Inverse Spectral Theory

Inverse Spectral Theory

This is volume 130 in PURE AND APPLIED MATHEMATICS H . Bass, A. Borel, J. Moser, and S.-T. Yau, editors Paul A. Smith and Samuel Eilenberg, founding editors A list of titles in this series appears at the end of this volume.

Inverse Spectral Theory Jiirgen Poschel Universitat Bonn Mathematisches Institut 0-5300Bonn Federal Republic of Germany

Eugene Trubowitz Mathematik ETH-Zentrum CH-8092 Zurich Switzerland

ACADEMIC PRESS, INC. Harcourt Brace Jovanovich, Publishers

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Copyright 0 1987 by Academic Press, Inc. All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage and retrieval system, without permission in writing from the publisher.

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United Kingdom Edition published by ACADEMIC PRESS INC. (LONDON) LTD. 24-28 Oval Road, London NWI 7DX

Library of Congress Cataloging-in-Publication Data Poschel, Jurgen. Inverse spectral theory. ) (Pure and applied mathematics ; Bibliography: p. Includes index. 1. Spectral theory (Mathematics) I. Trubowitz, Eugene. 11. Title. 111. Series: Pure and applied . mathematics (Academic Press) ; QA3.P8 510 s [515.7'222] 86-47801 [QA3201 ISBN 0-12-563040-9 (alk. paper)

87 88 89 90 9 8 7 6 5 4 3 2 1 Printed in the United States of America

Contents

Preface

xii

Chapter I Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6

A Fundamental Solution The Dirichlet Problem The Inverse Dirichlet Problem Isospectral Sets Explicit Solutions Spectra

1 25 49 67 87 107

Appendix A Appendix B Appendix C Appendix D Appendix E Appendix F Appendix G

Continuity, Differentiability and Analyticity Some Calculus Manifolds Some Functional Analysis Three Lemmas on Infinite Products Gaussian Elimination Numerical Computations

125 141 149 159 165 169 177

References Index Index of Notations

185 187 191 V

This Page Intentionally Left Blank

Preface

This book is based on lectures given during the winter semester of 1980 at the Courant Institute of Mathematical Sciences in New York and during the winter semester of 1981 at the Eidgenossische Technische Hochschule in Zurich. The present work, the result of a collaborative effort, incorporates many extensions and improvements. We have made a strenuous attempt to keep the discussion self-contained and as simple as possible to make it easily accessible to a general mathematical audience. On the other hand, the approach is novel, some of the material is new, and the book may therefore interest the expert reader. In 1836 both Sturm and Liouville ([St], [Li]) published articles in the same volume of the Journal de Mathkmatique concerning boundary value problems for the differential equation (1)

- y"

+ q(x)y = l y ,

0 Ix

I1.

Here, l is a complex parameter and q is a real-valued function which we shall assume is square-integrable over the unit interval [0, 11. Sturm and Liouville asked whether there exist nontrivial solutions of equation (1) satisfying boundary conditions of the form y(0) cos a

+ y'(1) sin

01

=0

y(1) cos /3

+ y'(1) sin /3

=0

where a,/3 are real numbers between 0 and n.

Vii

Inverse Spectral Theory

viii

A complex number A. is called an eigenvalue of q and a,p if the boundary value problem ( 1 ) and (2) can be solved. The corresponding nontrivial solutions are called eigenfunctions of q and a,/3 for 1.The collection of all eigenvalues is the spectrum of the boundary value problem. As is well known, a comprehensive spectral theory of general ordinary and partial differential operators has been developed that has its roots in the simple example discussed above. These sophisticated techniques give us insight into the nature of spectra and the behavior of eigenfunctions for a wide variety of problems. At this time, there is no general inverse spectral theory-even the simplest cases require considerable ingenuity for their resolution. In the present context let us consider two specific questions. Fix a and p. First we ask to what extent is the coefficient function q determined by the spectrum of the associated boundary value problem? Secondly, is it possible to characterize all sets of numbers that arise as the spectrum of some 4 for the fixed boundary conditions corresponding to a,p? It may come as a mild surprise that both questions have complete answers. The boundary conditions (2) are self-adjoint in the usual sense that for any pair of functions satisfying them

( f , Qg) where Q = - (d2/dx2) + q(x) and ( f , g) = jA f(x)g(x) dx. There is another (Qf, g)

=

class of self-adjoint boundary conditions for (1). Namely, (3)

where a, b, c, d a r e real with ad - bc = 1 and k is an arbitrary real number. In fact, every self-adjoint boundary condition for (1) is of the form (2) or (3). The same questions can be posed for the boundary conditions (3). They can also be answered. None of these inverse problems can really be posed unless a class of coefficients q is specified in advance. We take Lz = Lfc[O,11, the Hilbert space of all real-valued square-integrable functions on [0, 11. It is possible to work with other spaces of functions, or even measures, but the basic ideas and constructions are clearest for L2, as with Fourier series. For example, geometry is simpler in a Hilbert space than in a general Banach space: the gradient of a differentiable function f ( q ) on L2 is defined, and therefore the normal field to the level set f ( q ) = c is easily visualized-this picture will be important for us in Chapter 4.

Preface

ix

It is possible to systematically answer the questions raised above for all these different boundary conditions (see, for example, [IT], [IMT], and [MT]), but it is a large task. Our intentions for this book are by contrast rather modest. We shall discuss a single example-the Dirichlet problem of finding solutions to (1) satisfying the Dirichlet boundary conditions (4)

y(0) = 0 ,

y(1)

=

0.

This apparently naive case already exhibits many interesting features and displaysthebasic techniquesinatransparent form. It isashort story ofitsown. There is a simple physical interpretation of the inverse Dirichlet problem. The displacement u = u(x, t ) , 0 Ix IL , of a freely vibrating inhomogeneous stiring of length L and variable mass density p(x) > 0 satisfies the wave equation p(x)ut, = uxx

and the boundary conditions

u(L, t ) = 0.

u(0, t ) = 0 ,

A periodic vibration of the form u = y(x)(acos o f

+ b sin of),

with frequency o,is called a pure tone. Separating variables, y must satisfy (5)

y"

+ 02p(x)y = 0

and (6)

y(0) = 0 ,

y ( L ) = 0.

From this point of view it is reasonable to ask: how many ways can mass be distributed along a string to produce a given set of frequencies, and how can one tell when a set of numbers is the set of frequencies of an actual vibrating string? Liouville made the observation (see, for example, [MW], p. 51) that ( 5 ) and (6) can be transformed into (1) and (4) when p is twice continuously differentiable. Hence, the inverse Dirichlet problem can be interpreted as posing natural questions about the pure tones of strings. The necessary facts about the Dirichlet boundary value problem are derived in the first two chapters. Most of them are well known. However, a few results may be unfamiliar since we emphasize that various quantities, such as fundamental solutions, eigenvalues and eigenfunctions, are themselves analytic functions of the coefficient q. This perspective is essential for us.

X

Inverse Spectral Theory

Beginning with Chapter 3, the inverse Dirichlet problem is formulated and in due course completely solved. First, we answer the question: to what extent is a coefficient function determined by its Dirichlet spectrum? This is done by describing the set of all coefficients with the same, given, Dirichlet spectrum, the so-called isospectral set. Isospectral sets turn out to be infinite dimensional manifolds with many special properties. In the second part we address the problem of characterizing all sequences of Dirichlet eigenvalues. To make this problem more tangible, imagine shifting one of the eigenvalues in the spectrum of a function p to the right, or the left, by a small amount. Is the new sequence still realizable as the Dirichlet spectrum for some other coefficient q, or does it violate some internal constraint that is satisfied by an actual sequence of eigenvalues? A full characterization is given in Chapter 6 . It is necessary to point out that our treatment of this part of inverse spectral theory is very personal and consequently does not depend on or develop other established approaches and contains few references to published material. We refer to [CS] for an extensive bibliography and historical information, and to Borg [Bo], a student of Beurling, who wrote the first important paper on the subject, as well as to Gelfand and Levitan [GL] who were the first to discover a method for constructing coefficients with a given spectral measure. We recommend that the reader wishing to get a quick overall impression read the first few pages of each chapter and then skim the statements of the theorems. In this volume we have presented our subject in isolation and have not explored several important connections with other areas of mathematics. Were we to continue and write a second volume we would likely discuss, among other topics, the periodic problem and the relationships to classical mechanics, algebraic geometry, and the Korteweg de Vries equation of fluid dynamics. Finally, it is a pleasure to thank P. Deift; H.P. McKean, T. Nanda, C. Tomei and especially E. Isaacson, J. Moser, and J. Ralston for their help. This book is dedicated to my father on the occasion of his seventy-fifth birthday. Zurich, 1986

E. Trubowitz

1

A Fundamental Solution

The purpose of this chapter is to solve the initial value problem for the differential equation (1)

-y”

+ q(x)y = Ay,

0 Ix

I1.

Here, A E C, the complex numbers, and q E L$ = Li[O, 11, the Hilbert space of all complex valued, square integrable functions on [0, 11.’ The plan is to construct solutions y ~ ( xA, , q) and yz(x, 1,q) of equation (1) satisfying the initial conditions Yl(0, A , q) = Yi(0,A , Q) = 1

Yi(0, A , 4) = Y2(0,1,4)= 0, and to show that they are a fundamental solution. That is, any other solution y of equation (1) can be written as a linear combination of these two

solutions, namely, Y ( X ) = Y(O)Yl(X) + Y’(O)Yz(X).

An analysis of this fundamental solution will then give us information about the behavior of all solutions of (1).

’

Starting with Chapter 2, we restrict our attention to real valued functions q. However, it is important for us to develop the basic theory on the complex Hilbert space 156. See for example the proof of Theorem 3.1.

I

Inverse Spectral Theory

2

Starting with the next chapter, we are going to use y1 and y2 to study boundary value problems. It is a basic principle of the theory of linear ordinary differential equations that boundary value problems can be solved by first constructing appropriate solutions to the initial value problem. This principle is illustrated at the beginning of Chapter 2, where the Dirichlet problem for equation (1) is considered. As the notation indicates, we regard yl and y2 as functions of all three variables x , A and q. This point of view is essential for the applications we have in mind. In fact, we are going to construct y1 andy2 by expanding them as power series in q. It will be shown that they are entire functions on C x L t . See Appendix A for the basic facts about analysis on Banach spaces, and in particular analytic maps. Before we continue, we must explain what we mean by a solution of equation (1). After all, the coefficient q is in L i , so that the equation only makes sense almost everywhere. By definition, a function y is a solution of equation ( l ) , if it is continuously differentiable, y’ is absolutely continuous, and the equation holds almost everywhere.2 A common method for solving differential equations is to look for solutions in the form of a power series. We are going to apply this method to equation (1). To see what is involved, we first consider the simple problem of finding the solution of -u” = Au,

0

Ix I1,

with the initial conditions u’(0) = 0.

u(0) = 1,

Notice that u(x, A) = y ~ ( xA, , 0). Suppose that u is given by a power series in A: u(x, A) =

c

Un(X)Afl.

nrO

’Recall that an absolutely continuous function u on [0, 11 has a derivative u’ almost everywhere, which is integrable and satisfies u(x) = u(0) +

1:

U’(f)df.

Conversely, if v is an integrable function on [0, 11, then u(x) = jS v ( t ) df is absolutely continuous, and u‘ = v almost everywhere.

A Fundamental Solution

3

The zeroth coefficient is the solution to -u" uo(0) = 1, &(O) = 0. Thus,

=

l u for I

=

0, satisfying

uo(x) = 1. Formally, differentiating the power series two times with respect to x , using the differential equation and equating coefficients, we find that - U: = U n - 1 ,

The initial conditions for

Un

n 2 1.

are

un(0) = 0,

uA(0) = 0 ,

since u(0) = 1 + En, 1 un(0)In= 1 and u'(0) = Integrating twice, rx

n

2 1,

En, I

uA(0)l" = 0 for all A.

rr

so that by induction

Consequently,

which is no surprise at all. The point of this calculation is that the coefficients U n , n L 1, are determined iteratively by solving the inhomogeneous form of the differential equation -u" = A M for l = 0. We are going to imitate the last paragraph and construct y1 and y2 as power series in q. Judging by the above, a necessary prerequisite is the solution of the inhomogeneous form of equation (1) for q = 0.

Lemma 1. Let f

E

L6 and a, b E C. The unique solution of - y " = l y - f(x),

0 5 x 5 1

satisfying y(0) = a,

y'(0) = b

Inverse Spectral Theory

4

is y(x) = u cos f i x

"sinfi(x - t) f ( t ) dt.

+b

o

f

i

We will also use the notation a(x) = c o s a x , Then the solution reads

Proof of Lemma 1. Consider the integral Y~(x)=

s:

SX(X- t ) f ( t )dt.

By the addition theorem for the sine function,

s:

Y~(x)= Since af and

s:

a ( t ) f ( t )dt - CX(X) a ( t ) f ( t )dt-

af are integrable, y f is absolutely continuous.

yj(x) = a(x)

s:

a ( t ) f ( t ) d t+

s:

Hence,

fi(t)f(t)dt

almost everywhere, and consequently everywhere, because the right hand side is continuous in x . It ~ follows from another differentiation that yf is a particular solution of -y" = Ay - f(x) with

Jm = 0, Since a and

Uj(0) = 0.

a solve the homogeneous equation, Ax) = m ( x ) + M x ) + Y f W

is a solution of -y" = Ay - f(x) satisfying the correct initial conditions. If u is absolutely continuous and u is continuous with u' = u almost everywhere, then u is continuously differentiable, and u' = u everywhere. For, u(x) = u(0) + 16 u ' ( f )d f = u(0) + u ( t ) dt and thus u' = u everywhere by the fundamental theorem of calculus.

A Fundamental Solution

5

To prove uniqueness, suppose that JJ is another solution of the inhomogeneous differential equation satisfying the same initial conditions. Then the difference u = y - JJ satisfies -u" = Au,

u'(0) = 0.

u(0) = 0,

It follows that u vanishes identically, hence y = 9.

H

We are now ready to construct y1 and y 2 . We begin with y1. Suppose that yl(x, A , q ) is given as a power series in q. That is, Yl(X, A, 4 ) = CO(X, A)

+

Cn(x, A 4) = Cn(x, A 41

.

c

nz

1

Cn(X, I

, q),

where qn)

lq

,='" =qn=q'

and G ( x , A, 41, ..., q n ) is a bounded, multi-linear symmetric form on LE x x LE for each x and A. Thus, C1 is linear in q, C2 is quadratic in q , and so on. The zeroth order term is determined by setting q = 0. This gives Co(x, A) = cos f i x .

Proceeding as before, we formally differentiate the power series for y1 two times with respect to x , use ~ the differential equation and equate terms, which are homogeneous of the same degree in q. We obtain the differential equation - C i = ACn - QCn-1,

n

2 1.

The initial conditions are

Cn(0,A, q ) = 0,

n

CA(0, A, q ) = 0,

since y l ( 0 ) = 1 + 1, 1 Cn(0)= 1 and yi(0) = It follows from Lemma 1 that

1

2 1,

CA(0) = 0 for all q.5

4Here, one must not differentiate with respect to q using the chain rule. q is a variable in its own right. Consider, for example, the function

f i x , Q)

=x

s:

q(t) dt.

'Replacing q by tq, the sums become power series in t with coefficients Cn(O,I , q) and CA(0, I , q) respectively. These power series vanish identically in t , so all coefficients must vanish.

Inverse Spectral Theory

6

Hence,

and

2

cx(t1)

II [sx(ti+l- ti)q(ti)l dtl dt2.

i= 1

Proceeding by induction, (2) C f l kA, 4 ) =

i

0s I 1

n

i

... i t , , ,

II

Ch(f~) =x

[sA(ti+l

-

ti)q(ti)] dtl

dtn

i= 1

for n 2 1. Thus, we have written Y l k A, 4) = a ( x ) +

c Cfl(X,A, 4)

IlZl

as a formal power series in q. Such expansions were already known to Hermann Weyl [We]. The same reasoning applies to y2(x, A , q) and yields Y2(X,

1,4) = S d X ) +

c

n Z l

where r

"

S f l ( X , A, 41,

A Fundamental Solution

Problem I .

Show that, for n

L

1,

r

n

and

o s rI

5

... 5 r,,

,= x ti i II1

ti+^

- ti)(q(ti) -

I ) dtl

* * a

dtn.

=

[Hint: Use that yj(X, A , q) = yj(X90 , q - A ) for j

= 1,2.]

So far, these are just formal calculations. We now show that these series converge to genuine solutions of equation (1). Below, the standard notations

=

Sa

f(t)E(t)dt,

Ilf II =

rn

are used for the inner product and norm on LE.

Theorem 1. The formal power series f o r y l ( x ,I , q) and yz(x, A , q), whose coefficients are given by (2) and ( 3 ) respectively, converge uniformly on bounded subsets of [0, 11 x C x Lg to the unique solutions of -Y" + q(x)y = I y ,

0

Ix I 1,

satisfying the initial conditions YI(0,A, 4) = Yi(0,I , 4) = 1 Yi(0,I , 4)

=

YZ(0,A, 4 ) = 0.

Moreover, they satisfy the integral equations

a

Inverse Spectral Theory

Proof. The proof is given for y1. It is the same for y 2 . Recall the elementary inequalities

and, for 0

Ix I1,

The nth term in the expansion of y1 is then majorized by r

n

The value of the integral in the last line does not change under permutation of tl , .,., tn . Moreover, the union of all the permuted regions of integration is [O,x]". It follows that

by the Schwarz inequality. Thus, we obtain the estimate

This shows that the formal power series for yl converges uniformly on bounded subsets of [0, 11 x C x LE to a continuous function. Summing the majorants we obtain the estimate for y1.

A Fundamental Solution

9

Next, we observe that

since the uniform convergence allows us to interchange summation and integration. This verifies the integral equation for y l . It follows just as in the proof of Lemma 1 that y1 is a solution of -y" = Ay - q(x)y satisfying the correct initial conditions. To prove uniqueness, suppose that j3 is another solution of equation (1) satisfying the same initial conditions as y l . By Lemma 1,

J'i(x) = C X ( X ) + The difference

II

5:

SA(X -

t)q(t)Yl(t)dt.

= y1 - 91 then satisfies V(X) =

5:

SA(X

- t)q(t)u(t)dt,

hence

by the Schwarz inequality with c = )1q1)2maxo 1 Is?(t)l. It follows that the nonnegative function e-"jXg lu(t)I2 dt has a nonpositive derivative on [0, 11. Since it vanishes at zero, it must vanish identically on [0, 11. Thus, u = 0, proving uniqueness. H

10

Inverse Spectral Theory

Certainly, Theorem 1 is familiar. We recast its well known proof by Picard iteration as the method of power series in order to emphasize the dependence of y1 and yz on q. For q = 0, we have Yl = c o s a x ,

yz

=

sin f i x ~

f l .

These are entire function of A , because

This can be generalized. Analyticity Properties.

(a) For each x E [0, 11,

YAX, A , 4)s

uj'(x,A , 4)s

j

=

1,2

are entire functions on C x L i . They are real valued on IR x L,$. (b) The solution Y A * , A , q),

j = 192

is analytic as a map from C x L i into HZ. For k 2 0 an integer, H:= H:[O, 11 denotes the Hilbert space of all complex valued functions on [0, 11, which have k derivatives in L2.6Hk is the subspace of all real valued functions in H:. In particular, L i = HZ and L,$ = H$. Proof of Analyticity Properties. power series expansion is c n

=

5

0 L I1 L ... s t,, 1 = x

(a) We consider y l . The nth term in its n

n

i= 1

i= 1

cx(t1)

II sx(ti+l - ti) II q(ti) dtl ... dtn.

The first two factors in the integrand are entire functions of A . For each x , this term therefore is continuously differentiable in I and q everywhere on 6For example,fin L i belongs to Hd, if the difference quotients ( f ( x + h) strong limit in L ; .

- f ( x ) ) / hhave

a

A Fundamental Solution

11

C x L ; , hence is an entire function of A and q. By the uniform convergence of the sum of all these terms on bounded subsets of [0, 1 1 x C x LE, yl is also an entire function of A and q for each x . By the integral equation of Theorem 1, y;(x,A , q) = - fi sin f i x

+

i:

cos f i ( x - t)q(t)yl(t,A, q)dt.

This shows that for each x , the derivative y; is also continuously differentiable hence entire in A and q. By the differential equation, Yl(X,

La =

Y l k

A, q),

so yl and yl are real valued for real A and q. (b) Again, consideryl . Each term Cnis in fact continuously differentiable, hence analytic, when considered as a map from C x LE into CC, the Banach space of all complex valued continuous functions on [0, 11 with the usual supremum norm. The same is then true for yl by uniform convergence, and for yf by the integral equation. It follows that yl and yf are also analytic as maps from C x L; into LE, since the supremum norm is stronger than the L2-norm. By the differential equation,

Yl

= (q -

A)Yl

is analytic in the same sense. This shows that y1 is analytic as a map from w c x LE into H:. The Wronskian of two differentiable functions f and g is the function

We have the very important Wronskian Identity.

That is,'

' S L ( 2 , C)is the group of all 2 x 2 complex matrices with determinant 1 ,

12

Inverse Spectral Theory

Proof. We have "3

Yzl' = (YlYi - YiYz)' = YlY5 - y!yz = Y l ( 4 - A)Yz

- (4 - A)YlY2

=o almost everywhere, hence Yzl(X) =

"9

[Yl

Y

Yzl(0) = 1

everywhere by continuity. It is now possible to generalize Lemma 1 .

Theorem 2. Let f E L$ and a, b E C. Then there exists a unique solution of the inhomogeneous equation -y"

+ q(x)y = Ay - f ( x ) ,

0

Ix I1 ,

satisfying y(0) = a,

y'(0) = b.

It is given by

Proof. The proof is the same as that of Lemma 1. One only has to replace

SAW - t ) = sx(x)cx(t)- sx(t)cx(x) by yl(x)uz(t)- ul(x)yz(t)and make use of the Wronskian identity. Corollary 1. Every solution of equation ( 1 ) is uniquely given by Y ( X ) = Y(O)Yl(X) + U'(0)Yz(X).

Moreover, ifasolution hasa double root in [0, 11, then it vanishes identically. The initial value problem for equation ( 1 ) is now solved.

A Fundamental Solution

13

froof. The first statement is immediate. The second is easily verified. Suppose

Then $ ;(); identity.

= 0, because the 2

x 2-matrix is nonsingular by the Wronskian

W

Comparing Theorem 2 with Lemma 1, y1 and yz appear as generalizations of the trigonometric functions cos f i x and sin f i x / f i . It is indeed often helpful to think of them this way. For large I , this observation can be made quantitative by improving on the bound in Theorem 1. These asymptotic estimates will be used over and over again. Theorem 3 (Basic Estimates).

On [0,11 x C x LE,

lyl(x, A , q ) - c o s f i x l

1

I-exp(lImfilx

lfil

+ 11q11 fi)

Proof. Clearly, lYl(X, A , q ) - cos fix1 Each term Cn,n by

2

=c

na1

ICn(X, I , 411.

1, contains the factor sx(x - tn),which may be bounded

Otherwise, we proceed as in the proof of Theorem 1 to obtain the estimate

14

Inverse Spectral Theory

Summing these majorants and including the term for n = 0, we obtain the first inequality. The second one is proven by bounding both the factors ~ ~ ( t l ) and SA(X- t,) occurring in S n for n 1 1 as above. Differentiating the integral equation for yl one finds

s:

yi(x, A, q) = - \/si sin \/six +

cos \/si(x - t)q(t)yl(t, A, q) dt.

Using the estimate of Theorem 1 and the inequality Icx(t)I

I

exp(1Im \/silt),

proving the third inequality. To prove the fourth, show that

W

and mimic the last paragraph.

Problem 2. The entire function&) but for no E < 0, sup

If(z)I

is of order o < =

00

if for every E

> 0,

O(ero+').

IzI = r

It is of type

5

< 00 of order o if for every E > 0, but for no E < 0, SUP

If(z)l

=

~(e('+~)~").

IzI = r

Showthattheorderandtypeofyjandyj',j= 1,2,withrespecttoAiso = 7 = x respectively.

t,

The proof of Theorem 3 implicitly shows that y1(x, A) = cos \/six +

N

c C,(X, A) + 0

n=l

As it stands, this estimate is not a very useful one since Cn is rather complicated, even for large IAI. A small improvement is possible by splitting

A Fundamental Solution

15

terms apart using trigonometric identities. For example, using the identity 2 sin u cos b = sin@ 6) + sin@ - b), we obtain

+

Cl(X, A, 4)= -

s';

f i o

sin f i ( x - t) cos f i t q(t)dt j I q ( f ) d f+

-

s i n f l ( x - 2t)q(t)dt.

But in general, one can't do any better. The situation is quite different, when q has one or more derivatives. Then we may decompose C,, and refine its pieces by partial integration. If q has one derivative in L i , then

+

21;

cos f i ( x - 2t)q'(t)dt,

where the last term is o(exp(l1m filx)/lIl) as II) --* 0 3 . ~ If q has a second derivative in L? , we can integrate by parts once more and replace the third term by sin f i x (q'(x) + q'(0)) + 8 fi3

-~

&3."

sin f i ( x - Zt)q"(t)dt.

Now, the last term is o(exp(l1m filx)/lfl13) as (131

Problem 3.

-+

03. 8

Show that for u in L i and a > 0,

:l

u ( t ) sin f i t dt = o(etImJxlo)

for IA I --* 03. Note that A need not tend to infinity along the positive real axis. [Hint: First, prove the statement for continuously differentiable functions, using partial integration. Then write u = (u - u) + u, where u is C', and u - u has small L2-norm.] 'See the next Problem.

'That is,

Inverse Spectral Theory

16

In the expansions above, terms are ordered by inverse powers of fi,and each term (besides the error term) is a product of expressions in A and q alone, if we ignore the dependence on x. Thus, it is natural, assuming q to be smooth, to split the first Ncoefficients Cnup in this way and to collect terms of the same order in l / f i . The result of this procedure is a formal asymptotic expansion of YI to order N in inverse powers of fi with these special coefficients, There are more efficient ways to generate this expansion, see for example [Wa]. But in any case, it is a real task to actually calculate the coefficients. Luckily, we don’t have to, because the estimates of Theorem 3 meet all of our needs. The next theorem is included to give the expansions for twice differentiable functions.

Theorem 4. For q E El:,

+ O(exp(l1m filx))

1fii3

and

where

Proof. We have already seen that

A Fundamental Solution

17

By the same kind of calculation,

=z

+ zz.

