A Topological Introduction to Nonlinear Analysis [3ed.] 3319117939, 978-3-319-11793-5, 978-3-319-11794-2

Table of contents :
Front Matter....Pages i-x
Front Matter....Pages 1-1
The Topological Point of View....Pages 3-7
Ascoli–Arzela Theory....Pages 9-17
Brouwer Fixed Point Theory....Pages 19-23
Schauder Fixed Point Theory....Pages 25-31
The Forced Pendulum....Pages 33-42
Equilibrium Heat Distribution....Pages 43-47
Generalized Bernstein Theory....Pages 49-53
Front Matter....Pages 55-55
Brouwer Degree....Pages 57-61
Properties of the Brouwer Degree....Pages 63-69
Leray–Schauder Degree....Pages 71-76
Properties of the Leray–Schauder Degree....Pages 77-86
The Mawhin Operator....Pages 87-92
The Pendulum Swings Back....Pages 93-100
Front Matter....Pages 101-101
A Retraction Theorem....Pages 103-107
The Fixed Point Index....Pages 109-112
The Tubular Reactor....Pages 113-118
Fixed Points in a Cone....Pages 119-124
Eigenvalues and Eigenvectors....Pages 125-135
Front Matter....Pages 137-137
A Separation Theorem....Pages 139-141
Compact Linear Operators....Pages 143-154
Front Matter....Pages 137-137
The Degree Calculation....Pages 155-163
The Krasnoselskii–Rabinowitz Bifurcation Theorem....Pages 165-178
Nonlinear Sturm–Liouville Theory....Pages 179-189
More Sturm–Liouville Theory....Pages 191-203
Euler Buckling....Pages 205-210
Back Matter....Pages 211-240

Citation preview

Robert F. Brown

A Topological Introduction to Nonlinear Analysis Third Edition

Robert F. Brown

A Topological Introduction to Nonlinear Analysis Third Edition

Robert F. Brown Santa Monica, CA, USA

ISBN 978-3-319-11793-5 ISBN 978-3-319-11794-2 (eBook) DOI 10.1007/978-3-319-11794-2 Springer Cham Heidelberg New York Dordrecht London Library of Congress Control Number: 2014952328 Mathematics Subject Classification (2010): 47H10, 55M20, 47H11, 34B15, 34C23, 34B60, 55M25, 47J10, 34B24, 54H25 © Springer International Publishing Switzerland 1993, 2004, 2014 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. Exempted from this legal reservation are brief excerpts in connection with reviews or scholarly analysis or material supplied specifically for the purpose of being entered and executed on a computer system, for exclusive use by the purchaser of the work. Duplication of this publication or parts thereof is permitted only under the provisions of the Copyright Law of the Publisher’s location, in its current version, and permission for use must always be obtained from Springer. Permissions for use may be obtained through RightsLink at the Copyright Clearance Center. Violations are liable to prosecution under the respective Copyright Law. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. While the advice and information in this book are believed to be true and accurate at the date of publication, neither the authors nor the editors nor the publisher can accept any legal responsibility for any errors or omissions that may be made. The publisher makes no warranty, express or implied, with respect to the material contained herein. Printed on acid-free paper Springer is part of Springer Science+Business Media (www.birkhauser-science.com)

Preface

Nonlinear analysis is a remarkable mixture of topology (of several different types), analysis (both “hard” and “soft”), and applied mathematics. Mathematicians with a correspondingly wide variety of interests should become acquainted with this important, rapidly developing subject. But it’s a BIG subject. You can feel it: just weigh in your hand Eberhard Zeidler’s Nonlinear Functional Analysis and Its Applications (Zeidler, 1986). It’s heavy, as a 900 page book must be. Yet this is no encyclopedia; the preface accurately describes its “: : : very careful selection of material : : : .” And what you are holding in your hand is Part I of a five-part work. So how do you get started learning nonlinear analysis? Zeider’s book has a first page, and some people are quite content to begin right there. For an alternative, the bibliography in Zeidler (1986), which is 42 pages long, contains exposition as well as research results: monographs that explain greater or lesser portions of the subject to a variety of audiences. In particular Deimling (1985) covers much of the material of Zeidler’s book. Then what’s different about the exposition in this book? My answer is in three parts: this book is (i) topological, (ii) goal-oriented, and (iii) a model of its subject. The next three little paragraphs explain what each of these means. (i) As the title states, this is a topological book (though it’s not a book of topology). I’m a topologist and, as I’ve studied nonlinear analysis, I’ve become impressed by the extent to which the subject rests, in a strikingly simple and natural way, on basic topological ideas. These ideas come from general (point-set) topology, from metric space topology and, in the form of classical homology theory, from algebraic topology as well. It’s possible to disguise, or even to replace to some extent, the substantial topological content of this subject, but that won’t happen in this book. On the contrary, we’ll make sure our analysis rests on a secure base of carefully expounded topology. (ii) The goal of this book has a name: the Krasnoselski–Rabinowitz bifurcation theorem. By the time you finish this book you will know what this beautiful result says, understand why it is true, and, through a single but very striking instance, get some idea of how it is applied. You can come to this book with little specific preparation beyond the undergraduate real analysis level. Yet by the end of its v

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Preface

relatively few pages you will see how, in the late twentieth century (ca. 1970), we gained a new understanding of an eighteenth-century model of a column collapsing under excessive weight. (iii) Beyond its power and elegance, the Krasnoselski–Rabinowitz theorem has another virtue that made it irresistible as a topic for this book: the structure of its proof and this application is itself a model of the interplay of topological and analytic ideas that is characteristic of much of nonlinear analysis. The topological ingredients for the proof come from all the branches I mentioned: a separation theorem for compact topological spaces from general topology, Ascoli–Arzela theory from metric space topology, and the Leray–Schauder degree from algebraic topology. A key step in the proof is a calculation formula for the Leray–Schauder degree which, in turn, depends on a substantial topic in functional (“soft”) analysis: the spectral theory of compact linear operators on Banach spaces. The classical “hard” analysis comes into play once we have the relatively abstract bifurcation theorem and want to use it to study the ordinary differential equation problem that models the column buckling. As a curtain raiser to the relatively extensive discussions that lead us to the Krasnoselski–Rabinowitz theorem, I’ll show you a simpler and more classical tool from the nonlinear analyst’s toolbox: the Schauder fixed point theorem, along with a rather easily understood application of it. This is also a model of nonlinear analysis: the topological topics of the Ascoli–Arzela theorem and fixed point theory are applied, with the help of some elementary but clever calculus, to investigate the equilibrium distribution of heat in a rod. Jean Mawhin’s eloquent argument in Mawhin (1988) that much of nonlinear analysis could be illustrated in the context of the forced pendulum suggested some quite direct applications of the two main tools of this book, the Schauder fixed point theorem and the Leray–Schauder degree. In particular, the reader can see a demonstration of the usefulness of the degree before being introduced to bifurcation theory. This book was born at a conference at the University of Montreal organized by Andrzej Granas in 1983 where the talks, especially those of Ronald Guenther, Roger Nussbaum, and Paul Rabinowitz, made nonlinear analysis accessible. UCLA gave me the opportunity to communicate what I was learning about this subject, and to refine these notes, through specialized courses I taught in 1984, 1987, and 1992. The students and colleagues who attended these courses or talked to me about my plans helped me in many ways. I thank especially Joseph Bennish, Jerzy Dydak, Massimo Furi, Reiner Martin, and PierLuigi Zezza. The first time I taught about topology and nonlinear analysis, my late colleague Earl Coddington faithfully attended my lectures and didn’t seem to think it was ridiculous for a topologist to try to present analysis from his own point of view. The fact that this book was written is a consequence of Earl’s encouragement. Santa Monica, CA, USA

Robert F. Brown

About the Third Edition

This new edition presented an opportunity to expand the topological tool kit for nonlinear analysis and its applications. In addition to the Schauder fixed point theorem and the Leray–Schauder degree to which the rest of the book is devoted, the all-new Part III presents the fixed point index and applies this powerful and flexible tool both to a classical mathematical topic, the Krein–Rutman theory of functional analysis, and to the theory of the tubular reactor of chemical engineering. The bifurcation theory of Paul Rabinowitz, which contributed greatly to his qualifications for winning the Birkhoff Prize of the American Mathematical Society and the Society for Industrial and Applied Mathematics, is still the goal of this book and it now occupies Part IV. That part depends only on the Leray–Schauder degree theory of Part II, so a reader whose primary interest is in bifurcation theory can omit Part III or return to it to learn another important consequence of the work of Leray and Schauder. In addition to expanding the book by the presentation of the fixed point index, I have made many revisions throughout the text in an effort to improve the exposition. As I depended for guidance from Jean Mawhin in preparing the second edition, so this edition has benefitted from the very substantial help and encouragement I received from Roger Nussbaum. I am greatly indebted to both of them. Santa Monica, CA, USA September 2014

Robert F. Brown

vii

Contents

Part I Fixed Point Existence Theory 1

The Topological Point of View . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Outline of the Approximation Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Outline of the Topological Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3 4 6

2

Ascoli–Arzela Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

9

3

Brouwer Fixed Point Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

19

4

Schauder Fixed Point Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

25

5

The Forced Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

33

6

Equilibrium Heat Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

43

7

Generalized Bernstein Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

49

Part II Degree Theory 8

Brouwer Degree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

57

9

Properties of the Brouwer Degree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

63

10

Leray–Schauder Degree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

71

11

Properties of the Leray–Schauder Degree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

77

12

The Mawhin Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

87

13

The Pendulum Swings Back . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

93

Part III Fixed Point Index Theory 14

A Retraction Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

15

The Fixed Point Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 ix

x

Contents

16

The Tubular Reactor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

17

Fixed Points in a Cone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

18

Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

Part IV Bifurcation Theory 19

A Separation Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

20

Compact Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

21

The Degree Calculation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

22

The Krasnoselskii–Rabinowitz Bifurcation Theorem . . . . . . . . . . . . . . . . . 165

23

Nonlinear Sturm–Liouville Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179

24

More Sturm–Liouville Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191

25

Euler Buckling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

A

Singular Homology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.1 Construction of Homology Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.2 Basic Properties of Homology. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.3 Homology of Spheres and Brouwer’s Theorem . . . . . . . . . . . . . . . . . . . . .

B

Additivity and Product Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217

C

Bounded Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223

D

Metric Implies Paracompact . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229

211 211 213 214

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 Table of Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237

Part I

Fixed Point Existence Theory

Chapter 1

The Topological Point of View

This book is about the topological approach to certain topics in analysis, but what does that really mean? Starting with the “epsilon–delta” definitions of limits in calculus, analysis makes extensive use of topological ideas and techniques. Thus the issue is not whether analysis requires topology, but rather how central a role the topological material plays. Rather than attempt the hopeless task of defining precisely what I mean by the topological point of view in analysis, I’ll illustrate it by outlining two proofs of a well-known theorem about the existence of solutions to ordinary differential equations. In the first proof, the key step is the construction of a sequence of approximate solutions whose limit is the required solution. In the second proof, a general topological theorem about the behavior of self-maps of linear spaces implies the existence of the solution. The two proofs have several features in common, including their dependence on a substantial topological result, but I trust that even my (intentionally) very sketchy treatment will make it clear how basic the differences are in the ways that the two arguments reach the same conclusion. Here’s the theorem. Theorem 1.1 (Cauchy–Peano Existence Theorem). Given a function f W R2 ! R which is continuous in a neighborhood of a point .x0 ; y0 / 2 R2 , there exist ˛ > 0 and a solution to the initial-value problem y 0 D f .x; y/

y.x0 / D y0

on the interval Œx0 ˛; x0 C˛. That is, there exists a function  W Œx0 ˛; x0 C˛ ! R such that .x0 / D y0 and  0 .x/ D f .x; .x// for all x in the interval. The two proofs produce the number ˛ in the same way. Since f is continuous in a neighborhood of .x0 ; y0 / 2 R2 , there exists a > 0 such that if .x; y/ 2 R2 with jx  x0 j  a and jy  y0 j  a, then f is continuous at .x; y/. Let Q be the square in the plane consisting of such points, that is,

© Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__1

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1 The Topological Point of View

Q D f.x; y/ 2 R2 W jx  x0 j  a

jy  y0 j  ag

and

and choose M > 1 such that M  jf .x; y/j for all .x; y/ 2 Q. Then set ˛ D Ma . Notice how the definition of ˛ depended on some familiar topology. A neighborhood means an open set and therefore the euclidean topology of the plane gives us an open disc about .x0 ; y0 / on which f is continuous. We choose a small enough to fit the square Q inside the open disc. We know that the set of values jf .x; y/j for .x; y/ 2 Q is bounded, and therefore M exists, because Q is closed and bounded, that is, compact, so by a standard result its image under the continuous function jf j is a compact subset of the line and therefore bounded. Another feature the two proofs have in common is that they make use of the fact that the fundamental theorem of calculus gives us, as an equivalent form of the initial-value problem, the integral equation Z x y.x/ D y0 C f .t; y.t //dt: x0

That is, a function  W Œx0  ˛; x0 C ˛ ! R is a solution to the initial-value problem if and only if it is a solution to the integral equation. The remaining common feature is that substantial topological result I referred to earlier: the Ascoli–Arzela theorem. I’ll indicate in both proofs where and how this theorem is used, but in neither case is it necessary to state the result itself. However, I’ll present a detailed discussion and proof of this theorem in the next chapter because the Ascoli–Arzela theorem will play a crucial role throughout the entire book.

1.1 Outline of the Approximation Proof For each integer n  1, choose ın > 0 small enough so that jx  xj < ın and jy  yj < ın implies jf .x; y/  f .x; y/j
0, an -net S in a metric space X is a subset of X with the property that every point of X is within  of some point of S . To say the same thing more precisely, for x 2 X , define B.xI / D fy 2 X W d.x; y/ < g; where d is the metric of X , and say that a subset S of X is an -net if X is the union of all B.sI / for s 2 S . A metric space X is totally bounded (some people prefer the term precompact) if given an  > 0, there is a finite -net for X . For instance, every

© Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__2

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2 Ascoli–Arzela Theory

bounded subset of Rn has this property. Notice that a subset of a totally bounded set is also totally bounded. Furthermore, it is not hard to see that if a subset of a space is totally bounded, so is the closure of that subset. It is easy to demonstrate that if X is a compact metric space, then it is totally bounded: given  > 0, form the cover fBx D B.xI /gx2X of X and, for a finite subcover fBxj g, the set fxj g is an -net for X . The first step on the road toward the Ascoli–Arzela theorem is to show that, in complete metric spaces, the converse is also true. Theorem 2.1. Let X be a complete metric space. If X is totally bounded, then it is compact. Proof. Suppose that the complete, totally bounded space X is not compact, in other words, that there is an open cover of X that has no finite subcover. We’ll see that this leads to a contradiction. The existence of the cover will allow us to build a sequence fxn g in X with the following two properties: (1) d.xn ; xnC1 / < 2nC1 ; (2) the subset B.xn I 2nC1 / cannot be covered by a finite subfamily of the cover. Before constructing the sequence, let’s see why it proves the theorem. First of all, condition (1) implies that the sequence is Cauchy. To verify the Cauchy property, suppose we are given  > 0, then choose natural numbers m and n with, say, m  n and large enough so that 2mC2 < , then d.xm ; xn / <  because d.xm ; xn / 

n1 X kDm

d.xk ; xkC1 / 

n1 X

2kC1  2mC2 :

kDm

Let x be the limit of fxn g and let U be a member of the cover that contains x. Choose  small enough so that B.xI 2/  U . Then choose n large enough so that 2nC1 <  and also, by convergence of the sequence, d.xn ; x/ < . But that tells us that B.xn I 2nC1 /  U so B.xn I 2nC1 / certainly can be covered by a finite subfamily of the cover, in fact by the single member U , contrary to (2). Thus once we construct the sequence fxn g we know it will lead us to a contradiction, as we had hoped, so we now construct it—inductively. To identify x1 , use the total boundedness of X to construct a finite 1-net fy1 ; y2 ; : : : ; ym g. If each B.yj I 1/ could be covered by a finite subcover of the cover, we would have a finite subcover that covered all of X ; therefore there is a yj such that B.yj I 1/ cannot be so covered and we set x1 equal to that yj . Supposing that fx1 ; x2 ; : : : ; xn g have been found which satisfy (1) and (2), we need to produce xnC1 . To do that, we note that B.xn I 2nC1 / is totally bounded since X is and concentrate now on the subspace to locate xnC1 , essentially in the same way we found x1 . The subspace has a finite 2n -net fy1 ; y2 ; : : : ; ym g. Since (2) holds by assumption, that says B.xn I 2nC1 / cannot be covered by any finite subfamily of the given cover. Therefore there is a yj such that B.yj I 2n / has the same property and we let xnC1 D yj , which gives us property (2), and of course (1) holds because this all takes place within B.xn I 2nC1 /. t u

2 Ascoli–Arzela Theory

11

It often happens that we don’t need a set within a space to be compact itself, but we do want the set to lie in a compact subset of the space. When a subset S in a space X is contained in a compact subset K of X , then S is called a relatively compact subset of X . Since K is closed in X , it contains the closure S of S , as a closed subset, and thus S is itself compact. Therefore an equivalent definition of a relatively compact subset is one whose closure is compact. Theorem 2.1 gives us a lot of relatively compact subsets of complete metric spaces. If S is a totally bounded subset of a complete metric space X , then S is also totally bounded and it is complete because it is closed in X . Thus, by Theorem 2.1, we have Corollary 2.2. A totally bounded subset S of a complete metric space X is relatively compact. It follows that bounded subsets of euclidean spaces are relatively compact, but we knew that anyway. What we really want is a nice way to identify some relatively compact subsets of another class of complete metric spaces which we describe next. Let E be a metric space with metric denoted by d and let u W E ! R be a real-valued function which, for the moment, need not be continuous. However, we do want u to be bounded, that is, u.E/ must be a bounded subset of the reals. Then we can define the supremum norm (known familiarly as the sup norm, pronounced “soup”) kuk of u to be the least upper bound (also known as the supremum) of the set fju.x/j W x 2 Eg. The set B.E/ of all bounded real-valued functions on E is a metric space with distance between functions u and v given by ku  vk. Another name for kuk is the uniform norm and that emphasizes the fact that convergence of a sequence of functions in B.E/ is what real analysis texts call uniform convergence of the sequence. The space B.E/ is complete because it inherits that property from the reals. What I mean by this is that if you start with a sequence fun g in B.E/ that is Cauchy, then for each x 2 E the sequence of reals un .x/ is Cauchy and therefore has a limit we’ll call u.x/. This defines a function u and that function is bounded, so the limit is still in B.E/. The space we are really interested in, though, is C.E/, the bounded, continuous real-valued functions on E. Since convergence in B.E/ is uniform convergence, the limit of a sequence of continuous functions in B.E/ must also be continuous. In more topological language, C.E/ is a closed subset of the space B.E/. Therefore, C.E/ is also a complete metric space. The main result of this section identifies a useful class of relatively compact subsets of C.E/. By Corollary 2.2, since C.E/ is complete, all totally bounded subsets will be relatively compact. However, total boundedness of a set of realvalued functions isn’t an easy hypothesis to check, so we’ll replace it by two other hypotheses. One of the hypotheses is boundedness: a set A in C.E/ is bounded if there is a number ˇ such that kuk < ˇ for all functions u 2 A. To state the other hypothesis, a set A in C.E/ is said to be equicontinuous at x 2 E if given  > 0, there exists ıx > 0 such that if x; y 2 E with d.x; y/ < ıx , then ju.x/  u.y/j <  for all u 2 A. (Note that the functions u are real-valued and we are using the usual

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2 Ascoli–Arzela Theory

notion of distance in R when we write ju.x/  u.y/j.) Call the set A equicontinuous if it is equicontinuous at all x 2 E. Surprisingly, these hypotheses are reasonably convenient to check, and that’s how we’ll find relatively compact subsets of C.E/ when E itself is a compact space (a hypothesis we haven’t used up to this point) because of the following. Theorem 2.3 (Ascoli–Arzela Theorem). Let E be a compact metric space. If A is an equicontinuous, bounded subset of C.E/, then A is relatively compact. Proof. The idea is to prove that A is totally bounded by starting with a given  and use it in conjunction with the definition of equicontinuity to produce a finite -net (of functions, remember) for A. Specifically, for each x 2 E we use ıx from that definition, chosen so that if d.x; y/ < ıx then ju.x/  u.y/j < 4 for all u 2 A. This gives us a cover fB.xI ıx /g of E and here is where we use the compactness of the space E, to extract a finite subcover which we denote by fB.xj I ıj /g for j D 1; 2; : : : ; n. (In a sense, we want the finite set fxj g to substitute for the points of E.) For some xj (just one of them) look at the set of points fu.xj /g for all u 2 A, which we denote by AŒxj . The set AŒxj  is bounded in R since we assumed that A is a bounded subset of C.E/, so there is a finite 4 -net fz1 .xj /; z2 .xj /; : : : ; zk.j / .xj /g for the set AŒxj . This just means that given u 2 A, there is some zs .xj / for which ju.xj /  zs .xj /j < 4 and we want the net to substitute in some way for AŒxj . Let  D f1 ; 2 ; : : : ; n g denote an n-tuple of integers with the property 1  s  k.s/. For each , let v be a function in A such that jv .xj /  zj .xj /j
0, we will show that ı D ˇ will establish all the equicontinuity required. In other words, for any x 2 Œa; b, we will see that if jx  yj < ı, then .j / .j / jun .x/  un .y/j <  for any n and any j  k. The reason is the mean value theorem which implies that ˇ ˇ ˇ u.j / .x/  u.j / .y/ ˇ n ˇ ˇ n C1/ .c/j < ˇ ˇ D ju.j ˇ n ˇ ˇ xy for some c 2 Œa; b. The Ascoli–Arzela theorem (specifically its Corollary 2.5) gives us a function g in C Œa; b which is the limit of a subsequence of fun g. The derivatives of the functions in the subsequence are still bounded and equicontinuous, so they also have a subsequence which converges to a function in C Œa; b that we will call g . If we differentiate the functions in that subsequence, we can repeat the argument to obtain a continuous function g . We continue in this way until we have the function g . To avoid some potentially disastrous notational problems replace the original sequence, using only those terms that were used in the last step (which are a subset of those used in every previous one), but still call the sequence fun g. In this way, we have constructed functions g ; g ; g ; : : : ; g such .j / that un ! g for j D 0; 1; 2; : : : ; k, where the arrow indicates uniform convergence. Letting u D g , we will show that g D u.j / for j  k, which implies that u 2 C k Œa; b and also that fun g converges to u in that space, that is, we can make kun  ukk as small as we want by choosing n big enough. For 1  j  k, use the fundamental theorem of calculus to write Z x 1/ / .j 1/ .x/ D u.j .a/: u.j n n .t / dt C un a

Taking the limit, on the left-hand side we have g .x/ and on the other side we get Z a

x

g .t / dt C g .a/:

By uniqueness of the limit, we see that g is the derivative of g .

t u

We now have a useful tool for extracting convergent sequences, although I’m afraid you’ll have to wait a while to see it employed. The difficulty is that you can’t do much with it alone, so we’ll have to build more machinery in order to get a good payoff from all the work we did in this section. On the other hand, a lot of the concepts we met here will turn up again. To conclude the chapter, here is the proof I promised earlier. Proof of Theorem 2.4 (A metric space is compact if and only if every sequence has a convergent subsequence). If we assume we have a sequence fxn g in a compact space X , it’s not hard to extract a convergent subsequence; here’s how to do it.

2 Ascoli–Arzela Theory

17

If for each x 2 X we could find x > 0 such that B.xI x / contained only a finite subset of fxn g, then a finite subcover of fB.xI x /g would fail to cover all of the sequence, so it could hardly cover X . Thus there exists x 2 X such that B.xI n1 / contains infinitely many points of the sequence, so we can choose a different point xj.n/ 2 B.xI n1 / in the sequence for each n and that subsequence will do the job. The proof in the other direction uses the ideas of this chapter. We will show that if X is a space in which every sequence has a convergent subsequence (called a sequentially compact space) and X is metric, with metric d , then X is complete and totally bounded, so it is compact by Theorem 2.1. We can get completeness out of the way very quickly. Take a Cauchy sequence fxj g in the sequentially compact metric space X , so there is a subsequence fxjk g converging to some point x, but we want the entire sequence to converge to x. Given  > 0, there exists N1 > 0 such that jk  N1 implies d.xjk ; x/ < 2 . By the Cauchy property, there exists N2 > 0 such that j; jk  N2 implies d.xj ; xjk / < 2 . Therefore, if j is greater than both N1 and N2 , we get d.xj ; x/ < . To prove total boundedness and finish this chapter at last, we use a contrapositive argument. That is, we suppose that X is not totally bounded so there exists  > 0 for which X has no finite -net and show that X is therefore not sequentially compact, by constructing a sequence in X that has no convergent subsequence. Let any point of X be x1 , but choose x2 so it is not within  of x1 , where the  is the one for which there is no finite net. If there were no such x2 , that would mean x1 by itself was an -net for X . Choose x3 to be a point of X at a distance of more than  from both x1 and x2 . Continuing in this way, we choose xn 2 X that is in the complement of the union of the B.xj I / for j D 1; 2; : : : ; n  1. The next point of the sequence has to exist since otherwise the previous points would constitute a finite -net for X . By construction, no subsequence of fxn g converges. t u

Chapter 3

Brouwer Fixed Point Theory

A topological space Y has the fixed point property, abbreviated fpp, if every map (continuous function) f W Y ! Y has a fixed point f .y/ D y for some y 2 Y . The fixed point property is a topological property in the sense that it is preserved by homeomorphisms. That is, it’s easy to see that if a space Y has the fpp and Z is homeomorphic to Y , then Z also has the fpp. The fixed point theorem quoted in Chap. 1 as the key to the topological proof of the Cauchy–Peano theorem states that a compact, convex subset of a normed linear space has the fpp. We’ll prove that theorem and more in Chap. 4. The proof is accomplished in two steps: first prove a finite-dimensional fixed point theorem, and then generalize to normed linear spaces. This chapter will be devoted to the first of these steps. Let’s look at euclidean n-space Rn as a very special kind of normed linear space. Its elements are ordered n-tuples x D .x1 ; x2 ; : : : ; xn / of real numbers and Rn is given the structure of a real vector space with the usual vector addition and scalar multiplication along with the inner (dot) product x  y D .x1 ; x2 ; : : : ; xn /  .y1 ; y2 ; : : : ; yn / D x1 y1 C x2 y2 C    C xn yn : The norm of x is defined by jxj D

q

x12 C x22 C    C xn2 D

p

x  x:

The unit ball in Rn , denoted by B n , consists of all x 2 Rn with jxj  1. The usual euclidean distance between x and y in Rn is given by jx  yj. The fixed point theorem of Chap. 1 is a generalization of the following famous result. Theorem 3.1 (Brouwer Fixed Point Theorem). The unit ball B n has the fpp. I’m not going to prove Brouwer’s theorem at this point. If you are a fan of algebraic topology, you believe the result is true and can probably remember how to prove it. If you aren’t, I’m betting you’d rather not start off your journey through topology © Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__3

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20

3 Brouwer Fixed Point Theory

and analysis by scaling a wall of homology theory. So I’ll just assume the theorem. However, if it bothers you to generalize a result that you have not seen proved, by all means pay a visit to Appendix A. There you will find the basic definitions of singular homology theory, a summary of the properties of this theory, and the necessary computations and argument to establish Brouwer’s theorem in a reasonably selfcontained manner. The purpose of the appendix is not only to present a proof of the Brouwer theorem but also to exhibit the basic facts of homology theory, because we will need to refer to them later. So if you don’t check up on Brouwer now, you should still make a mental note that an outline of homology theory can be found in Appendix A. The traditional form of Brouwer’s result, that we quoted as Theorem 3.1, is not general enough for our purposes; we want something that more closely resembles the statement about compact convex subsets that we used in Chap. 1. Here it is. Theorem 3.2 (Generalized Brouwer Fixed Point Theorem). A compact, convex subset Q of Rn has the fpp. The rest of this section will be devoted to a proof of Theorem 3.2. The argument that it is a consequence of the traditional Brouwer theorem depends on a simple fact about the fpp. A subset A of a space X is a retract of X if there is a retraction of X onto A, that is, a map  W X ! A such that .x/ D x for all x 2 A. The fact is Lemma 3.3. If A is a retract of a space X and X has the fpp, then A also has the fpp. Proof. There is a retraction  W X ! A by hypothesis and we let i W A ! X be inclusion. Now given any map g W A ! A, define f W X ! X to be the composition 

g

i

f W X ! A ! A ! X then f .x/ D x for some x 2 X . But the image of f is contained in A, so .x/ D x and therefore g.x/ D x. t u The boundedness of the compact set Q in Theorem 3.2 means there exists r > 0 such that Q is contained in the n-ball Brn defined to be the set of all x 2 Rn with jxj  r. Certainly Brn has the fpp because it is homeomorphic to the unit ball B n , which the Brouwer Theorem 3.1 assures us has the fpp. By the end of this chapter, we will have demonstrated that Q is a retract of Brn , and that will prove the generalized form of the theorem. For each x 2 Rn , there is a unique point qx of Q that is “closest” to x in the sense that jx  qx j < jx  qj for any q of Q other than qx . The fact that at least one such qx exists is immediate from the compactness of Q: the function Dx W Q ! R defined by Dx .q/ D jx  qj is continuous and thus attains its minimum. To see why qx is unique, suppose q and q 0 are any two points of Q for which jx  qj D jx  q 0 j and assume for now that the points x; q, and q 0 are noncollinear. Let q 00 D 12 q C 12 q 0 , then x; q, and q 00 determine a plane in which those three points form a right triangle, so jx  q 00 j < jx  qj by the Pythagorean theorem.

3 Brouwer Fixed Point Theory

21

x Q q q  q

Fig. 3.1 Uniqueness of qx

See Fig. 3.1. If x; q, and q 0 are collinear, the argument is even easier. Notice that the convexity of Q is crucial to the uniqueness of the closest point. We need to improve our understanding of the relationship between x and that closest point qx in Q, beginning with the following, possibly mysterious looking, technical lemma. Lemma 3.4. If x 2 Rn and z 2 Q, a convex subset, then .z  qx /  .x  qx /  0: Proof. For t  0, define a non-negative, real-valued function by .t/ D jx  .qx C t .z  qx //j2 then  is differentiable from the right at t D 0. The convexity of Q tells us that Œqx C t .z  qx / 2 Q so the definition of qx implies that  is minimized at t D 0 and therefore  0 .0/  0. Now  0 .t / D 2.x  Œqx C t .z  qx //  ..z  qx // and therefore  0 .0/ D 2.x  qx /  .z  qx /  0:

t u

The lemma isn’t really mysterious at all if you look at the plane determined by x, z, and qx . In a plane, we know that the dot product of vectors depends on the cosine of the angle between them, so Lemma 3.4 just tells us that the angle between z  qx and x  qx is obtuse. Figure 3.2 indicates what this lemma implies about closest points: a shrinking of distance occurs when we replace points in Rn by their closest points in Q. The next result makes this statement precise.

22

3 Brouwer Fixed Point Theory

x

y

qx Q

qy

Fig. 3.2 Shrinking the distance

Lemma 3.5. For all x; y 2 Rn , and qx ; qy their closest points in the compact, convex set Q, jqx  qy j  jx  yj: Proof. We apply the previous lemma, first taking z D qy , so we have .qy  qx /  .x  qx /  0; then with z D qx , and y replacing x: .qx  qy /  .y  qy /  0: Adding these inequalities produces .qy  qx /  Œ.x  qx / C .qy  y/  0 which we rewrite as .qy  qx /  Œ.qy  qx / C .x  y/  0 and apply the distributive law jqy  qx j2 C .qy  qx /  .x  y/  0: By the Schwarz inequality, jqy  qx j2  .y  x/  .qy  qx /  jy  xjjqy  qx j and we finish by dividing by jqy  qx j.

t u

3 Brouwer Fixed Point Theory

23

Defining  W Brn ! Q by .x/ D qx produces a retraction since certainly .q/ D q for q 2 Q and  is continuous by Lemma 3.5. We have completed the proof of the generalized Brouwer Theorem 3.2: a compact, convex subset Q of euclidean space Rn really does have the fpp.

Chapter 4

Schauder Fixed Point Theory

The purpose of this chapter is to extend the Brouwer fixed point theory of maps on euclidean spaces to results about maps on normed linear spaces in general. Then, in the next chapter, we will combine the Ascoli–Arzela theory with this material to draw conclusions about fixed points of maps, specifically on those C k spaces we discussed in Chap. 2. Let X be a (real) normed linear space and F D fx1 ; x2 ; : : : ; xn g a finite subset of X . Then con.F /, the convex hull of F , is defined by con.F / D

8 n 0, there exist a finite subset F of X and a map P W K ! con.F /, called the Schauder projection, such that d.P .x/; x/ <  for all x 2 K. Proof. As the set F D fx1 ; : : : ; xm g, we take a finite -net for the compact set K. For i D 1; : : : ; m, define functions i W K ! R by letting i .x/ D   d.x; xi / if x 2 B.xi I / and i .x/ D 0 otherwise. See Fig. 4.1. Defining .x/ to be the sum of all the numbers i .x/, we see that .x/ > 0 for all x 2 K, since F is an -net. Define the Schauder projection by

fi(x) x

å

xi B(xi;å)

Fig. 4.1 Definition of i .x/

X

4 Schauder Fixed Point Theory

27

P .x/ D

m X i .x/

.x/

iD1

xi

which is continuous because the i are. The distance computation is    m  m m X  X i .x/ X i .x/ i .x/       x/ x x D .x d.P .x/; x/ D  i i     .x/ .x/ .x/ iD1



m X i .x/ iD1

.x/

i D1

iD1

kxi  xk
0 such that kxk D r implies f .x/ 6D x for all  > 1. See Fig. 4.2. In particular, since kxk D kxk, it is clear that f satisfies the Leray–Schauder boundary condition if kxk D r implies kf .x/k  r, and it is in this form that the condition most frequently makes its appearance. Notice that the condition imposes no restriction on how large r can be. Many functions that arise in interesting analytic problems turn out to satisfy “growth conditions” which, as kxk becomes large enough, prevent kf .x/k from growing at the same rate, so that eventually kxk  kf .x/k, which produces the Leray–Schauder condition in this special form.

{λ x : λ > 1}

x f (x) r 0

X

Fig. 4.2 The Leray-Schauder boundary condition

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4 Schauder Fixed Point Theory

The result that replaces the requirement in Schauder’s theorem, of a convex set carried to itself, by the Leray–Schauder boundary condition is really just a corollary of the Schauder theorem. But it is referred to often enough to merit a name of its own. Theorem 4.6 (Leray–Schauder Alternative). Let f W X ! X be a completely continuous map of a normed linear space and suppose f satisfies the Leray– Schauder boundary condition; then f has a fixed point. Proof. The Leray–Schauder condition gives us r > 0 such that kxk D r implies f .x/ 6D x for all  > 1. For the closed, convex set C required by the Schauder theorem, we use C D Br D fx 2 X W kxk  rg that is, the ball in X of radius r. If we restrict the given map f to Br , we have a map we write as f jBr W Br ! X . The map f jBr is compact because Br is bounded, but there is no reason to expect f jBr to map Br back into itself. In order to modify f for the purpose of getting the image into Br , we will define a map  W X ! Br and use f  D f jBr W Br ! Br . The map  is a retraction of X onto Br , which, remember, means that it is the identity on the subset Br , and its definition for x with kxk  r is .x/ D

r x: kxk

We know that f .Br / is contained in a compact set, call it K, then f  .Br /  .K/, which is compact since  is continuous, so f  is also a compact map. Therefore, the hypotheses of the Schauder fixed point theorem are satisfied (I told you this was really a corollary) and there exists x 2 Br such that f  .x/ D x. We claim that x has the property f .x/ 2 Br so f  .x/ D .f .x// D f .x/ and thus x is in fact the required fixed point of f . The argument that f .x/ 2 Br depends on the Leray– Schauder boundary condition. Suppose f .x/ 62 Br , that is, kf .x/k > r; therefore x D f  .x/ D .f .x// D

r f .x/: kf .x/k

Now, on the one hand, this last equation implies     r  f .x/ kxk D  Dr kf .x/k and, on the other hand, the equation can be rewritten as f .x/ D

kf .x/k x D x r

4 Schauder Fixed Point Theory

31

where  > 1 because we assumed kf .x/k > r. Thus the Leray–Schauder condition is precisely the hypothesis we need for the statement f .x/ 62 Br to lead to a contradiction and we can conclude that f .x/ D x. t u This result is called an “alternative” because it states that either the equation f .x/ D x for  > 1 has solutions with kxk arbitrarily large, or else the equation f .x/ D x must have a solution. However, neither condition excludes the other.

Chapter 5

The Forced Pendulum

The rest of Part I will be devoted to demonstrating the usefulness of two of the tools we have developed: the Schauder fixed point theorem and a compactness property of C k -spaces that is a consequence of the Ascoli–Arzela theorem. We used information from the Ascoli–Arzela and Schauder theories in Chap. 1, to prove the Cauchy– Peano theorem by topological methods. In this chapter, we will illustrate the use of these tools by showing how they establish the existence of solutions to a differential equation problem that comes up in the study of the pendulum. Figure 5.1 concerns one of the classical topics of elementary physics: the free simple pendulum. A wire of length ` is attached at the point A and an object is placed at the other end of the wire. We’ll think of the object as a ball in order to have something specific in mind. But we’ll concentrate all the weight of the ball at one point so that we can describe the motion of the ball as the motion of the point we have called P . The ball can swing back and forth in the x; y-plane of the figure, in which the origin has been placed at the rest position of the ball. That’s all there is to a simple pendulum. What makes it “free” is the assumption that the only force acting on the pendulum is that of gravity, which we can think of as the weight of the ball at the end of the wire. We can express that weight as mg where m is the mass of the ball and g is the gravitational constant (32 ft/s2 ). We are assuming that the wire has so little mass compared to that of the ball attached to it that we can ignore the weight of the wire itself. One of the fundamental ideas in physics is called the principle of conservation of energy which states that T C V D E; where T stands for kinetic energy, V for potential energy, and E, the total energy, is a constant. In the case of the pendulum, the kinetic energy depends on the mass of the ball and on how rapidly the arc length of the path of the ball, called s in the figure, is changing. Specifically, the kinetic energy is given by © Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__5

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5 The Forced Pendulum

Fig. 5.1 The free simple pendulum

T D

m 2



ds dt

2 :

The potential energy of the pendulum is just the weight of the ball times the distance the ball must fall to reach the rest position. Thus, when the ball is at the location P D .x; y/, the potential energy is V D mgy: The path of the ball lies on the circle of radius ` with center at A so, by a familiar formula, its arc length is s D ` where, as in the figure, is the angle the wire makes with the vertical axis. We can also relate the height y to that angle because cos D

`y `

and therefore y D `.1  cos /. Putting all this together, we’ll write the principle of conservation of energy for the free simple pendulum as an equation involving the angle , which will vary with the motion of the ball over time, and some constants: m 2 ` 2



d dt

2 C mg`.1  cos / D E:

5 The Forced Pendulum

35

When we differentiate that equation, we get m`2

d d 2 d C mg` sin D 0: dt dt 2 dt

The equation simplifies to the following equation concerning the angle the wire makes with the vertical axis: g d 2 C sin D 0: 2 dt ` There is an important variant of the free simple pendulum, called the forced pendulum, in which the pendulum is not free, that is, subject only to gravity, but instead it is subject to an outside force. That force, described by a function of time called the forcing term, replaces the zero on the right-hand side of the last equation. A particularly significant type is forcing occurs when the pendulum is made to vibrate according to the equation d 2 g C sin D c sin !t; dt 2 ` where c and ! > 0 are constants. Since this forcing term is an odd periodic map, we will look for odd periodic maps as solutions to the equation. It will be easier to describe the procedure for constructing such a map if we first rewrite the equation in a simpler and more general form like this: y 00 C a sin y D e; where we conveniently forget that the solution y represents an angle , we set a D g` and we allow the forcing term to be any odd T -periodic map e D e.t /. Instead of trying to solve the equation for a function y D y.t / defined for all real numbers t, we will seek a solution only for t in the interval Œ0; T2 . In particular, suppose we can find a map yW Œ0; T2  ! R such that y.0/ D y. T2 / D 0 and y.t / satisfies y 00 .t / C a sin y.t / D e.t / for all t 2 Œ0; T2 . We can extend y to the interval Œ T2 ; T2  by setting y.t / D y.t / for  T2  t  0 and we’ll get a continuous function because y.0/ D 0. This is a solution to the differential equation on Œ T2 ; T2  because the sine and e are odd functions. Periodicity then lets us complete the solution as follows. Cover the real line by the intervals fŒ T2 C.k1/T; T2 CkT g for all integers k and, for t 2 Œ T2 C.k1/T; T2 CkT , set y.t/ D y.t kT /. Notice that writing t D kT Ct0 for some integer k and t0 in the interval Œ T2 ; T2 , we have t D .k/T C .t0 / where t0 lies in the same interval and therefore y.t / D y.t0 / D y.t0 / D y.t / so we do obtain an odd function on R in this way. The continuity of the solution on all of R depends on the fact that y. T2 / D 0 because, if t D T2 C kT so that y.t/ D y. T2 /, then we can also write t D T2 C ..k C 1/  1/T in which case the definition gives y.t/ D y. T2 / D y. T2 /.

