (xk , xk+l ) + ε for all k ≥ n0 , l ≥ n0 and j ≥ k + n0 . Then, we have: (x j − q, xk − xk+l ) = (x j , xk ) − (x j , xk+l ) + (q, xk+l − xk ) ≥ (x j , xk ) − (x j−l , xk ) − ε + (q, xk+l − xk ) for all k ≥ n0 , l ≥ n0 and j ≥ l + k + n0 . Thus multiplying by an, j , we get: an, j (x j − q, xk − xk+l ) ≥ (an, j x j , xk ) − (an, j−l x j−l , xk ) + (an, j−l − an, j )(x j−l , xk ) − an, j ε + (an, j q, xk+l − xk )

First Order Evolution Equations

59

for k ≥ n0 , l ≥ n0 and j ≥ l + k + n0 . But we have: ∞



l−1



|an, j−l − an, j | ≤

∑ ∑ |an, j−l+i − an, j−l+i+1 |

j=l+k+n0 i=0

j=l+k+n0

l−1

=



|an, j−l+i − an, j−l+i+1 |

∑ ∑

i=0 j=l+k+n0 l−1 ∞



∑ ∑ |an, j − an, j+1 |

i=0 j=0

= lσn → 0 as n → +∞. Hence for fixed k ≥ n0 and l ≥ n0 , summing up over j and letting n = nr → +∞, by using the strong regularity of an, j , we get: 0 ≥ (q, xk ) − (q, xk ) − ε + (q, xk+l − xk ) = (q, xk+l − xk ) − ε Now letting l → +∞ and then k → +∞, we get: lim supn→+∞ (q, xn ) ≤ lim infn→+∞ (q, xn ) + ε, and since ε > 0 was arbitrary, we conclude that limn→+∞ (xn , q) exists, thus q ∈ C. Now the result follows by applying Theorem 4.9.1. Finally we have: ∞

lim (xn , p) = lim ( ∑ an, j x j , p)

n→+∞

n→+∞

j=0

= kpk2 

which completes the proof of the corollary.

Before stating our main result, we need the following lemma from [BRE-BRO]. Lemma 4.9.5 If {an, j } is a strongly regular summation method, then {bn, j } defined by: bn,0 = ∑∞j=0 a2n, j and bn,r = 2 ∑∞j=0 an, j an, j+r for r ≥ 1, is also a strongly regular summation method. Proof. We have ∞



r=0

j=0

∑ bn,r = ( ∑ an, j )2 = 1

j Now, an, j ≤ an,0 + ∑i=1 |an,i − an,i−1 | ≤ an,0 + σn for ∞ ∞ 2 ∑ j=0 an, j ≤ ∑ j=0 an, j (σn + an,0 ) = σn + an,0 → 0; 2 ∑∞j=0 an, j an, j+r ≤ 2(σn + an,0 ) → 0. Finally we have ∞



all j ≥ 0, hence bn,0 = and for r ≥ 1, bn,r =



∑ |bn,r+1 − bn,r | = |bn,1 − bn,0 | + 2 ∑ | ∑ an, j (an, j+r+1 − an, j+r )|

r=0

r=1 j=0 ∞ ∞

≤ |bn,1 − bn,0 | + 2 ∑ an, j ∑ |an, j+r+1 − an, j+r | j=0

r=1

60

Nonlinear Evolution and Difference Equations of Monotone Type ≤ bn,1 − bn,0 + 2σn → 0 

This completes the proof. Now we state and prove our main result.

Theorem 4.9.6 (i) Let xn be a sequence in H satisfying (4.59). Then the sequence yn = ∑∞j=0 an, j x j converges strongly in H to the unique element p ∈ C ∩ conv(ωw {xn }). If, moreover, limn→+∞ kxn k exists, then p is the asymptotic center of xn . (ii) If xn satisfies (x j , x j+r ) ≤ (xi , xi+r ) + ε(i, j − i) ∀i, r ≥ 0, and j ≥ i with

lim ε(i, m) = 0,

i,m→+∞

(4.60) then in addition p is the element of minimum norm in K, i.e. p = PK 0. Proof. (i) We already know from Corollary 4.9.4 that yn converges weakly to p ∈ C ∩ conv(ωw {xn }) and moreover (xn , p) → kpk2 . Therefore, to prove the strong convergence of yn , all we need to show is that : lim supn→+∞ kyn k2 ≤ kpk2 . Let M = supn≥0 kxn kand let {bn,r } denote the strongly regular summation method introduced in Lemma 4.9.5. Let ε > 0 given and choose n0 so that (x j , x j+r ) ≤ (xk , xk+r ) + ε for all k ≥ n0 , r ≥ n0 , and j ≥ k + n0 . Let k ≥ n0 fixed. We have: ∞



kyn k2 = ∑ ∑ an,i an, j (xi , x j )

i=0 j=0 ∞ ∞ ∞ = a2n, j kx j k2 + 2 an, j an, j+r (x j , x j+r ) j=0 j=0 r=1 ∞ ∞ = a2n, j kxk k2 + a2n, j (kx j k2 − kxk k2 ) j=0 j=0 ∞ ∞



∑∑





+2 ∑

∑ an, j an, j+r (xk , xk+r )

+2 ∑

∑ an, j an, j+r ((x j , x j+r ) − (xk , xk+r ))

j=0 r=1 ∞ ∞ j=0 r=1 ∞



= (xk , ∑ bn,r xk+r ) + ∑ a2n, j (kx j k2 − kxk k2 ) r=0 ∞ ∞

+2 ∑

j=0

∑ an, j an, j+r ((x j , x j+r ) − (xk , xk+r ))

j=0 r=1 ∞

≤ (xk , ∑ bn,r xk+r ) + M 2 bn,0 + ε r=0



∑ bn,r

r=n0

k+n0 −1 n0 −1

+2

∑ ∑ an, j an, j+r ((x j , x j+r ) − (xk , xk+r ))

j=0

r=1

First Order Evolution Equations

61



≤ (xk , ∑ bn,r xk+r ) + M 2 bn,0 + 4M 2 r=0

sup 0≤ j≤2n0 +k

a2n, j + ε.

Now since by Lemma 4.9.5, {bn,r } is a strongly regular summation method, for k ≥ n0 fixed, we deduce from Corollary 4.9.4 that ∑∞ r=0 bn,r xk+r converges weakly to p. Hence letting k ≥ n0 fixed, and n → +∞ in the above inequality, we get: lim supn→+∞ kyn k2 ≤ (xk , p) + ε for all k ≥ n0 . Now letting k → +∞, we get lim supn→+∞ kyn k2 ≤ kpk2 + ε, which implies the desired result since ε > 0 was arbitrary. (ii) Now assume that (4.60) holds and let us show that p = PK 0. Let zn = PKn 0; since Kn+1 ⊂ Kn , kzn k is nondecreasing, hence convergent. We have (0 − zn , zm − zn ) ≤ 0 for all m ≥ n. Hence kzn k2 ≤ (zn , zm ) for all m ≥ n, which implies kzn − zm k2 = kzn k2 − 2(zn , zm ) + kzm k2 ≤ kzm k2 − kzn k2 → 0, as m, n → +∞. zn converges strongly to some z ∈ K, and we have (0−z, y−z) ≤ 0 for all y ∈ K; hence kzk2 ≤ (y, z) ≤ kykkzk, which implies that kzk ≤ kyk for all y ∈ K, thus z = PK 0. Since p ∈ K, we have kzk ≤ kpk. Let ε > 0 given; we have: kn

kn

i=n

i=n

∑ αi = 1 such that k ∑ αi xi k2 ≤ kzn k2 + ε

∀n > 0 ∃kn ≥ n ∃αi ≥ 0, choosing n0 as in (i), we have: kn

kn kn

i=n

i=n j=n

k ∑ αi xi+l k2 = ∑ ∑ αi α j (xi+l , x j+l ) kn kn

≤ ∑ ∑ αi α j (xi , x j ) + ε i=n j=n kn

= k ∑ αi xi k 2 + ε i=n

for all n ≥ n0 and l ≥ n0 . Therefore we get: kn





kn

i=n

l=0

l=0

i=n





kn

i=n

k ∑ αi ∑ aN,l xi+l k2 = k ∑ (aN,l ∑ αi xi+l )k2 ≤ ( ∑ aN,l ) ∑ aN,l k ∑ αi xi+l k2 = ≤

l=0

l=0



kn

l=0

i=n

∑ aN,l k ∑ αi xi+l k2 ∞

kn

n0 −1

kn

l=0

i=n

l=0

i=n

∑ aN,l (k ∑ αi xi k2 + ε) + ∑ aN,l k ∑ αi xi+l k2

62

Nonlinear Evolution and Difference Equations of Monotone Type kn

≤ k ∑ αi xi k2 + ε + i=n

≤ kzn k2 + 2ε + M 2

n0 −1

kn

∑ aN,l k ∑ αi xi+l k2

l=0 n0 −1

∑ aN,l

i=n

for n ≥ n0 ,

l=0

where we have used Cauchy-Schwarz inequality for the first inequality above. Now for n ≥ n0 fixed, letting N → +∞, we have: ∞

∑ aN,l xi+l → p

for all i = n, · · · , kn

l=0

Hence we get kpk2 ≤ kzn k2 + 2ε for all n ≥ n0 . Letting n → +∞, we obtain kpk ≤ kzk, since ε > 0 was arbitrary; now since p ∈ K, by uniqueness of PK 0, we get p = z = PK 0, completing the proof of the theorem.  The following theorem is an immediate consequence of Theorem 4.9.6 Theorem 4.9.7 Let {xn } be any sequence in H satisfying the following condition: limi→+∞ (xi , xi+m ) = αm exists uniformly in m ≥ 0. Then sn = n1 ∑n−1 i=0 xi converges strongly to the asymptotic center of {xn }. Similarly, we have the following analogous result for curves. Theorem 4.9.8 Let {u(t)} be any curve in H satisfying the following condition: R limt→+∞ (u(t), u(t + h)) = α(h) exists uniformly in h ≥ 0. Then σT = T1 0T u(t)dt converges strongly to the asymptotic center of {u(t)}. The following result which follows from Theorem 4.9.8, is the main result of this section for the solutions to (4.57). Theorem 4.9.9 If u is a weak solution of the system (4.57) on every interval [0, T ], and satisfies limt→+∞ (u(t), u(t + h)) = α(h) exists uniformly in h ≥ 0, then σT = 1 RT T 0 u(t)dt converges strongly to the asymptotic center of the curve u(t).

4.10

STRONG CONVERGENCE OF SOLUTIONS

In this section, with additional assumptions on the monotone operator A or the convex function ϕ when A = ∂ ϕ, we study the strong convergence of solutions to (4.57). By Theorem 4.7.2 and Proposition 4.7.3, we only need to study the strong convergence of solutions to (4.58). In the following theorem, in addition to the strong convergence, we also determine the rate of convergence of the solutions. Theorem 4.10.1 Suppose that u(t) is a solution to (4.58), where A is α-strongly monotone, and p is the unique element of A−1 (0), then ku(t) − pk = O(e−αt ).

First Order Evolution Equations

63

Proof. Multiplying both sides of (4.58) by u(t) − p, and using the α-strong monotonicity of A, we get: 1 d ku(t) − pk2 ≤ −αku(t) − pk2 . 2 dt

(4.61)

Since by (4.5), ku(t) − pk is non-increasing (take v ≡ p ∈ A−1 (0)), if ku(t0 ) − pk = 0 for some t0 > 0, then ku(t) − pk = 0, for all t ≥ t0 and the result follows. Otherwise, dividing both sides of (4.61) by ku(t) − pk2 , we get: d ln ku(t) − pk2 ≤ −2α. dt Integrating from 0 to T , it follows that: ln ku(T ) − pk2 − ln ku(0) − pk2 ≤ −2αT.

(4.62)

Therefore: ku(T ) − pk ≤ ku(0) − pke−αT , 

which yields the theorem.

Theorem 4.10.2 Assume that A = ∂ ϕ, where ϕ : H → (−∞, +∞] is a proper, convex and lower semi-continuous function such that D(ϕ) = −D(ϕ) and ϕ(x) − ϕ(0) ≥ α(ϕ(−x) − ϕ(0)), ∀x ∈ D(ϕ)

(4.63)

where α is some positive real number. Then there exists a minimum point p of ϕ such that u(t) converges strongly to p. Proof. Without loss of generality we may assume that ϕ(0) = 0. Also without loss of generality we may assume that 0 < α ≤ 1. In fact we have: 0 = ϕ(0) = ϕ(x + (−x)/2) ≤ 1/2ϕ(x) + 1/2ϕ(−x) which implies that ϕ(x) ≥ −ϕ(−x). This inequality together with (4.63) implies that: ϕ(x) ≥ Max{αϕ(−x), −ϕ(−x)}. Therefore ϕ(x) ≥ 0, for all x ∈ D(ϕ). This shows that if (4.63) holds for α > 1, it also holds for any 0 < β < α, and in particular for any 0 < β ≤ 1. Therefore we may assume 0 < α ≤ 1. Now, for each x ∈ D(ϕ), by (4.63) and the convexity of ϕ, we have: 1 α x+ (−αx)) 1+α 1+α α 1 ≤ ϕ(x) + ϕ(−αx) 1+α 1+α α α ≤ ϕ(x) + ϕ(−x) 1+α 1+α α 1 ≤ ϕ(x) + ϕ(x) 1+α 1+α = ϕ(x).

ϕ(0) = ϕ(

64

Nonlinear Evolution and Difference Equations of Monotone Type

This implies that 0 is a minimum point of ϕ, or 0 ∈ A−1 (0). By Lemma 4.3.1, we have d ϕ(u(t)) = (u0 (t), −u0 (t)) = −ku0 (t)k2 ≤ 0, a.e. t > 0. dt Hence the function ϕ(u(t)) is non-increasing. Then by (4.63), the convexity of ϕ, and the sub-differential inequality, for s ≥ t, we get: ϕ(u(t)) ≥ ϕ(u(s)) ≥ αϕ(−u(s)) ≥ ϕ(−αu(s))  du ≥ ϕ(u(t)) + (t), αu(s) + u(t) , a.e. t ∈]0, s[ dt

(4.64)

For fixed s > 0, define g : [0, s] → R by 1 α g(t) = (1 + α)(ku(t)k2 − ku(s)k2 ) − ku(t) − u(s)k2 2 2 Then g is absolutely continuous and g(s) = 0. Moreover, it follows from (4.64) that for 0 < t ≤ s,  g0 (t) = u0 (t), αu(s) + u(t) ≤ 0, a.e. t ∈]0, s[ Therefore g(t) ≥ g(s) = 0, ∀t ∈]0, s[. In other words, (1 + α)(ku(t)k2 − ku(s)k2 ) ≥ αku(t) − u(s)k2 , ∀ t ∈]0, s[, which implies that ku(t)k2 is non-increasing and also u(t) converges strongly to some p, which by Theorem 4.8.1 belongs to Argminϕ. 

4.11

QUASI-CONVEX CASE

Consider the following non-homogeneous evolution system ( −x0 (t) = ∇φ (x(t)) + f (t), x(0) = x0 ∈ H.

(4.65)

where φ is a quasi-convex function as defined below. When f (t) ≡ 0, Goudou and Munier [GOU-MUN] proved the weak convergence of solutions to (4.65) to a critical point of φ . They also proved the strong convergence of solutions to (4.65) with additional assumptions on φ . In this section, we  study (4.65) with the condition f ∈ L1 (0, +∞); H . We prove the weak and strong convergence of solutions to (4.65) to a critical point of φ . These results extend the similar classical results on the asymptotic behavior of non-homogeneous gradient systems associated with convex functions which have been also extended to nonsmooth convex functions (see Theorem 4.8.1 of this chapter as well as [BRU]). A function φ : H → R is said to be quasi-convex if φ (λ x + (1 − λ )y) ≤ max{φ (x), φ (y)}, ∀x, y ∈ H, ∀λ ∈ [0, 1],

First Order Evolution Equations

65

or equivalently every sub-level set of φ is convex. A differentiable function φ on H is quasi-convex if φ (x) ≥ φ (y) ⇒ (∇φ (x), y − x) ≤ 0. A function φ is called pseudo-convex if the following condition is satisfied: If φ (y) > φ (x), then there exists β (x, y) > 0 and 0 < δ (x, y) ≤ 1, such that: φ (y) − φ (tx + (1 − t)y) ≥ tβ (x, y) for all t ∈ (0, δ (x, y)). A differentiable function φ is pseudo-convex if and only if φ (x) > φ (y) ⇒ (∇φ (x), y − x) < 0. Obviously convexity implies pseudo-convexity and pseudo-convexity implies quasiconvexity. We refer the reader to the interesting book by Cambini and Martein [CAM-MAR] for the definitions, properties and illustrative examples of convexity and its extensions. Throughout this section, we assume that φ : H → R is a continuously differentiable quasi-convex function with Argminφ , ∅, and ∇φ Lipschitz continuous on bounded subsets of H. The results are applicable even to one dimensional differential equations (where of course weak and strong convergence coincide). Consider the following nonlinear differential equation ( 2x(t) 2(t+1)6 2 −x0 (t) = ((x(t)) 2 +1)2 + (t+1)3 − ((t+1)4 +1)2 , x(0) = 1, which is in the form (4.65) with φ (x) =

x2 x2 +1

and f (t) =

2 (t+1)3

6

2(t+1) − ((t+1) 4 +1)2 . One

1 can easily verify that x(t) = (t+1) 2 is a solution which converges to zero as t → +∞, as predicted by Theorem 4.11.3.

Lemma 4.11.1 Suppose that x(t) is a solution to (4.65). If Argminφ , ∅, then limt→+∞ kx(t) − xk exists for each x ∈ Argminφ . Proof. Since x ∈ Argminφ , we have φ (x) ≤ φ (x(t)) for all t ≥ 0. By the quasiconvexity of φ , we have  ∇φ (x(t)), x − x(t) ≤ 0. Therefore  d kx(t) − xk2 = 2 x0 (t), x(t) − x dt  = 2 − ∇φ (x(t)) − f (t), x(t) − x ≤ 2k f (t)kkx(t) − xk.

(4.66)

First we prove that x(t) is bounded. By contradiction if x(t) is unbounded, there is an increasing sequence tn → +∞ such that kx(tn ) − xk → +∞, and kx(tn+1 ) − xk > 2kx(tn ) − xk, and kx(s) − xk < kx(tn+1 ) − xk, ∀s ∈ (tn ,tn+1 ). To show the existence of such a sequence, let f (t) = kx(t) − xk. f is continuous, and by contradiction it is unbounded. Define a sequence tn in this way: f (tn ) = maxn≤t≤n+1 f (t). Then define

66

Nonlinear Evolution and Difference Equations of Monotone Type

the subsequence n j as follows. Let n1 = 1, and n j+1 = min{k > n j : f (tk ) > f (tn j )}. Since f is unbounded, the sequence n j exists. By the definition of tn j , we have f (tk ) ≤ f (tn j ) < f (tn j+1 ), ∀n j < k < n j+1

(4.67)

Therefore f (tn j ) is unbounded, because f (tk ) is unbounded and since f (tn j ) is increasing, then f (tn j ) → +∞ as j → +∞. Now, choose a subsequence of tn j , which we again denote it by tn j such that f (tn j+1 ) ≥ 2 f (tn j ). Obviously this subsequence satisfies f (tk ) ≤ f (tn j+1 ), ∀n j ≤ k ≤ n j+1 by (4.67). Now, integrating (4.66) from tn j to tn j+1 , we get kx(tn j+1 ) − xk2 − kx(tn j ) − xk2 ≤ 2kx(tn j+1 ) − xk

Z tn j+1

k f (t)kdt.

tn j

Dividing both sides of the above inequality by kx(tn j+1 ) − xk, we get 3 kx(tn j ) − xk ≤ 2 2

Z tn j+1

k f (t)kdt.

tn j

We get a contradiction when j → +∞, because 0+∞ k f (t)kdt < +∞. Therefore x(t) is bounded. Let M := supt≥0 kx(t) − xk. Now integrating (4.66) from s to t > s, we get Z R

t

kx(t) − xk2 − kx(s) − xk2 ≤ 2M

k f (τ)kdτ.

s

Taking limsup as t → +∞ and liminf as s → +∞, we get that limt→+∞ kx(t) − xk exists.  Lemma 4.11.2 Suppose that x(t) is a solution to (4.65) and Argminφ , ∅, then lim φ (x(t)) exists. Proof. Since ∇φ is bounded on bounded subsets of H, by Equation (4.65) and Lemma 4.11.1, we have  d φ (x(t)) = ∇φ (x(t)), x0 (t) dt  = ∇φ (x(t)), −∇φ (x(t)) − f (t)  = −k∇φ (x(t))k2 − ∇φ (x(t)), f (t) ≤ k∇φ (x(t))kk f (t)k = k∇φ (x(t)) − ∇φ (x)kk f (t)k ≤ Lkx(t) − xkk f (t)k ≤ LMk f (t)k, where M = supt≥0 kx(t) − xk, L is the Lipschitz constant of ∇φ and x is a critical  point of φ . Now since f ∈ L1 (0, +∞); H , a similar proof as in Lemma 4.11.1 shows that limt→+∞ φ (x(t)) exists. 

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67

Theorem 4.11.3 Suppose that x(t) is a solution to (4.65). If Argminφ , ∅, then there is x ∈ H such that x(t) * x as t → +∞ and ∇φ (x) = 0. Proof. We consider the following two cases: 1) limt→+∞ φ (x(t)) = inf φ . Since x(t) is bounded by Lemma 4.11.1, there is a sequence {tn } and x˜ ∈ H such that x(tn ) * x, ˜ as n → +∞. By Lemma 4.11.2 and the weak lower semi-continuity of φ (by Mazur’s lemma), we have φ (x) ˜ ≤ lim inf φ (x(tn )) = lim φ (x(t)) = inf φ . n→+∞

t→+∞

Therefore x˜ ∈ Argminφ , which implies by Lemma 4.11.1 and Opial’s lemma [OPI], that x(t) * x ∈ Argminφ . 2) limt→+∞ φ (x(t)) > inf φ . Then there exist r > 0, t0 > 0 and x˜ ∈ Argminφ such that for all t ≥ t0 and every y ∈ B¯ r (x), ˜ φ (y) ≤ φ (x(t)). Now by the quasi-convexity of φ ,  ∇φ (x(t)) we have y − x(t), ∇φ (x(t)) ≤ 0. If ∇φ (x(t)) , 0, then letting y = x˜ + r k∇φ (x(t))k , we get:  rk∇φ (x(t))k ≤ x(t) − x, ˜ ∇φ (x(t)) 1 d ≤− kx(t) − xk ˜ 2 + Mk f (t)k, 2 dt where M = supt≥0 kx(t) − xk. ˜ By (4.66), this inequality  is obviously satisfied also 1 (0, +∞); H . By (4.65), this implies that if ∇φ (x(t)) = 0. Therefore ∇φ (x(t)) ∈ L  x0 (t) ∈ L1 (0, +∞); H . Therefore by the same argument as in Lemma 4.11.1, there is x ∈ H such that x(t) → x. Now by the continuity of ∇φ , ∇φ (x(t)) → ∇φ (x). Since ∇φ (x(t)) ∈ L1 (0, +∞); H , there exists a sequence tn → +∞ such that ∇φ (x(tn )) → 0. Therefore ∇φ (x) = 0.  Remark 4.11.4 Suppose that the assumptions of Theorem 4.11.3 are satisfied and φ is pseudo-convex. By the pseudo-convexity of φ , if ∇φ (x) = 0, then x ∈ Argminφ . In fact, suppose to the contrary that ∇φ (x) = 0 but x < Argminφ . Then there is an x0 ∈ H such that φ (x0 ) < φ (x). The pseudo-convexity of φ then implies that x0 − x, ∇φ (x) < 0, which is a contradiction. Therefore if φ is pseudo-convex, then x(t) * x ∈ Argminφ . Theorem 4.11.5 Suppose that x(t) is a solution to (4.65). If Argminφ , ∅ and either one of the following conditions is satisfied: a) x < Argminφ , where x is a weak cluster point of x(t), b) int(Argminφ ) , ∅, then x(t) → x and x is a critical point of φ . Proof. a) Suppose that x(tn ) * x < Argminφ . Then lim φ (x(t)) = lim inf φ (x(tn )) ≥ φ (x) > inf φ

t→+∞

n→+∞

Now the result follows from Case (2) in the proof of Theorem 4.11.3. b) If int(Argminφ ) , ∅, then there exist x˜ ∈ Argminφ , r > 0 and t0 > 0 such that

68

Nonlinear Evolution and Difference Equations of Monotone Type

for all t ≥ t0 and every y ∈ B¯ r (x), ˜ φ (y) ≤ φ (x(t)). Then by the quasi-convexity of φ ,  ∇φ (x(t)) y − x(t), ∇φ (x(t)) ≤ 0. Now if ∇φ (x(t)) , 0, then letting y = x˜ + r k∇φ (x(t))k , we have:  rk∇φ (x(t))k ≤ x(t) − x, ˜ ∇φ (x(t))  = x(t) − x, ˜ −x0 (t) − f (t) 1 d ≤− kx(t) − xk ˜ 2 + Mk f (t)k, 2 dt where M = supt≥0 kx(t) − xk. ˜ By (4.66), this inequality  is obviously satisfied also 1 (0, +∞); H . By (4.65), this implies that if ∇φ (x(t)) = 0. Therefore ∇φ (x(t)) ∈ L  x0 (t) ∈ L1 (0, +∞); H . Therefore by the same argument as in Lemma 4.11.1, there is x ∈ H such that x(t) → x as t → +∞. On the other hand, ∇φ (x(tn )) → 0 for a sequence tn → +∞ as n → +∞. Then the continuity of ∇φ implies that ∇φ (x) = 0.  Theorem 4.11.6 Suppose that x(t) is a solution to (4.65) and f (t) ≡ 0. If φ is even, then there is x ∈ H such that x(t) → x as t → +∞, where ∇φ (x) = 0. Proof. It follows from the proof of Lemma 4.11.2 that φ (x(t)) is non-increasing. Therefore for all t ≥ s, φ (x(t)) ≤ φ (x(s)). Then the quasi-convexity of φ implies that  ∇φ (x(s)), x(t) − x(s) ≤ 0. Since φ is even, φ (−x(t)) = φ (x(t)). Therefore again by the quasi-convexity of φ , for each t ≥ s, we get:  − x(t) − x(s), ∇φ (x(s)) ≤ 0. (4.68) Summing up the last two inequalities, we get:  x(s), ∇φ (x(s)) ≥ 0 ⇒

d kx(s)k2 ≤ 0. ds

Therefore kx(t)k is non-increasing. By (4.68) for each t ≥ s, we have   d x(t) + x(s), x0 (s) ≤ 0 ⇒ kx(s)k2 ≤ −2 x(t), x0 (s) . ds Integrating this inequality from s to t, we get:  kx(t)k2 ≤ kx(s)k2 − 2kx(t)k2 + 2 x(t), x(s) , ∀t > s From the above inequality, we get:  kx(t) − x(s)k2 = kx(t)k2 + kx(s)k2 − 2 x(t), x(s) ≤ 2(kx(s)k2 − kx(t)k2 ) → 0, as t, s → +∞. Therefore x(t) is a Cauchy net. So x(t) converges strongly to x ∈ H, and ∇φ (x) = 0, by Theorem 4.11.3. 

First Order Evolution Equations

69

Definition 4.11.7 Let f : H → (−∞, +∞] be proper, then f is said to be uniformly quasi-convex with modulus η : [0, +∞) −→ [0, +∞) if η is increasing, vanishes only at 0, and (∀ x, y ∈ dom f , ∀α ∈ (0, 1)) f (αx + (1 − α)y) + α(1 − α)η(kx − yk) 6 max{ f (x), f (y)} Example 4.11.8 We define f and η as follows:  2  x √ f (x) = 4 −x − 3   +∞

x > −1 −4 6 x 6 −1 x < −4

and η(x) =

x2 , ∀x ∈ [0, +∞). 16 + x2

f is not convex but it is uniformly quasi-convex with modulus η. Theorem 4.11.9 Suppose that x(t) is a solution to (4.65). If Argminφ , ∅ and φ is uniformly quasi-convex with modulus η, then x(t) converges strongly to the unique element of Argminφ . Proof. Let x˜ be the unique element of Argminφ , then φ (x) ˜ ≤ φ (x(t)). The uniform quasi-convexity of φ shows that  x˜ − x(t), ∇φ (x(t)) ≤ −η(kx(t) − xk) ˜  =⇒ 0 ≤ η(kx(t) − xk) ˜ ≤ x(t) − x, ˜ ∇φ (x(t))  −1 d = x(t) − x, ˜ −x0 (t) − f (t) ≤ kx(t) − xk ˜ 2 + Mk f (t)k, 2 dt where M = supkx(t) − xk. ˜ Now integrating both sides of the above inequality, we t>0

get: 0≤

Z +∞

η(kx(t) − xk)dt ˜

0

1 1 ≤ kx(0) − xk ˜ − lim kx(t) − xk ˜ +M t→∞ 2 2

Z +∞

k f (t)kdt < +∞.

0

Since lim kx(t) − xk ˜ exists and η is an increasing function which vanishes only at 0, t→∞

it follows that lim kx(t) − xk ˜ = 0. Hence, we conclude that x(t) → x. ˜ t→∞



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Nonlinear Evolution and Difference Equations of Monotone Type

REFERENCES BAR. V. Barbu, Nonlinear semigroups and Differential Equations in Banach Spaces, Noordhoff, Leyden, 1976. BAR-PRE. V. Barbu and Th. Precupanu, Convexity and Optimization in Banach Spaces, Editors Academiei, Buchrest, 1986 (and D. Reidel Publishing Company). BAI. J. B. Baillon, Un th´eor`eme de type ergodique pour les contractions non lin´eaires dans un espace de Hilbert, C. R. Acad. Sci. Paris 280 (1975), A1511–A1514. BRE. H. Br´ezis, “Op´erateurs Maximaux Monotones et Semi-groupes de Contractions dans les Espaces de Hilbert”, North-Holland Mathematics studies, Vol. 5, North-Holland Publishing Co., Amsterdam-London (1973). BRE-BRO. H. Br´ezis and F. E. Browder, Nonlinear ergodic theorems, Bull. Amer. Math. Soc. 82 (1976), 959–961. BRO. F.E. Browder, Nonexpansive nonlinear operators in a Banach space. Proc. Nat. Acad. Sci. USA 54, 1041–1044 (1965). BRU. R. E. Bruck, Asymptotic convergence of nonlinear contraction semigroups in Hilbert space, J. Funct. Anal. 18 (1975), 15–26. CAM-MAR. A. Cambini and L. Martein, Generalized Convexity and Optimization, Lecture Notes in Economics and Mathematicals Systems, 616. springer-Verlag, Berlin, 2009. DJA1. B. Djafari Rouhani, Ergodic theorems for nonexpansive sequences in Hilbert spaces and related problems, Ph.D. Thesis, Yale University, Part I, pp. 1–76 (1981). DJA2. B. Djafari Rouhani, Asymptotic behaviour of quasi-autonomous dissipative systems in Hilbert spaces, J. Math. Anal. Appl. 147 (1990), 465–476. DJA3. B. Djafari Rouhani, Asymptotic behaviour of almost nonexpansive sequences in a Hilbert space, J. Math. Anal. Appl. 151 (1990), 226–235. DJA4. B. Djafari Rouhani, An ergodic theorem for sequences in a Hilbert space, Nonlinear Anal. Forum, 4 (1999), 33–48. EDE. M. Edelstein, The construction of an asymptotic center with a fixed-point property, Bull. Amer. Math. Soc. 78 (1972), 206–208. GOH. D. G¨ohde, Zum Prinzip der kontraktiven Abbildung. Math. Nachr. 30 (1965), 251–258. GOU-MUN. X. Goudou and J. Munier, The gradient and heavy ball with friction dynamical systems: the quasiconvex case, Math. Program., Ser. B 116 (2009), 173–191. KHA-MOH. H. Khatibzadeh and V. Mohebbi, Non-homogeneous Continuous and Discrete Gradient Systems: The Quasi-convex Case, Bull. Iran. Math. Soc. KIR. W.A. Kirk, A fixed point theorem for mappings which do not increase distances. Amer. Math. Monthly 72 (1965), 1004–1006. LUC. R. Lucchetti, Convexity and well-posed problems. CMS Books in Mathematics/Ouvrages de Math´ematiques de la SMC, 22. Springer, New York, 2006. MIY. I. Miyadera, Nonlinear ergodic theorems for semigroups of Non-lipschitzian mappings in Banach Spaces II, Math J. Okayama, 43 (2001),123–135. MOR. G. Morosanu, Nonlinear Evolution Equations and Applications, Editura Academiei (and D. Reidel Publishing Company), Bucharest, 1988. OPI. Z. Opial, Weak convergence of the sequence of successive approximations for nonexpasive mappings, Bull. Amer. Math. Soc. 73 (1967), 591–597. ROC. R. T. Rockafellar, On the maximal monotonicity of subdifferential mappings, Pacific Math. J. 33 (1970), 209–216. WIT. R. Wittmann, Mean ergodic theorems for nonlinear operators, Proc. Amer. Math. Soc. 108 (1990), 781–788.

Order Evolution 5 Second Equations 5.1

INTRODUCTION

In this chapter, the existence and asymptotic behavior of solutions to the second order evolution equation of monotone type is studied. Let A : D(A) ⊂ H → H be a maximal monotone operator. Consider the following second order evolution equation associated to A: ( p(t)u00 (t) + q(t)u0 (t) ∈ Au(t), a.e. on(0, +∞) (5.1) u(0) = u0 ∈ D(A), supt≥0 ku(t)k < +∞ In the first section, we prove the existence of solutions to (5.1) with suitable assumptions on p(t) and q(t). The first results in this direction were proved by Barbu [BAR1, BAR2, BAR3] for the case p(t) ≡ 1 and q(t) ≡ 0, and by V´eron [VER1, VER2] for more general cases. In Section 2, we state the existence theorem of V´eron from [VER2]. Sections 3 and 4 are devoted to the nonhomogeneous case of the evolution equation with p(t) ≡ 1 and q(t) ≡ 0. In these sections, we consider the existence of solutions for a two point boundary value problem on an interval, as well as an existence result for a second order nonhomogeneous evolution equation on the positive axis. In Section 5, we consider the evolution equation ( u00 (t) ∈ Au(t) + f (t), a.e. t ∈ (0, +∞) (5.2) u(0) = u0 ∈ D(A), supt≥0 ku(t)k < +∞ where f is periodic with period T > 0. We prove the existence of a periodic solution to (5.2) and the weak convergence of solutions to a periodic solution. The results are due to Bruck [BRU1, BRU2]. In Section 6, the semigroup generated by (5.2) in the homogeneous case (i.e. when f ≡ 0) is considered and the square root of the maximal monotone operator generating this semigroup is defined. Sections 7 and 8 are devoted to the asymptotic behavior of solutions to (5.1) for the homogeneous and nonhomogeneous cases.

5.2

EXISTENCE AND UNIQUENESS OF SOLUTIONS

The main aim of this section is to study the existence of solutions to (5.1) with the following assumptions: p ∈ W 2,∞ (0, +∞), q ∈ W 1,∞ (0, +∞)

(5.3)

∃α > 0, such that ∀t ≥ 0, p(t) ≥ α.

(5.4)

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Nonlinear Evolution and Difference Equations of Monotone Type

Also to show the uniqueness if moreover: Z +∞ R t − 0

e

q(s) ds p(s)

dt = +∞

(5.5)

0

Let ρ(t) = exp this section. 5.2.1

R t q(s)−p0 (s) 0

p(s)

 ds . Theorems 5.2.1 and 5.2.11 are the main results of

THE STRONGLY MONOTONE CASE

In this subsection we consider the existence of solutions to (5.1), when the maximal monotone operator A is strongly monotone. Theorem 5.2.1 Let A : D(A) ⊂ H → H be a maximal monotone and β -strongly monotone operator such that 0 ∈ A(0). Then for each u0 ∈ D(A), there exists a unique continuously differentiable function u ∈ H 2 (0, +∞; H) that satisfies ( p(t)u00 (t) + q(t)u0 (t) ∈ Au(t), a.e. on [0, +∞) (5.6) u(0) = u0 , u(t) ∈ D(A), a.e. on [0, +∞) If u and v are solutions associated to the initial values u0 and v0 respectively, then ku(t) − v(t)k ≤ ku0 − v0 k.

(5.7)

The function kuk on [0, +∞) is nonincreasing. Moreover u ∈ Hρ2 (0, +∞; H). Proof. First we prove the uniqueness. If φ is a differentiable function, then we have 1 (pρφ )0 = pφ 0 + qφ ρ Let w = u − v. To show the inequality (5.7), by assumption, for almost all t ∈ [0, +∞) we have:  (pρw0 )0 (t), w(t) − β ρ(t)kw(t)k2 ≥ 0 (5.8) Since limt→+∞ kw(t)k = 0, then by contradiction, there is t1 > 0 such that kw(t1 )k > kw(0)k and kw(t1 )k = max0≤t t1 , since (w, w0 )(t1 ) = 12 dtd kw(t1 )k2 = 0, we get: ρ(t)p(t)(w, w0 )(t) ≥

Z t t1

ρ(t)p(t)kw0 (t)k2 + β

Z t

ρ(t)kw(t)k2 dt > 0,

t1

which is a contradiction with limt→+∞ kw(t)k = limt→+∞ kw0 (t)k = 0. Since 0 ∈ A0, the solution corresponding to the initial value 0 is 0. Therefore by (5.7) the function kuk is nonincreasing.  Now to prove the existence, we approximate u by a function uT which is the solution to a similar problem on a finite interval [0, T ]. To prove the existence we first need the following lemmas.

Second Order Evolution Equations

73

Lemma 5.2.2 The operator B defined by ( D(B) = {u ∈ H 2 (0, T ; H) s.t. u(0) = u0 ; u0 (T ) = 0} (Bu)(t) = −(ρ pu0 )0 (t)

(5.9)

is maximal monotone on L2 (0, T ; H). Proof. The monotonicity follows easily by integrating by parts. Let’s prove the maximality. We need to show that for a given f ∈ L2 (0, T ; H) there is u ∈ D(B) such that u + λ Bu = f for fixed λ > 0. Let H˜ 1 (0, T ; H) be the subset of H 1 (0, T ; H) consisting of the functions that are zero at 0. This is a Hilbert space with the same inner product. In this space, let the bilinear form bλ for λ > 0 be defined by Z T

bλ (φ , ψ) =

Z T

(φ , ψ)(t)dt + λ 0

ρ(t)p(t)(φ 0 (t), ψ 0 (t))dt

(5.10)

0

˜ T ; H) Then bλ is continuous, and µ > 0 exists such that for each φ ∈ H(0, bλ (φ , φ ) ≥ µkφ k2H(0,T ˜ ;H) ˜ T ; H) onto its dual in By Lax-Milgram Theorem, bλ defines an isomorphism of H(0, the following sense: ˜ T ; H))0 , ∃ψ ∈ H(0, ˜ T ; H) ∀L ∈ (H(0, such that ˜ T ; H), bλ (ψ, φ ) = L(φ ). ∀φ ∈ H(0, L2 (0, T ; H),

For f ∈ let g = f − u0 . Then g can be identified as an element of R ˜ T ; H))0 because the mapping φ 7→ 0T (g, φ )(t)dt is continuous on H(0, ˜ T ; H). (H(0, ˜ T ; H) exists such that for each φ ∈ H(0, ˜ T ; H) we have Therefore u˜ ∈ H(0, Z T

bλ (u, ˜ φ) =

(g, φ )(t). 0

Therefore in the sense of vectorial distributions, we have: u˜ − λ (ρ pu˜0 )0 = g

(5.11)

and since g ∈ L2 (0, T ; H), we have u˜00 ∈ L2 (0, T ; H). By choosing φ to be a function of class C1 that is 0 at zero, since (5.11) is satisfied almost everywhere, we deduce that: Z T 0

Z T

(g, φ )(t)dt = bλ (u, ˜ φ) =

Z T

(u, ˜ φ )dt + λ 0

ρ(t)p(t)(u˜0 , φ 0 )(t)dt

0

− λ ρ(T )p(T )(u˜0 , φ )(T ) Since this relation holds for each φ , then comparing it to (5.10), we conclude that u˜0 (T ) = 0. So the function u = u+u ˜ 0 belonges to D(B) and the relation u+λ Bu = f is satisfied. 

74

Nonlinear Evolution and Difference Equations of Monotone Type

¯ = D(A), and Now let A¯ = A − β I. Then A¯ is maximal monotone in H, D(A) ¯ ¯ ¯ 0 ∈ A0. Let J¯λ and Aλ be respectively the resolvent and Yosida approximation of A. Lemma 5.2.3 There is a unique function uλ ,T that is twice continuously differentiable and belongs to H 3 (0, T ; H) such that ( p(t)u00λ ,T (t) + q(t)u0λ ,T (t) − β uλ ,T (t) = A¯ λ (uλ ,T (t)), 0 ≤ t ≤ T (5.12) uλ ,T (0) = u0 , u0λ ,T (T ) = 0 Proof. For simplicity, we denote u = uλ ,T . Using the operator B, (5.12) can be written as: (Bu)(t) + ρ(t)A¯ λ u(t) + β ρu(t) = 0 (5.13) Define the operator Uλ on L2 (0, T ; H) by (Uλ u)(t) = (ρ A¯ λ u)(t) + β (ρu)(t). It is easy to see that Uλ is maximal monotone, strongly monotone and Lipschitzian on L2 (0, T ; H). By the previous lemma, Uλ +B is surjective and u is the unique element in (Uλ + B)−1 (0). By the regularity of u and the Lipschitz property of A¯ λ , we can get (5.12) and u ∈ H 3 (0, T ; H). The uniqueness is proved in a similar way as in the previous theorem, by showing that kuk is nonincreasing.  In order to pass to the limit, we now need some estimates on the norm of uλ ,T in H 2 (0, T ; H) ∩ Hρ2 (0, T ; H) where kuk2L2 ((0,+∞);H) :=

R +∞

ku(t)k2 ρ(t)dt.

0

ρ

Lemma 5.2.4 There exists a constant M, independent of λ and T , such that if uλ ,T is the solution to (5.12), then ( (i) kuλ ,T kH 2 (0,T ;H) ≤ M (5.14) (ii) kuλ ,T kHρ2 (0,T ;H) ≤ M Proof. Again for simplicity we denote uλ ,T = u. Using the monotonicity of A¯ λ , for each t ∈ [0, T ] we have: p(t)(u00 , u)(t) + q(t)(u0 , u)(t) − β ku(t)k2 ≥ 0 Integrating the above inequality by parts on the interval [0, T ], we get: −

Z T

p(t)ku0 (t)k2 dt − β

Z T

0

0

Since (u0 , u)(t) = Z T

α 0

1 d 2 2 dt ku(t)k

ku0 (t)k2 dt + β

ku(t)k2 dt + [p(u0 , u)]T0 +

(q − p0 )(t)(u0 , u)(t)dt ≥ 0

0

≤ 0, and u0 (T ) = 0 and p(t) ≥ α, we get:

Z T 0

Z T

ku(t)k2 ≤ −p(0)(u0 , u0 (0)) + k

q − p0 kL∞ ku0 k2 2

Second Order Evolution Equations

75

Set wh (t) = u(t + h) − u(t). Using the monotonicity of A¯ λ and Equation (5.12), we get: (p(t + h)u00 (t + h) − p(t)u00 (t), wh (t)) + (q(t + h)u0 (t + h) − q(t)u0 (t), wh (t)) − β kwh (t)k2 ≥ 0 Dividing both sides of the above inequality by h2 , and taking the limit as h → 0, since u ∈ H 3 (0, T ; H), we get: p(t)(u000 , u0 )(t) + (q + p0 )(u00 , u0 )(t) + (q0 (t) − β )ku0 (t)k2 ≥ 0

(5.15)

Integrating by parts the above inequality on the interval [0, T ], we get: −

Z T 0

p(t)ku00 (t)k2 dt + [p(u00 , u0 )]T0 +

 q 0 2 T Z T q0 (t) ku k 0 + ( − β )ku0 (t)k2 dt ≥ 0 2 2 0

Therefore Z T

α

ku00 (t)k2 dt ≤ k

0

q0 − β kL∞ 2

Z T

ku0 (t)k2 dt −

0

q(0) 0 ku (0)k2 − p(0)(u00 (0), u0 (0)) 2

Clearly u00 (0) can be estimated directly from (5.12). Then q(0) 0 q(0) 0 ku (0)k2 − p(0)(u00 (0), u0 (0)) = −β (u0 , u0 (0))+ ku (0)k2 −(A¯ λ u0 , u0 (0)) 2 2 ¯ by Theorem 3.4.1, we have: and since u0 ∈ D(A), −

kA¯ λ u0 k ≤ kA¯ 0 u0 k Therefore: Z T

α 0

ku00 (t)k2 dt ≤ k

q0 − β kL∞ 2

Z T 0

ku0 (t)k2 dt + M1 ku0 (0)k2 + N1 ,

where M1 and N1 are independent of λ and T . We have therefore obtained the following estimates  RT 2 0 2  (i) R0 ku(t)k dt ≤ A0 + B0 ku (0)k , (5.16) (ii) 0T ku0 (t)k2 dt ≤ A1 + B1 ku0 (0)k2 ,  R T 00  2 0 2 (iii) 0 ku (t)k dt ≤ A2 + B2 ku (0)k , where Ai , Bi , i = 0, 1, 2, are positive constants independent of λ and T . Since ku0 (0)k2 = −2

Z T

(u00 , u0 )(t)dt,

0

then by using the Cauchy-Shwarz inequality, and (5.16) (ii) and (iii), we get: ku0 (0)k4 ≤ 4(A1 + B1 ku0 (0)k2 )(A2 + B2 ku0 (0)k2 ). This inequality implies the boundedness of ku0 (0)k independently of λ and T , from which we get (5.14) (i). (5.14) (ii) is proved in a similar way. See [VER2] for details. 

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Nonlinear Evolution and Difference Equations of Monotone Type

Lemma 5.2.5 {uλ }λ >0 is a Cauchy net in H 1 (0, T ; H). Proof. We know that A¯ λ uλ ∈ A¯ J¯λ uλ . Now replacing J¯λ uλ by its value obtained from I−J¯ the equality A¯ λ = λ λ replaced in (5.12), we obtain: ¯ + λ β )uλ (t) − λ p(t)u00 (t) − λ q(t)u0 (t)]. p(t)u00λ (t) + q(t)u0λ − β uλ (t) ∈ A[(1 λ λ ¯ we get: Now letting x = λ uλ − µuµ and w = uλ − uµ , by the monotonicity of A,  (ρ pw0 )0 (t) − β ρ(t)w(t), w(t) + β x(t) − p(t)x00 (t) − q(t)x0 (t) ≥ 0, t ∈ [0, T ] Integrating the above inequality by parts on [0, T ], and using the given boundary conditions, we get: Z T

ρ(t)p(t)kw0 (t)k2 dt + β

0



Z T

Z T

ρ(t)kw(t)k2 dt

0

 (ρ pw0 )0 (t) − β ρ(t)w(t), β x(t) − p(t)x00 (t) − q(t)x0 (t) dt.

0

By Lemma 5.2.4, w remains bounded in H 2 (0, T ; H). Therefore there exists N independent of λ and µ such that kxkH 2 (0,T ;H) ≤ (λ + µ)N. By developing the integral on the right hand side, we see that there exists K independent of λ and µ such that Z T

0

2

Z T

ρ(t)p(t)kw (t)k dt + 0

β ρ(t)kw(t)k2 dt ≤ (λ + µ)K.

0

This completes the proof of the lemma, since the norms in L2 (0, T ; H) and Lρ2 (0, T ; H) are equivalent.  Proposition 5.2.6 For each u0 ∈ D(A), there exists a unique continuously differentiable function uT ∈ H 2 (0, T ; H) that satisfies ( p(t)u00T (t) + q(t)u0T (t) ∈ AuT (t), a.e. on [0, T ) (5.17) uT (0) = u0 , u0T (T ) = 0, uT (t) ∈ D(A), a.e. on [0, T ) If uT and vT are the solutions to (5.17) associated respectively to u0 and v0 , then kuT (t) − vT (t)k ≤ ku0 − v0 k, ∀ t ∈ [0, T ]

(5.18)

The function kuT k is nonincreasing and there exists a constant M independent of T such that ( kuT kHρ2 (0,T ;H) ≤ M (5.19) kuT kH 2 (0,T ;H) ≤ M

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77

Proof. Unicity follows from (5.18) whose proof is similar to the one given in Theorem 5.2.1. Let uT be the limit of uλ ,T in H 1 (0, T ; H), which exists by Lemma 5.2.5. Since u00λ ,T converges to u00T in D 0 (0, T ; H), it follows from Lemma 5.2.4 that u00λ ,T converges weakly to u00T in L2 (0, T ; H), as λ tends to zero. Moreover, since k(u0λ − u0µ )(t)k2 = −2

Z T t

(u0λ − u0µ , u00λ − u00µ )(t)dt

it follows that the net (uλ )λ >0 converges in C1 ([0, T ]; H). Let A¯ be the canonical extension of A¯ to the space L2 (0, T ; H). A¯ is maximal monotone in L2 (0, T ; H)(see [BRE]). By (5.12), in L2 (0, T ; H) we have: pu00λ + qu0λ − β uλ ∈ A¯J¯λ uλ

(5.20)

and kJ¯λ uλ − uT k ≤ λ kA¯ λ uλ k + kuλ − uT k. By Lemma 5.2.4, uλ is bounded in H 2 (0, T ; H), hence Aλ uλ also is bounded in L2 (0, T ; H). Therefore, passing to the limit in (5.20) as λ → 0, and since A¯ is demiclosed, we get: ( ¯ T (t) p(t)u00T (t) + q(t)u0T (t) − β uT (t) ∈ Au (5.21) 0 uT (0) = u0 , uT (T ) = 0, uT (t) ∈ D(A), a.e. on [0, T ] 

which proves the proposition.