Using the identity 2 sin a sin b = - cos(a by parts,

+ b) + cos(a - b) and integration

Another partial integration shows that

zz = 0(exp(l1m "1x1)

iat3

Collecting terms yields the expansion for y1. The expansion for yz follows from similar manipulations. H Problem 4. Theorem 4 can be improved. For instance, the error term in the expansion of y2 can be replaced by

Inverse Spectral Theory

18

We now investigate the dependence of y l and y2 on q. By construction, they are continuous in the strong topology on LE. It turns out, they are even compact. That is, continuous with respect to the weak topology."

Theorem 5. If the sequence q m converges weakly to q in L ; , then YAX,1,q m )

+

YAX,1,q),

j = 1,2

uniformly on bounded subsets of [0, 11 x C. In other words, yl and y2 are uniformly compact on bounded subsets of [0, 11 x C. This compactness property will ultimately be used to prove existence theorems.

Proof. Consider yl. Suppose the sequence qmconverges weakly to q. By the principle of uniform boundedness,

If A is any bounded subset of [0, 11 x C,then

uniformly on A . The second sum converges to zero as N tends to infinity. Therefore, it is enough to show that each term Cn(X, 1,q m ) converges to Cn(X, A, q) uniformly on A . 10

See Appendix A for compact mappings in general.

19

A Fundamental Solution

Fix n 2 1, and consider the function

A) = C n ( x , A, q m )

Am(X,

- Cn(x,

A, 4).

We can write

where n

and ll{...)is the indicator function of the set (. .). The inner product is taken in the space L$([O,11"). We have to show that s;p

[Am(X,

A)[

-+

as m

0

+

00.

By continuity, the supremum of I A m ( is attained at some point (Xm, Am) in the compact closure of A for each m.So equivalently we have to show that IAm(Xm, Am)[

as m + m .

0

--*

Suppose this is not the case. Passing to a subsequence we can assume that converges to (x*,A*), while

(Xm, A m )

[ A m ( X m , Am)l 2

6 > 0.

By the bounded convergence theorem, Pxm,XmstrOngPx*sA*

in Li([O, 11"). On the other hand, n

n

in Li([O, 11"). This is easily checked by integrating against the monomials t f l .-.I$", which span a dense subspace in Li([O, 11"). It follows that"

IA m ( x m

9

Am)

I

+

0,

which is a contradiction. "In general, if fm

+

f strongly and g, (fm,gm)

since (S," - f , g,)

+

=

(fm

-+

g weakly in a Hilbert space, then

-f*gm)

+


+

Cf,g)$

0 by the Schwarz inequality and the uniform boundedness of the gm

Inverse Spectral Theory

20

This proves the theorem for y1. The proof for y2 is of course the same. The functions yl and y2 are analytic in q. In particular, they are continuously differentiable in q. So, what are their derivatives and gradients? Recall that the derivative of a mapf: E + F between two Banach spaces E and F a t a point xis a bounded linear map from E into F,which we denote by dxf.12 Moreover, if E is a Hilbert space and F the real or complex line, then, by the Riesz representation theorem, there is a unique element af/ax in E , such that

for all u in E. This element is the gradient off at x. To get an idea what the gradients of y1 andy2 look like, we make a formal calculation. Differentiating both sides of the equation

with respect to q in the direction u in LE, we obtain

Formally, interchanging differentiation with respect to x and q, the first term becomes -(dqyj(u))". The initial values of dqyj(V) both vanish, since the initial values of yj are independent of q. Thus, Theorem 2 yields

and consequently

where l [ o , ~ ]is the indicator function of the interval [0, XI. This is easy to make rigorous. "See Appendix A for the definition of derivative.

21

A Fundamental Solution

Theorem 6. For j = 1,2,

The gradients are jointly continuous with respect to x , A , q.

Proof. (a) If q is continuous, then yj is twice continuously differentiable in x, and interchanging differentiation with respect to x and q is allowed. That is,

dqY"j(u) = (dqYj(u)Y'. Therefore, the result

dqyj(u) =

5:

Yj(t)[Yl(t)Yz(X)- Y I ( X ) Y Z ( ~ ) I ~ ( ~ dt )

of the preceding calulations indeed holds for continuous functions q. Now, each side of this equation is a continuous function of q in LE , and the equation holds on a dense subset of LE,namely the continuous functions. Hence it must hold in general. The formula for the gradient follows. Differentiating both sides with respect to x,

dq Y

X ~ =)

s:

Yj(t)[rl(t)ri(x)- ~ i ( x ) ~ z (u(t) t ) ldt

for continuous q. Again, this holds in general by continuity in q. (b) follows from the identity .Yj(x, 1

which implies

+ E , 4) = Yj(x, A , - E ) ,

Inverse Spectral Theory

22

Thus, ayj/aA is just the directional derivative of yj at q in the direction of the constant function - 1. Same for dy,!/aA. Part (a) of Theorem 6 can be written more compactly as

Problem 5. To interpret the last identity, fix x and A, and consider the map

from L i into SL(2, C). The tangent space to SL(2, C) at y0 is the three complex dimensional space of matrices of the form YoX, where X is in d(2, C), the Lie algebra of all complex 2 x 2-matrices with trace zero. Show that the tangent vector to the curve Y(x, A, q + t v ) at t = 0 is given by the above identity.

Problem 6. Prove directly that

I I Y ~ ( x , A, 4 + v ) - yj(x, A, 4 ) - Lq(u)II = ~ ( l l v l l ) , where Ldu)

=

:1

Y ~ ( ~ ) [ Y I ( ~ )-YY ~I ( (x )xY ~) ( ~ ) I u dt.

[Hint: Observe that yj(x, A, q + v) satisfies an inhomogeneous equation with inhomogeneity uyj(x, A, q + v ) , and apply Theorem 2.1

Problem 7. Are there critical points of the functions yl and yz on L;? That is, regard x and A as fixed, and determine the set of q s for which the gradient ayj/aq(t) vanishes identically in t. One interesting feature of Theorem 6 is that the derivative o f y l andy2 with respect to q is expressed in terms of products of y~ and y2. This phenomenon will occur many times. We end this chapter with a remark about such products.

A Fundamental Solution

Theorem 7.

23

(a) For each (A, q) E C x L;, the functions Yf ,YlY2 Yf 9

are linearly independent over [O, 11. (b) Let q E Cc![O,11, and L

=

d q(x)dx

d + dx -q(x)

-

Iff, g are two solutions of equation (1) f o r the same A , then

u:

r:

YIY2 Y: (Yf)’ (YlY2)’ (Y:)’ = 2YlYi Y l Y i + YiY2 2Y2Yi (Y:)” (YIY2)N (Yt)” 2(N2 2YiYi 2(YiI2 YlY2

Y:

=2

almost everywhere by the Wronskian identity. Hence, y : , ~ linearly independent on [0, 11. (b) follows by direction calculation.

1

~y 2 t

are ,

Problem 8. (a) Check that the map

*(a2 + b2 + c2 + d 2 ) *(a2 - b2 + c2 - d 2 ) ab + cd *(a2 + b2 - c2 - d 2 ) *(a2 - b2 - c2 + d 2 ) ab - cd ac + bd ac - bd ad + bc is a homomorphism of SL(2, C) into S0(1,2, C), the group of all 3 x 3matrices with complex elements and determinant 1, which preserve the quadratic form x 2 - y 2 - z 2 .

Inverse Spectral Theory

24

(b) Let 4 = (41,42) be any fundamental solution of equation ( 1 ) with [ + I , 421 = 1. Set @+ =

rt<4: + 49, H4: - 49,41421.

Verify the identity

Rg@+ = mg+. (c) Prove that there exists a g e SL(2, C) such that (@I, @j)

where

(@I,

@z,

@3)

=

Qg+.

= 6ij,

1 Ii , j I 3,

2

The Dirichlet Problem

This chapter is devoted to the simplest boundary value problem for the differential equation (1)

-y” + q(x)y

= ly,

0

4

x 5 1.

Here, l E C and q E L2 = Li[O, 11, the Hilbert space of all real valued, square integrable functions on [0, 11. We ask whether there are nontrivial solutions of equation (1) satisfying the Dirichlet boundary condition

y(0) = 0,

y(1) = 0.

This is the Dirichlet problem. A complex number l is called a Dirichlet eigenvalue of q if the Dirichlet problem can be solved. The corresponding nontrivial solutions are called eigenfunctions of q for I t . The collection of all eigenvalues of q is its Dirichlet spectrum. From a more abstract point of view, the Dirichlet problem belongs to the theory of self-adjoint operators. Precisely, the unbounded operator Q, ,defined by

25

Inverse Spectral Theory

26

for f in the dense subspace D of all functions in Hc[O, 11 that vanish at 0 and 1, is self-adjoint on the domain D.’ It can be shown that the spectrum of Q coincides with the Dirichlet spectrum of q, and so we may draw some conclusions about the Dirichlet problem from general facts about selfadjoint operators. For instance, the Dirichlet spectrum must be an unbounded sequence of real numbers. However, abstract operator theory is not helpful in answering many of the questions that interest us. For this reason, we shall follow Sturm’s classical approach [St] which may be called the “method of shooting. ” Intuitively speaking, the solution y2 is shot from the left end point of the interval with unit velocity. At x = 1, it reaches the height yz(1, A). The idea is to vary the parameter A so that the terminal height is 0. This point is illustrated in Figure 1. At the moment, the terminal velocityyz’(1,A) does not concern us. It becomes important in Chapter 3. The “shooting method” is easy to implement. If p is a zero ofyz( 1, A), then yz(x, p) is a nontrivial solution of equation (1) satisfying the Dirichlet boundary conditions. Hence, p is a Dirichlet eigenvalue of q. Conversely, suppose p is a Dirichlet eigenvalue of q with eigenfunction y(x). Then Y ( X ) = Y’(O)YZ(X,

Figure 1. ‘That is, for all f,g E D , (Qf,g) = ( f , Q g ) ,

and for all g

E

L;, the condition

f*;,:&-, l ( Q f y g ) l implies that g is in D. See [RS].

< O0

The Dirichlet Problem

21

by Corollary 1.1, hencey2( 1, p ) = 0 by the boundary conditions. We see that the Dirichlet spectrum of q is identical with the zero set of the entire function y2(l,A, q). From now on, we will therefore not distinguish between them. The spectrum of a finite dimensional linear map is the zero set of its characteristic polynomial. By analogy, we can consider y2( 1, A , q) as the “characteristic polynomial” of the linear operator - d2/dx2+ q with Dirichlet boundary conditions. For q = 0, the Dirichlet spectrum is the infinite sequence

n’, 4 n 2 ,9n2, ..., n 2 n 2 , ..., since y2( 1, A, 0) = sin fi/fi.The “Counting Lemma” below shows that there are infinitely many eigenvalues for every q. It also gives a first rough estimate of their location, to be refined later on. First, a simple technical lemma.

Lemma 1. If Iz - nnl 2 n/4 for all integers n, then

< 4lsin zl. Proof. Write z = x + iy with real x,y. Since JsinzJis even and periodic with period T I , it suffices to prove the lemma for 0 Ix I7112 and IzI 2 7114. We have lsin z12 = cosh’y - cos’x. For n/6

I

x

I

7712,

cos’x

I

$ I$ cosh’y

for all real y. For 0 Ix Id 6 , the assumption IzI 2 n/4 implies y 2 2 ( 7 ~ / 4 )-~ x2 2 &n2 2 and hence cosh’y

L

1

+ y 2 2 4 L 0 cos’x

as before. Thus, in both cases we have IsinzI2 2 $cosh2y > &e21Yl, and the result follows. The estimate of Lemma 1 is not optimal, but chosen for the sake of convenience. In general, there is a positive constant cg for each 6 > 0 such that

< calsinzl provided Iz - nnl > 6 for all n.

Inverse Spectral Theory

28

In the next lemma, we admit complex valued functions q. This is useful, for instance in the proof of Theorem 3.1. Lemma 2 (The Counting Lemma). Let q E Lf and N > 2e1I4l1be an integer. Then y2( 1, A, q) has exactly N roots, counted with multiplicities, in the open half plane ReA < (N + +)’n’, and f o r each n > N , exactly one simple root in the egg shaped region

There are no other roots.

and let K > N be another integer. Consider the Proof. Fix N > 2e11411, contours

+ i)n, R e f l = ( N + i)n, IflI

=

(K

and 71

Ifl-nnl=-, 2

n>N.

See Figure 2. By Lemma 1, the estimate eltrnfi1< 4lsinflI

holds on all of them. Therefore, by the Basic Estimate for y z , Yd1, A) - -

fl

also holds on them. It follows that yz(1, A) does not vanish on these contours. Hence, by Rouche’s theorem, yz(1, A) has as many roots, counted with multiplicities, as sin fl/flin each of the bounded regions and the remaining

29

The Dirichlet Problem

I-plane

I

1

I

1

Figure 2.

unbounded region. Since sin fi/fihas only the simple roots n2n2,n and since K > N can be chosen arbitrarily large, the lemma follows.

L

1,

Returning to real valued functions q , we have

Theorem 1. The Dirichlet spectrum of q in L i is an infinite sequence of real numbers, which is bounded from below and tends to + m.

Proof. We only have to prove reality, the rest follows from the Counting Lemma. Suppose A is a Dirichlet eigenvalue of q with eigenfunction y(x). Then -y”

+ q(x)y = Ay.

Conjugating the equation,

-p + q(x)P = xy, since q is real. Multiplying the first equation by 7, the second by y and taking the difference, we obtain [ y , jq’ = y p

- y”y= (A

- X)(yI2,

30

Inverse Spectral Theory

hence, by integration,

The left hand side vanishes, while the integral does not. Therefore - X = 0, that is, A is real. Of course, this is the standard argument, used to prove that eigenvalues of self-adjoint operators are real.

A

It has been shown that every eigenfunction for an eigenvalue A is a multiple of y2(x, A). The geometric multiplicity of A, which is the maximal number of linearly independent eigenfunctions for A , is therefore 1 . On the other hand, its algebraic multiplicity, defined as its order as a root of y2(l, A), might very well be larger. This, however, is not the case. We will often use the abbreviated notation = WdA.

-

Theorem 2.

If A is a Dirichlet eigenvalue of q in L2, then 32(1, A ) ~ i ( lA) ,

=

s:

&t,

= Ilu2(.,

A) d t

4112> 0.

In particular, y2( 1, A) # 0. Thus, all roots of y2( 1 , A) are simple.

Proof. Let yields2

y2 =

yz(x, A). Differentiating equation (1) with respect to A

-32” + q(x132 = yz + A 3 2 . Multiplying this equation by y 2 , the original equation by 3 2 and taking the difference, we obtain 2 Y2 =

Y232 - 32Y2

= [32,y21’.

Hence

’To avoid the problem of interchanging x- and I-derivative, first assume q is continuous and argue as in the proof of Theorem 1.6.

The Dirichkt Problem

31

since yz(0, A) and )'2(0,A) vanish for all A, and yz(1, A) vanishes for a since y2 is real for Dirichlet eigenvalue I. The integral is equal to 11 y2( , real A.

-

Problem 1. Prove the identity of Theorem 2 by starting from

and using Theorem 1.6. It follows from the preceeding results that the Dirichlet spectrum of every q in L2is a sequence of real numbers p d q ) < P2(4) <

***

satisfying pn(q) = n2n2

+ O(n).

To each eigenvalue we associate a unique eigenfunction gn normalized by

= g n ( X , q)

By Theorem 2,

-

Y ~ XPn) ,

J h ( 1 , PnjYi(l9 P n ) ' where the argument q has been suppressed on the right hand side. In particular, gn(x,0) = *sin

nnx.

Let us investigate the Dirichlet eigenvalues as functions defined on L2.

Theorem 3. Pn, n 2 1 , is a compact, real analytic function on L2. Its gradient is

32

Inverse Spectral Theory

Since p,, is real analytic, g,, is also a real analytic function of q by the expression above and the Analyticity Properties of Chapter 1.

Proof. To verify compactness, suppose the sequence q m , m weakly to q. By the principle of uniform boundedness, 11q11 I SZP Ilq.mll IM

Let N > 2e‘,

E

L

1, converges

< 03.

> 0, and consider the intervals

I,,= [ A E

m: [ A - pn(q)l < E l ,

1 I n I N.

If E is sufficiently small, then these intervals are all disjoint and contained in the half line (- 00, ( N + $)’n2) by the Counting Lemma. Moreover, y2(1, A , q) changes sign on each of them, since pn(q) is a simple root. As m tends to infinity, the functions y2(l, A , q m ) converge to yz(1, A , q) uniformly on I, u ... u IN by Theorem 1.5. Hence, for sufficiently large m, they also change sign on I I , ..., IN, so they must all have at least one root in each of these intervals. But there are only N roots on the whole half line (- m, ( N + +)’n’) by the Counting Lemma. Therefore, yz(1, A , q m ) has exactly one root in each interval I,,, 1 I n IN , which must be the nth eigenvalue of qm. Consequently, lp(n(qm) - pun(q)J< E ,

1 In IN ,

for all sufficiently large m. It follows that M q m )

+

pun(4)

form + 00, since N a n d E > 0 were arbitrary. Thus, p,, is a compact function of q. To prove real analyticity, fix p in L2. By Theorem 2, Yz(l9

Pu,(P),P )

f

0.

So the implicit function theorem applies, and there exists a unique continuous function P,, ,defined on some small neighborhood U C L2 of p , such that Y2(1,P n ( 4 h

4) = 0,

Prim = Pn(P)

on U. Furthermore, f i n is real analytic. On the other hand, p,, is also a continuous function on U satisfying yz(1, p,,(q), q) = 0. Therefore,

Pn(d

= PrI(4)

on U by uniqueness, and so p,, is real analytic.

The Dirichlet Problem

33

To calculate the gradient observe that

since yz(1, pn) = 0 and y,(l)yl(l) = 1 by the Wronskian identity. Hence,

Here is another, perhaps more intuitive proof of the last identity. Differentiating both sides of the differential equation of gn in the direction u we obtain - d q g X ~ )+ qdqgn(tr) + ugn = pndqgn(u) + dqpn(V)gn.

If q is continuous, then gn is twice continuously differentiable, and we may interchange differentiation with respect to x and q to obtain - (dqgn(u))" + qdqgn(u) + vgn = pn dqgn(u)

+ dqpn(u)gn.

Multiplying both sides by gn and integrating we find (Qdqgn(u1,gn)

where Q

=

-d2/dx2

+

(d, u> = pn(dqgn(v), gn) + dqpUn(V)

+ q. The first term equals

(dqgn(v), Qgn) = pn(dqgn(u), gn).

Hence dqpn(u) = < g i , u > ,

and

However, both sides are continuous functions of q, and the continuous q are dense in Lz. So this identity holds in general.

34

Problem 2.

Inverse Spectral Theory

Show that Ipn(q) - pn(p)l 5 114 - Pll-9

n 2 1,

if q , p are in L", the space of all real valued, essentially bounded functions on [0, 11. [Hint: Write pn(q) - pn(P)

=

1;

g p n ( t q + (1 - t ) p )dt.1

We make a small digression to illustrate the usefulness of Theorem 3 . Consider the restriction of the function p n , n L 1, to the sphere 11q11 = 1 in L2. The ball 11q11 I 1 is compact in the weak topology of L2,3and p,,is a compact function on L2. Hence, pn attains its maximum and minimum on 11q11 5 1. However, we have pn(q

+ C)

= pn(q)

+c

for real c by the differential equation. It follows that the maximum and minimum must be attained on the unit sphere 11q11 = 1, for otherwise they would not be extremal. Consequently, p n has at least two critical points on the unit sphere in L2. Let q be a critical point of pn on 11q11 = 1. The gradient of p n must be proportional to the unit normal at q. That is,

where

By (*), q has exactly n + 1 double roots in [0, 11 including 0 and 1, and does not change sign. Moreover, q is infinitely often differentiable. For, g," is always continuous, hence so is q . But then g," is in C ' , hence so is q , ad in finitum. The critical point q also satisfies a differential equation. Apply the operator L of Theorem 1.7 to (*) to obtain -2pnq'

+ 3qq' - +q'" = 0.

'Recall that the sphere 11q11 = 1 is not compact.

35

The Dirichlet Problem

Integration, multiplication by q’ and another integration yields the identity -p(nq

2

+ +q3- $(q’l2= cq

with some constant c # 0. The substitution q differential equation

(**I

= 2u

+ fpn then leads to the

(u’)= ~ 4(u - el)(u - e2)(u - e3)

where e l , e2, e3 are complex numbers whose sum is zero. For distinct e l , e2, e3, (**) is the differential equation for the Weierstrass @-function. It is possible to carry this analysis further using properties of the Weierstrass @-function. But we shall not. The Counting Lemma gave us a first rough estimate of the location of the Dirichlet eigenvalues. This estimate is now refined. Let P2 denote the Hilbert space of all real sequences (a1,a2, ...) such that C a: is finite. More generally, for k L 0, Pi denotes the Hilbert space of all real sequences (a1,a2, ...) such that

C

(nkanl2<

m.

n a 1

In analogy to the notation O ( l / n ) ,we use the notation $(n) for an arbitrary sequence of numbers which is an element of Pi’. For instance, a n

=

/jn

+ p2(n)

is equivalent to an = P n

Theorem 4.

+ I,,

C

I: <

03.

n a 1

For q in L2,

and

gXx, q ) = f i n n cos rrnx

+ 0(1).

These estimates hold uniformly on bounded subsets of [0, I ] x L2.

36

Inverse Spectral Theory

Proof. We “iterate” estimates on pn and g,, , thereby sharpening them. All estimates hold uniformly on bounded subsets of [O, 11 x Lz. Let p,, = p,,(q). By the Counting Lemma,

By the Basic Estimate for y2,

Using the identity 2 sin’ax = 1 - cos 2ax, one calculates

It follows that

Thus the preliminary estimate

is obtained. Our estimate of p,, may now be improved. By Theorem 3,

(3)

The Dirichlet Problem

31

Since (2) holds uniformly for tq, 0

It I1,

this implies

pn = n2n2 + 0(1)

or equivalently

a

=

nn

+

.(:),

and in turn gn(x) = *sin

nnx

+o

(3 -

.

Inserting the last estimate of gn into (3) and using the identity 2 sin’ux = 1 - cos 2ux again, we obtain pn -

n2n2

=

j: (1

-

cos2nnx

+0

1

= j o q ( t ) d f1

=

i o q ( t ) d t + t2(n),

since (cos 2nnx, q ) are the square summable Fourier cosine coefficients of q . Finally, we estimate gA . By the Basic Estimates,

= cos nnx

+ o(:).

This proves the theorem. It will also be important to have asymptotic estimates of the squares of the eigenfunctions and the products ~ ( x4,) = YI(X,P~)YZ(X, Pn),

n

1.

38

Inverse Spectral Theory

Corollary 1. For q in L2, 2

gn =

d -gi dx

=

1 - cos 2nnx

+O

2nn sin 2nnx

+ O(1)

and 1 an = -sin2nnx 2nn

d -an = cos 2nnx dx

+0

+0

All estimates hold uniformly on bounded subsets of [0, 11 x L2.

Proof. The estimates of gx and (d/dx)gi follow from Theorem 4 by straightforward calculations. Also, = nn + O(l/n) and the Basic Estimates for y1 and y2 yield

a

.YI(X, pn)

=

(3

+O

cos nnx

-

and

1

= -sin 2nnx

an = y1y2 h

2nn

= cos2nnx

(9

+0

-

.

+0

The Dirichlet Problem

Problem 3.

(a) Show that the estimate of Theorem 4 can be improved to pn(q) =

(b)

39

n

71 2 2

+

s:

q(t)dt - ( cos Z Z ~ X ,4)

+ P!(n).

The function q is odd on [0, 11, if g(l - x ) = - q ( x ) . Show that for

odd q, pn(q) =

n2n2 + ~ ? ( n ) .

(c) Show that for q in the Sobolev space H6 with

1; q(t) dt

= 0,

The characteristic polynomial of an n x n-matrix is equal to the product

where A,, ...,A n are the eigenvalues of the matrix. There is an analogous representation for yz(1, A , q). Theorem 5. For q in L2,

Proof. We havepn = n2n2 + O(1) by Theorem 4. It follows from Lemma E.2 that the infinite product

is an entire function of A, which satisfies

p(A)

=

(

sin dX 1 + 0 -(0; n))

-

a

uniformly on the circles 1A1 = rn = (n + +)'n2 for n large enough. Its roots are precisely the Dirichlet eigenvalues of q , so the quotient p(A)/yz(l, A) is also an entire function. By the Basic Estimate for y2 and Lemma 1, sin dX

Y 2 ( l , A) =-

dX

+

o('~;'p'>

Inverse Spectral Theory

40

uniformly for 111 = r n . Hence,

for 111

=

rn, that is,

as n + 03. If follows from the maximum principle that the difference vanishes identically. Hence, p(A) = yz(1, A). If the functions yz( 1, A, q) and yz(1, A, p ) are equal, then clearly P n ( q ) = p n ( p )for all n L 1. That is, q and p have the same Dirichlet spectrum. Now we know that the converse is also true. In other words, all the information in y2( 1, A) is already contained in the Dirichlet spectrum Pun, n L 1. Here are two useful consequences of the product formula. Corollary 2.

Proof. Part (a) follows from Lemma E.3. The first identity in (b) follows from (a), while the second one is a consequence of the first one and Theorem 2 . Problem 4. Prove Theorem 5 , using Hadamard's theorem [Ti]. [Hint: Observe that y2(l, A) is an entire function of order i, so that

41

The Dirichlet Problem

with

c1 = c o n ms1

-.m2n2 Pm

Now divide by sin fi

fi and let I tend to

--03

A = n m2n2 m2n2

to obtain c1

-

msl

=

11.

The rest of this chapter is devoted to properties of the eigenfunctions. Theorem 6. (a) For every q in L2,the eigenfunction g n ,n 1 1, has exactIy n + 1 roots in [0, 11. They are aN simple. (b) If q is even, then g n is even when n is odd and odd when n is even.

4.

Even and odd mean even and odd about the point The task of counting the roots of gn is simplified by the following “deformation lemma”:

Lemma 3. Let h t , 0 It I1 , be a famiIy of real valued functions on a 5 x Ib, which is jointly continuously differentiable in t and x. Suppose that f o r every t , ht has a finite number of roots in [a, b ] , aI1 of which are simple, and has boundary values that are independent o f t . Then ho and hl have the same number of roots in [a, b ] . Intuitively speaking, the roots of hr in the interior of [a,b] move as t moves, but they can never collide or split, since they are all simple. Therefore, their number is independent of t.

Proof of Lemma 3. Suppose for convenience that ht(a) = 0 = ht(b),

0

It I1 .