36

5 The Forced Pendulum

We have reduced the problem of the forced pendulum to the following: y 00 C a sin y D e   T D 0; y.0/ D y 2 where eW Œ0; T2  ! R is a given map and a is a nonzero constant. The boundary condition is a standard one in the theory of differential equations known as the Dirichlet boundary condition. Notice that the differential equation is nonlinear because sin y is a nonlinear function of the solution y. Thus our problem is a secondorder nonlinear Dirichlet boundary value problem and that is the kind of problem that we will be concerned with throughout the rest of Part I. With later applications in mind, it will be convenient to place our forced pendulum problem in a broader context of second-order Dirichlet problems, namely: y 00 D f .t; y; y 0 / y.0/ D y.1/ D 0; where f D f .t; p; q/W Œ0; 1  R  R ! R is continuous. For the forced pendulum with e an odd T -periodic function, and the function f is defined by f .t; p; q/ D

T2 .e. T2 4

t /  a sin p/:

Once we have found out enough about the general problem to show that a solution y to this problem exists, we will return to the forced pendulum and demonstrate that the problem has a solution for all constants a ¤ 0 and all such maps e because, letting u.t / D y. 2t /, it’s easy to see that u00 D e  a sin u and certainly u.0/ D T u. T2 / D 0. Thus, if we start with any odd T -periodic map e as the forcing term, the construction above will give us an odd T -periodic solution of the forced pendulum problem on all of R. Since we will be using fixed point theory, we need to characterize a solution to the boundary value problem as a fixed point of a map. I’ll first do it informally, in hopes of convincing you that this is a very natural way to describe the problem. Let’s focus on the differential equation y 00 D f .t; y; y 0 /, but think of the left-hand side not as the second derivative of a particular function y but rather as the result of performing an operation on any function, that is, finding its second derivative. The operation can be described as a function, which we’ll call L, that takes a function v D v.t /, defined for 0  t  1, to its second derivative, that is, Lv D v00 . We need to be more specific about the function L, in the sense of saying something about its domain and range. We have already required a function v in the domain of L to be defined on the interval Œ0; 1 but notice also that it doesn’t make any sense to write Lv D v00 unless v has a second derivative. And that isn’t sufficient since the differential equation can now be written as Ly D f .t; y; y 0 /, and we are told that the

5 The Forced Pendulum

37

function f is continuous, so we must also require that Ly be continuous. Thus, with regard to the domain of the function L, we are only interested in functions v whose second derivatives not only exist but are continuous. In the language we developed in Chap. 2, we want v to be in the linear space C 2 Œ0; 1. Since the image Lv D v00 is therefore a continuous function, the range of L is the linear space C Œ0; 1. We will need to modify this statement slightly in a moment, but for now let’s write L as a function L W C 2 Œ0; 1 ! C Œ0; 1. I called the function L, as most people do, because it is certainly a linear function. Now let’s suppose that L was not just a linear function but actually an isomorphism with a (continuous) inverse L1 . Then we could rewrite the differential equation Ly D y 00 D f .t; y; y 0 / as y D L1 f .t; y; y 0 /. This really would be a fixed point problem because we could define a function S by setting S.y/ D L1 f .t; y; y 0 / and then a solution to the differential equation would be described by y D S.y/, that is, a fixed point of S . I can’t really get away with all that for two reasons. The way I described L, it doesn’t have an inverse. Also, even if we proved that there exists y 2 C 2 Œ0; 1 for which S.y/ D y, that wouldn’t really solve the boundary value problem we stated since there would be no reason to suppose the function y would satisfy the Dirichlet boundary condition y.0/ D y.1/ D 0. Fortunately, the solution to the second problem solves the first one as well. Let C02 Œ0; 1 be the subset of C 2 Œ0; 1 consisting of the functions that satisfy the Dirichlet condition and now let L W C02 Œ0; 1 ! C Œ0; 1 be the restriction of L to the subset. Certainly C02 Œ0; 1 is a subspace of C 2 Œ0; 1 and L is still a linear function. But L W C02 Œ0; 1 ! C Œ0; 1 is more; it is a linear isomorphism; there is a continuous inverse function L1 W C Œ0; 1 ! C02 Œ0; 1. I’m about to tell you how L1 is defined, but let’s be sure to notice that a function y 2 C02 Œ0; 1 for which S.y/ D L1 f .t; y; y 0 / D y really does solve the boundary value problem because not only does y 00 D f .s; y; y 0 / but, since y is in the subspace, we know that y.0/ D y.1/ D 0 as well. If we write Lv D v00 D w where w 2 C Œ0; 1 then, in order to define L1 so that 1 L .w/ D v, we need to solve the equation v00 D w for v. The equation v00 D w for a given function w is a linear differential equation and there is classical theory available for dealing with it. A differential equation can be written as Dy D f where D is a differential operator and f is a function. The differential equation Dy D 0 is a called the homogeneous equation corresponding to the original inhomogeneous equation Dy D f . If y1 and y2 are two solutions to Dy D f and the operator D is linear, then y1  y2 is also a solution to the homogeneous equation. Turning that around, if you know all the solutions to the homogeneous equation and just one solution to the inhomogeneous linear differential equation, then you know the general solution to the original equation because you can just add that one solution to the general solution to the homogeneous equation. In our case, the corresponding homogeneous equation is v00 D 0 and we know that its solutions are all the linear functions, which I’ll write this way: vC .t / D c1 C c2 t . Now what we need to do is figure out one solution to v00 D w in order to find all of them.

38

5 The Forced Pendulum

To find that one solution, we can use a classical technique called variation of parameters. Replacing the constant parameters c1 and c2 in the general solution to the homogeneous equation, that is, vC .t / D c1 C c2 t , by functions (variable parameters) u1 .t / and u2 .t /, we will look for a solution to v00 D w of the form vP .t / D u1 .t / C u2 .t /  t: Since v0P .t / D u01 .t / C u2 .t / C u02 .t /  t , it is clear that, if we can solve the equations u01 .t /Cu02 .t /t D 0 and u02 .t / D w.t / for u1 and u2 , then vP .t / will be the solution we wanted. All we need is the fundamental theorem of calculus to solve those equations and we find that Z t .t  s/w.s/ ds vP .t / D 0

and therefore that the general solution to v00 D w is Z

t

v.t / D vC .t / C vP .t / D c1 C c2 t C

.t  s/w.s/ ds:

0

Now we need to remember that the reason we were interested in v00 D w is that we are trying to invert L W C02 Œ0; 1 ! C Œ0; 1, so we require L1 .w/ to lie in the subspace of C 2 Œ0; 1 determined by the Dirichlet condition. In terms of the differential equation v00 D w, this means that the general solution we just found must satisfy the boundary conditions v.0/ D v.1/ D 0. Those conditions tell us that c1 D 0 and that Z

1

c2 D

.s  1/w.s/ ds

0

and thus they give us a formula for L1 . The following result presents that formula in a traditional and convenient way, in terms of a function GW Œ0; 1  Œ0; 1 ! R, called a Green’s function. Theorem 5.1. The inverse L1 W C Œ0; 1 ! C02 Œ0; 1 of Lv D v00 such that v.0/ D v.1/ D 0 is defined by L1 w.t / D

Z

1

G.s; t /w.s/ ds; 0

where  G.s; t / D

.t  1/s; for 0  s  t  1 t .s  1/; for 0  t  s  1:

5 The Forced Pendulum

39

Proof. Substituting for c2 in the general solution to v00 D w gives us 1

Z

L w.t / D

1



Z

t

.s  1/w.s/ ds t C

0

.t  s/w.s/ ds:

0

which we can also write in the form L1 w.t / D

Z

t

Z .t  1/sw.s/ds C

0

1

t .s  1/w.s/ds:

t

The first integral is defined for 0  s  t  1 and the second for 0  t  s  1 just R1 as in the definition of G.s; t /, so L1 w.t / D 0 G.s; t /w.s/ds. t u From this explicit description of L1 we get the following useful information. Corollary 5.2. For all w 2 C Œ0; 1 we have kL1 .w/k2 

13 kwk: 4

Proof. Notice that G.s; t /  0 for all .s; t / 2 Œ0; 1  Œ0; 1 and that G reaches its minimum on that set when s D t so 1=4  G.s; t /  0. Let v D L1 .w/; then jv.t /j D jL1 .w/.t /j 

Z 0

1

jG.s; t /j  kw.t /k ds  14 kwk

Observing that v0 .t / D

Z 0

t

Z sw.s/ ds C

1

.s  1/w.s/ ds;

t

we see that jv0 .t /j  2kwk. Thus, from the definition of the norm of C 2 Œ0; 1, we have kL1 .w/k2 D kvk C kv0 k C kv00 k 

13 kwk: 4

t u

Corollary 5.2 tells us that the linear function L1 takes bounded subsets of C Œ0; 1 to bounded subsets of C 2 Œ0; 1. Therefore L1 W C Œ0; 1 ! C02 Œ0; 1 is a continuous function, by a standard result that is proved in Appendix C as Theorem C.1. Next let’s take a closer look at the right-hand side of our differential equation y 00 D f .t; y; y 0 /. Recall that f D f .t; p; q/W Œ0; 1  R  R ! R is a continuous function. We described the left-hand side of the equation in terms of a function L of functions defined by L.v/ D v00 and we can express the other side in terms of a function of functions also. Given a function u.t / defined for 0  t  1, let w.t / be the function defined by w.t / D f .t; u.t /; u0 .t //. For the function w to make sense,

40

5 The Forced Pendulum

certainly there must be a derivative u0 of the function u we used in the definition. If we require that the derivative be a continuous function, as we will, then we can conclude that w is also continuous since f is continuous by hypothesis. Now the “right-hand side function” we want, that we will call F , is defined by F .u/ D w. In other words, F .u/.t / D f .t; u.t /; u0 .t // for all t 2 Œ0; 1. Since u is assumed to have a continuous first derivative and we have seen that this makes w continuous, then F W C 1 Œ0; 1 ! C Œ0; 1. The function F is called the superposition (or Nemitski) function (or, more commonly in analysis, operator) corresponding to f . Theorem 5.3. If f D f .t; p; q/W Œ0; 1  R  R ! R is continuous, then the superposition operator F W C 1 Œ0; 1 ! C Œ0; 1 defined by F .u/.t / D f .t; u.t /; u0 .t // is also continuous. Proof. We must show that F is continuous at u 2 C 1 Œ0; 1. Thus, given  > 0, we must find ı > 0, depending on u and , such that kF .u/  F .w/k <  for all w 2 C 1 Œ0; 1 with ku  wk1 < ı. Let r  1 be such that r  kuk1 and consider f restricted to the compact set Sr D Œ0; 1  Œ2r; 2r  Œ2r; 2r. The uniform continuity of f on Sr implies that there exists ı 0 > 0 such that if .t; p; q/; .t 0 ; p 0 ; q 0 / 2 Sr with jt  t 0 j < ı 0 ; jp  p 0 j < ı 0 and jq  q 0 j < ı 0 , then jf .t; p; q/  f .t 0 ; p 0 ; q 0 /j < . We will see that if ı is the smaller of ı 0 and r, then it will have the required property. Notice that ku  wk1 < ı implies that kwk1 < 2r because of the way we defined r and ı and the triangle inequality which give us kwk1 < ı C kuk1 . So jw.t /j < 2r and jw0 .t /j < 2r which tells us that both .t; u.t /; u0 .t // and .t; w.t /; w0 .t // are in Sr . We also have ku  wk1 < ı  ı 0 implying ju.t /  w.t /j < ı 0 and ju0 .t /  w0 .t /j < ı 0 for all t 2 Œ0; 1. By the definition of ı 0 , we conclude that jf .t; u.t /; u0 .t //  f .t; w.t /; w0 .t //j <  for all t , which means that kF .u/  F .w/k < . t u We would now like to write the problem: y 00 D f .t; y; y 0 / y.0/ D y.1/ D 0 as that of finding y 2 C02 Œ0; 1 such that L.y/ D F .y/ or rather, since L has an inverse, as the fixed point problem y D L1 F .y/. But we have to be careful about writing our boundary value problem by equating L and F . The domain of the superposition operator F is the linear space C 1 Œ0; 1, but the function L is restricted to the subspace C02 Œ0; 1. We can adjust our formulation of the problem by defining j W C02 Œ0; 1 ! C 1 Œ0; 1 to be inclusion and writing the problem this way: find y 2 C02 Œ0; 1 such that L.y/ D Fj.y/. Since that’s the same as writing y D L1 Fj.y/ we now see that what we need, to solve the boundary value problem, is a fixed point of the function S defined to be the composition j

F

L1

S W C02 Œ0; 1 ! C 1 Œ0; 1 ! C Œ0; 1 ! C02 Œ0; 1

5 The Forced Pendulum

41

Theorem 5.4. Let S W C02 Œ0; 1 ! C02 Œ0; 1 be defined by S.y/ D L1 Fj.y/ where j W C02 Œ0; 1 ! C 1 Œ0; 1 is inclusion, F is the superposition operator of a continuous function, and L1 is the inverse of the second derivative operator with respect to the Dirichlet boundary condition; then S is completely continuous. Proof. Recall from Chap. 2 (specifically, the consequence of the Ascoli–Arzela theorem that we called Theorem 2.7) that the inclusion of C 2 Œ0; 1 in C 1 Œ0; 1 is a completely continuous map. It follows that the inclusion j W C02 Œ0; 1 ! C 1 Œ0; 1 is also completely continuous since a bounded subset B of C02 Œ0; 1 is bounded in C 2 Œ0; 1 as well, so B is indeed relatively compact in C 1 Œ0; 1. We remarked after Corollary 5.2 that L1 is continuous and F is continuous by Theorem 5.3. Therefore S D L1 Fj is completely continuous by Theorem 2.6. t u The problem y 00 D f .t; y; y 0 / y.0/ D y.1/ D 0 does not have a solution for all functions f . For instance, the general solution to the differential equation y 00 D .y 0 /2 is y.t/ D  ln.t C c/ C d and there are no values of c and d for which this function will satisfy the Dirichlet boundary condition. Thus, in the case of f .t; p; q/ D q 2 , the function S D L1 Fj , for F the corresponding superposition operator, does not have a fixed point. On the other hand, for the forced pendulum equation, where we have f W Œ0; 1  R  R ! R defined by  2  f .t; p; q/ D T4 e. T2 t /  a sin p ; we can show that S D L1 Fj does have a fixed point because the hypotheses of the Schauder fixed point theorem (4.4) are satisfied. We can find a closed bounded and convex subset C of C02 Œ0; 1 such that S maps C to C . Since C is bounded and we have just demonstrated that S is completely continuous, the restriction of S to C is a compact map and therefore it has a fixed point. We will have no difficulty finding C because f has the property that it has a bounded image, that is, there exists ˇ  0 such that jf .t; p; q/j  ˇ for all .t; p; q/ 2 Œ0; 1  R  R. To be specific, jf .t; p; q/j D

T2 4

 ˇ T ˇ ˇe. t /ˇ C ja sin pj  2

T2 4

.kek C jaj/ D ˇ:

Since f has bounded image, so also does its superposition operator F W C 1 Œ0; 1 ! C Œ0; 1.

42

5 The Forced Pendulum

Theorem 5.5. If f W Œ0; 1  R  R ! R has bounded image, then the Dirichlet boundary value problem y 00 D f .t; y; y 0 / y.0/ D y.1/ D 0 has a solution. Proof. Let ˇ be the bound for f , then for C we will use the ball in C02 Œ0; 1 of radius 13 ˇ, that is, 4 C D fu 2 C02 Œ0; 1W kuk2 

13 ˇg: 4

It is clear that C is closed, bounded, and convex and we will see that S.C /  C . Then the Schauder fixed point theorem (4.4) can be applied to conclude that S D L1 Fj has a fixed point and therefore the problem has a solution. In fact S maps all of C02 Œ0; 1 into C because kF .j.u//k  ˇ and then, by 5.2, we have kS.u/k2 D kL1 .F .j.u///k2 

13 kF .j.u//k 4

D

13 ˇ: 4

t u

It was easy to use the Schauder fixed point theorem for the forced pendulum equation because the function f satisfied a very strong property: it has a bounded image. As we will see in Chap. 7, when in a Dirichlet boundary value problem the function f does not satisfy such a convenient condition, the application of fixed point theory can be somewhat more challenging. We’ve already discussed how you can construct an odd T -periodic solution to the forced pendulum equation by solving the corresponding Dirichlet boundary value problem, so, applying Theorem 5.5, we have Theorem 5.6. Let eW R ! R be an odd T -periodic map, for some T > 0. Then the forced pendulum equation y 00 C a sin y D e has an odd T -periodic solution yW R ! R.

Chapter 6

Equilibrium Heat Distribution

In the next chapter, we will see the Ascoli–Arzela and Schauder theories used once again, to demonstrate the existence of solutions to a type of problem in the theory of ordinary differential equations that is quite different from what we encountered in studying the forced pendulum. The purpose of the present chapter is to present an illustration of how problems like those discussed in the next chapter come up. Although a single application is hardly sufficient to justify the entire theoretical apparatus, I hope that you will be willing to take my word for it that there are more where this one came from. Furthermore, by following the steps from the mathematical model of the physical problem to the abstract mathematics, you will be able to see why the differential equation problem has the particular form that it does. If I didn’t show you something of the sort, you would have every right to suspect that the somewhat complicated hypotheses of the mathematical problem were constructed just so that the abstract theory we’ve been developing could be applied to it. In this application, we suppose we have a metal rod that is to be heated, and that the temperature at the two ends of the rod is somehow held constant throughout the heating process. After a while, assuming the heating process never changes, the rod should reach a state where the temperature at each point stays constant over time, though the temperature may vary greatly from place to place on the rod. This equilibrium temperature distribution along the rod is believed to satisfy a certain ordinary differential equation. As a test of the validity of that mathematical description of the physical process, we want to know that the given equation actually has a solution. That solution is the function which gives the temperature, at equilibrium, at every location on the rod. The theoretical mathematics of previous chapters will not give us a formula for this temperature function. However, it assures us that the function exists, and therefore if, for instance, we use numerical methods to approximate the values of the function, we have reason to believe that these values correspond to the physical problem, at least to the extent that the mathematics describes it accurately. © Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__6

43

44

6 Equilibrium Heat Distribution

Now let’s discuss the situation in rather more detail. We will assume that the rod is made of a single material, not varying in different parts of the rod. We require that a location on the rod, at least insofar as it concerns our temperature distribution, can be specified by a single direction. It may be that the rod, or wire, is very thin so we can imagine it as one-dimensional. Or perhaps the rod reacts uniformly in terms of temperature in all but one direction. In any case, we wish to think of the rod as a (horizontal) line interval. A location on the rod will thus be determined entirely by the distance s from the left-hand endpoint. We might as well choose the distance unit so that our rod is of length one and thus 0  s  1. We assume, just for convenience in describing it, that the entire rod starts out at the same temperature as its environment. The rod is then heated by some process such as microwave heating, radioactive decay, absorption of radiation, or spontaneous chemical reaction. The important property of the heating process is that it should not change significantly over a long time, certainly much longer than the time it takes for the experiment. Let y be the temperature function for the rod. For a while after the experiment begins, we should think of y as a function of two variables: y D y.s; t/ where s gives the location on the rod and t the time from the beginning of the experiment, since we would expect the rod to heat up over a period of time. We assume in addition that the temperature is not allowed to change at all at the ends of the rod. This would be accomplished by insulating the ends of the rod, by keeping them in temperature controlled reservoirs, or some other system depending on the nature of the heating process. If we assign temperature units so that the entire rod is at a temperature we call 0 at the start, in other words y.s; 0/ D 0 for all locations s, then the condition on the ends states that y.0; t / D y.1; t / D 0 for all time t . In the mathematical description of the heating of the rod, the function y.s; t / is required to satisfy a certain partial differential equation that I will not write down. Instead, let’s accept the fact that under reasonable mathematical conditions, which reflect the physical process, after some time has elapsed the function y no longer changes as a function of time at any location, so now we have an “equilibrium” temperature function y D y.s/ which depends only on the location s on the rod. The condition on the ends of the rod has not changed, that is, as we now write it, y.0/ D y.1/ D 0. From an analysis of the partial differential equation, it can be shown that the equilibrium temperature function y D y.s/ satisfies an ordinary differential equation of the following form: .ky 0 /0 C q.s; y/ D 0: In addition, the mathematical modeling of the physical process tells us something about the functions k and q. The function k is the thermal conductivity of the rod. Since we assumed the material of the rod was uniform, the conductivity depends only on the temperature, in other words k D k.y/. Furthermore, k.y/ > 0 for all temperatures y. Also, the (approximate) form of the function q D q.s; y/ is

6 Equilibrium Heat Distribution

45

known; it may be assumed to be linear in the temperature y, specifically, q.s; y/ D R.s/y C S.s/ where R.s/ < 0 and S.s/ > 0 for all s. Now we can start to analyze the differential equation from an entirely mathematical point of view. Keeping in mind that k is a function of y, we apply the chain rule to .ky 0 /0 C q.s; y/ D 0 and obtain ky 00 C k 0 .y 0 /2 C q.s; y/ D 0: Solving for y 00 and substituting for q, we write the equation in the form 1 y 00 D  Œk 0 .y 0 /2 C Ry C S  D f .s; y; y 0 /: k The significant point is that even though we replaced q by a linear approximation with respect to y, the function f is still nonlinear because it is quadratic in y 0 . The boundary condition y.0/ D y.1/ D 0 is the same Dirichlet condition that we encountered in the previous chapter. In order to apply familiar analytic methods to this model, it was traditional to assume that the thermal conductivity of the material of the rod did not depend on its present temperature, in other words k.y/ D k0 , a constant. In that case, the awkward quadratic term vanishes and you have a classical linear Dirichlet boundary value problem in ordinary differential equations: given continuous functions R D R.s/ and S D S.s/ for 0  s  1 and a constant k0 > 0, find a function y D y.s/ for 0  s  1 satisfying y 00 D 

1 .Ry C S / k0

y.0/ D y.1/ D 0: But in fact thermal conductivity k.y/ is not constant and, even for temperatures not too far from the ambient temperature, a more reasonable approximation is a positive, monotone decreasing function of y. Using the topological methods we discussed in previous chapters, we do not need to assume constant conductivity in order to conclude that the differential equation problem has a solution. Thus we will work with the nonlinear function 1 1 f .s; y; y 0 / D  Œk 0 .y 0 /2 C q.s; y/ D  Œk 0 .y 0 /2 C Ry C S : k k So we again have a second-order nonlinear Dirichlet boundary value problem y 00 D f .s; y; y 0 / y.0/ D y.1/ D 0

46

6 Equilibrium Heat Distribution

where, this time, I’ll write f D f .s; u; p/W Œ0; 1  R  R ! R. For the heat distribution problem we have f .s; u; p/ D 

1 Œk 0 .u/p 2 C R.s/u C S.s//: k.u/

with R.s/ < 0 and S.s/ > 0 for all 0  s  1 and k.u/ > 0 for all u. We will need to understand how the function f behaves at a point of the interval where the value of the equilibrium heat function y is as large as it can become. This will certainly not happen at an endpoint of the interval, so the derivative of y must vanish at the maximum. In terms of the function f , we are asking what can be said about f .s; u; 0/ D 

1 ŒR.s/y C S.s/: k.u/

We should notice that, since R.s/ < 0, when juj is large enough we will have u.R.s/u C S.s// D R.s/u2 C S.s/u < 0 because R and S are bounded on Œ0; 1. The condition k.u/ > 0 then implies that uf .s; u; 0/ > 0 for juj sufficiently large. To be a bit more precise, we conclude that there is a positive number M with the property that if juj > M , then uf .s; u; 0/ > 0. Next I’ll write f in a little simpler way: f .s; u; p/ D 

1 Œk 0 .u/p 2 C q.s; u/ k.u/

because I’m not now concerned with the form of the function q. The triangle inequality then tells us that ˇ ˇ ˇ 0 ˇ ˇ k .u/ ˇ 2 ˇ q.s; u/ ˇ ˇp C ˇ ˇ jf .s; u; p/j  ˇˇ ˇ k.u/ ˇ: k.u/ ˇ We know how f behaves for juj > M , at least when p D 0. We also want to know what happens to f for the other values of u, that is, for M  u  M . The continuous functions k.u/ and k 0 .u/ are bounded on the closed interval ŒM; M  and, in particular, k.u/ > 0 is bounded away from zero there. Furthermore, q.s; u/ is continuous and therefore bounded for u in the same interval, since 0  s  1. What we have found out is that there exist positive numbers A and B such that ˇ 0 ˇ ˇ k .u/ ˇ ˇ ˇ ˇ k.u/ ˇ < A;

ˇ ˇ ˇ q.s; u/ ˇ ˇ ˇ ˇ k.u/ ˇ < B

6 Equilibrium Heat Distribution

47

for 0  s  1 and M  u  M . Therefore, jf .s; u; p/j < Ap 2 C B for these s and u and any p. If we allow ourselves to remember that the variable u is replacing y, which stands for the solution to the differential equation y 00 D f .s; y; y 0 /, then we should be pleased that we now know something about the right-hand side of the equation, at least when y 0 vanishes or when y is within a certain range. We will see just how useful this information is in the next chapter. We conclude this chapter by stating the, now abstract, differential equation problem we will treat in that chapter. The problem is to find a function y D y.s/ W Œ0; 1 ! R satisfying y 00 D f .s; y; y 0 / y.0/ D y.1/ D 0 where f D f .s; u; p/ W Œ0; 1  R  R ! R is a continuous function with the following two properties: (1) there exists M > 0 such that juj > M implies uf .s; u; 0/ > 0; (2) there exist A; B > 0 such that if 0  s  1 and juj  M , then jf .s; u; p/j < Ap 2 C B for all p. We will prove in the next chapter that there is a function y D y.s/ that satisfies the boundary condition and, for every s 2 Œ0; 1, it and its derivatives satisfy the relationship y 00 .s/ D f .s; y.s/; y 0 .s//. In other words, under these hypotheses on f , the nonlinear Dirichlet boundary value problem has a solution.

Chapter 7

Generalized Bernstein Theory

The title of this chapter refers to the fact that, by the use of topological methods, Granas, Guenther, and Lee Granas et al. (1978) were able to extend the classical boundary value theory of Bernstein Bernstein (1912). This chapter is based on their work. As we saw at the end of the last chapter, our goal is to demonstrate the existence of a solution y D y.s/ W Œ0; 1 ! R to the second-order boundary value problem y 00 D f .s; y; y 0 / y.0/ D y.1/ D 0 when f D f .s; u; p/ W Œ0; 1  R  R ! R is a continuous, but in general nonlinear, function with the properties: (1) there exists M > 0 such that juj > M implies uf .s; u; 0/ > 0; (2) there exist A; B > 0 such that if 0  s  1 and juj  M , then jf .s; u; p/j < Ap 2 C B for all p. Just as in Chap. 5, I’ll write this boundary value problem as the equation Ly D Fj.y/. Remember that j W C02 Œ0; 1 ! C 1 Œ0; 1 is the inclusion of the subspace of functions that satisfy the Dirichlet condition, F is the superposition operator corresponding to f , and LW C02 Œ0; 1 ! C Œ0; 1 is the second derivative operator. Making use of the inverse L1 from Theorem 5.1, then from Theorem 5.4 we know that the function S defined as the composition j

F

L1

S W C02 Œ0; 1 ! C 1 Œ0; 1 ! C Œ0; 1 ! C02 Œ0; 1 is completely continuous. We’ll still prove that the boundary value problem has a solution by proving that S must have a fixed point. But the argument has to be different than the one we used in Chap. 5 because it’s no longer true that f must have a bounded image, so we can’t use the Schauder fixed point theorem the way we did in Theorem 5.5. © Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__7

49

50

7 Generalized Bernstein Theory

For instance f .s; u; p/ D u C p 2 satisfies the conditions from Chap. 6 with, say, M D 1, but its image is all of R. Instead, the completely continuous function S has a fixed point because it satisfies the Leray–Schauder boundary condition, so the Leray–Schauder Alternative (4.6) that we obtained from the Schauder theorem will assure us that S has a fixed point. That is, we have to prove that there exists r > 0 such that if u 2 C02 Œ0; 1 with kuk2 D r, then S.u/ 6D u for all  > 1. What we’ll actually prove is: if S.u/ D u for some  > 1, then kuk2 < r. Then the contrapositive is actually a little stronger than we need: if kuk2  r, then S.u/ 6D u for all  > 1. Curiously, the argument depends on reversing the process we carried out above to turn a boundary value problem into a fixed point problem. Suppose we have u 2 C02 Œ0; 1 such that u D S.u/ D L1 Fj.u/, which is the same as L.u/ D 1 Fj.u/, for some  > 1. We can write this as the boundary value problem: find y 2 C 2 Œ0; 1 such that y 00 D f .s; y; y 0 / y.0/ D y.1/ D 0 where f .s; y; y 0 / D

1 f .s; y; y 0 / 

for some  > 1. We will prove that if this problem does have a solution y, then the solution must have the further property kyk2 < r, for a number r that we will determine. A result of this kind is called an “a priori estimate,” that is, an estimate of the norm of a solution before we attempt to establish that any solutions actually exist. So here is what we will prove: Theorem 7.1. Suppose f D f .s; u; p/ W Œ0; 1  R  R ! R is a continuous function with the properties: (1) there exists M > 0 such that juj > M implies uf .s; u; 0/ > 0; (2) there exist A; B > 0 such that if 0  s  1 and juj  M , then jf .s; u; p/j < Ap 2 C B for all p. There exists r > 0 such that if y D y.s/ W Œ0; 1 ! R is a solution to y 00 D f .s; y; y 0 / y.0/ D y.1/ D 0 where f .s; y; y 0 / D 1 f .s; y; y 0 / for some  > 1, then kyk2 < r. Proof. Recall that kuk2 D kuk C ku0 k C ku00 k, that is, kuk2 is the sum of the sup norms of u and its first two derivatives. We will find numbers M0 , M1 , and M2 , independent of , such that if y is a solution, then kyk < M0 , ky 0 k < M1 , and ky 00 k < M2 ; then we can set r D M0 CM1 CM2 . We claim that hypothesis (1) forces the solution y to have the property kyk < M so we can just let M0 D M . To see why this is true, notice that jy.s/j cannot attain its maximum at s D 0 or s D 1,

7 Generalized Bernstein Theory

51

since then the Dirichlet boundary condition would imply y was the zero function, which is certainly contrary to hypothesis (1). So we can assume that the maximum is at some other value s0 . The function 12 y 2 .s/ also has a maximum at s D s0 , so its second derivative at that point is nonpositive. This is interesting because, making use of the assumption that y solves the differential equation, we have ˇ d 2  1 2 ˇˇ 0 2 00 .s/ y ˇ D .y .s0 // C y.s0 /y .s0 / D y.s0 /f .s0 ; y.s0 /; 0/ ds 2 2 s0 which implies uf .s0 ; u; 0/  0 for u D y.s0 /. Hypothesis (1) then tells us that juj  M so jy.s0 /j  M which implies kyk  M , as we claimed. In order to find the bound M1 on ky 0 k, divide the interval Œ0; 1 into a finite number of subintervals Œ;  such that y 0 is of constant nowhere-zero sign on .; / and at least one of y 0 ./ and y 0 . / is zero. We can distinguish four cases depending on whether y 0 .s/ > 0 or y 0 .s/ < 0 on .; / and whether y 0 ./ D 0 or y 0 . / D 0. In every case, we can show that r 0

jy .s/j  M1 D

B 4AM  1/ .e A

and the argument is pretty much the same in each case. So let’s illustrate it with a single case: y 0 .s/ > 0 on .; / and y 0 . / D 0. See Fig. 7.1. Since we now know that jy.s/j  M , hypothesis (2) and the requirement  > 1 tell us that for a solution y, we have jy 00 .s/j D jf .s; y.s/; y 0 .s/j < A.y 0 .s//2 C B: Certainly y 00 .s/  jy 00 .s/j so y 00 .s/ < A.y 0 .s//2 C B

Fig. 7.1 The case y.s/ > 0 and y. / D 0

52

7 Generalized Bernstein Theory

which we rewrite as 

y 00 .s/  1: A.y 0 .s//2 C B

Since we are assuming y 0 .s/ > 0, then 2Ay 0 .s/ < 0, so when we multiply the last inequality by it we have 2Ay 0 .s/y 00 .s/  2Ay 0 .s/: A.y 0 .s//2 C B This implies the corresponding relationship when we integrate Z s



2Ay 0 . /y 00 . / d  A.y 0 . //2 C B

Z



2Ay 0 . / d

s

to obtain ˇ

ˇ

ˇ ˇ ln.A.y 0 . //2 C B/ˇˇ  2Ay. /ˇˇ : s

s

Since y 0 . / D 0, this reduces to ln.B/  ln.A.y 0 .s/2 C B/  2A.y. /  y.s// and thus 

A.y 0 .s//2 C B ln B

  2A.y. /  y.s//  4AM

because jy.s/j  M for all s implies y. /  y.s/  2M . Solving the inequality, we show that M1 should take the stated value, that is, .y 0 .s//2 

Be 4AM  B A

which implies r 0

jy .s/j 

B 4AM  1/: .e A

The final step of the proof is the easiest. If we restrict the continuous function f D f .s; u; p/ to the compact set where 0  s  1;

M0  u  M0 ;

M1  p  M1

7 Generalized Bernstein Theory

53

then we know that there exists M2 > 0 with jf .s; u; p/j < M2 . Now that we know jy.s/j < M0 and jy 0 .s/j < M1 for all s, we conclude that jy 00 .s/j D jf .s; y.s/; y 0 .s//j < jf .s; y.s/; y 0 .s//j < M2 : t u The a priori estimate of Theorem 7.1 has completed the proof that S satisfies the hypotheses of the Leray–Schauder Alternative and consequently has a fixed point. Thus, we have proved that the boundary value problem has a solution. In particular, we know that the mathematical model of the equilibrium heat distribution of the rod 1 y 00 D f .s; y; y 0 / D  Œk 0 .y 0 /2 C Ry C S  k y.0/ D y.1/ D 0; with R.s/ < 0 and S.s/ > 0 for all 0  s  1 and k.u/ > 0 for all u, does have a solution.

Part II

Degree Theory

Chapter 8

Brouwer Degree

The main technical tool of this second part of the book, and one of the most useful topological tools in analysis, is the Leray–Schauder degree. The setting for the Leray–Schauder degree is, in general, infinite-dimensional normed linear spaces. In the first part of the book, before proving the Schauder fixed point theorem for maps of such spaces, we studied the corresponding finite-dimensional setting, that is, euclidean spaces. We proved the finite-dimensional version of Schauder’s theorem, the Brouwer fixed point theorem, and then used the Schauder projection to extend to the infinite-dimensional version. The finite-dimensional version of Leray–Schauder degree is called Brouwer degree and, like Brouwer’s fixed point theorem, its context is euclidean space. In this chapter, I will present the Brouwer degree and, in the next chapter, I will demonstrate certain properties of it. These properties are the ones we will need in order to extend to the Leray–Schauder degree theory in the following chapter, again moving to infinite-dimensional spaces with the aid of the Schauder projection lemma. Just as we did not explore the many topological implications of the Brouwer fixed point theorem in the first part, here we will not be concerned with studying the Brouwer degree for its own sake, but instead we will proceed as efficiently as possible to the more general theory. The definition of the Brouwer degree requires some tools from homology theory. I’m going to begin this chapter by discussing those tools, so you may find it helpful to look at the summary of homology theory in Appendix A before going on. The cartesian product of topological pairs is defined by .X; A/  .Y; B/ D .X  Y; .X  B/ [ .A  Y //: We will be especially interested in topological pairs of the form .X; X  A/ and will make use of the easily verified fact that the definition of cartesian product preserves this form, that is, .X; X  A/  .Y; Y  B/ D .X  Y; .X  Y /  .A  B//: © Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__8

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8 Brouwer Degree

In particular, .Rp ; Rp  0/  .Rq ; Rq  0/ D .RpCq ; RpCq  0/ where in each case 0 represents the origin in the corresponding euclidean space. By assuming that we can always satisfy a very mild topological condition (“excisive couple”) that certainly holds for any topological pair we will deal with in our work here, and therefore I won’t even state, we can make use of a tool of algebraic topology called the cross product. The cross product is a homomorphism  W Hp .X; A/ ˝ Hq .Y; B/ ! HpCq ..X; A/  .Y; B// which takes z ˝ z0 to an element we write as z  z0 . I won’t describe the construction of the cross product, but instead refer you to a standard algebraic topology text such as Hatcher (2002) or Spanier (1966), where you will also find the (quite easy) verification of three important properties. Two of them are quickly stated. The cross product is associative: .z  z0 /  z00 D z  .z0  z00 /. Furthermore, it preserves integer products, that is, az  bz0 D ab.z  z0 / for integers a and b. In fact, when the homology groups are all infinite cyclic, the cross product may be identified with integer multiplication. The third property concerns maps f W .X; A/ ! .W; C / and g W .Y; B/ ! .Z; D/ which have a well-defined cartesian product f  g W .X; A/  .Y; B/ ! .W; C /  .Z; D/: The naturality property of the cross product states that, at the level of homology, the cartesian product of maps preserves cross products, as follows. Consider elements z 2 Hp .X; A/ and z0 2 Hq .Y; B/ with images f .z/ 2 Hp .W; C / and g .z0 / 2 Hq .Z; D/. The naturality property of the cross product states that .f  g/ W HpCq ..X; A/  .Y; B// ! HpCq ..W; C /  .Z; D// takes z  z0 to the cross product of the respective images, that is, .f  g/ .z  z0 / D f .z/  g .z0 /: We will use the cross product to select the generators of certain homology groups that we need to define the Brouwer degree. One set of generators, the n , will generate the groups Hn .Rn ; Rn  0/ which, as I’ll now show you, are infinite cyclic. Actually, we will calculate the groups Hn .S n ; S n  0/, but these are isomorphic to Hn .Rn ; Rn  0/ by excision. For n > 1, we just use part of the exact sequence of the pair, specifically Hn .S n  0/ ! Hn .S n / ! Hn .S n ; S n  0/ ! Hn1 .S n  0/:

8 Brouwer Degree

59

Since the space S n  0 is contractible and therefore the corresponding groups are trivial, the exactness of the sequence implies that the middle homomorphism is an isomorphism. We prove in Appendix A that the groups Hn .S n / are infinite cyclic, so we conclude that the groups Hn .S n ; S n  0/ for n > 1 are infinite cyclic as well. It is an immediate consequence of the definitions that if X is a pathwise connected space, then H0 .X / is infinite cyclic and if A is a pathwise connected subset of the pathwise connected X , then H0 .X; A/ D 0. To calculate H1 .S 1 ; S 1  0/, let’s look at this part of the exact sequence of a pair: H1 .S 1  0/ ! H1 .S 1 / ! H1 .S 1 ; S 1  0/ ! H0 .S 1  0/ ! H0 .S 1 / ! H0 .S 1 ; S 1  0/: Since H0 .S 1 ; S 1 0/ D 0, the next to last arrow is an epimorphism of infinite cyclic groups and thus an isomorphism. That makes the homomorphism with domain H1 .S 1 / an epimorphism, and since it is a monomorphism, Appendix A again completes the argument. Now that we know H1 .R1 ; R1  0/ is infinite cyclic, we choose one of the two possible generators once and for all and call it 1 . For the generator 2 of H2 .R2 ; R2  0/, we will use 1  1 and, in general, the chosen generator of Hn .Rn ; Rn  0/ will be n D 1  n1 . Since all the groups we consider are infinite cyclic, we can think of the cross product as multiplication of integers, so

n will generate its cyclic group because 1 and n1 have the same property. The associativity of the cross product implies that p  q D pCq . The other generators that we’ll need, the n , generate the infinite cyclic groups Hn .S n /. They are constructed out of the n in the following way. Let j W .Rn ; Rn  0/ ! .S n ; S n  0/ be inclusion then, as I mentioned, j W Hn .Rn ; Rn  0/ ! Hn .S n ; S n  0/ is an isomorphism by excision. Let k W Hn .S n / ! Hn .S n ; S n  0/ be the inclusion-induced homomorphism in the exact sequence of the pair that we showed above is an isomorphism (with a bit of extra work in the case n D 1). The definition is n D k1 j . n /

60

8 Brouwer Degree

and that certainly produces generators because we used only isomorphisms in the definition. Now that we have very carefully selected generators for all these various infinite cyclic groups, we are ready for the Brouwer degree. The Brouwer degree is defined in the following situation. We have an open set U in a euclidean space Rn and a map f W U ! Rn , where U denotes the closure of U . Let F D f 1 .0/. Call a subset A of U admissible if A is compact and A \ @U D ;, where @U D U  U is the boundary of U . We will require as part of the setting of the Brouwer degree that F is an admissible subset of U . Think of S n as Rn [ 1 so that U is a subset of S n . Inclusion induces a homomorphism k W Hn .S n / ! Hn .S n ; S n  F / in the exact sequence of the pair .S n ; S n  F /. The excision property of homology theory tells us that the inclusion j W .U; U  F / ! .S n ; S n  F / induces an isomorphism of homology. We can therefore identify an element 0n 2 Hn .U; U  F / by setting j1 k .n / D 0n where n 2 Hn .S n / is that generator that we just chose so carefully. We don’t know anything about the group Hn .U; U  F /, but we can be sure that 0n is nontrivial, provided only that F is nonempty, for the following reason. In this portion of the exact sequence of .S n ; S n  F /, i

k

Hn .S n  F / ! Hn .S n / ! Hn .S n ; S n  F /; the inclusion-induced homomorphism i can be factored as i W Hn .S n  F / ! Hn .S n  fxg/ ! Hn .S n / for any x 2 F . Since Hn .S n fxg/ D 0, we see that i is the trivial homomorphism, and therefore, by exactness, k is a monomorphism so n 6D 0 implies 0n 6D 0. The definition of F tells us that f maps U  F to Rn  0, so there is a homomorphism f W Hn .U; U  F / ! Hn .Rn ; Rn  0/: The Brouwer degree, denoted by deg.f; U /, of f on U is the integer defined by f .0n / D deg.f; U / n where n 2 Hn .Rn ; Rn 0/ is the generator chosen with the aid of the cross product.

8 Brouwer Degree

61

The only arbitrary choice we made in the entire definition of deg.f; U / was to decide which generator of H1 .R1 ; R1  0/ would be chosen as 1 . Suppose we had chosen the other generator, which in terms of this notation we would call  1 . By the properties of the cross product, in place of n we would use .1/n n to generate Hn .Rn ; Rn  0/. Checking out the definitions shows us that the 0n would similarly be replaced by .1/n 0n . Since f .0n / D deg.f; U /  n certainly implies that .1/n f .0n / D deg.f; U /  .1/n n , we can see that the definition of the Brouwer degree is independent of that single choice.

