Minimax Systems and Critical Point Theory

Martin Schechter Minimax Systems and Critical Point Theory Birkhäuser Boston • Basel • Berlin Martin Schechter Math...

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Martin Schechter

Minimax Systems and Critical Point Theory

Birkhäuser Boston • Basel • Berlin

Martin Schechter Mathematics Department University of California Irvine, CA 92697-3875 [email protected]

ISBN 978-0-8176-4805-3 DOI 10.1007/ 978-0-8176-4902-9

e-ISBN 978-0-8176-4902-9

Library of Congress Control Number: 2009928827 Mathematics Subject Classification (2000): 35J20, 35J65, 47J30, 49J35, 58E05 © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009 All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Birkhäuser Boston, c/o Springer Science+Business Media, LLC, 233 Spring Street, New York, NY 10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks, and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. Printed on acid-free paper Birkhäuser Boston is part of Springer Science+Business Media (www.birkhauser.com)

BS D To my wife, Deborah, our children, our grandchildren (twenty four so far) our great grandchildren (six so far) and our extended family. May they all enjoy many happy years.

Contents Preface

xi

1 Critical Points of Functionals 1.1 Introduction . . . . . . . . . . 1.2 Extrema . . . . . . . . . . . . 1.3 Palais–Smale sequences . . . . 1.4 Cerami sequences . . . . . . . 1.5 Linking sets . . . . . . . . . . 1.6 Previous definitions of linking 1.7 Notes and remarks . . . . . .

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1 1 2 2 3 3 4 5

2 Minimax Systems 2.1 Introduction . . . . . . . 2.2 Definitions and theorems 2.3 Linking subsets . . . . . 2.4 A variation . . . . . . . 2.5 Weaker conditions . . . . 2.6 Some consequences . . . 2.7 Notes and remarks . . .

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3 Examples of Minimax Systems 3.1 Introduction . . . . . . . . . . . . . . . . 3.2 A method using homeomorphisms . . . . 3.3 A method using metric spaces . . . . . . 3.4 A method using homotopy-stable families 3.5 Examples of linking sets . . . . . . . . . 3.6 Various geometries . . . . . . . . . . . . 3.7 A sandwich theorem . . . . . . . . . . . 3.8 Notes and remarks . . . . . . . . . . . .

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17 17 17 19 19 21 24 26 29

4 Ordinary Differential Equations 4.1 Extensions of Picard’s theorem . . . . . . . . . . . . . . . . . . . . . 4.2 Estimating solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Extending solutions . . . . . . . . . . . . . . . . . . . . . . . . . . .

31 31 33 34

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viii

Contents 4.4 4.5

The proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . An important estimate . . . . . . . . . . . . . . . . . . . . . . . . . .

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5 The Method Using Flows 5.1 Introduction . . . . . 5.2 Theorem 2.4 . . . . . 5.3 Theorem 2.12 . . . . 5.4 Theorem 2.14 . . . . 5.5 Theorem 2.21 . . . .

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39 39 39 41 44 45

6 Finding Linking Sets 6.1 Introduction . . . . . 6.2 The strong case . . . 6.3 The remaining proofs 6.4 Notes and remarks .

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51 51 52 54 56

7 Sandwich Pairs 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Notes and remarks . . . . . . . . . . . . . . . . . . . . . . . . . . .

57 57 58 61

8 Semilinear Problems 8.1 Introduction . . . . . . 8.2 Bounded domains . . . 8.3 Some useful quantities 8.4 Unbounded domains . 8.5 Further applications . . 8.6 Special cases . . . . . 8.7 The proofs . . . . . . . 8.8 Notes and remarks . .

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63 63 63 69 71 75 80 81 83

9 Superlinear Problems 9.1 Introduction . . . . . . 9.2 The main theorems . . 9.3 Preliminaries . . . . . 9.4 Proofs . . . . . . . . . 9.5 The parameter problem 9.6 The monotonicity trick 9.7 Notes and remarks . .

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85 85 86 88 89 92 97 104

10 Weak Linking 10.1 Introduction . . . . 10.2 Another norm . . . 10.3 Some examples . . 10.4 Some applications . 10.5 Notes and remarks

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107 107 108 112 114 126

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Contents

ix

11 Fuˇc´ık Spectrum: Resonance 11.1 Introduction . . . . . . . 11.2 The curves . . . . . . . . 11.3 Existence . . . . . . . . 11.4 Notes and remarks . . .

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12 Rotationally Invariant Solutions 12.1 Introduction . . . . . . . . . . . . . 12.2 The spectrum of the linear operator . 12.3 The nonlinear case . . . . . . . . . 12.4 Notes and remarks . . . . . . . . .

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13 Semilinear Wave Equations 13.1 Introduction . . . . . . . . . . . . . 13.2 Convexity and lower semi-continuity 13.3 Existence of saddle points . . . . . 13.4 Criteria for convexity . . . . . . . . 13.5 Partial derivatives . . . . . . . . . . 13.6 The theorems . . . . . . . . . . . . 13.7 The proofs . . . . . . . . . . . . . . 13.8 Notes and remarks . . . . . . . . .

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149 149 149 152 155 156 158 158 161

14 Type (II) Regions 14.1 Introduction . . . . . . . 14.2 The asymptotic equation 14.3 Local estimates . . . . . 14.4 The solutions . . . . . . 14.5 Notes and remarks . . .

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15 Weak Sandwich Pairs 15.1 Introduction . . . . . 15.2 Weak sandwich pairs 15.3 Applications . . . . . 15.4 Notes and remarks .

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177 177 178 184 190

16 Multiple Solutions 16.1 Introduction . . . . . . . . 16.2 Two examples . . . . . . . 16.3 Statement of the theorems 16.4 Some lemmas . . . . . . . 16.5 Local linking . . . . . . . 16.6 The proofs . . . . . . . . . 16.7 Notes and remarks . . . .

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191 191 191 192 194 206 208 209

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x 17 Second-Order Periodic Systems 17.1 Introduction . . . . . . . . . 17.2 Proofs of the theorems . . . 17.3 Nonconstant solutions . . . . 17.4 Notes and remarks . . . . .

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213 213 215 221 226

Bibliography

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Index

241

Preface Many problems in science involve the solving of differential equations or systems of differential equations. Moreover, many of these equations and systems come from variational considerations involving mappings (called functionals) into the real number system. As a simple example, consider the problem of finding a solution of (1)

u (x) = f (x, u(x)),

x ∈ I = [a, b],

under the conditions (2)

u(a) = u(b) = 0.

Assume that the function f (x, t) is continuous in I × R. The corresponding functional is b (3) G(u) = [(u )2 + 2F(x, u)] d x, a

where F(x, t) :=

t

f (x, s) ds.

0

It is easy to show that u(x) is a solution of the problem (1), (2) if and only if it satisfies (4)

G (u) = 0.

Thus, in such cases, solutions of the equations or systems are critical points of the corresponding functional. As a result, anyone who is interested in obtaining solutions of the equations or systems is also interested in obtaining critical points of the corresponding functionals. The latter problem is the subject of this book. The classical way of obtaining critical points was to search for maxima or minima. This is possible if the functional is bounded from above or below. However, when this is not the case, there is no organized way of finding critical points. Linking theory is an attempt to “level the playing field,” i.e., to find a substitute for semiboundedness. It finds a pair of subsets A, B of the underlying space that allow the functional to have the same advantages as semibounded functionals if the subsets separate the functional. They separate a functional G if (5)

sup G < inf G. A

B

xii

Preface

This is the theme of the book [122], which records much of the work of researchers on this approach up to that time. There are several methods of obtaining linking sets (some will be outlined later in Chapters 3 and 6). The purpose of the present volume is to unify some of these approaches and to study results and applications that were obtained since the publication of [122]. The underlying theme is to consider minimax systems depending on a set A. These are collections K of subsets such that if the functional G satisfies (6)

sup G < inf sup G, K ∈K K

A

then it has the same advantages as a semibounded functional. We show that the main approaches to linking can be combined by using minimax systems. In Chapters 1 and 2, we define minimax systems and show what they can accomplish. We consider several variations and generalizations that are useful in applications. In Chapter 3 we describe some methods used by researchers to obtain critical points of functionals, and we show that these results are contained in the theorems of Chapter 2. Various geometries are considered. In particular, the sandwich theorem of [143], [109], and [108] is generalized. This considers the following situation when N is a finite-dimensional subspace of a Hilbert space E and M = N ⊥ . If G is a functional on E such that G is bounded from below on M and bounded from above on N, then G has the advantages of a semibounded functional. If both subspaces are infinitedimensional, it may be necessary to impose additional assumptions on the functional G. This is discussed in Chapter 15. In Chapter 4 we prove some theorems concerning differential equations in abstract spaces. These results are needed in proving the theorems of Chapter 2. In Chapter 5 we give the proofs of the theorems of Chapter 2 using the results of Chapter 4. In Chapter 6 we search for linking subsets. We believe in principle that we have found most, if not all, of them, at least in an abstract way. We produce two criteria for subsets to satisfy, one slightly stronger than the other. The weaker criterion is necessary for linking, while the stronger is sufficient. In Chapter 7 we ask the question: Is there anything that can be done if one cannot find linking subsets that separate the functional? A surprising answer is yes. We show that there are subsets A, B that produce the same effect when they do not separate the functional, i.e., when they satisfy just the opposite: (7)

−∞ < inf G ≤ sup G < ∞. B

A

We call such sets a sandwich pair. The reason is that the subspaces described in the sandwich theorem of Chapter 3 form a sandwich pair. We describe criteria for obtaining sandwich pairs. In Chapter 8 we describe applications of the theories presented in Chapters 1–7. Some are more involved than those usually found in the literature. Other applications are given in Chapters 9–17.

Preface

xiii

In Chapter 9 we describe superlinear problems of the form (8)

−u = f (x, u), x ∈ ;

u = 0 on ∂,

where ⊂ Rn is a bounded domain whose boundary is a smooth manifold and f (x, t) ¯ × R. The problem is called superlinear if f (x, t) does is a continuous function on not satisfy an inequality of the form | f (x, t)| ≤ C(|t| + 1),

x ∈ , t ∈ R.

Almost all researchers who studied the superlinear problem make the assumption that there are constants μ > 2, r ≥ 0 such that (9)

0 < μF(x, t) ≤ t f (x, t),

|t| ≥ r.

This is a very convenient assumption, but it excludes a large number of superlinear problems. An assumption that is much weaker is: Either F(x, t)/t 2 → ∞ as t → ∞ or F(x, t)/t 2 → ∞ as t → −∞. We shall show that under this condition the boundary value problem (10)

−u = β f (x, u), x ∈ ;

u = 0 on ∂,

has a nontrivial solution for almost every positive β. We require a stronger hypothesis to show that the statement is true for any particular β. A fact of life which plagues researchers is that we can verify that subsets link only if at least one of them is contained in a finite dimensional manifold. In Chapter 10 we deal with this problem by adding an assumption to the functional. We try to make this assumption as weak as possible. Applications are given. Chapters 11 and 14 deal with the Fuˇc´ık spectrum. They deal with the situation concerning the semilinear problem (11)

Au = f (x, u),

u ∈ D(A),

in which f (x, t)/t

→ a a.e. as t → −∞ → b a.e. as t → +∞,

and (12)

Au = bu + − au − ,

u ± = max{±u, 0},

has a nontrivial solution. We call the set of those (a, b) ∈ R2 for which (12) has nontrivial solutions the Fuˇc´ık spectrum of A. When (a, b) ∈ , it is more difficult to

xiv

Preface

solve (11) than when (a, b) ∈ / . Chapter 11 deals with solving (11) when (a, b) ∈ . Moreover, not all (a, b) ∈ / are alike; some cause more difficulty than others. Chapter 14 deals with such points. Chapters 12 and 13 are concerned with the multidimensional wave equation (13) (14)

u ≡ u t t − u = p(x, t, u), u(x, t) = 0,

t ∈ R, x ∈ ∂.

In Chapter 12 we study rotationally invariant solutions. In Chapter 13 we study the more general situation. In Chapter 15 we discuss systems of equations for which the method of sandwich pairs would be ideal. However, sandwich pairs are plagued by the same problem as linking subsets, namely, that they cannot be verified as sandwich pairs unless at least one of the sets is contained in a finite-dimensional manifold. We show in Chapter 15 that they can both be infinite-dimensional if we add an assumption on G that is satisfied in most applications. In Chapter 16 we show that in many situations, nonlinear boundary-value problems have multiple solutions. Chapter 17 deals with second-order periodic systems of the form (15)

−x(t) ¨ = ∇x V (t, x(t)),

where (16)

x(t) = (x 1 (t), . . . , x n (t)).

We give several sets of conditions that imply the existence of solutions and conditions which imply the existence of nonconstant solutions. Martin Schechter Irvine, California TVSLB O

Chapter 1

Critical Points of Functionals 1.1 Introduction Many problems arising in science and engineering call for the solving of the Euler–Lagrange equations of functionals, i.e., equations equivalent to G (u) = 0,

(1.1)

where G(u) is a C 1 -functional (usually representing the energy) arising from the given data. By this we mean that functions are solutions of the Euler–Lagrange equations of G iff they satisfy (1.1). (For various definitions of the derivative G (u) of G, cf., e.g., [127].) Solutions of (1.1) are called critical points of G. Thus, solving the Euler–Lagrange equations is tantamount to finding critical points of the corresponding functional. As an illustration, the equation −u(x) = f (x, u(x)) is the Euler–Lagrange equation of the functional 1 2 G(u) = ∇u − F(x, u(x)) d x 2 on an appropriate space, where (1.2)

F(x, t) =

t

f (x, s) ds,

0

and the norm is that of L 2 . The variational approach to solving differential equations and systems has its roots in the calculus of variations. The original problem was to minimize or maximize a given functional. The approach was to obtain the Euler– Lagrange equations of the functional, solve them, and show that the solutions provided the required minimum or maximum. This worked well for one-dimensional problems. However, when it came to higher dimensions, it was recognized quite early that it was M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_1, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009

2

1. Critical Points of Functionals

more difficult to solve the Euler–Lagrange equations than it was to find minima or maxima of functionals. Consequently, the approach was abandoned for many years. Eventually, when nonlinear partial differential equations and systems arose in applications and people were searching for solutions, they began to check if the equations and systems were the Euler–Lagrange equations of functionals. If so, a natural approach is to find critical points of the corresponding functionals. The problem is that there is no uniform way of finding them.

1.2 Extrema The usual approach to finding critical points was to look for maxima or minima. Global extrema are the easiest to find, but they can exist only if the functional is semibounded. For instance, if the continuously differential functional G is bounded from below, then we can find a minimizing sequence {u k } such that (1.3)

G(u k ) → a = inf G > −∞.

If this series converges or has a convergent subsequence, we have a minimum. However, we have no guarantee that a subsequence will converge. Moreover, a minimizing sequence provides very little structure that can help one find a convergent subsequence. Indeed, one can find simple examples of functionals that are bounded below and have no minimum. For instance, the function y(x) = e x has no minimum on R even though it is bounded from below.

1.3 Palais–Smale sequences On the other hand, if the functional G is bounded from below, it can be shown that there is a sequence satisfying (1.4)

G(u k ) → a,

G (u k ) → 0.

(Actually, one can do better; see below.) If the sequence has a convergent subsequence, this will produce a minimum. Such a sequence is called a Palais–Smale (PS) sequence. The advantage of having a PS sequence instead of a minimizing sequence is that the additional information G (u k ) → 0 helps one show that a PS sequence has a convergent subsequence more readily than one can show that a minimizing sequence has a convergent subsequence.

1.4. Cerami sequences

3

1.4 Cerami sequences Actually, it can be shown that a C 1 -functional G that is bounded from below has a Cerami sequence, a sequence satisfying (1.5)

(1 + u k )G (u k ) → 0

G(u k ) → a,

(cf. Theorem 3.21 and its corollary). A Cerami sequence says more. There are examples for which one can show that a Cerami sequence has a convergent subsequence, while one cannot do the same for the corresponding PS sequence.

1.5 Linking sets When the functional is not semibounded, there is no clear way of obtaining critical points. In general, one would like to determine when a functional has PS or Cerami sequences. That is, one would like to find situations that give one the same advantages that one has in the case of semibounded functionals. An idea that has been very successful is to find appropriate sets that separate the functional. By this we mean the following: Definition 1.1. Two sets A, B separate the functional G(u) if (1.6)

a0 := sup G < b0 := inf G. A

B

We would like to find sets A and B such that (1.6) will imply (1.7)

∃ u : G(u) ≥ b0 , G (u) = 0.

This is too much to expect, since even semiboundedness alone does not imply the existence of a critical point. Consequently, we weaken our requirements and look for sets A, B such that (1.6) implies the existence of a PS sequence (1.4) with a ≥ b0 . This leads to Definition 1.2. We shall say that the set A links the set B if (1.6) implies (1.4) with a ≥ b0 for every C 1 -functional G(u). Of course, (1.4) is a far cry from (1.7), but if, e.g., the sequence (1.4) has a convergent subsequence, then (1.4) implies (1.7). Whether or not (1.4) implies (1.7) is a property of the functional G(u). We state this as Definition 1.3. We say that G(u) satisfies the PS condition if (1.4) always implies (1.7). The usual way of verifying this is to show that every sequence satisfying (1.4) has a convergent subsequence (there are other ways).

4

1. Critical Points of Functionals All of this leads to

Theorem 1.4. If G satisfies the PS condition and is separated by a pair of linking sets, then it has a critical point satisfying (1.7). This theorem cannot be applied until one identifies linking sets and functionals that satisfy the PS condition. Fortunately, they exist. We shall describe many of the known linking sets later in the book. Among other things, we shall discuss when a0 := sup G ≤ b0 := inf G

(1.8)

B

A

implies the existence of a PS sequence or a Cerami sequence. If there do not exist linking sets that separate a functional, all is not lost. There exist sets that produce PS or Cerami sequences when they do not separate a functional. That is, we shall find sets such that (1.9)

−∞ < b0 := inf G, B

a0 := sup G < ∞ A

implies the existence of a PS sequence or a Cerami sequence.

1.6 Previous definitions of linking In the earlier versions of linking, A is a compact set and is the “boundary” of a manifold S. They say that A links a set B if A∩B =φ and every continuous map ϕ from S to E that equals the identity on A must satisfy ϕ(S) ∩ B = φ. The underlying theorem is Theorem 1.5. If the set A links the set B and (1.6) holds, then there is a Palais–Smale sequence satisfying (1.4) with a ≥ b0 . There are distinct disadvantages of this definition. First, it requires A to be compact. Second, it requires A to be the “boundary” of a manifold S. Third, linking depends on the manifold S. Finally, there is no possibility of symmetry in infinite-dimensional spaces, i.e., if A links B, then B cannot link A according to this definition. Several methods have been used to circumvent these shortcomings and extend the definition of linking to cover cases when the old definition does not apply. Some of them are described in Chapter 3. The purpose of this volume is to present a uniform approach that includes most, if not all, theories of linking. It employs minimax systems and is presented in Chapter 2. The outline of this book is as follows. In Chapter 5 we give the proofs of the theorems of Chapter 2. We use methods of solving differential equations in Banach

1.7. Notes and remarks

5

spaces described in Chapter 4. In Chapter 6 we show that our uniform method is almost identical to the general Definition 1.2 given above. In Chapter 7 we consider the case when there are no linking sets that separate the functional. We find pairs of sets (called sandwich pairs) that produce PS sequences (1.4) when (1.6) fails. In Chapters 8, 9, 11, 14, 16, and 17 we give some applications to partial differential equations of the methods presented in Chapters 1–7. In Chapters 10, 12, 13, and 15 we are concerned with the situation when A and B are both infinite-dimensional.

1.7 Notes and remarks For an excellent review of critical point theory and linking, we refer to [96]. Previous definitions of linking theory are described in [8]. Other definitions of linking will be discussed in Chapters 2 and 3.

Chapter 2

Minimax Systems 2.1 Introduction We have described how the variational approach to solving nonlinear problems eventually leads to the search for critical points of related functionals. In the case of semibounded functionals, one can look for extrema. Otherwise, one is forced to use other methods. As we mentioned, linking provides a useful tool. There are several approaches to linking. In this book we unify these approaches, providing one theory that works for all of them. The idea is as follows. For each A ⊂ E, one wishes to find a collection K of sets K with the following properties. 1. If G ∈ C 1 (E, R) satisfies (2.1)

a0 := sup G < a := inf sup G, K ∈K K

A

then there is a sequence {u k } ⊂ E such that (1.4) holds. 2. If (2.2)

a0 ≤ sup G, K \A

K ∈ K,

then there is a sequence {u k } ⊂ E such that (1.4) holds. 3. If A, B ⊂ E satisfy (2.3)

A ∩ B = φ,

B ∩ K = φ,

K ∈ K,

and (2.4)

a0 := sup G < b0 := inf G, A

B

then there is a sequence {u k } ⊂ E such that (1.4) holds.

M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_2, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009

8

2. Minimax Systems 4. If

(2.5)

g K = {v ∈ K \A : G(v) ≥ a0 } = φ,

K ∈ K,

then there is a sequence {u k } ⊂ E such that (1.4) holds. In the next section we determine collections K that have these properties. Moreover, we shall show that all of these collections produce not only PS sequences (1.4), but also Cerami sequences (1.5) as well. Proofs of the theorems of this chapter will be given in Chapter 5.

2.2 Definitions and theorems We begin by studying C 1 -functionals on a Banach space E. Definition 2.1. We shall say that a map ϕ : E → E is of class if it is a homeomorphism onto E, and both ϕ, ϕ −1 are bounded on bounded sets. Definition 2.2. For A ⊂ E, we define (A) = {ϕ ∈ : ϕ(u) = u, u ∈ A}. Definition 2.3. For a nonempty set A ⊂ E, we define a nonempty collection K = K(A) of subsets K ⊂ E to be a minimax system for A if it has the following property: ϕ(K ) ∈ K,

ϕ ∈ (A), K ∈ K.

Every nonempty set has a minimax system. We have Theorem 2.4. Let K be a minimax system for a nonempty subset A of E, and let G(u) be a C 1 -functional on E. Define (2.6)

a := inf sup G, K ∈K K

and assume that a is finite and satisfies (2.7)

a > a0 := sup G. A

Let ψ(t) be a positive, nonincreasing, locally Lipschitz continuous function on [0, ∞) such that ∞ (2.8) ψ(r ) dr = ∞. 0

Then there is a sequence {u k } ⊂ E such that (2.9)

G(u k ) → a,

G (u k )/ψ( u k ) → 0.

Definition 2.5. We shall call σ (t) ∈ C([0, T ] × E, E), T > 0, a flow if σ (0) = I and for each t ∈ [0, T ], σ (t) is a homeomorphism of E onto itself.

2.3. Linking subsets

9

Theorem 2.6. Let K be a minimax system for a nonempty set A and let G(u) be a C 1 functional on E such that (2.10)

g K = {v ∈ K \A : G(v) ≥ a0 } = φ,

K ∈ K,

and a = inf sup G < ∞.

(2.11)

K

K

Assume that for any b > a, K ∈ K and flow σ (t) satisfying (2.12)

G(σ (t)v) < b,

v ∈ K , 0 < t ≤ T,

G(σ (t)u) < a,

u ∈ A, 0 < t ≤ T,

and (2.13)

there is a K˜ ∈ K such that (2.14)

K˜ ⊂

σ (t)A ∪ σ (T )[E b ∪ K ],

t ∈[0,T ]

where (2.15)

E α = {u ∈ E : G(u) ≤ α}.

Then the conclusions of Theorem 2.4 hold.

2.3 Linking subsets We now show how linking can play a major role in finding critical points. Definition 2.7. We shall say that a set A in E links a set B ⊂ E relative to a minimax system K for A if (2.16)

A∩B =φ

and (2.17)

B ∩ K = φ,

K ∈ K.

We shall say that A links B [mm] if there is a minimax system K for A such that A links B relative to K. Theorem 2.8. Let K be a minimax system for a nonempty set A, and assume that there is a subset B of E such that A links B relative to K. Assume that G ∈ C 1 (E, R) satisfies (2.18)

a0 := sup G < b0 := inf G A

B

and that the quantity a given by (2.6) is finite. Then, for each positive, nonincreasing, locally Lipschitz continuous function ψ(t) on [0, ∞) satisfying (2.8), there is a sequence {u k } ⊂ E such that (2.9) holds.

10

2. Minimax Systems

Definition 2.9. A subset A of a Banach space E links a subset B of E if, for every G ∈ C 1 (E, R) bounded on bounded sets and satisfying (2.18) there are a sequence {u k } ⊂ E and a constant a such that (2.19)

b0 ≤ a < ∞

and (1.4) holds. Definition 2.10. A subset A of a Banach space E links a subset B of E strongly if, for every G ∈ C 1 (E, R) bounded on bounded sets and satisfying (2.18) and each positive, nonincreasing, locally Lipschitz continuous function ψ(t) on [0, ∞) satisfying (2.8), there is a sequence {u k } ⊂ E such that (2.9) and (2.19) hold. Theorem 2.11. If A links B [mm], then it links B strongly. We note that Theorem 1.5 is a simple consequence of Theorem 2.4. In fact, we can let K be the collection K = {ϕ(S) : ϕ ∈ C(S, E), ϕ(u) = u, u ∈ A}. It is easily checked that K is a minimax system for A. Moreover, if A links B in the old sense, then it links B [mm]. We can now apply Theorem 2.4. In the following theorems, we allow a = a0 . Theorem 2.12. Let K be a minimax system for a nonempty set A, and assume that A links a subset B of E relative to K. Let G be a C 1 -functional satisfying (2.10), (2.11), and (2.20)

a0 := sup G ≤ b0 := inf G. B

A

Assume that for any b > a, K ∈ K, and flow σ (t) satisfying (2.12), (2.13) and such that (2.21)

σ (t)A ∩ B = φ,

t ∈ [0, T ],

there is a K˜ ∈ K such that (2.14) holds. Then the conclusions of Theorem 2.4 hold. Moreover, if a0 = a, then there is a sequence satisfying (1.4) and (2.22)

d(u k , B) → 0,

k → ∞.

Theorem 2.13. Let K be a minimax system for a nonempty set A satisfying (2.23)

K \A = φ,

K ∈ K.

Let G ∈ C 1 satisfy (2.10) and (2.11). Assume that for any K ∈ K and flow σ (t) satisfying (2.12), (2.13), one has S(K ) ∈ K, where (2.24)

S(u) = σ (d(u, A))u,

u ∈ E.

Assume also that for each K ∈ K, there is a ρ > 0 such that G(u) is Lipschitz continuous on (2.25)

K ρ = {u ∈ K : d(u, A) ≤ ρ}.


2.4. A variation

11

Theorem 2.14. Let K be a minimax system for a nonempty set A, and assume that there is a subset B of E such that A links B relative to K. Assume that for any K ∈ K and flow σ (t) satisfying (2.12), (2.13), one has S(K ) ∈ K, where S(u) is given by (2.24). Let G be a C 1 -functional satisfying (2.11) and (2.20). Assume also that for each K ∈ K there is a ρ > 0 such that G(u) is Lipschitz continuous on K ρ given by (2.25). Then the conclusions of Theorem 2.4 hold. Corollary 2.15. Let K be a minimax system for a nonempty set A, and let G ∈ C 1 satisfy (2.10) and (2.11). Assume that for any b > a, K ∈ K, and flow σ (t) satisfying (2.12), (2.13), and (2.21) for B = {u ∈ E\A : G(u) ≥ a0 },

(2.26)

there is a K˜ ∈ K satisfying K˜ ⊂ σ (T )E b . Then the conclusions of Theorem 2.4 hold.

2.4 A variation Theorem 2.16. Let K be a minimax system for a nonempty subset A of E, and let G(u) be a C 1 -functional on E. Define (2.27)

b := sup inf G, K ∈K K

and assume that b is finite and satisfies (2.28)

b < b0 := inf G. A

Then the conclusions of Theorem 2.4 hold with a replaced by b. We also have the following. Let A = {0}. Then a minimax system for A can be constructed as follows. K consists of the boundaries of all bounded, open sets ω containing 0. That K is a minimax system for A follows from the fact that 0 ∈ ϕ(ω) and ∂ϕ(ω) = ϕ(∂ω) for all such sets and ϕ ∈ (A). We have Theorem 2.17. Let b be defined by (2.27). Assume that G(0) < b < ∞. Then there is a sequence satisfying (2.29)

G(u k ) → b,

G (u k )/ψ( u k ) → 0.

Theorem 2.18. Assume that b < ∞ and that there is an open set ω0 containing 0 such that G(0) = inf G. ∂ω0

Then there is a sequence satisfying (2.29). We also have the counterpart of these theorems for the quantity a given by (2.6). Theorem 2.19. Assume that −∞ < a < G(0). Then the conclusions of Theorem 2.4 hold.

12

2. Minimax Systems

Theorem 2.20. Assume that a > −∞ and that there is an open set ω0 containing 0 such that G(0) = sup G. ∂ω0


2.5 Weaker conditions We now turn to the question as to what happens if some of the hypotheses of Theorem 2.12 do not hold. We are particularly interested in what happens when (2.20) is violated. In this case we let B := {v ∈ B : G(v) < a0 }.

(2.30) Note that

B = φ iff a0 ≤ b0 .

We assume that B = φ. Let ψ(t) be a positive, nonincreasing function on [0, ∞) satisfying the hypotheses of Theorem 2.4 and such that R+α ψ(t) dt (2.31) a0 − b0 < α

for some finite R ≤

d

:=

d(B ,

A), where α = d(0, B ). We assume d > 0. We have

Theorem 2.21. Let G be a C 1 -functional on E and A, B ⊂ E be such that A links B [mm] and (2.32)

−∞ < b0 ,

a < ∞.

Assume that for any b > a, K ∈ K and flow σ (t) satisfying (2.12), (2.13), (2.21), there is a K˜ ∈ K satisfying (2.14). Under the hypotheses given above, for each δ > 0, there is a u ∈ E such that (2.33)

G (u) < ψ(d(u, B )).

b0 − δ ≤ G(u) ≤ a + δ,

We can also consider a slightly different version of Theorem 2.21. We consider the set (2.34)

A := {u ∈ A : G(u) > b0 },

and we note that A = φ iff a0 ≤ b0 . We assume that ψ satisfies the hypotheses of Theorem 2.4 and R+β (2.35) a0 − b0 < ψ(t) dt β

holds for some finite R ≤ d := d(A , B), where β = d(0, A ). Assume that A = φ. We have

2.6. Some consequences

13

Theorem 2.22. Assume that for any b < b0 , K ∈ K, and flow σ (t) satisfying (2.36)

G(σ (t)v) > b,

(2.37)

G(σ (t)v) > b0 ,

v ∈ K , t > 0, v ∈ B, t > 0,

and σ (t)B ∩ A = φ,

(2.38)

t ∈ [0, T ],

there is a K˜ ∈ K such that K˜ ⊂

(2.39)

σ (t)B ∪ σ (T )[E b ∪ K ],

t ∈[0,T ]

where E α = {u ∈ E : G(u) ≥ α}.

(2.40) If B links A [mm] and (2.41)

−∞ < b0 ,

a 0, there is a u ∈ E such that (2.42)

b0 − δ ≤ G(u) ≤ a + δ,

G (u) < ψ(d(u, A )).

2.6 Some consequences We now discuss some methods that follow from Theorems 2.21 and 2.22. Let {Ak , Bk } be a sequence of pairs of subsets of E. Let Kk be a minimax system for Ak . For G ∈ C 1 (E, R), let (2.43)

ak0 = sup G,

bk0 = inf G, Bk

Ak

and (2.44)

ak = inf sup G. K ∈K k K

We assume ak < ∞ for each k. We define (2.45)

Bk := {v ∈ Bk : G(v) < ak0 },

(2.46)

Ak := {u ∈ Ak : G(u) > bk0 }

(2.47)

dk := d(Ak , Bk ),

dk := d(Ak , Bk ).

14

2. Minimax Systems

We have Theorem 2.23. Assume that Ak links Bk [mm] for each k, that dk → ∞ as k → ∞,

(2.48)

and for each k, there is a positive, nonincreasing function ψk (t) on [0, ∞) satisfying the hypotheses of Theorem 2.4 and such that Rk +αk (2.49) ak0 − bk0 < ψk (t)dt, αk

where αk = d(0, Bk ) and Rk ≤ dk . Then, under the hypotheses of Theorem 2.21, there is a sequence {u k } ⊂ E such that bk0 − (1/k) ≤ G(u k ) ≤ ak + (1/k)

(2.50) and

G (u k ) ≤ ψk (d(u k , Bk )).

(2.51)

Theorem 2.24. Assume that Bk links Ak [mm] for each k, that dk → ∞ as k → ∞,

(2.52)

and that, for each k, there is a positive, nonincreasing function ψk (t) on [0, ∞) satisfying the hypotheses of Theorem 2.4 and such that Rk +βk ψk (t)dt, (2.53) ak0 − bk0 < βk

Ak )

dk .

where βk = d(0, and Rk ≤ is a sequence {u k } ⊂ E such that (2.54)

Then, under the hypotheses of Theorem 2.22, there

bk0 − (1/k) ≤ G(u k ) ≤ ak + (1/k)

and (2.55)

G (u k ) ≤ ψk (d(u k , Ak )).

We combine the proofs of Theorems 2.23 and 2.24. Proof. For each k, take Rk equal to dk or dk , as the case may be. We may assume that bk0 < ak0 for each k. Otherwise the conclusions of the theorems follow from Theorem 2.4. We can now apply Theorems 2.21 and 2.22 for each k to conclude that there is a u k ∈ E such that bk0 − (1/k) ≤ G(u k ) ≤ ak + (1/k), and either or as the case may be.

G (u k ) < ψk (d(u k , Bk )) G (u k ) < ψk (d(u k , Ak )),


15

2.7 Notes and remarks The concept of a minimax system is related to that of [74] (cf. Section 3.4). The first situation that allowed a0 = a appeared in [75]. The first to consider the case ψ(t) = 1/(1 + |t|) was [39]. The first to consider the general case of arbitrary nonincreasing ψ was [107]. The theorems of Section 2.4 are due to [107]. The theorems of Sections 2.2, 2.3, 2.5, and 2.6 are due to [134].

Chapter 3

Examples of Minimax Systems 3.1 Introduction In this chapter we present some linking methods which are special cases of linking with respect to minimax systems. We consider three useful methods and show that they provide minimax systems. We then consider examples and applications.

3.2 A method using homeomorphisms One linking method can be described as follows. Let E be a Banach space and let be the set of all continuous maps = (t) from E × [0, 1] to E such that 1. (0) = I , the identity map. 2. For each t ∈ [0, 1), (t) is a homeomorphism of E onto E and −1 (t) ∈ C(E × [0, 1), E). 3. (1)E is a single point in E and (t)A converges uniformly to (1)E as t → 1 for each bounded set A ⊂ E. 4. For each t0 ∈ [0, 1) and each bounded set A ⊂ E, (3.1)

sup { (t)u + −1 (t)u } < ∞. 0≤t≤t0 u∈A

We have Theorem 3.1. Let G be a C 1 -functional on E, and let A be a subset of E. Assume that (3.2)

a0 := sup G < a := inf sup G((s)u) < ∞. A

∈

0≤s≤1 u∈A

Let ψ(t) be a positive, nonincreasing, locally Lipschitz continuous function on [0, ∞) satisfying (2.8). Then there is a sequence {u k } ⊂ E such that (2.9) holds. In particular, there is a Cerami sequence satisfying (1.5). M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_3, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009

18

3. Examples of Minimax Systems

Proof. In this case, we let K be the collection of sets of the form (3.3)

K = {(t)u : t ∈ [0, 1], u ∈ A, ∈ }.

To show that this is a minimax system, let ϕ be a mapping in (A), and let (t) be in . Define the mapping 1 (t)u = ϕ ◦ (t) ◦ ϕ −1 u,

u ∈ E.

Then 1 (t) ∈ . Moreover, 1 (t)u = ϕ ◦ (t)u,

u ∈ A.

Consequently, 1 (t)A = ϕ({(t)u : u ∈ A}), showing that the collection K is a minimax system for A. Since (3.2) now becomes (2.7), the result follows. Definition 3.2. We say that A links B [hm] if A, B are subsets of E such that A∩ B = φ and, for each (t) ∈ , there is a t ∈ (0, 1] such that (t)A ∩ B = φ. We have Corollary 3.3. If A links B [hm], then it links B [mm]. We also have Theorem 3.4. Let A, B be subsets of E such that A links B [hm], and let G be a C 1 -functional on E, satisfying (3.4)

a0 := sup G ≤ b0 := inf G. B

A

Assume that (3.5)

a := inf sup G((s)u) ∈

0≤s≤1 u∈A

is finite. Let ψ(t) be a positive, nonincreasing, locally Lipschitz continuous function on [0, ∞) satisfying (2.8). Then there is a sequence {u k } ⊂ E such that (2.9) holds. In particular, there is a Cerami sequence satisfying (1.5). Proof. The theorem follows from Theorem 2.12. If (3.6)

K = {(t)u : t ∈ [0, 1], u ∈ A}

for some ∈ , and σ (t) is any flow, let σ (2T s), ˜ (3.7) (s) = σ (T )(2s − 1),

0 ≤ s ≤ 12 , 1 2

< s ≤ 1.

It is easily checked that ˜ ∈ . Moreover, (3.8) K˜ ⊂ σ (t)A ∪ σ (T )K . t ∈[0,T ]

Hence, (2.14) is satisfied.

3.3. A method using metric spaces

19

3.3 A method using metric spaces Another general procedure is described in Mawhin–Willem [96] and Brezis–Nirenberg [28] as follows. One finds a compact metric space and selects a closed subset ∗ of such that ∗ = φ, ∗ = . One then picks a map p∗ ∈ C(, E) and defines A ={ p ∈ C(, E) : p = p∗ on ∗ }, a = inf max G( p(ξ )). p∈A ξ ∈

They assume (A) For each p ∈ A, maxξ ∈ G( p(ξ )) is attained at a point in \ ∗ . They then prove Theorem 3.5. Under the above hypotheses, there is a sequence satisfying (1.4). We shall prove Theorem 3.6. Under the same hypotheses, let ψ(t) be a positive, nonincreasing, locally Lipschitz continuous function on [0, ∞) satisfying (2.8). Then there is a sequence {u k } ⊂ E such that (2.9) holds. In particular, there is a Cerami sequence satisfying (1.5). Proof. If a > a0 , this follows from Theorem 2.4. In fact, we can take A = p∗ ( ∗ ) and K as the collection of sets K = { p(ξ ) : ξ ∈ , p ∈ A}. If ϕ ∈ (A), then ϕ( p(ξ )) = ϕ( p∗ (ξ )) = p ∗ (ξ ),

ξ ∈ ∗ .

Thus, ϕ( p(ξ )) ∈ K. Hence, K is a minimax system for A, and we can apply Theorem 2.4 to come to the desired conclusion. Now assume that a0 = a. Since each K ∈ K is compact, the same is true of K ρ given by (2.25) for each ρ > 0. Moreover, if σ (t) ∈ C(R+ × E, E) is such that σ (0) = I, one has S(K ) ∈ K, where S( p(ξ )) = σ (d( p(ξ ), A)) p(ξ ),

ξ ∈ .

The result follows from Theorem 2.14.

3.4 A method using homotopy-stable families Another approach is found in Ghoussoub [74]. The author considers a closed subset A of E, and a collection K of compact subsets of E having the property that if K ∈ K and ψ ∈ C([0, 1] × E, E) satisfies ψ(0)u = u,

u ∈ E,

20


and ψ(t)u = u,

t ∈ [0, 1], u ∈ A,

then ψ(1)K ∈ K. He calls such a collection a homotopy-stable family with extended boundary A. He proves Theorem 3.7. If K is a homotopy-stable family with extended closed boundary A, B is a closed subset of E satisfying (1.1), and G is a C 1 -functional that satisfies sup G ≤ a ≤ inf G, B

A

where the quantity a is given by (2.6), then there is a PS sequence satisfying (1.4). We shall prove Theorem 3.8. Under the same assumptions, for each function ψ(t) satisfying the hypotheses of Theorem 2.4, there is a sequence satisfying (2.9). In particular, there is a Cerami sequence satisfying (1.5). Proof. It is easy to show that such a collection K is indeed a minimax system. Indeed, if K ∈ K and ϕ ∈ (A), then ψ(t)u = tϕ(u) + (1 − t)u satisfies the stipulations above, and consequently ϕ(K ) ∈ K. Since each of the members of K is compact, it follows that every C 1 -functional is Lipschitz continuous on some set K ρ defined by (2.25) for ρ > 0 sufficiently small. Moreover, if σ (t) ∈ C(R+ × E, E) is such that σ (0) = I, one has S(K ) ∈ K, where S(u) = σ (d(u, A))u,

u ∈ E.

This follows from the fact that S(t)u = σ (td(u, A))u is in ∈ C([0, 1] × E, E) and satisfies S(0)u = u, u ∈ E, and S(t)u = u,

t ∈ [0, 1], u ∈ A.

Consequently, S(K ) = S(1)K ∈ K by hypothesis. It now follows that the conclusion of Theorem 2.14 holds. Note. It is not required to have a satisfy a ≤ inf G. B

If it does, one obtains additional information as described in [74].

3.5. Examples of linking sets

21

3.5 Examples of linking sets We now discuss various subsets of a Banach space E with respect to linking. First we have Proposition 3.9. [122] Let A, B be two closed, bounded subsets of E such that E\A is path connected. If A links B [hm], then B links A [hm]. The next proposition gives a very useful method of checking the linking of two sets. Proposition 3.10. [122] Let F be a continuous map from E to Rn , and let Q ⊂ E be such that F0 = F| Q is a homeomorphism of Q onto the closure of a bounded open subset of Rn . If p ∈ , then F0−1 (∂) links F −1 ( p) [hm]. Proposition 3.11. [122] If H is a homeomorphism of E onto itself and A links B [hm], then H A links H B [hm]. The following examples were given in [122]. Example 1. Let B be an open set in E, and let A consist of two points e1 , e2 with ¯ Then A links ∂ B [hm]. ∂ B links A [hm] as well if ∂ B is bounded. e1 ∈ B and e2 ∈ / B. Example 2. Let M, N be closed subspaces such that dim N < ∞ and E = M ⊕ N. Let B R = {u ∈ E : u < R}

(3.9)

and take A = ∂ B R ∩ N, B = M. Then A links B [hm]. Example 3. We take M, N as in Example 2. Let w0 = 0 be an element of M, and take A

=

{v ∈ N : v ≤ R} ∪ {sw0 + v : v ∈ N, s ≥ 0, sw0 + v = R},

B

=

∂ Bδ ∩ M, 0 < δ < R.

Then A and B link each other [hm]. Example 4. Take M, N as before and let v 0 = 0 be an element of N. We write N = {v 0 } ⊕ N . We take A B

= =

{v ∈ N : v ≤ R} ∪ {sv 0 + v : v ∈ N , s ≥ 0, sv 0 + v = R}, {w ∈ M : w ≥ δ} ∪ {sv 0 + w : w ∈ M, s ≥ 0, sv 0 + w = δ},

where 0 < δ < R. Then A links B [hm]. Example 5. This is the same as Example 4 with A replaced by A = ∂ B R ∩ N. Example 6. Let M, N be as in Example 2. Take A = ∂ Bδ ∩ N, and let v 0 be any element in ∂ B1 ∩ N. Take B to be the set of all u of the form u = w + sv 0 , w ∈ M,

22


satisfying any of the following: (a) w ≤ R, s = 0 (b) w ≤ R, s = 2R0 (c) w = R, 0 ≤ s ≤ 2R0 , where 0 < δ < min(R, R0 ). Then A and B link each other [hm]. Example 7. Let M, N be as in Example 2. Let v 0 be in ∂ B1 ∩ N and write N = {v 0 } ⊕ N . Let A = ∂ Bδ ∩ N, Q = B¯ δ ∩ N, and B = {w ∈ M : w ≤ R} ∪ {w + sv 0 : w ∈ M, s ≥ 0, w + sv 0 = R}, where 0 < δ < R. Then A and B link each other [hm]. Example 8. Let M, N be closed subspaces of E, one of which is finite-dimensional, and such that E = M ⊕ N. If B R := {u ∈ E : u < R}, then M ∩ ∂ B R links N [hm] for each R > 0. Example 9. Let M, N be closed subspaces of E such that E = M ⊕ N, with one of them being finite-dimensional. Let w0 be an element of M\{0}, and let 0 < δ < r < R. Take A

B

=

{v ∈ N : δ ≤ v ≤ R} ∪ {sw0 + v : v ∈ N, s ≥ 0, sw0 + v = δ}

∪

{sw0 + v : v ∈ N, s ≥ 0, sw0 + v = R},

=

∂ Br ∩ M, 0 < δ < r < R.

Then A and B link each other [hm]. Example 10. Let M, N be closed subspaces of E such that E = M ⊕ N, with one of them being finite-dimensional. Let w0 be an element of M\{0}, and let 0 < r < R, A B

= =

{w ∈ M : w = R}, {v ∈ N : v ≥ r } ∪ {u = v + sw0 : v ∈ N, s ≥ 0, u = r }.

Then A links B [hm].

3.5. Examples of linking sets

23

Example 11. Let M, N be as in Example 2. Take A = ∂ Bδ ∩ N, and let v 0 be any element in ∂ B1 ∩ N. Take B to be the set of all u of the form u = w + sv 0 ,

w ∈ M,

satisfying any of the following: (a) s = 0 (b) s = 2R0 where 0 < δ < R0 . Then A links B [hm]. Example 12. Let M, N be as in Example 2. Take A = ∂ Bδ ∩ N, and let v 0 be any element in ∂ B1 ∩ N. Take B to be the set of all u of the form u = w + sv 0 ,

w ∈ M,

satisfying any of the following: (a) w ≤ R, s = 0, (b) w = R, s > 0, where 0 < δ < ∞. Then A links B [hm]. Example 13. Let M be a closed subspace of a Hilbert space E with complement N ⊕ {v 0 }, where v 0 is an element in E having unit norm, and let δ be any positive number. Let ϕ(t) ∈ C 1 (R) be such that 0 ≤ ϕ(t) ≤ 1,

ϕ(0) = 1,

and ϕ(t) = 0,

|t| ≥ 1.

Let F(v + w + sv 0 ) = w + [s + δ − δϕ( v 2 /δ 2 )]v 0 ,

v ∈ N, w ∈ M, s ∈ R.

Assume that one of the subspaces M, N is finite-dimensional. Take A = [M ⊕ {v 0 }] ∩ ∂ B R and B = {v + r v 0 : v ∈ N, r = δϕ( v 2 /δ 2 )}. Then A links B [hm] provided 0 < δ < R. Proposition 3.12. [122] If K is any subset of a bounded open set ⊂ E, then ∂ links K [hm].

24


3.6 Various geometries We now apply the theorems of the preceding sections to various geometries in Banach space. As before, we assume that G ∈ C 1 (E, R) and that ψ satisfies the hypotheses of Theorem 2.4. Theorem 3.13. Assume that there is a δ > 0 such that G(0) ≤ α ≤ G(u),

(3.10)

u ∈ ∂ Bδ ,

and that there are a R0 < ∞ and a ϕ0 ∈ ∂ B1 such that G(Rϕ0 ) ≤ γ ,

(3.11)

R > R0 .

Then, for each function ψ(t) satisfying the hypotheses of Theorem 2.4, there is a sequence {u k } ⊂ E such that (3.12)

G(u k ) → c,

G (u k )/ψ( u k ) → 0.

α ≤ c ≤ γ,

Proof. We take A = {0, Rϕ0 }, B = ∂ Bδ . Then A = {Rϕ0 }. Note that a given by (2.6) is finite for each R since a R ≤ max G(r ϕ0 ). 0≤r≤R

We apply Theorem 2.24. We note that in each case aR ≤ γ ,

(3.13)

R > 0,

since the mapping (s)u = (1 − s)u

(3.14) (which is in ) satisfies (3.15)

G((s)u) ≤ γ ,

0 ≤ s ≤ 1, u ∈ A.

˜ This implies (3.13). We replace ψ(t) with ψ(t) = ψ(t + δ), which also satisfies the hypotheses of Theorem 2.4. By Theorem 2.24, we can find a sequence satisfying (3.16)

α − (1/k) ≤ G(u k ) ≤ γ + (1/k),

˜ G (u k )/ψ(d(u k , B)) → 0.

This implies (3.12), since u ≤ d(u, B) + δ. Theorem 3.14. Let M, N be closed subspaces of E such that (3.17)

E = M ⊕ N,

M = E,

N = E,

with (3.18)

dim M < ∞ or dim N < ∞.

3.6. Various geometries

25

Let G ∈ C 1 (E, R) be such that (3.19)

G(v) ≤ γ ,

v ∈ ∂ B R ∩ N, R > R0 ,

and G(w) ≥ α,

(3.20)

w ∈ M.

Then, for each function ψ(t) satisfying the hypotheses of Theorem 2.4, there is a sequence {u k } ⊂ E such that (3.21)

G(u k ) → c,

α ≤ c ≤ γ,

G (u k )/ψ(d(u k , M)) → 0.

Proof. This time we take A and B as in Example 2 above. Thus, A links B [hm]. Again, a R given by (2.6) is finite for each R since a R ≤ max G(u). u∈ B¯ R ∩N

Again we see that we can apply Theorem 2.24 to conclude that the desired sequence exists. Theorem 3.15. Let M, N be as in Theorem 3.14, and let G ∈ C 1 (E, R) satisfy (3.22)

G(v) ≤ α,

v ∈ N,

(3.23)

G(w) ≥ α,

w ∈ ∂ Bδ ∩ M,

(3.24)

G(sw0 + v) ≤ γ ,

s ≥ 0, v ∈ N, sw0 + v = R > R0 ,

for some w0 ∈ ∂ B1 ∩ M, where 0 < δ < R0 . Then, for each function ψ(t) satisfying the hypotheses of Theorem 2.4, there is a sequence {u k } ⊂ E such that (3.12) holds. Proof. Here we take A, B as in Example 3 above. Thus, A and B link each other [hm]. Here A = {sw0 + v : s ≥ 0, v ∈ N, sw0 + v = R}. Again, for each R, the quantity a given by (2.6) is finite since a R ≤ max G, Q

where Q = {sw0 + v : s ≥ 0, v ∈ N, sw0 + v ≤ R}. We now apply Theorem 2.24 to conclude that the desired sequence exists. Theorem 3.16. Let M, N be as in Theorem 3.14, and let v 0 ∈ ∂ B1 ∩ N. Take N = {v 0 } ⊕ N . Let G ∈ C 1 (E, R) be such that (3.25) (3.26)

G(v) ≤ γ , G(w) ≥ α,

(3.27)

G(sv 0 + w) ≥ α,

v ∈ ∂ B R ∩ N, R > R0 , w ∈ M, w ≥ δ, s ≥ 0, w ∈ M, sv 0 + w = δ,

where 0 < δ < R0 . Then, for each function ψ(t) satisfying the hypotheses of Theorem 2.4, there is a sequence {u k } ⊂ E such that (3.12) holds.