Other cases are handled analogously. Fix t i n [0, 11. By assumption, the roots of ht are simple. Therefore, we can place an interval around each of them, so that ht changes sign once on the intervals in the interior, hi is bounded away from zero on all the intervals, and in addition, ht is bounded away from zero on the complementary intervals. See Figure 3.

42

Inverse Spectral Theory

Figure 3.

By continuity, h, and h: behave in exactly the same way on these intervals, if s is sufficiently close to t. Consequently, h, has exactly the same number of roots as h r , when s is close to t . This argument applies to all t in [0, 11, so the number of roots is independent of t. In particular, it is the same for ho and hl . H

Proof of Theorem 6. (a) Since g,(x, q) does not vanish identically, its roots are all simple by Corollary 1.1. Their number is finite, since otherwise they would cluster at a multiple root. Now consider the deformation hr of gn given by h r ( ~= ) gn(X, tq),

0 It, x I1.

Clearly, for each I , hr vanishes at 0 and 1 and has only a finite number of roots in [0, 11 all of which are simple. Thus we may apply Lemma 3, and it follows that hl = gn has as many roots in [0, 11 as ho = a s i n nnx. Hence, g, has exactly n + 1 roots in [0, 11. (b) Let q be even. That is, q( 1 - x) = q(x) for 0 I x I 1. Substituting 1 - x for x in -S i

+ q(x)gn = pngn

9

we see that gn( 1 - x) is another eigenfunction of q forp,, with norm 1. Hence, gn(x) = - sgn gXl)gn(l -

XI.

Since sgn gXl) = (- 1)" by Corollary 2, the claim follows.

H

One of the main objectives of classical Sturm-Liouville theory is to generalize the Fourier sine expansion. The idea is to replace the sines by eigenfunctions of a boundary value problem for a second order equation.

The Dirichlet Problem

43

This was a precursor of the abstract spectral theorem for self-adjoint operators. For the Dirichlet problem, the result is

Theorem 7. For each q in Lz, the sequence gn(x, d,

n2 1

is an orthonormal basis for L2.

=

0.

Since pm # pn for m # n, this implies (gmvgn) = 0,

m

f

n.

Since also (gn, gn) = 1 by construction, the functions gn, n 2 1, are orthonormal. It remains to show that this system is complete in Lz. We do this by comparing the vectors gn with the vectors

en = \/z sin nnx,

n

L

1,

which do form an orthonormal basis of Lz. Consider the linear operator A on Lz defined by

Af=

C

(f,en)gn.

n r 1

A is well defined, since ( f , e n ) , n L 1, is an Pz-sequence and gn, n 2 1, is an orthonormal basis. A is an isometry, since IIAfII’ =

1

n.? 1

I(f,

=

IIfII’

by the orthogonality of the gn and Parseval’s identity. In particular, A is one-to-one. Also, by Theorem 4,

Inverse Spectral Theory

44

a,

so A - I is Hilbert-Schmidt and hence compact by Theorem D. 1. It follows

from the Fredholm Alternative, Theorem D.2, that A is onto L2 and has a bounded inverse. Therefore, gn, n 1 1, is an orthonormal basis for L2. There are more elementary proofs of the completeness of the Dirichlet eigenfunctions, which do not appeal directly to the Fredholm Alternative. See for instance Birkhoff and Rota [BR], who also make a number of interesting historical remarks. The method used here, however, is easy to generalize. Roughly speaking, any sequence of vectors, which is linearly independent and “sufficiently close” to some orthonormal basis, is itself a basis. Theorem D.3 makes this statement precise. It will be very useful in the sequel. Actually, we will never need Theorem 7, but include it as a prototype for other completeness theorems. For us, it will be much more important to settle independence and completeness questions about squares of eigenfunctions, for example in the proof of Theorem 3.2. The next theorem is essential for this and many other purposes. Recall that

Theorem 8. For m , n 2 1,

It is only a slight exaggeration to say that Theorem 8 is the basis of almost everything else we are going to do.

The Dirichlet Problem

45

Proof. (a) Integration by parts yields

lo 1

=

This clearly vanishes for m

=

gmgn[gm gnl dx.

n. If m # n, then p m # p,, and we can use

[gm, gnl’ = ( p m - punlgmgn to obtain

(b) Again, by partial integration,

If rn

=

n, then y2 is a multiple of g,, hence [ y z ,g,]

=

1,

using the Wronskian identity. (c) is proven in the same manner.

=

0, and we get

Inverse Spectral Theory

46

For q

=

0, the vectors 1,

n L 1, are a basis for L2[0,11. Theorem 8 and the asymptotic estimates of Corollary 1 make it possible to prove the same statement for every q . First, let us specify the notions of linear independence and basis. A sequence of vectors u1 , u 2 , ... in a Hilbert space is linearly independent, if none of the vectors un , n 1 1, is contained in the closed linear span of all the other vectors urn, m # n. One might try to define a stronger notion of linear independence by requiring U n to lie outside the weak closure of the span of urn, rn # n. However, if a sequence of vectors converges weakly, then the arithmetic means of some subsequence converge strongly to the same limit by the theorem of Banach-Saks [Ba]. Thus, the weak and strong closure of a linear space are the same, and nothing is gained. A sequence of vectors u l , u 2 , ... is a basis for a Hilbert space H, if there , ...) such that the correexists a Hilbert space A of sequences a = ( a ] a2, spondence a+

C

anvn

n z1

is a linear isomorphism between A and H.4 Theorem 9. A t every point in L2, the vectors 2

1,g, - 1,

n

2 1

are linearly independent. So are the vectors

The two sequences are mutually perpendicular, and together are a basis for L' . Precisely, ((,v)+

d

C t n z d + V O + nCa 1 Vn(g,'

n z1

is a linear isomorphism between pf x R x

P

- 1)

and L2.

4 A linear isomorphism between Banach spaces is an isomorphism between vector spaces, which is continuous and has a continuous inverse.

41

The Dirichlet Problem

It follows from Theorem 6 that at an even point in L2,the vectors 1, g,' - 1, n 2 1, are a basis for the subspace of even functions in L2, and the vectors (d/dx)g,', n 1 1, are a basis for the subspace of odd functions.

Proof. The vector g,' - 1, n 2 1, is not in the closed linear span of the vectors 1, g& - 1, m # n, since

but

by Theorem 8. Similar arguments apply to all other vectors. The two sequences are mutually perpendicular, since

for all m ,n again by Theorem 8. By Corollary 1, 2

gn - 1 =

- ~ 0 ~ 2 n n+x 0

1=1 1 d -g,' = sin27rnx 27rn dx

+o

We have shown that the vectors on the left are linearly independent. The vectors on the right without the error terms are (up to an irrelevant factor an orthonormal basis of L2, and the error terms are square summable. Therefore, Theorem D.3 applies. It follows that the vectors on the left are a basis with coefficients in x R x ?. From this, the last statement of the theorem follows.

a)

It is useful to write Theorem 9 in a slightly different form.

Inverse Spectral Theory

48

Corollary 3. A t every point q in L2, the subspaces

are perpendicular and closed. Their direct sum is all of L2.In particular, they are the odd and even subspaces respectively when q is even.

Problem 5. (a) Show that log N Conclude that 1 is in the closed linear span of gn', n L 1. (b) Show that the vectors gn', (d/dx)gn',n 1 1, span L2 (that is, finite linear combinations are dense) and are linearly independent, but not a basis. The basic facts about the Dirichlet eigenvalues and eigenfunctions are now proven. We are ready to approach the inverse Dirichlet problem.

3

The Inverse Dirichlet Problem

In the last chapter we introduced the Dirichlet spectrum associated to a function q in L2 and derived some of its properties. We found that the Dirichlet eigenvalues p,, , n 2 1, form a strictly increasing sequence of real numbers satisfying p,, =

n2n2 + c

+ P(n),

where c = 1; q(t) dt. It is natural to ask whether these conditions actually characterize all possible Dirichlet spectra. For instance, the sequence n 2 n 2 ,n 1 1, is the Dirichlet spectrum for q = 0. Suppose the first eigenvalue 71' is replaced by any number p1 below the second eigenvalue 4n2. Is the modified sequence still the Dirichlet spectrum of some q in L2? Perhaps, pl has to be chosen in a special way? It is also natural to ask to what extent a point p in Lz is determined by its Dirichlet spectrum. For instance, are there any functions q in L2 besides q = 0 with Dirichlet eigenvalues pun= nzn2,n 1 1 ? More ambitiously, what does the isospectral set M(P) = 14 E L2:P"(4) = Pn(P), n 1 1)

of all functions q with the same Dirichlet spectrum as p look like?

49

50

Inverse Spectral Theory

We have come to the inverse Dirichlet problem. It has two parts. First, describe the isospectral sets M ( p ) for all p in L2. Second, characterize all sequences of real numbers which arise as the Dirichlet spectrum of some q in L2. These problems will occupy us in this and the next three chapters. Surprisingly, they both have complete solutions.

Problem 1. Calculate the spectrum of the eigenvalue problem iy'

+ q(x)y = l y ,

0

I

x

I

1

Y(0) = Y ( 1 )

for q in L2. Make a list of all possible spectra and decide when two qs have the same spectrum. We begin by reformulating the inverse Dirichlet problem in more geometrical terms. For this purpose it is convenient to introduce the space S of all real, strictly increasing sequences 0 = ( ( T I , 0 2 , ...) of the form =

(T,

where s E R and 6 =

n2n2 + s

+ en,

nL 1

(al,62,...) E p . By Theorem 2.4, the map 4

+ P(4)

= (P1(4),P2(4),

...)

from q to its sequence of Dirichlet eigenvalues sends L2 into S. Characterizing spectra is equivalent to determining the image of p , and isospectral sets are the fibers of p , that is, W P ) = P-l(P(P)) =

( 4 E L2:P ( 4 ) = P(P)I.

Thus, solving the inverse Dirichlet problem amounts to analyzing the map p . For this purpose, observe that the correspondence between (T and (s, 6) is a one-to-one map between S and an open subset of IR x p . We shall identifv S with this open subset, and refer to (s,6) as the standard coordinate system on S. This identification allows us to do analysis on S as if it were an open subset of a Banach space.' In these coordinates the map p becomes 4 'See Appendix C, Example 3.

+

([41,P(4))-

The Inverse Dirichlet Problem

51

where [41 =

s:

q(t)dt

is the mean value of q, and p(q) = ( f i n ( @ , n 1 1 ) is the sequence defined by

P-n(Q)= P&)

-

n2n2 - [ql.

Theorem 1. ,u is a real analytic map from L2into S. Its derivative at q is the linear map from L2 into R x P given by dqr(((u)= ([vl, (gn' - 1, v > , n 2 1 ) .

Proof. Fixp in L2,and let N b e an integer. By the analyticity of the Dirichlet eigenvalues, there exists a complex neighborhood UN c LE of p such that P I , ...,p,v extend to analytic functions on UN and remain simple roots of y2(l, I ) . We show that such a neighborhood can be chosen independently of N. That is, all eigenvalues extend analytically to some fixed complex neighborhood of p . Choose N > 2elllll, and let V c LE be a complex neighborhood of p such that 2ellqll < N,

q

E

V.

By the Counting Lemma, there is a unique, simple root jin(q)of yz(1, I , q) in the egg-shaped region 1 6 - nnl < n / 2 for all n > N a n d q E V. Moreover, for real q, P"(4) = Punk?),

q

E

V fL li *

It follows from the implicit function theorem that f i n is an analytic extension of pun to V . Thus, all eigenvalues P I ,p2, ... extend analytically to the complex neighborhood U = UN n V of p , remaining simple roots of yz(1, A). There is a similar analytic extension of

to U,since 32(1, pun)# 0 and yi( 1, p,,) # 0 on U by construction. Proceeding word for word as in the proof of Theorem 2.4, one finds p n ( q ) = n2n2 + [q] - (cos 2nnx, q )

+o

52

Inverse Spectral Theory

uniformly on U. Hence, the extension of the map p(q) = ([q],p(q))is a bounded map from U into the complexification of R x P , all of whose coefficients are analytic. By Theorem A.3, the map p itself is analytic on U. In addition, its derivative is given by the derivatives of its components, and we have

by Theorem 2.3. The preceding argument applies to any point p in L2. Hence, p is real analytic on L ~ . It is easy to see that the m a p p is not globally one-to-one. Let * denote the reflection across the subspace of all even functions in L2. That is, q*(x)

= q(l -

x),

Ix I1.

0

Lemma 1. p(q*) = p(q). Loosely speaking, the Dirichlet spectrum cannot distinguish left from right.

Proof. One easily verifies that the reflected eigenfunction g?, n 1 1, is a Dirichlet eigenfunction of q* with eigenvaluepu,(q).It has exactly n + 1 roots in [0, 11. Therefore, p n ( q ) = pn(q*),

by Theorem 2.6.

n2 1

H

A small, but enlightening step towards understanding the behavior of p is to calculate its derivative at q = 0. It is the linear map from L2 into R x P given by u

since at q

=

-+

( [ u ] , -(cos2nnx, u ) , n

0,

ap,

-=

a4

g,'(x, 0) - I

= 2sin2nnx - 1 = -cos2nnx.

2

l),

The Inverse Dirichlet Problem

53

The kernel of dop is the subspace of all functions in L2 whose Fourier cosine coefficients vanish. Hence, it is the space

u = (q E L2:q*

= -q]

of all odd functions. Its orthogonal complement is the space E = (4 E L2: q*

= 4)

of all even functions, and dop is a linear isomorphism, in fact an isometry, between E and IR x Pz by the elementary theory of Fourier series. To exploit this observation, let p~ be the restriction of p to E. By the last paragraph,

is a boundedly invertible map between E and R x p. It follows from the inverse function theorem that p~ is a real analytic isomorphism between a neighborhood of 0 in E and a neighborhood of p(0) = (nzn2,n 1 1) in S . Hence, any small but arbitrary perturbation of the sequence n2n2,n 1 1, is the Dirichlet spectrum of an even function. In particular,

...

p1, 4n2,9n2,

is a sequence of Dirichlet eigenvalues, whenever pl is sufficiently close to n2. This analysis gives a local solution to one of the questions raised at the beginning of this chapter. Still, we want to know what happens when p1 is moved far to the left of n2 or very close to 4nZ. Such global questions will be answered in Chapter 6. We can extract some more information. Using the splitting Lz = E 0 U, write p(q) = p(e, u),

=

e+

u.

The derivative of this map with respect to e at (0,O)is an isometry of E and IR x P2. By the implicit function theorem, there exists a real analytic map u e(u) between a neighborhood of 0 in U and a neighborhood of 0 in E such that +

pu(e(u),u ) = ~(01,

do) = 0.

The graph of e in E 0 U is contained in M(0) and is homeomorphic to its projection onto U. Consequently, there is an infinite dimensional set of functions with the same Dirichlet spectrum as q = 0.

54

Inverse Spectral Theory

Figure 1.

Problem 2.

(a) Show that the map 4

+

( P ( 4 ) , t ( 4 - Q*))

from L2 to S x U is a coordinate system in a neighborhood of q = 0. [Hint: Write Lz = E 0 U. The map becomes (e, u ) (p(e,u), u ) . Compute the derivative at 0 and use the inverse function theorem.] (b) Show that in this coordinate system, the map from q to its Dirichlet spectrum is the projection of S x U onto S. (c) Conclude from (a) and (b) that for each e in a sufficiently small neighborhood of 0 in E, the space M(e) contains a set homeomorphic to an open subset of U. -+

Problem 3. The restriction of p to the space U of odd functions is even: = p(-q),

4E

u.

Show that on U the Taylor expansion of p,, at 0 is given by pn(q) = n2n2

+ 2(-

l)n+1n2nZ * ~ 2 ( 1 nzn2, , q) + 0(11q11~),

where SZ is the second order term in the power series expansion of y2 given in Theorem 1 . 1 . [Hint: By evenness, Pn(d

=

nzn2 + mz(a) + O(1141I4)1

Theorem 2. p~ is a local real analytic isomorphism at every point in E.

55

The Inverse Dirichlet Problem

Proof. We show that the derivative of

,UE is boundedly invertible everywhere on E. Then the result follows from the inverse function theorem. In the standard coordinates on S, the derivative of p~ is the linear map from E into IR x I’ given by

u

+

UUI,

, , n

L

1).

At every point in E, the vectors 2

1 , g n - 1,

n

L

1

are a basis for E by Theorem 2.9. Therefore, dqpE is boundedly invertible at every point in E. Each eigenvalue p n , n 2 1, is a compact function on L2 by Theorem 2 . 3 . On the other hand, p~ is open by the preceding theorem and therefore not compact. It turns out that p~ is one-to-one not only locally, but even globally. In other words, an even function is uniquely determined by its Dirichlet spectrum. This important fact was first proved by Borg [Bo]. A simplified proof was given by Levinson [Le]. We give another proof. Theorem 3.

is one-to-one on E.

Before presenting the proof, we make a slight digression into complex analysis. Lemma 2. Let f be a meromorphic function in the plane. If

for an unbounded sequence of positive real numbers rn, then the sum of the residues o f f is zero. In particular, i f the residues o f f are all nonnegative, then they are all zero. The residues are not necessarily absolutely summable, so their “sum” may depend on the order in which they are added together. We define the sum of the residues, by the Cauchy residue theorem, as the limit of the partial sums

56

Inverse Spectral Theory

Proof of Lemma 2. The sum of the residues o f f is, by definition,

But

so the sum is zero. Lemma 2 is an analogue of the fact that the sum of the residues of a meromorphic 1-form on a compact Riemann surface is zero. Since the plane is not compact, a growth condition is necessary. For instance, the residues of f i / s i n fi are 2(- l ) ' n ' ~ ~n, 2 1, and their sum is not even defined.

Proof of Theorem 3. Suppose q a n d p are even functions withp(q) = p(p). We have to show that q = p. Consider the meromorphic function

- [(YZ(X, I , 4 ) - Y Z k

I

9

P)l[(Yz(l - x , I , q ) - YZ(1 - x , I , PI1 YZ(1, A 4)

It has simple poles at pun, n 2 1, otherwise it is regular. As a consequence of Theorem 2.6, Y2(X, pun) = (- l)"+'YZ(l

- x , pun),

n

2

1,

for q and p. It follows that the residue of our function at p,,is

using (b) of Corollary 2.2. We now show that our function satisfies the hypothesis of Lemma 2 for r, = (n + &)2n2.By Theorem 1.3, its numerator is bounded from above by a constant multiple of JXlx elIm JXl(1 - x )

elImJXI

The Inverse Dirichlet Problem

T o bound the denominator from below we observe that

for

1 8e11q11. Therefore, by Lemma 2.1,

1 ellm&I

2 --

4

1elImfil

- --

)a)

8

lfi)

1 elImJxl 2 --

8

la1

for [A1 = r n and n sufficiently large. The quotient of the two bounds is = o( 1/ r n ) as required. O(1 So Lemma 2 applies, and it follows that

n 1 1.

YZ(X,Pn 9 4) = YZ(X, Pn 9 PI,

Hence, q a n d p not only have the same Dirichlet spectrum, but also the same eigenfunctions. This is actually more than we need to know. If only, say, ut(x, PI , 4) = Y ~ XP I, ,PI, then qy2 = P1(4)Y2

+ r4 = Pl(P)Y2 + yz

by the differential equation and therefore q By the last two theorems,

PE

=p

= py2

almost everywhere.

H

is a one-to-one real analytic map from E into

S, whose inverse is also real analytic. Thus, P E is a real analytic isomorphism

between E and an open subset of S. We will see in Chapter 6 that P E maps onto S.

Problem 4. (Another proof of Theorem 3): (a) Let ZIand 2 2 be the solutions of -y"

satisfying

+ q(x)y = Ay,

0

Ix I1,

58

Inverse Spectral Theory

Show that

ZdX,

I,4 ) = -Y2U

-

x , A, 4 * ) .

(b) Suppose q a n d p are even functions with the same Dirichlet spectrum. Show that

where x and I have been suppressed. [Hint: Write down the inverse and multiply out. Check that the entries of the result are entire functions of I bounded on circles of radius (n + i)’n2. Then use the maximum principle.] (c) Conclude from (b) that two even functions with the same Dirichlet spectrum are equal.

Problem 5. (Still another proof of Theorem 3 following Borg [Bo]): (a) Suppose q and p have the same Dirichlet spectrum. Show that’ ( 4 - P I gn(q)gn(P)) = 0,

n

2

1.

[Hint: Cross multiply the differential equations for g n ( q ) ,g n ( p ) and then subtract and integrate.] (b) Suppose q and p are even. Show that the sequence 1 , gn(q)gn(p) - 1, n I1 is a basis for E. [Hint: To prove independence, calculate the inner product between these vectors and

as in the proof of Theorem 2.8. Otherwise imitate the proof of Theorem 2.9.1 (c) Conclude from (a) and (b) that two even functions with the same Dirichlet spectrum are equal. [Hint: ( q - p , 1 ) = 0 by the asymptotics of the /.In .] The map from q to its Dirichlet spectrum is one-to-one on E , but, as we have seen, not on all of L2. Additional data are necessary to determine q . * W e frequently suppress x and write g,(q) for gn(x, q ) to simplify notation.

59

The Inverse Dirichlet Problem

We introduce the quantities

where the logarithm will be useful later on. Equivalently, Kn(q) =

log I Yi(l9 Pn)I

= log (- l ) n Y i ( l ,p n )

by Corollary 2.2. The numbers K I , ~ 2 ..,. are essentially the “terminal velocities” mentioned at the beginning of Chapter 2. The following theorem lists the basic properties of the K,,.

Theorem 4. Each K,, ,n 1 1 , is a compact, real analytic function on L2 with asymptotic behavior Kn(q) =

1 -(sin 2nn

=

&n).

2nnx, q )

+0

Its gradient is

=

1 -sin2nnt 2nn

+0

The error terms are uniform on bounded subsets of [0, 11 x L2. Recall that an = yl(x, pn)y2(x,pn).

Proof. yi(1, A, q ) and p n ( q )are compact, real analytic functions of ( I , q) and q , respectively. Since yi(1, p n ( q ) ,q) never vanishes, Kn(q) =

1% (- l ) n Y i ( l ,P n ( 4 ) ~4 )

is also compact and real analytic. By the chain rule,

Inverse Spectral Theory

60

Applying Theorem 1.6,

and

Inserting these identities into the expression for a K , , / a q and using

one obtains

1 2nn

+0

= -sin2nnt

The last line follows with Corollary 2.1. Finally, since ~ ~ ( =0 0,)

=

=

so 1

(ksin2nnt

1 -(sin2nnx,q) 2nn

+0

+0

Problem 6. (a) Show that for q in L2, actually

1 2nn

Kn(q) = -(sin

2nnx, q )

+ &n).

(b) Show that q is in the Sobolev space H' if and only if (cos 2nnx, q ) = t:(n) (sin 2nnx, q ) where c

= (q(0) -

q(1))/2n.

C =-

n

+ &n),

The Inverse Dirichlet Problem

61

[Hint: The Fourier sine coefficients of the odd function x - i are

-d2/nn, n

2 1.1

(c) Show that pn(q) =

n2n2 + 141 + R(n)

if and only if q is in H' and c = (q(0) - q(1))/47r2. [Hint: Recall Problem 2.3 and integrate by parts.] The gradients of Kn and p,,satisfy a set of simple relations, which will be essential later and are a direct consequence of Theorem 2.8.

Lemma 3. For m, n 1 1,

Proof. For instance, by Theorem 2.8,

By Theorem 4,the map 4

-+

K(4)

=

(Ki(q), Kz(Q), ...)

from q to its sequence of K-values maps Lz into the Hilbert space &?of all real sequences ( = ((1, ( 2 , ...) satisfying C n r 1 n2(i < 03. Combining K and p, we obtain a map 4

-+

(K

x co(4) = (K(d, p(q))

Inverse Spectral Theory

62

from L2 into the product space l?? x S . This map will play a central role in solving the inverse Dirichlet problem. As a first step we show that the K , are sufficient supplementary data to the Dirichlet eigenvalues to determine q uniquely. Theorem 5.

Proof.

K

x

p is one-to-one on L2

Suppose that

K(q) =

~ ( pand ) p(q) = p ( p ) . We have to show that

4 =P.

We vary the proof of Theorem 3 and consider the meromorphic function

-

[Y2(X,

A, 4) - Y 2 ( X , A,P)l[Y2(1 - x, 1,q*) - YZ(1 - x, A,P*)l , Y 2 ( l , A, a)

which has simple poles at p, , n 1 1. Our assumption K ( q ) = ~ ( pimplies )

Yi(l,Pn,q)= ~ i ( l , ~ n , P ) Moreover, by reflecting x into 1 - x ,

and similarly for p , since both sides are solutions of -y” + q*y = p,y with the same initial data at x = 1. Therefore, the residue of our function at pn is

Y ~ ( xpun, p)12 L 0. Pn 4)Yi(l Pfl9 PI

[ Y Z ( X , Pn 1 4) )’2(1,

9

3

9

We can also verify the hypothesis of Lemma 2 for r, = ( n + *)’7r2 by the same estimates as in the proof of Theorem 3. The argument is then completed as before. H While I( does not change upon reflecting q by Lemma 1, K changes sign. From this and the preceding uniqueness result we obtain a characterization of the even functions. Lemma 4. K(q*) = - K W

In particular, q is even i f and only i f

K(q) =

0.

The Inverse Diriehlet Problem

Proof.

63

In the proof of Theorem 5 we noticed that

On the left we can replace pn(q) by pn(q*) by Lemma 1 , hence

This holds for all n z 1 , so K ( q * ) = - K ( q ) . In particular, if q is even, then q = q*, and consequently K ( q ) = 0. Conversely, suppose K ( q )

=

0. Then

K(q*) = - K ( d

= K(q)

P ( 4 * ) = 1((4),

the second line being the content of Lemma 1 . Since the map K x p is one-toone, we obtain q* = q. That is, q is even. W Next we study the analytic properties of the map K x p. It is not very surprising that this map is a local real analytic isomorphism at every point. More remarkably, its derivative can be inverted explicitly. It is convenient to introduce some notation. For n 2 1 , set

We have V, = 4nn sin 2nnx

+ O(1)

Wn = -2cos2nnx

+0

(3 -

uniformly on bounded subsets of [O, 13 x L2 by Corollary 2.1.