Chapter 9

Properties of the Brouwer Degree

This chapter is devoted to the properties of the Brouwer degree that we will need in order to extend it to the Leray–Schauder degree. In all that follows, we assume that U is an open subset of Rn and that we have a map f W U ! Rn such that F D f 1 .0/ is admissible in U , that is, compact and disjoint from @U , so the Brouwer degree deg.f; U / is well defined. The properties of the degree are given names for easy identification; the terminology I’m using for this purpose is pretty much standard. Some of the properties will carry over to the infinite-dimensional case and others are needed in order to make the transition to that more general setting. The following simple lemma will be helpful in verifying some of those properties. Lemma 9.1. Let W be an open subset of U and let G be an admissible subset of W containing F D f 1 .0/. Let 00n 2 Hn .W; W  G/ be defined by analogy with 0n 2 Hn .U; U  F / and let ı be the integer defined by .f jW / .00n / D ı n , where .f jW / W Hn .W; W  G/ ! Hn .Rn ; Rn  0/ is induced by the restriction of f to W . Then ı D deg.f; U /. Proof. See Fig. 9.1. The inclusion h W .W; W  G/ ! .U; U  F / induces a homomorphism that is easily shown to take 00n to 0n , since all the homomorphisms involved are induced by inclusions. See Fig. 9.2. We have f h D .f jW / W .W; W  G/ ! .Rn ; Rn  0/ so .f jW / .00n / D f .0n / D deg.f; U / n t u

© Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__9

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64

9 Properties of the Brouwer Degree

F

G W U

Fig. 9.1 Subsets of U

Fig. 9.2 Diagram for Lemma 9.1

Theorem 9.2 (Solution Property). If deg.f; U / 6D 0, then f .x/ D 0 for some x 2 U. Proof. We will prove the contrapositive, so suppose f .x/ 6D 0 for all x 2 U . That means F D ; and therefore Hn .S n ; S n  F / D 0, so certainly 0n D 0 in this case and thus f .0n / D 0. t u Theorem 9.3 (Normalization Property). If U contains the origin and i W U ! Rn is the inclusion of U in Rn , then deg.i; U / D 1. Proof. Let I W Rn ! Rn denote the identity map, then checking back through the definitions we find that 0n D n in this case because the construction of 0n from n is precisely the inverse of the definition of n based on n . Therefore deg.I; Rn / D 1 and we have proved the proposition in the case that U D Rn . For the general case, we just apply Lemma 9.1 with F D G D 0 and U replacing Rn . t u

9 Properties of the Brouwer Degree

65

The next property is a trivial consequence of Lemma 9.1, but we include it here because it is sometimes convenient to be able to refer to it by name. Theorem 9.4 (Excision Property). Let W be an open subset of U containing F , then deg.f; U / D deg.f jW; W / where f jW denotes the restriction of f to W .

t u

Another consequence of Lemma 9.1 is Theorem 9.5 (Homotopy Property). Let H W U  Œ0; 1 ! Rn be a homotopy such that D fx 2 U W H.x; t/ D 0 for some

tg

is an admissible subset of U . Defining ht W U ! Rn by ht .x/ D H.x; t/, then deg.h0 ; U / D deg.h1 ; U /: Proof. Let 00n 2 Hn .U; U  / as in Lemma 9.1 with W D U and G D . 00 Noting that h1 0 .0/  , and defining ı0 by h0 .n / D ı0 n , we can see that Lemma 9.1 implies ı0 D deg.h0 ; U /. In the same way, we have ı1 D deg.h1 ; U / where h1 .00n / D ı1 n . Since h0 and h1 are homotopic, they induce the same homomorphism of homology, so ı0 D ı1 . t u The next two proofs require a somewhat more strenuous application of homology theory. In keeping with my general philosophy that even a topological introduction to nonlinear analysis shouldn’t make anyone unhappy just because they are not a topology fan, I’ve exiled these arguments to Appendix B. Consequently, in the process of finding out about the Brouwer degree, you won’t have to trip over a couple of large (though actually not very difficult) commutative diagrams. You can look up the proofs in Appendix B now, you can save the proofs for later when you’ve seen these properties used, or you can skip the Appendix altogether and take my word for it that the next two results are true. Theorem 9.6 (Additivity Property). Let U1 and U2 be disjoint open subsets of U such that F  U1 [ U2 and let fj denote the restriction of f to Uj , then deg.f; U / D deg.f1 ; U1 / C deg.f2 ; U2 /: Theorem 9.7 (Product Property). Let f W U ! Rm and g W V ! Rn be maps such that F D f 1 .0/ is admissible in U and G D g 1 .0/ is admissible in V , then for f  g W U  V ! RmCn , the degree deg.f  g; U  V / is well defined and deg.f  g; U  V / D deg.f; U /deg.g; V /:

66

9 Properties of the Brouwer Degree

The next property is crucial to extending the finite-dimensional Brouwer degree to the infinite-dimensional setting of the Leray–Schauder degree. Theorem 9.8 (Reduction of Dimension Property). Write Rn D Rm  Rnm and let nm W Rn ! Rnm be projection onto the last n  m coordinates. Let U be an open subset of Rn and let f W U ! Rn be a map such that nm f D nm and F D f 1 .0/ (which must be in Rm  f0g) is admissible in U . Let Um D U \ .Rm  f0g/, then, identifying Rm  f0g with Rm deg.f; U / D deg.f jUm ; Um / where f jU m W U m ! Rm is the restriction of f . Proof. Since U is open in the product topology on Rm  Rnm D Rn , we may cover U by subsets of U of the form Vi  Wj where Vi is open in Rm and S Wj is openTin Rnm . Take a finite subcover of the compact set F and set V D Vi , W D Wj where Vi 0 appears in the definition of V if and only if some Vi 0  Wj is in the finite cover of F and Wj 0 appears in the definition of W if and only if some Vi  Wj 0 is in the finite cover of F . See Fig. 9.3. It is clear that V  W is a subset of U and we claim that F  V  W so that deg.f; U / D deg.f; V  W / by the excision property. Note that since F  Rm  f0g, then if Vi  Wj 0 is in the finite cover of F it must be that Wj 0 contains 0 2 Rnm , so W has the same property. Since every point of F is in some Vi 0  Wj where Vi 0  V , the claim is established. Consider g D .f jV /  iW W V  W ! Rm  Rnm D Rn where, as in the statement of the theorem, Rm  f0g is identified with Rm and f jV is defined accordingly, and iW denotes the inclusion. Since W contains 0 2 Rnm , the product, normalization, and excision properties imply that

Fig. 9.3 The cover of F

9 Properties of the Brouwer Degree

67

deg.g; V  W / D deg..f jV /  iW ; V  W / D deg.f jV; V /  deg.iW ; W / D deg.f jV; V /  1 D deg.f jV; V / D deg.f jUm ; Um /: On the other hand, we will apply the homotopy property to show that deg.g; V  W / D deg.f; V W / and that will complete the proof. The homotopy H W V  W  Œ0; 1 ! Rn is defined by H.x; t/ D tf .x/ C .1  t /g.x/ so it just traces out the line segment in Rn between f .x/ and g.x/. Thinking of V  W as a subset of Rm  Rnm D Rn , we write its points as pairs x D .v; w/. The hypothesis nm f D nm means that f .v; w/ D .v0 ; w/ for some v0 2 Rm , whereas, by definition, g.v; w/ D ..f jV /.v/; w/ D .v00 ; w/. It follows that H has the form H.x; t/ D H..v; w/; t / D t .v0 ; w/ C .1  t /.v00 ; w/ D .tv0 C .1  t /v00 ; w/: See Fig. 9.4. Therefore if H.x; t/ D H..v; w/; t / D 0 2 Rn then w D 0 2 Rm . But we can identify g with f on Rm  f0g which tells us that the only x for which H.x; t/ D 0 is that for which f .x/ D 0. In other words, in the notation of Theorem 9.5, D F . Since F is compact and lies in V  f0g, it is admissible in V W and the homotopy property may be applied to conclude that deg.g; V W / D deg.f; V  W /. t u {0} × Rn−m

W

(v,0)

(v,m) V

Rm × {0}

{0} × Rn−m Rn

f(v,w)

H((v,w),t)

g(v,w) f(v,0) Rm × {0}

Fig. 9.4 The maps f and g

68

9 Properties of the Brouwer Degree

Fig. 9.5 The homotopy of f

Theorem 9.9. Let f W R ! R be a map and U D .a; b/ an open interval. If f .a/ < 0 and f .b/ > 0 then deg.f; U / D 1. If f .a/ > 0 and f .b/ < 0 then deg.f; U / D 1. Proof. Suppose f .a/ < 0 and f .b/ > 0. If a < 0 < b, then the homotopy property can be applied to the “straight-line homotopy” H W Œa; b  Œ0; 1 ! R defined by H.s; t/ D tf .s/ C .1  t /s between f and the inclusion since H.a; t/ < 0 and H.b; t/ > 0 for all t 2 Œ0; 1 and H 1 .0/ is a closed subset of .a; b/. Therefore deg.f; U / D deg.g; U / D deg.i; U / D 1 by the normalization property. If 0 < a < b, let V be the open interval .a; b/ and define W V ! R by setting .s/ D f .a/ < 0 for a  s  a and .s/ D f .s/ for a  s  b. Then, by the excision property, deg.f; U / D deg.; U / D deg.; V / D 1. The argument when a < b < 0 is similar. Now suppose f .a/ > 0 and f .b/ < 0. Choose some c > b and define intervals U1 D .b; c/ and U2 D .a; c/. Set W U 2 ! R again equal to f on Œa; b and on Œb; c, let  be the linear map such that .b/ D f .b/ and .c/ D f .a/ (see Fig. 9.5). By the additivity property we have deg.; U2 / D deg.; U / C deg.; U1 /. Another straight-line homotopy H.s; t/ will connect  to the constant map at f .a/ > 0 in such a way that H.a; t/ D H.c; t/ D f .a/ > 0 for all t, so we can use the homotopy property yet again to conclude that deg.; U2 / D 0 by the normalization property (notice how Theorem 9.2 was proved). Since  is identical to f on U and .b/ < 0 while .c/ > 0, we can take advantage of what we found out in the first part of this proof to calculate that

9 Properties of the Brouwer Degree

69

0 D deg.; U2 / D deg.f; U / C deg.; U1 / D deg.f; U / C 1 and therefore deg.f; U / D 1.

t u

The proof of Theorem 9.9 may have impressed you as pretty elaborate for a statement about maps on the real line. But keep in mind that all we really are working with are the properties of the Brouwer degree, and the only ones that give us actual numbers are the existence and normalization properties, so we had to find a way to express the problem in a form in which we could apply those properties. A function f W X ! X with the property f .f .x// D x is called an involution. I’ll use that term for our final property because it is a statement about a particular involution. Theorem 9.10 (Involution Property). Let In W Rn ! Rn be defined by In .x/ D x, then deg.In ; Rn / D .1/n : Proof. Since we can write In D I1  I1      I1 W Rn ! Rn D R1  R1      R1 ; by the product property it is sufficient to note that, by the previous result, deg.I1 ; R1 / D deg.I1 ; U / D 1 taking, say, U D .1; 1/.

t u

Chapter 10

Leray–Schauder Degree

The objective of Leray–Schauder degree theory is the same as that of the fixed point theory of the first part of the book. We want to demonstrate that if certain hypotheses are satisfied, then we can conclude that a map f has a fixed point, that is, that f .x/ D x. If the hypotheses are of the right type, we can hope to verify them in settings that arise in analysis and conclude that an analytic problem has a solution because we’ve managed to describe its solutions as fixed points. A major difference between Leray–Schauder theory and what we studied previously is the local nature of our new theory. A fixed point theorem generally states the existence of a fixed point somewhere in the domain of a map defined on an entire space. Degree theory, as in the last chapter, is concerned with a map defined on U , the closure of an open set U . Leray–Schauder theory seeks conditions that imply that the map has a fixed point specifically in U . However, Brouwer degree theory wasn’t about fixed points at all. The existence property doesn’t produce fixed points but instead zeros, that is, solutions to the equation f .x/ D 0 2 Rn . There’s no mystery about the connection between zeros and fixed points. In a euclidean space, or any other linear space, if, given a function f , we define a map g by g.x/ D x  f .x/, then solving g.x/ D 0, where 0 is the additive identity of the linear space, is equivalent to finding a fixed point of f . The map g is in the form g D I  f , where I is the identity map. Our degree theory will be specifically for maps that are written I  f so that a nonzero degree will produce a solution to .I  f /.x/ D 0 and therefore a fixed point of f . In this chapter, degree theory takes place in the setting of a map f W U ! X where U is still an open set but now a subset of a normed linear space X instead of a euclidean space Rn . In place of euclidean distance, we have the distance between points x and x 0 defined by kx  x 0 k in terms of the norm of X . We will add an important hypothesis to the map f itself, namely, that it is a compact map. Recall that this means that there is a compact set K in X that contains f .U /. In the euclidean setting of the previous chapter, I didn’t need to require compactness and that is fortunate because, in general, I  f isn’t a compact map. © Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__10

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An important consequence of the compactness of f W U ! X is described by the following result. For a subset C of U , let r.C / D inffkx  f .x//k W x 2 C g where inf denotes the infimum (greatest lower bound) of the set. Lemma 10.1. If f W U ! X is a compact map and C is a closed subset of U such that f .x/ 6D x for all x 2 C , then r.C / > 0. Proof. We will prove that if r.C / D 0, then f must have a fixed point in C , contrary to the hypothesis. Assuming that r.C / D 0, there is a sequence fxn g in C such that limn!1 kxn  f .xn /k D 0. The sequence ff .xn /g lies in a compact set K, so it has a convergent subsequence, we’ll still call it ff .xn /g, converging to some x0 2 K, which will turn out to be our fixed point. For the subsequence, we still have limn!1 kxn  f .xn /k D 0 so the corresponding subsequence of fxn g also converges to x0 . By requiring that C be closed, we assure that x0 2 C . Now the continuity of f completes the argument because it implies f .x0 / D f . lim xn / D lim f .xn / D x0 : n!1

n!1

t u

A particularly important closed subset of U is its boundary @U D U  U , so we will abbreviate r.@U / as r and specify the closed set C in r.C / otherwise. The key to moving from the finite-dimensional world of the Brouwer degree to the more general Leray–Schauder degree is the same result we used to extend the Brouwer fixed point theorem to the Schauder fixed point theorem, namely the Schauder projection lemma (Theorem 4.2). Specifically, we will make use of the following evident consequence of that lemma. Lemma 10.2. Let X be a normed linear space and let K be a compact subset of X . Given  > 0, there exists a finite-dimensional subspace X of X , the span of a finite -net for K, and a map P W K ! X such that kP .x/  xk <  for all x 2 K. Now if we start with a compact map f W U ! X , a compact set K containing f .U / and  > 0, we have the composition f

P

P f W U ! f .U /  K ! X : If we assume f has no fixed points on @U (which will correspond to the map I  f having no zeros on that set for the purposes of Brouwer degree theory), then for  small enough, the composition P f also has no fixed points on @U . In fact, we will take advantage of the following stronger result. Lemma 10.3. Suppose f .x/ 6D x for all x 2 @U . If  < for all x 2 @U .

r 2

then kx  P f .x/k 

r 2

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Proof. This is a consequence of the approximation property of P from the previous lemma and the triangle inequality as follows: kx  P f .x/k D k.x  f .x// C .f .x/  P f .x//k  kx  f .x/k  kf .x/  P f .x/k r r r D : 2 2

t u

By Theorem C.2 of Appendix C, the space X is homeomorphic to a euclidean space, though the dimension of the euclidean space may vary with the value of . We suppose that f has no fixed points on @U and we will be interested only in  small enough so that, by the lemma, kx  P f .x/k  2r for x 2 @U . Notice that P f .U / is contained in P .K/, which is a compact set and that the set of fixed points of P f is thus a closed subset of P .K/ and therefore also compact. Let I W X ! X and I W X ! X be the identity maps. Let U D U \ X and define f W U ! X to be the restriction of P f to U . The compactness of the set of fixed points of f is equivalent to the statement that .I  f /1 .0/ is compact. Furthermore, I  f has no zeros on @U ; in fact we know more: k.I  f /.x/  0k 

r 2

for all x 2 @Un . In particular, .I  f /1 .0/ is admissible in U . The Leray– Schauder degree deg.I  f; U / of I  f on U is defined by deg.I  f; U / D deg.I  f ; U / where the symbol on the right-hand side is the Brouwer degree and we require  < 2r . I was careful to establish that the conditions for the definition of the Brouwer degree were satisfied, so the definition of Leray–Schauder degree makes sense to that extent. But, unlike the definition of the Brouwer degree, we made a number of choices in defining the Leray–Schauder degree that, potentially, could change the value of the degree. The main result of this chapter will be a theorem showing that the Leray–Schauder degree is independent of those choices. The proof will make use of the following fact about compact maps that we’ll use in the next chapter too. Lemma 10.4. Let X be a normed linear space and Y  X any subset. Suppose that a; bW Y ! X are compact maps, then the straight-line homotopy H W Y Œ0; 1 ! X defined by H.x; t/ D t a.x/ C .1  t /b.x/ is a compact map and the set F ix.H / D fx 2 Y W H.x; t/ D x for some t 2 Œ0; 1g is compact.

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Proof. Since a and b are compact maps, there are compact subsets K and L of X such that a.Y /  K and b.Y /  L. Define qW K  L  Œ0; 1 ! X by q.k; `; t/ D t k C .1  t /`. Then q is continuous so Q D q.K  L  Œ0; 1/ is a compact subset of X and therefore H is a compact map because Q contains H.Y  Œ0; 1/. Since x 2 F ix.H / implies that x D H.x; t/ 2 Q, it follows that F ix.H /, a closed subset of Q, is compact. t u Now we’ll prove that we get the same value for the Leray–Schauder degree deg.I  f; U / no matter which choices we made in defining it. But first, let’s review what’s been done so that we can agree about what the conditions are under which we need to establish the proof. We have an open subset U of a normed linear space X and a map f W U ! X such that f .U / is contained in a compact set K and it has the property that f .x/ 6D x for all x 2 @U . The number r D inffkf .x/  xk W x 2 @U g is therefore positive. We choose a positive number  < 2r and a map P W K ! X , to a finite-dimensional subspace of X , that moves each point of K less than . Let I W X ! X be the identity map, and set U D U \X . Define a map f W U ! X by the composition i

f

P

f W U ! U ! K ! X where i is inclusion. We defined the Leray–Schauder degree of I  f on U by setting deg.I  f; U / D deg.I  f ; U /, the Brouwer degree. Theorem 10.5. The definition of the Leray–Schauder degree deg.I  f; U / is independent of the choices made: the compact set K containing f .U /, the positive number , the subspace X , and the map P W K ! X , provided only that  < 2r . Proof. We assume we have compact sets K1 and K2 containing f .U / and two positive numbers 1 and 2 , both smaller than 2r . For j D 1; 2 let Pj W Kj ! Xj be maps to finite-dimensional subspaces of X that move no point of Kj more than j . We thus have maps fj W Uj ! Xj , for j D 1; 2, defined by ij

f

Pj

fj W Uj ! U ! Kj ! Xj where ij W Uj D U \ Xj ! U is inclusion. Letting Ij W Xj ! Xj be the identity maps, we have to prove that the corresponding Brouwer degrees are equal, that is, deg.I1  f1 ; U1 / D deg.I2  f2 ; U2 / because either side of the equation could be used to define the Leray–Schauder degree deg.I  f; U /. Let X 0 be a finite-dimensional subspace of X that contains X1 [ X2 . Choosing a basis for X1 , we can extend it to a basis for X 0 , so, by

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Theorem C.2, we may think of X1 as a euclidean space Rm embedded as the subspace Rm  f0g in the euclidean space X 0 D Rn for some n  m. We denote the identity map of X 0 by I 0 . Letting U 0 D U \ X 0 , we have the inclusion i 0 W U 0 ! U . We also have the inclusion i10 W X1 ! X 0 . We define a map f10 W U 0 ! X 0 as the composition i0

f

P1

i10

f10 W U 0 ! U ! K1 ! X1 ! X 0 : It is important to note that although we defined f10 as a map into X 0 , in fact the image of U 0 under f10 lies in the subspace X1 . Thus, if we represent a point x 2 U 0  X 0 D Rn D Rm  Rnm in the form x D .v; w/ where v 2 Rm and w 2 Rnm , then we can write f10 .x/ D f10 .v; w/ D .v ; 0/ and therefore .I 0  f10 /.v; w/ D .v; w/  .v ; 0/ D .v00 ; w/ for some v00 2 Rm . In other words, for nm W Rn ! Rnm projection on the last n  m coordinates, we have nm .I 0  f10 / D nm . Therefore, by the reduction of dimension property of the Brouwer degree (Theorem 9.8), deg.I 0  f10 ; U 0 / D deg..I 0  f10 /jU1 ; U1 / D deg.I1  f1 ; U1 / because clearly the restriction of f10 to U1 is f1 . We can repeat the argument for j D 2, so, to prove that deg.I1  f1 ; U1 / D deg.I2  f2 ; U2 /, we just need to establish that deg.I 0  f10 ; U 0 / D deg.I 0  f20 ; U 0 /: That should be a lot easier because the trouble with comparing the degrees of the Ij  fj is that they are maps of different spaces, but both I 0  fj0 take U 0 into X 0 . In fact, the argument is nothing but the homotopy property, this time using the straight-line homotopy H W U 0  Œ0; 1 ! X 0 defined by H.x; t/ D t .I 0  f10 /.x/ C .1  t /.I 0  f20 /.x/: Let h W U 0  Œ0; 1 ! X 0 be the straight-line homotopy defined by h.x; t/ D .t /f10 .x/ C .1  t /f20 .x/: The maps fj0 are compact because their images lie in the compact sets Pj .Kj /, so F ix.h/ D fx 2 Y W h.x; t/ D x for some t 2 Œ0; 1g is compact by Lemma 10.4. Now notice that F ix.h/ is also the set of points in U 0

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with the property that H.x; t/ D 0 for some t as in Theorem 9.5, which was there called . Thus, one of the hypotheses of Theorem 9.5 has been satisfied, so, to apply the homotopy property, it remains to verify the other hypothesis: H.x; t/ 6D 0 for x 2 @U 0 . It is at this last step in the argument that we will finally make use of the hypothesis that the j < 2r . Think of H mapping not just into X 0 but into the normed linear space X . Then since x 2 @U 0 implies that x is in @U we compute that kH.x; t/  .I  f /.x/k D ktx  tf10 .x/ C x  tx  .1  t /f20 .x/  x C f .x/k  tkf .x/  f10 .x/k C .1  t /kf .x/  f20 .x/k D tkf .x/  P1 f .x/k C .1  t /kf .x/  P2 f .x/k r  t 1 C .1  t /2  : 2 t On the other hand, k.I  f /.x/k  r by the definition of r, so kH.x; t/k  2r . u If f W U ! X is a compact map and the normed linear space X is finitedimensional, let Ba D fx 2 X W kxk  ag where a is large enough so that Ba contains all points of X within 2r of f .U /. For any  < 2r , choose X D X so I is the identity map on X and U D U . Furthermore, we can let P be the inclusion of Ba in X , so f D f . Theorem 10.5 assures us that we can make all these choices and still obtain the Leray–Schauder degree, and therefore in this case that degree corresponds exactly to the Brouwer degree. To summarize: Theorem 10.6. If f W U ! X is a compact map with no fixed points on @U and X is finite-dimensional, then the Leray–Schauder degree deg.I  f; U / is the Brouwer degree of I  f on U . t u

Chapter 11

Properties of the Leray–Schauder Degree

You may have noticed that once the Brouwer degree was defined and its properties established, we used it in the next chapter only in a sort of formal way. In defining the Leray–Schauder degree we needed to know that there was a well-defined integer, called the Brouwer degree, represented by the symbol deg.I  f ; U /, but we did not have to specify how that integer was defined. Furthermore, and this is the point I want to emphasize, in the proof that the Leray–Schauder degree is well defined, which is really a theorem about the Brouwer degree, all we needed to know about that degree was two of its properties: homotopy and reduction of dimension. In this chapter, I will list and demonstrate properties of the Leray–Schauder degree, properties which, basically, are consequences of the corresponding properties of the Brouwer degree. Again all we will need to know about the Brouwer degree is its existence and properties. As in the previous chapter, no homology groups will ever appear. Then, once we have established these properties of the Leray–Schauder degree, that’s all we’ll need to know about that degree for the rest of the book. In other words, after this chapter we can forget how the degree was defined. It may seem rather an abstract way to do mathematics: to forget how a concept was defined and use only certain of its properties. But, on the other hand, in this way the concept becomes a tool that we can use very easily because it consists merely of a symbol and some relatively simple statements about how the symbol behaves. Since this particular tool comes from algebraic topology, but we want to use it in analysis, the abstract approach actually makes the Leray–Schauder degree neutral enough so that we are not distracted by its topological background. Throughout this chapter, we assume we have a map f W U ! X satisfying the hypotheses of the previous chapter so that the Leray–Schauder degree deg.I f; U / is defined, that is, U is the closure of an open subset of the normed linear space X and f is a compact map and it has no fixed points on the boundary of U . Just as with the Brouwer degree, the very first property we want to verify is the one that tells us what the purpose of the theory is

© Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__11

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Theorem 11.1 (Existence Property). If deg.I  f; U / 6D 0 then f .x/ D x for some x 2 U . Proof. The corresponding property of the Brouwer degree is the solution property (Theorem 9.2). We established that property by proving the contrapositive of the statement and we will do the same for this property. That is, we will assume that f .x/ 6D x for all x 2 U and prove that the Leray–Schauder degree is zero. That assumption and the compactness of f imply by Lemma 10.1 that r.U / D inffkx  f .x/k W x 2 U g > 0: Choose  < 12 r.U /, then it is easy to check that kx  fn .x/k 

r.U / 2

for all x 2 U and therefore, in particular, .I  f /.x/ 6D 0 for x 2 U  D U \ X . Since certainly r.U /  r.@U / D r, we conclude that deg.I  f ; U / D 0, by the contrapositive of the solution property, and therefore deg.I  f; U / D 0. t u The normalization property has exactly the same statement as the corresponding property of the Brouwer degree and follows so easily from it that I’ll omit the proof. The statement of the normalization property makes sense in the Leray–Schauder setting because it concerns the inclusion i of U in X and we can write i D I  f where f is the constant map taking U to the additive identity 0, so f is certainly a compact map. Theorem 11.2 (Normalization Property). Let i W U ! X be inclusion. If 0 2 U , then deg.i; U / D 1. Theorem 11.3 (Product Property). Let f W U ! X and g W V ! Y be maps and define f  g W U  V ! X  Y by .f  g/.x; y/ D .f .x/; g.y//. If f and g satisfy the hypotheses of the Leray–Schauder degree, then so does f g W U  V ! X Y and deg.IXY  .f  g/; U  V / D deg.IX  f; U /deg.IY  g; V /: Proof. For the metric ı of X  Y we may use ı..x; y/; .x 0 ; y 0 // D minfkx  x 0 kX ; ky  y 0 kY g; where k  kX and k  kY are the norms of the corresponding linear spaces. By hypothesis, there exist compact subsets Kf and Kg of X and Y , respectively, such that f .U /  Kf and g.V /  Kg . Let r be the smaller of the positive numbers inffjjf .x/  xjjX W x 2 @U g and inffjjg.y/  yjjY W y 2 @V g

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79

so that ı..f  g/.x; y/; .x; y//  r for all .x; y/ 2 @.U  V /. Choosing some  < 4r , by Lemma 10.2 we have maps Pf W Kf ! X and Pg W Kg ! Y , where X and Y are finite-dimensional subspaces of X and Y , respectively, such that kPf .x/  xkX <  for all x 2 Kf and kPg .y/  ykY <  for all y 2 Kg . It follows that ı..Pf  Pg /.x; y/; .x; y//
0, where r 0 D inffkx  H.x; t/k W x 2 @U; t 2 Œ0; 1g: 0

Take  < r2 and let P W K ! X 0 be a map to a finite-dimensional subspace of X such that kP .x/  xk <  for all x 2 K. Let I 0 W X 0 ! X 0 be the identity map and define U 0 D U \ X 0 . Define H  W U 0  Œ0; 1 ! X 0 by H  .x; t / D x  PH.x; t / D x  P ht .x/: We can think of H  as a homotopy between I 0  P h0 and I 0  P h1 , as maps from U 0 to X 0 . It is certainly clear that r 0 < rt D inffkx  ht .x/k W x 2 @U 0 g for any t 2 Œ0; 1 and so in particular r 0 < r0 . Furthermore, h0 is a compact map because h0 .U 0 /  K. We didn’t put any restrictions on the compact set K in the development of the theory (such as making it equal to the closure of the image), and we proved independence of the choice of K in Theorem 10.5. Therefore, the definition of Leray–Schauder degree permits us to write deg.I  h0 ; U / D deg.I 0  P h0 ; U 0 /: The same reasoning implies deg.I  h1 ; U / D deg.I 0  P h1 ; U 0 /„ so, if the homotopy H  satisfies the hypotheses of Theorem 9.5, the homotopy property of the Brouwer degree, we are through. Setting D fx 2 U 0 W H  .x; t / D 0 for some t 2 Œ0; 1g

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81

we have to prove that is admissible in U 0 , that is, we must demonstrate that is compact and disjoint from @U 0 . The second condition is a consequence of the inequality r 0 < rt that I pointed out earlier in the proof, so it comes down to establishing the compactness of . From the definition of H  , it is clear that consists of the points x 2 U 0 such that PH.x; t / D x for some t 2 Œ0; 1. So suppose we take a sequence f.xj ; tj /g 2 U 0  Œ0; 1 that is in , that is, each pair .xj ; tj / in the sequence has the property PH.xj ; tj / D xj . If we can prove that the sequence has a convergent subsequence, then Theorem 2.4 will complete the proof. By the compactness of Œ0; 1 there is a convergent subsequence of ftj g converging to some t 0 2 Œ0; 1. We’ll continue to use the j subscripts but understand that we have taken a subset so that ftj g is actually a convergent sequence. The sequence fH.xj ; tj /g is in the compact set K and, taking a subsequence, we may write limj !1 H.xj ; tj / D x 0 . The continuity of P then implies limj !1 PH.xj ; tj / D P .x 0 /, but now the assumption that PH.xj ; tj / D xj tells us that a subsequence of fxj g converges to P .x 0 /. Thus, by restricting the subscripts, we have proved limj !1 .xj ; tj / D .P .x 0 /; t 0 / and so we have succeeded in extracting a convergent subsequence from the sequence in . t u In Theorem 11.7, we required that the homotopy H W U  Œ0; 1 ! X be a compact map: there is a compact set K containing its image. It follows that each ht W U ! X is also a compact map, since its image is also in K. You might wonder if the converse is true. That is, if H is continuous and each ht is a compact map, does it follow that H is a compact map? To see that it doesn’t, take U D X D R and define 1  jx  t12 j; if jx  t12 j  1t and 0 < t  1 H.x; t/ D t 0; otherwise: The image of each ht is Œ0; 1t  for t > 0 and h0 is the zero function, so all the ht are compact. See Fig. 11.1. The only place we have to be concerned about the continuity of H is at t D 0, but given x0 > 0 we just have to note that H.x0 ; t / D 0 for t > 0 small enough so that t 2 < .1  t /x0 . Maps f; g W U ! X into a linear space are always homotopic because we can use the linear structure to define a straight-line homotopy H.x; t/ D tf .x/ C .1  t /g.x/. We proved in Lemma 10.4 that if f and g are compact maps, so also is the homotopy H . Thus we have the following useful consequence of the homotopy property. Theorem 11.8 (Straight-Line Homotopy Property). If f; g W U ! X are compact maps such that tf .x/ C .1  t /g.x/ 6D x for all x 2 @U and t 2 Œ0; 1, then deg.I  f; U/ D deg.I  g; U/:

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11 Properties of the Leray–Schauder Degree

graph of ht

1 t

1 t2

–

1

1

1

t

t2

t2

+

1 t

Fig. 11.1 Each ht is a compact map

We’ll use the straight-line homotopy right away, to prove a property of the Leray– Schauder degree that corresponds to the reduction in dimension property of the Brouwer degree (Theorem 9.8). The hypotheses are a bit different than in that theorem. However, if you look back at how Theorem 9.8 was used in the proof of Theorem 10.5, you’ll see that there we were working with maps of Rn of the form I  f where f mapped an open subset of Rn into a factor Rm  f0g of Rn D Rm  Rnm . That’s the sort of map f that this next property is about, except that the linear spaces are not necessarily finite-dimensional. Let X D X1  X2 be a product of normed linear spaces. The additive identity in X will be called 0 as usual, but I’ll put subscripts on the corresponding elements of X1 and X2 . Write x 2 X in the form x D .x1 ; x2 / where x1 2 X1 and x2 2 X2 . Then the metric topology on X induced by the norm kxk D k.x1 ; x2 /k D

q kx1 k21 C kx2 k22

defined in terms of the norms kk1 and kk2 of X1 and X2 is equivalent to the product topology obtained from the metric topologies of those factors. Suppose f W X1  X2 D X ! X is a map such that f .X /  X1  f02 g. Then the restriction of f to X1  f02 g gives us the map f1 W X1 ! X1 that is characterized by f .x1 ; 02 / D .f1 .x1 /; 02 /. For U an open subset of X , let U1 be the open subset of points x1 2 X1 such that .x1 ; 02 / 2 U . If f has no fixed points on @U then f1 has none on @U1 , and if f is a compact map, so also is its restriction to X1  f02 g and therefore f1 is also compact. Theorem 11.9 (Factor Property). If X D X1  X2 is a product of normed linear spaces, U is an open subset of X , and if f W U ! X1  f02 g  X is a compact map with no fixed points on the boundary of U , then

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83

deg.I  f; U/ D deg.I1  f1 ; U1 /; where I1 is the identity map of X1 . Proof. Define gW X ! X by g.x/ D g.x1 ; x2 / D f .x1 ; 02 / D .f1 .x1 /; 02 /: Since the image of g is contained in the image of the compact map f , then g is also compact. Noting that the equation x D tf .x/ C .1  t /g.x/ can have solutions only in X1  f02 g and that f .x1 ; 02 / D g.x1 ; 02 /, we find that there are no solutions on @U because, by hypothesis, the map f has no fixed points on @U . Thus the straightline homotopy property implies that deg.I  f; U / D deg.I  g; U /. The set ˚ D .I  f /1 .0/ D .I  g/1 .0/ is a compact subset of U lying in X1  f0g. Taking a cover of ˚ by subsets of U in the product topology S and then aTfinite subcover fVi  Wj g where Vi  X1 and Wj  X2 , let V D Vi and W D Wj (compare the proof of Theorem 9.8), then we see that ˚  V  W  U and also that ˚  V  f02 g  U1  f02 g. Thus, by the excision property Theorem 11.5, deg.I  g; U / D deg.I  g; V  W /. By its definition, g D f1  02 W X1  X2 ! X1  X2 where 02 here denotes the constant map. Therefore, the previously established product, excision, and normalization properties complete the proof because deg.I  g; V  W / D deg.I1  f1 ; V /  deg.I2  02 ; W / D deg.I1  f1 ; U1 /: t u The characteristic behavior of tools that are developed by means of algebraic topology is that they don’t distinguish between things that are homotopic. For instance, homotopic maps induce the same homology homomorphisms. The homotopy property of the Leray–Schauder degree of Theorem 11.7 reflects its algebraic topology source. We will see that this property is the key to the effective use of the degree in analysis. However, it often happens that in an analytic problem, it is not obvious that we are dealing with a homotopy at all. Rather than presenting us with a homotopy H W U  Œ0; 1 ! X , we will find that the problem has given us a map H W W ! X where W lies in X  R. The difficulty lies in the fact that although W lies in a cartesian product of spaces, unlike U  Œ0; 1 it has no such structure itself. Nevertheless, if H W W ! X is sufficiently well behaved, we can treat it like a homotopy and prove invariance of degree in an appropriate sense. Thus, our job in the final result of this chapter will be to extend the homotopy property to this setting, to find out what we must require of this map H and in what sense we can deduce homotopy invariance. Then, when we meet such a map H W W ! X later on, we will be prepared with a topological tool that is precisely in the form in which we need it.

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Let W be an open subset of X  R and define 1 W X  R ! X to be projection on the first factor. For t 2 R, define Wt D 1 .W \ .X  ftg//; in other words, Wt D fx 2 X W .x; t / 2 W g: We note that Wt is open in X and we denote the boundary of Wt by @Wt . Given H W W ! X , define ht W W t ! X by ht .x/ D H.x; t/. Theorem 11.10 (Generalized Homotopy Property). Let T D Œt1 ; t2  be a closed interval in R and let W be an open subset of X  T , where X is a normed linear space. If H W W ! X is a compact map such that ht .x/ 6D x for all x 2 @Wt and all t 2 T , then deg.I  ht ; Wt / is independent of t 2 T . Proof. Figure 11.2 illustrates the hypotheses: for instance we require that H.x; t/ 6D x, but we permit H.x 0 ; t2 / D x 0 . Let ˚ D f.x; t / 2 W W H.x; t/ D xgI then ˚ is compact because it is a closed subset of the compact set K  T , where K is a compact subset of X containing H.W /. Let ˚t D ˚ \ .X  ft g/, which is closed in ˚ and hence compact. By hypothesis, 1 .˚t / is contained in Wt for each t . Take t such that t1 < t < t2 ; the argument that follows must be modified slightly if t D t1 or t D t2 . Since W is open in X  T , which gets its topology from the product topology on X R, given .x;  / in ˚ , there exist ıx > 0 and an open subset

X

X

W W t1

t

t2

R

(x ′,t 2) (x, t )

Fig. 11.2 An open subset of X  T

x

11 Properties of the Leray–Schauder Degree

85

Fig. 11.3 Product neighborhood of ˚

Vx of X such that .x;  / 2 Vx  Œ  ıx ;  C ıx  and this set, in turn, is contained in W . From the cover fVx  Œ  ıx ;  C ıx g of ˚ we extract a finite subcover fVj  Œ  ıj ;  C ıj g. Letting V be the union of the Vj and ı be the minimum of the ıj , we can see that ˚ is contained in V  Œ  ı ;  C ı , which is a subset of W . See Fig. 11.3. Now comes the main step of the proof, which will permit us to apply the homotopy property we proved as Theorem 11.7, in order to obtain the generalization. The fact we will make use of is that V  Œ  ı ;  C ı  contains not only ˚ but also ˚t for all t sufficiently near  as well. That is, there exists  > 0 such that jt   j   implies ˚t  V  Œ  ı ;  C ı . Let A D .X  V /  f g, then this closed subset of W  T is disjoint from the compact set ˚ . See Fig. 11.4. It is easy to verify that the distance between these sets, that is, ı.˚; A / D inffk.x; t /  .x 0 ;  /k W .x; t / 2 ˚; .x 0 ;  / 2 A g is greater than zero. Now suppose that .x; t / 2 X  T such that jt  j <  D ı.˚; A / but x 62 V . Certainly ı..x; t /; A / D inffk.x; t /  .x 0 ;  /k W .x 0 ;  / 2 A g <  so .x; t / 62 ˚ . The contrapositive statement establishes the main step: if .x; t / 2 ˚ and jt   j < , then x 2 V . Now we notice that the homotopy property of

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11 Properties of the Leray–Schauder Degree

X

X

Ar

Φr

Φ

Ar r

t1 Φ

Φ

t2

R

Vr

Φr

Ar Fig. 11.4 The set ˚

Theorem 11.7 didn’t make use of any property of Œ0; 1 other than the fact that it is a closed interval in R (since the same is true of the homotopy property of the Brouwer degree established in Theorem 9.5), so we can apply Theorem 11.7 to the homotopy H 0 W V  Œ  ;  C  ! X; the restriction of H to that cartesian product. The image of H 0 is certainly in K and the definition of  assures us that H.x; t/ 6D x if x 2 @V , so the hypotheses are satisfied and we conclude that deg.I  h0t ; V / D deg.I  h0 ; V /: In other words, deg.I  h0t ; V / is constant for all t between    and  C . Also for these t , the fact that ˚t  V  Œ  ı ;  C ı  tells us that deg.I  h0t ; V / D deg.I  ht ; Wt / by the excision property. Thus, for each  2 T , we can find an open interval in T containing  on which the value of deg.I  ht ; Wt / is constant. Since T is connected and the Leray–Schauder degree is an integer, we conclude that deg.I  ht ; Wt / is the same for all t 2 T . t u

Chapter 12

The Mawhin Operator

In Chap. 5 we saw that if eW R ! R is an odd, T -periodic function, then the forced pendulum equation with forcing term e, that is, y 00 C a sin y D e always has a solution yW R ! R that is also an odd, T -periodic function. Remember how we constructed that solution: we built it up out of copies of a solution yW Œ0; T2  ! R to the Dirichlet boundary value problem corresponding to that differential equation, that is, with the boundary condition y.0/ D y. T2 / D 0. In the next chapter, we’ll take up the forced pendulum again, but this time only assuming that the forcing term e.t / is a T -periodic function. We’ll still build our solution yW R ! R by pasting together solutions to a boundary value problem. The construction is even simpler than in Chap. 5: find a solution y0 W Œ0; T  ! R and for t D kT C t0 where 0  t0  T , let y.t/ D y0 .t0 /. Certainly y will be a T -periodic function, provided it’s well defined. Since we can write t D kT also as t D .k  1/T C T , it must be that y0 .0/ D y0 .T /. (Restricting t0 to Œ0; T / would eliminate that potential ambiguity from the definition, but you still need the condition y0 .0/ D y0 .T / in order to make y continuous.) In fact, for our solution yW R ! R, we want a function with a continuous derivative and, for that purpose, we also need to know that y00 .0/ D y00 .T /. To summarize, a solution y0 to the boundary value problem y 00 C a sin y D e .P /

y.0/  y.T / D y 0 .0/  y 0 .T / D 0

gives us a T -periodic solution yW R ! R to the forced pendulum equation. The boundary conditions we labelled as .P / are called periodic boundary conditions. Notice that they specify that the solution takes the same value at 0 as at T but, unlike the Dirichlet condition y.0/ D y. T2 / D 0, they don’t tell us what © Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__12

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that value is. In Chap. 5 we defined a subspace C02 Œ0; T2  of C 2 Œ0; T2  consisting of the functions that satisfy the Dirichlet condition. The periodic boundary conditions give us a different subspace CP2 Œ0; T  D fu 2 C 2 Œ0; T W u.0/  u.T / D u0 .0/  u0 .T / D 0g: In Chap. 5, we wrote the differential equation in the form u00 D e  a sin u and viewed the left-hand side as the function LW C02 Œ0; T2  ! C Œ0; T2  defined by Lu D u00 . We found that L was an isomorphism and we used L1 to express the Dirichlet boundary problem as a fixed point problem. However, with respect to the periodic boundary conditions, the function LW CP2 Œ0; T  ! C Œ0; T  defined by Lu D u00 is not an isomorphism. Take u to be any constant function, then it lies in CP2 Œ0; T , but of course Lu D u00 D 0 and thus L has a nontrivial kernel. Although LW CP2 Œ0; T  ! C Œ0; T  is not an isomorphism, it turns out to be the next best thing, called a Fredholm operator of index zero. We will see in this chapter that the solutions to problems of the form Lu D Z.u/ that involve linear functions L of that kind can still be expressed as solutions to fixed point problems even though there is no inverse L1 available. That won’t mean that we’ll always have a solution to the forced pendulum equation with the periodic boundary conditions, as we did for the Dirichlet problem of Chap. 5. But under an appropriate condition we will be able to apply the Leray–Schauder degree to a fixed point problem in order to find solutions to the forced pendulum problem with periodic forcing term. So another goal of this chapter is to find out how to calculate the Leray–Schauder degree in a setting that will include the forced pendulum problem that we are interested in. Suppose LW X ! Y is a linear function where X and Y are normed linear spaces, L is continuous, and the image of L is closed in Y . Then N.L/, the null space of L, is the linear subspace of X consisting of x such that Lx D 0. Denoting the range of L by R.L/ D fy 2 Y W Lx D y for some x 2 X g, the cokernel of L is the quotient space CokerL D Y =R.L/. Thus the elements of CokerL are equivalence classes of points of Y with respect to the relation that y1 and y2 are equivalent if y1 y2 D Lx for some x 2 X . Denoting the equivalence class of y 2 Y by Œy, the definitions Œy C Œy 0  D Œy C y 0  and rŒy D Œry for r a real number make CokerL a linear space. A continuous linear function LW X ! Y is said to be a Fredholm operator if N.L/ and CokerL are both finite-dimensional linear spaces and R.L/ is closed in Y . The index of a Fredhold operator is the dimension of its null space minus the dimension of its cokernel, so index zero, which is the only case we will be interested in, means that the linear spaces N.L/ and CokerL are of the same finite dimension. Thus, an isomorphism may be viewed as a Fredholm operator of index zero for which the null space and the cokernel are both of dimension zero. The procedure for converting the equation Lu D Z.u/ into a fixed point problem depends on a tool of functional analysis called a projection, so that’s what we next

12 The Mawhin Operator

89

need to find out about. Since we can add elements of a linear space X and also multiply its elements by real numbers, that allows us to form linear combinations a1 x1 Ca2 x2 C  Cam xm of a finite set S D fx1 ; x2 ; : : : ; xm g of elements of X . The set of all such linear combinations is the span of S , denoted span.S /; it is a linear subspace of X . The set S is said to be linearly independent if the only representation of 0 2 X as an element of span.S / is the obvious one, where a1 D a2 D    D am D 0. If S is a linearly independent set, then it is called a basis for the linear space span.S /. If S is an infinite subset of X we can form all the linear combinations of all the finite subsets of S and produce a linear subspace that we will still call span.S /. The set S is said to be linearly independent if every finite subset of it is and, in that case, S is called a basis for span.S /. If S is a basis for a linear space X , it’s easy to see that each x 2 X can be expressed as a finite linear combination of elements of S in only one way. Theorem 12.1. Let X be a linear space and s a linearly independent subset of X , then s can be extended to a basis S for X . Proof. Let S be the set of all linearly independent subsets of X that contain that given set s, then set inclusion imposes a partial ordering on the elements of S. For every totally ordered subset of S there is an upper bound, namely, the union of all the elements of X that lie in members of the subset. Zorn’s lemma, a standard result from set theory, then tells us that there is a maximal element S in the set S and we claim that S is the basis for X that we need. Since S is in S, it is a linearly independent set and it contains s so we just have to show that span.S / D X . That’s where maximality comes in. If we took some x 2 X that isn’t in S and, adding it to every finite subset of S , we still had linearly independent sets, then S [ fxg would also be in S and contain S , which wouldn’t be maximal in S after all. So, for each x 2 X not in S , there is a finite set fx1 ; x2 ; : : : ; xm g in S so that you can write a0 x C a1 x1 C a2 x2 C    am xm D 0 in such a way that not all the aj D 0. It cannot be that a0 D 0 since, in that case, fx1 ; x2 ; : : : ; xm g would be a finite subset of S that wasn’t linearly independent. Solving the equation for x shows that x 2 span.S /. t u A projection is a linear function P W X ! X on a linear space with the property that P 2 D P , that is, P .P x/ D P x for all x 2 X . Given any subspace E of a linear space X , Theorem 12.1 allows us to define a projection P W X ! X whose range is E in the following way. It’s obvious what to do if E D f0g so take s to be any one nonzero element of E, then s is certainly a linearly independent set, and extend it to a basis SE for E. After that, use Theorem 12.1 once more to extend SE to a basis S for all of X . Writing each x 2 X as its unique linear combination of a finite number of elements of S and separating out the terms that involve members of SE , we write x D y C z where y 2 E and define P x D y. If we want a projection QW X ! X for which it is the null space of Q that is E, we can do that by setting Q D I  P for I the identity function and P the projection that we just defined.

