IRMA Lectures in Mathematics and Theoretical Physics 4 Edited by Vladimir G. Turaev
Institut de Recherche Mathe´matique Avance´e Universite´ Louis Pasteur et CNRS 7 rue Rene´ Descartes 67084 Strasbourg Cedex France
IRMA Lectures in Mathematics and Theoretical Physics
1 2 3
Deformation Quantization, Gilles Halbout (Ed.) Locally Compact Quantum Groups and Groupoids, Leonid Vainerman (Ed.) From Combinatorics to Dynamical Systems, Fre´de´ric Fauvet and Claude Mitschi (Eds.)
Three Courses on Partial Differential Equations Editor Eric Sonnendrücker
≥
Walter de Gruyter · Berlin · New York
Editor Eric Sonnendrücker De´partement de Mathe´matique, Universite´ Louis Pasteur, 7, rue Rene´ Descartes, 67084 Strasbourg Cedex, France, e-mail:
[email protected] Series Editor Vladimir G. Turaev Institut de Recherche Mathe´matique Avance´e (IRMA), Universite´ Louis Pasteur ⫺ C.N.R.S., 7, rue Rene´ Descartes, 67084 Strasbourg Cedex, France, e-mail:
[email protected] Mathematics Subject Classification 2000: 35-02; 35R30, 35R60, 74B99 Key words: Partial differential equations, elasticity, waves, random media, parameter identification, inverse problems
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Library of Congress Cataloging-in-Publication Data Three courses on partial differential equations / editor, Eric Sonnendrücker. p. cm. ⫺ (IRMA lectures in mathematics and theoretical physics ; 4) Includes bibliographical references. ISBN 3-11-017958-X (alk. paper) 1. Differential equations, Partial. I. Sonnendrücker, Eric. II. Series. QA377.T526 2003 515⬘.353⫺dc22 2003062470
ISBN 3-11-017958-X Bibliographic information published by Die Deutsche Bibliothek Die Deutsche Bibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data is available in the Internet at ⬍http://dnb.ddb.de⬎. 쑔 Copyright 2003 by Walter de Gruyter GmbH & Co. KG, 10785 Berlin. All rights reserved, including those of translation into foreign languages. No part of this book may be reproduced 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. Printed in Germany. Cover design: I. Zimmermann, Freiburg. Depicted on the cover is the Strasbourg Cathedral. Typeset using the authors’ TEX files: I. Zimmermann, Freiburg. Printing and binding: Hubert & Co. GmbH & Co. KG, Göttingen.
Preface
Partial differential equations (PDEs) have been playing a major role in the modeling of many phenomena in physics and engineering for a long time. The realm of applications is growing fast, in particular in economics and life sciences. Numerical simulation of the phenomena that are investigated with the aim of predicting their behavior is thus linked to a good understanding of the mathematical properties of the PDEs involved and the derivation of efficient and robust numerical methods to solve them. Therefore a good training in partial differential equations is essential for mathematicians interested in such applications. Following traditional introductory graduate courses on different topics of partial differential equations ranging from theoretical to numerical aspects that were given during the year 2001 (Diplôme d’Études Approfondies de mathématiques) at the University Louis Pasteur in Strasbourg, a special week consisting of five advanced lectures was proposed to the students. These lectures were the following. • Michel Chipot (University of Zürich): On some questions in elasticity. • Patrick Ciarlet (ENSTA, Paris): The finite element method in non-smooth computational domains. • Josselin Garnier (University Paul Sabatier, Toulouse): Waves in random media. • Otared Kavian (University of Versailles Saint-Quentin): Introduction to parameter identification. • François Murat (University Pierre et Marie Curie, Paris): Homogenisation. This volume contains enhanced versions of the lectures of M. Chipot, J. Garnier and O. Kavian. These three lectures all address very important and interesting, but very different aspects motivated by several applications in partial differential equations. Michel Chipot investigates equilibrium positions of several disks rolling on a wire. He is looking in particular for existence of, uniqueness of, and the exact position for an equilibrium. Josselin Garnier considers problems arising from acoustics and geophysics where waves propagate in complicated media, the properties of which can only be described statistically. He shows, in particular, that when the different scales presented in the problem can be separated, there exists a deterministic result. Otared Kavian is interested in so-called inverse problems, where one or several parameters of a PDE need to be determined, by, using for example, measurements on the boundary of the domain. The question that arises naturally is what information is necessary to determine the unknown parameters. This lecture answers to this questions in different settings.
vi
Preface
To conclude, I would like to thank all the people who made this special week on partial differential equations possible and very interesting: Olivier Debarre who suggested that I organize it, the five lecturers, Michel Chipot, Patrick Ciarlet, Josselin Garnier, Otared Kavian, and François Murat, who all gave very clear and interesting lectures. And, of course, thanks to all the students who came from the University of Strasbourg and from several other universities in France to attend these lectures. Many thanks also to those who made this volume possible, first of all the three contributors, but also Vladimir Turaev, editor of this series, the referees and the staff at Walter de Gruyter. Strasbourg, September 10, 2003
E. Sonnendrücker
Table of Contents
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v Michel Chipot Elastic deformations of membranes and wires . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Josselin Garnier Scattering, spreading, and localization of an acoustic pulse by a random medium systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 Otared Kavian Lectures on parameter identification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
Elastic deformations of membranes and wires Michel Chipot Institut für Mathematik, Angewandte Mathematik Winterthurerstrasse 190, 8057 Zürich, Switzerland email:
[email protected]
Abstract. When we let roll freely balls on an elastic membrane they will stop after some time in a position that we will call equilibrium position. We would like to determine such equilibrium and see if it is unique. Since this three dimensional problem is not so easy to handle we look also to the two dimensional analogue of disks rolling on a wire. 2000 Mathematics Subject Classification: 35J85, 49J40, 49J45
1 Introduction Consider a rigid planar frame . On this frame suppose that is spanned an elastic membrane. When we drop a ball on the membrane, this ball will dig its nest in the elastic film and roll for some time before to reach some equilibrium position (see Figure 1). The same phenomenon will occur if we let roll several balls on the membrane. This is this equilibrium position that we would like to investigate from the point of view of existence, uniqueness, location. In the case of one single ball the problem was studied in [4]. The symmetry of the heavy ball considered played a crucial rôle there. When one passes to a system formed by two different balls symmetry is lost and results are getting more challenging. Very few are in fact available and this is the reason why we are lead to consider the deformation of an elastic wire by one or several disks and more generally by two dimensional rigid bodies – like for instance a square. Then direct computations can be performed. As we will see some unexpected situations might then occur that could help to understand the three dimensional problem. For instance, in the case of two identical disks sitting on a wire, the equilibrium position reached is not always the obvious one of the two disks hanging at the same level in the middle of the wire. Tension could force them to tilt as shown on Figure 2 below when their weight become too high – see [3]. A similar loss of symmetry can occur in the case of a square for instance – see [2].
2
Michel Chipot
Figure 1. The case of a membrane
Figure 2. The case of two heavy disks on a wire
We will follow the following steps. First we will consider the case of a two dimensional object supported by a wire. Addressing first the case of a disk we will give detailed computations which will help to understand the main ingredients necessary to address the problem of two disks. Then we will consider the equilibrium of a square able to move freely on a wire. In the next section we will attack the question of two different disks. Finally we will treat the case of one ball rolling on an elastic membrane, then the case of several. All along we will point out the many open problems that are related to the different cases.
2 The case of a single two dimensional solid on a wire 2.1 The case of one disk on an elastic wire Let us denote by the interval (0, 1). is the undeformed configuration of a one dimensional elastic wire spanned between the two points 0 and 1 of the real axis. We will denote by 0x the horizontal real axis and by 0h the vertical one that we will suppose oriented downward (see Figure 3). Letting roll a heavy disk on this wire spanned between 0 and 1 we would like to determine its equilibrium position. Of
Elastic deformations of membranes and wires
3
course – for symmetry reasons – we are expecting that the disk will stabilize in the middle of the wire at a level such that the tension of the elastic support balances the weight of the disk. (We suppose an ideal situation where the two dimensional disk can be maintained vertically on the wire). However, we need yet to introduce the right energy considerations allowing us to show that. ζ
0
1
x
h C h Figure 3. One single disk on a wire
Let G be the weight of the disk and r its radius. In order to allow the disk to fit entirely between the points 0, 1 we will assume 1 . (2.1) 2 Then, clearly, the position of the disk in space is completely determined by the position of its center r<
C = (ζ, h) ∈ × R.
(2.2)
C being fixed the lower points of the disk can be described by the function ψ ζ h (x) = ψ(x) = h + r 2 − (x − ζ )2 ∀ x ∈ B(ζ, r),
(2.3)
where B(ζ, r) = (ζ − r, ζ + r) denotes the open ball of center ζ and radius r – see the figure below. An admissible deformation of the wire is a “smooth” function u such 0 ζ −r
ζ
x ζ +r
h
r 2 − (x − ζ )2
(x, ψ(x)) Figure 4. The function ψ
4
Michel Chipot
that u(0) = u(1) = 0,
u(x) ≥ ψ(x)
for x ∈ B(ζ, r),
(2.4)
i.e. a function u describing a reasonable state for the wire – that has to be attached at the end points of the interval (0, 1) and has to be below the disk. (This is the rôle of the inequality (2.4) – recall that the vertical axis is directed downward). Now, in order to develop analytical arguments we have to choose u in some appropriate space. We will suppose u ∈ H01 ()
(2.5)
H01 () = { v ∈ L2 () | v ∈ L2 (), v(0) = v(1) = 0 }.
(2.6)
where
(We refer to [6], [13], [17] for an introduction to Sobolev spaces.) Note that in our analysis we consider as “admissible”, deformations for which the wire can be loose – i.e. not spanned or not being tight as in Figure 3 for example. In a case of a nonhomogeneous wire the equilibrium position of the wire could be such that the disk is not supported by straight lines as one can see using the computation that we are going to develop now. Let us suppose the wire to be homogeneous. Then, the center of the disk being located at C = (ζ, h), the energy of an admissible deformation – i.e. a function u such that u ∈ H01 (),
u(x) ≥ ψ ζ h (x)
a.e. x ∈ B(ζ, r)
is given by – if ux denotes the derivative of u in x – 1 u2 (x) dx − Gh. E(u; ζ, h) = σ 2 x
(2.7)
(2.8)
The integral part of this energy corresponds to the elastic energy of the system, the other part is a potential energy, σ is some constant of elasticity characteristic of the material. Nature is careful and tends to adopt behaviors that minimize energy. In other words among all possible admissible deformations the one corresponding to the equilibrium of the disk on the wire will be the one minimizing E(u; ζ, h). So we have to find an eventual triple (u0 , ζ0 , h0 ) such that (ζ0 , h0 ) ∈ (r, 1 − r) × R, u0 admissible (i.e. satisfying (2.7)), (2.9) E(u0 ; ζ0 , h0 ) ≤ E(u; ζ, h) ∀ (ζ, h) ∈ (r, 1 − r) × R, ∀ u admissible. (We choose ζ ∈ (r, 1 − r) in such a way that the disk cannot touch the end points 0 and 1). Remark 2.1. The problem (2.9) is an unusual problem of the calculus of variations. The class in which we are looking for a minimizer is neither convex nor compact.
Elastic deformations of membranes and wires
5
One could consider a nonhomogeneous elastic wire. In this case the energy to minimize would be 1 σ (x)u2x (x) dx − Gh (2.10) E(u; ζ, h) = 2 where σ is some coefficient of elasticity. Similarly we could assume the disk to be nonhomogeneous and have a potential energy of the type G(x, y)y dx dy (2.11) − D
where the integration is taking place on the disk D of center (ζ, h) and radius r. The notation for G(x, y) – the density of weight – is clear. The analysis that we develop below is widely open in the cases (2.10), (2.11). Note that in the case of (2.11) the center of mass or the position of the disk around its center are additional parameters of the problem. We leave that to the curiosity of the reader. Let us now consider the problem (2.9). Dividing the energy (2.8) by σ it is clear that at the expense of rescaling G we can always assume that σ =1
(2.12)
This is what we will do from now on. Somehow – even if this is wrong in the nonhomogeneous case – our intuition tells us that the wire should be tight i.e. adopt a position made by straight lines outside of the contact zone of the disk. This is what we are about to see now. For that notice that if (u0 , ζ0 , h0 ) is a solution to (2.9) then it holds that:
⇐⇒
E(u0 ; ζ0 , h0 ) ≤ E(u; ζ0 , h0 ) 1 1 2 u (x) dx ≤ u2 (x) dx 2 0 0x 2 x
∀ u admissible
(2.13)
∀ u admissible.
(2.14)
To describe the set of admissible deformations, for C = (ζ, h) fixed let us introduce K(ζ, h) = { u ∈ H01 () | u(x) ≥ ψ ζ h (x) a.e. x ∈ B(ζ, r) }.