To prove the main result of this section (Theorem 5.2.1), we need to pass to the limit as T → +∞, and for this, we need the following lemmas. Lemma 5.2.7 Let xL be the solution to the following equation on [0, L]. ( p(t)xL00 (t) + q(t)xL0 (t) − β xL (t) = 0, on [0, L] xL0 (0) = 0, xL (L) = 2ku0 k

(5.22)

Then xL ∈ H 3 (0, L) and kxL kH 3 (0,L) ≤ M where M is independent of L. xL is a positive nondecreasing function, and for L0 fixed, we have: lim

L≥L0 ,L→+∞

kxL kH 2 (0,L0 ) = 0.

Proof. Let y = xL (L − t), p(t) ˜ = p(L − t), and q(t) ˜ = −q(L − t). Then we get: ( 00 (t) + qy p(t)y ˜ ˜ 0 (t) − β y(t) = 0 on [0, L] (5.23) y(0) = 2ku0 k, y0 (L) = 0 By Lemmas 5.2.3 and 5.2.4 (with H = R and A¯ = 0), the first three claims in the lemma are proved. Since the embedding of H 3 (0, L) into H 2 (0, L) is compact, there exist a function x ∈ H 3 (0, +∞) and a subnet of xL such that for each L0 , we have lim kxL − xkH 2 (0,L0 ) = 0. The function t 7→ x(t) is positive and nondecreasing as the limit of such functions, and since it belongs to L2 (0, +∞), it must be the constant zero. Now a classical unicity argument finishes the proof of the lemma. 

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Nonlinear Evolution and Difference Equations of Monotone Type

Lemma 5.2.8 Suppose that φ is a function in H 2 (0, L; H) that satisfies the following relations: ( φ (0) = 0, kφ (L)k ≤ 2ku0 k (5.24) p(t)(φ 00 , φ )(t) + q(t)(φ 0 , φ )(t) − β kφ (t)k2 ≥ 0 a.e. on [0, L]. Then kφ k is a nondecreasing function that is bounded above by the solution xL of the Equation (5.22). Proof. If φ (L) = 0, then after integrating by parts the relation (5.24) on [0, L], it follows immediately that φ is the constant zero. Otherwise, there exists t1 ∈ (0, L) such that kφ (t1 )k ≥ kφ (L)k. Without loss of generality, we may assume that kφ (t1 )k is maximal and therefore (φ , φ 0 )(t1 ) = 0. Integrating by parts on [t1 ,t] with t > t1 , we get: 1 d ρ(t)p(t) kϕ(t)k2 = ρ(t)p(t)(φ , φ 0 )(t) 2 dt ≥β

Z t t1

ρ(s)kφ (s)k2 ds +

Z t

ρ(s)p(s)kφ 0 (s)k2 ds ≥ 0

t1

which is a contradiction since dtd kϕ(t)k2 < 0 for t near t1 . By replacing L with an arbitrary t, it follows that kϕk is nondecreasing, and the first claim is proved. Suppose that (t0 , L) is the interval where kφ (t)k > 0. The function kφ k belongs to 2,1 Wloc (t0 , L) and we have: (kφ k2 )0 = 2(φ , φ 0 ) = 2kφ k(kφ k)0 ⇒ kφ 0 k ≥ |(kφ k)0 | (kφ k2 )00 = 2kφ 0 k2 + 2(φ , φ 00 ) = 2(kφ k0 )2 + 2kφ k(kφ k)00 and therefore (φ , φ 00 ) ≤ kφ k(kφ k)00 a.e. on (t0 , L). This implies that on (t0 , L) we have: p(t)kφ k00 (t) + q(t)kφ k0 (t) − β kφ (t)k ≥ 0. Let w = kφ k − xL . Then w satisfies the following relations: ( 2,1 w ∈ Wloc (t0 , L), w(t0 ) ≤ 0, w(L) ≤ 0 00 p(t)w (t) + q(t)w0 (t) − β w(t) ≥ 0, a.e. on (t0 , L).

(5.25)

Now by contradiction, assume that w(t1 ) > 0 for some t0 < t1 < L, where without loss of generality, we may assume that w(t1 ) is maximal. Then multiplying the relation (5.25) by w(t) for t > t1 , and integrating by parts on [t1 ,t], with a similar reasoning as above, we get a contradiction. Therefore w(t) ≤ 0, and this proves the theorem.  Proposition 5.2.9 For each fixed L, the net (uT )T ≥L defined in Proposition 5.2.6 is Cauchy in C1 ([0, L]; H).

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79

Proof. Suppose that uT and uT 0 are solutions on [0, T ] and [0, T 0 ] to the Equation (5.17) with the same initial value u0 . Suppose that T 0 > T and let L and L0 be such that 0 < L < L0 < T < T 0 . The function φ = uT − uT 0 satisfies the hypothesis of Lemma 5.2.8 on [0, L0 ], and therefore kφ k is bounded above by the solution xL0 to (5.22) on [0, L0 ]. By Lemma 5.2.7, for L fixed, we have limL0 →+∞ kφ (t)k = 0 uniformly on [0, L]. Moreover Z L 0

ρ(t)p(t)kw0 (t)k2 dt + β

Z L

ρ(t)kφ (t)k2 dt ≤ ρ(L)p(L)(φ , φ 0 )(L)

0

and since by Lemma 5.2.4, φ 0 (L) is bounded, we deduce that the net {uT }T >L is Cauchy in H 1 (0, L; H). Since u00T remains bounded in L2 (0, T ; H), by a similar proof as in Proposition 5.2.6, we conclude that this net is Cauchy in C1 ([0, L]; H), for each fixed L.  Now we complete the proof of Theorem 5.2.1: Proof. We have already proved the uniqueness. To prove the existence, by Proposition 5.2.9 and Lemma 5.2.4, there exists a function u belonging to H 2 (0, +∞; H) and Hρ2 (0, +∞; H), which is the limit of the net (uT ) in C1 ([0, L]; H), and moreover w − lim u00T = u00 in L2 (0, L; H) for each L > 0. Similar to the proof of Proposition 5.2.6, it follows that u(t) ∈ D(A), a.e. on (0, +∞), and that u is the solution to (5.6). This completes the proof of the theorem.  5.2.2

THE NON STRONGLY MONOTONE CASE

In this subsection, we assume that the assumptions (5.3), (5.4) and (5.5) are satisfied, and the operator A is maximal monotone and 0 ∈ A0, but A is not strongly monotone, and prove the existence of the solution to (5.1). Proposition 5.2.10 For every u0 ∈ D(A), there is a unique continuously differentiable function uT such that u00T ∈ L2 (0, T ; H) and satisfies the Equation (5.17). The function kuT k is nonincreasing and there exists M independent of T such that ( ku0 kH 1 (0,T ;H) ≤ M ku0 kHρ1 (0,T ;H) ≤ M

(5.26)

Proof. The proof of the uniqueness is similar to the proof of Proposition 5.2.6. We prove the existence. For δ > 0, let uδ be the unique solution in H 2 (0, T ; H) of the equation ( uδ (0) = u0 , u0δ (T ) = 0, uδ (t) ∈ D(A), a.e. on[0, T ] p(t)u00δ (t) + q(t)u0δ (t) − δ uδ (t) ∈ Auδ (t), a.e. on [0, T ]

(5.27)

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Nonlinear Evolution and Difference Equations of Monotone Type

(the existence of uδ follows from Proposition 5.2.6). By Remarks 2.6 and 2.7 from [VER2], there exist M independent of T and δ ≤ δ0 , such that ( ku0δ kH 1 (0,T ;H) ≤ M (5.28) ku0δ kHρ1 (0,T ;H) ≤ M and moreover the function kuδ k is nonincreasing. For each γ and δ ≤ δ0 , consider the solutions uγ and uδ of (5.27), and set w = uγ − uδ . By the monotonicity of A, we have: Z Z T



T

ρ(t)p(t)kw0 (t)k2 dt +

0

0

ρ(t)(δ uδ − γuγ , w)(t)dt ≥ 0.

Therefore by using (5.28), we get kw0 k2L2 (0,T ;H) ≤ K(γ +δ ). Now by letting γ, δ → 0, ρ

the proof is completed in a similar way as in Proposition 5.2.6.



Theorem 5.2.11 For each u0 ∈ D(A), there is a unique continuously differentiable function u, bounded on R+ , such that u0 ∈ H 1 (0, +∞; H) and u satisfies (5.6) and (5.7). Moreover kuk is nonincreasing and u0 ∈ Hρ1 (0, +∞; H). Proof. We first prove the uniqueness. Suppose that u1 and u2 are two solutions to (5.6) with the same initial value u0 , and let w = u1 − u2 . By the monotonicity of A, we have: Z t ρ(t)p(t)(w0 , w)(t) ≥ ρ(s)p(s)kw0 (s)k2 ds ≥ 0. 0

Therefore the function kw(t)k is nondecreasing. By contradiction, assume that w is not identically zero and let (T, +∞) be the interval where it is different from zero. Then after some computations (see [VER2]), we get the following inequality: R s q(τ) 0 exp(− 0 p(τ) dτ)ds ku1 − u2 kL∞ R n , R s q(τ) 0 exp(− 0 p(τ) dτ)ds Rt

kw(t)k ≤

∀t ∈ [T, n].

For fixed t, letting n → +∞ in the above inequality, it follows from (5.5) that w(t) = 0, ∀t ≥ T . Therefore T = +∞, and the unicity follows. For the existence, suppose that L < T < T 0 and w = uT − uT 0 , where uT and uT 0 are solutions to (5.17)(see Proposition 5.2.6) on [0, T ] and [0, T 0 ] respectively. We have: ( p(t)(w00 , w)(t) + q(t)(w0 , w)(t) ≥ 0, a.e. on [0, T ] (5.29) w(0) = 0, kw(T )k ≤ 2ku0 k. Then with a similar argument as in the proof of the unicity, we get: Rt R s q(τ)  0 exp − 0 p(τ) dτ ds kw(t)k ≤ 2ku0 k R T R s q(τ)  0 exp − 0 p(τ) dτ ds which converges uniformly to zero on every compact interval, as T tends to infinity. It is easily shown that w0 is bounded in H 1 (0, T ; H); then a similar argument as

Second Order Evolution Equations

81

in Proposition 5.2.6 shows that the net (uT )T >L is Cauchy in H 1 (0, L; H) for every L, and hence also in C1 ([0, L]; H). The proof is completed in a similar way as in Theorem 5.2.1. 

5.3

TWO POINT BOUNDARY VALUE PROBLEMS

In this section we consider the boundary value problem of the form: ( u00 (t) ∈ Au(t) + f (t), a.e. on (0, T ) u(0) = a, u(T ) = b

(5.30)

Definition 5.3.1 The function u : [0, T ] → H is called a solution of the two point boundary value problem (5.30) if u ∈ W 2,2 (0, T ; H), and u(t) ∈ D(A) for almost every t ∈ (0, T ), and u satifies the boundary conditions u(0) = a and u(T ) = b, and for almost every t ∈ (0, T ), u00 (t) ∈ Au(t) + f (t). We first study the existence problem for the case a, b ∈ D(A). Theorem 5.3.2 If A : D(A) ⊂ H → H is a maximal monotone operator in a Hilbert space H, a, b ∈ D(A) and f ∈ L2 (0, T ; H), then the problem (5.30) has a unique solution u ∈ W 2,2 (0, T ; H). If u1 and u2 are solutions to (5.30), then t 7→ ku1 (t) − u2 (t)k2 is a convex function and ku1 (t) − u2 (t)k ≤ max{ku1 (0) − u2 (0)k, ku1 (T ) − u2 (T )k} Z T 0

1 γ(t)ku01 (t) − u02 (t)kdt ≤ (ku1 (0) − u2 (0)k2 + ku1 (T ) − u2 (T )k2 ) 2

(5.31) (5.32)

where t ∈ [0, T ], and γ(t) = min{t, T − t} Proof. We start with the proof of uniqueness. Suppose that u1 and u2 are two solutions to (5.30). Then by the monotonicity of A, we get (ku1 − u2 k2 )00 ≥ 2ku01 − u02 k2 a.e. on (0, T ).

(5.33)

Therefore the function t 7→ ku1 (t) − u2 (t)k2 is convex on [0, T ], and (5.31) is satisfied. If u1 and u2 satisfy the boundary conditions, then the uniqueness follows. Multiplying (5.33) by γε (t) = min{t − ε, T − t − ε} for t ∈ (0, T ), and ε ∈ (0, T2 ) constant, and integrating from t = ε to t = T − ε, we get Z T −ε

2 ε

T −ε Z γε ku01 (t) − u02 (t)k2 dt ≤ γε (ku1 − u2 k2 )0 ε −

T 2

(ku1 (t) − u2 (t)k2 )0 dt

ε

Z T −ε

+

T 2

(ku1 (t) − u2 (t)k2 )0 dt

82

Nonlinear Evolution and Difference Equations of Monotone Type

or Z T −ε

2

γε ku01 (t) − u02 (t)k2 dt ≤ γε (T − ε)(ku1 − u2 k2 )0 (T − ε)

ε

− γε (ε)(ku1 − u2 k2 )0 (ε) + ku1 − u2 k2 (ε) + ku1 − u2 k2 (T − ε) Letting ε → 0+ , and using Fatou’s lemma, we get (5.32). Now we prove the existence result. Consider the following problem ( u00λ = Aλ uλ + f , a.e. on (0, T ) uλ (0) = a, uλ (T ) = b

(5.34)

We show that this problem has a unique solution uλ ∈ W 2,2 (0, T ; H) that converges uniformly on [0, T ] to a solution u of (5.30). We replace (5.34) with a boundary value problem with zero boundary conditions. To this end, we replace the function uλ with vλ = uλ − h where h(t) = Tt b + TT−t a, t ∈ [0, T ]. Then the above problem becomes ( v00λ − λ1 vλ = λ1 h − λ1 Jλ (vλ + h) + f , a.e. on (0, T ) (5.35) vλ (0) = vλ (T ) = 0. For all α ∈ L2 (0, T ; H), the problem ( v00λ − λ1 vλ = λ1 h − λ1 Jλ (α + h) + f , a.e. on (0, T ) vλ (0) = vλ (T ) = 0,

(5.36)

has a unique solution vαλ ∈ W 2,2 (0, T ; H). Now consider the operator C : L2 (0, T ; H) → L2 (0, T ; H) defined by Cα = vαλ . We show that C is a contraction. β

For α, β ∈ L2 (0, T ; H), let vαλ , vλ ∈ W 2,2 (0, T ; H) be the corresponding solutions to (5.36). We rewrite (5.36) in the form: vαλ =

1 1 1 ( I + B)−1 Jλ (α + h) − (I + λ B)−1 h − ( I + B)−1 f λ λ λ

(5.37)

where Jλ is defined by (Jλ v)(t) = Jλ v(t) for all v ∈ L2 (0, T ; H), and the operator B is defined by Bu = −u00 , D(B) = {u ∈ W 2,2 (0, T ; H), u(0) = u(T ) = 0} We observe that B is linear and maximal monotone on L2 (0, T ; H). Therefore −B generates a C0 semigroup of linear nonexpansive mappings on L2 (0, T ; H). Moreover for each µ > 0 and each u ∈ L2 (0, T ; H)  (µI + B)−1 u (t) =

T αt sinh(α − ) α sinh α T

Z t

sinh 0

αs u(s)ds T

Second Order Evolution Equations

83

T αt + sinh α sinh α T where α = T



Z T

sinh(α −

t

αs )u(s)ds, T

µ. Since (Bu, u) ≥ ckuk2 , ∀u ∈ D(B)

we deduce that k(µI + B)−1 u − (µI + B)−1 (v)k ≤

1 ku − vk, µ > 0. µ +c

By the nonexpansiveness of Jλ and (5.37) one obtains 1 1 1 k( I + B)−1 Jλ (α + h) − ( I + B)−1 Jλ (β + h)k λ λ λ 1 ≤ kα − β k 1+λc

kCα −Cβ k =

Therefore C is a contraction on L2 (0, T ; H) and hence has a unique fixed point α ∈ L2 (0, T ; H) that clearly is vαλ . Hence the problem (5.35) and consequently (5.34) has a unique solution uλ ∈ W 2,2 (0, T ; H). Now we prove that uλ → u in C([0, T ]; H) where u satisfies (5.30). To this end, we multiply (5.34) by u00λ and integrate from 0 to T , then we get: Z T 0

Z T T Z T ku00λ k2 dt = (Aλ uλ , u0λ ) 0 − ((Aλ uλ )0 , u0λ )dt + ( f , u00λ )dt. 0

(5.38)

0

We note that (Aλ uλ )0 exists for almost every t ∈ (0, T ), because the function t 7→ Aλ uλ (t) is Lipschitz continuous. By the monotonicity of Aλ , we have:  (Aλ uλ )0 , u0λ ≥ 0, a.e. on (0, T ). Since (Aλ b, u0λ (T ))−(Aλ a, u0λ (0)) =

Z T t 0

T −t 1 ( Aλ b+ Aλ a, u00λ )dt + (Aλ b−Aλ a, b−a) T T T

Equality (5.38) implies that  Z T Z 1 1 ku00λ k2 dt ≤ T 2 kA0 ak + T 2 kA0 bk + ( 0

0

1 + (kA0 ak + kA0 bk)kb − ak T

T

Z 1 k f k2 dt) 2 ( 0

T

1

ku00λ k2 dt) 2 (5.39)

This together with (5.34) implies the boundedness of {u00λ } and {Aλ uλ } in L2 (0, T ; H). Now we prove that uλ is a Cauchy sequence in C([0, T ]). Taking into account (u0λ − u0µ , uλ − uµ )0 = (u00λ − u00µ , uλ − uµ ) + ku0λ − u0µ k2

84

Nonlinear Evolution and Difference Equations of Monotone Type

almost everywhere on (0, T ), and using (5.34) and x = λ Aλ x + Jλ x, we get: Z T 0

ku0λ − u0µ k2 dt = − −

Z T 0

Z T 0

(Aλ uλ − Aµ uµ , λ Aλ uλ − µAµ uµ )dt (Aλ uλ − Aµ uµ , Jλ uλ − Jµ uµ )dt

Now the monotonicity of Aλ and the boundedness of Aλ uλ in L2 (0, T ; H) imply that: Z T

ku0λ − u0µ k2 dt ≤ K(λ + µ).

(5.40)

kuλ (t) − uµ (t)k2 ≤ K 0 (λ + µ)

(5.41)

0

Therefore K, K 0

where are positive constants. Then uλ is uniformly convergent on [0, T ] to a function u and u0λ → u0 in L2 (0, T ; H). Since u00λ is bounded in L2 (0, T ; H), and u00λ → u00 in D 0 (0, T ; H), we conclude that u00λ * u00 in L2 (0, T ; H). Moreover, we have the strong convergence of u0λ → u0 in C(0, T ; H) and Jλ uλ = λ Aλ uλ + uλ → u in C([0, T ]; H). Since A¯ (the canonical extension of A) is demiclosed (by Proposition 3.3.4), and Jλ uλ → u and u00λ * u00 ¯ in L2 (0, T ; H), letting λ → 0+ in (5.34) rewritten in the form u00λ − f ∈ A(J λ uλ ), we conclude that u ∈ D(A) and u is a solution to (5.30).  Remark 5.3.3 The solution u to problem (5.30) can be approximated by the unique solution uλ of the problem (5.34), because uλ → u in C([0, T ]; H). Moreover u0λ → u0 in C([0, T ]; H) and u00λ * u00 in L2 (0, T ; H). In the sequel, we find an estimation for u00 in a weighted function space. Suppose that γ : [0, T ] → R is the function defined by γ(t) = min{t, T − t} for t ∈ [0, T ]. Suppose that Lγ2 (0, T ; H) is the function space L2 (0, T ; H) with the weighted function γ. This means that the inner product of Lγ2 (0, T ; H) is defined by Z T

(u, v)γ =

γ(t)(u(t), v(t))dt 0

and the norm | · |γ is defined by |u|2γ =

Z T

γ(t)ku(t)k2 dt.

0

Similarly we define the weighted space Lγ23 (0, T ; H). Let (·, ·)γ 3 and | · |γ 3 be respectively the inner product and the norm for this space. We observe that L2 (0, T ; H) ⊂ Lγ2 (0, T ; H) ⊂ Lγ23 (0, T ; H) and

1

1

3

3

u ∈ Lγ2 (0, T ; H) ⇔ t 2 (T − t) 2 u(t) ∈ L2 (0, T ; H) u ∈ Lγ23 (0, T ; H) ⇔ t 2 (T − t) 2 u(t) ∈ L2 (0, T ; H)

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85

Proposition 5.3.4 With the assumptions of Theorem 5.3.2, if u ∈ W 2,2 (0, T ; H) is a solution to (5.30), then |u00 |γ 3 ≤ | f |γ 3 + 3|u0 |γ (5.42) Proof. Again we consider the approximate problem (5.34), and multiply the equation by Aλ uλ , and use the estimate (u0λ , Aλ uλ )0 ≥ (u00λ , Aλ uλ ) to get (u0λ , Aλ uλ )0 ≥ kAλ uλ k2 + ( f , Aλ uλ )

(5.43)

Multiplying (5.43) by γ 3 and integrating on [0, T ] we get: Z T 0

Z T

f 1 γ 3 kAλ uλ + k2 dt − 2 4

Z T

γ 3 k f k2 dt ≤

0

γ 3 (u0λ , Aλ uλ )0 dt

0

(5.44)

Integrating by parts, and using the equality γ(0) = γ(T ) = 0, we find: Z T 0

f 1 γ kAλ uλ + k2 dt − 2 4 3

Z T

≤3

γ 3 k f k2 dt

0

Z T 0

3 + 2

γku0λ k2 dt

Z T 0

T

1 Z

1 f γ 3 kAλ uλ + k2 dt 2 2

2

0

γku0λ k2 dt

1 Z 2

T

1

γ 3 k f k2 dt) 2

0

Therefore Z T 0

1 f 1 γ kAλ uλ + k2 dt 2 ≤ 2 2 3

Z T

3

2

1 2

γ k f k dt) + 3

Z T

0

0

γku0λ k2 dt

1 2

Since Aλ uλ = u00λ − f we get: Z T 0

γ 3 ku00λ k2 dt

1 2



Z T 0

γ 3 k f k2 dt

1 2

Z T

+3 0

γku0λ k2 dt

1 2

(5.45)

Letting λ → 0+ , and using the convergence u00λ * u00 in Lγ23 (0, T ; H) and u0λ → u0 in Lγ2 (0, T ; H) which follows from Remark 5.3.3, we get (5.42).



Theorem 5.3.5 If A : D(A) ⊂ H → H is maximal monotone on H and a, b ∈ D(A) and f ∈ L2 (0, T ; H), then the problem (5.30) has a unique solution u ∈ C([0, T ]; H) ∩ 2,2 Wloc (0, T ; H). Moreover, u00 ∈ Lγ23 (0, T ; H) and u0 ∈ Lγ2 (0, T ; H). Proof. The proof of uniqueness is similar to the proof of Theorem 5.3.2. We prove the existence. Since a, b ∈ D(A), there are sequences an and bn in D(A) such that an → a and bn → b in H. By Theorem 5.3.2 the problem ( u00n ∈ Aun + f (5.46) un (0) = an , un (T ) = bn

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Nonlinear Evolution and Difference Equations of Monotone Type

has a unique solution un ∈ W 2,2 (0, T ; H). Inequality (5.31) for un and um shows that un is a Cauchy sequence in C([0, T ]; H), therefore a function u ∈ C([0, T ]; H) exists such that un → u in C([0, T ]; H) and u(0) = a and u(T ) = b. By (5.32), u0n → u0 in Lγ2 (0, T ; H). Moreover since f ∈ L2 (0, T ; H) ⊂ Lγ23 (0, T ; H), Proposition 5.3.4 2,2 implies the boundedness of u00n in Lγ23 (0, T ; H). Therefore u ∈ Wloc (0, T ; H) and u0 ∈

Lγ2 (0, T ; H) and u00 ∈ Lγ23 (0, T ; H) and u00n * u00 in Lγ23 (0, T ; H). Suppose that ε ∈ (0, T2 ) and A is the operator defined by A u = {v ∈ L2 (ε, T − ε; H), v(t) ∈ Au(t) a.e. on (0, T )}. A is maximal monotone and therefore demiclosed. By rewriting the Equation (5.46) in the form u00n − f ∈ A un and noting that un → u and u00n * u00 in L2 (ε, T − ε; H), by taking the limit as n → +∞, we observe that u satisfies the equation u00 ∈ Au + f almost everywhere on (ε, T − ε). Since this holds for each ε ∈ (0, T2 ), u is a solution to (5.30). We have already shown that u satisfies the boundary conditions and u ∈ 2,2 C([0, T ]; H) ∩Wloc (0, T ; H) and u0 ∈ Lγ2 (0, T ; H) and u00 ∈ Lγ23 (0, T ; H). 

5.4

EXISTENCE OF SOLUTIONS FOR THE NONHOMOGENEOUS CASE

Theorem 5.4.1 Suppose that A is a maximal monotone operator in H and f ∈ 2 (0, +∞; H) and b ∈ D(A) is an initial value such that Problem (5.2) for b has Lloc 2,2 at least a solution v ∈ C([0, ∞); H) ∩ Wloc (0, ∞; H). Then Problem (5.2) has a so2,2 lution u ∈ C([0, ∞); H) ∩ Wloc (0, ∞; H) for each initial data a ∈ D(A). Moreover 1 3 t 2 u0 ∈ L2 (0, δ ; H) and t 2 u00 ∈ L2 (0, δ ; H) for each δ > 0. If u, v ∈ C([0, ∞); H) ∩ 2,2 Wloc (0, ∞; H) are two bounded solutions of u00 ∈ Au + f , then the function t 7→ ku(t) − v(t)k2 is convex and nonincreasing and Z ∞ 0

1 tku0 (t) − v0 (t)k2 dt ≤ ku(0) − v(0)k2 2

(5.47)

Proof. Suppose that v is a solution of (5.2) corresponding to the initial data v(0) = b. 2,2 If u ∈ C([0, ∞); H) ∩ Wloc (0, ∞; H) is a solution of u00 ∈ Au + f , since t 7→ ku(t) − 2 v(t)k is convex, we infer that sup ku(t)k ≤ max{ku(0)k, ku(T )k} + M

(5.48)

t∈[0,T ]

where M = 2 supt≥0 kv(t)k. Step 1. Consider the case a ∈ D(A). By Theorem 5.3.2, for every positive integer n, there is a unique solution un ∈ W 2,2 (0, n; H) to the problem: ( u00n (t) ∈ Aun (t) + f (t) a.e. on (0, n) un (0) = un (n) = a

Second Order Evolution Equations

87

Suppose that 0 < T < m ≤ n. Similar to the above, we can easily see that 2ku0n (t) − u0m (t)k2 ≤ (kun (t) − um (t)k2 )00 a.e. on (0, n) If u and v are two solutions to the Equation (5.2), then since t 7→ ku(t) − v(t)k2 is convex and bounded, we conclude that ku(t) − v(t)k is nonincreasing. Therefore d 2 dt ku(t) − v(t)k ≤ 0. Now multiplying both sides of the above inequality by t, and integrating by parts on the interval [0, T ], we get: Z T

2

tku0 (t) − v0 (t)k2 dt ≤

Z T

0

t(ku(t) − v(t)k2 )00 dt

0 T d d ku(t) − v(t)k2 |T0 − ku(t) − v(t)k2 dt dt 0 dt Z T d ≤− ku(t) − v(t)k2 dt dt 0 = |u(0) − v(0)k2 − ku(T ) − v(T )k2

Z

=t

≤ ku(0) − v(0)k2 which implies (5.47). By Theorem 5.3.2, one can show easily that 2,2 u ∈ Wloc (0, ∞; H) and verifies the equation u00 ∈ Au + f on (0, T ) for each T > 0. Then u is a solution of u00 ∈ Au + f and since kun (t)k ≤ kak + M for each t ∈ [0, n], it also satisfies the boundary conditions in (5.2). Step 2. We prove the theorem for a ∈ D(A). In this case an ∈ D(A) exists such that an → a. Let un be the solution of (5.2) with un (0) = an . Since t 7→ kun (t) − vn (t)k2 is convex and bounded, it is nonincreasing. Therefore: kun (t) − um (t)k ≤ kan − am k which shows that un converges uniformly on [0, +∞) to a function 2,2 u ∈ C([0, +∞); H) with u(0) = a. By Theorem 5.3.2, u ∈ Wloc (0, +∞; H), and it 00 is a solution to u ∈ Au + f . 

5.5

PERIODIC FORCING

Theorem 5.5.1 Suppose that A : D(A) ⊂ H → H is maximal monotone and f ∈ 2 (0, +∞; H) is periodic with period T > 0. Let u ∈ C(0, +∞; H) ∩W 2,2 (0, +∞; H) Lloc loc be a solution to (5.2) and un (t) = u(t + nT ), t ∈ [0, +∞), n ∈ N. Then there exists 2,2 a T -periodic solution w ∈ C(0, +∞; H) ∩Wloc (0, +∞; H) to the Equation (5.2) with the following properties: (i) u(t) − w(t) * 0 in H as t → +∞. 1 (ii) t 4 (u0 (t) − w0 (t)) → 0 in H as t → +∞. (iii) u00n * w00 in L2 (0, T ; H) as n → +∞. Proof. First we prove the weak convergence of (un (t))n in H for each t ∈ [0, T ]. The existence of a solution u which follows from Theorem 5.4.1 implies the existence of

88

Nonlinear Evolution and Difference Equations of Monotone Type

2,2 a unique solution v ∈ C([s, ∞); H) ∩Wloc (s, ∞; H) to the problem: ( v00 (t) ∈ Av(t) + f (t), a.e. t ∈ (s, ∞) v(s) = a

(5.49)

for every a ∈ D(A). Then we define an evolution system U(t, s) : D(A) → D(A) by U(t, s)a = v(t), 0 ≤ s ≤ t < +∞. By the uniqueness of the solution, we have U(t, s)U(s, r) = U(t, r), 0 ≤ r ≤ s ≤ t

(5.50)

Theorem 5.4.1 implies that kU(t, s)a −U(t, s)bk ≤ ka − bk, a, b ∈ D(A). This means that U(t, s) is nonexpansive. The T -periodicity of f implies that U(t + T, s + T ) = U(t, s). This together with (5.50) implies that U(t + nT,t) = U(t + T,t)n , n ∈ N. Then we have un (t) = u(t + nT ) = U(t + nT,t)u(t) = U(t + T,t)n u(t),

(5.51)

Therefore (un (t))n is an orbit of the nonexpansive mapping U(t +T,t). The boundedness of this sequence implies that this mapping has a fixed point (see [DJ1,DJ2,DJ3]). In particular, for t = 0 we obtain the existence of a T -periodic solution w1 for problem (5.2). By (5.47) we have Z ∞ 0

tku0 (t) − w01 (t)k2 dt ≤

1 |u(0) − w1 (0)k2 2

Therefore ku0n − w01 kL2 (0,T ;H) ≤ (2nT )

−1 2

ku(0) − w1 (0)k

(5.52)

Then u0n → w01 in L2 (0, T ; H). Since un (t) − un+1 (t) = [un (t) − w1 (t)] − [un (t + T ) − w1 (t + T )] =

Z t+T t

[w01 (s) − u0n (s)]ds

we obtain un+1 (t) − un (t) → 0 as n → ∞. By the nonlinear ergodic theorem (see [DJ1,DJ2, DJ3]), this together with (5.51) implies that (un (t))n is weakly convergent to a fixed point of U(t + T,t). Suppose that w(t) = w − limn→+∞ un (t), t ≥ 0. Then clearly w is periodic with period T . As we can see, the inequality (5.52) implies the convergence u0n → w01 in L2 (0, T ; H). Since by Theorem 2.4 of [BRU2], (u00n )n is

Second Order Evolution Equations

89

bounded in L2 (0, T ; H), we conclude that w ∈ W 2,2 (0, T ; H), w is a solution to (5.2), u0n → w0 and u00n * w00 in L2 (0, T ; H). Therefore (iii) is proved. Since u(t) − w(t) = un (0) − w(0) +

Z t

[u0 (s) − w0 (s)]ds

nT

for nT ≤ t < nT + T , and un (0) * w(0) and Z

Z t 0

1

[u (s) − w0 (s)]ds ≤ T 2

T

ku0n − w0 k2 dt

0

nT

1 2

→0

as n → +∞, this concludes (i). To prove (ii), (5.47) implies that Z t+T

α(t) =

sku0 (s) − w0 (s)k2 ds → 0,

(5.53)

t

as t → +∞. Moreover since u00n * w00 in L2 (0, T ; H), a constant c > 0 exists such that Z t+T

ku00 (s) − w00 (s)k2 ds ≤ c2 , t ≥ T

(5.54)

t 1

Let Pt = {s ∈ [t,t + T ]; ku0 (s) − w0 (s)k > α(t) 4 t 1

|Pt | < α(t) 2 t

−1 4

}. From (5.53)

−1 2

(5.55)

where |A| denotes the Lebesgue measure of A. By (5.54) 1

ku0 (t) − w0 (t)k ≤ ku0 (s) − w0 (s)k + c(s − t) 2 , s ∈ [t,t + T ]. 1

−1 2

1

−1 4

As a consequence of (5.55) we can choose s ∈ [t,t + α(t) 2 t 1

ku0 (t) − w0 (t)k ≤ α(t) 4 t

−1 4

+ cα(t) 4 t

] \ Pt . Therefore



which completes the proof of Part (ii).

5.6

SQUARE ROOT OF A MAXIMAL MONOTONE OPERATOR

We consider problem (5.2) with f (t) ≡ 0 and x ∈ D(A). By Theorem 5.4.1, Equation (5.1) with p(t) ≡ 1 and q(t) ≡ 0 has a unique solution that we denote by u(t). The nonexpansiveness of the trajectory of u(t) implies that the mapping defining u(t), which we denote by S 1 (t) : R × D(A) → D(A) defined by S 1 (t)x := u(t) is a nonex2

2

pansive semigroup, which is called the square root semigroup. Assume that A01 is the 2

infinitesimal generator of S 1 (t) and A 1 is the unique maximal monotone extension 2

2

2

2

of A01 . In the linear case (A 1 )2 = A. A 1 is also called the square root of A. 2

90

Nonlinear Evolution and Difference Equations of Monotone Type

Theorem 5.6.1 [APR1] The nonlinear nonexpansive semigroup {S 1 (t); t ≥ 0} has 2 the following properties: (i) S 1 (t) maps D(A) to D(A) for all t > 0. 2

(ii) For each x ∈ D(A), the function t 7→ S 1 (t)x is continuously differentiable on 2 (0, +∞) and d S 1 (t)x + A01 S 1 (t)x = 0, ∀t > 0 dt 2 2 2 (iii) If x ∈ D(A) and y ∈ A−1 (0), then Z ∞

tk 0

and k where

d+ dt

d 1 S 1 (t)xk2 dt ≤ kx − yk2 dt 2 2

d+ 1 S 1 (t)xk ≤ kx − yk, ∀t > 0 dt 2 t

is the right derivative.

(iv) The second derivative

d2 S (t)x dt 2 12

Z ∞ 0

t 3k

exists for almost every t ∈ (0, +∞) and

d2 3 S 1 (t)xk2 dt ≤ kx − yk2 2 dt 2 2

(v) If x ∈ D(Ao1 ), the function t 7→ S 1 (t)x is continuously differentiable on [0, +∞) and t

5.7

1 2

2

2

d2 S (t)x dt 2 12

∈ L2 (0, +∞; H).

ASYMPTOTIC BEHAVIOR

In this section, we study the asymptotic behavior of solutions to the following evolution equation: ( p(t)u00 (t) + r(t)u0 (t) ∈ Au(t) + f (t), (5.56) u(0) = x, supt≥0 ku(t)k < +∞ where A is a maximal monotone operator. We consider appropriate conditions on the real-valued functions p and r, and the function f : (0, +∞) → H, as well as the maximal monotone operator A. This section contains four subsetions. In the first one, we study the ergodic convergence of solutions, which is the convergence of the net of means of the solutions on the interval [0, T ], as T → +∞. The second subsection is devoted to the weak convergence of solutions to a zero of the maximal monotone operator A. In the third subsection, with suitable assumptions on p, r and f , or additional conditions on the monotone operator A, we prove the strong convergence of solutions to (5.56). The last subsection is devoted to an important special case of maximal monotone operators which are subdifferentials of proper, convex and lower semi-continuous functions. In this subsection, we investigate conditions by which

Second Order Evolution Equations

91

ergodic, weak or strong convergence of solutions to a minimum point of the convex function is obtained. First we recall V´eron’s theorem on the existence of solutions to (5.56) for the homogeneous case f ≡ 0. Theorem 5.7.1 Let A be a maximal monotone operator in H. Assume that f ≡ 0, and suppose that p and r are two real valued functions satisfying p ∈ W 2,∞ (0, +∞), r ∈ W 1,∞ (0, +∞)

(5.57)

∃α > 0, such that ∀t ≥ 0, p(t) ≥ α.

(5.58)

 Then (5.56) has a solution u such that u0 and u00 belong to L2 (0, +∞); H . The solution is unique if moreover Z +∞ R t − 0

e

r(s) ds p(s)

dt = +∞

(5.59)

0

Throughout this section, we always assume that the conditions (5.57) and (5.58) on p and r, which by V´eron’s theorem guarantee the existence of solutions to (5.56) are satisfied. We also concentrate on the homogeneous case of (5.56) (i.e. we assume that f ≡ 0). Throughout this section, we denote M := Max{kpkW 2,∞ , krkW 1,∞ , supt≥0 ku(t)k}, uh (t) := u(t + h) − u(t), and σT (u) := 1 RT T 0 u(t)dt. 5.7.1

ERGODIC CONVERGENCE

In this subsection, with the following assumptions on p and r, and without assuming A−1 (0) , ∅, we prove the almost weak convergence of solutions to (5.56) to a zero of A. With the above assumptions on the functions p and r, we also show that the existence of solutions to (5.56) implies that A−1 (0) , ∅, providing a converse to the existence theorem for the solutions to (5.56). Definition 5.7.2 We say that a sequence {xn } in a Banach space X is almost weakly convergent to z ∈ X if 1n ∑n−1 i=0 xk+i * z uniformly in k, as n → +∞. Lemma 5.7.3 Suppose that u(t) is a solution to (5.56) and f ≡ 0. If q ∈ H satisfies the following inequality:  p(t)u00 (t) + r(t)u0 (t), u(t) − q ≥ 0, then ku(t)−qk is nonincreasing or eventually increasing (i.e. there exists t0 > 0 such that ku(t) − qk is increasing for t ≥ t0 ); therefore there exists limt→+∞ ku(t) − qk. Proof. It follows from the assumption that p(t)

d2 d ku(t) − qk2 + r(t) ku(t) − qk2 dt 2 dt

92

Nonlinear Evolution and Difference Equations of Monotone Type   d 0 u (t), u(t) − q + 2r(t) u0 (t), u(t) − q dt   = 2p(t) u00 (t), u(t) − q + 2p(t)ku0 (t)k2 + 2r(t) u0 (t), u(t) − q  = 2 p(t)u00 (t) + r(t)u0 (t), u(t) − q + 2p(t)ku0 (t)k2 ≥ 0 = 2p(t)

R t r(s)

ds

Dividing both sides of the above inequality by p(t) and multiplying by e 0 p(s) , we get: r(s)  d R0t p(s) ds d e ku(t) − qk2 ≥ 0. (5.60) dt dt We consider two cases. If dtd ku(t) − qk2 ≤ 0 for each t > 0, then ku(t) − qk is nonincreasing. Otherwise, there exists t0 > 0 such that dtd ku(t) − qk2|t=t0 > 0. Integrating (5.60) from t0 to t, we get for each t ≥ t0 , R t r(s)

e

0 p(s) ds

R t0 d ku(t) − qk2 ≥ 2e 0 dt

r(s) ds p(s)

 u0 (t0 ), u(t0 ) − q > 0.

Then

d ku(t) − qk2 > 0, ∀t ≥ t0 , dt which shows that ku(t) − qk is eventually increasing.



Corollary 5.7.4 If A−1 (0) , ∅, q ∈ A−1 (0) and f ≡ 0, then the conclusions of Lemma 5.7.3 hold. Proof. If q ∈ A−1 (0), then by the monotonicity of A, q satisfies the inequality in Lemma 5.7.3.  Theorem 5.7.5 Let u be a solution to (5.56). Assume that p, r satisfy the assumptions (5.57) and (5.58). If either one of the following conditions hold: i) r and p0 are monotone, ii) r0 (t) ≥ p00 (t), iii) p00 (t) ≥ r0 (t), R then σT (uh ) := T1 0T u(t + h)dt * c ∈ A−1 (0) as T → +∞, uniformly for h ≥ 0. Moreover, c is the asymptotic center of the curve (u(t))t≥0 . Proof. By the monotonicity of A, we have  p(t)u00 (t) + r(t)u0 (t), u(t) − u(s + h) ≥  p(s + h)u00 (s + h) + r(s + h)u0 (s + h), u(t) − u(s + h) . Integrating from s = 0 to s = T and dividing by T , we get:  p(t)u00 (t) + r(t)u0 (t), u(t) − σT (uh ) Z  1 T d 0 ≥ p(s + h) u (s + h), u(t) − u(s + h) ds T 0 ds

Second Order Evolution Equations

93

1 T r(s + h) d − ku(t) − u(s + h)k2 ds T 0 2 ds  1 = p(T + h) u0 (T + h), u(t) − u(T + h) T  p0 (T + h) 1 − p(h) u0 (h), u(t) − u(h) + ku(t) − u(T + h)k2 T 2T Z 1 0 1 T 00 − p (h)ku(t) − u(h)k2 − p (s + h)ku(t) − u(s + h)k2 ds 2T 2T 0 r(T + h) r(h) − ku(t) − u(T + h)k2 + ku(t) − u(h)k2 2T 2T Z 1 T 0 + r (s + h)ku(t) − u(s + h)k2 ds. (5.61) 2T 0 Z

By (5.57) and (5.58), the third, fourth, sixth, and seventh terms on the right hand side of (5.61) converge to zero as T → +∞. Since we have: ku0 (t + h) − u0 (t)k = k ≤

Z t+h

Z

t t+h

u00 (s)dsk

ku00 (s)kds

t



√ h||u00 ||L2 ((0,+∞);H)

0 it follows that u0 , and hence  also ku k, is uniformly 0 continuous on (0, +∞). Then 0 2 since u ∈ L (0, +∞); H , it follows that limt→+∞ ku (t)k = 0. This implies that the first and second terms on the right hand side of (5.61) converge to zero, uniformly for h ≥ 0. Now assume that (i) holds. We will show that the fifth and the last terms on the right hand side of (5.61) are bigger than expressions tending to zero as T → +∞. If p00 (t) ≤ 0, then:



1 2T

Z T

p00 (s + h)ku(t) − u(s + h)k2 ds ≥ 0.

(5.62)

0

Otherwise, p00 (t) ≥ 0 and by the boundedness of u, we have: −

1 2T

Z T

p00 (s + h)ku(t) − u(s + h)k2 ds ≥ 4M 2

0

1 0 (p (h) − p0 (T + h)) → 0 (5.63) 2T

as T → +∞, uniformly for h ≥ 0. If r0 (t) ≥ 0, then 1 2T

Z T

r0 (s + h)ku(t) − u(s + h)k2 ds ≥ 0.

(5.64)

0

Otherwise, r0 (t) ≤ 0 and we get: 1 2T

Z T 0

r0 (s + h)ku(t) − u(s + h)k2 ds ≥ 4M 2

1 (r(T + h) − r(h)) → 0 2T

(5.65)

94

Nonlinear Evolution and Difference Equations of Monotone Type

as T → +∞, uniformly for h ≥ 0. Suppose q is a weak cluster point of σT (uh ). Then for any two sequences Tn and hn of positive real numbers such that Tn → +∞ and σTn (uhn ) * q, by replacing T by Tn and h by hn in (5.61)–(5.65), and letting n → +∞, we get:  p(t)u00 (t) + r(t)u0 (t), u(t) − q ≥ 0 (5.66) If p(t) and r(t) satisfy condition (ii), then we get again (5.66) from (5.61)–(5.65). If p(t) and r(t) satisfy condition (iii), then we have 1 T 00 − p (s + h)ku(t) − u(s + h)k2 ds 2T 0 Z 1 T 0 + r (s + h)ku(t) − u(s + h)k2 ds 2T 0 Z 1 T 0 = (r (s + h) − p00 (s + h))ku(t) − u(s + h)k2 ds 2T 0 Z 1 T 0 ≥ 4M 2 (r (s + h) − p00 (s + h))ds 2T 0 4M 2 = (r(T + h) − r(h) − p0 (T + h) + p0 (h)), 2T Z

which converges to zero as T → +∞. Then we get again (5.66) from (5.61) to (5.65). Now by Lemma 5.7.3, there exists limt→+∞ ku(t) − qk. If c is another weak cluster point of σT (uh ), then there exists limt→+∞ (ku(t) − qk2 − ku(t) − ck2 ). This implies that there exists limt→+∞ (u(t), c − q), then there exists limT →+∞ (σT (uh ), c − q) uniformly for h ≥ 0. It follows that (c, c − q) = (q, c − q). Therefore c = q, and hence σT (uh ) * c as T → +∞, uniformly for h ≥ 0, which shows the almost weak convergence of u(t) to c as t → +∞. Now we prove that c ∈ A−1 (0). Let y ∈ Ax. By the monotonicity of A, we have:    x − u(t), y = x − u(t), y − Au(t) + x − u(t), Au(t)  ≥ x − u(t), p(t)u00 (t) + r(t)u0 (t) Integrating from t = 0 to T , dividing by T , and letting T → +∞, by a similar proof as above we get: (x − c, y) ≥ 0. Now the maximality of A implies that c ∈ A−1 (0). Finally, we show that c is the asymptotic center of the curve u(t). Let x ∈ H, with x , c. Then:  ku(t) − xk2 = ku(t) − ck2 + kx − ck2 + 2 u(t) − c, c − x . Integrating from 0 to T , and dividing by T , then taking limsup when T → +∞, since σT * c and limn→+∞ ku(t) − ck exists, we get: lim sup ku(t) − xk2 ≥ lim sup ku(t) − ck2 + kx − ck2 t→+∞

t→+∞

> lim sup ku(t) − ck2 . t→+∞

Hence c is the asymptotic center of u(t) as desired. The proof is now complete.



Second Order Evolution Equations 5.7.2

95

WEAK CONVERGENCE

In this subsection, we prove the weak convergence of solutions to (5.56) to a zero of A. The proof is based on Lorentz’ Tauberian theorem which states that an almost weakly convergent sequence {xn } is weakly convergent if and only if xn+1 − xn * 0 as n → +∞. Theorem 5.7.6 Let u be a solution to (5.56) with f ≡ 0. If (5.57), (5.58) and the assumptions of Theorem 5.7.5 are satisfied, then u(t) * c ∈ A−1 (0) as t → +∞, where c is the asymptotic center of the curve (u(t))t≥0 .  Proof. Since u0 ∈ L2 (0, +∞); H , then u is asymptotically regular (i.e. u(t + h) − u(t) → 0, as t → +∞, ∀h ≥ 0). Now the result follows from Theorem 5.7.5 and G. G. Lorentz’ Tauberian condition for almost convergence (see [LOR]).  Remark 5.7.7 The conclusions of Theorems 5.7.5 and 5.7.6 still hold if the assumptions (i), (ii) or (iii) in Theorem 5.7.5 are satisfied only for large enough t(i.e. for t ≥ t0 ). Remark 5.7.8 Since u0 and u00 belong to L2 (0, +∞; H), then as stated in the proof  of Theorem 5.7.5, we have that limt→+∞ ku0 (t)k = 0. Since p, r ∈ L∞ (0, +∞); H , then pu00 + ru0 ∈ L2 (0, +∞); H . Therefore there is a sequence tn → +∞ such that p(tn )u00 (tn ) + r(tn )u0 (tn ) → 0 as n → +∞. Now the demiclosedness of A shows that A−1 (0) , ∅. However, the weak convergence of u(t) cannot be directly derived from this fact, and we need the arguments in Theorems 5.7.5 and 5.7.6. Example 5.7.9 Consider the following second order evolution equation: ( 0 q(t)u0 (t) ∈ Au(t) a.e. on R+ u(0) = u0 , supt≥0 ku(t)k < +∞

(5.67)

 where A is a maximal monotone operator and q(t) ∈ W 2,∞ (0, +∞); H . Then (5.57) and condition (ii) of Theorem 5.7.5 are satisfied. (5.58) is satisfied if q(t) ≥ α > 0, ∀t ≥ 0. Then u(t) converges weakly to a zero of A as t → +∞. In this case, condition (5.59) is equivalent to the following condition: Z ∞ 1 0

q(t)

= +∞.