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In order to use projections as a topological tool, we need them to be continuous. Theorem 12.2. Let LW X ! Y be a Fredholm operator of index zero between normed linear spaces, then there are continuous projections P W X ! X with R.P / D N.L/ and QW Y ! Y where N.Q/ D R.L/. A proof of continuity in Theorem 12.2 would require concepts from functional analysis that we don’t need for any other purpose in this book (see [Brown and Page (1970), Exercise 8, page 339]), so we’ll omit it. When we take up the forced pendulum problem in the next chapter, we won’t use Theorem 12.2 anyway because we will explicitly define the projections and it will be clear that they are continuous. A useful property of a projection P is that N.P / \ R.P / D 0. The reason is that if x 2 N.P / \ R.P / then we have x D P x 0 for some x 0 2 X , but then 0 D P x D P 2 x 0 D P x 0 D x. Let LW X ! Y be a Fredholm operator of index zero, let P W X ! X be a projection with R.P / D N.L/ and let LP denote the restriction of L to the null space of the projection. A point in the null space of LP would lie in N.P / \ R.P / so LP is one-to-one. Moreover, the range of LP is all of R.L/ because for any y 2 R.L/ we can take any x such that Lx D y and then notice that x P x 2 N.P / while we still have L.x  P x/ D LP .x  P x/ D y. Therefore LP W N.P / ! R.L/ is an isomorphism. We will need the inverse of LP in the main construction of this chapter, but, in order to use it, we must impose an additional property on our normed linear spaces X and Y . A Banach space is a normed linear space in which sequences that are Cauchy with respect to the distance defined on it by its norm converge. That is, a Banach space is a complete normed linear space. A closed subspace of a Banach space is also a Banach space. The null space N.P / and, because L is Fredholm operator, R.L/ as well are closed subspaces of X and Y , respectively. Thus, if X and Y are Banach spaces, then LP is an isomorphism of Banach spaces. The open mapping theorem (Theorem C.6) therefore implies that L1 P W R.L/ ! N.P / is continuous. Let LW X ! Y be a Fredholm operator of index zero, where X and Y are Banach spaces. The subspaces N.L/ and CokerL are of the same finite dimension, so Theorem C.2 implies that they are both isomorphic to the same euclidean space and therefore there is an isomorphism that we’ll call W CokerL ! N.L/. Let ˘ W Y ! CokerL be the map that takes y to its equivalence class Œy. Now, given a (nonlinear in general) continuous function ZW X ! Y , we will define a function that we will call the Mawhin operator M W X ! X corresponding to the functions L and Z as follows: M.x/ D P x C ˘ Z.x/ C L1 P .I  Q/Z.x/: To see that M is well defined, notice that the property Q2 D Q implies that y  Qy 2 N.Q/ D R.L/ for any y 2 Y . Since X and Y are Banach spaces, we know that L1 P is continuous and, assuming Theorem 12.2, then all the other functions that make up the definition of the Mawhin operator are continuous, so we conclude that the Mawhin operator M is continuous also.

12 The Mawhin Operator

91

We will see that the Mawhin operator converts the problem of solving the equation Lx D Z.x/ into a fixed point problem when L is not invertible, but it is Fredholm of index zero. Before I prove that, we’ll look at a related family of operators. Lemma 12.3. Let L; ZW X ! Y be continuous functions where L is a Fredholm operator of index zero. If x 2 X is a fixed point of Mt .x/ D P x C ˘ Z.x/ C L1 P .I  Q/.Q C t .I  Q//Z.x/ for some t 2 Œ0; 1, then QZ.x/ D 0 and Lx D t Z.x/. Proof. If Mt .x/ D x, then ˘ Z.x/ C L1 P .I  Q/.Q C t .I  Q//Z.x/ D .I  P /x: The image of L1 P is the null space of P and x  P x 2 N.P /, so ˘ Z.x/ 2 N.P / also. Now the image of  is N.L/ D R.P /, so that implies ˘ Z.x/ 2 N.P / \ R.P / D 0. Since  is an isomorphism, that gives us ˘ Z.x/ D 0 as well, which means Z.x/ is in the range of L. That, in turn, shows us that QZ.x/ D 0. Then the equation comes down to just this: 1 1 1 L1 P t Z.x/ D .I  P /x D LP LP .I  P /x D LP .Lx  LP x/ D LP Lx

and therefore t Z.x/ D Lx.

t u

Notice that if t D 1 then we have the Mawhin operator M corresponding to the functions L and Z above. Theorem 12.4. Let L; ZW X ! Y be continuous functions where L is a Fredholm operator of index zero and let M W X ! X be the corresponding Mawhin operator. Then Lx D Z.x/ if and only if M.x/ D x. Proof. Lemma 12.3 tells us that M.x/ D x implies Lx D Z.x/, so it remains to verify the converse. Assuming Z.x/ D Lx, we can substitute it into the definition of the Mawhin operator 1 M.x/ D P x C ˘ Lx C L1 P .I  Q/Lx D P x C LP Lx

because R.L/ D N.˘ / D N.Q/. Noting that LP x D 0 and x  P x 2 N.P /, we rewrite the formula for M.x/ some more 1 M.x/ D P x C L1 P L.x  P x/ D P x C LP LP .x  P x/ D x

to show that x is a fixed point of M .

t u

Now that we know that fixed points of M will give us solutions to Lx D Z.x/, we will want to use the Leray–Schauder degree to show that M has fixed points.

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The next result gives us a good tool for this purpose because it allows us to calculate that degree by computing the degree of a map on N.L/, which is finite-dimensional. Thus, by Theorem 10.6, we will be able to use what we found out in Chap. 9 about computing the Brouwer degree. Theorem 12.5. Let X and Y be Banach spaces and let L; ZW X ! Y be continuous functions where L is a Fredholm operator of index zero and Z is completely continuous. Let M W X ! Y be the corresponding Mawhin operator. Suppose U is a bounded open subset of X such that Lx ¤ t Z.x/ for all x 2 @U and 0 < t  1 and QZ.x/ ¤ 0 for all x 2 @U \ N.L/, then deg.I  M; U/ D deg.m; U \ N.L// where m D ˘ ZjN.L/W N.L/ ! N.L/. Proof. The function Mt .x/ D P x C ˘ Z.x/ C L1 P .I  Q/.Q C t .I  Q//Z.x/ that we introduced in Lemma 12.3 defines a homotopy of the Mawhin operator M1 D M . The restriction of the homotopy to U  Œ0; 1 is compact because Z is completely continuous, U is bounded and the range of P is finite-dimensional. If x is a fixed point of M0 , then, by Lemma 12.3, Lx D 0Z.x/ D 0, so x 2 N.L/, and also QZ.x/ D 0 by that same lemma. A hypothesis then tells us that this fixed point cannot be in the boundary of U . Another hypothesis is designed to guarantee that no fixed point of Mt .x/ for 0 < t  1 can lie in @U . Thus the hypotheses of the homotopy property of the Leray–Schauder degree (Theorem 11.7) are satisfied and we can conclude that deg.I  M; U / D deg.I  M0 ; U / where, since .I  Q/Q is the constant function at 0, the definition reduces to M0 .x/ D P x C ˘ Z.x/. Thus M0 maps all of X to the subspace N.L/ and therefore by the factor property (Theorem 11.9) we have deg.I  M0 ; U / D deg.I  M0 jN.L/; U \ N.L// where both I denote the appropriate identity functions. Then, noting that P x D x for all x 2 N.L/, we see that I  M0 jN.L/ D I  P  ˘ ZjN.L/ D ˘ ZjN.L/ D m:

t u

More information about the Mawhin operator can be found in Gaines and Mawhin (1977), which was the source for most of this chapter.

Chapter 13

The Pendulum Swings Back

In the previous chapter, we returned to the equation of the forced pendulum y 00 C a sin y D e; assuming now that the forcing term eW R ! R is a continuous T -periodic function, but not necessarily an odd function as it was in Chap. 5. Recall that a D g` where g is the gravitational constant and ` is the length of the pendulum, and therefore a > 0. I pointed out that we could build T -periodic solutions to the equation by solving the corresponding periodic boundary value problem y 00 C a sin y D e y.0/  y.T / D y 0 .0/  y 0 .T / D 0: But, in contrast to the boundary value problem that we studied in Chap. 5, this problem doesn’t always have a solution. If the forcing term is too big, as measured by its average value, compared to the constant a, then there can’t be any solutions to the boundary value problem. For u 2 C Œ0; T , we’ll use this notation for the average value: .u/ D

1 T

Z

T

u.t / dt: 0

Theorem 13.1. If there is a solution yW Œ0; T  ! R to the differential equation y 00 C a sin y D e that satisfies y 0 .0/ D y 0 .T /, then j.e/j  a. Proof. Writing the equation as e.t /  a sin y.t/ D y 00 .t / and integrating we have

© Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__13

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13 The Pendulum Swings Back

Z

T

Z

T

e.t / dt  a

0

Z

T

sin y.t/ dt D

0

y 00 .t / dt D y 0 .T /  y 0 .0/ D 0

0

so ˇZ ˇ T j.e/j D ˇˇ

T 0

ˇ ˇZ ˇ ˇ ˇ e.t / dt ˇ D a ˇˇ

T 0

ˇ Z ˇ ˇ sin y.t/ dt ˇ  a

T

j sin y.t /j dt  aT

0

which completes the argument.

t u

The purpose of this chapter is to show you that, when we put a sufficiently strong restriction on the forcing term, we do get solutions to the boundary value problem. One feature of the differential equation y 00 C a sin y D e that we should notice is that if we find one solution y0 W Œ0; T  ! R, then we have an infinite number of solutions. The reason is just that if we define yk .t / D y0 .t / C k2 , then the lefthand side of the equation doesn’t change. But we don’t think of all those solutions as really different from each other. However, using the techniques we worked out in Chap. 12, we’ll also prove that there are at least two solutions that are truly distinct because they do not differ by a constant. In the previous chapter we introduced the space of functions satisfying the periodic boundary conditions CP2 Œ0; T  D fu 2 C 2 Œ0; T W u.0/  u.T / D u0 .0/  u0 .T / D 0g: Writing the forced pendulum equation as u00 D e  a sin u, we represented the left-hand side as Lu D u00 where LW CP2 Œ0; T  ! C Œ0; T . The superposition operator F W C Œ0; T  ! C Œ0; T  is defined by F .u/.t / D e.t /  a sin u.t /. Defining j W CP2 Œ0; T  ! C Œ0; T  to be the inclusion, let Z D Fj and then we can write the equation in the form we studied in Chap. 12: Lu D Z.u/. Since the definition of CP2 Œ0; T  incorporated the periodic boundary conditions, we can see that the solutions to Lu D Z.u/ are the solutions to our boundary value problem. The way we will demonstrate that the periodic boundary value problem has a solution, under suitable conditions, is by showing that the hypotheses of Theorem 12.5 are satisfied and that deg.I  M; U / D deg.m; U \ N.L// ¤ 0, for the right choice of U . Therefore the Mawhin operator M has a fixed point, so we have a solution by Theorem 12.4. The continuity of the Mawhin operator, and consequently the proof of Theorem 12.5, required that the normed linear spaces be complete. The following result will give us all the completeness we need, not only for this chapter but for the rest of the book. Theorem 13.2. For each k  0, the space C k Œa; b is a Banach space. Proof. The space C 0 Œa; b D C Œa; b is complete because, by the completeness of the reals, a Cauchy sequence in C Œa; b is pointwise convergent and thus, since Œa; b is compact, it is uniformly convergent. Let fun g be a Cauchy sequence in C k Œa; b, then, from the definition of the C k norm, we see that, for each j D 0; 1; : : : ; k, the

13 The Pendulum Swings Back

95

.j /

sequence of derivatives fun g is Cauchy and thus it has a limit g in C Œa; b. As in the proof of Theorem 2.7, the fundamental theorem of calculus implies that g is the derivative of g . Therefore, if we let u D g , then g D u.j / . Given .j /  > 0, for each j D 0; 1; : : : ; k, let Nj be such that n > Nj implies kun  u.j / k <  , and then, for n > maxj Nj , we have kun  ukk < , that is, fun g converges to kC1 u in C k Œa; b. t u To apply the theory of the Mawhin operator, the next thing we need to show is that LW X ! Y defined by Lu D u00 is a Fredholm operator of index zero from X D CP2 Œ0; T  to Y D C Œ0; T . The general solution to u00 D 0 is a linear function and the boundary condition u.0/ D u.T / implies its slope is zero, so the null space of L is the one-dimensional vector space of constant real-valued functions. The following result describes the range of L. Lemma 13.3. If y 2 C Œ0; T , then y is in the range of LW CP2 Œ0; T  ! C Œ0; T  defined by Lu D u00 if, and only if, .y/ D 0. Proof. The fundamental theorem of calculus tells us that if u00 D y and u0 .0/ D u0 .T /, then T.y/ D 0. On the other hand, suppose y 2 C Œ0; T  with .y/ D 0. Since the general solution to u00 D y is u.t / D a C bt C

Z tZ

s

y.r/ dr ds 0

0

then 

1 u.t / D  T

Z 0

T

Z

s 0

 y.r/ dr ds t C

Z tZ

s

y.r/ dr ds 0

0

is a particular solution with the property u.0/ D u.T / D 0. The condition .y/ D 0 is then precisely what we need to conclude that u0 .0/ D u0 .T / and thus that u 2 CP2 Œ0; T . t u Lemma 13.4. The function LW CP2 Œ0; T  ! C Œ0; T  defined by Lu D u00 is a Fredholm operator of index zero. Proof. If we view  as a function W C Œ0; T  ! R, then we can see that it is continuous because j.y/j  kyk. Lemma 13.3 tells us that R.L/ D 1 .0/, so we have R.L/ closed in C Œ0; T . Besides that, it tells us that if y; z 2 C Œ0; T  and y  z 2 R.L/, then .y/ D .z/, so we can define a linear isomorphism  between C okerL and N.L/ by sending the equivalence class Œy to the constant function that takes the value .y/ on Œ0; T . t u We know from the general results we discussed in the previous chapter that there are continuous projections P W CP2 Œ0; T  ! CP2 Œ0; T  and QW C Œ0; T  ! C Œ0; T  with R.P / D N.L/ and N.Q/ D R.L/. As I promised there, in this specific case we have natural choices for P and Q. For x 2 CP2 Œ0; T , take P x to be the constant

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13 The Pendulum Swings Back

map whose value is x.0/. The average value function can be used to determine Q: for y 2 C Œ0; T , let Qy be the constant function whose value is .y/. It is easy to show that the functions P and Q are continuous. We will make use of the fact that, for u 2 CP2 Œ0; 1, we have QZ.u/ D .e  a sin u/ D .e/  a.sin u/: Now we need to find out about the function m D ˘ ZjN.L/ of Theorem 12.5. Since the domain of m will be in N.L/, we can start with a constant function uW Œ0; T  ! R where u.t / D u0 for all t 2 Œ0; T . Now ˘ takes y 2 C Œ0; T  to the equivalence class in C okerL containing y which, by Lemma 13.3, consists of all functions in C Œ0; 1 whose average value is .y/. Thus we can identify ˘y with the real number .y/. We defined the isomorphism W C okerL ! N.L/ in the proof of Lemma 13.4: it takes the equivalence class Œy to the constant function with value .y/. Since u is the constant function at u0 , the average value of y D Z.u/ D e  a sin u is .y/ D .e/  a sin u0 and that number describes the equivalence class in C okerL that contains u. We conclude that m.u/ D ˘ Z.u/ D ..e/  a sin u0 /: We can describe m more simply by identifying constant functions with the corresponding real numbers: then the map mW R ! R is defined by m.t / D a sin t  .e/: The key to the application of the Leray–Schauder degree, via the Mawhin operator, to the forced pendulum problem is the choice of the open set U  CP2 Œ0; T . In order to define U , we will need a fact about functions in u 2 C 2 Œ0; T , namely that the numbers kuk; ku0 k, and ku00 k cannot take on arbitrary values, but they are instead related in the following way. Lemma 13.5. If u 2 C 2 Œ0; T  such that kuk  A and ku00 k < B, then r 0

ku k < ˚.A; B/ D 2

A2 C AB: T2

Proof. Suppose ju0 .t /j reaches its maximum at t D b. We can assume that u0 .b/ is positive because, if it isn’t, we can use the function u in place of u and if we show that k  u0 k < ˚.A; B/ then we are done since the same A and B can be used. We’ll suppose for now that u0 .b/ > 2A and take care of the case 0 < u0 .b/  2A at the end T T 2A 0 0 of the proof. Now u .b/ > T implies that there exists t 2 Œ0; T  with u .t /  2A T since otherwise we’d have Z u.T /  u.0/ D

T

0

Z

u .t / dt > 0

0

T

2A dt D 2A T

13 The Pendulum Swings Back

97

but kuk  A implies ju.t1 /  u.t2 /j  2A for all t1 ; t2 2 Œ0; T . Thus the set .u0 /1 . 2A / is nonempty and, since it’s compact, there is a point a in it whose T distance from b is a minimum. Assume for convenience that a < b, then 0


2A , T

then for all t 2 Œ0; T  we have r

0

0

2A 2 / < 4AB T

ju .t /j  u .b/ < 2

A2 C AB D ˚.A; B/: T2

If 0 < u0 .b/  2A , then ku0 k < ˚.A; B/ because 2A < ˚.A; B/. Thus, for all T T 0 values of u .b/ > 0, we have shown that ku0 k < ˚.A; B/. u t We will use the set  that consists of these three open intervals:  D f.

3 3 3 3  ; /; . ; /; . ; /g 2 2 2 2 2 2

and, for .˛; ˇ/ 2 , we define U.˛;ˇ/ D fu 2 CP2 Œ0; T W ˛ < u.t / < ˇ and ju00 .t /j < 2aI for all t 2 Œ0; T g: Theorem 13.6. The set U.˛;ˇ/ for .˛; ˇ/ 2  is a bounded open subset of CP2 Œ0; T . Proof. If u 2 U.˛;ˇ/ , then by Lemma 13.5, kuk2 D kuk C ku0 k C ku00 k
˛. Define  D minfˇ  umax ; umin  ˛; 2a  ku00 kg > 0: If v 2 CP2 Œ0; T  such that kv  uk2 < , then kv  uk <  and ku00  v00 k <  and it follows that v 2 U.˛;ˇ/ . This proves that U.˛;ˇ/ is open. t u Lemma 13.7. If j.e/j < a, then QZ.u/ D .e/  a.sin u/ ¤ 0 for all u 2 @U.˛;ˇ/ \ N.L/ with .˛; ˇ/ 2 . Proof. If u 2 @U.˛;ˇ/ \ N.L/, then the value of the constant function u is either ˛ or ˇ, where .˛; ˇ/ comes from the set . In either case, j.sin u/j D 1 so QZ.u/ D .e/ ˙ a ¤ 0 since j.e/j < a. t u We know from 13.1 that, to solve the boundary value problem, we must at least assume that j.e/j  a, so the hypothesis of Lemma 13.7 is close to the weakest one possible. We’ll use a much stronger hypothesis to obtain the next result. Somewhat weaker hypotheses than that one have been shown to be sufficient to establish the existence of solutions to the boundary value problem that arises from the forced pendulum equation, but it would take us beyond the scope of this book to explore that topic. In Chap. 5, it was easy to use the Schauder fixed point theorem because we were concerned with a problem of the form y 00 D f .t; y.t /; y 0 .t // where the function f W Œ0; 1  R  R ! R had a bounded image. In the present boundary value problem that arises from a forced pendulum equation, the only difference is in the boundary conditions, so we still have f with bounded image because now jf .t; s; p/j D je.t /  a sin sj  je.t /j C a  kek C a: Lemma 13.8. If kek < a, then Lu ¤ t Z.u/ for u 2 @U.˛;ˇ/ , where .˛; ˇ/ 2  and 0 < t  1. Proof. If u is a solution to Lu D t Z.u/, that means u00 .s/ D t .e.s/ C a sin u.s// for all s 2 Œ0; T  and therefore ju00 .s/j < t.kek C a/ < 2a. Consequently, if u 2 @U.˛;ˇ/ is a solution to Lu D t Z.u/, then there must exist s0 2 Œ0; T  such that u.s0 / D ˛ or u.s0 / D ˇ. If u.s0 / D ˛ then u attains its minimum at s0 and therefore u00 .s0 /  0. Now .˛; ˇ/ 2  implies that ˛ is either 3 or 2 so sin u.s0 / D 1. If u is a solution 2 to Lu D t Z.u/ then, in particular, u00 .s0 / D t .e.s0 /  a sin u.s0 // D t .e.s0 /  a/ < 0 Because, by hypothesis, e.s/ < a for all s 2 Œ0; T . Similarly, if u.s0 / D ˇ, then we are at the maximum and u00 .s0 /  0. Now ˇ is either  or 3 so sin u.s0 / D 1. 2 2 But then if u is a solution to Lu D t Z.u/, that would imply u00 .s0 / D t .e.s0 /Ca/ > 0 since e.t / > a for all t . t u

13 The Pendulum Swings Back

99

Remember that m.t / D a sin t  .e/, so assuming that j.e/j < a gives us m.

3  / D m. / D a  .e/ < 0 2 2

and m.

3 / D m. / D a  .e/ > 0: 2 2

We use Theorem 9.9 to calculate that deg.m; U. 3 ;  / \ N.L// D deg.m; U. ; 3 / \ N.L// 2

2

2

2

D deg.m; U. 3 ; 3 / \ N.L// D 1: 2

2

By putting together the general results from Chap. 12 with all the specific information about the forced pendulum equation that we obtained in this chapter, we can prove our main result, which comes from Mawhin’s paper Mawhin (1982). Theorem 13.9. If eW R ! R is a T -periodic function and kek < a, then there are T -periodic solutions y0 ; y1 W R ! R to the forced pendulum equation y 00 C a sin y D e such that y0  y1 is not a constant function. Proof. For the Mawhin operator M that comes from the periodic boundary value problem of the forced pendulum equation, we’ve satisfied the hypotheses of Theorem 12.5, under the condition kek < a, for the open sets U.˛;ˇ/ with .˛; ˇ/ 2 . We’ve just shown that deg.m; U.˛;ˇ/ \ N.L// D 1 for all .˛; ˇ/ 2 , so, by Theorem 12.5, we know that deg.I  M; U. 3 ;  / / D deg.I  M; U. ; 3 / / 2

2

2

2

D deg.I  M; U. 3 ; 3 / / D 1: 2

2

Then Theorem 12.4 tells us in particular that there is a solution y0 to the problem that lies in U. 3 ;  / so 3 < y0 .t / <  for all t 2 Œ0; T . It also tells us that there 2 2 2 2 is a solution y 2 U. ; 3 / , but it could be that y.t/ D y0 .t / C 2 for all t and that’s 2 2 not really a different solution, as I pointed out before. However, we can get more information by using Theorem 11.4, the additivity property of the Leray-Schauder degree: deg.I  M; U. 3 ; 3 / / D deg.I  M; U. 3 ;  / / C deg.I  M; U. ; 3 / / 2

2

2

2

/ C deg.I  M; U.  2 ;2/

2

2

100

13 The Pendulum Swings Back

/ D 1, so there is a solution y1 2 U  . and therefore deg.I  M; U.  . 2 ;2/ 2 ;2/ If y1  y0 were a constant function, the differential equation would then imply sin y0 .t / D sin y1 .t / for all t, so that constant would have to be a multiple of 2 . But the only possible constant value is 2 and that would place y1 in U. ; 3 / rather 2 2

. Thus there are two solutions that do not differ by a constant. t u than U.  2 ;2/

I used the Mawhin operator M for the forced pendulum equation y 00 Ca sin y D e with periodic boundary conditions because the linear operator Lu D u00 is Fredholm of index zero, but it isn’t invertible with respect to those boundary conditions, as it is with the Dirichlet condition. A different approach would have been to write the forced pendulum equation as y 00  y D e  a sin y  y because the linear function Lu D u00  u can be inverted with respect to the periodic boundary conditions. That would have given us a fixed point problem of the form y D L1 Fj.y/ as in Chap. 5. However, notice that for the right-hand side of the equation, the corresponding function f W Œ0; T   R  R ! R is f .t; s; p/ D e.t /  a sin s  s which no longer has as its image a bounded subset of R. The bounded image property greatly simplified the discussion in this chapter, just as it did in Chap. 5, and that’s why it was worthwhile to introduce Mawhin’s theory—because I wanted to preserve that property. A more realistic model of the pendulum would have taken into account the damping effect of friction on the motion of the pendulum. The classical equation for the damped free pendulum is y 00 C cy 0 C a sin y D 0 where c is a constant. If the right-hand side is replaced by a periodic forcing function e.t /, then we have the damped and forced pendulum equation that can still be analyzed by the methods we’ve been using. I’ve limited this chapter to the undamped case just to keep things a little simpler. There is considerable literature devoted to “pendulum-like equations,” in which the constant c in the damping term is replaced by a function g.y/, so there’s a lot more you can find out about this topic if you want to go beyond what’s in this book.

Part III

Fixed Point Index Theory

Chapter 14

A Retraction Theorem

Let’s begin with an example. Inside the linear space C Œ0; 1 of maps u W Œ0; 1 ! R is the positive cone of C Œ0; 1, written CC Œ0; 1, which is the set of all u 2 C Œ0; 1 such that u.t /  0 for all t. The positive cone is obviously convex and it is a closed subset of C Œ0; 1 with respect to the sup norm because if u0 .t0 / < 0, so u0 … CC Œ0; 1, then u … CC Œ0; 1 if kuu0 k < ju0 .t0 /j. There is a retraction  W C Œ0; 1 ! CC Œ0; 1 defined by setting .u/.t / D u.t / if u.t /  0 and .u/.t / D 0 if u.t / < 0. The function  is continuous because k.u/.t /  .u0 /.t /k  ku.t /  u0 .t /k. We will learn in this chapter that there is a retraction of a normed linear space onto any closed and convex subset of it. Recall that if X is a normed linear space, then d W X  X ! R defined by d.x1 ; x2 / D kx1  x2 k is a metric on X . For each x0 2 X and  > 0 we have the epsilon-neighborhood of x0 , defined by B.x0 I / D fx 2 X W d.x; x0 / < g: The open sets of the metric space topology on X are the unions of the epsilonneighborhoods. We call an open set that contains x0 2 X a neighborhood of x0 . It will be important that the epsilon-neighborhoods are convex subsets of X , so let’s verify it: if x1 ; x2 2 B.x0 I / and 0  t  1, then kx0  .tx1 C .1  t /x2 /k D kt .x0  x1 / C .1  t /.x0  x2 /k  tkx0  x1 k C .1  t /kx0  x2 k < : The notion of the distance between points extends to the distance between subsets of a metric space this way: if P; Q  X then d.P; Q/ D inffd.p; q/ W p 2 P; q 2 Qg:

© Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__14

103

104

14 A Retraction Theorem

Lemma 14.1. Let U be an open subset of a metric space X . The function dU W X ! R defined by dU .x/ D d.x; X n U / is continuous. Proof. To prove continuity at x0 2 X , first suppose x0 … U so dU .x/ D 0. Given  > 0, let x 2 B.x0 I /, then dU .x/ < d.x; x0 / <  and it remains to prove continuity on U . For x0 2 U and  > 0, we will see that if d.x0 ; x 0 / < 12 , then jdU .x0 /  dU .x 0 /j < . Choose x … U such that d.x0 ; x / < dU .x0 / C 12 , then dU .x 0 / < d.x 0 ; x /  d.x 0 ; x0 / C d.x0 ; x /
0 such that if y 2 B, then kyk   so jy.s/j   for all s 2 Œ0; 1 and thus r.y.s// 2 r.Œ; /. Since r is continuous, there exists

> 0 such that jr.y.s//j  for all y 2 B and s 2 Œ0; 1. We then have ˇZ ˇ jf .y/.t /j D ˇˇ

1 0

ˇ Z ˇ G.t; s/r.y.s// ds ˇˇ 

1

jG.t; s/r.y.s//j ds  

0

so kf .y/k   for all y 2 B and we have proved that the set f .B/ is bounded. To prove that f .B/ is equicontinuous at t0 2 Œ0; 1, suppose we are given  > 0, then, by the uniform continuity of G, there exists ı > 0 such that if jt  t0 j < ı, then jG.t; s/  G.t0 ; s/j <  for all s 2 Œ0; 1. Therefore, if jt  t0 j < ı, then ˇZ ˇ jf .y/.t /  f .y/.t0 /j D ˇˇ Z

1 0

 0

1

Z

1

G.t; s/r.y.s// ds  0

ˇ ˇ G.t0 ; s/r.y.s// ds ˇˇ

jG.t; s/  G.t0 ; s/jjr.y.s//j ds < :

16 The Tubular Reactor

117

Since jt  t0 j < ı implies jf .y/.t /  f .y/.t0 /j <  for all y 2 B, we have shown that f .B/ is an equicontinuous subset of C Œ0; 1. t u So far the mathematics of the tubular reactor problem looks pretty much like that of the forced pendulum problem of Chap. 5, but that is about to change. A fixed point of the map f is a continuous function y W Œ0; 1 ! R, but now we need to remember that y.t/ represents the temperature of the liquids in the reactor at location t . Those liquids cannot be frozen, so we should only consider solutions y such that y.t /  0 (Celsius) for all t. We will denote the subset of C Œ0; 1 of such “positive” functions by CC Œ0; 1, which was the positive cone that we introduced at the start of Chap. 14. We will restrict the domain of f to positive functions so that any fixed point will be a possible steady state for the reactor problem. It is important to note that the restriction of a completely continuous function is completely continuous, so that hypothesis for f is still fulfilled. In order to establish the existence of fixed points of the function f , and thus solutions to the boundary value problem, we need to show that f satisfies the hypotheses of the theorems of the next chapter. The function f will have the required properties provided that the reaction rate function r is nice enough. The following theorem tells us how r has to behave. Theorem 16.2. Let f W CC Œ0; 1 ! C Œ0; 1 be defined by Z

1

f .y/.t / D

G.t; s/r.y.s// ds 0

where r W Œ0; 1/ ! R is continuous and 

( G.t; s/ D

exp 1;

ts ˇ



; for 0  t  s  1 for 0  s  t  1:

Assuming that there exist 0 < d < a < ae 1=ˇ D z such that the function r.x/ has the following properties: (i) if x  z then r.x/  0, (ii) if x  d then r.x/ < d , (iii) if a  x  z then r.x/ > a.ˇ  ˇe 1=ˇ /1 , it follows that the function f .y/ has the following properties: 1. 2. 3. 4.

if kyk  z then f .y/.t /  0 for all 0  t  1, if kyk  d then kf .y/k < d , if a  y.t /  z then inf0t1 f .y/.t / > a, if kf .y/k > z then inf0t1 f .y/.t / > az kf .y/k.

Proof. Since G.t; s/  0 for all s; t 2 Œ0; 1, it follows from the definition of f .y/ that hypothesis (i) implies conclusion (1). For each s 2 Œ0; 1, the function G.t; s/ is a nondecreasing function of t, so f .y/.t / is also nondecreasing. Therefore

118

16 The Tubular Reactor

inf0t1 f .y/.t / D f .y/.0/ and kf .y/k D f .y/.1/. For (2) we use hypothesis (ii): if kyk  d then Z kf .y/k D f .y/.1/ D

Z

1

1

G.1; s/r.y.s// ds < d 0

G.1; s/ ds D d:

0

To prove (3), since a  y.t/  z, then hypothesis (iii) gives us Z inf f .y/.t / D f .y/.0/ D

0t1

1

e s=ˇ r.y.s// ds

0

> a.ˇ  ˇe 1=ˇ /1

Z

1

e s=ˇ ds D a:

0

Conclusion (4) holds because kf .y/k > z D ae 1=ˇ implies Z inf f .y/.t / D f .y/.0/ D

0t1

e

1=ˇ

1

e s=ˇ r.y.s// ds

0

Z

1

r.y.s// ds 0

D e 1=ˇ f .y/.1/ D e 1=ˇ kf .y/k D

a kf .y/k: z

t u

In order to obtain solutions to the boundary value problem that can be applied to the tubular reactor problem, we had to leave the setting of maps of normed linear spaces. Instead, the map is defined on a closed, convex subset of such a space. In the next chapter we will use the fixed point index to obtain fixed point information about such maps. The conclusions of Theorems 16.1 and 16.2 will then furnish the hypotheses that will allow us to apply those abstract results to the map whose fixed points are solutions to the boundary value problem that represent steady state processes in the tubular reactor.

Chapter 17

Fixed Points in a Cone

In this chapter, we will use the fixed point index to show that functions that satisfy certain conditions must have at least two or at least three fixed points. That will lead us to the conclusion that, for some reaction rate functions r, the chemical reaction in the tubular reactor that we discussed in the previous chapter must have at least two or, under an additional condition, at least three steady state temperature functions y.t/ for the chemical reaction. Our first, quite general, preliminary result can be thought of as a refinement of the Schauder fixed point theorem (Theorem 4.4), since this lemma, along with Theorem 15.3, the existence property of the fixed point index, implies that theorem. Lemma 17.1. Let A be a closed convex subset of a normed linear space and let f W A ! A be a compact map, then i nd.f; A/ D 1. Proof. Note that A is an open subset of itself and there is no boundary, so the compactness of f implies that i nd.f; A/ is well defined. Choose y0 2 A, then a homotopy H W A  Œ0; 1 ! A can be defined by H.y; t/ D .1  t /.y/ C ty0 since A is convex. The straight-line homotopy property and normalization property of the fixed point index imply that i nd.f; A/ D i nd.h0 ; A/ D i nd.h1 ; A/ D i nd.x0 ; A/ D 1:

t u

The functions we are interested in will be defined on a subset of a normed linear space X . A closed, convex subset C of X is a cone if y 2 C implies ay 2 C for all real a  0 and y 2 C implies y … C . For z > 0, let Cz D fy 2 C W kyk  zg. Let f W Cz ! C be a compact map and U  Cz an open set such that f .y/ ¤ y for all y 2 @U , then we have the fixed point index i nd.f; U / of Chap. 15.

© Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__17

119

120

17 Fixed Points in a Cone

Define ˛ W CC Œ0; 1 ! Œ0; 1/ by ˛.u/ D inf0t1 u.t /, so obviously ˛.u/  kuk. The function ˛ is continuous because, given  > 0, if ku1  u2 k <  and ˛.uj / D uj .tj / for j D 1; 2, then ˛.u2 /  u2 .t1 / < u1 .t1 / C  D ˛.u1 / C  and ˛.x1 / < ˛.u2 / C  similarly, so j˛.u1 /  ˛.u2 /j < . Moreover, the function ˛ is concave on the convex set CC Œ0; 1, that is, ˛.u1 C .1  /u2 /  ˛.u1 / C .1  /˛.u2 / for 0    1. The reason is that if we set u.t / D u1 .t / C .1  /u2 .t / and write ˛.u/ D inf0t1 u.t / D u.t0 /, then the inequality holds because u1 .t0 /  inf0t1 u1 .t / and u2 .t0 /  inf0t1 u2 .t /. In the general setting of a compact map f W Cz ! C , where C is a cone in a normed linear space X , we will assume we have a continuous, concave function ˛ W C ! Œ0; 1/ such that ˛.y/  kyk. We will use ˛ to define some useful subsets of Cz . Given constants 0 < a < b  z, we define Sb .˛  a/ D fy 2 Cb W ˛.y/  ag and Sz .˛  a/  Cz similarly. The fact that ˛ is a continuous function implies that Sb .˛  a/ is closed since its complement is open. Furthermore, since ˛ is a concave function, Sb .˛  a/ is a convex set because ˛.y1 /  a and ˛.y2 /  a implies ˛.y1 C .1  /y2 /  ˛.y1 / C .1  /˛.y2 /  a C .1  /a D a: We will denote the interior of Sb .˛  a/ by Sb .˛ > a/ because it consists of the y 2 Cb such that ˛.y/ > a. Our multiple fixed point results will make use of the following fixed point index calculation. Lemma 17.2. Let f W Cz ! C be a compact map. Suppose there exist 0 < a < b  z such that (1) (2) (3) (4)

Sb .˛ > a/ ¤ ; If y 2 Sb .˛  a/ then ˛.f .y// > a If y 2 Sz .˛  a/ then kf .y/k  z If y 2 Sz .˛  a/ and kf .y/k > b then ˛.f .x// > a

Then the fixed point index of f on Sz .˛  a/ is defined and i nd.f; Sz .˛  a// D 1. Proof. From (1) we choose y0 2 Sb .˛ > a/. Define H W Sz .˛  a/  Œ0; 1 ! Cz by H.y; t/ D .1  t /f .y/ C ty0 . This homotopy is well defined by hypothesis (3) and the convexity of Cz . Moreover, since f is a compact map, then H is also a

17 Fixed Points in a Cone

121

compact map by Lemma 10.4. Suppose H.y; t/ D y for some 0  t  1. We claim that ˛.y/ > a and therefore H has no fixed points on the boundary of Sz .˛  a//, since .y; t / a boundary point implies that ˛.y/ D a. To establish the claim, first suppose that H.y; t/ D y and kf .y/k  b, then kyk D kH.y; t/k D k.1  t /f .y/ C ty0 k  .1  t /kf .y/k C tky0 k  b since y0 2 Sb .˛ > a/  Cb . Therefore, ˛.f .y// > a by hypothesis (2) and so by the concavity of ˛ we have ˛.y/ D ˛.H.y; t// D ˛..1  t /f .y/ C ty0 /  .1  t /˛.f .y// C t ˛.y0 / > a: Now if H.y; t/ D y and kf .y/k > b, then by hypothesis (4) we again have ˛.f .y// > a so ˛.y/ > a which establishes our claim. Thus, the hypotheses of Theorem 15.7, the homotopy property of the fixed point index, are satisfied and ind.ht ; Sz .˛  a// is well defined and constant for all 0  t  1. Theorem 15.4, the normalization property, completes the computation because ind.f; Sz .˛  a// D ind.h0 ; Sz .˛  a// D ind.h1 ; Sz .˛  a// D ind.y0 ; Sc .˛  a// D 1:

t u

Our application to the tubular reactor model will depend on two fixed point results. The first furnishes us not only with three fixed points but also with rather precise information on the location of those fixed points, and that is information that we will need to prove the subsequent theorem. Theorem 17.3. Suppose f W Cz ! Cz is a compact map and there exist 0 < d < a < b  z such that (1) (2) (3) (4)

Sb .˛ > a/ ¤ ; If y 2 Sb .˛  a/ then ˛.f .y// > a If kyk  d then kf .y/k < d If y 2 Sz .˛  a/ and kf .y/k > b then ˛.f .y// > a

Then each of the disjoint open sets U1 D fy 2 Cz W kyk < d g; U2 D Sz .˛ > a/ and U3 D Cz n .U 1 [ U 2 / contains a fixed point of f . Proof. The property ˛.y/  kyk implies that if y 2 U 1 , then ˛.x/  d < a, so the closures of U1 and U2 are disjoint. Since hypothesis (3) states that f .U 1 /  U1 , Lemma 17.1 implies that i nd.f; U1 / D 1, so Theorem 15.3, the existence property of the index, tells us that there is a fixed point y1 of f in U1 . The hypotheses of this theorem are the same as those of Lemma 17.2 except for hypothesis (3) of Lemma 17.2 and that is satisfied because f maps Cz to itself. Therefore ind.f; U2 / D 1 by Lemma 17.2, so f also has a fixed point y2 in U2 . Since f is a self-map of Cz , Lemma 17.1 also implies that ind.f; Cz / D 1. We have disjoint open

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17 Fixed Points in a Cone

sets U1 ; U2 , and U3 D Cz n .U 1 [ U 2 / satisfying the hypotheses of Theorem 15.6, the additivity property of the fixed point index, so 1 D ind.f; Cz / D ind.f; U1 / C ind.f; U2 / C ind.f; U3 / D 2 C ind.f; U3 / and therefore ind.f; U3 / D 1 so there is a third fixed point y3 , in U3 .

t u

The next result differs from the previous one in an important way: it does not require that f be a self-map of Cz . Theorem 17.4. Suppose f 0 < d < a  z such that (1) (2) (3) (4)

W Cz ! C is a compact map and there exist

Sz .˛ > a/ ¤ ;, if y 2 Sz .˛  a/ then ˛.f .y// > a, if kyk  d then kf .y/k < d , if kf .y/k > z then ˛.f .y//  az kf .y/k,

then f has at least two fixed points. Proof. From hypothesis (3), Lemma 17.1 again gives us a fixed point y1 of f in Cd , so the problem is to establish the existence of a fixed point of f somewhere else in Cz . We will make use of the theorem we have just proved, but, to do so, we need a self-map g of Cz . Define a retraction  W C ! Cz by ( .y/ D

x; z x; kyk

if kyk  z if kyk  z

and set g D f . Let S be a subset of Cz . Since f is a compact map, there exists a compact subset K of C that contains f .S /. Then g.S / D f .S /  .K/ which is compact, so we conclude that g is also a compact map. We want to apply Theorem 17.3 to g, so we will show that all its hypotheses are satisfied, with b D z. Hypothesis (1) of this theorem implies the same for Theorem 17.3, since it doesn’t depend on the map, hypothesis (3) holds because d < z, and hypothesis (4) of Theorem 17.3 is vacuously satisfied for g when b D z because then kg.y/k  z D b. Thus, all that is left for us to do is to verify hypothesis (2) of Theorem 17.3 for the map g, that is, show that if y 2 Sz .˛  a/, then ˛.g.y// > a. If kf .y/k  z, then ˛.g.y// D ˛.f .y// > a by our present hypothesis (2), so suppose kf .y/k > z. Then by the concavity of ˛ we have z f .y// kf .y/k z z f .y// C .1  /0/ D ˛. kf .y/k kf .y/k

˛.g.y// D ˛.