Theorem 6 . K x p is a local real analytic isomorphism at every point in L2. Moreover, the inverse of dq(K x p) is the linear map from e: x R x e2 onto L2 given by (dq(K X P))-'(<$V ) =

C

n r I

LVn

+ vo + C

n>l

V n Wn

-

Inverse Spectral Theory

64

Proof. p is real analytic on L2 by Theorem 1. We have to show that also is real analytic. Fix p in L2. We showed in the proof of Theorem 1 that there exists a and g; extend as analytic complex neighborhood U of p , to which all functions of q, satisfying the estimates of Theorem 2.4. Therefore, K

Kn =

log(- l)",,Y,$(l,Pn),

2

1

and

are also analytic on U,sinceyi(1, p n ) does not vanish. Moreover, imitating the proof of Theorem 4, we have

uniformly on U.It follows that K extends to a bounded map from U into the complexification of f f , all of whose components are analytic. Hence, K is analytic on U by Theorem A.3. The preceding argument applies to any point p . Therefore, the map K is analytic on L'. By Theorem A.3, the derivative of K x p, in the standard coordinates on S , is the linear map from L2 into ff x R x given by

We show that this map is boundedly invertible. Then, by the inverse function theorem, K x p is a local real analytic isomorphism. By Theorem 4 and Corollary 2.1, aKn

2nn-

a4

=

sin2nnx

+0

1=1 _ - -cos2nnx+ 0 a4

These vectors are linearly independent, since, by Lemma 3, for each one of them there is another vector which is perpendicular to all vectors but the given one. Moreover, the vectors on the right, without the error terms, are an

65

The Inverse Dirichlet Problem

a),

orthonormal basis of L2 (up to an irrelevant factor and the error terms are square summable. Thus, Theorem D.3 applies, and the above map, with aKn/dq replaced by 2nn aKn/aq, is a linear isomorphism between L2 and a? x IR x a?. Hence, the original map is a linear isomorphism between L2 and P: x IR x P2. Now consider the inverse. By the asymptotics of K, and Wn,

-

=

C

tnK,

n.z 1

converges in L2when 4 E

+ vo + C

n

~

and q E R x P2. By Lemma 3,

hence d9(K x

This proves the theorem.

v

n a l

P)(U)

=

(r,v).

W

As a special case we obtain the inverse of the derivative of p~ . By Theorems 2 and 3 , this map is a real analytic isomorphism between E and an open subset of s. ~ the linear map from Corollary 1. For q in E , the inverse of d 9 p is IR x a? = T,(,)S onto E 2: T,E given by (dqpE)-l(V)= v o

+ n a 1 VnK .

In particular, the sequence 1, Wn, n 2 1, is a basis for E. It is implicit that the functions Wn are even at an even point q .

Proof. By the last theorem, K x p is a real analytic isomorphism between L2 and an open subset of P: x IR x a?. In particular, by Lemma 4, the linear subspace E is mapped isomorphically to the intersection of this open set with the linear space 0 x IR x a?. It follows that p i 1 is the restriction of ( K x p)-' to this intersection, and (dqp$' is the restriction of (dq(K x p))-' to the linear space 0 x IR x a?. This proves the identity. The second statement is immediate, since (d,pE)-' is a linear isomorphism.

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4

Isospectral Sets

How does the isospectral set

lie inside L2? Is it connected? Bounded? Is it a submanifold? In this chapter, we shall describe M ( p ) by answering questions of this kind. Our basic intuition about isospectral sets comes from a simple picture in which we view M ( p ) as the intersection of infinitely many hypersurfaces. Namely,

n

M ( ~ ) = Mn(p), na1

where Mn(P) = 14 E L2:Pn(4)

=

Pn(P))

is the set of all functions in L2 with the same nth Dirichlet eigenvalue as p .

67

68

Inverse Spectral Theory

Each of these sets is a real analytic submanifold of L2 of codimension one, since the gradients

never vanish identically. We saw at the end of Chapter 2 that the gradients gn', n L 1, are linearly independent at every point in L2. This result now tells us that every finite intersection M d p ) n ... n Mn(P)

is a real analytic submanifold of L2, whose normal space at q is spanned by the vectors' g:(q),

* **9

gk).

Hence there is reason to conjecture that the infinite intersection M ( p ) =

On,I Mn(p) is also a real analytic submanifold of L2, whose normal space at q is the closed linear span of the gradients gn'(q), n 2 1. The linear independence of the gradients, however, is not enough to guarantee this. A simple counterexample is given in Appendix C. More has to be done. We will show that at every point q on M ( p ) , the derivative dqp is a linear isomorphism between the orthogonal complement of its kernel and the tangent space to S at p ( p ) . Then p ( p ) is a regular value of the map p, and by the regular value theorem in Appendix C, M ( p ) is a real analytic submanifold of L2. Moreover, its tangent space at q is T q W p ) = kerdqp,

the kernel of dqp, and its normal space at q is NqM(p) = ker' dqp,

its orthogonal complement. As we will see in a moment, they are the closed linear spans of the vectors (d/dx)g:, n 1 1, and g i , n 2 1, respectively. Let us verify our intuitive picture. To simplify notation, set

uo =

1

u n = g n ' - 1,

n L 1,

' We frequently suppress x, and for example write g,(q) for gn(x,q) to simplify notation.

69

lsospectral Sets

and, as in Chapter 3,

Vn

=

d 2-gg,', dx

n 2 1.

These vectors depend on x and 4. We write ~7

C

=

tlnun

?I20

and

v, = n C2 1 LV",

r

where q E IR x ? and E P?. By the asymptotics of g,' - 1 and (d/dx)gi, these sums converge in L2. By Corollary 2.3,

( U v :q

E

(Vt:

r

m

x P2],

E

Pa

are perpendicular, closed linear subspaces, whose direct sum is L2.

Theorem 1. (a) For allp in L2,M ( p ) is a real analytic submanifold of L2 lying in the hyperplane of all functions with mean [ p ] . (b) A t every point 4 in M ( p ) , the normal space is N,M(P) = (Uv(4):tl

E

m

r

E

x P2L

and the tangent space is G W P ) = (W):

61.

Proof. Let 4 be a point in M(p). The derivative o f p at 4 is the linear map from L2 into IR x P2 given by dqp(w) = (( un, w > ,n 2 0). We have the splitting

L2 = ker,

0 ker;,

where ker, is the kernel of d,p. We show that the restriction of d,p to ker; is boundedly invertible. Then, p ( p )is a regular value of p , and by the regular value theorem, M ( p ) is a real analytic submanifold of L2. Moreover, TQM(p) = ker, and N,M(p) = ker; .

70

Inverse Spectral Theory

We have kert

=

[U,, c =

1

V n U n : r,~E IR x P2 ,

nzo

since, by the form of the derivative, the vectors Un are perpendicular to ker,, any vector perpendicular to them belongs to kerq, and the fight hand side is closed by Corollary 2.3. By the way, the sequence Un, n L 0, is a basis of kert . With respect to this basis, the restriction of dqp to ker; is represented by the matrix operator

acting on R x P2. By Corollary 2.1,

Using the Bessel inequality, we have

and similarly for the other terms. This shows that D - Zis Hilbert-Schmidt, hence compact. D is also one-to-one, since the Un are a basis of k e r t . It follows from the Fredholm alternative that D is boundedly invertible, as was to be proven. We have shown that M ( p )is a real analytic submanifold with normal space

.

N q M ( p ) = kert

=

(U,,(q):q

E

R x P2).

Its tangent space is the orthogonal complement of ker t ,that is,

T,M(p) = ker,

= [Q(q):

t E P?)

by Corollary 2.3. Finally, for q in M ( p ) ,

by Theorem 2 . 4 . So M ( p ) lies in the hyperplane of all functions with mean [ p ] .

Isospectral Sets

71

Problem 1. Show that a tangent vector V can be expanded as V=

C (V,an)h,

n r 1

and a normal vector U as

Problem 2. Show that for all real A ,

Verify the identities

Our picture of intersecting hypersurfaces is primitive, but quickly led us to a good understanding of the local properties of M ( p ) . There is another, more global picture of isospectral sets : we view M ( p ) as a horizontal slice of the real analytic coordinate system K x p on L2constructed in Chapter 3. Namely, M ( p ) is the set of all points whose p-coordinate is p ( p ) . We have, almost at once, an alternate version of Theorem 1.

Theorem l*. (a) For allp in L2,M ( p ) is a real analytic submanifold of L2,lying in the hyperplane of all functions with mean [ p ] . (b) A t every point q in M ( p ) , the normal space is Nqiu(p) = {u,:9 E R x

e2),

and the tangent space is T,M(p) = 16:

r

E

P?).

(c) K is a global coordinate system on M ( p ) . Its derivative dqK is an isomorphism between T,M(p) and T,(,,P? = P?, which is given by

dqK(6) =

r.

12

Inverse Spectral Theory

Figure 1.

In more detail, K is a real analytic isomorphism between M ( p ) , as a submanifold of L', and its image K ( M ( ~ )an ) , open subset of P:.

Proof. M ( p ) is a real analytic submanifold, and K is a global coordinate system on it, by the properties of the K x p-map and the definition of submanifold in Appendix C. Its tangent space is

by Corollary 2.3. Finally, d*K(I/E)=

by Lemma 3.3.

r

Isospectral Sets

13

Every isospectral set is a real analytic submanifold of L2 isomorphic to an open subset of p:. This more than answers one of the questions raised at the beginning of this chapter. We now go on to answer the other questions. Let M ( p ) be an isospectral submanifold, and let q be a point on it. Every tangent vector to M ( p ) at q is of the form

V,=

c

na

rnKl 1

<

with uniquely determined coefficients = ( ( I , &, ...) in p:. These coefficients in turn uniquely determine a tangent vector at every other point on M ( p ) by the same expression, since the K, are globally defined. In this way, every tangent vector V,at a given point determines a globally defined vectorfield on M ( p ) , which we denote by the same symbol. The space of these vectorfields is isomorphic to any tangent space & M ( p )and hence isomorphic to p:. We are going to study the solution curves of V,. By definition, a curve

on M ( p ) is a solution curve of the vectorfield V , with initial value q, if

and 4'(q) = q. We implicitly assume that the curve is differentiable in t, and that a < 0 c b. See Appendices B and C for solution curves of vectorfields in general. The local existence and uniqueness of such solution curves is immediate. In the K-coordinate system on M(p),the vectorfield V, becomes the constant vectorfield by Theorem l*.Consequently, any solution curve in this coordinate system is a straight line, and we must have

<

Applying the real analytic map K - ' , we obtain an expression for 4'(q, 6).It follows that +'(q, V,) is an onalytic function o f t , ( and q, and is defined as long as the straight line K ( q ) + t( remains inside the open set x(M(p)). The local existence and uniqueness of @(q, V,) may also be established without reference to the special coordinates K , by looking at M ( p )alone. The argument is outlined in the following Problem.

Inverse Spectral Theory

74

Problem 3 (Another local existence and uniqueness proof). Show directly that V, is a real analytic vectorfield on M ( p ) , in fact, on all of L2. Apply the local existence and uniqueness theorem in Appendix C and conclude that there is a unique real analytic solution curve of Q for every initial value on M(Pb2 [Hint: Fix in P:. One can show as in the proof of Theorem 3.1 that

<

V,

=

4nn sin 2nnx

+ O(1)

uniformly on a complex neighborhood of every point in L2. Thus the sum C C n V , consists of two terms, one independent of q, and one converging uniformly on every such neighborhood. Apply Theorem A.2 to conclude that V, is real analytic for each (1. Our immediate goal is to show that each solution curve exists for all time. This is done in the usual manner by deriving an a priori bound for 11#(q, V,)II. First. two technical lemmas.

Proof. Using the differential equation and Theorem 2.2 we find

'This is not enough to show, however, that d'(q, V,) is also real analytic in <. For the stronger result, one must show that V , is real analytic in q and <.

lsospectral Sets

Thus, by the definition of

Kn

and

=

an,

46, sinh K n .

We notice that the functions 6, are positive on L2 and have the asymptotic behavior

by Corollary 2.2. Moreover, by the same Corollary, they are constant on every isospectral set, since

depends only on the Dirichlet spectrum of q ,

Proof. For 6'(q) = &(q, &), we have

Inverse Spectral Theory

76

by Lemma 1 . To obtain the derivative at time t # 0, replace q in the above expression by 4'(q) and make use of the fact that 4'+'(q) = 4'(4'(q)).Then

=4

C

t n d n ( q ) sinh ( K n ( q )

n r 1

+ ttn),

since Gn(d'(q)) = 6 n ( q ) and ~n(4'(4))= K n ( q ) + tr. The right hand side converges uniformly on bounded intervals of time, because tnsn

sinh ( K n

+ trn) = O ( t n S n ( K n + t c n ) ) =

for bounded t , and C n'ti summation sign to obtain

O(n2t,2)

< 00. We may therefore integrate under the

Theorem 2. For every q on M ( p ) and every ( in f i , the solution curve #(q, 6)exists for all time. This together with the identity of Lemma 2 implies that the set M ( p ) is unbounded.

<

Proof. Fix q in M = M ( p ) and in f i . Theorem 2 is equivalent to the statement that the straight line K ( q ) + tc never leaves the open set K ( M ) . Suppose, to the contrary, it does so for some positive t , say. Then there exists t* > 0 such that K ( q ) + t*< lies on the boundary of K ( M ) ,while the segment K ( q ) + tt, 0 c t c t * , lies in its interior. By Lemma 2,

Hence we can choose a sequence

tn

converging to t* from below such that

+'"(q) converges weakly to some point q* . This point lies on M with K(q*) =

K(q)

+ t*r,

77

Isospectral Sets

since the punand K,, are compact functions on L2. It follows that lies in the interior of K ( M ) .That is a contradiction. Problem 4. That is,

K(q)

+ t*<

Show that the flows of any two vectorfields V, and &commute.

dm&, GI, V,) = dr(dJS(49V,), &I for all s, t. Using the notation of Appendix B, this can be written more succinctly as

@i.@i= @.@, where @c, @( denote the flows of the vectorfields V,, & respectively. We can now define the exponential map exp, at a point q in M(p). This is the map from T,M(p) into M ( p ) given by exp,(Q) = 674, &)

Theorem 3. For all q in M ( p ) , the exponential map exp, is a real analytic isomorphism between T,M(p) = 6 and M ( p ) . It satisfies K(exPq(&)) = K ( 4 ) +

r.

It follows that M ( p ) is connected and simply connected.

Proof.

The identity K(exPq(&)) =

K ( d

+

r

follows immediately from the definition of exp, and the fact that & is the constant vectorfield 4 in the K-coordinates on M(p). This identity shows that the K-coordinate system maps M ( p ) onto 8, and that exp, is the inverse of the translated map r + K(r) - K ( q ) . The latter is a real analytic isomorphism between M ( p )and 6by Theorem I*, so the exponential map is a real analytic isomorphism between 6 2: T,M(p)and M ( p ) . Corollary 1. On each isospectral manifold M ( p ) , there is a unique even point. It is closer to the origin of L2 than any other point on M(p).

78

Inverse Spectral Theory

Proof. Consider the point

e

=

expd-V,d

on M ( p ) . We have K(e) =

W)- K ( P ) = 0,

so e is even by Lemma 3.4. It is the only even point on M ( p ) ,since the map p~ is one-to-one by Theorem 3.3. It is closer to the origin in L2 than any other

point on M ( p ) , because by Lemma 2, IlexP4vE)II =

(

lle1I2 + 8

c an(cosh
nzO

1/2

- 1))

'llell

for all

< # 0, and expe maps onto M ( p ) .

Problem 5 (Another proof of Corollary 1). Consider the differentiable function p on M = M ( p ) defined by P ( 4 ) = 1141l2.

(a) Show that p attains its minimum on M . [Hint: Pick a minimizing sequence, show that it is bounded, extract a weakly converging subsequence, and use the compactness of the functionspun .] (b) Suppose q in M is a critical point for p. That is, d,p I T,M vanishes. Show that q is even. (c) Conclude from (a) and (b) that there is a necessarily unique even point e on M , and that it is closer to the origin of L2 than any other point on M . (d) Prove directly, without using the exponential map, that M is connected. [Hint: Minimize p over each component of M.]

Problem 6. (a) Show that an even function belongs to the Sobolev space H ' , if M(e) contains an odd function. [Hint: Use Problems 2.3 and 3.6.1 (b) Conclude that the generic isospectral set does not contain an odd function. Let us summarize the results obtained so far in the following simple picture. Every isospectral set intersects the space E of all even functions in a unique point e. Hence,

L ~ [ o11 , =

U M(e),

eeE

Isospectral Sets

19

Figure 2.

and we may think of isospectral sets as fibers of a fibration of Lz over the base E. On each fiber, we have a global coordinate system K , which maps it onto 6 . Its inverse is the exponential map expe.

Problem 7. Let e be the even point on M = M(p). (a) Show that the intersection of M with the sphere of radius r centered at the origin of Lz is analytically isomorphic to S, =

< E P::

8

c G,(e)(cosh(,

- 1) =

rz

-

))el)’].

n z l

(b) Show that S, is empty if r < Ilell, contains just the point 0 if r = llell, and is the boundary of a bounded convex body in P: if r > ))e)). (c) Conclude from (a) and (b) that the intersection of M with any sphere centered at the origin is connected and simply connected.

Problem 8. Let e be the even point on M points q and r in M by

=

M(p). Define the sum of two

80

Inverse Spectral Theory

Show that e is the identity element, and that every point q has an inverse 04. Show that

oq

= q*,

the reflection of q across the even subspace. Another way to see how M ( p )lies inside L2 is to study its projections onto linear subspaces. A natural choice is L2[0, i ]or L2[i, 1 1 . For, the even point in M ( p ) is uniquely determined by its projection onto either one of these subspaces. One might therefore expect that these projections are well behaved at least in a small neighborhood of this even point. Also, the restriction of a function q to the left half interval, say, determines "half" of q. Loosely speaking, the Dirichlet spectrum determines a different "half" of q . It may be hoped that they together determine q uniquely. So decompose L2as the direct sum

L2 = L2[0,+] @ L2[*,11 of square integrable functions of the left and right half intervals, and consider the natural projections

ICL

I

RR

1

Theorem 4. For allp in L2, the projection RL is a reul analytic isomorphism between M ( p ) and an open subset of L2[0, i ] . Of course, an analogous result holds for the projection RR . It follows that every function on the left half interval has at most one extension to a function on the whole interval that lies in M ( p ) .This fact was discovered by Hochstadt and Liebermann [HL]. The main ingredient of the proof of Theorem 4 is

Lemma 3. (a) A t every point in M ( p ) , the sequence of vectors

RLK, i]. is a basis of L2[0,

n r l

Isospectral Sets

81

(b) Let q and r be any two points in M ( p ) . Then the sequence of vectors 1, nL(gfl(q)gfl(r) - 11,

n

2 1

is a basis for L2[0,i]. (c) Analogous statements hold, when I I L is replaced by I I R in (a) and (b).

Proof of Lemma 3. (a) Fix a point q in M ( p ) , and consider the vectors Vn = Vn(q).We have 1

-Vn

2nn

= 2sin2nnx

+0

by Corollary 2.1. The sequence of trigonometric functions 2 sin 2nnx, n 1 1, is an orthonormal basis for L2[0,i],and the error terms are square summable. Therefore, the statement follows from Theorem D.3, provided the vectors IILV,are either linearly independent over [0, i]or span L2[0, i]. Each time we have applied Theorem D.3 it was possible to verify independence directly. Here, however, this is difficult to do. To prove the linear independence of a set of functions we always needed to know their boundary values in order t o construct a biorthogonal set. See for example the proofs of Theorem 2.8 and 2.9. But in the present situation, nothing is known about the values of IILKat i.Therefore, we are going to show that the sequence of vectors I I L V, , n 2 1, spans L ~ [ o3, . Suppose there is a vector u in L2[0, +], which is perpendicular to all the vectors I I L Vn. Then the entire function

has a root at each Dirichlet eigenvalue p n , n 2 1, of q. We show that f vanishes identically, and subsequently that u vanishes identically. The roots of y2(l, A) coincide with the Dirichlet spectrum of q and are simple. Therefore, the quotient f(A)/yz(1, A) is entire. By Theorem 1.3,

82

Inverse Spectral Theory

for all A, so that, by Problem 1.3,

for all A. Here it is important that we integrate only up to hand, referring to the proof of Theorem 3.3,

-.(

YZ(1,A) on all sufficiently large circles 1A1

=

t . On the other

‘““I) F

(n + t)2n2.Thus,

on the same circles. It follows from the maximum principle that this quotient and thus the function f (12) vanishes identically. Now let g be the even extension of nLq to [0, 11, and let p,,,n 1 1, be its Dirichlet eigenvalues. Since YZ(X,

1-41= Y Z ( X , A,

a,

05x 5

t,

and since f vanishes in particular at p,,, we have U l n L V , ,

n 2 1

for V, = V,(q).On [0, 11, the V, are all odd by Theorem 2.6. Hence, if we extend u to an odd function a on [0,1], then also

aLK,

n2l.

However, since g i s even, the V, are a basis of the space of all odd functions on [0, 11 by Theorem 2.9. Consequently, 0 and so u must be zero. This shows that the vectors ~ L V ,n, 2 1, span L2[0,8], and the proof of (a) is complete. (b) is proven analogously. Here one needs to show that the vectors 1, nr(gn(q)gn(r) - 11,

n

1

1

83

Isospectral Sets

span L2[0,*]. Suppose they do not. Then there exists a function u in L2[0,$1 which is perpendicular to all of them. It follows as in the proof of (a) that the function

1

1/2

f(A)

=

0

~ ( X ) Y Z ( X A, , ~ ) Y z ( xA, , r) dx

vanishes not only at the eigenvalues of q and r, but vanishes identi~ally.~ Now extend nLq to a function w on [0, 11 by setting

w

= q

on [O,$],

w

=

r* on [+, 11.

Then w* is an extension of nLr to [0, 11, and so, for 0 YdX,

Ix I$,

A, 4) = Y 2 ( X , A, w), Y 2 ( X , A, r) = Y z ( X , A , w*).

Since f vanishes in particular at p,(w) perpendicular to all the vectors

=

p,(w*), n L 1 , it follows that u is

1 , nr(g,(w)g,(w*) - 11,

n

2

1.

The vectors g,(w)g,( w*)are all even on [0, 11. Hence the even extension ii of u is perpendicular on [0, 11 to all the vectors l,g,(w)g,(w*) - 1 ,

n 2 1.

But these vectors are a basis for the space E of all even functions by the next Problem. Consequently, ii and so u must be zero. This shows that the vectors 1 , nr(g,(q)g,(r) - 1) span L2[0, (c) The analogous statements for the projection on the right half interval follow from (a) and (b) by reflection.

t].

Probfem 9. Show that at every point q in L2,the vectors 1 , g,(q)g,(q*) - 1 , n L 1 , are a basis for E . [Hint: These vectors are even since g,(q*) = (- l)"gt(q). To prove their independence, calculate the inner product between these vectors and

as in the proof of Theorem 2.8. Otherwise imitate the proof of Theorem 2.9.1 'The proof is actually easier than before because now

84

Inverse Spectral Theory

Proof of Theorem 4. The restriction RL I M of n~ to M = M ( p ) is clearly real analytic, and its derivative at q is the restriction of dqnL to T,M. It is the linear map from T,M into L2[0, 31 such that

K-nLK,

n L 1.

Since the vectors ~ L Vare , a basis for L2[0, 31 by Lemma 3, this derivative is boundedly invertible, and n~ I M is a local real analytic isomorphism by the inverse function theorem. We show that nL 1 M i s one-to-one. Suppose q and r are two points on M which agree on the left half interval. Cross multiplying the differential equations for g n ( q ) and g n ( r ) and taking their difference, we get

0 = ( 4 - r, g n ( q k n ( r 1 ) = ( n R ( q - r), n R ( g n ( q ) g n ( r ) ) )

for all n

L

1. Moreover,

0 = [41 - [rl = (q -

r, 1)

= (nR(q -

Part (b) of Lemma 3 for the projection n R ( q - r)

Hence, q = r.

r), 1). ZR

=

now implies that

0.

H

Problem 10. (a) Show that the projection of M ( p )into L2[$, $1 0L2[$, 11 is a local real analytic isomorphism at the event point. [Hint: Use the fact that at an even point, the functions g;, n L 1, are a basis for the space E of even functions.] (b) What about the projection into L z [ f ,$1 ? Problem 11. Fix u in L2[0, t ] and consider the function pu on L2defined by Pu(4) =

[

1/2

(q - ~ ) ~ d x .

JO

Show that u lies in the projection of M ( p ) into L2[0, 31 if and only if the restriction of pu to M ( p ) has a critical point. Show that such a critical point is necessarily unique.

85

Isospectral Sets

The question remains whether M ( p ) ever projects onto L2[0,*]. The answer is, it never does. In other words, there always is a function in L2[0,$1 which can not be extended in L2[0, 11 to a function in M ( p ) . Such functions can be written down explicitly. Fix q in M ( p ) and n 2 1. As will be seen in Chapter 5 , a5 t QO, the curve #(q, V,) = &(q) converges in L2[0, *] to the function +

but tends to infinity in L 2 [ i ,I]. It follows from Theorem 4 that &(q) is not contained in the range of the projection of M ( p ) into L2[0,$1, but lies on its boundary. For otherwise, &(q) would be an interior point. In this case, applying the inverse of the isomorphism n~ I M ( p ) ,

6x4) = nL ' ( n L ( 9 m ) ) must be bounded in L2[0, 11 for t

+

QO

by continuity. This is a contradiction.

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5

Explicit Solutions

In the last century an ordinary differential equation was “solved” when all its solutions could be expressed as finite or perhaps infinite combinations of known functions. For example, as a power series whose coefficients satisfy a manageable recurrence relation. The work of Poincare and others, however, led t o the realization that this is generally impossible. As a consequence, qualitative and geometric methods were developed, and the search for explicit solutions was all but given up. In 1965 it therefore came as a surprise that certain interesting nonlinear evolution equations can be solved in closed form. One well known example is the Korteweg-de Vries equation describing the propagation of water waves in a canal. This discovery renewed interest in explicit solutions, and quite a number of such “integrable systems” were subsequently found. We shall see that the differential equation

on L2 is also “integrable”. Exploiting an observation that goes back to Darboux [Dal], we shall derive a closed formula for its solution curves. For

87

Inverse Spectral Theory

88

the vectorfields Vn

= 2(d/dx)g:

this formula reads

+((q,v,) = 4 - 2

d2 dx

log m , t , 41,

where &(x, t , q) = 1

+ (e'

c

- 1)

g&, q ) ds.

In addition, there is a closed formula for the exponential map exp, between P: = T , M ( p ) and M ( p ) . Namely,

where 0 is an infinite matrix with entries &(x,

t, q) = 6ij + (eti -

1)

:1

gi(s, q)gj(s,q ) ds.