(5.68)

For example, q(t) = sint + 2 satisfies all of the above conditions. Example 5.7.10 Consider bounded solutions to the following ordinary differential equation: 3 00 2 0 u + u = u3 , u(0) = 1. 2 t +1 One can easily verify that the solution is given by u(t) = zero as t → +∞, as predicted by Theorem 5.7.6.

1 t+1 ,

which converges to

96

Nonlinear Evolution and Difference Equations of Monotone Type

5.7.3

STRONG CONVERGENCE

In this subsection, we prove the strong convergence of solutions to (5.56) with f ≡ 0, to a zero of the maximal monotone operator A. In particular, as in the previous subsection, we show that the existence of solutions implies that A−1 (0) , ∅. We assume that p and r satisfy (5.57) and (5.58), and then by Theorem 5.7.1 we have  u0 , u00 ∈ L2 (0, +∞); H . Lemma 5.7.11 Let u be a solution to (5.56). Then lim inft→+∞ p(t) dtd kv(t, h)k2 ≤ 0 and lim supt→+∞ r(t)kv(t, h)k2 ≤ 0, where v(t, h) = u(t + h) − u(t). Proof. Assume by contradiction that lim inft→+∞ p(t) dtd kv(t, h)k2 ≥ c > 0. Since 0 < α ≤ p(t) ≤ M, integrating from t = s to T , we get: c (T − s). M Letting T → +∞, we get a contradiction since u is bounded. On the other hand, kv(T, h)k2 − kv(s, h)k2 >

|r(t)|kv(t, h)k2 = |r(t)|k Z

Z t+h t t+h

≤ Mh(

u0 (s)dsk2

ku0 (s)k2 ds) → 0

t

 as t → +∞, since u0 ∈ H 1 (0, +∞); H .



Theorem 5.7.12 Let u(t) be a solution to (5.56) with f ≡ 0. Assume that the following conditions (i) and (ii) are satisfied: (i)

R ∞ − R0s r(τ) dτ 2p(τ) 0

e

ds < +∞,

R ∞ R t − Rst

r(τ)



p(τ) (ii)M1 := 0 [ 0 e R(s)ds]1/2 dt < +∞, where R(s) := Max{0, supt≥s r0 (t)}. Then u(t) → p ∈ H, as t → +∞.

Proof. Take v(t) = u(t + h) − u(t). By the monotonicity of A and (5.56), we get:    p(t) v00 (t), v(t) + p(t + h) − p(t) u00 (t + h), v(t) +    r(t) v0 (t), v(t) + r(t + h) − r(t) u0 (t + h), v(t) ≥ 0   1 d2 ⇒ p(t) 2 kv(t)k2 + p(t + h) − p(t) u00 (t + h), v(t) 2 dt   1 d + r(t) kv(t)k2 + r(t + h) − r(t) u0 (t + h), v(t) ≥ 0. 2 dt Integrating by parts from s to t, we get: 1 d 1 d 1 p(t) kv(t)k2 − p(s) kv(s)k2 − 2 dt 2 ds 2

Z t s

p0 (τ)

d kv(τ)k2 dτ+ dτ

Second Order Evolution Equations

97

Z t

  1 1 p(τ + h) − p(τ) u00 (τ + h), v(τ) dτ + r(t)kv(t)k2 − r(s)kv(s)k2 − 2 2 s Z Z t   1 t 0 r (τ)kv(τ)k2 dτ + r(τ + h) − r(τ) u0 (τ + h), v(τ) dτ ≥ 0 2 s s By Lemma 5.7.11, there is a sequence tn → +∞ such that  lim p(tn ) v(tn ), v0 (tn ) ≤ 0. n→+∞

Now replacing t by tn in the above inequality, and letting n → +∞, we get: 1 d 1 p(s) kv(s)k2 + r(s)kv(s)k2 ≤ 2 ds 2 Z Z ∞   1 ∞ 0 d − p (τ) kv(τ)k2 dτ + p(τ + h) − p(τ) u00 (τ + h), v(τ) dτ 2 s dτ s Z Z ∞   1 ∞ 0 2 − r (τ)kv(τ)k dτ + r(τ + h) − r(τ) u0 (τ + h), v(τ) dτ. 2 s s Dividing by h2 and letting h → 0, by an application of Fatou’s Lemma, it follows from the assumptions that: p(s)

d 0 ku (s)k2 + r(s)ku0 (s)k2 ≤ ds

Z ∞

r0 (τ)ku0 (τ)k2 dτ ≤ R(s)

Z ∞

s

ku0 (τ)k2 dτ.

s

(5.69)

Then: R s r(τ)

Rs d 0 ku (s)k2 e t0 ds

r(τ) dτ p(τ)

Rs r(s) 0 + ku (s)k2 e t0 p(s)

r(τ) dτ p(τ)

e t0 p(τ) ≤ p(s)



Z ∞

R(s)

ku0 (τ)k2 dτ.

s

Therefore: R s r(τ)

Rs d ku0 (s)k2 e t0 ds

r(τ) dτ p(τ)



e t0 p(τ) ≤ p(s)



Z ∞

R(s)

ku0 (τ)k2 dτ.

s

Integrating from t0 to t with respect to s, we get: 0

2

0

ku (t)k ≤ ku (t0

R

R r(τ) − tt p(τ) dτ 0

= ku0 (t0 )k2 e

R r(τ) − tt p(τ) dτ 0

r(τ)

− t dτ )k2 e t0 p(τ)

+

+

Z t

e

1 α

α Z t

e t0

t0

R s r(τ)

e t0

p(τ)



Z ∞

R(s)

 ku0 (τ)k2 dτ ds

s

Z ∞ R r(τ)  − st p(τ) dτ R(s) ku0 (τ)k2 dτ ds. s

Therefore: R r(τ) − tt 2p(τ) dτ 0

ku0 (t)k ≤ ku0 (t0 )ke

1  + √ α

Z t

Z ∞ R r(τ) 1 − st p(τ) dτ R(s) ku0 (τ)k2 dτ)ds 2 . s

(e t0

(5.70)

98

Nonlinear Evolution and Difference Equations of Monotone Type

Hence: ku(T 0 ) − u(T )k ≤

Z T0 T

1 +√ α Assume

u(Tn0 ) *

ku0 (t)kdt ≤ ku0 (t0 )k

Z T0 Rt − t 0

e

r(τ) dτ 2p(τ)

dt

T

Z T0 Z t T

Z ∞ R r(τ) 1 − st p(τ) dτ (e R(s) ku0 (τ)k2 dτ)ds 2 dt. t0 s

p. Then we get: ku(T ) − pk ≤ ku0 (t0 )k

1 +√ α

Z ∞ Z t T

t0

(e

Z ∞ Rt − t 0

e

r(τ) dτ 2p(τ)

dt

T

Z ∞ R r(τ) 1 − st p(τ) dτ R(s) ku0 (τ)k2 dτ)ds 2 dt. s

Given ε > 0, choose t0 big enough such that: s∞ ku0 (τ)k2 dτ ≤ ε 2 , ∀s ≥ t0 . Then we have: Z ∞ R s r(τ) M1 ε − dτ ku(t) − pk ≤ ku0 (t0 )k e t0 2p(τ) ds + √ . (5.71) α t R

Therefore lim supt→+∞ ku(t) − pk ≤ u(t) → p as t → +∞.

M √1 ε . α

Since ε > 0 is arbitrary, we conclude that 

Corollary 5.7.13 Let u(t) be a solution to (5.56) with f ≡ 0. Assume that r(t) ≥ 0. In addition, assume that (i) and the following stronger condition (ii0 ) is satisfied: R s r(τ)



(ii0 )M10 := 0+∞ e 0 p(τ) R(s)ds < +∞. Then u(t) → p ∈ A−1 (0) as t → +∞. R

r(τ) r(τ) r(τ) Proof. First of all, since: − st p(τ) dτ = −2 0t 2p(τ) dτ + 0s p(τ) dτ, it is clear that (i) 0 and (ii ) imply (ii). Therefore by Theorem 5.7.12, u(t) → p ∈ H as t → +∞. Now since r(t) ≥ 0, it follows from (i), (ii0 ) and (5.70) that ku0 (t)k → 0 as t → +∞. Since u00 ∈ L2 ((0, +∞); H), there is a sequence tn → +∞ such that u00 (tn ) → 0 as n → +∞. Since p(t) and r(t) are bounded, the closedness of A implies that p ∈ A−1 (0). 

R

R

R

Corollary 5.7.14 Let u(t) be a solution to (5.56) with f ≡ 0. Assume that r0 (t) ≤ 0. If R ∞ − R0s r(τ) dτ 2p(τ)

(i) holds, then u(t) → p ∈ A−1 (0) as t → +∞ and ku(t) − pk = O(

t

e

ds).

Proof. Since r0 (t) ≤ 0, then (i) implies that r(t) ≥ 0, and R(s) = 0 so that (ii0 ) is clearly satisfied. The conclusion follows now from Corollary 5.7.13. The rate of convergence is concluded by (5.71), because M1 = 0.  4 . Then the assumptions of Corollary Example 5.7.15 Let p(t) ≡ 1, r(t) = t+1 5.7.14 are satisfied; therefore u(t) → p ∈ A−1 (0) as t → +∞. Moreover, the proof of Theorem 5.7.12 shows that we have the following rate of convergence: 1 |u(t) − p| = O( t+1 ).

Second Order Evolution Equations

99

Example 5.7.16 Let p(t) ≡ 1 and r(t) = 1 − e−t ≥ 0. Then r0 (t) = e−t > 0 and r0 (t) → 0 as t → +∞. In this case, (i) and (ii) are satisfied, but (ii0 ) does not hold. Therefore, it follows from Theorem 5.7.12 that u(t) → p ∈ H, as t → +∞. Problem 5.7.17 Say for p(t) ≡ 1, is it possible to get the strong convergence of u(t) by assuming that limsupt→+∞ r0 (t) ≤ 0, and is there any relationship between this condition and condition (ii) in Theorem 5.7.12? 5.7.4

SUBDIFFERENTIAL CASE

In this subsection, we consider the evolution Equation (5.56) with f (t) ≡ 0 when the monotone operator A is the subdifferential ∂ ϕ of a proper, convex and lower semicontinuous function ϕ : H → (−∞, +∞]. We prove a weak convergence theorem with suitable assumptions on p(t) and r(t), as well as a strong convergence theorem with additional assumptions on ϕ. Lemma 5.7.18 Suppose that u(t) is a solution to (5.56) with f ≡ 0. Then for each p ∈ A−1 (0), limt→+∞ ku(t) − pk2 exists and lim inft→+∞ dtd ku(t) − pk2 ≤ 0. In addition, if either (5.59) is satisfied or A is strongly monotone, then ku(t) − pk2 is nonincreasing. Proof. The existence of limt→+∞ ku(t) − pk2 follows from Lemma 5.7.3. By contradiction, assume that lim inft→+∞ dtd ku(t)− pk2 > 0. Then there exists t0 > 0 and λ > 0, such that for each t ≥ t0 , d ku(t) − pk2 ≥ λ . dt Integrating from t = t0 to t = T , we get ku(T ) − pk2 − ku(t0 ) − pk2 ≥ λ T − λt0 . Letting T → +∞, we deduce that u is not bounded, a contradiction. If in addition (5.59) is satisfied, assume that ku(t) − pk is eventually increasing. Then there exists t0 > 0 such that (u0 (t0 ), u(t0 ) − p) > 0. Integrating (5.60) from t0 to t, then dividing R t r(s)

both sides by e

0 p(s) ds

2

, and integrating from t = t0 to t = T , we get: 2

R t0 r(s)

ku(T ) − pk − ku(t0 ) − pk ≥ 2e

0

p(s)

ds

Z u (t0 ), u(t0 ) − p 0

T

R r(s) − 0t p(s) ds dt.

e

t0

Letting T → +∞, we obtain a contradiction to Assumption (5.59). This implies that ku(t) − pk is nonincreasing. Finally, assume that A is strongly monotone, and let p ∈ A−1 (0). Then we have  p(t)u00 (t) + r(t)u0 (t), u(t) − p ≥ β ku(t) − pk2 . This implies that p(t)

d2 d ku(t) − pk2 + r(t) ku(t) − pk2 ≥ 2β ku(t) − pk2 . dt 2 dt

100

Nonlinear Evolution and Difference Equations of Monotone Type

Suppose to the contrary that ku(t) − pk is increasing for t ≥ T0 > 0. Let K (resp. M) be an upper bound for p(t) (resp. |r(t)|). Integrating both sides of this inequality from t = T0 to t = T , we get: Z T

2β ku(t) − pk2 dt T0    Z T r(t) d d 2 0 2 ≤K ku(T ) − pk − 2 u (T0 ), u(T0 ) − p + ku(t) − pk dt dT T0 p(t) dt    M  d 2 0 2 2 ≤K ku(T ) − pk − 2 u (T0 ), u(T0 ) − p + ku(T ) − pk − ku(T0 ) − pk . dT α Since ku(t) − pk is increasing for t ≥ T0 > 0, we have: 2β ku(T0 ) − pk2 (T − T0 ) ≤    M  d K ku(T ) − pk2 − 2 u0 (T0 ), u(T0 ) − p + ku(T ) − pk2 − ku(T0 ) − pk2 . dT α Taking lim inf as T → +∞ of both sides in the above inequality, by the first part of this Lemma we deduce that u(t) is unbounded, a contradiction.  In the following, we prove a mean ergodic theorem when A is the subdifferential of a proper, convex and lower semicontinuous function. Theorem 5.7.19 Suppose that u(t) is a solution to (5.56) with f ≡ 0 and A = ∂ ϕ, where ϕ : H → (−∞, +∞] is a proper, convex and lower semicontinuous function. If R (5.59) is satisfied, then σT := T1 0T u(t)dt * p ∈ A−1 (0), as T → +∞. Proof. By the subdifferential inequality and Equation (5.56), we get for each p ∈ A−1 (0),  ϕ(u(t)) − ϕ(p) ≤ p(t)u00 (t) + r(t)u0 (t), u(t) − p r(t) d p(t) d 2 ku(t) − pk2 + ku(t) − pk2 2 dt 2 2 dt R t r(s) r(s)  p(t) − R0t p(s) ds d ds d = e e 0 p(s) ku(t) − pk2 . 2 dt dt ≤

Let K be an upper bound for p(t) 2 . Integrating the above inequality from t = 0 to t = T , and using integration by parts, we get: Z T

(ϕ(u(t)) − ϕ(p))dt

0 T r(t) d  d ku(T ) − pk2 − 2(u0 (0), u(0) − p) + ku(t) − pk2 dt dT 0 p(t) dt Z T   r(t) d ≤ K − 2 u0 (0), u(0) − p + ku(t) − pk2 dt (5.72) 0 p(t) dt

≤K

Z

Second Order Evolution Equations

101

(the second inequality holds by Lemma 5.7.18). Let R be an upper bound for |r(t)|, which exists by (5.57). Since ku(t) − pk is nonincreasing (by Lemma 5.7.18) we get from (5.72), lim sup T →+∞

1 T

Z T

 ϕ(u(t)) − ϕ(p) dt

0

K T r(t) d ≤ lim sup ku(t) − pk2 dt T →+∞ T 0 p(t) dt  −KR 1 ≤ lim sup ku(T ) − pk2 − ku(0) − pk2 = 0 α T →+∞ T Z

(5.73)

Since p ∈ A−1 (0) and A = ∂ ϕ, p is a minimum point of ϕ. Convexity of ϕ implies, 0 ≤ ϕ(σT ) − ϕ(p) ≤

1 T

Z T

ϕ(u(t))dt − ϕ(p).

0

Taking the lim sup as T → +∞ in the above inequality we get by (5.73) lim sup ϕ(σT ) ≤ ϕ(p). T →+∞

Assume that σTn * q for some sequence {Tn } converging to +∞ as n → +∞. Since ϕ is lower semicontinuous, we have lim inf ϕ(σTn ) ≥ ϕ(q). n→+∞

Therefore ϕ(p) ≥ lim sup ϕ(σT ) ≥ lim inf ϕ(σTn ) ≥ ϕ(q). T →+∞

n→+∞

Hence, q ∈ A−1 (0) and by Lemma 5.7.18 limt→+∞ ku(t) − qk2 exists. Now if p is another weak cluster point of σT , then limt→+∞ (ku(t) − pk2 − ku(t) − qk2 ) exists. It follows that limt→+∞ (u(t), p − q) exists, hence limT →+∞ (σT , p − q) exists. This implies that p = q and therefore σT * p ∈ A−1 (0), as T → +∞.  Proposition 5.7.20R Let u(t) be a solution to (5.56) with f ≡ 0. Assume that (5.57) and (5.58) hold. If 0∞ R(t)dt < +∞, then limt→+∞ ϕ(u(t)) exists. Proof. By Lemma 5.7.3, (5.56) and (5.69), we have:   d ϕ u(t) = ∂ ϕ(u(t)), u0 (t) dt  = p(t)u00 (t) + r(t)u0 (t), u0 (t) 1 d = p(t) ku0 (t)k2 + r(t)ku0 (t)k2 2 dt Z 1 ∞ 0 1 ≤ r (s)ku0 (s)k2 ds + r(t)ku0 (t)k2 2 t 2

102

Nonlinear Evolution and Difference Equations of Monotone Type 1 ≤ R(t) 2

Z ∞ t

1 ku0 (s)k2 ds + r(t)ku0 (t)k2 . 2

Therefore: Z ∞ Z 0   1 Z T0 1 T 0 2 ϕ u(T ) − ϕ u(T ) ≤ R(t) ku (s)k dsdt + r(t)ku0 (t)k2 dt 2 T 2 T t Z ∞  Z T 0  1 ≤ ku0 (t)k2 dt R(t)dt + M 2 T T   R This implies that: lim supt→+∞ ϕ u(t) ≤ ϕ u(T) + C T∞ ku0 (s)k2 ds, for some 0 2 constant C. Now  since u ∈ L (0,+∞); H , letting T → +∞, we get: lim supt→+∞ ϕ u(t) ≤ lim inft→+∞ ϕ u(t) , which completes the proof of the proposition.  0

Remark 5.7.21 In Proposition 5.7.20, if r(t) ≤ 0 and r0 (t) ≤ 0, then ϕ(u(t)) is nonincreasing. Lemma 5.7.22 Suppose that u(t) is a solution to (5.56) with f ≡ 0, and q ∈ H. Then lim inft→+∞ t dtd ku(t) − qk2 ≤ 0. Proof. Assume by contradiction that lim inft→+∞ t dtd ku(t) − qk2 > 0. Then There exist t0 > 0 and c > 0 such that for each t > t0 , we have t dtd ku(t) − qk2 ≥ c > 0. Dividing both sides by t and then integrating from t = t0 to T , we get ku(T ) − qk2 − ku(t0 ) − qk2 ≥ c(ln T − lnt0 ). Since u(t) is bounded, we get a contradiction by letting T → +∞.



Theorem 5.7.23 Let u(t) be a solution to (5.56) with f ≡ 0. Suppose that the assumptions of Proposition 5.7.20 are satisfied. Then u(t) converges weakly to an element of A−1 (0), as t → +∞. Proof. By Remark 5.7.8, we know that A−1 (0) , ∅. Let q ∈ A−1 (0). By the subdifferential inequality and (5.56), we get   ϕ u(t) − ϕ(q) ≤ p(t)u00 (t) + r(t)u0 (t), u(t) − q ≤

d2 1 d 1 p(t) 2 ku(t) − qk2 + r(t) ku(t) − qk2 2 dt 2 dt

(5.74)

Integrating from t = 0 to T , we get: Z T 0

1 − 2

Z T 0

  p(T ) d ϕ(u(t)) − ϕ(q) dt ≤ ku(T ) − qk2 − p(0) u0 (0), u(0) − q 2 dT Z d 1 T d p0 (t) ku(t) − qk2 + r(t) ku(t) − qk2 dt dt 2 0 dt

Second Order Evolution Equations

103

By Lemma 5.7.3, ku(t) − qk is nonincreasing or eventually increasing. If ku(t) − qk is nonincreasing, then Z T

  ϕ(u(t)) − ϕ(q) dt ≤ −p(0) u0 (0), u(0) − q + M(ku(0) − qk2 − ku(T ) − qk2 ).

0

(5.75) If ku(t) − qk is eventually increasing, then there exists t0 such that for each t ≥ t0 , d dt ku(t) − qk > 0. Integrating (5.74) from t = t0 to T , we get: Z T t0

1 − 2

Z T t0

  p(T ) d ϕ(u(t)) − ϕ(q) dt ≤ ku(T ) − qk2 − p(t0 ) u0 (t0 ), u(t0 ) − q 2 dT Z 1 T d d 0 2 p (t) ku(t) − qk + r(t) ku(t) − qk2 dt dt 2 t0 dt

Therefore: Z T t0

  M d ϕ(u(t)) − ϕ(q) dt ≤ ku(T ) − qk2 − p(t0 ) u0 (t0 ), u(t0 ) − q 2 dT + M(ku(T ) − qk2 − ku(t0 ) − qk2 ).

(5.76)

Taking liminf of (10.14) and (8.85) as T → +∞ , by Lemma 5.7.18, we get Z ∞

 ϕ(u(t)) − ϕ(q) dt < +∞.

t0

  Therefore lim inft→+∞ ϕ u(t) ≤ ϕ(q). By Proposition 5.7.20, limt→+∞ ϕ u(t) = ϕ(q) = Min{ϕ(z); z ∈ H}. If u(tn ) * s as n → +∞, then ϕ(s) ≤ lim inf ϕ(u(tn )) = lim ϕ(u(t)) = ϕ(q). n→+∞

t→+∞

Hence s ∈ A−1 (0), and therefore by Corollary 5.8.8, there exists limt→+∞ ku(t) − sk. Now the proof is completed by a similar argument as in Theorem 5.7.5.  Theorem 5.7.24 Let u(t) be a solution to (5.56) with f ≡ 0. Suppose (5.57) and (5.58) hold and that r(t) ≤ 0 and r0 (t) ≤ 0. Let ϕ : H →] − ∞, +∞] be a proper, convex and lower semicontinuous function satisfying the following conditions: D(ϕ) = −D(ϕ), and  ϕ(x) − ϕ(0) ≥ a(kxk) ϕ(−x) − ϕ(0) , ∀x ∈ D(ϕ), where a : R+ → (0, 1) is a continuous function. Then u(t) → q ∈ A−1 (0) as t → +∞, which is a minimum point of ϕ. Proof. By Remark 5.7.8, we know that A−1 (0) , ∅, and therefore ϕ has a minimum point. Without loss of generality, we may assume that ϕ(0) = 0 and 0 is a minimum point of ϕ. For t ≤ s, by the assumptions, Proposition 5.7.20 and Remark 5.7.21, we get:     ϕ u(t) ≥ ϕ u(s) ≥ a(ku(s)k)ϕ − u(s) + 1 − a(ku(s)k) ϕ(0)

104

Nonlinear Evolution and Difference Equations of Monotone Type  ≥ ϕ − a(ku(s)k)u(s)   ≥ ϕ u(t) + ∂ ϕ(u(t)), −a(ku(s)k)u(s) − u(t)   = ϕ u(t) − p(t)u00 (t) + r(t)u0 (t), a(ku(s)k)u(s) + u(t) .

Therefore:  p(t)u00 (t) + r(t)u0 (t), a(ku(s)k)u(s) + u(t) ≥ 0. Let:   g(t) = 1 + a(ku(s)k) ku(t)k2 − ku(s)k2 − a(ku(s)k)ku(t) − u(s)k2 then  g0 (t) = 2 u0 (t), u(t) + a(ku(s)k)u(s) ,  g00 (t) = 2 u00 (t), u(t) + a(ku(s)k)u(s) + 2ku0 (t)k2 . Hence p(t)g00 (t) + r(t)g0 (t) ≥ 0. Now the same argument as in Lemma 5.7.3, with ku(t) − qk2 replaced by g(t), shows that g(t) is either nonincreasing or eventually increasing. Since r(t) ≤ 0, condition (5.59) is satisfied. Therefore by Lemma 5.7.18, we conclude that g(t) is nonincreasing. Then g(t) ≥ g(s) = 0 for t ≤ s. It follows that: 1 + a(ku(s)k) (ku(t)k2 − ku(s)k2 ) a(ku(s)k) 2 < (ku(t)k2 − ku(s)k2 ), ∀s ≥ t a(ku(s)k)

ku(t) − u(s)k2 ≤

By Corollary 5.8.8, there exists lims→+∞ ku(s)k. If ku(s)k → 0 as s → +∞, then u(s) → 0 and this yields the theorem. Otherwise, if ku(s)k → r > 0, from the continuity of a, we have lims→+∞ a(ku(s)k) = a(lims→+∞ ku(s)k) = a(r) > 0. Therefore {u(t)} is a cauchy sequence in H, hence u(t) → q as t → +∞, and q ∈ A−1 (0) by Theorem 5.7.23. 

5.8

ASYMPTOTIC BEHAVIOR FOR SOME SPECIAL NONHOMOGENEOUS CASES

In this section, we study the weak and strong convergence of solutions to (5.56) when p(t) ≡ 1 and r(t) ≡ c, where c is a constant real number. Our investigations in this section include the nonhomogeneous case. We consider the following second order nonhomogeneous evolution equation ( u00 (t) + cu0 (t) ∈ Au(t) + f (t) a.e. on R+ (5.77) u(0) = u0 , supt≥0 ku(t)k < +∞ where A is a maximal monotone operator, and f : R+ → H satisfies the following condition: Z ∞

There exists t0 > 0 such that

t0

tk f (t) − f∞ kdt < +∞,

(5.78)

Second Order Evolution Equations

105

for some f∞ ∈ H. By replacing f (t) by f (t) − f∞ and A by A + f∞ , we may assume without loss of generality that f∞ = 0. We study the asymptotic behavior of (5.77) in two subsections: first when the constant c is non-positive, and then in the next subsection when c is positive. 5.8.1

CASE C ≤ 0

In this case, (5.77) satisfies the assumptions in V´eron’s existence Theorem (Theorem 5.7.1) when f ≡ 0. Lemma 5.8.1 Suppose f : [0, +∞) → R is bounded from above and absolutely continuous on every compact subinterval, then lim inft→+∞ f 0 (t) ≤ 0. Proof. To get a contradiction, assume that lim inft→+∞ f 0 (t) > 0. Then there exist λ > 0 and t0 > 0 such that for each t ≥ t , f 0 (t) ≥ λ . Integrating this inequality from RT 0 0 t0 to T , we get: f (T ) − f (t0 ) = t0 f (t)dt ≥ λ (T − t0 ). Letting T → +∞, we get a contradiction.  Lemma 5.8.2 Assume f : [0, +∞) → R is bounded from above and f and f 0 are absolutely continuous on every compact subinterval. If f 00 (t) + c f 0 (t) ≥ −Mg(t)

(5.79)

with g(t) ≥ 0, and g satisfying t+∞ tg(t)dt < +∞, for some t0 > 0, then f (T2 ) ≤ 0 f (T1 ) + η(T1 ), ∀T2 ≥ T1 ≥ t0 , where limT1 →+∞ η(T1 ) = 0, and limt→+∞ f (t) exists. R

Proof. Multiplying both sides of (5.79) by ect , we get d ct 0 (e f (t)) ≥ −Mect g(t) dt Integrating from t = S to t = T , we get ecT f 0 (T ) − ecS f 0 (S) ≥ −M

Z T

ect g(t)dt ≥ −MecS

S

Z T

g(t)dt S

This implies that ecS f 0 (S) ≤ ecT f 0 (T ) + MecS

Z T

g(t)dt S

Taking lim inf as T → +∞, by Lemma 5.8.1, we get f 0 (S) ≤ M

Z ∞

g(t)dt S

Integrating the above inequality from S = T1 to S = T2 , and applying Fubini’s Theorem, we obtain f (T2 ) ≤ f (T1 ) + M

Z T2

Z ∞

dS T1

g(t)dt S

106

Nonlinear Evolution and Difference Equations of Monotone Type

≤ f (T1 ) + M = f (T1 ) + M

Z +∞ T1 Z ∞

≤ f (T1 ) + M

Z t

T

T1

g(t)dt S

dt Z 1∞

= f (T1 ) + M

Z ∞

dS

g(t)dS T1

(t − T1 )g(t)dt

Z +∞

tg(t)dt T1

= f (T1 ) + η(T1 ) where η(T1 ) = M limT →+∞ f (T ).

R +∞ T1

tg(t)dt → 0 as T1 → +∞. This implies that there exists 

Theorem 5.8.3 Assume that u is a solution to (5.77) and f (t) satisfies the condition R (5.78), then σT := T1 0T u(t)dt converges weakly as T → +∞ to some p ∈ L := {q ∈ H : limn→+∞ ku(t) − qk exists}, which is also the asymptotic center of the curve (u(t))t≥0 . Proof. We show that the curve u is almost nonexpansive. Let h ≥ 0 be fixed. By the monotonicity of A, we get from (5.77) that   d2u d2u du du (t + h) − (t), u(t + h) − u(t) + c (t + h) − (t), u(t + h) − u(t) dt 2 dt 2 dt dt  ≥ f (t + h) − f (t), u(t + h) − u(t) ≥ −k f (t + h) − f (t)kku(t + h) − u(t)k M ≥ − k f (t + h) − f (t)k 2 M M ≥ − k f (t + h)k − k f (t)k 2 2 where M := supt≥0 ku(t)k. This implies that 1 d2 c d ku(t + h) − u(t)k2 + ku(t + h) − u(t)k2 2 dt 2 2 dt ≥ ku0 (t + h) − u0 (t)k2 M M − k f (t + h)k − k f (t)k 2 2 From Lemma 5.8.2 we conclude that there exists limT →+∞ ku(T + h) − u(T )k and R u(t) is almost nonexpansive. By Theorem 4.5.23, we get: 1t 0t u(s)ds * p as t → +∞, where p is the asymptotic center of u and limt→+∞ ku(t) − pk2 exists.  Lemma 5.8.4 Assume that u is a solution to (5.77), and f satisfies (5.78). If p is the asymptotic center of u, then we have:  d2u du (t) + c (t) − f (t) + f∞ , u(t) − p ≥ 0, for a.e. t ∈ R+ . dt 2 dt

(5.80)

Second Order Evolution Equations

107

Proof. Without loss of generality, we assume that f∞ = 0. Let u be a solution to (5.77), and let t ∈ R+ be fixed so that u(t) satisfies (5.77). Then by the monotonicity of A we have: u00 (t) + cu0 (t) − f (t), u(t) −

1 T

Z T

 u(s)ds

t0

Z 1 T

 u00 (t) + cu0 (t) − f (t) − u00 (s) − cu0 (s) + f (s), u(t) − u(s) T t0  + u00 (s) + cu0 (s) − f (s), u(t) − u(s) ds Z  1 T 00 ≥ u (s) + cu0 (s) − f (s), u(t) − u(s) ds T t0 Z  1 Td 0 ≥ u (s), u(t) − u(s) + ku0 (s)k2 T t0 ds  c d − ku(t) − u(s)k2 − k f (s)kku(t) − u(s)k ds 2 ds   1 0 ≥ u (T ), u(t) − u(T ) − u0 (t0 ), u(t) − u(t0 ) T  1 c c − ku(t) − u(T )k2 − ku(t) − u(t0 )k2 T 2 2 Z 1 T − k f (s)kku(t) − u(s)kds T t0 =

Letting T → +∞, by Lemma 5.8.1, we get:  −1 d u00 (t) + cu0 (t) − f (t), u(t) − p ≥ lim sup ku(t) − u(T )k2 ≥ 0 T →+∞ 2T dT 

This proves the lemma.

Theorem 5.8.5 Let u be a solution to (5.77) and f satisfies (5.78). Then u(t) * p as t → +∞, where p is an element of A−1 (− f∞ ), as well as the asymptotic center of u. Proof. First we show that u is asymptotically regular. For all h ≥ 0 we have ku(t + h) − u(t)k ≤

Z t+h t

ku0 (s)kds ≤

√ Z h

t+h

1 ku0 (s)k2 ds 2 .

t

From (5.80), we deduce that ku0 (s)k2 ≤

 c d d 0 u (s), u(s) − p + ku(s) − pk2 + Mk f (s) − f∞ k ds 2 ds

where M := supt≥0 ku(t) − pk. Integrating on [t,t + h], we get Z t+h t

  ku0 (s)k2 ds ≤ u0 (t + h), u(t + h) − p − u0 (t), u(t) − p

108

Nonlinear Evolution and Difference Equations of Monotone Type c c + ku(t + h) − pk2 − ku(t) − pk2 2 2 Z t+h

+M t

k f (s) − f∞ kds

Since there exists lim ku(t + h) − u(t)k2

t→+∞

it is sufficient to prove that lim inf ku(t + h) − u(t)k2 = 0 t→+∞

By Assumption (5.78) and Lemma 8.116, we get Z t+h

lim inf ku(t + h) − u(t)k2 ≤ h lim inf ku0 (s)k2 ds t→+∞ t→+∞ t    ≤ lim inf u0 (t + h), u(t + h) − p − u0 (t), u(t) − p t→+∞

c c lim ku(t) − pk2 + lim ku(t + h) − pk2 t→+∞ 2 2 t→+∞   1 d = lim inf ku(t + h) − pk2 − ku(t) − pk2 ≤ 0 2 t→+∞ dt −

Therefore lim ku(t + h) − u(t)k = 0

t→+∞

By Theorem 4.5.25, u(t) * p where p is the asymptotic center of u. Now we prove that p ∈ A−1 (− f∞ ). Suppose that [x, y] ∈ A. By the monotonicity of A and Equation (5.77), we get: x−

1 T

Z T t0

 1ZT  u(t)dt, y + f∞ = x − u(t), y + f∞ dt T t0 Z  1 T ≥ x − u(t), u00 (t) + cu0 (t) − f (t) + f∞ dt T t0 Z  c d 1 Td ≥ x − u(t), u0 (t) − ku(t) − xk2 T t0 dt 2 dt  − k f (t) − f∞ kku(t) − xk dt   1 = x − u(T ), u0 (T ) − x − u(t0 ), u0 (t0 ) T  c c − ku(T ) − xk2 + ku(t0 ) − xk2 2 2 Z M T − k f (t) − f∞ kdt T t0

where M = supt≥t0 ku(t) − xk. Taking limsup as T → +∞ and using Lemma 5.8.1, we get (x − p, y + f∞ ) ≥ 0. Now the maximality of A implies that p ∈ A−1 (− f∞ ). 

Second Order Evolution Equations

109

Theorem 5.8.6 Assume that the operator A in (5.77) is strongly monotone, and f satisfies (5.78). Let u be a solution to (5.77). Then u(t) converges strongly as t → +∞ to the asymptotic center p of u. Proof. Without loss of generality, we assume that f∞ = 0. Assume (y2 − y1 , x2 − x1 ) ≥ αkx2 − x1 k2 , for all [xi , yi ] ∈ A, i = 1, 2 and for some α > 0. Then by using the strong monotonicity of A, and a similar proof as in Lemma 5.8.4, we get: u00 (t) + cu0 (t) − f (t), u(t) −

1 T

Z T

 u(s)ds

t0

1 T ≥α ku(t) − u(s)k2 ds T t0  1 d 1 0 − ku(t) − u(T )k2 − u (t0 ), u(t) − u(t0 ) 2T dT T  1 c c 2 − ku(t) − u(T )k − ku(t) − u(t0 )k2 T 2 2 Z 1 T − k f (s)kku(t) − u(s)kds T t0 Z

Taking limsup as T → +∞ in the above inequality and using Lemma 5.8.1, Theorem 5.8.3 and Theorem 5.8.5, we get: Z  1 T u00 (t) + cu0 (t) − f (t), u(t) − p ≥ α lim inf ku(t) − u(s)k2 ds T →+∞ T t0

≥ α lim inf ku(t) − u(s)k2 s→+∞

≥ αku(t) − pk2 . Now integrating both sides of the above inequality on [ζ ,t] where ζ > t0 , we get Z t

α

2

 u0 (s), u(s) − p − ku0 (s)k2 ds  c d ku(s) − pk2 − f (s), u(s) − p ds + 2 ds   ≤ u0 (t), u(t) − p − u0 (ζ ), u(ζ ) − p Z t c c + ku(t) − pk2 − ku(ζ ) − pk2 + M k f (s)kds 2 2 ζ

ku(s) − pk ds ≤

ζ

Z t d ζ

where M := supt≥0 ku(t) − pk. We define F(t) := α ζt ku(s) − pk2 ds. Integrating the above inequality on [T, 2T ] and dividing by T , since F is nondecreasing, we get: R

F(T ) ≤

1 T

Z 2T

F(t)dt T

≤ M(ζ ) +

1 T

Z 2T T

 u0 (t), u(t) − p dt

110

Nonlinear Evolution and Difference Equations of Monotone Type

 1 ku(2T ) − pk2 − ku(T ) − pk2 2T  R where M(ζ ) := − u0 (ζ ), u(ζ ) − p − 2c ku(ζ ) − pk2 + M ζ+∞ k f (s)kds. This implies that Z = M(ζ ) +

+∞

ku(s) − pk2 ds < +∞

ζ

Hence lim inft→+∞ ku(t)− pk2 = 0. Since there exists limt→+∞ ku(t)− pk2 , the proof of the theorem is now complete.  Now we prove the strong convergence of solutions to (5.77) without even the monotonicity assumption on A, but assuming A to satisfy some positivity condition and the following condition (b): (x2 , y1 ) + (x1 , y2 ) + b(kx1 k, kx2 k){(x1 , y1 ) + (x2 , y2 )} ≥ 0,

(5.81)

where [x1 , y1 ] ∈ A, [x2 , y2 ] ∈ A and b : R+ × R+ → R+ is a continuous function. Condition (5.81) is a weaker version of the following condition (a): |(x1 , y2 ) + (x2 , y1 )| ≤ a(kx1 k, kx2 k){(x1 , y1 ) + (x2 , y2 )},

(5.82)

where [[x1 , y1 ] ∈ A, [x2 , y2 ] ∈ A and a : R+ × R+ → R+ is a continuous function. The condition (5.82) has been applied by Mitidieri [MIT] to prove the strong convergence of solutions to (5.77) when c = 0 in the homogeneous case ( f ≡ 0). Since these conditions are no longer invariant by translation of A with f∞ , we will assume in our next theorem that f satisfies (5.78) with f∞ = 0. Theorem 5.8.7 Assume that the (not necessarily monotone) operator A in (5.77) satisfies (y, x) ≥ 0, ∀[x, y] ∈ A, as well as condition (b), and f satisfies (5.78) with f∞ = 0. Let u be a solution to (5.77). Then u(t) converges strongly as t → +∞ to the asymptotic center p of u. Proof. For h ≥ 0, let M := max{1, supt≥0 b(ku(t)k, ku(t + h)k)} and H(t) :=

 M M ku(t)k2 + ku(t + h)k2 + u(t), u(t + h) 2 2

By the assumption on A, we have   H 00 (t) = M u00 (t), u(t) + Mku0 (t)k2 + M u00 (t + h), u(t + h) + Mku0 (t + h)k2   + u00 (t), u(t + h) + u00 (t + h), u(t)) + 2(u0 (t), u0 (t + h)    = M u00 (t) + cu0 (t) − f (t), u(t) − Mc u0 (t), u(t) + M f (t), u(t)  + Mku0 (t)k2 + M u00 (t + h) + cu0 (t + h) − f (t + h), u(t + h)   − Mc u0 (t + h), u(t + h) + M f (t + h), u(t + h) + Mku0 (t + h)k2    + u00 (t) + cu0 (t) − f (t), u(t + h) − c u0 (t), u(t + h) + f (t), u(t + h)

Second Order Evolution Equations

111

  + u00 (t + h) + cu0 (t + h) − f (t + h), u(t) − c u0 (t + h), u(t)   + f (t + h), u(t) + 2 u0 (t), u0 (t + h)    ≥ −Mc u0 (t), u(t) + M f (t), u(t) − Mc u0 (t + h), u(t + h)    + M f (t + h), u(t + h) − c u0 (t), u(t + h) + f (t), u(t + h)   − c u0 (t + h), u(t) + f (t + h), u(t)  Mc d Mc d d ≥− ku(t)k2 − ku(t + h)k2 − c u(t), u(t + h) 2 dt 2 dt dt − Mk f (t)kku(t)k − Mk f (t + h)kku(t + h)k − k f (t)kku(t + h)k − k f (t + h)kku(t)k  d = −c H(t) − k f (t)k Mku(t)k + ku(t + h)k dt  − k f (t + h)k Mku(t + h)k + ku(t)k ≥ −Kk f (t)k − Kk f (t + h)k − cH 0 (t) where K := 2M supt≥0 ku(t)k. This implies that H 00 (t) + cH 0 (t) ≥ −Kk f (t)k − Kk f (t + h)k a.e. on R+ From Lemma 5.8.2, we deduce that H(t) ≤ H(s) + η(s) ∀t ≥ s ≥ t0 where lims→+∞ η(s) = 0. This implies that   u(t), u(t + h) ≤ u(s), u(s + h) + η(s)  M + ku(s)k2 − ku(t)k2 + ku(s + h)k2 − ku(t + h)k2 2 Multiplying (5.77) by u(t), and using the positivity of A, we get   u00 (t) + cu0 (t), u(t) ≥ f (t), u(t) This implies that exists

d2 ku(t)k2 dt 2

(5.83)

(5.84)

K + c dtd ku(t)k2 ≥ − M k f (t)k. By Lemma 5.8.2, there

lim ku(t)k2

t→+∞

From (5.83), we obtain ku(s + h) − u(s)k2 ≤ ku(t + h) − u(t)k2 + 2η(s)  + (M + 1) ku(s)k2 − ku(t)k2 + ku(s + h)k2 − ku(t + h)k2 . This implies that for all δ > 0 there exists t1 > 0 such that for all t ≥ s ≥ t1 and h ≥ 0, we have ku(s + h) − u(s)k2 ≤ ku(t + h) − u(t)k2 + δ (5.85)

112

Nonlinear Evolution and Difference Equations of Monotone Type

From (5.84) we get: ku0 (t)k2 ≤

d 0 c d (u (t), u(t)) − ku(t)k2 + Mk f (t)k dt 2 dt

Now integrating from s to s + h and with a similar proof to that in Theorem 5.8.5, we conclude that u(t) is asymptotically regular. From (5.85), by letting t → +∞ and using the asymptotic regularity of u, we get: ku(s + h) − u(s)k ≤ δ

∀s ≥ t0 , ∀h ≥ 0.

Therefore (u(t))t≥0 is a Cauchy net in H, and hence u(t) → p as t → +∞.



Now we state the following two Corollaries. Corollary 5.8.8 Assume that the operator A in (5.77) is monotone and satisfies the Condition (b), and f satisfies (5.78) with f∞ = 0. If u is a solution to (5.77), then u(t) converges strongly as t → +∞ to the asymptotic center p of u. Proof. Let [x1 , y1 ], [x2 , y2 ] ∈ A. By Condition (5.81), we have: −(x1 , y2 ) − (x2 , y1 ) ≤ b(kx1 k, kx2 k){(x1 , y1 ) + (x2 , y2 )} On the other hand, by the monotonicity of A, we have: (x1 , y2 ) + (x2 , y1 ) ≤ (x1 , y1 ) + (x2 , y2 ) Adding the above two inequalities, we get: (x1 , y1 ) + (x2 , y2 ) ≥ 0. This implies that A is positive. Then the result follows from Theorem 5.8.7.



Corollary 5.8.9 Let u be a solution to (5.77) with A = ∂ ϕ, where ϕ is a proper, convex and lower semicontinuous function on H satisfying the following Condition (a1 ): D(ϕ) = −D(ϕ) and ϕ(x) − ϕ(0) ≥ a(kxk) ϕ(−x) − ϕ(0) , ∀x ∈ D(ϕ), where a : R+ → (0, +∞) is a continuous function. Assume that f satisfies (5.78) with f∞ = 0. Then u(t) converges strongly as t → +∞ to the asymptotic center p of u. Proof. We prove that the assumption on ϕ implies the condition (b), and then the result follows from Corollary 5.8.8. Without loss of generality we may assume that ϕ(0) = 0. From the convexity of ϕ and the assumption, we have: a((kxk) 1 x+ (−a(kxk)x)) 1 + a(kxk) 1 + a(kxk) a(kxk) 1 ≤ ϕ(x) + ϕ(−a(kxk)x) 1 + a(kxk) 1 + a(kxk) a(kxk) a(kxk) ≤ ϕ(x) + ϕ(−x) 1 + a(kxk) 1 + a(kxk)

0 = ϕ(0) = ϕ(

Second Order Evolution Equations

113

a(kxk) 1 ϕ(x) + ϕ(x) 1 + a(kxk) 1 + a(kxk) = ϕ(x) ≤

Now suppose that [x, y], [x, ˆ y] ˆ ∈ ∂ ϕ. By the subdifferential inequality, we have: 0 ≤ ϕ(x) ≤ (y, x) Again by the subdifferential inequality and the convexity of ϕ, we get:  ϕ(x) − a(kxk)ϕ(−x) ˆ ≤ ϕ(x) − ϕ − a(kxk)xˆ  ≤ y, x + a(kxk)xˆ Therefore

1 1 ϕ(x) − ϕ(−x) ˆ ≤ (x, y) + (x, ˆ y) a(kxk) a(kxk)

(5.86)

and ϕ(x) ˆ − a(kxk)ϕ(−x) ˆ ≤ ϕ(x) ˆ − ϕ(−a(kxk)x) ˆ ≤ (y, ˆ xˆ + a(kxk)x) ˆ Hence

1 1 ϕ(x) ˆ − ϕ(−x) ≤ (x, ˆ y) ˆ + (x, y) ˆ a(kxk) ˆ a(kxk) ˆ

(5.87)

Adding up (5.86) and (5.87), by the assumption on ϕ, we get 1 1 (x, y) + (x, ˆ y) ˆ + (x, ˆ y) + (x, y) ˆ ≥0 a(kxk) a(kxk) ˆ 1 Therefore the condition (b) is satisfied by taking b(kxk, kxk) ˆ = Max{ a(kxk) , a(k1xk) ˆ }. 