17 Fixed Points in a Cone

123

z z ˛.f .y// C .1  /˛.0/ kf .y/k kf .y/k z ˛.f .y//:  kf .y/k 

Hypothesis (4) of this theorem then completes the verification of the hypotheses of Theorem 17.3 for the map g because ˛.g.y// 

z z a ˛.f .y// > kf .y/k D a: kf .y/k kf .y/k z

Now we can apply Theorem 17.3 to the map g and, in particular, conclude that there is a fixed point y3 of g in the set U3 D Cz n .U 1 [ U 2 / where U1 D fy 2 Cz W kyk < d g and U2 D Sz .˛ > a/. Since y3 … U1 , then y3 ¤ y1 and y3 … U 2 imply ˛.y3 / < a. Suppose kf .y3 /k > z, then hypothesis (4) and the inequality between ˛.g.y// and ˛.f .y// that we just established would imply a > ˛.y3 / D ˛.g.y3 // 

z z a ˛.f .y3 //  kf .y3 /k D a: kf .y3 /k kf .y3 /k z

Therefore y3 is also a fixed point of f because kf .y3 /k  z and thus y3 D g.y3 / D f .y3 / D f .y3 /. t u In the previous chapter we discussed the tubular reactor model in which the stable states for the reaction correspond to the solutions y 2 CC Œ0; 1 to the boundary value problem ˇy 00 .t /  y 0 .t / C r.y.t // D 0;

ˇ>0

. / ˇy 0 .0/  y.0/ D 0;

y 0 .1/ D 0

We are now ready to demonstrate that this problem has multiple solutions, provided that the reaction rate function r has the properties we introduced in the previous chapter. In that chapter, we proved that the fixed points of the function f W CC Œ0; 1 ! C Œ0; 1 defined by Z

1

f .y/.t / D

G.t; s/r.y.s// ds 0

where r W Œ0; 1/ ! R is continuous and 

( G.t; s/ D

exp 1;

ts ˇ



; for 0  t  s  1 for 0  s  t  1:

are the solutions to the boundary value problem (*).

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17 Fixed Points in a Cone

Theorem 17.5. If there exist 0 < d < a < ae 1=ˇ D z such that the reaction rate function r W Œ0; 1/ ! R has the following properties: (i) if x  z then r.x/  0, (ii) if x  d then r.x/ < d , (iii) if a  x  z then r.x/ > a.ˇ  ˇe 1=ˇ /1 , then the function f W CC Œ0; 1 ! C Œ0; 1 whose fixed points are the solutions to the boundary value problem (*) has at least two fixed points. If, in addition, f .Cz /  Cz , then there are at least three fixed points. Proof. We will apply Theorems 17.3 and 17.4 to the map f W Cz Œ0; 1 ! CC Œ0; 1 which is the restriction of f to Cz Œ0; 1 D fy 2 CC Œ0; 1 W kyk  zg. Since Cz Œ0; 1 is bounded, this restricted map is compact by Theorem 16.1. The range of the restriction is the cone CC Œ0; 1 by hypothesis (i) because G.s; t /  0 for all s; t 2 Œ0; 1. The properties of the map r are the hypotheses of Theorem 16.2 and the conclusions of that theorem are hypotheses (2), (3), and (4) of Theorem 17.4. Hypothesis (1) of Theorem 17.4, that Sz .˛ > a/ is nonempty, is satisfied by, for instance, the constant function y.t/ D ae 1=2ˇ . Therefore f has at least two fixed points. If f W Cz ! Cz , the hypotheses of Theorem 17.3 are satisfied with b D z D ae 1=ˇ because if kf .y/k > z, then conclusion (3) of Theorem 16.2 implies that ˛.f .y// > a which is the requirement of hypothesis (4) of Theorem 17.3. Thus there are at least three fixed points in this case. t u Most of the content of both this chapter and the previous one comes from the paper Leggett and Williams (1979) of Leggett and Williams. In Williams and Leggett (1982), the same authors made a detailed study of the tubular reactor model for the most common reaction rate function, which is called the Arrhenius reaction rate function: 

k r.x/ D p.q  x/ exp 1Cx

 ;

choosing q D 1:1 and k D 10 as what they called “physically interesting” values of those constants. In that paper, they identify a range of values of the constants ˇ and p for which the boundary value problem has at least three positive solutions and thus there are at least three steady states for the chemical process.

Chapter 18

Eigenvalues and Eigenvectors

This chapter describes a quite different sort of application of the fixed point index; it is an application to mathematics itself. For any function f W X ! X on a linear space, the definitions from linear algebra can be extended to call a real number  an eigenvalue of f if f .x/ D x for some x ¤ 0 in X , and then x is an eigenvector for the eigenvalue . We will use the fixed point index to establish the existence of eigenvectors in the setting of nonlinear functions on normed linear spaces. Then that nonlinear result will furnish us with a tool to prove that certain linear functions on such spaces have a specific, very important, eigenvalue. Thus our area of application will be classical “linear” functional analysis. Let C be a cone in a normed linear space X . Notice that if c1 ; c2 2 C , then so is c1 C c2 , since 2c1 and 2c2 are in C and c1 C c2 is the midpoint of the line segment connecting them in the convex set C . A cone induces a partial ordering on X by defining x 0  x if x 0  x 2 C . Call f W X ! X order-preserving if x 0  x implies f .x 0 /  f .x/. The map f is homogenous (of degree one) if f .tx/ D tf .x/ for all t 2 R. So if f is homogenous, then f .0/ D 0. To illustrate those definitions, let X be the plane and C the subset of points .x1 ; x2 / 2 X such that x1  0 and x2  0. An example of an order-preserving linear (and therefore homogenous) function f W X ! X is given by f .x/ D Ax where A D Œaij  is a non-negative two-by-two matrix, that is, one with all aij  0. For a nonlinear example of an order-preserving, homogenous function, let f .x/ D f .x1 ; x2 / D .x1 C x2 ; .x13 C x23 /1=3 /: A function f W X ! X can be composed with itself to define a sequence of functions ff m g by setting f 1 .x/ D f .x/ and then f m .x/ D f .f m1 .x//. The set ff m .x/ W m  1g is called the orbit of x by f or, more briefly, the f -orbit of x. Now that we have collected all our definitions, we can find out what the theorem says about eigenvectors of a nonlinear self-map of a cone.

© Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__18

125

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18 Eigenvalues and Eigenvectors

Theorem 18.1. Let C be a cone in a Banach space X and let f W C ! C be a completely continuous order-preserving homogenous function. Suppose there is u 2 C that has an unbounded f -orbit. Then there is x0 2 C with kx0 k D 1 and t0  1 such that f .x0 / D t0 x0 , that is, x0 is an eigenvector of f and the corresponding eigenvalue is t0 . Proof. Suppose f .x/ D x for some x ¤ 0 in C , then, since f is homogenous, f .sx/ D sx for all s  0. In particular, letting x0 D sx where s D 1=kxk, we see that f .x0 / D t0 x0 with t0 D 1. Thus we have something to prove only in the case that f .x/ ¤ x for all x 2 C n f0g. It may seem strange to then use fixed point theory if we already know that there is only that single fixed point x D 0, but that is just what we will do. We have the fixed point index ind.f; U / of Chap. 15 available for U a bounded subset of C because C is a closed and convex subset of the normed linear space X , the map f is completely continuous, and thus a compact map of U , and no relatively open subset of C will have any fixed points on its boundary. Letting U D fx 2 C W kxk < 1g, we will compute i nd.f; U /. Then, assuming that there is no eigenvector x0 such that kx0 k D 1 with eigenvalue t0  1, we will compute ind.f; U / once more. But this time we will obtain a different answer for ind.f; U /, which implies that such an eigenvector must exist after all. Since the restriction of f to U is a compact map, by Lemma 10.1 there exists r > 0 such that kx  f .x/k  r for all x 2 @U . By hypothesis we have u 2 C with an unbounded f -orbit. Let u0 D

r u 2kuk

so ku0 k D r=2 and, since f is homogenous, f m .u0 / D

r f m .u/ 2kuk

and therefore the f -orbit of u0 is also unbounded. Define g W C ! C by g.x/ D f .x/ C u0 and also define H W U  Œ0; 1 ! C by H.x; t/ D f .x/ C t u0 , a homotopy between f and g. Suppose x 2 @U , then kH.t; x/  xk D k.f .x/ C t u0 /  xk  kf .x/  xk  kt u0 k 

r 2

so H.x; t/ ¤ x for all x 2 @U and t 2 Œ0; 1, and therefore ind.f; U / D ind.g; U / by Theorem 15.7, the homotopy property of the fixed point index. We will show that g has no fixed points in U . Suppose x 2 U is a fixed point, that is, g.x / D f .x / C u0 D x . Since x 2 C and thus f .x / 2 C , we conclude that x  u0 . Moreover, since f is order-preserving, x D f .x /  f .u0 / and thus, by induction, x  f m .u0 / for all m  1. We define a sequence ff mj .u0 /g as follows. Let m1 D 1 and then let mj be the smallest integer such

18 Eigenvalues and Eigenvectors

127

that kf mj .u0 /k  kf mj 1 .u0 /k, which must exist because the sequence ff m .u0 /g is unbounded. We have constructed the sequence ff mj .u0 /g to have the property mj 0/ . The map f is kf mj .u0 /k  kf k .u0 /k for all k  mj . Let amj D kff mj .u .u0 /k homogenous, so we may write amj D

f mj .u0 / Df kf mj .u0 /k



 f mj 1 .u0 / : kf mj .u0 /k

Since kf mj 1 .u0 /k  kf mj .u0 /k and f is completely continuous, then the sequence famj g has a convergent subsequence which, to avoid complicating the notation any further, we’ll still call famj g. Let w be the limit of that sequence, then kwk D 1. We showed that x  f mj .u0 /, so x =kf mj .u0 /k  amj 2 C which is closed in X , so, if there is a limit to the sequence fx =kf mj .u0 /k  amj g, it must lie in C . But   x x  amj D lim  lim amj D 0  w D w … C: lim n!1 kf mj .u0 /k n!1 kf mj .u0 /k n!1 This contradiction establishes our claim that g has no fixed points in U and thus, by Theorem 15.3, the existence property of the index, ind.f; U / D ind.g; U / D 0. Now suppose there is no point x0 2 @U and t0  1 such that f .x0 / D t0 x0 . This time we define H W U  Œ0; 1 ! C by H.x; s/ D sf .x/. There are no x 2 @U such that H.x; s/ D sf .x/ D x since otherwise f .x/ D 1s x and 1s  1, so, for 0 the constant map to 0, Theorems 15.7 and 15.4, the homotopy and normalization properties, imply a different value for ind.f; U / because then ind.f; U / D ind.0; U / D 1: Thus f does have an eigenvector x0 such that kx0 k D 1 with the corresponding eigenvalue t0  1. t u In order to apply this theorem to functional analysis, we need some information from that branch of mathematics. The symbol X will still represent a normed linear space. It is traditional in functional analysis to call a linear function a linear operator, so we’ll use that language too. A linear operator T W X ! X is said to be bounded if, for any bounded subset B of X , the image T .B/ is also bounded. Denote by L.X / the set of all bounded linear operators from X to itself. By Theorem C.1, a linear operator is bounded if and only if it is continuous in the metric topology of X obtained from the norm. The norm of T 2 L.X / is denoted by kT k and defined by kT k D supfkT xk W x 2 X; kxk D 1g where sup means the supremum (least upper bound) of a set. The norm is well defined because the set fx 2 X W kxk D 1g is bounded. It’s not hard to show that,

128

18 Eigenvalues and Eigenvectors

with respect to this norm, the set of operators L.X / is itself a normed linear space. Since, for x 6D 0, the linearity of T 2 L.X / implies that kT xk 1 x Dk T xk D kT . /k kxk kxk kxk we have the following equivalent definition of the norm: kT k D supf

kT xk W x 6D 0g kxk

which gives us the useful inequality kT xk  kT kkxk that is true for all x 2 X . The set L.X / is obviously closed under composition; moreover, that useful inequality gives us another inequality. Lemma 18.2. If S; T 2 L.X /, then kS T k  kS k  kT k: Proof. All we have to do is apply the inequality twice: k.S T /xk D kS.T x/k  kS kkT xk  kS kkT kkxk because then, for all x ¤ 0 we get k.S T /xk  kSkkT k: kxk

t u

By an eigenvalue of T 2 L.X / we still mean a real number  such that T x D x for some x ¤ 0. The next result relates the eigenvalues of T to the norms of its iterates, that is, the linear operators T k 2 L.X / where T 1 x D T x and T k x D T .T k1 x/. Lemma 18.3. If  is an eigenvalue of T 2 L.X /, then jj  kT k k1=k for all k  1. Proof. Let x ¤ 0 be an eigenvector for  so T x D x and therefore, since T is linear and hence homogenous, T k x D k x for all k  1. Then jjk kxk D kk xk D kT k xk  kT k kkxk by the useful inequality. Dividing by kxk ¤ 0 completes the argument.

t u

18 Eigenvalues and Eigenvectors

129

Define r.T /, the spectral radius of T , by r.T / D inffkT k k1=k W k  1g which, the lemma tells us, is an upper bound for the absolute values of the eigenvalues of T . The following result gives us another description of the spectral radius. Theorem 18.4. For T 2 L.X /, the sequence fkT k k1=k g converges and its limit is r.T /. Proof. Since r.T / D inffkT k k1=k exists m  1 such that kT m k1=m kT m k  kT km < .r.T / C /m . For n 0  qn  m  1, and there exists pn Lemma 18.2, we find that

W k  1g then, given  > 0, there < r.T / C . By Lemma 18.2 we have  1, divide m into n with remainder qn , so  0 such that n D pn m C qn . Again using

kT n k D kT pn mCqn k  k.T m /pn kkT qn k  kT m kpn kT kqn < .r.T / C /mpn kT kqn : Thus, by Lemma 18.3,  r.T /  kT n k1=n < .r.T / C /mpn =n kT kqn =n D .r.T / C /

kT k r.T / C 

qn =n

because mpn =n D 1  qn =n. Since qn < m for all n  1, then limn!1 qn =n D 0 and therefore, given  > 0, there exists N such that n > N implies 

Choosing  D

 , r.T /C



kT k r.T / C 

qn =n < 1 C :

we see that if n > N then

kT k r.T / C 

qn =n 0, then there exist u; v 2 C such that x D u  v and kuk C kvk < 1 C . Now since jT xj D jT u  T vj  jT uj C jT vj D kT ukC C kT vkC  kT kC kuk C kT kC kvk D kT kC .kuk C kvk/ < kT kC C kT kC for all  > 0, it must be that jT xj  kT kC for all x 2 CQ such that jxj D 1 and therefore jT j  kT kC . t u

18 Eigenvalues and Eigenvectors

131

Although CQ is a linear space that is a subset of the normed linear space X , it is not a subspace as a normed linear space because its norm is not the norm of X . However, we will prove that if X is a Banach space, then CQ is also a Banach space, that is, it is complete with respect to its own norm. Our proof will make use of a characterization of completeness in terms of series of elements of CQ rather than sequences. Let fxn g be a sequence in a normed linear space X . The corresponding series P1 g and the nD1 xn is, as usual, the sequence of partial sums fsn D x1 C    C xnP 1 series converges if and only if the sequence fs g is convergent. The series n nD1 xn P1 is absolutely convergent if the series nD1 kxn k of real numbers is convergent. Lemma 18.6. A normed linear space X is a Banach space if and only if every absolutely convergent series in X is convergent. Proof. Suppose that the normed linear space X has the property that absolutely convergent series converge and let fxn g be a Cauchy sequence in X , so, given  > 0, there exists N such that m; n  N implies kxm  xn k < . In particular, there exists N1 such that m; n  N1 implies kxm  xn k < 21 . Suppose we have chosen Nj such that m; n  Nj implies kxm  xn k < 2j , then choose Nj C1 > Nj such that m; n  Nj C1 implies kxm  xn k < 2.j C1/ . Let y1 D xN1 and, for j > 1, let yj D xNj  xNj 1 , then k X

kyj k D kxN1 k C kxN2  xN1 k C    kxNk  xNk1 k < kxN1 k C

j D1

k X

2.j 1/

j D2

and therefore 1 X j D1

kyj k < kxN1 k C

1 X

2.j 1/ D kxN1 k C 1;

j D2

P is absolutely convergent and hence convergent. Since the partial that is, 1 nD1 yj P Pk sum of the series 1 nD1 yj is j D1 yj D xNk we have proved that the subsequence fxNj g of fxn g converges and we write limj !1 xNj D x. Now, given  > 0, choose k so that 2k < . There exists M such that Nj > M implies kxNj  xk < 2.kC1/ . Choose j  k C 1 such that if Nj > M , then kxn  xk  kxn  xNj k C kxNj  xk < 2j C 2.kC1/  2k <  and therefore the Cauchy sequence fxn g converges to x. We have proved that X is a Banach space. P To prove the converse, suppose X is aPBanach space and that 1 nD1 xn is an absolutely convergent series, so the series 1 kx k converges. Therefore, by the n nD1 Cauchy criterion, given  > 0, there exists N such that n > m  N implies

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18 Eigenvalues and Eigenvectors

kxm k C    C kxn k < . The sequence of partial sums fsn D x1 C    C xn g is Cauchy because if n > m  N then ksn  sm k D kxm C    C xn k  kxm k C    C kxn k <  converges because X is complete, and that means the absolutely and therefore fsn gP convergent series 1 t u nD1 xn also converges. Lemma 18.7. If C is a cone in a Banach space X , then the space .CQ ; j  j/ is a Banach space. P1 P1 Q Proof. We will P show that if nD1 xn is a series in C such that nD1 jxn j Q is complete by Lemma 18.6. converges, then 1 x converges and therefore C n nD1 By the definition of the norm j  j, for each n there exist un ; vn 2 C such that xn D un  vn and kun k C kvn k < 2jxn j. Therefore 1 X

kun k C

nD1

1 X

kvn k  2

nD1

1 X

jxn j < 1

nD1

P1 P kvn k converge.P Since .X; k  k/ is a Banach so both series 1 nD1 kun k and nD1P 1 1 space, Lemma 18.6 tells us that u and nD1 n nD1 vn converge. Writing P1 u D u then, since the sequence of partial sums lies in the closed subset C nD1 n P of X , we conclude that u 2 C and we write 1 v D v 2 C as well. Now, given nD1 n PN P  > 0, there exists N such that ku  nD1 un k < =2 and kv  N nD1 vn k < =2. Letting x D u  v completes the proof because jx 

N X

xn j D j.u  v/ 

nD1

N X

.un  vn /j

nD1

D j.u 

N X

un /  .v 

nD1

 ju 

N X nD1

D ku 

N X nD1

and therefore

P1 nD1

xn D x.

N X

vn //j

nD1

un j C jv 

N X

vn j

nD1

un k C kv 

N X

vn k < 

nD1

t u

We need one more preliminary result in order to prove our main theorem. It is a special case of the “uniform boundedness principle” of functional analysis.

18 Eigenvalues and Eigenvectors

133

Lemma 18.8. Let X be a Banach space and let f W X ! X be a continuous linear operator. If the sequence fkf m k W m  1g is unbounded, then there exists x0 2 X which has an unbounded f  orbit. Proof. We will prove the contrapositive: if fkf m .x/k W m  1g is bounded for each x 2 X , then the sequence fkf m k W m  1g is bounded. For each n  1, let Xn D fx 2 X W kf m .x/k  n for all mg: Since f is continuous, then the Xn are closed subsets of X . We are assuming that S n1 Xn D X . The space X is a complete metric space so the Baire category theorem (Theorem C.5) implies that some X` has a nonempty interior. Therefore there exists x0 2 X` and  > 0 such that if kx0  xk   then kf m .x/k  ` for all m  1. Let u 2 X such that kuk D 1. Using the linearity of f , we find that 1 1 kf m .u/k D kf m . x0 C u  x0 /k   1 1  kf m . x0 C u/k C kf m . x0 /k   2` 1 D .kf m .x0 C u/k C kf m .x0 /k/    because kx0  .x0 C u/k D . Therefore, for all m  1 we have the bound 2` t u :  Back in Chap. 2, I mentioned that the meaning of “compact” for linear functions was quite different from the way the same word was used in the nonlinear setting. Specifically, a linear operator T W X ! X is said to be compact if for every bounded subset B of X , the image T .B/ is relatively compact, that is, the closure of T .B/ in X is compact. By Theorem 2.4, an equivalent definition is that T is a compact linear operator if, for every bounded sequence fxn g in X , the sequence fT xn g contains a convergent subsequence. Thus “compact” for linear operators means the same as “completely continuous” for functions in general. In particular, since compact linear operators are bounded operators, then they are continuous by Theorem C.1. Here is the theorem about compact linear operators in which we apply the fixed point index to functional analysis by means of Theorem 18.1. kf m k D supfkf m .u/k W kuk D 1g 

Theorem 18.9. Let C be a cone in a Banach space X and let T W X ! X be a compact linear operator such that T .C /  C . If rC .T / > 0, then rC .T / is an eigenvalue of the restriction of T to C , that is, there exists x  2 C n f0g such that T x  D rC .T /x  .

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18 Eigenvalues and Eigenvectors

Proof. We will use Theorem 18.1 to construct a sequence of positive real numbers fn g, converging to rC .T /, that are eigenvalues of T with eigenvectors xn 2 C such that kxn k D 1. Assuming that we have that sequence, then T xn D n xn implies xn D T . 1n xn /. The sequence f 1n xn g is bounded because limn!1 n D rC .T / > 0 so there is some N such that n > N implies n > rC .T /=2 and therefore k

1 1 2 1 xn k D kxn k D  : n n n rC .T /

The linear operator T is compact so the sequence T . 1n xn / D fxn g contains a convergent subsequence and, simplifying the notation in the usual way, we write limn!1 xn D x  . Now, given  > 0 there exists N such that n > N implies jn  rC .T /j < =2, so n < rC .T / C =2, and also that kxn  x  k
0, we can choose n0 such that n10 < rC .T /. Let sn D rC .T / 

1 >0 n C n0

and define Sn D s1n T , which is still a compact linear operator that takes C to itself and so, by linearity, it takes CQ to itself as well. Since kSnk k1=k D s1n kT k k1=k then rC .Sn / D s1n rC .T /. We will apply Theorem 18.1 to the linear operator Sn to find

18 Eigenvalues and Eigenvectors

135

xn 2 C such that kxn k D 1 and Sn xn D tn xn for some tn  1. Then we let n D sn tn so T xn D n xn . Therefore n is an eigenvalue of T , which by Lemma 18.2 implies that rC .T / 

1 < sn  sn tn D n  rC .T / n

and thus limn!1 n D rC .T /. Our task then is to verify that the hypotheses of Theorem 18.1 are satisfied by the function f D Sn on the Banach space CQ . Since f is a compact linear operator, it is a completely continuous function. The linearity of f implies that it is homogenous. If x  x 0 in the order induced by the cone, that is x  x 0 2 C , then since f maps C to itself, we have f .x  x 0 / D f .x/  f .x 0 / 2 C and thus Sn is orderpreserving. There is just one more hypothesis to verify: the existence of u 2 C with an unbounded f -orbit. Since rC .T / > 0, we use Lemma 18.5 to conclude that rCQ .f / D rC .f / D

1 1 rC .T / > rC .T / > 1: sn rC .T /  n1

If the sequence fjf n jg were bounded with jf n j  ˇ for all n, then rCQ .f / D inffjf n j1=n W n  1g  inffˇ 1=n W n  1g  1: Thus fjf n jg is unbounded and, by Lemma 18.8, there exists y 2 CQ such that fjf n .y/jg is unbounded. Write y D u  v for some u; v 2 C , then jf n .y/j  kf n .u/k C kf n .v/k, so at least one of the sequences fkf n .u/kg or fkf n .v/kg is unbounded. Thus the hypotheses of Theorem 18.1 are satisfied and the proof is complete. t u Theorem 18.9 is a variant of an important theorem of functional analysis called the Krein-Rutman theorem. That theorem requires that the cone C be total, which means that the Banach space CQ is dense in X . Then, for a compact linear operator T W X ! X on the Banach space X that has the properties T .C /  C and rC .T / > 0, the theorem states that rC .T / D r.T / and, as in Theorem 18.9, that the spectral radius r.T / is an eigenvalue of T . Moreover, r.T / is also an eigenvalue of the dual linear operator T  on the dual space X  of linear real-valued functions on X . I didn’t go into all that because it would have taken us too far into the world of functional analysis for an introductory book about nonlinear analysis. However, it’s important to point out that not only can the Krein–Rutman theorem itself be proved by applying the fixed point index via Theorem 18.1, but that a significantly more general result can be obtained by using the fixed point index. In that version, the linear operator T may not be compact, but instead it need only satisfy a substantially weaker hypothesis. This generalization is due to Nussbaum who gives a careful exposition of it in Nussbaum (1985), which was my source for the less ambitious result presented in this chapter.

Part IV

Bifurcation Theory

Chapter 19

A Separation Theorem

Our mathematical travels take us next to a quite different part of topology. It is called general topology or point-set topology and it seeks to discover properties of topological spaces that hold for very broad classes of spaces. A substantial part of the subject concerns spaces that are not necessarily metric and this will be the case of the theorem we will discuss here; it holds for any compact topological space. When we come to apply the theorem, it will be in the setting of metric spaces, because all the spaces that arise in the analytic context are metric. However, a proof that made use of a metric would be unnecessarily complicated; since the property has nothing to do with distance, there is no point in bringing it up. The first step is a lemma that does not even require topology because it is a property of families of abstract sets. Lemma 19.1. T Let S be a set and let fF˛TW ˛ 2 Ag be a family of sets indexed by a set A. Then ˛2A F˛  S if and only if ˛2A .F˛  S / D ;. Proof. We Tprove both parts of the statement T by establishing their contrapositives. Suppose ˛2A .F˛  S / 6D ; and let xT 2 ˛2A .F˛  S /. Therefore x 2 F˛ for all ˛ 2 A, but x 62 S which tells us that ˛2A F˛ is not T entirely contained in S . For the other contrapositive, there has to be a pointTx 2 ˛2A F˛ that is not in S . That means x 2 F˛  S for all ˛ 2 A, so the set ˛2A .F˛  S / is not empty since it contains x. t u Our first fact about compact spaces is immediate from the definition. Lemma 19.2. Let X be a compact space and let F T D fF˛ W ˛ 2 Ag be a family of closed subsets of X indexed by a set A such that T˛2A F˛ D ;. There exists a finite subset Fn D fF˛1 ; F˛2 ; : : : ; F˛n g of F such that F˛j , the intersection of all the sets in Fn , is empty. The next lemma, on the other hand, is the main step in proving the result we require. It concerns subsets of a space X that are both open and closed in X . Such a subset is said to be clopen in X . © Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__19

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Lemma 19.3. Let X be a compact set and let x0 be a point of X . Let F D fF˛ W ˛ 2 T Ag be the family of all clopen subsets of X that contain x0 . Then the set Y D ˛2A F˛ is a connected subset of X . Proof. Suppose Y D C [ D where C and D are disjoint open (and therefore clopen) subsets of Y . Since x0 is in Y it’s in one of the subsets, let’s take it to be in C . We’ll then prove that D must be empty. That will prove the connectedness of Y because if it were disconnected, by definition, there would be a way to express Y as C [ D where C and D were disjoint, open, nonempty subsets of Y . The set Y is an intersection of closed subsets of X , so it is closed also, and that makes its closed subsets C and D closed in X as well. A compact space is normal, so we can find disjoint open subsets U and V of X containing C and D, respectively. Define F˛ D F˛  .U [ V /. By Lemma 19.1, Y D

\

F˛ D C [ D  U [ V

˛2A

T implies ˛2A F˛ D ;. Since F˛ is closed in X , we can apply Lemma 19.2 and conclude that there exists a finite set fF˛1 ; F˛2 ; : : : ; F˛n g such that their intersection T F˛j is empty. Using 19.1 once again, and recalling how F˛ was defined, we see T T that F˛j  U [ V . Now F˛j is a finite intersection T of clopen sets, so it is itself clopen, and it certainly contains x0 , so it must be that F˛j D Fˇ , for some Fˇ 2 F, the family of all such sets. The set Fˇ \ U is open and it contains x0 . We claim that Fˇ \ U is also a closed subset of X . Once the claim is established, the proof is over because then Fˇ \ U D F 2 F and yet F \ V D ; so Y \ V D ; by the definition of Y . But that implies Y \ D D ; and since Y D C [ D, it can only be that D D ;. To prove that Fˇ \ U is closed, take a sequence fxi g in Fˇ \ U which converges to a point x 2 X . Now x 2 Fˇ because Fˇ is closed. But, furthermore, recall that Fˇ  U [ V , which puts x either in U or in V , since these sets are disjoint. If x were in the open set V it could hardly be the limit of a sequence in Fˇ \ U , so x 2 U . t u We are now ready to prove the result from point-set topology that we will need later. It concerns a compact space X that contains disjoint closed sets P and Q such that no connected subset of X intersects both of them. The hypothesis implies in particular that X must be disconnected since otherwise X itself would be a connected set intersecting both P and Q. The theorem says that we can conclude more: while there may be many ways to write the space X as a union of disjoint clopen nonempty sets, there is at least one such union that “separates” P from Q, in the sense that each is contained in a different clopen set. Here is the precise statement. Theorem 19.4 (Separation Theorem). Let P and Q be disjoint closed subsets of a compact space X . If there is no closed connected subset of X that intersects both P and Q, then there exist disjoint closed subsets KP and KQ of X such that P  KP , Q  KQ , and X D KP [ KQ .

19 A Separation Theorem

141

T Proof. Choose x0 2 P and, as in Lemma 19.3, define Y D ˛2A F˛ where F D fF˛ W ˛ 2 Ag is the family of all clopen subsets of X that contain x0 . Since Y \P 6D ;, because it contains x0 , and Y is connected by the lemma, the hypothesis of the T theorem implies that Y \ Q D ;. We may write this fact in the form ˛2A .F˛ \ Q/ D ;. Both F˛ and Q are closed in X, so Lemma 19.2 tells us that there exists a T finite subset fF ; F ; : : : ; F g of F such that .F \Q/ D ;. Or, to put it another ˛ ˛ ˛ ˛ 1 2 n j T T way, . F˛j / \ Q D ;. The set F is clopen and contains x0 so, just as in the ˛ Tj proof of Lemma 19.3, we can write F˛j D Fˇ for some Fˇ 2 F. To summarize, starting with an arbitrary point x0 2 P , we found Fˇ 2 F containing x0 and disjoint from Q. Thus we can find a cover fFˇ .x/g of P consisting of clopen S subsets of X disjoint from Q. Let fFˇ .xj /g be a finite subcover and let KP D Fˇ .xj / which is a clopen subset of X containing P and disjoint from Q. Choosing KQ D X KP completes the proof. t u The compactness of X played a crucial role in the proof of this separation theorem; a noncompact space need not have such a separation property. For example, let Jn be the line segment in the plane connecting . n1 ; 0/ and . n1 ; 1/, let P D .0; 0/ and Q D .0; 1/, and define X to be the subset of the plane that is the union of all the Jn for n  1 together with P and Q. See Fig. 19.1. Although P and Q are closed subsets of X such that no connected subset of X contains both, for any decomposition X D C [ D of X into disjoint clopen subsets, one of C or D must contain both P and Q. To see why, suppose C contains P ; then, since C is open in the plane, there is an open disc in C containing P and that disc will intersect the Jn for all n greater than some N . The Jn are connected, so if some Jn intersects C , it must be contained in C . In particular, all the points . n1 ; 1/ for n > N , and consequently their limit point Q, are in the closed set C .

1

1

Q

• • •

• • •

Jn

P 0

1/n

J3

1/3

J2

1/2

Fig. 19.1 A noncompact counterexample

J1

1

Chapter 20

Compact Linear Operators

Recall that L.X / is the normed linear space of bounded linear operators T W X ! X . An operator T 2 L.X / is called regular if it is invertible, that is, if there exists S 2 L.X / such that S T D T S D I (the identity operator). As you would expect, we’ll call S the inverse of T and write it as T 1 . By Lemma 18.2 we have 1 D kI k D kT 1 T k  kT 1 kkT k and therefore kT 1 k  kT k1 . Results from Appendix C tell us that, when X is a Banach space, the regular operators in L.X / are the linear homeomorphisms from X to itself because bounded linear operators are continuous by Theorem C.1 and, if X is Banach, they are open by the open mapping theorem (C.6). Thus, if T 2 L.X / has an inverse, then that inverse is continuous. In contrast to what we know about a linear transformation of a finite-dimensional vector space to itself, for a linear operator on an infinite-dimensional linear space X , neither of the hypotheses one-to-one or onto implies the other. For some examples, we’ll again use the Hilbert space l2 of sequences .x1 ; x2 ; x3 ; : : : / such P 2 that 1 j D1 xj < 1: Define S; T 2 L.l2 / by S.x1 ; x2 ; x3 ; : : : / D .0; x1 ; x2 ; x3 ; : : : / and T .x1 ; x2 ; x3 ; : : : / D .x2 ; x3 ; : : : /: It’s clear that S is one-to-one but not onto whereas T is onto, but it is not one-to-one. Throughout the rest of this chapter, the space X is assumed to be Banach. Then, by Theorem C.4, the normed linear space L.X / is also a Banach space. Lemma 20.1. If T 2 L.X / such that kT k < 1, then I  T is regular.

© Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__20

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20 Compact Linear Operators

Proof. P Let T j denote the composition of the operator T with itself j times. By the j series 1 j D1 T we mean the sequence fT; T C T 2 ; T C T 2 C T 3 ; : : : g of partial sums, which are operators in L.X /. Lemma 18.2 implies that 1 X

kT j k 

j D1

1 X

kT kj

j D1

and the series converge because we assumed kT kP < 1. The convergence of the series j of norms of operators means that the sequence 1 j D1 T is absolutely convergent and therefore it converges in the Banach space L.X / by Lemma 18.6. It follows that S DIC

1 X

Tj

j D1

is a well-defined operator in L.X /. To see that S D .I  T /1 , notice that S is the limit of the sequence of operators Sn D I C T C T 2 C    C T n and that .I  T /Sn D I  T nC1 : Think of I  T as an operator on L.X /, defined by composing. Then I  T W L.X / ! L.X / is a continuous function, and therefore .I  T /S D .I  T /. lim Sn / D lim .I  T /Sn D lim I  T nC1 D I: t u n!1 n!1 n!1 Theorem 20.2. Let G denote the set of regular elements of L.X /; then G is an open subset of the Banach space L.X /. Proof. Given an operator S 2 G, we’ll see that the neighborhood U D fT 2 L.X / W kS  T k
kT k then  is not in  .T /. The reason is that jj > kT k implies k 1 T k < 1 so I  1 T is regular by Lemma 20.1, and therefore a linear homeomorphism. But then .I 

1 T / D I  T 

is also a linear homeomorphism. Thus  .T / is bounded in R. Now define a function F W R ! L.X / by F ./ D I  T and notice that F is continuous. Theorem 20.2 states that the subset G of regular operators is open in L.X /, but by definition  .T / D R  F 1 .G/, so it is closed in R. t u Denote by K.X / the set of compact linear operators T W X ! X , that is, the ones with the property that if B  X is bounded then T .B/ is relatively compact. Certainly a compact linear operator is bounded, and in fact K.X / is actually a linear subspace of L.X /. To see why, take S; T 2 K.X / and real numbers a; b and let fxn g be any bounded sequence in X . We can simplify the notation by writing again as fxn g the subsequence such that fS xn g is convergent. Then taking fxnk g to be a subsequence of that fxn g for which fT xnk g converges, we

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still have fS xnk g convergent since it’s a subsequence of a convergent sequence. Therefore f.aS C bT /xnk g is a convergent sequence, and we have shown that aS C bT 2 K.X /. Theorem 20.4. If X is a Banach space, then K.X / is a closed subspace of L.X / and therefore also a Banach space. Proof. Let fTn g be a sequence in K.X / converging to a bounded linear operator T ; we must prove that T is a compact linear operator. Given  > 0, there exists N such that n  N implies kT  Tn k < 3 . Now let D be the unit ball in X , that is, x 2 X is in D if and only if kxk  1. Then, in particular, we have that x 2 D and n  N implies kT x  Tn xk < 3 . Since TN 2 K.X /, then TN .D/ is relatively compact and therefore, as we recall from Chap. 2, that makes TN .D/ totally bounded. So we can find fx1 ; x2 ; : : : ; xm g in D such that fTN x1 ; TN x2 ; : : : ; TN xm g is an 3 -net for TN .D/. That same finite set fx1 ; x2 ; : : : ; xm g also gives us an -net for T .D/. To see why, take any x 2 D, then there exists k so that kTN x  TN xk k < 3 and for that xk 2 D we have kT x  T xk k  kT x  TN xk C kTN x  TN xk k C kTN xk  T xk k < : We’ve demonstrated that T .D/ is totally bounded and therefore, since we assume X is complete, T .D/  C for some compact set C  X by Corollary 2.2. But then we can use the linearity of T to show that T .B/ is relatively compact for any bounded subset B of X , and therefore T 2 K.X /, as follows. For a > 0, define a W X ! X to be scalar multiplication, that is, a.x/ D ax, then T .aD/ D aT .D/  a.C /, a compact set. Since B bounded means B  aD for some a > 0, then T .B/  a.C /. t u All we were able to say about the spectrum of an operator in L.X / in general was that it was a compact subset of R. But if the operator is in K.X /, its spectrum is a very special kind of compact subset of R. I will devote the rest of this chapter to showing you what kind of a subset it is. We’ll begin with another of those wonderful results, like the Schauder projection lemma, that are elementary and easy to prove, yet very, very useful. It is an infinitedimensional analogue of the following observation about euclidean space. Let V be a vector subspace of Rn with V 6D Rn . Suppose x 2 Rn is a unit vector orthogonal to V , that is, kxk D 1 and x  v D 0 for all v 2 V ; then inffkx  vk W v 2 V g D 1: We can interpret the conclusion as stating that the distance from x to the vector space V is 1. To prove it, just apply the Pythagorean theorem to the plane determined by x and v 2 V , when v 6D 0 of course. See Fig. 20.1. A vector subspace V of Rn is closed in Rn . By Corollary C.3, if V is a finitedimensional linear subspace of a normed linear space X , then V is closed in X , even if X itself is not finite-dimensional. However, that property cannot be assumed

20 Compact Linear Operators

147

Fig. 20.1 Reisz’s lemma

for linear subspaces of a normed linear space in general, so we will add it to our hypotheses. But, in this infinite-dimensional setting, the conclusion of the theorem is still weaker than in the euclidean result. Here it is. Theorem 20.5 (Riesz’s Lemma). Let V be a closed linear subspace of a Banach space X , with V 6D X . Given  > 0, there exists x  2 X such that kx  k D 1 and inffkx   vk W v 2 V g  1  : Proof. Choose any y 2 X  V ; then since V is closed in X , we can find an open ball about y disjoint from V . It follows that if we set ˛ D inffky  vk W v 2 V g then ˛ > 0. On the other hand, since ˛ is an infimum, there exists v 2 V such that ky  v k 

1 ˛ 1

The proof now just consists of defining x D

y  v ky  v k

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20 Compact Linear Operators

and checking that it works. For any v 2 V we have     y  v    v kx   vk D .ky  v k/1 ky  v k    ky  v k    D .ky  v k/1 y  v  .ky  v k/v D .ky  v k/1 ky  v0 k where v0 D v C .ky  v k/v: Since V is a vector space, we see that v0 2 V and therefore ky  v0 k  ˛. Consequently, kx   vk D .ky  v k/1 ky  v0 k 1  ˛ D 1  : t u ˛ The only value of  that we will use when we apply Riesz’s lemma will be  D 12 . That is, for V a proper, closed subspace of a Banach space X , there exists x  2 X with kx  k D 1 such that  .ky  v k/1 ˛ 

inffkx   vk W v 2 V g 

1 : 2

Lemma 20.6. If T 2 K.X /, then the null space N.I  T / of the linear operator I  T is finite-dimensional. Proof. This time let D be the unit ball in N.I  T /. Note that x 2 N.I  T / just means T x D x, so D D T .D/ is relatively compact and therefore totally bounded. For small  > 0, take a finite -net for D and let V be the span of the net, which is closed in N.I  T / because it is finite-dimensional (Corollary C.3). Suppose N.I  T / were not finite-dimensional; then certainly V 6D N.I  T /. Therefore, by Riesz’s lemma there exists x  2 N.I  T / with kx  k D 1, so x  2 D, such that kx  vk  12 for all v 2 V . But, for  < 12 , this statement contradicts the assumption that V is the span of an -net for D, so N.I  T / must be finite-dimensional. t u We can also say something significant about the range R.I  T / of I  T , but first we must prove a technical lemma. Lemma 20.7. Suppose T 2 K.X /. There exists M > 0 with the following property. Given y 2 R.I  T /, there is a point x 2 X such that .I  T /x D y and kxk  M kyk. Proof. There is nothing to prove if y is 0, so we’ll assume it’s not. Let’s suppose that no such positive number M exists and see how that leads us to a contradiction.