26


Proof. We take A, B as in Example 5 above. Thus, A links B [hm]. As before, we note that a R < ∞ for each R. Hence, (3.12) holds by Theorem 2.24.

3.7 A sandwich theorem We now discuss a very useful theorem that allows one to consider functionals that are bounded from below on one subspace and bounded from above on another with no correlation between the bounds. This provides such functionals with the same advantages as those that are semibounded. One drawback is the requirement that one of the subspaces be finite-dimensional. This condition will be removed in Chapter 15 if we assume that the functional satisfies more than the mere continuity of its derivative. Theorem 3.17. Let N be a closed subspace of a Hilbert space E and let M = N ⊥ . Assume that at least one of the subspaces M, N is finite-dimensional. Let G be a C 1 -functional on E such that m 0 := inf G(w) = −∞

(3.28)

w∈M

and m 1 := sup G(v) = ∞.

(3.29)

v∈N

Let ψ(t) be a positive, nonincreasing, locally Lipschitz continuous function on [0, ∞) such that (2.8) holds. Then there are a constant c ∈ R and a sequence {u k } ⊂ E such that (3.30)

G(u k ) → c,

m0 ≤ c ≤ m1,

G (u k )/ψ( Pu k ) → 0,

where P is the (orthogonal) projection onto M. Proof. We may assume dim N < ∞; otherwise, we can consider −G in place of G. Let A be the set ∂ B R ∩ N, and take B = M, where R > 0 is arbitrary. Then A links B [hm] by Example 2 above. We now apply Theorem 2.21. Note that a0 ≤ m 1 , m 0 = b0 , and (2.31) holds for R sufficiently large. We also note that a R ≤ sup G ≤ m 1 B R ∩N

by taking (s)u = (1 − s)u, u ∈ E. Hence, by Theorem 2.21, for each δ > 0, there is a u ∈ E such that m 0 − δ ≤ G(u) ≤ m 1 + δ,

G (u) < ψ(d(u, B )).

Since this is true for each δ > 0, we obtain the desired conclusion.

3.7. A sandwich theorem

27

An immediate consequence is Corollary 3.18. Under the hypotheses of Theorem 3.17, there are a constant c ∈ R and a sequence {u k } ⊂ E such that (3.31)

G(u k ) → c,

m 0 ≤ c ≤ m 1,

(1 + Pu k )G (u k ) → 0,

where P is the (orthogonal) projection onto M. The following is a consequence of Theorem 2.23. Theorem 3.19. Under the hypotheses of Theorem 3.17, for any sequence {Rk } ⊂ R+ such that Rk → ∞, there are a constant c ∈ R and a sequence {u k } ⊂ E such that (3.32)

G(u k ) → c,

m 0 ≤ c ≤ m 1,

(Rk + u k ) G (u k ) ≤

m1 − m0 . ln(4/3)

Proof. We may assume dim N < ∞. Let Ak be the set ∂ B Rk ∩ N, and take Bk = M. Then, for each k, Ak links Bk [hm] by Example 2 above. We now apply Theorem 2.23. Note that αk = Rk and ak0 ≤ m 1 , m 0 = bk0 . Take ψk (t) =

m1 − m0 . [2Rk + t] ln(4/3)

Since Rk + d(u, Bk ) ≥ u , we see that (3.32) holds for each k. The following is another consequence of Theorem 2.23. Theorem 3.20. Let N be a closed subspace of a Hilbert space E, and let M = N ⊥ . Assume that at least one of the subspaces M, N is finite-dimensional. Let G be a C 1 -functional on E such that (3.33)

m 0 := sup inf G(v + w) = −∞ v∈N w∈M

and (3.34)

m 1 := inf sup G(v + w) = ∞. w∈M v∈N

Then, for any ε > 0 and any sequence {Rk } ⊂ R+ such that Rk → ∞, there are a constant c ∈ R and a sequence {u k } ⊂ E such that (3.35)

G(u k ) → c,

m 0 − ε ≤ c ≤ m 1 + ε, m 1 − m 0 + 3ε . (Rk + u k ) G (u k ) ≤ ln(5/4)

Proof. We may assume dim N < ∞. Let ε > 0 be given. Then there is a u ε = v ε + wε such that m 0 − ε < inf G(v ε + w), sup G(v + wε ) < m 1 + ε. w∈M

v∈N

28


Note that αk = Rk and ak0 ≤ m 1 , m 0 = bk0 . Take ψk (t) =

m 1 − m 0 + 3ε . [3Rk + t] ln(5/4)

Note that

2Rk

m 1 − m 0 + 2ε
−∞, E

with (3.39)

G(u k ) → b0 ,

G (u k )/ψ( u k ) → 0.


29

Proof. We refer to Theorem 2.24. We take a sequence of points such that G(v k ) > a0 − (1/k) and a sequence {Rk } such that Rk > 2 v k and ck = 2 k

Rk +2βk

−1 ψ(t) dt

→ 0,

k → ∞,

2βk

where βk = v k . We then take ψk (t) = ck ψ(t + βk ) in Theorem 2.24. Then (2.53) holds. We used the fact that ∂ B Rk +βk links {v k } for each k (Theorem 3.12). The conclusion follows since u ≤ d(u, v k ) + βk . In the second case, we replace G with −G. Corollary 3.22. If (3.36) holds, then there is a sequence satisfying (3.40)

G(u k ) → a0 ,

(1 + u k )G (u k ) → 0.

If (3.38) holds, there is a sequence satisfying (3.41)

G(u k ) → b0 ,

(1 + u k )G (u k ) → 0.

3.8 Notes and remarks The results of Section 3.2 are from [136], [114], and [120] (cf. also [122]). Sections 3.5 and 3.6 are from [122]. The results of Section 3.7 come from [143], [109], [108], [129], and [132].

Chapter 4

Ordinary Differential Equations 4.1 Extensions of Picard’s theorem In proving the theorems of Chapter 2, we shall make use of various extensions of Picard’s theorem in a Banach space. Some are well known, and some are of interest in their own right. All of them will be used in proving the theorems of Chapter 2. Theorem 4.1. Let X be a Banach space, and let B0 = {x ∈ X : x − x 0 ≤ R0 } and I0 = {t ∈ R : |t − t0 | ≤ T0 }. Assume that g(t, x) is a continuous map of I0 × B0 into X such that (4.1)

g(t, x) − g(t, y) ≤ K 0 x − y ,

x, y ∈ B0 , t ∈ I0 ,

and g(t, x) ≤ M0 ,

(4.2)

x ∈ B0 , t ∈ I 0 .

Let T1 be such that (4.3)

T1 ≤ min(T0 , R0 /M0 ),

K 0 T1 < 1.

Then there is a unique solution x(t) of (4.4)

d x(t) = g(t, x(t)), dt

|t − t0 | ≤ T1 ,

x(t0 ) = x 0 .

Lemma 4.2. Let γ (t) and ρ(t) be continuous functions on [0, ∞), with γ (t) nonnegative and ρ(t) positive and nondecreasing. Assume that ∞ T dτ (4.5) > γ (s) ds, u 0 ρ(τ ) t0 M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_4, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009

32

4. Ordinary Differential Equations

where t0 < T and u 0 ≥ 0 are given numbers. Then there is a unique solution of (4.6)

u (t) = γ (t)ρ(u(t)),

t ∈ [t0 , T ), u(t0 ) = u 0

that is positive in [t0 , T ) and depends continuously on u 0 . Proof. One can separate variables to obtain u t dτ W (u) ≡ = γ (s) ds. u 0 ρ(τ ) t0 The function W (u) is differentiable and increasing in R, is positive in (u 0 , ∞), depends continuously on u 0 , and satisfies

∞

W (u) → L =

u0

dτ > ρ(τ )

T

γ (s) ds

as u → ∞.

t0

Thus, for each t ∈ [t0 , T ), there is a unique u ∈ [u 0 , ∞) such that t W (u) = γ (s) ds. t0

Hence, u = W −1

t

γ (s) ds

t0

is the unique solution of (4.6) and depends continuously on u 0 . Lemma 4.3. Let γ (t) and ρ(t) be continuous functions on [0, ∞), with γ (t) nonnegative and ρ(t) positive and nondecreasing. Assume that

u0

(4.7) m

dτ > ρ(τ )

T

γ (s) ds,

t0

where t0 < T and m < u 0 are given numbers. Then there is a unique solution of (4.8)

u (t) = −γ (t)ρ(u(t)),

t ∈ [t0 , T ), u(t0 ) = u 0 ,

which is ≥ m in [t0 , T ) and depends continuously on u 0 . Proof. One can separate variables to obtain u0 t dτ W (u) ≡ = γ (s) ds. ρ(τ ) u t0 The function W (u) is differentiable and decreasing in R, is positive in [m, u 0 ], depends continuously on u 0 , and satisfies W (u) → L =

u0 m

dτ > ρ(τ )

T t0

γ (s) ds

as u → m.

4.2. Estimating solutions

33

Thus, for each t ∈ [t0 , T ), there is a unique u ∈ [m, u 0 ] such that t W (u) = γ (s) ds. t0

Hence, u = W −1

t

γ (s) ds

t0

is the unique solution of (4.8) and depends continuously on u 0 .

4.2 Estimating solutions Theorem 4.4. Assume, in addition to the hypotheses of Theorem 4.1, that (4.9)

g(t, x) ≤ γ (t)ρ( x ),

x ∈ B0 , t ∈ I0 ,

where γ (t) and ρ(t) satisfy the hypotheses of Lemma 4.2 with T = t0 + T1 . Let u(t) be the positive solution of (4.10)

u (t) = γ (t)ρ(u(t)), t ∈ [t0 , T ), u(t0 ) = u 0 ≥ x 0

provided by that lemma. Then the unique solution of (4.4) satisfies (4.11)

x(t) ≤ u(t),

t ∈ [t0 , T ).

Proof. Assume that there is a t1 ∈ [t0 , T ) such that u(t1 ) < x(t1 ) . For ε > 0, let u ε (t) be the solution of (4.12)

u (t) = [γ (t) + ε]ρ(u(t)),

t ∈ [t0 , T ), u(t0 ) = u 0 .

By Lemma 4.2, a solution exists for ε > 0 sufficiently small. Moreover, u ε (t) → u(t) uniformly on any compact subset of [t0 , T ). Let w(t) = x(t) − u ε (t). Then we may take ε sufficiently small so that w(t0 ) ≤ 0,

w(t1 ) > 0.

Let t2 be the largest number in [t0 , t1 ) such that w(t2 ) = 0 and w(t) > 0,

t ∈ (t2 , t1 ].

For h > 0 sufficiently small, we have w(t2 + h) − w(t2 ) > 0. h

34


Consequently,

D + w(t2 ) ≥ 0.

But (4.13)

D + w(t2 ) = D + x(t2 ) − u ε (t2 ) ≤ x (t2 ) − u ε (t2 ) = g(t2 , x(t2 )) − [γ (t2 ) + ε]ρ(u ε (t2 )) ≤ γ (t2 )ρ( x(t2 ) ) − [γ (t2 ) + ε]ρ(u ε (t2 )) = − ερ(u ε (t2 )) < 0.

This contradiction proves the theorem.

4.3 Extending solutions Theorem 4.5. Let g(t, x) be a continuous map from R× H to H , where H is a Banach space. Assume that for each point (t0 , x 0 ) ∈ R × H, there are constants K , b > 0 such that (4.14) g(t, x)− g(t, y) ≤ K x − y ,

|t −t0 | < b, x − x 0 < b, y − x 0 < b.

Assume also that (4.15)

g(t, x) ≤ γ (t)ρ( x ),

x ∈ H, t ∈ [t0 , TM ),

where TM ≤ ∞, and γ (t), ρ(t) satisfy the hypotheses of Lemma 4.2. Then, for each x 0 ∈ H and t0 > 0, there is a unique solution x(t) of the equation (4.16)

d x(t) = g(t, x(t)), dt

t ∈ [t0 , TM ), x(t0 ) = x 0 .

Moreover, x(t) depends continuously on x 0 and satisfies (4.17)

x(t) ≤ u(t),

t ∈ [t0 , TM ),

where u(t) is the solution of (4.6) in that interval satisfying u(t0 ) = u 0 ≥ x 0 . Before proving Theorem 4.5, we note that the following is an immediate consequence. Corollary 4.6. Let V (y) be a locally Lipschitz continuous map from H to itself satisfying (4.18)

V (y) ≤ C(1 + y ),

y ∈ H.

Then, for each y0 ∈ H, there is a unique solution of (4.19)

y (t) = V (y(t)),

t ∈ R+ , y(0) = y0 .

4.4. The proofs

35

4.4 The proofs We now give the proof of Theorem 4.5. Proof. By Theorems 4.1 and 4.4, there is an interval [t0 , t0 + m], m > 0, in which a unique solution of (4.20)

d x(t) = g(t, x(t)), dt

t ∈ [t0 , t0 + m], x(t0 ) = x 0

exists and satisfies (4.21)

t ∈ [t0 , t0 + m],

x(t) ≤ u(t),

where u(t) is the unique solution of (4.22)

u (t) = γ (t)ρ(u(t)),

t ∈ [t0 , TM ), u(t0 ) = u 0 = x 0 .

Let T ≤ TM be the supremum of all numbers t0 + m for which this holds. If t1 < t2 < T, then the solution in [t0 , t2 ] coincides with that in [t0 , t1 ], since such solutions are unique. Thus, a unique solution of (4.20) satisfying (4.21) exists for each t0 < t < T. Moreover, we have t2

x(t2 ) − x(t1 ) =

g(t, x(t)) dt. t1

Consequently, x(t2 ) − x(t1 ) ≤

t2

g(t, x(t)) dt

t1

≤

t2

γ (t)ρ( x(t) ) dt

t1

≤

t2

γ (t)ρ(u(t)) dt

t1

= u(t2 ) − u(t1 ). Assume that T < TM . Let tk be a sequence such that t0 < tk < T and tk → T . Then, x(tk ) − x(t j ) ≤ u(tk ) − u(t j ) → 0. Thus, {x(tk )} is a Cauchy sequence in H. Since H is complete, x(tk ) converges to an element x 1 ∈ H. Since x(tk ) ≤ u(tk ), we see that x 1 ≤ u(T ). Moreover, we note that x(t) → x 1 as t → T. To see this, let ε > 0 be given. Then there is a k such that x(tk ) − x 1 < ε,

u(T ) − u(tk ) < ε.

36


Then, for tk ≤ t < T, x(t) − x 1 ≤ x(t) − x(tk ) + x(tk ) − x 1 ≤ u(t) − u(tk ) + x(tk ) − x 1 < 2ε. We define x(T ) = x 1 . Then we have a solution of (4.20) satisfying (4.21) in [0, T ]. By Theorem 4.1, there is a unique solution of (4.23)

d y(t) = g(t, y(t)), dt

y(T ) = x 1

satisfying y(t) ≤ u(t) in some interval |t − T | < δ. By uniqueness, the solution of (4.23) coincides with the solution of (4.20) in the interval (T − δ, T ]. Define t0 ≤ t < T,

z(t) = x(t), z(T ) = x 1 , z(t) = y(t),

T < t ≤ T + δ.

This gives a solution of (4.20) satisfying (4.21) in the interval [t0 , T + δ), contradicting the definition of T. Hence, T = TM . We also have the following. Theorem 4.7. Let g(t, x) be a continuous map from R× H to H , where H is a Banach space. Assume that for each point (t0 , x 0 ) ∈ R × H, there are constants K , b > 0 such that (4.24) g(t, x)− g(t, y) ≤ K x − y ,

|t −t0 | < b, x − x 0 < b, y − x 0 < b.

Assume also that g(t, x) satisfies (4.15), where TM ≤ ∞, and γ (t), ρ(t) satisfy the hypotheses of Lemma 4.2 with ρ(t) nondecreasing. Then, for each x 0 ∈ H and t0 ∈ R, there is a unique solution x(t) of (4.16) depending continuously on x 0 and satisfying (4.25)

x(t) ≥ u(t),

t ∈ [t0 , TM ),

where u(t) is the solution of (4.8) in that interval.

4.5 An important estimate Lemma 4.8. Let ρ, γ satisfy the hypotheses of Lemma 4.3, with ρ locally Lipschitz continuous. Let u(t) be the solution of (4.8), and let h(t) be a continuous function satisfying t γ (r )ρ(h(r )) dr, t0 ≤ s < t < T, h(t0 ) ≥ u 0 . (4.26) h(t) ≥ h(s) − s

Then (4.27)

u(t) ≤ h(t),

t ∈ [t0 , T ).

4.5. An important estimate

37

Proof. Assume that there is a point t1 in the interval such that h(t1 ) < u(t1 ). Let y(t) = u(t) − h(t),

t ∈ [t0 , T ).

Then y(t0 ) ≤ 0 and y(t1 ) > 0. Let τ be the largest point < t1 such that y(τ ) = 0. Then y(t) > 0,

(4.28)

t ∈ (τ, t1 ].

Moreover, by (4.8) and (4.26), we have t (4.29) y(t) ≤ − γ (s)[ρ(u(s)) − ρ(h(s))] ds ≤ L τ

t τ

y(s) ds,

where L is the Lipschitz constant for ρ at u(τ ) times the maximum of γ in the interval. Let w(t) = Then Consequently,

t

τ

y(s) ds.

[e−Lt w(t)] = e−Lt [y(t) − Lw(t)] ≤ 0, e−Lt w(t) ≤ e−Lτ w(τ ) = 0,

t ∈ [τ, t1 ].

Hence, y(t) ≤ Lw(t) ≤ 0, contradicting (4.28). This completes the proof.

t ∈ [τ, t1 ].

t ∈ [τ, t1 ],

Chapter 5

The Method Using Flows 5.1 Introduction In this chapter we give the proofs of the theorems of Chapter 2. They rely on the theorems for ordinary differential equations in abstract spaces developed in Chapter 4.

5.2 Theorem 2.4 We begin with the proof of Theorem 2.4. Proof of Theorem 2.4. First, we note that if the theorem were false, there would be a δ > 0 and a ψ satisfying the hypotheses of Theorem 2.4 such that ||G (u) ≥ ψ( u )

(5.1) when

u ∈ Q = {u ∈ E : |G(u) − a| ≤ 3δ}.

(5.2)

Reduce δ so that 3δ < a − a0 . Since G ∈ C 1 (E, R), for any θ ∈ (0, 1), there is a locally Lipschitz continuous mapping Y (u) of Eˆ = {u ∈ E : G (u) = 0} into E such that (5.3)

Y (u) ≤ 1,

θ G (u) ≤ (G (u), Y (u)),

u ∈ Eˆ

(cf., e.g., [120]). Let Q0

=

{u ∈ E : |G(u) − a| ≤ 2δ},

Q1

=

{u ∈ E : |G(u) − a| ≤ δ},

Q2

=

E\Q 0 ,

η(u)

=

d(u, Q 2 )/[d(u, Q 1 ) + d(u, Q 2 )].


40

5. The Method Using Flows

It is easily checked that η(u) is locally Lipschitz continuous on E and satisfies ⎧ ⎪ u ∈ Q1, ⎪ ⎨η(u) = 1, (5.4) η(u) = 0, u ∈ Q¯ 2 , ⎪ ⎪ ⎩η(u) ∈ (0, 1), otherwise. Let ρ(t) = 1/ψ(t). Then ρ is a positive, nondecreasing, locally Lipschitz continuous function on [0, ∞) such that ∞ dτ (5.5) =∞ ρ(τ ) 0 by (2.8). Let W (u) = −η(u)Y (u)ρ( u ). Then W (u) ≤ ρ( u ),

u ∈ E.

By (5.5), for each u ∈ E, there is a unique solution of (5.6)

σ (t) = W (σ (t)),

t ∈ R+ , σ (0) = u

(cf. Theorem 4.5). We have (5.7)

d G(σ (t)u)/dt

=

−η(σ (t)u)(G (σ (t)u), Y (σ (t)u))ρ( σ (t)u )

≤

−θ η(σ ) G (σ ) ρ( σ )

≤

−θ η(σ ).

Thus, G(σ (t)u) ≤ G(u), G(σ (t)u) ≤ a0 ,

t ≥ 0,

u ∈ A, t ≥ 0,

and (5.8)

σ (t)u = u,

u ∈ A, t ≥ 0.

This follows from the fact that G(σ (t) u) ≤ a0 < a − 3δ,

u ∈ A, t ≥ 0.

Hence, η(σ (t) u) = 0 for u ∈ A, t ≥ 0. This means that σ (t) u = 0,

u ∈ A, t ≥ 0,

and σ (t) u = σ (0) u = u,

u ∈ A, t ≥ 0.

5.3. Theorem 2.12

41

Let E α = {u ∈ E : G(u) ≤ α}.

(5.9) There is a T > 0 such that

σ (T )E a+δ ⊂ E a−δ .

(5.10)

In fact, we can take T > 2δ/θ. To see this, let u be any element in E a+δ . If there is a t1 ∈ [0, T ] such that σ (t1 )u ∈ / Q 1 , then G(σ (T )u) ≤ G(σ (t1 )u) < a − δ by (5.7). Hence, σ (T )u ∈ E a−δ . On the other hand, if σ (t)u ∈ Q 1 for all t ∈ [0, T ], then η(σ (t)u) = 1 for all such t, and (5.7) yields (5.11)

G(σ (T )u) ≤ G(u) − θ T < a − δ.

Hence, (5.10) holds. Now, by (2.6), there is a K ∈ K such that K ⊂ E a+δ .

(5.12)

As we saw, σ (T ) ∈ (A). Let K˜ = σ (T )(K ). Then K˜ ∈ K by definition. But sup G = sup G(σ (T )u) < a − δ, K˜

u∈K

which contradicts (2.6), proving the theorem. Next, we prove Theorem 2.8. Proof. By (2.6) and (2.17), we see that b0 ≤ a. This implies (2.7) in view of (2.18). The result now follows from Theorem 2.4. Theorem 2.11 follows obviously from Theorem 2.8.

5.3 Theorem 2.12 Now we give the proof of Theorem 2.12. Proof. If a0 < a, the conclusion follows from Theorem 2.4. Assume that a0 = a. If there did not exist a sequence satisfying (2.9), then there would be positive numbers δ, θ, T such that 2δ < θ T and (5.1) holds whenever u ∈ Q, where Q is given by (5.2). Since a = a0 , we see by (2.6), (2.17), and (2.20) that b0 = a. Define Q 0 , Q 1 , Q 2 , and η(u) as before and let σ (t) be the flow generated by the mapping (5.6). Let u be any / Q 1 , then element in E a+δ . If there is a t1 ≤ T such that σ (t1 )u ∈ (5.13)

G(σ (t1 )u) < a − δ.

42


On the other hand, if σ (t)u ∈ Q 1 for all t ∈ [0, T ], then η(σ (t)u) ≡ 1 in [0, T ] and (5.14)

G(σ (t)u) ≤ G(u) − θ t ≤ a + δ − θ t,

t ∈ [0, T ].

Thus we have G(σ (T )u) < a − δ.

(5.15)

Since b0 := inf B G = a, this shows that σ (T )E a+δ ∩ B = φ.

(5.16) Moreover,

G(σ (t)u) ≤ a − θ t,

(5.17)

u ∈ A, t ∈ [0, T ].

It therefore follows that (2.12), (2.13), and (2.21) hold for b = a +δ. Let K ∈ K satisfy (5.12). Then, by hypothesis, there is a K˜ ∈ K satisfying (2.14) with b = a + δ. Now (2.21), (5.14), (5.16), and (5.17) imply that K˜ ∩ B = φ, contradicting (2.17). To prove the last statement, we take ψ(t) = ρ(t) ≡ 1. Still assuming a0 = a, we note that if there did not exist a sequence satisfying both (1.4) and (2.22), then there would be positive numbers , δ, T such that δ < T and (5.1) holds whenever u ∈ Q = {u ∈ E : d(u, B) ≤ 4T, |G(u) − a| ≤ 3δ}. Let Q0

=

{u ∈ E : d(u, B) ≤ 3T, |G(u) − a| ≤ 2δ},

Q1

=

{u ∈ E : d(u, B) ≤ 2T, |G(u) − a| ≤ δ}.

Since a = b0 , we see that Q 1 = φ. Define Q 0 , Q 1 , Q 2 , and η(u) as before and let σ (t) be the flow generated by the mapping W (u) = −η(u)Y (u), with everything now with respect to the new sets Q j . In this case (5.18)

W (u) ≤ 1.

Let u be any element in E a+δ . If there is a t1 ≤ T such that σ (t1 )u ∈ / Q 1 , then either (5.19)

G(σ (t1 )u) < a − δ

or (5.20) Since

d(σ (t1 )u, B) > 2T. σ (t)u − σ (t )u ≤ |t − t |

5.3. Theorem 2.12

43

by (5.18), (5.20) implies d(σ (t)u, B) > T,

(5.21)

0 ≤ t ≤ T.

On the other hand, if σ (t)u ∈ Q 1 for all t ∈ [0, T ], then (5.22)

G(σ (T )u) ≤ G(u) − 2T ≤ a + δ − 2δ = a − δ.

Thus, either we have G(σ (T )u) < a − δ

(5.23)

or (5.21) holds. Since b0 = a, this shows that σ (T )E a+δ ∩ B = φ.

(5.24) We also note that (5.25)

σ (t)A ∩ B = φ,

0 ≤ t ≤ T.

To see this, we observe, by (5.7), that

t

G(σ (t)u) ≤ a0 − 2

η(σ (τ )u)dτ,

u ∈ A.

0

If σ (t)u ∈ B, we must have G(σ (t)u) ≥ b0 ≥ a0 . The only way this can happen is if η(σ (τ )u) ≡ 0,

0 ≤ τ ≤ t.

But this implies σ (τ )u ∈ Q¯ 2 for such τ , and this in turn implies either G(σ (τ )u) < a − δ,

0 ≤ τ ≤ t,

d(σ (τ )u, B) > 2T,

0 ≤ τ ≤ t.

or In either case we cannot have σ (t)u ∈ B. Thus, (5.25) holds. Now, by (2.6), there is a K ∈ K such that K ⊂ E a+δ ,

(5.26)

and there is a K˜ ∈ K such that (2.14) holds. By (5.25), K˜ ∩ B = φ, contradicting the fact that A links B [mm]. This completes the proof of the theorem. Next, we give the proof of Theorem 2.6. Proof. Define

B= v∈

K \A : G(v) ≥ a0 .

K ∈K

By (2.23), A links B relative to K, and (2.20) holds. Apply Theorem 2.12.

44


5.4 Theorem 2.14 Now we give the proof of Theorem 2.14. Proof. We assume that a = a0 and follow the proof of Theorem 2.4. In this case we cannot take 3δ < a − a0 and we cannot conclude that (5.8) holds. Because of this, it does not follow that σ (T )K ∈ K for all K ∈ K. Let K ∈ K be such that (5.27)

a ≤ sup G < a + δ, K \A

and let K ρ be given by (2.25). Note that (5.28)

a0 ≤ sup G, K \A

K ∈ K,

by (2.20). Since G(u) is Lipschitz continuous on K ρ and a = a0 , there is a constant C such that (5.29)

G(u) ≤ a + Cd(u),

u ∈ Kρ ,

where d(u) = d(u, A). Pick T > 2δ/θ and T > ρC/θ. As in the proof of Theorem 2.4, we have (5.30)

σ (T )K ⊂ E a−δ .

Let ζ (t) be a nondecreasing, continuous function on R+ satisfying 0, t = 0, (5.31) ζ(t) = T, t ≥ ρ, and (5.32)

T t/ρ < ζ(t) < T,

0 < t < ρ.

Note that the flow σ˜ (t) = σ (ζ(t)) satisfies (2.12) and (2.13). Consider the map (5.33)

Su = σ (ζ(d(u)))u,

u ∈ K.

Consequently, by hypothesis, S(K ) ∈ K. Since d(u) = 0 for u ∈ A, and σ (0)u = u, we have Su = u, u ∈ A. If u ∈ K \K ρ , then Su = σ (T )u ∈ E a−δ . If u ∈ K ρ \A and Su ∈ E a−δ , then η(σ (t)u) = 1,

0 ≤ t ≤ ζ(d(u)).

Consequently, (5.34)

G(Su) ≤ G(u) − θ ζ (d(u)) ≤ a + Cd(u) − θ T d(u)/ρ < a.

5.5. Theorem 2.21

45

Thus, G(Su) < a,

u ∈ K \A,

contradicting (5.28). This completes the proof of the theorem. Now, we prove Theorem 2.13. Proof. Let B be given by (2.26). Then A links B relative to the system K. Now apply Theorem 2.14.

5.5 Theorem 2.21 We can now prove Theorem 2.21. Proof. We may assume that a = a0 . Otherwise, by Theorem 2.4, a sequence (2.9) ˜ exists with ψ replaced by ψ(t) = ψ(t + α). Since ψ˜ satisfies the hypotheses of Theorem 2.4 and d(u, B ) ≤ u + α, for each δ > 0, we can find a u ∈ E such that a − δ ≤ G(u) ≤ a + δ,

˜ G (u) < ψ( u ) ≤ ψ(d(u, B )),

which certainly implies (2.33). If the conclusion of the theorem were not true, there would be a δ > 0 such that ψ(d(u, B )) ≤ G (u)

(5.35) would hold for all u in the set (5.36)

Q = {u ∈ E : b0 − 3δ ≤ G(u) ≤ a + 3δ}.

By reducing δ if necessary, we can find θ < 1, T > 0 such that (5.37)

a0 − b0 + δ < θ T,

T ≤

R+α δ+α

ψ(s) ds.

Thus, by Lemma 4.3, if u(t) is the solution of (4.6) with ρ(t) = 1/ψ(t), γ = 1, t0 = 0, and u 0 = R, then u(t) ≥ δ, t ∈ [0, T ]. Let (5.38)

Q 0 = {u ∈ Q : b0 − 2δ ≤ G(u) ≤ a + 2δ},

(5.39)

Q 1 = {u ∈ Q : b0 − δ ≤ G(u) ≤ a + δ},

and (5.40)

Q 2 = E\Q 0 ,

η(u) = d(u, Q 2 )/[d(u, Q 1 ) + d(u, Q 2 )].

46


As before, we note that η satisfies (5.4). There is a locally Lipschitz continuous map Y (u) of Eˆ = {u ∈ E : G (u) = 0} into itself such that (5.41)

Y (u) ≤ 1,

u ∈ Eˆ

θ G (u) ≤ (G (u), Y (u)),

(cf., e.g., [120]). Let σ (t) be the flow generated by W (u) = −η(u)Y (u)ρ(d(u, B )),

(5.42)

˜ where ρ(τ ) = 1/ψ(τ ). Since W (u) ≤ ρ(d(u, B )) ≤ ρ( u ) ˜ = 1/ψ( u ) and is + locally Lipschitz continuous, σ (t) exists for all t ∈ R in view of Theorem 4.5. Since t W (σ (τ )u)dτ, (5.43) σ (t)u − u = 0

we have

t

σ (t)u − σ (s)u ≤

ρ(d(σ (r )u, B )) dr.

s

If v ∈ B , we have h(s) = d(σ (s)u, B ) ≤ σ (s)u − v ≤ σ (t)u − v +

t

ρ(d(σ (r )u, B )) dr,

s

which implies

t

h(s) ≤ h(t) +

(5.44)

ρ(h(r )) dr.

s

We also have (5.45) d G(σ (t)u)/dt

=

(G (σ ), σ ) = −η(σ )(G (σ ), Y (σ ))ρ(d(σ, B ))

≤

−θ η(σ ) G (σ ) ρ(d(σ, B ))

≤

−θ η(σ )ψ(d(σ, B ))ρ(d(σ, B ))

=

−θ η(σ )

in view of (5.35) and (5.41). Now suppose u ∈ E a+δ is such that there is a t1 ∈ [0, T ] for which σ (t1 )u ∈ / Q 1 . Then G(σ (t1 )u) < b0 − δ, since we cannot have G(σ (t1 )u) > a + δ for such u by (5.45). But this implies (5.46)

G(σ (T )u) < b0 − δ.

On the other hand, if σ (t)u ∈ Q 1 for all t ∈ [0, T ], then T G(σ (T )u) ≤ G(u) − θ dt ≤ a − θ T < b0 − δ 0

5.5. Theorem 2.21

47

by (5.37). Thus, (5.46) holds for u ∈ E a+δ and, in particular, for any u ∈ A. By the definition (2.6) of a, there is a K ∈ K such that sup G < a + (δ/2).

(5.47)

K

Note that σ (t)u = v ∈ B for u ∈ A and t ∈ [0, T ]. For v ∈ B , this follows from (5.44) and the fact that h(t) = d(σ (t)u, B ) ≥ u(t) ≥ δ,

t ∈ [0, T ],

in view of Lemma 4.8. If v ∈ B\B , we have, by (5.45), t G(σ (t)u) ≤ a − θ η(σ (τ )u)dτ < a, t > 0, 0

unless η(σ (τ )u) = 0 for 0 ≤ τ ≤ t. But this would mean that η(σ (τ )u) ∈ Q¯ 2 in view of (5.4). But then we would have G(u) ≤ b0 −2δ since we cannot have G(u) ≥ a +2δ. Thus, G(σ (t)u) < a. This shows that (5.48)

B ∩ σ (t)A = φ,

0 ≤ t ≤ T.

Taking b = a + (δ/2), (2.14) tells us that there is a K˜ ∈ K such that (5.49) K˜ ⊂ σ (t)A ∪ σ (T )[E b ∪ K ]. t ∈[0,T ]

But (5.46), (5.47), and (5.48) imply that K˜ ∩ B = φ, contradicting the fact that A links B [mm]. We also give the proof of Theorem 2.22. Proof. Again, we may assume that a = a0 . We interchange A and B and consider the ˜ functional G(u) = −G(u). Then a˜ 0 = sup G˜ = − inf G = −b0 < ∞ B

B

and

b˜0 = inf G˜ = − sup G = −a0 > −∞. A

A

Moreover, a˜ 0 − b˜0 = a0 − b0 < where

R+β β

ψ(t) dt,

R ≤ d = d(A , B).

48


Since

˜ < a˜ 0 }, A = {u ∈ A : G(u)

we can apply Theorem 2.21 to conclude that for each δ > 0, there is a u ∈ E such that (5.50)

˜ ≤ a˜ 0 + δ, b˜0 − δ ≤ G(u)

G˜ (u) < ψ(d(u, A )).

This implies (2.42). Theorems 2.24 and 2.19 follow from the same arguments used in the proof of Theorem 2.4 (in the case of Theorem 2.24 we substitute −G for G). Now we give the proof of Theorem 2.18. Proof. If the conclusion of the theorem were not true, then there would exist ε > 0 and ψ(r ) ∈ such that (5.51)

G (u) ≥ ψ( u )

whenever (5.52)

|G(u) − b| ≤ 3ε.

Let Q = {u ∈ E : |G(u) − b| ≤ 2ε}, Q 1 = {u ∈ E : |G(u) − b| ≤ ε}, Q 2 = B\Q, and η(u) = d(u, Q 2 )/[d(u, Q 1 ) + d(u, Q 2 )]. Note that η(u) is locally Lipschitz continuous and satisfies ⎧ ⎪ u ∈ Q1, ⎪ ⎨η(u) = 1, (5.53) η(u) = 0, u ∈ Q¯ 2 , ⎪ ⎪ ⎩η(u) ∈ (0, 1), otherwise. Let Y (u) be a locally Lipschitz continuous pseudo-gradient for G satisfying (5.54)

G (u)Y (u) ≥ α G (u) ,

Y (u) ≤ 1,

for some α > 0. Let W (u) = η(u)Y (u). Then W (u) is a locally Lipschitz continuous mapping of E into itself such that (5.55)

G (u)W (u) ≥ 0,

W (u) ≤ 1.

Let σ (t)u be the solution of the initial-value problem (5.56)

dσ (t)u/dt = W (σ (t)u),

σ (0)u = u.

5.5. Theorem 2.21

49

The solution of (5.56) exists for every u ∈ E and t ≥ 0 by Theorem 4.5. By (5.55), σ (t)u − u ≤ t,

(5.57) and by (5.54) and (5.56), (5.58)

d G(σ (t)u)/dt = G (σ (t)u)W (σ (t)u) ≥ αη(σ (t)u) G (σ (t)u) .

This implies that G(σ (t1 )u) ≤ G(σ (t2 )u),

(5.59)

0 ≤ t1 < t2 .

By hypothesis, G(u) ≥ b,

(5.60)

u ∈ ∂ω0 .

Let M = sup u

(5.61)

u∈∂ω0

and let T be given by T = d(0, ∂ω0 )/2.

(5.62) Take ε > 0 so small that

2ε < α

(5.63)

T +M

ψ(t) dt.

M

By (5.57) and (5.62), σ (t)u = 0,

(5.64)

u ∈ ∂ω0 , 0 ≤ t ≤ T.

If u ∈ ∂ω0 , but u ∈ Q 1 , we must have G(u) > b + ε

(5.65) since we cannot have

G(u) < b − ε

(5.66) by (5.60). Thus, by (5.59), (5.67)

G(σ (t)u) ≥ G(u) > b + ε,

u ∈ ∂ω0 , u ∈ Q 1 , 0 ≤ t ≤ T.

On the other hand, if u ∈ ∂ω0 ∩ Q 1 , let t1 be the largest number not greater than T such that σ (t)u ∈ Q 1 for 0 ≤ t ≤ t1 . If t1 < T, then for δ > 0 sufficiently small, G(σ (t1 + δ)u) ≥ G(σ (t1 )u) ≥ b

50


and since σ (t1 + δ)u ∈ Q 1 , we must have G(σ (t1 + δ)u) > b + ε. Consequently, (5.68)

G(σ (T )u) > b + ε.

If t1 = T , then by (5.51), (5.53), (5.57), (5.58), (5.61),and (5.63),

T

G(σ (T )u) − G(u) ≥

G (σ (t)u) dt

0

T

≥α

T

≥α

ψ( σ (t)u ) dt

0

ψ( u + t) dt

0 T

≥α

ψ(M + t) dt

0

=α

T +M

ψ(r )dr

M

> 2ε. Thus, (5.68) holds as well in this case by (5.60). Consequently, (5.68) holds for all u ∈ ∂ω0 . Let ωT be the set of points σ (T, u) where u ∈ ω0 . Then ωT is a bounded, open set in E with ∂ωT consisting of those points of the form σ (T )u, u ∈ ∂ω0 by (5.57) and the continuous dependence of σ (T )u on u. Since 0 is in Q 2 and η ≡ 0 there, we see that σ (T )0 = 0 by the uniqueness of solutions of (5.56). Thus, 0 = σ (T )0 ∈ ωT . Thus, ∂ωT ∈ K, and (5.69)

G(u) > b + ε,

u ∈ ∂ωT ,

by (5.68). But (5.69) contradicts (2.27). This completes the proof. In proving Theorem 2.20, we merely replace G by −G and follow the proof of Theorem 2.18.

Chapter 6

Finding Linking Sets 6.1 Introduction As we saw in Chapter 3, there are several sufficient conditions that imply that a set A links a set B in the sense of Definition 2.9. Our goal is to find all subsets that link according to this definition. At the present, we are very close to achieving this goal. In the present chapter we define two relationships that are close to each other. The stronger one is sufficient for linking, while the weaker one is necessary. We use the following maps. Definition 6.1. We shall say that a map ϕ : E → E is of class if it is a homeomorphism onto E and both ϕ, ϕ −1 are bounded on bounded sets. If, furthermore, ϕ, ϕ −1 ∈ C 1 (E; E), we shall say that ϕ ∈ U . Definition 6.2. We shall say that a bounded set A is chained to a set B if A ∩ B = φ and (6.1)

inf ϕ(x) ≤ sup ϕ(x) ,

x∈B

ϕ ∈ U .

x∈A

Definition 6.3. We shall say that a bounded set A is strongly chained to a set B if A ∩ B = φ and (6.1) holds for every ϕ ∈ . Definition 6.4. For A ⊂ E, we define −1 KU (A) = ϕ (B R ) : ϕ ∈ U , R > sup ϕ(u) , u∈A

where B R = {u ∈ E : u < R}.


52

6. Finding Linking Sets We shall show

Theorem 6.5. If a bounded set A is strongly chained to a set B, then A links B. Theorem 6.6. Let G be a (C 2 ∩CU )-functional on E, and let A, B be nonempty subsets of E such that A is bounded and chained to B with (6.2)

a0 := sup G < b0 := inf G. B

A

Let (6.3)

a :=

inf

sup G.

K ∈KU ( A) K

Then there is a sequence {u k } ⊂ E such that (6.4)

G(u k ) → a, G (u k ) → 0.

Theorem 6.7. If E is a Hilbert space and A links B, then it is chained to B.

6.2 The strong case Definition 6.8. For A ⊂ E, we define (A) = {ϕ ∈ : ϕ(u) = u, u ∈ A} and U (A) = {ϕ ∈ U : ϕ(u) = u, u ∈ A}. The following are easily proved. Lemma 6.9. ϕ1 , ϕ2 ∈ ⇒ ϕ1 ◦ ϕ2 ∈ . ϕ1 , ϕ2 ∈ U ⇒ ϕ1 ◦ ϕ2 ∈ U . ϕ1 , ϕ2 ∈ (A) ⇒ ϕ1 ◦ ϕ2 ∈ (A). ϕ1 , ϕ2 ∈ U (A) ⇒ ϕ1 ◦ ϕ2 ∈ U (A). Lemma 6.10.

ϕ ∈ iff ϕ −1 ∈ . ϕ ∈ U iff ϕ −1 ∈ U . ϕ ∈ (A) iff ϕ −1 ∈ (A). ϕ ∈ U (A) iff ϕ −1 ∈ U (A).

Definition 6.11. For A ⊂ E, we define K(A) = ϕ −1 (B R ) : ϕ ∈ , R > sup ϕ(u) . u∈A

6.2. The strong case

53

Lemma 6.12. If K ∈ K(A) and σ ∈ (A), then σ (K ) ∈ K(A). Proof. There is a ϕ ∈ such that K = ϕ −1 (B R ) with R > supu∈A ϕ(u) . Let ϕ˜ = ϕ ◦ σ −1 . Then ϕ˜ ∈ (Lemma 6.9). Now, ϕ˜ −1 = σ ◦ ϕ −1 . Hence,

σ (K ) = σ [ϕ −1 (B R )] = ϕ˜ −1 (B R ).

Moreover, ϕ(u) ˜ = ϕ(u),

u ∈ A.

Hence, sup ϕ(x) ˜ = sup ϕ(x) < R. x∈A

x∈A

Therefore, σ (K ) ∈ K(A). Corollary 6.13. K(A) is a minimax system for A. Lemma 6.14. If A is strongly chained to B and K ∈ K(A), then K ∩ B = φ. Proof. There is a ϕ ∈ such that K = ϕ −1 (B R ) with R > supu∈A ϕ(u) . Since A is strongly chained to B, we have inf ϕ(x) ≤ sup ϕ(x) < R.

x∈B

x∈A

Hence, there is a v ∈ B such that ϕ(v) ∈ B R . Thus, v = ϕ −1 ϕ(v) ∈ ϕ −1 (B R ). Corollary 6.15. If A is strongly chained to B, then A links B relative to K(A). Corollary 6.16. If A is strongly chained to B, then A links B strongly. Proof. Theorem 2.11. We can now give the proof of Theorem 6.5. Proof. Let G be a (C 1 ∩ CU )-functional on E, and let A, B be nonempty subsets of E such that A is strongly chained to B and (6.2) holds. Then (6.5)

a :=

inf

sup G

K ∈K( A) K

is finite. By Lemma 6.14, K ∩ B = φ for all K ∈ K(A). Hence, a ≥ b0 . Since a0 < a, then the theorem follows from Corollary 6.16.

54

6. Finding Linking Sets

6.3 The remaining proofs Lemma 6.17. If A is chained to B and K ∈ KU (A), then K ∩ B = φ. Proof. There is a ϕ ∈ U such that K = ϕ −1 (B R ) with R > supu∈A ϕ(u) . Since A is chained to B, we have inf ϕ(x) ≤ sup ϕ(x) < R.

x∈B

x∈A

Hence, there is a v ∈ B such that ϕ(v) ∈ B R . Hence, v = ϕ −1 ϕ(v) ∈ ϕ −1 (B R ). Lemma 6.18. If K ∈ KU (A) and σ ∈ U (A), then σ (K ) ∈ KU (A). We can now give the proof of Theorem 6.6. Proof. Let G be a (C 2 ∩ CU )-functional on E, and let A, B be nonempty subsets of E such that A is chained to B and (6.2) holds. Then a given by (6.3) is finite. By Lemma 6.17, K ∩ B = φ for all K ∈ KU (A). Hence, a ≥ b0 . If (6.4) were false, there would exist a positive constant δ such that 3δ < a − b0 and G (u) ≥ 3δ

(6.6) whenever (6.7)

u ∈ Q = {u ∈ E : |G(u) − a| ≤ 3δ}.

Let Q0

= {u ∈ E : |G(u) − a| ≤ 2δ},

Q1

= {u ∈ E : |G(u) − a| ≤ δ},

Q2

=

η(u)

E\Q 0 ,

= d(u, Q 2 )/[d(u, Q 1 ) + d(u, Q 2 )].

It is easily checked that η(u) is locally Lipschitz continuous on E and satisfies η(u) = 1, u ∈ Q 1 ; η(u) = 0,

u ∈ Q¯ 2 ;

0 < η(u) < 1, otherwise.

Consider the differential equation (6.8)

σ (t) = W (σ (t)),

t ∈ R, σ (0) = u,

where (6.9)

W (u) = −η(u)G (u)/ G (u) .

Since G ∈ C 2 (E, R), the mapping W is locally Lipschitz continuous on the whole of E and is bounded in norm by 1. Hence, by Corollary 4.6, (6.8) has a unique solution

6.3. The remaining proofs

55

for all t ∈ R. Let us denote the solution of (6.8) by σ (t) u. The mapping σ (t) is in CU1 (E × R, E) and is called the flow generated by (6.9). Note that (6.10)

d G(σ (t)u)/dt

=

(G (σ (t)u), σ (t)u)

=

−η(σ (t)u) G (σ (t)u)

≤

−2δη(σ (t)u).

Thus, G(σ (t)u) ≤ G(u), G(σ (t)u) ≤ a0 ,

t ≥ 0,

u ∈ A, t ≥ 0,

and σ (t)u = u,

u ∈ A, t ≥ 0.

Again, this follows from the fact that G(σ (t) u) ≤ a0 ≤ b0 < a − 3δ,

u ∈ A, t ≥ 0.

Hence, η(σ (t) u) = 0 for u ∈ A, t ≥ 0. This means that σ (t) u = 0,

u ∈ A, t ≥ 0,

and σ (t) u = σ (0) u = u,

u ∈ A, t ≥ 0.

Let E α = {u ∈ E : G(u) ≤ α}.

(6.11)

We note that there is a T > 0 such that σ (T )E a+δ ⊂ E a−δ .

(6.12)

(In fact, we can take T = 1.) Let u be any element in E a+δ . If there is a t1 ∈ [0, T ] such that σ (t1 )u ∈ / Q 1 , then G(σ (T )u) ≤ G(σ (t1 )u) < a − δ by (6.10). Hence, σ (T )u ∈ E a−δ . On the other hand, if σ (t)u ∈ Q 1 for all t ∈ [0, T ], then η(σ (t)u) = 1 for all t, and (6.10) yields (6.13)

G(σ (T )u) ≤ G(u) − 2δT ≤ a − δ.

Hence, (6.12) holds. Now, by (6.3), there is a K ∈ KU (A) such that K ⊂ E a+δ .

(6.14)

Note that σ (T ) ∈ U (A). Let K˜ = σ (T )(K ). Then K˜ ∈ KU (A) by Lemma 6.12. But sup G = sup G(σ (T )u) < a − δ, K˜

u∈K

which contradicts (6.3), proving the theorem.

56

6. Finding Linking Sets Now we give the proof of Theorem 6.7.

Proof. Assume that ϕ ∈ U does not satisfy (6.1), and let G(u) = ϕ(u) 2 . Then, by the definition of the class U , G ∈ CU1 (E, R), sup G(A) < inf G(B), and G has no critical level a ≥ infx∈B G(x) > 0. To show the latter, assume that there is a sequence u k satisfying (6.15)

G(u k ) → a,

G (u k ) → 0.

Then we have, for any bounded sequence v k ∈ E, (6.16)

(ϕ (u k )v k , ϕ(u k )) → 0.

Let v k := (ϕ (u k ))−1 (ϕ(u k )). Then (ϕ (u k )v k , ϕ(u k )) = G(u k ) → a. However, the sequence v k is bounded: G(u k ) → a implies that ϕ(u k ) is bounded, which implies that u k , and consequently that ϕ (u k ))−1 and v k are also bounded. Hence, G (u k ) → 0 implies G(u k ) → 0, showing that a = 0. Thus, A does not link B.

6.4 Notes and remarks The results of this chapter are from [138]. For related material cf. [137] and [156].

Chapter 7

Sandwich Pairs 7.1 Introduction In this chapter we discuss the situation in which one cannot find linking sets that separate a functional G, i.e., satisfy (7.1)

a0 := sup G ≤ b0 := inf G. B

A

Are there weaker conditions that will imply the existence of a PS sequence (7.2)

G(u k ) → a,

G (u k ) → 0?