5.8.2

THE CASE C > 0

In this subsection, we consider (5.77) with c > 0 and f satisfing (5.78). We prove the strong convergence of solutions to (5.77) with c > 0, for a general maximal monotone operator A. Lemma 5.8.10 Assume u is a solution to (5.77). Then there exists p ∈ H such that: (

d2u du (t) + c (t) − f (t), u(t) − p) ≥ 0, for a.e. t ∈ R+ dt 2 dt

(5.88)

Proof. Let u be a solution to (5.77), and let t ∈ R+ be fixed so that u(t) satisfies (5.77). Then by the monotonicity of A, we get: u00 (t) + cu0 (t) − f (t), u(t) −

1 T

Z T t0

 u(s)ds

114

Nonlinear Evolution and Difference Equations of Monotone Type Z  1 T 00 ≥ u (s) + cu0 (s) − f (s), u(t) − u(s) ds T t0 Z  1 Td 0 = u (s), u(t) − u(s) + ku0 (s)k2 T t0 ds  c d − ku(t) − u(s)k2 − f (s), u(t) − u(s) ds 2 ds   1 0 ≥ u (T ), u(t) − u(T ) − u0 (t0 ), u(t) − u(t0 ) T  1ZT c c 2 2 − ku(t) − u(T )k + ku(t) − u(t0 )k − Mk f (s)kds 2 2 T t0

where M := sups≥t0 ku(t) − u(s)k. Integrating the above inequality from T = ξ to T = T 0 and dividing by T 0 , we get u00 (t) + cu0 (t) − f (t), u(t) −

1 T0

R t0

Z T0

σT dT +

0

u(s)ds T0

Z T0 dT 

T  1 0 ≥ 0 ku(t) − u(T )k2 − u (t0 ), u(t) − u(t0 ) T ξ 2T dT T Z  c c 1 T − ku(t) − u(T )k2 + ku(t) − u(t0 )k2 − Mk f (s)kds dT 2T 2T T t0 ξ

ξ

Z 0 1 T  −1 d

There exists a sequence {Tn0 } such that 1 Tn0

Z T0 n

σT dT * p

ξ

as n → +∞. Replacing T 0 by {Tn0 } and letting n → +∞, after integration by parts we get:  u00 (t) + cu0 (t) − f (t), u(t) − p ≥ 0 where p is a weak cluster point of

1 R T0 T 0 ξ σT dT .



Theorem 5.8.11 Let u be a solution to (5.77). Then u(t) → p as t → +∞, where p ∈ A−1 (− f∞ ). Proof. Without loss of generality we assume f∞ = 0. Let h > 0 be fixed. Then from (5.77) and the monotonicity of A, we have   u00 (t + h) − u00 (t), u(t + h) − u(t) + c u0 (t + h) − u0 (t), u(t + h) − u(t)  ≥ f (t + h) − f (t), u(t + h) − u(t) ≥ −k f (t + h) − f (t)kku(t + h) − u(t)k ≥ −Mk f (t + h) − f (t)k ≥ −Mk f (t + h)k − Mk f (t)k

Second Order Evolution Equations

115

where M = supt≥0 ku(t + h) − u(t)k. Then d2 ku(t + h) − u(t)k2 + c dt 2

Z t

 ku(s + h) − u(s)k2 ds ≥ −2Mk f (t + h)k − 2Mk f (t)k

0

First we show that

(5.89) Z t

ku(s + h) − u(s)k2 ds

0

is bounded from above. From (5.89), we get ku(s + h) − u(s)k2 ≤

Z s+h

ku0 (r)kdr

2

≤h

Z s+h

s

ku0 (r)k2 dr

s

Z h s+h  d 2 d ku(r) − pk2 + c ku(r) − pk2 2 2 s dr dr  − f (r), u(r) − p dr h d d ≤ ku(s + h) − pk2 − ku(s) − pk2 2 ds ds



+ cku(s + h) − pk2 − cku(s) − pk2 + M

Z s+h

k f (r)kdr

s

where M = supt≥t0 ku(t) − pk. Integrating the above inequality on [0,t], we get Z t 0

h ku(s + h) − u(s)k2 ds ≤ [ku(t + h) − pk2 − ku(h) − pk2 2 − ku(t) − pk2 + ku(0) − pk2 Z t

+c

ku(s + h) − pk2 ds − c

Z t

0

|u(s) − p|2 ds

0

Z t

+M

Z s+h

ds 0

k f (r)kdr]

s

h ≤ [ku(t + h) − pk2 + ku(0) − pk2 2 Z t+h

+c

ku(s) − pk2 ds − c

Z h

ku(s) − pk2 ds

0

t

Z t

Z t+h

+ M(

rk f (r)kdr + 0

hk f (r)kdr)] t

≤ R < +∞ Therefore ku(t + h) − u(t)k2 + c

Z t

ku(s + h) − u(s)k2 ds

0

is bounded from above, hence by (5.89) and Lemma 5.8.2, we have ku(t + h) − u(t)k2 + c

Z t 0

ku(r + h) − u(r)k2 dr



116

Nonlinear Evolution and Difference Equations of Monotone Type ≤ku(s + h) − u(s)k2 + c

Z s

ku(r + h) − u(r)k2 dr

0

Z +∞

+ 4M

rk f (r)kdr

(5.90)

s

Hence ku(t + h) − u(t)k2 ≤ ku(s + h) − u(s)k2 + 4M

Z +∞

rk f (r)kdr

(5.91)

s

This implies that u is almost nonexpansive. From (5.90) we also get: Z t

c

ku(r + h) − u(r)k2 dr ≤ ku(s + h) − u(s)k2 − ku(t + h) − u(t)k2

s

Z +∞

+ 4M

rk f (r)kdr s

Then integrating (5.91) on [s,t], and using the above inequality, we get: cku(t + h) − u(t)k2 (t − s) − 4Mc

dr s

ku(s + h) − u(s)k2 + 4M

Z +∞

Z t

r0 k f (r0 )kdr0 ≤

r

Z +∞

rk f (r)kdr s

Then we obtain ku(s + h) − u(s)k2 4M + c(t − s) t −s Z +∞ 4M + rk f (r)kdr c(t − s) s

ku(t + h) − u(t)k2 ≤

Z +∞

rk f (r)kdr s

Hence u(t) is a Cauchy net in H. Therefore u(t) converges strongly as t → +∞, to some p ∈ H. Now we prove that 0 ∈ A−1 (− f∞ ). Suppose that [x, y] ∈ A. By the monotonicity of A and (5.77) we get   x − u(t), y ≥ x − u(t), u00 (t) + cu0 (t) − f (t)  c d  d ≥ x − u(t), u0 (t) − kx − u(t)k2 − x − u(t), f (t) dt 2 dt Integrating the above inequality on [t0 , T ] and dividing by T , we get Z T  T − t0 1 (x, y + f∞ ) − u(t)dt, y + f∞ T T t0   1 1 ≥ x − u(T ), u0 (T ) − x − u(t0 ), u0 (t0 ) T T c c 2 − kx − u(T )k + kx − u(t0 )k2 2T 2T

Second Order Evolution Equations K − T

Z T t0

117

k f (t) − f∞ kdt

where K = supt≥t0 ku(t) − xk. Letting T → +∞, by Lemma 5.8.10 we obtain (x − p, y + f∞ ) ≥ 0. Now the maximality of A implies that 0 ∈ A−1 (− f∞ ).  Remark 5.8.12 In the homogeneous case ( f (t) ≡ 0), by (5.89) g(t) := ku(t + h) − u(t)k2 + c

Z t

ku(s + h) − u(s)k2 ds

0

is convex. Since g(t) is bounded from above, then g(t) is nonincreasing and g0 (t) ≤ 0. If u is nonconstant, we get:  d ln ku(t + h) − u(t)k2 ≤ −c dt ⇒ ln ku(t + h) − u(t)k2 − ln ku(h) − u(0)k2 ≤ −ct ⇒ ku(t + h) − u(t)k2 ≤ e−ct ku(h) − u(0)k2 c

⇒ ku0 (t)k ≤ ku0 (0)ke− 2 t and ku(t) − pk = lim ku(t) − u(T )k T →+∞

= lim k

Z T

T →+∞

≤ lim

t T

Z

T →+∞ t

u0 (s)dsk

ku0 (s)kds

≤ ku0 (0)k lim

Z T

T →+∞ t

c

e− 2 s ds

c c 2 = − ku0 (0)k lim (e− 2 T − e− 2 t ) T →+∞ c 2 0 − 2c t = ku (0)ke c c

c

Therefore ku0 (t)k = O(e− 2 t ) and ku(t) − pk = O(e− 2 t ).

REFERENCES APR1. N. C. Apreutesei, Nonlinear second order evolution equations of monotone type and applications, Pushpa Publishing House, Allahabad, India, 2007. BAR1. V. Barbu, Nonlinear semigroups and Differential Equations in Banach Spaces, Noordhoff, Leyden, 1976.

118

Nonlinear Evolution and Difference Equations of Monotone Type

BAR2. V. Barbu, Sur un probl`eme aux limites pour une classe d’´equations diff´erentielles non lin´eaires abstraits du deuxi`eme ordre en t, C. R. Acad. Sci. Paris S´er. A–B 274 (1972), A459–A462. BAR3. V. Barbu, A class of boundary problems for second order abstract differential equations, J. Fac. Sci. Univ. Tokyo Sect. IA Math. 19 (1972), 295–319. BAI. J. B. Baillon, Un th´eor`eme de type ergodique pour les contractions non lin´eaires dans un espace de Hilbert, C. R. Acad. Sci. Paris 280 (1975), A1511–A1514. BRE. H. Br´ezis, “Op´erateurs Maximaux Monotones et Semi-groupes de Contractions dans les Espaces de Hilbert”, North-Holland Mathematics studies, Vol. 5, North-Holland Publishing Co., Amsterdam-London, (1973). BRU1. R. E. Bruck, Asymptotic convergence of nonlinear contraction semigroups in Hilbert space, J. Funct. Anal. 18 (1975), 15–26. BRU2. R. E. Bruck, Periodic forcing of solutions of a boundary value problem for a second order differential equation in Hilbert space, J. Math. Anal. Appl. 76 (1980), 159–173. DJ1. B. Djafari Rouhani, Ergodic theorems for nonexpansive sequences in Hilbert spaces and related problems, Ph.D. Thesis, Yale University, Part I, pp. 1–76 (1981). DJ2. B. Djafari Rouhani, Asymptotic behaviour of quasi-autonomous dissipative systems in Hilbert spaces, J. Math. Anal. Appl. 147 (1990), 465–476. DJ3. B. Djafari Rouhani, Asymptotic behaviour of almost nonexpansive sequences in a Hilbert space, J. Math. Anal. Appl. 151 (1990), 226–235. DJ4. B. Djafari Rouhani, An ergodic theorem for sequences in a Hilbert space, Nonlinear Anal. Forum, 4 (1999), 33–48. DJ-KH1. B. Djafari Rouhani and H. Khatibzadeh, Asymptotic behavior of solutions to some homogeneous second-order evolution equations of monotone type, 2007, Article ID 72931, 8 pages. DJ-KH2. B. Djafari Rouhani and H. Khatibzadeh, Asymptotic behavior of bounded solutions to a class of second order nonhomogeneous evolution equations, Nonlinear Anal. 70 (2009), 4369–4376. DJ-KH3. B. Djafari Rouhani and H. Khatibzadeh, Asymptotic behavior of bounded solutions to a nonhomogeneous second order evolution equation of monotone type, Nonlinear Anal. 71 (2009), e147–e159. DJ-KH4. B. Djafari Rouhani and H. Khatibzadeh, Asymptotic behavior of bounded solutions to some second order evolution systems, Rocky Mountain J. Math. 40 (2010), 1289–1311. DJ-KH5. B. Djafari Rouhani and H. Khatibzadeh, A strong convergence theorem for solutions to a nonhomogeneous second order evolution equation, J. Math. Anal. Appl. 363 (2010), 648–654. DJ-KH6. B. Djafari Rouhani and H. Khatibzadeh, A note on the strong convergence of solutions to a second order evolution equation, J. Math. Anal. Appl. 401 (2013), 963– 966. DJ-KH7. B. Djafari Rouhani and H. Khatibzadeh, Asymptotic behavior for a general class of homogeneous second order evolution equations in a Hilbert space, Dynam. Systems Appl. 24 (2015), 1–15. EDE. M. Edelstein, The construction of an asymptotic center with a fixed-point property, Bull. Amer. Math. Soc. 78 (1972), 206–208. LOR. G. G. Lorentz, A contribution to the theory of divergent sequences, Acta. Math. 80 (1948), 167–190.

Second Order Evolution Equations

119

MIT. E. Mitidieri, Some remarks on the asymptotic behaviour of the solutions of second order evolution equations, J. Math. Anal. Appl. 107 (1985), 211–221. VER1. L. V´eron, Probl`emes d’´evolution du second ordre associ´es a` des op´erateurs monotones, C. R. Acad. Sci. Paris S´er. A 278 (1974), 1099–1101. VER2. L. V´eron, Equations d’´evolution du second ordre associ´ees a` des op´erateurs maximaux monotones, Proc. Roy. Soc. Edinburgh Sect. A 75 (1975/76), 131–147.

Ball with Friction 6 Heavy Dynamical System 6.1

INTRODUCTION

In this brief chapter we consider the following second order evolution equation ( x00 (t) + γx0 (t) + A(x(t)) 3 0, x(0) = x0 , x0 (0) = x1

a.e. t ≥ 0

(6.1)

where γ > 0 is a positive constant, and A is a maximal monotone operator. When A = ∇ϕ, where ϕ is a convex and differentiable function, this system models the rolling of a heavy ball on the slid of the convex function ϕ with the friction impact γ > 0. For general maximal monotone operators, (6.1) does not necessarily have a solution. By assuming the existence of a solution, to the best of our knowledge, the study of the asymptotic behavior of the solution is still an open problem. Only for some special monotone operators, the asymptotic behavior is well-known. In this chapter, we study these special cases where the solution to (6.1) is weakly convergent. We refer the reader to [ALV-ATT1, ALV-ATT2, ATT-ALV1, ATT-GOU-RED] for more details about (6.1) when A = ∇ϕ and its various variants.

6.2

MINIMIZATION PROPERTIES

The following theorem was proved in [ALV]. Theorem 6.2.1 Suppose that ϕ ∈ C1 (H; R) is convex and inf ϕ > −∞. If u ∈ C2 (0, +∞; H) is a solution to (6.1), then u0 ∈ L2 (0, +∞; H), u0 (t) → 0 as t → +∞, and lim ϕ(u(t)) = inf ϕ. t→+∞

Moreover, if Argminϕ , ∅, then there exists p ∈ Argminϕ such that u(t) * p as t → +∞. Proof. Fix x ∈ H, and define the following auxiliary function ψ(t) := 12 ku(t) − xk2 . Since u is a solution to (6.1), it follows that ψ 00 + γψ 0 = h∇ϕ(u), x − ui + ku0 k2 . Now the convexity of ϕ yields ψ 00 + γψ 0 ≤ ϕ(x) − ϕ(u) + ku0 k2 .

(6.2)

122

Nonlinear Evolution and Difference Equations of Monotone Type

Let:

1 E(t) := ϕ(u(t)) + ku0 (t)k2 . 2 Then we can rewrite the last inequality as: 3 ψ 00 + γψ 0 ≤ ϕ(x) − E(t) + ku0 k2 2 Since E(t) is nonincreasing because E 0 (t) ≤ 0, then given t > 0, for all τ ∈ [0,t] we have: 3 ψ 00 (τ) + γψ 0 (τ) ≤ ϕ(x) − E(t) + ku0 (τ)k2 2 γτ Multiplying both sides of the above inequality by e , and then integrating from 0 to θ , we get: 1 3 ψ 0 (θ ) ≤ e−γθ ψ 0 (0) + (1 − e−γθ )[ϕ(x) − E(t)] + γ 2

Z θ

e−γ(θ −τ) ku0 (τ)k2 dτ

0

Integrating once more on [0,t], we obtain: 1 1 ψ(t) ≤ ψ(0) + (1 − e−γt )ψ 0 (0) + 2 (γt − 1 + e−γt )[ϕ(x) − E(t)] + h(t), γ γ where h(t) :=

3 2

Z tZ θ 0

(6.3)

e−γ(θ −τ) ku0 (τ)k2 dτdθ

0

Since E(t) ≥ ϕ(u(t)), for t > 1γ , (6.3) gives: 1 1 1 (γt − 1 + e−γt )ϕ(u(t)) ≤ ψ(0) + (1 − e−γt )ψ 0 (0) + 2 (γt − 1 + e−γt )ϕ(x) + h(t) γ2 γ γ Dividing both sides of the above inequality by γ12 (γt − 1 + e−γt ) and letting t → +∞, we get: γ lim sup ϕ(u(t)) ≤ ϕ(x) + lim sup h(t) (6.4) t→∞ t→+∞ t Let’s show that h(t) remains bounded as t → +∞. By Fubini’s theorem, we have: h(t) =

3 2

Z tZ t 0

e−γ(θ −τ) ku0 (τ)k2 dθ dτ =

τ

3 2γ

Z t

ku0 (τ)k2 (1 − e−γ(t−τ) )dτ.

0

From the equality E 0 = −γku0 k2 , it follows that: 1 0 2 ku k + ϕ(u) + γ 2

Z t 0

ku0 (τ)k2 dτ = E0 ,

Heavy Ball with Friction Dynamical System therefore

Z t

ku0 (τ)k2 dτ ≤

0

123

E0 − inf ϕ < +∞. γ

Then u0 ∈ L2 (0, +∞; H), and 3 t 0 h(t) ≤ ku (τ)k2 dτ 2γ 0 Z 3 ∞ 0 ≤ ku (τ)k2 dτ < +∞. 2γ 0 Z

On the other hand, since E(·) is nonincreasing and bounded from below by inf ϕ, it converges as t → +∞. If limt→+∞ E(t) > inf ϕ, then limt→+∞ ku0 (t)k > 0, because by (6.4), limt→+∞ ϕ(u(t)) = inf ϕ, contradicting the fact that u0 ∈ L2 . Therefore, limt→+∞ E(t) = inf ϕ, hence u0 (t) → 0 as t → +∞. Now we prove the weak convergence of u(t) when Argminϕ , ∅. To this end, we use Opial’s lemma [OPI]. Since it follows from (6.3) that u(t) is bounded, then there exists xˆ ∈ H such that u(tk ) * xˆ for some sequence tk → +∞. Since ϕ is convex and continuous, then it is lower semicontinuous. Hence ϕ(x) ˆ ≤ lim inf ϕ(u(tk )) = lim ϕ(u(t)) = inf ϕ, t→+∞

k→+∞

which shows that xˆ ∈ Argmin ϕ. By Opial’s lemma, we only need to show that ∀z ∈ Argminϕ, lim ku(t) − zk exists. t→+∞

For this, fix z ∈ Argminϕ and define v(t) := 21 ku(t) − zk2 . The following lemma provides a sufficient condition on [v0 ]+ , the positive part of the derivative, in order to ensure the convergence of v. Lemma 6.2.2 Let θ ∈ C1 ([0, ∞); R) be bounded from below. If [θ 0 ]+ ∈ L1 ([0, ∞); R), then θ (t) converges as t → +∞. Proof. Set w(t) := θ (t) −

Z t 0

[θ 0 (τ)]+ dτ

Since w(t) is bounded from below and w0 (t) ≤ 0, then w(t) converges as t → +∞, and consequently θ (t) converges as t → +∞.  On account of this result, it is sufficient to prove that [v0 ]+ belongs to L1 (0, ∞). Of course, to obtain information on v0 we shall use the fact that u(t) is a solution to (6.1). Due to the optimality of z, it follows from (6.2) that v00 + γv0 ≤ ku0 k2

(6.5)

Lemma 6.2.3 If ω ∈ C1 ([0, ∞); R) satisfies the differential inequality ω 0 (τ) + γω(τ) ≤ g(τ), ∀τ ∈ [0, ∞), with γ > 0 and g

∈ L1 ([0, ∞); R),

then [ω]+

∈ L1 ([0, ∞); R).

(6.6)

124

Nonlinear Evolution and Difference Equations of Monotone Type

Proof. We may assume that g ≥ 0, for if not, we replace g by |g|. Multiplying both sides of (6.6) by eγτ and integrating on [0,t], we get: ω(t) ≤ e−γt ω(0) +

Z t

e−γ(t−τ) g(τ)dτ

0

Therefore [ω(t)]+ ≤ e−γt [ω(0)]+ + and Fubini’s theorem gives

Z t

e−γ(t−τ) g(τ)dτ

0

R ∞ R t −γ(t−τ) R g(τ)dτdt = 1γ 0∞ g(τ)dτ < +∞. 0 0e



Recall that we already proved ku0 k2 ∈ L1 ([0, ∞); R). Now the proof of the theorem is completed by using Lemma 6.2.3 and (6.5).  Recently, the asymptotic behavior of (6.1) with nonconstant friction was studied in [MAY, BAL-MAY]. For other recent work and developments on the asymptotic behavior of various versions of (6.1) see [BEN-HAR, CAB-ENG-GAD, CAB-FRA].

REFERENCES ALV. F. Alvarez, On the minimizing property of a second order dissipative system in Hilbert spaces, SIAM J. Control Optim. 38 (2000), 1102–1119. ALV-ATT1. F. Alvarez and H. Attouch, Convergence and asymptotic stabilization for some damped hyperbolic equations with non-isolated equilibria, ESAIM Control Optim. Calc. Var. 6 (2001), 539–552. ALV-ATT2. F. Alvarez and H. Attouch, An inertial proximal method for maximal monotone operators via discretization of a nonlinear oscillator with damping, Set-Valued Anal. 9 (2001), 3–11. ATT-ALV1. H. Attouch and F. Alvarez, The heavy ball with friction dynamical system for convex constrained minimization problems, Optimization (Namur, 1998), 25–35, Lecture Notes in Econom. and Math. Systems, 481, Springer, Berlin, 2000. ATT-GOU-RED. H. Attouch, X. Goudou and P. Redont, The heavy ball with friction method, I. The continuous dynamical system: Global exploration of the local minima of a real-valued function by asymptotic analysis of a dissipative dynamical system, Commun. Contemp. Math. 2 (2000), 1–34. BAL-MAY. M. Balti and R. May, Asymptotic for the perturbed heavy ball system with vanishing damping term, Evol. Equ. Control Theory 6 (2017), 177–186. BEN-HAR. I. Ben Hassen and A. Haraux, Convergence and decay estimates for a class of second order dissipative equations involving a non-negative potential energy, J. Funct. Anal. 260 (2011), 2933–2963. CAB-ENG-GAD. A. Cabot, H. Engler and S. Gadat, On the long time behavior of second order differential equations with asymptotically small dissipation, Trans. Amer. Math. Soc. 361 (2009), 5983–6017.

Heavy Ball with Friction Dynamical System

125

CAB-FRA. A. Cabot and P. Frankel, Asymptotics for some semilinear hyperbolic equations with non-autonomous damping, J. Differential Equations, 252 (2012), 294–322. MAY. R. May, Long time behavior for a semilinear hyperbolic equation with asymptotically vanishing damping term and convex potential, J. Math. Anal. Appl. 430 (2015), 410–416. OPI. Z. Opial, Weak convergence of the sequence of successive approximations for nonexpasive mappings, Bull. Amer. Math. Soc. 73 (1967), 591–597.

Part III Difference Equations of Monotone Type

Order Difference 7 First Equations and Proximal Point Algorithm 7.1

INTRODUCTION

Consider the following first order evolution equation of monotone type ( −u0 (t) ∈ Au(t) + f (t), u(0) = u0 ∈ D(A)

(7.1)

that was studied in Chapter 4. By the backward Euler discretization u0 (t) =

u(t) − u(t − h) + o(h), h

in (7.1), we get the following difference inclusion ( un−1 − un ∈ cn Aun + fn u(0) = u0 Using the resolvent operator, (7.2) can be written as follows: ( un = JcAn (un−1 − fn ) u(0) = u0 .

(7.2)

(7.3)

In (7.2) and (7.3), we can take u0 ∈ H, and not necessarily in D(A). The existence of the sequence {un } in (7.2) or equivalently (7.3) follows from the surjectivity of the resolvent operator JλA (see Theorem 3.3.2 of Chapter 3). In this chapter, we prove some results on the boundedness, mean convergence, as well as weak and strong convergence of solutions to (7.2), similar to the results of Chapter 4 for (7.1). The resolvent Equation (7.3) is called the proximal point algorithm, and it was first introduced by Rockafellar [ROC] to approximate a zero of a maximal monotone operator. Rockafellar [ROC] proved that when A−1 (0) , ∅, lim infn→+∞ cn > 0 and ∑∞ n=1 k f n k < +∞, then the sequence {un } converges weakly to a zero of the maximal monotone operator A. Br´ezis and Lions [BRE-LIO] and Lions [LIO] continued the study of the convergence of the proximal point algorithm, and proved the weak convergence of the mean and some results on the weak convergence of the sequence ∞ 2 {un }. They proved that when A−1 (0) , ∅, ∑∞ n=1 cn = +∞ and ∑n=1 k f n k < +∞, the sequence {un } converges weakly to a zero of A. Clearly this condition is weaker than

130

Nonlinear Evolution and Difference Equations of Monotone Type

the one used by Rockafellar. Before Rockafellar, the proximal point algorithm was studied by Martinet [MAR] for convex functions. If f : H → (−∞, +∞] is a proper, convex and lower semicontinuous function with Argmin f , ∅, by Theorem 2.4.3 there exists a sequence {un } satisfying ( un = Argminy∈H { f (y) + 2c1n ky − un−1 k2 }, n ≥ 1 (7.4) u0 ∈ H. Martinet proved that the sequence {un } given by (7.4) converges weakly to a minimum point of f . It is easy to check that (7.4) is equivalent to (7.3) when the maximal monotone operator A is equal to ∂ f . This chapter is devoted to the asymptotic behavior of the sequence {un } given by (7.3). Section 2 describes sufficient conditions for the boundedness of the sequence {un }. In Section 3, we consider the periodic forcing case. In Section 4, we investigate the weak and strong convergence of the sequence {un } generated by (7.2) when the error sequence { fn } is summable (i.e. ∑∞ n=1 k f n k < +∞). The convergence analysis of the proximal point algorithm with a non-summable error sequence is the subject of Section 5. Finally, in Section 6 we u −u − f study the rate of convergence. Throughout the chapter we denote n−1 cn n n by Aun .

7.2

BOUNDEDNESS OF SOLUTIONS

It is a well-known result (see also Section 4 of this chapter) that “{un } is bounded if ∞ and only if A−1 (0) , ∅ provided that ∑∞ n=1 cn = +∞ and ∑n=1 k f n k < +∞”. In this very short section, we prove that the coercivity condition on the operator A implies the boundedness of the sequence {un } without assuming A−1 (0) , ∅. The following theorem is also proved by Boikanyo and Morosanu [BOI-MOR] with stronger conditions on { fn } and {cn }. Definition 7.2.1 The (possibly multivalued) operator A in H is said to be coercive 0) if there exists y0 ∈ H such that limkzk→+∞ (w,z−y kz−y k = +∞, ∀[z, w] ∈ A. 0

Theorem 7.2.2 Let A be a coercive maximal monotone operator. If the sequence { kcfnn k } is bounded, then for each u0 ∈ H, the sequence {un } generated by (7.3) is bounded. Proof. Let C > 0 be such that for each n ≥ 1,

k fn k cn

< C. By coerciveness of A,

0) there exist K > 0 and y0 ∈ H such that for all [z, w] ∈ A, with kzk > K, (w,z−y kz−y0 k > C. Suppose that there exists n such that kun+1 − y0 k > K. From (7.2), we have

un − un+1 − fn+1 ∈ cn+1 Aun+1 . Multiplying both sides of (7.5) by

un+1 −y0 kun+1 −y0 k ,

we get

cn+1C + kun+1 − y0 k ≤ kun − y0 k + k fn+1 k.

(7.5)

First Order Difference Equations and Proximal Point Algorithm

131

This implies that kun+1 − y0 k ≤ kun − y0 k + cn+1 (

k fn+1 k −C) < kun − y0 k, cn+1

for each n ≥ 0 such that kun+1 − y0 k > K. It follows that for all n ≥ 0, kun+1 − y0 k ≤ max{ku0 − y0 k, K}. Hence the sequence {un } is bounded.

7.3



PERIODIC FORCING

In this section we prove the discrete version of Theorem 4.3.3 for the first order evolution Equation (7.1). But as it is shown in the following, the discrete version holds true for general maximal monotone operators. Theorem 7.3.1 Suppose that A is a maximal monotone operator in H. If (7.2) has a solution {un }n≥1 satisfying lim infn→+∞ ksn k < +∞, where sn = n1 ∑n−1 k=0 ukN , and { fn }n≥1 and {cn }n≥1 are periodic with period N, then there exists an N-periodic solution {wn }n≥1 of (7.2) such that un − wn * 0 as n → +∞. Moreover, any two periodic solutions differ by an additive constant. Proof. Let x ∈ H, and let m ≥ 0 be an integer. Since A is maximal monotone, there is a unique solution to: ( ui−1 − ui ∈ ci Aui + fi , i > m (7.6) um = x. For n ≥ m, we define the operators Q(n, m) : H → H by Q(n, m)x := un . Let vn be a solution of (7.6) with vm = y. The monotonicity of A implies that (un−1 − vn−1 , un − vn ) − kun − vn k2 ≥ 0. Therefore kun − vn k ≤ kun−1 − vn−1 k. Hence by definition of Q(n, m), it follows that Q(n, m) is nonexpansive. The nonexpansiveness of Q(n, m) also shows the uniqueness of the solution to (7.6) with um = x. By the uniqueness of the solution, we have Q(n, m)Q(m, k) = Q(n, k) for n ≥ m ≥ k, and by the periodicity, we also have: Q(n + N, m + N) = Q(n, m)

132

Nonlinear Evolution and Difference Equations of Monotone Type

for n ≥ m. It follows that Q(m + N, m)n = Q(m + nN, m). In particular, taking m = 0, it follows that {ukN }k≥0 is a nonexpansive sequence in H. Therefore, by the ergodic theorem proved in [ROU], we deduce that sn converges weakly in H, and the limit is a fixed point of Q(N, 0). This shows the existence of a periodic solution to (7.2), which is therefore bounded. Hence all solutions to (7.2) are bounded. Suppose that {un } is a bounded solution of (7.2) with u0 = x and let {vn } be a periodic solution of (7.2) with v0 = y. Let zn := un − vn . From (7.2) we have: zn−1 − zn ∈ cn (Aun − Avn ). Multiplying the above inclusion by zn , we get (zn−1 − zn , zn ) ≥ 0. Hence kzn k2 ≤ (zn−1 , zn )

(7.7)

which implies that kzn k is non increasing, therefore it has a limit. We also get: (zn−1 − zn , zn − zn−1 ) + (zn−1 − zn , zn−1 ) ≥ 0 Therefore kzn − zn−1 k2 ≤ (zn−1 − zn , zn−1 ) Summing up the above inequality from n = 1 to m and letting m → +∞, and using (7.7), we get that limn→+∞ kzn k exists, and: +∞

+∞

∑ kzn − zn−1 k2 ≤ ∑ (zn−1 − zn , zn−1 )

n=1

(7.8)

n=1 +∞

=

∑ (kzn−1 k2 − (zn , zn−1 ))

(7.9)

n=1 +∞



lim kzn k2 < +∞. ∑ (kzn−1 k2 − kzn k2 ) = kz0 k2 − n→+∞

(7.10)

n=1

Since {vn } is a periodic solution of (7.6), for each m ≥ 0, we have (n+1)N−1

um+nN − um+(n+1)N =



(um+i − vm+i − (um+i+1 − vm+i+1 )) → 0

i=nN

as n → +∞. Let xn := Q(m + nN, 0)u0 = um+nN

(7.11)

First Order Difference Equations and Proximal Point Algorithm

133

Then {xn }n≥1 is a nonexpansive sequence, which is asymptotically regular (i.e. xn+1 − xn → 0, as n → +∞), by (7.11). It follows from Theorem 5.7.6 of Chapter 4 that: um+nN * wm (7.12) as n → +∞. Since {vn }n≥1 is a periodic solution of (7.2), we have lim (um+nN − um−1+nN − (vm − vm−1 ))

n→+∞

= lim (um+nN − vm+nN − (um−1+nN − vm−1+nN )) n→+∞

=0 Therefore, limn→+∞ (um+nN − um−1+nN ) = (vm − vm−1 ) exists, and from (7.12), we get: lim (um+nN − um−1+nN ) = wm − wm−1 n→+∞

This implies that wm − vm = w0 − v0 = constant, showing that any two periodic solutions differ by an additive constant. By (7.2), we have [um+nN , c1m (um−1+nN − um+nN − fm )] ∈ A. Since by Theorem 3.3.4, A is demiclosed, by letting n → +∞, we get 1 [wm , (wm−1 − wm − fm )] ∈ A cm Therefore wm ∈ D(A), and wm−1 − wm − fm ∈ cm Awm . Since wn is N-periodic un − wn = un − uk+mN + uk+mN − wk+mN + wk+mN − wn = un − wn − (uk+mN − wk+mN ) + uk+mN − wk = zn − zk+mN + uk+mN − wk where k + mN ≤ n < k + (m + 1)N and zn = un − wn . Now from (7.12), we have uk+mN − wk * 0 as m → +∞, and from (7.8), we get: i=k+(m+1)N−1

kzn − zk+mN k ≤



kzi − zi−1 k → 0

i=k+mN

as n → +∞. This shows that un − wn * 0, as n → +∞, and completes the proof of the theorem. 

7.4

CONVERGENCE OF THE PROXIMAL POINT ALGORITHM

As was mentioned in the introduction, the first convergence result for the proximal point algorithm was proved by Rockafellar. Then Br´ezis and Lions [BRE-LIO] and Lions [LIO] improved his result. All the previous authors studied the convergence of (7.3) by assuming that A−1 (0) , ∅. In this chapter, we prove weak and strong convergence results without assuming A−1 (0) , ∅. First we recall an elementary lemma without proof.

134

Nonlinear Evolution and Difference Equations of Monotone Type

Lemma 7.4.1 Suppose that {an }n≥1 and {bn }n≥1 are nonnegative real sequences and ∑+∞ n=1 bn < +∞. If an+1 ≤ an + bn , for all n ≥ 1, then there exists limn→+∞ an . We start with a weak ergodic theorem which was proved by Lions (see [LIO]), by assuming that A−1 (0) , φ (see also [MOR], Theorem 3.1, p.139). We denote: wn := (∑nk=1 ck )−1 ∑nk=1 ck uk . Theorem 7.4.2 Assume that {un }n≥1 given by (7.2) is bounded, ∑+∞ n=1 cn = +∞ and +∞ k f k < +∞. Then the sequence {w } converges weakly as n → +∞ to an ele∑n=1 n n ment p ∈ A−1 (0), which is also the asymptotic center of {un }. Proof. By the monotonicity of A, we have (Aun , uk ) + (Auk , un ) ≤ (Aun , un ) + (Auk , uk ), for all k, n ≥ 1. Multiplying both sides of the above inequality by ck cn and using (7.2), we get (un−1 − un − fn , ck uk ) + (uk−1 − uk − fk , cn un ) ≤ ck (un−1 − un − fn , un ) + cn (uk−1 − uk − fk , uk ). Assume that wm j * p as j → +∞. Summing up both sides of the above inequality mj from k = 1 to k = m j , dividing by ∑k=1 ck , and letting j → +∞, we get (un−1 − un − fn , un − p) ≥ 0

(7.13)

This implies that kun − pk ≤ kun−1 − pk + k fn k. By Lemma 7.4.1, there exists limn→+∞ kun − pk. If q is another cluster point of {wn }n≥1 , limn→+∞ kun − qk exists too. It follows that there exists limn→+∞ 21 (kun − pk2 − kun − qk2 ) and hence limn→+∞ (un , p − q) exists. Therefore limn→+∞ (wn , p − q) exists. This implies that (q, p − q) = (p, p − q). Hence p = q and therefore wn * p ∈ H as n → +∞. Now suppose that [x, y] ∈ A. By the monotonicity of A, we have n

n

k=1 n

k=1 n

k=1 n

k=1 n

k=1 n

k=1 n

(y, x − wn ) = ( ∑ ck )−1 ∑ ck (y, x − uk ) = ( ∑ ck )−1 ∑ ck [(y − Auk , x − uk ) + (Auk , x − uk )] ≥ ( ∑ ck )−1 ∑ (uk−1 − uk − fk , x − uk ) 1 1 ≥ ( ∑ ck )−1 ∑ ( |uk − x|2 − |uk−1 − x|2 − | fk ||uk − x|) 2 2 k=1 k=1

First Order Difference Equations and Proximal Point Algorithm



135

n n −(∑nk=1 ck )−1 |u0 − x|2 − ( ∑ ck )−1 ∑ | fk ||uk − x|. 2 k=1 k=1

Letting n → +∞, it follows that (y, x − p) ≥ 0. Then, by the maximality of A, we have 0 ∈ A(p) as desired. For x ∈ H, we have kun − pk2 = kun − xk2 − kxk2 + kpk2 − 2(un , p − x). Multiplying both sides of this equality by cn , summing up from n = 1 to n = m, dividing by ∑m n=1 cn , and letting m → +∞, we obtain lim kun − pk2 = lim sup kun − xk2 − kx − pk2

n→+∞

n→+∞

< lim sup kun − xk2 , if x , p. n→+∞

Hence p is the asymptotic center of the sequence {un }, and the proof of the theorem is now complete.  Remark 7.4.3 The proof of Theorem 7.4.2 shows that for fn ≡ 0 and {un } given by (7.2), the sequence {wn } actually converges weakly as n → +∞ to some p ∈ A−1 (0), if and only if lim infn→+∞ kwn k < +∞ (without assuming the boundedness of {un }). Therefore in this case, {un } is bounded if and only if lim infn→+∞ kwn k < +∞. Problem 7.4.4 However, in the nonhomogeneous case where fn , 0, we do not know whether in Theorem 7.4.2, the boundedness of {un } can be replaced by lim infn→+∞ kwn k < +∞. Now we show the weak convergence of {un }. Remark 7.4.5 By Br´ezis and Lions (see [BRE-LIO], Remark 14, p. 344), the weak (resp. strong) convergence of {un } given by (7.2) in the homogeneous case where fn ≡ 0 implies the weak (resp. strong) convergence of {un } in the nonhomogeneous case fn , 0, if ∑+∞ n=1 k f n k < +∞. Therefore for the study of weak (resp. strong) convergence of {un }, we may assume without loss of generality that fn ≡ 0. 2 Theorem 7.4.6 Assume that {un }n≥1 is given by (7.2), ∑+∞ n=1 cn = +∞ and +∞ ∑n=1 k fn k < +∞. Then {un } converges weakly as n → +∞ to an element p ∈ A−1 (0), which is also the asymptotic center of {un }, if and only if lim infn→+∞ kwn k < +∞.

Proof. Necessity being obvious, let’s prove the sufficiency. By Remark 7.4.5, we assume without loss of generality that fn ≡ 0. From (7.2) we have un−1 − p ∈ un − p + cn Aun where p is the weak limit of {wn }, which exists by Remark 7.4.3, since the assumption on {cn }n≥1 implies that ∑+∞ n=1 cn = +∞. Squaring this inclusion and taking into account (7.2) and (7.13) we obtain

136

Nonlinear Evolution and Difference Equations of Monotone Type c2n kAun k2 ≤ kun−1 − pk2 − kun − pk2 .

Summing up this inequality from n = 1 to m and letting m → +∞, we get +∞

lim kun − pk2 < +∞ ∑ c2n kAun k2 ≤ ku0 − pk2 − n→+∞

n=1

since by Theorem 7.4.2, limn→+∞ kun − pk2 exists. By assumption on {cn } we deduce that lim infn→+∞ kAun k2 = 0. On the other hand, since A is monotone, we have (Aun−1 − Aun , un−1 − un ) ≥ 0. From (7.2) when fn ≡ 0, we get (Aun−1 − Aun , cn Aun ) ≥ 0. Hence kAun k2 ≤ kAun−1 k2 . Then limn→+∞ kAun k = 0. Now assume that un j * q as j → +∞. Then by the monotonicity of A, we have (Aun − Aun j , un − un j ) ≥ 0. Hence we get (Aun , un − q) ≥ 0. From (7.2) when fn ≡ 0, we deduce that kun − qk ≤ kun−1 − qk. Thus limn→+∞ kun − qk2 exists, and by a similar proof as in Theorem 7.4.2, we conclude that: un * p as n → +∞.  In our next two theorems, we show the strong convergence of {un }. +∞ 2 Theorem 7.4.7 Assume that (I + A)−1 is compact, ∑+∞ n=1 cn = +∞ and ∑n=1 k f n k < +∞. Then the sequence {un } given by (7.2) converges strongly as n → +∞ to an element p ∈ A−1 (0), which is also the asymptotic center of {un }, if and only if lim infn→+∞ kwn k < +∞.

Proof. Necessity being obvious, we prove the sufficiency. Again by Remark 7.4.3, we assume without loss of generality that fn ≡ 0. From the proof of Theorem 7.4.6 it follows that limn→+∞ kAun k = 0, and un * p as n → +∞. Therefore the sequence {un + Aun } is bounded, and hence the compacity of (I + A)−1 implies that {un } has a strongly convergent subsequence {un j } to p. Now, by the monotonicity of A, we have (Aun − Aun j , un − un j ) ≥ 0, ∀ n ≥ 1.

First Order Difference Equations and Proximal Point Algorithm

137

Letting j → +∞, we get (Aun , un − p) ≥ 0, ∀ n ≥ 1. Now the proof of Theorem 7.4.6 shows that limn→+∞ kun − pk2 exists, which implies that un → p as n → +∞, as desired.  +∞ Theorem 7.4.8 Assume that A is strongly monotone, ∑+∞ n=1 cn = +∞ and ∑n=1 k f n k < +∞. Then the sequence {un } given by (7.2) converges strongly as n → +∞ to an element p ∈ A−1 (0), which is also the asymptotic center of {un }, if and only if lim infn→+∞ kwn k < +∞.

Proof. Necessity being obvious, we prove the sufficiency. Again by Remark 7.4.5, we assume without loss of generality that fn ≡ 0. By the proof of Theorem 7.4.2 (see also Remark 7.4.3), we know that {wn } converges weakly as n → +∞ to p ∈ A−1 (0), and limn→+∞ kun − pk2 exists. Since A is strongly monotone and {un } is given by (7.2), we have αcn kun − pk2 ≤ (un−1 − un , un − p) 1 1 ≤ kun−1 − pk2 − kun − pk2 . 2 2 Summing up both sides of this inequality from n = 1 to m, and letting m → +∞, we get +∞

∑ cn kun − pk2 < +∞.

n=1

Since ∑+∞ n=1 cn = +∞, this implies that: lim inf kun − pk2 = 0, n→+∞

and hence un → p as n → +∞, as desired.

7.5



CONVERGENCE WITH NON-SUMMABLE ERRORS

Suppose that un is a sequence given by (7.3). By taking xn−1 = un−1 − fn and en = fn+1 in (7.3), we get xn = (I + cn A)−1 xn−1 + en Rockafellar proposed the following open problem: “does the sequence xn converge weakly to a zero of A if the error sequence {en } is non-summable?” In this section, we give a partial affirmative answer to this problem and study the convergence of the sequence {un } in (7.3), without summability assumptions on the error sequence { fn }. First we prove an elementary lemma. Lemma 7.5.1 Suppose that {an } and {bn } are positive sequences such that ∑m an n=1 an ∑+∞ n=1 bn = +∞ and limn→+∞ bn = 0, then limm→+∞ ∑m bn = 0. n=1

138

Nonlinear Evolution and Difference Equations of Monotone Type

Proof. Let ε > 0. By the assumption there is n0 > 0 such that for each n ≥ n0 , an < εbn . Now n

∑ 0 an + ε ∑m ∑m n=n0 bn n=1 an < n=1 m m ∑n=1 bn ∑n=1 bn n ∑ 0 an < n=1 +ε → ε ∑m n=1 bn as m → +∞. The result follows now since ε > 0 is arbitrary.



Theorem 7.5.2 Let {un } be a bounded sequence generated by (7.3). If kcfnn k → 0 as −1 n → +∞ and ∑+∞ n=1 cn = +∞, then A (0) , φ . Moreover, every weak cluster point −1 of wn belongs to A (0). Proof. Suppose that y ∈ A(x). By the monotonicity of A, we have k

(y, x − wk ) = ( ∑ cn )−1 n=1 k

= ( ∑ cn )−1 n=1 k

≥ ( ∑ cn )−1 n=1 k

≥ ( ∑ cn )−1 n=1



k

∑ cn (y, x − un )

n=1 k

∑ cn [(y − Aun , x − un ) + (Aun , x − un )]

n=1 k

∑ (un−1 − un − fn , x − un )

n=1 k

1

n=1

−(∑kn=1 cn )−1 2

1

∑ ( 2 kun − xk2 − 2 kun−1 − xk2 − k fn kkun − xk) k

ku0 − xk2 − ( ∑ cn )−1 n=1

k

∑ k fn kkun − xk.

n=1

Since {wk } is bounded, it has a subsequence {wk j } such that wk j * p ∈ H as j → +∞; substituting k by k j in the above inequality and letting j → +∞, by Lemma 7.5.1 and our assumptions, we get (y, x − p) ≥ 0. Then, by the maximality of A, we have p ∈ A−1 (0) as desired.



In the following two theorems, we show that the weak ω−limit set of the bounded sequence {un } generated by (7.3) is a subset of A−1 (0). Theorem 7.5.3 Let {un } be a bounded sequence generated by (7.3) and 2

∞ k fn k 2 ∑+∞ n=1 cn = +∞. If ∑n=1 c2n < +∞, and A−1 (0).

k fn k c2n

→ 0 as n → +∞, then ωw (un ) ⊂

First Order Difference Equations and Proximal Point Algorithm

139

Proof. By the monotonicity of A and (7.3), we have (Aun−1 − Aun , Aun +

fn ) ≥ 0. cn

This implies that kAun k2 ≤ (Aun−1 , Aun ) + (Aun−1 − Aun ,

fn ) cn

1 1 1 1 ≤ kAun−1 k2 + kAun k2 − kAun − Aun−1 k2 + kAun−1 − Aun k2 2 2 2 2 1 k fn k2 + . 2 c2n Then kAun k2 ≤ kAun−1 k2 +

k f n k2 . c2n

(7.14)

2

k fn k Since ∑∞ n=1 c2n < +∞, there exists limn→+∞ kAun k = l. Let L = supn≥1 kAun k. Theorem 7.5.2 shows that A−1 (0) , φ . Let p ∈ A−1 (0). By the monotonicity of A and (7.3), we get (un−1 − un − fn , un − p) ≥ 0.

This implies that kun − un−1 k2 ≤ kun−1 − pk2 − kun − pk2 + Mk fn k, where M := 2 supn≥1 kun − pk. Summing up from n = 1 to k, by (7.3), we get k

k

fn

∑ c2n kAun + cn k2 ≤ ku0 − pk2 − kuk − pk2 + M ∑ k fn k.

n=1

n=1

It follows that k

∑ c2n (kAun k2 +

n=1

k k fn k k fn k2 2 2 − 2kAu k ) ≤ ku − pk − ku − pk + M n 0 k ∑ k fn k. c2n cn n=1

Hence k

k

k

∑ c2n kAun k2 ≤ 2L ∑ cn k fn k + ku0 − pk2 − kxk − pk2 + M ∑ k fn k.

n=1

n=1

n=1

By (7.14), we get k

kAuk k2 ≤ kAun k2 +

k fi k2 . 2 i=n+1 ci



140

Nonlinear Evolution and Difference Equations of Monotone Type

Then k

kAuk k2

k

∑ c2n ≤

n=1

k

k k f i k2 + 2L ∑ c2 ∑ cn k fn k + ku0 − pk2 − kuk − pk2 i i=n+1 n=1

∑ c2n

n=1

k

+ M ∑ k fn k. n=1

From the assumptions and Lemma 7.5.1, it follows that Auk → 0 as k → +∞. If unk * q as k → +∞, then the demiclosedness of A implies that q ∈ A−1 (0).  Theorem 7.5.4 Let {un } be a bounded sequence generated by (7.3) and A = ∂ ϕ, where ϕ : H → (−∞, +∞] is a proper, convex and lower semicontinuous function. If ∞ ∑∞ n=1 cn = +∞, ∑n=1

k fn k2 cn

< ∞ and

k fn k cn

→ 0 as n → +∞, then ωw (un ) ⊂ A−1 (0).

Proof. By the subdifferential inequality and (7.3), we get ci (ϕ(ui ) − ϕ(ui−1 )) ≤ ci (∂ ϕ(ui ), ui − ui−1 ) = (ui−1 − ui − fi , ui − ui−1 ) 1 1 ≤ −kui − ui−1 k2 + kui − ui−1 k2 + k fi k2 2 2 1 ≤ k fi k2 . 2

(7.15)

Dividing both sides of (7.15) by ci and summing up from i = n + 1 to k, we obtain ϕ(uk ) ≤ ϕ(un ) +

1 k k f i k2 ∑ ci . 2 i=n+1

(7.16)

By Theorem 7.5.2, we know that A−1 (0) , φ . Let p ∈ A−1 (0). By the subdifferential inequality and (7.3), we have cn (ϕ(un ) − ϕ(p)) ≤ (un−1 − un − fn , un − p) 1 1 ≤ ( kun−1 − pk2 − kun − pk2 ) + Mk fn k, 2 2 where M := supn≥1 kun − pk. Summing up from n = 1 to k, we obtain k

1

1

k

∑ cn (ϕ(un ) − ϕ(p)) ≤ 2 ku0 − pk2 − 2 kuk − pk2 + M ∑ k fn k.

n=1

n=1

(7.17)

First Order Difference Equations and Proximal Point Algorithm

141

From (7.16), we get: k

(ϕ(uk ) − ϕ(p)) ∑ cn ≤ n=1

k 1 k k fi k2 1 1 c + ku0 − pk2 − kuk − pk2 n ∑ ∑ 2 n=1 i=n+1 ci 2 2 k

+ M ∑ k fn k. n=1

From the assumptions and Lemma 7.5.1, it follows that: ϕ(uk ) − ϕ(p) → 0 as k → +∞. If uk j * q as j → +∞, then ϕ(q) ≤ lim inf j→+∞ ϕ(uk j ) = ϕ(p). Hence q ∈ A−1 (0).  Remark 7.5.5 Theorems 7.5.3 and 7.5.4 show that if A−1 (0) is a singleton (which happens if, for example, A is strictly monotone in Theorem 7.5.3 or ϕ is strictly convex in Theorem 7.5.4), then un * p, where p is the unique element of A−1 (0). Remark 7.5.6 Although Theorems 7.5.3 and 7.5.4 do not imply the weak convergence of un to p ∈ A−1 (0) unless when A is strictly monotone or ϕ is strictly convex, they improve the errors in the proximal point algorithm. They show that the error sequence {k fn k} can go to infinity, provided that a suitable assumption on cn holds. Lemma 7.5.7 Assume that {yn } is a positive real sequence satisfying the following inequality: bn yn ≤ yn−1 − yn + an , (7.18) where {bn } and {an } are positive sequences; then the following hold: i) If { abnn } is bounded, then the sequence {yn } is bounded. ii) If limn→+∞ abnn = 0, then there exists limn→+∞ yn . iii) If limn→+∞ abnn = 0 and ∑+∞ n=1 bn = +∞, then yn → 0 as n → +∞. Proof. i) First we prove the boundedness of {yn }. There exists B such that for all n ≥ 1. If yn > B, then yn ≤ yn−1 + bn (

≤ B,

an an − yn ) < yn−1 + bn ( − B) ≤ yn−1 . bn bn

It follows that: yn ≤ max{y0 , B}. ii) For each ε > 0 there is m0 > 0 such that for each m ≥ m0 , argument for all k > m ≥ m0 , we have yk ≤ max{yk−1 ,

an bn

am bm

< ε. By the above

ak am+1 ak } ≤ · · · ≤ max{ym , , · · · , } ≤ max{ym , ε} ≤ ym + ε. bk bm+1 bk

Then, there exists limn→+∞ yn . iii) We divide both sides of (7.18) by bn and take liminf as n → +∞. It suffices to show that lim infn→+∞ b1n (yn−1 − yn ) ≤ 0. Suppose to the contrary that lim inf n→+∞

1 (yn−1 − yn ) > λ > 0. bn

142

Nonlinear Evolution and Difference Equations of Monotone Type

Then, there exists n0 such that for each n ≥ n0 , b1n (yn−1 − yn ) > λ . Multiplying both sides of this inequality by bn , summing up from n = n0 to m and letting m → ∞, we get a contradiction.  Theorem 7.5.8 Let {un } be the sequence generated by (7.3) and A be a maximal | fn | monotone and strongly monotone operator. If ∑∞ n=1 cn = +∞ and cn → 0 as n → +∞, then un → p as n → +∞, where p is the unique element of A−1 (0). Proof. Theorem 7.2.2 implies the boundedness of {un }. Now, by Theorem 7.5.2, A−1 (0) is nonempty. Suppose that p is the single element of A−1 (0). By the strong monotonicity of A and (7.3), we get (un−1 − un − fn , un − p) ≥ αcn kun − pk2 . It follows that 2αcn kun − pk2 ≤ kun−1 − pk2 − kun − pk2 + 2Mk fn k, where M := supn≥1 kun − pk. The theorem is now proved by using the assumptions and Lemma 7.5.7(iii). 