(2.15)
Then, following (2.14), u0 is solution of a minimization problem that we make more precise in the proposition below: Proposition 2.1. Let (ζ, h) ∈ (r, 1 − r) × R be fixed. Then, there exists a unique u = u(ζ, h) solution to u ∈ K(ζ, h), (2.16) 1 1 2 ux (x) dx ≤ vx2 (x) dx ∀ v ∈ K(ζ, h). 2 2
6
Michel Chipot
Moreover, u is the solution of the so called Variational Inequality u ∈ K(ζ, h), ux (vx − ux ) dx ≥ 0 ∀ v ∈ K(ζ, h).
(2.17)
Proof. Let us denote by | · |1,2 the quantity defined as 1/2 u2x (x) dx . |u|1,2 =
(2.18)
It is well known – see [9], [18] – that this defines a Hilbert norm on H01 () which is a Hilbert space when equipped with it. We now leave to the reader to show that K(ζ, h) = K is a closed convex subset of H01 ().
(2.19)
Then, (2.16) can be also written as u ∈ K(ζ, h),
|u − 0|1,2 ≤ |v − 0|1,2
∀ v ∈ K(ζ, h),
(2.20)
i.e. u is the projection of 0 on the closed convex set K. It is well known see for in-
K u 0
v
Figure 5. The projection of 0 on a closed convex set
stance [8] – that a solution u to (2.20) exists and is unique. Moreover, u is equivalently solution to – see Figure 5 u ∈ K,
(0 − u, v − u)1,2 ≤ 0
∀ v ∈ K,
(2.21)
associated to (2.18) and defined where ( ·, ·)1,2 denotes the scalar product on by (u, v)1,2 = ux vx dx. (2.22) H01 ()
It is then clear that (2.21) is nothing but (2.17). This completes the proof of the proposition. We can now describe exactly what is the solution to (2.16) or (2.17). We have:
7
Elastic deformations of membranes and wires
Proposition 2.2. For (ζ, h) ∈ (r, 1 − r) × R the solution u of (2.16) or (2.17) is given by the following function: h ≤ −r
h > −r
1
0
1
y1
0
x1 ζ h C X1
Figure 6. The solution u
i.e. if h ≤ −r if h > −r
u ≡ 0, u(x) = xψx (x1 ) = ψ(x) = (x − 1)ψx (y1 )
(2.23) for x ∈ [0, x1 ], for x ∈ [x1 , y1 ], for x ∈ [y1 , 1],
(2.24)
where ψx denotes the derivative of the function ψ = ψ ζ h and x1 , y1 the coordinates of the points of contact of the tangents to the disk issued from 0 and 1 – see Figure 6. Proof. Suppose that h ≤ −r. Then, by (2.3) ψ ζ h (x) ≤ 0 and u = 0 belongs to K(ζ, h) and satisfies (2.17). This completes the proof in this case. Note that this is the case when the disk does not touch the wire. Suppose now that h > −r. Then, the function u defined by (2.24) belongs to K. Moreover, it holds that: x1 y1 ux (vx − ux ) dx = ux (vx − ux ) dx + ux (vx − ux ) dx
0
+
x1 1
(2.25) ux (vx − ux ) dx.
y1
Writing down in each intervals ux (vx − ux ) = (ux (v − u))x − uxx (v − u), it comes after integration
x
y ux (vx − ux ) dx =ux (v − u) 01 +ux (v − u) x1 1 y1
1 − ψxx (v − ψ) dx + ux (v − u) y . x1
1
8
Michel Chipot
(We used the fact that uxx = 0 on (0, x1 ), (y1 , 1)). Since u, v vanish on the boundary of and since u is clearly a C 1 -function it comes y1 ux (vx − ux ) dx = − ψxx (v − ψ) dx ∀ v ∈ K(ζ, h). (2.26)
x1
Since ψxx ≤ 0, v ≥ ψ on (x1 , y1 ) it comes u ∈ K(ζ, h), ux (vx − ux ) dx ≥ 0
∀ v ∈ K(ζ, h),
(2.27)
and u defined by (2.24) is solution to (2.16), (2.17). This completes the proof of the proposition. Remark 2.2. The variational inequality (2.17) is called an obstacle problem (see [5], [15], [19]). The set [x1 , y1 ] = { x ∈ | u(x) = ψ ζ h (x) }
(2.28)
is called the “coincidence set” of this obstacle problem. One should also notice that it holds that uxx ≤ 0 in , uxx = 0 in \ [x1 , y1 ], u=ψ in [x1 , y1 ].
(2.29)
(h is directed downward.) Let us now come back to our problem (2.9). Due to (2.14) and Proposition 2.1 we see that if a solution exists then u0 = u0 (ζ0 , h0 ) and it holds that E(u(ζ, h); ζ, h) ≤ E(u; ζ, h) Thus, for u = u(ζ, h) let us set 1 e(ζ, h) = 2
∀ u ∈ K(ζ, h).
(2.30)
u2x dx − Gh.
(2.31)
Due to (2.30), the problem (2.9) reduces to minimize e(ζ, h) on (r, 1 − r) × R, i.e. to a minimization problem in two dimensions. However, the set where we are minimizing is not compact and we do not know yet if the function e = e(ζ, h) is continuous. This is what we would like to address now. For that let us find the expressions of x1 , y1 in term of ζ, h. We have
Elastic deformations of membranes and wires
9
Lemma 2.1. Let us assume h > −r and the notation of Figure 6. Then it holds that r 2 ζ + rh h2 + ζ 2 − r 2 , (2.32) x1 = x1 (ζ, h) = ζ − h2 + ζ 2 r 2 (1 − ζ ) + rh h2 + (1 − ζ )2 − r 2 y1 = y1 (ζ, h) = ζ + . (2.33) h2 + (1 − ζ )2 In particular, x1 , y1 are differentiable. Proof. Since 0X1 and X1 C are orthogonal it holds that
⇐⇒ ⇐⇒
x1 (x1 − ζ ) + ψ(x1 )(ψ(x1 ) − h) = 0, x1 (x1 − ζ ) + {h + r 2 − (x1 − ζ )2 } r 2 − (x1 − ζ )2 = 0, ζ (x1 − ζ ) + r 2 + h r 2 − (x1 − ζ )2 = 0.
Setting X = ζ − x1 we get
− ζ X + r 2 + h r 2 − X2 = 0
⇐⇒
h2 (r 2 − X 2 ) = (r 2 − ζ X)2 ,
(ζ X − r 2 ) · h ≥ 0.
(2.34)
Expanding the square in the first equation it comes: X2 (ζ 2 + h2 ) − 2r 2 ζ X + r 4 − r 2 h2 = 0 ⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒
2r 2 ζ X r 4 − r 2 h2 + =0 ζ 2 + h2 ζ 2 + h2 2 (r 4 − r 2 h2 )(ζ 2 + h2 ) − r 4 ζ 2 r 2ζ + =0 X− 2 ζ + h2 (ζ 2 + h2 )2 2 r 2 h2 {ζ 2 + h2 − r 2 } r 2ζ = X− 2 ζ + h2 (ζ 2 + h2 )2 r 2 ζ ± rh ζ 2 + h2 − r 2 X= . ζ 2 + h2
(2.35)
X2 −
(2.36)
(Recall that ζ ∈ (r, 1 − r) so that the term under the square root symbol is positive.) Going back to (2.34) we see that ζ X = r 2 ± h r 2 − X 2 , (ζ X − r 2 ) · h ≥ 0 so that X is the largest root of the equation (2.35) for h > 0 and the smallest for h < 0. Thus X = ζ − x1 is given by (2.36) with the sign +. This leads to (2.32). The proof of (2.33) is similar changing ζ into 1 − ζ and this is left to the reader.
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Michel Chipot
In order to minimize the function e(ζ, h) we first study its differentiability. We have then: Theorem 2.1. The function e = e(ζ, h) defined by Equation (2.31) is differentiable on (r, 1 − r) × R and it holds that: • For h ≤ −r ∂e (ζ, h) = 0, ∂ζ
∂e (ζ, h) = −G, ∂h
(2.37)
1 ∂e (ζ, h) = {ψx2 (y1 ) − ψx2 (x1 )}, ∂ζ 2 ∂e (ζ, h) = ψx (x1 ) − ψx (y1 ) − G. ∂h
(2.38)
• For h > −r
Here x1 , y1 are the points given by (2.32), (2.33) and ψx denotes the derivative in x of the function ψ. Proof. a) A formal proof. The rigorous proof of (2.38) being tedious let us give first a formal proof of it. We restrict ourselves to the first formula of (2.38). Due to (2.31) – if every differentiation under the integral sign is justified – we have 1 ∂ ∂e 2 (ζ, h) = u dx = ux uxζ dx (2.39) ∂ζ 2 ∂ζ x with an obvious notation for uxζ . Now, clearly, we have u = u(ζ, h) = 0
at 0, 1
∀ ζ, h.
(2.40)
(Recall that u = u(ζ, h) is the solution of the variational inequality (2.17)). Thus it holds that uζ = 0
∀ ζ, h.
at 0, 1
(2.41)
Integrating the last integral of (2.39) by parts we get 1 ∂e (ζ, h) = − uxx uζ dx. ∂ζ 0 Due to (2.29) we obtain then ∂e (ζ, h) = − ∂ζ
y1
ψxx ψζ dx. x1
According to (2.3) we have ψζ = −ψx
(2.42)
Elastic deformations of membranes and wires
so that (2.42) becomes ∂e (ζ, h) = ∂ζ
y1
ψxx ψx dx =
x1
11
1 {ψx (y1 )2 − ψx (x1 )2 } 2
which is the first formula (2.38). b) The rigorous proof. First, due to (2.23), (2.31) for h ≤ −r we have e(ζ, h) = −Gh and thus (2.37) holds in this case. Let us assume that h > −r. Due to (2.24), (2.31) we have 1 y1 2 1 1 2 1 x1 2 ux dx + u dx + u dx − Gh e(ζ, h) = 2 0 2 x1 x 2 y1 x 1 y1 2 1 1 ψx dx + (1 − y1 )ψx2 (y1 ) − Gh. = x1 ψx2 (x1 ) + 2 2 x1 2 Let us set A= Applying the chain rule we get ∂A 1 ∂x1 (ζ, h) = − ψx2 (x1 ) + ∂ζ 2 ∂ζ
1 2
y1
x1
y1 x1
ψx2 dx.