Our answer is yes, and we find pairs of subsets such that the absence of (7.1) produces a PS sequence. We have Definition 7.1. We shall say that a pair of subsets A, B of a Banach space E forms a sandwich if, for any G ∈ C 1 (E, R), the inequality (7.3)

−∞ < b0 := inf G ≤ a0 := sup G < ∞ B

A

implies that there is a sequence satisfying (7.4)

G(u k ) → c,

b0 ≤ c ≤ a0 ,

G (u k ) → 0.

Unlike linking, the order of a sandwich pair is immaterial; i.e., if the pair A, B forms a sandwich, so does B, A. Moreover, we allow sets forming a sandwich pair to intersect. One sandwich pair has been studied in Chapter 3. In fact, Theorem 3.17 tells us that if M, N are closed subspaces of a Hilbert space E, and M = N ⊥ , then M, N form a sandwich pair if one of them is finite-dimensional. Until recently, only complementing subspaces have been considered. The purpose of the present chapter is to show that other sets can qualify as well. Infinite-dimensional sandwich pairs will be considered in Chapter 15. M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_7, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009

58

7. Sandwich Pairs

7.2 Criteria In this section we present sufficient conditions for sets to qualify as sandwich pairs. We have Proposition 7.2. If A, B is a sandwich pair and J is a diffeomorphism on the entire space having a derivative J satisfying (7.5)

J (u)−1 ≤ C,

u ∈ E,

then JA, JB is a sandwich pair. Proof. Suppose G ∈ C 1 satisfies (7.6)

−∞ < b0 := inf G ≤ a0 := sup G < ∞. JB

JA

G 1 (u) = G(J u),

u ∈ E.

Let Then (7.7)

−∞ < b0 := inf G = inf G(J u) = inf G 1 JB

J u∈ J B

B

≤ a0 := sup G = sup G(J u) = sup G 1 < ∞. JA

J u∈ J A

A

Since A, B form a sandwich pair, there is a sequence {h k } ⊂ E such that (7.8)

G 1 (h k ) → c,

b0 ≤ c ≤ a0 ,

G 1 (h k ) → 0.

If we set u k = J h k , this becomes (7.9)

G(u k ) → c,

b0 ≤ c ≤ a0 ,

G (u k )J (h k ) → 0.

In view of (7.5), this implies G (u k ) → 0. Thus, J A, J B is a sandwich pair. Proposition 7.3. Let N be a closed subspace of a Hilbert space E with complement M = M ⊕{v 0 }, where v 0 is an element in E having unit norm, and let δ be any positive number. Let ϕ(t) ∈ C 1 (R) be such that 0 ≤ ϕ(t) ≤ 1,

ϕ(0) = 1,

and ϕ(t) = 0,

|t| ≥ 1.

Let (7.10) F(v + w + sv 0 ) = v + [s + δ − δϕ( w 2 /δ 2 )]v 0 ,

v ∈ N, w ∈ M, s ∈ R.

Assume that one of the subspaces M, N is finite-dimensional. Then A = N = N ⊕ {v 0 }, B = F −1 (δv 0 ) form a sandwich pair.

7.2. Criteria

59

Proof. Define J (v + w + sv 0 ) = v + w + [s + δ − δϕ( w 2 /δ 2 )]v 0 ,

v ∈ N, w ∈ M, s ∈ R.

Then J is a diffeomorphism on E with its inverse having a derivative satisfying (7.5). Moreover, J A = N and J B = M +δv 0 . Hence, J A, J B form a sandwich pair as long as one of them is finite-dimensional (Theorem 3.17). We now apply Proposition 7.2. Theorem 7.4. Let N be a finite dimensional subspace of a Banach space E. Let F be a Lipschitz continuous map of E onto N such that F = I on N and F(g) − F(h) ≤ K g − h ,

(7.11)

g, h ∈ E.

Let p be any point of N. Then A = N, B = F −1 ( p) form a sandwich pair. Proof. Let G be a C 1 -functional on E satisfying (7.3), where A, B are the subsets of E specified in the theorem. If the theorem is not true, then there is a δ > 0 such that G (u) ≥ 3δ

(7.12) whenever

b0 − 3δ ≤ G(u) ≤ a0 + 3δ.

(7.13)

Since G ∈ C 1 (E, R), there is a locally Lipschitz continuous mapping Y (u) of ˆ E = {u ∈ E : G (u) = 0} into E such that Y (u) ≤ 1, and

ˆ u ∈ E,

(G (u), Y (u)) ≥ 2δ

whenever u satisfies (7.13) (for the construction of such a map, cf., e.g., [112]). Let Q0

= {u ∈ E : b0 − 2δ ≤ G(u) ≤ a0 + 2δ},

Q1

= {u ∈ E : b0 − δ ≤ G(u) ≤ a0 + δ},

Q2

=

η(u)

E\Q 0 ,

= ρ(u, Q 2 )/[ρ(u, Q 1 ) + ρ(u, Q 2 )].

It is easily checked that η(u) is locally Lipschitz continuous on E and satisfies η(u) = 1, u ∈ Q 1 ;

η(u) = 0, u ∈ Q 2 ;

0 < η(u) < 1, otherwise.

Consider the differential equation (7.14)

σ (t) = W (σ (t)),

t ∈ R, σ (0) = u ∈ N,

where W (u) = −η(u)Y (u).

60

7. Sandwich Pairs

The mapping W is locally Lipschitz continuous on the whole of E and is bounded in norm by 1. Hence, by Theorem 4.5, (7.13) has a unique solution for all t ∈ R. Let us denote the solution of (7.13) by σ (t)u. The mapping σ (t) is in C(E × R, E) and is called the flow generated by W (u). Note that (7.15)

d G(σ (t)u)/dt

=

(G (σ (t)u), σ (t)u)

=

−η(σ (t)u)(G (σ (t)u), Y (σ (t)u))

≤

−2δη(σ (t)u).

Let E α = {u ∈ E : G(u) ≤ α}. I claim that there is a T > 0 such that (7.16)

σ (T )E a0 +δ ⊂ E b0 −δ .

In fact, we can take T > (a0 − b0 + δ)/2δ. Let u be any element in E a0 +δ . If there is a t1 ∈ [0, T ] such that σ (t1 )u ∈ / Q 1 , then G(σ (T )u) ≤ G(σ (t1 )u) < b0 − δ by (7.15). Hence, σ (T )u ∈ E b0 −δ . On the other hand, if σ (t)u ∈ Q 1 for all t ∈ [0, T ], then η(σ (t)u) = 1 for all t, and (7.15) yields G(σ (T )u) ≤ G(u) − 2δT ≤ a0 − 2δT ≤ b0 − δ. Hence, (7.16) holds. Let be a bounded open subset of N containing the point p such that (7.17)

ρ(∂, p) > K T + δ,

where ρ is the distance in E and K is the constant in (7.11). If v ∈ ∂, then v − p ≤ v − Fσ (t)v + Fσ (t)v − p . Then (7.18)

Fσ (t)v − p > K T + δ − t K > 0,

since Fσ (t)v − v ≤ K

t

v ∈ ∂, 0 ≤ t ≤ T,

σ (s)v ds ≤ K t.

0

Let H (t) = Fσ (t). Then H (t) is a continuous map of into N for 0 ≤ t ≤ T. Moreover, H (t)v = p for v ∈ ∂ by (7.18). Hence, the Brouwer degree d(H (t), , p) is defined. Consequently, d(H (T ), , p) = d(H (0), , p) = d(I, , p) = 1.


61

This means that there is a v ∈ such that Fσ (T )v = p. But then

σ (T )v ∈ F −1 ( p) = B.

This is not consistent with (7.16). Hence, A, B form a sandwich pair.

7.3 Notes and remarks The material of this chapter comes from [135].

Chapter 8

Semilinear Problems 8.1 Introduction In the present chapter we study several nonlinear boundary value problems that arise frequently in applications and illustrate the techniques described in the book.

8.2 Bounded domains We assume that is a bounded domain in Rn with boundary ∂ sufficiently regular so that the Sobolev inequalities hold and the embedding of H m,2() in L 2 () is compact (cf., e.g., [1]). Let A be a self-adjoint operator on L 2 (). We assume that A ≥ λ0 > 0 and that C0∞ () ⊂ D := D(A1/2 ) ⊂ H m,2() for some m > 0, where C0∞ () denotes the set of test functions in (i.e., infinitely differentiable functions with compact supports in ) and H m,2() denotes the Sobolev space. If m is an integer, the norm in H m,2 () is given by ⎛ u m,2 := ⎝

(8.1)

⎞1/2 D μ u 2 ⎠

.

| μ|≤m

Here D μ represents the generic derivative of order |μ| and the norm on the right-hand side of (8.1) is that of L 2 (). We shall not assume that m is an integer. Let q be any number satisfying (8.2)

2

≤ q ≤ 2n/(n − 2m),

2m < n

2

≤ q < ∞,

n ≤ 2m

and let f (x, t) be a Carathéodory function on × R. This means that f (x, t) is continuous in t for a.e. x ∈ and measurable in x for every t ∈ R. We make the following assumption. M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_8, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009

64

8. Semilinear Problems

(A) The function f (x, t) satisfies | f (x, t)| ≤ V (x)q (|t|q−1 + W (x)) and f (x, t)/V (x)q = o(|t|q−1 ) as |t| → ∞, where V (x) > 0 is a function in L q () such that V u q ≤ C u D ,

(8.3)

u ∈ D,

and W is a function in L ∞ (). Here u q :=

1/q |u(x)| d x q

and u D := A1/2 u .

(8.4)

With the norm (8.4), D becomes a Hilbert space. Define t f (x, s) ds F(x, t) := 0

and (8.5)

G(u) := u 2D − 2

F(x, u) d x.

It is readily shown that G is a continuously differentiable functional on the whole of D (cf., e.g., [122]). Since the embedding of D in L 2 () is compact, the spectrum of A consists of isolated eigenvalues of finite multiplicity: 0 < λ0 < λ1 < · · · < λ < · · · . Let λ , > 0, be one of these eigenvalues. We assume that the eigenfunctions of λ are in L ∞ () and that the following hold: (8.6)

2F(x, t) ≤ λ t 2 + W1 (x),

x ∈ , t ∈ R

for some W1 (x) ∈ L 1 (R), (8.7)

λ t 2 ≤ 2F(x, t),

|t| ≤ δ,

for some δ > 0, (8.8)

νt 2 ≤ 2F(x, t),

x ∈ , t ∈ R,

for some ν > λ−1 , (8.9)

H (x, t) := 2F(x, t) − t f (x, t) ≤ C(|t| + 1),

8.2. Bounded domains

65

and (8.10)

σ (x) := lim sup H (x, t)/|t| < 0

a.e.

|t |→∞

We wish to obtain a solution of (8.11)

Au = f (x, u),

u ∈ D.

By a solution of (8.11) we shall mean a function u ∈ D such that (8.12)

(u, v) D = ( f (·, u), v),

v ∈ D.

If f (x, u) is in L 2 (), then a solution of (8.12) is in D(A) and solves (8.11) in the classical sense. Otherwise we call it a weak (or semi-strong) solution. We have Theorem 8.1. Under the above hypotheses, (8.13)

Au = f (x, u),

u∈D

has at least one nontrivial solution. Proof. Let N denote the subspace of L 2 () spanned by the eigenfunctions of A corresponding to the eigenvalues λ0 , . . . , λ , and let M = N ⊥ ∩ D. Thus, D = M ⊕ N. This time we take G(u) = 2 F(x, u) d x − u 2D , the negative of (8.5). We are therefore looking for solutions of G (u) = 0. Let N be the set of those functions in N that are orthogonal to E(λ ). It is spanned by those eigenfunctions corresponding to λ0 , . . . , λ−1 . Let v 0 be an eigenfunction of λ with norm 1. Let M1 = M ⊕ E(λ ) {v 0 }. We can write E = M1 ⊕ {v 0 } ⊕ N . Consider the mapping F(v + w + sv 0 ) = w + [s + ρ − ρϕ( v 2 /ρ 2 )]v 0 ,

v ∈ N, w ∈ M1 , s ∈ R,

where ϕ satisfies the hypotheses of Proposition 7.3 and ρ > 0 is to be chosen. We take A = M1 ⊕ {v 0 },

B = F −1 (ρv 0 ).

By Proposition 7.3, A, B form a sandwich pair. For v ∈ N, we write v = v + y, where v ∈ N and y ∈ E(λ ). Since E(λ ) is finite-dimensional and contained in L ∞ (), there is a ρ > 0 such that (8.14)

y D ≤ ρ implies y ∞ ≤ δ/2,

where δ is given by (8.7). Thus, if (8.15)

v D ≤ ρ

and

|v(x)| ≥ δ,

66


then

δ ≤ |v(x)| ≤ |v (x)| + |y(x)| ≤ |v (x)| + δ/2.

Hence,

|v(x)| ≤ 2|v (x)|

¯ satisfying (8.15). Thus by (8.7) holds for all x ∈ 2 G(v) ≥ λ v dx − 2 {|V v|q + |V q v|W }d x − v 2D |v|δ

≥

λ v 2 − λ

≥

λ v

≥

λ v 2 − v 2D − C v D λ q−2 − 1 − C v D v 2D . λ−1

≥

2

|v|>δ

− v 2D

v 2 d x − v 2D − C

−C

2|v |>δ

2|v |>δ

{|V v |q + δ 1−q |V v |q }d x

{|V v |q + δ 1−q |V v |q + δ 2−q |v |q }d x

q

To see this, note that when (8.15) holds, we have |V v| ≤ 2|V v | and |V q v| ≤ V q |v| Moreover,

|v|q−1 ≤ δ 1−q V q |2v |q . δ q−1

v q ≤ C v m,2 ≤ C v D

by the Sobolev inequality and the embedding of D in H m,2(). From this we see that there are positive constants , ρ such that G(v) ≥ v 2D ,

v D ≤ ρ, v ∈ N .

Moreover, this shows that for each positive ρ1 ≤ ρ, (8.16)

G(v) ≥ 1 ,

v D = ρ1 , v ∈ N ,

for some positive 1 unless there is a solution of Ay = λ y = f (x, y),

y ∈ E(λ )\{0}

(cf. [122]). Since such a solution would solve (8.13), we may assume that (8.16) holds. Since v 2D ≤ λ v 2 , v ∈ N, and λ+1 w 2 ≤ w 2D ,

w ∈ M,

we have, by (8.6), G(w) ≤ λ w 2 + B1 − w 2D ≤ B1 ,

w ∈ A,

8.2. Bounded domains where B1 =

67

W1 (x) d x. Moreover, (8.8) implies G(v ) ≥ (ν − λ−1 ) v 2 ,

v ∈ N.

Hence, there is an ε > 0 such that G(v) ≥ ε,

v ∈ B.

In view of these inequalities, we can now apply Proposition 7.3 to conclude that there is a sequence {u k } ⊂ D such that G(u k ) → c,

(8.17)

ε ≤ c ≤ B1 ,

Let ρk = u k D . If ρk → ∞, then G(u k ) = 2

(8.18) and

(G (u k ), u k )/2 =

G (u k ) → 0.

F(x, u k ) d x − ρk2 → c

f (x, u k )u k d x − ρk2 = o(ρk ).

Hence,

H (x, u k ) d x = o(ρk ).

Let u˜k = u k /ρk . Then u˜ k D = 1. Thus, there is a renamed subsequence such that u˜ k → u˜ weakly in D, strongly in L 2 (), and a.e. in . By (8.9) and (8.10), lim sup H (x, u k ) d x/ρk ≤ lim sup[H (x, u k )/|u k |]|u˜ k | d x

=

σ (x)|u| ˜ d x.

Since σ (x) < 0 a.e. in , the last two statements imply that u˜ ≡ 0. However, we see from (8.18) that 2

F(x, u k ) d x/ρk2 → 1,

while (8.6) implies

lim sup 2

F(x, u k ) d x/ρk2

≤ λ

u˜ 2 d x,

showing that u˜ ≡ 0. This contradiction tells us that the ρk must be bounded. We can now apply Theorem 3.4.1 of [122] to conclude that there is a u ∈ D satisfying (8.19)

G(u) = c,

G (u) = 0.

Since c ≥ ε > 0, we see that u = 0, and the proof is complete.

68

8. Semilinear Problems The proof of Theorem 8.1 implies

Corollary 8.2. If λ is a simple eigenvalue, then hypothesis (8.6) in Theorem 8.1 can be weakened to (8.20)

2F(x, t) ≤ λ+1 t 2 + W1 (x),

x ∈ , t ∈ R,

for some W1 (x) ∈ L 1 (R). Remark 8.3. The proof of Theorem 8.1 is much simpler if = 0. In this case N = {0} and (8.14) immediately implies (8.16). The rest of the proof is unchanged. We now show that we can essentially reverse the inequalities (8.6)–(8.10) and obtain the same results. In fact, we have Theorem 8.4. Equation (8.13) has at least one nontrivial solution if we assume > 0 and (8.21)

λ t 2 ≤ 2F(x, t) + W1 (x),

x ∈ , t ∈ R,

for some W1 (x) ∈ L 1 (R), (8.22)

2F(x, t) ≤ λ t 2 ,

|t| ≤ δ,

for some δ > 0, (8.23)

2F(x, t) ≤ νt 2 ,

x ∈ , t ∈ R,

for some ν < λ+1 , (8.24)

H (x, t) ≥ −C(|t| + 1),

x ∈ , t ∈ R,

and (8.25)

lim inf H (x, t)/|t| > 0 |t |→∞

a.e.

Proof. In this case we take G to be the functional (8.5). We take A = N, N {v 0 }, and consider the mapping F(v + w + sv 0 ) = v + [s + δ − δϕ( w 2 /δ 2 )]v 0 ,

N =

v ∈ N , w ∈ M, s ∈ R,

where ϕ satisfies the hypotheses of Proposition 7.3. By (8.21), we have G(v) ≤ B1 ,

v ∈ N.

For w ∈ M1 , we write w = w + y, where w ∈ M and y ∈ E(λ ). Then (8.22) implies G(w) ≥ ε1 ,

w D = ρ, w ∈ M1 ,

unless (8.13) has a nontrivial solution. Hence, by the argument given in the proof of Theorem 8.1 we have a sequence satisfying (8.17). If u˜ k and u˜ are as in the proof of

8.3. Some useful quantities

69

Theorem 8.1, then (8.24), (8.25) imply that u˜ ≡ 0 as in that proof. However, (8.18) implies F(x, u k ) d x/ρk2 → 1, 2

while (8.23) implies

lim sup 2

F(x, u k ) d x/ρk2 ≤ ν

u˜ 2 d x,

showing that u˜ ≡ 0. This contradiction proves the theorem as in the case of Theorem 8.1. The proof of Theorem 8.4 implies Corollary 8.5. If λ is a simple eigenvalue, then hypothesis (8.21) in Theorem 8.4 can be weakened to (8.26)

λ−1 t 2 ≤ 2F(x, t) + W1 (x),

x ∈ , t ∈ R, for some W1 (x) ∈ L 1 (R).

8.3 Some useful quantities We now show how we can improve the results of the last section. For each fixed k, let Nk denote the subspace of D := D(A1/2 ) spanned by the eigenfunctions corresponding to λ0 , . . . , λk , and let Mk = Nk⊥ ∩ D. Then D = Mk ⊕ Nk . We define (8.27)

αk := max{(Av, v) : v ∈ Nk , v ≥ 0, v = 1},

where v denotes the L 2 ()-norm of v. We assume that A has an eigenfunction ϕ0 of constant sign a.e. on corresponding to the eigenvalue λ0 . Next we define for a ∈ R (8.28)

γk (a) := max{(Av, v) − a v − 2 : v ∈ Nk , v + = 1}

and (8.29)

k (a) := inf{(Aw, w) − a w− 2 : w ∈ Mk , w+ = 1},

where u ± = max{±u, 0}. We take any integer ≥ 0 and let N denote the subspace of L 2 () spanned by the eigenspaces of A corresponding to the eigenvalues λ0 , λ1 , . . . , λ . We take M = N ⊥ ∩ D, where D = D(A1/2 ). We assume that F(x, t) satisfies (8.30)

a1 (t − )2 + γ (a1 )(t + )2 − W1 (x) ≤ 2F(x, t) ≤ a2 (t − )2 + ν(t + )2 ,

x ∈ , t ∈ R,

for numbers a1 , a2 satisfying α < a1 ≤ a2 , where W1 is a function in L 1 () and ν < (a2 ). We also assume that (8.31)

2F(x, t) ≤ λ+1 t 2 ,

|t| ≤ δ

70


for some δ > 0, W ∈ L 2 (),

(8.32)

| f (x, t)| ≤ C|t| + W (x),

(8.33)

f (x, t)/t → α± (x) a.e. as t → ±∞,

and the only solution of Au = α+ (x)u + − α− (x)u −

(8.34) is u ≡ 0. We have

Theorem 8.6. Under the above hypotheses, (8.13) has a nontrivial solution. Proof. By (8.28), (8.35)

v 2D ≤ a1 v − 2 + γ (a1 ) v + 2 ,

v ∈ N,

and by (8.29), we have (8.36)

a2 w− 2 + (a2 ) w+ 2 ≤ w 2D ,

w ∈ M.

Hence, G(v) ≤ B1 ,

(8.37)

v ∈ N.

Since ν < (a2 ), we see by continuity that there is an ε > 0 such that a2 . ν < (1 − ε) 1−ε Hence, (8.38)

a2 ν G(w) ≥ ε w 2D + (1 − ε) − w+ 2 1−ε 1−ε ≥ ε w 2D ,

w ∈ M,

by (8.30). As in the proof of Theorem 8.1, we note that the following alternative holds: Either (a) there is an infinite number of eigenfunctions y ∈ E(λ )\{0} such that (8.39)

Ay = f (x, y) = λ y,

or (b) for each ρ > 0 sufficiently small, there is an ε > 0 such that (8.40)

G(w) ≥ ε,

w D = ρ, w ∈ M .

8.4. Unbounded domains

71

Since option (a) solves our problem, we may assume that option (b) holds. Let v 0 ∈ E(λ ), and let F be the mapping (7.10). Take A = N, B = F −1 (δv 0 ). By (8.37), (8.38), and (8.40) we see that (7.3) holds with b0 > 0 and a0 = B1 . By Proposition 7.3, we can conclude that there is a sequence {u k } ⊂ D such that (8.41)

G(u k ) → c,

b 0 ≤ c ≤ B1 ,

Thus, (8.42)

G (u k ) → 0.

G(u k ) = u k 2D − 2

F(x, u k )d x → c

and (8.43)

(G (u k ), v) = 2(u k , v) D − 2( f (u k ), v) → 0,

v ∈ D.

If ρk = u k D → ∞, let u˜ k = u k /ρk . Then u˜ k D = 1. Thus, there is a renamed subsequence such that u˜ k → u˜ weakly in D, strongly in L 2 (), and a.e. in . Hence, 2 2 G(u k )/ρk = u˜ k D − 2 F(x, u k )d x/ρk2 → 0.

Since (8.44)

|F(x, u k )|/ρk2 ≤ C(|u˜ k (x)|2 + W3 (x)/ρk2 ),

W3 ∈ L1 (),

2 in L 1 () and by (8.30), and the right-hand side of (8.44) converges to C|u(x)| ˜

(8.45)

2F(x, u k (x))/ρk2 → α+ (x)(u˜ + )2 + α− (x)(u˜ − )2

a.e.,

we see that the convergence in (8.45) is not only pointwise a.e., but also in L 1 (). Since u˜ k D = 1, (8.44) implies {α+ (u˜ + )2 + α− (u˜ − )2 }d x = 1. (8.46)

Also,

(G (u k ), v)/ρk = 2(u˜ k , v) D − 2( f (u k ), v)/ρk → 0

for each v ∈ D. This implies (u, ˜ v) ˜ D = (α+ u˜ + − α− u˜ − , v),

v ∈ D.

Consequently, u˜ is a solution of (8.34). By hypothesis, u˜ ≡ 0. But this contradicts (8.46). Hence, ρk ≤ C. The theorem now follows from Theorem 3.4.1 of [122].

8.4 Unbounded domains Now we allow the domain ⊂ Rn to be unbounded. Assume hypothesis (A), and assume that (8.47)

H (x, t) = 2F(x, t) − t f (x, t) ≥ −W3 (x) ∈ L 1 (),

x ∈ , t ∈ R,

72


and H (x, t) → ∞ a.e. as |t| → ∞.

(8.48) We have

Theorem 8.7. Assume that the spectrum of A consists of isolated eigenvalues of finite multiplicity 0 < λ0 < λ1 < · · · < λk < · · · ,

(8.49)

and let be a nonnegative integer. Take N to be the subspace of D spanned by the eigenspaces of A corresponding to the eigenvalues λ0 , λ1 , . . . , λ . We take M = N ⊥ ∩ D. Assume that there are functions W1 , W2 ∈ L 1 () and numbers a1 , a2 such that α < a1 ≤ a2 and (8.50) a1 (t − )2 + γ (a1 )(t + )2 − W1 (x) ≤ 2F(x, t) ≤ a2 (t − )2 + (a2 )(t + )2 + W2 (x),

x ∈ , t ∈ R,

and that (8.47) and (8.48) hold. Then (8.13) has at least one solution. Proof. First, we note that (8.51)

sup G ≤ B1 , N

inf G ≥ −B2 , M

Bj =

W j (x)d x.

To see this, note that by (8.28) we have v 2D ≤ a1 v − 2 + γ (a1 ) v + 2 ,

(8.52)

v ∈ N.

By (8.29) we have (8.53)

a2 w− 2 + (a2 ) w+ 2 ≤ w 2D ,

w ∈ M.

Hence, G(v) ≤ B1 ,

v ∈ N,

and G(w) ≥ −B2 ,

w ∈ M,

by (8.50). By Theorem 3.19, we conclude that for any sequence Rk → ∞, there is a sequence {u k } ⊂ D such that (8.54)

G(u k ) → c,

−B2 ≤ c ≤ B1 ,

(Rk + u k D ) G (u k ) ≤

In particular, we have (8.55)

u k 2D

−2

F(x, u k ) d x → c

B1 + B2 . ln(4/3)

8.4. Unbounded domains and

73

u k 2D − ( f (·, x k ), u k ) ≤ K .

(8.56) Consequently,

H (x, u k ) d x ≤ K .

(8.57)

If ρk = u k D → ∞, let u˜ k = u k /ρk . Then u˜ k D = 1. Consequently, there is a renamed subsequence such that u˜ k → u˜ weakly in D, strongly in L 2 (), and a.e. in . In view of (8.55), we have (8.58) 1−2 F(x, u k )/ρk2 d x → 0.

But by (8.50), we have + 2 2 2 2F(x, u k )/ρk2 ≤ a2 (u˜ − k ) + (a2 )(u˜ k ) + W2 (x)/ρk .

(8.59)

In the limit this implies 1 ≤ a2 u˜ − 2 + (a2 ) u˜ + 2 . This shows that u˜ ≡ 0. Let 0 be the subset of on which u˜ = 0. Then |u k (x)| = ρk |u˜ k (x)| → ∞,

(8.60)

If 1 = \0 , then we have (8.61) H (x, u k ) d x =

0

+

1

x ∈ 0 .

≥

0

H (x, u k ) d x −

1

W1 (x) d x → ∞.

This contradicts (8.57), and we see that ρk = u k D is bounded. Once we know that the ρk are bounded, we can apply Theorem 3.4.1 of [122] to obtain the desired conclusion. Remark 8.8. It should be noted that the crucial element in the proof of Theorem 8.7 was (8.56). If we had been dealing with an ordinary Palais–Smale sequence, we could only conclude that u k 2D − ( f (·, u k ), u k ) = o(ρk ), which would imply only that

H (x, u k )d x = o(ρk ).

This would not contradict (8.61), and the argument would not go through.

74

8. Semilinear Problems We also have

Theorem 8.9. The conclusion of Theorem 8.7 holds if in place of (8.47), (8.48) we assume that H (x, t) ≤ W1 (x) ∈ L 1 (),

(8.62)

x ∈ , t ∈ R,

and H (x, t) → −∞ a.e. as |t| → ∞.

(8.63)

Proof. We use (8.62) and (8.63) to replace (8.61) with (8.64) H (x, u k ) d x = + ≤ H (x, u k ) d x +

0

1

0

1

W1 (x) d x → −∞.

We then proceed as before. As a specific example of an operator A satisfying the hypotheses of Theorems 8.7 and 8.9, let g(x) be a measurable function satisfying g(x) ≥ c0 > 0,

x ∈ Rn ,

for some positive constant c0 . We consider the problem (8.65)

−u(x) + g(x)2 u(x) = f (x, u(x)),

x ∈ Rn .

We define the operator A on L 2 = L 2 (Rn ) by u ∈ D(A) and Au = f if u ∈ D = H 1,2 = H 1,2(Rn ) and (u, v) D = (∇u, ∇v) + (gu, gv) = ( f, v),

v ∈ H 1,2.

We assume that g(x) ∈ L 2loc and that multiplication by g −1 is a compact operator from H 1,2 to L 2 . It follows that A is a self-adjoint operator on L 2 that is bijective. Moreover, A−1 is a compact operator on L 2 . It follows that the spectrum of A consists of isolated eigenvalues of finite multiplicity satisfying (8.49). Thus, A satisfies the hypotheses of Theorems 8.7 and 8.9. Solutions of (8.11) satisfy (8.65). Hence, Theorems 8.7 and 8.9 produce weak solutions of (8.65). We summarize this as Theorem 8.10. Let g(x) be a function satisfying the conditions described above. Then there exists a sequence of eigenvalues for the equation (8.66)

−u(x) + g(x)2 u(x) = λu(x),

x ∈ Rn ,

satisfying (8.49). Let f (x, t) be a Carathéodory function satisfying hypothesis (A) for = Rn , and assume that (8.47), (8.48), and (8.50) hold for some > 0. Then (8.65) has at least one solution.

8.5. Further applications

75

Remark 8.11. We could have assumed a1 ≤ lim inf 2F(x, t)/t 2 ≤ lim sup 2F(x, t)/t 2 ≤ a2 t →−∞

t →−∞

and γ (a1 ) ≤ lim inf 2F(x, t)/t 2 ≤ lim sup 2F(x, t)/t 2 ≤ (a2 ) t →∞

t →∞

in place of (8.50). Remark 8.12. This theorem generalizes results of several authors, including [10], [46], [71], [98], and [104], with various conditions on the function g(x) to ensure that the spectrum of (8.66) is discrete. We guarantee it by assuming that multiplication by g −1 is a compact operator from H 1,2 to L 2 . A sufficient condition for this is given in [106]. Since g −1 is bounded, a simple sufficient condition is that for each constant b > 0, (8.67)

m{x ∈ Rn : |x − y| < 1, g(x) < b} → 0 as |y| → ∞.

Remark 8.13. If we choose a1 = λ and a2 = λ+1 , inequality (8.50) reduces to (8.68)

λ t 2 − W1 (x) ≤ 2F(x, t) ≤ λ+1 t 2 + W2 (x),

x ∈ Rn , t ∈ R.

By choosing a1 , a2 to be different values, we allow a wider range of possibilities for F(x, t).

8.5 Further applications Now we look for solutions of (8.13) under different conditions. Let A be a self-adjoint operator on L 2 (). We assume that A ≥ λ0 > 0 and that C0∞ () ⊂ D := D(A1/2 ) ⊂ H m,2() for some m > 0. Let q ∗ be given by q∗

= =

2n/(n − 2m), ∞,

2m < n n ≤ 2m,

and let f (x, t) be a Carathéodory function on × R. We assume the following. I. There are positive functions V (x), V0 (x), V1 (x), W (x), W1 (x), r0 (t), r1 (t) and constants q, q1 such that the following hold. (a) 2 < q < q ∗ , 1 < q1 < q ∗ . (b) Multiplication by V or W is a bounded operator from D to L 2 (), multiplication by V is compact from D to L q−1 (K ) for each compact subset K of , and W is in L q (). Multiplication by V0 is compact from D to L q (), V0 is locally bounded, and (8.69)

| f (x, st)| ≤ V0 (x)(|V (x)s|q−1 + W (x)q−1 )tr0 (t),

s ∈ R, t ≥ 1.

76

8. Semilinear Problems (c) Multiplication by V1 is a compact operator from D to L q1 (), W1 is in L 1 (),

and (8.70)

|H (x, st)| ≤ (|V1 (x)s|q1 + W1 (x))r1 (t),

s ∈ R, t ≥ 1.

(d) There is a ψ ∈ such that (8.71)

ψ(t) ≤ tr0 (t),

tψ(t) ≤ r1 (t),

t ≥ 1,

and (8.72)

ri (t) → ri

as t → ∞,

i = 0, 1,

with r0 < ∞

(8.73) and

r1 ≤ ∞.

(8.74)

II. There are positive constants α, β such that (8.75)

2F(x, t) ≤ λ0 (1 − α)t 2 ,

t 2 < β.

III. (a) If r0 = 0, it is assumed that there is a measurable function γ (x, a) on × R such that (8.76)

f (x, st)/tr0 (t) → γ (x, a) a.e. as s → a, t → ∞.

(b) If r0 = 0 and r1 = 0, it is assumed that there is a measurable function (x, a) on × R such that (8.77)

H (x, st)/r1(t) → (x, a) a.e. as s → a, t → ∞.

If r1 = ∞, then (8.77) is required to hold only for a = 0. IV. If r0 = 0 and r1 = 0, then we assume that there does not exist a function u ∈ D\{0} and b > 0 such that 1 (Au, u) = 1, 2 r0−1 Au = γ (x, u) a.e.,

(8.78) (8.79) (8.80)

a(u) :=

(x, u) d x = −b/r1 .

V. If r0 = 0 and r1 = ∞, we assume that all solutions u ≡ 0 of (8.79) are nonzero a.e. We have the following.


77

Theorem 8.14. Under hypotheses I to V, there is a solution of Au = f (x, u)

(8.81)

in D. If there is a u 0 ∈ D\{0} such that G(u 0 ) ≤ 0,

(8.82) where

G(u) = a(u) −

(8.83)

F(x, u) d x,

then (8.81) has a nonzero solution. Proof. Under hypothesis I(b), it is easily checked that G(u) given by (8.83) is continuously Fréchet differentiable with f (x, u)vd x, u, v ∈ D. (8.84) (G (u), v) = 2a(u, v) −

Also by hypothesis I(b), V u 2q + V0 u 2q + W u 2q ≤ Ca(u),

(8.85)

u ∈ D.

Thus, by (8.75), (8.86)

1 G(u) ≥ αa(u) + (1 − α)a(u) − λ0 (1 − α) u 2 2 −C |V0 u|(|V u|q−1 + β (1−q)/2|W u|q−1 ) d x u 2 >β

≥ a(u)(αC1 a(u)(q/2)−1). Hence, there are positive constants ρ, δ such that (8.87)

G(u) ≥ ρ,

a(u) = δ 2 .

We consider two cases. Case 1. G(u) ≥ 0 for all u ∈ D. In this case 0 is a minimum point and we must have G (0) = 0. By (8.84), we see that 0 is a solution of (8.81). Case 2. There is a u 0 ∈ D\{0} such that (8.82) holds. In this case the hypotheses of Theorem 2.18 are satisfied. In particular, there is a sequence {u k } ⊂ D such that (8.88)

G(u k ) → b

and (8.89)

G (u k )/ψ(rk ) → 0,

78


where b is given by (8.80), tk2 = a(u k ), and ψ(t) ∈ is the function satisfying (8.71). We shall show that this sequence satisfies a(u k ) ≤ C.

(8.90)

If so, we can find a renamed subsequence that converges weakly in D to a function u and such that V0 u k → V0 u in L q () and a.e. in while V u k → V u in L q−1 (K ) for each compact subset K of . By (8.84) and (8.89), (8.91)

2a(u k , v) − ( f (x, u k ), v) → 0,

v ∈ D.

If v ∈ C0∞ (), then, by hypothesis I(b), | f (x, u k )v| ≤ C(|V (x)u k | + W (x)q−1 )|V0 v|, and the right-hand side converges in L 1 () to C(|V u|q−1 + W q−1 )|V0 u|. Thus, (8.92)

v ∈ C0∞ ().

( f (x, u k ), v) → ( f (x, u), v),

Since u k → u weakly in D, we have (8.93)

v ∈ C0∞ (),

2a(u, v) = ( f (·, u), v),

which shows that u is a solution of (8.81). I claim that u = 0. To see this, note that b > 0 by (8.87). Moreover, by (8.70), |H (x, u k )| ≤ C(|V1 u k |q1 + W1 ), and the right-hand side converges in L 1 () to C(|V1 u k |q1 + W1 ). Since H (x, u k ) → H (x, u) a.e., we have

H (x, u k ) d x →

But for any u ∈ D, (8.94)

H (x, u) d x.

1 (G (u), u) = G(u) + 2

H (x, u) d x.

Hence, (8.88) and (8.89) imply that H (x, u) d x = −b = 0.

Thus, u = 0 since H (x, 0) ≡ 0.


79

It therefore remains only to prove (8.90). Assume that tk → ∞, and let u˜ k = u k /tk . Then a(u˜ k ) = 1

(8.95)

and there is a renamed subsequence {u˜ k } that converges weakly in D to a function u˜ and such that V0 u˜ k → V0 u˜ in L q () and a.e. in while V u k → V u˜ in L q−1 (K ) for each compact subset K of . By (8.69), | f (x, u k )|/tk r0 (tk ) ≤ (|V u˜ k |q−1 + W q−1 )|V0 u˜ k |. Thus, (8.96)

q−1 q−1 f (x, u k ) d x /tk r0 (tk ) ≤ ( V u˜ k q + W q ) V0 u˜ k q .

Moreover, by (8.71) and (8.89), (8.97)

2a(u˜ k )/r0 (tk ) − ( f (x, u k )/tk r0 (tk ), u˜ k ) → 0.

This shows that u(x) ˜ ≡ 0. Otherwise, we would have V0 u˜ k → 0 in L q () and consequently the left-hand side of (8.96), which is equal to the second term of (8.97), will converge to 0. But this would mean that a(u˜ k ) → 0, contradicting (8.95). If r0 = 0, we obtain another contradiction, for the second term in (8.97) is bounded by (8.96) while the first becomes infinite by (8.95) if tk → ∞. Hence, r0 = 0 implies (8.90), and the proof is complete in this case. It remains to consider the case when r0 = 0. By (8.69), | f (x, u k )v|/tk r0 (tk ) ≤ (|V u˜ k |q−1 + W q−1 )|V0 v|. When v ∈ C0∞ (), the right-hand side converges in L 1 () to (|V u| ˜ q−1 + W q−1 )|V0 v| by hypothesis II(b). By (8.76), the left-hand side converges a.e. to γ (x, u). ˜ Thus, f (x, u k )v(x)d x/tk r0 (tk ) → γ (x, u)v(x) ˜ dx (8.98)

for each v ∈ (8.99)

C0∞ ().

By (8.71) and (8.89),

2a(u˜ k /t0 (tk ), v) − ( f (x, u k )/tk r0 (tk ), v) → 0

for each v ∈ D. Thus, by (8.72) and (8.98), 2a(u/r ˜ 0 , v) = (γ (x, u), v),

v ∈ C0∞ ().

This shows that u˜ is a solution of (8.79). By (8.70), (8.100)

|H (x, u k )|/r1 (tk ) ≤ |V1 u˜ k |q1 + W1 ,

80


and the right-hand side converges in L 1 () to the function ˜ q1 + W1 |V1 u| by hypothesis II(c). If r1 = 0, this produces a contradiction, for by (8.88), (8.89), and (8.94), H (x, u k ) d x/r1(tk ) (8.101)

converges to −b/r1. Since b = 0, this means that expression (8.101) becomes infinite. Thus, the proof is complete for the case when r1 = 0. It therefore remains to consider the case when r0 = 0, r1 = 0. If r1 = ∞, the integrand of (8.101) converges a.e. to (x, u) ˜ by hypothesis III(b). If r1 = ∞, this convergence takes place only for those points x where u(x) ˜ = 0. But by hypothesis V, u(x) ˜ = 0 a.e. since it is a solution of (8.79). Hence, the convergence is a.e. in all cases, and expression (8.101) converges to (x, u) ˜ d x. (8.102)

Since it also converges to −b/r1 by (8.88), (8.89) and (8.94), we see that u˜ is a solution of (8.80) as well as (8.79). As before, (8.97) holds by (8.71) and (8.89). In view of (8.76), this implies that ˜ u). ˜ 2/r0 = (γ (x, u), If we combine this with (8.79), we see that u˜ satisfies (8.78) as well as (8.79) and (8.80). A contradiction is now provided by hypothesis IV. Hence, (8.90) holds, and the proof is complete.

8.6 Special cases In this section we present some consequences of Theorem 8.14. Theorem 8.15. Assume that (8.103)

|H (x, t)| ≤ W (x) ∈ L 1 (),

(8.104)

H (x, t) → H± (x) as t → ±∞ a.e.,

(8.105)

2t −2 F(x, t) → b± (x) as t → ±∞ a.e.,

(8.106) and that

| f (x, t)| ≤ W1 (x) ∈ L 1 (),

|t| ≤ 1,

(8.107) u>0

H+ (x) d x +

u 2 such that b1/q , W 1/q are compact operators from D to L q (), where b(x) = max |b± (x)|, and that solutions of (8.108) that are not identically zero a.e. are never zero a.e. Then (8.81) has a solution u ∈ D\{0}. Theorem 8.16. If we replace (8.103), (8.104) in Theorem 8.15 by (8.109)

|H (x, t)| ≤ V (x)|t|γ + W (x),

0 < γ < 2,

and (8.110)

H (x, t)/|t|γ → H± (x) as t → ±∞ a.e.,

the theorem will hold if γ (8.111) H± (x)|u| d x + u>0

u 0

for all u ∈ D\{0} satisfying (8.108), and V 1/q is a compact operator from D to L q (). In this case the requirement that solutions of (8.108) that are not identically zero a.e. are never zero a.e. can be removed. Theorem 8.17. Assume that there is a number γ such that 0 ≤ γ < 1 and (8.112)

| f (x, t)| ≤ V (x)|t|γ + W (x),

with V 1/q , W 1/q compact operators from D to L q (), where q > 2. Then (8.81) has a solution u ∈ D\{0}. Theorem 8.18. Assume that (8.113)

| f (x, t)| ≤ V (x)|t| + W (x)

and (8.114)

f (x, t)/t → b± (x) as t → ±∞ a.e.,

with b1/q , V 1/q , and W 1/q compact operators from D to L q (), where b(x) = max |b ± (x)|, and q > 2. Assume in addition that (8.108) has no nontrivial solutions. Then (8.81) has a solution u ∈ D\{0}.

8.7 The proofs In this section we give the proofs of Theorems 8.15 to 8.18. First, we give the Proof of Theorem 8.15. We apply Theorem 8.14. We take r0 (t) = r1 (t) = 1, ψ(t) = t −1 . Inequality (8.70) is satisfied with q1 = q, V1 arbitrary, and W1 = W. Since (8.115)

∂(Ft −2 )/∂t = −2t −3 H (x, t),

82


we have 1 (8.116)

F(x, t) =

2 2 b+ (x)t 1 2 2 b− (x)t

+ F0 (x, t),

t > 0,

+ F0 (x, t),

t < 0,

where (8.117)

F0 (x, t) =

∞

−3 H (x, s) ds, t s t −2t 2 −∞ s −3 H (x, s) ds,

2t 2

t > 0, t < 0.

Thus, (8.118)

∞ b+ (x)t + 4t ( t ) − 2H (x, t)/t, f (x, t) = t b− (x)t − 4t ( −∞ ) − 2H (x, t)/t

t > 0, t < 0.


| f (x, t)| ≤ b(x)|t| + 4W |t|−1 ,

|t| ≥ 1.

This combined with (8.106) gives (8.69) with V0 = V = b1/q . Moreover, by (8.119), (8.120)

f (x, st)/t → ±b±

as s → a, t → ∞.

Thus, (8.79) becomes (8.108). Moreover, by (8.104), the left-hand side of (8.80) is the left-hand side of (8.107). Hence, (8.107) assures us that (8.80) cannot hold for any solution u ∈ D of (8.79). Thus, all of the hypotheses of Theorem 8.14 are satisfied. Proof of Theorem 8.16. In this case we take r0 (t) = 1, r1 (t) = t γ . Since r1 = ∞, the right-hand side of (8.80) vanishes. Thus, strict inequality in (8.111) is needed to guarantee that no solutions of (8.78)–(8.80) exist. Moreover, we do not require hypothesis V for this case. Proof of Theorem 8.17. Here we take r0 (t) = t γ −1 and r1 (t) = t γ +1 . Note that by (8.112), (8.121)

|F(x, t)| ≤ p−1 V (x)|t| p + W (x)|t|,

where p = 1 + γ . Thus, by (8.2), 1 1 3 + (8.122) |H (x, t)| ≤ V (x)|t| p + W (x)|t|. 2 p 2 Since r0 = 0 and r1 = ∞, all of the hypotheses of Theorem 8.14 hold. Proof of Theorem 8.18. Now we take r0 (t) = 1, r1 (t) = t 2 . By (8.113), |F(x, t)| ≤

1 V (x)t 2 + W (x)|t| 2


83

and |H (x, t)| ≤

(8.123)

1 3 V (x)t 2 + W (x)|t|, 2 2

Moreover, by (8.114), (8.124)

2F(x, t)/t 2 → b± (x) a.e. as t → ±∞.

Consequently, by (8.2), H (x, t)/t 2 → 0 a.e. as t → ±∞. Thus r (x, a) → 0 a.e. and (8.80) is always satisfied. Thus, we must assume that (8.79) has no solutions in D\{0}.

8.8 Notes and remarks The material in this chapter is from [120], [110], [107], and [132].

Chapter 9

Superlinear Problems 9.1 Introduction Consider the problem (9.1)

−u = f (x, u), x ∈ ;

u = 0 on ∂,

where ⊂ Rn is a bounded domain whose boundary is a smooth manifold, and f (x, t) ¯ × R. This semilinear Dirichlet problem has been studied is a continuous function on by many authors. It is called sublinear if there is a constant C such that | f (x, t)| ≤ C(|t| + 1),

x ∈ , t ∈ R.

Otherwise, it is called superlinear. Beginning in [7], almost all researchers studying the superlinear problem assumed (a1 ) There are constants c1 , c2 ≥ 0 such that | f (x, t)| ≤ c1 + c2 |t|s , where 0 ≤ s < (n + 2)/(n − 2) if n > 2. (a2 ) f (x, t) = o(|t|) as t → 0. (a3 ) There are constants μ > 2, r ≥ 0 such that (9.2)

0 < μF(x, t) ≤ t f (x, t),

where

t

F(x, t) =

|t| ≥ r,

f (x, s) ds.

0

They proved Theorem 9.1. Under hypotheses (a1 )–(a3 ), problem (9.1) has a nontrivial weak solution. M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_9, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009

86

9. Superlinear Problems

The condition (a3 ) is convenient, but it is very restrictive. In particular, it implies that there exist positive constants c3 , c4 such that (9.3)

F(x, t) ≥ c3 |t|μ − c4 ,

x ∈ , t ∈ R.

Although this condition is weaker, it still eliminates many superlinear problems. A much weaker condition that implies superlinearity is (a3 ) Either F(x, t)/t 2 → ∞ as t → ∞ or F(x, t)/t 2 → ∞ as t → −∞. The purpose of the present chapter is to explore what happens when (a3 ) is replaced with (a3 ). Surprisingly, we find the following to be true. Theorem 9.2. Under hypotheses (a1 ), (a2 ), (a3 ), the boundary-value problem (9.4)

−u = β f (x, u), x ∈ ;

u = 0 on ∂,

has a nontrivial solution for almost every positive β. We generalize this theorem and present some variations below.

9.2 The main theorems We consider the boundary-value problems described in Sections 8.2 and 8.4 under assumption (A) stipulated there. We allow the domain to be unbounded. We also assume (B) The point λ0 is an isolated simple eigenvalue with a bounded eigenfunction ϕ0 (x) = 0 a.e. in . (C) There is a δ > 0 such that 2F(x, t) ≤ λ0 t 2 , where (9.5)

|t| ≤ δ, x ∈ ,

t

F(x, t) :=

f (x, s)ds.

0

(D)

There is a function W (x) ∈ L 1 () such that either W (x) ≤ F(x, t)/t 2 → ∞ as t → ∞,

x ∈

or W (x) ≤ F(x, t)/t 2 → ∞ as t → −∞,

x ∈ .

(The function W (x) need not be positive.) (E) There are constants μ > 2, C ≥ 0 such that [μF(x, t) − t f (x, t)]/(t 2 + 1) ≤ C,

t ∈ R, x ∈ .

9.2. The main theorems

87

We shall prove Theorem 9.3. Under the above hypotheses, the problem Au = f (x, u),

(9.6)

u∈D

has at least one nontrivial solution. We also have Theorem 9.4. If we replace hypothesis (E) with (E ) The function H (x, t) := t f (x, t) − 2F(x, t)

(9.7)

is convex in t. then problem (9.6) has at least one nontrivial solution. As we noted, problem (9.6) is called sublinear if f (x, t) satisfies | f (x, t)| ≤ C(|t| + 1),

(9.8)

x ∈ , t ∈ R.

Otherwise, it is called superlinear. Hypothesis (D) requires (9.6) to be superlinear. If we drop hypothesis (E) completely, then we are able to prove the following theorems. Theorem 9.5. If we replace hypotheses (C), (D) with (C ) There are a δ > 0 and a λ˜ > λ0 such that ˜ 2, 2F(x, t) ≥ λt and (D )

|t| ≤ δ, x ∈

there is a function W (x) ∈ L 1 () such that W (x) ≥ P(x, t) → −∞ as |t| → ∞,

x ∈ ,

where (9.9)

1 P(x, t) := F(x, t) − λ0 t 2 . 2

and drop hypothesis (E), then problem (9.6) has at least one nontrivial solution. We also have Theorem 9.6. Assume that (A)–(D) hold. Then, for almost every β ∈ (0, 1), the equation (9.10)

Au = β f (x, u)

has a nontrivial solution. In particular, the eigenvalue problem (9.10) has infinitely many solutions.

88


Theorem 9.7. If we replace hypothesis (C) in Theorem 9.6 with (C ) There are a δ > 0 and a λ˜ ≤ λ0 such that ˜ 2, 2F(x, t) ≤ λt and (D) with (D ) Either

or

|t| ≤ δ, x ∈ .