7.6

RATE OF CONVERGENCE

G¨uler [GUL] computed the rate of convergence of ϕ(un ) to ϕ(p) as n → +∞, where {un } is generated by (7.3) with A = ∂ ϕ and p is a minimum point of ϕ, provided that un → p as n → +∞. In this section, we give a simple proof to G¨uler’s result without assuming that un → p, which does not always occur as it has been shown by G¨uler [GUL]. First we prove the following elementary lemma. Lemma 7.6.1 Suppose that {an } and {bn } are two positive real sequences such that {an } is nonincreasing and convergent to zero, and ∑+∞ n=1 an bn < +∞; then (∑nk=1 bk )an → 0 as n → +∞. Proof. By the assumptions on {an } and {bn }, for each m > k, we have: m

k

m

am ( ∑ bn ) ≤ am ( ∑ bn ) + n=1

n=1



an bn .

n=k+1

Taking limsup as m → +∞, we get: m

lim sup am ( ∑ bn ) ≤ m→+∞

n=1

The lemma is now proved by letting k → +∞.

+∞



an bn .

n=k+1



Theorem 7.6.2 Suppose that {un } is a bounded sequence generated by (7.3) with fn ≡ 0 and A = ∂ ϕ, where ϕ is a proper, convex and lower semicontinuous function. n −1 If ∑+∞ n=1 cn = +∞, then ϕ(un ) − ϕ(p) = o((∑i=1 ci ) ), where p is a minimum point of ϕ.

First Order Difference Equations and Proximal Point Algorithm

143

Proof. Summing up (7.17) from n = 1 to k, we get: k

1

∑ cn (ϕ(un ) − ϕ(p)) ≤ 2 ku0 − pk2 .

n=1

By the proof of Theorem 7.5.4, ϕ(un ) − ϕ(p) is nonincreasing and convergent to 0. Now, the theorem follows by using Lemma 7.6.1. 

REFERENCES BOI-MOR. O.A. Boikanyo, G. Morosanu, Modified Rockafellar’s algorithm, Math. Sci. Res. J. 13 (2009), 101–122. BRE-LIO. H. Br´ezis and P. L. Lions, Produits infinis de resolvantes, ´ Israel J. Math. 29 (1978), 329–345. ROU. B. Djafari Rouhani, Ergodic theorems for nonexpansive sequences in Hilbert spaces and related problems, Ph.D. Thesis, Yale University, Part I (1981), pp. 1–76. ROU-KHA1. B. Djafari Rouhani and H. Khatibzadeh, On the proximal point algorithm, J. Optim. Theory Appl. 137 (2008), 411–417. ROU-KHA2. B. Djafari Rouhani and H. Khatibzadeh, Existence and asymptotic behaviour of solutions to first and second-order difference equations with periodic forcing. J. Difference Equ. Appl. 18 (2012), 1593–1606. GUL. O. G¨uler, On the convergence of the proximal point algorithm for convex minimization. SIAM J. Control Optim. 29 (1991), 403–419. KHA. H. Khatibzadeh, Some remarks on the proximal point algorithm. J. Optim. Theory Appl. 153 (2012), 769–778. LIO. P.L. Lions, Une m´ethode it´erative de r´esolution d’une in´equation variationnelle, Israel J. Math. 31 (1978), 204–208. MOR. G. Morosanu, ‘Nonlinear Evolution Equations and Applications’, Editura Academiei Romane (and D. Reidel publishing Company), Bucharest, 1988. MAR. B. Martinet, R´egularisation d’in´equations variationnelles par approximations successives. Rev. Franc¸aise Informat. Recherche Op´erationnelle 4 (1970), 154–158. ROC. R.T. Rockafellar, Monotone operators and the proximal point algorithm, SIAM J. Control Optimization 14 (1976), 877–898.

Order Difference 8 Second Equations 8.1

INTRODUCTION

Consider the following second order evolution equation of monotone type (

p(t)u00 (t) + r(t)u0 (t) ∈ Au(t) + f (t), u(0) = u0 , supt≥0 ku(t)k < +∞

(8.1)

that we studied in Chapter 5. Using the following difference quotients, u0 (t) =

u(t) − u(t − h) + o(h), h

and u00 (t) =

u(t + h) − 2u(t) + u(t − h) + o(h) h2

to approximate the first and second derivatives of u in (8.1), we get the following difference inclusion ( un+1 − (1 + θn )un + θn un−1 ∈ cn Aun + fn u0 = x, supn≥0 kun k < +∞,

(8.2)

where cn and θn are two positive real sequences, fn is a sequence in H and x ∈ H. In this chapter, we prove some results on the existence, periodicity and asymptotic behavior of solutions to (8.2), similar to the results of Chapter 5 for (8.1). Section 2 is devoted to the study of existence and uniqueness of solutions to (8.2). In Section 3, we consider the periodic forcing case. We prove the existence of a periodic solution if cn , θn and fn are periodic. Also, we show that each solution asymptotically converges to a periodic solution. In Section 4, the continuous dependence on initial data is studied. In Section 5, ergodic, weak and strong convergence of solutions to (8.2) for the homogeneous case ( fn ≡ 0) is investigated. In Section 6, we consider the subdifferential case, that is when the monotone operator A is the subdifferential of a proper, convex and lower semicontinuous function. Section 7 is devoted to the asymptotic behavior of solutions to (8.2) for the non-homogeneous case. Finally in Section 8, we present some applications of these results to convex optimization.

146

8.2

Nonlinear Evolution and Difference Equations of Monotone Type

EXISTENCE AND UNIQUENESS

In this section, we prove the existence and uniqueness of solutions to the following second order difference equations ( uNi+1 − (1 + θi )uNi + θi uNi−1 ∈ ci AuNi + fi , 1 ≤ i ≤ N (8.3) uN0 = a, uNN+1 = b, and

( ui+1 − (1 + θi )ui + θi ui−1 ∈ ci Aui + fi , u0 = a, supi≥1 kui k < +∞,

i≥1

(8.4)

where a, b ∈ H, θi , ci > 0, fi ∈ H for i ≥ 1 and A : D(A) ⊂ H → H is maximal monotone. Define the sequence ai as follows. a0 = 1,

ai =

1 , i≥1 θ1 · · · θi

(8.5)

Let HaNi := H × · · · × H endowed with the inner product N

h(ui )1≤i≤N , (vi )1≤i≤N i = ∑ ai (ui , vi )

(8.6)

i=1

for each (ui )1≤i≤N , (vi )1≤i≤N ∈ H N . It is obvious that HaNi is a Hilbert space and it is algebraically and topologically isomorphic with H N . Define the operator B by B((ui )1≤i≤N ) = (−ui+1 + (1 + θi )ui − θi ui−1 )1≤i≤N

(8.7)

D(B) := {(ui )1≤i≤N ∈ H N , u0 = a, uN+1 = b}

(8.8)

Proposition 8.2.1 The operator B is maximal monotone in HaNi . Proof. (8.7) can be rewritten as: 1 B((ui )1≤i≤N ) = −( (ϕi+1 − ϕi )1≤i≤N ) ai

(8.9)

where ϕi = ai−1 (ui − ui−1 ). If we take (ui )1≤i≤N , (vi )1≤i≤N ∈ D(B), ϕi = ai−1 (ui − ui−1 ), ψi = ai−1 (vi − vi−1 ), we have hB((ui )1≤i≤N ) − B((vi )1≤i≤N ), (ui − vi )1≤i≤N i N

= − ∑ (ϕi+1 − ϕi − ψi+1 + ψi , ui − vi ) i=1

N

N

= ∑ ai kui+1 − ui − vi+1 + vi k2 − ∑ (ϕi − ψi , ui+1 − ui − vi+1 + vi ) i=1

i=1

Second Order Difference Equations

147

N

− ∑ (ϕi+1 − ϕi − ψi+1 + ψi , ui+1 − vi+1 ) i=1 N

= ∑ ai kui+1 − ui − vi+1 + vi k2 i=1 N

+ ∑ [(ϕi − ψi , ui − vi ) − (ϕi+1 − ψi+1 , ui+1 − vi+1 )] i=1 N

= ∑ ai kui+1 − ui − vi+1 + vi k2 + ku1 − v1 k2 ≥ 0 i=1

This proves the monotonicity of B. To prove the maximal monotonicity of B, it is sufficient to show that for each (hi )1≤i≤N ∈ H N , there exists a sequence (ui )1≤i≤N ∈ H N that satisfies ( ui+1 − (2 + θi )ui + θi ui−1 = hi , 1 ≤ i ≤ N (8.10) u0 = a, uN+1 = b We are going to look for solutions to (8.10) of the form ui = vi + αi x + βi y,

1≤i≤N

(8.11)

where x, y ∈ H and vi , αi and βi are respective solutions to the following problems: ( vi+1 − (2 + θi ) + θi vi−1 = hi , 1 ≤ i ≤ N (8.12) v0 = 0, v1 = 0

and

( αi+1 − (2 + θi )αi + θi−1 αi−1 = 0, 1 ≤ i ≤ N α0 = 0, α1 = c > 0,

(8.13)

( βi+1 − (2 + θi )βi + θi−1 βi−1 = 0, 1 ≤ i ≤ N βN+1 = 0, βN = −c

(8.14)

We note that ui given by (8.11) satisfies Equation (8.10) for each x, y ∈ H. In order for (b−vN+1 ) the boundary conditions u0 = a and uN+1 = b to be satisfied, we choose x = αN+1 and y = βa ; then the problem (8.10) has a solution and consequently B is maximal 0 monotone in HaNi .  Now we prove the existence result for (8.3). Theorem 8.2.2 Assume that A : D(A) ⊂ H → H is a maximal monotone operator in H such that 0 ∈ D(A), and ci > 0, θi > 0 and fi ∈ H for 1 ≤ i ≤ N are given sequences. Then for each a, b ∈ H, the finite scheme (8.3) has a unique solution (ui )1≤i≤N ∈ D(A)N .

148

Nonlinear Evolution and Difference Equations of Monotone Type

Proof. The operator A ((ui )1≤i≤N ) = (c1 v1 , · · · , cN vN ) with vi ∈ Aui for 1 ≤ i ≤ N is maximal monotone in H N . Let Aλ be the Yosida approximation of A and Aλ be the Yosida approximation of A . Then Aλ ((ui )1≤i≤N ) = (c1 Aλ u1 , · · · , cN Aλ uN ) Since B is maximal monotone in HaNi (see Proposition 8.2.1), and Aλ is maximal monotone and everywhere defined, then B + Aλ is maximal monotone in HaNi (see Theorem 3.3.5); hence R(ωI + B + Aλ ) = HaNi for each λ , ω > 0 (see Theorem 3.3.2). Then for a given sequence ( fi )1≤i≤N ∈ HaNi , there exists a sequence (uλi ω )1≤i≤N ∈ H N , such that: ( ω − (1 + θ )uλ ω + θ uλ ω = c A uλ ω + ωuλ ω + f , uλi+1 1≤i≤N i i i i−1 i λ i i i (8.15) λ ω λ ω u0 = a, uN+1 = b We show that uλi ω is bounded respect to λ and ω, then we take the limit in (8.15) as λ → 0 and ω → 0. Multiplying (8.15) by ai uλi ω and summing up from i = 1 to i = N, we get: N

N

∑ ai (uλi+1ω − uλi ω , uλi ω ) − ∑ ai θi (uλi ω − uλi−1ω , uλi ω )

i=1

i=1

N

N

N

= ∑ ci ai (Aλ uλi ω , uλi ω ) + ω ∑ ai kuλi ω k2 + ∑ ai ( fi , uλi ω ). i=1

i=1

i=1

Without loss of generality, we may assume that 0 ∈ A0. Otherwise, we consider ˜ i = Aui − Ao 0 instead of A (where Ao x is the least minimum norm element of Ax Au which was defined in Chapter 3), and f˜i = fi + ci Ao 0 instead of fi . Since ai θi = ai−1 , from the monotonicity of Aλ we conclude that: N

N

ω λω λω λω λω ω ∑ ai kuλi ω k2 ≤ ∑ [ai (uλi+1 − uωλ i , ui ) − ai−1 (ui − ui−1 , ui−1 )] i=1

i=1

N

N

ω 2 − ∑ ai−1 kuλi ω − uλi−1 k − ∑ ai ( fi , uλi ω ) i=1

Therefore

i=1

N

N

ω 2 ω ∑ ai kuλi ω k2 + aN kuλNω k2 + ∑ ai−1 kuλi ω − uλi−1 k i=1

i=1

N

1

N

1

≤ aN kbkkuλNω k + kakkuλ1 ω k + kak2 + ( ∑ ai k fi k2 ) 2 ( ∑ ai kuλi ω k2 ) 2 i=1

(8.16)

i=1

One can see that N 1 1 kuλk ω k ≤ √ ( ∑ ai kuλi ω k2 ) 2 , 1 ≤ k ≤ N ak i=1

(8.17)

Second Order Difference Equations

149

By using (8.16) and (8.17) one can write N

N

N

1

ω 2 ω ∑ ai kuλi ω k2 + ∑ ai−1 kuλi ω − uλi−1 k ≤ K1 ( ∑ ai kuλi ω k2 ) 2 + K2 i=1

i=1

(8.18)

i=1

where K1 , K2 > 0 are constants independent of λ and ω. On the other hand, k

k

ω ω kuλk ω k = ∑ (kuλi ω k − kuλi−1 k) + kak ≤ ∑ kuλi ω − uλi−1 k + kak i=1

i=1

Therefore k

k ak ω 2 )( ∑ ai−1 kuλi ω − uλi−1 k ) + 2kak2 ak a i=1 i−1 i=1

ak kuλk ω k2 ≤ 2( ∑

Summing up from k = 1 to k = N, we get: N

N

∑ ak kuλk ω k2 ≤ K3 ( ∑ ai−1 kuλi ω − uλi−1ω k2 ) + K4

(8.19)

i=1

k=1

Combining (8.19) with (8.18) we get N

N

1

∑ ai−1 kuλi ω − uλi−1ω k2 ≤ K5 ( ∑ ai−1 kuλi ω − uλi−1ω k2 ) 2 + K6

i=1

(8.20)

i=1

which implies that N

∑ ai−1 kuλi ω − uλi−1ω k2 ≤ K7

(8.21)

i=1

where K5, K6 and K7 are constants independent of λ and ω. By (8.19) and (8.17) we conclude that: N

∑ ai kuλi ω k2 ≤ K8 ,

kuλi ω k ≤ K9 , 1 ≤ i ≤ N

(8.22)

i=1

and by (8.15) we get: kAλ uλi ω k ≤ K10 , 1 ≤ i ≤ N

(8.23)

HaNi

We prove the strong convergence of in as λ → 0. To this aim, we subtract µω (8.15) with µ from (8.15) with λ , then we multiply the difference by ai (uλi ω − ui ), and sum up from i = 1 to i = N. We conclude that: uλi ω

N µω

µω

∑ ai (uλi+1ω − ui+1 − uλi ω + ui

µω

, uλi ω − ui )

i=1 N

µω

− ∑ ai θi (uλi ω − ui i=1

µω

µω

ω − uλi−1 + ui−1 , uλi ω − ui )

150

Nonlinear Evolution and Difference Equations of Monotone Type N

N

= ∑ ai ci (Aλ uλi ω − Aµ ui , uλi ω − ui ) + ω ∑ ai kuλi ω − ui k2 µω

µω

µω

i=1

(8.24)

i=1

Suppose that M1 and M2 are the left and right hand side of (8.24). By using the initial conditions in (8.15) and the well known relation uλi ω = Jλ uλi ω + λ Aλ uλi ω , 1 ≤ i ≤ N

(8.25)

we conclude that: N

M1 = −aN kuλNω − uN k2 − ∑ ai−1 kuλi ω − ui µω

µω

ω − uλi−1 + ui−1 k2 µω

(8.26)

i=1

and N µω

µω

M2 = ∑ ai ci (Aλ uλi ω − Aµ ui , Jλ uλi ω − Jµ ui ) i=1 N

µω

µω

+ ∑ ai ci (Aλ uλi ω − Aµ ui , λ Aλ uλi ω − µAµ ui ) i=1

N

+ ω ∑ ai kuλi ω − ui k2 . µω

i=1

Since A is monotone and Aλ uλi ω ∈ A(Jλ uλi ω ) we have: N

M2 ≥ ∑ ai ci (λ kAλ uλi ω k2 + µkAµ ui k2 ) µω

i=1

N

N

−(λ + µ) ∑ ai ci (Aλ uλi ω , Aµ ui ) + ω ∑ ai kuλi ω − ui k2 µω

i=1

µω

(8.27)

i=1

By using (8.26) and (8.27) in (8.24) and with the aid of (8.23) we obtain: N

N

ω ∑ ai kuλi ω − ui k2 + ∑ ai−1 kuλi ω − ui µω

i=1

µω

ω − uλi−1 + ui−1 k2 ≤ K11 (λ + µ) (8.28) µω

i=1

ω ) are strongly convergent sequences as Then we conclude that (uλi ω ) and (uλi ω − uλi−1 λ ω ω λ → 0. Let ui → ui as λ → 0. Passing to subsequences if necessary, suppose that λ ω → uω as λ → 0 in Aλ uλi ω * wω i as λ → 0 in H. By (8.25), we conclude that Jλ ui i H. Since A is maximal monotone in H (and therefore strong-weak closed), by taking the limit in the inclusion Aλ uλi ω ∈ A(Jλ uλi ω ), we get:

uω i ∈ D(A),

ω wω i ∈ Aui

Therefore taking the limit in (8.15) as λ → 0, we conclude that: ( ω ω ω ω uω i+1 − (1 + θi )ui + θi ui−1 ∈ ci Aui + ωui + f i , 1 ≤ i ≤ N ω ω u0 = a, uN+1 = b

(8.29)

(8.30)

Second Order Difference Equations

151

λω Since kuω i k = limλ →0 kui k, (8.22) implies that

kuω i k ≤ K9

(8.31)

ω The next step of the proof consists of showing the strong convergence of (uω i − ui−1 ) as ω → 0. Considering the Equation (8.30) for δ , ω > 0, and subtracting them, then δ multiplying both sides by ai (uω i − ui ) and summing up from i = 1 to i = N, we get the following equality: N

∑ ai (uωi+1 − uδi+1 − uωi + uδi , uωi − uδi )

i=1 N

δ ω δ ω δ − ∑ ai θi (uω i − ui − ui−1 + ui−1 , ui − ui ) i=1

N

N

δ ω δ ω δ ω δ = ∑ ai ci (vω i − vi , ui − ui ) + ∑ ai (ωui − δ ui , ui − ui ) i=1

(8.32)

i=1

ω δ δ where vω i ∈ Aui and vi ∈ Aui . Since the left hand side can be written in the form N δ 2 ω δ ω δ 2 M1 = −aN kuω N − uN k − ∑ ai−1 kui − ui − ui−1 + ui−1 k i=1

then (8.32) implies that N

N

∑ ai−1 kuωi − uδi − uωi−1 + uδi−1 k2 ≤ − ∑ ai ci (vωi − vδi , uωi − uδi )

i=1

i=1

N

N

2 δ 2 ω δ − ∑ ai [ωkuω i k + δ kui k ] + (ω + δ ) ∑ ai (ui , ui ) i=1

i=1

By the monotonicity of A and the boundedness of find that

uω i

which was shown in (8.31), we

N

∑ ai−1 kuωi − uδi − uωi−1 + uδi−1 k2 ≤ K12 (ω + δ )

(8.33)

i=1 ω ) (uω − u i i−1 ω

Therefore is a Cauchy net and hence strongly convergent in H as ω → 0. On the other hand, passing to a subsequence if necessary, we may assume that uω i is weakly convergent. Suppose that uω i * ui as ω → 0. By rewriting the Equation (8.30) in the form ω ω ω ω ω uω i+1 − ui − θi (ui − ui−1 ) ∈ ci Aui + ωui + f i , 1 ≤ i ≤ N

(8.34)

and taking the limit as ω → 0, we conclude that ui ∈ D(A) and ui is a solution to (8.3). Now we prove the uniqueness. If (ui )1≤i≤N and (vi )1≤i≤N are solutions to (8.3) and wi = ui − vi , then the monotonicity of A implies that N

N

∑ ai (wi+1 − wi , wi ) − ∑ ai−1 (wi − wi−1 , wi ) ≥ 0

i=1

i=1

(8.35)

152

Nonlinear Evolution and Difference Equations of Monotone Type

Since wN+1 = w0 = 0, we get N

∑ ai−1 kwi − wi−1 k2 + aN kwN k2 ≤ 0

(8.36)

i=1



which shows the uniqueness of the solution.

Next we are going to investigate the existence of solutions to (8.4). In each of the two cases θn ≥ 1 and 0 < θn < 1, we prove that if (8.4) has a solution for an initial value, then it has a solution for every other initial value. Then we prove the existence of solutions when A−1 (0) , ∅. First we prove a lemma. Lemma 8.2.3 If the following assumption on θn ∞

k

1

∑ hk = +∞,

1 θ ...θ k i=1 i

where hk = ∑

k=1

(8.37)

is satisfied, then every positive and bounded sequence an satisfying an ≤

1 θn an+1 + an−1 1 + θn 1 + θn

(8.38)

is nonincreasing. Proof. (8.38) implies that an is nonincreasing or eventually increasing. Suppose to the contrary that an is eventually increasing. Then there exists m > 0 such that for all i > m, ai > ai−1 . Form (8.38) we obtain θi ≤

ai+1 − ai , ai − ai−1

then θi · · · θk ≤

ak+1 − ak ai − ai−1

∀i > m

∀i > m

and m

k k k 1 1 ai − ai−1 ak − am 1 + ∑ ≥ ∑ ≥ ∑ = . θ · · · θ θ · · · θ θ · · · θ a − a a k k k k k+1 − ak i=m+1 i i=m+1 i i=m+1 k+1 i=1 i

hk = ∑ Then: ∞

1

m

∞ m ak+1 − ak ak+1 − ak ak+1 − ak 1 + ∑ ≤∑ + λ k=1 ak − am k=m+1 ak − am k=1 ak − am

∑ hk ≤ ∑

k=1





(ak+1 − ak )

k=m+1

m

=

ak+1 − ak 1 + (l − am+1 ) < +∞ λ k=1 ak − am



where 0 < λ = am+1 − am ≤ ak − am for all k > m, and l = limk→+∞ ak . This is a contradiction. 

Second Order Difference Equations

153

Theorem 8.2.4 Suppose that A : D(A) ⊂ H → H is a maximal monotone operator such that 0 ∈ D(A), and fi ∈ H, ci > 0 and θi ∈ (0, 1) for all i ≥ 1 such that (8.37) is satisfied. If the problem (8.4) has a solution for some u0 = b ∈ H, then for each initial value u0 = a ∈ H, it has a unique solution. If (ui )i≥1 and (vi )i≥1 are two solutions to (8.4) with u0 = a and v0 = b, then kui − vi k is nonincreasing and kui − vi k ≤ ka − bk. Proof. Let b ∈ H be such that the problem ( wi+1 − (1 + θi )wi + θi wi−1 ∈ ci Awi + fi , i ≥ 1 w0 = b, supi≥1 kwi k < +∞

(8.39)

has a solution (wi )i≥1 , with wi ∈ D(A), i ≥ 1. If a ∈ H is an arbitrary element, then we show that problem (8.4) with u0 = a has a solution (ui )i≥1 with ui ∈ D(A), i ≥ 1. By Theorem 8.2.2 the problem ( uNi+1 − (1 + θi )uNi + θi uNi−1 ∈ ci AuNi + fi , 1 ≤ i ≤ N (8.40) uN0 = uNN+1 = a has a unique solution (uNi )1≤i≤N ∈ D(A)N . We show that the limit ui = limN→+∞ uNi exists uniformly with respect to i belonging to a finite set of natural numbers, and the limit is a solution of (8.4). Let yi = uNi − wi for 1 ≤ i ≤ N. Since A is monotone, by substituting wi in (8.4) and subtracting from (8.40), then multiplying the difference by yi , we conclude that: (yi+1 − (1 + θi )yi + θi yi−1 , yi ) ≥ 0

(8.41)

so

1 θi kyi+1 k + kyi−1 k (8.42) 1 + θi 1 + θi Since the right hand side of the above inequality is a convex combination, we get: kyi k ≤

kuNi k ≤ max{ky0 k, kyN+1 k} + kwi k ≤ kak + K

(8.43)

K = sup{kwi k, i ≥ 0}

(8.44)

where Inequality (8.41) also implies that (yi+1 − yi , yi ) ≥ θi (yi − yi−1 , yi ) therefore θi kyi − yi−1 k2 ≤ (yi+1 − yi , yi ) − θi (yi − yi−1 , yi−1 ), 1 ≤ i ≤ N. Multiplying the above inequalities for 1 ≤ i ≤ N − 1 respectively by θN · · · θi+1 , and then adding them up from i = 1 to i = N, we get: N

∑ θN · · · θi+1 θi kyi − yi−1 k2 ≤ (yN+1 − yN , yN ) − θN · · · θ1 (y1 − y0 , y0 )

i=1

(8.45)

154

Nonlinear Evolution and Difference Equations of Monotone Type

For each fixed N0 , (8.44) and (8.45) imply the existence of M > 0, independent of N, such that N0

∑ θN · · · θi+1 θi kuNi − uNi−1 k2 ≤ M

(8.46)

i=1

We show that for each N0 , the sequence (uNi )1≤i≤N0 is a Cauchy sequence with respect to N. Suppose that N0 < N1 < N2 are natural numbers and zi = uNi 1 − uNi 2 , 0 ≤ i ≤ N1 + 1. By (8.40) (zi+1 − zi , zi ) ≥ θi (zi − zi−1 , zi ), 1 ≤ i ≤ N1 which implies the inequality θi kzi − zi−1 k2 ≤ (zi+1 − zi , zi ) − θi (zi − zi−1 , zi−1 ), 1 ≤ i ≤ N1

(8.47)

Multiplying (8.47) by θk θk−1 · · · θi+1 (where k ∈ {1, · · · , N1 }, 1 ≤ i ≤ k − 1) and then summing up from i = 1 to i = k, we get: k

∑ θk · · · θi+1 θi kzi − zi−1 k2 ≤ (zk+1 − zk , zk ),

1 ≤ k ≤ N1

(8.48)

1 ≤ k ≤ N1

(8.49)

i=1

because z0 = 0. Therefore k

1

∑ θk · · · θi+1 θi kzi − zi−1 k2 ≤ 2 (kzk+1 k2 − kzk k2 ),

i=1

This shows that the sequence (kzk k)k is nondecreasing. Moreover k

k

kzk k = ∑ [kzi k − kzi−1 k] ≤ ∑ kzi − zi−1 k i=1

i=1

Therefore k

k 1 )( ∑ θk · · · θi+1 θi kzi − zi−1 k2 ), 1 ≤ k ≤ N1 i=1 θk · · · θi+1 θi i=1

kzk k2 ≤ ( ∑

(8.50)

Using (8.49) we get: 1 kzk k2 ≤ hk (kzk+1 k2 − kzk k2 ), 1 ≤ k ≤ N1 2 Summing up from k = N0 to k = N1 , and using (8.43), we get: N1

∑ k=N0

1 1 kzk k2 ≤ kzN1 +1 k2 ≤ C hk 2

(8.51)

Second Order Difference Equations

155

where C is a positive constant. Since (kzk k) is nondecreasing, for i ∈ {1, · · · , N0 } we can write N1 N1 1 1 kzi k2 ( ∑ )≤ ∑ kzk k2 ≤ C (8.52) h h k k k=N k=N 0

0

Therefore kuNi 1 − uNi 2 k2 ≤

C 1 (∑Nk=N 0

1 hk )

, 1 ≤ i ≤ N0

(8.53)

N1 1 Since by assumption ∑∞ k=1 hk = +∞, limN1 →+∞ ui = ui exists uniformly with respect to i belonging to every finite subset of natural numbers. The operator A is maximal monotone in H and therefore demiclosed. By letting N → +∞ in (8.40), we conclude that ui satisfies (8.4). To prove the uniqueness, suppose that (ui )i≥1 and (vi )i≥1 are two solutions of (8.4) with u0 = a and v0 = b, and let qi = ui − vi . Subtracting the corresponding equations for ui and vi , and multiplying by qi , by the monotonicity of A we get: 1 θi kqi k ≤ kqi+1 k + kqi−1 k, i ≥ 1 (8.54) 1 + θi 1 + θi

By Lemma 8.2.3, kqi k is nonincreasing and therefore kui − vi k is nonincreasing. Therefore kui − vi k ≤ ka − bk. Choosing a = b, we get the uniqueness and the proof is complete.  1 k Theorem 8.2.5 Suppose that 0 < θn < 1 and ∑∞ k=1 h = +∞ where hk = ∑i=1 θ

If { fn } is a sequence in H such that tion (8.4) has a unique solution.

k fn k ∑+∞ n=1 n θ1 ···θn

k

< +∞ and

A−1 (0) , φ ,

1 k ···θi

.

then Equa-

Proof. Suppose that p ∈ A−1 (0) or 0 ∈ A(p). By Theorem 8.2.4, it is sufficient to show that the problem (8.4) has a solution for u0 = p ∈ A−1 (0). Then it will have a unique solution for all x ∈ H. Therefore it suffices to prove the existence of a solution to the following problem: ( un+1 − (1 + θn )un + θn un−1 ∈ cn Aun + fn (8.55) u0 = p ∈ A−1 (0), supn≥1 kun k < +∞ We define B(x) = A(x + p), and vn = un − p. Then (8.55) is equivalent to the following problem: ( vn+1 − (1 + θn )vn + θn vn−1 ∈ cn Bvn + fn (8.56) v0 = 0, supn≥1 kvn k < +∞ By Theorem 8.2.2, the following equation has a solution: ( vNn+1 − (1 + θn )vNn + θn vNn−1 ∈ cn BvNn + fn , 1≤n≤N N N v0 = vN+1 = 0

(8.57)

156

Nonlinear Evolution and Difference Equations of Monotone Type

Multiplying both sides of this equation by vNn , and using the fact that B is monotone and 0 ∈ B(0), we get: (vNn+1 , vNn ) − (1 + θn )kvNn k2 + θn (vNn−1 , vNn ) ≥ ( fn , vNn ) ⇒ (1 + θn )kvNn k ≤ kvNn+1 k + θn kvNn−1 k + k fn k ⇒ an (kvNn k − kvNn+1 k) − an−1 (kvNn−1 k − kvNn k) ≤ an−1 where an =

1 θ1 ···θn .

k fn k θn

Summing up from n = r to n = N, we get: N

kvNr k − kvNr−1 k ≤

k fn k

∑ θ1 · · · θn

n=r

Summing up both sides of the above inequality from r = 1 to r = k, we get r=k N

kvNk k − kvN0 k ≤

k fn k

∑ ∑ θ1 · · · θn

r=1 n=r r=N N

⇒ kvNk k ≤

k fn k



k fn k

∑ ∑ θ1 ...θn < ∑ n θ1 · · · θn < +∞.

r=1 n=r

n=1

Now let N0 < N1 < N2 , and set zn = vNn 1 − vNn 2 , 0 ≤ n ≤ N1 + 1. On the other hand, by (8.57), we get: (zn+1 − (1 + θn )zn + θn zn−1 , zn ) ≥ 0 ⇒ (zn+1 − zn , zn ) − θn (zn − zn−1 , zn−1 ) ≥ θn kzn − zn−1 k2 Multiplying the above inequality by θn+1 · · · θk where k ∈ {1, · · · , N} and then summing up from n = 1 to n = k, we get: k

1

∑ θn · · · θk kzn − zn−1 k2 ≤ (zk+1 − zk , zk ) ≤ 2 (kzk+1 k2 − kzk k2 )

n=1

In addition k

k

kzk k = ∑ (kzi k − kzi−1 k) ≤ ∑ kzi − zi−1 k i=1

i=1

hence k 1 )( ∑ θi · · · θk kzi − zi−1 k2 ) i=1 θi · · · θk i=1 k

kzk k2 ≤ ( ∑ By (8.58)

1 kzk k2 ≤ hk (kzk+1 k2 − kzk k2 ) 2

(8.58)

Second Order Difference Equations

157

Summing up from k = N0 to k = N1 , we get N1

1 1 kzk k2 ≤ kzN1 +1 k2 ≤ C hk 2

∑ k=N0

where C is a positive constant. Since kzk k is nondecreasing, for 1 ≤ i ≤ N0 we have: kzi k2 (

N1



k=N0

N1 1 1 )≤ ∑ kzk k2 ≤ C hk h k=N k

then kuNi 1 − uNi 2 k2 ≤

0

C 1 1 (∑Nk=N ) 0 hk

,

1 ≤ i ≤ N0

Hence, vi = limn→+∞ vni exists. The result now follows from the fact that A is demiclosed.  Theorem 8.2.6 Let the sequences ci > 0, θi ≥ 1 and fi ∈ H be given. Suppose that A : D(A) ⊂ H → H is a maximal monotone operator with 0 ∈ D(A). If (8.4) has a solution for an initial value u0 = b ∈ H, then it has a unique solution for every initial value u0 = a ∈ H. If (ui )i≥1 and (vi )i≥1 are two solutions with u0 = a and v0 = b, then the sequence (kui − vi k) is nonincreasing and kui − vi k ≤ ka − bk. Proof. Suppose that b ∈ H is given such that the difference inclusion ( wi+1 − (1 + θi )wi + θi wi−1 ∈ ci Awi + fi , i ≥ 1 w0 = b, supi≥1 kwi k = K < +∞

(8.59)

admits a solution (wi )i≥1 . Let a ∈ H be arbitrarily chosen. We prove that (8.4) has a unique solution with initial condition u0 = a. By Theorem 8.2.2, there exists a unique solution (uNi )1≤i≤N ∈ D(A)N for the auxiliary problem ( uNi+1 − (1 + θi )uNi + θi uNi−1 ∈ ci AuNi + fi , 1 ≤ i ≤ N (8.60) uN0 = uNN+1 = a We show that limN→+∞ uNi = ui exists uniformly on every finite set of natural numbers, and that (ui )i≥1 is a solution to (8.4). Similar to the proof of Theorem 8.2.4 we find that kuNi k ≤ kak + 2K, ∀N ≥ 1, 1 ≤ i ≤ N. (8.61) If N0 < N1 < N2 are natural numbers and zi = uNi 1 − uNi 2 , 0 ≤ i ≤ N1 + 1, then subtracting the corresponding equations for uNi 1 and uNi 2 , and multiplying the difference by zi , we get: (zi+1 − (1 + θi )zi + θi zi−1 , zi ) ≥ 0, 1 ≤ i ≤ N1 (8.62) which implies that (zi+1 − zi , zi ) − θi (zi − zi−1 , zi−1 ) ≥ θi (kzi k − kzi−1 k)2 , 1 ≤ i ≤ N1

158

Nonlinear Evolution and Difference Equations of Monotone Type

Multiplying by θi+1 · · · θk , where 1 ≤ i ≤ k − 1 and 1 ≤ k ≤ N1 and summing up from i = 1 to i = k, we get: k

1

∑ θi θi+1 · · · θk (kzi k − kzi−1 k)2 ≤ 2 (kzk+1 k2 − kzk k2 ),

1 ≤ k ≤ N1

(8.63)

i=1

Now we sum from k = N0 to k = N1 . Taking into account (8.61) and θi ≥ 1, ∀i ≥ 1, this implies that N0

(N1 − N0 + 1) ∑ (kzi k − kzi−1 k)2 ≤ C

(8.64)

i=1

On the other hand, we see that N0

(N1 − N0 + 1) ∑ (kzi k − kzi−1 k)2 ≥ i=1

=

N1 − N0 + 1 N0 ( ∑ (kzi k − kzi−1 k))2 N0 i=1 N1 − N0 + 1 kzN0 k2 N0

(8.65)

By (8.64) and (8.65) we conclude that kzN0 k2 ≤

CN0 N1 − N0 + 1

(8.66)

The inequality (8.63) also implies that (kzk k) is a nondecreasing sequence, and so for 1 ≤ i ≤ N0 we have CN0 kuNi 1 − uNi 2 k ≤ N1 − N0 + 1 Therefore limN→+∞ uNi = ui , exists uniformly with respect to i belonging to every finite set of natural numbers. By taking the limit as N → +∞ in (8.60), we conclude that (ui )i≥1 is a solution to (8.4). The proof of the last part of the theorem is similar to the proof of Theorem 8.2.4.  Theorem 8.2.7 Suppose that θn ≥ 1 and { fn } is a sequence in H such that k fn k −1 ∑+∞ n=1 n θn < +∞ and A (0) , φ , then Equation (8.4) has a unique solution. Proof. Suppose that p ∈ A−1 (0) or 0 ∈ A(p). By Theorem 8.2.6, it suffices to show that the problem (8.4) has a solution for u0 = p ∈ A−1 (0). Then it will have a unique solution for every initial value x ∈ H. Therefore it is sufficient to prove the existence of a solution for the following difference inclusion: ( un+1 − (1 + θn )un + θn un−1 ∈ cn Aun + fn (8.67) u0 = p ∈ A−1 (0), supn≥1 kun k < +∞ We define B(x) = A(x + p), and vn = un − p. Then Equation (8.67) is equivalent to the following equation: ( vn+1 − (1 + θn )vn + θn vn−1 ∈ cn Bvn + fn (8.68) v0 = 0, supn≥1 kvn k < +∞

Second Order Difference Equations

159

By Theorem 8.2.2, the following equation has a solution: ( vNn+1 − (1 + θn )vNn + θn vNn−1 ∈ cn BvNn + fn , 1≤n≤N vN0 = vNN+1 = 0

(8.69)

We multiply both sides of the above equation by vNn . Since B is monotone and 0 ∈ B(0), we get: (vNn+1 , vNn ) − (1 + θn )kvNn k2 + θn (vNn−1 , vNn ) ≥ ( fn , vNn ) ⇒ (1 + θn )kvNn k ≤ kvNn+1 k + θn kvNn−1 k + k fn k ⇒ an (kvNn k − kvNn+1 k) − an−1 (kvNn−1 k − kvNn k) ≤ an−1 where an =

1 θ1 ···θn .

k fn k θn

Summing up from n = r to n = N, we get: N

ar−1 (kvNr k − kvNr−1 k) ≤

∑ an−1

n=r

N k fn k k fn k ≤ ar−1 ∑ θn n=r θn

Summing up both sides of the above inequality from r = 1 to r = k, we get: r=k N

kvNk k − kvN0 k ≤

k fn k r=1 n=r θn

∑∑

r=N N

⇒ kvNk k ≤

N ∞ k fn k k fn k k fn k = n < ∑ ∑ θn ∑ θn ∑ n θn < +∞. r=1 n=r n=1 n=1

Now let N0 < N1 < N2 , and set zn = vNn 1 − vNn 2 , 0 ≤ n ≤ N1 + 1. On the other hand by (8.69) we get: (zn+1 − (1 + θn )zn + θn zn−1 , zn ) ≥ 0 ⇒ an (zn+1 − zn , zn ) − an−1 (zn − zn−1 , zn−1 ) ≥ an−1 kzn − zn−1 k2 Summing up from n = 1 to n = k, we get: k

∑ an−1 kzn − zn−1 k2 ≤ ak (zk+1 − zk , zk ) ≤

n=1

ak (kzk+1 k2 − kzk k2 ) 2

Since the sequence {ak } is nonincreasing, we deduce that: k

1

∑ kzn − zn−1 k2 ≤ 2 (kzk+1 k2 − kzk k2 )

n=1

Summing up (8.70) from k = N0 to N1 , we get: N0 ∞ 1 k fn k 2 (N1 − N0 + 1) ∑ kzn − zn−1 k2 ≤ kzN1 +1 k2 ≤ 2( ∑ n ) . 2 θn n=1 n=1

(8.70)

160

Nonlinear Evolution and Difference Equations of Monotone Type

On the other hand N1 − N0 + 1 N1 − N0 + 1 N0 kzN0 k2 = [ ∑ (kzi k − kzi−1 k)]2 N0 N0 i=1 ≤

N1 − N0 + 1 N0 N0 ∑ kzi − zi−1 k2 N0 i=1 N0

= (N1 − N0 + 1) ∑ kzi − zi−1 k2 i=1



k fn k 2 ) . ≤ 2( ∑ n θn n=1 Therefore kzN0 k2 ≤

∞ 2N0 k fn k 2 (∑ n ) . N1 − N0 + 1 n=1 θn

Since by (8.70), kzn k is nondecreasing, this implies that vi = limn→+∞ vni exists uniformly with respect to i belonging to every finite set of natural numbers. Then the result follows because A is demiclosed. Now we prove the uniqueness. If un and vn are two solutions to (8.4) with u0 = v0 = x, then by the monotonicity of A and Lemma 8.2.3, kun − vn k is nonincreasing. This implies uniqueness. 

8.3

PERIODIC FORCING

In this section, we assume that cn , θn and fn in the difference inclusion (8.4) are periodic, and we prove the existence of a periodic solution, as well as the weak convergence of any bounded solution to a periodic solution. This section contains a theorem from [ROU-KHA4]. Also in [APR3, POF-REI] the reader can find some special cases of some theorems similar to the following theorem. Theorem 8.3.1 Suppose that A : D(A) ⊂ H → H is a maximal monotone operator with 0 ∈ D(A), and cn , θn and fn are periodic with period N > 0 such that either (8.37) is satisfied or θn ≥ 1. If (8.4) has a solution un for an initial value, then it has a periodic solution wn with period N, and un − wn * 0 as n → +∞. Moreover, any two periodic solutions differ by an additive constant. Proof. Let x ∈ H, and let m ≥ 0 be an integer. Since A is maximal monotone, there is a unique solution to: ( ui+1 − (1 + θi )ui + θi ui−1 ∈ ci Aui + fi , i > m um = x. For n ≥ m, we define the operators Q(n, m) : H → H by Q(n, m)x := un .

Second Order Difference Equations

161

Let vn be a solution to (8.4) with vm = y. The monotonicity of A implies that: (un+1 − vn+1 − (1 + θn )(un − vn ) + θn (un−1 − vn−1 ), un − vn ) ≥ 0. Therefore (1 + θn )kun − vn k ≤ kun+1 − vn+1 k + θn kun−1 − vn−1 k. By Lemma 8.2.3 and (8.37), we get: kun − vn k ≤ kun−1 − vn−1 k. Hence from the definition of Q(n, m), it follows that Q(n, m) is nonexpansive. The Nonexpansiveness of Q(n, m) also shows the uniqueness of the solution to (8.4) with um = x. By the uniqueness of the solution, we have: Q(n, m)Q(m, k) = Q(n, k) for n ≥ m ≥ k, and by the periodicity, we also have: Q(n + N, m + N) = Q(n, m) for n ≥ m. It follows that Q(m + N, m)n = Q(m + nN, m). In particular, taking m = 0, it follows that {ukN }k≥0 is a nonexpansive sequence in H. Therefore, by the ergodic theorem proved in [DJA1], we deduce that the sequence sn = 1n ∑n−1 k=0 ukN converges weakly in H, and the limit is a fixed point of Q(N, 0). This shows the existence of a periodic solution to (8.4), which is therefore bounded. Hence all solutions to (8.4) are bounded. Suppose that un and vn are respectively a bounded and a periodic solution to (8.4). Set zn = un − vn . Then by (8.4) and the monotonicity of A, we get  zn+1 − (1 + θn )zn + θn zn−1 , zn ≥ 0 This implies that kzn+1 − zn k2 ≤ (zn+1 − zn , zn+1 ) − θn (zn − zn−1 , zn ) 1 1 θn θn θn 1 = kzn+1 − zn k2 + kzn+1 k2 − kzn k2 − kzn − zn−1 k2 − kzn k2 + kzn−1 k2 2 2 2 2 2 2 Since by Lemma 8.2.3, kzn k is nonincreasing, we get: kzn − zn−1 k2 ≤ kzn−1 k2 − kzn k2 2 Summing up from n = 1 to m and then letting m → +∞, we get ∑+∞ n=1 kzn+1 − zn k < +∞. Since {vn } is a periodic solution of (8.4), for each m ≥ 0, we have (n+1)N−1

um+nN − um+(n+1)N =



i=nN

(um+i − vm+i − (um+i+1 − vm+i+1 )) → 0

(8.71)

162

Nonlinear Evolution and Difference Equations of Monotone Type

as n → +∞. Let xn := Q(m + nN, 0)u0 = um+nN Then {xn }n≥1 is a nonexpansive sequence, which is asymptotically regular by (8.71). It follows from [DJA2, DJA3] that: um+nN * wm

(8.72)

as n → +∞. Since {vn }n≥1 is a periodic solution of (8.4), we have lim (um+nN − um−1+nN − (vm − vm−1 ))

n→+∞

= lim (um+nN − vm+nN − (um−1+nN − vm−1+nN )) = 0 n→+∞

Therefore, limn→+∞ (um+nN − um−1+nN ) = (vm − vm−1 ) exists, and from (8.72), we get: lim (um+nN − um−1+nN ) = wm − wm−1 n→+∞

This implies that wm − vm = w0 − v0 = constant, showing that any two periodic solutions differ by an additive constant, as we show below that wn is a periodic solution to (8.4). By (8.4) we have [um+nN ,

1 (um+1+nN − (1 + θm+nN )um+nN + θm+nN um−1+nN − fm )] ∈ A. cm

Since A is demiclosed, by letting n → +∞, we get: [wm ,

1 (wm+1 − (1 + θm )wm + θm wm−1 − fm )] ∈ A cm

Therefore wm ∈ D(A), and wm+1 − (1 + θm )wm + θm wm−1 − fm ∈ cm Awm , which shows that wn is a periodic solution to (8.4). Since wn is N-periodic, we have: un − wn = un − uk+mN + uk+mN − wk+mN + wk+mN − wn = un − wn − (uk+mN − wk+mN ) + uk+mN − wk = zn − zk+mN + uk+mN − wk where k + mN ≤ n < k + (m + 1)N and zn = un − wn . Now from (8.72), we have uk+mN − wk * 0 as m → +∞, and from kzn − zn−1 k → 0, we get: i=k+(m+1)N−1

kzn − zk+mN k ≤



kzi − zi−1 k → 0

i=k+mN

as n → +∞. This shows that un − wn * 0, as n → +∞, and completes the proof of the theorem. 