1 ∂y1 . ψx ψxζ dx + ψx (y1 )2 2 ∂ζ
(2.43)
(2.44)
(2.45)
(2.46)
(Recall that by (2.32), (2.33) x1 and y1 are differentiable and so is ψ). Since by (2.3) ψxζ = −ψxx , (2.46) can be written 1 1 ∂x1 1 1 ∂y1 ∂A = − ψx2 (x1 ) + ψx (x1 )2 − ψx (y1 )2 + ψx (y1 )2 . ∂ζ 2 ∂ζ 2 2 2 ∂ζ
(2.47)
Going back to (2.44) and differentiating e we obtain ∂x1 ∂ψx (x1 ) ∂A 1 ∂e (ζ, h) = ψx2 (x1 ) + x1 ψx (x1 ) + ∂ζ 2 ∂ζ ∂ζ ∂ζ ∂ψ ∂y 1 1 x (y1 ) + (1 − y1 )ψx (y1 ) − ψx (y1 )2 2 ∂ζ ∂ζ ∂ψx (x1 ) 1 + ψx (x1 )2 = x1 ψx (x1 ) ∂ζ 2 1 ∂ψx (y1 ) , − ψx (y1 )2 + (1 − y1 )ψx (y1 ) 2 ∂ζ
(2.48)
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Michel Chipot
by (2.47). We remark then that x1 ψx (x1 ) = ψ(x1 )
and
(y1 − 1)ψx (y1 ) = ψ(y1 ),
(2.49)
hold for any (ζ, h). Thus differentiating in ζ these equalities we obtain ∂x1 ∂x1 ∂ψx (x1 ) + x1 = ψζ (x1 ) + ψx (x1 ) , ∂ζ ∂ζ ∂ζ ∂y1 ∂ψx (y1 ) ∂y1 + (y1 − 1) = ψζ (y1 ) + ψx (y1 ) . ψx (y1 ) ∂ζ ∂ζ ∂ζ ψx (x1 )
Thus it holds ∂ψx (x1 ) = ψζ (x1 ) = −ψx (x1 ), x1 ∂ζ
(y1 − 1)
(2.50) (2.51)
∂ψx (y1 ) = ψζ (y1 ) = −ψx (y1 ). ∂ζ
Reporting this to (2.48) we obtain the first formula of (2.38). To get the second formula we proceed similarly. We have first by the chain rule and (2.45) y1 ∂A 1 ∂x1 ∂y1 1 ∂A = ψx ψxh dx + ψx (y1 )2 (ζ, h) = − ψx (x1 )2 + . (2.52) ∂h ∂h 2 ∂h 2 ∂h x1 Since ψxh = ψhx = 0 it comes 1 1 ∂x1 ∂y1 ∂A = − ψx (x1 )2 + ψx (y1 )2 . ∂h 2 ∂h 2 ∂h Differentiating e given by (2.44) in the h direction we obtain ∂x1 ∂ψx (x1 ) ∂A 1 ∂e (ζ, h) = ψx2 (x1 ) + x1 ψx (x1 ) + ∂h 2 ∂h ∂h ∂h ∂y1 ∂ψx (y1 ) 1 2 + (1 − y1 )ψx (y1 ) −G − ψx (y1 ) 2 ∂h ∂h ∂ψx (x1 ) ∂ψx (y1 ) + (1 − y1 )ψx (y1 ) −G = x1 ψx (x1 ) ∂h ∂h by (2.53). Differentiating then (2.49) with respect to h we obtain
(2.53)
(2.54)
∂x1 ∂x1 ∂ψx (x1 ) + x1 = ψh (x1 ) + ψx (x1 ) , ∂h ∂h ∂h ∂y1 ∂ψx (y1 ) ∂y1 + (y1 − 1) = ψh (y1 ) + ψx (y1 ) ψx (y1 ) ∂h ∂h ∂h ψx (x1 )
and thus ∂ψx ∂ψx (x1 ) (2.55) = ψh (x1 ) = 1, (y1 − 1) (y1 ) = ψh (y1 ) = 1. ∂h ∂h Using that in (2.54) leads to the second formula of (2.38). Now, clearly when h > −r, h → −r, ψx (x1 ), ψx (y1 ) → 0 and thus these partial derivatives of e are continuous in (r, 1 − r) × R. The differentiability of e follows and the proof of the theorem is complete. x1
13
Elastic deformations of membranes and wires
We can now solve the problem (2.9). We have Theorem 2.2. The problem (2.9) admits a unique solution given by
1 G G2 ζ0 = , h0 = −r 1+ , u0 = u(ζ0 , h0 ) 2 4 4
(2.56)
where u(ζ0 , h0 ) is the solution of the variational inequality (2.17) given by (2.24). Proof. First let us remark that, as we could expect, the disk reaches its equilibrium when centered in the middle of the wire. As observed below (2.31) the problem (2.9) reduces to minimize the function e(ζ, h) on (r, 1 − r) × R. Let us consider first h > −r. Then, due to (2.38) we have ∂e (ζ, h) < 0 ∂ζ
∀ζ <
1 , 2
∂e (ζ, h) > 0 ∂ζ
∀ζ >
1 . 2
(2.57)
Thus in this case the minimum of e(ζ, h) is clearly achieved for ζ = 21 . For h≤ −r we have e(ζ, h) = −Gh
(2.58)
i.e. e is independent of ζ and of course the minimum of e is achieved also for ζ = ζ0 = 21 . Thus, we have now reduced the problem to minimize e( 21 , h) over R. For h > −r one has 1 ∂e 1 d e ,h = , h = ψx (x1 ) − ψx (y1 ) − G = 2ψx (x1 ) − G (2.59) dh 2 ∂h 2 since for symmetry reasons ψx (y1 ) = −ψx (x1 ). Thus, we have 1 d e ,h < 0 for ψx (x1 ) < dh 2 1 d e ,h > 0 for ψx (x1 ) > dh 2
G , 2 G 2
(2.60)
and a unique minimum is achieved on (−r, +∞) when it holds that ψx (x1 ) = −ψx (y1 ) =
G . 2
(2.61)
Since e is continuous for h = −r, this minimum is also the minimum of e( 21 , h) on the whole real line. This completes the proof of existence and uniqueness of a unique minimizer (u0 , ζ0 , h0 ). To determine h0 , if we set G = tang θ 2
(2.62)
14
Michel Chipot
then we have (see Figure 7) G G π − θ = x1 − r cos θ, h0 = x1 − r sin 2 2 2 1 π − θ = − r sin θ. x1 = ζ − r cos 2 2
x1
0
1 2
(2.63) (2.64)
1
θ
θ
Figure 7. The equilibrium position of a single disk
From this it follows that h0 =
G G − r sin θ − r cos θ 4 2
(2.65)
with (by (2.62)) sin θ = Since
G cos θ. 2
G2 cos2 θ, 1 = sin2 θ + cos2 θ = 1 + 4
we obtain 1 cos θ = 1+
G2 4
,
sin θ =
G 1 · 2 1+
G2 4
and by (2.65)
G G2 h0 = −r 1+ . 4 4 This completes the proof of the theorem.
(2.66)
15
Elastic deformations of membranes and wires
2.2 The case of one square on a wire In this section we would like to address the case of a square moving freely on an elastic wire. We will not provide all the details of the analysis in this case referring the interested reader to [2], but we would like to show that a surprising behaviour can happen in this case. The setting is the same as in the previous section. We suppose that in its undeformed configuration the elastic wire occupies the domain = (0, 1). We denote by 2r the length of the sides of the square and by G its weight. If Q denotes the closed set of the points occupied by the square we will always assume that Q ⊂ × R.
(2.67)
We will denote by P = (ζp , hp ) the barycenter of the square – see Figure 8. For ζp
0
hp
1
x
P A
h
B Figure 8. A square on a wire
u ∈ H01 () we set Cu = { (x, h) ∈ × R | h ≤ u(x) }.
(2.68)
(Recall that a function in H01 () is a continuous function in this case – see [6]). Then, for a position Q of the square, an admissible deformation of the wire is a function u in H01 () such that Q ⊂ Cu .
(2.69)
The energy of an admissible deformation is given by the same formula as in the preceding section. Then we would like to find 1 Inf u2x (x) dx − GhpQ (2.70) (Q,u) 2 when Q is located in the strip × R and when u ∈ H01 () is such that (2.69) holds. Q hp denotes the second coordinate of the barycenter P of the square occupying the
16
Michel Chipot
position Q. More precisely we would like to show that the problem (2.70) admits a solution which is unique up to symmetry. As for the case of a disk we need the square to be able to fit inside the interval (0, 1) and in order to insure that we suppose r<
1 . 4
(2.71)
In (2.70) we are minimizing over a set of isometric squares and admissible deformations so that the situation seems slightly unusual. As we have done in the case of a disk we are going to reformulate the problem as a minimization problem on a subset of R4 . Suppose that the position of the square Q is fixed in the strip × R. Denote by IQ = x (Q) the interval of projection of Q on the x-axis ( x is the orthogonal projection on the x-axis). For x ∈ IQ set ψQ (x) = Sup{ y | (x, y) ∈ Q}.
(2.72)
Then, clearly, the function ψQ is a function describing the lower border of Q. Then, with this notation, the constraint (2.69) can be written u ∈ CQ = { u ∈ H01 () | u(x) ≥ ψQ (x) on IQ }. It is clear if (Q0 , u0 ) is a minimizer (2.70) that it holds that 1 1 2 u dx ≤ u2 dx ∀ u ∈ CQ0 , u0 ∈ CQ0 , 2 0x 2 x i.e. u0 is solution of the variational inequality u0 ∈ CQ0 , u0x (vx − u0x ) dx ≥ 0
∀ v ∈ CQ0 .
(2.73)
(2.74)
(2.75)
(It is easy to see that CQ0 is a closed convex set of H01 (). As in the case of a disk, u0 is the projection of 0 on CQ0 – see Proposition 2.1.) Thus, for Q satisfying (2.67) we denote by u = uQ the unique solution to u ∈ CQ , (2.76) ux (vx − ux ) dx ≥ 0 ∀ v ∈ CQ .
It is then clear that (2.70) is equivalent to 1 u2Qx dx − GhpQ , Inf Q⊂×R 2
(2.77)
i.e., we minimize on Q only. Now, a square is perfectly determined by the position of two of its vertices – for instance by A, B as in Figure 8. A being the further left
Elastic deformations of membranes and wires
17
vertex and B the further down. If we set A = (xA , yA ),
B = (xB , yB )
(2.78)
yA ≤ yB
(2.79)
we can always assume that 0 < xA < xB < 1,
(recall that the vertical axis is directed downward). Now, the two points A, B are not able to move freely. They always have to fulfill the constraint (xA − xB )2 + (yA − yB )2 = 4r 2 .
(2.80)
If we choose A, B satisfying (2.79), (2.80) the square Q is perfectly determined and we have 1 u2 dx − GhpQ = e(xA , yA , xB , yB ) (2.81) 2 Qx i.e. the energy that we have to minimize is a function of the points belonging to the subset of R4 defined by (2.79), (2.80). Then, using a compactness argument we can show Theorem 2.3. The problems (2.70), (2.77) admit a minimizer. Proof. As seen above, it is enough to minimize (2.81). We refer the reader to [2] for the details of this minimization. Now when it comes to find minimizers for (2.70), (2.77) the first guess is to think that the square will find its equilibrium by sitting horizontally in the middle of the wire. It is true for small values of the weight of the square. Surprisingly this holds no more when G becomes large. In other words, the horizontal position of the square becomes unstable when the tension of the wire increases. We have indeed Theorem 2.4. Assume that
√ G ≤ 2( 3 − 1)
(2.82)
then the problems (2.70), (2.77) admit a unique solution given by a square centered in the middle of the wire (see Figure 9) with 1 1 G 1 −r . (2.83) xA = − r, xB = + r, yA = yB = 2 2 2 2 Assume that
√ 2( 3 − 1) < G < 2
(2.84)
18
Michel Chipot
√ G ≤ 2( 3 − 1)
G 2
A
B
Figure 9. The square sitting in the middle
then the problems (2.70), (2.77) admit two solutions. One is given by A, B satisfying xA = yA = xB = yB =
√ r 1 2 G2 + 4G − 8 −√ , 2+ 2 G G2 + 4G − 4 r G −√ G + G2 + 4G − 8 , 2 4 G + 4G − 4 √ r 1 2 G2 + 4G − 8 +√ , 2− 2 G G2 + 4G − 4 r G −√ 2 − G2 + 4G − 8 . 2 4 G + 4G − 4
(2.85) (2.86) (2.87) (2.88)
The other solution is obtained by reflection with respect to the vertical axis x = (see Figure 10).
G 2
A B
G 2
G 2
G 2
√ 2( 3 − 1) < G < 2
1 2
C B
Figure 10. The two disks are slightly tilted
Assume that √ 2≤G≤2 7
(2.89)
19
Elastic deformations of membranes and wires
then the problems (2.70), (2.77) admit a unique minimizer given by a square centered in the middle of the wire with A, B given by 1 √ G 1 √ 1 G √ G xA = − 2r, yA = − 2r , xB = , yB = + 2r 1− (2.90) 2 2 2 2 4 2 (see Figure 11).
√ 2≤G 7
G 2
A B Figure 11. The square is centered again
Assume now that
√ 2 7
(2.91)
then the problems (2.70) or (2.77) admit two symmetric solutions – see Figure 12.
G 2
G 2
C
A C B
A B
Figure 12. The square tilts again for large G
Proof. We refer to [2]. Let us just mention that the proof relies on Lagrange multipliers technique. In the case of (2.91) it is also possible to determine A and B exactly. However, the expressions are complicated. They are given in [1], [2].
20
Michel Chipot
2.3 Some open questions As we mentioned in Remark 2.1 one can consider the case of an inhomogeneous wire and the case of an inhomogeneous two dimensional body sitting above it: for instance a disk, a square... In the homogeneous case – for an object or a wire – one can consider the case of a rectangle which should be a variant of the case of a square. More challenging is the case of an ellipse. Indeed in this case, depending of the weight of it the equilibrium positions should be like in Figure 13 for light weights and like in Figure 14 for large weights, i.e., when the ellipse is light then it should be lying with its long axis horizontal and on the contrary with its long axis vertical when the weight is large. (Note that the scales of the figures are not the same.) Of course the example of the square tells us that in between or for very large weights a tilted position could also occur. We leave these issues to the reader’s curiosity. 0
1
0
1
Figure 13. The case of an ellipse with light weight Figure 14. The case of an ellipse with large weight
Another variant of the above problem is to consider an energy related to the arc length of the deformation (which would correspond to the mean curvature operator in higher dimensions). In this case one would minimize 1 1 + a(x)u2x (x) dx − Gh (2.92) E= 2 and all our analysis has to be revisited. Of course the notation of equilibrium refers also to an evolution process, i.e. the equilibrium reached for instance by the disk corresponds to the final state of its motion when we drop it on the wire and let him roll freely. This has to be established mathematically and in particular the evolution problem has to be defined and considered together with its asymptotic behaviour.
21
Elastic deformations of membranes and wires
3 The case of two disks on an elastic wire 3.1 Formulation of the problem We consider two solid disks of different sizes and weights allowed to roll freely on an elastic wire – see Figure 15. 0 x1 y1
1
x2 y2 ζ1
x
ζ2 u
h1
C1
C2
X1
h2
Y2
Y1 X2
h Figure 15. Two disks rolling on a wire
We denote by r1 , r2 the different radii of these disks and by G1 , G2 their respective weights. In its undeformed state the elastic wire is supposed to occupy the interval = (0, 1). Letting roll the two disks on it, they will reach an equilibrium position that we would like to determine. First, as we did before for one single disk we notice that the position of the disks is completely determined by the location of their centers C1 = (ζ1 , h1 ),
C2 = (ζ2 , h2 ).