F(x, Rϕ0 ) d x/R 2 → ∞ as R → ∞

F(x, −Rϕ0 ) d x/R 2 → ∞ as R → ∞.

then (9.10) has a nontrivial solution for almost every β ∈ (0, λ0 /λ˜ ). Corollary 9.8. If we replace hypothesis (C ) in Theorem 9.7 with (C ) F(x, t)/t 2 → 0 uniformly as t → 0. then (9.10) has a nontrivial solution for almost every β ∈ (0, ∞).

9.3 Preliminaries Define (9.11)

G(u) := u 2D − 2

F(x, u)d x.

Under hypothesis (A), it is known that G is a continuously differentiable functional on the whole of D. In fact, the following were proved in [122, pp. 56–58]. Proposition 9.9. Under hypothesis (A), F(x, u(x)) and v(x) f (x, u(x)) are in L 1 () whenever u, v ∈ D. Proposition 9.10. G(u) has a Fréchet derivative G (u) on D given by (9.12)

(G (u), v) D = 2(u, v) D − 2( f (·, u), v).

Proposition 9.11. The derivative G (u) given by (9.12) is continuous in u. Theorem 9.12. Under hypotheses (A)–(C), the following alternative holds: Either (a) there is an infinite number of y(x) ∈ D(A) \ {0} such that

(9.13)

Ay = f (x, y) = λ0 y

or (b) for each ρ > 0 sufficiently small, there is an ε > 0 such that

(9.14)

G(u) ≥ ε,

u D = ρ.

9.4. Proofs

89

9.4 Proofs We now give the proof of Theorem 9.3. Proof. We take (9.15)

G(u) =

u 2D

−2

F(x, u)d x.

Under our hypotheses, Propositions 9.9–9.11 apply, and (9.16)

(G (u), v) = 2(u, v) D − 2( f (·, u), v),

u, v ∈ D.

By Theorem 9.12, we see that there are positive constants ε, ρ such that (9.17)

G(u) ≥ ε,

u D = ρ,

unless (9.18)

Au = λ0 u = f (x, u),

u ∈ D \ {0},

has a solution. This would give a nontrivial solution of (9.6). We may therefore assume that (9.17) holds. Next, we note that G(±Rϕ0 )/R 2 = ϕ0 2D − 2 {F(x, ±Rϕ0 )/R 2 ϕ02 }ϕ02 d x → −∞ as R → ∞

by hypothesis (D), depending on which part of (D) is assumed, since ϕ0 = 0 a.e. Since G(0) = 0 and (9.17) holds, we can now apply Theorem 3.13 to conclude that there is a sequence {u k } ⊂ D such that G(u k ) → c ≥ ε, Then (9.19)

G (u k ) → 0.

G(u k ) = ρk2 − 2

F(x, u k )d x → c

and (9.20)

(G (u k ), u k ) = 2ρk2 − 2( f (·, u k ), u k ) = o(ρk ),

where ρk = u k D . Assume that ρk → ∞, and let u˜ k = u k /ρk . Since u˜ k D = 1, there is a renamed subsequence such that u˜ k → u˜ weakly in D, strongly in L 2 (), and a.e. in . By (9.19), 2F(x, u k ) 2 u˜ k d x → 1. u 2k Let 1 = {x ∈ : u(x) ˜ = 0},

2 = \ 1 .

90


Then

2F(x, u k ) 2 u˜ k → ∞, u 2k

x ∈ 1 ,

by hypothesis (D). If 1 has positive measure, then 2F(x, u k ) 2 2F(x, u k ) 2 u˜ k d x ≥ u˜ k d x + [−W (x)] d x → ∞. u 2k u 2k 1 2 Thus, the measure of 1 must be 0, i.e., we must have u˜ ≡ 0 a.e. Moreover, 2F(x, u k ) − μu k f (x, u k ) 2 u˜ k d x → 1 − μ. u 2k But by hypothesis (E), lim sup

u 2k + 1 2 2F(x, u k ) − μu k f (x, u k ) 2 u ˜ ≤ lim sup C u˜ k = 0, k u 2k u 2k

which implies that 1 − μ ≤ 0, contrary to our assumption. Hence, the ρk are bounded. We can now follow the usual procedures to obtain a weak solution of (9.6) satisfying G(u) = c ≥ ε (cf., e.g., [122, p. 64]). Since G(0) = 0, we see that u = 0. This completes the proof. We postpone the proof of Theorem 9.4 until the next section. In proving Theorem 9.5, we shall make use of Lemma 9.13. Under hypothesis (C ), there is an α = 0 such that G(αϕ0 ) < 0. Proof. We can assume that ϕ0 D = 1.

(9.21) Thus,

G(αϕ0 ) = α 2 − 2

F(x, αϕ0 ) d x ≤ α 2 − λ˜ α 2 ϕ0 (x)2 d x |αϕ0 (x)|δ

V q (|αϕ0 |q + |αϕ0 |)

q ≤ α − λ˜ α 2 ϕ0 2 + C|α|q V ϕ0 q 2

˜ 0 ) + C |α|q−2 ]. ≤ α 2 [1 − (λ/λ This can be made negative by taking α sufficiently small. Lemma 9.14. Under hypothesis (D ), (9.22)

G(u) → ∞

as u D → ∞.

9.4. Proofs

91

Proof. Suppose there is a sequence {u k } ⊂ D such that ρk = u k → ∞ and G(u k ) ≤ K . Write u k = wk + αk ϕ0 ,

u˜ k = u k /ρk ,

w˜ k = wk /ρk ,

α˜ k = αk /ρk ,

where wk ⊥ϕ0 . If λ1 > λ0 is the next point in the spectrum of A, then λ1 w 2 ≤ w 2D ,

w⊥ϕ0 .

Thus, G(u k ) =

u k 2D

− λ0 u k − 2 2

P(x, u k ) d x

λ0 wk 2D − 2 P(x, u k ) d x 1− λ1 λ0 2 W (x) d x. ≥ 1− wk D − 2 λ1

≥

The only way this would not converge to ∞ is if wk D is bounded. But then w˜ k D → 0 and |α˜ k | → 1. Since u˜ k D = 1, there is a renamed subsequence such that u˜ k → u˜ weakly in D, strongly in L 2 (), and a.e. in . Since w˜ = 0 and |α| ˜ = 1, we have u(x) ˜ = αϕ ˜ 0 (x) = 0 a.e. Hence, |u k (x)| = ρk |u˜ k (x)| → ∞ a.e. Consequently, P(x, u k ) d x → −∞,

showing that G(u k ) → ∞. This completes the proof. We can now give the proof of Theorem 9.5. Proof. Let m = inf G. D

Then there is a sequence {u k } ⊂ D such that G(u k ) → m. In view of Lemma 9.14, we must have u k D ≤ C. Thus, there is a renamed subsequence such that u k → u weakly in D, strongly in L 2loc (), and a.e. in . Now, F(x, u) d x G(u) = u 2D − 2

=

u k 2D

− 2([u k − u], u) D − u k − u 2D − 2 F(x, u k ) d x + 2 [F(x, u k ) − F(x, u)]d x ≤ G(u k ) − 2([u k − u], u) D + 2 [F(x, u k ) − F(x, u)]d x.

92


From our hypotheses, it follows that F(x, u k ) d x → F(x, u) d x

(cf., e.g., [122, p. 64]). We therefore have in the limit G(u) ≤ m, from which we conclude that G(u) = m and G (u) = 0. Hence, u is a weak solution of (9.6). We see from Lemma 9.13 that m < 0. Since G(0) = 0, we see that u = 0. This completes the proof.

9.5 The parameter problem In this section we shall give the proofs of Theorems 9.4, 9.6, and 9.7. They will be based on the following result proved in the next section. Let E be a reflexive Banach space with norm · , and let A, B be two closed subsets of E. Suppose that G ∈ C 1 (E, R) is of the form G(u) := I (u) − J (u), u ∈ E, where I, J ∈ C 1 (E, R) map bounded sets to bounded sets. Define G λ (u) = λI (u) − J (u),

λ ∈ ,

where is an open interval contained in (0, +∞). Assume one of the following alternatives holds. (H1 ) I (u) ≥ 0 for all u ∈ E and either I (u) → ∞ or |J (u)| → ∞ as u → ∞. (H2 ) I (u) ≤ 0 for all u ∈ E and either I (u) → −∞ or |J (u)| → ∞ as u → ∞. Furthermore, we suppose that (H3 ) a0 (λ) := sup A G λ ≤ b0 (λ) := inf B G λ , for any λ ∈ . We let be the set of mappings (t) ∈ C(E × [0, 1], E) described in Chapter 3. We have Theorem 9.15. Assume that (H1 ) or (H2 ) holds together with (H3 ). (1) If A links B [hm] and A is bounded, then, for almost all λ ∈ , there exists u k (λ) ∈ E such that supk u k (λ) < ∞, G λ (u k (λ)) → 0, and G λ (u k (λ)) → a(λ) := inf

sup

∈ s∈[0,1],u∈A

G λ ((s)u),

k → ∞.

Furthermore, if a(λ) = b0 (λ), then dist(u k (λ), B) → 0, k → ∞. (2) If B links A[hm] and B is bounded, then, for almost all λ ∈ , there exists v k (λ) ∈ E such that supk v k (λ) < ∞, G λ (v k (λ)) → 0, and G λ (v k (λ)) → b(λ) := sup

inf

∈ s∈[0,1],v∈B

G λ ((s, v)),

k → ∞.

Furthermore, if a0 (λ) = b(λ), then dist(v k (λ), A) → 0, k → ∞.

9.5. The parameter problem

93

We shall also need the following extension of Theorem 9.12. Theorem 9.16. Let λ be a parameter satisfying 1 < λ ≤ K < ∞. Under hypotheses (A)–(D), for each ρ > 0 sufficiently small (not depending on λ), we have (9.23) G λ (u) := λ u 2D − 2 F(x, u)d x ≥ (λ − 1)ρ 2 , u D = ρ.

If we replace hypothesis (C) with hypothesis (C ), assuming 1 < λ˜ /λ0 < λ ≤ K < ∞, then we have λ˜ (9.24) G λ (u) ≥ λ − ρ 2 , u D = ρ. λ0 Proof. Let λ1 > λ0 be the next point in the spectrum of A, and let N0 denote the eigenspace of λ0 . We take M = N0⊥ ∩ D. By hypothesis (B), there is a ρ > 0 such that y D ≤ ρ ⇒ |y(x)| ≤ δ/2,

y ∈ N0 .

Now suppose u ∈ D satisfies u D ≤ ρ

(9.25)

|u(x)| ≥ δ

and

for some x ∈ . We write u = w + y,

(9.26)

y ∈ N0 .

w ∈ M,

Then for those x ∈ satisfying (9.25) we have δ ≤ |u(x)| ≤ |w(x)| + |y(x)| ≤ |w(x)| + (δ/2). Hence, |y(x)| ≤ δ/2 ≤ |w(x)|,

(9.27) and consequently,

|u(x)| ≤ 2|w(x)|

(9.28)

for all such x. Now we have by (9.26) and (9.28), u2d x − C G λ (u) ≥ λ u 2D − λ0 |u|δ

(|V u|q + V q |u|)d x

≥

λ u 2D − λ0 u 2 − C

≥

(λ − 1) y 2D + λ w 2D − λ0 w 2 − C

|u|>δ

|V u|q d x |V w|q d x 2|w|>δ

94


in view of the fact that y 2D = λ0 y 2 and (9.28) holds. Thus, by (8.3), λ0 q−2 w 2D , u D ≤ ρ. − C w D (9.29) G λ (u) ≥ (λ − 1) y 2D + λ − λ1 We take ρ > 0 to satisfy 1−

λ0 > C ρ q−2 . λ1

This gives λ0 q−2 −C ρ − λ + 1 w 2D G λ (u) ≥ (λ − 1)ρ + λ − λ1 2

≥ (λ − 1)ρ 2 ,

u D = ρ.

Hence, (9.23) holds. To prove (9.24) under hypothesis (C ), let σ = λ˜ /λ0 and = (σ, K ). Under hypothesis (C ), we have in place of (9.29) λ˜ q−2 2 − C w D w 2D , u D ≤ ρ. (9.30) G λ (u) ≥ (λ − σ ) y D + λ − λ1 We take ρ > 0 to satisfy σ−

λ˜ > C ρ q−2 . λ1

Consequently,

λ˜ q−2 G λ (u) ≥ (λ − σ )ρ + λ − −C ρ − λ + σ w 2D λ1 2

≥ (λ − σ )ρ 2 , u D = ρ. This gives (9.24), and the proof is complete. We now turn to the proofs of Theorems 9.6 and 9.7. We prove the latter first. We shall prove Theorem 9.7 by applying Theorems 9.15 and 9.16. ˜ 0 , K > 1 is a finite number, Proof. We take E = D, = (σ, K ), where σ = λ/λ and I (u) = u 2D , J (u) = 2 F(x, u) d x.

For the purpose of this application, it is sufficient to know that the sets A± = [0, ±Rϕ0 ],

B = {x ∈ D : x D = ρ}

link each other if R > ρ (cf., e.g., [120]). In our case hypothesis (H1 ) is satisfied. We now check that (H3) holds. We observe that G λ (u) = 0

9.5. The parameter problem

95

is equivalent to (9.10) with β = 1/λ. Now, at least one of the expressions J (±Rϕ0 )/R 2 = 2 F(x, ±Rϕ0 ) d x/R 2 → ∞ as R → ∞

by hypothesis (D ). Hence, for R sufficiently large, one of the inequalities 2 2 G λ (±Rϕ0 )/R ≤ K ϕ0 D − 2 {F(x, ±Rϕ0 )/R 2 ≤ 0

holds. Thus, a0 (λ) ≤ 0,

λ ∈ .

Moreover, it follows from Theorem 9.16 that (9.24) holds. Hence, b0 (λ) ≥ (λ − σ )ρ 2 ,

λ ∈ .

This shows that hypothesis (H3 ) holds. We can now apply Theorem 9.15 to conclude that for almost all λ ∈ , there exists u k (λ) ∈ D such that supk u k (λ) < ∞, G λ (u k (λ)) → 0, and G λ (u k (λ)) → a(λ) ≥ b0 (λ). Once it is known that the sequence {u k } is bounded, we can apply the usual theory to conclude that there is a solution of G λ (u) = 0,

G λ (u) = a(λ)

(cf., e.g., [122, p. 64]). Moreover, from the definition, we see that a(λ) ≥ (λ − σ )ρ 2 . Hence, the equation G λ (u) = 0 has a nontrivial solution for almost every λ ∈ . This is equivalent to (9.10) having a nontrivial solution for almost every β ∈ (K −1 , σ −1 ). Since K was arbitrary, the result follows. To prove Theorem 9.6, it suffices to take λ˜ = λ0 and show that hypothesis (D) implies hypothesis (D ). To see this, we note that F(x, ±Rϕ0 ) 2 F(x, ±Rϕ0 ) d x/R 2 = ϕ0 d x → ∞ R 2 ϕ02 by hypothesis (D) and the fact that ϕ0 (x) = 0 a.e. To prove Corollary 9.8, we let ε be any positive number. By hypothesis (C ), there is a δ > 0 such that F(x, t)/t 2 ≤ ε, |t| ≤ δ, x ∈ . By Theorem 9.7, (9.10) has a nontrivial solution for a.e. β ∈ (0, λ0 /ε). Since ε was arbitrary, the result follows. We now give the proof of Theorem 9.4.

96


Proof. By Theorem 2.4, for each arbitrary K > 1, and a.e. λ ∈ (1, K ), there exists u λ such that G λ (u λ ) = 0, G λ (u λ ) = a(λ) ≥ (λ − 1)ρ 2 . Choose λn → 1, λn > 1. Then there exists u n such that G λn (u n ) = 0,

G λn (u n ) = a(λn ) ≥ a(1) ≥ b0 (1).

By Theorem 3.4, we may assume that b0 (1) ≥ ε > 0. Therefore, 2F(x, u n ) d x ≤ c. 2 u n D Now we prove that {u n } is bounded. If u n D → ∞, let wn = u n / u n D ; then wn → w weakly in D, strongly in L 2 (), and a.e. in . We now have two cases: Case 1: w = 0 in D. We get a contradiction as follows: 2F(x, u n ) 2F(x, u n ) c≥ d x = |wn |2 d x 2 2 u u n D n 2F(x, u n ) ≥ |wn |2 d x − W1 (x) d x → ∞. u 2n w =0 w=0 Case 2: w = 0 in D. We define tn ∈ [0, 1] by G λn (tn u n ) = max G λn (tu n ). t ∈[0,1]

For any c > 0 and w¯ n = cwn , we have F(x, w¯ n ) d x → 0

(cf., e.g., [122, p. 64]). Thus, 2

G λn (tn u n ) ≥ G λn (cwn ) = c λn − 2

F(x, w¯ n ) d x ≥ c2 /2

for n large enough. That is, limn→∞ G λn (tn u n ) = ∞ and (G λn (tn u n ), u n ) = 0. Therefore, f (x, tn u n )tn u n − 2F(x, tn u n ) d x G λn (tn u n ) = =

H (x, tn u n ) d x → ∞.

By hypothesis (E ), G λn (u n ) =

H (x, u n ) d x ≥

H (x, tn u n ) d x → ∞.

9.6. The monotonicity trick

97

But G λn (u n ) = a(λn )

≤ ≤
0 such that (9.32)

n (s)u ≤ k0

whenever

G λ (n (s)u) ≥ a(λ) − (λn − λ).

In fact, by the definition of a(λn ), there exists n ∈ such that (9.33)

sup s∈[0,1],u∈A

G λ (n (s)u) ≤

sup s∈[0,1],u∈A

G λn (n (s)u) ≤ a(λn ) + (λn − λ).

If G λ (n (s)u) ≥ a(λ) − (λn − λ) for some u ∈ A, s ∈ [0, 1], then, by (9.31) and (9.33), we have that (9.34)

I (n (s)u)

= ≤ ≤

G λn (n (s)u) − G λ (n (s)u) λn − λ a(λn ) + (λn − λ) − a(λ) + (λn − λ) λn − λ a (λ) + 3,

and it follows that (9.35)

J (n (s)u)

= λn I (n (s)u) − G λn (n (s)u) ≤

λn (a (λ) + 3) − G λ (n (s)u)

≤

λn (a (λ) + 3) − a(λ) + (λn − λ)

≤

2λ(a (λ) + 3) − a(λ) + λ.

98


On the other hand, by (H1 ), (9.31), and (9.33), (9.36)

J (n (s)u)

=

λn I (n (s)u) − G λn (n (s)u)

≥

−G λn (n (s)u)

≥

−(a(λn ) + (λn − λ))

≥

−(a(λ) + (λn − λ)(a (λ) + 2))

≥

−a(λ) − λ|a (λ) + 2|.

Combining (9.34)–(9.37) and (H1 ), we see that there exists k0 (λ) := k0 (depending only on λ) such that (9.32) holds. First, we consider the case of a(λ) > b0 (λ). For each n, we define (9.37)

Q n (λ) := {u ∈ E : u ≤ k0 + 1, |G λ (u) − a(λ)| ≤ λn − λ}.

We claim that Q n (λ) = φ and that inf{ G λ (u) : u ∈ Q n (λ)} → 0

(9.38)

as n → ∞. By the definition of a(λ), there exists (s0 , u 0 ) ∈ [0, 1] × A such that G λ (n (s0 )u 0 ) ≥ a(λ) − (λn − λ). Then, by (9.32), n (s0 )u 0 ≤ k0 (λ). It follows that n (s0 )u 0 ∈ Q n (λ). Therefore, Q n (λ) = φ. Next, we want to prove (9.38). If this were not so, there would exist a positive ε < [a(λ) − b0 (λ)]/3 such that G λ (u) ≥ 3ε for all u ∈ Q n (λ). We take n so large that (a (λ) + 2)(λn − λ) ≤ ε, λn − λ ≤ ε. By Lemma 2.10.1 of [122], we may construct a locally Lipschitz continuous map Yλ of Eˆ such that 1. Yλ (u) ≤ 1,

ˆ ∀u ∈ E,

2. (G λ (u), Yλ (u)) ≥ 2ε, 3. (G λ (u), Yλ (u)) ≥ 0,

∀u ∈ Q n (λ), ˆ ∀u ∈ E.

Consider the initial boundary-value problem: dσ (t)u = −Yλ (σ (t)u), dt

σ (0, u) = u.

By Theorem 4.5, there exists a unique continuous solution σ (t)u such that G λ (σ (t)u) is nonincreasing in t. Define σ (2s)u, 0 ≤ s ≤ 1/2, ˜ (s)u := σ (1)n (2s − 1)u, 1/2 ≤ s ≤ 1. Then it is easy to check that ˜ ∈ . We want to prove that (9.39)

˜ ≤ a(λ) − (λn − λ), G λ ((s)u)

∀(s, u) ∈ [0, 1] × A,


99

which provides the desired contradiction. Choose any u ∈ A. If 0 ≤ s ≤ 1/2, by (H3 ), we get ˜ G λ ((s)u)

(9.40)

=

G λ (σ (2s)u)

≤

G λ (u)

≤

a0 (λ)

≤

b0 (λ)

0, T > 0, we define ¯ Q(ε, T, λ) :={u ∈ E : u ≤ k0 (λ) + 4 + d A , |G λ (u) − a(λ)| ≤ 3ε, d(u, B) ≤ 4T }. ¯ We claim that Q(ε, T, λ) = φ. By (9.33), we may choose n so large that (9.46)

sup s∈[0,1],u∈A

G λ (n (s)u) ≤

sup s∈[0,1],u∈A

G λn (n (s)u) ≤ a(λ) + 3ε.

Since A links B [hm], there exists (s0 , u 0 ) ∈ [0, 1] × A such that n (s0 )u 0 ∈ B. Hence dist(n (s0 )u 0 , B) = 0 and (9.47)

G λ (n (s0 )u 0 ) ≥ b0 (λ) = inf G λ = a(λ) > a(λ) − (λn − λ) ≥ a(λ) − 3ε. B

¯ T, λ). By (9.32), n (s0 )u 0 ≤ k0 . Hence, n (s0 )u 0 ∈ Q(ε, Next, we prove that (9.48)

¯ inf{ G λ (u) : u ∈ Q(ε, T, λ)} = 0

for ε, T sufficiently small. If not, there would exist δ > 0, ε1 > 0, and T1 ∈ (0, 1) such that (9.49)

G λ (u) ≥ 3δ

¯ 1 , T1 , λ). for u ∈ Q(ε


101

Define (9.50)

Q¯ ∗ (ε1 , T1 , λ) := {u ∈ E : u ≤ k0 + 4 + d A , dist(u, B) ≤ 4T1 , a(λ) − (λn − λ) ≤ G λ (u) ≤ a(λ) + 3ε1}.

¯ 1 , T1 , λ). Let n be By (9.46) and (9.47), Q¯ ∗ (ε1 , T1 , λ) = φ and Q¯ ∗ (ε1 , T1 , λ) ⊂ Q(ε so large that λn − λ ≤ ε1 , (a (λ) + 2)(λn − λ) < ε1 and λn − λ < δT1 . Similarly, we may construct a locally Lipschitz continuous map Y¯λ of Eˆ such that ˆ ∀u ∈ E,

1. Y¯λ (u) ≤ 1,

2. (G λ (u), Y¯λ (u)) ≥ 2δ, 3. (G λ (u), Y¯λ (u)) ≥ 0,

¯ 1 , T1 , λ), ∀u ∈ Q(ε ˆ ∀u ∈ E.

Define (9.51) Q 1 := {u ∈ E : u ≤ k0 + 2 + d A , |G λ (u) − a(λ)| ≤ 2ε1, dist(u, B) ≤ 3T1 }. ¯ 1 , T1 , λ). Moreover, the distance As in the proof of (9.38), Q 1 = φ and Q 1 ⊂ Q(ε ¯ 1 , T1 , λ) is positive. Thus, we may choose a Lipschitz continubetween Q 1 and E\ Q(ε ¯ 1 , T1 , λ). ous map γ from E into [0, 1] that equals 1 on Q 1 and vanishes outside Q(ε Consider the following initial boundary-value problem: d(σ1 (t)u) = −γ (σ1 )Y¯λ (σ1 ), dt

σ1 (0)u = u.

Let σ1 (t)u be the unique continuous solution. Then we have that d G λ (σ1 (t)u) ≤ −2δγ (σ1 (t)u) ≤ 0. dt

(9.52) First, we note that

σ1 (t)u ∈ B,

(9.53)

s ∈ [0, T1 ], u ∈ A.

For u ∈ A, by (9.52), we have that (9.54)

G λ (σ1 (t)u) ≤ G λ (u) ≤ a0 (λ) ≤ b0 (λ) = a(λ),

and (9.55)

G λ (σ1 (t)u)

=

G λ (u) +

≤ for all t ∈ [0, T1 ].

G λ (u) −

t 0 t 0

d G λ (σ1 (θ )u) dθ dθ 2δγ (σ1 (θ )u)dθ

∀t ∈ [0, T1 ],

102


If (9.53) were not true, then there would be a t0 ∈ [0, T1 ] such that σ1 (t0 )u ∈ B. Then G λ (σ1 (t0 )u) ≥ a(λ) = b0 (λ) = inf B G λ . By (9.52)–(9.56), we see that t0 2δγ (σ1 (θ )u)dθ = 0. 0

Hence, γ (σ1 (θ )u) = 0 for θ ∈ [0, t0 ]; i.e., σ1 (θ )u ∈ Q 1 ∀θ ∈ [0, t0 ]. Therefore, one of the following three cases occurs:

(9.56)

σ1 (θ )u > k0 + 2 + d A ;

or (9.57)

|G λ (σ1 (θ )u) − a(λ)| > 2ε1 ;

or (9.58)

dist(σ1 (θ )u, B) > 3T1 .

Inequality (9.56) cannot hold, since (9.59)

σ1 (θ, u) − σ1 (θ )u ≤ |θ − θ |

and σ1 (θ )u ≤ σ1 (0)u + T1 ≤ d A + 1,

∀θ ∈ [0, t0 ].

If (9.57) holds, then G λ (σ1 (θ )u) < a(λ) − 2ε1 . Hence, σ1 (θ )u ∈ B. Evidently, (9.58) implies that σ1 (θ )u ∈ B. Therefore, (9.53) is true. Next, we note that (9.60)

σ1 (T1 )n (2s − 1)u ∈ B,

u ∈ A, s ∈ [1/2, 1].

For any fixed u ∈ A and s ∈ [1/2, 1], we divide the proof into two cases. Case 1: If σ1 (θ )n (2s − 1)u ∈ Q 1 for all θ ∈ [0, T1 ], by (9.52) and (9.32), we have that G λ (σ1 (T1 )n (2s − 1)u) = G λ (n (2s − 1)u) + ≤ G λ (n (2s − 1)u) −

T1 0 T1

d G λ (σ1 (θ )n (2s − 1)u) dθ dθ 2δγ (σ1 (θ )n (2s − 1)u)dθ

0

= G λ (n (2s − 1)u) − 2δT1 ≤ a(λ) − 2δT1 + (a (λ) + 2)(λn − λ), which implies that σ1 (T1 )n (2s − 1)u ∈ B since a(λ) = b0 (λ).


103

Case 2: If there exists t0 ∈ [0, T1 ] such that σ1 (t0 )n (2s − 1)u ∈ Q 1 , then one of the following alternatives holds: either σ1 (t0 )n (2s − 1)u > k0 + 2 + d A ;

(9.61) or

|G λ (σ1 (t0 )n (2s − 1)u) − a(λ)| > 2ε1 ;

(9.62) or

dist(σ1 (t0 )n (2s − 1)u, B) > 3T1 .

(9.63)

Assume that (9.61) holds. If σ1 (T1 )n (2s − 1)u ∈ B, then b0 (λ) = a(λ) ≤ G λ (σ1 (T1 )n (2s − 1)u) ≤ G λ (n (2s − 1)u). By (9.32), n (2s − 1)u ≤ k0 . Furthermore, since σ1 (t0 )n (2s − 1)u − σ1 (0)n (2s − 1)u ≤ t0 , it follows that σ1 (t0 )n (2s − 1)u ≤ k0 + t0 ≤ k0 + 1, which contradicts (9.61). Hence, σ1 (T1 )n (2s − 1)u ∈ B. Assume that (9.62) holds. Note, by (9.32), that G λ (σ1 (t0 )n (2s − 1)u)

≤

G λ (σ1 (0)n (2s − 1)u)

=

G λ (n (2s − 1)u)

≤

a(λ) + ε1 .

Therefore, (9.62) implies that G λ (σ1 (T1 )n (2s − 1)u) ≤ G λ (σ1 (t0 )n (2s − 1)u) ≤ a(λ) − 2ε1. It follows that σ1 (T1 )n (2s − 1)u ∈ B since a(λ) = b0 (λ). Assume (9.63) holds. Note that σ1 (t)u − σ1 (t )u ≤ |t − t |. It therefore follows that σ1 (t)n (2s − 1)u − w ≥ σ1 (t0 )n (2s − 1)u − w − |t − t0 | for all w ∈ B, t ∈ [0, T1 ]. Hence, dist(σ1 (t)n (2s − 1)u, B) ≥ 2T1 for all t ∈ [0, T1 ]. In particular, σ1 (T1 )n (2s − 1)u ∈ B. Hence, (9.60) holds in this case as well. Define σ1 (2sT1 )u, 1∗ (s)u := σ1 (T1 )n (2s − 1)u,

0 ≤ s ≤ 1/2, 1/2 ≤ s ≤ 1.

104


Then 1∗ ∈ . Moreover, by (9.53) and (9.60), 1∗ (s)A ∩ B = φ for all s ∈ [0, 1]. This gives the contradiction that completes the proof. Now, we consider conclusion (1) with the second alternative (H2 ). For this case, the map λ → a(λ) is nonincreasing and a (λ) = da(λ)/dλ exists for almost all λ > 0. Therefore, we consider those λ where a (λ) exists. We choose λn ∈ (0, λ) ∩ and λn → λ as n → ∞. Then (9.31) is still true for n large enough. We also prove that there exists a n ∈ , k0 = k0 (λ) > 0 such that (9.64)

n (s)u ≤ k0 if G λ (n (s, u)) ≥ a(λ) − (λ − λn ).

In fact, by the definition of a(λn ), there exists a n ∈ such that (9.65)

sup s∈[0,1],u∈A

G λ (n (s)u)

≤

sup s∈[0,1],u∈A

G λn (n (s)u)

≤

a(λn ) + (λ − λn )

≤

a(λ) + (a (λ) − 2)(λn − λ).

If G λ (n (s)u) ≥ a(λ) − (λ − λn ), then by (9.31), (9.64), and (9.65), (9.66)

I (n (s, u))

= ≥ ≥

G λn (n (s)u) − G λ (n (s)u) λn − λ a(λn ) − a(λ) + 2(λ − λn ) λn − λ a (λ) − 3.

On the other hand, by (H2 ), (9.31), (9.65), (9.66), (9.67)

J (n (s)u)

= λn I (n (s)u) − G λn (n (s)u) ≥

−λ|a (λ) − 3| − λ − a(λn )

≥

−λ|a (λ) − 3| − λ − λ|a (λ) − 1| − a(λ)

and (9.68)

J (n (s)u) ≤ −G λn (n (s)u) ≤ −G λ (n (s)u) ≤ −a(λ) + λ.

Hence, (9.66)–(9.68) imply that n (s)u ≤ k0 = k0 (λ), a constant depending only on λ. The rest is similar to the proof under the assumption (H1 ); we omit the details. Finally, conclusion (2) can be proved immediately by interchanging A and B, replacing G λ by −G λ , and using conclusion (1).

9.7 Notes and remarks Problem (9.6) has been studied by many people. The vast majority of results obtained concern sublinear problems. Much less has been proved for the superlinear case. In [7] the basic assumption was (9.69)

0 < μF(x, t) ≤ t f (x, t),

|t| ≥ r,


105

for some μ > 2 and r ≥ 0. This is a very convenient hypothesis since it readily achieves mountain pass geometry as well as satisfaction of the Palais–Smale condition. However, it is a severe restriction; it strictly controls the growth of f (x, t) as |t| → ∞. Almost every author discussing superlinear problems has made this assumption. We have been able to weaken assumption (9.69) considerably, but not to our complete satisfaction. We assume either that μF(x, t) − t f (x, t) ≤ C(t 2 + 1),

|t| ≥ r,

for some μ > 2 and r ≥ 0 or that (9.7) is convex in t. These allow much more freedom for the function f (x, t). But they do not allow as much freedom as we would like. We were able to weaken it much further and assume only hypothesis (D). However, we paid a heavy price for this generalization: We were only able to solve (9.10) for almost all positive values of β. The method (called the monotonicity trick), which allowed us to solve (9.10) for almost all values of β in some interval, was first introduced by Struwe [148] for minimization problems. It was applied by Jeanjean [76] and others for various types of problems. The material of this chapter comes from [141] and [139]. See also [49].

Chapter 10

Weak Linking 10.1 Introduction As we noted, a subset A of a Banach space E links a subset B of E if, for every G ∈ C 1 (E, R) satisfying (10.1)

a0 := sup G ≤ b0 := inf G, B

A

there are a sequence {u k } ⊂ E and a constant c such that (10.2)

b0 ≤ c < ∞

and (10.3)

G(u k ) → c,

G (u k ) → 0.

We also saw that there are several criteria that imply that a set A links a set B. However, all of them require that at least one of the sets A, B be contained in a finite-dimensional manifold. It is not clear if it is possible for A to link B if neither is contained in such a manifold. For instance, if E = M ⊕ N, where M, N are closed, infinite-dimensional subspaces of E and B R is the ball centered at the origin of radius R in E, it is unknown if the set A = M ∩ ∂ B R links B = N. (If either M or N is finite-dimensional, then A does link B; cf. Example 2 of Section 3.4.) Very little is known for the infinitedimensional case. Unfortunately, this situation arises in some important applications, including Hamiltonian systems, the wave equation, and elliptic systems, to name a few. The purpose of the present chapter is to study linking when both M and N are infinite-dimensional and G has some additional continuity property. The property we have chosen is that of weak-to-weak continuity: Definition 10.1. Let E be a Banach space. We shall call a functional G ∈ C 1 (E, R) weak-to-weak continuously differentiable if, for each sequence (10.4)

u k → u weakly in E,


108

10. Weak Linking

there exists a renamed subsequence such that (10.5)

G (u k ) → G (u) weakly.

We restrict our attention to Hilbert space and prove the following. Theorem 10.2. Let E be a separable Hilbert space, and let G be a weak-to-weak continuously differentiable functional on E. Let N be a closed subspace of E, and let Q be a bounded, convex, open subset of N containing the point p. Let F be a continuous map of E onto N such that (a) F| Q = I, and (b) for each finite-dimensional subspace S of E containing p such that F S = {0}, there is a finite-dimensional subspace S0 = {0} of N containing p such that (10.6)

v ∈ Q¯ ∩ S0 ,

w ∈ S ⇒ F(v + w) ∈ S0 .

(The restriction F S = {0} is made in case p = 0.) Set A = ∂ Q, B = F −1 ( p). If (10.7)

a1 = sup G < ∞ Q¯

and (10.1) holds, then there is a sequence {u k } ⊂ E such that (10.2), (10.3) hold and c ≤ a1 . If F does not satisfy (a), but the restriction F0 of F to Q¯ is a homeomorphism of Q¯ onto the closure of a convex, open subset ⊂ N, then this can replace (a). Moreover, most, if not all, sets A, B known to link when one of the subspaces M, N is finite-dimensional will link now as well. This leads to the following. Definition 10.3. A subset A of a Banach space E links a subset B weakly if, for every G ∈ C 1 (E, R) that is weak-to-weak continuously differentiable and satisfies (10.1), there are a sequence {u k } ⊂ E and a constant c such that (10.2) and (10.3) hold. Thus we have Corollary 10.4. Under the hypotheses of Theorem 10.2, the set A = ∂ Q links the set B = F −1 ( p) weakly. Theorem 10.2 will be proved in the next section.

10.2 Another norm In this section we prove Theorem 10.2 by introducing a different norm that is equivalent to the weak topology on bounded sets.

10.2. Another norm

109

Proof. Assume that there is no such sequence. Then there is a positive number δ > 0 such that G (u) ≥ 2δ

(10.8) whenever u belongs to the set (10.9)

Eˆ = {u ∈ E : b0 − 2δ ≤ G(u) ≤ a1 + 2δ}.

Since E is separable, we can norm it with a norm |u|w satisfying (10.10)

|u|w ≤ u ,

u ∈ E,

and such that the topology induced by this norm is equivalent to the weak topology of E on bounded subsets of E. This can be done as follows. Let {ek } be an orthonormal basis for E. We then set |u|2w =

∞ |(u, ek )|2 k=1

k2

.

˜ Let We denote E equipped with this norm by E. E = {u ∈ E : G (u) = 0}. For u ∈ E , let h(u) = G (u)/ G (u) . Then, by (10.8), (10.11)

(G (u), h(u)) ≥ 2δ,

ˆ u ∈ E.

Let T = (a1 − b0 + 4δ)/δ, B R = {u ∈ E : u < R}, (10.12)

R = sup u + T, Q

ˆ Bˆ = B¯ R ∩ E. ˆ there is an E˜ neighborhood W (u) of u such that For each u ∈ B, (10.13)

(G (v), h(u)) > δ,

ˆ v ∈ W (u) ∩ B.

Otherwise, there would be a sequence {v k } ⊂ Bˆ such that (10.14)

|v k − u|w → 0

and

(G (v k ), h(u)) ≤ δ.

Since Bˆ is bounded in E, v k → u weakly in E and (10.5) implies that (10.15)

(G (v k ), h(u)) → (G (u), h(u)) ≤ δ

110

10. Weak Linking

in view of (10.14). This contradicts (10.11). Let B˜ be the set Bˆ with the inherited ˜ It is a metric space, and W (u) ∩ B˜ is an open set in this space. Thus, topology of E. ˜ u ∈ B, ˜ is an open covering of the paracompact space B˜ (cf., e.g., [112]). {W (u) ∩ B}, Consequently, there is a locally finite refinement {Wτ } of this cover. For each τ, there is an element u τ such that Wτ ⊂ W (u τ ). Let {ψτ } be a partition of unity subordinate to this covering. Each ψτ is locally Lipschitz continuous with respect to the norm |u|w and consequently with respect to the norm of E. Let ˜ (10.16) Y (u) = ψτ (u)h(u τ ), u ∈ B. Then Y (u) is locally Lipschitz continuous with respect to both norms. Moreover, (10.17) Y (u) ≤ ψτ (u) h(u τ ) ≤ 1 and (10.18)

(G (u), Y (u)) =

ψτ (u)(G (u), h(u τ )) ≥ δ,

ˆ u ∈ B.

ˆ let σ (t)u be the solution of For u ∈ Q¯ ∩ E, (10.19)

σ (t) = −Y (σ (t)),

t ≥ 0, σ (0) = u.

ˆ Moreover, it is continuous in (u, t) Note that σ (t)u will exist as long as σ (t)u is in B. with respect to both topologies. ˆ we cannot have σ (t)u ∈ Bˆ and G(σ (t)u) > b0 − δ Next we note that if u ∈ Q¯ ∩ E, for 0 ≤ t ≤ T : for by (10.18), (10.19), (10.20)

d G(σ (t)u)/dt = (G (σ ), σ ) = −(G (σ ), Y (σ )) ≤ −δ

ˆ Hence, if σ (t)u ∈ Bˆ for 0 ≤ t ≤ T , we would have as long as σ (t)u ∈ B. (10.21)

G(σ (T )u) − G(u) ≤ −δT = −(a1 − b0 + 4δ).

Thus, we would have G(σ (T )u) < b0 − 4δ. On the other hand, if σ (s)u exists for ˆ To see this, note that 0 ≤ s < T, then σ (t)u ∈ B. t Y (σ (s)u)ds. (10.22) u − σ (t)u = z t (u) := 0

By (10.17), (10.23)

z t (u) ≤ t.


σ (t)u ≤ u + t < R.

ˆ We can now conclude that for each u ∈ Q¯ ∩ E, ˆ there is a t ≥ 0 such Thus, σ (t)u ∈ B. that σ (s)u exists for 0 ≤ s ≤ t and G(σ (t)u) ≤ b0 − δ. Let (10.25)

Tu := inf{t ≥ 0 : G(σ (t)u) ≤ b0 − δ},

ˆ u ∈ Q¯ ∩ E.

10.2. Another norm

111

Then σ (t)u exists for 0 ≤ t ≤ Tu and Tu < T . Moreover, Tu is continuous in u. Define σ (t)u, 0 ≤ t ≤ Tu , σˆ (t)u = Tu ≤ t ≤ T, σ (Tu )u, ˆ For u ∈ Q¯ \ E, ˆ define σˆ (t)u = u, 0 ≤ t ≤ T . Then σˆ (t)u is for u ∈ Q¯ ∩ E. continuous in (u, t), and ¯ (10.26) G(σˆ (T )u) ≤ b0 − δ, u ∈ Q. Let ¯ 0 ≤ t ≤ T. ϕ(v, t) = F σˆ (t)v, v ∈ Q, Then ϕ is a continuous map of Q¯ × [0, T ] to N. Let ¯ t ∈ [0, T ]}. K = {(u, t) : u = σˆ (t)v, v ∈ Q,

(10.27)

Then K is a compact subset of E˜ × R. To see this, let (u k , tk ) be any sequence in K . ¯ Since Q is bounded, there is a subsequence such Then u k = σ (tk )v k , where v k ∈ Q. that v k → v 0 weakly in E and tk → t0 in [0, T ]. Since Q¯ is convex and bounded, v 0 is in Q¯ and |v k − v 0 |w → 0. Since σˆ (t) is continuous in E˜ × R, we have u k = σˆ (tk )v k σˆ (t0 )v 0 = u 0 ∈ K . ˆ Each u 0 ∈ B has a neighborhood W (u 0 ) in E˜ and a finite-dimensional subspace S(u 0 ) ˆ Since σˆ (t)u is continuous in (u, t), for such that Y (u) ⊂ S(u 0 ) for u ∈ W (u 0 ) ∩ B. ˜ each (u 0 , t0 ) ∈ K there are a neighborhood W (u 0 , t0 ) ⊂ E×R and a finite-dimensional subspace S(u 0 , t0 ) ⊂ E such that zˆ t (u) ⊂ S(u 0 , t0 ) for (u, t) ∈ W (u 0 , t0 ), where t ˆ Y (σˆ (s)u)ds, u ∈ E, (10.28) zˆ t (u) := u − σˆ (t)u = 0 ˆ 0, u ∈ E. Since K is compact, there are a finite number of points (u j , t j ) ∈ K such that K ⊂ W = ∪W (u j , t j ). Let S be a finite-dimensional subspace of E containing p and all ¯ we have zˆ t (v) ∈ S. Then, the S(u j , t j ) and such that F S = {0}. Then, for each v ∈ Q, by assumption (b) of Theorem 10.2, there is a finite dimensional-subspace S0 = {0} of N containing p such that F(v − zˆ t (v)) ∈ S0 for all v ∈ Q¯ ∩ S0 . We note that ϕ(u, t) maps Q¯ ∩ S0 × [0, T ] into S0 . For t in [0, T ], let ϕt (v) = ϕ(v, t). Then (10.29)

ϕt (v) = p,

v ∈ ∂(Q ∩ S0 ) = ∂ Q ∩ S0 , 0 ≤ t ≤ T,

for, if ϕ(v, t) = p, then σˆ (t)v ∈ F −1 ( p) = B. This implies G(σˆ (t)v) ≥ b0 by (10.1). But (10.20) and (10.1) imply that G(σˆ (t)v) < b0 for t > 0. Since p ∈ ∂ Q by hypothesis, we obtain a contradiction. Thus, (10.29) holds. Consequently, the Brouwer degree d(ϕt , Q ∩ S0 , p) is defined. Since ϕt is continuous, we have (10.30)

d(ϕT , Q ∩ S0 , p) = d(ϕ0 , Q ∩ S0 , p) = d(I, Q ∩ S0 , p) = 1.

Hence, there is a v ∈ Q such that F σˆ (T )v = p. Consequently, σˆ (T )v ∈ F −1 ( p) = B. In view of (10.1), this implies G(σˆ (T )v) ≥ b0 , contradicting (10.26). This completes the proof.

112

10. Weak Linking

10.3 Some examples The following are examples of sets that link weakly. Example 1. Let M, N be closed subspaces such that E = M ⊕ N (both can be infinitedimensional). Let B R = {u ∈ E : u < R} and take A = ∂ B R ∩ N, B = M. Then A links B weakly. To see this, take = ¯ For u ∈ E, we write B R ∩ N, Q = . u = v + w,

(10.31)

v ∈ N, w ∈ M,

and take F to be the projection Fu = Pu = v. For any finite-dimensional subspace S of E such that F S = {0}, take S0 = P S. Since F| Q = I and M = F −1 (0), we see by Theorem 10.2 that A links B weakly. Example 2. We take M, N as in Example 1. Let w0 = 0 be an element of M, and take A

=

{v ∈ N : v ≤ R} ∪ {sw0 + v : v ∈ N, s ≥ 0, sw0 + v = R},

B

=

∂ Bδ ∩ M, 0 < δ < R.

Then A links B weakly. Again, we may assume that w0 = 1. Let Q = {sw0 + v : v ∈ N, s ≥ 0, sw0 + v ≤ R}. Then A = ∂ Q in N˜ = N ⊕ {w0 }. If u is given by (10.31), we define Fu = v + w w0 . Then F is a continuous map of E onto N˜ , F| Q = I, and B = F −1 (δw0 ). For any finite-dimensional subspace S of E, take S0 = P S ⊕{w0 }. We can now apply Theorem 10.2 to conclude that A links B weakly. Example 3. Take M, N as before and let v 0 = 0 be an element of N. We write N = {v 0 } ⊕ N . We take A

= {v ∈ N : v ≤ R} ∪ {sv 0 + v : v ∈ N , s ≥ 0, sv 0 + v = R},

B

= {w ∈ M : w ≥ δ} ∪ {sv 0 + w : w ∈ M, s ≥ 0, sv 0 + w = δ},

where 0 < δ < R. Then A links B weakly. To see this, we let Q = {sv 0 + v : v ∈ N , s ≥ 0, sv 0 + v ≤ R} and reason as before. For simplicity, we assume that v 0 = 1, that E is a Hilbert space, and that the splitting E = N ⊕ {v 0 } ⊕ M is orthogonal. If (10.32)

u = v + w + sv 0 ,

v ∈ N , w ∈ M, s ∈ R,

10.3. Some examples we define F(u)

113

=

2 2 v + s + δ − δ − w v0 ,

w ≤ δ

=

v + (s + δ)v 0 ,

w > δ.

Note that F| Q = I while F −1 (δv 0 ) is precisely the set B. For any finite-dimensional subspace S of E containing (δv 0 ), take S0 = F S ⊕ {δv 0 }. Hence, we can conclude via Theorem 10.2 that A links B weakly. Example 4. This is the same as Example 3 with A replaced by A = ∂ B R ∩ N. The proof is the same with Q replaced by Q = B¯ R ∩ N. Example 5. Let M, N be as in Example 1. Take A = ∂ Bδ ∩ N, and let v 0 be any element in ∂ B1 ∩ N. Take B to be the set of all u of the form u = w + sv 0 ,

w ∈ M,

satisfying any of the following: (a) w ≤ R, s = 0; (b) w ≤ R, s = 2R0 ; (c) w = R, 0 ≤ s ≤ 2R0 , where 0 < δ < min(R, R0 ). Then A links B weakly. To see this, take N = {v 0 } ⊕ N . Then any u ∈ E can be written in the form (10.32). Define R0 w , |s − R0 | v 0 + v F(u) = R0 − max R and Q = B¯ δ ∩ N. Then F ∈ C(E, N) and F| Q = I . Moreover, A = F −1 (0). For any finite-dimensional subspace S of E such that F S = {0}, take S0 = P S ⊕ {v 0 }. Hence, A links B by Theorem 10.2. Example 6. Let M, N be as in Example 1. Let v 0 be in ∂ B1 ∩ N and write N = {v 0 } ⊕ N . Let A = ∂ Bδ ∩ N, Q = B¯ δ ∩ N, and B = {w ∈ M : w ≤ R} ∪ {w + sv 0 : w ∈ M, s ≥ 0, w + sv 0 = R}, where 0 < δ < R. Then A links B weakly. To see this, write u = w + v + sv 0 , w ∈ M, v ∈ N , s ∈ R and take F(u) = (c R − max{c w + sv 0 , |c R − s|})v 0 + v , where c = δ/(R − δ). Then F is the identity operator on Q, and F −1 (0) = B. For any finite-dimensional subspace S of E such that F S = {0}, take S0 = P S ⊕ {v 0 }. Apply Theorem 10.2.

114

10. Weak Linking

10.4 Some applications In this section we apply Theorem 10.2 to semilinear boundary-value problems. Let be a domain in Rn and let A be a self-adjoint operator in L 2 () having 0 in its resolvent set.Thus, there is an interval (a, b) in its resolvent set satisfying a < 0 < b. Let f (x, t) be a Carathéodory function on × R such that (10.33)

| f (x, t)| ≤ V (x)2 |t| + W (x)V (x),

x ∈ , t ∈ R,

and f (x, t)/t → α± (x)

(10.34)

as t → ±∞,

where V, W ∈ L 2 (), and multiplication by V (x) > 0 is a compact operator from D = D(|A|1/2) to L 2 (). Let ∞ a M= d E(λ)D, N = d E(λ)D, −∞

b

where {E(λ)} is the spectral measure of A. Then M, N are invariant subspaces for A and D = M ⊕ N. If (10.35) α(u, v) = (α+ u + − α− u − )vd x, α(u) = α(u, u),

then we assume that (10.36) (10.37)

α(v) ≥ (Av, v), (Aw, w) ≥ α(w),

v ∈ N, w ∈ M.

We also assume that the only solution of (10.38)

Au = α+ u + − α− u −

is u ≡ 0, where u ± = max{±u, 0}. We have Theorem 10.5. Under the above hypotheses, there is at least one solution of (10.39)

Au = f (x, u),

u ∈ D(A).

Proof. Let (10.40)

a(u, v) = (Au, v),

a(u) = (Au, u),

u, v ∈ D,

and (10.41)

G(u) = a(u) − 2

F(x, u) d x,

u ∈ D,

10.4. Some applications

115

where

t

F(x, t) =

(10.42)

f (x, s)ds.