Second Order Difference Equations

8.4

163

CONTINUOUS DEPENDENCE ON INITIAL CONDITIONS

Consider the following second order difference equation: ( ui+1 − (1 + θi )ui + θi ui−1 ∈ ci Aui , i ≥ 1 u0 = a, supi≥1 kui k < +∞,

(8.73)

where a ∈ H and A is a maximal monotone operator in H, for which we already proved the existence and uniqueness of the solution. Our aim in this section is to show that the function that associates to the initial data {a, A} the solution (ui )i≥1 to (8.73) is continuous. The results of this section are from [APR-APR1] and [APRAPR2] to which we refer for some of the proofs. First we recall a similar theorem for (8.3) from [APR-APR1], to which we refer for its proof. Theorem 8.4.1 Suppose that (uNi )1≤i≤N and (unN i )1≤i≤N are respectively solutions to the bilocal problems ( uNi+1 − (1 + θi )uNi + θi uNi−1 ∈ ci AuNi + fi , 1 ≤ i ≤ N (8.74) uN0 = a, uNN+1 = b and ( nN nN n nN n unN i+1 − (1 + θi )ui + θi ui−1 ∈ ci A ui + f i , 1 ≤ i ≤ N unN unN 0 = an , N+1 = bn

(8.75)

where A and An are maximal monotone operators in H with 0 ∈ D(A) ∩ D(An ), a, b, an , bn , fi , fin ∈ H, ci > 0, θi > 0, 1 ≤ i ≤ N. If an → a, bn → b, fin → fi and An converges to A in the sense of resolvent, (that is (I + λ An )−1 x → (I + λ A)−1 x, ∀λ > 0, ∀x ∈ H), then N lim unN i = ui , 1 ≤ i ≤ N n→+∞

In this section, we prove a similar result for (8.4). Let un = (uni )i≥1 be the solution to the following difference inclusion: ( uni+1 − (1 + θi )uni + θi uni−1 ∈ ci An uni , i ≥ 1 (8.76) un0 = an , supi≥1 kuni k = Cn < +∞, n ≥ 1 where An : D(An ) ⊂ H → H is a sequence of maximal monotone operators in H and an ∈ H. Theorem 8.4.2 Suppose that A : D(A) ⊂ H → H and An : D(An ) ⊂ H → H are maximal monotone operators in H such that 0 ∈ D(A) ∩ D(An ) and 0 ∈ A0 ∩ An 0. Also suppose that a, an ∈ H, ci > 0, and θi > 0, i ≥ 1 is a sequence such that i



0 < θi < 1,

1 ∑ hi = +∞, i=1

where hi =

1

∑ θ j · · · θi

j=1

(8.77)

164

Nonlinear Evolution and Difference Equations of Monotone Type

or θi ≥ 1. un

(8.78)

(uni )i≥1

Assume that u = (ui )i≥1 and = are respectively solutions to (8.73) and (8.76). If an → a in H and An → A in the sense of resolvent, then uni → ui as n → +∞, uniformly on every finite subset of integers. Proof. We are going to estimate the solutions of problems (8.73) and (8.76) by the solutions to the following problems on the finite set {1, · · · , N}. ( uNi+1 − (1 + θi )uNi + θi uNi−1 ∈ ci AuNi , 1 ≤ i ≤ N (8.79) uN0 = uNN+1 = a, and

( nN nN n nN unN i+1 − (1 + θi )ui + θi ui−1 ∈ ci A ui , 1 ≤ i ≤ N nN nN n u0 = uN+1 = a , n ≥ 1

(8.80)

By Theorem 8.2.2, we know that these problems have unique solutions uN = (uNi )1≤i≤N ∈ D(A)N n N unN = (unN i )1≤i≤N ∈ D(A ) .

Moreover, we can prove the following auxiliary result.



Lemma 8.4.3 the sequence Cn in (8.76) is bounded in R. Proof. Since 0 ∈ D(An ) and 0 ∈ An 0 for each n ≥ 1, uni ≡ 0 is the unique solution of problem (8.76) with un0 = 0. Then Theorems 8.2.4 and 8.2.6 imply that kuni k ≤ kan k, ∀n, i ≥ 1 where un = (uni )i≥1 is the solution to (8.76) with initial data un0 = an . Therefore Cn = supi≥1 kuni k is bounded in R.  n By Theorems 8.2.6 and 8.2.4, uNi → ui and unN i → ui as N → +∞, uniformly for i belonging to every finite subset of the integers. The following lemma is in Lemma 2.2 of [APR-APR2]. n Lemma 8.4.4 With the same assumptions as in Theorem 8.4.1, limN→+∞ unN i = ui exists for each n ≥ 1, uniformly on every finite subset of the integers. Moreover, for each integer N0 ≥ 1 and 1 ≤ i ≤ N0 , if θi satisfies either (8.77) or (8.78), then the following respective estimates hold:

2kan k n q kunN − u k ≤ i i ∑Nk=N0 or n kunN i − ui k ≤

1 hk

2N0 kan k (N − N0 + 1)

(8.81)

(8.82)

Also, similar estimates hold for the respective solutions u and uN to problems (8.73) and (8.74).

Second Order Difference Equations

165

Proof. (for Theorem 1.4.2). Suppose that N is a fixed integer. We have nN N N kuni − ui k ≤ kuni − unN i k + kui − ui k + kui − ui k, 1 ≤ i ≤ N, n ≥ 1

Suppose that 1 ≤ N0 ≤ N. Then applying Lemma 8.4.4 to both problems in (8.79), we get: 2kan k + 2kak N kuni − ui k ≤ q + kunN (8.83) i − ui k, 1 ≤ i ≤ N0 ∑Nk=N0 h1k if we assume (8.77) for θn , and we get: kuni − ui k ≤

2N0 (kan k + kak) N + kunN i − ui k, 1 ≤ i ≤ N0 N − N0 + 1

(8.84)

if θn satisfies (8.78). For N fixed, by Theorem 3.1 of [APR-APR1], we have unN i → uNi in H as n → +∞, for 1 ≤ i ≤ N. By taking limsup as n → +∞ in (8.83) and (8.84) we obtain either 4kak lim sup kuni − ui k ≤ q n→+∞ ∑Nk=N0 h1k or lim sup kuni − ui k ≤ n→+∞

4N0 kak , 1 ≤ i ≤ N0 N − N0 + 1

respectively, depending on whether condition (8.77) or (8.78) is satisfied by θn . In both cases, the result follows by letting N → +∞. 

8.5

ASYMPTOTIC BEHAVIOR FOR THE HOMOGENEOUS CASE

In this section, we concentrate on the homogeneous case when fn ≡ 0. We study the convergence of solutions to (8.4) and their weighted averages to a zero of the maximal monotone operator A, with some suitable assumptions on the parameters cn and θn . Throughout this section, we denote by M = supn≥0 kun k, an = (θ1 · · · θn )−1 , with a0 = 1, and Aun , the element 8.5.1

un+1 −(1+θn )un +θn un−1 cn

in H.

WEAK ERGODIC CONVERGENCE

In this subsection, we study the asymptotic behavior of the weighted averages of solutions to (8.4). We introduce three types of weighted averages, and prove the weak or strong convergence of each of them to a zero of the maximal monotone operator A. Moreover, we show that the existence of a solution to (8.4) implies that A−1 (0) , ∅, provided that the appropriate conditions on the parameters cn and θn hold. We denote wn := (∑ni=1 ci )−1 (∑ni=1 ci ui ), zn := (∑ni=1 ai ci )−1 ∑ni=1 ai ci ui , n ∞ −1 σn := (∑ni=1 ∑∞ k=i ak ck ) (∑i=1 ∑k=i ak ck uk ). We will need the following elementary lemmas in order to prove our main results.

166

Nonlinear Evolution and Difference Equations of Monotone Type

Lemma 8.5.1 Let {xi } and {yi } be two sequences of real numbers. Then, k

k

∑ (∆xi )yi = xk+1 yk+1 − x1 y1 − ∑ xi+1 (∆yi ),

i=1

i=1

where ∆xi = xi+1 − xi . The following lemma was proved in Chapter 7 (see Lemma 1.6.1). We recall it for the convenience of the reader. Lemma 8.5.2 Let {an } and {bn } be two sequences of real positive numbers. If {an } n is nonincreasing and convergent to zero and ∑+∞ n=1 an bn < +∞, then (∑k=1 bk )an → 0 as n → +∞. −1 Lemma 8.5.3 Let {an } be a sequence of positive numbers with ∑+∞ n=1 an = + ∞. If {bn } is a bounded sequence, then lim infn→+∞ an (bn+1 − bn ) ≤ 0, and lim infn→+∞ an (bn − bn+1 ) ≤ 0.

Proof. It suffices to show that lim infn→+∞ an (bn+1 − bn ) ≤ 0. The second inequality is proved in a similar way. Suppose to the contrary that there is an integer n0 > 0 and λ > 0 such that for each integer n ≥ n0 , we have an (bn+1 − bn ) > λ > 0. Then dividing by an and summing up from n = n0 to ∞, we get a contradiction.  Lemma 8.5.4 Let {un } be a solution to (8.4). Then an−1 kun − un−1 k is either nonincreasing or eventually increasing. Proof. From the monotonicity of A, we have (Aui+1 − Aui , ui+1 − ui ) ≥ 0,

∀i ≥ 1.

By (8.4), we get 1 θi+1 1 (ui+2 − ui+1 , ui+1 − ui ) − kui+1 − ui k2 − kui+1 − ui k2 ci+1 ci+1 ci +

θi (ui − ui−1 , ui+1 − ui ) ≥ 0, ci

∀i ≥ 1.

It follows that 1 θi 1 θi+1 kui+1 − ui k − kui − ui−1 k ≤ kui+2 − ui+1 k − kui+1 − ui k, ci ci ci+1 ci+1

(8.85)

for all i ≥ 1. If {ai−1 kui − ui−1 k} is not nonincreasing, then there exists j ≥ 1 such that a j ku j+1 − u j k > a j−1 ku j − u j−1 k. Then (8.85) and the identity an θn = an−1 imply that the sequence {ai−1 kui − ui−1 k}i≥ j is increasing.  Lemma 8.5.5 Suppose that ui is a solution to (8.4) and p ∈ A−1 (0). Then kun − pk is nonincreasing or eventually increasing. Moreover, if ∑+∞ n=1 θ1 · · · θn = +∞, then kun − pk and an−1 kun − un−1 k are nonincreasing and an−1 kun − un−1 k converges to zero as n → +∞.

Second Order Difference Equations

167

Proof. From the monotonicity of A and (8.4), we get  ui+1 − (1 + θi )ui + θi ui−1 , ui − p ≥ 0.

(8.86)

This implies that kui+1 − pk2 −kui − pk2 +θi (kui−1 − pk2 −kui − pk2 ) ≥ kui+1 −ui k2 +θi kui −ui−1 k2 ≥ 0. (8.87) If kui − pk is not nonincreasing, there is j > 0 such that ku j − pk < ku j+1 − pk. Then by (8.87), the sequence {kui − pk}i≥ j+1 is increasing. For the second part of the lemma, multiplying (8.87) by ai , we get ai−1 kui − ui−1 k2 ≤ ai (kui+1 − pk2 − kui − pk2 ) − ai−1 (kui − pk2 − kui−1 − pk2 ). Summing up from i = k to m, we get: m

∑ ai−1 kui − ui−1 k2 ≤ am (kum+1 − pk2 − kum − pk2 ) − ak−1 (kuk − pk2 − kuk−1 − pk2 ). i=k

Taking liminf when m → +∞, by our assumption and Lemma 8.5.3, lim inf am (kum+1 − pk2 − kum − pk2 ) ≤ 0, m→+∞

then we get: ∞

∑ ai−1 kui − ui−1 k2 ≤ ak−1 (kuk−1 − pk2 − kuk − pk2 ).

(8.88)

i=k −1 2 2 (8.88) implies that {kuk − pk2 } is nonincreasing and ∑∞ i=1 ai−1 ai−1 kui − ui−1 k < +∞. The assumption on {θi } implies that lim infi→+∞ ai−1 kui − ui−1 k = 0. By Lemma 8.5.4, ai−1 kui − ui−1 k is nonincreasing and therefore limi→+∞ ai−1 kui − ui−1 k = 0. 

Theorem 8.5.6 Let {un }n≥1 be a solution to (8.4). Assume that ∑+∞ n=1 cn = +∞, ∑nk=1 |θk −θk−1 | ∞ −1 → 0 as n → +∞. Then A (0) , ∅ and ∑n=1 θ1 · · · θn = +∞ and ∑n c k=1 k

wn * p ∈ A−1 (0). In particular, the latter condition holds if either one of the following two conditions is satisfied: 1) ∑∞ n=1 |θn − θn−1 | < +∞ |θ −θ | 2) limn→+∞ n cnn−1 = 0. Proof. By the monotonicity of A and (8.4), we have for all k, n ≥ 1,   un+1 − (1 + θn )un + θn un−1 , ck uk + uk+1 − (1 + θk )uk + θk uk−1 , cn un   ≤ ck un+1 − (1 + θn )un + θn un−1 , un + cn uk+1 − (1 + θk )uk + θk uk−1 , uk . (8.89)

168

Nonlinear Evolution and Difference Equations of Monotone Type

Summing up the above inequality from k = 1 to m, and using the technique of Lemma 8.5.1, we get: m

un+1 − (1 + θn )un + θn un−1 , ∑ ck uk



k=1 m



∑ ck

  un+1 − (1 + θn )un + θn un−1 , un − um+1 − u1 , cn un

k=1

+ θm (um , cn un ) − θ0 (u0 , cn un ) m

− ∑ (θk − θk−1 )(uk−1 , cn un ) + k=1

cn [kum+1 k2 − ku1 k2 − θm kum k2 2

m

+ θ0 ku0 k2 + ∑ (θk − θk−1 )kuk−1 k2 ]. k=1

Suppose that wm j * p. Then dividing both sides by ∑m k=1 ck , substituting m by m j and letting j → +∞ in the above inequality, we get (8.86). Now the proof of Lemma 8.5.5 implies that kun − pk is nonincreasing. If q is another cluster point of wn , then kun − qk is nonincreasing too. Therefore limn→+∞ (un , p − q) exists. This implies that limn→+∞ (wn , p − q) exists. Then (p, p − q) = (q, p − q), which implies that p = q. Therefore wn * p. Now we prove that p ∈ A−1 (0). Let [x, y] ∈ A. By the monotonicity of A, we have: (y, uk − x) ≤ (Auk , uk − x). Multiplying the above inequality by ck and summing up from k = 1 to m, by applying Lemma 8.5.1, we get: m

1

∑ ck (y, uk − x) ≤ 2 [(kum+1 − xk2 − ku1 − xk2 )

k=1

m

− ∑ θk (kuk − xk2 − kuk−1 − xk2 )] k=1

1 θm ≤ (kum+1 − xk2 − ku1 − xk2 ) − kum − xk2 2 2 m−1 θ1 1 + ku0 − xk2 + ∑ (θk+1 − θk )kuk − xk2 2 2 k=1 1 θ1 1 m−1 ≤ kum+1 − xk2 + ku0 − xk2 + ∑ (θk+1 − θk )kuk − xk2 . 2 2 2 k=1 Dividing both sides of the above inequality by ∑m k=1 ck , and letting m → +∞, we get (y, p − x) ≤ 0, for all [x, y] ∈ A. Since A is maximal monotone, we conclude that p ∈ A−1 (0).



Second Order Difference Equations

169

Theorem 8.5.7 Assume that {un }n≥1 is a solution to (8.4) and either one of the following two conditions holds: 1) θn ≥ 1 an 2) 0 < θn < 1, ∑+∞ n=1 θ1 · · · θn = +∞ and ∑n a c → 0 as n → +∞. k=1 k k

−1 If ∑+∞ n=1 an cn = +∞, then A (0) , ∅. Moreover, zn converges weakly as n → +∞ to −1 some p ∈ A (0).

Proof. Multiplying both sides of (8.89) by ak and summing up from k = 1 to m, we get: m

un+1 − (1 + θn )un + θn un−1 , ∑ ak ck uk



k=1

+ am (um+1 − um ) − a0 (u1 − u0 ), cn un



m

≤ ( ∑ ak ck ) un+1 − (1 + θn )un + θn un−1 , un



k=1

1 + cn [am (kum+1 k2 − kum k2 ) − a0 (ku1 k2 − ku0 k2 )]. 2 Dividing both sides by ∑m k=1 ak ck , we get: un+1 − (1 + θn )un + un−1 , zm



m

≤ ( ∑ ak ck )−1 a0 (u0 − u1 , cn un ) k=1 m

( ∑ ak ck )−1 am (um − um+1 , cn un ) k=1

+ un+1 − (1 + θn )un + θn un−1 , un



1 m + ( ∑ ak ck )−1 cn a0 (ku0 k2 − ku1 k2 ) 2 k=1 1 m + ( ∑ ak ck )−1 cn am (kum+1 k2 − kum k2 ) 2 k=1

(8.90)

Suppose that zm j * p. Substituting m by m j in the above inequality, and letting j → +∞, we get (8.86). By the proof of Lemma 8.5.5, this implies that: kun − pk ≤ kun−1 − pk. If q is another cluster point of zn , then there exists limn→+∞ kun − qk. This implies that there exists limn→+∞ (zn , p − q), then (p, p − q) = (q, p − q), hence p = q. Therefore zn * p. Now let [x, y] ∈ A. By the monotonicity of A, we get (y, uk − x) ≤ (Auk , uk − x). Multiplying by ak ck and summing up from k = 1 to m, we get m

∑ ak ck (y, uk − x) k=1

170

Nonlinear Evolution and Difference Equations of Monotone Type m



m

∑ ak (uk+1 − uk , uk − x) − ∑ ak θk (uk − uk−1 , uk − x) k=1 m



k=1

1 1 m ak (kuk+1 − xk2 − kuk − xk2 ) − ∑ ak−1 (kuk − xk2 − kuk−1 − xk2 ) ∑ 2 k=1 2 k=1

1 1 1 m−1 = am kum+1 − xk2 − a1 ku1 − xk2 + ∑ (ak − ak+1 )kuk+1 − xk2 2 2 2 k=1 1 1 1 m − ku0 − xk2 + am kum − xk2 + ∑ (ak−1 − ak )kuk − xk2 . 2 2 2 k=1

(8.91)

If θn ≥ 1, (8.91) implies that m

3

∑ ak ck (y, uk − x) ≤ 2 (M + kxk)2 .

k=1

Dividing by ∑m k=1 ak ck and letting m → +∞, we get (y, p − x) ≤ 0. If 0 < θn < 1 we have m

∑ ak ck (y, uk − x) ≤ (M + kxk)2 am . k=1

Dividing by ∑m k=1 ak ck and letting m → +∞, we get (y, p − x) ≤ 0. of A, we have p ∈ A−1 (0), which is therefore nonempty.

By the maximality 

Theorem 8.5.8 Assume that {un }n≥1 is a solution to (8.4) with either θn ≥ 1, or an 0 < θn < 1, ∑+∞ → 0 as n → +∞. If ∑+∞ +∞ n=1 θ1 · · · θn = +∞ and n n=1 nan cn = a c ∑i=1 ∑k=i k k

+∞ +∞, then A−1 (0) , ∅. Moreover, if ∑+∞ n=1 an cn < +∞ and ∑n=1 nan cn = +∞, then σn −1 converges weakly as n → +∞ to some p ∈ A (0).

Proof. Multiplying both sides of (8.89) by ak and summing up from k = i to m, we get m

un+1 − (1 + θn )un + θn un−1 , ∑ ak ck uk



k=i

+ am (um+1 − um ) − ai−1 (ui − ui−1 ), cn un



m

≤ ( ∑ ak ck ) un+1 − (1 + θn )un + θn un−1 , un



k=i

 1  + cn am (kum+1 k2 − kum k2 ) − ai−1 (kui k2 − kui−1 k2 ) . 2 Taking lim inf when m → +∞, and using the first inequality in Lemma 8.5.3 for the right hand side and the second for the left hand side, we get: +∞

un+1 − (1 + θn )un + θn un−1 , ∑ ak ck uk k=i



Second Order Difference Equations

171

≤ ai−1 (ui − ui−1 , cn un ) ∞

+ ( ∑ ak ck ) un+1 − (1 + θn )un + θn un−1 , un



k=i

1 + cn ai−1 (kui−1 k2 − kui k2 ). 2

(8.92)

From Lemma 8.5.1, if θn ≥ 1, we get: m

∑ ai−1 [(ui , cn un ) − (ui−1 , cn un )]

i=1

m

= am (um , cn un ) − a0 (u0 , cn un ) + ∑ (ai−1 − ai )(ui , cn un ) i=1 2

2

≤ −a0 (u0 , cn un ) + cn M am + cn M (a0 − am ) ≤ 2cn M 2 < +∞,

(8.93)

and if 0 < θn < 1, we have: m

∑ ai−1 [(ui , cn un ) − (ui−1 , cn un )] ≤ cn M2 a0 + cn M2 am + cn M2 (am − a0 )

i=1

= 2cn M 2 am

(8.94)

If θn ≥ 1, by Lemma 8.5.1, we get: m

m

∑ ai−1 (kui−1 k2 − kui k2 ) = a0 ku0 k2 − am kum k2 + ∑ (ai − ai−1 )kui k2

i=1

i=1

2

≤ ku0 k < +∞,

(8.95)

and if 0 < θn < 1, we have: m

∑ ai−1 (kui−1 k2 − kui k2 ) ≤ ku0 k2 − am kum k2 + M2 (am − a0 )

i=1

≤ M 2 am .

(8.96)

∞ Summing up (8.92) from i = 1 to m, and then dividing by ∑m i=1 ∑k=i ak ck , if θn ≥ 1 by (8.93) and (8.95), we get m ∞

m +∞

un+1 − (1 + θn )un + θn un−1 , ( ∑ ∑ ak ck )−1 ∑ ∑ ak ck uk i=1 k=i

≤ un+1 − (1 + θn )un + θn un−1 , un



m ∞ 1 + cn ( ∑ ∑ ak ck )−1 (ku0 k2 + 4M 2 ), 2 i=1 k=i

i=1 k=i



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Nonlinear Evolution and Difference Equations of Monotone Type

and if 0 < θn < 1, by (8.94) and (8.96), we get m +∞

m ∞

un+1 − (1 + θn )un + θn un−1 , ( ∑ ∑ ak ck )−1 ∑ ∑ ak ck uk i=1 k=i

≤ un+1 − (1 + θn )un + θn un−1 , un



i=1 k=i



m ∞ 1 + cn ( ∑ ∑ ak ck )−1 (M 2 am + 4M 2 am ) 2 i=1 k=i m ∞  5 = cn ( ∑ ∑ ak ck )−1 M 2 am + un+1 − (1 + θn )un + θn un−1 , un . 2 i=1 k=i

Assume σm j * p. Substituting m by m j and letting j → +∞, we get (8.86). By the proof of Lemma 8.5.5, kun − pk is nonincreasing. If q is another cluster point of σn , then there exists limn→+∞ kun − qk. This implies that there exists limn→+∞ (σn , p − q), then (p, p − q) = (q, p − q), hence p = q. Therefore σn * p as n → +∞. Now we prove that p ∈ A−1 (0). Let [x, y] ∈ A. By the monotonicity of A, we have: (y, uk − x) ≤ (Auk , uk − x). Multiplying by ak ck and then summing up from k = i to m, and using the identity ak θk = ak−1 , we get: m

∑ ak ck (y, uk − x) k=i

 1 m  1 m ak kuk+1 − xk2 − kuk − xk2 − ∑ ak θk kuk − xk2 − kuk−1 − xk2 ∑ 2 k=i 2 k=i   1 = am kum+1 − xk2 − kum − xk2 − ai−1 kui − xk2 − kui−1 − xk2 . 2 Since ∑∞ k=1 ak ck < +∞, then taking liminf when m → +∞, and using Lemma 8.5.3, we get: ∞  1 ∑ ak ck (y, uk − x) ≤ 2 ai−1 kui−1 − xk2 − kui − xk2 . k=i ≤

∞ Summing up from i = 1 to m, and dividing by ∑m i=1 ∑k=i ak ck , we get:

(y, σm − x) m  1 m ∞ ≤ ( ∑ ∑ ak ck )−1 ∑ ai−1 kui−1 − xk2 − kui − xk2 2 i=1 k=i i=1 m   1 m ∞ = ( ∑ ∑ ak ck )−1 a0 ku0 − xk2 − am kum − xk2 + ∑ (ai − ai−1 )kui − xk2 . 2 i=1 k=i i=1

If θn ≥ 1, then letting m → +∞, we get (y, x − p) ≥ 0, for all [x, y] ∈ A. Otherwise (i.e. if 0 < θn < 1), (y, σm − x)

Second Order Difference Equations

173

m   1 m ∞ ≤ ( ∑ ∑ ak ck )−1 ku0 − xk2 + (M + kxk)2 ∑ (ai − ai−1 ) 2 i=1 k=i i=1

  1 m ∞ = ( ∑ ∑ ak ck )−1 ku0 − xk2 + (M + kxk)2 (am − a0 ) . 2 i=1 k=i Now letting m → +∞, we get again (y, x − p) ≥ 0, for all [x, y] ∈ A. Since A is maximal monotone, we obtain p ∈ A−1 (0).  8.5.2

STRONG ERGODIC CONVERGENCE

In this short subsection, we show that when the maximal monotone operator A is odd, the weighted average zn converges strongly to a zero of A. Proposition 8.5.9 Suppose that un is a solution to (8.4) and A−1 (0) , ∅. 1) If 0 < θn < 1 and ∑+∞ n=1 θ1 · · · θn = +∞, then limn→+∞ nkun − un−1 k = 0. 2) If θn ≥ 1, then lim nan kun − un−1 k = 0. Proof. 1) Let p ∈ A−1 (0). From the monotonicity of A and (8.4), we get (8.86). By Lemma 8.5.5, this implies that kun − pk is nonincreasing. Now we have: kui+1 − pk2 −kui − pk2 +θi (kui−1 − pk2 −kui − pk2 ) ≥ kui+1 −ui k2 +θi kui −ui−1 k2 ≥ 0. (8.97)

Multiplying (8.97) by ai , we get ai−1 kui − ui−1 k2 ≤ ai (kui+1 − pk2 − kui − pk2 ) − ai−1 (kui − pk2 − kui−1 − pk2 ). Summing up from i = k to m, we get: m

∑ ai−1 kui −ui−1 k2 ≤ am (kum+1 − pk2 −kum − pk2 )−ak−1 (kuk − pk2 −kuk−1 − pk2 ). i=k

Taking lim inf when m → +∞, by our assumption and Lemma 8.5.3, we have: lim inf am (kum+1 − pk2 − kum − pk2 ) ≤ 0. m→+∞

Therefore we get: ∞

∑ ai−1 kui − ui−1 k2 ≤ ak−1 (kuk−1 − pk2 − kuk − pk2 ). i=k

Since {ai } is increasing, we have ∞

∑ kui − ui−1 k2 ≤ kuk−1 − pk2 − kuk − pk2 . i=k

Summing up from k = 1 to ∞, we get: ∞

∑ kkuk − uk−1 k2 ≤ ku0 − pk2 − l(p)2 ≤ ku0 − pk2 < +∞, k=1

(8.98)

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Nonlinear Evolution and Difference Equations of Monotone Type

where l(p) = limk→+∞ kuk − pk. Therefore limk→+∞ kuk − uk−1 k2 = 0. Since by Lemma 8.5.5, ak−1 kuk −uk−1 k is nonincreasing and ak is increasing then kuk −uk−1 k is nonincreasing. Now, by Lemma 8.5.2 (∑nk=1 k)kun − un−1 k2 → 0, then nkun − un−1 k → 0 as n → +∞. 2) Summing up (8.98) from k = 1 to ∞, we get ∞





∑ ∑ ai−1 kui − ui−1 k2 ≤ ∑ ak−1 (kuk−1 − pk2 − kuk − pk2 ). k=1 i=k

k=1

Therefore since ak ≤ 1, we have ∞



∑ kak−1 kuk − uk−1 k2 ≤

∑ (kuk−1 − pk2 − kuk − pk2 )

k=1

k=1

≤ ku0 − pk2 − lim kuk − pk2 k→+∞

2

≤ ku0 − pk < +∞ k 2 ∞ 2 2 Hence ∑∞ k=1 ak ak kuk+1 − uk k = ∑k=1 kak kuk+1 − uk k < +∞, because by Lemma 8.5.5, ak kuk+1 − uk k is nonincreasing. Since it is also convergent to zero, by Lemma 8.5.2, we conclude that: n

n

k 2 )an kun+1 − un k2 → 0 a k=1 k

( ∑ k)a2n kun+1 − un k2 ≤ ( ∑ k=1

as n → +∞. Therefore nan kun+1 − un k → 0 as n → +∞.



Theorem 8.5.10 Suppose that {un } is a solution to (8.4) with θn ≥ 1 and the mono−1 tone operator A is odd. If ∑∞ n=1 an cn = +∞, then zn → p ∈ A (0) as n → +∞. Proof. Since A is monotone and odd,   4 un , A(un ) = un − (−un ), A(un ) − A(−un ) ≥ 0. Therefore (8.86) is satisfied with p = 0. Hence kun k is nonincreasing. Since A is monotone and odd, we have: ±[(ui , Au j ) + (u j , Aui )] ≤ (ui , Aui ) + (u j , Au j ). Multiplying both sides by ai a j ci c j and using (8.4), we get:  ± (ai (ui+1 − ui ) − ai−1 (ui − ui−1 ), a j c j u j )  + (a j (u j+1 − u j ) − a j−1 (u j − u j−1 ), ai ci ui ) a jc j ≤ {ai (kui+1 k2 − kui k2 ) − ai−1 (kui k2 − kui−1 k2 )} 2 ai ci + {a j (ku j+1 k2 − ku j k2 ) − a j−1 (ku j k2 − ku j−1 k2 )}. 2

(8.99)

Second Order Difference Equations

175

Summing up from i, j = k to n, we get n   ± 2 (an (un+1 − un ) − ak−1 (uk − uk−1 ), ∑ ai ci ui ) i=k n

≤ ( ∑ ai ci ){an (kun+1 k2 − kun k2 ) − ak−1 (kuk k2 − kuk−1 k2 )}.

(8.100)

i=k

For any fixed k, we have n n   lim zn − ( ∑ ai ci )−1 ∑ ai ci ui = 0.

n→+∞

i=k

(8.101)

i=k

By (8.100) and (8.101), since by Proposition 8.5.9, limn→+∞ an kun+1 − un k = 0, we get 1 lim sup |(uk − uk−1 , wn )| ≤ (kuk−1 k2 − kuk k2 ). (8.102) 2 n→+∞ On the other hand, for any k < n n

|(uk − un , zn )| = |(

(ui − ui−1 ), zn )|



i=k+1 n





|(ui − ui−1 , zn )|.

i=k+1

By taking limsup, we get n

lim sup |(uk − un , zn )| ≤ lim sup n→+∞



|(ui − ui−1 , zn )|

n→+∞ i=k+1 +∞





lim sup |(ui − ui−1 , zn )|

i=k+1 n→+∞ +∞



1 1 ( kui−1 k2 − kui k2 ) 2 2 i=k+1



1 1 = kuk k2 − l 2 , 2 2 where l = limn→+∞ kun k. Therefore limk→+∞ lim supn→+∞ |(uk − un , zn )| = 0. The rest of the proof is exactly similar to the proof of Lemma 1 in [BRU1] (see also Theorem 3.3, pp. 147–149 in [MOR]). Finally, p ∈ A−1 (0) by the first part of Theorem 8.5.7.  Theorem 8.5.11 Suppose that {un } is a solution to (8.4) with 0 < θn < 1 and lim infn→+∞ θ1 · · · θn > 0. If the monotone operator A is odd and ∑∞ n=1 an cn = +∞, then zn → p ∈ A−1 (0) as n → +∞. Proof. The assumption on θn implies that an is bounded and ∑+∞ n=1 θ1 · · · θn = +∞. Therefore by Proposition 8.5.9, we have an kun+1 − un k → 0. Now the result follows by a proof similar to that of Theorem 8.5.10. Moreover, p ∈ A−1 (0) by the second part of Theorem 8.5.7. 

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Nonlinear Evolution and Difference Equations of Monotone Type

8.5.3

WEAK CONVERGENCE OF SOLUTIONS

In this section, we study the weak convergence of un to a zero of A. Theorem 8.5.12 Suppose that un is a solution to (8.4) with θn ≥ 1 and A−1 (0) , ∅. −1 If lim infn→+∞ ∑kn i=1 an+i cn+i > 0, for some positive integer k, then un * p ∈ A (0). Proof. By Proposition 8.5.9, limn→+∞ nan kun+1 − un k = 0. Assume that un j * p. kn

Without loss of generality, assume that ∑i=1j cn j +i ≥ α2 > 0, ∀ j ≥ 1. Then, an |un+1 − un | ≤ εnn , where limn→+∞ εn = 0. By the monotonicity of A, we have: (Auk − Aun j +i , uk − un j +i ) ≥ 0. Multiplying by ck an j +i cn j +i , and summing up from i = 1 to kn j and dividing by kn

(∑i=1j an j +i cn j +i )−1 , by (8.4) we get: kn j kn j  um+1 − (1 + θm )um + θm um−1 , ( ∑ an j +i cn j +i )−1 ∑ an j +i cn j +i un j +i + i=1

kn j

i=1

kn j

( ∑ an j +i cn j +i )−1 ∑ an j +i (un j +i+1 − un j +i ) − an j +i−1 (un j +i − un j +i−1 ), cm um i=1



i=1

kn j

≤ (um+1 − (1 + θm )um + θm um−1 , um ) + cm ( ∑ an j +i cn j +i )−1 i=1

kn j





 an j +i (un j +i+1 − un j +i ) − an j +i−1 (un j +i − un j +i−1 ), un j +i .

(8.103)

i=1

On the other hand kn j

kn j

( ∑ an j +i cn j +i )−1 ∑ an j +i (un j +i+1 − un j +i ) − an j +i−1 (un j +i − un j +i−1 ), cm um i=1



i=1

kn j

= ( ∑ an j +i cn j +i )−1 a(k+1)n j (u(k+1)n j +1 − u(k+1)n j ) − an j (un j +1 − un j ), cm um



i=1

kn j

  ≤ cm M( ∑ an j +i cn j +i )−1 a(k+1)n j ku(k+1)n j +1 − u(k+1)n j k + an j kun j +1 − un j k i=1

εn j 2 ε(k+1)n j ≤ ( + ) → 0, α (k + 1)n j nj

(8.104)

as j → +∞. Also, kn j

kn j

( ∑ an j +i cn j +i )−1 ∑ an j +i (un j +i+1 − un j +i ) − an j +i−1 (un j +i − un j +i−1 ), un j +i i=1

i=1



Second Order Difference Equations kn j

= ( ∑ an j +i cn j +i )−1 i=1

 kn j



177

an j +i (un j +i+1 − un j +i ) − an j +i−1 (un j +i − un j +i−1 ), un j +i − un j



i=1

kn j

+ ∑ (an j +i (un j +i+1 − un j +i ) − an j +i−1 (un j +i − un j +i−1 ), un j )]. i=1

But kn j kn j  ( ∑ an j +i cn j +i )−1 ∑ an j +i (un j +i+1 − un j +i ) − an j +i−1 (un j +i − un j +i−1 ), un j → 0 i=1

i=1

as j → +∞, in a similar way as in (8.104). And, kn j

kn j

( ∑ an j +i cn j +i )−1 ∑ an j +i (un j +i+1 − un j +i ) − an j +i−1 (un j +i − un j +i−1 ), un j +i − un j i=1



i=1

kn j

kn

j 1 ≤ ( ∑ an j +i cn j +i )−1 ∑ an j +i (kun j +i+1 − un j k2 − kun j +i − un j k2 ) 2 i=1 i=1  − an j +i−1 (kun j +i − un j k2 − kun j +i−1 − un j k2 )

kn

1 j ≤ ( ∑ an j +i cn j +i )−1 a(k+1)n j (ku(k+1)n j +1 − un j k2 − ku(k+1)n j − un j k2 ) 2 i=1 kn

1 j ≤ ( ∑ an j +i cn j +i )−1 a(k+1)n j ku(k+1)n j +1 − u(k+1)n j kku(k+1)n j +1 + u(k+1)n j − 2un j k 2 i=1 kn j

≤ 2M( ∑ an j +i cn j +i )−1 a(k+1)n j ku(k+1)n j +1 − u(k+1)n j k i=1

≤ 4M

ε(k+1)n j α(k + 1)n j

→ 0,

(8.105)

as j → +∞. Then letting j → +∞ in (8.103), we get:  um+1 − (1 + θm )um + θm um−1 , um − p ≥ 0. Therefore (8.86) holds, and by the proof of Lemma 8.5.5, kun − pk is nonincreasing. Now by a similar proof as in Theorem 8.5.6, it follows that un * p as n → +∞. Finally, we need to show that p ∈ A−1 (0). By Part 2 of Proposition 8.5.9, nan kun − un−1 k → 0 as n → +∞. Then: nan cn kAun k = nan kun+1 − (1 + θn )un + θn un−1 k ≤ nan kun+1 − un k + nan−1 kun − un−1 k → 0 as n → +∞. Therefore for each ε > 0 there exists n0 > 0 such that for every n ≥ n0 , nan cn kAun k ≤ ε. Then (n j + i)an j +i cn j +i kAun j +i k ≤ ε for each j ≥ j0 . This implies that kn j

kn j

ε n i=1 j + i

∑ cn j +i an j +i kAun j +i k ≤ ∑

i=1

178

Nonlinear Evolution and Difference Equations of Monotone Type kn j

ε n i=1 j

≤∑

= εk, ∀ j ≥ j0 . Therefore lim infn→+∞ kAun k = 0. Now, the demiclosedness of A implies that p ∈ A−1 (0).  Theorem 8.5.13 Suppose that un is a solution of (8.4) with 0 < θn < 1 and A−1 (0) , kn ∅. If ∑+∞ n=1 θ1 · · · θn = +∞ and lim infn→+∞ ∑i=1 cn+i > 0, for some positive integer k, −1 then un * p ∈ A (0). Proof. By Proposition 8.5.9, limn→+∞ nkun+1 − un k = 0. Assume that un j * p. kn

Without loss of generality, assume that ∑i=1j cn j +i ≥ α2 > 0, ∀ j ≥ 1. Given ε > 0, there is an integer n0 such that kun+1 − un k ≤ εn , ∀n ≥ n0 . Then we have: kn j

kn j

( ∑ cn j +i )−1 ∑ cn j +i kun j +i − un j k i=1

i=1

kn j

kn j

i

≤ ( ∑ cn j +i )−1 ∑ cn j +i ∑ kun j +l − un j +l−1 k i=1

i=1

kn j

l=1

kn j

i

1 l=1 n j + l

≤ ε( ∑ cn j +i )−1 ∑ cn j +i ∑ i=1

i=1

kn j

kn j

≤ ε( ∑ cn j +i )−1 ∑ cn j +i i=1

≤ε

i=1

Z n j +i dx

x

nj

Z (k+1)n j dx

x

nj

= ε ln(k + 1), ∀ j ≥ j0 .

(8.106)

Also kn j

kn j

( ∑ cn j +i )−1 ∑ kun j +i+1 − un j +i k ≤

2ε α



2ε α

i=1

i=1

kn j

1

∑ nj +i+1

i=1 kn j

1

∑ nj

i=1

2kε = . α

(8.107)

Similarly kn j

kn j

( ∑ cn j +i )−1 ∑ kun j +i − un j +i−1 k ≤ i=1

i=1

2kε , ∀ j ≥ j0 . α

(8.108)

Second Order Difference Equations

179

By the monotonicity of A, we have: kn j

kn j

( ∑ cn j +i )−1 ∑ un j +i+1 − (1 + θn j +i )un j +i + θn j +i un j +i−1 , cm um i=1



i=1

kn j

kn j

+ ( ∑ cn j +i )−1 um+1 − (1 + θm )um + θm um−1 , ∑ cn j +i un j +i i=1



i=1

kn j

kn j

≤ ( ∑ cn j +i )−1 cm ∑ un j +i+1 − (1 + θn j +i )un j +i + θn j +i un j +i−1 , un j +i i=1



i=1

 + um+1 − (1 + θm )um + θm um−1 , um . Letting j → +∞, by (8.104), (8.105) and (8.106), we get (um+1 − (1 + θm )um + θm um−1 , um − p). ≥ 0

(8.109)

By the proof of Lemma 8.5.5, the sequence kum − pk is nonincreasing. Then a similar proof as in Theorem 8.5.6 shows that un * p as n → +∞. Now we show that p ∈ A−1 (0). We know that nkun+1 − un k → 0 as n → +∞. Therefore ncn kAun k = nkun+1 − (1 + θn )un + θn un−1 k ≤ nkun+1 − un k + nθn kun − un−1 k → 0 as n → +∞, because 0 < θn < 1. Therefore for each ε > 0 there exists n0 > 0 such that for every n ≥ n0 , ncn kAun k ≤ ε. Then (n j + i)cn j +i kAun j +i k ≤ ε for each j ≥ j0 . This implies that kn j

kn j

kn

j ε ε ∑ cn j +i kAun j +i k ≤ ∑ n j + i ≤ ∑ n j = εk, ∀ j ≥ j0 . i=1 i=1 i=1

Therefore lim infn→+∞ kAun k = 0. Now the demiclosedness of A implies that p ∈ A−1 (0).  Corollary 8.5.14 The conclusions of Theorems 8.5.12 and 8.5.13 hold without assuming that A−1 (0) , ∅, if we assume the additional condition ∑n |θ −θ | limn→+∞ k=1∑n k c k−1 = 0. k=1 k

Proof. Since the condition on {cn } in Theorems 8.5.12 and 8.5.13 implies that −1  ∑∞ n=1 cn = +∞, then we know from Theorem 8.5.6 that A (0) , ∅. ( 1, if n is even Example 8.5.15 Let cn = 1 Then {cn } satisfies the condition in , if n is odd. n2 Theorems 8.5.12 and 8.5.13 with k = 1, and either θn ≡ 1 or 0 < θn < 1.

180

Nonlinear Evolution and Difference Equations of Monotone Type

8.5.4

STRONG CONVERGENCE OF SOLUTIONS

In this section, with additional assumptions on the maximal monotone operator A, we study the strong convergence of the sequence un to a zero of A. Theorem 8.5.16 Assume that A is strongly monotone, and let {un }n≥1 be a solution to (8.4), 1) If ∑∞ n=1 nan cn = +∞, and θn ≥ 1, then un converges strongly as n → +∞ to some 1 p ∈ A−1 (0) and kun − pk = o((∑nk=1 kak ck )− 2 ). ∞ ∞ 2) If ∑n=1 ncn = +∞, 0 < θn < 1, ∑n=1 θ1 · · · θn = +∞ and ∑m ama c → 0 as n → k=1 k k

+∞, then un converges strongly as n → +∞ to some p ∈ A−1 (0) and kun − pk = 1 o((∑nk=1 kck )− 2 ). ∞ Proof. In both cases (1) and (2), we have ∑∞ n=1 nan cn = +∞. If ∑n=1 an cn = +∞, then by the strong monotonicity of A, and Theorem 8.5.7 with p = weak − limn→+∞ zn , we get:

(Aun , un − p) = lim inf(Aun , un − zk ) ≥ lim inf αkun − zk k2 ≥ αkun − pk2 . k→+∞

k→+∞

am Similarly, if ∑∞ → 0, we have n=1 an cn < +∞, since ∑m k=1 ak ck with p = weak − limn→+∞ σn , by Theorem 8.5.8, we get:

am +∞ ∑m i=1 ∑k=i ak ck

→ 0, then

(Aun , un − p) = lim inf(Aun , un − σk ) ≥ lim inf αkun − σk k2 ≥ αkun − pk2 . k→+∞

k→+∞

Multiplying both sides of the above inequality by an cn and using (8.4), we get: αan cn kun − pk2 ≤ an (kun+1 − pk2 − kun − pk2 ) + an−1 (kun−1 − pk2 − kun − pk2 ). If θn ≥ 1, then +∞

∑ nan cn kun − pk2 < +∞.

(8.110)

n=1

Hence lim infn→+∞ kun − pk2 = 0. Since kun − pk is nonincreasing, then un → p as n → +∞. On the other hand, by (8.110) and Lemma 8.5.2, since kun − pk is nonin1 creasing and converges to zero, we get kun − pk = o((∑nk=1 kak ck )− 2 ). Otherwise, in the second case 0 < θn < 1, since an is increasing, then αcn kun − pk2 ≤ (kun+1 − pk2 − kun − pk2 ) − (kun − pk2 − kun−1 − pk2 ). Therefore

+∞

∑ ncn kun − pk2 < +∞

n=1

and by Lemma 8.5.2, kun − pk = o((∑nk=1 kck )

−1 2

).



Theorem 8.5.17 Assume that un is a solution to (8.4), and A satisfies the following condition (weaker than the strong monotonicity of A): ku − vk ≥ αkx − yk,

∀u ∈ Ax, ∀v ∈ Ay

Second Order Difference Equations

181

If either one of the following conditions holds: 1) θn ≥ 1 and lim supn→+∞ nan cn > 0, 2) 0 < θn < 1, ∑+∞ n=1 θ1 · · · θn = +∞ and lim supn→+∞ ncn > 0, then A−1 (0) , ∅ and un → p ∈ A−1 (0) as n → +∞. −1 Proof. In both cases ∑∞ n=1 nan cn = +∞. Theorem 8.5.8 implies that A (0) , ∅. Let p = weak − limn→+∞ zn or σn , as in Theorems 8.5.7 and 8.5.8. It follows from (8.4) that:

αnan cn kun − pk ≤ nan cn kAun k = nan kun+1 − (1 + θn )un + θn un−1 k ≤ nan kun+1 − un k + nan−1 kun − un−1 k

(8.111)

By Part 2 of Proposition 8.5.9 for θn ≥ 1, nan cn kun − pk → 0 as n → +∞. Therefore un → p as n → +∞. Otherwise if 0 < θn < 1, from (8.111), we have αncn kun − pk ≤ nkun+1 − un k + nkun − un−1 k. The result follows now from Part 1 of Proposition 8.5.9.



Theorem 8.5.18 Assume that (I + A)−1 is compact, and let {un }n≥1 be a solution to (8.4). If either 2 2 1) θn ≥ 1, and ∑∞ n=1 nan cn = +∞, or ∞ 2 2) 0 < θn < 1, ∑n=1 θ1 · · · θn = +∞, ∑m ama c → 0 as m → ∞ and ∑∞ n=1 ncn = +∞, k=1 k k

then un converges strongly as n → +∞ to some p ∈ A−1 (0). −1 Proof. 1) If ∑∞ n=1 an cn = +∞, then by Theorem 8.5.7, A (0) , ∅. Otherwise ∞ −1 (0) , ∅. By (8.4), we have na c = +∞ and by Theorem 8.5.8, we have A ∑n=1 n n

cn Aun + (1 + θn )un = un+1 + θn un−1 . Subtracting (1 + θn )p from both sides of the above equation, where p ∈ A−1 (0), and then multiplying by an , we get cn an Aun + (an + an−1 )(un − p) = an (un+1 − p) + an−1 (un−1 − p). Squaring both sides and using the monotonicity of A, we get: c2n a2n kAun k2 ≤ (a2n + an an−1 )(kun+1 − pk2 − kun − pk2 ) + (a2n−1 + an an−1 )(kun−1 − pk2 − kun − pk2 ).

(8.112)

Since θn ≥ 1 (note that in this case an ≤ 1 and an ≤ an−1 ) and kun − pk is nonincreasing, we have c2n a2n kAun k2 ≤ 2a2n (kun+1 − pk2 − kun − pk2 ) − 2a2n−1 (kun − pk2 − kun−1 − pk2 ).

182

Nonlinear Evolution and Difference Equations of Monotone Type

Summing up from n = k to infinity, we get +∞

lim 2a2n (kun+1 − pk2 − kun − pk2 ) − 2a2k−1 (kuk − pk2 − kuk−1 − pk2 ) ∑ c2n a2n kAun k2 ≤ n→+∞

n=k

= 2a2k−1 (kuk−1 − pk2 − kuk − pk2 ).

Now summing up from k = 1 to +∞, we obtain +∞ +∞



∑ ∑ c2n a2n kAun k2 ≤ 2 ∑ a2k−1 (kuk−1 − pk2 − kuk − pk2 ) k=1 n=k

k=1 ∞

≤ 2 ∑ (kuk−1 − pk2 − kuk − pk2 ) < +∞. k=1

This implies that ∞

∑ na2n c2n kAun k2 < +∞.

n=1

Therefore lim infn→+∞ kAun k = 0. Let un j be a subsequence of {un } such that lim j→+∞ kAun j k = 0. Then {un j + Aun j } is bounded. Since (I + A)−1 is compact, there exists a subsequence of {un j }, denoted again by {un j }, such that un j → q. By the monotonicity of A, we have (Aun − Aun j , un − un j ) ≥ 0. Letting j → +∞, we get (Aun , un − q) ≥ 0. This implies that kun − qk is nonincreasing and hence un → q = p. ∞ 2 2 2 2) The assumption ∑∞ n=1 ncn = +∞ implies that ∑n=1 nan cn = +∞. A similar ar−1 −1 gument as in Part 1 implies that A (0) , ∅. Let p ∈ A (0). Now the assumption 0 < θn < 1 (note that in this case an−1 < an and an > 1), together with (8.112) implies that c2n a2n kAun k2 ≤ (an + an−1 ){an (kun+1 − pk2 − kun − pk2 ) − an−1 (kun − pk2 − kun−1 − pk2 )} ≤ 2a2n (kun+1 − pk2 − kun − pk2 ) + 2an an−1 (kun−1 − pk2 − kun − pk2 ) ≤ 2a2n (kun+1 − pk2 − kun − pk2 ) + 2a2n (kun−1 − pk2 − kun − pk2 ). Then c2n kAun k2 ≤ 2(kun+1 − pk2 − kun − pk2 ) + 2(kun−1 − pk2 − kun − pk2 ). 2 2 This implies that ∑+∞ n=1 ncn kAun k < +∞, and the rest of the proof is similar to Part 1 of the proof. 