(3.1)
If u is an admissible deformation of the wire – we will come back to this point later – the energy of the configuration is given by 1 u2 dx − G1 h1 − G2 h2 . (3.2) E= 2 x Our goal in studying this one dimensional problem is to get some insight on what happens in higher dimension that is to say for instance for two balls rolling on a membrane. First, we will assume that both disks are located in the strip × R, that is to say, that ζi ∈ (ri , 1 − ri ),
i = 1, 2.
(3.3)
Next, we will suppose that (ζ1 − ζ2 )2 + (h1 − h2 )2 ≥ (r1 + r2 )2 i.e. we will not allow the disks to interpenetrate each other.
(3.4)
22
Michel Chipot
The lower part of each disk can be described by the function ψi = ψ ζi hi (x) = hi + ri2 − (x − ζi )2 , i = 1, 2,
(3.5)
defined on B(ζi , ri ) = (ζi − ri , ζi + ri ). Then, an admissible deformation is a function u such that the disks are located below u – note that we suppose the vertical direction h oriented downward (see Figure 15). More precisely an admissible deformation is a function u such that u ∈ K = K(C1 , C2 ) = { v ∈ H01 () | v(x) ≥ ψi (x) on B(ζi , ri ) i = 1, 2 }. (3.6) Then the problem we have to consider is to find 1 Inf u2x dx − G1 h1 − G2 h2 (u,C1 ,C2 ) 2
(3.7)
where u belongs to K and C1 , C2 the center of the disks are satisfying the constraints (3.3), (3.4). We would like to show first that one can reformulate the problem (3.7) as a minimization problem in R4 . Assume that we have found an element (u0 , C10 , C20 ) = (u0 , ζ10 , h01 , ζ20 , h02 ) minimizing (3.7). Then, clearly, u0 minimizes 1 v 2 dx − G1 h01 − G2 h02 2 x on the set K 0 = K 0 (ζ10 , h01 , ζ20 , h02 ) = { v ∈ H01 () | v(x) ≥ ψi (x) on B(ζi0 , ri ), i = 1, 2 } i.e. since −G1 h01 − G2 h02 is independent of v, u0 is the solution of the variational inequality 0 0 u ∈ K , u0x (vx − u0x ) dx ≥ 0 ∀ v ∈ K 0 .
Thus, for ξ = (ζ1 , h1 , ζ2 , h2 ) ∈ R4 we introduce u = u(ξ ) the solution of the variational inequality 1 u ∈ K(ξ ) = { v ∈ H0 () | v(x) ≥ ψi (x) on B(ζi , ri ), i = 1, 2 }, (3.8) ux (vx − ux ) dx ≥ 0 ∀ v ∈ K(ξ ),
and we set for u = u(ξ ) 1 e(ξ ) = e(ζ1 , h1 , ζ2 , h2 ) = 2
u2x dx − G1 h1 − G2 h2 .
(3.9)
Elastic deformations of membranes and wires
The problem (3.7) is then clearly equivalent to: find ξ 0 = (ζ10 , h01 , ζ20 , h02 ) satisfying (3.3), (3.4) such that e(ξ 0 ) ≤ e(ξ ) ∀ ξ = (ζ1 , h1 , ζ2 , h2 ) satisfying (3.3), (3.4).
23
(3.10)
Thus, our problem is reformulated as a problem of minimization on R4 . Note that the set of points where we are minimizing – see (3.3) – is not compact and even if e is continuous, as before, some work has to be done to obtain the existence of a minimizer. First – and without loss of generality – we will assume all along r1 ≥ r2 .
(3.11)
Moreover, we will suppose that when one of the disks is sitting on the wire we can always fit the other one on one side or the other. In other words we will suppose 2r1 + r2 <
1 . 2
(3.12)
3.2 Preparatory results Let us denote by Di = Di (Ci , ri ), i = 1, 2 the closed disk of center Ci = (ζi , hi ) and radius ri . Due to our assumption (3.3) the two disks are located in the strip (0, 1) × R. Denote then by U the convex hull in R2 of D1 ∪ D2 ∪ [0, 1] × (−∞, 0]. Then for x ∈ [0, 1] set u(x) = Sup{ y | (x, y) ∈ U }.
(3.13)
One has: Proposition 3.1. The function u defined by (3.13) coincides with the solution u = u(ξ ) = u(ζ1 , h1 , ζ2 , h2 ) of the variational inequality (3.8). Proof. In the case D1 , D2 ⊂ (0, 1) × (−∞, 0] one has u=0 and u = 0 satisfies clearly (3.8) in this case. Let us assume now that we are in the case of Figure 15 – which is somehow the generic case. u = u(ξ ) is the function shown on the picture. Then it holds – with the notation of Figure 15 – for any v ∈ K(ξ ): 1 1 ux (vx − ux ) dx = (ux (v − u))x − uxx (v − u) dx, (3.14) 0
0
where the second integral is understood as the sum of the integrals on (0, x1 ), (x1 , y1 ), (y1 , x2 ), (x2 , y2 ), (y2 , 1) intervals on which u is C 2 . The first integral of the second
24
Michel Chipot
side of (3.14) vanishes. Indeed this follows after summing the equalities: x1
x 1
(ux (v − u))x dx = ux (v − u) = ux (v − u)(x1 ), 0 0 y1
y 1
(ux (v − u))x dx = ux (v − u) = ux (v − u)(y1 ) − ux (v − u)(x1 ), . . . x1
x1
...... (We use the fact that u is C 1 , v − u = 0 on {0, 1}.) Thus, from (3.14) it follows that 1 y1 y2 ux (v − ux ) dx = − ψ1,xx (v − ψ1 ) dx − ψ2,xx (v − ψ2 ) dx (3.15) x1 x2 0 ≥0 ∀ v ∈ K(ξ ). (It is due to the fact that uxx = 0 outside (xi , yi ) and the fact that ψixx ≤ 0, v ≥ ψi on (xi , yi ), i = 1, 2). This completes the proof in this case. In any other case one proceeds with the same arguments splitting the integral into the different intervals where u is of class C 2 . Note that this works also when the disks are interpenetrating each other. The details are left to the reader. Remark 3.1. What we call above the other cases is for instance when only one disk is touching the membrane, or when the small one is located below the large one – see below Figure 17. Remark 3.2. We can define the coincidence sets Ii = { x ∈ (0, 1) | u(x) = ψi (x) },
i = 1, 2.
(3.16)
Then, the solution u satisfies clearly −uxx ≥ 0 uxx = 0 u = ψi
in (0, 1), in (0, 1) \ I1 ∪ I2 , on Ii , i = 1, 2.
(3.17) (3.18) (3.19)
We can now show: Proposition 3.2. The function e = e(ζ1 , h1 , ζ2 , h2 ) defined by (3.9) is continuous on the set of points ξ = (ζ1 , h1 , ζ2 , h2 ) ∈ (r1 , 1 − r1 ) × R × (r2 , 1 − r2 ) × R. Proof. As already mentioned in the previous proposition we can relax the assumption (3.4) in the definition of e since u = u(ξ ) is perfectly defined when the disks are penetrating each other. Due to the definition (3.9) the continuity of e will follow if we can show that the mapping u(ξ )2x (x) dx ξ →
Elastic deformations of membranes and wires
25
is continuous. (u(ξ )x (x) denotes the derivative of u = u(ξ ) at the point x). But when ξ → ξ we have u(ξ )x → u(ξ )x uniformly on each interval where u(ξ ) is C 2 . It follows that 2 u(ξ )x dx → u(ξ )2x dx
and the continuity of e is proved. We now determine precisely the coincidence sets I1 , I2 . First we have Proposition 3.3. In the case of Figure 15 it holds that: r12 ζ1 + r1 h1 h21 + ζ12 − r12 x1 = ζ1 − , h21 + ζ12 r1 y1 = ζ1 − 2 {(h2 − h1 ) d 2 − (r1 − r2 )2 − (r1 − r2 )(ζ2 − ζ1 )}, d r2 x2 = ζ2 − 2 {(h2 − h1 ) d 2 − (r1 − r2 )2 − (r1 − r2 )(ζ2 − ζ1 )}, d r22 (1 − ζ2 ) + r2 h2 h22 + (1 − ζ2 )2 − r22 y2 = ζ2 + h22 + (1 − ζ2 )2 where d = (h2 − h1 )2 + (ζ2 − ζ1 )2 .
(3.20) (3.21) (3.22)
(3.23)
Proof. The first and the last formulae above are obtained exactly as in (2.32), (2.33). To obtain the second formula of (3.21) let us enlarge Figure 15 – see Figure 16 C1 τ r1
t
t
Y1 θ C2 r2 X2 Figure 16. The blow-up of part of Figure 15
26
Michel Chipot
We have then clearly y1 = ζ1 − r1 sin t,
x2 = ζ2 − r2 sin t.
(3.24)
Moreover ⇒
t =τ −θ sin t = sin τ cos θ − sin θ cos τ.