The infinite determinant is defined later on as a limit of finite dimensional determinants. There are two steps in the derivation of these formulae. First, the solution curves of the vectorfields V, are determined. Then they are combined. Along the way, a general method is developed that will be used again in the next chapter. The heart of the matter is the following algebraic fact.

Lemma 1. Pick a real number p , and let g be a nontrivial solution of

-yrr + qy

(1)

for A

= p.

=

Iy

Iff is a nontrivial solution of ( 1 ) f o r I

# p,

then

is a nontrivial solution of

(

)

-y"+ q - 2 7 l o g g y for the same A . Also, f o r A

= p,

dx d2

Iy

the general solution of (2) is given by

1( a + b

g

=

:'1

g2(s)ds) ,

where a and b are arbitrary constants. In particular, l/g is a solution.

Explicit Solutions

89

If g has roots in [0, 11, then equation (2) is understood to hold between them. Notice that the sign of g is unimportant for the logarithmic derivative (d/dx) log g = g’/g.

Proof. Lemma 1 can be established by direct calculation. However, there is a more systematic proof based on the observation that the second order operator - ( d 2 / d x 2 )+ 4 - p can be written as the product of two first order operators. This observation, sometimes attributed to d’Alembert, is traditionally referred to as the “method of reduction”. Let A = g(d/dx)(l/g)and A* = -(l/g)(d/dx)g. The operator A* is the formal adjoint of A , if g and p are real. Using the differential equation for g we find’

It follows that equation (1) can be rewritten as (1‘)

A*Ay = (A - p ) y .

Similarly,

so that equation (2) becomes (2‘)

AA*y =

(A - p)y.

The lemma is now easy to prove. Applying A to both sides of equation (1’) we obtain AA*Ay =

(A - p)Ay.

So, if y is a solution of (l’), then Ay is a solution of (2’). Consequently, if # p , then

f is a solution of (1) for A

‘For some insight into this identity see the next Problem.

90

Inverse Spectral Theory

is a solution of ( 2 ) . Moreover, Af is a nontrivial solution when f is nontrivial. Otherwise, the Wronskian [ g ,f ] vanishes identically, which implies 0

=

k,fl’ = ( P

- A1g.f.

But then A = p , since g and f are nontrivial. To prove the second statement, suppose h is a solution of equation ( 2 ) for A = p. Then

d l d AA*h = - g - - - g h dx g 2 dx

= 0,

or

d -gh dx

=

bg2

for some constant 6 . The result follows from another integration.

Problem I . (a) Show that the most general real, formally self adjoint second order ordinary differential operator is of the form d d -p (x ) - + r(x). dx dx (b) Deduce from (a), by calculating the leading coefficients, that A*A and AA* are both of the form - ( d 2 / d x 2 )+ r(x). Here, as in the proof of Lemma 1 , A = g ( d / d x ) ( l / g ) ,where g is a nontrivial solution of ( l ) , and A* = - ( l / g ) ( d / d x ) gthe , formal adjoint of A for real g . (c) Derive the identities

d 2 g” + dx2 g

A*A = - -

- 9

from Ag = 0 and A * ( l / g ) = 0 respectively. Use the differential equation for g to show that ( 1 ) and (2) are equivalent to ( 1 ‘ ) and (2’) respectively. Lemma 1 was known to Gaston Darboux. A proof appeared in his 1882 Comptes rendus note “Sur une proposition relative aux equations liniaires” . In article 408 of his “Theory of Surfaces” [Da2] Darboux calls it a “curious theorem of analysis” and writes: “This proposition, which is easy to verify directly, evidently permits one to associate to every equation of the form ( l ) , that one knows how to integrate for all values of I ,

Explicit Solutions

91

an infinite sequence of differential equations of the same form that one also knows how to integrate for all values of the parameter 1. Each passage from one equation to the next introduces two new arbitrary constants’; in general, the successive equations differ more and more from the initial form and become more and more complicated. There are, however, exceptional cases in which the form of the equation is preserved, when one chooses the particular solutions appropriately.”

Actually, Darboux did not express the new differential equation in terms of the second logarithmic derivative of g. Instead he wrote the equation in the form

-y”

+g

(J -

y

= (p

-

A)y.

It is the addition law of the logarithm that makes iteration of Lemma 1 practical.

Lemma 2. Pick real numbersp and v. Let g be a nontrivialsolution of (1) f o r A = p and h a nontrivial solution of ( 2 )f o r A = v. Iff is a nontrivial solution of (1) for A # p , v, then

is a nontrivial solution of (3)

for the same A . Also, l / h is a nontrivial solution of ( 3 )f o r A

=

v.

Equation (3) is understood to hold between the roots of gh.

Proof.

Apply Lemma 1 to equation (2)with h in place of g and place o f f .

f ] / g in

[gl

We are ready to determine the solution curves of the vectorfield V,.

Theorem 1. The solution curve of the vectorfield Vn = 2(d/dx)gi with initial value q on M ( p ) is given by

d2 log e,cx, t , 41, dx

+r(q, v,) = 4 - 27

’The initial values of g [the authors].

- 00

< t < 00,

Inverse Spectral Theory

92

where &(x, t , q ) = 1

+ (e' - 1)

1

1

g&) ds.

Moreover, gj(x, 4') = e6Jn'/2 where 4'

=

4'(q, V,) and gj = g j ( X , 4) for j

2 1.

Observe that

Proof. Fix q in M ( p ) and a positive integer n. By Lemma 1, h

=

gn L(1 + c[Ig,'(s)ds)

is a solution of equation (2) for A = pn(q) and g = gn(x,q ) . Here, c is a real parameter. The idea of the proof is to apply Lemma 2 to this solution h of equation (2) and the solutions

g=gn,

j#n

f=gj9

of equation (1). Set = gh = gnh = 1

+c

ix

gi(s) ds.

0

For c >

-

is strictly positive on [0, 11. For,

1, the function

0

IIXg,'(s)dsI1, 0

and so

0

Ix I1

Explicit Solutions

93

Set d2

qc =

- 2~logOn,..

For c > - 1, the function qc belongs to L2, since On,. is strictly positive. Furthermore, for j 2 1, let3

Once again, we dropped the arguments xa nd q. In the second line the identity [gn, gj] = (pn - p j ) Jigngjds is used. Notice that

By Lemma 2, the function gj, c , j 2 1, is a genuine nontrivial solution of -Y” + qcY = pj(q)Y. Moreover, it vanishes at 0 and 1. Therefore, p,(q) is a Dirichlet eigenvalue for q c . In fact, gj,c has the same number of roots in [0, 11 as gj.0 = gj by Lemma 2.3. It follows that g,,c is a constant multiple of t h e j t h normalized eigenfunction of q c , and consequently that pj(qc) = pj(i(4) for allj 2 1. That is, qc has the same Dirichlet spectrum as q for all c > - 1. We determine the normalized eigenfunctions of qc . Squaring the second expression for gj,= and using = cg,‘, one finds after a short calculation that

Hence, I;g:.ds

=

1 1-l + c 8Jn. -

1 ~

1

+ C8jn

’Keep in mind that qc and g,,= also depend on the index n, which is suppressed for simplicity.

Inverse Spectral Theory

94

Also, g,!,c(0)= gj(0) is positive for all j

1. Therefore,

L

gj(qc) = . J l + c s , , g j , c

for a l l j L 1 . So far we have constructed a path q c ,c > - 1, that lies on M ( p ) .It satisfies

= Kj(q) - S j n

log(1

+ c),

j

L

1.

On the other hand, the solution curve &(q, K) is the unique path on M ( p ) that satisfies

K j ( 4 ' ( q , K)) Thus, if we choose c

= Kj(q)

= e-' - 1

> - 1,

Kj(qc) = Kj(q)

+ Sjnt, - 03

j L 1.

< t < 03, then

+ Sjn t ,

j 2 1.

Moreover, = 1

+ (e-'

= e-'( 1

- 1 ) ( 1 - $:g:ds)

+ (e'

= e-'O,,(x,

- 1)

1:

g,' ds)

t).

Therefore,

4'(4, K)

= qc =q - 2

d2 log e- '&(x, t ) dx

= q -2

d2 log &(x, t ) . dx

The proof of the first part of Theorem 1 is complete. The second part is obtained by a straightforward calculation.

95

Explicit Solutions

Problem 2 (Proof of Theorem 1 by direct calculation). (a) Show that tyi(q) = q - 2 ( d 2 / d x 2 )log 8, is a differentiable curve in L ~ . (b) Show that tyz(q) = q and -tyi(q) d = -2--log8, d2 d dt dx2 dt

= 2-hn, d 2

dx

where hn

=

er / 2 g. 8, ’

and gn, 6, are evaluated at q . (c) Show that

To this end check that (1 1 - h: + Wi(q)hn = Pn(q)hn ; (2) h, has exactly n + 1 roots in [0, 11 including 0 and 1

(3) (4)

hn has positive derivative at 0 h, has norm one in L2.

[Hint: For the last point use the identity hi = ( e r / e r- 1)(1/8,)’.] (d) Conclude that t y i is a solution curve of the vectorfield V, with initial value q , hence t y i = &(q, V,) by uniqueness. The proof of Theorem 1 is not difficult, but at the same time, not particularly transparent. We now try t o make the whole approach seem more “natural” in hindsight by exposing an underlying “trick”. First a general remark. Consider two operatorsA and B on a Hilbert space. In finite dimensions, A B and BA always have the same spectrum. This is easy to see. If A is invertible, then det(AB - A) = det(A-’) det(AB - A) det(A)

=

det(BA - A).

Invertible matrices are dense, so the identity det(AB - A) = det(BA - A) holds in general. Thus, A B and BA have the same characteristic polynomial and hence the same spectrum. In infinite dimensions, A B and BA have the same spectrum away from zero, provided they are bounded. That is, 0 may belong to the spectrum of

Inverse Spectral Theory

96

A B , but not to the spectrum of BA, or vice versa. This is also easy to see. One shows, by multiplying out, that for each A # 0 in the resolvent set of A B , the operator B(AB

-

A)-lA

-

I

is a two sided inverse of BA - A , so that A also belongs to the resolvent set of BA. The opposite inclusion holds by symmetry. Thus, A B and BA have the same resolvent set away from zero, hence also the same spectrum away from zero. In a special case, there is a generalization to unbounded operators. If A is a closed operator with adjoint A * , then A * A and AA* have the same spectrum away from zero [De, DeT]. These observations suggest a way of constructing families of functions with the same Dirichlet s p e c t r ~ m . ~ Pick a function q in L2 and let g,, m 2 1, be its Dirichlet eigenfunctions. Fix n L 1. As in the proof of Lemma 1, the operator - (d2/dx2) + q - p, may be factored as A * A , where

d l A =g,--, dx g n

and

1 d A*=--g, dxgn

is its formal adjoint. Interchanging the factors, one may expect that

has the same “Dirichlet spectrum” as A*A with the possible exception of 0. Formally, this is so. The functions Ag,, m # n, are “Dirichlet eigenfunctions” of AA* : they are nontrivial and vanish at 0 and 1 by 1’Hospital’s rule. On the other hand, by Lemma 1 the general solution of AA*f = 0 is unbounded at 0 or 1 unless it vanishes identically. So, 0 is not a “Dirichlet eigenvalue”. Unfortunately, the coefficient in AA* does not belong to L2, and the “eigenfunctions” Agm are not defined at the roots of g,. T o overcome these difficulties and construct functions with the same spectrum as q, let us factor once more. In the hope of restoring 0 as an eigenvalue, choose a solution

4 W e thank P. Deift for helpful discussions on this point.

Explicit Solutions

91

of AA*f = 0, and write AA*

=

B*B,

where B = h(d/dx)(l/h) and B* = - (l/h)(d/dx)h. Once again, permuting the factors, the “Dirichlet spectrum” of

ought to be the same away from zero. Indeed, for appropriate choices of a and b, everything turns out alright, as we have seen in the proof of Theorem 1. In particular,

_1 -h

gn

a

+ bjtgids

is a genuine solution of BB*f = 0, when the numerator is positive, so that 0 is in fact a Dirichlet eigenvalue. The process of factoring and permuting the factors is familiar. Consider for example the QR-algorithm well known to numerical analysts. Here, a non-singular, real symmetric matrix A is factored as A

=

QR,

where Q is an orthogonal matrix, and R = Q-’A is an upper triangular matrix. Q is the matrix, whose columns are obtained by orthogonalizing the columns of A from left to right. Interchanging Q and R , one obtains a new matrix Ai

=

R Q = Q-lAQ,

which is similar to A . Repeating this process inductively, one writes An = QnRn, An+’ = RnQn

=

Qi’AnQn.

One can show that the matrices An converge to a diagonal matrix D and the products QQI Qn to an orthogonal matrix U such that 1..

U-’AU

=

D.

This algorithm provides a numerically efficient way of determining the spectrum of a symmetric matrix.

Inverse Spectral Theory

98

The explicit formulas of Theorem 1 lend themselves to numerical calculations. They are particularly simple for the initial value q = 0, where the Dirichlet eigenfunctions are just trigonometric functions :

Sample programs in FORTRAN for this case are given in Appendix G . They were used to produce the figures in this and the next chapter. We plot the points on the solution curves of 6 and VZ with initial value 0 for

(*I

t = k.0.5,

k

=

0 , ..., 5 .

That is, we increment the first and second K-coordinate respectively in steps of 0.5 starting at q = 0.

&(O, VI) with t as in (*)

Explicit Solutions

99

$'(O,

V ) with I as in

(*)

V

100

Inverse Spectral Theory

g2

along $40,VZ)

It is a consequence of Lemma 4.2 that the functions &(q, V,) become unbounded as t becomes unbounded. A quick look at the plots suggests that this happens in a very specific way. As t + 03, these functions seem to “explode” at the right hand endpoint of the interval [0, 11, while they appear to converge to a finite limit on every interval [0, 1 - E ] strictly contained in [0, 11. This is the case. Fix q in L2, and let g, be its nth eigenfunction, n 2 1. The formula of Theorem 1 can be rewritten as d2

r$‘(q, V,) = q - 2-1og

dx

The integral j: g,’ ds is bounded away from zero as long as xis bounded away from 1 , whereas at 1 it tends to zero like a power of 1 - x. Consequently, as t 03, the curve &(q, V,) converges to the function +

4,’

d2 dx

= q - 2-log

5

1

g:ds

X

in L2[0,1 - E ] for every 0 < E < 1 , but tends to infinity in L2[1 every 0 < E < 1.

E,

11 for

Explicit Solutions

101

By the way, this completes the argument given at the end of Chapter 4 showing that the projection of any isospectral set into L2[0,i ] is never onto. Similarly, as t -, - 00, the same curve converges to the function

in L2[c,11 but diverges in Lz[O,E] for every 0 < E < 1.

Problem 3. Let q be in Lz and let gn be its nth eigenfunction, n 9 m = 4 4 4 , vn). are independent of t . (a) Show that the roots of g,(&q)) Fix two roots 0 5 a < b I 1 of g,,. Verify the identity (b)

L

1. Set

[Hint: Proceed as in the proof of Lemma 4.1 and use the identity following Theorem 1.1 (c) Check that ~~

and conclude that

(d) Suppose that b < 1. By the remark above, the curve &q) has a limit q,’ in L2[a,b ] as t -+ 00. Show that

In particular, if e is even and n is even, show that

Combining the explicit solutions for thevectorfields V,, it is now possible to obtain the formula for exp,(Q) stated at the beginning of this chapter. For E ,:P consider the infinite matrix

<

102

Inverse Spectral Theory

where

Bij(x, t, 4)

= 6ij

+

(e‘l

- 1)

1:

gi(s, q)gj(s, 4 ) ds.

We define the determinant of 0 as the limit of the determinants of its principal minors. That is, det 0

=

lim det O‘“’,

n-m

where = (Bi,)isi,,sn.

This simple definition of det 0 agrees with the invariantly defined Fredholm determinant of 0. For the argument see the next Problem.

Theorem 2. For q in L2 and exp,(&) = q

t in P:, -

d2 27 log det 0 ( x , t, q). dx

It is implicit in the statement of the theorem that the determinant of 0 always exists and never vanishes. Ifp is given, then, at least in principle, one can obtain all the eigenfunctions of p and “write down” the determinant of 0 at p. Hence, Theorem 2 provides an explicit description of the set

M(P)

=

(exp,(W:

t E I:!

of all functions with the same Dirichlet spectrum as p.

Proof. Let 4n = expq(K,,d,

n 2 1,

where r n t = ((1, ..., t,,,0, ...). The truncations r,< converge to the vectors Knc converge to in L2. It follows that 4n+

exp,(&)

strongly in L2 by the continuity of the exponential map. On the other hand, we have q n = exPq,-,(tnK) =

4%-

1,

Vn),

n 2 1,

< in P:,

so

Explicit Solutions

103

with q o = q, since q n - 1 and q n differ only in their nth K-coordinate by Applying Theorem 1 to q o , ...,q n - I ,

tn.

In Appendix F, the product on the right is identified with the determinant of the nth minor of 0 by Gaussian elimination. That is, n

n

dk(X, ( k

,q k -

1)

=

det @'"'(X,

k = l

t, 4).

Hence,

d2 dx

q n = qo - 2 7 log det @("'(X,

<, q).

Also, at x = 0, det 0'") has value exp(Ci= I t k ) and a vanishing first derivative. So integrating twice with respect to x and exponentiating we obtain

The convergence of the qn and the absolute convergence of z k , 1 ( k imply that det converges uniformly on [0, 11 to a positive function. Consequently, det 0 always exists, never vanishes, and the formula for exp,(fi) holds as stated.

Corollary 1. For q in L2 and ( in P?, &(q,

For q

=

5) = q

d2 dx

- 2710gdet

O ( x , t t , q).

0, the formula of Theorem 2 becomes

expo(fi) = - 2

d2 log 0 ( x , t, 0), dx

104

Inverse Spectral Theory

where O(x, <, 0) = Z +

((&

- 1)

5

1

2 sin(nis) sin(njs) X

This formula is used to plot some points on M(0). In the figure captions only the nonzero <-coordinates are displayed.

100

.. 6

100

Explicit Solutions

loo

105

1

(i=-k.f,(2=2,0,k=O

loo

,..., 6

1

M

Inverse Spectral Theory

106

Problem 4 (The Fredholm determinant of 0). Let A be a linear operator on a Hilbert space. If A is trace class, then the Fredholm determinant of I + A is defined by the identity det(Z + A ) =

tr(Ak(A)). krO

where Ak(A)is the kth exterior power of A . See [RS] for the details as well as for the definition of the trace class norm 11 * 11 1 and the Hilbert Schmidt norm II 112. We will use the following facts proven in [RS]. The determinant of I + A is a continuous function of A with respect to the trace class norm. If B , C are both Hilbert Schmidt, then BC is trace class, and llBC(1I IllBll2llCllz. (a) Write 0 - I as the product of the two matrices

and

Show that both matrices are Hilbert Schmidt on Pz. It follows that 0 - Z is trace class on Pz, hence the determinant of 0 according to the above definition exists. [Hint: Use that

=

11 1[ x , 11gill~

by Theorem 2.7 and the Parseval identity.] (b) Let C'"), n 2 1, be the principal minors of C so that BC'") = 0'")- I. Show that the C(")converge to C in the Hilbert Schmidt norm. Conclude that BC'"' = 0'")- Z converges to 0 - Z in the trace class norm. It follows that the usual, finite dimensional determinant of 0'")converges to the Fredholm determinant of 0 by continuity. 'Observe that

!&-Il e'

-

1

=

0.

6

Spectra

At the beginning of Chapter 3 it was pointed out that the Dirichlet spectrum of a function q in L2 belongs to the space S of all real, strictly increasing sequences a = (a,,0 2 , ...) of the form on = n2n2 + s

+ Gn,

n 2 1,

where s E R and 6 = ( G I , & , ...) E P2. This observation led us to recast the problem of characterizing spectra as the problem of determining the image of the map p from L2 into S . The principal result of this chapter is that p maps onto S . In fact, we shall show that p maps the even subspace E of L2 onto S . The strategy here is to construct a special sequence of vectorfields on E . The nth vectorfield is chosen so that the Dirichlet eigenvalues pj, j # n, are kept fixed by the associated flow while p,, is moving. In combination, they allow us to shift the spectrum into any desired position, and to reach every point on S . Such vectorfields are easy to construct. Let I n , n 2 1, be the constant vectorfield on S given by 1,

= (O,Bn1n,m 2

1)

in the standard coordinate system on S . These vector fields have the desired properties on S . It therefore suffices to pull them back to vectorfields on E

107

108

using 3.1,

Inverse Spectral Theory

,LIE.

That is, we consider the vectorfields (dgpuE)-l(Un).By Corollary

=

WXX, 4 ) .

More generally, one may consider the vectorfields w,=qo+

c qnw,

qEmxP2

nz 1

on E . In the p-coordinate system, W, becomes the constant vectorfield q. Consequently, its solution curves $'(q, W,) become straight lines:

,u(d'(q, w,))= P ( 4 ) + w . It follows that &(q, W,) is a real analytic function of t , q, q and is defined as long as the straight line p(q) + tq remains inside the open set p ( E ) of S . So far, the vectorfields W, and the vectorfields V, discussed in Chapter 4 seem to have similar properties. In contrast to the latter, however, the solution curves of W, are not defined for all time. For example, the curve @(q,Wn), n L 1 , is characterized by p(d'(q, w,))= P(4)

+ tl n

*

Clearly, t is restricted by the condition' pun-I(q)

< P n ( 4 ) + t < p(n+l(q),

since p, + t is confined by its neighboring eigenvalues. Our immediate goal is to show that &(q, W,) exists for all t that satisfy this necessary condition. That is, the solution curves exist for the maximal time interval possible. The proof of this important fact is not quite straightforward. One would like to proceed as in Chapter 4 and obtain an a priori bound for ((qY(q,Wn)II over any closed subset of this maximal interval. This is possible, but requires several additional concepts that have not been developed. Instead, the technique of Chapter 5 is applied, and the solution curves are written down explicitly. It is convenient to introduce some notation and state a technical lemma. For n L 1 , set

Spectra

109

+ qy = Ay

This is the unique solution of -y” conditions wn(0, A )

=

1,

Wn(1, A)

=

satisfying the boundary

yl(l,~n)

for all A outside the Dirichlet spectrum of q. As a function of A , Wn has poles at A = pcm,m # n, but a removable singularity at A = pn, since the limit

exists by I’Hospital’s rule and Theorem 2.2. Furthermore, write zn(x, 4) = Y Z ( X , pn(q), 4)

for the unnormalized Dirichlet eigenfunctions of q. In this chapter, they are easier t o use than the usual normalized eigenfunctions gn .

Lemma 1. For each q in L2, the function an(X,

A, 4) = [Wn, znl,

is strictly positive on [0, 11 x

(pn-I(q),

n 2 1

pun+I(q)).

Proof.2 The function O n is continuous on [0, 11 X

(pun-1, pn+l). Suppose it vanishes, say, on [0, 11 x [ p n , pn+l). The case of a root in [O, 11 X (pn-1, pun] is handled analogously. Consider the set

[A E [ p n , p n + l ) : O n > 0 on This set is not empty, since Un(X,

pun) = [UI,YZI

[O, 11 x [pun,A l l .

I

= 1 i = ~ n

for all 0 Ix I1 by the Wronskian identity. Hence it has a least upper bound X, which is strictly less than pn+l by assumption. Now consider the function On = O n ( X , 1)on [0, l].3 By continuity, On (x )2 0,

0

Ix I1,

’For helping us with this proof we thank J . Moser and T. Nanda. ’Here, - is a convenient local notation, not to be confused with complex conjugation

Inverse Spectral Theory

110

and by construction, fi,(X) = 0 for somexin [0, I]. However, for all I , and in particular for 1,we have O n ( 0 , I ) = Yl(0, I)zXO) = 1, On(1,

A)

=

Yl(1,p,,)zxl) = 1.

Hence, X must lie in the interior of [0, 11. That is, 0 < X < 1. wnor, pun)

1

I

X

0

1

Figure 1.

By the preceding discussion, X. Hence,

0

=

f i n has

a local minimum at the interior point

a;(X)

= (A

- Pn) - Wi)n(X)zn(X),

where W,,= w,,(x, 1).Consequently, iV,,(X) or z,,(X) vanishes, since X # p,,. But the roots of W,,and I,,are all simple, so the equation

= W,,z:,- iVAZn

implies that W,,(X) and z,(X) must both vanish. Hence, by Taylor's formula,

W,,(x) = ( x Z,(X)

=

-

X) * r(x)

( x - X) s(x)

Spectra

111

in a neighborhood of X with nonvanishing continuous functions r, s, since iVn and zn are continuously differentiable.4 It follows that ~ A ( x=)

(I- pn)

*

(X

-

XI' * ~ ~ x ) s ( > x )o

in a punctured neighborhood of X. But this implies that f i n is strictly monotone near Xand therefore changes sign. This contradicts the fact that Xis a local minimum of a n . Theorem 1. The solution curve +'(q, W,) of the vectorfield W, with initial value q in E is given by

The proof of this theorem also provides the unnormalized eigenfunctions along +r = +'(q, Wn). They are Zn

zn(x, +'I

=O n .1

and zj(x, 6') = zj - t ;wn ' [ ' Z j ( S ) Z n ( S ) ds, On,'

where w,, I , con,r are the functions are evaluated at q.

j Z n,

0

,

W n W n evalutated

at pn

+ t , q, and Z j , Zn

Proof of Theorem 1. Fix q in E and a positive integer n. Write W n , and W n .1 for the functions wn and conrespectively evaluated at L = p, + t and q. The function w,,' is a solution of the equation

- y"

(1)

for I

= pn

+ qy = I y

+ t, so by Lemma 5.1,

40bserve that in general w. and z,,are not twice continuously differentiable.

Inverse Spectral Theory

112

is a solution of the equation

for the same A . The idea now is to apply Lemma 5.2 to this solution h of equation (2) and the solutions

f = zj,

g =z ~ ,

j f n

of equation (1). By Lemma 1, 0n.1

=

is strictly positive on [0, 11 for p n - l

gh

=

znh

< p n + t < p n + l .Hence, d2

qr = q - 2 7 log O n , t

dx

belongs to L2.So do the functions

and, for j # n,

Here, the identity [ Z j , z n ] = (pj - p n ) J$ z j z n d S was used. Now, by Lemma 5.2, the functions Z j , l , j 1 1, are genuine solutions of the equation -y”

+ qry

=

Ay

for A = pj + t s j n . They vanish at 0 and 1 and have the same number of roots in [0, 11 as z j , ~= Z j by Lemma 2.3. In fact, checking the x-derivative at 0 we find that Zj,t

=

Zj(X9qt)

113

Spectra

for all j 1 1. Hence, p;(qr) =

+

ta;n,

as required. It remains to check that the functions

Kj

j 2 1

d o not change along qr. Indeed,

= K;W

for a l l j 2 1, using the fact that It follows that

On,r

=

1 at 0 and 1.