In the following theorem we prove the strong convergence of solutions without additional assumptions on the monotone operator but we assume A−1 (0) , ∅. Theorem 8.5.19 Assume that un is a solution to (8.4), A−1 (0) , ∅, and ∑∞ n=1 θ1 . . . θn 0, 2) lim supn→+∞ cn > 0, then un → p as n → +∞, where p ∈ A−1 (0); moreover, kun − pk = O(∑∞ k=n θ1 . . . θk ).

Second Order Difference Equations

183

Proof. If kui − pk is nonincreasing, from (8.87), we get kui − ui−1 k2 ≤ kui−1 − pk2 − kui − pk2 .

(8.113)

Otherwise, by Lemma 8.5.5, kui − pk is eventually increasing and by (8.87), we obtain kui+1 − ui k2 ≤ kui+1 − pk2 − kui − pk2 , (8.114) for large i. Summing up (10.14) and (8.114) from i = 1 to ∞, we get ∞

∑ kui+1 − ui k2 < +∞.

(8.115)

i=1

On the other hand, multiplying the inequalities (10.14) and (8.114) by i and taking liminf, by Lemma 8.5.3, we get: lim inf ikui+1 − ui k2 = 0.

(8.116)

i→+∞

Summing up (8.85) from n = k to n = m − 1, we obtain 1 1 θk θm kuk+1 − uk k − kum+1 − um k ≤ kuk − uk−1 k − kum − um−1 k. (8.117) ck cm ck cm √ Taking liminf as m → +∞, if lim infm→+∞ mcm > 0 then, by (8.116), we get kuk+1 − uk k ≤ θk kuk − uk−1 k.

(8.118)

If lim sup cm > 0, then there exists a subsequence cm j > c0 > 0. Substituting m by m j in (8.117) and letting j → +∞, by (9.1.2), we get again (8.118). For each n > m, (8.118) implies n−1

kun − um k = k

n−1

n−1

∑ (uk+1 − uk )k ≤ ∑ kuk+1 − uk k ≤ ku1 − u0 k ∑ θ1 . . . θk . k=m

k=m

(8.119)

k=m

It follows that un is Cauchy, and therefore un → p ∈ H. Now we prove that p ∈ A−1 (0). Suppose that [x, y] ∈ A. By the monotonicity of A and (8.4), we have: (x − un , y) = (x − un , y − Aun ) + (x − un , Aun ) 1 ≥ (x − un , cn Aun ) cn 1 = (x − un , un+1 − (1 + θn )un + θn un−1 ) cn 1 θn = (x − un , un+1 − un ) − (x − un , un − un−1 ) cn c √ √n − n n ≥ ( √ kun+1 − un k − √ kun−1 − un k)(M + kxk), ncn ncn

184

Nonlinear Evolution and Difference Equations of Monotone Type

√ where M := supn≥0 kun k. If lim infn→+∞ ncn > 0, from the above inequality and by (8.116), we get (x − p, y) ≥ 0. If lim supn→+∞ cn > 0, there is a subsequence cn j ≥ c0 > 0; substituting n j for n in the above inequality, letting j → +∞, then using (9.1.2) we get (x − p, y) ≥ 0. Since A is maximal monotone then p ∈ A−1 (0). Now by letting n → +∞ in (8.119), we get ∞

kum − pk ≤ ku0 − u1 k

∑ θ1 · · · θk . k=m

This implies that kum − pk

8.6

= O(∑∞ k=m θ1 · · · θk ).



SUBDIFFERENTIAL CASE

In this section, we consider an important special case where the maximal monotone operator A is the subdifferential of a proper, convex and lower semicontinuous function ϕ. We study the weak convergence of the sequence given by (8.4) to a minimum point of ϕ, as well as the rate of convergence of ϕ(un ) to the minimum value of ϕ, which is important in optimization problems. We also prove the strong convergence of un with additional assumptions on ϕ. In this section we also concentrate on the homogeneous case i.e. fn ≡ 0. Theorem 8.6.1 Suppose that un is a solution to (8.4) with A = ∂ ϕ, where ϕ : H → (−∞, +∞] is a proper, convex and lower semicontinuous function and A−1 (0) , ∅. −1 If θn ≥ 1 and ∑+∞ n=1 ncn an = +∞, then un * p ∈ A (0), which is a minimum point n −1 of ϕ; moreover, ϕ(un ) − ϕ(p) = o((∑i=1 ici ai ) ). Proof. By the subdifferential inequality, Lemma 8.5.5 and (8.4), we have ϕ(ui ) − ϕ(ui−1 ) ≤ (∂ ϕ(ui ), ui − ui−1 ) 1 = (ui+1 − (1 + θi )ui + θi ui−1 , ui − ui−1 ) ci 1 ≤ (ai kui+1 − ui k − ai−1 kui − ui−1 k)kui − ui−1 k ci ai ≤ 0. (8.120) Let p ∈ A−1 (0). Using again the subdifferential inequality and (8.4), we get: ai ci (ϕ(ui ) − ϕ(p)) ≤ ai (ui+1 − (1 + θi )ui + θi ui−1 , ui − p) 1 ≤ ai (kui+1 − pk2 − kui − pk2 ) 2  − ai−1 (kui − pk2 − kui−1 − pk2 ) . Summing up from i = k to m, taking liminf when m → +∞, and then again summing up from k = 1 to +∞, by Lemmas 8.5.3 and 8.5.5, since ai ≤ 1, we obtain: +∞

1

∑ ici ai (ϕ(ui ) − ϕ(p)) ≤ 2 kx − pk2 < +∞.

i=1

(8.121)

Second Order Difference Equations

185

From the assumption in the theorem, it follows that lim infi→+∞ (ϕ(ui ) − ϕ(p)) = 0. Since by (8.120), ϕ(ui ) is nonincreasing, then limi→∞ ϕ(ui ) = ϕ(p). If ui j * q as j → +∞, then lim inf j→+∞ ϕ(ui j ) ≥ ϕ(q). Hence ϕ(p) = limi→+∞ ϕ(ui ) ≥ ϕ(q). This implies that q ∈ A−1 (0). Now the weak convergence of un follows from Opial’s Lemma [OPI]. The rate of convergence of ϕ(un ) to ϕ(p) follows directly from (8.121) and Lemma 8.5.2.  Theorem 8.6.2 Suppose that un is a solution to (8.4) with A = ∂ ϕ, where ϕ : H → (−∞, +∞] is a proper, convex and lower semicontinuous function and A−1 (0) , ∅. If +∞ −1 0 < θn < 1, ∑∞ n=1 θ1 · · · θn = +∞ and ∑n=1 ncn = +∞, then un * p ∈ A (0), which n −1 is a minimum point of ϕ; moreover, ϕ(un ) − ϕ(p) = o((∑i=1 ici ) ). Proof. By (8.120) and Lemma 8.5.5, {ϕ(ui )} is nonincreasing. Let p ∈ A−1 (0). Using again the subdifferential inequality and (8.4), we get:  ai ci (ϕ(ui ) − ϕ(p)) ≤ ai ui+1 − (1 + θi )ui + θi ui−1 , ui − p 1 ≤ ai (kui+1 − pk2 − kui − pk2 ) 2  − ai−1 (kui − pk2 − kui−1 − pk2 ) . Summing up from i = k to m, and taking liminf when m → +∞, by Lemma 8.5.3, we get +∞



ak−1 ∑ ci (ϕ(ui ) − ϕ(p)) ≤ ∑ ai ci (ϕ(ui ) − ϕ(p)) i=k

i=k

1 ≤ ak−1 (kuk−1 − pk2 − kuk − pk2 ) 2 (where in the first inequality we used the assumption that 0 < θi < 1 which implies that ai is increasing). Summing up again from k = 1 to +∞, we obtain +∞

∑ ici (ϕ(ui ) − ϕ(p)) < +∞.

(8.122)

i=1

The assumption in the theorem implies that lim infi→+∞ (ϕ(ui ) − ϕ(p)) = 0. Since by (8.120), ϕ(ui ) is nonincreasing then limi→∞ ϕ(ui ) = ϕ(p). If ui j * q as j → +∞, then lim inf j→+∞ ϕ(ui j ) ≥ ϕ(q). Hence ϕ(p) = limi→+∞ ϕ(ui ) ≥ ϕ(q). This implies that q ∈ A−1 (0). The rest of the proof is similar to that of Theorem 8.6.1.  Theorem 8.6.3 Assume that un is a solution to (8.4) with A = ∂ ϕ, where ϕ : H → (−∞, +∞] is a proper, convex and lower semicontinuous function on H satisfying the following conditions: D(ϕ) = −D(ϕ), ϕ(x) − ϕ(0) ≥ a(kxk)(ϕ(−x) − ϕ(0)), ∀x ∈ D(ϕ),

(8.123)

186

Nonlinear Evolution and Difference Equations of Monotone Type

where a : R+ → (0, 1) is a continuous function. If either one of the following two assumptions is satisfied a) θn ≥ 1 and ∑+∞ n=1 ncn an = +∞, +∞ b) 0 < θn < 1, ∑∞ n=1 θ1 · · · θn = +∞ and ∑n=1 ncn = +∞, then un converges strongly as n → +∞ to some p ∈ A−1 (0), which is a minimum point of ϕ. Proof. Without loss of generality, we may assume that ϕ(0) = 0. By the proof of Theorem 4.10.2 in Chapter 4, we have ϕ(0) ≤ ϕ(x), ∀x ∈ D(ϕ). Then 0 is a minimum point of ϕ, and 0 ∈ (∂ ϕ)−1 (0). Hence by Lemma 8.5.5, kun k is nonincreasing. By the subdifferential inequality and (8.4), we obtain ci (ϕ(ui ) − ϕ(0)) ≤ ci (∂ ϕ(ui ), ui ) = ui+1 − (1 + θi )ui + θi ui−1 , ui



= (ui+1 , ui ) − (1 + θi )kui k2 + θi (ui−1 , ui ) 1 ≤ [kui+1 k2 − kui k2 + θi (kui−1 k2 − kui k2 )]. 2 Now (a) implies (8.121) and (b) implies (8.122) with p = 0. By (8.120), and Lemma 8.5.5, {ϕ(un )} is nonincreasing. On the other hand, by (8.121) and (8.122) and our hypothesis on ci , lim infi→+∞ ϕ(ui ) = 0. Therefore limi→+∞ ϕ(ui ) = 0. For m ≤ n,we define gm := (1 + a(kun k))(kum k2 − kun k2 ) − a(kun k)kun − um k2 . We obtain gm − gm−1 = (1 + a(kun k))(kum k2 − kum−1 k2 ) − a(kun k)(kum − un k2 − kum−1 − un k2 ) = (1 + a(kun k))(um − um−1 , um + um−1 ) − a(kun k)(um − um−1 , um + um−1 − 2un ) = kum k2 − kum−1 k2 + 2a(kun k)(um − um−1 , un ) Similarly, for m < n, we have: gm − gm+1 = kum k2 − kum+1 k2 + 2a(kun k)(um − um+1 , un ). On the other hand, since ϕ is convex, by (8.4) and the assumption on ϕ, for m ≤ n, we get ϕ(um ) ≥ ϕ(un ) ≥ a(kun k)ϕ(−un ) ≥ ϕ(−a(kun k)un )  ≥ ϕ(um ) + ∂ ϕ(um ), −a(kun k)un − um  1 = ϕ(um ) + (1 + θm )um − um+1 − θm um−1 , a(kun k)un + um . cm

Second Order Difference Equations

187

This implies that θm (um − um−1 , a(kun k)un + um ) + (um − um+1 , a(kun k)un + um ) ≤ 0.

(8.124)

From (8.124), we get: (1 + θm )gm − gm+1 − θm gm−1 = θm (gm − gm−1 ) + gm − gm+1 = θm kum k2 − θm kum−1 k2 + 2θm a(kun k)(um − um−1 , un ) + kum k2 − kum+1 k2 + 2a(kun k)(um − um+1 , un ) ≤ θm kum k2 − θm kum−1 k2 + kum k2 − kum+1 k2 − 2θm (um − um−1 , um ) − 2(um − um+1 , um ) = −θm kum k2 − kum k2 − θm kum−1 k2 − kum+1 k2 + 2θm (um−1 , um ) + 2(um+1 , um )  = −θm kum k2 + kum−1 k2 − 2(um−1 , um )  − kum k2 + kum+1 k2 − 2(um+1 , um ) ≤ 0. Hence θm (gm − gm−1 ) ≤ gm+1 − gm .

(8.125)

It is obvious that gn = 0. We prove that gn−1 ≥ 0. Since 0 < a(kun k) < 1, and {un } is nonincreasing, we get gn−1 = (1 + a(kun k))(kun−1 k2 − |un k2 ) − a(kun k)kun−1 − un k2 ≥ kun−1 k2 − kun k2 − kun−1 − un k2 = kun−1 k2 − kun k2 − kun−1 k2 − kun k2 + 2(un−1 , un ) = 2(un , un−1 − un )  cn 1 = 2 un , ∂ ϕ(un ) + (un − un+1 ) θn θn 2 cn = (un , un − un+1 ) + 2 (∂ ϕ(un ), un ) θn θn 2 2 ≥ kun k2 − (un , un+1 ) θn θn 2 1 1 2 ≥ kun k − kun k2 − kun+1 k2 θn θn θn 1 = (kun k2 − kun+1 k2 ) ≥ 0 θn It follows from (8.125) that gm ≥ 0 for all m ≤ n. This implies that kum − un k2 ≤

1 + a(kun k) (kum k2 − kun k2 ) a(kun k)

188

Nonlinear Evolution and Difference Equations of Monotone Type


0, from the continuity of a, we have lim a(kun k) = a( lim kun k) = a(r) > 0.

n→+∞

n→+∞

Therefore {un } is a Cauchy sequence in H, hence un → p ∈ A−1 (0) as n → +∞. 

8.7 8.7.1

ASYMPTOTIC BEHAVIOR FOR THE NON-HOMOGENEOUS CASE MEAN ERGODIC CONVERGENCE

In this section, we study the weak convergence of the weighted average vn := (∑nk=1 θck )−1 (∑nk=1 θck uk ), to a zero of the monotone operator A for the nonhomogek k neous case. Throughout this section, we denote the element c1n (un+1 − (1 + θn )un + θn un−1 − fn ) in H by Aun . We consider the following assumptions on θn , cn and fn : θn is a sequence of positive real numbers satisfying one of the following assumptions: T1 ) θn ≥ 1 and nonincreasing. T2 ) θn is a monotone sequence, and θ := limn→+∞ θn , 0, 1, and +∞. T3 ) θn is a monotone sequence with 0 < θn < 1, and ∑+∞ n=1 (1 − θn ) < +∞. cn is a positive sequence satisfying one of the following assumptions: cn C1 ) ∑∞ n=1 θn = +∞. C2 ) lim infn→+∞ cn > 0. Finally fn is a sequence in H satisfying one of the following assumptions: F1 )∑∞ n=1 k f n k < +∞. k fn k F2 ) ∑∞ n=1 n θn < +∞. Theorem 8.7.1 Assume that {un }n≥1 is a solution to (8.4). If θn , cn and { fn }n≥1 satisfy the assumptions (T1 ), (C1 ) and (F2 ), respectively, then vn converges weakly as n → +∞ to some p ∈ A−1 (0), which is also the asymptotic center of {un }n≥1 . Proof. For all k, n ≥ 0, by the monotonicity of A, we have (Aun , uk ) + (Auk , un ) ≤ (Aun , un ) + (Auk , uk ). Multiplying both sides by cn ck and using by (8.4), we get (un+1 − (1 + θn )un + θn un−1 − fn , ck uk ) + (uk+1 − (1 + θk )uk + θk uk−1 − fk , cn un ) ≤ ck (un+1 − (1 + θn )un + θn un−1 − fn , un )

Second Order Difference Equations

189

+ cn (uk+1 − (1 + θk )uk + θk uk−1 − fk , uk ).

(8.126)

Dividing both sides of the above inequality by θn and rearranging the terms, we obtain cn cn un ) + θk (uk−1 − uk , un ) θn θn ck ck fn cn ≤ (un+1 − un , un ) − (un+1 − un , uk ) + ( , ck uk ) + ( fk , un ) θn θn θn θn fn cn cn − ck ( , un ) − ( fk , uk ) + ck (un−1 − un , un ) + (uk+1 − uk , uk ) θn θn θn cn + θk (uk−1 − uk , uk ). θn (un−1 − un , ck uk ) + (uk+1 − uk ,

cn Summing both sides of this inequality from n = 1 to m and dividing by ∑m n=1 θn , we get: m

cn −1 ) (u0 − um , ck uk ) θ n=1 n

(∑

m

cn −1 m cn  ) ∑ un n=1 θn n=1 θn

+ uk+1 − uk , ( ∑

m

cn −1 m cn  ) ∑ un n=1 θn n=1 θn

+ θk uk−1 − uk , ( ∑ m

 cn −1 m 1 ) ∑ un+1 − un , un − uk n=1 θn n=1 θn

≤ ck ( ∑ m

cn −1 m fn ) ( ∑ , ck uk ) n=1 θn n=1 θn

+(∑

m

cn −1 m cn  ) ∑ un n=1 θn n=1 θn

+ fk , ( ∑ m

cn −1 m fn ) ∑ ( , un ) − ( fk , uk ) n=1 θn n=1 θn

− ck ( ∑ m

cn −1 m 1 1 ) ∑ [ kun−1 k2 − kun k2 ] 2 n=1 θn n=1 2

+ ck ( ∑

+ (uk+1 − uk , uk ) + θk (uk−1 − uk , uk ). Since {vm } is bounded, there exists a subsequence {vm j } of {vm } such that vm j * p; substituting m j by m in the above inequality and letting j → +∞, we get: (uk+1 − uk , p) + θk (uk−1 − uk , p) mj

j→+∞

m

cn −1 j 1 ) ∑ (un+1 − un , un − uk ) n=1 θn n=1 θn

≤ lim sup ck ( ∑

190

Nonlinear Evolution and Difference Equations of Monotone Type mj

m

cn −1 j k fn k ) (∑ )ck kuk k + ( fk , p) n=1 θn n=1 θn

+ lim ( ∑ j→+∞

+ (uk+1 − uk , uk ) − ( fk , uk ) + θk (uk−1 − uk , uk ) mj

m

cn −1 j 1 ) ∑ [(un+1 − uk , un − uk ) − kun − uk k2 ] θ n n=1 n=1 θn

= lim sup ck ( ∑ j→+∞

+ ( fk , p) + (uk+1 − uk , uk ) − ( fk , uk ) + θk (uk−1 − uk , uk ) mj

m

cn −1 j 1 ) ∑ [kun+1 − uk k2 − kun − uk k2 ] θ 2θ n n n=1 n=1

≤ lim sup ck ( ∑ j→+∞

+ ( fk , p) + (uk+1 − uk , uk ) − ( fk , uk ) + θk (uk−1 − uk , uk ) mj

m

cn −1 j 1 1 ) ∑[ kun+1 − uk k2 − kun − uk k2 ] θ 2θ 2θ n n n+1 n=1 n=1

≤ lim ck ( ∑ j→+∞

+ ( fk , p) + (uk+1 − uk , uk ) − ( fk , uk ) + θk (uk−1 − uk , uk ) = ( fk , p) + (uk+1 − uk , uk ) − ( fk , uk ) + θk (uk−1 − uk , uk ) (where in the last inequality we used the assumption T1 ). Therefore, we proved the following inequality: (uk+1 − uk , p) + θk (uk−1 − uk , p) ≤ (uk+1 − uk , uk ) + θk (uk−1 − uk , uk ) + ( fk , p) − ( fk , uk ). (8.127) This implies that θk kuk − uk−1 k2 ≤ (p, uk − uk+1 ) + θk (p, uk − uk−1 ) + ( fk , p) + (uk+1 − uk , uk ) − ( fk , uk ) + θk (uk−1 − uk , uk−1 ) = (uk+1 − uk , uk − p) − θk (uk − uk−1 , uk−1 − p) + ( fk , p) − ( fk , uk ). Dividing both sides of this inequality by θk and using the polarization identity, we get: 1 [(uk+1 − p, uk − p) − kuk − pk2 ] θk − [(uk − p, uk−1 − p) − kuk−1 − pk2 ] 1 − ( fk , uk − p) θk 1 1 ≤ kuk+1 − pk2 − kuk − pk2 2θk 2θk 1 1 − kuk − pk2 − kuk−1 − pk2 2 2 1 fk 2 + kuk − uk−1 k + kuk−1 − pk2 + k kkuk − pk. 2 θk

kuk − uk−1 k2 ≤

Second Order Difference Equations

191

Since {θk }k≥1 is nonincreasing, we deduce that: kuk − uk−1 k2 ≤

1 1 kuk+1 − pk2 − kuk − pk2 θk+1 θk

+ kuk−1 − pk2 − kuk − pk2 + Mk

fk k, θk

where M := 2 supk≥0 kuk − pk. Summing up this inequality from k = 1 to +∞, we get: +∞

∑ kuk − uk−1 k2 < +∞.

(8.128)

k=1

From (8.127), we have  un+1 − (1 + θn )un + θn un−1 − fn , un − p ≥ 0, for all n ≥ 1. It follows that 1 (kun+1 − pk − kun − pk) θn k fn k + θn 1 ≤ (kun+2 − pk − kun+1 − pk) θn θn+1 k fn+1 k k fn k + + θn θn+1 θn ≤ ··· 1 ≤ (kun+m+1 − pk − kun+m − pk) θn · · · θn+m

kun − pk − kun−1 − pk ≤

n+m

+

∑ k=n



k fk k θn · · · θk

1 kun+m+1 − un+m k θn · · · θn+m n+m

+

∑ k=n

k fk k θk

1 ≤ kun+m+1 − un+m k θn n+m k fk k +∑ , k=n θk since θn ≥ 1 for all n ≥ 1. Letting m → +∞, and using (8.128), we obtain +∞

kun − pk − kun−1 − pk ≤

fk

∑ k θk k.

k=n

(8.129)

192

Nonlinear Evolution and Difference Equations of Monotone Type

By Assumption (F2 ), this implies that limn→+∞ kun − pk exists. If q is another cluster point of vn , then there exists limn→+∞ kun − qk. This implies that there exists limn→+∞ (kun − pk2 − kun − qk2 ) and therefore limn→+∞ (un , q − p) exists. It follows that there exists limn→+∞ (vn , p − q), then (p, p − q) = (q, p − q). Therefore p = q, and vn converges weakly to p, as n → +∞. Now we prove that p ∈ A−1 (0). Let [x, y] ∈ A. Then: n

ck −1 n ck ) ∑ (y, x − uk ) k=1 θk k=1 θk

(y, x − vn ) = ( ∑ n

ck −1 n ) ∑ (Auk , x − uk ) k=1 θk k=1

≥ (∑ n

ck −1 n  −1 ) ∑ (uk+1 − x, uk − x) k=1 θk k=1 θk

= (∑

 1 + θk fk kuk − xk2 − (uk−1 − x, uk − x) + ( , uk − x) θk θk n ck −1 n  1 1 ≥ (∑ ) ∑ kuk − xk2 − kuk+1 − xk2 2θk+1 k=1 θk k=1 2θk 1 1 fk  + kuk − xk2 − kuk−1 − xk2 − Mk k 2 2 θk n ck −1  1 1 = (∑ ) ku1 − xk2 − kun+1 − xk2 2θ1 2θn+1 k=1 θk +

n 1 1 fk + kun − xk2 − ku0 − xk2 − M ∑ k k] → 0, 2 2 θ k k=1

as n → +∞, so that we get (y, x − p) ≥ 0 for all [x, y] ∈ A. Since A is maximal monotone, it follows that p ∈ A−1 (0). Finally we prove that p is the asymptotic center of {un }. For x ∈ H, we have kun − pk2 = kun − xk2 + kx − pk2 + 2(un − x, x − p). Multiplying both sides of this equality by θcnn , summing up from n = 1 to n = m, cn dividing by ∑m n=1 θn , and letting m → +∞, we get: lim kun − pk2 ≤ lim sup kun − xk2 − kx − pk2

n→+∞

n→+∞

< lim sup kun − xk2 , if x , p, n→+∞

completing the proof of the theorem.



Theorem 8.7.2 Assume that un is a solution to (8.4). If (T2 ), (C1 ) and (F1 ) are satisfied, then vn * p ∈ A−1 (0), and p is the asymptotic center of {un }.

Second Order Difference Equations

193

Proof. First, assume that θn is nonincreasing. Let vm j * p. By a similar computation as in Theorem 8.7.1, we get: (uk+1 − uk , p) + θk (uk−1 − uk , p) mj

m

cn −1 j 1 1 ) ∑[ kun+1 − uk k2 − kun − uk k2 ] θ 2θ 2θ n n+1 n=1 n n=1

≤ lim sup ck ( ∑ j→+∞

+ ( fk , p) + (uk+1 − uk , uk ) − ( fk , uk ) + θk (uk−1 − uk , uk ).

(8.130)

(In the last inequality, we used the fact that {θn } is nonincreasing). Therefore we obtain (8.127). Again, by a similar computation as in Theorem 8.7.1, we get: 1 1 1 1 kuk+1 − pk2 − kuk − pk2 − kuk − pk2 − kuk−1 − pk2 2θk 2θk 2 2 1 fk 2 2 + kuk − uk−1 k + kuk−1 − pk + k kkuk − pk. 2 θk

kuk − uk−1 k2 ≤

Since {θk }k≥1 is nonincreasing, we deduce that 1 fk kuk − pk2 +kuk−1 − pk2 −kuk − pk2 +2Mk k, θk+1 θk θk (8.131) where M := supk≥0 kuk − pk. Then we get (8.128) and (8.129). It follows that: kuk −uk−1 k2 ≤

1

kuk+1 − pk2 −

Mk fn k + kun+1 − pk2 − kun − pk2 ≥ θn (kun − pk2 − kun−1 − pk2 ) ≥ θn kun − pk2 − θn−1 kun−1 − pk2

(8.132)

Summing up (8.132) from n = k to n = m, we get: m

θk−1 kuk−1 − pk2 − kuk − pk2 ≥ θm kum − pk2 − kum+1 − pk2 − M ∑ k fn k. (8.133) n=k

First, suppose that θ > 1. Taking lim sup when m → +∞ and then lim inf when k → +∞ in (8.133), and using (F1 ), we get: (θ − 1) lim inf kun − pk2 ≥ (θ − 1) lim sup kun − pk2 . n→+∞

n→+∞

Hence there exists limn→+∞ kun − pk. If θ < 1, then from (8.133), we get: m

kum+1 − pk2 − θm kum − pk2 ≥ kuk − pk2 − θk−1 kuk−1 − pk2 − M ∑ k fn k (8.134) n=k

Now, first taking liminf when m → +∞ and then limsup when k → +∞ in (8.134), we get: (1 − θ ) lim inf kun − pk2 ≥ (1 − θ ) lim sup kun − pk2 , (8.135) n→+∞

n→+∞

194

Nonlinear Evolution and Difference Equations of Monotone Type

which implies again that there exists limn→+∞ kun − pk. If q is another cluster point of vn , then there exists limn→+∞ kun − qk. This implies that there exists limn→+∞ (kun − pk2 − kun − qk2 ) and therefore limn→+∞ (un , q − p) exists. It follows that there exists limn→+∞ (vn , p − q), hence (p, p − q) = (q, p − q), and thus p = q. Therefore vn * p as n → +∞. Now we prove that p ∈ A−1 (0). Let [x, y] ∈ A; then: n

ck −1 n ck ) ∑ (y, x − uk ) k=1 θk k=1 θk

(y, x − vn ) = ( ∑ n

ck −1 n ) ∑ (Auk , x − uk ) k=1 θk k=1

≥ (∑ n

ck −1 n  −1 1 + θk ) ∑ (uk+1 − x, uk − x) + kuk − xk2 θ θ θk k k=1 k k=1  fk − (uk−1 − x, uk − x) + ( , uk − x) θk n ck −1 n  1 1 ≥ (∑ ) ∑ kuk − xk2 − kuk+1 − xk2 θ 2θ 2θ k k+1 k=1 k k=1 1 1 fk  + kuk − xk2 − kuk−1 − xk2 − (M + kx − pk)k k 2 2 θk n ck −1  1 1 ≥ (∑ ) ku1 − xk2 − kun+1 − xk2 2θ1 2θn+1 k=1 θk = (∑

 1 1 M + kx − pk n + kun − xk2 − ku0 − xk2 − k fk k → 0. ∑ 2 2 θ k=1

(8.136)

By letting n → +∞, we get (y, x − p) ≥ 0 for all [x, y] ∈ A. Since A is maximal monotone, this completes the proof for the case {θn } nonincreasing. Now suppose that {θn } is nondecreasing. In the above proof, the assumption {θn } nonincreasing was used in (8.130), (8.131), (8.132) and (8.136). In the last inequality of (8.130), if {θn } is nondecreasing, then: m

mj

cn −1 j 1 ) ∑ [kun+1 − uk k2 − kun − uk k2 ] θ 2θ n n n=1 n=1

lim sup ck ( ∑ j→+∞

+ ( fk , p) + (uk+1 − uk , uk ) − ( fk , uk ) + θk (uk−1 − uk , uk ) mj

m

cn −1 j 1 1 ) ∑[ kun+1 − uk k2 − kun − uk k2 ] θ 2θ 2θ n n n+1 n=1 n=1

≤ lim sup ck ( ∑ j→+∞

mj

m

cn −1 j 1 1 ) ∑( − )kun+1 − uk k2 θ 2θ 2θ n n n+1 n=1 n=1

+ lim sup ck ( ∑ j→+∞

+ ( fk , p) + (uk+1 − uk , uk ) − ( fk , uk ) + θk (uk−1 − uk , uk ).

Second Order Difference Equations

195

1 Since θ1n − θn+1 > 0, and un is bounded, (8.127) follows. In (8.131), if {θn } is nondecreasing, we get:

1 1 1 1 kuk+1 − pk2 − kuk − pk2 + ( − )kuk+1 − pk2 θk+1 θk θk θk+1 1 + kuk−1 − pk2 − kuk − pk2 + 2M k fk k θk

kuk − uk−1 k2 ≤

Since {θn } is nondecreasing and un is bounded, we get (8.128) by summing up the above inequality from n = 1 to ∞. In (8.132) too {θn } nonincreasing was used. When {θn } is nondecreasing, from (8.129) we get: Mk fn k + kun+1 − pk2 − kun − pk2 ≥ θn (kun − pk2 − kun−1 − pk2 ) ≥ θn kun − pk2 − θn−1 kun−1 − pk2 + M 2 (θn−1 − θn ) Now the rest of the proof for the weak convergence of vn is similar to the case where θn is nonincreasing, because limn→∞ θn = θ . In the last two inequalities of (8.136) too, we used {θn } nonincreasing. If {θn } is nondecreasing, we get: n

ck −1 n  −1 1 + θk ) ∑ (uk+1 − x, uk − x) + kuk − xk2 − (uk−1 − x, uk − x) θ θ θ k k k k=1 k=1  fk + ( , uk − x) θk n n  ck 1 1 ≥ ( ∑ )−1 ∑ kuk − xk2 − kuk+1 − xk2 θ 2θ 2θ k k k+1 k=1 k=1 (∑

1 1 1 1 − )kuk+1 − xk2 + kuk − xk2 − kuk−1 − xk2 2θk+1 2θk 2 2 fk  − (M + kx − pk)k k → 0, θk +(

since un is bounded and θ 1 − θ1 < 0. Finally, a similar proof as in Theorem 8.7.1 k+1 k shows that p is the asymptotic center of the sequence {un }. The proof is now complete.  Theorem 8.7.3 Assume that un is a solution to (8.4). If (T3 ), (C1 ) and (F2 ) are satisfied, then vn * p ∈ A−1 (0), and p is the asymptotic center of {un }. Proof. Again (8.129) holds, and therefore summing (8.132) from n = k to n = m, and letting m → +∞, we get: kuk − pk2 ≤ (1 − θ ) lim inf kun − pk2 + θk−1 kuk−1 − pk2 n→+∞



+ M ∑ k fi k + M 2 (θ − θk−1 ). i=k

196

Nonlinear Evolution and Difference Equations of Monotone Type

Since θ = 1, we get: ∞

kun − pk2 ≤ θn−1 kun−1 − pk2 + M ∑ k fi k + M 2 (1 − θn−1 ) i=n

2

≤ kun−1 − pk + εn , with ∑∞ n=1 εn < +∞, by (T3 ) and (F2 ). It follows that there exists limn→+∞ kun − pk. The rest of the proof is similar to that of Theorem 8.7.1.  8.7.2

WEAK CONVERGENCE OF SOLUTIONS

Theorem 8.7.4 Let {un }n≥1 be a solution to (8.4). Assume that (T2 ) and (F1 ) are satisfied, and cn ≥ c > 0 for all n ≥ 1. Then un * p ∈ A−1 (0) as n → +∞. Proof. Dividing both sides of (8.126) by cn and using the Cauchy-Schwarz inequality, we get: (uk+1 − uk , un ) + θk (uk−1 − uk , un ) ck θn ck ≤ kun+1 − un kkuk k + kun−1 − un kkuk k cn cn ck ck + k fn kkuk k + ( fk , un ) + (un+1 − un , un ) cn cn ck ck + θn (un−1 − un , un ) + k fn kkun k cn cn + (uk+1 − uk , uk ) + θk (uk−1 − uk , uk ) − ( fk , uk ). cn Since in this case also ∑+∞ n=1 θn = +∞, by (8.128), we have kun − un−1 k → 0 as n → +∞; assume that un j * p. Replacing n by n j in the above inequality and letting j → +∞, we get:

(uk+1 − uk , p) + θk (uk−1 − uk , p) ≤ (uk+1 − uk , uk ) + θk (uk−1 − uk , uk ) + ( fk , p) − ( fk , uk ) This implies (8.129). By the same proof as in Theorem 8.7.2, we deduce that limn→+∞ kun − pk exists, and then un * p ∈ A−1 (0) as n → +∞.  Theorem 8.7.5 Assume that un is a solution to (8.4). If (T2 ), (C2 ) and (F1 ) are satisfied, then un * p ∈ A−1 (0). cn Proof. Since in this case, we have ∑+∞ n=1 θn = +∞, then (8.128) holds, and therefore, we have kun − un−1 k → 0 as n → +∞. Hence kAun k → 0 as n → +∞. Assume that un j * p; then we get (8.129). Now the same proof as in Theorem 8.7.2 shows that limn→+∞ kun − pk exists. The rest of the proof is now similar to that of Theorem 8.7.2. 

Theorem 8.7.6 Assume that un is a solution to (8.4). If (T3 ), (C2 ) and (F2 ) are satisfied, then un * p ∈ A−1 (0).

Second Order Difference Equations

197

Proof. The proof of Theorem 8.7.3 shows that (8.129) holds here. Now the rest of the proof is similar to that of Theorem 8.7.1.  8.7.3

STRONG CONVERGENCE OF SOLUTIONS

Theorem 8.7.7 Assume that the operator A in (8.4) is strongly monotone, and let {un }n≥1 be a solution to (8.4). If (T1 ), (C1 ) and (F2 ) are satisfied, then un converges strongly as n → +∞ to some p ∈ A−1 (0). Proof. Assume that (y2 − y1 , x2 − x1 ) ≥ αkx2 − x1 k2 , for all [xi , yi ] ∈ A, i = 1, 2, and for some α > 0. Then by using the strong monotonicity of A, a similar proof as in Theorem 8.7.1 gives the following inequalities: k

ci −1 k ci ) ( ∑ ui )) i=1 θi i=1 θi

(Aun , un − vk ) = (Aun , un − ( ∑ k

ci −1 k ci ) ∑ (Aun , un − ui ) i=1 θi i=1 θi

= (∑ k

 ci −1 k ci  ) ∑ (Aun − Aui , un − ui ) + (Aui , un − ui ) i=1 θi i=1 θi

= (∑ k

ci −1 k ci ) ∑ αkun − ui k2 θ i i=1 i=1 θi

≥ (∑

k

ci −1 k 1 ) ∑ (ui+1 − (1 + θi )ui + θi ui−1 − fi , un − ui ) i=1 θi i=1 θi

+ (∑ k

ci −1 k ci ) ∑ kun − ui k2 θ i i=1 i=1 θi

= α( ∑ k

ci −1 k 1  ) ∑ (ui+1 − un , un − ui ) i=1 θi i=1 θi

+ (∑

 + (1 + θi )kun − ui k2 + θi (ui−1 − un , un − ui ) − ( fi , un − ui ) k

ci −1 k ci ) ∑ kun − ui k2 θ i i=1 i=1 θi

≥ α( ∑ k

ci −1 k  1 1 ) ∑ − kui+1 − un k2 − kui − un k2 θ 2θ 2θ i i i i=1 i=1

+ (∑

(1 + θi ) 1 1 kun − ui k2 − kui−1 − un k2 − kui − un k2 θi 2 2  k fi k − kui − un k θi +

k

ci −1 k ci ) ∑ kun − ui k2 θ i=1 i i=1 θi

≥ α( ∑

198

Nonlinear Evolution and Difference Equations of Monotone Type k

ci −1 k 1 1 ) ∑[ kun − ui k2 − kun − ui+1 k2 θ 2θ 2θ i i i+1 i=1 i=1

+ (∑

1 1 k fi k + kui − un k2 − kui−1 − un k2 − kui − un k ]. 2 2 θi (where in the last inequality, we used the assumption that {θn } is nonincreasing). Letting k → +∞, by Theorem 8.7.2 and the Cauchy-Schwarz inequality, we get: k

ci −1 k ci ) ∑ kun − ui k2 i=1 θi i=1 θi

(Aun , un − p) ≥ α lim inf( ∑ k→+∞

k

ci −1 k ci ) ∑ kun − ui k]2 i=1 θi i=1 θi

≥ α[lim inf( ∑ k→+∞

≥ α[lim inf kun − vk k]2 ≥ αkun − pk2 . k→+∞

Multiplying both sides of (8.137) by

cn θn

(8.137)

and using (8.4), we get:

1 cn (un+1 − (1 + θn )un + θn un−1 − fn , un − p) ≥ α kun − pk2 . θn θn It follows that α

cn 1 (1 + θn ) kun − pk2 ≤ (un+1 − p, un − p) − kun − pk2 θn θn θn fn + (un−1 − p, un − p) − ( , un − p) θn 1 1 ≤ kun+1 − pk2 − kun − pk2 2θn+1 2θn 1 1 fn + kun−1 − pk2 − kun − pk2 + Mk k, 2 2 θn

where M := supn≥0 kun − pk. Summing up both sides of this inequality from n = 1 to m and letting m → +∞, we get +∞

cn

∑ θn kun − pk2 < +∞.

n=1

This implies that lim inf kun − pk2 = 0. n→+∞

Since by Theorem 8.7.2, limn→+∞ kun − pk exists, then un → p ∈ A−1 (0) as n → +∞.  Theorem 8.7.8 Assume that {un }n≥1 is a solution to (8.4), (I + A)−1 is compact and 2

cn { fn }n≥1 satisfies (F2 ). If ∑+∞ n=1 (1+θn )2 = +∞, and (T1 ) or (T3 ) is satisfied, then un converges strongly as n → +∞ to some p ∈ A−1 (0).

Second Order Difference Equations

199

cn Proof. Since in this case ∑+∞ n=1 θn = +∞, by Theorems 8.7.1 and 8.7.3 we already know that vn converges weakly as n → +∞ to some p ∈ A−1 (0), and limn→+∞ kun − pk exists. From (8.4), we have

un+1 − p + θn (un−1 − p) ∈ cn Aun + fn + (1 + θn )(un − p).

(8.138)

Squaring both sides of (8.138), we obtain kun+1 − pk2 + θn2 kun−1 − pk2 + 2θn (un+1 − p, un−1 − p) = c2n kAun k2 + k fn k2 + (1 + θn )2 kun − pk2 + 2cn (Aun , fn ) + 2cn (1 + θn )(Aun , un − p) + 2(1 + θn )( fn , un − p). Dividing both sides of the above inequality by (1 + θn )2 and using (8.4) and (8.129), we get: c2n 1 θn kAun k2 ≤ kun+1 − pk2 + kun−1 − pk2 − kun − pk2 (1 + θn )2 1 + θn 1 + θn 2kun+1 − pk 2θn kun−1 − pkk fn k fn 2 + k fn k + +k k (1 + θn )2 (1 + θn )2 1 + θn 1 ≤ (kun+1 − pk2 − kun − pk2 ) 1 + θn θn fn fn + (kun−1 − pk2 − kun − pk2 ) + 4Mk k + k k2 1 + θn θn θn 1 1 θn−1 ≤ kun+1 − pk2 − kun − pk2 + kun−1 − pk2 1 + θn+1 1 + θn 1 + θn−1 θn fn fn − kun − pk2 + 4Mk k + k k2 . 1 + θn θn θn Summing up both sides of the above inequality from n = 1 to m and letting m → +∞, we get: +∞ c2 ∑ (1 + nθn )2 kAun k2 < +∞. n=1 Therefore lim infn→+∞ kAun k = 0. It follows that there exists a subsequence {un j } of {un } such that lim j→+∞ kAun j k = 0. Then {un j + Aun j } is bounded. Since (I + A)−1 is compact, there exists a subsequence of {un j }, denoted again by {un j }, such that un j → q. By the monotonicity of A, we have (Auk − Aun j , uk − un j ) ≥ 0. Letting j → +∞, we get (Auk , uk − q) ≥ 0. Multiplying both sides of this inequality by ck , we get (8.129) with p replaced by q. Then we conclude that limn→+∞ kun − qk exists. Therefore un → q ∈ A−1 (0). 

200

Nonlinear Evolution and Difference Equations of Monotone Type

In the following theorem we assume that A−1 (0) is nonempty. Definition 8.7.9 Let P : H → A−1 (0) be the metric projection onto the (closed and convex) zero set of A. The operator A is said to satisfy the “convergence condition” if [xn , yn ] ∈ A, kxn k ≤ K, kyn k ≤ K for a positive constant K, and limn→+∞ (yn , xn − Pxn ) = 0 imply that limn→+∞ (xn − Pxn ) = 0. Theorem 8.7.10 Suppose that A satisfies the “convergence condition” and {un }n≥1 is a solution to (8.4). If the conditions (T1 ), (C1 ) and (F2 ) are satisfied, then un converges strongly to a zero of A. Proof. We have: cn (Aun , un − Pun ) θn un+1 − un fn =( + un−1 − un − , un − Pun ) θn θn 1 1 = (un+1 − Pun+1 , un − Pun ) + (Pun+1 − Pun , un − Pun ) θn θn 1 2 − kun − Pun k + (un−1 − Pun−1 , un − Pun ) θn fn + (Pun−1 − Pun , un − Pun ) − kun − Pun k2 − ( , un − Pun ) θn 1 1 1 ≤ kun+1 − Pun+1 k2 + kun − Pun k2 − kun − Pun k2 2θn 2θn θn 1 1 fn + kun−1 − Pun−1 k2 + kun − Pun k2 − kun − Pun k2 + k kkun − Pun k 2 2 θn 1 1 1 1 ≤ ( kun+1 − Pun+1 k2 − kun − Pun k2 ) + (kun−1 − Pun−1 k2 2 θn+1 θn 2 f n − kun − Pun k2 ) + Mk k θn

0≤

where M := supn≥0 kun − Pun k. Summing up the above inequality from n = 1 to m and letting m → +∞, we get: +∞

cn

∑ θn (Aun , un − Pun ) < +∞

n=1

It follows that lim infn→+∞ (Aun , un − Pun ) = 0; hence by the convergence condition, there exists a subsequence {unk } of {un } such that limk→+∞ (unk − Punk ) = 0. By a similar proof as in Theorem 8.7.1, we have: +∞

kun+1 − Pun+1 k ≤ kun+1 − Pun k ≤ kun − Pun k +



i=n+1

k

fi k θi

Second Order Difference Equations

201

Thus, limn→+∞ kun − Pun k exists, and we have un − Pun → 0 as n → +∞. Since kun+m − un k ≤ kun+m − Pun k + kun − Pun k +∞

≤ 2kun − Pun k +



(i − n)k



ik

i=n+1 +∞

≤ 2kun − Pun k +

i=n+1

fi k θi

fi k→0 θi

as n → +∞, uniformly in m ≥ 0, we conclude that {un } converges strongly to a zero of A.  Theorem 8.7.11 Let un be a solution to (8.4). If (T2 ) holds with limn→+∞ θn := θ < ci k fn k n 1, lim supn→+∞ cn > 0 and ∑∞ n=1 (∑i=1 θ i ) cn < +∞, then un → p as n → +∞, where p ∈ A−1 (0). Proof. By the monotonicity of A, we have (Aun+1 − Aun , un+1 − un ) ≥ 0 By (8.4), we get: 1

θn+1 1 kun+1 − un k2 − kun+1 − un k2 cn+1 cn+1 cn θn fn+1 fn + (un − un−1 , un+1 − un ) − ( − , un+1 − un ) ≥ 0. cn cn+1 cn (un+2 − un+1 , un+1 − un ) −

This implies that θn 1 θn+1 1 kun+1 − un k − kun − un−1 k ≤ kun+2 − un+1 k − kun+1 − un k cn cn cn+1 cn+1 fn+1 fn +k − k. cn+1 cn Summing up the above inequality from n = k to n = m − 1, we get: 1 1 θk θm kuk+1 − uk k − kum+1 − um k ≤ kuk − uk−1 k − kum − um−1 k ck cm ck cm m fn + 2 ∑ k k. c n n=k cn Since lim supn→+∞ cn > 0, we have ∑∞ n=1 θn = +∞; on the other hand, for large n, and we have: ∞ ∞ n k fn k cn k fn k ci k fn k =∑ n ≤ ∑ (∑ i ) < +∞. n θ θ c θ cn n n=1 n=1 n=1 i=1 ∞



k fn k θn



k fn k θn

202

Nonlinear Evolution and Difference Equations of Monotone Type

k fn k This implies that ∑∞ n=1 θn < +∞, and hence (F1 ) holds. In a similar way as in Theorem 8.7.4, (8.128) holds, and we have limn→+∞ kun+1 − un k = 0. On the other hand, there exists a subsequence cn j of cn and ε > 0 such that cn j ≥ ε for all j ≥ 1. Substituting m by n j in the above inequality, and letting j → +∞, we get: ∞

kuk+1 − uk k ≤ θk kuk − uk−1 k + 2ck

fn

∑ k cn k.

n=k

This implies that k

kuk+1 − uk k ≤ λ k ku1 − u0 k + ∑ λ k−i ci ai i=1

fn where θ < λ < 1 and ai = 2 ∑∞ n=i k cn k. Then we get: n−1

kun − um k = k

n−1

∑ (uk+1 − uk )k ≤ ∑ kuk+1 − uk k k=m

k=m n−1



k

∑ (λ k ku1 − u0 k + ∑ λ k−i ci ai ) i=1

k=m

= ≤

λm −λn 1−λ λm −λn 1−λ

n−1

ku1 − u0 k +

k=m n−1

ku1 − u0 k +

k

ci ai i i=1 λ

∑ λk ∑ k

ci ai . i i=1 θ

∑ λk ∑ k=m

Since by assumption we have: k

∞ ∞ n ci ai ci ∞ fn ci fn ≤ 2 ∑ i ∑ k k = 2 ∑ ( ∑ i )k k < +∞, i θ θ c θ cn n n=i i=1 i=1 n=1 i=1



it follows that un is Cauchy, and therefore un → p ∈ H. Now we prove that p ∈ A−1 (0). Suppose that [x, y] ∈ A. By the monotonicity of A and (8.4), we have: (x − un , y) = (x − un , y − Aun ) + (x − un , Aun ) ≥

1 (x − un , cn Aun ) cn

1 (x − un , un+1 − (1 + θn )un + θn un−1 − fn ) cn 1 θn 1 = (x − un , un+1 − un ) − (x − un , un−1 − un ) − (x − un , fn ) cn cn cn −1 θn ≥ kx − un kkun+1 − un k − kx − un kkun−1 − un k cn cn k fn k − kx − un k. cn =

Second Order Difference Equations Let’s show that

k fn k cn

→ 0 as n → +∞. Since

203 cn θn


0, then (Λ2 ) is satisfied and α ∞ αn2 kun − un−1 k2 ≤ 2 ∑ αn kun − un−1 k2 < +∞, 2 λ n=1 λn+1 n=1 ∞



Discrete Nonlinear Oscillator Dynamical System and the Inertial

215

by (α1 ) and (α2 ). Then (α5 ) is also satisfied.