0

From (10.33) it is readily verified that G is continuously differentiable on D and (10.43)

(G (u), v)/2 = a(u, v) − ( f (u), v),

u, v ∈ D,

where we write f (u) in place of f (x, u) (cf., e.g., [112]). We note that G (u) satisfies (10.5) on E = D : for, if u k → u weakly in D, then a(u k , v) → a(u, v) for each v ∈ D, and there is a renamed subsequence such that V u k → V u in L 2 (). Since V (x) > 0 in , there is a renamed subsequence such that u k → u a.e. in . Thus, f (x, u k )v → f (x, u)v

(10.44)

a.e.

Since | f (x, u k )v| ≤ |V u k V v| + W |V v| and the right-hand side converges to |V uV v|+W |V v| in L 1 (), we see that ( f (u k ), v) → ( f (u), v). Thus, G (u) satisfies (10.5). I claim that (10.45)

G(v) → −∞ as v D → ∞,

v ∈ N,

(10.46)

G(w) → ∞ as w D → ∞,

w ∈ M,

where u D = |A|1/2u . To prove (10.45), let {v k } be a sequence such that ρk = v k D → ∞, and let v˜k = v k /ρk . Then v˜k D = 1, and there is a renamed subsequence that converges weakly in D and a.e. in to a function v˜ ∈ D (again using the properties of V ). I claim that F(x, v k )d x/ρk2 → α(v) ˜ (10.47) 2

and (10.48)

˜ = ( v ˜ 2D − 1) − ( v ˜ 2D + α(v)) ˜ ≤ 0. G(v k )/ρk2 → −1 − α(v)

To see this, note that 2F(x, v k ) ρk2

=

2F(x, v k ) v k2

v˜k2 −→ α+ (v˜ + )2 + α− (v˜ − )2 a.e.

Moreover, by (10.33), ˜2 |F(x, v k )|/ρk2 ≤ C(|V v˜k |2 + W |V v˜k |/ρk ) −→ C|V v| in L 1 (). This implies (10.47) and (10.48). Now, the only way the right-hand side of (10.48) can vanish is if (10.49)

v ˜ D=1

116

10. Weak Linking

and (Av, ˜ v) ˜ = α(v). ˜

(10.50) Let

(v) = (Av, v) − α(v). Then (v) ≤ 0,

v ∈ N.

If (v) ˜ = 0, then v˜ is a maximum point for on N. This means that (v) ˜ = 0, i.e., that (Av, ˜ v) − α(v, ˜ v) = 0, v ∈ N. Thus, v˜ is a solution of (10.38). However, this contradicts (10.49). Hence, the righthand side of (10.48) must be negative. This proves (10.45). The proof of (10.46) is similar. Note that (10.46) implies b0 = inf G > −∞, M

for if G(wk ) → b0 , {wk } ⊂ M, then wk D ≤ C by (10.46). But then (10.33) implies that G is bounded on bounded sets in D. From (10.45) we see that there is an R such that (10.1) holds with A = N ∩ ∂ B R , B = M, Q = N ∩ B R , and F = P, the orthogonal projection of D onto N, p = 0. If S is a finite-dimensional subspace of D such that P S = {0}, let S0 = P S. Clearly, (10.6) is satisfied. We can now apply Theorem 10.2 to conclude that there is a sequence {u k } ⊂ D such that (10.51)

G(u k ) → c,

b0 ≤ c ≤ a1 ,

G (u k ) → 0.

Assume first that ρk = u k D → ∞, and write u˜ k = u k /ρk . Then u˜ k D = 1. Consequently, there is a renamed subsequence such that u˜ k → u˜ weakly in D and a.e. in (using the properties of V ). Since | f (x, u k )v|/ρk ≤ |V u˜ k V v| + W |V v|/ρk and the right-hand side converges in L 1 (), we see by (10.34) that (10.52) Thus,

( f (u k ), v)/ρk → α(u, ˜ v). (G (u k ), v)/2ρk → a(u, ˜ v) − α(u, ˜ v),

v ∈ D.

Since this limit vanishes by (10.51), we see that u˜ is a solution of (10.38). Hence, u˜ ≡ 0 by hypothesis. Let u˜ k = v˜k + w˜ k , where v˜k ∈ N, w˜ k ∈ M. Arguments similar to that given in the proof of (10.47) show that ( f (u k ), v˜k )/ρk → α(u, ˜ v) ˜ and ( f (u k ), w˜ k )/ρk → α(u, ˜ w), ˜


117

where u˜ = v˜ + w. ˜ Since v˜k 2D + w˜ k 2D = 1, we have for a renamed subsequence (G (u k ), v˜k )/2ρk

→ γ1 − α(u, ˜ v), ˜

(G (u k ), w˜ k )/2ρk

→ γ2 − α(u, ˜ w), ˜

where γ2 + γ1 = 1. By (10.51), both of these expressions must vanish. But this cannot happen if u˜ ≡ 0. Thus, the ρk must be bounded, and consequently there is a renamed subsequence of {u k } such that u k → u weakly in D and a.e. in . As before, this implies ( f (u k ), v) → ( f (u), v) for v ∈ D. Hence, (G (u k ), v) → (G (u), v),

v ∈ D.

By (10.51), this limit must vanish for each v in D. We conclude that G (u) = 0 and that u is a solution of (10.39). As another application, let ⊂ Rn and let A ≥ λ0 > 0 be a self-adjoint operator on L 2 () with compact resolvent. Let F(x, s, t) be a function on × R2 such that (10.53)

f (x, s, t) = ∂ F/∂t,

g(x, s, t) = ∂ F/∂s

are Carathéodory functions satisfying (10.54)

| f (x, s, t)| + |g(x, s, t)| ≤ C(|s| + |t| + 1),

s, t ∈ R,

and (10.55)

f (x, s, t)/ρ → α(x)˜s + β(x)t˜,

(10.56)

g(x, s, t)/ρ → γ (x)˜s + δ(x)t˜

as ρ 2 = s 2 + t 2 → ∞, s/ρ → s˜ , t/ρ → g. ˜ We wish to solve the system (10.57)

Av = f (x, v, w),

Aw = g(x, v, w),

v, w ∈ D(A).

We assume that the only solution of (10.58)

Av = αv + βw,

Aw = γ v + δw

is v ≡ w ≡ 0 in . If λ0 is an eigenvalue of A, we assume that corresponding eigenfunctions are = 0 a.e. on . Finally, we assume that (10.59)

β + γ − α − δ ≤≡ 2λ0 ,

β + γ + α + δ ≤≡ 2λ0 .

118

10. Weak Linking

We have Theorem 10.6. Under the above assumptions, system (10.57) has a solution. Proof. Let D = D(A1/2 ), E = D × D. Then D becomes a Hilbert space with norm given by (10.60)

u 2E = (Av, v) + (Aw, w),

We define (10.61)

u = (v, w) ∈ D.

G(u) = (Av, w) −

F(x, v, w) d x,

u ∈ D.

In view of (10.54), G ∈ C 1 (D, R) and (10.62)

(G (u), h) = (Av, h 2 ) + (Aw, h 1 ) − ( f (u), h 2 ) − (g(u), h 1 )

for u = (v, w) ∈ E, h = (h 1 , h 2 ) ∈ E, and we write f (u), g(u) in place of f (x, v, w), g(x, v, w), respectively. It follows from (10.62) that u ∈ E is a solution of (10.57) if u satisfies G (u) = 0.

(10.63)

Let N denote the subspace of E consisting of those u = (v, w) for which w = −v and let M consist of those for which w = v. Then M = N ⊥ . We note that (10.64)

G(v, −v) → −∞ as v D → ∞

and (10.65)

G(w, w) → +∞ as w D → ∞.

To prove (10.64), let {v k } ⊂ D be such that ρk = v k D → ∞, and take v˜k = v k /ρk . Since v˜k D = 1, there is a renamed subsequence such that v˜k → v˜ weakly in N, strongly in L 2 (), and a.e. in such that 2G(v k , −v k )/ρk2 = −2 v˜k 2D − 2 (10.66) F(x, v k , −v k )d x/ρk2

→ −2 − =−

[γ − δ − α + β]v˜ 2 d x

[2λ0 + β + γ − α − δ]v˜ 2 d x + 2λ0 v ˜ 2 − 2.

˜ 2 = 1. But this would mean that In view of (10.59), this is negative unless λ0 v v˜ ∈ E(λ0 ), the eigenspace of λ0 . If λ0 is not an eigenvalue, then we conclude immediately that (10.67)

lim sup G(v, −v)/ v 2D < 0

v D →∞


119

and (10.64) holds. If λ0 is an eigenvalue, then v˜ = 0 a.e. by hypothesis. But then the last integral in (10.66) must be positive by (10.59). Again, this implies (10.67) and (10.64). A similar argument implies (10.65). Now that we have (10.64) and (10.65), we know that (10.1) holds for A = N ∩ ∂ B R and B = M when R is sufficiently large. If we let F be the projection of E onto N, we can take Q = N ∩ B R and p = 0. We can conclude from Theorem 10.2 that there is a sequence {u k } ⊂ E satisfying (10.2), (10.3). I claim that (10.68)

ρk = u k E ≤ C.

To see this, assume that ρk → ∞, and let u˜ k = u k /ρk . Then there is a renamed subsequence such that u˜ k → u˜ weakly in E, strongly in L 2 (), and a.e. in . If h = (h 1 , h 2 ) ∈ E, then, by (10.62), (10.69) (G (u k ), h)/ρk = (Av˜k , h 2 )+(Aw˜ k , h 1 )−( f (u k ), h 2 )/ρk −(h(u k ), h 1 )/ρk , where u k = (v k , wk ). Taking the limit and applying (10.54) – (10.56) we see that u˜ = (v, ˜ w) ˜ is a solution of (10.58). By hypothesis, u˜ ≡ 0. On the other hand, there is a renamed subsequence such that v˜k D → a, w˜ k D → b with a 2 + b2 = 1. Moreover, (10.70)

(G (u k ), (v˜k , −v˜k ))/ρk = − 2 v˜k 2D + ( f (u k ), v˜k )/ρk − (g(u k ), v˜k )/ρk → − 2a 2 + [α + β − γ − δ]v˜ 2 d x

and (10.71)

(G (u k ), (w˜ k , w˜ k ))/ρk =2 w˜ k 2D − ( f (u k ), w˜ k )/ρk − (g(u k ), wk )/ρk →2b2 − [α + β + γ + δ]w˜ 2 d x.

By (10.2), both these limits are 0. Since a and b cannot both vanish, the same is true of v˜ and w. ˜ Hence, u˜ ≡ 0, and this contradiction shows that (10.68) does not hold. We can now use the usual procedures to show that {u k } has a renamed subsequence converging weakly in E to a function u, strongly in L 2 (), and a.e. in . Then this limit is a solution of (10.63) and consequently of (10.57). For another application, let A, B be positive, self-adjoint operators on L 2 () with compact resolvents, where ⊂ Rn . Let F(x, v, w) be a function on × R2 such that (10.72)

f (x, v, w) = ∂ F/∂v,

g(x, v, w) = ∂ F/∂w

are Carathéodory functions satisfying (10.73)

| f (x, v, w)| + |g(x, v, w)| ≤ C0 (|v| + |w| + 1),

v, w ∈ R,

120

10. Weak Linking

and (10.74)

f (x, t y, tz)/t → α+ (x)v + − α− (x)v − + β+ (x)w+ − β− (x)w− ,

(10.75)

g(x, t y, tz)/t → γ+ (x)v + − γ− (x)v − + δ+ (x)w+ − δ− (x)w−

as t → +∞, y → v, z → w, where a ± = max(±a, 0). We wish to solve the system (10.76)

Av = − f (x, v, w),

(10.77)

Bw = g(x, v, w).

Let λ0 (μ0 ) be the lowest eigenvalue of A(B). We assume that the only solution of (10.78)

−Av = α+ v + − α− v − + β+ w+ − β− w− ,

(10.79)

Bw = γ+ v + − γ− v − + δ+ w+ − δ− w−

is v = w = 0. We have Theorem 10.7. Assume that eigenfunctions of λ0 are = 0 a.e. on , α± (x) ≥≡ −λ0 ,

(10.80)

x ∈ ,

and (10.81)

2F(x, 0, t) ≤ μ0 t 2 + W (x),

x ∈ , t ∈ R,

where W (x) ∈ L 1 (). Then the system (10.76), (10.77) has a solution. Proof. Let D = D(A1/2 ) × D(B 1/2 ). Then D becomes a Hilbert space with norm given by (10.82)

u 2D = (Av, v) + (Bw, w),

We define (10.83)

u = (v, w) ∈ D.

G(u) = b(w) − a(v) − 2

F(x, v, w) d x,

u ∈ D,

where (10.84)

a(v) = (Av, v),

b(w) = (Bw, w).

Then G ∈ C 1 (D, R) and (10.85)

(G (u), h)/2 = b(w, h 2 ) − a(v, h 1 ) − ( f (u), h 1 ) − (g(u), h 2 ),

where we write f (u), g(u) in place of f (x, v, w), g(x, v, w), respectively. It is readily seen that the system (10.76), (10.77) is equivalent to (10.86)

G (u) = 0.


121

We let N be the set of those (v, 0) ∈ D and M the set of those (0, w) ∈ D. Then M, N are orthogonal closed subspaces such that D = M ⊕ N.

(10.87) If we define (10.88)

Lu = 2(−v, w),

u = (v, w) ∈ D

then L is a self-adjoint, bounded operator on D. Also, G (u) = Lu + c0 (u),

(10.89) where (10.90)

c0 (u) = −(A−1 f (u), B −1 g(u))

is compact on D. This follows from (10.73) and the fact that A and B have compact resolvents. Now, by (10.81), 2 W (x) d x, (0, w) ∈ M. (10.91) G(0, w) ≥ b(w) − μ0 w −

Thus,

inf G ≥ −

(10.92)

M

W (x)d x.

On the other hand, (10.80) implies (10.93)

sup G → −∞ as R → ∞. N∩∂ B R

To see this, let (v k , 0) be any sequence in N such that ρk2 = a(v k ) → ∞. Then F(x, v k , 0)d x/ρk2 , (10.94) G(v k , 0)/ρk2 = −a(v˜k ) − 2

where v˜k = v k /ρk . Note that a(v˜k ) = 1. Thus, there is a renamed subsequence v˜k → v˜ weakly in N, strongly in L 2 (), and a.e. in such that 2 G(v k , 0)/ρk → − 1 − {α+ (v˜ + )2 + α− (v˜ − )2 }d x

=−

{(λ0 + α+ )(v˜ + )2 + (λ0 + α− )(v˜ − )2 }d x

+ λ0 v ˜ 2 − 1. This is negative unless λ0 v ˜ 2 = 1. Since a(v) ˜ ≤ 1, this would mean that v˜ ∈ E(λ0 ), the eigenspace of λ0 . Thus, v˜ = 0 a.e. by hypothesis. But then the integral cannot vanish by (10.80). Hence, (10.95)

lim sup G(0, v)/a(v) < 0, a(v)→∞

122

10. Weak Linking

and (10.93) holds. Since N is an invariant subspace of L, we can apply Theorem 10.2 to conclude that there is a sequence {u k } ⊂ D such that (10.2) and (10.3) hold. Let u k = (v k , wk ). I claim that ρk2 = a(v k ) + b(wk ) ≤ C.

(10.96)

To see this, assume that ρk → ∞, and let u˜ k = u k /ρk . Then there is a renamed subsequence such that u˜ k → u˜ weakly in D, strongly in L 2 (), and a.e. in . If h = (h 1 , h 2 ) ∈ D, then (10.97) ˜ k , h 2 ) − 2a(v˜k , h 1 ) − 2( f (u k ), h 1 )/ρk − 2(g(u k ), h 2 )/ρk . (G (u k ), h)/ρk = 2b(w Taking the limit and applying (10.73) – (10.75), we see that u˜ = (v, ˜ w) ˜ is a solution of (10.78) (10.79). Hence, u˜ = 0 by hypothesis. On the other hand, since a(v˜k ) + b(w˜ k ) = 1, there is a renamed subsequence such that a(v˜k ) → a, ˜ b(w˜ k ) → b˜ with a˜ + b˜ = 1. Thus, by (10.74), (10.75), and (10.85), (10.98) (G (u k ), (v˜k , 0))/2ρk = −a(v˜k ) − ( f (u k ), v˜k )/ρk → −a˜ − (α+ v˜ + − α− v˜ − + β+ w˜ + − β− w˜ − )vd ˜ x

and (10.99) (G (u k ), (0, w˜ k ))/2ρk = b(w˜ k ) − (g(u k ), w˜ k )/ρk → b˜ − (γ+ v˜ + − γ− v˜ − + δ+ w˜ + − δ− w˜ − )wd ˜ x.

Thus, by (10.42), (10.100)

a˜ = −

and (10.101)

b˜ =

(α+ v˜ + − α− v˜ − + β+ w˜ + − β− w˜ − )vd ˜ x

(γ+ v˜ + − γ− v˜ − + δ+ w˜ + − δ− w˜ − )wd ˜ x.

Since one of the two numbers a, ˜ b˜ is not zero, we see that we cannot have u˜ ≡ 0. This contradiction proves (10.96). Once this is known, we can use the usual procedures to show that there is a renamed subsequence such that u k → u in D, and u satisfies (10.86). Corollary 10.8. In Theorem 10.7 we can replace (10.80), (10.81) with (10.102)

2F(x, v, 0) ≥ −λ0 v 2 − W (x),

provided eigenfunctions of μ0 are = 0 a.e. in and (10.103)

δ± (x) ≤≡ μ0 .

x ∈ , v ∈ R,


123

Proof. We interchange the roles of M and N in the proof of Theorem 10.7. Theorem 10.9. If λ0 is simple and the eigenfunctions of λ0 and μ0 are bounded and = 0 a.e. in , we can replace (10.81) in Theorem 10.7 with (10.104)

2F(x, t, 0) ≥ −λ1 t 2 ,

(10.105)

2F(x, s, t) ≤ μ0 t − λ0 s 2 ,

t ∈ R,

2

|t| + |s| ≤ δ,

where λ1 is the next eigenvalue of A and δ > 0. Moreover, system (10.76), (10.77) has a nontrivial solution. Proof. Let N be the orthogonal complement of N0 = {ϕ0 } in N. Then N = N ⊕ N0 and (10.106) G(v , 0) = −a(v ) − 2

F(x, v , 0) d x ≤ −a(v ) + λ1 v 2 ≤ 0,

v ∈ N,

and (10.107)

G(v, 0) → −∞ as a(v) → ∞

by (10.80) [cf. (10.93)]. Let M0 be the subspace of M spanned by the eigenfunctions of B corresponding to μ0 , and let M be its orthogonal complement in M. Since N0 and M0 are contained in L ∞ (), there is a positive constant ρ such that (10.108)

a(y) ≤ ρ 2 ⇒ y ∞ ≤ δ/4,

y ∈ N0 ,

(10.109)

b(h) ≤ ρ ⇒ h ∞ ≤ δ/4,

h ∈ M0 ,

2

where δ is the number given in (10.105). If (10.110)

a(y) ≤ ρ 2 ,

b(w) ≤ ρ 2 ,

|y(x)| + |w(x)| ≥ δ,

we write w = h + w , h ∈ M0 , w ∈ M , and (10.111)

δ ≤ |y(x)| + |w(x)| ≤ |y(x)| + |h(x)| + |w (x)| ≤ (δ/2) + |w (x)|.

Thus, (10.112)

|y(x)| + |h(x)| ≤ δ/2 ≤ |w (x)|

and (10.113)

|y(x)| + |w(x)| ≤ 2|w (x)|.

124

10. Weak Linking

Now, by (10.105) and (10.113), G(y, w) =b(w) − a(y) − 2

F(x, y, w)d x

{μ0 w2 − λ0 y 2 }d x

≥b(w) − a(y) − − c0

|y|+|w|δ

≥b(w) − a(y) − μ0 w 2 + λ0 y 2 − c1

|w |4 d x

2|w |>δ

≥b(w ) − μ0 w 2 − c2 b(w )2 μ0 ≥ 1− − c2 b(w ) b(w ) μ1 when a(y) ≤ ρ 2 ,

b(w) ≤ ρ 2 ,

where μ1 is the next eigenvalue of B after μ0 . If we reduce ρ accordingly, we can find a positive constant ν such that (10.114)

G(y, w) ≥ νb(w ),

a(y) ≤ ρ 2 , b(w) ≤ ρ 2 .

I claim that either (a) (15.48), (15.49) has a nontrivial solution or (b) there is an > 0 such that G(y, w) ≥ ,

(10.115)

a(y) + b(w) = ρ 2 .

To see this, suppose (10.115) did not hold. Then there would be a sequence {yk , wk } such that a(yk ) + b(wk ) = ρ 2 and G(yk , wk ) → 0. If we write wk = wk + h k , wk ∈ M , h k ∈ M0 , then (10.114) tells us that b(wk ) → 0. Thus, a(yk ) + b(h k ) → ρ 2 . Since N0 , M0 are finite-dimensional, there is a renamed subsequence such that yk → y in N0 and h k → h in M0 . By (10.108) and (10.109), y ∞ ≤ δ/4 and h ∞ ≤ δ/4. Consequently, (10.105) implies 2F(x, y, h) ≤ μ0 h 2 − λ0 y 2 .

(10.116) Since (10.117) we have (10.118)

G(y, h) = b(h) − a(y) − 2

F(x, y, h) d x = 0,

{2F(x, y, h) + λ0 y 2 − μ0 h 2 }d x = 0.


125

In view of (10.116), this implies 2F(x, y, h) ≡ μ0 h 2 − λ0 y 2 .

(10.119)

For ζ ∈ C0∞ () and t > 0 small, we have (10.120)

2[F(x, y + tζ, h) − F(x, y, h)]/t ≤ −λ0 [(y + tζ )2 − y 2 ]/t.

Taking t → 0, we have f (x, y, h)ζ ≤ −λ0 yζ.

(10.121)

Since this is true for all ζ ∈ C0∞ (), we have f (x, y, h) = −λ0 = −Ay.

(10.122) Similarly, (10.123)

2[F(x, y, h + tζ ) − F(x, y, h)]/t ≤ μ0 [(h + tζ )2 − h 2 ]/t,

and, consequently, (10.124)

g(x, y, h)ζ ≤ μ0 hζ

and (10.125)

g(x, y, h) = μ0 h = Bh.

We see from (15.89) and (10.125) that (10.76), (10.77) has a nontrivial solution. Thus, we may assume that (10.115) holds. But then we can combine it with (10.106) and (10.107) to conclude from Theorem 10.2 that there is a sequence {u k } ⊂ D such that (10.3) holds with c ≥ . Arguing as in the proof of Theorem 10.7, we see that there is a u ∈ D such that G(u) = c, G (u) = 0. Since c = 0 and G(0) = 0, we see that u = 0, and the proof is complete. Theorem 10.10. Assume that λ0 is simple and the eigenfunctions corresponding to λ0 and μ0 are bounded and = 0 a.e. in . Assume (10.80), (10.105) and (10.126)

2F(x, 0, t) ≤ μ(x)t 2 ,

x ∈ , t ∈ R,

where (10.127)

μ(x) ≤≡ μ0 ,

x ∈ .

Then (10.76), (10.77) has a nontrivial solution. Proof. As in the proof of Theorem 10.7, (10.80) implies (10.93). As in the proof of Theorem 10.9, (10.105) implies (10.115) for y ∈ N0 , w ∈ M unless (10.76), (10.77) has a nontrivial solution. Thus, we may assume that (10.115) holds. Next we note that there is a ν > 0 such that (10.128)

G(0, w) ≥ νb(w),

w ∈ M.

126

10. Weak Linking

Assuming this for the moment, we see that inf G ≥ ε1 > 0,

(10.129)

B

where (10.130)

B = {w ∈ M : b(w) ≥ ρ 2 } ∪ {u = (sϕ0 , w) : s ≥ 0, w ∈ M},

and 1 = min{, νρ 2 }. By (10.93), there is an R > ρ such that sup G ≤ 0,

(10.131)

A

where A = N ∩ ∂ B R . We can now apply Theorem 10.2 to conclude that there is a sequence in D satisfying (10.3). As in the proof of Theorem 10.7, this leads to a solution of (10.86) satisfying G(u) = c ≥ 1 . Hence, u = 0, and we have a nontrivial solution of (10.76), (10.77). It therefore remains only to prove (10.128). Clearly, ν ≥ 0. If ν = 0, then there is a sequence {wk } ⊂ M such that G(0, wk ) → 0,

(10.132)

b(wk ) = 1.

Thus, there is a renamed subsequence such that wk → w weakly in M, strongly in N, and a.e. in . Consequently, (10.133) G(0, wk ) ≥ 1 − μ(x)wk2 d x ≥ [μ0 − μ(x)]wk2 d x → 0

and (10.134)

1=

μ(x)w2 d x ≤ μ0 w 2 ≤ b(x) ≤ 1,

which means that we have equality throughout. It follows that we must have w ∈ E(μ0 ), the eigenspace of μ0 . Since w ≡ 0, we have w = 0 a.e. But (10.135) [μ0 − μ(x)]w2 d x = 0

implies that the integrand vanishes identically on , and consequently μ(x) ≡ μ0 , violating (10.127). This establishes (10.128) and completes the proof of the theorem.

10.5 Notes and remarks The concept of weak linking was begun in [19] and [78]. Further work in this direction was carried out in [106]. The main thrust was to consider the case when G can be written in the form (10.136)

G(u) =

1 (Lu, u) + b(u) 2


127

where L is a self-adjoint operator and b (u) is weakly continuous and E is a Hilbert space. With this method, very few sets have been found to link. The results of [19], [78], and [122] do not require E to be separable. However, they require G to be of the form (10.136) with b (u) compact. In [78] the compactness of b (u) is replaced by the assumption that it be weak-to-weak continuous, and b(u) itself is required to be bounded from below and be weakly lower semicontinuous. A theorem is given in [78] that requires only (10.5) but also requires G to be τ -upper semi-continuous (the τ -topology is specially constructed to accommodate the splitting of E into subspaces). In all of the results mentioned only two examples of linking sets A, B are given. The presentation given here is from [117], [118], [140]. It has several advantages. It does not require G to be of the form (10.136) [which indeed satisfies (10.5)]. It does not require hypotheses on each of an exhausting sequence of finite-dimensional subspaces. Moreover, all sets A, B known to link when one of the subspaces M, N is finite-dimensional will link now as well.

Chapter 11

Fuˇc´ık Spectrum: Resonance 11.1 Introduction Let be a bounded domain in Rn and let A ≥ λ0 > 0 be a self-adjoint operator on L 2 () with compact resolvent and eigenvalues (11.1)

0 < λ0 < λ1 < · · · < λk < · · · .

If f (x, t) is a Carathéodory function on × R, then the semilinear problem (11.2)

Au = f (x, u),

u ∈ D(A)

is said to have asymptotic resonance at infinity if (11.3)

f (x, t)/t → λk as |t| → ∞,

where λk is one of the eigenvalues of A. Interest in resonant problems began with the pioneering work of Landesman and Lazer [79] and has continued until the present because such problems are more difficult to solve than nonresonant problems. The reason is that for |u(x)| large, (11.2) approximates the eigenvalue problem Au = λk u

(11.4)

with its inherent instabilities. Even if (11.3) does not occur, but (11.5)

f (x, t)/t

→ a a.e. as t → −∞ → b a.e. as t → +∞,

one encounters the same difficulties if (11.6)

Au = bu + − au − ,

u ± = max{±u, 0}

has a nontrivial solution. In fact, the difficulties are compounded because there is no eigenspace with which to work in this case. We call the set of those (a, b) ∈ R2 for M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_11, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009

130

11. Fuˇc´ık Spectrum: Resonance

which (11.6) has nontrivial solutions the Fuˇc´ık spectrum of A. For each , it was shown in [111, 122] that in the square [λ−1 , λ+1 ]2 there are decreasing curves C,1 , C,2 (which may coincide) passing through the point (λ , λ ) such that all points above or below both curves in the square are not in . We denote the lower curve by C,1 and the upper curve by C,2 . Further discussions concerning the Fuˇc´ık spectrum can be found in Chapter 14. In the present chapter we shall study resonance problems with respect to the Fuˇc´ık spectrum. We shall allow (a, b) to be on either of the curves C, j in Q = (λ−1 , λ+1 )2 . We do not presently consider the case when (a, b) is a point between the curves (when there are points in in that region). Such points will be discussed in Chapter 14. We have Theorem 11.1. Let f (x, t) be a Carathéodory function such that | f (x, t)| ≤ C(|t| + 1),

(11.7)

x ∈ , t ∈ R,

and (11.6) has a nontrivial solution with (a, b) ∈ C,1 . Assume that λ−1 t 2 ≤ 2F(x, t) + W1 (x),

(11.8)

x ∈ , t ∈ R,

and H (x, t) ≤ W0 (x),

(11.9) where Wi (x) ∈ L 1 (),

x ∈ , t ∈ R,

(11.10)

t

F(x, t) :=

f (x, s)ds,

0

and H (x, t) := 2F(x, t) − t f (x, t).

(11.11) Assume further that

H (x, t) → H± (x) a.e. as t → ±∞

(11.12) and

(11.13) u>0

H+ (x)d x +

u0

H+ (x)d x +

x ∈ , t ∈ R,

u B1 u=0

for all nontrivial solutions u of (11.6). Then (11.2) has a solution.

11.2. The curves

131

Note that (11.13) is automatically satisfied if H (x, t) → −∞ a.e. as |t| → ∞

(11.17)

and (11.16) is automatically satisfied if (11.18)

H (x, t) → +∞ a.e. as |t| → ∞.

11.2 The curves In this section we shall describe the curves C,1 and C,2 and discuss some of their properties. For each fixed positive integer , we let N denote the subspace of E spanned by the eigenfunctions corresponding to λ0 , . . . , λ , and we let M = N⊥ ∩ D, where D = D(A1 /2). Then D = M ⊕ N . For a, b ∈ R, we define (11.19)

I (u, a, b) = (Au, u) − a u − 2 − b u + 2 ,

(11.20)

M (a, b) = inf sup I (v + w, a, b),

(11.21)

m (a, b) = sup inf I (v + w, a, b),

w∈M w =1

v∈N v =1

u ∈ D,

v∈N

w∈M

(11.22)

ν (a) = sup{b : M (a, b) ≥ 0},

(11.23)

μ (a) = inf{b : m (a, b) ≤ 0}.

We have Lemma 11.3. [111, 122] In the square Q the functions μ (a), ν−1 (a) have the following properties: (a) μ (λ ) = ν−1 (λ ) = λ ; (b) μ (a), ν−1 (a) are continuous and strictly decreasing; (c) (a, μ (a)) and (a, ν−1 (a)) are in ; (d) if (a, b) ∈ Q and b > μ (a), then (a, b) ∈ / ; (e) if (a, b) ∈ Q and b < ν−1 (a), then (a, b) ∈ / . It follows from Lemma 11.3 that ν−1 (a) ≤ μ (a) in Q . If ν−1 (a) < μ (a), this lemma does not cover those points in Q for which (11.24)

ν−1 (a) < b < μ (a).

We let C,1 be the lower curve b = ν−1 (a) and we take C,2 to be the upper curve b = μ (a). We shall need

132


Lemma 11.4. [111, 122] If (a, b) ∈ Q , w ∈ M−1 , v 1 , v 2 ∈ N−1 , and (11.25)

I (w + v j , a, b) ⊥ N−1 ,

j = 1, 2,

then v 1 = v 2 . Lemma 11.5. [111, 122] I f (a, b) ∈ Q , v ∈ N , w1 , w2 ∈ M , and (11.26)

I (v + w j , a, b) ⊥ M ,

j = 1, 2,

then w1 = w2 . Lemma 11.6. [111, 122] If (a, b) ∈ Q , then, for each w ∈ M−1 , there is a v 0 ∈ N−1 such that (11.27)

I (w + v 0 , a, b) = sup I (w + v, a, b). v∈N−1

Lemma 11.7. [111, 122] If (a, b) ∈ Q , then, for each v ∈ N , there is a w0 ∈ M such that (11.28)

I (v + w0 , a, b) = inf I (v + w, a, b). w∈M

Lemma 11.8. If (a, b) ∈ C,1 , then I (u, a, b) ≥ 0,

(11.29)

u ∈ S,1 ,

where (11.30)

S,1 = {u ∈ E : I (u, a, b) ⊥ N−1 }.

Proof. If (a, b) ∈ C,1 , then b = ν−1 (a). Thus, by (11.22), M−1 (a, b) = 0.

(11.31) This implies by (11.20) that (11.32)

sup I (v + w, a, b) ≥ 0,

v∈N−1

w ∈ M−1 .

Let u 0 be any element in S,1 . Then u 0 = w0 + v 0 , w0 ∈ M−1 , v 0 ∈ N−1 , and (11.33)

I (w0 + v 0 , a, b) ⊥ N−1

by (11.30). In view of Lemma 11.6, there is a v 1 ∈ N−1 such that (11.34)

I (w0 + v 1 , a, b) = sup I (w0 + v, a, b). v∈N−1

From this and (11.32), it follows that (11.35)

I (w0 + v 1 , a, b) ⊥ N−1

11.2. The curves

133

and I (w0 + v 1 , a, b) ≥ 0.

(11.36)

But (11.33) and (11.35) imply via Lemma 11.4 that v 1 = v 0 . Thus, I (u 0 , a, b) ≥ 0, and the lemma is proved. Lemma 11.9. If (a, b) ∈ C,2 , then I (u, a, b) ≤ 0,

(11.37)

u ∈ S,2 ,

where S,2 = {u ∈ E : I (u, a, b) ⊥ M }.

(11.38)

Proof. If (a, b) ∈ C,2 , then b = μ (a). Consequently, by (11.23), m (a, b) = 0.

(11.39) By (11.21), this implies (11.40)

inf I (v + w, a, b) ≤ 0,

w∈M

v ∈ N .

Let u 0 be any element in S,2 and write u 0 = v 0 + w0 with v 0 ∈ N , w ∈ M . Then (11.41)

I (v 0 + w0 , a, b) ⊥ M .

By Lemma 11.7, there is a w1 ∈ M such that (11.42)

I (v 0 + w1 , a, b) = inf I (v 0 + w, a, b). w∈M


I (v 0 + w1 , a, b) ⊥ M

and (11.44)

I (v 0 + w1 , a, b) ≤ 0.

If we now apply Lemma 11.5, we see that w1 = w0 . Thus, I (u 0 , a, b) ≤ 0, and the proof is complete. Lemma 11.10. N−1 ∩ ∂ B R links S,1 [hm] for each R > 0.

134


Proof. We suppress the subscript − 1. Let P be the orthogonal projection of D onto N, and define (11.45)

F(u) = P I (u),

u ∈ D.

I claim that the restriction F0 of F to N is a homeomorphism of N onto N. If this is so, then the image of N ∩ B¯ R under F0 is the closure of a bounded, open set. Moreover, this open set contains the point 0 since F0 (0) = 0 and 0 is an interior point of N ∩ B R . Proposition 3.10 can then be used to show that N ∩ ∂ B R links F −1 (0) = S,1 . Clearly, F0 is a continuous map of N into itself. It is surjective. To see this, let h be any element of N and take (11.46)

G 0 (v) = I (v, a, b) − (h, v).

I claim that (11.47)

G 0 (v) → −∞ as v → ∞,

v ∈ N.

Assuming this for the moment, we see that G 0 (v) has a maximum on N. At a point of maximum we have G 0 (v) ⊥ N, producing a solution of P I (v, a, b) = h.

(11.48) Moreover, if (11.49)

P I (v 0 , a, b) = h 0 ,

P I (v 1 , a, b) = h 1 ,

then (11.50)

(h, v) = (Av, v) + a(v 1− − v 0− , v) − b(v 1+ − v 0+ , v)

where h = h 1 − h 0 , v = v 1 − v 0 . Thus, (11.51)

(b − λ−1 )(v 1+ − v 0+ , v) − (a − λ−1 )(v 1− − v 0− , v), + [λ−1 v 2 − (Av, v)] = −(h, v).

Since a, b > λ−1 , the left-hand side of (11.51) is positive for all v ≡ 0. Consequently, there is a δ > 0 such that δ v 2 ≤ h v

(11.52)

showing not only that F0 is injective but that F0−1 is continuous. To prove (11.47), let {v k } ⊂ N be a sequence such that ρk2 = (Av k , v k ) → ∞, and let v˜k = v k /ρk . Then v˜k 2D = (Av˜k , v˜k ) = 1 and there is a subsequence such that v˜k → v˜ in D. We have (11.53)

G 0 (v k )/ρk2

=

I (v˜k , a, b) − (h, v˜k )/ρk

→

I (v, ˜ a, b)

=

[ v ˜ 2D − λ−1 v ˜ 2] +(λ−1 − a) v˜ − 2 +(λ−1 − b) v˜ + 2 .

11.3. Existence

135

Since v ˜ D = 1, the right-hand side of (11.53) is negative. This gives (11.47) and completes the proof of the lemma. The following can be proved in the same way. Lemma 11.11. M ∩ ∂ B R links S,2 [hm] for each R > 0.

11.3 Existence In this section we give the proofs of Theorems 11.1 and 11.2. Let (11.54)

p(x, t) = f (x, t) + at − − bt + ,

t ± = max{±t, 0},

and

t

P(x, t) =

(11.55)

p(x, s) ds. 0

Under hypothesis (11.7), it is readily checked [122] that the functional F(x, u) d x = I (u, a, b) − 2 P(x, u) d x (11.56) G(u) = u 2D − 2

is in C 1 on D with (11.57)

(G (u), v) = 2(u, v) D − 2( f (·, u), v) = (I (u, a, b), v) − 2( p(·, u), v)

and (11.58)

(I (u, a, b), v)/2 = (u, v) D + a(u − , v) − b(u + , v).

From (11.5) and (11.54) we see that (11.59)

2P(x, t)/t 2 → 0 as |t| → ∞.

p(x, t)/t → 0,

This and the fact that (11.60)

∂(F/t 2 )/∂t = ∂(P/t 2 )/∂t = −t −3 H

imply that (11.61)

P(x, t)

=

t

2 t

=

−t 2

∞

s −3 H (x, s)ds, t

−∞

s −3 H (x, s)ds,

t >0 t < 0.

136


Proof of Theorem 11.1. By (11.9) and (11.61), we have 2P(x, t) ≤ W0 (x),

(11.62)

x ∈ , t ∈ R.

In view of Lemma 11.8, this implies (11.63) G(u) ≥ −2 P(x, u)d x ≥ −B0 = − W0 (x)d x,

u ∈ S,1 .

I claim that G(v) → −∞ as v → ∞,

(11.64)

v ∈ N−1 .

Assume this for the moment. Then there is an R > 0 sufficiently large that (3.4) holds with A = N−1 ∩ ∂ B R and B = S,1 . Moreover, for this choice of A and B, A links B [hm] by Lemma 11.10. We can now apply Theorem 3.4 to conclude that there is a sequence {u k } ⊂ D such that (1 + u k D )G (u k ) → 0,

G(u k ) → c,

(11.65)

and from (11.56) we readily estimate c by −B0 ≤ c ≤ B1 .

(11.66) From (11.65) we find

(11.67)

I (u k , a, b) − 2

P(x, u k )d x →c,

I (u k , a, b) − ( p(·, u k ), u k ) →0,

(11.68) (11.69)

(I (u k , a, b), v) − 2( p(·, u k ), v) →0,

v ∈ D.

Assume that (11.70)

ρk = u k D → ∞,

and let u˜ k = u k /ρk . Then u˜ k D = 1. Thus, there is a renamed subsequence such that u˜ k → u˜ weakly in D, strongly in L 2 (), and a.e. in . As a consequence, (11.59) and (11.68) imply (11.71)

1 = a u˜ − 2 + b u˜ + 2 .

This shows that u˜ ≡ 0. Moreover, (11.69) implies (11.72)

I (u, ˜ a, b) = 0.

Thus, u˜ is a solution of (11.5) and satisfies (11.73)

I (u, ˜ a, b) = 0.

11.3. Existence

137

If we combine this with (11.71), we see that u ˜ D = 1. Consequently, u˜ k → u˜ strongly in D. Combining (11.67) and (11.68), we obtain H (x, u k ) d x → −c. (11.74)

˜ > 0}, 0 = {x ∈ : u(x) ˜ = 0}. Then u k (x) → ±∞ for Let ± = {x ∈ : ±u(x) x ∈ ± . By (11.9) and (11.13), −c = lim H (x, u k ) d x

≤

+

H+ (x) d x

+ +

H− (x) d x

−

0

W0 (x) d x

< − B1 , contradicting (11.66). Thus, (11.70) does not hold. Once we know that the sequence {u k } is bounded in D, we can use standard techniques [122] to obtain a solution of G(u) = c,

(11.75)

−B0 ≤ c ≤ B1 ,

G (u) = 0,

which is a solution of (11.2). It thus remains to prove (11.64). Let {v k } ⊂ N−1 be a sequence such that ρk = v k D → ∞. Let v˜k = v k /ρk . Then v˜k D = 1 and there is a renamed subsequence such that v˜k → v˜ in D and a.e. in . Since |F(x, t)| ≤ C(t 2 + |t|)

(11.76) by (11.7), we have

|F(x, v k )/ρk2 | ≤ C(v˜k2 + |v˜k |/ρk )

(11.77) and, consequently, (11.78)

2

F(x, v k )d x/ρk2 → a v˜ − 2 + b v˜ + 2

by (11.6). This means that G(v k )/ρk2

=

v˜k 2D − 2

F(x, v k )d x/ρk2

→ v ˜ 2D − a v˜ − 2 − b v˜ + 2 ≤

(λ−1 − a) v˜ − 2 + (λ−1 − b) v˜ + 2 .

Since both a and b are greater than λ−1 , this will always be positive since v˜ ≡ 0. Thus, (11.64) holds, and the proof of Theorem 11.1 is complete.

138


Proof of Theorem 11.2. By (11.15) and (11.61) we have −2P(x, t) ≤ W0 (x),

(11.79)

x ∈ , t ∈ R.

Consequently, G(u) ≤ −2

(11.80)

P(x, u) d x ≤ B0 ,

u ∈ S,2 .

Moreover, I claim that G(w) → ∞ as w → ∞,

(11.81)

w ∈ M .

Assuming this, we note that there is an R > 0 sufficiently large that sup G ≤ inf G,

(11.82)

B

A

where A = M ∩ ∂ B R and B = S,2 . This is not quite (3.4). To correct the situation, we let G 1 = −G. Then (11.82) becomes sup G 1 ≤ inf G 1 .

(11.83)

A

B

We know from Lemma 11.11 that A links B [hm]. Consequently, we may apply Theorem 3.4 to conclude that there is a sequence {u k } ⊂ D such that (11.65) holds with G replaced by G 1 . Now c satisfies the estimates inf G 1 ≤ c ≤ sup G 1

(11.84)

B

M∩B R

or inf G ≤ −c ≤ sup G.

(11.85)

M∩B R

B

By (11.14), (11.56), and (11.80), we see that c satisfies (11.66) as well. Thus, for the given sequence, (11.65), (11.67)–(11.69) hold with c replaced by −c, while c satisfies (11.66). If we assume that (11.70) holds, we can reason as in the proof of Theorem 11.1 that there is a renamed subsequence of {u˜ k } converging in D to a function u˜ ∈ D and a.e. in . Moreover, the function u˜ satisfies (11.72) and (11.73). Also, (11.74) holds with c replaced by −c. But then (11.15) and (11.16) imply (11.86) H (x, u k )d x ≥ H+ (x)d x + H−(x)d x − W0 (x) d x > B1 . c = lim

+

−

0

This contradiction shows that the ρk are bounded, and we can now employ standard techniques to obtain a solution of (11.87)

G(u) = c1 ,

This produces the desired result.

−B1 ≤ c1 ≤ B0 , G (u) = 0.


139

It therefore remains only to prove (11.81). Let {wk } ⊂ M be a sequence such that ρk = wk D → ∞. Let w˜ k = wk /ρk . Then we have w˜ k D = 1, and there is a renamed subsequence such that w˜ k → w˜ weakly in D, strongly in L 2 (), and a.e. in . Thus, (11.88) 2 F(x, wk )d x/ρk2 → a w˜ − 2 + b w˜ + 2 .

Consequently, G(wk )/ρk2 → 1 − a w˜ − 2 − b w˜ + 2 ≥ 1 − w ˜ 2D + (λ+1 − a) w˜ − 2 + (λ+1 − b) w˜ + 2 , which is positive since both a and b are less than λ+1 . This completes the proof of the theorem.

11.4 Notes and remarks The presentation given here is from [124]. Further work was done in [102].

Chapter 12

Rotationally Invariant Solutions 12.1 Introduction In this chapter (and the next) we study periodic solutions of the Dirichlet problem for the semilinear wave equation. In this chapter we study radially symmetric solutions for the problem u := u t t − u = f (t, x, u),

(12.1) (12.2) (12.3)

u(t, x) = 0,

t ∈ R, x ∈ B R ,

t ∈ R, x ∈ ∂ B R ,

u(t + T, x) = u(t, x), t ∈ R,

x ∈ BR,

where (12.4)

B R = {x ∈ Rn : |x| < R}.

In this case we have f (t, x, u) = f (t, |x|, u),

x ∈ BR.

Our basic assumption is that the ratio R/T is rational. Thus, we can write (12.5)

8R/T = a/b,

where a, b are relatively prime positive integers. We show that (12.6)

n ≡ 3

(mod(4, a))

implies that the linear problem corresponding to (12.1)–(12.3) has no essential spectrum. If (12.7)

n≡3

(mod(4, a)),

then the essential spectrum of the linear operator consists of precisely one point: (12.8)

λ0 = −(n − 3)(n − 1)/4R 2 .


142

12. Rotationally Invariant Solutions

This shows that the spectrum has at most one limit point. We can then consider the nonlinear case (12.9)

f (t, r, s) = μs + p(t, r, s),

where μ is a point in the resolvent set, r = |x|, and (12.10)

| p(t, r, s)| ≤ C(|s|θ + 1),

s ∈ R,

for some number θ < 1. Our main theorem is Theorem 12.1. If (12.6) holds, then (12.1)–(12.3) has a weak rationally invariant solution. If (12.7) holds and λ0 < μ, assume, in addition, that p(t, r, s) is nondecreasing in s. If μ < λ0 , assume that p(t, r, s) is nonincreasing in s. Then (12.1)–(12.3) has a weak rotationally invariant solution. Our proof of this theorem will make use of Theorem 10.2. For the definition of essential spectrum, cf., e.g., [126] and [106].

12.2 The spectrum of the linear operator In proving Theorem 12.1, we shall need to calculate the spectrum of the linear operator applied to periodic rotationally symmetric functions. Specifically, we shall need Theorem 12.2. Let L 0 be the operator (12.11)

L 0 u = u t t − u rr − r −1 (n − 1)u r

¯ satisfying applied to functions u(t, r ) in C ∞ () u t (T, r ) = u t (0, r ),

(12.12)

u(T, r ) = u(0, r ),

(12.13)

u(t, R) = u R (t, 0) = 0,

0 ≤ r ≤ R,

t ∈ R,

where = [0, T ] × [0, R]. Then L 0 is symmetric on L 2 (, ρ), where ρ = r n−1 . Assume that 8R/T = a/b, where a, b are relatively prime integers [i.e., (a, b) = 1]. Then L 0 has a self-adjoint extension L having no essential spectrum other than the point λ0 = −(n − 3)(n − 1)/4R 2 . If n ≡ 3 (mod(4, a)), then L has no essential spectrum. If n ≡ 3 (mod(4, a)), then the essential spectrum of L is precisely the point λ0 . Proof. Let ν = (n − 2)/2 and let γ be a positive root of Jν (x) = 0, where Jν is the Bessel function of the first kind. Set (12.14)

ϕ(r ) = Jν (γ r/R)/r ν .

Then (12.15)

ϕ + (n − 1)ϕ /r = (x 2 Jν + x Jν − ν 2 Jν )/r ν+2 = −γ 2 Jν /R 2 .

12.2. The spectrum of the linear operator

143

If ψ(t, r ) = ϕ(r )e2πikt /T ,

(12.16) then

L 0 ψ = [(γ /R)2 − (2πk/T )2 ]ψ.

(12.17)

Let γ j be the j th positive root of Jν (x) = 0, and set ψ j k (t, r ) = r −ν Jν (γ j r/R)e2πikt /T .

(12.18)

Then ψ j k (t, r ) is an eigenfunction of L 0 with eigenvalue λ j k = (γ j /R)2 − (2πk/T )2 .

(12.19)

It is easily checked that the functions ψ j k , when normalized, form a complete orthonormal sequence in L 2 (, ρ). We shall show that the corresponding eigenvalues (12.19) are not dense in R. It will then follow that L 0 has a self-adjoint extension L with spectrum equal to the closure of the set {λ j k } (cf., e.g., [126]). Now (12.20)

γ j = β j − (μ − 1)/8β j + O(β −3 j ) as β j → ∞,

where βj = π

(12.21)

1 1 , j+ ν− 2 4

μ = 4ν 2

(cf., e.g., [157]). Thus, λ j k R2

=

[β j − τk − (μ − 1)/8β j + O(β −3 j )] ·[β j + τk − (μ − 1)/8β j + O(β −3 j )]

=

β 2j − τk2 − (μ − 1)/4 + O(β −2 j )

where τk = 2kπ R/T . (We may assume k ≥ 0.) Now 1 1 (12.22) β j − τk = π j + ν − − ak/4b = π[(4 j + n − 3)b − ak]/4b. 2 4 Since the expression in the brackets is an integer, we see that either β j = τk or |β j − τk | ≥ π/4b.

(12.23) Thus, (12.24)

lim λ j k = −(μ − 1)/4R 2 = λ0

j,|k|→∞ β j =τk

144


and lim |λ j k | = ∞.

(12.25)

j,|k|→∞ β j =τ k

If n − 3 is not a multiple of (4, a), then (12.26)

β j − τk = π((4 j + n − 3) − ak/b)/4

can never vanish. To see this, note that if (b, k) = b, then ak/b is not an integer. Hence, β j = τk . If b = (b, k), then (12.27)

(n − 3) = ak − 4 j

∀ j, k = k/b.