In particular, the curve is defined as long as pn-1 < pn proof of Theorem 1 is complete.

+ t < pn+l. The

As an illustration some points on the curves $‘(O, W I )and 1$~(0, WZ)are plotted along with their first normalized eigenfunction.

4‘(0, W , ) where I increases from 0 so that ,u, approaches p2

114

Inverse Spectral Theory

I 100

1

d~~(0, Wz) where f increases from 0 so that pu~approaches p3

I

115

Spectra

Problem 1. Show that, with the notation as in the remark following Theorem 1 ,

and

[Hint: Observe that

and recall Theorem 2.2.1 It is now possible to answer another question raised at the beginning of Chapter 3 . Calculating the solution curve of WI with initial value q = 0 we find that d2 -2-log dx2

[sin\/;;(l

-

- x) - s i n G x , sin nx sin 6

is the unique even function with Dirichlet spectrum p1, 4n2,9n2,...

for all p~ strictly less than 4n2.Hence, any such sequence with p1 < p2(0) arises as the Dirichlet spectrum of a function in L2.

Theorem 2. p maps E onto S. Consequently, p~ is a real analytic isomorphism between E and S. is a real analytic isomorphism between Proof. If p maps E onto S , then E and S , since p . ~is a local real analytic isomorphism by Theorem 3.2 and is globally one-to-one by Theorem 3.3. To prove that p maps onto S , let o be an arbitrary point in S . In the standard coordinates on S , we have (T = (s, 6). It suffices to consider the case s = 0. For, if there is a q in E such that P(4) = (0,

a),

Inverse Spectral Theory

116

then p(q

+ s)

= (s,

a),

and q + s is clearly contained in E . So let s = 0. Consider the modified sequences ON =

(PI,

..., p N , ( T N + I ,

...1,

where p n = n2n2,1 In IN. The sequences oNconverge to p ( 0 ) as N tends to infinity, so for sufficiently large N,they must be contained in the open image of the map p. Hence, if N is sufficiently large, (p1,

...,pN,

(TN+I,

...) = p ( q )

for some q in E . That is, the tale of any point in S with s = 0 is attained. It remains t o shift the first N eigenvalues of q to P I , ..., p ~ This . is easily done using the flows of Theorem 1, since only finitely many eigenvalues need to be adjusted. However, care must be taken t o avoid the crossing of eigenvalues. To be on the safe side, first shift p1, ...,p~ t o the far left, then move them into the desired positions beginning with p N . 5 We thus have a complete characterization of all possible Dirichlet spectra. Corollary 1. The sequence (T = ( a l , 0 2 , ...) is the Dirichlet spectrum of some function in L2 ifand only i f it is real, strictly increasing, and of theform ( T ~=

n2n2 + s

+ P2(n),

where s is a real number. Also, Theorem 3.7 can be sharpened: Corollary 2.

K

x p is a real analytic isomorphism between L2 and P: x S .

There is also a closed formula for pi': S

-+

E.

It is the counterpart of the formula for exp, which was derived in Chapter 5 . 'More elegantly, one can shift the eigenvalues into position moving each of them only once. See the Proof of Theorem 3.

117

Spectra

For the sake of convenience we restrict pi' to the subset

so = (a = (s,a):s = 0 ) c

s.

The general case follows from a translation. For a in SO we set

The convergence of the infinite product follows from Lemma E. 1, since

and similarly for the second factor

Theorem 3. For a in S O , p,E'(a)

= -2

d2 log(ll(o) det Q(x, a)), dx

where Q is the infinite matrix whose elements are given by (- 1)' - cos

o i j ( X , 0) =

f o r i,j

L

6sin G

sin Ja,

1

sin njx x ,nj

1, and the determinant of Q is defined as in Chapter 5.6

Observe that

where p u n n, 2 1, are the Dirichlet eigenvalues of q

oii(x, 0)

=

[ w i ,z;]

lq=o

x = a/

=

0. In particular,

= W I ( X , 090)

in the notation of Lemma 1. It is implicit in the statement of the theorem that the product n(a)det Q(x, a) always exists and never vanishes. It is possible that ll vanishes, in which case the determinant of Q is infinite. This happens 6This is an unpublished result of J. Ralston and E. Trubowitz.

118

Inverse Spectral Theory

precisely when at = j 2 n 2for some i # j . Otherwise, det SZ is finite, does not vanish, and we equivalently have d2 dx

p(El(a) = - 2 7 log det

(3)

SZ(x,a).

This formula holds on a relatively open dense subset of SO.As to the Fredholm determinant of C2 see Problem 3 below.

Proof of Theorem 3.

Fix a in SO,and let

p ( n = 2n n2 ,

n z l

be the Dirichlet eigenvalues of q = 0. The idea is to apply the flows of the vectorfields WI, W Z ,... one after the other t o shift p l , p z , ... into the positions a l , a2, ... and to combine the closed expressions provided by Theorem 1. In general, however, we can not move the eigenvalues in their natural succession. For example, if a1 > p z , then we can not begin by moving pl to a], since eigenvalues must not cross. Instead, we have t o move them in an order depending on a as follows. By the asymptotic behavior of the sequences in S there exists an integer N such that Ipn - afll <

I P ~ - aml

for all n , m

L

N and n # m.

Hence, if we can move the first Neigenvalues into position, then we can continue t o move the remaining eigenvalues in their natural order. To this end, we show that for any two strings P I < L(2 <

***

< a2 <

--*

a1

< Pn < an

of length n there is a way t o shift the first sequence into the second preserving order and moving each element once. We show this by induction on n. For strings of length 1 there is nothing t o prove. So suppose the statement is true for strings of length n 2 1 . If p n + l < a n + ]then , first shift this element to its new position to get pi

< p2 <

***

< p n < an+i,

then shift the remaining elements using the induction hypotheses. If pn+l 2 afl+1,interchange these two steps. You first get a1

< a2 < ... < an < / . I n + ] ,

then you can move p f l + l to

a,,+l.This

completes the induction.

Spectra

119

In effect there exists a finite permutation j l ,j 2 , ... of the natural numbers 1, 2, ...' such that the sequences on =

(d, a;,

...)

with

all belong to S O . These sequences on converge to

pfl

0

in S . Hence the functions

= pEI(an),

n

L

1

converge to q = pE1(a)

in L2 by the continuity of p i 1 . On the other hand, we have

pn

=

4In(pn-1,W,,),

t n = o j , - pjn

with PO = 0, since p n and p n -1 differ only in their jnth eigenvalue. Applying Theorem 1 t o p o , ...,P , , - ~ ,

In Appendix F it is shown that for all sufficiently large n, (4)

fi

k=l

Ojk(x, f k

,pk- 1)

where

'That is, j , , = n for all sufficiently large n

=

n'"'(0)det ~ ( " ' ( X ,0 )

Inverse Spectral Theory

120

and the sign is determined by the sign of the permutationjl, ...,j,. Hence,

d2 dx

Pn = - 2 7 log

x

n'")(a)det Q'")(x, a).

The left hand side of (4) has value 1 and a vanishing first derivative at 0. So integrating twice with respect to x and exponentiating we obtain

=

f

n'")(a)det Q'")(x, a) = ex,(

-

jx

(x - t)pn(t)dz).

0

It follows from the convergence of the p n that the left hand side converges uniformly on [0, 11 to a positive function. The product ll'") itself converges to ll. So also the determinant of Q'") converges to a finite value, if ll # 0, otherwise it tends to infinity. In any event, going to the limit we obtain q = -2

d2 log(n(a) det Q(x, a)). dx

Here are some image points of the map p2'. In the captions, pn = n 2 712 , n 2 1, are the eigenvalues of q = 0.

100-

UI

is fixed at a value between p , and p ~ while ,

a1

tends towards p3; otherwise, un = pn

121

Spectra

100

As before, but now u? tends towards p4

100

0 1 ,U Z ,u3

tend towards p ~p3, , p4 respectively at different rates; otherwise, un = pn

Inverse Spectral Theory

122

We mention a natural generalization of the preceding results. Consider the sets

of all functions in L2 with the same K-values as a given functionp. As a special case we have

K(0) = E by Lemma 3.4. These sets are counterparts of the isospectral sets M(p). K(p) may be viewed as a vertical slice of the real analytic coordinate system K x p on L2. So each such set is a real analytic submanifold of L2, and p is a global coordinate system on it mapping into I? x P2.

Problem 2. (a) Show that with q also the straight line q -00 < c < 00, is contained in K(p). [Hint: Use that y2(x, A, q + c) = y2(x, A - c, q).] (b) Show that

+ c,

(c) Show that the derivative of p is a linear isomorphism between &K@) and T,(,,S = IR x P2 given by

Theorem 2 generalizes to

Theorem 2*. For allp in L2 the restriction p ~ @of) p to K(p) is a real analytic isomorphism between K(p) and S. Making use of the last Problem the proof is essentially the same as for Theorem 2. Observe that the vectorfields Wn are tangent to K(p), so any solution curve with initial value in K(p) stays in that set. There is also a closed formula for p i & ) . Restricting this map to the subset sp

=

(a= (s, 6 ) : s

=

[p]) c

s

123

Spectra

it is given by

where WO,P)=

n

0;- j i j p ; - Oj

*-

;>izl 0;- O j p ; - / l ;

SZ is the infinite matrix with elements

and pn, n L 1, are the Dirichlet eigenvalues of p . The proof is the same as that of Theorem 3.

Problem 3 (The Fredholm determinant of a). Fix p in L2, let pn, n 2 1, be its Dirichlet eigenvalues, and let u be in S p . For the notion of the Fredholm determinant and some related facts, see Problem 5.4. (a) Verify the identity

Show that the three matrices

define Hilbert-Schmidt operators on P2, provided that O; # p; for i # j . Conclude that in this case, SZ - Z is trace class on P2, hence the Fredholm determinant of SZ is well defined. (b) Imitate the argument of Problem 5.3 to show that the finite dimensional determinants of a'"' converge to the Fredholm determinant of SZ as

n

+

03.

This Page Intentionally Left Blank

A

Continuity, Differentiability and Analyticity

We discuss the notions of continuity, differentiability and especially analyticity for maps between Banach spaces. Let U be an arbitrary subset of a Banach space E , and let F be another Banach space. A map

f:U-+F is continuous on U,if it maps strongly convergent sequences in U into strongly convergent sequences in F. That is, if Xn x strongly in U,then f ( x n ) f ( x ) strongly in F. This is the familiar notion of a continuous map. However, in an infinite dimensional setting, continuous maps lack many of the useful properties one would like to have. For instance, they need not be bounded on closed bounded subsets of E. This is due to the fact that the closed unit ball is no longer compact in the strong topology. -+

+

Problem 1. Construct a continuous, unbounded function with bounded support. [Hint: Choose a sequence of unit vectors xn such that ((xn- x,(( L 6 > 0 for m # n. For each n, construct a function f n using Urysohn’s lemma

125

Inverse Spectral Theory

126

which is equal to n at x,, and vanishes outside IIx - x,ll 2 6/3. Add these functions together.] A stronger notion of continuity, which rules out such behavior, is obtained by generalizing the notion of a compact operator familiar from linear functional analysis. A map f: U -,F is compact on U,if it maps weakly convergent sequences in U into strongly convergent sequences in F. That is, if x,, -,x weakly in U,then f(x,,) -,f ( x ) strongly in F. A strongly convergent sequence is also weakly convergent, so every compact map is continuous. But not every continuous map is compact.

Problem 2. Show that the function u

3

-+

u2(t)dt 0

is continuous on Li[O, 11, but not compact. The closed unit ball, and more generally every closed, bounded and convex subset of a reflexive Banach space is weakly compact regardless of the dimension. For this reason, compact maps are much nicer than merely continuous maps. For instance, the norm of a compact function on a closed, bounded, convex subset is bounded and attains its maximum. Moreover, the function itself is uniformly continuous. We now turn to differentiable maps. Let U C E b e open. As usual, the map

f:U-F is differentiable at x E U,if there exists a bounded linear map

d x f :E -,F such that

I l f @ + h) - f W

- dxf(h)II = o(llhll).

That is, for every E > 0 there is a 6 > 0 such that 11 f ( x + h ) - f ( x ) d,f(h)((Iellhll for all h with llhll < 6. The linear map d x f is uniquely determined, and is called the derivative o f f at x . The map f is differentiable on U,if it is differentiable at each point in U. In this case, the derivative is a map from U into the Banach space L(E, F ) of all bounded linear maps from E into F, denoted by df. If this map is continuous, then f is continuously differentiable, or of class C ' , on U.

Continuity, Differentiability and Analyticity

127

We frequently encounter complex valued differentiable functions f , which are defined on open subsets of LE[O, 11. In this case, dqf is a bounded linear functional on Li?[O, 11. It follows from the Riesz representation theorem that there is a unique element af/aq in L;[O, 11 such that

for all u E L;[O, 11. The function af/aq is the gradient off at q. Our notation imitates the usual finite dimensional notation.

Problem 3.

Suppose f is differentiable on U C E . Show that

d d, f ( h ) = -f(x d& 1

+ Eh)

= lim - ( f (x &+O&

+ Eh) - f (x))

for all x E Uand h E E . The right hand side is called the directionalderivative o f f at x in the direction h. Conversely, suppose the limit

exists locally uniformly inxand h. That is, suppose for each x in Uand 6 > 0 there exists a neighborhood V of x and an EO > 0 such that

1a

<

I

( f ( T + Eh) - f ( 0 )- 6c(h) < 6

IE~

for all in V , llhll < 1 and < EO. Show that f is continuously differentiable on U,and that d, f = 6,. [Hint: To prove additivity of 6, in h write

6x(h + k) - 6x(h) - 6x(k) =

1

lim - ( f ( x + Eh

&+O&

+ ~ k +) f(x)

- f(x

+ eh) - f(x + ~k)).

Add and subtract terms to rewrite the last expression as a combination of four different difference quotients at the point 4 = x + E(h + 412.1 'The gradient is defined in the same way for the complexification of any real Hilbert space.

Inverse Spectral Theory

128

Here are several examples of differentiable maps.

Example 1. A bounded linear map L : E on E, and d,L = L for all x .

+

F k continuously differentiable

Example2. The functionf: q jd q ( t )d t is continuously differentiable on Li[O, 11. Its derivative is given by +

1

1

dqf(u) =

u(t) dt

= (u,

0

hence af/aq

=

1 >,

1.

Example 3. The function g : q derivative is

Sb q2(t)d t

+

is also C' on Li[O, 11, its

Consequently, ag/aq = 4.

Example 4. Let M(n, C) be the Hilbert space of all complex n x n-matrices with inner product ( A , B ) = C ; , j a;j6;,,where a;,, b;jare the entries of A and B respectively. The determinant function det : A

-+

det(A)

is C' on M(n, C), since it is a polynomial in the entries of A . The gradient of det at A is the adjoint of A ,

which is the matrix of cofactors a;' may also be written as

a det -

-

= (-

l)'+'det(a&

#,,/#;

of A . This

det(A)A-'.

aA

in case A is invertible. For the proof, let 1 Ij In, and apply dAdet to a matrix U,which is nonzero only in t h e j t h column. Expanding the determinant with respect to

Continuity, Differentiability and Analyticity

129

this column you get dA det(U) =

d dc

- det(A

Ir-O

+ EU)

By linearity, this holds also for general U. Hence,

a det/aA

=

adj(A).

Problem 4. Let A,, 1 I j In, denote t h e j t h column of a matrix A . Show that the directional derivative of det in the direction U is given by n

dAdet(U)

=

C

det(A1,

..., U j , ..., An)

j = 1

=

det(A) tr(A-'U),

where the last term applies only to invertible matrices A .

Example 5. Let CY be a 1-form on R" with smooth coefficients, and let C'(T', IR") be the space of all continuously differentiable loops in R", that is, all periodic continuously differentiable maps from the real line into R" with period 1. Then the map loopa: y

+

17CY

=

1

1

Y*CY

0

is continuously differentiable on C'(T', Rn), and its derivative is the linear map c

d7loopa: tp

-+

l7

where i+ denotes interior multiplication by

i+da, tp : i+da =

dol(w, .).

130

Inverse Spectral Theory

This can be seen as follows. By the calculus of differential forms, or by a direct computation using coordinates,

=

1' ( y + sw)*(iGda + diGa) ds. J O

The integral of ( y that loop,(y

+ sty)*di+aover [0, 11 vanishes by periodicity. It follows

+ w ) - loop,(y)

-

[(y

+ w ) * a - y*a - y*i$da]

ri ri

[(y JO

.\o

+ sty)*iGda - y*i+da]ds.

Higher derivatives are defined inductively. Iff: U F i s differentiable on U , and if its derivative d f : U L(E, F ) is also differentiable on U , then f is twice differentiable. Its second derivative is a map -+

-+

d2f:U

+

L2(E,F)

from U into the Banach space Lz(E,F ) of all bounded, bilinear maps from E x E into F. The map is of class C2, if d 2 f is continuous. In general, f is p times differentiable on U , p 2 1, if

dPf = d(dp-'f): U

+

Lp(E,F)

exists as a map from Uinto the Banach space Lp(E,F ) of all boundedp-linear x E ( p times) into F. It is of class C p ,if d P fis continuous. maps from E x Finally, f is smooth, or of class C", if it is of class C p for all p 1 1 . I f f is C p , then actually d f f is a symmetric bounded p-linear map from E x x E into F. This is a restatement of the classical theorem about interchanging the order of partial differentiation. For a proof, see for instance DieudonnC [Di] or Lang [Lal]. We are not going to list and prove all the elementary properties of differentiable maps and their derivatives. They can be found in the standard references. However, because it is used so often, we mention the chain rule.

Continuity, Differentiability and Analyticity

131

If f : U -,V and g : V - W are two C’-maps, then the composite map g f : U + W is also C ’ , and 0

dx(g o f 1 = d!(x)g* dxf is its derivative at x. Let us also recall Taylor’s formula with integral remainder. Iff is p times continuously differentiable, p 1 1, and if the segment x + th, 0 I t I 1, is contained in U , then

f(X

+ h) = f ( x ) + dxf(h) +

+

3 . .

1

(P

-

I)!

dxP-’f(k, ...,h)

The integral remainder is bounded by ( 1 1 hllp/p!)sup0 special case is the “mean value theorem”, where p = 1:

I

Ildf+thf

11 .* A

IIf(x) - f(Y)II 5 IIx - Yll o yIldtx+(l-t)uf l II. If, in addition, f is smooth, and the integral remainder converges to 0 as p tends to infinity uniformly in some ball llhll < r, then f admits a Taylorseries expansion

f ( x + h) =

1

C -d!f(h, ...,h) krok!

at x in this ball. The integral in Taylor’s formula is the familiar Riemann integral of a continuous, Banach space valued function on [0, 11. T o recall its definition, let G be a Banach space. For a step map s: [0, I ] G , one sets +

where 0 = to c tl < < t,, = 1 is a partition of [0, 11 such that s(t) = si for ti- 1 < t < t i . For two step maps s and s’,

’Here and in similar cases, Ildpfll is understood to denote the operator norm of d p f . That is, for L in LP(E,F),

Inverse Spectral Theory

132

It follows that

is well defined for any map g : [0, 11 G , which is the uniform limit of step maps Sm . In particular, the integral is defined for continuous maps g. All the usual properties of an integral hold. For example, +

and

L

g(t) d t =

5

1

Lg(t)dt

0

for every L E G * , the dual space of G. The multiple Riemann integral of continuous Banach space valued maps on [0, 11” is constructed in the same way. The order of integration may be interchanged. We finally turn to analytic maps. Suppose fiU-F is a map from an open subset U of a complex Banach space E into a complex Banach space F. By definition, f is analytic on U,if it is continuously differentiable on U.This is the straightforward generalization of Riemann’s notion of an analytic function of one complex variable.

Example 6. Let n L 1, and A E L:(E, F ) , the Banach space of all bounded, n-linear symmetric maps from E x ... x E into F. Then P(x) = A(x, ...,x), the evaluation of A on the diagonal, is analytic on E. The function P i s called a homogeneous, F-valued polynomial of degree n on E. We also use the notation P = a to indicate the linear map A from which P is obtained.

Example 7. Let P n , n 2 0, be homogeneous polynomials of degree n on E , where POis simply a constant. If the “power series”

C

PAXI

n20

converges absolutely and uniformly on an open subset U C E , then it defines an analytic function on U.

Continuity, Differentiability and Analyticity

133

Problem 5. (a) Let A E L:(E, F ) , and let P = a be the corresponding homogeneous polynomial. Verify the polarization identity

(b) Show that there is a one-to-one correspondence between bounded symmetric n-linear maps on E and homogeneous polynomials of degree n on E . It is convenient to introduce another notion of analyticity. The map f: U F is weakly analytic on U , if for each x E U , h E E and L E F*, the function +

z

+

L f ( x + zh)

is analytic in some neighborhood of the origin in C in the usual sense of one complex variable. The radius of weak analyticity off at x is the supremum of all r 2 0 such that the above function is defined and analytic in the disc IzI < 1 for all L E F* and h E E with llhll < r. It is easy to see that the radius r of weak analyticity at x is equal to the distance p of x to the boundary of U. For, r Ip by definition. On the other hand, if L and h are given with llhll < p, then the function z L f ( x + zh) is well defined on the disc IzI < 1 and analytic in some neighborhood of each point in it, since f is weakly analytic on all of U . Consequently, this function is analytic on IzI < 1, and so also r 1 p. It follows that r = p. The notion of a weakly analytic map is weaker than that of an analytic map. For instance, every globally defined, but unbounded linear operator is weakly analytic, but not analytic. Remarkably, a weakly analytic map is analytic, if, in addition, it is locally bounded, that is, bounded in some neighborhood of each point of its domain of definition. +

Theorem 1. Let f: U F be a map from an open subset U of a complex Banach space E into a complex Banach space F. Then the following three statements are equivalent. +

f is analytic on U. ( 2 ) f is locally bounded and weakly analytic on U. ( 3 ) f is infinitely often differentiable on U , and is represented by its Taylor series in a neighborhood of each point in U. (1)

134

Inverse Spectral Theory

A prerequisite for the proof is a version of

Cauchy’s Formula. Suppose f is weakly analytic and continuous on U. Then, f o r every x E U and h E E ,

where r is the radius of weak analyticity o f f at x .

Proof. Fix x in U,and let r > 0 be the radius of weak analyticity off at x . Then the open ball of radius r aroundxis contained in U.For every h E E , the integral

is well defined, since f is continuous and ll4-hll < r, 14- Then, for every L E F*,

= Lf(x

ZI > 0 for 14-1 = p.

+ zh)

by the usual Cauchy formula. Since this holds for all L , the statement follows. rn

Proof of Theorem I . (1) * ( 2 ) Suppose f is analytic. A differentiable map is continuous, and a continuous map is locally bounded, so f is locally bounded. Furthermore, for every L E F * , the composite map L f ( x + zh) is continuously differentiable in z wherever it is defined. Thus, f is weakly analytic. (2) * (1) Suppose f is locally bounded and weakly analytic. We first show that f is continuous. Fix x E U and choose r > 0 so small that

By the usual Cauchy formula,

135

Continuity, Differentiability and Analyticity

where IlLll denotes the operator norm of L . This estimate holds for all L in F* uniformly for lz\ < and llhl\ < r. Consequently,

1"

+ zh) - f(x)ll

~

2M

Z

for IzI c -1- and llhll < r. From this, the continuity off follows. Now, f being weakly analytic and continuous, Cauchy's formula applies, and

for IzI c 1 and llhll < r. It follows thatfhas directional derivatives in every direction h, namely &(h)

=

1

lim - ( f ( x + zh) - f ( x ) )

2-0

In fact, this limit is uniform in

z

I\<

- XI( < r / 2 and llhll < r / 2 , since

t.

for IzI c It follows from Problem 3, thatfis continuously differentiable, hence analytic on U. (1) * (3) Suppose f is analytic on U.Fix x in U and r > 0 such that

For h in E and n 2 0, define

where p > 0 is chosen sufficiently small. The integral is independent of p as long as 0 < p Ir/llhll, sincefis analytic.

Inverse Spectral Theory

136

For instance, PO@)= f ( x ) and A ( h ) = d,f(h). We show that P,(h) is the nth directional derivative o f f in the direction h. First of all, Cauchy's formula and the expansion

yield

for llhll < r. Choosingp = r/llhll for h # 0, the norm of the right hand side is bounded by

Consequently,

f ( x + h) =

" 1 c -+(h) ,=on.

for llhll < r. Moreover, the sum converges uniformly in every ball llhll < P < r. We now show that each P, is a homogeneous polynomial of degree n. That is, there exists a bounded symmetric n-linear map A, such that P, = an,the polynomial associated to A, (see Example 6 ) . Consider the map A, defined by

where E > 0 is sufficiently small, say E < min r/llhill. For every L E F * , the map

..., z,)

( ~ 1 ,

+

L f ( x + zlhl

+

-

0

.

+ Znhn)

is analytic in a neighborhood of the origin in C". Hence, by the usual Cauchy formula for n complex variables,

Continuity, Differentiability and Analyticity

137

It follows that A nis linear and symmetric in all arguments. A n is also bounded by a straightforward estimate. Finally, using Cauchy’s formula again,

LAn(h,..., h) = =

(2-

L f ( x + zh)l 0

LPn(h)

for all L , and therefore

An(h, ..., h) = An(h) = Pn(h), as we wanted to show. Thus, on the ball of radius r around x , the map f is represented by a power series, which converges uniformly on every smaller ball around x . It is a basic fact that such a map is infinitely often differentiable. In particular, we have

d:f = An for all n 2 0. (3) = (1) is trivial.

Problem 6. Let co be the Banach space of all sequences z = ( a ,ZZ,...) of complex numbers tending to zero, normed by IzI = SuPnz I I Z n l . For n 2 1, define P,,on co by Pn(Z) = ZI

Zn.

(a) Show that each Pn is a homogeneous polynomial of degree n on (b) Show that

C

nS 1

CO.

Pn(z)

converges uniformly on every ball with a radius smaller than 1 to a function F which is analytic on CO. (c) Show that F is unbounded on every closed ball of radius 1.