Clearly
d 2 − (r1 − r2 )2 cos θ = , d ζ2 − ζ1 cos τ = . d
r1 − r2 , sin θ = d h2 − h 1 , sin τ = d
(3.25) (3.26)
The formulae (3.21), (3.22) follow. The coincidence set I1 (see (3.16)) might be not connected when r2 < r1 and when the small disk is located below the large one – see Figure 17. Then one can show: 0
x1
y1
ζ1 x1
ζ2
1 y1
x
h C1 X1
X1
Y1 Y1 C2
X2 = (x2 , ψ2 (x2 ))
(y2 , ψ2 (y2 )) = Y2
Figure 17. The case of the little disk below the large one
Proposition 3.4. Assume that we are in the case of Figure 17. It holds: r12 ζ1 + r1 h1 h21 + ζ12 − r12 , x1 = ζ1 − h21 + ζ12
(3.27)
Elastic deformations of membranes and wires
y1 = ζ1 +
r12 (1 − ζ1 ) + r1 h1 h21 + (1 − ζ1 )2 − r12 h21 + (1 − ζ1 )2
,
r1 {(h − h )(r − r ) + (ζ − ζ ) d 2 − (r1 − r2 )2 }, 2 1 1 2 2 1 d2 r1 y1 = ζ1 − 2 {(ζ2 − ζ1 ) d 2 − (r1 − r2 )2 − (h2 − h1 )(r1 − r2 )}, d r2 x2 = ζ2 − 2 {(h2 − h1 )(r1 − r2 ) + (ζ2 − ζ1 ) d 2 − (r1 − r2 )2 }, d r2 y2 = ζ2 − 2 {(ζ2 − ζ1 ) d 2 − (r1 − r2 )2 − (h2 − h1 )(r1 − r2 )}. d x1 = ζ1 −
27
(3.28) (3.29) (3.30) (3.31) (3.32)
Proof. It is clear that the formulae (3.27), (3.28) are just (3.20) and (3.23). To prove the other equalities let us enlarge the bottom part of Figure 17 – see Figure 18. As before we have x1 = ζ1 − r1 sin t,
x2 = ζ2 − r2 sin t
C1 t X1
Y1
τ
θ
θ
X2
Y2
Figure 18. The blow-up of Figure 17
with π = t + τ +
π 2
− θ i.e. t = θ − τ + π2 . So we have
sin t = cos(θ − τ ) = cos θ cos t + sin θ sin τ with sin θ , cos θ, sin τ , cos τ given by (3.25), (3.26). The formulae (3.29), (3.31) follow. The last formulae are obtained the same way. We can now turn to the differentiability of e. We have first:
28
Michel Chipot
Proposition 3.5. Let ξ = (ζ1 , h1 , ζ2 , h2 ) be a point of R4 such that the configuration is the one of Figure 15. Then e is differentiable at ξ and it holds: ∂e 1 2 2 (ξ ) = {ψ1,x (y1 ) − ψ1,x (x1 )}, ∂ζ1 2 1 2 ∂e 2 (ξ ) = {ψ2,x (y2 ) − ψ2,x (x2 )}, ∂ζ2 2 ∂e (ξ ) = ψ1,x (x1 ) − ψ1,x (y1 ) − G1 , ∂h1 ∂e (ξ ) = ψ2,x (x2 ) − ψ2,x (y2 ) − G2 , ∂h2
(3.33) (3.34) (3.35) (3.36)
where xi , yi are given by (3.20)–(3.23). Proof. 1. Formal proof. Let us for instance prove (3.35). Differentiating under the integral sum, it comes – see (3.9) ∂e (ξ ) = ux uxh1 dx − G1 . ∂h1 Integrating by parts we derive easily (see (3.18), (3.19)) ∂e (ξ ) = − uxx uh1 dx − G1 ∂h1 =− ψ1,xx ψ1,h1 dx − ψ2,xx ψ2,h1 dx − G1 I1 I2 y1 =− ψ1,xx dx − G1 x1
and the result follows. The other cases can be treated similarly. 2. Rigorous proof. We do it for (3.33). With the notation of Figure 15 we have 1 y1 2 1 2 1 2 1 2 (x1 )x1 + ψ1,x (x) dx − ψ1,x (y1 )y1 + ψ2,x (x2 )x2 e = ψ1,x 2 2 x1 2 2 1 y2 2 1 2 + ψ (x) dx + (1 − y2 )ψ2,x (y2 ) − G1 h1 − G2 h2 . 2 x2 2,x 2 Due to Proposition 3.3 the xi , yi are differentiable. Thus, taking the derivative in the direction ζ1 – it is easy to see that every differentiation we will make is justified – we obtain: ∂e 1 2 (ξ ) = ψ1,x (x1 )x1,ζ1 + x1 ψ1,x (x1 )(ψ1,x (x1 ))ζ1 ∂ζ1 2 1 2 1 2 (y1 )y1,ζ1 − ψ1,x (x1 )x1,ζ1 + + ψ1,x 2 2
29
Elastic deformations of membranes and wires
y1
+
x1
1 2 ψ1,x ψ1,xζ1 dx − ψ1,x (y1 )y1,ζ1 − y1 ψ1,x (y1 )(ψ1,x (y1 ))ζ1 2
1 2 + ψ2,x (x2 )x2,ζ1 + x2 ψ2,x (x2 )(ψ2,x (x2 ))ζ1 2 1 2 1 2 + ψ2,x (y2 )y2,ζ1 − ψ2,x (x2 )x2,ζ1 2 2 y2 1 2 + ψ2,x ψ2,xζ1 dx + (1 − y2 )ψ2,x (y2 )(ψ2,x (y2 ))ζ1 − ψ2,x (y2 )y2,ζ1 . 2 x2 Using the cancellations it comes (note that ψ2 , y2 are independent of ζ1 ) ∂e (ξ ) = x1 ψ1,x (x1 )(ψ1,x (x1 ))ζ1 ∂ζ1 y1 + ψ1,x ψ1,xζ1 dx − y1 ψ1,x (y1 )(ψ1,x (y1 ))ζ1
(3.37)
x1
+ x2 ψ2,x (x2 ))(ψ2,x (x2 ))ζ1 . Since ψ1,x (y1 ) = ψ2,x (x2 ) we have ψ2 (x2 ) − ψ1 (y1 ) = ψ2,x (x2 )x2 − ψ1,x (y1 )y1 . Differentiating in ζ1 we get ψ2,x (x2 )x2,ζ1 − ψ1,ζ1 (y1 ) − ψ1,x (y1 )y1,ζ1 = ψ2,x (x2 )x2,ζ1 + (ψ2,x (x2 ))ζ1 x2 − (ψ1,x (y1 ))ζ1 y1 − ψ1,x (y1 )y1,ζ1 . Thus we obtain −ψ1,ζ1 (y1 ) = (ψ2,x (x2 ))ζ1 x2 − (ψ1,x (y1 ))ζ1 y1 . Since ψ2,x (x2 ) = ψ1,x (y1 ), ψ1,x = −ψ1,ζ1 it follows from (3.37) that y1 ∂e (ξ ) = x1 ψ1,x (x1 )(ψ1,x (x1 ))ζ1 − ψ1,x ψ1,xx dx + ψ1,x (y1 )2 . ∂ζ1 x1 Using now the equality ψ1 (x1 ) = x1 ψ1,x (x1 ) and differentiating we obtain ψ1,ζ1 (x1 ) + ψ1,x (x1 )x1,ζ1 = ψ1,x (x1 )x1,ζ1 + x1 (ψ1,x (x1 ))ζ1 i.e. −ψ1,x (x1 ) = x1 (ψ1,x (x1 ))ζ1 . Reporting in (3.38) we obtain 1 ∂e (ξ ) = −ψ1,x (x1 )2 − ∂ζ1 2
y1
x1
2 (ψ1,x )x dx + ψ1,x (y1 )2
(3.38)
30
Michel Chipot
and the result follows by integration. Note that the differentiability follows since the partial derivatives that we obtain are clearly continuous. The other cases can be handled the same way. In the case of Figure 17 one has Proposition 3.6. Let ξ = (ξ1 , h1 , ξ2 , h2 ) be a point of R4 such that the configuration is the one of Figure 17. Then e is differentiable at ξ and it holds: 1 2 ∂e 2 2 2 (ξ ) = {ψ1,x (x1 ) − ψ1,x (x1 ) + ψ1,x (y1 ) − ψ1,x (y1 )}, ∂ζ1 2 1 2 ∂e 2 (ξ ) = {ψ2,x (y2 ) − ψ2,x (x2 )}, ∂ζ2 2 ∂e (ξ ) = ψ1,x (x1 ) − ψ1,x (x1 ) + ψ1,x (y1 ) − ψ1,x (y1 ) − G1 , ∂h1 ∂e (ξ ) = ψ2,x (x2 ) − ψ2,x (y2 ) − G2 . ∂h2
(3.39) (3.40) (3.41) (3.42)
Proof. It is identical to that of Proposition 3.5 and we leave the details to the reader. Remark 3.3. In the case of Figure 15 and when y2 → x2 one has
∂e ∂ζ2 (ξ )
→ 0,
→ −G2 . This corresponds to the case where the second disk does not touch the wire. Similarly when one passes from Figure 18 to Figure 15 the derivatives change smoothly in such a way that the function e is of class C 1 on its domain of definition – i.e. for every ξ such that ξ ∈ (r1 , 1 − r1 ) × R × (r2 , 1 − r2 ) × R. ∂e ∂h2 (ξ )
3.3 Existence result In order to show that the infimum (3.7) is achieved we will restrict the problem on a compact set by showing that only some configurations has to be considered in the minimization process (3.7). So, let us show Theorem 3.1. Under the assumptions (3.3), (3.11), (3.12) there exists (u, C1 , C2 ) achieving the infimum (3.7). Proof. We split our arguments in several steps. Step 1: We can assume h1 > −r1 , or h2 > −r2 . If not we have h1 ≤ −r1 , h2 ≤ −r2 – i.e. both disks are located above the rest position of the wire and u = u(ξ ) = 0. Thus one has e = −G1 h1 − G2 h2 ≥ G1 r1 + G2 r2 .
(3.43)
If one considers only the disk D1 on the wire it will take an equilibrium position with h1 > −r1 and an energy < G1 r1 (see for instance Section 2). Thus for D1 in this
31
Elastic deformations of membranes and wires
position and D2 at the level h2 = −r2 – i.e. in such a way that it does not touch the membrane which is possible by (3.12) – the energy e of the configuration will be e < G1 r1 + G2 r2 which contradicts (3.43). Step 2: We can assume or consider only the situation h1 > −r1 , h2 > −r2 . Indeed from step 1 we can assume hi > −ri for i = 1 or 2. Assume that for j = i it holds hj ≤ −rj . Due to (3.12)we can move Dj on one side or another of Di until it
1 x
0 C1
h Figure 19. The two disks have to touch the wire
touches the wire and thus is such that hj > −rj . This process will lower the energy since u will be unchanged but hj will increase. We can go down until the two disks touch each other (see Figure 19). Step 3: We can assume −ri ≤ hi ≤ H where H is some constant. Indeed first we know that we can assume hi > −ri , i = 1, 2. Then, by the Poincaré Inequality there exists a constant c such that c 1 u2x dx − G1 h1 − G2 h2 ≥ u2 dx − G1 h1 − G2 h2 . e= 2 2 Choosing hi ≥ 0 will cause u to be greater than hi – see (3.5) – on B(ζi , ri ). In other words we will have c e ≥ π(r12 h21 + r22 h22 ) − G1 h1 − G2 h2 . 2 Selecting ||h||∞ = Max{|h1 |, |h2 |} large enough we then force e to be larger than the infimum (3.7) and a contradiction. One can thus assume hi ≤ H , i = 1, 2 where H is some constant.
32
Michel Chipot
Step 4: Analysis of the configuration of Figure 15. Suppose that hi > −ri , i = 1, 2. Suppose that one of the disks is not touching the wire. Then moving it down will cause a decrease in its energy – see step 2. We can thus assume that both disks are touching the wire. Then we would like to show that in a case like in Figure 15 it is not necessary to consider all the configurations but only the ones where the disks are in contact – i.e. the ones for which the constraint (3.4) is saturated. Indeed consider a case where (ζ1 − ζ2 )2 + (h1 − h2 )2 > (r1 + r2 )2 . (3.44) ∂e ∂e ∂e ∂e We claim that ∇e = ∂ζ , , , = 0. Indeed, if not, from (3.33), (3.34) we 1 ∂h1 ∂ζ2 ∂h2 derive 2 2 2 2 ψ1,x (y1 ) = ψ1,x (x1 ) = ψ2,x (x2 ) = ψ2,x (y2 ).
By (3.35), (3.36) – recall that ψ1,x (y1 ) = ψ2,x (x2 ) – this forces ψ1,x (x1 ) = −ψ2,x (y2 ) =
G1 + G2 . 2
∂e ∂e 2 which renders ∂h = 0 or ∂h = 0 But then ψ1,x (y1 ) = ψ2,x (x2 ) = ± G1 +G 2 1 2 impossible. Thus one of the derivatives of e is not equal to 0 and moving one disk around will cause a decrease in energy. One can thus assume that (3.4) is saturated. Due to the symmetry of the problem we can assume without loss of generality that
ζ1 ≤ ζ2 .
(3.45)
So, consider a configuration of the type of Figure 15 where the two disks are in contact. Suppose that h1 , h2 are maintained fixed and move the two disks horizontally in order to search the minimum of energy in this situation – i.e. assume ζ2 = ζ1 + cst = ζ1 + γ . Due to (3.33), (3.34) it holds that d e(ζ1 , h1 , ζ1 + γ , h2 ) = dζ1
(3.46)
∂e ∂e + (ζ1 , h1 , ζ1 + γ , h2 ) ∂ζ1 ∂ζ2
1 2 2 {ψ (y2 ) − ψ1,x (x1 )}. 2 2,x Thus, the minimum will be achieved for =
−ψ2,x (y2 ) = ψ1,x (x1 ). This imposes that the two disks are tangent to the lines with slope λ = ψ1,x (x1 ) = −ψ2,x (y2 ) (see Figure 20). Thus it holds that ζ1 ≤
1 ≤ ζ2 . 2
(3.47)
33
Elastic deformations of membranes and wires
We obtain x1 y1
0
ζ1 21 ζ2 y2
1
x2
h
t θ
σ
λ
Figure 20
1 − (r1 + r2 ) ≥ δ + r1 , 2 1 ζ2 = ζ1 + ζ2 − ζ1 ≤ + (r1 + r2 ) ≤ 1 − r2 − δ 2 ζ1 = ζ2 + ζ1 − ζ2 ≥
where δ = 21 − (2r1 + r2 ) > 0 by (3.11), (3.12). So, we can assume in this case due to step 3 that ζ = (ζ1 , h1 , ζ2 , h2 ) belongs to a compact of R4 . Step 5: The case of the configuration of Figure 18. Of course this configuration has to be considered only when r2 < r1 . Then, arguing as in the preceding step – i.e. assuming h1 , h2 fixed and ζ1 , ζ2 such that (3.46) holds we have – see (3.39), (3.40) d e(ζ1 , h1 , ζ1 + γ , h2 ) dζ1 ∂e ∂e + = (ζ1 , h1 , ζ1 + γ , h2 ) ∂ζ1 ∂ζ2 1 2 2 2 2 2 2 (x1 ) − ψ1,x (x1 ) + ψ1,x (y1 ) − ψ1,x (y1 ) + ψ2,x (y2 ) − ψ2,x (x2 )} = {ψ1,x 2 1 2 2 (y1 ) − ψ1,x (x1 )}. = {ψ1,x 2 Thus the minimum is achieved in this case for −ψ1,x (y1 ) = ψ1,x (x1 )
34
Michel Chipot
i.e. for ζ1 = 21 . It holds then that 1 1 − r2 , + r2 ⊂ (r2 , 1 − r2 ) ζ2 ∈ 2 2 and we can only consider ξ = (ζ1 , h1 , ζ2 , h2 ) for ξ belonging to a compact subset of R4 . Step 6: End of the proof. One minimizes e, a continuous function, (see Proposition 3.2) on a compact subset of R4 . Thus the minimum is achieved. This completes the proof.
3.4 Research of the minima of energy We consider first the case where the disks are like on Figure 20. We have seen there that a minimum can be reached only when ψ1,x (x1 ) = −ψ2,x (y2 ) = λ. Now, letting ζ1 , ζ2 fixed and moving the disks vertically we see that a minimum can be achieved only when ∂e ∂e + = 2λ − G1 + G2 = 0 ∂h1 ∂h2
⇐⇒
λ = (G1 + G2 )/2 = G.