Thus, in this case we always have β j = τk and |λ j k | → ∞ as j, k → ∞. On the other hand, if n ≡ 3(mod(4, a)), then there are an infinite number of positive integers j, k such that n − 3 = ak − 4 j.

(12.28)

Hence, the point λ0 is a limit point of eigenvalues. Consequently, it is in σe (L). This completes the proof.

12.3 The nonlinear case We now turn to the problem of solving (12.1)–(12.3). If one is searching for rotationally invariant solutions, the problem reduces to (12.29)

Lu = f (t, r, u),

u ∈ D(L)

where L is the self-adjoint extension of the operator L 0 given in Theorem 12.2. Under the hypotheses of that theorem, the spectrum of L is discrete. We assume that (12.30)

f (t, r, s) = μs + p(t, r, s),

where μ is a point in the resolvent set of L and p(t, r, s) is a Carathéodory function on × R such that (12.31)

| p(t, r, s)| ≤ C(|s|θ + 1),

s ∈ R,

for some number θ < 1. We have Theorem 12.3. Let f (t, r, s) satisfy (12.30) and (12.31), and assume the hypotheses of Theorem 12.2. If (12.32)

n ≡ 3 (mod(4, a)),

make no further assumptions. If (12.33)

n≡3

(mod(4, a))

and λ0 < μ, assume that p(t, r, s) is nondecreasing in s. If (12.33) holds and μ < λ0 , assume that p(t, r, s) is nonincreasing in s. Then (12.29) has at least one weak solution.

12.3. The nonlinear case

145

Proof. Since μ is in the resolvent set of L, there is a δ > 0 such that |λ j k − μ| ≥ δ

(12.34)

∀ j, k,

where the λ j k are given by (12.19). Each u ∈ L 2 (, ρ) can be expanded in the form (12.35) u= α j k ψ j k (t, r ), where the ψ j k are given by (12.18). Let N0 be the subspace of those u ∈ L 2 (, ρ) for which α j k = 0 if β j = τk (cf. the proof of Theorem 12.2). For u ∈ N0 , α j k ψ j k (t, r ), (12.36) u= [0]

where summation is taken over those j, k for which β j = τk . Let E be the subspace of L 2 (, ρ) consisting of those u for which (12.37) u 2E = |λ j k − μ||α j k |2 is finite. With this norm, E becomes a separable Hilbert space. Note that E ⊂ D(|L|1/2 ) and the embedding of E N0 into L 2 (, ρ) is compact [we use (12.25) for this purpose]. Let P(t, r, u)ρdtdr, u ∈ E, (12.38) G(u) = ([L − μ]u, u) − 2

where

s

P(t, r, s) =

(12.39)

p(t, r, σ )dσ,

0

and then scalar product is that of L 2 (, ρ). C 1 -functional on E with (12.40)

One checks readily that G is a

(G (u), v)/2 = ([L − μ]u, v) − ( p(u), v),

u, v ∈ E,

where we write p(u) in place of p(t, r, u). This shows that u is a weak solution of (12.29) iff G (u) = 0. Let N be the subspace of E spanned by the ψ j k corresponding to those λ j k < μ and let M denote the subspace of E spanned by the rest. Thus, M = N ⊥ in E. Assume first that N ∩ N0 = {0}. Then (12.41) u 2E = (μ − λ j k )|α j k |2 , u ∈ N. Thus,

G(v)

= − v 2E − 2

P(t, r, v)ρdtdr

≤

− v 2E

≤

− v 2E + C ( v 1+θ + v ) → −∞,

+C

(|v|1+θ + |v|)ρdtdr v E → ∞, v ∈ N.

146


If w ∈ M, (12.42)

G(w) ≥ δ w 2 − C( w 1+θ + w ) ≥ −K ,

w ∈ M.

We can now make use of Theorem 10.2. If Q is a large ball in N, then sup G ≤ inf G.

(12.43)

M

∂Q

By Example 1 of Section 10.3, ∂ Q links M weakly. Moreover, if {u k } ⊂ E is a sequence such that v k = Pu k → v = Pu weakly on N and wk = (I − P)u k → w = (I − P)u strongly in M, where P is the projection of E onto N, then {u k } has a renamed subsequence that converges strongly in L 2 (, ρ). The reason is that {v k } has such a subsequence because the embedding of E N0 in L 2 (, ρ) is compact. Thus, G (u n ) → G (u) weakly in E. Hence, all of the hypotheses of Theorem 10.2 are satisfied, and we can conclude that there is a sequence {u k } satisfying (10.3). Write u k = v k + wk + yk , where v k ∈ N, wk ∈ M N0 , yk ∈ N0 . Then (G (u k ), v k )/2 = ([L − μ]u k , v k ) − ( p(u k ), v k )

(12.44) and, consequently, (12.45)

v k 2E ≤ G (u k ) v k E /2 + C v k ( u k θ + 1)

in view of (12.31) and (12.37). Similarly, (12.46)

wk 2E ≤ G (u k ) wk E /2 + C wk ( u k θ + 1).

If N0 = {0}, then it follows from (12.45) and (12.46) that u k E is bounded and consequently there is a renamed subsequence that converges weakly in E and strongly in L 2 (, ρ) to a function u. Thus, G (u k ) → G (u) weakly. But G (u k ) → 0. Consequently G (u) = 0 and the proof for this case is complete. If N0 = {0}, we note that (12.47)

yk 2E ≤ G (u k ) yk ) E /2 + C yk ||( u k θ + 1)

as well. Again, this together with (12.45) and (12.46) implies that u k || E is bounded and has a renamed subsequence that converges weakly in E and such that u k = v k +wk converges strongly in L 2 (, ρ). Now (12.48)

(G (u k ), yk − y)/2 =

([L − μ](yk − y), yk − y) −( p(u k ) − p(u k + y), yk − y) +( p(u k + y) − p(u), yk − y) +([L − μ]y, yk − y),

where yk → y weakly in E and L 2 (, ρ) and u k → u weakly in E and strongly in L 2 (, ρ). By hypothesis (12.49)

( p(u k ) − p(u k + y), yk − y) ≥ 0


147

if μ > λ0 . Moreover, ( p(u k

(G (u k ), yk − y) →

0,

+ y) − p(u), yk − y) →

0,

and (12.50)

([L − μ]y, yk − y) → 0.

Hence (12.51)

yk − y 2E ≤ o(1),

k → ∞.

This shows that yk → y in E, and the proof proceeds as before. If λ0 > μ, we apply Theorem 12.1 to −G(u) and come to the same conclusion. In this case the inequality in (12.49) is reversed. This completes the proof.

12.4 Notes and remarks Many authors have studied the one-dimensional periodic-Dirichlet problem for the semilinear wave equation u t t − u x x = p(t, x, u), u(t, x) = 0,

t ∈ R, x ∈ (0, π),

t ∈ R, x = 0, x = π,

u(t + 2π, x) = u(t, x),

t ∈ R, x ∈ (0, π)

A basic problem in this one dimensional case is that the null space N of the linear part u = u t t − u x x is infinite dimensional. On the other hand, has a compact inverse on the orthogonal complement of N. In contrast to this, the higher dimensional periodicDirichlet problem for the semilinear wave equation (13.1)–(13.2) has the additional difficulty that does not have a compact inverse on N ⊥ . In fact, it has a sequence of eigenvalues of infinite multiplicities stretching from −∞ to ∞. This is a serious complication that causes all of the methods used to solve the one-dimensional case to fail. Recently some authors have examined the radially symmetric counterpart of (13.1)–(13.3) that was considered in this chapter (cf. [146], [15],[14],[23],and [121]). It is assumed that the function f (t, x, u) is radially symmetric in x. This allows one to reduce the problem to u t t − u rr − r −1 (n − 1)u r = f (t, r, u), u(2π, r ) = u(0, r ),

u t (2π, r ) = u t (0, r ),

u(t, R) = u R (t, R) = 0,

0 ≤ r ≤ R,

t ∈ R.

This is much more difficult than the one-dimensional problem for the wave equation, but the techniques used in solving it cannot be used to solve the n-dimensional problem for the wave operator when the region and functions are not radially symmetric. This will be addressed in Chapter 13.

Chapter 13

Semilinear Wave Equations 13.1 Introduction In this chapter we shall consider the higher-dimensional periodic-Dirichlet problem for the semilinear wave equation (13.1)

u ≡ u t t − u = p(x, t, u),

(x, t) ∈ ,

t ∈ R, x ∈ ∂(0, π)n ,

(13.2)

u(x, t) = 0,

(13.3)

u(x, t + 2π) = u(x, t),

(x, t) ∈ ,

where = (0, π)n × (0, 2π). Here, x = (x 1 , . . . , x n ) ∈ Rn and (0, π)n = {x ∈ Rn : 0 < x k < π, 1 ≤ k ≤ n}. In studying this problem, we shall make use of the theory of saddle points.

13.2 Convexity and lower semi-continuity A set M is called convex if (1−t)w0 +tw1 ∈ M whenever w0 , w1 ∈ M and 0 ≤ t ≤ 1. Let M be a convex subset of a Hilbert space E, and let G be a functional (realvalued function) defined on M. We call G convex on M if G((1 − t)w0 + tw1 ) ≤ (1 − t)G(w0 ) + t G(w1 ),

w0 , w1 ∈ M, 0 ≤ t ≤ 1.

We call it strictly convex if the inequality is strict when 0 < t < 1, w0 = w1 . G(v) is called upper semi-continuous (u.s.c.) at w0 ∈ M if wk → w0 ∈ M implies G(w0 ) ≥ lim sup G(wk ). It is called lower semi-continuous (l.s.c.) if the inequality is reversed and lim sup is replaced by lim inf. We have M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_13, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009

150

13. Semilinear Wave Equations

Lemma 13.1. If M is closed, convex, and bounded in E and G is convex and l.s.c. on M, then there is a point w0 ∈ M such that G(w0 ) = min G.

(13.4)

M

If G is strictly convex, then w0 is unique. In proving Lemma 13.1, we shall make use of Lemma 13.2. If u k u in E, then there is a renamed subsequence such that u¯ k → u, where u¯ k = (u 1 + · · · + u k )/k.

(13.5)

Proof. We may assume that u = 0. Take n 1 = 1, and inductively pick n 2 , n 3 , . . . , so that 1 1 |(u nk , u n1 )| ≤ , . . . , |(u nk , u nk−1 )| ≤ . k k This can be done since (u n , u n j ) → 0 as n → ∞,

1 ≤ j ≤ k.

Since u k ≤ C for some C, we have u¯ k 2

=

⎡ ⎤ j −1 k k ⎣ u j 2 + 2 (u i , u j )⎦ /k 2 ⎡

≤

j =1

j =1 i=1

⎤ j k 1 ⎣kC 2 + 2 ⎦ /k 2 j j =1 i=1

≤

(C 2 + 2)/k → 0.

Lemma 13.3. If G(u) is convex and l.s.c. on E, and u k u, then G(u) ≤ lim inf G(u k ). Proof. Let L = lim inf G(u k ). Then there is a renamed subsequence such that G(u k ) → L. Let ε > 0 be given. Then (13.6)

L − ε < G(u k ) < L + ε

13.2. Convexity and lower semi-continuity

151

for all but a finite number of k. Remove a finite number and rename the subsequence so that (13.6) holds for all k. Moreover, there is a renamed subsequence such that u¯ k → u by Lemma 13.2, where u¯ k is given by (13.5). Thus, ⎛ ⎞ k 1 G(u) ≤ lim inf G(u¯ k ) = lim inf G ⎝ u j⎠ k j =1

≤

lim inf

1 k

k

G(u j ) ≤ lim inf

j =1

1 · k(L + ε) = L + ε. k

Since ε was arbitrary, we see that G(u) ≤ L, and the proof is complete. A subset M ⊂ E is called weakly closed if u ∈ M whenever there is a sequence {u k } ⊂ M converging weakly to u in E. A weakly closed set is closed in a stronger sense than an ordinary closed set. It follows from Lemma 13.2 that Lemma 13.4. If M is a closed, convex subset of E, then it is weakly closed in E. Proof. Suppose {u k } ⊂ M and u k u in E. Then, by Lemma 13.2, there is a renamed subsequence such that u¯ k → u, where u¯ k is given by (13.5). Since M is convex, each u¯ k is in M. Since M is closed, we see that u ∈ M. We can now give the proof of Lemma 13.1. Proof. Let α = inf G. M

(At this point we do not know if α = −∞.) Let {wk } ⊂ M be a sequence such that G(wk ) → α. Since M is bounded, we see that there is a renamed subsequence such that wk w0 . Since M is closed and convex, it is weakly closed (Lemma 13.4). Hence, w0 ∈ M. By Lemma 13.3, G(w0 ) ≤ lim inf G(wk ) = α. Since G(w0 ) ≥ α, we see that (13.4) holds. So far, we have only used the convexity of G. We use the strict convexity to show that w0 is unique. If there were another element w1 ∈ M such that G(w1 ) = α, then we would have 1 1 1 w0 + w1 < [G(w0 ) + G(w1 )] = α, G 2 2 2 which is impossible from the definition of α. This completes the proof. We also have Lemma 13.5. If M is closed and convex, G is convex, is l.s.c., and satisfies (13.7)

G(u) → ∞ as u → ∞,

u∈M

(if M is unbounded), then G is bounded from below on M and has a minimum there. If G is strictly convex, this minimum is unique.

152


Proof. If M is bounded, then Lemma 13.5 follows from Lemma 13.1. Otherwise, let u 0 be any element in M. By (13.7), there is an R ≥ u 0 such that G(u) ≥ G(u 0 ),

u ∈ M, u ≥ R.

By Lemma 13.1, G is bounded from below on the set M R = {w ∈ M : w ≤ R} and has a minimum there. A minimum of G on M R is a minimum of G on M. Hence, G is bounded from below on M and has a minimum there.

13.3 Existence of saddle points We say that (v 0 , w0 ) is a saddle point of G if (13.8)

G(v, w0 ) ≤ G(v 0 , w0 ) ≤ G(v 0 , w),

v ∈ N, w ∈ M.

We now present some sufficient conditions for the existence of saddle points. Let M, N be closed, convex subsets of a Hilbert space, and let G(v, w) : M × N → R be a functional such that G(v, w) is convex and l.s.c. in w for each v ∈ N, and concave and u.s.c. in v for each w ∈ M. If M is unbounded, assume also that there is a v 0 ∈ N such that (13.9)

G(v 0 , w) → ∞ as w → ∞,

w ∈ M.

If N is unbounded, assume that there is a w0 ∈ M such that (13.10)

G(v, w0 ) → −∞ as v → ∞,

v ∈ N.

[If M is bounded, then (13.9) is automatically satisfied; the same is true for (13.10) when N is bounded.] We have Theorem 13.6. Under the above hypotheses, G has at least one saddle point. Proof. Assume first that M, N are bounded. Then, for each v ∈ N, there is at least one point w where G(v, w) achieves its minimum (Lemma 13.1). Let J (v) = min G(v, w). w∈M

Since J (v) is the minimum of a family of functionals that are concave and u.s.c., it is also concave and u.s.c. In fact, if v t = (1 − t)v 0 + tv 1 ,

t ∈ [0, 1],

then ˆ + t min G(v 1 , w), ˆ G(v t , w) ≥ (1 − t) min G(v 0 , w) w∈M ˆ

w∈M ˆ

w ∈ M.

13.3. Existence of saddle points

153

Since this is true for each w ∈ M, we see that J (v t ) ≥ (1 − t)J (v 0 ) + t J (v 1 ).

(13.11)

Similarly, if v k → v ∈ N, then we have J (v k ) ≤ G(v k , w),

w ∈ M.

Thus, lim sup J (v k ) ≤ lim sup G(v k , w) ≤ G(v, w),

w ∈ M.

Since this is true for each w ∈ M, we have lim sup J (v k ) ≤ inf G(v, w) = J (v).

(13.12)

w∈M

Therefore, J (v) is concave and u.s.c. Consequently, J (v) has a maximum point v¯ satisfying J (v) ≤ J (v), ¯ v∈N (Lemma 13.1). In particular, we have J (v) ¯ = min G(v, ¯ w) ˆ ≤ G(v, ¯ w),

(13.13)

w∈M ˆ

w ∈ M.

Let v be an arbitrary point in N, and let v θ = (1 − θ )v¯ + θ v,

0 ≤ θ ≤ 1.

Since G is concave in v, we have G(v θ , w) ≥ (1 − θ )G(v, ¯ w) + θ G(v, w). Consequently, J (v) ¯ ≥ J (v θ ) = G(v θ , wθ ) ≥ (1 − θ )G(v, ¯ wθ ) + θ G(v, wθ ) ≥ (1 − θ )J (v) ¯ + θ G(v, wθ ), where wθ is any point in M such that G(v θ , wθ ) = min G(v θ , w). w∈M

This gives (13.14)

J (v) ¯ ≥ G(v, wθ ),

v ∈ N, 0 < θ ≤ 1.

Let {θk } be a sequence converging to 0, and let v k = v θk , wk = wθk . Then v k → v. ¯ Since M is bounded, there is a renamed subsequence such that wk w. ¯ Since (1 − θ )G(v, ¯ wθ ) + θ G(v, wθ ) ≤ G(v θ , wθ ) ≤ G(v θ , w),

w ∈ M,

154


we have ¯ wk ) + θk J (v) ≤ G(v k , w), (1 − θk )G(v,

w ∈ M.

In the limit this gives G(v, ¯ w) ¯ ≤ G(v, ¯ w),

w∈M

(cf. Lemma 13.3). Since J (v) ¯ ≥ G(v, wk ), we have G(v, w) ¯ ≤ J (v) ¯ ≤ G(v, ¯ w),

v ∈ N, w ∈ M,

in view of (13.13) and (13.14). Take v = v¯ and w = w. ¯ Then G(v, ¯ w) ¯ ≤ J (v) ¯ ≤ G(v, ¯ w), ¯ showing that G(v, ¯ w) ¯ = J (v) ¯ and G(v, w) ¯ ≤ G(v, ¯ w) ¯ ≤ G(v, ¯ w),

v ∈ N, w ∈ M.

Thus, (v, ¯ w) ¯ is a saddle point. Now, we remove the restriction that M, N are bounded. Let R be so large that v 0 < R, w0 < R. The sets M R = {w ∈ M : w ≤ R},

N R = {v ∈ N : v ≤ R}

are closed, convex, and bounded. By what we have already proved, there is a saddle point (v¯ R , w¯ R ) such that (13.15)

G(v, w¯ R ) ≤ G(v¯ R , w¯ R ) ≤ G v¯ R , w),

v ∈ NR , w ∈ MR .

In particular, we have G(v 0 , w¯ R ) ≤ G(v¯ R , w¯ R ) ≤ G(v¯ R , w0 ). Since G(v 0 , w) is convex, is l.s.c., and satisfies (13.9), it is bounded from below on M (Lemma 13.5). Thus, G(v 0 , w¯ R ) ≥ A > −∞. Similarly, G(v, w0 ) is bounded from above. Hence, G(v¯ R , w0 ) ≤ B < ∞. Combining these with (13.15), we have A ≤ G(v 0 , w¯ R ) ≤ G(v¯ R , w¯ R ) ≤ G(v¯ R , w0 ) ≤ B. By (13.9) and (13.10), the sequences {v¯ R }, {w¯ R } are bounded. Hence, there are renamed subsequences such that v¯ R v, ¯ w¯ R w¯ as R → ∞

13.4. Criteria for convexity

155

and G(v¯ R , w¯ R ) → λ as R → ∞. In view of (13.15), we have in the limit G(v, w) ¯ ≤ λ ≤ G(v, ¯ w),

v ∈ N, w ∈ M.

This shows that λ = G(v, ¯ w), ¯ and (v, ¯ w) ¯ is a saddle point. The theorem is completely proved.

13.4 Criteria for convexity If G is a differentiable functional on a Hilbert space E, there are simple criteria that can be used to verify the convexity of G. We have Theorem 13.7. Let G be a differentiable functional on a closed, convex subset M of E. Then G is convex on E iff it satisfies any of the following inequalities for u 0 , u 1 ∈ M : (13.16)

(G (u 0 ), u 1 − u 0 ) ≤G(u 1 ) − G(u 0 )

(13.17)

(G (u 1 ), u 1 − u 0 ) ≥G(u 1 ) − G(u 0 ),

(13.18)

(G (u 1 ) − G (u 0 ), u 1 − u 0 ) ≥0.

Moreover, it will be strictly convex iff there is strict inequality in any of them when u 0 = u 1 . Proof. Let u t = (1 − t)u 0 + tu 1 , 0 ≤ t ≤ 1, and ϕ(t) = G(u t ). If G is convex, then (13.19)

G(u t ) ≤ (1 − t)G(u 0 ) + t G(u 1 )

or (13.20)

ϕ(t) ≤ (1 − t)ϕ(0) + tϕ(1),

0 ≤ t ≤ 1.

In particular, the slope of ϕ at t = 0 is less than or equal to the slope of the straight line connecting (0, ϕ(0)) and (1, ϕ(1)). Thus, ϕ (0) ≤ ϕ(1) − ϕ(0), and this is merely (13.16). Reversing the roles of u 0 , u 1 produces (13.17). We obtain (13.18) by subtracting (13.16) from (13.17). Conversely, (13.18) implies ϕ (t) − ϕ (s) = (G (u t ) − G (u s ), u 1 − u 0 ) = (G (u t ) − G (u s ), u t − u s )/(t − s) ≥ 0, Thus,

ϕ (t) ≥ ϕ (s),

0 ≤ s < t ≤ 1.

0 ≤ s ≤ t ≤ 1,

which implies (13.20). Since this is equivalent to (13.19), we see that G is convex. If G is strictly convex, we obtain strict inequalities in (13.16)–(13.18), and strict inequalities in any of them implies strict inequalities in (13.20) and (13.19).

156


Corollary 13.8. Let G be a differentiable functional on a closed, convex subset M of E. Then G is concave on E iff it satisfies any of the following inequalities for u 0 , u 1 ∈ M : (13.21)

(G (u 0 ), u 1 − u 0 ) ≥ G(u 1 ) − G(u 0 ),

(13.22)

(G (u 1 ), u 1 − u 0 ) ≤ G(u 1 ) − G(u 0 ),

(13.23)

(G (u 1 ) − G (u 0 ), u 1 − u 0 ) ≤ 0.

Moreover, it will be strictly concave iff there is strict inequality in any of them when u 0 = u 1 . Proof. Note that G(u) is concave iff −G(u) is convex.

13.5 Partial derivatives Let M, N be closed subspaces of a Hilbert space H satisfying H = M ⊕ N. Let G(u) be a functional on H. We can consider “partial” derivatives of G in the same way we considered total derivatives. We keep w = w0 ∈ M fixed and consider G(u) as a functional on N, where u = v + w0 , v ∈ N. If the derivative of this functional exists at v = v 0 ∈ N, we call it the partial derivative of G at u 0 = v 0 + w0 with respect to v ∈ N and denote it by G N (u 0 ). Similarly, we can define the partial derivative G M (u 0 ). We have Lemma 13.9. If G exists at u 0 = v 0 + w0 , then G M (u 0 ) and G N (u 0 ) exist and satisfy (13.24)

(G (u 0 ), u) = (G M (u 0 ), w) + (G N (u 0 ), v),

v ∈ N, w ∈ M.

Proof. By definition,

Therefore, and But and

G(u 0 + u) = G(u 0 ) + (G (u 0 ), u) + o( u ),

u ∈ H.

G(u 0 + v) = G(u 0 ) + (G (u 0 ), v) + o( v ),

v∈N

G(u 0 + w) = G(u 0 ) + (G (u 0 ), w) + o( w ), G(u 0 + v) = G(u 0 ) + (G N (u 0 ), v) + o( v ),

w ∈ M. v∈N

G(u 0 + w) = G(u 0 ) + (G M (u 0 ), w) + o( w ),

w ∈ M.

In particular, we have (G (u 0 ) − G N (u 0 ), v) = o( v ) as v → 0, Thus,

v ∈ N.

(G (u 0 ) − G N (u 0 ), tv) = o(|t|) as |t| → 0

13.5. Partial derivatives

157

for each fixed v ∈ N. This means that (G (u 0 ) − G N (u 0 ), v) = Hence, Similarly,

o(|t|) → 0 as t → 0. t

(G (u 0 ), v) = (G N (u 0 ), v),

v ∈ N.

(G (u 0 ), w) = (G M (u 0 ), w),

w ∈ M.

These two identities combine to give (13.24). Lemma 13.10. Under the hypotheses of Lemma 13.9, assume that G is differentiable on H , convex on M, and concave on N. Then, (13.25)

G(u) − G(u 0 ) ≤ (G N (u 0 ), v − v 0 ) + (G M (u), w − w0 ), u = v + w, u 0 = v 0 + w0 , v, v 0 ∈ N, w, w0 ∈ M.

If G is either strictly convex on M or strictly concave on N (or both), then one has strict inequality in (13.25) when u = u 0 . Proof. This follows from Theorem 13.7 and its corollary. In fact, we have G(u) − G(u 0 ) = G(u) − G(v + w0 ) + G(v + w0 ) − G(u 0 ) ≤ (G (u), w − w0 ) + (G (u 0 ), v − v 0 ). Apply Lemma 13.9. We also have Lemma 13.11. Under the hypotheses of Lemma 13.9, if G (u 0 ) exists and u 0 = v 0 +w0 is a saddle point, then G (u 0 ) = G M (u 0 ) = G N (u 0 ) = 0. Proof. By definition, G(v + w0 ) ≤ G(u 0 ) ≤ G(v 0 + w),

v ∈ N, w ∈ M.

Since v 0 is a maximum point on N, we see that G N (u 0 ) = 0. Since w0 is a minimum point on M, we have G M (u 0 ) = 0 for the same reason. We then apply Lemma 13.9. Corollary 13.12. Under the hypotheses of Lemma 13.10, if G is either strictly convex on M or strictly concave on N (or both), then G has exactly one saddle point. Proof. This follows from inequality (13.25).

158


13.6 The theorems In solving problem (13.1)–(13.3) we take p(x, t, ξ ) to be a Carathéodory function on × R which is 2π-periodic in t and satisfies (13.26)

θ− ξ 2 ≤ ξ [ p(x, t, ξ1 ) − p(x, t, ξ0 )] ≤ θ+ ξ 2 ,

ξ = ξ1 − ξ2 ∈ R.

Let μ = (μ1 , . . . , μn ),

μ j ≥ 0,

μ j ∈ Z,

μ2 = |μ|2 =

μ2j .

Let σ () be the set of integers of the form λ = μ2 − k 2 , and let ρ() = R\σ (). We shall prove Theorem 13.13. If [θ− , θ+ ] ⊂ ρ(),

(13.27)

then there is a unique weak solution of (13.1)–(13.3). Theorem 13.14. If (θ− , θ+ ) ⊂ ρ(),

(13.28)

and there are constants β± such that θ− ≤ β− ≤ β+ ≤ θ+ , [β− , β+ ] ⊂ ρ() and (13.29)

β− ≤ lim inf 2P(x, t, ξ )/ξ 2 ≤ lim sup 2P(x, t, ξ )/ξ 2 ≤ β+ |ξ |→∞

|ξ |→∞

uniformly in , where

ξ

P(x, t, ξ ) =

p(x, t, s)ds, 0

then (13.1)–(13.3) has at least one weak solution. Theorems 13.13 and 13.14 are proved in the next section.

13.7 The proofs In this section we present the proof of Theorems 13.13 and 13.14. Let x = (x 1 , . . . , x n ), ϕμ (x) = sin μ1 ξ1 · · · sin μn x n /(π/2)n/2 . Note that (ϕμ , ϕν ) = δμν , where the scalar product is that of L 2 ([0, π]n ). We take ek (t) = eikt /(2π)1/2,

t ∈ [0, 2π].

Then (ek , e ) = δk ,

13.7. The proofs

159

where the scalar product is that of L 2 ([0, 2π]). If (13.30) u= αμk ϕμ (x)ek (t), then u ≡ u t t − u =

(μ2 − k 2 )αμk ϕμ (x)ek (t).

Let γ be a fixed number in (θ− , θ+ ) and define P(x, t, u), G(u) = ∇u 2 − u t 2 − 2

u ∈ E,

where the norm is that of L 2 (), and E is the set of all u ∈ L 2 () of the form (13.30) such that |μ2 − k 2 − γ | · |αμk |2 < ∞. u 2E = It is easily checked that G ∈ C 1 (E, R) and (G (u), v)/2 = (u, v) − ( p(u), v),

u, v ∈ E,

where we write p(u) in place of p(x, t, u). Let m + be the smallest point of σ () above γ and m − the largest point of σ () below γ . Let M = {u ∈ E : αμk = 0 when μ2 − k 2 < γ }, N = {u ∈ E : αμk = 0 when μ2 − k 2 > γ }. We have Lemma 13.15.

θ+ − γ (G (v + w1 ) − G (v + w0 ), w)/2 ≥ 1 − w 2E , m+ − γ

where w = w1 − w0 ∈ M, v ∈ N. Proof. The left-hand side equals (w, w) − ( p(v + w1 ) − p(v + w0 ), w) ≥ w 2E − (θ+ − γ ) w 2 θ+ − γ ≥ 1− w 2E . m+ − γ

Lemma 13.16.

γ − θ− v 2E , (G (v 1 + w) − G (v 0 + w), v)/2 ≤ − 1 − γ − m−

where v = v 1 − v 0 ∈ N, w ∈ M.

160


Proof. Now the left-hand side equals (v, v) − ( p(v 1 + w) − p(v 0 + w), v) ≤ − v 2E + (γ − θ− ) v 2 γ − θ− ≤− 1− v 2E . γ − m−

We can now give the proof of Theorem 13.14. Proof. By (13.28) and Lemmas 13.15 and 13.16, G(v +w) is convex in w and concave in v. Moreover, (13.31)

G(w) → ∞ as w E → ∞,

w ∈ M,

G(v) → −∞ as v E → ∞,

v ∈ N.

and (13.32) To see this, let wj =

( j)

αμk ϕμ ek

be a sequence in M such that ρ j = w j E → ∞. Take w˜ j = w j /ρ j , and let ε > 0 be such that β+ < m + − ε. Then there is a constant K such that 2P(x, t, ξ )/ξ 2 < β+ + ε,

|ξ | > K .

Hence, 2

P(x, t, w j )/ρ 2j ≤

|w j |>K

(β+ + ε)w˜ 2j +

|w j |≤K

P(x, t, w j )/ρ 2j

β+ + ε + C/ρ 2j ≤ m+ − γ since w˜ j E = 1. Therefore, β+ + ε − γ G(w j )/ρ 2j ≥ 1 − − C/ρ 2j . m+ − γ Consequently,

β+ + ε − γ > 0. lim inf G(w j )/ρ 2j ≥ 1 − m+ − γ

This proves (13.31). A similar argument proves (13.32). Thus, G(v + w) is convex in w, is concave in v and satisfies (13.31), (13.32). The theorem now follows from Theorems 13.6, 13.7, Corollary 13.8, and Lemma 13.11.


161

Next, we prove Theorem 13.13. Proof. First, we note that the hypotheses of Theorem 13.13 imply those of Theorem 13.14. In fact, we can take β± = θ± , for we have P(x, t, ξ ) =

ξ

s[ p(x, t, s) − p(x, t, 0)]

0

ds + p(x, t, 0)ξ. s

Hence, θ− ξ 2 ≤ 2P(x, t, ξ ) − 2 p(x, t, 0)ξ ≤ θ+ ξ 2 by (13.26). Thus, there exists a weak solution of (13.1)–(13.3). To show that it is unique, we note that for fixed v ∈ N, G(v + w) is strictly convex in w. Hence, if there were two distinct points such that G (u 0 ) = G (u) = 0, we would have by Lemma 13.10 both G(u) < G(u 0 ) and G(u 0 ) < G(u), an impossibility.

13.8 Notes and remarks The problem (13.1)–(13.3) has been considered by Smiley [147] and Mawhin [90]. They assume β0 ξ 2 ≤ ξ [ p(x, t, ξ1 ) − p(x, t, ξ0 )],

ξ = ξ1 − ξ0 ∈ R,

and | p(x, t, ξ1 ) − p(x, t, ξ0 )| ≤ β1 |ξ |, where 0 < β1 and < β0 . This implies 0 < β0 ≤ β1 < 1 and β1 − β0 < 1 (1 − β ). Since ρ() consists only of integers, our results are extensions of their 0 2 results. Theorems 13.13 and 13.14 were proved in [125]. β12

Chapter 14

Type (II) Regions 14.1 Introduction The Fuˇc´ık spectrum described in Chapter 11 arises in the study of semilinear elliptic boundary-value problems of the form Au = f (x, u),

(14.1)

where A is a selfadjoint operator having compact resolvent on L 2 (), ⊂ Rn , and ¯ × R such that f (x, t) is a Carathéodory function on (14.2)

f (x, t)/t

→

a a.e. as t → −∞

→

b a.e. as t → +∞.

If |u(x)| is large, then (14.1) approximates the equation (14.3)

Au = bu + − au − ,

where u ± = max{±u, 0}. It was first noticed by Fuˇc´ık [68] that (14.3) plays an important part in the study of (14.1) when (14.2) holds. Fuˇc´ık studied the problem (14.4)

−u = bu + − au − in (0, π),

u(0) = u(π) = 0

and showed that there is a substantial difference in the solvability of (14.1) if (14.3) has nontrivial solutions. We now call the set of those (a, b) ∈ R2 for which (14.3) has nontrivial solutions the Fuˇc´ık spectrum of A. Fuˇc´ık showed that for (14.4), consists of a sequence of decreasing curves passing through the points (k 2 , k 2 ), k = 1, 2, . . . , with one or two curves emanating from each of these points. Points not on these curves are not in . He also noticed that there are two different types of regions between the curves, namely, (I)

regions between curves passing through consecutive points (k 2 , k 2 ), ([k + 1]2 , [k + 1]2 )


164 and (II)

14. Type (II) Regions

regions between curves passing through the same point (k 2 , k 2 ).

In the type (I) regions one can solve (14.5)

−u = bu + − au − + p(x),

x ∈ (0, π), u(0) = u(π) = 0,

for arbitrary p(x) ∈ L 2 (0, π), while this is not so for regions of type (II). No complete description of for the general case (14.1) has been found. If (14.6)

0 < λ0 < λ1 < · · · < λk < · · ·

are the eigenvalues of A, it was shown in [111] and [122] that in the square [λ−1 , λ+1 ]2 , there are decreasing curves C,1 , C,2 (which may coincide) passing through the point (λ , λ ) such that all points above or below both curves in the square are not in [and are of type (I)] while points on the curves are in . The status of points between the curves (when they do not coincide) is unknown in general. However, it was shown in [72] that when λ is a simple eigenvalue, points between the curves are not in . On the other hand, C. and W. Margulies [91] have shown that there are boundary-value problems for which many curves in emanate from a point (λ , λ ) when λ is a multiple eigenvalue. (Clearly, these curves are contained in the region between the curves C,1 and C,2 .) As expected, the boundary-value problem (14.1) is more readily solved when (a, b) is in a region of type (I) (i.e., on the same side of both curves C1 , C2 ). For such points, the following was proved in [111] and [122]. Theorem 14.1. Assume that f (x, t) is of the form (14.7)

f (x, t) = bt + − at − + p(x, t),

where p(x, t) satisfies (14.8)

| p(x, t)| ≤ V (x)1−σ |t|σ + W (x),

with 0 ≤ σ < 1 and V, W ∈ L 2 (). Then (14.1) has a solution. In particular, (14.9)

Au = bu + − au − + p(x)

has a solution for each p ∈ L 2 (). However, when it comes to points on the curves C,1 , C,2 , no such theorem holds. In Chapter 11 we addressed this issue and presented sufficient conditions for the existence of solutions of (14.1) when (a, b) is on either C,1 or C,2 . [Needless to

14.1. Introduction

165

say, much more is required of p(x, t) in Theorems 11.1 and 11.2 than in Theorem 14.1.] In the present chapter we consider the situation when (a, b) is in the region between the curves C,1 , C,2 (when they do not coincide). As we saw earlier, such points may or may not belong to . We shall be concerned with those points in this region that are not in . Even in the case when none of these points is in (as in the case of a simple eigenvalue), we cannot prove a theorem as comprehensive as Theorem 14.1. However, we can prove the theorems below. We assume that (14.10)

| p(x, t)| ≤ C(|t| + 1),

x ∈ , t ∈ R,

and that E(λ ), the subspace of eigenfunctions corresponding to λ , is contained in L ∞ () (i.e., the eigenfunctions corresponding to λ are bounded). We define t (14.11) F(x, t) = f (x, s) ds 0

and (14.12)

t

P(x, t) =

p(x, s) ds. 0

We have Theorem 14.2. Let (a, b) be a point in Q := (λ−1 , λ+1 )2 that lies below the upper curve C,2 . Assume (14.2), (14.7), and (14.13)

f (x, t1 ) − f (x, t0 ) >λ−1 (t1 − t0 ),

t0 < t1 , x ∈ ,

(14.14)

2P(x, t) ≤W1 (x) ∈ L 1 (),

(14.15)

2F(x, t) ≤λ+1 t 2 ,

(14.16)

λ t ≤2F(x, t), 2

x ∈ , t ∈ R,

x ∈ , t ∈ R, |t| < δ for some δ > 0.

If (a, b) is not in , then (14.1) has a nontrivial solution. Theorem 14.3. Let (a, b) be a point in Q \ that lies above the lower curve C,1 . Assume (14.2), (14.7), and (14.17) (14.18) (14.19) (14.20)

2P(x, t) ≥ − W1 (x) ∈ L 1 (), f (x, t1 ) − f (x, t0 ) 0.

Then (14.1) has a nontrivial solution. It should be noted that these theorems hold even when λ is a multiple eigenvalue. In proving the theorems, we examine the functional 2 F(x, u)d x, u ∈ D = D(A1/2 ), (14.21) G(u) = u D − 2

166


where u D = A1/2u , and search for solutions of G (u) = 0. Even though (a, b) is in a region of type (II), we can find a manifold S on which G is bounded from below. Unfortunately, S has a boundary that cannot be linked with a subspace, and one must search for another manifold that links the boundary of S and on which G is bounded from above. Once this is achieved, we can apply the theorems of Chapter 2 to obtain a Palais–Smale sequence. We then show that there is a convergent subsequence due to the fact that (a, b) is not in . In Theorem 14.3 we reverse the procedure, finding a manifold on which G is bounded from above and then searching for a linking set on which it is bounded from below.

14.2 The asymptotic equation In this section we show how information concerning (14.3) affects the solvability of (14.1). For each fixed positive integer , we let N denote the subspace of D = D(A1/2 ) spanned by the eigenfunctions of A corresponding to the eigenvalues λ0 , λ1 , . . . , λ , and we let M = N⊥ ∩ D. Then D = M ⊕ N . For (a, b) ∈ R2 , we define (14.22)

u 2D =(Au, u),

(14.23)

I (u, a, b) = u 2D − a u − 2 − b u + 2 ,

(14.24)

M (a, b) = inf

(14.25)

m (a, b) = sup

w∈M w D =1

v∈N v D =1

u ∈ D,

sup I (v + w, a, b),

v∈N

inf I (v + w, a, b),

w∈M

(14.26)

ν (a) =sup{b : M (a, b) ≥ 0},

(14.27)

μ (a) = inf{b : m (a, b) ≤ 0}.

It follows from Lemma 11.3 that ν−1 (a) ≤ μ (a) in Q . We let C,1 be the (lower) curve b = ν−1 (a) and we let C,2 be the (upper) curve b = μ (a). Thus, by (14.27), if (a, b) ∈ Q is below the curve C,2 , then m (a, b) > 0.

(14.28)

Consequently, there is a y0 ∈ N0 = E(λ ) such that (14.29)

sup

inf I (v + w + y0 , a, b) > 0.

v∈N−1 w∈M

Note that I (v + w + y0 , a, b) is strictly convex and lower semi-continuous in w ∈ M and strictly concave and continuous in v ∈ N−1 because (a, b) ∈ Q . Moreover, I (w + y0 , a, b) → ∞ as w D → ∞,

w ∈ M.

To see this, let w D = ρ → ∞. Then, for u = (w + y0 )/ρ, we have I (u, a, b) ≥ 1 − a u − 2 − b u + 2 → 1 − a w˜ − 2 − b w˜ + 2 > 0,

14.2. The asymptotic equation

167

where w ˜ ≤ 1, since a, b < λ+1 . Similarly, I (v + y0 , a, b) → ∞ as v D → ∞,

v ∈ N.

Hence we may apply Theorem 13.6 to conclude that there are unique v 0 ∈ N−1 , w0 ∈ M such that (14.30)

I (v 0 + w0 + y0 , a, b) = =

sup

inf I (v + w + y0 , a, b)

inf

sup I (v + w + y0 , a, b).

v∈N−1 w∈M w∈M v∈N−1

In particular, this shows that y0 = 0, for otherwise v 0 = 0, w0 = 0 would satisfy (14.30) and the uniqueness would violate (14.29). We also see from (14.29) and (14.30) that (14.31)

sup I (v + w + y0 , a, b) > 0,

w ∈ M .

v∈N−1

Moreover, it follows that there is a continuous map θ from M−1 to N−1 such that (14.32)

θ (w) ≤ C w D ,

(14.33)

θ (sw) = sθ (w),

w ∈ M−1 , s ≥ 0,

and v˜ = θ (w) is the only solution of I (w + v, ˜ a, b) ⊥ N−1 ,

(14.34)

w ∈ M−1 ,

and (14.35)

I (w + v, ˜ a, b) = max I (w + v, a, b), v∈N−1

w ∈ M−1 .

Thus, (14.31) implies (14.36)

I (w + sy0 + θ (w + sy0 ), a, b) ≥ 0,

w ∈ M , s ≥ 0.

Another way of writing (14.36) is I (u, a, b) ≥ 0,

(14.37)

u ∈ S2

where (14.38)

S2 , = {u ∈ D : I (u, a, b) ⊥ N−1 , (u, y0 ) ≥ 0}.

Thus, we have proved Lemma 14.4. If (a, b) ∈ Q is below the upper curve C,2 , then there is a y0 ∈ E(λ )\{0} such that (14.37) holds for all u ∈ S2 , where S2 is given by (14.38).

168


In the same vein, if (a, b) ∈ Q is above the lower curve C1 , then there is a y1 ∈ N0 such that (14.39)

inf

sup I (v + w + y1 , a, b) < 0.

w∈M v∈N−1

Using the same reasoning as before, we have y1 = 0 and (14.30) holds with y0 replaced by y1 . This implies inf I (v + w + y1 , a, b) < 0,

(14.40)

w∈M

v ∈ N−1 .

Now we use the fact that there is a continuous map τ from N to M such that (14.41)

τ (v) D ≤ C v ,

(14.42)

τ (sv) = sτ (v),

v ∈ N , s ≥ 0,

and w˜ = τ (v) is the unique solution of I (v + w, ˜ a, b) ⊥ M ,

(14.43)

v ∈ N ,

and I (v + w, ˜ a, b) = inf I (v + w, a, b),

(14.44)

w∈M

v ∈ N .

Thus, (14.40) implies (14.45)

I (v + sy1 + τ (v + sy1 ), a, b) ≤ 0,

v ∈ N−1 , s ≥ 0,

or (14.46)

I (u, a, b) ≤ 0,

u ∈ S1 ,

where (14.47)

S1 = {u ∈ D : I (u, a, b) ⊥ M , (u, y1 ) ≥ 0}.

We therefore have Lemma 14.5. If (a, b) ∈ Q lies above the lower curve C,1 , then there is a y1 ∈ E(λ )\{0} such that (14.46) holds for all u ∈ S1 , where S1 is given by (14.47).

14.3 Local estimates In proving Theorems 14.2 and 14.3, we shall make use of the functional 2 (14.48) G(u) = u D − 2 F(x, u) d x = I (u, a, b) − 2 P(x, u) d x

for functions u ∈ D. When (14.10) holds, it is easily checked that G ∈ C 1 (D, R) and that u ∈ D is a solution of (14.1) if, and only if, it satisfies (14.49)

G (u) = 0.

In this section we shall examine some properties of G under the hypotheses of the theorems. First, we have

14.3. Local estimates

169

Lemma 14.6. If (14.16) holds, then for each ρ > 0 sufficiently small, there is a positive ε such that (14.50)

G(v + y) ≤ −ε v 2 ,

v ∈ N−1 , y ∈ E(λ ), v + y ≤ ρ.

Proof. Since E(λ ) ⊂ L ∞ (), there is a ρ > 0 such that y D ≤ ρ implies |y(x)| ≤ δ/2,

(14.51)

y ∈ E(λ ),

where δ is the constant given in (14.16). Let w = v + y, where v ∈ N−1 , y ∈ E(λ ). If w D ≤ ρ

(14.52)

|w(x)| ≥ δ,

and

then δ ≤ |w(x)| ≤ |v(x)| + |y(x)| ≤ |v(x)| + δ/2.

(14.53)

Consequently, (14.52) implies |y(x)| ≤ δ/2 ≤ |v(x)|

(14.54) and

|w(x)| ≤ 2|v(x)|.

(14.55) By (14.10),

|F(x, t)| ≤ C(t 2 + |t|).

(14.56) Thus, (14.57)

G(w)

≤

w 2D

− λ

w dx + C 2

|w|δ

− λ v + 4C (1 + δ 2

≤ v 2D − λ v 2 + C

|w|>δ

(w2 + |w|)d x

(w2 + |w|)d x

−1

)

v2d x 2|v|>δ

|v|σ d x

λ σ −2 1− v 2D , + C v D λ−1

where σ > 2. If we take ρ sufficiently small, this implies (14.50). Lemma 14.7. If (14.20) holds, then for each ρ > 0 sufficiently small, there is an ε > 0 such that (14.58)

G(w + y) ≥ ε w 2 ,

w ∈ M , y ∈ E(λ ), w + y ≤ ρ.

170


The proof of Lemma 14.7 is similar to that of Lemma 14.6 and is omitted. Lemma 14.8. Let θ be a continuous map from M−1 to N−1 satisfying (14.32), and define (14.59)

H (v + w) = v + w + θ (w),

v ∈ N−1 , w ∈ M−1 .

If (14.16) holds and G is given by (14.48), then we have the following alternative: Either (a) for each ρ > 0 sufficiently small, there is an ε > 0 such that G(H v) ≤ −ε,

(14.60)

v ∈ N ∩ ∂ Bρ ,

or (b) there is a y ∈ E(λ )\{0} such that Ay = λ y = f (x, y).

(14.61)

Proof. For v ∈ N , write v = v + y, where v ∈ N−1 and y ∈ E(λ ). Then H v = v + θ (y),

(14.62)

H v ≤ C v .

Thus, by Lemma 14.6, for each ρ > 0 sufficiently small, there is an ε > 0 such that (14.63)

G(H v) ≤ −ε v + θ (y) 2 ,

v ∈ N ∩ Bρ .

In particular, G(H v) ≤ 0 for such v. Assume that option (a) does not hold. Then there is a sequence v k = v k + yk , v k ∈ N−1 , yk ∈ E(λ ) such that G(H v k ) → 0,

(14.64)

v k = ρ.

Since H v k = v k + θ (u k ), we see from (14.63) that v k + θ (yk ) → 0. Moreover, there is a renamed subsequence such that yk → y0 in E(λ ). By continuity, θ (yk ) → θ (y0 ) and v k → −θ (y0 ). If y0 = 0, then v k → 0, contradicting the fact that v k = ρ. Thus, y0 = 0 and H v k → y0 . This implies G(y0 ) = 0.

(14.65)

If ρ > 0 is such that (14.51) holds, then (14.16) implies that (14.66)

λ y0 (x)2 ≤ 2F(x, y0(x)),

x ∈ .

But (14.65) says (14.67)

{λ y0 (x)2 − 2F(x, y0 (x))}d x = 0.

This together with (14.66) implies that (14.68)

λ y0 (x)2 ≡ F(x, y0 (x)),

x ∈ .

Let ζ (x) be any function in C0∞ (). Then, for t > 0 sufficiently small, (14.69)

t −1 λ [(y0 + tζ )2 − y02 ] ≤ t −1 2[F(x, y0 + tζ ) − F(x, y0 )].

14.3. Local estimates

171

Taking the limit as t → 0, we have (14.70)

λ y0 (x)ζ (x) ≤ f (x, y0 )ζ(x),

x ∈ .

From this we conclude that (14.71)

λ y0 (x) ≡ f (x, y0 (x)),

x ∈ .

Since y0 ∈ E(λ ), this implies that y0 satisfies (14.61). Lemma 14.9. Let τ be a continuous map from N to M such that (14.41) holds and define (14.72)

H (v + w) = v + w + τ (v),

v ∈ N , w ∈ M .

If F(x, t) satisfies (14.20), then the following alternative holds: Either (a) for each ρ > 0 sufficiently small, there is an ε > 0 such that (14.73)

G(H w) ≥ ε,

w ∈ M−1 ∩ ∂ Bρ ,

or (b) there is a y ∈ E(λ )\{0} such that (14.61) holds. Proof. The proof is similar to that of Lemma 14.8. If w = w +y, w ∈ M , y ∈ E(λ ), then H w = w + τ (y), H w D ≤ C w D and (14.74)

G(H w) ≥ ε w + τ (y) 2 ,

w ∈ M−1 ∩ Bρ ,

for ρ sufficiently small by Lemma 14.7. If (a) does not hold and wk = wk + yk , wk ∈ M , yk ∈ E(λ ), wk ∈ ∂ Bρ , and G(H wk ) → 0, then wk + τ (yk ) → 0, yk → y1 , τ (yk ) → τ (y1 ), and wk → −τ (y1 ) for a renamed subsequence. If y1 = 0, then wk → 0, contradicting the fact that wk D = ρ. Now H wk → y1 and G(y1 ) = 0.

(14.75)

If ρ is sufficiently small, (14.51) holds and (14.76)

2F(x, y1(x)) ≤ λ y1 (x)2 ,

x ∈ ,

by (14.20). This together with (14.75) implies that (14.77)

2F(x, y1 (x)) ≡ λ y1 (x)2 ,

x ∈ .