Theorem 2. Let f n be a sequence of analytic maps on U C E , which converges uniformly to a map f . Then f is also analytic. Proof. Given x in U, choose r > 0 so that the sequence f n converges uniformly on the ball of radius r around x . Then, for every L E F* and every h E E, llhll < r, the function

z

-,Lf(x + zh)

Inverse Spectral Theory

138

is, on IzI < 1, the uniform limit of the analytic functions z Lfn(x + zh), hence is itself analytic. It follows that f is weakly analytic on U. The limit is locally bounded, since it is continuous. By Theorem 1, f is analytic on U. H -+

It is clear that a map from an open subset of C ninto C‘” is analytic if and only if all its m coordinate functions are analytic functions of their n arguments. This is not necessarily so in infinite dimensions, because a map may be discontinuous although every coordinate function is analytic. An additional assumption is required. Theorem 3. Let f ; U -, H be a map from an open subset U of a complex Banach space into a Hilbert space with orthonormal basis en, n 2 1. Then f is analytic on U if and only iff is locally bounded, and each “coordinate function” fn

=

< f ,e n ) :U - + Q)

is analytic on U. Moreover, the derivative off is given by the derivatives of its “coordinate functions” ; df(h) =

c df2h)en.

nzl

Proof. Let L E H*. By the Riesz representation theorem, there is a unique element Pin H such that L 4 = (4, P) for all 4 in H. Write

P=

C

n z 1

Anen,

and set m ...

P,,, =

C

Anen,

m 2 1.

n= I

Then L is the operator norm limit of the functionals Lm defined by L,n@= (4, Pm). That is,

as m 4 0 0 . Now, given x in U , choose r > 0 so that f is bounded on the ball of radius r around x. Fix h in the complex Banach space containing CJ with ((h((< r.

Continuity, Differentiability and Analyticity

139

On IzJ < 1, the functions rn

z

-+

Lrrtf(x

+ Zh) = C A n fn(x + ~ h ) , n=l

m

2

1

are analytic by hypotheses and tend uniformly t o the function

r

+

Lf(x

+ rh),

sincef is bounded. Hence that function is also analytic on IzI < 1. This shows that f is weakly analytic and locally bounded. By Theorem 1, the function f is analytic. Conversely, i f f is analytic, then of course, f is locally bounded, and each coordinate function is analytic. Finally, i f f is analytic, then d, f(h) exists and is a n element of H , hence can be expanded with respect to the orthonormal basis en, n 2 1. Its nth coefficient is (d,f(h), en> = &(f, e n ) ( h ) = dxfn(h) by the chain rule, since ( * , en>is a linear function. Thus dxf(h)

=

c dxfn(h)en

n 2 I

as was to be proven. Finally, we introduce the notion of a real analytic map. Let E , F be real Banach spaces, let CE, C F b e their complexifications, and let U c E be open. A map f:U+F is real analytic on U , if for each point in U there is a neighborhood V c CE and an analytic map g: V -+ CF, such that f = g

on

UnV.

It follows that a real analytic map can be expanded into a Taylor series with real coefficients in a ball at each point. The converse is also true. For more information about analytic maps, we refer the reader for example to [Na] and [Din].

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B

Some Calculus

In this appendix the inverse and implicit function theorems are derived in the framework of Banach spaces of arbitrary dimension. Of course, these theorems are well known and can be found in every standard text book on analysis. They are included here because we use them so often. Also, the basic existence and uniqueness theorem for solution curves of vectorfields is reviewed. We will restrict ourselves to the analytic case, since this suffices for our applications. Also, the proofs are slightly simpler, since we can use the fact that the locally uniform limit of analytic maps is analytic. See Theorem A.2. First some terminology. Let E , F be Banach spaces and U open in E . The map

f:U + F is a (real)analytic isomorphism on U , if it is a homeomorphism between U and its image Vin F, and f,f-' are (real) analytic on U , Vrespectively. The map f is a local (real) analytic isomorphism at a in U , if its restriction to some neighborhood A of a is a [real] analytic isomorphism on A .

I41

142

Inverse Spectral Theory

Inverse Function Theorem. Let f : U F be an analytic map from an open subset U of a complex Banach space E into a complex Banach space F. Let a E U. If d, f is a linear isomorphism between E and F , then f is a local analytic isomorphism at a. If E, F are real Banach spaces, then the same holds, if “analytic” is replaced by “real analytic”. +

Proof. Replacing f by (daf ) - ’ * f , it suffices to consider the case where E = Fand daf

I,

=

the identity map. By the continuity of df we can choose r > 0 such that, in the operator norm,

lldxf - Ill

IIx

on

I

-

all

Ir.

Then, by applying the mean value theorem to f - id, (1)

IIf(xl)

-f(xd

- (XI -

XZ)II

5 SUP XE-

I

tllx1

lldxf - 1 1 111x1 - ~ -

~ 1 1

xzll

for 11x1 - all Ir a n d 11x2 - all Ir. It follows thatfis one-to-one on the ball IIx - all Ir. We now construct the inverse g o f f on the ball 11 y - bll I r/2, where b = f ( a ) . T o this end, we inductively define the maps

(2)

gk=gk-I+id-fOgk-l,

k = 1 , 2 ,...,

where go 3 a , the constant map with value a. Let us show by induction that all these maps are well defined on I\y - bll I r/2 and satisfy

r

SF, Obviously, for k

=

1,

gl

k = 1 , 2,

is well defined, and

... .

143

Some Calculus

F o r k > 1,

r using (1). Thus, the gk converge uniformly on IIy - bll Going to the limit in (2),

f

0

Ir / 2 to a continuous map g .

g = id.

It follows that f is a homeomorphism between an open neighborhood of a and the open ball Ily - 611 < r / 2 . Moreover, g is analytic on this ball by Theorem A.2, since the maps gk are analytic. Consequently, f is a local analytic isomorphism at a. It remains t o show that g is real analytic when f is. Let f b e the analytic extension off to some complex neighborhood of a. Then d,fis the identity on the complexification of E, hence p i s an analytic isomorphism on some complex neighborhood A' of a. On the other hand, f is also a homeomorphism on some real neighborhood A of a by the same arguments as before. Consequently, f - I = f-'on A n A'. Hence, f is real analytic, and f is a real analytic isomorphism at a.

-'

The implicit function theorem generalizes the inverse function theorem. Here, an analytic map f:UxV+G from the product of open subsets U,Vof two Banach spaces E , Frespectively into a Banach space G is considered. At a point (a, 6 ) in U x V the partial derivative o f f with respect to, say, the second coordinate is required to be

144

Inverse Spectral Theory

a linear isomorphism between F and G. Here, the partial derivative abf:

F-+ G

off at (a, 6 ) with respect to the second coordinate is obtained by fixing the first coordinate at a, so that f becomes an analytic map from V into G , whose derivative is taken at b. Implicit Function Theorem. spaces E , F respectively, let

Let U,V be open subsets of complex Banach f:UxV-.+G

be analytic, and let (a, b) E U x V. If the partial derivative off at (a, 6 ) with respect to b, abf: F + G , is a linear isomorphism, then there exists an open neighborhood A of a and a unique continuous map u : A V such that -+

f ( x ,u(x)) = f ( a , b),

u(a) = b

on A. Moreover, u is analytic on A . If E , F are real Banach spaces, then the same holds, if “analytic” is replaced by “real analytic”. Proof. We may replace f by ( a bf ) - ’ * f . Then f maps U x V into F, and abf is the identity map on F. Now consider the map cp from U x V into E x F defined by (x9

Y)

+

(x,f (x,Y ) ) .

Its derivative

at (a, 6 ) is a linear isomorphism on E , F, so cp is a local (real) analytic isomorphism at (a, b ) by the inverse function theorem. Its local inverse has the form

145

Some Calculus

and for c = f ( a , b) we have g(a, c) = b.

Hence, in a neighborhood A of a, we can define u by u(x) = g(x, c).

The map u is (real) analytic on A , proving existence. To prove uniqueness we first choose A to be connected and so small that q~ is a local isomorphism at every point (x, u(x)) with x in A . Now suppose u : A + V is another map such that f ( x , u(x)) = f ( a , b),

u(a) = b

on A . Let B be the subset of A where u and u agree. B is closed by continuity and contains a. The set B is also open. For, if x E B, then u(x) = u(x) and consequently qJ(x,u(x)) = qJ(x9 W ) ) .

Since is a homeomorphism in a neighborhood of ( x , u(x)) by construction, u and u must agree also in a neighborhood of x . It follows that B = A , since A is connected. This proves uniqueness. H Let U be an open subset of a real Banach space E . A vectorfield on U is simply a map X : U + E.

Geometrically, a vectorfield assigns a tangent, or “velocity” vector to each point in U. A solution curve of the vectorfield X with initial value a in U is a differentiable map

4: J - , U from an open interval J containing 0 into U such that

4(0)

= a

and

d -4(t) dt

=

X(4(t)),

t

E

J.

Thus, the solution curve passes through a at time t = 0, and its velocity agrees with the vectorfield at every point it passes through.

Inverse Spectral Theory

146

+

A solution curve depends on the time t , the initial value a and the vectorfield X . We often use the notations +‘(a) and +‘(a,X ) to make this dependence explicit. The local existence and uniqueness of solution curves is guaranteed by a simple condition. A vectorfield X i s locally Lipschitz on U , if for every point a in U there is a neighborhood A and a positive constant L such that

for all x , y in A . Every continuously differentiable vectorfield is locally Lipschitz by the mean value theorem, and every locally Lipschitz vectorfield is continuous and locally bounded.

Local Existence and Uniqueness Theorem. Suppose X isa locally Lipschitz vectorfield on U. Then,f o r every a E U , there exists an interval J containing 0 and a solution curve

4: J - t U of X with initial value a. This curve is unique on J. If X is real analytic, then is a real analytic function oft and a.

+

We skip the standard proof, which may be found for example in [Di, Lal]. As a matter of fact, we d o not need the theorem at all, but include it for the sake of completeness only. In all cases of interest t o us, the solution curves can be written down explicitly. Every point on a solution curve may be considered as its initial value if the time is translated accordingly. It follows that if two solution curves pass through the same point, then they must be identical on their common interval of definition. Therefore, every solution curve has at most one extension on each side, and to each initial value a one can associate a unique maximal interval J, on which the solution curve with initial value a is defined. J, is the union of the domains of all possible solution curves through a. The interval J, may be finite, infinite on one side or the whole real line. In Chapters 4 and 6 we encounter all three possibilities. The solution curves of the vectorfields Vn are defined for all time, those of the vectorfield Wl are defined on the half line (- 03, PI), and those of Wn, n > 1, are defined on the finite intervals ( ~ ~ - 1Pn+i). , Another important consequence of this observation is the

Some Calculus

147

Semi-Group-Property. Let 4 be a solution curve of a locally Lipschitz vectorfield X . Then, for every point a on it, #+"(a) = 4t(4s(a))

for all s and t for which both sides are defined.

Proof. As functions of t , both sides are solution curves of the vectorfield X with initial value @(a). By uniqueness, the two curves are equal. To shed some more light on the semi-group-property, suppose all solution curves of a vectorfield X on a domain D are defined for all time. Then, for every real t , we have a map @': D

-, D,

where @(x) is obtained by following the solution curve with initial value x up to time t . Clearly, @' = id0

by definition, and @'+S

= @t

0 @S

by the semi-group-property. Thus, we obtain a one-parameter-group of continuous' maps of D onto itself. In fact, each such map Qt is a homeosince morphism of D with inverse Qt o @ - t = @ - t

o

mt =

=

idD.

Such a one-parameter-group at,t E I?of , homeomorphisms of D is called a flow on D . Conversely, every flow on D which is differentiable in t arises from a vectorfield on D.Simply set X(x) = - @ ( x ) dt We leave the elementary proof to the reader. 'Continuity follows from the proof of the local existence and uniqueness theorem and the semi-group-property.

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C

Manifolds

In this appendix we present the basic facts about manifolds required to make our discussion of isospectral sets self contained. All the manifolds that are of interest to us arise as subsets of Hilbert spaces. For this reason, we shall only consider submanifolds of Banach spaces, and not develop the theory of manifolds in general. To begin with, recall the following two terms. A diffeomorphism between two open subsets of Banach spaces is a continuously differentiable homeomorphism between these two sets, whose inverse is also continuously differentiable. A splitting of a Banach space is a direct sum decomposition into two subspaces, which are both closed. By definition, a subset M of a Banach space E is a submanifold of E, if for every point x on M there is a coordinate system of the following kind. There is a diffeomorphism p:

u+v

between an open neighborhood U of x in E and an open subset Vof another Banach space F and a splitting F = Fh 0 F,, such that p(UnM) = V n F h . In short, a submanifold locally looks like a “horizontal slice” of a Banach space.

I49

150

Inverse Spectral Theory

Depending on the regularity of the coordinate systems, one obtains different classes of submanifolds. For example, a submanifold is real analytic, if for every point there is a real analytic coordinate system. All our manifolds are of that class.

Example 1. An open subset of a Banach space is a real analytic submanifold. Example 2. A closed linear subspace is a real analytic submanifold if and only if it can be complemented. That is, one can find another closed linear subspace so that the two form a splitting. This is always possible in a Hilbert space, where one can take the orthogonal complement. This is not always possible in a Banach space. There are closed linear subspaces of Banach spaces, which can not be complemented [Ru]. Example 3. The sequence space S introduced in Chapter 3 is not given as a subset of a Banach space. However, there is a one-to-one correspondence between S and an open subset of R x P2. This allows us to identify S with a real analytic submanifold. Example 4. The isospectral set MP)

=

(4E L2:P ( d = P(P)I

introduced in Chapter 3 is a real analytic submanifold of L2.For, by Theorem 3.7, the translated map

4

+

(K

x P)(d -

(K

x PMP)

defines a global, real analytic coordinate system, which maps M ( p ) into the horizontal subspace P? x (0) of P? x (R x P2) (after identifying S with R x P2).

Example 5. Let M be any compact connected submanifold of a Banach space E, and let a, be any coordinate system on M mapping into a Banach space F = Fh 0 F, . Then a,-' is a homeomorphism between a closed ball in Fh and a closed subset of M , that is, the intersection of a closed subset of E with M . This intersection is compact by assumption, hence also the ball is compact. This implies that Fh is finite dimensional, since every locally compact Banach space is finite dimensional [Ru]. It follows that every

Manifolds

151

compact connected submanifold is finite dimensional. Conversely, an infinite dimensional compact subset of E can never be a submanifold. To give an example, let Pi be the Hilbert space of all complex P2sequences z = (ZI, Z Z , ...) and let r n , n L 1, be a sequence of positive numbers such that

Consider the set

Topologically, T is the product of the infinitely many circles lznl = r n ,so T is an infinite dimensional torus. However, the radii of these circles tend to zero just fast enough to make T a compact subset of Pi. So T is not a submanifold. One can give other examples of this phenomenon. For instance, every finite intersection of the subsets

of P2 is a real analytic submanifold of P2. Their infinite intersection, however, is compact, hence not a submanifold. The notion of tangent space is of fundamental importance. Intuitively, the tangent space of a manifold M at a point x consists of the velocity vectors at x of all smooth curves on M which pass through x . The tangent space is a closed linear space which one thinks of being attached to the point x . For submanifolds of a Banach space this notion is easy to make precise. Let M be such a submanifold, let x be a point on M , and let p be a coordinate system around x mapping into a Banach space F = Fh @ F,,.The tangent space of M at x is the closed linear space

T,M

=

(d,p)-'(Fh).

' I f M is a connected submanifold, then one can show that the spaces Fh for different coordinate systems are isomorphic. The dimension of M is then by definition the dimension of any of these spaces.

Inverse Spectral Theory

152

Equivalently,

=

ker dx(7cu* (D),

where nu denotes the projection of F onto F, along Fh . We show that the definition of T,M is independent of the coordinate system chosen. Suppose v/ and (D are two coordinate systems around x mapping into F = Fh @ F, and G = Gh @ G , respectively. Then iy (D-' is a diffeomorphism between open subsets V c F and W c G such that 0

W n Gh =

I/

0

(D-'(Vn Fh).

It follows that

or

The opposite inclusion holds by symmetry, so we have equality. Hence the definition of T,M does not depend on the coordinate system. In a Hilbert space we can also consider the orthogonal complement N X M= (T,M)' of T,M. This is the normal space of M at x .

Example I . If U is an open subset of a Banach space El then T,U = E for all x in U. Example 2.

If M is a closed linear subspace, then T,M

=

A4 for all x in M .

Example 3. For all o in S , one has T , S = IR x P2. This is a particular instance of Example 1. Example 4. The tangent and normal spaces of M ( p ) are computed in the proof of Theorem 4.1 *.

Manifolds

153

Manifolds often arise as the solution set of some nonlinear equation, typically as the level set, or “fiber”, of a mapping between two Banach spaces. Not every such level set, however, is a submanifold, as the simple example ( ( x ,y ) :x2 - y 2 = 0)

of a subset in IR2 shows. It is therefore useful to have a simple sufficient criterion for a level set to be a submanifold.

Figure 1.

Let f : A -,F be a continuously differentiable map from an open subset A of a Banach space E into another Banach space F. A point c in F is a regular value o f f , if for every point x in the level set

Mc = ( X E A :f ( x ) =

C)

there exists a splitting E = Eh 0 E,, such that d, f I E,, the restriction of d, f to E,, is a linear isomorphism between E, and F. Note that c is a regular value when Mc is empty. If F = IR, then c is a regular value off if and only if d, f f 0 at every point x in M, . More generally, if F = R“, then c is a regular value off, if d, f has rank n (that is, is onto) at every point x in M,.

154

Inverse Spectral Theory

Regular Value Theorem. Suppose f : A F is a real analytic map from an open subset A of a Banach space E into another Banach space F. If c E F is a regular value o f f , then +

Mc

= (X E A

:f ( x ) =

C)

is a real analytic submanifold of E. Moreover, rXMc = kerd,f at every point x in Mc .

Proof. Without loss of generality we may assume that c = 0. By hypotheses, for every x in MO there is a splitting E = Eh @ E,, such that with respect to this splitting the second partial derivative L =

a,f:

E,, -,F

is a linear isomorphism between E,, and F. We may therefore define a map cp from A into E by setting v(xh xu) = 9

(xh 9

L-'f

(xh 9

xu)).

This map is real analytic. Moreover,

(*I

~ ( Mno U )= p(U) n Eh

for every neighborhood U of x in A . The derivative of cp at x is a linear isomorphism of E, since

By the inverse function theorem, cp is a real analytic isomorphism between an open neighborhood I/ of x and an open neighborhood V of cp(x). Hence, in view of (*), cp is a real analytic coordinate system around x . This shows that MOis a real analytic submanifold of E. Finally,

K M o = ker dx(n,, cp) 0

= ker

nu - d,cp

= ker L- Idxf =

ker d, f ,

since L-' is a linear isomorphism. Here, nu denotes the projection of E onto E,,.

155

Manifolds

Example 4. The isospectral 5et M ( p ) is the fiber of the real analytic map p : L2+ S over p ( p ) . The proof of Theorem 4.1 shows that every point in S is a regular value of p. Hence, M ( p ) is a real analytic submanifold of r?. Example 5. On !$, the functions fn(z)

n

= IznJ2,

2 1,

are real analytic, since IznI2= x,' + y,' is a polynomial in the real and imaginary part of zn = xn + iyn. Their gradients are given by

They do not vanish, in fact, they are all linearly independent at every point

z in !$ all of whose components do not vanish. It follows that every hypersurface T, = ( Z

= (z E

e$: lznl

is a real analytic submanifold of intersection

n

=

E G:Jn(z)

T,, = (z E

21 rn > o

= rn),

t i . The

G :ltnl

same is true for every finite

= r n , 1 In IN J

IsnsN

of them. However, their infinite intersection

T = nT,, n z I

is not a submanifold of f't, if C r,' < 00, since this set is compact. This example shows that the linear independence of gradients is not enough to guarantee that a level set is a submanifold.

Example 6. For p in L2and n 1 1, Mn(P) = (4E L'

=

pn(p))

is a level set of the real analytic function punon L2.Its derivative dqpn(v) =

is never identically zero. Thus, every real number is a regular value of p,,,

156

Inverse Spectral Theory

and Mn(p) is a real analytic submanifold of Lz for every p . Moreover,

&Mn(p) = ker dqpn = (u E

L 2 :(gn2(q), u )

=

0).

Clearly, NqMn(p)is the line spanned by g:(q). A vectorfield on a submanifold M of a Banach space E is a map

X:M+E such that

X(x) E Z M for every x E M . A solution curve of X with initial value a on M is a differentiable map

4:J-M from an open interval J containing 0 into M such that 4(0)

=

a and

for all t E J. The local existence and uniqueness theorem of Appendix B generalizes immediately to submanifolds. Here, a vectorfield is locally Lipschitz, if for every point a on the submanifold M there is a neighborhood A of a in E and a positive constant L such that

IIX(x) - X(Y)II 5 LIlx - YII for all x, y E A n M .

Local Existence and Uniqueness Theorem. Suppose X is a locally Lipschitz vectorfield on a submanifold M of a Banach space E. Then,f o r every a E M there exists a solution curve

4: J + M of X with initial value a. This curve is unique on J. If M and X are real analytic, then 4 is a real analytic function o f t and a. The vectorfield X i s real analytic, if there exists an open neighborhood I/’ of M and a real analytic map .%: U E such that .%I M = X . +

Manifolds

157

We omit the standard proof. In short, it consists of three simple steps. Using a local coordinate system, the vectorfield X is pushed forward to an open subset of a Banach space. The local existence and uniqueness theorem of Appendix B is applied to obtain its solution curves, which are then pulled back to the submanifold M . As a matter of fact, in Chapters 4 and 6 we proceed in just this way instead of applying the general theorem, because in both cases the situation is particularly simple. We have a global coordinate system which maps the given (complicated looking) vectorfield into a consrant vectorfield on a Banach space. The unique solution curves of the latter are straight lines, which are pulled back to the submanifold. This approach in addition yields the analytic dependence of these curves on time, initial value andvectorfield.

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D

Some Functional Analysis

In this appendix we collect some facts about compact operators and state the Fredholm Alternative in a form, which is useful for us. Let H be a Hilbert space. A linear operator T o n H i s compact, if it maps weakly converging sequences into strongly converging sequences. Equivalently, a compact operator T maps bounded subsets into relatively compact subsets of H.' Obviously, compact operators are bounded. An operator has finite rank, if its range is finite dimensional. Clearly, bounded operators with finite rank operators are compact. From these, more interesting compact operators are obtained by

Lemma 1.

The uniform limit of compact operators is compact.

Proof. Let T be the uniform limit of compact operators G .Let xm be a weakly converging sequence, and let x be its weak limit. Then, by the 'The latter is the usual definition of compactness. The equivalence to the former requires a short proof, which we forego, since it is standard.

159

Inverse Spectral Theory

160

principle of uniform boundedness,

hence

IITxm - Txll

IIITXm -

Zxnll

I(IT - T,ll

+ IIZxm - Zxll + 1 1 % ~

* M + IIT,xm - Zxll

-

T’l

+ IIT, - TI1 .M.

The first and third term can be made small by choosing n sufficiently large. Then the middle term can be made equally small by choosing m sufficiently large, since all T, are compact. It follows that

( ( T x-~ Txll

+

0

as m tends to infinity. That is, Txn converges strongly to Tx. Thus, T is compact. W Suppose H is a separable Hilbert space (as are all Hilbert spaces we encounter in this book). A linear operator T o n H i s Hilbert-Schmidt if

C

nz1

IITenIJ’ <

00

for some orthonormal basis e n , n L 1, of H . A Hilbert-Schmidt operator is bounded. To see this, write

x =

c xne,.

n r 1

Then llx((’ =

En> I

Ixn(’, and by the triangle and Schwarz inequality,

IITxll

= I

I I

Hence, T is bounded. Furthermore, we have

Theorem 1. If T is Hilbert-Schmidt, then T is compact.

Some Functional Analysis

Proof. Let e l , e2, ... be an orthonormal basis of H such that finite. For n L 1, let Tn be the linear operator defined by

161

C 11 Tern(('is

as n tends to infinity. Thus, T is the uniform limit of compact operators, hence compact by Lemma 1. W We frequently have to consider compact perturbations of the identity map. These are operators of the form Z - T, where I is the identity and T is compact. By slight abuse of terminology, we call them Fredholm operator^.^ In some respect, they resemble operators on finite dime.piona1 spaces: Theorem 2 (Fredholm Alternative). Suppose A is a Fredholm operator. Then the following three statements are equivalent. (1) A is boundedly invertible. (2) A is onto. ( 3 ) A is one-to-one. 'As usual, 11 TI1 = supllxll = I (1 Txll denotes the operator norm of T.

'Usually, this term is reserved for the much larger class of operators with finite index and coindex. See for instance [Ka].

Inverse Spectral Theory

162

Proof. (1) * (2) Obvious. (2) * (3) Suppose A = I - T is onto. The kernels of A", n = 0, 1, ..., form an increasing sequence of closed subspaces of H . Suppose no two of them are equal. Then we can choose unit vectors

x, We have x,

-+

E

ker(A") n ker(A"-')',

n

2 1.

0 weakly, hence Txn -0 strongly by compactness. Then also ( Txn - TXn-1 ,Xn)

+

0.

On the other hand, A maps ker(A") into ker(A"-'), and T = I - A maps ker(A") into itself. Therefore,

(Txn -

TXn-l,Xn> =

(Txn,xn) = (xn,xn>=

1,

a contradiction. Thus, we have ker(A") = ker(A"-') for some n L 1. That is, AA"-'x = 0 implies A"-'x = 0 for all An-' is onto for every n 1 1, it follows that A is one-to-one.

x. Since

(3) * (1) Suppose A = I - T is one-to-one. Suppose we have

0 strongly. Then we can choose a sequence of unit vectors X, such that Ax" Extracting a subsequence, Tx, converges to some x by compactness. Then also -+

strongly, hence llxll

=

1. On the other hand,

AX hence x

=

=

lim Ax,

=

0,

0, since A is one-to-one. This is a contradiction, and we must have

It follows that the pointwise inverse of A is bounded on its range rg(A), for

IIA-'Axll

=

llxll

5

s-'IIAxll

for all x. This further implies that rg(A) is closed.

Some Functional Analysis

163

It remains to show that rg(A)is all of H . Consider rg(A") for n = 0, 1, ... . This is a decreasing sequence of closed subspaces of H , since T" is compact, hence A" is Fredholm for all n 2 1. As before, one shows that

rg(A") = rg(A"-') must hold for some n 2 1. Hence, for each y there is an x such that Any = A"-'X.Since A" is one-to-one for every n 2 1, we have y = Ax. This shows that A is onto. Among other things, the Fredholm Alternative can be used to show that a sequence of vectors is a basis. The following theorem describes the general situation. Theorem 3. Let en, n 2 1, be an orthonormal basis of a Hilbert space H. Suppose dn , n 2 1, is another sequence of vectors in H that either spans or is linearly independent. If, in addition,

C

n r 1

Ildn - en]/' <

00,

then dn , n 1 1, is also a basis of H. Moreover, the map

x

+

( ( x ,dn), n

2 1)

is a linear isomorphism between H a n d P 2 .