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149

If there isn’t any such M , then we can find a sequence fyn g of nonzero elements of X with the property that .I  T /x D yn implies kxk > nkyn k. For each n, choose an xn such that .I  T /xn D yn . Since N.I  T / is a finite-dimensional subspace by the previous result, and therefore closed in X , then setting dn D inffkxn  vk W v 2 N.I  T /g we know that dn > 0. Choose vn 2 N.I  T / so that kxn  vn k  2dn and let zn D

xn  vn : kxn  vn k

Now kzn k D 1 and therefore by the compactness of T we may assume that the sequence fT .zn /g (really a subsequence) converges to some z 2 X . After that blizzard of definitions, let’s see what we can find out about z. We claim that z is also the limit of the corresponding sequence fzn g itself. That will imply that z 2 N.I T / since .I  T /z is the limit of fzn  T .zn /g. To prove the claim, notice that we defined zn so that .I  T /.kxn  vn kzn / D yn and therefore the defining property of yn tells us that kxn  vn k  kzn k > nkyn k: But then k.I  T /zn k D

1 1 1 kyn k < kzn k D kxn  vn k n n

so f.I  T /.zn /g converges to 0 and thus fzn g D f.I  T /.zn / C T .zn /g converges to 0 C z D z, as we claimed. We will arrive at a contradiction by showing that in fact fzn g does not converge to z. Unfortunately, this sends us off on another computation, this one very similar to the one we used to prove Riesz’s lemma. Here goes:     xn  vn   z kxn  vn k  kzn  zk D kxn  vn k    kx  v k n n   D xn  vn  .kxn  vn k/z D kxn  z0 k

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where z0 D vn C .kxn  vn k/z which is an element of N.I  T / because we proved that z was. Consequently, we know that kxn  vn k  kzn  zk  dn : Remember that we chose the vn so that kxn  vn k  2dn and thus dn 1  kxn  vn k 2 which implies kzn  zk 

1 2

for all n, so fzn g does not converge to z.

t u

That was the hard part. Now let’s get the payoff. Theorem 20.8. If T 2 K.X /, then R.I  T / is a closed subspace of X . Proof. Take a sequence fyn g in R.I  T / converging to some y 2 X ; we’ll see that y 2 R.I  T /. For each yn in the given sequence, we’ll use the previous result to choose xn such that .I  T /xn D yn and kxn k  M kyn k for the constant M > 0 from that lemma. Since the sequence fyn g is convergent it is bounded, and then the inequality makes the sequence fxn g bounded also. The operator T is compact, so we can assume the sequence fT .xn /g is convergent. Thus fxn g D fyn C T .xn /g is a sum of convergent sequences and therefore it converges to an element of X that we’ll call x. The function I  T is continuous, so .I  T /x is the limit of the sequence f.I  T /.xn /g D fyn g and therefore .I  T /x D y. t u Now we begin to collect some facts about the spectrum of a compact linear operator. Theorem 20.9. If T 2 K.X / and X is infinite-dimensional, then 0 2  .T /. Proof. Since 0I  T D T , the theorem just states that a compact linear operator cannot have an inverse. Suppose T did have an inverse T 1 in L.X /. Then T 1 is a continuous function by Theorem C.1 and therefore it maps convergent sequences to convergent sequences. Writing I D T 1 T , since T 2 K.X /, we would conclude that I 2 K.X / also. But then Lemma 20.6 would imply that X D N.I  I / was finite dimensional, contrary to the hypothesis, so T 1 cannot exist after all. t u

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151

Although 0 is in the spectrum of T , it might not be an eigenvalue of T . That’s what would happen if T is one-to-one but not onto. We have already seen a simple example of an operator, we called it S , on l2 that is one-to-one and not onto, but it won’t serve to illustrate Theorem 20.9 because it’s not compact. However, there are also compact operators on l2 that are one-to-one but not onto. One such operator is defined by setting 1 1 1 T .x1 ; x2 ; x3 ; : : : ; xn ; : : : / D .x1 ; x2 ; x3 ; : : : ; xn ; : : : /: 2 3 n Notice, for instance, that .1; 12 ; 13 ; : : : / 2 l2 is not in the image. The reason we know that this operator is compact is that we can write it as the limit of a sequence of compact operators fTn g in L.l2 /, and that makes T compact since K.X / is closed in L.X / by Theorem 20.4. The operators Tn are defined by 1 1 1 Tn .x1 ; x2 ; x3 ; : : : ; xn ; : : : / D .x1 ; x2 ; x3 ; : : : ; xn ; 0; : : : ; 0 : : : / 2 3 n and these are compact because their ranges are finite-dimensional. The spectrum of a compact operator T may contain a point, namely 0, that is not an eigenvalue of T , but the next result will show that all the other points in the spectrum of T 2 K.X / are eigenvalues of T . In other words,  .T / consists of 0 and the set of eigenvalues of T (of which 0 might be one). Thus, for an operator T 2 K.X / and  6D 0, we will see that there are two possibilities:  is an eigenvalue of T , so that .I  T /x D 0 has a nonzero solution x or I  T is regular, which implies that, given y 2 X , there is a unique solution to .I T /x D y. The equation .I  T /x D 0 is called the “homogeneous problem” (for the operator T ) and .I  T /x D y the “inhomogeneous problem.” The statement of the next theorem expresses an alternative regarding compact linear operators: for  6D 0, either the homogeneous problem has a nonzero solution or the corresponding inhomogeneous problem has a unique solution. This is the alternative referred to in the name of the theorem. Theorem 20.10 (Fredholm Alternative). If T 2 K.X /, and  2  .T / is nonzero, then  is an eigenvalue of T . Proof. Suppose that  6D 0 and  is not an eigenvalue of T ; we must prove that  is not in  .T /. That is, we must show that I  T is regular. Since  is not an eigenvalue, we know that the operator I  T is one-to-one. Then, by the open mapping theorem (C.6), all we have to demonstrate is that I T is onto. The range of I  T is closed because T 2 K.X / implies that 1 T is also compact, which tells us that I  1 T has a closed range by 20.8, and thus I  T D .I 

1 T/ 

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has the same property. The idea of the proof is to assume that R.I  T / 6D X and use Riesz’ lemma (Theorem 20.5) to produce a contradiction. Define a sequence of subspaces fXn g of X by setting X0 D X X1 D R.I  T / D .I  T /.X / and, in general, Xn D .I  T /.Xn1 /; which is a subspace of Xn1 , as you can show by an easy induction argument, and 20.8 implies that it is a closed subspace. We will also find that, for all n, the spaces Xn are distinct, but let’s put off that argument for a moment in order to see how we will use the Xn . By Riesz’ lemma, for each n there exists yn 2 Xn with kyn k D 1 and kyn  yk > 12 for all y 2 Xn . We will prove that the sequence fyn g cannot really exist, and that will establish what we need: R.I  T / D X . However, since certainly yn must not be in Xn , we had first better find out how we know the Xn are distinct. Our assumption that R.I T / 6D X means that X1 6D X0 . If the Xn are not all distinct, then there exists n such that Xn D XnC1 but Xj 6D Xj C1 for all j < n. Given w 2 Xn1 , set y D .I  T /.w/. Thus y 2 Xn D XnC1 , so there exists x 2 Xn with .I  T /.x/ D y. Furthermore, x is unique since I  T is one-toone, so x D w and we have shown that w 2 Xn which implies Xn1 D Xn , contrary to the definition of n. Now that we have justified the construction of the sequence fyn g, we notice that it is certainly bounded so the compactness of T implies that the sequence fT yn g converges. But I claim fT yn g cannot converge, and the claim will thus give us the contradiction we need. To establish the claim, we take positive integers m; n with n > m and write 1 1 .T ym  T yn / D ym  fyn C Œ.I  T /ym  .I  T /yn  g:   Look at the individual terms within the brackets. We note that yn 2 Xn1 and therefore .I T /yn 2 Xn , where both Xn1 and Xn are contained in Xm . Similarly, .I  T /ym 2 Xm , so since Xm is a linear space, we may write 1 .T ym  T yn / D ym  y  with y 2 Xm . By the definition of ym , we know that    1  .T ym  T yn / D kym  yk > 1 :   2

20 Compact Linear Operators

153

That implies kT ym  T yn k 

jj >0 2

for all m; n with n > m, which could not be true if the sequence fT yn g converged. t u The following well-known fact from linear algebra is not difficult to prove for X any linear space, so we’ll just state it. Lemma 20.11. If 1 ; 2 ; : : : n are distinct eigenvalues of T 2 L.X / then corresponding eigenvectors, that is, nonzero solutions to T xi D i xi , form a linearly independent subset of X . We can now describe the spectrum of a compact operator quite precisely. Theorem 20.12. If T 2 K.X /, then either  .T / is finite or it is a bounded sequence of real numbers converging to 0. Proof. We suppose  .T / contains a sequence fj g which converges to some nonzero number, and we set out to obtain a contradiction. Since  .T / is a compact subset of R by 20.3, that will prove the theorem. We may assume that the j are nonzero and therefore eigenvalues of T according to the Fredholm alternative; let fxj g be a sequence of eigenvectors corresponding to the j . Define Xn to be the span of fx1 ; x2 ; : : : ; xn g, which is a closed subspace of X because it is finitedimensional. Lemma 20.11 tells us that Xn1 , the span of fx1 ; x2 ; : : : ; xn1 g, is a proper subspace of Xn so, by Riesz’ lemma, there exists yn 2 Xn such that kyn k D 1 and kyn  vk  12 for all v 2 Xn1 . As in the proof of the Fredholm alternative (20.10), we can show that if n > m, then 1 1 T yn  T ym D yn  y n m where y 2 Xn1 . Therefore   1   T yn  1 T ym   1 :   2 m n On the other hand, for any m; n we have   ˇ ˇ 1  ˇ ˇ  T yn  1 T ym   ˇ 1  1 ˇ  kT yn k C 1 kT yn  T ym k:   ˇ ˇ   m n m n m Since fj g converges to a nonzero number, the sequence f 1j g is bounded and, choosing m; n large enough, we can make j 1n 

1 j m

as small as we wish. We may

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assume that the sequence fT yj g is convergent, since kyj k D 1 and T is compact, and hence it is bounded and we can make kT ym  T yn k small by choosing m and n sufficiently large. A previous inequality therefore implies that    1  T yn  1 T ym  < 1  2  m n for sufficiently large m and n, and that produces the required contradiction.

t u

Chapter 21

The Degree Calculation

In this chapter, we will use the spectrum of a compact linear operator to establish an important fact about the Leray–Schauder degree. As in the previous chapter, X is an infinite-dimensional Banach space. The binomial theorem writes an expression of the form .1 C x/n as a polynomial in x by using the operations of elementary algebra. These operations apply to T 2 L.X / as well, so the proof of the binomial theorem produces the same formula: .I  T /n D

n X

.1/r Crn T r D I 

rD0

n X .1/rC1 Crn T r rD1

where Crn D

nŠ : rŠ.n  r/Š

Define a linear operator Sn by setting Sn D

n X .1/nC1 Crn T r rD1

so that .I  T /n D I  Sn . Since K.X / is a linear subspace of L.X / and compositions of compact operators are compact, if T is compact so also is Sn . That means that everything we learned in Chap. 20 applies to the operator .I  T /n . For T 2 K.X /, let Nn denote the null space of .I  T /n D I  Sn ; then Nn is finitedimensional by Lemma 20.6. Notice that Nn is a subspace of NnC1 . Theorem 21.1. For each T 2 K.X /, there exists an integer such that Nn 6D NnC1 for n < and Nn D NnC1 for all n  .

© Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__21

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21 The Degree Calculation

Proof. Suppose Nn 6D NnC1 for all n. Since Nn is closed in NnC1 , by Riesz’ lemma,    for each n there exists xnC1 2 NnC1 with kxnC1 k D 1 and kxnC1  vk  12 for all  v 2 Nn . This gives us a bounded sequence fxs g so, since T is compact, fT xs g must contain a convergent sequence. But take any integers r; s with r < s and write T xs  T xr D xs  .I  T /xs C xr  .I  T /xr : Now .I  T /s1 ..I  T /xs / D .I  T /s xs D 0 so .I  T /xs 2 Ns1 and similarly .I  T /xr 2 Nr1  Nr  Ns1 . Thus we can write T xs  T xr D xs  x where x D .I  T /xs C xr  .I  T /xr 2 Ns1 : The property of xs would then imply that kT xs  T xr k  12 and fT xn g couldn’t contain a convergent sequence after all. The contradiction tells us that there must be some n for which Nn D NnC1 . But then couldn’t we have Nn D NnC1 for some n and yet Nr 6D NrC1 for some r > n? It’s not hard to show that the answer is no, as follows. Let be the minimum of the set of integers n such that Nn D NnC1 and suppose r > . For x 2 Nr , we have .I  T / C1 ..I  T /r 1 x/ D .I  T /r x D 0 so .I  T /r 1 x 2 N C1 D N : Consequently, 0 D .I  T / ..I  T /r 1 x/ D .I  T /r1 x which tells us that Nr D Nr1 .

t u

For T 2 K.X /, we define Rn to be the range of .I  T / , which is a closed subspace of X by Theorem 20.8. Of course RnC1 is a subspace of Rn . The Rn exhibit the same sort of stability behavior that we demonstrated for the Nn spaces in the previous result. The proof is very similar, so we’ll omit it. n

Theorem 21.2. For each T 2 K.X /, there exists an integer  such that Rn 6D RnC1 for n <  and Rn D RnC1 for all n  . Lemma 21.3. Let  be the integer defined by Theorem 21.2 and let m  1 be an integer; then Nm \ R D 0.

21 The Degree Calculation

157

Proof. Let z 2 Nm \ R . Choose n at least as large as both  and the integer of Theorem 21.1, so in particular, Nn D NnCm for all m. Since Rn D R , and z 2 R , there exists zn 2 X such that .I  T /n zn D z which implies .I  T /nCm zn D .I  T /m z D 0 because z 2 Nm . Thus zn is in NnCm D Nn and therefore z D 0.

t u

Theorem 21.4. For T 2 K.X /, let and  be the integers of Theorems 21.1 and 21.2, respectively; then D . Proof. Suppose  < and let x 2 NC1  N so .I  T / x D y ¤ 0 which tells us that y 2 R . However, .I  T /C1 y D .I  T /2C1 x D .I  T / .I  T /C1 x D 0 because .I  T /C1 x D 0. But that puts y in Nm \ R which makes y D 0 by Lemma 21.3. On the other hand, supposing that  > also leads us to a contradiction, in the following way. Let x 2 R be any point. Since R D R21 , then x D .I  T /21 y for some y 2 X . Consider .I  T /2 y 2 R2 D R ; then .I  T /2 y D .I  T / z for some z. Write z D .z  .I  T / y/ C .I  T / y: Now z  .I  T / y is in N1 because z  .I  T / y 2 N and  > implies N D N1 . Therefore .I  T /1 z D 0 C .I  T /21 y D x which means x 2 R1 . But we chose x to be any point of R so this would mean that R D R1 , contrary to the definition of  (see Theorem 21.2). t u The purpose of what we’ve been doing is to split X as a product of linear spaces in a useful way. Theorem 21.5. Let T 2 K.X / and let  be as in Theorem 21.2; then X D N ˚ R .D N ˚ R /. Proof. Since R D R2 , given x 2 X , we know that .I  T / x D .I  T /2 y for some y 2 X . That tells us that .I  T / Œx  .I  T / y D 0 so if we write x in the form x D .x  .I  T / y/ C .I  T / y

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21 The Degree Calculation

then we have expressed it as the sum of an element of N and an element of R . Setting m D  in Lemma 21.3 gives us N \ R D 0. u t A compact linear operator on X will preserve that splitting because Lemma 21.6. For all n, an operator T 2 K.X / takes Nn to Nn and Rn to Rn . Proof. The linearity of T implies that T Sn D T

n X .1/rC1 Crn T r D Sn T rD1

and therefore that T .I  T /n D .I  T /n T , from which the result follows easily. u t Letting  be a nonzero real number, we’re now going to use the subspaces from Theorem 21.5 to investigate the compact operator 1 T . For  the integer of Theorem 21.2 for this operator, that is, where the range of .I  1 T /n stabilizes, we define N ./ to be the null space of .I  1 T / , which is a finite-dimensional, and therefore closed, subspace of X . Defining R ./ to be the range of the same operator, Theorem 21.5 allows us to write X D N ./ ˚ R ./. Furthermore, a slight adjustment of the proof of Lemma 21.6 suffices to prove that T takes each of N ./ and R ./ to itself. Thus the operator T respects this splitting of the linear space X . The spectrum of T also respects this splitting in a sense that will be made precise by the next two results. We’ll simplify the notation a bit more by writing N ./ and R ./ as N and R , respectively. Denote the corresponding restrictions of T by T jN and T jR . Theorem 21.7. If  2  .T / and  ¤ 0, then the spectrum of T jN , the restriction of T to N D N ./, consists of the number  only, that is,  .T jN / D fg. Proof. We denote the identity map of N by I . In the case  D 1, we see that N1 consists of all x such that .I  1 T /x D 0 which is equivalent to T x D x. Therefore T jN1 D I1 and  is its one and only eigenvalue. Now suppose that   2. The eigenvectors of T corresponding to  are contained in N , so certainly  is an eigenvalue of T jN . We will show there are no others by proving that if  2 R is not zero, then the number    is not in the spectrum of T jN . Recalling the definition of the spectrum from the previous chapter, this means we have to prove that the operator .   /I  T jN has an inverse. Let S be the restriction of I  T to N , which maps N to itself. The reason we want to introduce S is that .   /I  T jN D .I  T  I /jN D S  I and S  I has an inverse for the following reason. By definition, S  D 0, so we can write

21 The Degree Calculation

159

  I D   I  S  D .I  S /.

 X

 j S j 1 /

j D1

by the usual sort of telescoping sum argument. We have shown that ..   /I  T jN /1 D .S  I /1 D 

 1 X j j 1  S   j D1

and thus that the spectrum of T jN is the single number .

t u

Theorem 21.8. For the splitting of X D N ˚R , there is a corresponding splitting of the spectrum, that is,  .T jR / D  .T /  fg. Proof. We wish to apply earlier results of this chapter to N . We may do so because N D N ./ is the null space of .I  1 T / D I  S where S D

 X 1 .1/rC1 Cr . T /r  rD1

which is in K.X / because T is. Thus N is finite-dimensional so the space R is infinite-dimensional and therefore 0 2  .T jR / by Theorem 20.9. By the Fredholm alternative, the rest of  .T / consists of eigenvalues of T . Let ˛ be an eigenvalue of T not equal to either 0 or  and let x be a corresponding eigenvector. Then we claim x 2 R and therefore ˛ 2  .T jR / also. To prove it, notice that .I  T /x D .  ˛/x implies .I  T / x D .  ˛/ x and therefore that we can write x D .I 

1    T/ . / x:  ˛

The only eigenvalue of T not yet accounted for is . Let x now be an eigenvector of T corresponding to ; then certainly x 2 N1 ./ which is a subspace of N . The direct sum decomposition X D N ˚ R therefore tells us that x 62 R since it’s nonzero. Thus  is not an eigenvalue of T jR and consequently it is not in its spectrum. t u The multiplicity of  2  .T / is defined to be the dimension of N ./. Let  be a nonzero real number with the property that 1 is not an eigenvalue of T ; thus 1 is not in  .T /. Define H./ D f 2  .T / W  > 0

and

1 jj > j jg: 

See Fig. 21.1. Since we know that  .T / is compact and has no limit points other than, possibly, 0, we conclude that the set H./ is finite. Define ˇ./ to be the sum of the multiplicities of all those eigenvalues of T that lie in H./.

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21 The Degree Calculation

Fig. 21.1 The set H./

We need a splitting of the Banach space X in terms of . Lemma 21.9. Let T 2 K.X / and let  be a nonzero real number such that 1 is not in  .T /; then X D X1 ˚ X2 such that dim X1 D ˇ./, each of X1 and X2 is mapped to itself by T , and  .T jX1 / D H./ whereas  .T jX2 / D  .T /  H./. Proof. Let f1 ; 2 ; : : : ; k g be the eigenvalues of T that form the set H./. For each j , we’ll apply Theorem 21.2 to the compact operator 1j T to obtain an integer j with the property that R..I 

1 j

T /n / D R..I 

1 j

T /j / for all n  j . Then,

defining  to be the largest of all those j , we see that R..I  1 j

1 j

T / / D R..I 

T /j / for all j D 1; : : : ; k. Now we’ll simplify the notation by setting R.j / D

R..I 

1 j

T / / and also N.j / D N..I 

1 j

T / / for that same integer .

If we apply Theorem 21.5 to 11 T , we get a splitting X D N.1 / ˚ R.1 /. What we want to do is to continue to split off each of the N.j / for j 2 H./, one at a time, and their direct sum will be the space X1 . By 21.8, we have  .T jR.1 // D  .T /  f1 g. Lemma 21.6 implies that T .R.1 //  R.1 /, so it also implies that 1 T .R.1 //  R.1 /. We’ll use the operator 12 T jR.1 / to split the linear space 2 R.1 / by Theorem 21.5: R.1 / D N0 .2 / ˚ R0 .2 /: The linear space N0 .2 / consists of all x 2 R.1 / such that .I  12 T / .x/ D 0 for  sufficiently large. It is a subspace of N.2 /, the set of elements of X with the same property. We will show that, in fact, N0 .2 / D N.2 /. To do that, let x 2 N.2 / and, using the splitting X D N.1 / ˚ R.1 /, write x D a C b where a 2 N.1 / and b 2 R.1 /; then we’ll prove that a D 0 so x 2 R.1 / and therefore x 2 N 0 .2 /. By definition .I 

1  T / .a C b/ D 0 2

and thus .I 

1 1  T / .a/ D .I  T / .b/: 2 2

21 The Degree Calculation

But .I 

1 T / .a/ 2

161

2 N.1 / whereas .I  .I 

1 T / .b/ 2

2 R.1 / so, in particular,

1  T / .a/ D 0 2

which means that a 2 N.2 /. To complete the proof that a D 0, we’ll demonstrate that N.1 / \ N.2 / D 0, which we will need also to express X1 as a direct sum. Suppose there were a nonzero element x in N.1 / \ N.2 /. Certainly x cannot be an eigenvector for both 1 and 2 , so without loss of generality, we can assume that there is an m  2 such that .I  11 T /m x D 0, but y D .I 

1 m1 T/ x 6D 0: 1

We are assuming that x 2 N.2 /, so Lemma 21.6 implies that y 2 N.2 / also. But since .I 

1 1 m T / x D .I  T /y D 0 1 1

we find that y is an eigenvector of T corresponding to the eigenvalue 1 . That would give us 1 2  .T jN.2 //, contrary to Theorem 21.7, and this completes the argument that N.1 / \ N.2 / D 0 and thus that N0 .2 / D N.2 /. Consequently, we can write R.1 / D N.2 / ˚ R0 .2 / where, by Theorem 21.8,  .T jR0 .2 // D  .T /  f1 ; 2 g: Let’s carry the process one step further and, writing R0 .2 / D N0 .3 / ˚ R0 .3 /, we need to prove that N0 .3 / D N.3 /. Using the fact that N.1 / \ N.2 / D 0, this time we’ll write X D .N.1 / ˚ N.2 // ˚ R0 .2 /: Take x 2 N.3 / and write x D a C b where a 2 N.1 / ˚ N.2 / and b 2 R0 .2 /. The same sort of argument we just used demonstrates that a 2 N.3 / and then that .N.1 / ˚ N.2 // \ N.3 / D 0. Repeating this procedure for each element of H./, we end up splitting X as X D X1 ˚ X2 where X1 is the direct sum X1 D

k X

N.j /

j D1

so the dimension of the linear space X1 is the integer we called ˇ./. The operator T takes each of X1 and X2 to itself, because each step of the process respects the corresponding splitting, with  .T jX1 / D H./ and  .T jX2 / D  .T /  H./. u t

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21 The Degree Calculation

For a positive real number , we have the open -ball B D fx 2 X W kxk < g: We will be concerned now with the operator I  T on the closed ball B  . If T x D x for x 6D 0, that would make 1 an eigenvalue of T , so the restriction on the value of  assures us that the compact map T jB  has no fixed points on @B . Therefore, the Leray–Schauder degree deg.I  T; B / is well defined. The final, and principal, result of this chapter will compute this degree. Theorem 21.10. Let T 2 K.X / and let  6D 0 such that  > 0,

1 

62  .T /; then, for any

deg.I  T; B / D .1/ˇ./ : Proof. From Lemma 21.9 we have the splitting X D X1 ˚ X2 where X1 is of dimension ˇ./. Furthermore, for j D 1; 2 we have T .Xj /  Xj and therefore .I  T /.Xj /  Xj . In order to calculate deg.I  T; B / and thus prove the theorem, we will calculate the two Leray–Schauder degrees defined by setting dj D deg.I  T jXj ; Bj / where Bj D B \Xj . Here is the reason those computations will prove the theorem. Since 1 isn’t an eigenvalue of T , the only fixed point of T is 0, so by the excision property of the Leray–Schauder degree (Theorem 11.5), deg.I  T; B / D deg.I  T; B1  B2 /: Then the product property (Theorem 11.3) implies that deg.I  T; B1  B2 / D d1  d2 : Now, to calculate d1 , define f; g W B1 ! X1 by setting f .x/ D T x and g.x/ D 2x. We’ll show that tf .x/ C .1  t /g.x/ 6D x for all x 2 @B1 . If for some t 2 Œ0; 1 there was a solution to tf .x/ C .1  t /g.x/ D tT x C .1  t /2x D x in @B1 , then x 6D 0 and it would have the property Tx D which makes conclude that

2t1 t

2t  1 x t

an eigenvalue of T . Thus, since x 2 X1 , by Lemma 21.9, we 2t  1 2 H./ t

21 The Degree Calculation

163

which tells us, by the definition of H./, that 

 2t  1 >0 t

and

ˇ ˇ ˇ ˇ ˇ 2t  1 ˇ ˇ 1 ˇ ˇ ˇ ˇ ˇ ˇ t ˇ > ˇ  ˇ:

But that implies 2t  1 > t , so there are no such solutions and we may apply the straight-line homotopy property (Theorem 11.8) to conclude that d1 D deg.I  T; B1 / D deg.I  f; B1 / D deg.I  g; B1 / Now .I  g/.x/ D x  2x D x D I.x/ and X1 is a finite-dimensional linear space that we can identify with Rˇ./ . In the finite-dimensional setting, the Leray– Schauder degree is the Brouwer degree by Theorem 10.6, and therefore we may apply the involution property of that degree (Theorem 9.10), to conclude that d1 D deg.I; B1 / D deg.I; Rˇ./ / D .1/ˇ./ : The proof that d2 D 1 is similar: in this case we will demonstrate that the equation tT x C .1  t /0 D x has no nontrivial solutions x 2 X2 . If it did, then 1 2  .T jX2 / D  .T /  H./ t by Lemma 21.9. Since . then

1 t

1 1 / D > 0 t t

62 H./ implies ˇ ˇ ˇ ˇ ˇ 1 ˇ ˇ1ˇ ˇ ˇ 1. Therefore, by the homotopy (Theorem 11.7) and normalization (Theorem 11.2) properties of the Leray–Schauder degree, d2 D deg.I  T jX2 ; B2 / D deg.I; B2 / D 1: t u

Chapter 22

The Krasnoselskii–Rabinowitz Bifurcation Theorem

In the previous chapter, we used spectral theory to make a computation of Leray– Schauder degree. For this chapter, which presents the main result of the book, we’ll also need the separation theorem from point-set topology that we proved in Chap. 19. However, we first must introduce a hypothesis that permits us to apply the theory of compact linear operators in a more general, nonlinear, setting. In elementary calculus, if a function f W R ! R is differentiable at x0 , then the tangent line to the graph of f at the point .x0 ; f .x0 // is well defined and it serves as an approximation to f in a neighborhood of x0 . The tangent line is the graph of a linear function, call it L.x/, through that point and with slope f 0 .x0 /, so L.x/ D f 0 .x0 /x C Œf .x0 /  f 0 .x0 /x0 : The sense in which L approximates f near x0 can be made precise by noticing that the difference quotient definition of f 0 .x0 / implies that lim

jxx0 j!0

jf .x/  L.x/j D 0: jx  x0 j

We can reverse this observation and define f to be differentiable at x0 if there is a linear function L for which that limit exists and is zero. In that case, we would define the number f 0 .x0 / to be the slope of L. Besides whatever pedagogical benefit we might gain by using this approach to explain to our students what differentiability means, this definition invites us to make a very far-reaching generalization of the concept of differentiability. At least that’s what it suggested to Frechet, as we will now see. Let X be a Banach space and f W X ! X be a map. We only need Frechet’s idea in the somewhat specialized case that f .0/ D 0 and our concern is just at that point. The map f is said to be Frechet differentiable at 0 if there exists T 2 L.X / such that © Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__22

165

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22 The Krasnoselskii–Rabinowitz Bifurcation Theorem

lim

kxk!0

kf .x/  T xk D 0: kxk

Noticing that this definition agrees with the previous one in the case R D X and x0 D 0, we call T the Frechet derivative of f at 0. We next prove some results about the Frechet derivative that allow us to apply facts we learned earlier, about compact linear operators, and about the Leray– Schauder degree, to the map f , provided it satisfies the complete continuity hypothesis. Theorem 22.1. If f W X ! X is a completely continuous map of a Banach space such that f .0/ D 0 and f is Frechet differentiable at 0, then its Frechet derivative T W X ! X is a compact linear operator. Proof. We’ll prove the contrapositive, so we start out by assuming that T 62 K.X / and seek to demonstrate that f cannot be completely continuous. We suppose there’s a bounded subset B of X such that T .B/ isn’t relatively compact. That implies there is a sequence fxi g in B for which the sequence fT xi g contains no subsequence convergent in X . Basically, what we are going to show is that the sequence ff .xi /g has the same property, so f isn’t completely continuous. Since multiplying everything by a constant won’t change the situation, we will assume kxi k  1 for all i. The completeness of X tells us that fT xi g cannot be Cauchy so let’s assume there exists ı > 0 such that kT xi  T xj k > ı for all i 6D j . That’s not quite true, but it is for i and j big enough so it can only fail for a finite number of the xi and we can throw those away. The Frechet differentiability hypothesis allows xk us to make kf .x/T as small as we like by making kxk small enough. Specifically, kxk there exists  > 0 such that if kxk  , then kf .x/  T xk 

ı kxk: 3

What we’ll actually show is that the sequence ff .xi /g isn’t Cauchy. Since the sequence fxi g is bounded (specifically, what we need to know is that kxi k < ), which will demonstrate that f isn’t completely continuous. The argument just uses the triangle inequality: for all i 6D j we have kf .xi /  f .xj /k D k.T xi  T xj / C .f .xi /  T xi / C .f .xj / C T xj /k  kT xi  T xj k  kf .xi /  T xi k  kf .xj /  T xj k ı ı ı ı  ı  kxi k  kxj k  ı  2  D : 3 3 3 3

t u

22 The Krasnoselskii–Rabinowitz Bifurcation Theorem

167

Before we prove the other main result about the Frechet derivative, it’s convenient to get one detail out of the way: Lemma 22.2. If X is a Banach space and T 2 L.X / such that I  T is regular, then there exists a > 0 such that kx  T xk  akxk for all x 2 X . Proof. Any a will do if x D 0. The hypothesis tells us that .I  T /1 2 L.X / exists and, looking back at the definition of the norm of a bounded linear operator in Chap. 18, for x 6D 0 (and therefore .I  T /x 6D 0), we have k.I  T /1 .I  T /xk k.I  T /1 uk  sup D k.I  T /1 k: k.I  T /xk kuk u2X This implies kxk D k.I  T /1 .I  T /xk  k.I  T /1 k  k.I  T /xk and we can write kx  T xk  akxk by letting aD

1 : k.I  T /1 k

t u

Theorem 22.3. Suppose f W X ! X is a completely continuous map of a Banach space such that f .0/ D 0 and f is Frechet differentiable at 0 with Frechet derivative T 2 K.X /. If I  T 2 L.X / is regular, then there exists  > 0 such that for B D fx 2 X W kxk < g we have deg.I  f; B / D deg.I  T; B /: Proof. The first thing we should notice is that, when restricted to B , both f and T are compact in the sense of nonlinear analysis; that is, the image of B has compact closure, so, if they have no fixed points on the boundary of B , then the Leray– Schauder degrees in the conclusion are both defined. What we have to do now is prove that they both take the same value on the ball of radius . We’ll choose the

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22 The Krasnoselskii–Rabinowitz Bifurcation Theorem

number  by using the Frechet differentiability hypothesis as we used it in the proof of Theorem 22.1. Specifically, we choose  > 0 such that kxk   implies kf .x/  T xk
0 comes from Lemma 22.2; that is, kx  T xk  akxk for all x. The equality of the degrees follows from the straight-line homotopy property (Theorem 11.8) of the Leray–Schauder degree. All we have to prove is that kxk D  implies that tf .x/ C .1  t /T x 6D x and that’s the triangle inequality again: k.tf .x/ C .1  t /T x/  xk D kT x  x C t .f .x/  T x/k  kT x  xk  tkf .x/  T xk a a a  akxk  t kxk  kxk D  > 0: 2 2 2

t u

Bifurcation theory concerns maps G W R  X ! X , where X is a Banach space. The space R  X is itself a Banach space, with the norm of .; x/ 2 R  X given by k.; x/k D jj C kxk: We need to extend the idea of Frechet differentiability at 0 for maps G that have the property G.; 0/ D 0. The definition is: there exists T 2 L.X / such that given  > 0 and an interval Œ0 ; 1   R, there exists ı > 0 with the property that kxk < ı implies kG.; x/  T xk 0 such that G.; x/ 6D x for all .; x/ for which j  j <  and 0 < kxk < . In particular,  is not a bifurcation point for the solutions to G.; x/ D x. Proof. Since 1 is not in  .T /, the bounded operator . 1 I  T /1 exists and so also does .I  T /1 . For  we choose D

1 3k.I  T /1 k  kT k

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22 The Krasnoselskii–Rabinowitz Bifurcation Theorem

which we can assume is smaller than jj. We use the definition of the derivative to define  as follows: choose  small enough so that kG.; x/  T xk 1  kxk 3k.I  T /1 k for all .; x/ such that j  j <  and kxk  . Now suppose we did have .; x/ such that j  j <  and kxk < , and yet G.; x/ D x. Here’s what happens: kxk  k.I  T /1 k  kx  T xk D k.I  T /1 k  kG.; x/  T xk D k.I  T /1 k  k.G.; x/  T x/ C .  /T xk  k.I  T /1 k  kG.; x/  T xk C k.I  T /1 k  j  j  kT k  kxk  and therefore x D 0

2 kxk C k.I  T /1 k    kT k  kxk D kxk 3 3 t u

Corollary 22.5. If G W R  X ! X is completely continuous and Frechet differentiable at 0, then the set of bifurcation points is a discrete subset of R. Proof. Let T be the derivative of G. If  6D 0 is a bifurcation point, then Theorem 22.4 tells us that 1 2  .T /. By Theorem 20.12, the spectrum is either finite or a bounded sequence converging to 0. The previous theorem eliminates as possible bifurcation points all numbers except 0 and the reciprocals of nonzero members of  .T /, which form a discrete set no matter whether  .T / is finite or not. t u We need to remember from the previous chapter that the multiplicity of  2  .T / is the dimension of the null space of .I  1 T / for  big enough. So if the reciprocal of  ¤ 0 is in  .T /, by the multiplicity of 1 we mean the dimension of .I  T / . Theorem 22.6. Let X be a Banach space and let G W R  X ! X be completely continuous and Frechet differentiable at 0, with derivative T . Suppose  ¤ 0 is a real number such that 1 2  .T / and it is of odd multiplicity. Then  is a bifurcation point for the solutions to G.; x/ D x. Proof. Suppose  isn’t such a bifurcation point; then there is a neighborhood of .; 0/ in R  X that contains no nontrivial solutions to G.; x/ D x. Using the product topology, we can find positive numbers  and  so that W \ S D ; for W D Œ  ;  C   B  (see Fig. 22.2). Define G W X ! X by G .x/ D G.; x/.