This and (14.20) imply that (14.78)

t −1 2[F(x, y1 + tζ ) − F(x, y1 )] ≤ t −1 λ [(y1 + tζ )2 − y12 ]

for t small and ζ ∈ C0∞ (). This yields (14.79)

f (x, y1 ) ≡ λ y1 (x),

and y1 is a solution of (14.61).

x∈

172


14.4 The solutions We can now present the proof of Theorems 14.2 and 14.3. We begin with the proof of Theorem 14.2. Proof. We shall study G given by (14.48) and look for solutions of (14.49). By (14.13) we have for w ∈ M−1 , v 0 , v 1 ∈ N−1 , v = v 1 − v 0 , (14.80)

1 (G (w + v 1 ) − G (w + v 0 ), v) 2 = v 2D − ( f (w + v 1 ) − f (w + v 0 ), v) ≤ {λ−1 v 2 − [ f (x, w + v 1 ) − f (x, w + v 0 )]v}d x.

The right-hand side will be negative unless v ≡ 0 in view of (14.13). Thus, there is a unique solution θˆ (w) of G(w + θˆ (w)) = sup G(w + v)

(14.81)

v∈N−1

(Lemma 13.5). Moreover, θˆ is continuous and satisfies (14.32). Define (14.82)

H (v + w) = v + w + θˆ (w),

v ∈ N , w ∈ M−1 ,

which is continuous as well. Let (14.83)

A R = [M ∩ B¯ R ] ∪ {u = w + sy0 : w ∈ M , s ≥ 0, u = R}

where y0 is the element of E(λ )\{0} given in Lemma 14.4. By Example 7 of Section 3.5, A R links N ∩ ∂ Bρ [hm] whenever 0 < ρ < R. In view of Proposition 3.11, H A R links H [N ∩ ∂ Bρ ] [hm]. Let θ be the map described in Section 14.2 satisfying (14.32)–(14.36). Then, by (14.81), (14.84) G(w + sy0 + θˆ (w + sy0 ))

≥

G(w + sy0 + θ (w + sy0 ))

=

I (w + sy0 + θ (w + sy0 ), a, b) −2 P(x, w + sy0 + θ (w + sy0 ))d x

≥

−

≡

−B1

W1 (x) d x

for w ∈ M , s ≥ 0. Thus, (14.85)

G(u) ≥ −B1 ,

u ∈ H A R.

14.4. The solutions

173

We can do a bit better on that portion of A R contained in M . In fact, (14.16) implies G(w + sy0 + θˆ (w)) ≥ G(w) = w 2D − 2 (14.86) F(x, w) d x

≥

w 2D

≥

0

− λ+1 w

2

for w ∈ M . On the other hand, we see from Lemma 14.8 that G(u) ≤ −ε < 0,

(14.87)

u ∈ H [N ∩ ∂ Bρ ],

for ρ > 0 sufficiently small unless there is a y ∈ E(λ )\{0} satisfying (14.61). But in the latter case the theorem is proved. Thus, we may assume that (14.87) holds. We can now apply Theorem 3.15 to conclude that there is a sequence {u k } ⊂ D such that (14.88)

G(u k ) → c,

−∞ < c ≤ −ε,

G (u k ) → 0.

Thus u k 2D − 2

(14.89)

F(x, u k )d x → c

and (u k , v) D − ( f (u k ), v) → 0,

(14.90)

v ∈ D.

If ρk = u k D → ∞,

(14.91)

let u˜ k = u k /ρk . Then u˜ k D = 1, and there is a renamed subsequence such that u˜ k → u˜ weakly in D, strongly in L 2 (), and a.e. in . By (14.56), |F(x, u k )|/ρk2 ≤ C(u˜ 2k + |u˜ k |/ρk )

(14.92)

and the right-hand side converges to C u˜ 2 in L 1 (). Moreover, (14.93)

2F(x, u k )/ρk2 → b(u˜ + )2 + a(u˜ − )2 as k → ∞ a.e. in .

Hence, (14.94)

2

F(x, u k )d x/ρk2 → b u˜ + 2 + a u˜ − 2 .

But the left-hand side of (14.94) converges to 1 by (14.89). Thus, u˜ ≡ 0. Now, by (14.10), (14.95)

| f (x, u k )|/ρk ≤ C(|u˜ k | + ρk−1 )

174


and the right-hand side converges to C|u| ˜ in L 2 (). Since f (x, u k )/ρk → b u˜ + − a u˜ − a.e. in as k → ∞,

(14.96) we have

(u, ˜ v) D = (b u˜ + − a u˜ − , v),

(14.97)

v ∈ D,

by (14.90). This says that u˜ satisfies (14.3). Since (a, b) ∈ , we must have u˜ ≡ 0, contradicting the conclusion reached earlier. Thus, (14.91) cannot hold. But then standard arguments imply that (14.1) has a solution u satisfying G(u) = c ≤ −ε. This shows that u ≡ 0, and the proof is complete. Proof of Theorem 14.3. In this case we take A R = [N−1 ∩ B¯ R ] ∪ {u = v + sy1 : v ∈ N−1 , s ≥ 0, u = R},

(14.98)

where y1 is the element of E(λ )\{0} given by Lemma 14.5. By (14.18), there is a continuous map τˆ from N to M such that (14.41) holds (for τˆ ) and G(v + τˆ (v)) = inf G(v + w).

(14.99)

w∈M

If (14.100)

H (v + w) = v + w + τˆ (v),

v ∈ N , w ∈ M ,

then H is a homeomorphism of D onto itself. Thus, H A R links H [N ∩ ∂ Bρ ] for 0 < ρ < R (Example 7 of Section 3.5 and Proposition 3.11). Let τ be the map described in Section 14.2 satisfying (14.41)–(14.45). Then we have G(v + sy1 + τˆ (v + sy1 ))

≤

G(v + sy1 + τ (v + sy1 ))

=

I (v + sy1 + τ (v + sy1 ), a, b) −2 P(x, v + sy1 + τ (v + sy1 ))d x

≤

B1

for v ∈ N−1 , s ≥ 0. This implies G(u) ≤ B1 ,

(14.101)

u ∈ H A R.

Moreover, we have

G(v + τˆ (v))

≤

G(v) = v 2D − 2

≤

v 2D − λ−1 v 2 ≤ 0,

F(x, v) d x v ∈ N−1 .

Also, Lemma 14.9 implies that (14.102)

G(w) ≥ ε > 0,

w ∈ H [M−1 ∩ ∂ Bρ ],


175

for ρ > 0 sufficiently small unless there is a y ∈ E(λ )\{0} satisfying (14.61). The latter option implies the conclusion of the theorem. We may therefore assume that (14.102) holds. We can now apply Theorem 3.15 to conclude that there is a sequence {u k } ⊂ D such that (14.103)

G(u k ) → c,

ε ≤ c < ∞,

G (u k ) → 0.

We now follow the proof of Theorem 14.2 to conclude that there is a solution of (14.1) satisfying G(u) = c ≥ ε. This implies that u ≡ 0, and the proof is complete.

14.5 Notes and remarks Since the work of Fuˇc´ık , many authors have studied other problems of the form (14.1) when (14.2) holds (cf., e.g.,[29]–[30], [52], [54], [55], [58], [56], [63], [64], [72], [73], [80], [81], [82], [90], [105], [122]–[116] and the references quoted in them. In [72], [73], [33], [30] the problem (14.1) was considered for (a, b) ∈ Q when λ is a simple eigenvalue. If ϕ is a corresponding eigenfunction, then it was shown in [72] that for each s ∈ R, there are a unique function u s and a unique constant Cs (a, b) such that (14.104)

− Au s = bu + s − au s + Cs (a, b)ϕ,

(u s , ϕ) = s,

with Cs = Cs (a, b) = =

sC1 ,

s≥0

sC−1 ,

s < 0.

Clearly, (a, b) ∈ iff C1 C−1 = 0. The region where C1 > 0, C−1 > 0 is below both curves, the region C1 < 0, C−1 < 0 is above the curves, and the region C1 C−1 < 0 is between the curves. It is in this sense that problems for regions of type (II) were considered by these authors for simple eigenvalues. Theorems 14.2 and 14.3 were given in [116].

Chapter 15

Weak Sandwich Pairs 15.1 Introduction In Chapter 7 we discussed the situation in which one cannot find linking sets that separate the functional, i.e., satisfy (7.1). Are there sets such that the opposite of (7.1) will imply (3.31)? More precisely, are there sets A, B such that (7.3) implies that there is a sequence satisfying (7.4)? This was answered in the affirmative in Chapter 7. Such pairs exist. This has led to Definition 15.1. We say that a pair of subsets A, B of a Banach space E forms a sandwich, if, for any G ∈ C 1 (E, R), inequality (7.3) implies the existence of a PS sequence (7.4). It follows from Theorem 3.17 that M, N form a sandwich pair if one of them is finite-dimensional. (Note that m 0 ≤ m 1 .) This is a severe drawback in many applications. The purpose of the present chapter is to find a counterpart of sandwich pairs that deals with the case when both sets in the pair are infinite-dimensional. In order to do this, we required weak-to-weak continuous differentiability of the functional as we did in Theorem 15.2. We call such pairs weak sandwiches. The purpose of this chapter is to solve systems of equations of the form (15.1)

Av = f (x, v, w)

(15.2)

Bw = g(x, v, w),

where A, B are linear partial differential operators. The variational approach to solving such a system is to study a functional G(u) chosen so that the system is equivalent to (15.3)

G (u) = 0.

(In very many cases, such a functional can be found.) The sandwich theorem, Theorem 3.17, is very useful in dealing with equations or systems for which the corresponding functional is semibounded in one of the directions only on a subspace of finite M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_15, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009

178

15. Weak Sandwich Pairs

dimension. However, there are many systems for which this is not the case. On the other hand, the theorem is probably not true if both subspaces are infinite-dimensional. In the present chapter we shall show that the theorem is indeed true if we require more than mere continuous differentiability of the functional. The requirement we have chosen is present in many applications. It is the weak-to-weak continuous differentiability defined in Chapter 10 (cf. Definition 10.1). For such functionals, we have Theorem 15.2. Let N be a closed subspace of a Hilbert space E and let M = N ⊥ . Let G be a weak-to-weak continuously differentiable functional on E such that (15.4)

m 0 := inf G(w) = −∞ w∈M

and m 1 := sup G(v) = ∞.

(15.5)

v∈N

Then there are a constant c ∈ R and a sequence {u k } ⊂ E such that (15.6)

G(u k ) → c,

m 0 ≤ c ≤ m 1,

quG (u k ) → 0.

We shall prove Theorem 15.2 in the next section, where we introduce weak sandwich pairs. Applications will be given in Section 15.3.

15.2 Weak sandwich pairs We now introduce the corresponding definition for the case when both sets A, B are infinite-dimensional. Definition 15.3. We shall say that a pair of subsets A, B of a Banach space E forms a weak sandwich pair if, for any weak-to-weak continuously differentiable G ∈ C 1 (E, R), the inequality (15.7)

−∞ < b0 := inf G ≤ a0 := sup G < ∞ B

A

implies that there is a sequence {u k } satisfying (15.8)

G(u k ) → c,

b0 ≤ c ≤ a0 ,

G (u k ) → 0.

We have Theorem 15.4. Let E be a separable Hilbert space, let N be a closed subspace of E, and let p be any point of N. Let F be a Lipschitz continuous map of E onto N such that F| N = I, (15.9)

F(g) − F(h) ≤ K g − h ,

g, h ∈ E,

and, for each finite-dimensional subspace S of E containing p such that F S = {0}, there is a finite-dimensional subspace S0 = {0} of N containing p such that (15.10)

v ∈ S0 , w ∈ S ⇒ F(v + w) ∈ S0 .

Then A = N, B = F −1 ( p) form a weak sandwich pair.

15.2. Weak sandwich pairs

179

Proof. Assume that the theorem is false. Let G be a weak-to-weak continuously differentiable functional on E satisfying (15.7), where A, B are the subsets of E specified in the theorem, such that there is no sequence satisfying (15.8). Then there is a positive number δ > 0 such that G (u) ≥ 2δ

(15.11) whenever u belongs to the set (15.12)

Eˆ = {u ∈ E : b0 − 2δ ≤ G(u) ≤ a0 + 2δ}.

Since E is separable, we can norm it with a norm |u|w satisfying (15.13)

|u|w ≤ u ,

u ∈ E,

and such that the topology induced by this norm is equivalent to the weak topology of E on bounded subsets of E. This can be done as follows. Let {ek } be an orthonormal basis for E. We then set |u|2w =

∞ |(u, ek )|2 k=1

k2

.

˜ For u ∈ E, ˆ let h(u) = G (u)/ G (u) . We denote E equipped with this norm by E. Then, by (15.11), (15.14)

(G (u), h(u)) ≥ 2δ,

ˆ u ∈ E.

Let T = (a0 − b0 + 4δ)/δ, B R = {u ∈ E : u < R}, (15.15)

R = sup u + T,

ˆ Bˆ = B¯ R ∩ E, where is a bounded, open subset of N containing the point p such that (15.16)

ρ(∂, p) > K T + δ,

ˆ there is an E˜ neighborhood W (u) of u such and ρ is the distance in E. For each u ∈ B, that (15.17)

(G (v), h(u)) > δ,

ˆ v ∈ W (u) ∩ B.

Otherwise there would be a sequence {v k } ⊂ Bˆ such that (15.18)

|v k − u|w → 0

and

(G (v k ), h(u)) ≤ δ.

180


Since Bˆ is bounded in E, v k → u weakly in E and (10.5) implies that (15.19)

(G (v k ), h(u)) → (G (u), h(u)) ≥ 2δ

in view of (15.14). This contradicts (15.17). Let B˜ be the set Bˆ with the inherited ˜ It is a metric space, and W (u) ∩ B˜ is an open set in this space. Thus, topology of E. ˜ ˜ is an open covering of the paracompact space B˜ (cf., e.g., [77]). {W (u) ∩ B}, u ∈ B, Consequently, there is a locally finite refinement {Wτ } of this cover. For each τ, there is an element u τ such that Wτ ⊂ W (u τ ). Let {ψτ } be a partition of unity subordinate to this covering. Each ψτ is locally Lipschitz continuous with respect to the norm |u|w and consequently with respect to the norm of E. Let ˜ (15.20) Y (u) = ψτ (u)h(u τ ), u ∈ B. Then Y (u) is locally Lipschitz continuous with respect to both norms. Moreover, (15.21) Y (u) ≤ ψτ (u) h(u τ ) ≤ 1 and (15.22)

(G (u), Y (u)) =

ψτ (u)(G (u), h(u τ )) ≥ δ,

ˆ u ∈ B.

ˆ let σ (t)u be the solution of ¯ ∩ E, For u ∈ (15.23)

σ (t) = −Y (σ (t)),

t ≥ 0,

σ (0) = u.

ˆ Moreover, it is continuous in (u, t) Note that σ (t)u will exist as long as σ (t)u is in B. with respect to both topologies. ˆ we cannot have σ (t)u ∈ Bˆ and G(σ (t)u) > b0 −δ ¯ ∩ E, Next, we note that if u ∈ for 0 ≤ t ≤ T : for by (15.23), (15.22), (15.24)

d G(σ (t)u)/dt = (G (σ ), σ ) = −(G (σ ), Y (σ )) ≤ −δ

ˆ Hence, if σ (t)u ∈ Bˆ for 0 ≤ t ≤ T , we would have as long as σ (t)u ∈ B. (15.25)

G(σ (T )u) − G(u) ≤ −δT = −(a0 − b0 + 4δ).

Thus, we would have G(σ (T )u) < b0 − 4δ. On the other hand, if σ (s)u exists for ˆ To see this, note that 0 ≤ s < T, then σ (t)u ∈ B. t Y (σ (s)u)ds. (15.26) u − σ (t)u = z t (u) := 0

By (15.21), (15.27)

z t (u) ≤ t.


σ (t)u ≤ u + t < R.

15.2. Weak sandwich pairs

181

ˆ We can now conclude that for each u ∈ ˆ there is a t ≥ 0 such ¯ ∩ E, Thus, σ (t)u ∈ B. that σ (s)u exists for 0 ≤ s ≤ t and G(σ (t)u) ≤ b0 − δ. Let (15.29)

Tu := inf{t ≥ 0 : G(σ (t)u) ≤ b0 − δ},

ˆ ¯ ∩ E. u∈

Then σ (t)u exists for 0 ≤ t ≤ Tu and Tu < T . Moreover, Tu is continuous in u. Define σ (t)u, 0 ≤ t ≤ Tu , σˆ (t)u = σ (Tu )u, Tu ≤ t ≤ T, ˆ For u ∈ ˆ define σˆ (t)u = u, 0 ≤ t ≤ T . Then σˆ (t)u is continuous ¯ ∩ E. ¯ \ E, for u ∈ in (u, t), and G(σˆ (T )u) ≤ b0 − δ,

(15.30)

¯ u ∈ .

Let (15.31)

ϕ(v, t) = F σˆ (t)v,

¯ v ∈ ,

0 ≤ t ≤ T.

¯ × [0, T ] to N. Let Then ϕ is a continuous map of ¯ t ∈ [0, T ]}. K = {(u, t) : u = σˆ (t)v, v ∈ Q, Then K is a compact subset of E˜ × R. To see this, let (u k , tk ) be any sequence in K . ¯ Since Q is bounded, there is a subsequence such Then u k = σ (tk )v k , where v k ∈ Q. that v k → v 0 weakly in E and tk → t0 in [0, T ]. Since Q¯ is convex and bounded, v 0 is in Q¯ and |v k − v 0 |w → 0. Since σˆ (t) is continuous in E˜ × R, we have u k = σˆ (tk )v k σˆ (t0 )v 0 = u 0 ∈ K . Each u 0 ∈ Bˆ has a neighborhood W (u 0 ) in E˜ and a finite-dimensional subspace S(u 0 ) ˆ Since σˆ (t)u is continuous in (u, t), for each such that Y (u) ⊂ S(u 0 ) for u ∈ W (u 0 )∩ B. (u 0 , t0 ) ∈ K , there are a neighborhood W (u 0 , t0 ) ⊂ E˜ × R and a finite-dimensional subspace S(u 0 , t0 ) ⊂ E such that zˆ t (u) ⊂ S(u 0 , t0 ) for (u, t) ∈ W (u 0 , t0 ), where t ˆ Y (σˆ (s)u)ds, u ∈ E, (15.32) zˆ t (u) := u − σˆ (t)u = 0 ˆ 0, u ∈ E. Since K is compact, there are a finite number of points (u j , t j ) ⊂ K such that K ⊂ W = ∪W (u j , t j ). Let S be a finite-dimensional subspace of E containing p and all ¯ we have zˆ t (v) ∈ S. the S(u j , t j ) and such that F S = {0}. Then, for each v ∈ , By hypothesis, there is a finite-dimensional subspace S0 = {0} of N containing p such ¯ ∩ S0 . We note that ϕ(u, t) maps ¯ ∩ S0 × [0, T ] that F(v − zˆ t (v)) ∈ S0 for all v ∈ into S0 . For t in [0, T ], let ϕt (v) = ϕ(v, t). Then (15.33)

ϕt (v) = p,

v ∈ ∂( ∩ S0 ) = ∂ ∩ S0 ,

0 ≤ t ≤ T.

To see this, note that if v ∈ ∂, then v − p ≤ v − F σˆ (t)v + F σˆ (t)v − p .

182


Hence, (15.34)

F σˆ (t)v − p > K T + δ − t K > 0,

since F σˆ (t)v − v ≤ K

t

v ∈ ∂, 0 ≤ t ≤ T

σˆ (s)v ds ≤ K t.

0

Thus, (15.33) holds. Consequently, the Brouwer degree d(ϕt , ∩ S0 , p) is defined. Since ϕt is continuous, we have (15.35)

d(ϕT , ∩ S0 , p) = d(ϕ0 , ∩ S0 , p) = d(I, ∩ S0 , p) = 1.

Hence, there is a v ∈ such that F σˆ (T )v = p. Consequently, σˆ (T )v ∈ F −1 ( p) = B. In view of (15.7), this implies G(σˆ (T )v) ≥ b0 , contradicting (15.30). Thus, (15.8) holds, and the proof is complete. We can now give the proof of Theorem 15.2. Proof. We take A = N, B = M, p = 0, and F = PN , the projection onto N. If S is a finite-dimensional subspace such that F S = {0}, we take S0 = F S. All of the hypotheses of Theorem 15.4 are satisfied. Definition 15.5. Let E, F be Banach spaces. We shall call a map J ∈ C(E, F) weakto-weak continuous if, for each sequence (15.36)

u k → u weakly in E,

there exists a renamed subsequence such that (15.37)

J (u k ) → J (u) weakly in F.

We have Proposition 15.6. If A, B is a weak sandwich pair and J is a weak-to-weak continuous diffeomorphism on the entire space having a derivative J (u) depending compactly on u and satisfying (15.38)

J (u)−1 ≤ C,

u ∈ E,

then JA, JB is a weak sandwich pair. Proof. Let G be a weak-to-weak continuously differentiable functional on E satisfying (15.39)

−∞ < b0 := inf G ≤ a0 := sup G < ∞. JB

JA

G 1 (u) = G(J u),

u ∈ E.

Let

15.2. Weak sandwich pairs Then

183

(G 1 (u), h) = (G (J u), J (u)h).

If u k → u weakly, then there is a renamed subsequence such that J (u k ) → J (u).

J (u k ) → J (u) weakly; Hence,

(G 1 (u k ), h) → (G (J u), J (u)h),

and G 1 is weak-to-weak continuously differentiable. Moreover, (15.40)

−∞ < b0 := inf G = inf G(J u) = inf G 1 JB

J u∈ J B

B

≤ a0 := sup G = sup G(J u) = sup G 1 < ∞. JA

J u∈ J A

A

Since A, B form a weak sandwich pair, there is a sequence {h k } ⊂ E such that (15.41)

G 1 (h k ) → c,

b0 ≤ c ≤ a0 ,

G 1 (h k ) → 0.

If we set u k = J h k , this becomes (15.42)

G(u k ) → c,

b0 ≤ c ≤ a0 ,

G (u k )J (h k ) → 0.

In view of (15.38), this implies G (u k ) → 0. Thus, J A, J B is a sandwich pair. Proposition 15.7. Let N be a closed subspace of a Hilbert space E with complement M = M ⊕{v 0 }, where v 0 is an element in E having unit norm, and let δ be any positive number. Let ϕ(t) ∈ C 1 (R) be such that 0 ≤ ϕ(t) ≤ 1,

ϕ(0) = 1,

and ϕ(t) = 0,

|t| ≥ 1.

Let (15.43) F(v + w + sv 0 ) = v + [s + δ − δϕ( w 2 /δ 2 )]v 0 ,

v ∈ N, w ∈ M, s ∈ R.

Then A = N = N ⊕ {v 0 }, B = F −1 (δv 0 ) forms a weak sandwich pair. Proof. Define J (v + w + sv 0 ) = v + w + [s − δ + δϕ( w 2 /δ 2 )]v 0 ,

v ∈ N, w ∈ M, s ∈ R.

Then J is a diffeomorphism on E satisfying the hypotheses of Proposition 15.6. Moreover, A = J N and B = J [M + δv 0 ]. Since N and M + δv 0 form a weak sandwich pair by Theorem 15.2, we see that A, B also form a weak sandwich pair (Proposition 15.6).

184


15.3 Applications Let A, B be positive, self-adjoint operators on L 2 () with compact resolvents, where ⊂ Rn . Let F(x, v, w) be a Carathéodory function on × R2 such that (15.44)

f (x, v, w) = ∂ F/∂v,

g(x, v, w) = ∂ F/∂w

are also Carathéodory functions satisfying (15.45)

| f (x, v, w)| + |g(x, v, w)| ≤ C0 (|v| + |w| + 1),

v, w ∈ R,

and (15.46)

f (x, t y, tz)/t → α+ (x)v + − α− (x)v − + β+ (x)w+ − β− (x)w− ,

(15.47)

g(x, t y, tz)/t → γ+ (x)v + − γ− (x)v − + δ+ (x)w+ − δ− (x)w−

as t → +∞, y → v, z → w, where a ± = max(±a, 0). We wish to solve the system (15.48)

Av = − f (x, v, w)

(15.49)

Bw = g(x, v, w).

Let λ0 (μ0 ) be the lowest eigenvalue of A(B). We assume that the only solution of (15.50)

−Av = α+ v + − α− v − + β+ w+ − β− w− ,

(15.51)

Bw = γ+ v + − γ− v − + δ+ w+ − δ− w−

is v = w = 0. Our first result is Theorem 15.8. Assume (15.52)

2F(x, s, 0) ≥ −λ20 − W1 (x),

x ∈ , t ∈ R,

2F(x, 0, t) ≤ μ0 t 2 + W2 (x),

x ∈ , t ∈ R,

and (15.53)

where Wi (x) ∈ L 1 (). Then the system (15.48), (15.49) has a solution. Proof. Let D = D(A1/2 ) × D(B 1/2 ). Then D becomes a Hilbert space with norm given by (15.54)

u 2D = (Av, v) + (Bw, w),

We define (15.55)

u = (v, w) ∈ D.

G(u) = b(w) − a(v) − 2

F(x, v, w)d x,

where (15.56)

a(v) = (Av, v),

b(w) = (Bw, w).

u ∈ D,

15.3. Applications

185

Then G ∈ C 1 (D, R) and (15.57)

(G (u), h)/2 = b(w, h 2 ) − a(v, h 1 ) − ( f (u), h 1 ) − (g(u), h 2 ),

where we write f (u), g(u) in place of f (x, v, w), g(x, v, w), respectively. It is readily seen that the system (15.48), (15.49) is equivalent to G (u) = 0.

(15.58)

We let N be the set of those (v, 0) ∈ D and M the set of those (0, w) ∈ D. Then M, N are orthogonal closed subspaces such that D = M ⊕ N.

(15.59) If we define (15.60)

Lu = 2(−v, w),

u = (v, w) ∈ D,

then L is a self-adjoint, bounded operator on D. Also, G (u) = Lu + c0 (u),

(15.61) where (15.62)

c0 (u) = −(A−1 f (u), B −1 g(u))

is compact on D. This follows from (15.45) and the fact that A and B have compact resolvents. It also follows that G has weak-to-weak continuity, for if u k → u weakly, then Lu k → Lu weakly and c0 (u k ) has a convergent subsequence. Now, by (15.53), (15.63) G(0, w) ≥ b(w) − μ0 w 2 − W2 (x) d x, (0, w) ∈ M.

Thus, (15.64)

inf G ≥ − M

W (x) d x ≡ b0 .

On the other hand, (15.52) implies (15.65)

G(v, 0) ≤ −a(v) + λ0 v + 2

Thus, (15.66)

W1 (x) d x,

(v, 0) ∈ N.

sup G ≤ N

W1 (x) d x ≡ a0 .

We can now apply Theorem 15.2 to conclude that there is a sequence {u k } ⊂ D such that (15.8) holds. Let u k = (v k , wk ). I claim that (15.67)

ρk2 = a(v k ) + b(wk ) ≤ C,

186


for assume that ρk → ∞, and let u˜ k = u k /ρk . Then there is a renamed subsequence such that u˜ k → u˜ weakly in D, strongly in L 2 (), and a.e. in . If h = (h 1 , h 2 ) ∈ D, then (15.68) ˜ k , h 2 ) − 2a(v˜k , h 1 ) − 2( f (u k ), h 1 )/ρk − 2(g(u k ), h 2 )/ρk . (G (u k ), h)/ρk = 2b(w Taking the limit and applying (15.45)–(15.47), we see that u˜ = (v, ˜ w) ˜ is a solution of (15.50), (15.51). Hence, u˜ = 0 by hypothesis. On the other hand, since a(v˜k ) + b(w˜ k ) = 1, there is a renamed subsequence such that a(v˜k ) → a, ˜ b(w˜ k ) → b˜ with ˜ a˜ + b = 1. Thus, by (15.46), (15.47), and (15.57), (G (u k ), (v˜k , 0))/2ρk = −a(v˜k ) − ( f (u k ), v˜k )/ρk → −a˜ − (α+ v˜ + − α− v˜ − + β+ w˜ + − β− w˜ − )v˜ d x

and (G (u k ), (0, w˜ k ))/2ρk = b(w˜ k ) − (g(u k ), w˜ k )/ρk ˜ → b − (γ+ v˜ + − γ− v˜ − + δ+ w˜ + − δ− w˜ − )w˜ d x.

Thus, by (15.8), (15.69)

a˜ = −

and (15.70)

b˜ =

(α+ v˜ + − α− v˜ − + β+ w˜ + − β− w˜ − )vd ˜ x

(γ+ v˜ + − γ− v˜ − + δ+ w˜ + − δ− w˜ − )w˜ d x.

Since one of the two numbers a, ˜ b˜ is not zero, we see that we cannot have u˜ ≡ 0. This contradiction proves (15.67). Once this is known, we can use the usual procedures to show that there is a renamed subsequence such that u k → u in D, and u satisfies (15.58). Theorem 15.9. In addition, assume that the eigenfunctions of λ0 and μ0 are bounded and nonzero a.e. in and that there is a q > 2 such that (15.71)

w 2q ≤ Cb(w),

w ∈ M.

Assume (15.72)

2F(x, 0, t) ≤ μ(x)t 2 ,

x ∈ , t ∈ R,

where (15.73)

μ(x) ≤≡ μ0 ,

x ∈ ,

and for some δ > 0, (15.74)

2F(x, s, t) ≤ μ0 t 2 − λ0 s 2 ,

|t| + |s| ≤ δ.

Then the system (15.48), (15.49) has a nontrivial solution.

15.3. Applications

187

Proof. Let N be the orthogonal complement of N0 = {ϕ0 } in N, where ϕ0 is the eigenfunction of A corresponding to λ0 . Then N = N ⊕ N0 . Let M0 be the subspace of M spanned by the eigenfunctions of B corresponding to μ0 , and let M be its orthogonal complement in M. Since N0 and M0 are contained in L ∞ (), there is a positive constant ρ such that (15.75)

a(y) ≤ ρ 2 ⇒ y ∞ ≤ δ/4,

y ∈ N0 ,

(15.76)

b(h) ≤ ρ 2 ⇒ h ∞ ≤ δ/4,

h ∈ M0 ,

where δ is the number given in (15.74). If (15.77)

a(y) ≤ ρ 2 ,

b(w) ≤ ρ 2 , |y(x)| + |w(x)| ≥ δ,

we write w = h + w , h ∈ M0 , w ∈ M and (15.78)

δ ≤ |y(x)| + |w(x)| ≤ |y(x)| + |h(x)| + |w (x)| ≤ (δ/2) + |w (x)|.

Thus, |y(x)| + |h(x)| ≤ δ/2 ≤ |w (x)|

(15.79) and

|y(x)| + |w(x)| ≤ 2|w (x)|.

(15.80)

Now, by (15.74) and (15.80), G(y, w) = b(w) − a(y) − 2

F(x, y, w) d x

≥ b(w) − a(y) − − c0

{μ0 w2 − λ0 y 2 }d x

|y|+|w|δ

≥ b(w) − a(y) − μ0 w 2 + λ0 y 2 − c1

|w |q d x

2|w |>δ

≥ b(w ) − μ0 w 2 − c2 b(w )q/2 μ0 (q/2)−1 b(w ), ≥ 1− − c2 b(w ) μ1

a(y) ≤ ρ 2 , b(w) ≤ ρ 2 ,

where μ1 is the next eigenvalue of B after μ0 . If we reduce ρ accordingly, we can find a positive constant ν such that (15.81)

G(y, w) ≥ νb(w ),

a(y) ≤ ρ 2 , b(w) ≤ ρ 2 .

188


I claim that either (15.48), (15.49) has a nontrivial solution or there is an > 0 such that G(y, w) ≥ ,

(15.82)

a(y) + b(w) = ρ 2 .

To see this, suppose (15.82) did not hold. Then there would be a sequence {yk , wk } such that a(yk ) + b(wk ) = ρ 2 and G(yk , wk ) → 0. If we write wk = wk + h k , wk ∈ M , h k ∈ M0 , then (15.81) tells us that b(wk ) → 0. Thus, a(yk ) + b(h k ) → ρ 2 . Since N0 , M0 are finite-dimensional, there is a renamed subsequence such that yk → y in N0 and h k → h in M0 . By (15.75) and (15.76), y ∞ ≤ δ/4 and h ∞ ≤ δ/4. Consequently, (15.74) implies 2F(x, y, h) ≤ μ0 h 2 − λ0 y 2 .

(15.83) Since (15.84)

G(y, h) = b(h) − a(y) − 2

we have

F(x, y, h)d x = 0,

(15.85)

{2F(x, y, h) + λ0 y 2 − μ0 h 2 }d x = 0.

In view of (15.83), this implies 2F(x, y, h) ≡ μ0 h 2 − λ0 y 2 .

(15.86)

For ζ ∈ C0∞ () and t > 0 small, we have (15.87)

2[F(x, y + tζ, h) − F(x, y, h)]/t ≤ −λ0 [(y + tζ )2 − y 2 ]/t.

Taking t → 0, we see that f (x, y, h)ζ ≤ −λ0 yζ.

(15.88)

Since this is true for all ζ ∈ C0∞ (), we have f (x, y, h) = −λ0 y = −Ay.

(15.89) Similarly, (15.90)

2[F(x, y, h + tζ ) − F(x, y, h)]/t ≤ μ0 [(h + tζ )2 − h 2 ]/t,

and, consequently, (15.91)

g(x, y, h)ζ ≤ μ0 hζ

and (15.92)

g(x, y, h) = μ0 h = Bh.

15.3. Applications

189

We see from (15.89) and (15.92) that (15.48), (15.49) has a nontrivial solution. Thus, we may assume that (15.82) holds. Next, we note that there is an ε > 0 depending on ρ such that G(0, w) ≥ ε,

b(w) ≥ ρ > 0.

To see this, suppose that {wk } ⊂ M is a sequence such that G(0, wk ) → 0,

b(wk ) ≥ ρ.

If bk = b(wk ) ≤ C, this implies b(wk ) − μ0 wk 2 → 0 and

[μ0 − μ(x)]wk2 d x → 0

since G(0, w) ≥ b(w) − μ0 w 2 +

[μ0 − μ(x)]w2 d x,

w ∈ M.

If we write wk = wk + h k , wk ∈ M , h k ∈ M0 as before, then this tells us that b(wk ) → 0. Since M0 is finite-dimensional, there is a renamed subsequence such that h k → h. But the two conclusions above tell us that h = 0. Since b(h) ≥ ρ, we see that ε > 0 exists for any constant C. If the sequence {bk } is not bounded, we take 1/2 w˜ k = wk /bk . Then ˜ k )2 d x. G(0, wk )/bk ≥ b(w˜ k ) − μ0 w˜ k 2 + [μ0 − μ(x)](w Next, we note that there is a ν > 0 such that (15.93)

G(0, w) ≥ νb(w),

w ∈ M.

Assuming this for the moment, we see that (15.94)

inf G ≥ ε1 > 0, B

where (15.95) B = {w ∈ M : b(w) ≥ ρ 2 } ∪ {u = (sϕ0 , w) : s ≥ 0, w ∈ M, u D = ρ}, and ε1 = min{ε, νρ 2 }. By (15.66), there is an R > ρ such that (15.96)

sup G = a0 < ∞, A

where A = N. By Proposition 15.7, A, B form a weak sandwich pair. Moreover, G satisfies (15.7) with ε1 ≤ b0 . Hence, there is a sequence {u k } ⊂ D such that (15.8)

190


holds with c ≥ ε1 . Arguing as in the proof of Theorem 15.8, we see that there is a u ∈ D such that G(u) = c ≥ ε1 > 0, G (u) = 0. Since c = 0 and G(0) = 0, we see that u = 0, and we have a nontrivial solution of the system (15.48), (15.49). It therefore remains only to prove (15.93). Clearly, ν ≥ 0. If ν = 0, then there is a sequence {wk } ⊂ M such that G(0, wk ) → 0,

(15.97)

b(wk ) = 1.

Thus, there is a renamed subsequence such that wk → w weakly in M, strongly in L 2 (), and a.e. in . Consequently, (15.98) [μ0 − μ(x)]wk2 d x ≤ 1 − μ(x)wk2 d x ≤ G(0, wk ) → 0

and (15.99)

1=

μ(x)w2 d x ≤ μ0 w 2 ≤ b(w) ≤ 1,

which means that we have equality throughout. It follows that we must have w ∈ E(μ0 ), the eigenspace of μ0 . Since w ≡ 0, we have w = 0 a.e. But (15.100) [μ0 − μ(x)]w2 d x = 0

implies that the integrand vanishes identically on , and consequently μ(x) ≡ μ0 , violating (15.73). This establishes (15.93) and completes the proof of the theorem.

15.4 Notes and remarks Weak sandwich pairs were introduced in [133].

Chapter 16

Multiple Solutions 16.1 Introduction Nonlinear problems are characterized by the fact that very often solutions are not unique. However, it is sometimes quite difficult to show that more than one solution exists. In this respect, variational methods can be very helpful. We have chosen some examples that illustrate this point.

16.2 Two examples Let be a smooth, bounded domain in Rn , and let A be a self-adjoint operator on L 2 (). We assume that C0∞ () ⊂ D := D(|A|1/2 ) ⊂ H T ,2()

(16.1)

holds for some T > 0 (T need not be an integer), and the eigenvalues of A satisfy 0 < λ0 < λ1 < · · · < λk < · · · . Let f (x, t) be a Carathéodory function on × R. We assume that the function f (x, t) satisfies | f (x, t)| ≤ C(|t| + 1),

(16.2)

x ∈ , t ∈ R.

We shall prove Theorem 16.1. Assume that for some integers l < m, the following inequalities hold: (16.3)

t[ f (x, t1 ) − f (x, t0 )] ≤ at 2 ,

t j ∈ R, t = t1 − t0 ,

where a < λm+1 , (16.4)

a0 t 2 ≤ 2F(x, t) ≤ a1 t 2 ,

|t| < δ,


192

16. Multiple Solutions

for some δ > 0, with λl < a0 ≤ a1 < λl+1 , where t F(x, t) := f (x, s)ds, 0

and a2 t 2 − W1 (x) ≤ 2F(x, t),

(16.5)

|t| > K ,

for some K ≥ 0, where a2 > λm and W1 ∈ L 1 (). Then the equation Au = f (x, u),

(16.6)

u ∈ D,

has at least two nontrivial solutions. Theorem 16.2. Equation (16.6) will have at least two nontrivial solutions if we assume that for some integers l > m, the following inequalities hold: (16.7)

t[ f (x, t1 ) − f (x, t0 )] ≥ at 2 ,

t j ∈ R, t = t1 − t0 ,

where a > λm , a0 t 2 ≤ 2F(x, t) ≤ a1 t 2 ,

(16.8)

|t| < δ,

for some δ > 0, with λl < a0 ≤ a1 < λl+1 , 2F(x, t) ≤ a2 t 2 + W2 (x),

(16.9)

|t| > K ,

for some K ≥ 0, where a2 < λm+1 and W2 ∈ L 1 (). More comprehensive theorems will be presented in the next section. Proofs will be given in Section 16.6.

16.3 Statement of the theorems We use the notation a(u, v) = (Au, v),

a(u) = a(u, u),

u, v ∈ D.

We define (16.10)

u D := A1/2 u

and G(u) := u 2D − 2

F(x, u)d x.

It is known that G is a continuously differentiable functional on the whole of D e.g., [120]) and (G (u), v) D = 2(u, v) D − 2( f (u), v),

(cf.,

16.3. Statement of the theorems

193

where we write f (u) in place of f (x, u(x)). In connection with the operator A, the following quantities are very useful. For each fixed positive integer , we let N denote the subspace of D spanned by the eigenfunctions corresponding to λ0 , . . . , λ , and let M = N⊥ ∩ D. Then D = M ⊕ N . For real a, b, we define I (u, a, b) = (Au, u) − a u − 2 − b u + 2 , γ (a) = sup{I (v, a, 0) : v ∈ N , v + = 1}, (a) = inf{I (w, a, 0) : w ∈ M , w+ = 1}, F1 (w, a, b) = sup{I (v + w, a, b) : v ∈ N }, F2 (v, a, b) = inf{I (v + w, a, b) : w ∈ M }, M (a, b) = inf{F1 (w, a, b) : w ∈ M , w D = 1}, m (a, b) = sup{F2 (v, a, b) : v ∈ N , v D = 1}, ν (a) = sup{b : M (a, b) ≥ 0}, μ (a) = inf{b : m (a, b) ≤ 0}, where u ± (x) = max{±u(x), 0}. Our first result is Theorem 16.3. Assume that for some integers l < m, the following inequalities hold: (16.11)

t[ f (x, t1 ) − f (x, t0 )] ≤ a(t − )2 + b(t + )2 ,

t j ∈ R, t = t1 − t0 ,

where b < m (a), (16.12)

a0 (t − )2 + b0 (t + )2 ≤ 2F(x, t) ≤ a1 (t − )2 + b1 (t + )2 ,

|t| < δ,

for some δ > 0, with a0 , b0 < λl+1 , a1 , b1 > λl , b0 > μl (a0 ), and b1 < νl (a1 ), (16.13)

a2 (t − )2 + b2 (t + )2 − W1 (x) ≤ 2F(x, t),

|t| > K ,

for some K ≥ 0, where a2 , b2 < λm+1 , b2 > μm (a2 ), and W1 ∈ L 1 (). Then (16.6) has at least two nontrivial solutions. In contrast to this, we have Theorem 16.4. Equation (16.6) will have at least two nontrivial solutions if we assume that for some integers l > m, the following inequalities hold: (16.14)

t[ f (x, t1 ) − f (x, t0 )] ≥ a(t − )2 + b(t + )2 ,

t j ∈ R, t = t1 − t0 ,

where b > γm (a), (16.15)

a0 (t − )2 + b0 (t + )2 ≤ 2F(x, t) ≤ a1 (t − )2 + b1 (t + )2 ,

|t| < δ

for some δ > 0, with a0 , b0 < λl+1 , b0 > μl (a0 ) and a1 , b1 > λl , b1 < νl (a1 ), (16.16)

2F(x, t) ≤ a2 (t − )2 + b2 (t + )2 + W2 (x),

|t| > K ,

for some K ≥ 0, where a2 , b2 > λm , b2 < νm (a2 ), and W2 ∈ L 1 ().

194


It was shown in [114] that the functions γl , μl , νl−1 , l−1 all emanate from the point (λl , λl ) and satisfy l−1 (a) ≤ νl−1 (a) ≤ μl (a) ≤ γl (a) on their common domains. It would therefore give a weaker result if we assumed in Theorems 16.3 and 16.4 that b0 > γl (a0 ) and b1 < l (a1 ). However, the functions γl , l are defined on the whole of R, while the others are not. For cases in which the other functions are not defined, we state the following. Theorem 16.5. Theorems 16.3 and 16.4 remain true if we assume that (16.12) holds with b0 > γl (a0 ), and b1 < l (a1 ) for some a0 , a1 ∈ R.

16.4 Some lemmas The proofs of the theorems of the previous section will be based on a series of lemmas. Lemma 16.6. If b < l (a), then there is an > 0 such that I (w, a, b) ≥ w 2D ,

(16.17)

w ∈ Ml .

Proof. By the continuity of l , there is a t < 1 such that b/t < l (a/t). Then I (w, a/t, b/t) = w 2D −

a − 2 b + 2 w − w ≥ 0, t t

w ∈ Ml .

Therefore, I (w, a, b) = t

w 2D

a − 2 b + 2 − w − w + (1 − t) w 2D ≥ (1 − t) w 2D . t t

Lemma 16.7. If b > γl (a), then there is an > 0 such that I (v, a, b) ≤ − v 2D ,

(16.18)

v ∈ Nl .

Proof. By the continuity of γl , there is a t > 1 such that b/t > γl (a/t). Hence, I (v, a/t, b/t) = v 2D −

a − 2 b + 2 v − v ≤ 0, t t

v ∈ Nl ,

and I (v, a, b) = t

v 2D

a − 2 b + 2 − v − v + (1 − t) v 2D ≤ (1 − t) v 2D . t t

16.4. Some lemmas

195

Lemma 16.8. If (16.19)

t[ f (x, t1 ) − f (x, t0 )] ≤ a(t − )2 + b(t + )2 ,

t j ∈ R, t = t1 − t0 ,

then (16.20) (G (v + w1 ) − G (v + w0 ), w) ≥ 2I (w, a, b),

v, w j ∈ D, w = w1 − w0 .

Proof. We have ( f (x, v + w1 ) − f (x, v + w0 ), w) ≤ a w− 2 + b w+ 2 . Hence, (G (v + w1 ) − G (v+w0 ), w)/2 = w 2D − ( f (x, v + w1 ) − f (x, v + w0 ), w) ≥ I (w, a, b).

Lemma 16.9. Suppose there are closed subspaces X,Y of E such that E = X ⊕ Y, and (16.21) (G (x + y1 ) − G (x + y2 ), y1 − y2 ) ≥ m( y1 − y2 ),

x ∈ X, y1 , y2 ∈ Y,

for some function m(t) satisfying (a) m(t) > 0,

t > 0,

(b) there is a t0 > 0 such that

t0 0

m(t) dt < ∞, t

(c) m(t) → 0 i mpli es t and (d) 1 M

M 0

m(t) dt → ∞ t

as

t → 0,

M → ∞.

Then there exists a continuous function ϕ : X → Y satisfying (i) G(x + ϕ(x)) = min y∈Y G(x + y). ˜ (ii) The function G(x) = G(x + ϕ(x)) : X → R is of class C 1 and satisfies (16.22)

(G˜ (x), h) = (G (x + ϕ(x)), h),

h ∈ X.

196


Proof. For each x ∈ X, we define G x : Y → R by G x (y) = G(x + y). From (16.21), we have (G x (y1 ) − G x (y2 ), y1 − y2 ) ≥ m( y1 − y2 ). In particular, G x (y) is strictly convex (Theorem 13.7). Thus, for each x ∈ X, G x (y) has a unique critical point ϕ(x). To see this, note that (G x (y) − G x (0), y) ≥ m( y ), Thus,

y ∈ Y.

(G x (y), y) ≥ (G x (0), y) + m( y ).

Consequently, G x (y) − G x (0) =

d G x (t y) dt dt

1

1 (G (t y), t y) dt t x

0

=

1

0 1

1 [(G x (0), t y) + m( t y )] dt 0 t 1 1 m( t y ) dt = (G x (0), y) + 0 t y m(s) ds = (G x (0), y) + s 0

≥

→ ∞ as y → ∞. Hence, G x (y) has a unique minimum ϕ(x) (Lemma 13.5). Moreover, G x (ϕ(x)) = min G x (y) = min G(x + y). y∈Y

y∈Y

Therefore, ϕ(x) is the only element of Y such that (16.23)

0 = (G x (ϕ(x)), y) = (G (x + ϕ(x)), y),

y ∈ Y.

Note that ϕ is continuous. Suppose, on the contrary, that {x n } is a sequence in X such that x n → x ∈ X. Let tn = ϕ(x n ) − ϕ(x) . Since G is continuous, for each ε > 0 and n sufficiently large, we have |(G (x n + ϕ(x)), y)| < ε y ,

(16.24)

y ∈ Y.

Because of (16.21), we obtain (16.25)

(G (x n + ϕ(x)) − G (x n + ϕ(x n )), ϕ(x) − ϕ(x n )) ≥ m(tn ).

Since, by (16.23),

(G (x n + ϕ(x n )), y) = 0,

y ∈ Y,

16.4. Some lemmas

197

inequalities (16.24) and (16.25) imply m(tn )/tn < ε. Thus, m(tn )/tn → 0, and consequently, tn → 0. Hence, ϕ is continuous. This proves part (i). For t > 0 and h ∈ X, we have (16.26)

˜ + th) − G(x) ˜ G(x + th + ϕ(x + th)) − G(x + ϕ(x)) G(x = t t G(x + th + ϕ(x)) − G(x + ϕ(x)) ≤ t 1 (G (x + ϕ(x) + sth), h) ds. = 0

In a similar manner, we see that ˜ + th) − G(x) ˜ G(x + th + ϕ(x + th)) − G(x + ϕ(x)) G(x = t t G(x + th + ϕ(x + th)) − G(x + ϕ(x + th)) ≥ t 1 (G (x + ϕ(x + th) + sth), h) ds. ≥ 0

Therefore, since G and ϕ are continuous, we have lim

t →0+

˜ + th) − G(x) ˜ G(x = (G (x + ϕ(x)), h). t

This shows that G˜ has a continuous Gateaux derivative and hence is of class C 1 . From the above, we see that (16.27)

(G˜ (x), h) = (G (x + ϕ(x)), h),

h ∈ X.

Lemma 16.10. Let M, N be closed subspaces of a Banach space E such that dim N < ∞ and E = M ⊕ N. Let w0 = 0 be an element of M, and take K0 B

= {v ∈ N : v ≤ R} ∪ {sw0 + v : v ∈ N, s ≥ 0, sw0 + v = R}, = ∂ Bδ ∩ M, 0 < δ < R.

Let Q = {sw0 + v : v ∈ N, s ≥ 0, sw0 + v ≤ R},

198


Then K 0 = ∂ Q. Let ξ(u) be a continuous map from Q to E such that ξ(v) = v,

v ∈ N ∩ BR

and ξ(u) ≥ r > δ > 0,

u ∈ K0 ∩ ∂ BR .

Then ξ(Q) ∩ B = φ. Proof. Let P be the projection of E onto N via M. Then the conclusion of the lemma is true iff there is a u ∈ Q such that S(u) := Pξ(u) + ξ(u) w0 = δw0 . Let ξt (u) = (1 − t)ξ(u) + tu and St (u) = Pξt (u) + [(1 − t) ξ(u) + t u ]w0 . Then S0 = S and

S1 (u) = Pu + u w0 .