Proof. Define an operator A on H by AX =

C

(X,en)

dn.

n r1

A maps en into d,, . This operator is a compact perturbation of the identity, for

C

n r1

ll(A - I)enl12=

C

n r 1

Ildn - enllZ<

00,

so A - I is Hilbert-Schmidt and hence compact by Theorem 1. If the sequence d, is linearly independent, then A is also one-to-one: if A x = 0, then

( x , en) = 0,

n

2 1

by linear independence, and consequently x = 0, since the en are a basis. If, on the other hand, the sequence dn spans, then the range of A is dense in H.

Inverse Spectral Theory

164

It follows that A is onto, since the range of a compact perturbation of the identity is always closed [ R u ] . ~ So in both cases the operator A is boundedly invertible by the Fredholm alternative. Hence the d,, are also a basis. Finally, we have ( x , d n ) = ( X , A e n ) = ( A * x , en>.

The map x

+

( ( A * x , en>,n 2 1)

is a linear isomorphism between H a n d f", since A* is boundedly invertible, and the e,, are an orthonormal basis. 4This is not proven here, because this part of the theorem is used only once in the proof of Lemma 4.3.

E

Three Lemmas on Infinite Products

Lemma 1. (a) Suppose amn, m , n > 1, are complex numbers satisfying

Then

(b) In addition, if bn, n numbers, then

Proof.

L

1, is a square summable sequence of complex

(a) By assumption,

m#n

m#n

165

166

Inverse Spectral Theory

with some positive constant C. The sum on the right can be estimated by

c

m z l m#n

1

C

=

1m2 - n 2 )

lsms2n m#n

2

I- ( 1

n

1

1

1

+ + C

~Im - nl m n m>2nm2 -

n2

+ logn) + -n1.

Hence we obtain, with a different constant C',

Iexp(c'

(b) By the proof of (a), Schwarz inequality,

The claim follows.

Problem I . improved.

smfn

1

+ logn

larnnlIC(l

)

- 1

+ log n ) / n . Hence, by the

H

Show that the error term O(1og n / n ) in Lemma 1 can not be

167

Three Lemmas on Infinite Products

Recall the well known product expansion [Ah, Ti] sin fi -

--

fi

m r l

m2n2 - A m2n2

If in the numerator the numbers m2n2are replaced by complex numbers Zm with the same asymptotic behavior, then we obtain an entire function, which approximates sin fi/fion certain circles. This is the content of the following lemma.

Lemma 2.

Suppose z m ,m z 1, is a sequence of complex numberssuch that

z,,,

m2n2 + 0(1).

=

Then the infinite product

n -m n A Zm -

m r l

71

is an entirefunction of A, whose roots are precisely the Zm ,m 2 1. Moreover,

n mv n =A -sinfifi (1 + 0(lo, n)) Zm -

mzl

uniformly on the circles Proof.

1A1 = ( n + *)’n2.

By the uniform boundedness of

Zm -

m2n2for m 2 1,

tm -

m2n2 - A

converges uniformly on bounded subsets of C. Therefore, the infinite product converges to an entire function of A, whose roots are precisely z m , m 2 1. The quotient of the given product and sin fi/fiis the infinite product

n

mzl

On the circles

Zm -

A

m2n2 - A

’

[A1 = (n + s)2n2the uniform estimates m=n

Inverse Spectral Theory

168

hold. Then, by Lemma 1 ,

n

m r1

m2n2 Z m --A A = (I = 1

+ 0(;))(1+ o($)) + o(T) log n

uniformly on these circles. The last lemma is a variant of the preceding one.

Lemma 3. Suppose z m , m 2 1, is a sequence of complex numbers such that Zm =

Then, f o r each n

1

m2n2 + 0 ( 1 ) .

1

n

Z m - A

ma1 m#n

m71

2

2

is an entire function of A such that

uniformIyfor A = n2n2 + 0(1).

Proof. We only prove the last statement. By the product expansion for sin fi/fi,we have

The quotient of this and the given product is

n

ma1 m#n

Zm - A = 1+ m2n2- n2n2

+)

uniformly for A = n2nZ + 0(1)by Lemma 1.

log n

F

Gaussian Elimination

In Chapters 5 and 6 explicit formulas for the isomorphisms exp, and pEi are given, which only involve data at a single point q. In both cases the crucial ingredient of their derivation is an identity relating the determinant of an n x n-matrix to an n-fold product. In this appendix proofs of these identities are given. They are based on the familiar Gaussian elimination scheme. For our purposes it can be stated as follows.

Lemma 1 (The Gaussian Elimination Scheme). Let A = (a!) be an n x nmatrix. For 1 5 k < n , let

assuming that a k ' # 0for 1 Ik < n. Then A can be transformed into the upper triangular matrix

I69

Inverse Spectral Theory

170 0 -

-a?l a L a L

o a22 o o ...

...

0

0

gin

a23 ...

1 a2n

aB

2 ~3~

... ...

0

***

a,";

by elementary row transformations. Consequently, n

n a,kl

= detA.

k=l

More generally,

if

with real or complex numbers a;,&, then

Proof. Apply induction on n. The statement is true for 1 x 1-matrices. So assume it is true for ( n - 1) x ( n - 1)-matrices, n > 1. By assumption, aP1 # 0. In case (1) holds, multiplying the first row of A by - (a;l/aYl)and adding it to the ith row, 1 < i 5 n , we obtain the matrix

(3)

To the lower right ( n - 1) x ( n - 1)-matrix the induction hypotheses applies. Hence, A can be transformed into the matrix (2) by elementary row transformations. These transformations do not affect the determinant of a matrix, so

fi a$il

= detA.

k=l

' An elementary row transformation consists in adding a multiple of one row to another row.

Gaussian Elimination

171

In case (1*) holds, first multiply the matrix A from left and right with the diagonal matrices 1 a; a;

1

an.

respectively. You obtain a matrix

A

=

(riij)

such that

By the same row transformations as before, matrix (3). Hence,

fl

det A

A

is transformed into the

aj/# = a71det A ',

1ejan

where A ' denotes the lower right (n - 1) x (n - 1)-matrix in (3). To A' the induction hypotheses applies, and we have

n

de tA'

. .

ajpj

=

Zaiejan

n

ak'.

2skan

The claim follows. We apply the Gaussian elimination scheme to the matrix

0'"'

= (Oij)iai,jan

with elements 8ij

=

eij(x,

=

6ij

t, 4)

+ (e" - 1)

s:

gi(s, q)gj(S, 4)ds.

To this end we verify an identity of the form (1) relating the &functions of 40 =

4,

qk = +"(qk-l

3

I/k),

15

as they were defined in the proof of Theorem 5.2.

k

5

n,

lnverse Spectral Theory

172

Lemma 2 . For 1 I k

I

n and k

< i, j

In,

We drop the arguments x , <,which are fixed in the following discussion. Combining Lemma 2 with the Gaussian elimination scheme we obtain n

det 0'"'=

n n

dkk(qk-1)

k=l

n

=

o k ( x ,
qk- 1)

k=l

in the notation of chapter 5 as we wanted to show.

Proof of Lemma 2. By Theorem 5.1,

=gi--

g k Oki ekk

for i # k,where the functions on the right hand side are all evaluated at q k Hence, for k < i, j In, we have eij(qk)

=

6ij

4-

(eEi- 1 )

s:

gi(qk)&(qk)dS

We thank T . Nanda for his help in proving this lemma.

1

.

Gaussian Elimination

173

and

the second term becomes

=

!k%lxm

All the functions are evaluated at Qk-1, so the lemma is proven. Next we apply Lemma 1 to the matrix

where

See Chapter 6 for the definition of Wi and zj. We shall verify an identity of the form (1*) relating the co-functions of

to each other. Recall that in general the flows of the vectorfields W1 , ..., W, are applied in some permuted order instead of their natural order to avoid the crossing of eigenvalues. As a consequence the Gaussian elimination scheme has to be applied to the columns of Qcfll in the same permuted order. Equivalently, the columns of Qc")first have to be permuted properly so as to apply the scheme from left to right as usual. This permutation, however, affects only the sign of the determinant of Q("), which is of no importance to us. Thus, for the sake of simplicity, we may assume that the flows of WI, ..., W, are applied in their natural order.

Inverse Spectral Theory

174

Again, we dropped the arguments x , U , which are fixed in the following. Here, the more general form of Lemma 1 applies, with

It follows that

as we wanted to show.

Proof of Lemma 3. By definition,

where here and in the following, the functions w; are evaluated at 1 = without further mention. By the remark following Theorem 6.2,

U;

(4)

where the functions on the right hand side are evaluated at P k arguments, Vj(Pk)

=

w; -

is a genuine solution of -y"

gk ~

-

pk

wk

-[ W i , Z k ] ,

0; - p k a k k

1

. By the same

i #k

+ p k y = 1y for 1 = 0 ; . We claim that

'This lemma is an unpublished result of J. Ralston and E. Trubowitz.

Gaussian Elimination

175

To prove this it suffices to check the boundary values of u;. One calculates

and

using the Wronskian identity and the fact that the boundary values of are always 1 . By equation (4),

Okk

and therefore also

again by the Wronskian identity. It follows that the right hand side in ( 5 ) has the same values at 0 and 1 as W ; ( p k ) , namely 1 and yl(l, p ; , p k ) respectively. This proves ( 5 ) . We determine the Wronskian of U ; ( P k ) and Z j ( p k ) . An elementary but lengthy calculation shows that

To the third term apply the identity

Inverse Spectral Theory

176

which yields

It follows that

It remains to multiply both sides of this equation by

On the left hand side we obtain

Using corresponding identities on the right hand side we arrive at mij(Pk)

as was to be proven.

pk

- oi U k - p j

Qk

-

=-

H

Qi p k

-pj

G

Numerical Calculations

In this appendix the FORTRAN programs are described and listed that were used to produce the figures in Chapters 5 and 6. Since the actual plotting of these figures depends on the machine and software used, only those subroutines are given that produce the data to be plotted. To make them easy to read, virtually no attempts was made to optimize them with respect to speed or code. There are three subroutines KAPPA, MUE and EXPON. In each case, the output of the subroutine is a “function on [0, 11” given by a vector result of length len + 3, where the value of the function at the point

is stored in the k + 2nd element of this vector. Its first and last element are only used during intermediate calculations and do not contain meaningful data. KAPPA determines the function q = #(O, K), when the parameter j is zero, or its normalized eigenfunction gj, when j 2 1, according to the formulas of Theorem 5.1. Similarly, MUE determines the function q = 4‘(0, Wn), whenj = 0, or its normalized eigenfunction gj, whenj 2 1.

I77

Inverse Spectral Theory

178

This program makes use of the fact that q is always even, while gj is even when j is odd and odd when j is even. The syntax is

CALL KAPPA(resu/t, j , n ,t, vectorJen) and

CALL MUE(result,j,n ,s,vector, /en). Here, vector is a vector of length len + 3 used to store intermediate results, and - 1 < s < 1 is a rescaled time variable such that

t

=

( n + s)27r2-

n27r2.

See the program listings for the type of each parameter. EXPON either determines q = expo(K) or q = pil(s) for arguments s = (SI, s2, ...) having only finitely many coordinates S n l , ...,Snm

different from zero. In this case, the formulas of Theorem 5.3 and 6.2 are both of the form

d2

q = -2710gdetE, dx

where E is an rn x rn-matrix with entries eninj(x, sni,0) and U n i n , ( X , Sni) respectively. For the sake of simplicity, only the case 1 5 rn < 3 is supported. The syntax is

CALL EXPON(resu1t,n vec,svec,su bnarne, vector,array ,rn ,/en). Here, nvec and svec are vectors of length rn containing the indices nr , ...,nrn and the arguments sn,,...,snm respectively, subnarne is either THETA or OMEGA, the name of the subroutine calculating the entries of E, and vector and array are an auxiliary vector of length len + 3 and an auxiliary array of size rn x rn x (len + 3) respectively. In case subnarne is OMEGA, the arguments in svec must lie in the interval (- 1, 1) and are rescaled to (n; + sni)’n2- nfn’, 1 Ii 5 rn. See the program listing for the type of each parameter. KAPPA, MUE and EXPON call a number of other subroutines, which we only describe briefly. OMEGA calculates the u-function at 0 using the identity

Numerical Calculations

179

where t = I - pun.The program makes use of the fact that this function is even. LOGDER approximates the second logarithmic derivative

of a function u by the second difference expression

Here, Uk is the value of u at the kth point of a grid of width h. NORMAL normalizes a vector so that it has euclidian length 1. DET yields the determinant of an rn x rn-matrix. For the sake of simplicity, only the case 1 Irn I3 is supported. The calculation of the function w n is speeded up by writing wn(x, I , 0) = c o s f i x

+ c,,(I)sinfix

with

cfl(I) =

(- 1y - c o s f i

sinfi

The function COEFF yields the value of c,(I) for given n and fi.This value is passed as a parameter to the function W calculating w n . These programs were run on an IBM Personal Computer equipped with an 8087 math coprocessor. The plots were produced on a Hewlett Packard 7475 Plotter. C C C C

P L

SUBROUTINE KAPPA(V,J,N,T,TH, L) REAL*8 V(*),TH(*) CALL THETA(TH,N,N,T,L) IF (J.EQ.0) THEN CALL LOGDER(TH,V,L) ELSE IF (J.EQ.N) THEN FCT = EXP(T/2.) DO 1 K = 2 , L + 2 = REAL(K - 2)/L X V(K) = FCT*G(N,X)/TH(K) 1 CONTINUE ELSE CALL THETA(V,J,N,T,L) DO 2 K = 2 , L + 2 X = REAL(K-2)/L V(K) = G(J,X) - G(N,X)*V(K)/TH(K)

Inverse Spectral Theory

180

2 C

C C C C

C

C

C C

C

C

CONTINUE END IF RETURN END

SUBROUTINE MUE(V,J,N,S,OM,L) REAL*8 SUM,V(*),OM(*) REAL LAM,MU CALL OMEGA(OM,N,N,S,L) IF (J.EQ.0) THEN CALL LOGDER(OM,V,L) ELSE IF (J.EQ.N) THEN DO 1 K = 2 , (L+4)/2 = REAL(K-2)/L X V(K) = Z(N,X)/OM(K) 1 CONTINUE ELSE = 3.141593 PI RL = PI*(N+S) LAM = RL**2 = LAM-MU(N) T CO = COEFF(N,RL) H = 1./L SUM = 0. DO 2 K = 2 , (L+4)/2 = REAL(K-2)/L X SUM = SUM + Z(J,X)*Z(N,X) V(K) = Z(J,X) - T*W(CO,RL,X)*SUM*H/OM(K) 2 CONTINUE END IF SIG = 2*MOD(J,2)- 1 DO 3 K=(L+4)/2,L+2 V(K) = SIG*V(L+4-K) 3 CONTINUE CALL NORMAL(V,L) END IF RETURN END SUBROUTINE EXPON(V,N,S,ELEM,W,MAT,M,L) REAL*8 V(*),W(*),MAT(M,M,*),DET S(*) REAL INTEGER N(*) EXTERNAL ELEM DO 1 I = l , M

Numerical Calculations

181

DO 1 J = l , M CALL ELEM(W.N(I),N(J),S(I),L) . _ . . _ .. _ . DO 2 K = l , L + 3 MAT(I,J,K) = W(K) 2 CONTINUE 1 CONTINUE DO 3 K = l . L + 3 W(K) = DET(MAT(~,~ , K ) , M ) 3 CONTINUE CALL LOGDER(W, V, L) RETURN END ,

C C C

C C

C

C

C C

C

SUBROUTINE THETA(TH ,N,M,T,L) REAL*8 TH(*),SUM FCT = EXP(T)- 1. H = 1./L X = l.+H/2. SUM = - G(N,X)*G(M,X) IF (N.EQ.M) THEN DO 1 K = L + 3 , 1 , - 1 TH(K) = 1. + FCT*SUM*H SUM = SUM + G(N,X)**2 X = X-H 1 CONTINUE ELSE DO 2 K = L + 3 , 1 , - 1 TH(K1 = FCT*SUM*H SUM ’ = SUM + G(N,X)*G(M,X) = X-H X 2 CONTINUE END IF RETURN END SUBROUTINE OMEGA(OM,N,M,S,L) REAL *8 SUM ,OM(*) REAL LAM,MU = 3.141593 PI RL = PI*(N+S) LAM = RL**2 T = LAM-MU(N) CO = COEFF(N,RL) IF (N.EQ.M) THEN TRM = 1.

Inverse Spectral Theory

182

C C

C C

C C C

C C C C

C C C

ELSE TRM = T/(LAM-MU(M)) END IF H = 1./L X = -H/2. SUM = - W(CO,RL,X)*Z(M,X) DO 1 K = 1,(L +4)/2 OM(K) = TRM + T*SUM*H SUM = SUM + W(CO,RL,X)*Z(M,X) X = X+H 1 CONTINUE DO 2 K = (L + 4)/2,L + 3 OM(K) = OM(L + 4 - K) 2 CONTINUE RETURN END SUBROUTINE LOGDER(U ,V, L) REAL*8 UM,UN,UP,U(*),V(*) FCT = -2*(L**2) DO 1 K = 2 , L + 2 UM = U(K- 1) UN = U(K) UP = U ( K + l ) V(K) = FCT*((UP - 2*UN + UM)*UN ((UP-UM)/2)**2)/UN**2 CONTINUE RETURN END REAL*8 FUNCTION DET(A,M) REAL*8 A(M,M) IF (M.EQ. 1) THEN DET = A(1,l) ELSE IF (M.EQ.2) THEN DET = A( 1,1)*A(2,2) - A( 1,2)*A(2,1) ELSE IF (M.EQ.3) THEN DET = A(l, l)*(A(2,2)*A(3,3) - A(2,3)*A(3,2)) * - A(1,2)*(A(2,1)*A(3,3) - A(2,3)*A(3,1)) * + A(1,3)*(A(2,1)*A(3,2) - A(2,2)*A(3,1)) ELSE M > 3 not supported END IF END

Numerical Calculations

C C C C C C

C C

C C C

C C

C C

C C

SUBROUTINE NORMAL(V,L) REAL*8 V( *), SUM, DSQRT SUM = 0. DO 1 K = 2 , L + 2 SUM = SUM+V(K)**2 CONTINUE FCT = DSQRT(L/SUM) DO 2 K = 2 , L + 2 V(K) = FCT*V(K) CONTINUE RETURN END REAL FUNCTION MU(N) PI = 3.141593 MU = (PI*N)**2 END REAL FUNCTION G(N,X) DATA P1,ROOT /3.141593,1.4142 141 G = ROOT*SIN(PI*N*X) END REAL FUNCTION Z(N,X) PI = 3.141593 PIN = PI*N = SIN(PIN*X)/PIN Z END REAL FUNCTION W(CO,RL,X) RLX = RL*X = COS(RLX) + CO*SIN(RLX) W END REAL FUNCTION COEFF(N,RL) DATA PI,EPS /3.141593,0.01/ IF (ABS(RL - PI*N).GT.EPS) THEN COEFF = (1 - 2*MOD(N,2) - COS(RL))/SIN(RL) ELSE -~ COEFF = SIN(RL)/COS(RL) END IF END ~~

C

183

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References

[Dil [Din]

L. Ahlfors, Complex Analysis. McGraw-Hill, New York, 1979. V. Ambarzumian, “Uber eine Frage der Eigenwerttheorie”. Zeitschr. f. Phys. 53, 690-695, 1929. A. Balakrishnan, Applied Functional Analysis. Springer, New York, 1981. G. Borg, “Eine Umkehrung der Sturm-LiouvilleschenEigenwertaufgabe”. ActaMath. la, 1-96, 1946. G. Birkhoff, G. Rota, “On the completeness of Sturm-Liouville expansions”. Am. Math. Monthly 61, 835-851, 1960. K . Chadan, P. Sabatier, Inverse Problems in Quantum Scattering Theory. Springer, New York, 1977. G. Darboux, “Sur la representations sphkrique des surfaces”. Compt. Rend. 94, 1343-1345, 1882. G. Darboux, LeCons sur le Thdorie gdndrale des Surfaces et les applications gdomdtrique du calcul infinitesimal. Vol. 2, Gauthier Villars, Paris, 1915. B. Dahlberg, E. Trubowitz, “The inverse Sturm-Liouville problem 111”. Comm. Pure Appl. Math. 31, 255-267, 1984. P. Deift, “Applications of acommutation formula”. DukeMath. J. 45,267-310, 1978. P. Deift, E. Trubowitz, “Inverse scattering on the line”. Comm. Pure Appl. Math. 32, 121-251, 1979. J. Dieudonnt, Foundations of Modern Analysis. Academic Press, New York, 1969. S . Dineen, Complex Analysis in Locally Convex Spaces. North Holland, Amsterdam, 1981. I. Gel’fand, B. Levitan, “On the determination of a differential equation from its spectral function”. Amer. Math. SOC.Transl. 1(2), 253-304, 1955. Russian: Izv. Akad. Nauk SSSR 15, 309-360, 1951.

185

186

Inverse Spectral Theory

H. Hochstadt, B. Lieberman, “An inverse Sturm-Liouville problem with mixed given data”. SIAM J . Appl. Math. 34, 676-680, 1978. E. Isaacson, H. McKean, E. Trubowitz, “The inverse Sturm-Liouville problem 11”. Comm. Pure Appl. Math. 37, 1-11, 1984. E. Isaacson, E. Trubowitz, “The inverse Sturm-Liouville problem I ” . Comm. Pure Appl. Math. 36, 767-783, 1983. R. Jost, W. Kohn, “Equivalent Potentials”. Phys. Rev. Ser. 2 88, 382-385, 1952. S. Lang, Real Analysis. Addison-Wesley, Reading, 1969. S. Lang, Differentiable Manifolds. Addison-Wesley, Reading, 1972. N. Levinson, “The inverse Sturrn-Liouville problem”. Mat. Tidsskr. B . , 25-30, 1949. J. Liouville, “Memoire sur le developpement des fonctions ou parites de fonctions en series dont les divers termes sont assujettis a satisfaire a une mCme equation differentielles du second ordre contenant un pararnttre variable”. J. de Math. 1, 253-265, 1836; 2, 16-35 and 418-436, 1837. H. McKean, E. Trubowitz, ‘‘Hill’s operator and hyperelliptic function theory in the presence of infinitely many branch points”. Comm. Pure Appl. Math. 29, 143-226, 1976. L. Nachbin, “Topology on Spaces of Holornorphic Mappings”. Ergebnisse der Mathematik und ihrer Grenzgebiete 41. Springer, 1969. M. Reed, B. Simon, Methods of Modern Mathematical Physics, volume I and I V . Academic Press, New York, 1980 and 1978. W. Rudin, Functional Analysis. McGraw-Hill, New York, 1973. C. Sturrn, “Sur les equations differentielle lineares du second ordre”. J . de Math. 1, 106-186, 1836. E. Titchmarsh, The Theory of Functions. Clarendon Press, Oxford, 1932. W. Wasow, Asymptotic Expansions for Ordinary Differential Equations. Interscience, New York, 1965. H. Weyl, “Uber gewohnliche Differentialgleichungen mit Singularitaten und die zugehorigen Entwicklungen willkiirlicher Funktionen”. Math. Ann. 68.220-269.1910.

Index

A absolutely continuous 2 adjoint 128 algebraic multiplicity 30 analytic isomorphism 141 local 141 analytic map 132 real 139 weakly 133 analyticity properties 10

B Basic Estimates 13 basis 46 boundary condition 25 boundary value problem 25

C Cauchy’s formula 134 chain rule 130 compact 18, 126 map 126 operator 159 continuous 125 absolutely 2 continuously differentiable 126

coordinate system 149 Counting Lemma 28

D Deformation Lemma 41 derivative 20, 126 at a point 126 partial 144 determinant 102 Fredholm 106 diffeomorphism 149 differentiable 126 at a point 126 continuously 126 Dirichlet boundary condition 25 eigenfunction 25 eigenvalue 25 spectrum 25 E

elementary row transformation 170 even function 41, 5 3 existence and uniqueness theorem 146, 156 exponential map 77

187

188

Inverse Spectral Theory

F fiber 50, 153 fibration 79 finite rank operator 159 flow 147 Fredholm Alternative 161 determinant 106 operator 161 fundamental solution 1

locally bounded 133 Lipschitz 146

M

Gaussian elimination scheme 169 geometric multiplicity 30 gradient 20, 127

map analytic 132 compact 18, 126 continuous 125 differentiable 126 exponential 77 mean value theorem 131 method of shooting 26 minor 102 multiplicity 30

H

N

Hilbert-Schmidt operator 160 homogeneous polynomial 132

norm on L: 7 operator 131 normal space 152

G

I implicit function theorem 144 independent 46 initial value 73, 145, 156 integral 131 inverse Dirichlet problem 50 function theorem 142 isomorphism analytic 141 linear 46 local analytic 141 isospectral set 49

K Korteweg-de Vries equation 87

L

0

odd function 39, 41, 53 operator compact 159 Fredholm 161 norm 131 self adjoint 25

P partial derivative 144 Picard iteration 10 polarization identity 133 polynomial 132 power series 2, 5, 132 principal minor 102

Q

linear independence 46 isomorphism 46 Lipschitz 146 local analytic isomorphism 141 existence and uniqueness theorem 146, 156

QR-algorithm 97

R radius of weak analyticity 133 real analytic map 139 submanifold 150

Index regular value 153 theorem 154 Riemann integral 131 row transformation 170

189

T tangent space 151 Taylor’s formula 131 Taylor series expansion 131 trace class 106

S

V semi-group-property 147 shooting, method of 26 smooth 130 solution 2 curve 73, 145, 156 fundamental 1 splitting 149 submanifold 149

vectorfield 145, 156

W weakly analytic map 133 Weierstrass @-function 35 Wronskian 11 identity 11

This Page Intentionally Left Blank

Index of Notations

The symbols are listed in their order of appearance. 20 25 Qi

30 31 31 35 35 37 49

50

50 20, 127

51

191

192

Inverse Spectral Theory

L7, P n

51

an

107

4*

52

w,

108

E, u

53

9'(4,w,)

108

PE

53

Kn

59

K

61

K X P

61

v,

63

Wn

63

UII

68

4

69

v,

69

d'(9, Y )

73

122

an

74

130

exp,

77

130

9"

88

131

0

88

151

gij

88

152

108 109 109 117 117 117, 123 117, 123 122

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