9.4

SUBDIFFERENTIAL CASE

In this section, we establish the weak convergence of the sequence {un } generated by (9.4) to an element of A−1 (0), when A = ∂ ϕ. Lemma 9.4.1 Suppose that {un } is a bounded sequence generated by (9.4) and A = ∂ ϕ, where ϕ : H → (−∞, +∞] is a proper, convex and lower semicontinuous function. If the conditions +∞

+∞ αn2 kun − un−1 k2 ken k2 < +∞ and (E5 ) ∑ < +∞. λn+1 n=1 n=1 λn+1

(α7 ) ∑

are satisfied, then there exists limn→+∞ ϕ(un ). Proof. By the subdifferential inequality and (9.4), we get λn+1 (ϕ(un+1 ) − ϕ(un )) ≤ (λn+1 ∂ ϕ(un+1 ), un+1 − un ) = (un − un+1 + αn (un − un−1 ) + en , un+1 − un ) = −kun+1 − un k2 + (αn (un − un−1 ) + en , un+1 − un ) 1 ≤ kαn (un − un−1 ) + en k2 2 ≤ αn2 kun − un−1 k2 + ken k2 . Hence ϕ(un+1 ) ≤ ϕ(un ) +

αn2 ken k2 kun − un−1 k2 + . λn+1 λn+1

It follows from the assumptions (α7 ), (E5 ) that there exists limn→+∞ ϕ(un ).

(9.11) 

The following proposition is the subdifferential version of Proposition 9.3.2. Proposition 9.4.2 Let {un } be a bounded sequence generated by (9.4) and A = ∂ ϕ. If (α7 ), (E5 ), and the following conditions hold:  +∞  (Λ1 ) ∑n=1 λn = +∞, αn kun −un−1 k → 0, (α3 ) ∑+∞ (9.12) n=1 αn kun − un−1 k < +∞ or (α4 ) λn+1   ken k +∞ (E1 ) ∑n=1 ken k < +∞ or (E2 ) λ → 0, n+1

then lim ϕ(un ) = infx∈H ϕ, (∂ ϕ)−1 (0) , ∅ and ωw (un ) ⊂ (∂ ϕ)−1 (0). Proof. By Proposition 9.2.2, (∂ ϕ)−1 (0) , ∅. Assume that p ∈ (∂ ϕ)−1 (0); then we have: λn+1 (ϕ(un+1 ) − ϕ(p)) ≤ (λn+1 ∂ ϕ(un+1 ), un+1 − p)

216

Nonlinear Evolution and Difference Equations of Monotone Type = (un − un+1 + αn (un − un−1 ) + en , un+1 − p) = (un − p, un+1 − p) + (p − un+1 , un+1 − p) + (αn (un − un−1 ), un+1 − p) + (en , un+1 − p).

So, 1 1 λn+1 (ϕ(un+1 ) − ϕ(p)) ≤ kun − pk2 − kun+1 − pk2 + 2 2 αn kun − un−1 kkun+1 − pk + ken kkun+1 − pk.

(9.13)

Summing up both sides of (9.13) from n = 1 to n = k and dividing by ∑kn=1 λn+1 , then letting k → ∞, since by Lemma 9.4.1, limn→+∞ ϕ(un ) − ϕ(p) = l exists, then k

lim ( ∑ λn+1 )−1

k→+∞ n=1

k

lim (ϕ(un ) − ϕ(p)) = l. ∑ λn+1 (ϕ(un+1 ) − ϕ(p)) = k→+∞

n=1

It follows now from the assumptions that l = 0. Thus limk ϕ(uk ) = ϕ(p). Therefore, if un j * q, then ϕ(q) ≤ lim inf j ϕ(un j ) = ϕ(p), which implies that q ∈ (∂ ϕ)−1 (0). Hence ωw (un ) ⊂ (∂ ϕ)−1 (0).  The following theorem shows that in the special case A = ∂ ϕ, which is important from the optimization point of view, the conditions (Λ2 ) and (α5 ) in Theorem 9.3.4 can be replaced by the weaker conditions (Λ1 ) and (α7 ). Theorem 9.4.3 Let {un } be a bounded sequence generated by (9.4) and A = ∂ ϕ. Assume that the conditions (Λ1 ), (α1 ), (α2 ), (α7 ), (E1 ) and 2

ken k (E5 ) ∑+∞ < +∞ are satisfied. Then un * p ∈ (∂ ϕ)−1 (0) as n → ∞. n=1 λ n+1

Proof. By Proposition 9.2.2, A−1 (0) , ∅. Assume that p ∈ (∂ ϕ)−1 (0); then we have: λn+1 (ϕ(un+1 ) − ϕ(p)) ≤ (λn+1 ∂ ϕ(un+1 ), un+1 − p) = (un − un+1 + αn (un − un−1 ) + en , un+1 − p) = (p − un+1 , un+1 − p) + (un − p + αn (un − un−1 ) + en , un+1 − p) 1 1 ≤ − kun+1 − pk2 + kun − p + αn (un − un−1 ) + en k2 2 2 1 1 α2 = − kun+1 − pk2 + kun − pk2 + n kun − un−1 k2 2 2 2 1 + ken k2 2 + (un − p, en ) + αn (un − un−1 , un − p) + (αn (un − un−1 ), en )

Discrete Nonlinear Oscillator Dynamical System and the Inertial

217

1 1 αn ≤ kun − pk2 − kun+1 − pk2 + 3 kun − un−1 k2 2 2 2 αn αn 2 2 + kun − pk − kun−1 − pk + ken k2 2 2 + ken kkun − pk. So 1 1 αn λn+1 (ϕ(un+1 ) − ϕ(p)) ≤ kun − pk2 − kun+1 − pk2 + 3 kun − un−1 k2 2 2 2 α 2 2 + [kun − pk − kun−1 − pk ]+ 2 + ken k2 + ken kkun − pk. (9.14) On the other hand, by Lemma 9.4.1, there exists limn→+∞ (φ (un ) − φ (p)). Let limn→+∞ (φ (un ) − φ (p)) = l. Summing up both sides of (9.14) from n = 1 to n = k, dividing by ∑kn=1 λn+1 , and letting k → +∞, we get: k

lim ( ∑ λn+1 )−1

k→+∞ n=1

k

∑ λn+1 (φ (un ) − φ (p)) = l.

n=1

It follows now from Proposition 9.2.1 (b) and the assumptions on {λn }, {αn } and {en } that l = 0. Therefore limn→+∞ φ (un ) = φ (p). Now if un j * q, then ϕ(q) ≤ lim inf j ϕ(un j ) = ϕ(p), which implies that q ∈ (∂ ϕ)−1 (0). Hence ωw (un ) ⊂ (∂ ϕ)−1 (0). The rest of the proof is now similar to that of Theorem 9.2.4.  Remark 9.4.4 Obviously (Λ2 ) ⇒ (Λ1 ); on the other hand, by (α1 ), (α2 ) and (α5 ), we have: +∞

+∞ 2 1 αn kun − un−1 k2 1 +∞ 2 αn2 kun − un−1 k2 ≤ ( ) 2 ( ∑ αn kun − un−1 k2 ) 2 ∑ ∑ 2 λn+1 λn+1 n=1 n=1 n=1 1

+∞

1 αn2 kun − un−1 k2 1 +∞ ) 2 ( ∑ αn kun − un−1 k2 ) 2 < +∞. 2 λn+1 n=1 n=1

≤ α 2(∑

Then (α7 ) is satisfied. This shows that, in the subdifferential case and when en ≡ 0, the weak convergence of un is proved with weaker assumptions on the parameters, than in Theorem 9.3.4.

9.5

STRONG CONVERGENCE

In this final section, with additional assumptions on the maximal monotone operator A, we show the strong convergence of the bounded sequence {un } generated by (9.4) to a zero of A. Theorem 9.5.1 Assume that {un } is a bounded sequence generated by (9.4). If (I + A)−1 is a compact operator and the conditions (Λ2 ), (α1 ), (α3 ) and (E1 ) are satisfied, then un → p ∈ A−1 (0).

218

Nonlinear Evolution and Difference Equations of Monotone Type

Proof. By (9.10) and the assumptions, we get: lim infn kAun k = 0. Therefore there exists a subsequence {Aun j } of {Aun } such that kAun j k → 0 and {un j + Aun j } is bounded. Since (I + A)−1 is compact, {un j } has a strongly convergent subsequence (denoted again by {un j }) to some p ∈ H. The maximality of A implies that p ∈ A−1 (0). On the other hand, by part (b) of Proposition 9.2.1, limn kun − pk exists. Hence un → p ∈ A−1 (0).  Theorem 9.5.2 Let {un } be a bounded sequence generated by (9.4) and A be a maximal monotone and strongly monotone operator. If (Λ1 ), (α3 ) and (E1 ) are satisfied, then un → p, where p is the unique element of A−1 (0). Proof. By Proposition 9.2.2, A−1 (0) , ∅. Assume that p is the single element of A−1 (0). By the strong monotonicity of A and (9.4), we get: β λn+1 kun+1 − pk2 ≤ (un − un+1 + αn (un − un−1 ) + en , un+1 − p), for some β > 0. It follows that 2β λn+1 kun+1 − pk2 ≤ kun − pk2 − kun+1 − pk2 + 2αn kun − un−1 kkun+1 − pk + 2ken kkun+1 − pk. Now summing up both sides of the above inequality from n = 1 to n = k, dividing by ∑kn=1 λn+1 , and letting k → ∞, since by Proposition 9.2.1 (c), we know that limn→∞ kun − pk = l exists, it follows that the left hand side is equal to l. Using the assumptions and Lemma 9.1.1, it follows that the right hand side is equal to zero. Therefore l = 0, and the proof is now complete. 

REFERENCES ALV-ATT. F. Alvarez and H. Attouch, An inertial proximal method for maximal monotone operators via discretization of a nonlinear oscillator with damping, Set-Valued Anal. 9 (2001), 3–11. JUL-MAI. F. Jules, P. E. Maing´e, Numerical approach to a stationary solution of a second order dissipative dynamical system, Optimization 51 (2002), 235–255. KHA-RAN. H. Khatibzadeh and S. Ranjbar, Inexact inertial proximal algorithm for maximal monotone operators. An. S¸tiint¸. Univ. “Ovidius” Constant¸a Ser. Mat. 23 (2015), 133–146. MAI. P. E. Maing´e, Convergence theorems for inertial KM-type algorithms, J. Comput. Appl. Math. 219 (2008), 223–236.

Part IV Applications

Applications to 10 Some Nonlinear Partial Differential Equations and Optimization 10.1

INTRODUCTION

In this final chapter of the book, we present some applications of the first and second order evolution equations of monotone type to minimization problems, variational problems and partial differential equations.

10.2

APPLICATIONS TO CONVEX MINIMIZATION AND MONOTONE OPERATORS

Consider the convex minimization problem: Minx∈H ϕ(x),

(10.1)

where ϕ : H → (−∞, +∞] is a convex, proper and lower semicontinuous function. There are several methods for solving the above minimization problem. In this section, we consider some dynamic methods by using the solutions to first and second order evolution equations governed by the subdifferential of the convex function ϕ. Consider the first order system: ( −u0 (t) ∈ A(u(t)), t ∈ (0, +∞), (10.2) u(0) = u0 ∈ D(A). As we saw in Chapter 4 of the book, solutions to (10.2) are not necessarily weakly convergent for a general monotone operator A. Bruck [BRU] proved the weak convergence of the solutions for a demipositive operator A, which includes the case when A = ∂ ϕ, where ϕ is a proper, convex and lower semicontinuous function. In Chapter 5, we saw that the solutions to the following second order system: ( u00 (t) ∈ Au(t), t ∈ (0, +∞), (10.3) u(0) = u0 ∈ D(A), supt≥0 ku(t)k < +∞, converge weakly for a general maximal monotone operator A. Then (10.2) and (10.3) provide dynamical approaches for finding a minimizer of the convex function ϕ.

222

Nonlinear Evolution and Difference Equations of Monotone Type

From a numerical point of view, it is important to obtain strong convergence results, because weak open neighborhoods are unbounded in infinite dimensional Hilbert spaces. G¨uler [GUL1] for (10.2) and V´eron [VER1] for (10.3) showed that the strong convergence does not hold in general, unless if we impose some additional assumptions on the convex function ϕ or the monotone operator A. (see Chapter 5 of the book). But as we showed in Theorem 5.8.11 of Chapter 5 of the book, the solutions to the second order evolution equation ( u00 (t) + cu0 (t) ∈ Au(t), t ∈ (0, +∞), (10.4) u(0) = u0 ∈ D(A), supt≥0 ku(t)k < +∞, with c > 0 converge strongly to a minimum point of ϕ. Moreover by Remark 5.8.12 of Chapter 5, the solutions to (10.4) converge to a zero of A exponentially. Therefore (10.4) seems more suitable for approximating a minimizer of ϕ. This section contains three subsections. In the first subsection, we study the rate of convergence of the solutions to (10.2), (10.3) and (10.4) in the case when A = ∂ ϕ, to a minimum point of the proper, convex and lsc function ϕ. In the second subsection, we study the rate of convergence for strongly maximal monotone operators. Finally in the third subsection, we show how one can apply (10.3) or (10.4) to approximate a zero of the monotone operator A. 10.2.1

RATE OF CONVERGENCE

This section is devoted to the study and comparison of the rate of convergence of the solutions to (10.2), (10.3) and (10.4). G¨uler [GUL2] studied the gradient flow for (10.2) for the case that A = ∂ ϕ. He proved that ϕ(x(t)) − ϕ(p) = o( 1t ), provided that x(t) converges strongly to p, where p is a minimum point of ϕ. Here, we give a simpler proof for G¨uler’s result, without assuming the strong convergence of x(t). First we prove some elementary lemmas. Lemma 10.2.1 Let f : R+ → R+ be a bounded and differentiable function, then lim inft→+∞ t f 0 (t) ≤ 0. Proof. Suppose to the contrary that lim inft→+∞ t f 0 (t) ≥ λ > 0. Then, there exists t0 > 0 such that for all t ≥ t0 > 0, t f 0 (t) ≥ λ . Dividing by t and integrating from t = t0 to T , we get f (T ) − f (t0 ) ≥ λ (ln T − lnt0 ). By letting T → +∞, we get a contradiction.



Lemma 10.2.2 Let R f , g : R+ → R+ . Assume that f is nonincreasing and limt→+∞ f (t) = 0. If 0+∞ f (s)g(s)ds < +∞, then Z t

lim f (t)(

t→+∞

g(s)ds) = 0. 0

Some Applications to Nonlinear Partial Differential Equations

223

Proof. By assumption, for all 0 < t < T , we have Z T

f (T )(

Z t

g(s)ds) ≤ f (T )(

0

Z T

g(s)ds) + 0

f (s)g(s)ds. t

The result follows by taking limsup as T → +∞, and then liminf as t → +∞.



Theorem 10.2.3 Suppose that u(t) is a solution to (10.2), then ϕ(u(t)) − ϕ(p) = o( 1t ), where p is a minimum point of ϕ. Proof. By the subdifferential inequality and (10.2), we get: ϕ(u(t)) − ϕ(p) ≤ (−u0 (t), u(t) − p) =

−1 d ku(t) − pk2 . 2 dt

Integrating the above inequality from t = 0 to t = T , we get Z T 0

1 (ϕ(u(t)) − ϕ(p))dt ≤ (−ku(T ) − pk2 + ku(0) − pk2 ) 2 1 ≤ ku(0) − pk2 . 2

This implies that Z +∞

(ϕ(u(t)) − ϕ(p))dt < +∞.

(10.5)

0

Then lim inft→+∞ ϕ(u(t)) − ϕ(p) = 0. On the other hand, by Lemma 2.2, pp. 57–58 of [MOR], we get d ϕ(u(t)) = (∂ ϕ(u(t)), u0 (t)) = −ku0 (t)k2 ≤ 0. dt This shows that ϕ(u(t)) − ϕ(p) is nonincreasing. Therefore lim ϕ(u(t)) − ϕ(p) = 0.

t→∞

Now the theorem follows from (10.5) and Lemma 10.2.2.



The following theorem shows that the second order evolution Equation (10.3) provides a faster rate of convergence for ϕ(u(t)) to ϕ(p) than (10.2). Theorem 10.2.4 Suppose that u(t) is a solution to (10.3), then ϕ(u(t)) − ϕ(p) = o( t12 ), where p is a minimum point of ϕ. Proof. By (10.3), for all t, h > 0, we get u00 (t + h) − u00 (t) ∈ ∂ ϕ(u(t + h)) − ∂ ϕ(u(t)). Multiplying both sides of the above inclusion by u(t + h) − u(t), by the monotonicity 2 of ∂ ϕ, we get: dtd 2 ku(t + h) − u(t)k2 ≥ 0. Therefore ku(t + h) − u(t)k2 is convex and

224

Nonlinear Evolution and Difference Equations of Monotone Type

bounded. Then, ku(t +h)−u(t)k2 is nonincreasing i.e. for all t > s > 0 and h > 0, we have ku(t + h) − u(t)k2 ≤ ku(s + h) − u(s)k2 . Dividing both sides of this inequality by h2 and letting h → 0, we get: ku0 (t)k2 is nonincreasing. Now, by Lemma 2.2, pp. 57–58 of [MOR], we get d 1 d 0 ϕ(u(t)) = (∂ ϕ(u(t)), u0 (t)) = (u00 (t), u0 (t)) = ku (t)k2 ≤ 0. dt 2 dt

(10.6)

By the subdifferential inequality and (10.3), we get: ϕ(u(t)) − ϕ(p) ≤ (u00 (t), u(t) − p) =

1 d2 ku(t) − pk2 − ku0 (t)k2 . 2 dt 2

Multiplying both sides of the above inequality by t, integrating from t = 0 to t = T , and integrating by parts, we get: Z T 0

1 d 1 T d t(ϕ(u(t)) − ϕ(p))dt ≤ T ku(T ) − pk2 − ku(t) − pk2 dt 2 dT 2 0 dt 1 d 1 1 = T ku(T ) − pk2 − ku(T ) − pk2 + ku(0) − pk2 . 2 dT 2 2 Z

Taking liminf as T → +∞, by Lemma 10.2.1, we get Z ∞

t(ϕ(u(t)) − ϕ(p))dt < +∞.

(10.7)

0

Therefore lim inft→+∞ ϕ(u(t)) − ϕ(p) = 0. (10.6) implies that limt→+∞ ϕ(u(t)) − ϕ(p) = 0. Now the theorem follows from (10.7) and Lemma 10.2.2.  Lemma 10.2.5 If u(t) is a solution to (10.4), then d 0 ku (t)k + cku0 (t)k ≤ 0. dt Proof. Let h be a small positive constant, and denote y(t) := u(t + h) − u(t). By the monotonicity of A, we get (y00 (t), y(t)) + c(y0 (t), y(t)) ≥ 0. Therefore

1 d2 c d ky(t)k2 + ky(t)k2 ≥ ky0 (t)k2 ≥ 0. (10.8) 2 2 dt 2 dt By the proof of Theorem 5.8.11 of Chapter 5, ky(t)k is nonincreasing. If there exists t0 such that ky(t0 )k = 0, then ky(t)k = 0, ∀t ≥ t0 . Therefore, eventually Z t

c

ky(s)kds ≤ ky(r)k − ky(t)k

r

Otherwise ky(t)k > 0 for all t > 0. Then (10.8) implies that 1 d d d (2ky(t)k ky(t)k) + cky(t)k ky(t)k ≥ ky0 (t)k2 . 2 dt dt dt

(10.9)

Some Applications to Nonlinear Partial Differential Equations Then (

225

d d2 d ky(t)k)2 + ky(t)k 2 ky(t)k + cky(t)k ky(t)k ≥ ky0 (t)k2 . dt dt dt 0

(t),y(t)) 2 Since ky0 (t)k2 ≥ ( (y ky(t)k ) = ( dtd ky(t)k)2 , we get

ky(t)k(

d2 d ky(t)k + c ky(t)k) ≥ 0. 2 dt dt

If ky(t)k > 0, ∀t > 0, then d2 d ky(t)k + c ky(t)k ≥ 0. dt 2 dt Hence ky(t)k + c 0t ky(s)kds is a convex function. Since by Remark 5.8.12 of Chap−c ter 5, for p ∈ Argminϕ, ku(t) − pk ≤ 2c ku0 (0)ke 2 t , we get R

ky(t)k + c

Z t

ky(s)kds ≤ 2(M + kpk) + c

Z t

0

 ku(s + h) − pk + ku(s) − pk ds

0

8 ≤ 2(M + kpk) + ku0 (0)k < +∞, c where M = supt≥0 ku(t)k. Therefore, ky(t)k + c By a result in convex analysis, we have ky(t)k + c

Z t

ky(s)kds ≤ ky(r)k + c

0

Rt

Z r

0 ky(s)kds

is bounded from above.

ky(s)kds, ∀t ≥ r,

(10.10)

0

which implies (10.9). Dividing both sides of (10.9) by h and letting h → 0, by Fatou’s Lemma, we get: Z t

c r

Hence ku0 (t)k + c

Rt

Z t

ky(s)k ds h Z t ky(s)k ≤ c lim inf ds h→0 h r ky(r)k ky(t)k ≤ lim inf( − ) h→0 h h = ku0 (r)k − ku0 (t)k.

ku0 (s)kds = c

lim inf

r

0 0 ku (s)kds

h→0

is nonincreasing. Therefore

d 0 ku (t)k + cku0 (t)k ≤ 0, dt as desired.



Theorem 10.2.6 Assume that u(t) is a solution to (10.4); then ku0 (t)k = O(e−ct ) and ku(t) − pk = O(e−ct ), where p ∈ A−1 (0) is the strong limit of u(t), as t → +∞.

226

Nonlinear Evolution and Difference Equations of Monotone Type

Proof. By Lemma 10.2.5, we get d (ku0 (t)kect ) ≤ 0 ⇒ ku0 (t)k ≤ e−ct ku0 (0)k. dt Then ku0 (t)k = O(e−ct ). On the other hand, by Theorem 5.8.11 of Chapter 5, we get ku(t) − pk = lim ku(t) − u(T )k T →+∞

= lim k

Z T

T →+∞

Z

≤ lim

t T

T →+∞ t

u0 (s)dsk

ku0 (s)kds

≤ lim ku0 (0)k

Z T

T →+∞

e−cs ds

t

1 = e−ct ku0 (0)k. c Hence ku(t) − pk = O(e−ct ).



Theorem 10.2.7 Suppose that u(t) is a solution to (10.4); then ϕ(u(t)) − ϕ(p) = o(e−ct ), where p ∈ A−1 (0) is the strong limit of u(t), as t → +∞. Proof. By the subdifferential inequality and (10.4), we get: ϕ(u(t)) − ϕ(p) ≤ (u00 (t) + cu0 (t), u(t) − p) =

1 d2 1 d ku(t) − pk2 − ku0 (t)k2 + c ku(t) − pk2 . 2 dt 2 2 dt

Multiplying both sides of the above inequality by ect , we get ect (ϕ(u(t)) − ϕ(p)) ≤

1 d ct d (e ku(t) − pk2 ). 2 dt dt

Integrating the above inequality from t = 0 to t = T , we get Z T 0

1 d ect (ϕ(u(t)) − ϕ(p))dt ≤ ecT ku(T ) − pk2 − (u0 (0), u(0) − p) 2 dT c c ≤ e 2 T ku0 (T )ke 2 T ku(T ) − pk − (u0 (0), u(0) − p).

Letting T → +∞, by Theorem 10.2.6, we have Z +∞ 0

ect (ϕ(u(t)) − ϕ(p))dt ≤ ku0 (0)kku(0) − pk < +∞.

(10.11)

Some Applications to Nonlinear Partial Differential Equations

227

Then lim inft→+∞ ϕ(u(t)) − ϕ(p) = 0. On the other hand, by Lemma 2.2, pp. 57–58 of [MOR] and Lemma 10.2.5, we get d ϕ(u(t)) = (∂ ϕ(u(t)), u0 (t)) dt = (u00 (t) + cu0 (t), u0 (t)) 1 d 0 = ku (t)k2 + cku0 (t)k2 2 dt d = ( ku0 (t)k + cku0 (t)k)ku0 (t)k ≤ 0, dt if ku0 (t)k > 0, for all t > 0. But if there exists t0 such that ku0 (t0 )k = 0, since ku0 (t)k is nonincreasing (see the proof of Theorem 5.8.11 of Chapter 5), ku0 (t)k = 0 for all t ≥ t0 . Then d ϕ(u(t)) = (∂ ϕ(u(t)), u0 (t)) dt = (u00 (t) + cu0 (t), u0 (t)) 1 d 0 = ku (t)k2 + cku0 (t)k2 = 0 2 dt This shows that ϕ(u(t)) − ϕ(p) is nonincreasing. Therefore lim ϕ(u(t)) − ϕ(p) = 0.

(10.12)

t→∞



The theorem is proved by (10.11), (10.12) and Lemma 10.2.2. 10.2.2

STRONGLY MONOTONE CASE

One of the most important methods for solving (10.1), called the regularization method, is to consider a family of auxiliary problems: Minx∈H ϕα (x),

(10.13)

α 2 2 kxk ,

where ϕα (x) := ϕ(x) + α > 0. Obviously if ϕ is convex with Argminϕ , ∅, then ϕα is strongly convex and coercive, i.e. limkxk→+∞ ϕα (x) = +∞. By Theorem 2.4.3 of Chapter 2, ϕα has a unique minimum point that we denote by xα . It is well known in the literature, that xα converges strongly to a point x∗ as α → 0, which is an element of Argmin(ϕ) with minimum norm (in fact, xα = J 1 0, i.e. the 2α

1 resolvent of ϕ of order 2α at 0, see [TIK, MOR]). Therefore the solutions to (10.13) approximate the least norm solution of (10.1). On the other hand, as we mentioned in the introduction, solutions to (10.2), (10.3) and (10.4) converge to a minimum point of ϕ. This convergence, in both cases of (10.2) and (10.3) is weak in general (except for (10.4)), and is strong when ϕ satisfies some additional conditions such as being strongly convex or even (see Theorems 4.10.1 and 4.10.2 of Chapter 4 and Theorem 5.7.24 of Chapter 5). Now, we plan to study and compare the convergence rate of the solutions to (10.2), (10.3) and (10.4), when ϕ is strongly convex, or even more generally, when A is α-strongly monotone.

228

Nonlinear Evolution and Difference Equations of Monotone Type

Theorem 10.2.8 Suppose that u(t) is a solution to (10.2), where A is α-strongly monotone, and p is the unique element of A−1 (0), then ku(t) − pk = O(e−αt ). Proof. Multiplying both sides of (10.2) by u(t) − p, and using the α-strong monotonicity of A, we get: αku(t) − pk2 ≤

−1 d ku(t) − pk2 . 2 dt

(10.14)

Since ku(t) − pk is nonincreasing, if ku(t0 ) − pk = 0 for t0 > 0, then ku(t) − pk = 0, for all t ≥ t0 and the result holds. Otherwise, dividing both sides of (10.14) by ku(t) − pk2 , we get: d ln ku(t) − pk2 ≤ −2α. dt Integrating from 0 to T , we get: ln ku(T ) − pk2 − ln ku(0) − pk2 ≤ −2αT.

(10.15)

Therefore ku(T ) − pk ≤ ku(0) − pke−αT , 

which yields the theorem.

Theorem 10.2.9 Suppose that u(t) is a solution to (10.4) with c ≥ 0, where A is α−1 −βt strongly monotone, and qp is the unique element of A (0), then ku(t)− pk = o(e ), for each c < β < 2c +

c2 4

+ α, where c is the positive constant appearing in (10.4).

Proof. Multiplying both sides of (10.4) by u(t) − p, and using the α-strong monotonicity of A, we get: αku(t) − pk2 + ku0 (t)k2 ≤

1 d2 1 d ku(t) − pk2 + c ku(t) − pk2 . 2 2 dt 2 dt

(10.16)

If there exists t0 > 0 such that ku(t0 ) − pk = 0, then ku(t) − pk = 0 for all t ≥ t0 and the result holds. Now suppose that ku(t) − pk > 0 for each t > 0. From (10.16), we have d d d (ku(t)− pk ku(t)− pk)+cku(t)− pk ku(t)− pk ≥ ku0 (t)k2 +αku(t)− pk2 dt dt dt 2 d d d ⇒ ( ku(t) − pk)2 + ku(t) − pk 2 ku(t) − pk + cku(t) − pk ku(t) − pk dt dt dt ≥ ku0 (t)k2 + αku(t) − pk2 . 0

(t),u(t)−p) 2 Since ( dtd ku(t) − pk)2 ≤ ( (u ku(t)−pk ) ≤ ku0 (t)k2 , and ku(t) − pk > 0, we obtain

d2 d ku(t) − pk + c ku(t) − pk ≥ αku(t) − pk. dt 2 dt

(10.17)

Some Applications to Nonlinear Partial Differential Equations

Multiplying both sides of (10.17) by eβt for c < β < from 0 to T , we get: Z T

α 0

c 2

+

q

c2 4

229

+ α and integrating

d d ku(T ) − pk − ku(t) − pk|t=0 dT dt Z T d + (c − β ) eβt ku(t) − pkdt dt 0 ≤ C + (c − β )eβ T ku(T ) − pk

eβt ku(t) − pkdt ≤ eβ T

− (cβ − β 2 )

Z T

eβt ku(t) − pkdt,

0

for some constant C independent of T . Therefore R +∞ 2 βt 0 (α + cβ − β )e ku(t) − pkdt < +∞. Now the result follows from Lemma 10.2.2.  Remark 10.2.10 Theorems 10.2.8 and 10.2.9 show that the rate of convergence of solutions to (10.4) is much better than (10.2), especially if c is chosen large enough. Even if c = 0, the rate of convergence of (10.3) is better than (10.2), when 0 < α < 1. This advantage of (10.4) compared to (10.2), with respect to the rate of convergence, allows us to use (10.4) for the approximation of the unique minimum point of the strongly convex functions ϕα , which estimate a minimum point of ϕ, for sufficiently small α. 10.2.3

USING THE SECOND ORDER EVOLUTION EQUATION FOR APPROXIMATING A MINIMIZER

Even in R, a simple example shows that the equation ( u00 (t) + cu0 (t) ∈ Au(t), u(0) = u0 ,

(10.18)

has infinitely many solutions, and only one of them is bounded. In general, we don’t know of any method for finding this bounded solution. Each numerical method for solving (10.18) requires u0 (0) or some other additional condition, and for a given choice of u0 (0), the solution is not necessarily bounded. But by [APR2] (see also [APR1]), we know that if uT is a solution to the two point boundary value problem below ( u00T (t) + cu0T (t) ∈ AuT (t), (10.19) uT (0) = u0 , uT (T ) = vT on [0, T ], with vT = u0 , then uT converges uniformly to a solution u of (10.18) on each compact subinterval of (0, +∞), as T → +∞. Therefore we use (10.19) instead of (10.4) for approximating a zero of the monotone operator A, or a minimum point of ϕα . To this aim, we need to compute the solution uT at an appropriate point in the

230

Nonlinear Evolution and Difference Equations of Monotone Type

interval [0, T ]. To obtain a suitable point in the interval, let L > 0 fixed and L < S < T . By relation (2.7) of [APR2], we have kuT (t) − uS (t)k2 ≤

ecS kuT (S)k2 . 2 ln LS

(10.20)

By relation (2.4) of [APR2], there exists M > 0 independent of T and of t ∈ [0, T ] such that ect kuT (t)k2 ≤ M, ∀t ∈ [0, T ], ∀T > 0. (10.21) Therefore limT →+∞ uT (t) = u(t), uniformly on every bounded interval [0, L] ⊂ [0, +∞). The function u is in fact the solution of (10.4). Now letting T → +∞ in (10.20), by (10.21), we get ku(t) − uS (t)k2 ≤ √ In the above inequality, set t = L = S, then √ √ ku( S) − uS ( S)k2 ≤

M . 2 ln LS

M M √ = . 2 ln S ln S

(10.22)

(10.23)

√ √ The last √ inequality shows that when S is sufficiently large, uS ( S) is near u( S). Since u( S) converges √ to a zero of A (weakly when c = 0 and strongly when c > 0), as S → +∞, then uS ( S) provides an approximation to a zero of A when √ S is sufficiently large. By Theorem 10.2.9 and Remark 10.2.10, although u( S) approximates a zero of A with√a rate of convergence better than √ (10.2) for large S, but we can compute only uS ( S), which is an estimate to u( S)(by (10.23)). In addition, since kuS (t) − pk, where p is a zero of A, is convex on [0, S] (see Theorem 2.1.2 of [APR1]), we have kuS (t) − pk ≤ Max{kuS (S) − pk, kuS (0) − pk} = ku0 − pk, which shows the stability of the curve uS (t). To see an example and numerical implementation, the reader can consult [KHA-SHO].

10.3

APPLICATION TO VARIATIONAL PROBLEMS

Consider the second order evolution equation of the form ( u00 (t) ∈ ∂ ϕ(u(t)), u(0) = x, supt≥0 ku(t)k < +∞

(10.24)

where ϕ : H → (−∞, +∞] is a proper, convex and lower semicontinuous function. In this brief section, we show that the solution to (10.24) is exactly the minimizer of the functional Z +∞  1 0 F(v) = kv (t)k2 + ϕ(v(t)) dt (10.25) 2 0 on a suitable space.

Some Applications to Nonlinear Partial Differential Equations

231

Theorem 10.3.1 Let u(t), t ≥ 0 be a solution to (10.24). Then u is the minimizer of the functional F defined in (10.25) on the space 2,2 H = {v ∈ Wloc (R+ ; H);

dv ∈ L2 (R+ ; H) and v(0) = x} dt

Proof. By the subdifferential inequality for ϕ and the Equation (10.24), we get: ϕ(u(t)) − ϕ(v(t)) ≤ (u00 (t), u(t) − v(t)) d = (u0 (t), u(t) − v(t)) − ku0 (t)k2 + (u0 (t), v0 (t)) dt d 1 1 ≤ (u0 (t), u(t) − v(t)) − ku0 (t)k2 + kv0 (t)k2 dt 2 2 Integrating the above inequality from 0 to T , we get: Z T 0

T 1 1 (ϕ(u(t)) + ku0 (t)k2 )dt ≤ (ϕ(v(t)) + kv0 (t)k2 )dt 2 2 0 + (u0 (T ), u(T ) − v(T )) − (u0 (0), u(0) − v(0))

Z

Now the result follows by letting T go to infinity, because ku0 (T )k → 0 as T → +∞.  For more information, and the proof that the solution to a more general second order equation is the minimizer of a functional, we refer the interested reader to [APR1].

10.4

SOME APPLICATIONS TO PARTIAL DIFFERENTIAL EQUATIONS

The first and second order evolution equations have a lot of applications in partial differential equations. In this section we give only one concrete example for a first and a second order differential equation, as well as a difference equation that was discussed in this book. This section contains three subsections. In the first subsection, we give an example of a concrete partial differential equation that can be modeled by a first order nonlinear differential equation of monotone type. In the second subsection, we give an example of a second order differential inclusion associated to a maximal monotone operator. Finally in the last subsection, we give an application of the second order difference equation of monotone type discussed in Chapter 8. These examples were chosen from [MOR], [APR2, APR3](see also [APR1]). 10.4.1

A CONCRETE EXAMPLE OF THE FIRST ORDER EQUATION

Consider the nonlinear partial differential equation  2 ∂ u ∂u +   ∂t 2 − ∆u + β ( ∂t ) 3 f (t, x), (t, x) ∈ R × Ω u(t, x) = 0, (t, x) ∈ R+ × Γ   u(0, x) = u0 , ∂∂tu (0, x) = v0 , x ∈ Ω.

(10.26)

232

Nonlinear Evolution and Difference Equations of Monotone Type

where Ω ⊂ Rn is open and bounded with a sufficiently smooth boundary and β : D(β ) ⊂ R → R is maximal monotone with 0 ∈ β (0). Equation (10.26) models the vibration of an elastic membrane with fixed boundary in the presence of a friction force β ( ∂∂tu ), where f (t, x) on the right hand side represents an external force. The Hilbert space H = H01 (Ω) × L2 (Ω) is equipped with the inner product Z

(U1 ,U2 )H = Ω

Z

(∇u1 , ∇u2 )Rn dx +

v1 v2 dx Ω

for every U1 = [u1 , v1 ] and U2 = [u2 , v2 ]. By Proposition 1.3 of [MOR], β = ∂ j where j : R → (−∞, +∞] is proper, convex and lower semicontinuous, and since 0 ∈ β (0) one can write the function j in the form Z r

j(r) =

β 0 (s)ds.

0

Obviously, inf j = j(0) = 0. Now consider the function ϕ : H01 (Ω) → (−∞, +∞] defined by (R j(v(x))dx, if jov ∈ L1 (Ω) ϕ(v) = Ω +∞, otherwise Obviously ϕ is proper, convex and lower semicontinuous. By Proposition 3.5.1 of Chapter 3, the extension of ϕ to L2 (Ω) is also lower semicontinuous. The subdifferential β˜ of ϕ is defined by β˜ : D(β˜ ) ⊂ H01 (Ω) → H −1 (Ω) with z ∈ β˜ (v) ⇔ ϕ(v) − ϕ(w) ≤ (v − w, z)H 1 (Ω),H −1 (Ω) , ∀w ∈ H01 (Ω). 0

By Theorem 3.4.2 of Chapter 3, β˜ is maximal monotone. The Laplacian −∆ is an operator from H01 (Ω) to H −1 (Ω) defined by the bilinear form Z a(v1 , v2 ) = (∇v1 , ∇v2 )Rn dx, ∀v1 , v2 ∈ H01 (Ω) Ω

which is exactly the inner product of H01 (Ω). Therefore −∆ is precisely the canonical isomorphism from H01 (Ω) to H −1 (Ω), which is given by the Riesz representation Theorem (see Theorem 1.1.10 of Chapter 1). On the other hand, L2 (Ω) is isometrically isomorphic with its dual, and therefore algebrically and topologically we have H01 (Ω) ⊂ L2 (Ω) ⊂ H −1 (Ω). The first embedding is the canonical injection i : H01 (Ω) → L2 (Ω), and the second one is the adjoint map i∗ : L2 (Ω) → H −1 (Ω). For simplicity, we denote i∗ (ω) again by ω. By Rellich-Kondrachov theorem (see [ADA] p.143), the embedding i : H01 (Ω) → L2 (Ω) is compact, and by Schauder’s theorem (see [YOS] p.282), i∗ : L2 (Ω) → H −1 (Ω) is also compact. In other words, H01 (Ω) is compactly embedded in H −1 (Ω). Now define the operator A : D(A) ⊂ H → H by D(A) = {[p, q] ∈ H01 (Ω) × H01 (Ω); (−∆p + β˜ (q)) ∩ L2 (Ω) , ∅} A([p, q]) = {−q} × (−∆p + β˜ (q)) ∩ L2 (Ω), ∀[p, q] ∈ D(A).

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233

It can be shown that A is maximal monotone (see [MOR]). Then the problem can be written as a first order evolution equation associated to A in the form: ( dU dt + AU 3 F(t), 0 < t < +∞ U(0) = U0 where F(t) = [0, f (t, ·)] with f ∈ L1 (0, +∞; L2 (Ω)), and U0 = [u0 , v0 ] ∈ D(A). To see more examples, we refer the interested reader to [MOR]. 10.4.2

AN EXAMPLE OF A SECOND ORDER EVOLUTION EQUATION

Let H = L2 (Ω) where Ω ⊆ Rn is a bounded domain with smooth boundary Γ. Let j : R → (−∞, +∞] be a proper, convex and lower semicontinuous function, and β = ∂ j. We assume for simplicity that 0 ∈ β (0). Define n

∂ 2u 2 i=1 ∂ xi

Au = −∆u = − ∑ with D(A) = {u ∈ H 2 (Ω) ,

−∂ u (x) ∈ β (u(x)) , a.e. on Γ} ∂η

where ( ∂∂ηu (x)) is the outward normal derivative to Γ at x ∈ Γ. It is known that A = ∂ φ , where φ : L2 (Ω) → (−∞ , +∞], is the functional: ( R R 1 2 if u ∈ H 1 (Ω) and β (u) ∈ L1 (Γ) Ω |∇u| dx + Γ β (u(x))dσ , 2 φ (u) = +∞, otherwise. Consider the following equation  2 2 ∂ u  (t, x) + ∑i ∂∂ xu2 (t, x) = 0 a.e. on R+ × Ω   ∂t 2 i   ∂u − ∂ η (t, x) ∈ β u(t, x) a.e. on R+ × Γ   u(0, x) = u0 (x) a.e. on Ω   R 2  supt≥0 Ω u (t, x)dx < +∞. Then this is an example of a second order evolution equation. 10.4.3

AN EXAMPLE OF A SECOND ORDER DIFFERENCE EQUATION

Let H = L2 (Ω) where Ω ⊆ Rn is a bounded domain with smooth boundary Γ. Denote by A˜ the operator from W 1,p (Ω) into W −1,q (Ω)(where 1 + 1 = 1, p ≥ 2), given by 0

N

˜ v) = (Au,



Z

k=1 Ω

p

|

q

∂ u p−2 ∂ u ∂ v | dx, ∀u, v ∈ W01,p (Ω) ∂ xk ∂ xk ∂ xk

(10.27)

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Nonlinear Evolution and Difference Equations of Monotone Type

The operator A˜ coincides with the subdifferential of the convex and lower semicontinuous function Z 1 N ∂u p | dx (10.28) ϕ(u) = ∑ | p k=1 Ω ∂ xk Let A be the restriction of A˜ to D(A) = W01,p (Ω). This operator is maximal monotone in L2 (Ω). Moreover, A = ∂ ψ, where ψ : L2 (Ω) → (−∞, +∞], is defined by ( ϕ(u), u ∈ W01,p (Ω), ψ(u) = (10.29) +∞, otherwise The sequence of the following boundary value problems  ∂ ui p−2 ∂ ui ∂ N  ui+1 − (1 + θi )ui + θi ui−1 ∈ −ci ∑k=1 ∂ xk (| ∂ xk | ∂ xk ), ui (x) = 0, x ∈ ∂ Ω,  R  u0 (x) = a(x), ∀x ∈ Ω, supi≥1 Ω u2i (x)dx < +∞

x∈Ω (10.30)

where a ∈ L2 (Ω), ci > 0, θi > 0, ∀i ≥ 1, is an example of a second order difference equation of monotone type.

REFERENCES ADA. R. Adams and J. J. F. Fournier, Sobolev spaces. First Edition, Academic Press, New York 1975. Second edition. Pure and Applied Mathematics (Amsterdam), 140. Elsevier/Academic Press, Amsterdam, 2003. APR1. N. C. Apreutesei, Nonlinear second order evolution equations of monotone type and applications, Pushpa Publishing House, Allahabad, India, 2007. APR2. N. C. Apreutesei, Second-order differential equations on half-line associated with monotone operators, J. Math. Anal. Appl. 223 (1998), 472–493. APR3. N. C. Apreutesei, On a class of difference equations of monotone type, J. Math. Anal. Appl. 288 (2003), 833–851. BAR. V. Barbu, Nonlinear semigroups and Differential Equations in Banach Spaces, Noordhoff, Leyden, 1976. BRU. R. E. Bruck, Asymptotic convergence of nonlinear contraction semigroups in Hilbert space, J. Funct. Anal. 18 (1975), 15–26. DJA-KHA1. B. Djafari Rouhani, H. Khatibzadeh, A strong convergence theorem for solutions to a nonhomogeneous second order evolution equation, J. Math. Anal. Appl. 363 (2010), 648–654. GUL1. O. G¨uler, On the convergence of the proximal point algorithm for convex minimization, SIAM J. Control Optim. 29 (1991), 403–419. GUL2. O. G¨uler, Convergence rate estimates for the gradient differential inclusion, Optim. Methods Softw. 20 (2005), 729–735. KHA-SHO. H. Khatibzadeh and A. Shokri, On the first and second order strongly monotone dynamical systems and minimization problems, Optim. Methods and Soft. 30 (2015), 1303–1309. MOR. G. Morosanu, Nonlinear Evolution Equations and Applications, Editura Academiei (and D. Reidel Publishing Company), Bucharest, 1988.

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TIK. A. N. Tikhonov, V. Ya. Arsenin, Methods for the solution of ill-posed problems, Third edition, “Nauka”, Moscow, 1986. VER1. L. V´eron, Un exemple concernant le comportement asymptotique de la solution 2 born´ee de l’´equation ddt 2u ∈ ∂ ϕ(u), Monatsh. Math. 89 (1980), 57–67. YOS. K. Yosida, Functional analysis. Sixth edition. Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 123, Springer-Verlag, Berlin, New York, 1980.

Index A absolutely continuous 8, 11, 33, 39, 40, 64, 105 almost nonexpansive curve 33, 51 almost nonexpansive sequence 46, 53 almost orbit 54, 55 almost weakly convergent 91, 95 asymptotic behavior 33, 64, 71, 90, 104, 105, 121, 124, 130, 145, 165, 188, 207 asymptotic center 47, 49, 50, 52, 54, 57, 58, 60, 62, 92, 94, 95, 106–110, 112, 134–137, 188, 192, 195, 211

B best approximation theorem 5 boundary value problem 71, 81, 82, 229, 234 bounded linear functional 4, 5

C canonical extension 28, 29, 42, 77, 84 closed convex hull 13, 46 coercive 6, 130, 227 compact operator 217 continuous dependence on initial conditions 163 convergence condition 200 convex function 13, 14, 16–19, 26, 28, 47, 52, 62, 64, 65, 81, 91, 121, 130, 221, 222, 225, 229 convex set 7, 13, 14

D demiclosed 23, 37, 77, 84, 86, 133, 155, 157, 160, 162 discrete nonlinear oscillator 207 distributions 8–11, 73 dynamical system 121, 207

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Nonlinear Evolution and Difference Equations of Monotone Type

E epigraph 14 ergodic retraction 46 ergodic theorem 45, 46, 88, 100, 132, 134, 161, 208 Euler discretization 129 eventually increasing 91, 92, 99, 103, 104, 152, 166, 183 evolution system 64, 88

F Fenchel conjugate 19 Fenchel subdifferential 18 Fenchel-Moreau theorem 19 first order difference equation 129 first order evolution equation 33, 56, 129, 131, 233 friction 121, 124, 207, 232

G Gronwall type lemma 34

H heavy ball 121, 207 Hilbert space 3–11, 14, 18, 21, 46, 51, 56, 73, 81, 146, 222, 232 Homogeneous 33, 56, 64, 71, 91, 110, 117, 135, 145, 165, 184, 188

I inertial proximal algorithm 207 inner product 4, 5, 10, 11, 73, 84, 146, 232

L LP spaces 8 Lax-Milgram theorem 73 linear mapping 3, 4, 6 locally Lipschitz 14–16 lower semicontinuous 16, 25, 99, 100, 101, 103, 112, 123, 130, 140, 142, 145, 184, 185, 203, 207, 215, 221, 230, 232–234

M maximal monotone operator 21, 23–25, 28, 33, 37, 45, 55, 71, 72, 81, 86, 89, 90, 91, 95, 96, 104, 113, 121, 129, 130, 131, 147, 153, 157, 160, 163, 165, 173, 180, 184, 203, 207, 211, 217, 221, 222, 231

Index

Mazur theorem 16 metric projection 5, 200

N nonexpansive curve 33, 46, 51 nonexpansive operator 24, 33, 45, 53, 82, 89, 90, 131, 209 nonexpansive semigroup 45, 89, 90 nonexpansive sequence 45–47, 53, 132, 133, 161, 162 nonhomogeneous 33, 71, 86, 104, 135, 188 nonlinear partial differential equations 221 non-summable errors 137 Normed linear space 3, 4, 13

O Opial’s lemma 42, 44, 67, 123, 185 optimization 13, 17, 18, 21, 145, 184, 203, 216, 221 orthogonal projection 6

P periodic solution 33, 41, 42, 44, 71, 87, 88, 131–133, 145, 160–162 polarization identity 5, 48, 190 proximal point algorithm 18, 129, 130, 133, 141, 207 pseudo-convex function 65

Q quasi-convex function 64, 65

R rate of convergence 62, 98, 130, 142, 184, 185, 222, 223, 229, 230 reflexive Banach spaces 7, 18 resolvent 23, 24, 26, 28, 74, 129, 163, 164, 207, 209, 227 Riesz representation theorem 5, 232

S second order difference equation 145, 146, 163, 231, 233, 234 second order evolution equation 71, 95, 121, 145, 221–223, 229–231, 233 Sobolev spaces 5, 8, 9, 11 square root 71, 89 strictly convex 13, 14, 17, 47, 52, 141 strictly monotone 21, 141, 213 strong ergodic convergence 57, 173

239

240

Nonlinear Evolution and Difference Equations of Monotone Type

strong solution 33, 34, 37, 39, 45 strongly monotone 21, 62, 72, 74, 79, 99, 109, 137, 142, 180, 197, 218, 227, 228 strongly regular summation method 57, 59–61 symmetric 6

U uniformly quasi-convex 69

W weak cluster point 7, 42, 67, 94, 101, 114, 138, 208–210 weak convergence 6–8, 53–55, 57, 64, 71, 87, 90, 91, 94, 95, 99, 123, 129, 135, 141, 160, 176, 184, 185, 188, 195, 196, 208, 211, 213, 215, 217, 221 weak solution 34, 38–40, 53, 54, 62 weak topology 6, 7, 16 weakly asymptotically regular 50–52 weighted averages 165, 208, 209

Y Yosida approximation 23, 24, 26, 35, 74, 148