Clearly, there is a unique u ∈ Q such that S1 (u) = δw0 . The St map Q into E, but there is no point on ∂ Q that gets mapped into δw0 . If v ∈ N ∩ B R , then St (v) = v + v w0 , which cannot equal δw0 . If u ∈ ∂ Q\N, then [(1 − t) ξ(u) + t u ] ≥ r > δ. Thus, no such point can be mapped into δw0 . Hence, the Brouwer degree d(St , Q, δw0 ) is defined for t ∈ [0, 1], and d(S0 , Q, δw0 ) = d(S1 , Q, δw0 ) = 1. This completes the proof. Corollary 16.11. A = ξ(∂ Q) links B [mm]. Proof. Let K be the collection of all sets K = {η(Q) : η ∈ C(Q, E), η(u) = ξ(u), u ∈ ∂ Q}. It is easily checked that K is a minimax system. If ϕ ∈ (A), then ϕ(K ) ∈ K for K ∈ K. Since A ∩ B = φ and K ∩ B = φ for K ∈ K, we see that A links B [mm]. Lemma 16.12. If f (x, t) satisfies (16.19) and b < m (a), then there is a continuous map ϕ from Nm → Mm such that (16.28)

J (v) ≡ G(v + ϕ(v)) = min G(v + w) ∈ C 1 (Nm , R), w∈Mm

and (16.29)

J (v) = G (v + ϕ(v)),

v ∈ Nm .

v ∈ Nm ,

16.4. Some lemmas

199

Proof. In view of Lemmas 16.6 and 16.8, we have (G (v + w1 ) − G (v + w0 ), w) ≥ w 2D ,

w ∈ Mm .

We can now apply Lemma 16.9 to arrive at the conclusion. Lemma 16.13. If, in addition, a0 (t − )2 + b0 (t + )2 ≤ 2F(x, t),

(16.30)

|t| < δ,

for some δ > 0, with a0 , b0 < λl+1 , b0 > μl (a0 ), l ≤ m, then there are ε > 0, r > 0 such that J (v) ≤ −ε v 2D ,

(16.31)

v ∈ Nl ∩ Br ,

where Br = {u ∈ D : u D ≤ r }. Proof. Let q be any number satisfying (16.32)

2 < q ≤ 2n/(n − 2T ), 2T < n

(16.33)

2 < q < ∞, n ≤ 2T.

It follows from Theorem 13.6 that there is a continuous map τ : Nl → Ml such that τ (s v) = s τ (v),

(16.34) (16.35)

s ≥ 0,

I (v + τ (v), a0 , b0 ) = inf I (v + w, a0 , b0 ), w∈Ml

τ (v) D ≤ C v D ,

(16.36)

v ∈ Nl ,

v ∈ Nl .

Then, for u = v + τ (v), we have, by (16.2), [a0 (u − )2 + b0 (u + )2 − 2F(x, u)] d x J (v) ≤ G(u) ≤ I (u, a0 , b0 ) + ≤ F2l (v, a0 , b0 ) + C

|u|>δ

|u|>δ

|u|q d x ≤ m l (a0 , b0 ) v 2D + o( v 2D ) ≤ −ε v 2D

for r sufficiently small. Lemma 16.14. Assume that (16.37)

a(t − )2 + b(t + )2 − W1 (x) ≤ 2F(x, t),

|t| > K ,

for some K ≥ 0, where a, b < λm+1 , b ≥ μm (a), l ≤ m, and W1 ∈ L 1 (). Then there is a K 1 < ∞ such that (16.38)

J (v) ≤ K 1 .

If b > μm (a), then (16.39)

J (v) −→ −∞ as v D −→ ∞.

200


Proof. For u = v + w, v ∈ Nm , w ∈ Mm , we have G(u) ≤ I (u, a, b) + C |u|q d x + W1 (x) d x ≤ I (u, a, b) + K . |u| μm (a), then m(a, b) < 0. This proves (16.39). Lemma 16.15. If l < m and λl < a, b < λm+1 , then there are continuous functions ξ : Nm ∩ Ml → Nl , η : Nm ∩ Ml → Mm homogeneous of degree 1 and such that (16.40) I (ξ(y)+η(y)+ y, a, b) = sup inf I (v +w+ y, a, b) = inf sup I (v +w+ y, a, b) v∈Nl w∈Mm

w∈Mm v∈Nl

for y ∈ Nm ∩ Ml . Proof. Let L y (v, w) = I (v + w + y, a, b). Then L y is a strictly convex, lower semicontinuous functional in w ∈ Mm , and strictly concave and continuous in v ∈ Nl . By Theorem 13.6 and Corollary 13.12, for each y0 ∈ Nm ∩ Ml , there are unique elements v 0 = ξ(y0 ) ∈ Nl , w0 = η(y0 ) ∈ Mm such that (16.40) holds, i.e., that L y0 (v, w0 ) ≤ L y0 (v 0 , w0 ) ≤ L y0 (v 0 , w),

v ∈ Nl , w ∈ Mm .

The functions ξ, η are clearly homogeneous of degree 1. To prove continuity, let y j → y0 in Nl ∩ Mm , and let v j = ξ(y j ), w j = η(y j ). We note that the functions v j , w j are bounded in D. Otherwise, it is easy to show that I (v + w j + y j , a, b) −→ ∞ as j −→ ∞ for any v ∈ Nl , and I (v j + w + y j , a, b) −→ −∞ as j −→ ∞ for any w ∈ Mm . This would contradict (16.40). Thus there are renamed subsequences such that v j → v 1 , w j w1 in D. Since I (v + w j + y j , a, b) ≤ I (v j + w j + y j , a, b) ≤ I (v j + w + y j , a, b),

v ∈ Nl , w ∈ Mm ,

16.4. Some lemmas

201

we have in the limit I (v + w1 + y0 , a, b) ≤ I (v 1 + w1 + y0 , a, b) ≤ I (v 1 + w + y0 , a, b),

v ∈ Nl , w ∈ Mm ,

showing that v 1 = v 0 , w1 = w0 . Since this is true for any subsequence, the result follows. Lemma 16.16. If 2F(x, t) ≤ a1 (t − )2 + b1 (t + )2 ,

(16.41)

|t| ≤ δ,

for some δ > 0, with a1 , b1 > λl , b1 < νl (a1 ), l < m, then there are ε > 0, r > 0 such that J (y + ξ(y)) ≥ ε y 2D ,

(16.42)

y ∈ Nm ∩ Ml ∩ Br .

Proof. By Lemma 16.15 (16.43)

inf I (ξ(y) + y + w, a1 , b1 ) = inf sup I (v + y + w, a1 , b1 )

w∈Mm

w∈Mm v∈Nl

for y ∈ Nm ∩ Ml . Then, for y ∈ (Nm ∩ Ml ∩ Br )\{0}, J (ξ(y) + y) = G(ξ(y) + y + ϕ(ξ(y) + y)) ≥ I (ξ(y) + y + ϕ(ξ(y) + y), a1 , b1 ) − o( y 2D ) ≥ inf I (ξ(y) + y + w, a1 , b1 ) − o( y 2D ) w∈Mm

= inf sup I (v + y + w, a1 , b1 ) − o( y 2D ) w∈Mm v∈Nl

≥ inf Ml (a, b) y + w 2D − o( y 2D ) w∈Mm

= Ml (a, b) y 2D − o( y 2D ) ≥ ε y 2D .

Lemma 16.17. Assume (16.44)

t[ f (x, t1 ) − f (x, t0 )] ≥ a(t − )2 + b(t + )2 ,

t j ∈ R, t = t1 − t0 .

Then (16.45) (G (v 1 + w) − G (v 0 + w), v) ≤ 2I (v, a, b),

v j , w ∈ D, v = v 1 − v 0 .

202


Proof. We have ( f (x, v 1 + w) − f (x, v 0 + w), v) ≥ a v − 2 + b v + 2 . Hence, (G (v 1 + w) − G (v 0 +w), v)/2 = v 2D − ( f (x, v 1 + w) − f (x, v 0 + w), v) ≤ I (v, a, b). Lemma 16.18. If f (x, t) satisfies (16.44), and b > γm (a), then there is a continuous map ψ from Mm → Nm such that (16.46)

J (w) ≡ G(w + ψ(w)) = max G(v + w) ∈ C 1 (Mm , R), v∈Nm

w ∈ Mm ,

and J (w) = G (w + ψ(w)),

(16.47)

w ∈ Mm .

Proof. In view of Lemmas 16.7 and 16.17, we have (G (v 1 + w) − G (v 0 + w), v) ≤ − v 2D ,

v ∈ Nm .

We can now apply Lemma 16.9 to obtain the conclusion. Lemma 16.19. If, in addition, (16.48)

a0 (t − )2 + b0 (t + )2 ≤ 2F(x, t),

|t| < δ,

for some δ > 0, with a0 , b0 < λl+1 , b0 > μl (a0 ), l > m, then there are ε > 0, r > 0 such that (16.49)

J (y + η(y)) ≤ −ε y 2D ,

y ∈ Nl ∩ Mm ∩ Br .

Proof. For y ∈ Mm ∩ Nl , let u = y + η(y) ∈ Mm . By (16.2), J (u) = G(u + ψ(u)) ≤ I (u + ψ(u), a0 , b0 ) + o( u 2D ) ≤ sup I (u + v, a0 , b0 ) + o( u 2D ) v∈Nm

= I (y + η(y) + ξ(y), a0 , b0 ) + o( u 2D ) = sup inf I (y + v + w, a0 , b0 ) + o( u 2D ) v∈Nm w∈Ml

= sup F2l (y + v, a0 , b0 ) + o( u 2D ) v∈Nm

≤ sup m l (a0 , b0 ) y + v 2D + o( u 2D ) v∈Nm

≤ −ε y 2D for r sufficiently small.

16.4. Some lemmas

203

Lemma 16.20. If 2F(x, t) ≤ a1 (t − )2 + b1 (t + )2 ,

(16.50)

|t| ≤ δ,

for some δ > 0, with a1 , b1 > λl , b1 < νl (a1 ), l > m, then there are ε > 0, r > 0 such that J (w) ≥ ε w 2D ,

(16.51)

w ∈ Ml ∩ Br .

Proof. We recall from Theorem 13.6 that there is a continuous map θ : Ml → Nl such that (16.52)

θ (s w) = s θ (w),

s ≥ 0,

(16.53)

I (θ (w) + w, a1 , b1 ) = sup I (v + w, a1 , b1 ), v∈Nl

w ∈ Ml .

Thus, J (w) ≥ G(w + θ (w), a1 , b1 ) ≥ I (w + θ (w), a1 , b1 ) − o( w 2D ) = sup I (v + w, a1 , b1 ) − o( w 2D ) v∈Nl

= F1l (w, a1 , b1 ) − o( w 2D ) ≥ Ml (a1 , b1 ) w 2D − o( w 2D ) ≥ ε w 2D for r sufficiently small. Lemma 16.21. Assume that (16.54)

2F(x, t) ≤ a(t − )2 + b(t + )2 + W1 (x),

|t| > K ,

for some K ≥ 0, where a, b > λm , b ≤ νm (a), l ≥ m, and W1 ∈ L 1 (). Then there is a K 1 < ∞ such that (16.55)

J (w) ≥ −K 1 ,

w ∈ Mm .

If b < νm (a), then (16.56)

J (w) −→ ∞ as w D −→ ∞.

Proof. For u = v + w, v ∈ Nm , w ∈ Mm , we have G(u) ≥ I (u, a, b) − C |u|q d x − W1 (x)d x ≥ I (u, a, b) − K . |u| 0. This proves (16.56). Lemma 16.22. If (16.57)

a0 (t − )2 + b0 (t + )2 ≤ 2F(x, t),

|t| < δ,

for some δ > 0, with b0 > γl (a0 ), l ≤ m, then there are ε > 0, r > 0 such that J (v) ≤ −ε v 2D ,

(16.58)

v ∈ Nl ∩ Br ,

where Br = {u ∈ D : u D ≤ r }. Proof. Let q be any number satisfying 2 < q ≤ 2n/(n − 2T ), 2T < n 2 < q < ∞, n ≤ 2T. By (16.2), J (v) ≤ G(v) ≤ I (v, a0 , b0 ) + ≤ − v 2D + C

|v|>δ

|v|>δ

[a0 (v − )2 + b0 (v + )2 − 2F(x, v)] d x

|v|q d x

≤ − v 2D + o( v 2D ) ≤ −ε v 2D for r sufficiently small. Lemma 16.23. If (16.59)

2F(x, t) ≤ a1 (t − )2 + b1 (t + )2 ,

|t| ≤ δ,

for some δ > 0, with b1 < l (a1 ), l < m, then there are ε > 0, r > 0 such that (16.60)

J (v) ≥ ε v 2D ,

v ∈ Nm ∩ Ml ∩ Br .

16.4. Some lemmas

205

Proof. Let u = v + ϕ(v) ∈ Ml . Then J (v) = G(u) ≥ I (u, a1 , b1 ) + ≥ u 2D − C

|u|>δ

|u|>δ

[a0 (u − )2 + b0 (u + )2 − 2F(x, u)] d x

|u|q d x

≥ u 2D − o( u 2D ) ≥ v 2D − o( v 2D ) ≥ ε v 2D for r sufficiently small, since v D ≤ u D ≤ C v D .

Lemma 16.24. If a0 (t − )2 + b0 (t + )2 ≤ 2F(x, t),

(16.61)

|t| < δ,

for some δ > 0, with b0 > γl (a0 ), l ≥ m, then there are ε > 0, r > 0 such that J (w) ≤ −ε w 2D ,

(16.62)

w ∈ Nl ∩ Mm ∩ Br .

Proof. For w ∈ Mm ∩ Nl , let u = w + ψ(w) ∈ Nl . By (16.2), J (w) = G(w + ψ(w)) = G(u) ≤ I (u, a0 , b0 ) + [a0 (v − )2 + b0 (u + )2 − 2F(x, u)] d x ≤ − u 2D + C

|u|>δ

|u|>δ

|u|q d x

≤ − u 2D + o( u 2D ) ≤ −ε u 2D for r sufficiently small. Since w D ≤ u D ≤ C w D , the result follows. Lemma 16.25. If (16.63)

2F(x, t) ≤ a1 (t − )2 + b1 (t + )2 ,

|t| ≤ δ

for some δ > 0, with b1 < l (a1 ), l > m, then there are ε > 0, r > 0 such that (16.64)

J (w) ≥ ε w 2D ,

w ∈ Ml ∩ Br .

206


Proof. We have G(w) ≥ I (w, a1 , b1 ) + ≥

w 2D

−C

|w|>δ

|w|>δ

[a0 (u − )2 + b0 (w+ )2 − 2F(x, w)] d x

|w|q d x

≥ w 2D − o( w 2D ) ≥ w 2D − o( w 2D ) ≥ ε w 2D for r sufficiently small. Since J (w) = sup G(v + w) ≥ G(w), v∈Nl

the result follows.

16.5 Local linking The following theorem will also be used in the proofs of the theorems of Section 16.3. It is also of interest in its own right. Theorem 16.26. Let M, N be closed subspaces of a Hilbert space E such that 0 < dim N < ∞ and M = N ⊥ . Let G ∈ C 1 (E, R) satisfy the PS condition and G(v) ≤ 0,

v ∈ N ∩ BR,

G(w) ≥ 0,

w ∈ M ∩ BR,

for some R > 0. Assume that −∞ < α = inf G < 0. E

Then G has at least three critical points. Proof. Since G satisfies the PS condition, it has a minimum point satisfying G(u 0 ) = α. Clearly, 0 is also a critical point. Assume that there are no others. Then the set Eˆ = {u ∈ E : G (u) = 0} contains all points except u 0 and 0. If θ < 1, then there is a mapping Y (u) from Eˆ to E that is locally Lipschitz continuous and satisfies Y (u) ≤ 1,

θ G (u) ≤ (G (u), Y (u)),

ˆ u ∈ E.

For v ∈ N ∩ B R \{0}, let σ (t)v be the solution of σ (t) = −Y (σ (t)),

t ≥ 0, σ (0) = v.

16.5. Local linking

207

Then d G(σ (t)v)/dt = (G (σ ), σ ) = −(G (σ ), Y (σ )) ≤ −θ G (σ ) < 0 ˆ Note that σ (t)v is continuous in t and v for v = 0. For each as long as σ (t)v is in E. v ∈ N ∩ B R \{0}, there is a maximal interval 0 < t < Tv in which σ (t)v exists and satisfies G (σ (t)v) = 0 and G(σ (t)v) < 0. I claim that σ (t)v → u 0

(16.65)

as t → Tv .

To see this, suppose that tk → Tv . Then tk Y (σ (t)v) dt| ≤ |tk − t j | → 0. σ (tk )v − σ (t j )v ≤ | tj

Thus, σ (tk )v → h in E. By continuity,

G (σ (tk )v) → G (h).

If G (h) = 0, the solution can be continued beyond Tv , contrary to the way it was chosen. Thus, G (h) = 0, showing that h = u 0 . Consequently, σ (tk )v → u 0 . Since this is true for any such sequence, (16.65) holds. Note that Tv is continuous in v for v = 0. Define σˆ (t)v = σ (t)v,

(16.66)

= u0,

t < Tv ,

t ≥ Tv .

Let w0 be an element of M with unit norm, and take K 0 = {sw0 + v : v ∈ N, s ≥ 0, sw0 + v = R}. Let Q = {sw0 + v : v ∈ N, s ≥ 0, sw0 + v ≤ R}. Let ε > 0 be given, and let T > 0 satisfy (16.67)

ε2 Tv2 + ≤ 1, T2 R2

v ∈ N, v = ε.

Let ξ(u) be the continuous map from ∂ Q to E such that ξ(v) = v,

v ∈ N ∩ BR ,

and for u = sw0 + v ∈ K 0 , (16.68)

ξ(sw0 + v) = σˆ (T s/R)v, = u0,

v ≥ ε,

v < ε.

208


By (16.67) and (16.68), σˆ (T s/R)v = u 0 ,

v = ε.

Hence, ξ is continuous on ∂ Q. Moreover, G(ξ(u)) ≤ 0,

u ∈ ∂ Q.

In addition, ξ(u) ≥ r > 0,

u ∈ K0.

Let B = ∂ Bδ ∩ M,

0 < δ < r < R.

By Corollary 16.11, A = ξ(∂ Q) links B [mm]. Since a0 := sup G ≤ b0 := inf G,

(16.69)

B

A

we can apply Theorem 2.12 to conclude that (1.4) holds. If a > 0, this provides a third critical point by the PS condition. If a = 0, then there is a sequence satisfying (1.4) and d(u k , B) → 0,

(16.70)

k → ∞.

Since G satisfies the PS condition, there is a subsequence converging to a critical point on B. Again, this provides a third critical point.

16.6 The proofs We prove the theorems of Section 16.3. First, we prove Theorem 16.3. Proof. By Lemma 16.12, it suffices to show that J (v) has two nontrivial solutions. Now J is bounded from above by Lemma 16.14 and satisfies (PS) by (16.39). Moreover, J (v) < 0,

(16.71)

v ∈ Nl ∩ Br \{0},

by Lemma 16.13, and (16.72)

J (ξ(y) + y) > 0,

y ∈ Nm ∩ Ml ∩ Br \{0},

by Lemma 16.16. Thus, J has a positive maximum on Nm . We can now apply Theorem 16.26 and Lemma 16.9 to obtain the desired conclusion.


209

Similarly, we prove Theorem 16.4. Proof. By Lemma 16.18, it suffices to show that J (w) given by (16.46) has two nontrivial solutions. Now J is bounded from below by Lemma 16.21 and satisfies (PS) by (16.56). Moreover, (16.73)

J (w + η(w)) < 0,

w ∈ Nl ∩ Mm ∩ Br \{0},

by Lemma 16.19, and (16.74)

J (w) > 0,

w ∈ Ml ∩ Br \{0}

by Lemma 16.20. Thus, J has a negative minimum on Mm . We can now apply Theorem 16.26 and Lemma 16.9 to obtain the desired conclusion. Next, we prove Theorem 16.5. Proof. With reference to Theorem 16.3, we note that, by Lemma 16.12, it suffices to show that J (v) has two nontrivial solutions. Now J is bounded from above by Lemma 16.14 and satisfies (PS) by (16.39). Moreover, (16.75)

J (v) < 0,

v ∈ Nl ∩ Br \{0}

by Lemma 16.22, and (16.76)

J (v) > 0,

v ∈ Nm ∩ Ml ∩ Br \{0}

by Lemma 16.23. Thus J has a positive maximum on Nm . We can now apply Theorem 16.26 and Lemma 16.9 to obtain the desired conclusion. With respect to Theorem 16.4, we note that by Lemma 16.18, it suffices to show that J (w) given by (16.46) has two nontrivial solutions. Now J is bounded from below by Lemma 16.21 and satisfies (PS) by (16.56). Moreover, (16.77)

J (w) < 0,

w ∈ Nl ∩ Mm ∩ Br \{0},

by Lemma 16.24, and (16.78)

J (w) > 0,

w ∈ Ml ∩ Br \{0},

by Lemma 16.25. Thus, J has a negative minimum on Mm . We can now apply Theorem 16.26 and Lemma 16.9 to obtain the desired conclusion.

16.7 Notes and remarks In his studies of semilinear elliptic problems with jumping nonlinearities, Các [29]– [34] proved the following.

210


Theorem 16.27. Let be a bounded domain in Rn , n ≥ 2, with smooth boundary ∂. Let 0 < λ0 < λ1 < · · · < λk < · · · be the sequence of distinct eigenvalues of the eigenvalue problem −u = λu i n ,

(16.79)

u = 0 on ∂.

Let p(t) be a continuous function such that p(0) = 0 and p(t)/t −→ a

as t −→ −∞

p(t)/t −→ b

as t −→ +∞.

and Assume that for some k ≥ 1, we have a ∈ (λk−1 , λk ), b ∈ (λk , λk+1 ), and the only solution of −u = bu + − au − i n ,

(16.80)

u = 0 on ∂

is u ≡ 0, where u ± = max[±u, 0]. Assume further that p(s) − p(t) ≤ ν < λk+1 , s−t

(16.81)

s, t ∈ R, s = t.

Assume also that p (0) exists and satisfies p (0) ∈ (λ j −1 , λ j ) for some j ≤ k. Then −u = p(u) i n ,

(16.82)

u = 0 on ∂

has at least two nontrivial solutions. This theorem generalizes the work of Gallouët and Kavian [72], [73] which required λk to be a simple eigenvalue and the left-hand side of (16.81) to be sandwiched in between λk−1 and λk+1 and bounded away from both of them. Các proves a counterpart of the theorem in which the inequalities are reversed. In the present chapter we generalized this theorem and its reverse-inequality counterpart by not requiring p(t)/t to converge to limits at either ±∞ or ±0. Rather, we worked with the primitive t f (x, s)ds F(x, t) := 0

and bounded 2F(x, t)/t 2 near ±∞ and ±0 [we replaced p(t) with a function f (x, t) depending on x as well]. Our main assumptions were (16.83)

t[ f (x, t1 ) − f (x, t0 )] ≤ a(t − )2 + b(t + )2 ,

(16.84)

t j ∈ R, t = t1 − t0 ,

a0 (t − )2 + b0 (t + )2 ≤ 2F(x, t) ≤ a1 (t − )2 + b1 (t + )2 ,

for some δ > 0, (16.85)

a2 (t − )2 + b2 (t + )2 − W1 (x) ≤ 2F(x, t),

|t| > K ,

|t| < δ,


211

for some K > 0 and W1 ∈ L 1 (), where the constants a, a0, a1 , a2 , b, b0 , b1 , b2 are suitably chosen (they include the cases considered by Các). The advantage of such inequalities is that they do not restrict the expression 2F(x, t)/t 2 or f (x, t)/t to any particular interval. The results of this chapter come from [101] with changes in the proofs. Theorem 16.26 is from [28] with variations made in the proof. Lemma 16.9 is due to Castro [38].

Chapter 17

Second-Order Periodic Systems 17.1 Introduction In this chapter we study a general system of second-order differential equations, and we look for periodic solutions. We show that for several sets of hypotheses such systems can be solved by the methods used in the book. We consider the following problem. One wishes to solve −x(t) ¨ = ∇x V (t, x(t)),

(17.1) where

x(t) = (x 1 (t), · · · , x n (t))

(17.2)

is a map from I = [0, T ] to Rn such that each component x j (t) is a periodic function with period T, and the function V (t, x) = V (t, x 1 , . . . , x n ) is continuous from Rn+1 to R with ∇x V (t, x) = (∂ V /∂ x 1, . . . , ∂ V /∂ x n ) ∈ C(Rn+1 , Rn ).

(17.3)

For each x ∈ Rn , the function V (t, x) is periodic in t with period T. We shall study this problem under the following assumptions: 1. V (t, x) ≥ 0,

t ∈ I, x ∈ Rn .

2. There are constants m > 0, α ≤ 6m 2 /T 2 such that V (t, x) ≤ α,

|x| ≤ m, t ∈ I, x ∈ Rn .

3. There is a constant μ > 2 such that (17.4)

Hμ(t, x) ≤ W (t) ∈ L 1 (I ), |x|2

|x| ≥ C, t ∈ I, x ∈ Rn ,


214

17. Second-Order Periodic Systems and (17.5)

lim sup |x|→∞

Hμ (t, x) ≤ 0, |x|2

where Hμ (t, x) = μV (t, x) − ∇x V (t, x) · x.

(17.6)

4. There is a subset e ⊂ I of positive measure such that (17.7)

lim inf |x|→∞

V (t, x) > 0, |x|2

t ∈ e.

We have Theorem 17.1. Under the above hypotheses, system (17.1) has a solution. As a variant of Theorem 17.1, we have Theorem 17.2. The conclusion in Theorem 17.1 is the same if we replace hypothesis 2 with 2A. There is a constant q > 2 such that V (t, x) ≤ C(|x|q + 1),

t ∈ I, x ∈ Rn ,

and there are constants m > 0, α < 2π 2 /T 2 such that V (t, x) ≤ α|x|2 ,

|x| ≤ m, t ∈ I, x ∈ Rn .

We also have Theorem 17.3. The conclusion of Theorem 17.1 holds if we replace hypothesis 3 with 3A. There is a constant μ < 2 such that (17.8)

Hμ (t, x) ≥ −W (t) ∈ L 1 (I ), |x|2

|x| ≥ C, t ∈ I, x ∈ Rn ,

and (17.9)

lim inf |x|→∞

Hμ (t, x) ≥ 0. |x|2

And we have Theorem 17.4. The conclusion of Theorem 17.1 holds if we replace hypothesis 1 with 1A. 0 ≤ V (t, x) ≤ C(|x|2 + 1), and hypothesis 3 with

t ∈ I, x ∈ Rn .

17.2. Proofs of the theorems

215

3B. The function given by (17.10)

H (t, x) = 2V (t, x) − ∇x V (t, x) · x

satisfies (17.11)

H (t, x) ≤ W (t) ∈ L 1 (I ),

|x| ≥ C, t ∈ I, x ∈ Rn ,

and (17.12)

|x| → ∞, t ∈ I, x ∈ Rn .

H (t, x) → −∞,

Theorems 17.1–17.4 show the existence of solutions, which conceivably could be constants. The following theorems provide the existence of non-constant solutions. Theorem 17.5. If we replace hypothesis 4 in Theorem 17.1 with 4A. There are constants β > 2π 2 /T 2 and C such that V (t, x) ≥ β|x|2 ,

|x| > C, t ∈ I, x ∈ Rn ,

then system (17.1) has a nonconstant solution. As a variant of Theorem 17.5, we have Theorem 17.6. The conclusion in Theorem 17.5 is the same if we replace hypothesis 2 with hypothesis 2A. We also have Theorem 17.7. The conclusion of Theorem 17.5 holds if we replace hypothesis 1 with hypothesis 1A and hypothesis 3 with hypothesis 3B. We shall prove Theorems 17.1–17.7 in the next section. We use the linking method of Chapter 2.

17.2 Proofs of the theorems We now give the proof of Theorem 17.1. Proof. Let X be the set of vector functions x(t) described above. It is a Hilbert space with norm satisfying n x j 2H 1 . x 2X = j =1

We also write x 2 =

n j =1

where · is the L 2 (I ) norm.

x j 2 ,

216

17. Second-Order Periodic Systems

Let N = {x(t) ∈ X : x j (t) ≡ constant, 1 ≤ j ≤ n}, and M = N ⊥ . The dimension of N is n, and X = M ⊕ N. The following is known (cf., e.g., Proposition 1.3 of [95]). Lemma 17.8. If x ∈ M, then x 2∞ ≤ and

T x ˙ 2 12

T x . ˙ 2π

x ≤

Proof. It suffices to prove the lemma for continuously differential periodic functions. First, consider the case T = 2π. Using Fourier series, we have ∞

x=

(17.13)

αk ϕk ,

k=−∞

where αk = (x, ϕ¯ k ),

(17.14)

k = 0, ±1, ±2, . . . ,

and 1 ϕk (x) = √ eikx , 2π

(17.15)

k = 0, ±1, ±2, . . .

Thus, ∞

x 2 =

(17.16)

|αk |2 .

k=−∞

If x ⊥ M, then α0 = 0, and 2

x ≤

∞

|kαk |2 = x ˙ 2.

k=−∞

Moreover, x 2∞

1 ≤ 2π

2

∞

|αk |

k=−∞

∞ 1 ≤ |kαk |2 2π

k=−∞

2

∞ k=1

k −2

=

1 x ˙ 2 π 2 /3. 2π

This proves the lemma for the case T = 2π. Otherwise, we let y(t) = x(T t/2π). Then x 2 =

T y 2 , 2π

x ˙ 2=

2π y˙ 2 , T

x ∞ = y ∞ .

17.2. Proofs of the theorems

217

Thus, x 2 =

T T y 2 ≤ y˙ 2 = 2π 2π

and x ∞ = y ∞ ≤

T 2π

2 x ˙ 2

π π T T y˙ 2 = x ˙ 2= x ˙ 2. 6 6 2π 12

Note that it follows that x ∞ ≤ C x X , We define (17.17)

x ∈ X.

G(x) = x ˙ 2−2

V (t, x(t)) dt,

x ∈ X.

I

For each x ∈ X, write x = v + w, where v ∈ N, w ∈ M. For convenience, we shall use the following equivalent norm for X: x 2X = w ˙ 2 + v 2 . If x ∈ M and

12 2 m , T ≤ m, and we have by hypothesis 2 that V (t, x) ≤

x ˙ 2 = ρ2 = then Lemma 17.8 implies that x ∞ α. Hence, (17.18)

G(x)

≥ x ˙ 2−2

|x|<m

α dt

≥ ρ 2 − 2αT ≥ 0. We also note that hypothesis 1 implies G(v) ≤ 0,

(17.19)

v ∈ N.

Take A

=

∂ Bρ ∩ M,

B

=

N,

ρ2 =

12 2 m , T

where Bσ = {x ∈ X : x X < σ }. By Example 2 of Section 3.4, A links B. Moreover, if R is sufficiently large, (17.20)

sup[−G] ≤ 0 ≤ inf[−G]. A

B

218


Hence, we may apply Theorem 3.4 to conclude that there is a sequence {x (k) } ⊂ X such that (k) (k) 2 G(x ) = x˙ − 2 V (t, x (k) (t)) dt → c ≥ 0, (17.21) (17.22)

(G (x (k) ), z)/2 = (x˙ (k) , z˙ ) −

I

∇x V (t, x (k) ) · z(t) dt → 0,

z ∈ X,

I

and (17.23)

(G (x

(k)

), x

(k)

)/2 = x˙

(k) 2

∇x V (t, x (k) ) · x (k) dt → 0.

− I

If

ρk = x (k) X ≤ C,

then there is a renamed subsequence such that x (k) converges to a limit x ∈ X weakly in X and uniformly on I. From (17.22), we see that ˙ z˙ ) − ∇x V (t, x(t)) · z(t) dt = 0, z ∈ X, (G (x), z)/2 = (x, I

from which we conclude easily that x is a solution of (17.1). If ρk = x (k) X → ∞, let x˜ (k) = x (k) /ρk . Then x˜ (k) X = 1. Let x˜ (k) = w˜ (k) + v˜ (k) , where w˜ (k) ∈ M and v˜ (k) ∈ N. There is a renamed subsequence such that [x˜ (k) ]· → r and x˜ (k) → τ, where r 2 + τ 2 = 1. From (17.21) and (17.23), we obtain [x˜ (k) ]· 2 − 2 V (t, x (k) (t)) dt/ρk2 → 0 I

and [x˜ (k) ]· 2 −

I

∇x V (t, x (k) ) · x (k) dt/ρk2 → 0.

Thus, (17.24)

2 I

and

(17.25) I

V (t, x (k) (t)) dt/ρk2 → r 2

∇x V (t, x (k) ) · x (k) dt/ρk2 → r 2 .

Hence, (17.26) I

Hμ (t, x (k) (t)) dt/ρk2 →

μ 2

− 1 r 2.

17.2. Proofs of the theorems Note that

219

|x˜ (k) (t)| ≤ C x˜ (k) X = C.

If

|x (k) (t)| → ∞,

then, by hypothesis 3, lim sup

Hμ (t, x (k) (t))

≤ lim sup

ρk2

If

Hμ (t, x (k) (t)) (k) 2 |x˜ (t)| ≤ 0. |x (k) (t)|2

|x (k) (t)| ≤ C,

then

Hμ (t, x (k) (t)) → 0. ρk2

Hence,

lim sup I

Thus, by (17.26),

Hμ (t, x (k) (t)) dt/ρk2 ≤ 0.

μ 2

− 1 r 2 ≤ 0.

If r = 0, this contradicts the fact that μ > 2. If r = 0, then w˜ (k) → 0 uniformly in I by Lemma 17.8. Moreover, T |v˜ (k) |2 = v˜ (k) 2 → 1. Thus, there is a renamed subsequence such that v˜ (k) → v˜ in N with |v| ˜ 2 = 1/T. Hence, x˜ (k) → v˜ uniformly in (k) I. Consequently, |x | → ∞ uniformly in I. Thus, by hypothesis 4, lim inf I

V (t, x (k) (t)) dt/ρk2 ≥

lim inf e

V (t, x (k) (t)) (k) 2 |x˜ (t)| dt > 0. |x (k) (t)|2

This contradicts (17.24). Hence, the ρk are bounded, and the proof is complete. The proof of Theorem 17.2 is similar to that of Theorem 17.1 with the exception of inequality (17.18) resulting from hypothesis 2. In its place we reason as follows: If x ∈ M, we have, by hypothesis 2A, 2 2 G(x) ≥ x ˙ −2 α|x(t)| dt − C (|x|q + 1) dt |x|<m

≥

|x|>m

x ˙ 2 − 2α x 2 − C(1 + m 2−q + m −q )

≥

x ˙ 2 (1 − [2αT 2 /4π 2 ]) − C

≥

(1 − [αT /2π 2

2

]) x 2X

−C

|x|>m

|x|>m

|x|q dt q

I

x X dt

|x|q dt

220

17. Second-Order Periodic Systems ≥ =

(1 − [αT 2 /2π 2 ]) x 2X − C x X q−2 x 2X 1 − [αT 2 /2π 2 ] − C x X q

by Lemma 17.8. Hence, we have Lemma 17.9. G(x) ≥ ε x 2X ,

(17.27)

x X ≤ ρ, x ∈ M,

for ρ > 0 sufficiently small, where ε < 1 − [αT 2 /2π 2 ]. The remainder of the proof is essentially the same. In proving Theorem 17.3, we follow the proof of Theorem 17.1 until we reach (17.26). Then we reason as follows. If |x (k) (t)| → ∞, then lim inf

Hμ (t, x (k) (t)) ρk2

If

≥ lim inf

Hμ (t, x (k) (t)) (k) 2 |x˜ (t)| ≥ 0. |x (k) (t)|2

|x (k) (t)| ≤ C,

then

Hμ (t, x (k) (t)) ρk2

Hence,

lim inf I

Thus, by (17.26),

→ 0.

Hμ (t, x (k) (t)) dt/ρk2 ≥ 0. μ 2

− 1 r 2 ≥ 0.

If r = 0, this contradicts the fact that μ < 2. If r = 0, then w˜ (k) → 0 uniformly in I by Lemma 17.8. Moreover, T |v˜ (k) |2 = v˜ (k) 2 → 1. Hence, there is a renamed subsequence such that v˜ (k) → v˜ in N with |v| ˜ 2 = 1/T. Hence, x˜ (k) → v˜ uniformly in (k) I. Consequently, |x (t)| → ∞ uniformly in I. Thus, by Hypothesis 4, lim inf I

V (t, x (k) (t)) dt/ρk2 ≥

lim inf e

V (t, x (k) (t)) (k) 2 |x˜ (t)| dt > 0. |x (k) (t)|2

This contradicts (17.24). Hence, the ρk are bounded, and the proof is complete. In proving Theorem 17.4, we follow the proof of Theorem 17.1 until (17.26). Assume first that r > 0. Note that (17.21) and (17.23) imply that (17.28) H (t, x (k)(t)) dt → −c. I

17.3. Nonconstant solutions

221

On the other hand, by hypothesis 1A, we have 0 ← [x˜ (k) ]· 2 − 2 V (t, x (k) (t)) dt/ρk2 I

≥ [x˜ (k) ]· 2 − 2C

I

→ r 2 − 2C

(|x˜ (k) (t)|2 + ρk−2 ) dt

|x(t)| ˜ 2 dt. I

Thus, x(t) ˜ ≡ 0. Let 0 ⊂ I be the set on which x(t) ˜ = 0. The measure of 0 is positive. Moreover, |x (k) (t)| → ∞ as k → ∞ for t ∈ 0 . Hence, (k) (k) H (t, x (t)) dt ≤ H (t, x (t)) dt + W (t) dt → −∞ 0

I

I \0

by hypothesis 3A. But this contradicts (17.28). If r = 0, then w˜ (k) → 0 uniformly in I by Lemma 17.8. Moreover, T |v˜ (k) |2 = v˜ (k) 2 → 1. Thus, there is a renamed subsequence such that v˜ (k) → v˜ in N with |v| ˜ 2 = 1/T. Hence, x˜ (k) → v˜ uniformly in (k) I. Consequently, |x | → ∞ uniformly in I. Thus, by hypothesis 4, V (t, x (k) (t)) (k) 2 (k) 2 |x˜ (t)| dt > 0. lim inf V (t, x (t)) dt/ρk ≥ lim inf |x (k) (t)|2 I e This contradicts (17.24). Hence, the ρk are bounded, and the proof is complete.

17.3 Nonconstant solutions We now turn to the proofs of Theorems 17.5–17.7. First, we prove Theorem 17.5. Proof. As before, we define (17.29)

G(x) = x ˙ 2−2

V (t, x(t)) dt,

x ∈ X.

I

For each x ∈ X, write x = v + w, where v ∈ N, w ∈ M. As before, if x ∈ M and x ˙ 2 = ρ2 =

12 2 m , T

then Lemma 17.8 and hypothesis 2 imply that (17.30)

G(x)

≥ x ˙ 2−2

|x|<m

α dt

≥ ρ 2 − 2αT ≥ 0. Note that hypothesis 4A is equivalent to (17.31) for some constant C.

V (t, x) ≥ β|x|2 − C,

t ∈ I, x ∈ Rn ,

222


Next, let y(t) = v + sw0 , where v ∈ N, s ≥ 0, and w0 = (sin(2πt/T ), 0, . . . , 0). Then w0 ∈ M, and

w0 2 = T /2,

w˙ 0 2 = 2π 2 /T.

Note that y 2 = v 2 + s 2 T /2 = T |v|2 + T s 2 /2. Consequently, G(y) = s w˙ 0 − 2 2

V (t, y(t)) dt

2

I

≤ 2π 2 s 2 /T − 2β

|y(t)|2 dt + T C I

≤ 2π 2 s 2 /T − 2β( v 2 + T s 2 /2) + T C ≤ (2π 2 − βT 2 )s 2 /T − 2Tβ|v|2 + T C → −∞ as s 2 + |v|2 → ∞. We also note that hypothesis 1 implies G(v) ≤ 0,

(17.32)

v ∈ N.

Take A

=

{v ∈ N : v ≤ R} ∪ {sw0 + v : v ∈ N, s ≥ 0, sw0 + v = R},

B

=

∂ Bρ ∩ M, 0 < ρ < R,

where Bσ = {x ∈ X : x X < σ }. By Example 3 of Section 3.4, A links B. Moreover, if R is sufficiently large, (17.33)

sup G ≤ 0 ≤ inf G. B

A

Hence, we may apply Theorem 3.4 to conclude that there is a sequence {x (k) } ⊂ X such that (17.34) G(x (k) ) = x˙ (k) 2 − 2 V (t, x (k) (t)) dt → c ≥ 0, (17.35)

(G (x (k) ), z)/2 = (x˙ (k) , z˙ ) −

I

∇x V (t, x (k) ) · z(t) dt → 0, I

z ∈ X,


223

and (17.36)

(G (x (k) ), x (k) )/2 = x˙ (k) 2 −

∇x V (t, x (k) ) · x (k) dt → 0. I

If

ρk = x (k) X ≤ C,

then there is a renamed subsequence such that x (k) converges to a limit x ∈ X weakly in X and uniformly on I. From (17.35), we see that (G (x), z)/2 = (x, ˙ z˙ ) − ∇x V (t, x(t)) · z(t) dt = 0, z ∈ X, I

from which we conclude easily that x is a solution of (17.1). By (17.34), we see that G(x) ≥ c ≥ 0, showing that x(t) is not a constant, for if c > 0 and x ∈ N, then G(x) = −2 V (t, x(t)) dt ≤ 0. I

If c = 0, we know that d(x (k) , B) → 0 by Theorem 2.12. Hence, there is a sequence {y (k) } ⊂ B such that x (k) − y (k) → 0 in X. If v ∈ N, then (x, v) = (x − x (k) , v) + (x (k) − y (k) , v) → 0 since y (k) ∈ M. Thus, x ∈ M. If

ρk = x (k) X → ∞,

let x˜ (k) = x (k) /ρk . Then x˜ (k) X = 1. Let x˜ (k) = w˜ (k) + v˜ (k) , where w˜ (k) ∈ M and · v˜ (k) ∈ N. There is a renamed subsequence such that [w˜ (k) ] → r and v˜ (k) → τ, 2 2 where r + τ = 1. From (17.21) and (17.23) we obtain (k) · 2 [x˜ ] − 2 V (t, x (k) (t)) dt/ρk2 → 0 I

and [x˜ (k) ]· 2 −

I

∇x V (t, x (k) ) · x (k) dt/ρk2 → 0.

Thus, (17.37)

2 I

and

(17.38) I

V (t, x (k) (t)) dt/ρk2 → r 2

∇x V (t, x (k) ) · x (k) dt/ρk2 → r 2 .

224


Hence, (17.39) I

Hμ (t, x (k) (t)) dt/ρk2 →

μ 2

− 1 r 2.

By hypothesis 3, the left-hand side of (17.37) is ≥ (2β x (k) 2 − 4πC)/ρk2 → 2βτ 2 . Hence, r 2 ≥ 2βτ 2 = 2β(1 − r 2 ), showing that r2 ≥

2β > 0. 1 + 2β

Note that |x˜ (k) (t)| ≤ C x˜ (k) X = C. If |x (k) (t)| → ∞, then lim sup

Hμ (t, x (k) (t)) Hμ (t, x (k) (t)) (k) 2 ≤ lim sup |x˜ (t)| ≤ 0. |x (k) (t)|2 ρk2

If |x (k) (t)| ≤ C, then Hμ (t, x (k) (t)) ρk2 Hence,

lim sup I

→ 0.

Hμ (t, x (k) (t)) dt/ρk2 ≤ 0.

In view of (17.39), this implies that μ 2

− 1 r 2 ≤ 0,

contrary to hypothesis 3. Hence, the ρk are bounded, and the proof is complete. The proof of Theorem 17.6 is similar to that of Theorem 17.5 with the exception of inequality (17.18) resulting from hypothesis 2. In its place we reason as follows:


225

If x ∈ M, we have, by hypothesis 2A, 2 2 α|x(t)| dt − C G(x) ≥ x ˙ −2 |x|<m

≥

x ˙ 2 (1 − [2αT 2 /4π 2 ]) − C

≥

(1 − [αT 2 /2π 2 ]) x 2X − C

=

(|x|q + 1) dt

x ˙ 2 − 2α x 2 − C(1 + m 2−q + m −q )

≥

≥

|x|>m

2

2

]) x 2X

|x|>m

|x|>m

|x|q dt

|x|q dt q

I

x X dt q

(1 − [αT /2π − C x X q−2 1 − [αT 2 /2π 2 ] − C x X x 2X

by Lemma 17.8. Hence, we have Lemma 17.10. G(x) ≥ ε x 2X ,

(17.40)

x X ≤ ρ, x ∈ M,

for ρ > 0 sufficiently small, where ε < 1 − [αT 2 /2π 2 ] is positive. The remainder of the proof is essentially the same, but in this case c > ε > 0, obviating the need to consider the situation when c = 0. In proving Theorem 17.7, we follow the proof of Theorem 17.5. In particular, it follows that r > 0. Moreover, (17.21) and (17.23) imply that H (t, x (k)(t)) dt → −c. (17.41) I

On the other hand, by hypothesis 1A, we have (k) · 2 0 ← [x˜ ] − 2 V (t, x (k) (t)) dt/ρk2 I

≥ [x˜ (k) ]· 2 − 2C

I

→ r 2 − 2C

(|x˜ (k) (t)|2 + ρk−2 ) dt

|x(t)| ˜ 2 dt. I

˜ = 0. The measure of 0 is Hence, x(t) ˜ ≡ 0. Let 0 ⊂ I be the set on which x(t) positive. Moreover, |x (k) (t)| → ∞ as k → ∞ for t ∈ 0 . Thus, H (t, x (k)(t)) dt ≤ H (t, x (k)(t)) dt + W (t) dt → −∞ I

0

I \0

by hypothesis 4A. But this contradicts (17.41). Hence, the ρk are bounded, and the proof is complete.

226


17.4 Notes and remarks The periodic, nonautonomous problem x(t) ¨ = ∇x V (t, x(t))

(17.42)

has an extensive history in the case of singular systems (cf., e.g., Ambrosetti–Coti Zelati [2]). The first to consider it for potentials satisfying (17.3) were Berger and the author [21] in 1977. We proved the existence of solutions to (17.42) under the condition that V (t, x) → ∞ as |x| → ∞ uniformly for a.e. t ∈ I. Subsequently, Willem [159], Mawhin [93], Mawhin–Willem [95], Tang [151], [152], Tang–Wu [154], [153], Wu–Tang [160] and others proved existence under various conditions (cf. the references given in these publications). The periodic problem (17.1) was studied by Mawhin–Willem [96],[95], Long [88], Tang–Wu [155] and others (cf. the refernces quoted in them). Tang–Wu [155] proved the existence of solutions of problem (17.1) under the following hypotheses: (I)

V (t, x) → ∞ as

|x| → ∞

uniformly for a.e. t ∈ I, (II) ∃ a ∈ C(R+ , R+ ),

b ∈ L 1 (0, T, R+ )

such that |V (t, x)| + |∇V (t, x)| ≤ a(|x|)b(t) ∀x ∈ Rn and a.e. t ∈ [0, T ], and (III) ∃ 0 < μ < 2,

M >0

such that ∇V (t, x) · x ≤ μV (t, x) ∀ |x| ≥ M and a.e. t ∈ [0, T ]. Rabinowitz [103] proved existence under stronger hypotheses. In particular, he assumed (I ) ∃ constants a1 , a2 > 0, μ0 > 1, such that

V (t, x) ≥ a1 |x|μ0 + a2 ∀ x ∈ Rn and a.e. t ∈ [0, T ]

in place of (I), and

(III )

∃ 0 < μ < 2, M > 0

such that 0 < ∇V (t, x) · x ≤ μV (t, x) ∀ |x| ≥ M and a.e. t ∈ [0, T ] in place of (III). Mawhin–Willem [96] proved existence for the case of convex potentials, while Long [88] studied the problem for even potentials. They assumed that V (t, x) is subquadratic in the sense that ∃ a3 < (2π/T )2 and a4


227

such that |V (t, x)| ≤ a3 |x|2 + a4 ∀x ∈ Rn and a.e. t ∈ [0, T ]. Mawhin–Willem [95] also studied the problem for a bounded nonlinearity. Tang–Wu [155] also proved the existence of solutions if one replaces (I) with

T

V (t, x) dt → ∞ as |x| → ∞

0

and V (t, x) is γ − subadditive with γ > 0 for a.e. t ∈ [0, T ]. All of these authors studied only the existence of solutions. Here, we studied the problem under much weaker assumptions and showed the existence of nonconstant solutions. Little was done concerning nonconstant solutions of problem (17.1). For the homogeneous case, Ben Naoum–Troestler–Willem [18] proved the existence of a nonconstant solution. For the case T = 2π, Theorem 17.5, with substantially stronger hypotheses, was proved by Nirenberg (cf. Ekeland and Ghoussoub [60]). In place of hypothesis 2, they assumed V (t, x) ≤

3 , 2π 2

|x| ≤ 1, t ∈ R, x ∈ Rn .

In place of hypotheses 3 and 4, they assumed the superquadraticity condition V (t, x) > 0, Hμ (t, x) ≤ 0,

|x| ≥ C, t ∈ R, x ∈ Rn ,

for some μ > 2, which implies these hypotheses and V (t, x) ≥ C|x|μ − C , among other things.

x ∈ Rn , C > 0,

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Index B R , 21, 112 F(x, t), 64 , 129 K, 8 , 8 (A), 8 asymptotic resonance, 129 Cerami sequence, 3 convex, 149 critical points, 1 Euler–Lagrange equations, 1 examples of linking, 21 extensions of Picard’s theorem, 31 extrema, 2 flows, 8, 31 Fuˇc´ık spectrum, 130 iff=if, and only if, 155 l.s.c., 149 linking, 10 linking [hm], 18 linking sets, 3, 51 linking with respect to a minimax system, 9 local linking, 206 lower semi-continuous, 149 minimax system, 8 minimizing sequence, 2

multiple solutions, 191 ordinary differential equations, 31 Palais–Smale sequences, 2 paracompact, 110 partial derivative, 156 periodic-Dirichlet problem, 149 resonance Problems, 129 resonance problems with respect to the Fuˇc´ık spectrum, 130 rotationally invariant solutions, 141 saddle point, 152 sandwich pairs, 57 sandwich theorem, 26 second-order periodic systems, 213 semilinear problems, 63 semilinear wave equations, 149 separate a functional, 3 strictly convex, 149 strong linking, 10 superlinear problems, 85 type (II) regions, 163 u.s.c., 149 upper semi-continuous, 149 wave equation, 141 weak linking, 107 weak sandwich pairs, 177 weakly closed, 151

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