22 The Krasnoselskii–Rabinowitz Bifurcation Theorem

171

Fig. 22.2 Not a bifurcation point

Choose  and  small enough so that we can apply Theorem 22.3 to both G and GC to conclude that deg.I  G˙ ; B / D deg.I  . ˙ /T; B /: Since, by assumption, the only fixed point of G jB  for  2 Œ  ;  C  is 0, the homotopy property of the Leray–Schauder degree (Theorem 11.7) implies that deg.I  GC ; B / D deg.I  G ; B / and therefore deg.I  . C /T; B / D deg.I  .  /T; B /: We can calculate these degrees by Theorem 21.10: deg.I  . ˙ /T; B / D .1/ˇ.˙/ :

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22 The Krasnoselskii–Rabinowitz Bifurcation Theorem

Now 1 lies in one of the sets H.  / or H.  / but not in the other, and thus, since 1 is of odd multiplicity, we arrive at a contradiction because then .1/ˇ.C/ ¤ .1/ˇ./ : Therefore  is a bifurcation point after all.

t u

The next result is a major step in preparing for the Krasnoselskii–Rabinowitz bifurcation theorem. Its purpose is to bring the information from the separation theorem of Chap. 19 into our study of the solutions to the equation G.; x/ D x. The form of the statement is a bit unusual. It says that if, in the closure of the set S of nontrivial solutions, there is not a closed, connected set C that is either unbounded or connects two bifurcation points, then there is an open bounded set O in RX with some other properties. In the proof of the bifurcation theorem, we will establish that such a set O cannot exist and therefore, as a part of the conclusion to the theorem, we know that there must be a set C in S with at least one of the properties: it’s unbounded or it contains two bifurcation points. Theorem 22.7. Let G W R  X ! X be completely continuous and Frechet differentiable at 0, with derivative T , and let 1 2  .T / such that  is a bifurcation point for the solutions to G.; x/ D x. Suppose there is no closed connected subset C of S containing .; 0/ such that either (i) C is unbounded in R  X or (ii) C contains . ; 0/ for some bifurcation point  6D . Then there exists an open bounded subset O of R  X containing .; 0/ such that (a) @O \ S D ; and (b) there exists  > 0 for which .; 0/ 2 O implies j  j < 2 and  is the only bifurcation point in Œ  ;  C . Proof. Let C be the component, that is the maximal connected subset, of S that contains .; 0/. We claim that C is compact, even though S may not be. Hypothesis (i) implies that C is bounded in R  X . In particular, then, there exists an interval Œ1 ; 2  in R such that .; x/ 2 C implies 1    2 . Since G is completely continuous, the boundedness of C also tells us that G.C / is compact. But the definition of S implies that the restriction GjC is just the projection of C onto X so C is a subset of the compact set Œ1 ; 2   G.C /. See Fig. 22.3. Since components are closed subsets, the claim is established. Recalling Corollary 22.5, let  > 0 be less than the distance from  to any other bifurcation point and let J D R .  ;  C /. Now comes our chance to use hypothesis (ii). It tells us that J  0 is a closed subset of R  X disjoint from the compact set C , so the distance between them is a well-defined positive number; let ı be half that distance and denote the ı-neighborhood of C in RX by Uı .C /. See Fig. 22.4. In other words, .; x/ 2 Uı .C / if there exists .0 ; x 0 / 2 C such that k.; x/  .0 ; x 0 /k < ı. Define K D Uı .C / \ S;

22 The Krasnoselskii–Rabinowitz Bifurcation Theorem

173

Fig. 22.3 The component C

X Ud (Cm) Cm

J

m −h

m

m +h

J

R

Fig. 22.4 Construction of the set O

then K is compact by pretty much the same argument we used for C . Now C \ .@Uı .C / \ S / D ; but the separation theorem (19.4) permits us to say more. If there were a closed connected subset C of S intersecting both C and @Uı .C / \ S , then C [ C would be connected because the union of two connected sets with nonempty intersection

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22 The Krasnoselskii–Rabinowitz Bifurcation Theorem

is connected, contradicting the maximality of C . In particular, there is no such connected set in K, and the hypotheses of the separation theorem hold. Therefore, we may write K D A [ B where A and B are disjoint closed subsets of K with A containing C and B containing @Uı .C / \ S. Choose  > 0 so that a 2 A and b 2 B implies ka  bk > 2; then certainly  < ı. We take O to be the  -neighborhood of A in R  X . It is clear from the construction that O contains 2 no bifurcation points .; 0/ other than .; 0/ and that @O \ B D ;, so O has the required properties. t u Here it is at last: the main result of the book (Rabinowitz (1971)). Theorem 22.8 (Krasnoselskii–Rabinowitz Bifurcation Theorem). Let X be a Banach space and let G W R  X ! X be completely continuous and Frechet differentiable at 0, with derivative T . Suppose  is a real number such that 1 2  .T / and it is of odd multiplicity. Then there exists a maximal closed connected  subset C of S which contains .; 0/ and either C is unbounded in R  X or C contains . ; 0/ for some bifurcation point  6D . Proof. We will assume that C doesn’t exist and show that it leads to a contradiction. Since  is a bifurcation point by Theorem 22.6, then our assumption that C doesn’t exist gives us the hypotheses of Theorem 22.7 and therefore an open bounded set O that avoids .0 ; 0/ for any bifurcation point 0 other than  itself, and with the property @O \ S D ;. For each real number , define O D fx 2 X W .; x/ 2 Og: The sets O will play a central role in our proof. The first thing to notice about them is that since O is bounded, O D ; for  sufficiently far from . If .; 0/ 62 O, choose ./ > 0 such that k.; 0/  .0 ; x 0 /k > ./ for all .0 ; x 0 / 2 O. See Fig. 22.5. For .; 0/ 2 O but  6D , we choose ./ small enough so that k.; 0/  .0 ; x 0 /k > ./ for .0 ; x 0 / 2 S . See Fig. 22.6. Next, for G W X ! X defined by G .x/ D G.; x/, we have G.; x/ D x if and only if .I  G /.x/ D 0. For any r > 0, let B r D fx 2 X W kxk  rg: For each  6D , we consider the set O  B ./ which is open and bounded in X , and the restriction of G to it is certainly a compact map. Furthermore, there are no zeros of I  G on the boundary of O  B ./ , so the set of zeros in the interior is compact and thus the Leray–Schauder degree deg.I  G ; O  B ./ / is well defined. Looking back at Theorem 22.3, for  6D  such that .; 0/ 2 O, let’s be sure to choose ./ small enough so that deg.I  G ; B./ / D deg.I  T; B./ /:

22 The Krasnoselskii–Rabinowitz Bifurcation Theorem

175

Fig. 22.5 The distance ./ for  far from 

Fig. 22.6 The distance ./ for  near 

We’ll make good use of that at the end of the proof, but for now we need to show that deg.I  G ; O  B ./ / D 0 for all  6D . This claim is certainly true for  sufficiently far from  since then O D ; (recall the empty set property, Theorem 11.6). Now let 1 and 2 be real numbers such that .i ; 0/ 62 O and either both i are greater than  or both are less than . Let  be the closed interval with endpoints the i ; then W D O \ .  X / is open in   X . Noting that  2  implies O  B ./ D O , the generalized homotopy property of the Leray–Schauder degree (Theorem 11.10) tells

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22 The Krasnoselskii–Rabinowitz Bifurcation Theorem

X

W

O r(λ2)

r(λ1) m

λ1

Λ

λ2

R

Fig. 22.7 Independence of 

us that deg.I  G ; O  B ./ / is independent of the choice of  2 . Thus, we’ve established the claim for all  such that .; 0/ 62 O. See Fig. 22.7. We’ll choose 1 and 2 again, but this time so that .1 ; 0/ is in O and .2 ; 0/ isn’t, but still retaining the property that  is not contained in the interval   R with endpoints the i . We’ll also want 2 close enough to  so that no bifurcation point lies in . Therefore, there exists  > 0, smaller than both .1 / and .2 /, such that .  B  / \ S D ;: We’ll use the generalized homotopy property once again, this time with W D O \ .  .X  B  //: See Fig. 22.8. Thus we see that deg.I  G1 ; O1  B  / D 0 but then since there cannot be any zeros of I  G1 in B .1 /  B  , the excision property of the Leray– Schauder degree (Theorem 11.5) completes the proof of the claim that deg.I  G ; O  B ./ / D 0 for all  6D . Now we are ready for the argument that will lead to a contradiction. Choose real numbers  <  and  >  close enough to  so that .; 0/ and .; 0/ lie in O. We make use of the generalized homotopy property one more time, now with W D O \ .Œ;   X /, to conclude that deg.I  G ; O / D deg.I  G ; O /:

22 The Krasnoselskii–Rabinowitz Bifurcation Theorem

177

X O W

r(λ1) λ1 m

r(λ2)

R

λ2

W

Λ × Br

Fig. 22.8 A different W

X O

W

h

m

z

R

Fig. 22.9 Invariance of the degree

See Fig. 22.9. The first part of this proof consisted of establishing a claim which, together with the additivity property of the Leray–Schauder degree (Theorem 11.4), tells us that deg.I  G ; O / D deg.I  G ; O  B ./ / C deg.I  G ; B./ / D 0 C deg.I  G ; B./ /

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22 The Krasnoselskii–Rabinowitz Bifurcation Theorem

and, similarly, that deg.I  G ; O / D deg.I  G ; B./ /: Therefore we can write deg.I  G ; B./ / D deg.I  G ; B./ /: Now recalling the condition on ./ when .; 0/ 2 O, Theorem 22.3 implies that deg.I  T; B./ / D deg.I  T; B./ /: But, just as in the proof of Theorem 22.6, since 1 is of odd multiplicity and contained in exactly one of the sets H./ or H./, by Theorem 21.10, we have deg.I  T; B./ / D .1/ˇ./ ¤ .1/ˇ./ D deg.I  T; B./ / which contradicts the equality of these degrees that we just established and therefore proves the theorem. t u

Chapter 23

Nonlinear Sturm–Liouville Theory

In Chap. 25, we’ll apply the Krasnoselskii–Rabinowitz bifurcation theorem in a very specific way: to the Euler buckling problem. The buckling problem belongs to an important class of problems in ordinary differential equations called nonlinear Sturm–Liouville problems. To begin this chapter I’ll describe the Euler buckling problem and place it in that more general differential equation context. Then I’ll apply the bifurcation theorem to the general class of nonlinear Sturm–Liouville problems to obtain a tool that I’ll be able to use for the buckling problem. A buckling problem concerns a load placed on a vertical column. If the load is great enough the column will “buckle,” that is, be displaced from vertical. To set up the same problem in a mathematically convenient way, imagine the column as the interval Œ0;  in the x-axis of the plane. The endpoint at the origin is held fixed while a constant force P is applied to the other endpoint . ; 0/. In other words, instead of placing a load on a vertical column, we put the column on its side and push on one end. See Fig. 23.1. In Euler’s theory, the force is applied so that when the endpoint .0; / is moved by the force, it stays on the x-axis (in the more realistic picture, the top of the column is pushed straight down). Furthermore, and this is the really crucial assumption, the column is not allowed to twist into a third dimension. In the mathematical picture, this means that, after the force is applied, each point of the interval Œ0;  is moved only in a vertical direction. For each s 2 Œ0; , let y.s/ denote the y-coordinate of the point to which s is moved by the buckling. Since s D 0 doesn’t move and s D stays in the x-axis, we have y.0/ D y. / D 0. Thus the buckled column can be represented as a curve in the plane with endpoints on the x-axis. We will assume the buckled column is a smooth curve, so there is a tangent line to the curve at each point. Let .s/ be the angle between the tangent at the point to which s moves and the horizontal. The Euler–Bernoulli law states that P y.s/ D M 0 .s/

© Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__23

179

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23 Nonlinear Sturm–Liouville Theory

weight

weight

column

column

ground

ground

y

y

force 0

s (arc length)

π

0

s =π

Fig. 23.1 A buckled column

where M is a constant, called the “modulus of flexural rigidity” that is determined by experiments on the particular column. Since both P and M are constants, we’ll combine them by setting D

P M

where of course the choice of the name  for the constant is no coincidence. Thus the Euler–Bernoulli law has the form  0 .s/ D y.s/: approxiAs Fig. 23.2 indicates, for a small displacement s of s, the ratio y.s/ s mates the function sin..s// so, in the limit, y 0 .s/ D sin..s//. To take advantage of that observation, let’s differentiate the Euler–Bernoulli law and substitute, which gives us  00 D y 0 D  sin : The Euler–Bernoulli law tells us that  0 .0/ D  0 . / D 0. The Euler buckling problem therefore becomes the following problem in ordinary differential equations: find a continuous function  D .s/ W Œ0;  ! R and a real number  satisfying  00 D  sin   0 .0/ D  0 . / D 0: The second condition is called the Neumann boundary condition. The type of problem we are discussing is called an eigenvalue problem in differential equations

23 Nonlinear Sturm–Liouville Theory

181

Fig. 23.2 The Euler-Bernoulli law

because the constant  is analogous to an eigenvalue in the linear theory. (We will see that the analogy is not just a formal one.) Since we required  to be continuous, the right-hand side of the differential equation tells us that  00 is also continuous. Thus, in terms of the notation of Chap. 2, the function  must be in the space C 2 Œ0; . I said that the Euler buckling problem was an example of a nonlinear Sturm– Liouville problem, so now that we have the Euler problem expressed in a good mathematical form, let’s find out what Sturm–Liouville problems in general look like. The first step in the definition of such a problem requires two functions p; q W Œ0;  ! R where p is differentiable and p.s/ > 0 for all s 2 Œ0; . Sturm– Liouville theory uses these functions to define a linear operator L W C 2 Œ0;  ! C Œ0;  by setting Lu D .pu0 /0 C qu: Another ingredient of this type of problem is a function F W Œ0;   R3 ! R which is Frechet differentiable in the following sense. There exists a map a W R ! R with a.s/ > 0 p for all s such that, given  > 0 and an interval Œ0 ; 1 , there exists ı > 0 such that  2 C 2 < ı implies that jF .s; ; ; /  a.s/j 0; then since lim

!0

sin  D1 

we can find ı > 0 such that ˇ ˇ ˇ sin  ˇ j sin   j ˇ D jj  ˇ  1ˇˇ <  jj  for jj < ı and  2 Œ0 ; 1 . In the terms of the definition, the function a is given by a.s/ D 1 for all s. Let’s suppose we are given a nonlinear Sturm–Liouville eigenvalue problem to solve. That means not only that the functions p and q, and thus the operator L, have been specified, along with the function F , but also the constants ai and bi that determine which particular separated boundary conditions, Dirichlet or Neumann or some other, must be satisfied. Denote by C02 Œ0;  the subspace of C 2 Œ0;  consisting of the functions u W Œ0;  ! R such that a0 u.0/ C b0 u0 .0/ D 0 and a1 u. / C b1 u0 . / D 0 for the given values of the ai and bi . Now we’ll assume that, in the given problem, L and the boundary conditions are such that L W C02 Œ0;  ! C Œ0;  has an inverse. This was precisely the situation we met in Chaps. 5 through 7, where Lu D u00 was invertible with respect to the Dirichlet boundary conditions. As in the forced pendulum and equilibrium heat distribution problems, to obtain a solution to the differential equation Lu D F .s; u; u0 ; /, it is sufficient to obtain a solution to the fixed point problem u D L1 F .s; u; u0 ; /: There is, however, an important distinction between the formulations of the differential equation and the fixed point problem. The operator L involves a second derivative, but the function F , and therefore the fixed point problem, uses only the first derivative. Let X D C01 Œ0; , that is, X is the subspace of C 1 Œ0;  consisting of the functions u (with continuous first derivative) that satisfy the given (sBC). As before, making use of the function ˚ W R  X ! C Œ0;  defined by ˚.; u/.s/ D F .s; u.s/; u0 .s/; /: the fixed point problem can be written in the form u D L1 ˚.; u/. But this formulation isn’t quite satisfactory since the domain of ˚ is R  X whereas the image of L1 is C02 Œ0; . However, since C02 Œ0;  is a subspace of X D C01 Œ0; , we can let j be the inclusion of C02 Œ0;  in X and, defining G W R  X ! X to be the composition ˚

L1

j

G W R  X ! C Œ0;  ! C02 Œ0;  ! C01 Œ0;  D X;

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23 Nonlinear Sturm–Liouville Theory

a solution to the Sturm–Liouville problem can be obtained as a solution to G.; u/ D u. In terms of our earlier discussion, G D E˚ W R  X ! X where E D jL1 so from that discussion we know that G is Frechet differentiable at 0, with derivative T D L1 a. Let’s collect our thoughts. We are assuming that we have a nonlinear Sturm– Liouville eigenvalue problem, which we can abbreviate as Lu D F (sBC), for which L1 , the inverse of L with respect to the boundary conditions, exists. We will apply the bifurcation theory of the previous chapter to G D jL1 ˚ . Here’s the first thing we need to know. Theorem 23.1. The function G D jL1 ˚ W R  X ! X is completely continuous. Proof. As the norm of R  X we have k.; u/k D jj C kuk1 D jj C sup ju.s/j C sup ju0 .s/j: 0s

0s

Let Z be a bounded subset of R  X , so there exists M such that k.; u/k < M for all .; u/ 2 Z. In particular, then, .; u/ 2 Z tells us that  2 ŒM; M  and that ju.s/j < M , ju0 .s/j < M for all s 2 Œ0; . Consider the compact subset Q D Œ0;   ŒM; M   ŒM; M   ŒM; M  of Œ0;   R3 and, using the fact that F is continuous, let denote the maximum of jF .q/j for q 2 Q. We have seen that if .; u/ 2 Z, then .s; u.s/; u0 .s/; / 2 Q, so j˚.; u/.s/j D jF .s; u.s/; u0 .s/; /j  for all s 2 Œ0;  and therefore kF .; u/k  which tells us that ˚.Z/ is a bounded subset of C Œ0; . So we have shown that ˚ is a bounded map, that is, it takes bounded subsets to bounded subsets. The operator L1 is also bounded, because it is a continuous linear function (Theorem C.1). Thus, for Z a bounded subset of R  X , we know that L1 ˚.Z/ is bounded in C02 Œ0; . The form of the Ascoli–Arzela theorem stated as Theorem 2.7 tells us that the inclusion of C 2 Œ0;  into C 1 Œ0;  is completely continuous, so its restriction to C02 Œ0; , which we are calling j , has the same property. Therefore, for Z a bounded subset of R  X , we conclude that G.Z/ D j.L1 ˚.Z// is a relatively compact subset of X . t u In order to apply the Krasnoselskii–Rabinowitz theorem to nonlinear Sturm– Liouville problems, we will need to investigate the spectrum of T D L1 a. The eigenvalue equation T u D u, for  2 R, is thus L1 au D u. Recalling what L1 is, namely the inverse with respect to (sBC) of the operator L defined by

23 Nonlinear Sturm–Liouville Theory

185

Lu D .pu0 /0 C qu, we rewrite the eigenvalue problem as Lu D au where D 1 . To say it yet another way, an eigenvector u with respect to an eigenvalue  is a solution to the boundary value problem .pu0 /0 C qu D au .sBC /: Such a problem is an example of a classical, that is linear, Sturm–Liouville boundary value problem. It is called the linearized form of the nonlinear problem Lu D F (sBC). Let’s assume we have chosen a specific value for ; then the differential equation of the linearized problem can be written in the form pu00  p 0 u0 C .q  a/u D 0

(DE)

where p; q; a W Œ0;  ! R are maps, is a constant, and 0 denotes the zero function on Œ0; . Equation (DE) is a relatively elementary sort of problem: a linear secondorder ordinary differential equation. Our next step in analyzing the eigenvalues of T depends on standard facts about such equations. Theorem 23.2. The space of solutions to k2 u00 C k1 u0 C k0 u D 0; where k0 ; k1 ; k2 W Œ0;  ! R are maps with k2 ¤ 0, is two-dimensional. Proof. The Cauchy–Peano existence theorem that we discussed back in Chap. 1 can easily be extended to higher-order differential equations. If we choose initial conditions u. 2 / D u0 and u0 . 2 / D u1 , then there is a local solution at s D 2 . But, since the differential equation is linear, that solution is defined over all of Œ0; . Not only that, but the solution is unique, as can be shown by an argument in the spirit of the approximation proof of existence that we presented in Chap. 1. To demonstrate that the solution space is at least two-dimensional, we’ll use the solutions u1 and u2 that we get by choosing as our initial conditions u1 . 2 / D 1; u01 . 2 / D 0 and u2 . 2 / D 0; u02 . 2 / D 1, respectively. These two solutions are linearly independent because if we write c1 u1 .s/ C c2 u2 .s/ for constants c1 and c2 and evaluate that expression and its derivative at s D 2 , the initial conditions imply that c1 D c2 D 0. Now let u be any solution at all and set

v.s/ D u.s/  u. /u1 .s/  u0 . /u2 .s/ 2 2 which is also a solution to the linear differential equation because it’s a linear combination of solutions. Since v satisfies the initial conditions v.0/ D v0 .0/ D 0, uniqueness then tells us that v D 0 and therefore u D c1 u1 C c2 u2 , with c1 D u. 2 / and c2 D u0 . 2 /. t u

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23 Nonlinear Sturm–Liouville Theory

From Theorem 23.2 we can see that it’s important to know whether two solutions are linearly independent. Another standard result furnishes the tool we need to make that decision. Theorem 23.3. Solutions u1 and u2 to k2 u00 C k1 u0 C k0 u D 0; are linearly independent if, and only if, the wronskian wW Œ0;  ! R defined by w.s/ D W .u1 ; u2 /.s/ D u1 .s/u02 .s/  u2 .s/u01 .s/; has the property w.s/ ¤ 0 for all s 2 Œ0; . Proof. We’ll prove the theorem in the equivalent form: u1 and u2 are linearly dependent if and only if w has a zero. For each s, the value of the wronskian w.s/ is just the determinant of the matrix  u1 .s/ u2 .s/ : u01 .s/ u02 .s/



That determinant vanishes at some value s0 if and only if there is a nonzero vector .c1 ; c2 / in the null space of the corresponding linear transformation. Thus the solution u D c1 u1 Cc2 u2 D 0 by uniqueness because it satisfies the initial conditions u.s0 / D u0 .s0 / D 0. t u It can be shown that if w has a zero, then in fact it vanishes for all s, but we won’t need that. Looking back at how we obtained (DE), we can see that another way to write it is Lu  au D 0 and, recalling that D 1 , we divide by and obtain Lu  au D 0. Applying the linear operator L1 to this form of (DE), we find that a solution u to (DE) has the property .I  T /u D 0, which we can also write as .I  1 T /u D 0, so, in the terminology of Chap. 21, Theorem 23.2 has shown us that dimN1 ./ D dim.N.I  T //  2: Theorem 23.3 can tell us more because we have specific boundary conditions in mind, namely those we call (sBC). Lemma 23.4. If  is an eigenvalue of T D L1 a where L1 is the inverse of Lu D .pu0 /0 C qu with respect to (sBC), then dim.N.I  T // D 1. Proof. Since  is an eigenvalue of T , then dim.N.I  T //  1. We’ll see that, when we impose (sBC), the wronskian w.s/ D W .u1 ; u2 /.s/ must vanish at s D 0, so dim.N.I  T // < 2. If, for i D 1; 2, the solutions ui to (DE) satisfy (sBC), then, in particular, a0 ui .0/ C b0 u0i .0/ D 0:

23 Nonlinear Sturm–Liouville Theory

187

If a0 D 0, it’s obvious that w.0/ D 0. But if a0 6D 0 we obtain the same conclusion because we can then write ui .0/ D

b0 0 u .0/ a0 i

and therefore w.0/ D

b0 0 b0 u .0/u02 .0/  u01 .0/ u02 .0/ D 0: a0 1 a0 t u

In the proof of Lemma 23.4, we showed that if u1 and u2 are any functions in C 1 Œ0;  that satisfy (sBC), then W .u1 ; u2 /.0/ D 0. In fact, the proof only used the first half of (sBC). From the other half, that is, a1 ui . / C b1 u0i . / D 0, the same argument tells us that W .u1 ; u2 /. / D 0 as well. We’ll need both of these consequences of (sBC) quite soon. In Chap. 21, the multiplicity of an eigenvalue  of T was defined to be the dimension of N.I  T /n , where n is large enough so that N.I  T /n D N.I  T /nC1 . Theorem 23.5. Every eigenvalue  of T D L1 a is of multiplicity one. Proof. Suppose N.I  T /2 ¤ N.I  T /; then there exists w 2 N.I  T /2  N.I  T / with .I  T /w D x 2 N.I  T / where x ¤ 0. Define a linear operator W C 2 Œ0;  ! C Œ0;  by setting u D Lu  au D .pu0 /0 C qu 

1 au 

and observe that u D 0 if and only if u 2 N.I  T /. Now L.I  T /w D Lx D

1 ax 

so w D

1 ax: 

For u; v 2 C Œ0; , there is an “inner product,” written .u; v/, defined by Z .u; v/ D

u.s/v.s/ds: 0

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23 Nonlinear Sturm–Liouville Theory

The main step of this proof is to show that .u; v/ D .u; v/ for all u; v 2 C 2 Œ0;  that satisfy (sBC). To prove it, we just write from the definitions: Z u.s/v.s/  u.s/v.s/ds .u; v/  .u; v/ D Z

0

D

Œ.p.s/u0 .s//0 C q.s/u.s/  a.s/u.s/v.s/

0

 u.s/Œp.s/v0 .s//0 C q.s/v.s/  a.s/v.s/ds Z D  Œ.p.s/u0 .s//0 v.s/  .p.s/v0 .s//0 u.s/ds 0

then integrate by parts ˇ ˇ .u; v/  .u; v/ D  .p.s/u0 .s/v.s/  p.s/v0 .s/u.s//ˇˇ 0

Z

C

Œ.p.s/u0 .s//v0 .s/  .p.s/v0 .s//u0 .s/ds

0

ˇ ˇ D p.s/.u .s/v.s/  v .s/u.s//ˇˇ 0

0

ˇ ˇ D p.s/W .u; v/.s/ˇˇ

0

0

D p. /W .u; v/. / C p.0/W .u; v/.0/ D 0: In particular, .w; x/ D .w; x/ D .w; 0/ D 0: But, on the other hand, 1 1 .w; x/ D . ax; x/ D  

Z

ax.s/x.s/ ds > 0 0

because x ¤ 0. This contradiction proves that N.I  T /2 D N.I  T / and therefore  is of multiplicity one. t u Thus the Krasnoselskii–Rabinowitz bifurcation theorem produces a bifurcation at every eigenvalue of T D L1 a because they are all of odd multiplicity, namely, multiplicity one. But the bifurcation theorem tells us quite a bit more. In order

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to obtain a convenient statement for the application of the bifurcation theorem in this setting, I’ll write the linearized form of the nonlinear Sturm–Liouville problem Lu D F (sBC) as Lu D au (sBC) and suppose that  is an eigenvalue of this problem, by which I just mean there is a solution .; u/ to it, with u 6D 0, for that . If we rewrite Lu D au in the form 1 u D L1 au D T u  we see that 1 2  .T /. So looking back to the statement of Theorem 22.8, here’s what the Krasnoselskii–Rabinowitz theorem tells us about nonlinear Sturm– Liouville eigenvalue problems. Theorem 23.6. Let Lu D F .s; u; u0 ; / .sBC / be a nonlinear Sturm–Liouville eigenvalue problem where Lu D .pu0 /0 C qu is invertible with respect to the given separated boundary conditions (sBC). If  is an eigenvalue of the linearized form of the problem Lu D au .sBC / then  is a bifurcation point of the nonlinear problem and there exists a closed connected set C of solutions to that problem containing .; 0/ such that either C is unbounded in R  X or C contains . ; 0/ for some bifurcation point  6D .

Chapter 24

More Sturm–Liouville Theory

In Theorem 23.6, the bifurcation theorem for nonlinear Sturm–Liouville eigenvalue problems Lu D F (sBC) that concluded the previous chapter, a key hypothesis was the invertibility of Lu D .pu0 /0 C qu with respect to the given boundary conditions. Certainly a necessary condition for an operator to be invertible is that it be one-to-one. The nonlinear Sturm–Liouville eigenvalue problem that arises from Euler’s theory of the buckling of columns is u00 D  sin u u0 .0/ D u0 . / D 0: In this problem, Lu D u00 and the alert reader may have been bothered already by the fact that, with respect to the given (Neumann) boundary condition, the operator L W C02 Œ0;  ! C Œ0;  couldn’t possibly have an inverse since Lu D 0 for any constant function u. The way around this problem is to define operators L W C02 Œ0;  ! C Œ0; , for small  > 0, by L u D u00  u: W The operator L does have an inverse; see Ritger and Rose (1968). Define L1  C Œ0;  ! C02 Œ0;  by L1  v.t / D

Z

G.s; t /v.s/ds 0

where G is the Green’s function defined by ( G.s; t / D

1p  sin.  / p 1p  sin.  / p

p p cos. .t  // cos. s/; for 0  s  t  p p cos. t/ cos. .s  //; for 0  t  s  :

© Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__24

191

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24 More Sturm–Liouville Theory

Writing y D L1  v in the form Z t p p 1 cos. s/v.s/ ds y.t / D p Œcos. .t  // p  sin.  / 0 Z p p cos. .s  //v.s/ ds C cos. t/ t

we can see that Z t p p 1 cos. s/v.s/ ds y .t / D p Œsin. .t  // p  sin.  / 0 Z p p cos. .s  //v.s/ ds C sin. t/ 0

t

and it is evident that y 0 .0/ D y 0 . / D 0. Another differentiation shows us that y 00 .t / D y.t /  v.t /, so L y D v as required. Thus the form of the Krasnoselskii–Rabinowitz bifurcation theorem that I obtained as Theorem 23.6 will apply to the following modified Euler buckling problem: L u D u00  u D  sin u u0 .0/ D u0 . / D 0: We will see that the conclusions that we are able to draw about this problem by using Theorem 23.6 will extend to the actual Euler buckling problem as well. Since only the right-hand side of the differential equation is used in calculating the Frechet derivative, the linearized form of the modified Euler buckling problem is u00  u D u u0 .0/ D u0 . / D 0: Writing the differential equation more briefly as u00  . C /u D 0 the theory of second-order linear differential equations from Theorems 23.2 and 23.3 tells us that the general solution is p p u.s/ D c1 sin.  C s/ C c2 cos.  C s/: Since p p p p u0 .s/ D c1  C  cos.  C s/  c2  C  sin.  C s/

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then, from the Neumann conditions, u0 .0/ D 0 impliespthat c1 D 0 whereas up0 . / D 0 tell us that a nonzero solution exists only if sin.  C  / D 0. Setting  C  D k for k an integer, we see that, for all  > 0, the eigenvalues of the linearized problem are of the form  C  D k 2 , and the corresponding eigenvectors are the functions uk .s/ D cos.ks/. Before going on, let’s look more closely at the eigenfunction uk .s/ D cos.ks/. Its zeros in the interval Œ0;  are

3 .2k  1/ ; ;  ; : 2k 2k 2k Let’s notice that there are precisely k of them. Furthermore, all the zeros are simple, that is, the derivative of uk is nonzero at those points, because u0k .

r r / D k sin.k / D ˙k 6D 0 2k 2k

for r an odd integer. Theorem 23.6 tells us that the modified Euler buckling problem described by the differential equation L D  sin u has branches of nontrivial solutions Ck at the bifurcation points k D k 2  . We also found out in Theorem 23.6 that either Ck is unbounded in R  X or it contains another bifurcation point k  . I’ll show you that the second alternative, that Ck connects two bifurcation points, cannot in fact occur. Therefore, the branches of solutions from bifurcation points, that is, the components of the set of nontrivial solutions that contain bifurcation points in their closures, are unbounded. I will prove that this is true for any nonlinear Sturm– Liouville eigenvalue problem, but it has an especially graphic consequence for the Euler buckling problem, as you’ll see later on. The linearized form of the modified Euler buckling problem, u00  u D u u0 .0/ D u0 . / D 0 has eigenvalues k D k 2   and eigenfunctions uk .s/ D cos.ks/ with only simple zeros and exactly k zeros on the interval .0; /. The classical, linear Sturm–Liouville eigenvalue theory that I’ll now present, based on the exposition in Coddington and Levenson (1955), will tell us that this structure is characteristic of all such problems. I’ll begin that theory by writing the linear problem of Theorem 23.6 in this way: .pu0 /0 C .q  r/u D 0 cos ˛ u.0/  p.0/ sin ˛ u0 .0/ D 0 cos ˇ u. /  p. / sin ˇ u0 . / D 0:

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The hypotheses of Sturm–Liouville theory are that p; q and r are continuous on Œ0;  and p; q > 0, that is, that p.s/ > 0 and q.s/ > 0 for all s in Œ0; . When possible, I’ll use notation like p > 0 that avoids writing the independent variable. When an inequality holds only on a subinterval of Œ0; , I’ll tell you what the subinterval is. By multiplying the original boundary conditions, such as a0 u.0/ C b0 u.0/ D 0, by 1 if necessary, I can see to it that 0  ˛ < and 1 0 < ˇ  . If we choose the initial condition u.0/ D sin ˛; u0 .0/ D p.0/ cos ˛, then the first boundary condition is satisfied. Since the differential equation is linear, basic results from the theory of ordinary differential equations that I described in the previous chapter tell us that, for any  2 R, there is a unique solution u W Œ0;  ! R 1 to it that has the property u .0/ D sin ˛; u0 .0/ D p.0/ cos ˛. We want to find the values of  for which u will also satisfy the boundary condition at s D . Those numbers  are called the eigenvalues of the linear Sturm–Liouville problem and the nonzero functions u the corresponding eigenfunctions. The first step in finding the eigenvalues is to understand how u changes as we change . As  increases, let’s think of q  r as a family of functions that increases along with it. The form of q  r doesn’t matter for now, so I’ll simplify notation by replacing it by f and then our differential equation just looks like this, .pu0 /0 C f u D 0, and we’ll study how changes in f affect its solutions. Setting v D pu0 , the differential equation gives us v0 D f u so that, when we choose the initial condition u.0/ D sin ˛ and v.0/ D cos ˛, there is a unique solution .u; v/ to the linear first-order system. Next we’ll introduce polar coordinates: u D  sin and v D  cos . Then 2 D u2 C v2 D u2 C .pu0 /2 which vanishes at some value s D s0 only if u.s0 / D u0 .s0 / D 0. But uniqueness would then imply that u D 0, contrary to the conditions at s D 0. So  > 0 and u.s0 / D 0 can only occur when .s0 / is an integer multiple of . Differentiating the polar coordinate equations, we have these equations r0 sin C .cos / 0 D u0 D

v 1 D  cos p p

0 cos  .sin / 0 D v0 D f u D f sin : When we solve for 0 and 0 , here’s what we get 0 D . 0 D

1  f / sin cos p

1 cos2 C f sin2 : p

. /

I gave the second equation the name (*) because I’ll need to refer to it several times later on. The function reflects changes in f , as the following result shows us.

24 More Sturm–Liouville Theory

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Lemma 24.1. Suppose, for i D 1; 2, that ui D i sin i satisfy .pi u0i /0 C fi ui D 0 where p1  p2 > 0 and u1 .0/ D u2 .0/. If f2 > f1 on .0; , then 2 > 1 on the same interval. Proof. Setting D 2  1 , our first task will be to prove that  0. From (*) we get 0 D .f1 

1 1 1 /.sin2 2  sin2 1 / C .f2  f1 / sin2 2 C .  / cos2 2 : p1 p2 p1

(**)

When ¤ 0, we can set h D .f1 

  1 sin 2  sin 1 / /.sin 2 C sin 1 / p1

and then (**) implies that 0  h  0. The function h is nice enough so that Z H.s/ D

h.t/ dt s

is well defined and it has the property H 0 D h. Now .e H /0 D e H . 0  h /  0 which tells us that e H is a nondecreasing function. In particular, e H.s/ .s/  e H.0/ .0/ D 0 which implies  0. But the lemma requires more: that > 0 on .0; . Suppose that isn’t so, that is, there exists s0 > 0 such that .s0 / D 0. I claim that there must exist c > 0 such that D 0 on Œ0; c. If there were no such interval, that would mean there is a sequence fsn g in Œ0;  converging to zero with all .sn / > 0. Take some n so that sn < s0 ; then, since we showed that e H is a nondecreasing function, we have e H.s0 / .s0 /  e H.sn / .sn / > 0 which contradicts the property of s0 . But the existence of the interval Œ0; c on which vanishes also leads to a contradiction. The reason is that of course 0 D 0 on that interval so from (**) we get 0 D .f2  f1 / sin2 2 C .

1 1  / cos2 2 D 0 p2 p1

and, since f2 f1 > 0 and p12  p11  0 by hypothesis, both summands must be zero. Therefore on Œ0; c we have that 2 D k for some integer k and also p1 D p2 . The contradiction now comes from applying (*) to 2 . For the interval Œ0; c, the lefthand side of (*) is the derivative of a constant function but, when 2 D k , the right-hand side reduces to p12 , which is positive by hypothesis. t u

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Now let’s return to the Sturm–Liouville differential equation .pu0 /0 C 1 .q  r/ D 0 with the initial conditions u.0/ D sin ˛; u0 .0/ D p.0/ cos ˛. For each  2 R, we’ll call the unique solution to that initial value problem u D  sin  1 where  .0/ D ˛  0. Then (*) implies that 0 .s/ D p.s/ > 0 whenever  .s/ D k for some integer k, so  > 0 for all . Just as we did before, we’ll set v D pu0 D  cos  . The boundary condition that u must satisfy at s D in order for  to be an eigenvalue can be written cos ˇ u . /  sin ˇ v . / D 0. That’s the reason the boundary conditions were written in the form that they were. When we substitute to get this equation cos ˇ  sin  . / C sin ˇ  cos  . / D 0 and remember that  > 0, we see that it’s satisfied only if  . / D ˇ C k for some integer k. We arranged it that ˇ > 0 so, since  > 0, the only values of  that could be eigenvalues are those, which we’ll call k , for which k D ˇ C k for the integers k  0. In Theorem 24.4 we’ll show that these k , for all k  0, are the eigenvalues of the linear Sturm–Liouville problem. The function W R  Œ0;  ! R defined by .; s/ D  .s/ is continuous. For each c 2 Œ0; , define c W R ! R by c ./ D .; c/. Lemma 24.1 implies that each c , for c > 0, is a monotone increasing function. The proof of Theorem 24.4 depends on finding out a lot more about these functions c , and that’s what we’ll do in the next two lemmas. Lemma 24.2. For any c 2 .0; , the function c W R ! R has the property lim c ./ D 1:

!1

Proof. We are assuming that, in the differential equation .pu0 /0 C .q  r/ D 0, we have p; q > 0. Choose positive numbers P; Q and R such that p.s/  P and q.s/ > Q and r.s/  R for all s 2 Œ0; . The general solution to the constant coefficient linear differential equation P u00 C .Q  R/ D 0 is r u D k1 sin

! ! r Q  R Q  R s C k2 cos s : P P

Taking any c > 0, define uO  to be the unique solution to the constant coefficient equation subject to the initial conditions c c uO  . / D u . /; 2 2

c p.0/ 0 c uO 0 . / D u . /: 2 P  2

On the interval Πc2 ; c, the number n./ of zeros of uO  is approximately

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197

c 2

r

Q  R P

and therefore lim!1 n./ D 1. Writing the solution in polar coordinates as uO  D O sin O and remembering what I pointed out to you about zeros of solutions just before the proof of Lemma 24.1, we can see that there are n./ values s on Œ c2 ; c for which O .s/ in an integer multiple of . Equation (*) tells us that O .s/ D P1 > 0 at those values so n./ is the number of integers k for which O .s/ D k has a solution for some s 2 Œ c2 ; c. Keeping in mind that O . c2 /  0, we can see that lim!1 O .c/ D 1: Letting p1 D P be the constant function, p2 D p; f1 D Q  R and f2 D q  r, the hypotheses of Lemma 24.1 are satisfied (replacing the interval Œ0;  by Œ c2 ; c) so, in particular, we can conclude that O .c/ <  .c/ D c ./ to complete the argument. t u Lemma 24.3. For any c 2 .0; , the function c W R ! R has the property lim c ./ D 0:

!1

Proof. We know that c ./ D  .c/ > 0 so what we need to do is prove that, given  > 0, there exists 0 < 0 such that  < 0 implies c ./ < . Remembering that 0  ˛ < , we’ll make sure that  is small enough so that ˛ <  . Since p; q > 0, we can find positive numbers P and Q such that P < p.s/ and Q < q.s/ for all s in Œ0; . If   .s/   , then 0 < sin2   sin2 .s/ so from (*) we get 0 .s/
, then, by the mean value theorem, there would exist s0 in .0; c/ with 0 .s0 / D

 .c/   .0/

> c c

contrary to the definition of 0 . Theorem 24.4. The linear Sturm–Liouville eigenvalue problem .pu0 /0 C .q  r/u D 0 cos ˛ u.0/  p.0/ sin ˛ u0 .0/ D 0 cos ˇ u. /  p. / sin ˇ u0 . / D 0;

t u

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where 0  ˛ < and 0 < ˇ  , has a nonzero solution if and only if  D k from a sequence 0 < 1 < 2 <    < k < : : : with limk!1 k D 1. Moreover, if uk is a nonzero solution corresponding to k , then uk has exactly k zeros on .0; / and all of the zeros of uk are simple. Proof. If a solution u to .pu0 /0 C.q r/u D 0 had a non-simple zero at s0 2 Œ0; , then u.s0 / D u0 .s0 / D 0 which, by the uniqueness of solutions, would imply u D 0. For  2 R, let u D  sin  be a solution to the differential equation with the initial 1 condition u.0/ D sin ˛; u0 .0/ D p.0/ cos ˛ so we can take  .0/ D ˛. Defining W R ! R by ./ D  . /, Lemmas 24.1 through 24.3 tell us that is a monotone increasing function taking R onto .0; 1/. Noting that ˇ > 0, we see that k D 1 .ˇ C k / exists for all integers k  0 and k satisfies the condition for an eigenvalue that we discussed after the proof of Lemma 24.1. Let uk D k sin k 1 be the eigenfunction with uk .0/ D ˛; u0k .0/ D p.0/ cos ˛ corresponding to k . Then k has the properties k .0/ D ˛ < and k . / D ˇ C k > k , so, for each j D 1; : : : ; k, the equation k .s/ D j has exactly one solution on .0; /. Therefore uk has exactly k zeros on that interval. t u Our next task will be to prove that the solutions to nonlinear Sturm–Liouville eigenvalue problems exhibit the same behavior as the solutions to the linear problem described by Theorem 24.4. So we’re back to talking about nonlinear Sturm–Liouville eigenvalue problems Lu D F .s; u; u0 ; / .sBC / where Lu D .pu0 /0 C qu and about the linearized version Lu D au with the same boundary conditions. Recall that X D C01 Œ0;  where the subscript indicates that the functions in X must satisfy the separated boundary conditions a0 u.0/ C b0 u0 .0/ D 0 a1 u. / C b1 u0 . / D 0 with .a02 C b02 /.a12 C b12 / 6D 0: The norm on X is the C 1 norm obtained from the sup norm; that is, kuk1 D kuk C ku0 k:

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199

Jj Sj

0

π

Fig. 24.1 The monotone decreasing case

Define subsets Sk D fu 2 X W u has k zeros on .0; /; all the zeros of u are simpleg: Notice that the functions in Sk may have zeros at 0 or or both, in addition to the k zeros on the open interval. It is important that these additional zeros are required to be simple, just as the zeros on .0; / are. Theorem 24.4 states that in a linear Sturm–Liouville problem, the eigenfunction uk corresponding to the eigenvalue k is in Sk . The first step in obtaining the related result for nonlinear Sturm–Liouville problems is to find out more about the sets Sk , and that’s the purpose of the next two lemmas. Lemma 24.5. The set Sk is open in X . Proof. For u 2 Sk , let 1 be the minimum value of ju0 .sj /j for all zeros sj of u; then of course 1 > 0 by the definition of Sk . Take disjoint open intervals Jj in Œ0;  about the sj such that, for each j , for all s 2 J j either u0 .s/  21 , and u is monotone increasing there, or u0 .s/   21 and u is monotone decreasing. See Fig. 24.1. Deleting the Jj leaves a closed subset of Œ0;  on which u doesn’t vanish, so let 2 be the minimum of juj on that set. Letting  be the smaller of 21 and 22 , we claim that if v 2 X with ku  vk1 < , then v 2 Sk . Since ju.s/  v.s/j < , it’s clear that v.s/ cannot have any zeros outside the Jj and in fact u.s/ and v.s/ are either both positive or both negative. Furthermore, ju0 .s/  v0 .s/j <  forces v to be monotone increasing on Jj if u is and monotone decreasing there if u is monotone decreasing on the interval. Since for each zero on .0; / the function u changes sign, the same is true for v and it therefore has exactly one zero on each corresponding Jj . Now if u has a zero at s D 0 so also has v. The reason is that u 2 Sk and u.0/ D 0 would imply u0 .0/ 6D 0 since all zeros are simple. Since u is in X it must satisfy a0 u.0/ C b0 u0 .0/ D 0

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where a02 C b02 6D 0 so in the definition of X it must be that b0 D 0 and a0 6D 0. If v.0/ 6D 0, we would have a0 v.0/ C b0 v0 .0/ D a0 v.0/ 6D 0 which contradicts the assumption that v 2 X . Since v is still monotone near 0, there cannot be another zero in the interval Jj that contains 0. For the same reason, v. / D 0 if and only if u. / D 0, so v has exactly k zeros on .0; /, and thus v 2 Sk . t u Lemma 24.6. If u 2 X such that u 2 @Sk , then u has a nonsimple zero. Proof. If u had a finite set of zeros, all of them simple, it would belong to some Sm , which is open, and thus u 62 @Sk . If u had only simple zeros but there were an infinite number of them, the set of zeros would contain a subsequence that converges to a zero s0 , but then u0 .s0 / D 0 and s0 is a nonsimple zero. t u The only solution to a linear Sturm–Liouville problem that has a nonsimple zero is the trivial solution because of the uniqueness of solutions to linear initial value problems. The next result extends the same conclusion to solutions of nonlinear Sturm–Liouville problems. Lemma 24.7. If .; u/ is a solution to the nonlinear Sturm–Liouville problem Lu D F .s; u; u0 ; / .sBC / and u has a nonsimple zero, then u is the zero function. Proof. Since part of the hypotheses of a Sturm–Liouville problem is that F is Frechet differentiable at 0, we make use of that property when we define H W Œ0;   R3 ! R by H.s; ; ; / D F .s; ; ; /  a.s/ to conclude that jH.s; ; ; /j p  2 C 2 can be made arbitrarily small, for  2 Œ1 ; 2 , by choosing .; / close enough to .0; 0/. It follows from that property that the linear differential equation Lv D av C

H.s; w; w0 ; /w0 0 H.s; w; w0 ; /w vC v 02 2 w Cw w2 C w0 2

(DE)

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201

is well defined at w D 0. We know that no nontrivial solution to (DE) has a nonsimple zero. Now suppose u is a solution to the given nonlinear Sturm–Liouville problem, which can be written in the form Lu D au C H.s; u; u0 ; /: Rewrite that differential equation in the following sneaky way: Lu D au C

H.s; u; u0 ; /u0 0 H.s; u; u0 ; /u uC u: 02 2 u Cu u2 C u0 2

It now has the form of (DE) for the case w D u. Thus u is also a solution to a form of the linear problem and consequently has no nonsimple zeros. t u We continue to use the symbol S for the set of solutions to the nonlinear Sturm– Liouville problem Lu D F .s; u; u0 ; /, (sBC). Notice that the following is the first lemma that requires the hypothesis that L is invertible with respect to the boundary conditions of the problem. Lemma 24.8. Given that L is invertible with respect to the boundary conditions of the problem, there exists a neighborhood Nk of the bifurcation point .k ; 0/ in R  X such that if .; u/ 2 S \ Nk and u 6D 0, then u 2 Sk . Proof. Suppose there is no such neighborhood. That means there exists a sequence f.˛n ; un /g in S , converging to .k ; 0/, with the un nontrivial, for which un 62 Sk for all n. Writing the nonlinear problem using the function H that we introduced in the proof of the previous lemma, for each n, we have Lun D ˛n aun C H.s; un ; u0n ; ˛n / which we can rewrite, employing the linearity of L1 , in the form un L1 H.s; un ; u0n ; ˛n / un D ˛n L1 a C : kun k1 kun k1 kun k1

(&)

Recalling that the norm on X is the C 1 norm, let’s notice that u.s/2 C u0 .s/2  . sup ju.s/j/2 C . sup ju0 .s/j/2 s2Œ0; 

s2Œ0; 

 Œ sup ju.s/j C sup ju0 .s/j2 D kuk21 : s2Œ0; 

s2Œ0; 

Then from the definition of the norm of a bounded linear operator in Chap. 20, we can write for each value of s that

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24 More Sturm–Liouville Theory

jL1 H.s; un ; u0n ; ˛n /j jH.s; un ; u0n ; ˛n /j  kL1 k kun k1 kun k1 jH.s; un ; u0n ; ˛n /j  kL1 k p un .s/2 C u0n .s/2 and we recall from the proof of Lemma 24.7 that the definition of the Frechet derivative implies that the right-hand term goes to zero as the sequence fun g goes to 0. Therefore, the right-hand term of (&) also approaches 0 as n goes to infinity. Since L1 a D T was the Frechet derivative of a completely continuous function, by Theorem 22.1, it is a compact linear operator and therefore, because the sequence f kuunnk1 g is bounded, we have lim L1 a

n!1

un D w 2 X: kun k1

Now we use the fact that f˛n g converges to k to conclude that lim

n!1

un D k w: kun k1

Set v D k w; then by the continuity of L1 lim L1 a

n!1

un un D L1 a lim D L1 av: n!1 kun k1 kun k1

We have computed the same limit in two ways, and we find that L1 av D lim L1 a n!1

un v DwD kun k1 k

which we rewrite in the form Lv D k av: We can conclude from this that v is an eigenfunction for the linear eigenvalue problem Lu D au and therefore v 2 Sk . Since Sk is open in X by Lemma 24.5, it certainly must be that kuunnk1 2 Sk for n sufficiently large. But division by the constant kun k1 cannot change the set of zeros of un , so un 2 Sk for n sufficiently large, contrary to what we assumed at the start of the proof. Therefore, the neighborhood Nk does exist. t u Now we’re ready to prove the result for nonlinear Sturm–Liouville problems that corresponds to Theorem 24.4 and, happily, just about all the work has already been done.