(3.48)
In what follows we will assume that (3.48) holds so that G = (G1 + G2 )/2 is the slope of the two tangents (in absolute value) issued from the end points of the wire. Then, see Figure 20, the disks being tangent to these straight lines their position is uniquely determined by the parameter α = ψ1,x (y1 ) = ψ2,x (x2 ).
(3.49)
Let us then compute the energy of the configuration in terms of α. We will set T = tan
(3.50)
(see Figure 20). Then we have cos =
r1 − r2 , r1 + r2
√ 2 r1 r2 r1 − r2 2 1/2 sin = 1 − = . r1 + r2 r1 + r2
(3.51)
√ 1 = (r1 − r2 )/2 r1 r2 . tan
(3.52)
Thus, T = tan = Then, we can show
Elastic deformations of membranes and wires
35
Proposition 3.7. Assume that T ≤ 1/G. Then, in the case of Figure 20, the energy e = e(α) ˜ of the configuration is given by √ T +α 2 1 − αT 2 +G √ e(α) ˜ = r1 r2 −T ln[ 1 + α + α] + (α + G1 − G2 ) √ 1 + α2 1 + α2 2 r 1 + r2 r 1 + r2 G ln[ 1 + G2 + G] − + G 1 + G2 . + (3.53) 2 2 2 Remark 3.4. For T > 1/G i.e. tan < G, the small disk can fit under the large one. Proof of Proposition 3.7. Using the notation of Figure 20 we have 1 1 2 e = e(α) ˜ = u dx − G1 h1 − G2 h2 2 0 x 1 2 1 y1 2 1 = G x1 + ψ1,x (s) ds + α 2 (x2 − y1 ) (3.54) 2 2 x1 2 1 1 y2 2 ψ2,x (x) dx + G2 (1 − y2 ) − G1 h1 − G2 h2 . + 2 x2 2 We have different quantities to compute. First x1 = ζ1 − r1 sin θ
1 G with tan θ = G ⇒ cos θ = √ , sin θ = √ 1 + G2 1 + G2 r1 G ⇒ x1 = ζ1 − √ . (3.55) 1 + G2
This allows us to get h1 given by
h1 = Gx1 − r1 cos θ = Gζ1 − r1 1 + G2 . √ Since tan t = α we have sin t = α/ 1 + α 2 and r1 α , y1 = ζ1 − r1 sin t = ζ1 − √ 1 + α2 r2 α , x2 = ζ2 − r2 sin t = ζ2 − √ 1 + α2 r2 G . y2 = ζ2 + r2 sin θ = ζ2 + √ 1 + G2
(3.56)
(3.57) (3.58) (3.59)
Finally, for h2 we get
h2 = (1 − y2 )G − r2 cos θ = −ζ2 G + G − r2 1 + G2 .
We come back now to (3.54). Due to (3.5) we have ψi,x = −(x − ζi )/ ri2 − (x − ζi )2 ,
i = 1, 2
(3.60)
36
Michel Chipot
and thus 2 ψi,x = (x − ζi )2 /{ri2 − (x − ζi )2 }
=
ri2 ri2
− (x − ζi
)2
−1=
1−
1 x−ζi ri
2 − 1
, i = 1, 2.
(3.61)
Thus, for i = 1, 2, a simple computation gives 1 2
yi
xi
1 1 2 ψi,x (x) dx = − (yi − xi ) + 2 2 1 = − (yi − xi ) + 2 1 = − (yi − xi ) + 2
yi
xi
dx i 2 1 − ( x−ζ ri )
ri (yi −ζi )/ri dy/(1 − y 2 ) (3.62) 2 (xi −ζi )/ri ri + (xi − ζi ) ri ri + (yi − ζi ) − ln . ln 4 ri − (yi − ζi ) ri − (xi − ζi )
Using now (3.55), (3.57)–(3.59) we obtain 1 2
y1
x1
2 ψ1,x (x) dx =
r1 α 1 r1 G −√ √ 2 1 + α2 1 + G2 √ √ r1 1 + α2 − α 1 + G2 − G + ln √ − ln √ , 4 1 + α2 + α 1 + G2 + G
(3.63)
r2 α 1 r2 G −√ −√ 2 1 − α2 1 + G2 √ √ r2 1 + G2 + G 1 + α2 − α + ln √ − ln √ . 4 1 + G2 − G 1 + α2 + α
(3.64)
and 1 2
y2
x2
2 ψ2,x (x) dx =
Noting that √ 1 1 + x2 − x = −2 ln 1 + x 2 + x , = ln √ ln √ 1 + x2 + x ( 1 + x 2 + x)2
Elastic deformations of membranes and wires
37
we derive from (3.63), (3.64), (3.54) that (r1 − r2 ) ln 1 + α 2 + α e = e(α) ˜ =− 2 1 α r1 + r2 ln 1 + G2 + G + (r1 − r2 ) √ + 2 2 1 + α2 1 G 1 2 r1 G − (r1 + r2 ) √ + G ζ1 − √ 2 2 1 + G2 1 + G2 α 1 + α 2 ζ2 − ζ1 + (r1 − r2 ) √ 2 1 + α2 r2 G 1 − G 1 h1 − G 2 h2 + G2 1 − ζ2 − √ 2 1 + G2 r 1 + r 2 (r1 − r2 ) ln 1 + α 2 + α + ln 1 + G2 + G =− 2 2 1 2 + (r1 − r2 )α 1 + α 2 1 1 1 + G2 (ζ1 − ζ2 ) + α 2 (ζ2 − ζ1 ) + G2 2 2 2 1 − (r1 + r2 )G 1 + G2 − G1 h1 − G2 h2 . (3.65) 2 We turn now to the computation of h1 and h2 . We have – see Figure 20 σ = + t.
(3.66)
Thus tan σ =
tan + tan t T +α = . 1 − tan tan t 1 − αT
(3.67)
It follows that it holds cos2 σ =
1 1 (1 − αT )2 4r1 r2 (1 − αT )2 = = = T +α 2 (1 + α 2 )(1 + T 2 ) (r1 + r2 )2 1 + α 2 1 + tan2 σ 1 + ( 1−αT )
(recall (3.52)). Thus it comes √ 2 r1 r2 1 − αT cos σ = , √ r1 + r2 1 + α 2
√ 2 r1 r2 T + α sin σ = tan σ cos σ = . √ r1 + r 2 1 + α 2
(3.68)
We derive then √ T +α h2 − h1 = (r1 + r2 ) sin σ = 2 r1 r2 √ 1 + α2
(3.69)
√ 1 − αT ζ2 − ζ1 = (r1 + r2 ) cos σ = 2 r1 r2 √ . 1 + α2
(3.70)
and
38
Michel Chipot
From (3.56), (3.60) we have also
h1 + h2 = −G(ζ2 − ζ1 ) + G − (r1 + r2 ) 1 + G2 √ 1 − αT = −G2 r1 r2 √ + G − (r1 + r2 ) 1 + G2 . 1 + α2
(3.71)
Combining (3.69) and (3.71) we get √ √ 1 − αT G (r1 + r2 ) T +α − r1 r2 G √ + − 1 + G2 , h1 = − r1 r2 √ 2 2 1 + α2 1 + α2 √ √ 1 − αT G (r1 + r2 ) T +α − r1 r2 G √ + − 1 + G2 . h2 = r1 r2 √ 2 2 1 + α2 1 + α2
(3.72) (3.73)
Using (3.70), (3.72), (3.73) in (3.65) we obtain r 1 + r 2 √ e = e(α) ˜ = − r1 r2 T ln 1 + α 2 + α + ln 1 + G2 + G 2 2 √ √ α(1 + α ) √ 1 − αT 1 − αT + r1 r2 T √ − r1 r2 G2 √ + r1 r2 α 2 √ 2 2 1+α 1+α 1 + α2 2 √ G T +α 1 + − (r1 + r2 )G 1 + G2 + r1 r2 (G1 − G2 ) √ 2 2 1 + α2 √ 1 − αT + r1 r2 2G2 √ − G2 + G(r1 + r2 ) 1 + G2 2 1+α r 1 + r 2 √ ln 1 + G2 + G = − r1 r2 T ln 1 + α 2 + α + 2 √ √ T +α 1 − αT + r1 r2 (α + G1 − G2 ) √ + r1 r2 G2 √ 2 1+α 1 + α2 r1 + r2 G2 + G 1 + G2 − 2 2 which completes the proof of (3.53). We are now able to differentiate the energy to get: Proposition 3.8. Assume that T ≤ 1/G, then it holds that √ r1 r2 e˜ (α) = {α 3 −T α 2 +α[2−G2 −T (G1 −G2 )]−G2 T +G1 −G2 }. (3.74) (1 + α 2 )3/2 Proof. From (3.53) we derive √ T T +α α +1 + √ + e˜ (α) = r1 r2 − √ √ 2 2 1+α +α 1+α 1 + α2
39
Elastic deformations of membranes and wires
√ √ 1 + α 2 − (T + α)α/ 1 + α 2 + (α + G1 − G2 ) (1 + α 2 ) √ √ 2 2 2 −T 1 + α − (1 − αT )α/ 1 + α +G (1 + α 2 ) √ T +α 1−Tα T +√ + (α + G1 − G2 ) = r1 r2 − √ (1 + α 2 )3/2 1 + α2 1 + α2 (T + α) − G2 (1 + α 2 )3/2 √ r1 r2 {α 3 − T α 2 + α[2 − G2 − T (G1 − G2 )] − G2 T + G1 − G2 }. = (1 + α 2 )3/2 This completes the proof of the proposition. In what follows we will set P (α) = α 3 − T α 2 + α[2 − G2 − T (G1 − G2 )] − G2 T + G1 − G2 .
(3.75)
By simple computation we have P (−G) = −G3 − T G2 − G[2 − G2 − T (G1 − G2 )] − G2 T + G1 − G2 (3.76) = −2G(1 + GT ) + (G1 − G2 )(1 + GT ) = −2G2 (1 + GT ) < 0, and also P (G) = G3 − T G2 + G[2 − G2 − T (G1 − G2 )] − G2 T + G1 − G2 = 2G1 (1 − GT ).
(3.77)
3.4.1 The case of two disks of same size (T = 0). In this case we have r1 = r2 and thus, see (3.52), T = 0. The polynomial P above becomes P (α) = α 3 + α(2 − G2 ) + G1 − G2 .
(3.78)
3.4.1.1. The case of two identical disks (G1 = G2 ). This case was studied in [3] by a Lagrange multiplyer method. Here our approach is different. We have √ Theorem 3.2. If G ≤ 2 then the two disks have a unique equilibrium position (up √ to exchanging each other) for α = 0 – see Figure 21. If G > 2 then the two disks
G≤
Figure 21. The two disks at the same level
√ 2
40
Michel Chipot
have two equilibrium positions which are symmetric (see Figure 22). If we exchange each other this leads to four possible equilibria.
G>
√
2
Figure 22. The two disks tilted
Proof. We have by (3.78) P (α) = α[α 2 + 2 − G2 ].
(3.79)
If G2 ≤ 2 then α = 0 is the only root of P on [−G, G]. Thus since e˜ is decreasing on (−G, 0) increasing on (0, G), α = 0 is the only minimum of e˜ on (−G, G) and we are in the case of Figure 21. If now G2 > 2 then it holds that P (α) = α(α − G2 − 2)(α + G2 − 2). (3.80) √ e˜ has then two local minima at ± G2 − 2 which are also global minima for obvious symmetry reasons. We are in the case of Figure 22. This completes the proof. Remark 3.5. The breaking of the symmetry of Figure 21 is somehow a priori unexpected. However, when the weight of the disks is getting higher, – due to the tension of the wire – the equilibrium becomes unstable and the tilting of Figure 22 occurs. 3.4.1.2. The case of different disks in weight (G1 = G2 ). In this case we can show Theorem 3.3. Assume that G1 = G2 . Up to a symmetry the equilibrium position of the two disks is unique. Proof. By (3.78) we have P (α) = 3α 2 + 2 − G2 .
(3.81)
In the case where G2 ≤ 2 then P (α) ≥ 0 and P is increasing on (−G, G). Since P (−G) < 0, P (G) > 0 – see (3.76), (3.77) – P admits a unique 0 on (−G, G) which is the unique minimum of e˜ on (−G, G). This completes the proof in this case. In the case where G2 > 2 then it holds that √ √ G2 − 2 G2 − 2 P (α) = 3 α − α+ . (3.82) √ √ 3 3
Elastic deformations of membranes and wires
Let us set
√ α± = ± G2 − 2/ 3.
41
(3.83)
It is clear that P is increasing on (−G, α− ), decreasing on (α− , α+ ), increasing on (α+ , G). Thus if P (α− ) ≤ 0
or
P (α+ ) ≥ 0
(3.84)
then P changes its sign only once and e˜ has a unique minimum on (−G, G) – recall that P (−G) < 0, P (G) > 0. This completes the proof in this case. If now P (α− ) > 0
and
P (α+ ) < 0
(3.85)
then P admits three zeros on (−G, G) and e˜ admits two local minima on (−G, G). However, only one is a global minimum. This is due to the fact that the light disk will always sit above the more heavy one. Let us denote by α1 , α2 , α1 < α− < α+ < α2 the two points of (−G, G) where e˜ achieves its local minimum. Suppose that (3.86)
G2 < G1 .
Then, e˜ cannot achieve a global minimum in α2 where the second disk is in a position lower than the first one. Indeed suppose that 1 ˜ 2) = u2 − G1 h1 − G2 h2 . e1 = e(α 2 x Exchanging the two disks will produce a configuration with an energy 1 u2 − G2 h1 − G1 h2 e2 = 2 x so that e2 − e1 = (h1 − h2 )(G1 − G2 ) < 0
(h1 < h2 )
(3.87)
recall that h is directed downward. Thus, by (3.87), the energy e2 is smaller than e1 which renders impossible to have a minimum at α2 . Of course when G2 > G1 one would prove the same way that e˜ cannot have an absolute minimum in α1 . This completes the proof of the theorem. Remark 3.6. One has – see (3.78): 2 (G − 2)3/2 (G2 − 2)3/2 P (α± ) = ± − + G 1 − G2 √ √ 3 3 3 2 (G2 − 2)3/2 =± − + G 1 − G2 . √ 3 3
(3.88)
42
Michel Chipot
Thus (3.85) holds if we have 2 2 G − 2 3/2 G − 2 3/2 < G1 − G2 < 2 −2 3 3 2 3/2 G −2 ⇐⇒ |G1 − G2 | < 2 3
(3.89)
which can of course occur. For the case of different disks in size and weight we refer the reader to [11].
3.5 Further problems In this last section we would like to mention briefly several other issues which have yet to be investigated. To consider the case of two ellipses free to roll on an elastic wire would be probably a nightmare to study and we do not insist on this topic. A natural continuation of the problem of two disks would be the problem of n identical disks – (see [1] for very partial results). Then – in the case of three disks for instance – a new situation can occur: one disk can sit atop of the two others and a special investigation of the energy has to be made in this case (see Figure 23). Of course with more disks
Figure 23. A new situation in the case of three disks
the possible packing of the disks together can become more involved! We let this to the curiosity of the reader.
4 The case of a single ball on an elastic membrane In this section is a bounded domain of R2 with a smooth boundary . As we mentioned in our introduction is a horizontal frame on which an elastic membrane is spanned, is the undeformed position of the membrane. Letting roll freely on this membrane a heavy ball of weight G and radius r we are interested in finding the
Elastic deformations of membranes and wires
43
x2
x1
h Figure 24. The equilibrium position of a ball on a membrane
equilibrium position reached by the ball. As before in the case of a disk on a wire, the position of the ball is completely determined by the position of its center C = (ζ, h) = (ζ1 , ζ2 , h),
(4.1)
(recall that the vertical axis is oriented downward). The function describing the bottom part of the ball is now given by – see Figure 4 – (4.2) ψ = ψ ζ h (x) = h + r 2 − |x − ζ |2 , x ∈ B(ζ, r), where B(ζ, r) denotes the open ball in R2 of center ζ and radius r – i.e. B(ζ, r) = { x ∈ R2 | |x − ζ | < r }.
(4.3)
| · | is the euclidean norm in R2 defined by |z| = {z12 + z22 }1/2
∀ z = (z1 , z2 ) ∈ R2 .
(4.4)
For a position (ζ, h) of the center of the ball an admissible deformation is an element u of K = K(ζ, h) defined by K = K(ζ, h) = { v ∈ H01 () | v(x) ≥ ψ ζ h (x) a.e. x ∈ B(ζ, r) }.
(4.5)
It is easy to check that K(ζ, h) is a closed convex set of H01 (). In order to avoid the ball to be in contact to the frame we will consider only positions of the center such that dist(ζ, ) > r
(4.6)
dist(ζ, ) = Inf |ζ − z|.
(4.7)
r = { ζ ∈ | dist(ζ, ) > r }
(4.8)
where z∈
If we define
44
Michel Chipot
we will choose (ζ, h) ∈ r × R
(4.9)
and for such a point the energy of an admissible deformation u is given by 1 |∇u(x)|2 dx − Gh. E(u; ζ, h) = 2
(4.10)
∂u ∂u , ). Then to look (∇u denotes the usual gradient i.e. the vector of components ∂x 1 ∂x2 for an equilibrium position of the ball reduces to look for an element (u0 , ζ0 , h0 ) such that (ζ0 , h0 ) ∈ r × R, u0 ∈ K(ζ0 , h0 ), (4.11) E(u0 ; ζ0 , h0 ) ≤ E(u; ζ, h) ∀ (ζ, h) ∈ r × R, ∀ u ∈ K(ζ, h).
As observed in previous situations, if such a point u0 does exist then it holds that E(u0 ; ζ0 , h0 ) ≤ E(u; ζ0 , h0 )
∀ u ∈ K(ζ0 , h0 )
that is to say u0 satisfies u0 ∈ K(ζ0 , h0 ) and 1 1 2 |∇u0 | dx − Gh0 ≤ |∇u|2 dx − Gh0 ∀ u ∈ K(ζ0 , h0 ), 2 2 1 1 2 |∇u0 | dx ≤ |∇u|2 dx ∀ u ∈ K(ζ0 , h0 ). ⇐⇒ 2 2 (4.12) It is well known that H01 () is a Hilbert space when equipped with the norm
|∇u| = 2
|∇u|2 dx
1/2 .
(4.13)
Thus, by (4.12), u0 is the projection of 0 on the closed convex set K(ζ0 , h0 ) and also the solution of the variational inequality (see Proposition 2.1) u 0 ∈ K(ζ0 , h0 ), (4.14) ∇u0 · ∇(v − u0 ) dx ≥ 0 ∀ v ∈ K(ζ0 , h0 ).
(We denote with a dot the scalar product.) Thus, for (ζ, h) ∈ r × R we can introduce u = u(ζ, h) the solution to u ∈ K(ζ, h), (4.15) ∇u · ∇(v − u) dx ≥ 0 ∀ v ∈ K(ζ, h).
Elastic deformations of membranes and wires
45
(Such a solution exists and is unique – see Proposition 2.1.) For u = u(ζ, h) we can then define e by 1 |∇u|2 dx − Gh. (4.16) e(ζ, h) = 2 Due to (4.12) it is clear that if (u0 , ζ0 , h0 ) is a minimizer to (4.11) then (ζ0 , h0 ) is a minimizer of e on r × R. Conversely if (ζ0 , h0 ) is a minimizer of e on r × R, then for u0 solution to (4.14), (u0 , ζ0 , h0 ) is a minimizer to (4.11). Thus, we are now reduced to find a point (ζ0 , h0 ) such that (ζ0 , h0 ) ∈ r × R, (4.17) e(ζ0 , h0 ) ≤ e(ζ, h) ∀ (ζ, h) ∈ r × R, i.e. we are reduced to minimize e on r × R. It should be noticed that this later set is not compact, thus the problem requires some work in order to be solved. Remark 4.1. In this note we will restrict ourselves to the Dirichlet integral. One could consider also an energy of the type 1 1 + |∇u(x)|2 dx − Gh (4.18) E(u; ζ, h) = 2 which corresponds to a membrane of a soap film type. Of course in this case in order for the membrane not to brake we have to impose some limit to h. We refer the reader to [4] for details.
4.1 Some remarks on the obstacle problem In this section we would like to study more deeply the problem (4.15). This is a problem of obstacle type – i.e. the solution is forced to stay above an “obstacle” (see [5], [7], [8], [18]). However, this obstacle problem is not a classical one due to the fact that ψ = ψ ζ h – the so called obstacle – is not defined on the whole domain and moreover its gradient does not remain bounded. So, it requires some analysis. First we would like to show that the solution u of (4.15) is such that u ∈ W 2,∞ ().
(4.19)
(We refer the reader to [13], [17], [18] for notation and results on Sobolev spaces.) For that, let us first prove: Lemma 4.1 (Monotonicity of the solution with respect to the domain). Let ∗ ⊂ be an open subset of containing B(ζ, r). Let u∗ be the solution to 1 ∗ ∗ ∗ ∗ ζh u ∈ K = K (ζ, h) = { v ∈ H0 ( ) | u(x) ≥ ψ (x) a.e. x ∈ B(ζ, r) }, ∇u∗ · ∇(v − u∗ ) dx ≥ 0 ∀ v ∈ K ∗ . ∗
(4.20)
46
Michel Chipot
Let u be the solution to (4.15). Then if u∗ is supposed to be extended by 0 on \ ∗ it holds that u∗ (x) ≤ u(x) a.e. x ∈ .
(4.21)
Proof. If (·)+ denotes the positive part of a function we want to show that (u∗ − u)+ = 0.
(4.22)
For that, we remark first that u and u∗ are nonnegative. Let us show it for u, the proof for u∗ being similar. We remark that v = u+ ∈ K(ζ, h) and thus by (4.15) it holds that
∇u · ∇(u+ − u) dx ≥ 0.
(4.23)
(4.24)
If u− = (−u)+ denotes the negative part of u we have u = u+ − u− and (4.24) leads to − 0≤ ∇u · ∇u dx = − ∇u− · ∇u− dx, (4.25)
u−
= 0 i.e. u ≥ 0 a.e. in . This is what we wanted to prove (see [9]). It follows that first. Next, we notice that v = u∗ − (u∗ − u)+ ∈ H01 (∗ ).
(4.26)
(This is due to the fact that u ≥ 0 on the boundary of ∗ .) Now v = u∗ or u, it is clear that v ∈ K ∗ (ζ, h) and from (4.20) we derive ∇u∗ · ∇(u∗ − u)+ dx ≤ 0.
(4.27)
Now since v = u + (u∗ − u)+ = u or u∗ also, this is a suitable test function for (4.15) and we obtain −∇u · ∇(u∗ − u)+ dx ≤ 0.
(4.28)
Summing up (4.27) and (4.28) it comes ∗ ∗ + ∇(u − u)∇(u − u) dx = |∇(u∗ − u)+ |2 dx ≤ 0
which implies (4.22) and completes the proof of the lemma.
(4.29)
Elastic deformations of membranes and wires
47
Let us denote by the set = { x ∈ B(ζ, r) | u(x) = ψ ζ h (x) }.
(4.30)
is the coincidence set of the obstacle problem (4.15). For the time being this is a measurable subset of B(ζ, r). We are going to show that this is a closed subset of B(ζ, r). For that, let us denote by R the distance from ζ to the boundary of – i.e. R = dist(ζ, ) = Inf |ζ − γ |. γ ∈
(4.31)
Clearly for ζ ∈ r it holds that R > r.
(4.32)
∗ = B(ζ, R).
(4.33)
Let us set
Then we have: Lemma 4.2. Suppose that ζ ∈ r , h > −r. There is a unique s ∈ (0, r) such that s 1 f (s) = ln + 2 h r 2 − s 2 + r 2 − s 2 = 0. (4.34) R s Moreover, if ∗ is given by (4.33) the solution to (4.20) is given by ζh if = |x − ζ | ≤ s, ψ (x) 2 (4.35) u∗ (x) = −s √ ln if = |x − ζ | ≥ s. r 2 − s2 R In particular u∗ is radially symmetric, u∗ ∈ W 2,∞ (∗ )
(4.36)
and the coincidence set is given by ¯ s) ∗ = { x ∈ ∗ | u∗ (x) = ψ ζ h (x) } = B(ζ,
(4.37)
¯ where B(ζ, s) denotes the closed ball of center ζ and radius s. Proof. If f is defined by (4.34) – i.e. if we set 1 f () = ln + 2 {h r 2 − 2 + r 2 − 2 }, R
(4.38)
we have
1 2 1 −h f () = + 2 − 2 − 3 {h r 2 − 2 + r 2 − 2 } r 2 − 2 2 2 − 2r h 2h 2 = − − r − 2 < 0 for ∈ (0, r). 3 2 2 3 r −
(4.39)
48
Michel Chipot
This shows that f is decreasing. Moreover since lim f () = +∞,
→0
f (r) = ln
r <0 R
it is clear that there exists a unique s ∈ (0, r) such that (4.34) holds. Now (4.34) can also be written −s 2 s h + r 2 − s2 = √ (4.40) ln 2 2 R r −s which shows that u∗ given by (4.35) and ψ ζ h coincide for = s. Moreover, since u∗ and ψ are radially symmetric we have with obvious notation dψ du∗ −s = (s) = √ (s). d d r 2 − s2
(4.41)
Thus the function u∗ defined by (4.35) is a radially symmetric C 1 -function and thus belongs clearly to W 2,∞ (∗ ). Next, due to (4.39), (4.40) it holds that −s 2 −2 h + r 2 − 2 < <√ ∀ ∈ (s, r), (4.42) ln ln R R r 2 − s2 r 2 − 2 (the function → √ 2
2
r −2
is clearly increasing) and thus (4.37) holds. To see that u∗
given by (4.35) is solution to (4.20) one remarks first that (4.42) implies that u∗ ∈ K ∗ (ζ, h).