24 More Sturm–Liouville Theory

203

Theorem 24.9. Let Lu D F .s; u; u0 ; / .sBC / be a nonlinear Sturm–Liouville eigenvalue problem where Lu D .pu0 /0 C qu is invertible with respect to the given separated boundary conditions. Let k be the eigenvalue of the linearized form of the problem Lu D au .sBC / such that the corresponding eigenfunction has exactly k zeros on .0; /. If Ck is a branch of nontrivial solutions to the nonlinear problem containing .k ; 0/ then .; u/ 2 Ck with u 6D 0 implies that u 2 Sk ; that is, u has exactly k zeros on .0; /, and all the zeros of u are simple. It follows that Ck is unbounded. Proof. Lemma 24.8 assures us that Ck must intersect R  Sk . We’ll assume that not all nonzero u are in Sk for .; u/ 2 Ck and seek a contradiction. Recalling that Sk is open in X , the connectedness of Ck implies that there must be . ; u / 2 Ck where u is in @Sk . Lemmas 24.6 and 24.7 then tell us that u D 0 which makes  a bifurcation point. But, by Theorem 22.4, the bifurcation points must be eigenvalues of the linearized problem, so  D m for some m by Theorem 24.4. Lemma 24.8 states that there is a neighborhood Nm of . ; 0/ for which all solutions .; u/ with u 6D 0 have the property that u 2 Sm lie in R  Sm with m 6D k. However, . ; u / 2 R  @Sk would make Nm intersect R  Sk which establishes a contradiction. Now that we’ve established that .; u/ 2 Ck with u 6D 0 implies u 2 Sk , we understand why Ck cannot connect two eigenvalues, say k and m : because u cannot have both exactly k zeros on .0; / and also exactly m 6D k zeros. The Krasnoselskii–Rabinowitz bifurcation theorem (Theorem 23.6) then tells us that the Ck are unbounded. t u

Chapter 25

Euler Buckling

The Euler buckling problem is u00 D  sin u .E/ u0 .0/ D u0 . / D 0: Even though Lu D u00 isn’t invertible with respect to the given boundary condition, we can apply Theorem 24.9 to the modified problem u00  u D  sin u .E / u0 .0/ D u0 . / D 0 to prove Theorem 25.1. Let X D C01 Œ0; , the continuously differentiable functions that satisfy the boundary conditions. For each k D 1; 2; : : : , the integer k 2 is a bifurcation point for the Euler buckling problem (E) and the branch Ck of nontrivial solutions containing .k 2 ; 0/ is an unbounded subset of RX such that if .; u/ 2 Ck with u 6D 0, then u 2 Sk . That is, u has k zeros on .0; / and all the zeros are simple. Proof. Denote the component of solutions to .E / containing .k 2  ; 0/ by Ck . Let k be the limit set of the Ck for a given k; that is, it is the set of limits of all convergent sequences f. ; u /g where . ; u / 2 Ck and  ! 0. See Fig. 25.1. The first claim is that the elements of .; u/ 2 k are solutions to .E/. Since .; u/ 2 X , and therefore the boundary conditions are satisfied, what we have to show is that u00 D  sin u. We are assuming that .; u/ is the limit of a convergent, and therefore bounded, sequence f. ; u /g of solutions to .E /. Thus, we know that © Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2__25

205

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25 Euler Buckling

Fig. 25.1 Components as  goes to zero

u 00   sin u D u and also that there exists M > 0 such that ku k < M . We see that k  u 00   sin u k  M for all , so it must be that k  u00   sin uk D 0 and .; u/ does indeed solve .E/. It is clear that .k 2 ; 0/ 2 k , since it’s the limit of f.k 2  ; 0/g 2 Ck , but k contains much more; in fact it is an unbounded subset of R  X . To see why, 1=n let M  1 be an integer and for each positive integer n choose .n ; un / 2 Ck 1=n with k.n ; un /k1 D M , which exists because Ck is unbounded and connected. The Ascoli–Arzela theorem (2.3) tells us that there is a subsequence of f.n ; un /g (continue to denote the subscripts by n) that converges in the sup norm to .; u/ 2 X which, as we already have seen, is a solution to u00 D  sin u. We can make 1 ku00  un 00 k D k   sin u  . un  n sin un /k n arbitrarily small because the sequence fun g is bounded and the sine function is continuous. Thus, after integrating, we can have ku0  u0n k as small as we wish, and therefore we establish that f.n ; un /g converges in the C 1 norm to .; u/ 2 k . Since it is evident that k.; u/k1 D M , we now know that k is unbounded. Next, we’ll see that there has to be an unbounded component of k that contains .k 2 ; 0/. If that were not so, we could write k D .U \ k / [ .V \ k / where U and V are open subsets of R  X . Since the Ck are connected and unbounded, 1=n

1=n

1=n

.U \ Ck / [ .V \ Ck / 6D Ck

25 Euler Buckling

207

and we can find a bounded, and therefore convergent, sequence f.n ; un /g converging to .; u/ 2 k . But since the sequence lies in the closed set R  X  .U [ V /, so does its limit. Now I’ll use some of the lemmas we proved earlier to show you that if .; u/ 2 k with u 6D 0, then u 2 Sk . Theorem 24.9 states that if . ; u / 2 Ck with u 6D 0, then u 2 Sk , so since u is a limit, it is in the closure of Sk . I made a point of the fact that Lemmas 24.6 and 24.7 do not require the operator L to be invertible, so we can use them to conclude that u cannot be in @Sk . Now we turn at last to Ck itself. Clearly Ck contains that component of k that includes .k 2 ; 0/, so Ck is unbounded. Furthermore, since there are .; u/ 2 Ck with u 2 Sk , the connectedness of Ck together with Lemmas 24.6 and 24.7 imply that if .; u/ 2 Ck with u 6D 0 then u 2 Sk . t u Although Ck is unbounded in R  X , its behavior is severely constrained, as the following result demonstrates. Lemma 25.2. If .; u/ 2 Ck for any k  1, then kuk1  . C 1/jj: Proof. The functions u are in X D C01 Œ0; ; that is, they are continuously differentiable functions u W Œ0;  ! R such that u0 .0/ D u0 . / D 0. Let  2 .0; / be such that ju0 .s/j attains its maximum on Œ0;  at s D . Now we make use of the boundary condition u0 .0/ D 0 along with the fact that .; u/ is a solution to Euler’s differential equation to put a bound on ju0 .s/j, as follows: ˇZ ˇ ju0 .s/j  ju0 ./j D ˇˇ Z



 0

ˇ Z ˇ u00 . /d ˇˇ  Z

00

ju . /jd D

0



ju00 . /jd

0

j sin u. /jd  jj :

0

From Theorem 25.1, we know that u has k  1 zeros on .0; /, so choose one of them; call it ˛. Let ju.s/j reach its maximum at  2 Œ0;  and assume, without loss of generality, that ˛ < ; then using what we just found out about ju0 .s/j, we calculate that ˇ Z ˇZ ˇ ˇ   ˇ ˇ u0 . /d ˇ  ju0 . /jd ju.s/j  ju./  u.˛/j D ˇ ˇ ˇ ˛ ˛ Z Z ju0 . /jd  jj d D jj 2 :  0

0

The result follows by the definition of the C 1 norm.

t u

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25 Euler Buckling

Fig. 25.2 Bifurcation in the linearized problem

In Chap. 23, we defined D

P M

where P was the force applied to the column and M a constant “modulus of flexural rigidity” that depends on the physical characteristics of the column. As Fig. 25.2 suggests, the linearized form of the buckling problem, which was used to analyze the problem prior to Rabinowitz’s work, is an unrealistic model since it shows buckling when P D k 2 M exactly, but no buckling occurs if the force is increased slightly. (The linearized form is not useless, however, since it does imply that bifurcation must occur at the first eigenvalue.) Now we can see that the nonlinear problem exhibits a strikingly different behavior, as follows. Theorem 25.3. If  > k 2  1, then for each j  k, there is a solution .; uj / to the Euler buckling problem u00 D  sin u u0 .0/ D u0 . / D 0 for that value of  such that uj has exactly j zeros on .0; /, all of them simple. Proof. We will prove that each branch Ck contains a solution .; u/ for every  > k 2 and since u has exactly k zeros by Theorem 25.1, this will prove the theorem. See Fig. 25.3. We first establish the fact that if .; u/ 2 Ck with u not the zero function, then  > 0. From the differential equation we see that  D 0 implies u00 .s/ D 0 for all s, so u0 is a constant function. The boundary condition u0 .0/ D 0 then forces u

25 Euler Buckling

209

Fig. 25.3 Nonlinear bifurcation

Fig. 25.4 The solution .s/

to be a constant function, contrary to Theorem 25.1. Since Ck is connected, it must lie entirely in .0; 1/  X ; that is,  > 0. Now suppose there exists b > 0 such that .; u/ 2 Ck implies   b. By Lemma 25.2, we would have to conclude that .; u/ 2 Ck implies  2 Œ0; b and kuk1  . C 1/b, so that Ck is bounded, contrary to Theorem 25.1. t u Looking back to Chap. 23 once more, we should recall that a solution to the nonlinear Sturm–Liouville eigenvalue problem that we have been calling the Euler buckling problem does not describe the shape function y.s/ of the buckled column; instead, the solution is the function we called .s/, which measures the angle between the tangent line at y.s/ and the horizontal. As Fig. 25.4 indicates, a

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25 Euler Buckling

weight

column

ground

P> 9 Fig. 25.5 Buckling for k D 3

knowledge of .s/ does, however, show us what the buckling looks like as well. Thus, Theorem 25.3 tells us that if P > k 2 M, the column can buckle in k distinct ways, as demonstrated by Fig. 25.5 for the case k D 3.

Appendix A

Singular Homology

Sections A.1 and A.2 of this appendix present a good-sized piece of basic algebraic topology, summarized at approximately the speed of sound. It should, therefore, be approached with all the care you would normally use in the vicinity of any object moving at that speed. If you would like to see that reasonably self-contained proof of Brouwer’s fixed point theorem that I promised in Chap. 3, go directly to Sect. A.3, referring back to the previous material as the need arises.

A.1 Construction of Homology Groups Within euclidean space Rq we give names to some useful points, namely the origin, which we call E0 , and the standard vector space basis E1 ; E2 ; : : : ; Eq where Ej D .0; 0; : : : ; 0; 1; 0; : : : ; 0/; with the 1 in the j th location. These points determine a subset of Rq , called the standard q-simplex and denoted by q , which is defined by q D f

q X

j D0

tj Ej W

q X

tj D 1; tj  0g

j D0

Thus 1 is the segment E0 E1 in the real line, 2 is the triangle with vertices E0 ; E1 , and E2 in the plane, 3 is a tetrahedron, and so on (whatever that “so on” may mean, since most of us cannot visualize the rest of the ’s). Letting kq denote the subset Pq of all the points j D0 tj Ej 2 q for which tk D 0 produces copies of q1 in the boundary of q . Specifically, a homeomorphism i k W q1 ! kq can be defined

© Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2

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Fig. A.1 The standard 2-simplex

by sending Ej 2 Rq1 to Ej 2 Rq for j < k and to Ej C1 2 Rq otherwise, and extending linearly. See Fig. A.1. Let X be a topological space and let Sq .X / denote the q-th singular chain group, that is, the free abelian group generated by Pall maps from q to X . Thus the elements of Sq .X / are all the formal sums c D ˛ a˛ ˛ where the ˛ W q ! X are the maps from q to X and the a˛ are integers, all but a finite number of them zero. Let ˛k W kq ! X denote the restriction of ˛ . Given a map ˛ W q ! X , let q X .1/k .˛k i k / 2 Sq1 .X /; @q .˛ / D kD0

Extending linearly produces a homomorphism @q W Sq .X / ! Sq1 .X / for each q  0. If we start with a map  W qC1 ! X and first compose it with @qC1 and then apply @q to the formal sum @qC1 . / 2 Sq .X /, some combinatorial bookkeeping will show that all the integer coefficients cancel, that is, @q @qC1 . / D 0. Therefore I m@qC1 , the image of @qC1 , is a subgroup of the abelian group Ker@q , the kernel of @q . The quotient group Hq .X / D

Ker@q I m@qC1

is called the q-th homology group of X . For a map f W X ! Y between topological spaces and cD

X ˛

a˛ ˛ 2 Sq .X /

A.2 Basic Properties of Homology

213

define f# c D

X

a˛ .f ˛ / 2 Sq .Y /:

˛

Thus f induces a homomorphism f# W Sq .X / ! Sq .Y /. If c is in the kernel of @q W Sq .X / ! Sq1 .X / then it’s not hard to show that f# c is in the kernel of the corresponding homomorphism for the space Y . In that case, sending the equivalence class of Hq .X / containing c to the equivalence class of f# c in Hq .Y / defines a homomorphism f W Hq .X / ! Hq .Y /. The only thing left to do in the way of definitions, for now at least, is to extend homology from a topological space X to a topological pair .X; A/, which just means that X is a topological space and A is a subset of X . Then Sq .A/ is a subgroup of Sq .X / and we let Sq .X; A/ be the quotient group. Certainly @q takes Sq .X; A/ to Sq1 .X; A/ and we may define the q-th homology group of .X; A/ by Hq .X; A/ D

Ker@q W Sq .X; A/ ! Sq1 .X; A/ I m@qC1 W SqC1 .X; A/ ! Sq .X; A/

In the case that A D ;, the empty set, then Hq .X; ;/ may be identified with the homology group Hq .X / defined previously. A map f W .X; A/ ! .Y; B/ between topological pairs is a map f W X ! Y such that f .A/  B. Although there are a number of details you can worry about if you want to, it’s really not hard to show that starting with a formal sum c and defining the sum f# c just as before produces a well-defined homomorphism f W Hq .X; A/ ! Hq .Y; B/.

A.2 Basic Properties of Homology The following collection of theorems about homology groups will be the tools we need to prove Brouwer’s theorem without further waving of hands. Of course all the details can be found in standard algebraic topology texts such as Hatcher (2002) and Spanier (1966). Theorem A.1 (Functorial Property). If f W .X; A/ ! .X; A/ is the identity function, then f W Hq .X; A/ ! Hq .X; A/ is the identity homomorphism. If f W .X; A/ ! .Y; B/, g W .Y; B/ ! .Z; C / and h W .X; A/ ! .Z; C / are maps such that h D gf , then h D g f . For topological pairs .X; A/ and .Y; B/, a homotopy is a map F W .X Œ0; 1; A Œ0; 1/ ! .Y; B/. As usual, a homotopy F determines a family of maps ft for 0  t  1 by ft .x/ D F .x; t /. Theorem A.2 (Homotopy). If F W .X  Œ0; 1; A  Œ0; 1/ ! .Y; B/ is a homotopy, then f0 D f1 .

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For a topological pair .X; A/, we define inclusions i W A ! X and j W .X; ;/ ! .X; A/. Recall that Hq .X; ;/ can be identified with Hq .X /. Theorem A.3 (Exactness). For a topological pair .X; A/, there is a homomorphism @ W Hq .X; A/ ! Hq1 .A/, for each q  0, such that i

j

@

   ! Hq .A/ ! Hq .X / ! Hq .X; A/ ! Hq1 .A/ !    ! H0 .X; A/ ! 0 is an exact sequence of groups and homomorphisms, that is, the image of each homomorphism equals the kernel of the homomorphism to its right. Theorem A.4 (Excision). Let .X; A/ be a topological pair and let U be a subset of X whose closure is contained in the interior of A; then the homomorphism Hq .X  U; A  U / ! Hq .X; A/ induced by inclusion is an isomorphism for all q. Theorem A.5 (Dimension). Let X be the space consisting of a single point, then H0 .X / D Z, the group of integers, and Hq .X / D 0 for all q > 0.

A.3 Homology of Spheres and Brouwer’s Theorem The unit sphere S n consists of the points x D .x1 ; x2 ; : : : ; xnC1 / 2 RnC1 with n n x12 C x22 C    xn2 D 1. Write S n D Bn [ BC where Bn and BC consist of the points n x D .x1 ; x2 ; : : : ; xnC1 / 2 S with xnC1  0 and xnC1  0, respectively. Both subsets are homeomorphic to B n and their intersection is homeomorphic to S n1 . The proof of the Brouwer fixed point theorem depends on a fact about the homology groups of the spheres, namely (Theorem A.7) that Hn .S n / D Z, the integers. That information is also used in Chap. 8 in defining the Brouwer degree. We begin with the case n D 1. Theorem A.6. H1 .S 1 / D Z. Proof. A space X is path-connected if given any x; x 0 2 X , there is a path ! between them, that is, a map ! W I ! X with !.0/ D x and !.1/ D x 0 . The definition of homology is all you need to prove that if X and Y are path-connected spaces and f W X ! Y is a map, then f W H0 .X / ! H0 .Y / is an isomorphism of groups isomorphic to Z. It also shows you that H0 .S 0 / is the free abelian group 1 on two generators and that the kernel of i W H0 .S 0 / ! H0 .BC / is isomorphic 1 1 0 to Z. The exact sequence of .BC ; S / then implies that H1 .BC ; S 0 / D Z. But 1 H1 .BC ; S 0 / D H1 .S 1 ; B1 / by excision and the exact sequence of .S 1 ; B1 / produces an isomorphism between H1 .S 1 ; B1 / and H1 .S 1 /. t u Theorem A.7. Hn .S n / D Z for all n. Proof. A space X is contractible if the identity map on X is homotopic to the constant map that takes X to a point x0 2 X . The homotopy property tells us that a contractible space X has the same homology as that point so,

A.3 Homology of Spheres and Brouwer’s Theorem

215

by the dimension property, it is trivial except for H0 .X /, which is isomorphic to the integers Z. Letting n  2, the exact sequence of .S n ; Bn / produces an n isomorphism Hn .S n / ! Hn .S n ; Bn / and the exact sequence of .BC ; S n1 / n n1 n1 produces an isomorphism Hn .BC ; S / ! Hn1 .S /. Excision gives us an n isomorphism Hn .BC ; S n1 / ! Hn .S n ; Bn /. Therefore Hn .S n / D Hn1 .S n1 / and Theorem A.6 completes the argument. t u Proof of the Brouwer Fixed Point Theorem (B n has the fpp). For x 2 Rn not the origin, let .x/ D

x : jxj

Suppose there were a map f W B n ! B n with no fixed points. Then, for n  2, the following homotopy F W S n1 Œ0; 1 ! S n1 would show that S n1 is contractible, which is impossible by Theorem A.7:  F .x; t / D

.x  2tf .x//; for 0  t  12 ..2  2t /x  f ..2  2t /x//; for 12  t  1:

To see that F is well defined, observe that the definitions agree when t D 12 , that x  2tf .x/ cannot be the origin when t < 12 since jxj D 1 and for t  12 the assumption that f has no fixed points makes  well defined. Thus F would show that the identity map is homotopic to the constant map taking S n1 to .f .0//. The remaining case, n D 1, of the Brouwer theorem is an obvious consequence of the intermediate value theorem. t u

Appendix B

Additivity and Product Properties

We have some unfinished business from Chap. 9: the proof of two of the properties of the Brouwer degree. We will need two more tools from algebraic topology for these proofs. For the first tool, we start with a topological space X ; let U1 and U2 be disjoint open subsets and let A1 and A2 be subsets of the corresponding U ’s. For i D 1; 2, let hi W .Ui ; Ai / ! .U1 [ U2 ; A1 [ A2 / be the inclusions. On the singular chain groups, define h1# ˚ h2# W Sq .U1 ; A1 / ˚ Sq .U2 ; A2 / ! Sq .U1 [ U2 ; A1 [ A2 / by setting .h1# ˚ h2# /.c1 ; c2 / D h1# c1 C h2# c2 . We will also make use of a homomorphism that goes in the opposite direction. To define it, we first observe that since the standard q-simplex q is connected, the image of a map  W q ! U1 [U2 either lies entirely in U1 or entirely in U2 . We then have a homomorphism 1 W Sq .U1 [ U2 ; A1 [ A2 / ! Sq .U1 ; A1 / defined by letting 1 . / D  if  .q /  U1 and 1 . / D 0 otherwise, and extending linearly. Of course there’s a 2 as well. The homomorphism s W Sq .U1 [ U2 ; A1 [ A2 / ! Sq .U1 ; A1 / ˚ Sq .U2 ; A2 / is defined by X X X s. ˛ / D . 1 .˛ /; 2 .˛ //: ˛

˛

˛

© Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2

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B Additivity and Product Properties

It’s easy to check that s is the inverse of h1# ˚ h2# . Furthermore, it’s quite clear from the definitions that @s D s@ (with appropriate dimension subscripts of course) so s produces an isomorphism of homology s W Hq .U1 [ U2 ; A1 [ A2 / ! Hq .U1 ; A1 / ˚ Hq .U2 ; A2 /: Now what’s really interesting is the inverse of this, which is clearly h1 ˚ h2 . Since ˚ was defined to be the group operation, what this means, in effect, is that the disjoint union of subsets becomes, at the level of homology, the group operation. The other tool arises from a space X and an equivalence relation  defined on X . The symbol X= will represent the set of equivalence classes made into a topological space by the quotient topology. All this means is that there is the obvious projection of X onto the set X= and a set in X= is open if and only if its inverse image under the projection is open in X . An important special case occurs when all but one of the equivalence classes are degenerate, that is, consist of a single point of X . Thus the classes consist of a subset A of X , the nondegenerate equivalence class, and each point of X not in A. In this case, we will write X= as X=A. For instance, let Ip be the cartesian product of p copies of the unit interval Œ0; 1 and let @Ip be its boundary. Thus @Ip consists of all p-tuples .x1 ; x2 ; : : : ; xp / of numbers between 0 and 1 such that at least one of the xj is either 0 or 1. The quotient space Ip =@Ip is homeomorphic to the sphere S p . The collapsing map W S p  S q ! S pCq lends itself to a description using quotient spaces. Define an equivalence relation  on IpCq D Ip  Iq by representing points as pairs .x; y/, where x 2 Ip and y 2 Iq , and setting .x; y/.x 0 ; y 0 / if and only if one of the following is true: (1) both x and x 0 are in @Ip and y D y 0 or (2) x D x 0 and both y and y 0 are in @Iq . See Fig. B.1. Since the nondegenerate equivalence classes are of the form @Ip  fyg and fxg  @Iq , we see that IpCq = is homeomorphic to S p  S q . On the other hand, all those classes are certainly subsets of @IpCq , so we can define

W IpCq = ! IpCq =@IpCq D S pCq by identifying all of those nondegenerate equivalence classes inside the (larger) single nondegenerate class @IpCq . Another way to view S p D Ip =@Ip is by noting that since the set of degenerate equivalence classes is the interior of Ip , which is homeomorphic to the euclidean space Rp , we can write S p D Rp [ 1p where the point 1p corresponds to the nondegenerate class @Ip . Writing S p S q D .Rp [1p /.Rq [1q / D RpCq [.1p Rq /[.Rp 1q /[.1p 1q / the map sends RpCq to itself and identifies all of .1p  Rq / [ .Rp  1q / [ .1p  1q / to 1pCq in S pCq D RpCq [ 1pCq :

B Additivity and Product Properties

219

Fig. B.1 The collapsing map

Fig. B.2 The additivity property

Now we are ready to present the proofs of the two properties of the Brouwer degree that we didn’t prove in Chap. 9. We have an open set U in Rn and a map f W U ! Rn such that F D f 1 .0/ is admissible in U , that is, F is compact and disjoint from @U . Theorem 9.6 (Additivity Property). Let U1 and U2 be disjoint open subsets of U such that F  .U1 [ U2 / and let fj denote the restriction of f to Uj ; then deg.f; U / D deg.f1 ; U1 / C deg.f2 ; U2 /: Proof. By Lemma 9.1, we may replace U by U1 [ U2 in the definition of deg.f; U /. Therefore, the left-hand side of the diagram in Fig. B.2 shows n taken to deg.f; U / n . Let Fj D F \ Uj . Except for the map f and its restrictions, all

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Fig. B.3 The additivity subdiagram

the homomorphisms in Fig. B.2 are induced by inclusions. The diagram indicates the definition of 0n 2 Hn .U1 [ U2 ; U1 [ U2  F1 [ F2 /, 01n 2 Hn .U1 ; U1  F1 /, and 02n 2 Hn .U2 ; U2  F2 / and we can see that the right-hand side takes n to deg.f1 ; U1 / n C deg.f2 ; U2 / n . Therefore, in order to prove the additivity property, it is sufficient to prove that the diagram in Fig. B.2 commutes. The top triangle of the diagram commutes because the homomorphisms are inclusion-induced and the bottom diagram commutes by the definition of the fj , so it remains to verify that the subdiagram labelled (?) commutes. For this purpose, consider the diagram in Fig. B.3. The subdiagram labelled (&) commutes because all homomorphisms in it are induced by inclusions. The remaining subdiagram uses the homomorphism s that we just defined. Commutativity can be established directly from the definitions; thus the diagram of Fig. B.3 commutes, which implies that the diagram it induces in the corresponding homology groups also commutes. The commutativity of the diagram in Fig. B.2 now follows from the fact that s is the inverse of h1 ˚ h2 . u t Theorem 9.7 (Product Property). Let f W U ! Rm and g W V ! Rn be maps such that F D f 1 .0/ is admissible in U and G D g 1 .0/ is admissible in V ; then for f  g W U  V ! RmCn , the degree deg.f  g; U  V / is well defined and deg.f  g; U  V / D deg.f; U /deg.g; V / Proof. The degree is well defined because .f  g/1 .0/ D F  G. From Chap. 8, we know that the cross product from Hm .Rm ; Rm  0/ ˝ Hn .Rn ; Rn  0/ to HmCn .RmCn ; RmCn  0/ takes m ˝ n to mCn . But the cross product cannot be quite so accommodating in relating the ’s because it takes Hm .S m / ˝ Hn .S n / into HmCn .S m  S n / rather than into HmCn .S mCn /. That is why we need the collapsing map W S m  S n ! S mCn . We can view F  G as a subset of the part of S m  S n identified with Rm  Rn and, as well, as a subset of RmCn  S mCn , that is, where no collapsing under takes place. Therefore, subdiagram (a) of the diagram Fig. B.4 commutes, where k; k 0 , and k 00 all denote inclusion. Similarly, when we include U  V in Rm  Rn  S m  S n that subset is not collapsed by , so the effect is the same as if we included U  V in S mCn . Consequently, subdiagram (b) also commutes. The commutativity of the entire diagram in Fig. B.4 then follows by the naturality property of the cross product. The homomorphisms down the

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Fig. B.4 The product property

left-hand side of Fig. B.4 take m ˝ n to deg.f; U / m ˝ deg.g; V / n and those along the right-hand side take mCn to deg.f  g; U  V / mCn . Now take a look at Fig.B.5, which is the commutative diagram of Fig. B.4 in the special case that f is the identity map of Rm and g is the identity map of Rn . Since m  n D mCn according to Chap. 8, then Fig. B.5 tells us that

 .m  n / D mCn Therefore, the commutativity of the diagram of Fig.B.4, together with the fact that the cross product preserves integer products, implies that deg.f g; U V / mCn D deg.f; U / m deg.g; V / n D deg.f; U /deg.g; V / mCn which completes the argument.

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Fig. B.5 The homomorphism 

Appendix C

Bounded Linear Operators

Nonlinear analysis studies functions on normed linear spaces and, if all those functions were linear, the subject wouldn’t have that name. But some of the functions we study are linear and functional analysis is the branch of mathematics that can tell us a lot about such linear functions. There is substantial material from functional analysis in Chaps. 15 and 18, so, as background to those chapters and for several other topics, we need the facts about linear operators that I have collected here. Everything in this appendix can be found in textbooks of functional analysis. But those books are long and this appendix is short, so, if you don’t already know these things, you’ll find it a bit easier to locate them here. I’ll assume that you are looking up these facts to support what is in the text and therefore you’ll know from the text what the words and symbols mean. A linear operator T W X ! Y of normed linear spaces, with norms k  kX and k  kY , is defined by the properties T .x1 C x2 / D T x1 C T x2 for x1 ; x2 2 X and T ax D aT x for a real number a. To show that a linear operator is bounded, it’s enough to find ˇ > 0 such that kxkX  1 implies kT xkY < ˇ. The reason is that if S  X is a bounded subset, with x 2 S implying kxkX  ˛ for some ˛ > 0, then kT xkY D ˛kT . ˛1 x/kY < ˛ˇ and thus T .S / is also a bounded set. Theorem C.1. A linear operator T W X ! Y of normed linear spaces is bounded if and only if it is continuous in the topologies induced by the norms of those spaces. Proof. Suppose that T is continuous, so, in particular, given  D 1 we have ı > 0 such that kxkX < ı implies kT xkY < 1. We can use that ı to prove that T is a bounded linear operator because if kxkX  1 then kT xkY < 2ı since 2ı kT xkY D kT . 2ı x/kY < 1: On the other hand, suppose T is a bounded linear operator, so in particular there exists ˇ > 0 such that kxkX  1 implies kT xkY < ˇ. Given  > 0 define ı D ˇ . Not only is T continuous, it’s equicontinuous because for any x0 2 X if kx  x0 kX < ı, then the linearity of T gives us

© Springer International Publishing Switzerland 2014 R.F. Brown, A Topological Introduction to Nonlinear Analysis, DOI 10.1007/978-3-319-11794-2

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1 kT x  T x0 kY D kıT . .x  x0 //kY < ıˇ D : ı

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Theorem C.2. Let X be an n-dimensional normed linear space. Then X is linearly homeomorphic to Rn . Proof. Let fx1 ; x2 ; : : : ; xn g be a basis for X . Define T W Rn ! X by sending a D .a1 ; a2 ; : : : ; an / to T a D a1 x1 C a2 x2 C    C an xn : It is clear that T is a one-to-one linear function onto X so it’s fine algebraically, but we need to demonstrate that topologically it’s nice also, that is, show that T and its inverse are both continuous. By Theorem C.1 that’s the same as showing T and q T 1 are bounded functions, and we’ll start by proving that T is bounded. If kak D a12 C a22 C    C an2  1, then jaj j  1 for each j . All we need to remember is that the norm on X has to satisfy the triangle inequality so kT ak D ka1 x1 C    C an xn k  ja1 jkx1 k C    C jan jkxn k  kx1 k C    C kxn k: To show that T 1 is a bounded functions, we still look at T , but now on the unit sphere S n1 in Rn . We’ve just shown that T is continuous so T .S n1 / is a compact set, and it certainly doesn’t contain 0 2 X , so the set kT ak for a 2 S n1 has a 1 positive lower bound; call it b. For a 2 Rn a nonzero vector, we have kak a 2 S n1 1 so kT . kak a/k  b and therefore kT ak  bkak. Now we can see that T 1 is a bounded function because for any x 2 X with kxk  1 we let T 1 x D a and calculate that kT 1 xk D kak 

1 1 1 kT ak D kxk  : b b b

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Corollary C.3. If V is a finite-dimensional linear subspace of a normed linear space X , then V is a Banach space and consequently V is a closed subspace of X . Proof. From the proof of Theorem C.2 we have the linear function T 1 W V ! Rn . Let fxn g be a Cauchy sequence in V ; then it follows from Lemma 18.2 that kT 1 x  T 1 yk  kT 1 kkx  yk so fT 1 xn g is a Cauchy sequence in Rn , which is complete because the reals are. Letting a 2 Rn denote the limit of that sequence, the continuity of T assures us that xn converges to T a and therefore V is complete. It follows that V is closed in X because if x is in the closure of V , it is the limit of a sequence of points of V . But that convergent sequence is Cauchy, so it has a limit in V and that limit point must be x. t u

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Theorem C.4. If X is a Banach space, then L.X / is also a Banach space. Proof. Starting with a Cauchy sequence fTn g in L.X /, we need to produce T 2 L.X / to which the sequence converges. Since fTn g is Cauchy, given  > 0, we have N > 0 such that m; n  N implies 

k.Tm  Tn /xk  kTm  Tn k D sup Wx ¤ 0 < kxk 2 so kTm x  Tn xk 

 kxk 2

for all x 2 X . Once we choose x, that last inequality allows us to make kTm xTn xk as small as we like, so fTn xg is a Cauchy sequence in X and we define T x D limn!1 Tn x. The triangle inequality property of the norm makes sequences in X behave nicely with respect to sums and scalar products and that allows us to use the linearity of the Tn to show that T is linear because, for a; b 2 R and x; y 2 X , T .ax C by/ D lim Tn .ax C by/ D a lim Tn x C b lim Tn y D aT x C bT y: n!1

n!1

n!1

Choose x 2 X and make m big enough so that kT x  Tm xk < 2 kxk but also be sure to make m bigger than that number N above that didn’t depend on x. Now we can see that if n  N , then kT x  Tn xk  kT x  Tm xk C kTm x  Tn xk < kxk: Now suppose we take any x 2 X with kxk  1; then kT xk  kT x  TN xk C kTN xk <  C kTN k; which tells us that T 2 L.X /, while kT x  Tn xk   and that means limn!1 Tn D T . t u A set S in a metric space X is somewhere dense in X if there exist x0 2 X and 0 > 0 such that every point x 2 B.x0 I 0 /, the open ball of radius 0 centered at x0 , has the property B.xI / \ S ¤ ; for all  > 0. In other words, at a point x0 where S is somewhere dense in X , the ball B.x0 I 0 / lies in S. If there is no such point x0 , then S is nowhere dense in X . If S is nowhere dense in X , then X  S is dense in X , that is, the closure of X  S is all of X because a point of X that wasn’t in that closure would be a point where S is somewhere dense in X . Theorem S C.5 (Baire Category Theorem). If X is a complete metric space and XD 1 nD1 An , then at least one of the sets An is somewhere dense in X .

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Proof. I’ll assume all the An are nowhere dense in X and prove that it leads to a contradiction. be dense in X T and I’ll show you T The open sets Un D X An wouldSall 1 1 that makes 1 U also dense in X . But X D A implies n n nD1 nD1 nD1 Un D ;, so T there’s our contradiction. Take any x 2 X and  > 0, then, to prove 1 nD1 Un dense in X , we need to find a point x0 2 B.xI / that lies in all the Un . I’ll obtain x0 as the limit of a Cauchy sequence fxn g by constructing a sequence of balls fB.xn I n /g, N nC1 I nC1 /  Un \ B.xn I n /. To do where n < n1 , to have the property that B.x it, we can safely assume  < 1 and start with B.x1 I 1 / D B.xI /. Now supposing B.xn I n / has been chosen then, since Un is dense in X , the open set Un \ B.xn I n / is nonempty and we can take any point xnC1 in it and then pick any value of nC1 < 1 small enough so that the closed ball B.xnC1 I nC1 / also lies in the open set. If nC1 m  n, then xm 2 B.xn I n / and therefore the distance from xm to xn is less than n1 so the sequence is Cauchy and x0 D limn!1 xn exists. Now fix n and notice that xm 2 B.xnC1 I nC1 / for all m > n and, since that set is closed, we get x0 2 B.xnC1 I nC1 /  Un \ B.xn I n /  Un \ B.xI / for all n.

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Theorem C.6 (Open Mapping Theorem). A bounded linear operator T W X ! Y of the Banach space X onto the Banach space Y is an open mapping, that is, if U is open in X , then T .U / is open in Y . Proof. Given x0 2 U , we have to prove that there exists ı > 0 such that B.T x0 I ı/  T .U /. What I want to do is prove that there exists r such that T .B.0I r// contains the unit ball in Y , that is, the set of y 2 Y with kykY  1. That’s all we need in order to prove the theorem because there exists  > 0 with B.x0 I /  U and if y 2 B.T x0 I ı D r / then y 2 T .U / because y D T x where x 2 B.x0 I /. Here’s why: r .y  T x0 / is in the unit ball in Y so r .y  T x0 / D T x 0 for some x 0 2 B.0I r/ and, letting x D r x 0 C x0 , then T x D y and kx  x0 kX D k r x 0 kX < . The Baire category theorem will give us r 0 such that the closure of T .B.0I r 0 // will contain the unit ball of Y . It will then take a little additional effort to show that there exists r such that T .B.0I r// itself contains the unit ball. Since T maps onto 1 Y , we can apply Theorem C.5 to the union Y D [1 nD1 T .B.0I n// D [nD1 An to conclude that some Am is somewhere dense, that is, there exist y0 2 Y and  > 0 such that B.y0 I /  Am . We’ll set r 0 D m and verify that the unit ball of Y lies in Ar 0 . Take y in the unit ball and notice that y C y0 and y  y0 both lie in B.y0 I /. Adding points in Am and multiplying them by numbers less than one keep us in Am by the linearity of T . Therefore y D 12 .y C y0 / C 12 .y  y0 / 2 Am , so, using the linearity of T once more, y 2 Ar 0 . That almost did it, but we still need to find r such that Ar D T .B.0I r// contains the ball. I’ll represent a point y in the unit ball as the limit of a series y D P1unit1k 2 yk where yk 2 P Ar 0 . To do that, I’ll inductively define a sequence fyk g in kD1 Ar 0 with the property ky  nkD1 21k yk kY  2n . Since we already know y 2 Ar 0 , we can find y1 2 Ar 0 such that ky  y1 kY  12 . Assuming that y1 ; y2 ; : : : ; yn

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in Ar 0 have P been obtained with the required property, that property tells us that k2n .y  nkD1 21k yk /kY  1 so that point is in Ar 0 and therefore we can find ynC1 2 Ar 0 such that k2n .y 

n X

21k yk /  ynC1 kY 

kD1

1 2

r 0 //, there exists which completes the induction step. Since yk 2 Ar 0 D T .B.0IP 0 1k xk with kxk kX  r such that T xk D yk . Therefore the series 1 xk kX kD1 k2 0 converges, to a number no greater than 2r . TheP space X is complete and the series 1k is absolutely convergent, so, by Lemma 18.6, 1 xk converges, to a point kD1 2 I’ll call x. Since T is continuous and linear, we conclude that Tx D T.

1 X kD1

21k xk / D

1 X

21k yk D y:

kD1

P 1k We’ve seen that kxk is bounded by the limit of the series 1 xk kX , which kD1 k2 0 0 can’t be bigger than 2r , so, setting r D 2r , we have shown that the unit ball in Y lies in T .B.0I r//. t u

Appendix D

Metric Implies Paracompact

To start out, we need to discuss some mathematical foundations. A total order  on a set S is a binary relation such that if a; b 2 S then either a  b or b  a or both are true, in which case a D b, with the property that a  b and b  c implies a  c. A total order of S is a well-ordering if in any nonempty subset T of S there is a least element of T , that is, aT 2 T such that aT  b for all b 2 T . The statement that every set can be well ordered is accepted by most mathematicians because it is equivalent to the intuitively reasonable axiom of choice, so we won’t hesitate to use a well-ordering in this appendix. Suppose X is a metric space with metric d . For x 2 X , let Bn .x/ D fx 0 2 X W d.x; x 0 / < 21n g and, more generally, for A  X let Bn .A/ be the 21n -neighborhood of A, that is, Bn .A/ D fx 2 X W d.x; A/