(4.43)
Next we remark that u∗ is harmonic outside B(ζ, s). Indeed outside this ball we have if we use the Laplace operator in polar coordinates −s 2 d 2 u∗ 1 du∗ 1 1 ∗ u = =√ + − 2 + 2 = 0. (4.44) d2 d r 2 − s2 Thus we have, if we denote simply B(ζ, s) by B ∗ ∗ ∇u · ∇(v − u ) dx = ∇ · {(v − u∗ )∇u∗ } dx ∗ \B
∗ \B
(v − u∗ )
=− ∂B ∗ ( ∂u ∂n
∂u∗ dσ (x) ∂n
(4.45) ∀ v ∈ K∗
denotes the derivative in the direction of the outward unit normal n to ∂B the boundary of B, dσ is the surface area on ∂B – recall that v, u∗ vanish on the boundary of ∗ and u∗ is a C 1 -function). Similarly on B we have ∗ ∗ ∇u · ∇(v − u ) dx = ∇ · {(v − u∗ )∇u∗ } − (v − u∗ )u∗ dx B B (4.46) ∗ ∗ ∂u dσ (x) − (v − ψ)ψ dx. (v − u ) = ∂n ∂B B
Elastic deformations of membranes and wires
Summing up (4.45) and (4.46) leads to ∗ ∗ ∇u · ∇(v − u ) dx = − (v − ψ)ψ dx. ∗
49
(4.47)
B
Now arguing like in (4.44) we have ψ = with − dψ , = d r 2 − 2
1 dψ d 2ψ + 2 d d
d 2ψ (−)(−) 2 2 = − r − − /(r 2 − 2 ). d2 r 2 − 2
(4.48)
(4.49)
Thus −(r 2 − 2 ) − 2 r2 d 2ψ = = − . d2 (r 2 − 2 )3/2 (r 2 − 2 )3/2
(4.50)
Combining this with (4.48) we get ψ =
1 2 − 2r 2 −r 2 − = ≤ 0. (r 2 − 2 )3/2 (r 2 − 2 )3/2 r 2 − 2
(4.51)
Since for v ∈ K ∗ (ζ, h) we have v−ψ ≥0
on B(ζ, r),
we deduce from (4.47), (4.51) ∇u∗ · ∇(v − u∗ ) dx ≥ 0 ∗
∀ v ∈ K ∗ (ζ, h)
and thus u∗ is solution to (4.20). This completes the proof of the lemma. Let ζ be a point of r . Let u∗ be the function defined by (4.35) that we suppose to be extended by the bottom formula of (4.35) outside of B(ζ, R). Then, clearly, as seen in the preceding lemma we have u∗ ∈ W 2,∞ ().
(4.52)
Moreover: Lemma 4.3. Let ζ ∈ r , h > −r. Let u be the unique solution to (4.15) then u is also the unique solution to the variational inequality u ∈ Ku∗ = { v ∈ H01 () | v(x) ≥ u∗ (x) a.e. x ∈ }, (4.53) ∇u · ∇(v − u) dx ≥ 0 ∀ v ∈ Ku∗ .
Proof. It is clear that Ku∗ is a closed convex set of H01 (), so the existence and uniqueness of a solution to (4.53) follows the arguments that we developed before.
50
Michel Chipot
Let us, for the time being, denote by u the solution to (4.53). Since – see (4.43) – u∗ ∈ K ∗ (ζ, h) we have clearly u ∈ K(ζ, h) so that by (4.15) ∇u · ∇(u − u) dx ≥ 0. (4.54)
Next, due to Lemmas 4.1 and 4.2 we have u ≥ u∗ , and by (4.53) we derive
i.e.,
u ∈ Ku∗
∇u · ∇(u − u ) dx ≥ 0.
(4.55)
Summing now (4.54) and (4.55) we get |∇(u − u )|2 dx ≤ 0
i.e. u = u . This completes the proof of the lemma. Then we are now able to prove Proposition 4.1. Let ζ ∈ r , h > −r. Let u be the solution to (4.15) then it holds u ∈ W 2,∞ ().
(4.56)
= { x ∈ B(ζ, r) | u(x) = ψ ζ h (x) }
(4.57)
The coincidence set
is a closed subset of B(ζ, r) such that for some ε > 0 ⊂ B(ζ, r − ε).
(4.58)
Moreover, it holds that −u ≥ 0 in ,
(4.59)
−u = 0 in \
(4.60)
((4.59) means that u is a nonpositive measure in , (4.60) means that u is harmonic outside the coincidence set). Proof. By Lemma 4.3 u is solution to (4.53) with u∗ ∈ W 2,∞ (). It follows by well known results of regularity – see [5], [7], [8], [19] – that (4.56) holds. Moreover, see (4.42), we have ⊂ B(ζ, s)
(4.61)
which is (4.58). The fact that is closed follows from the fact that both u and ψ are continuous. Now let us consider ϕ ∈ D(),
ϕ ≥ 0,
(4.62)
Elastic deformations of membranes and wires
51
where D() denotes the space of infinitely differentiable functions with compact support in . It is clear that v = u + ϕ ∈ K(ζ, h) and v is a suitable test function for (4.15). Thus we have ∇u · ∇ϕ dx ≥ 0 ∀ ϕ ∈ D(), ϕ ≥ 0 (4.63) ⇐⇒ −u, ϕ ≥ 0 ∀ ϕ ∈ D(), ϕ ≥ 0, ·, · denotes the duality bracket between D () and D(), see [9]. This is (4.59). From this and (4.44) we deduce −(u − u∗ ) = −u ≥ 0 ∗
u−u ≥0
in \ B(ζ, s), ¯ in \ B(ζ, s).
(4.64) (4.65)
Thus, by the strong maximum principle – see [17] – we deduce that u > u∗
¯ in \ B(ζ, s).
(4.66)
¯ Since on B(ζ, s) we have u∗ = ψ,
(4.67)
it follows that = { x ∈ B(x, r) | u(x) = ψ(x) } = { x ∈ | u(x) = u∗ (x) }.
(4.68)
Let then ϕ be a function such that ϕ ∈ D( \ ).
(4.69)
\ = { x ∈ | u(x) > u∗ (x) }
(4.70)
The set
is of course an open subset. Since ϕ has compact support for ε small enough it holds that u ± εϕ ≥ u∗
in .
Thus, v = u ± εϕ are both test functions for (4.53) and we obtain ∇u · ∇ϕ dx ≥ 0 ∀ ϕ ∈ D( \ ), ±ε
(4.71)
(4.72)
i.e., −u = 0
in \
(4.73)
in the distributional sense or almost everywhere by (4.56). This completes the proof of the proposition.
52
Michel Chipot
4.2 Existence of an equilibrium As seen previously the minimization problem (4.11) reduces to (4.17) – i.e. to the minimization of the function e(ζ, h) defined by (4.16). This is what we would like to address now. Note that the preceding section provides us with some information about the function u used in the definition of e. First we would like to show that the function e is continuous on r × R. This will be a direct consequence of the following lemma: Lemma 4.4. Let (ζn , hn ) ∈ r × R be a sequence of points such that (ζn , hn ) → (ζ, h) ∈ r × R.
(4.74)
If un = u(ζn , hn ), u = u(ζ, h) denote the solutions to (4.15) we have un → u in H01 ().
(4.75)
Proof. Step 1: We claim that there exists n0 such that K(ζn , hn ) ∩ K(ζ, h) = ∅
(4.76)
n≥n0
i.e. there exists a function v belonging to all the K(ζn , hn ), n ≥ n0 and to K(ζ, h). For that set 1 δ = {dist(ζ, ) − r}. (4.77) 2 Then, for n ≥ n0 we have |ζn − ζ | < δ
(4.78)
and thus B(ζn , r) ⊂ B(ζ, r + δ) Since ψ ζn hn (x) = hn +
∀ n ≥ n0 .
(4.79)
r 2 − |x − ζn |2 ≤ hn + r
(4.80)
and since hn is bounded – since converging – there exists a constant H > 0 such that ψ ζn hn (x) ≤ hn + r ≤ H
∀ x ∈ B(ζn , r),
Then, one can clearly find a smooth function v such that H on B(ζ, r + δ), v= 0 outside B(ζ, r + 2δ),
∀ n ≥ n0 .
(4.81)
(4.82)
this function v belongs then to the set defined by (4.76) (we see that v ∈ K(ζ, h) by passing to the limit in (4.81)).
53
Elastic deformations of membranes and wires
Step 2: Uniform bound for un . Using (4.15) for v given by (4.82), we obtain ∇un · ∇(v − un ) dx ≥ 0 ∀ n ≥ n0
(4.83)
|∇un | 2 = ⇒ |∇un |2 dx ≤ ∇un · ∇v dx ≤ |∇un | |∇v|
2
2
2
where | · |2 denotes the L2 ()-norm, | · | the euclidean norm of the gradient. It follows that it holds that
|∇un | ≤ |∇v| ∀ n ≥ n0 (4.84) 2 2 and thus un is bounded independently of n. Thus there exists a subsequence of n – still denoted by n such that for some u∞ ∈ H01 () it holds that un u∞ in H01 (),
un → u∞ in L2 (),
un → u∞ a.e. in ,
(4.85)
(recall that H01 () is compactly imbedded in L2 ()). Step 3: We have u∞ = u. If we can show that u∞ = u then by uniqueness of this weak limit the whole sequence un will be converging toward u in H01 () weak. First, since un ∈ K(ζn , hn ), there exists a subset Un of measure 0 such that un (x) ≥ ψ ζn hn (x)
a.e. x ∈ B(ζn , r) \ Un .
(4.86)
Moreover there is a set V of measure 0 in such that un (x) → u∞ (x)
in \ V .
(4.87)
Set U = (∪n Un ) ∪ V .
(4.88)
Then U is a set of measure 0. Let x ∈ B(ζ, r) \ U . For n large enough we have x ∈ B(ζn , r) ∀ n and thus it holds that (4.89) un (x) ≥ ψ ζn hn (x) = hn + r 2 − |x − ζn |2 . Letting n → +∞ and using (4.87) we obtain u∞ (x) ≥ ψ ζ h (x)
(4.90)
i.e. u∞ ∈ K(ζ, h). Next, we consider u ∈ K(ζ, h) the solution to (4.15). We set for n large enough ε u∨ ψ ζn hn on B ζ, r − , 2 (4.91) wn = ε u , outside B ζ, r − 2
54
Michel Chipot
where ε is the real defined in (4.58) and ∨ denotes the maximum of two functions. We claim that for n large enough wn ∈ K(ζn , hn ), wn → u
(4.92)
in H01 ().
Indeed, for n large enough it holds that ε ε B ζ, r − ⊂ B ζn , r − ⊂ B(ζn , r) 2 3
(4.93)
(4.94)
so that the definition (4.91) makes sense. Moreover, since u and ψ = ψ ζ h are continuous, for some δ > 0 it holds that 2ε u − ψ ζ h ≥ δ > 0 on \ B ζ, r − . (4.95) 3 Now it is clear that on B(ζ, r − for n large enough it holds that
2ε 3 ),
ψ ζn hn converges uniformly toward ψ ζ h and thus
u − ψ ζn h n ≥ 0
2ε . on \ B ζ, r − 3
Denote then by α a smooth function such that 2ε α = 1 on B ζ, r − , α=0 3
ε outside B ζ, r − . 2
(4.96)
(4.97)
It is clear that it holds that wn = α(u∨ ψ ζn hn ) + (1 − α)u.
(4.98)
Recall that the maximum of two functions of H 1 belongs to H 1 (see [9]). Then from (4.94) we have that it holds that ε u∨ ψ ζn hn ∈ H 1 B ζ, r − (4.99) 2 and thus wn ∈ H01 (). Moreover, combining (4.96) and the definition of wn we have wn ∈ K(ζn , hn ).
(4.100)
It is easy to see that – see also (4.94) – ψ ζn hn → ψ ζ h
ε in H 1 B ζ, r − 2
(4.101)
u ∨ ψ ζ n hn → u
ε in H 1 B ζ, r − . 2
(4.102)
and thus (see [9])
Elastic deformations of membranes and wires
It follows from (4.98) that (4.93) holds. Thus from (4.15) we derive ∇un · ∇(wn − un ) dx ≥ 0 ∇un · ∇wn dx ≥ |∇un |2 dx. ⇒
Since un
55
(4.103)
u∞ in H01 (), taking the lim inf
of both sides of the inequality we obtain ∇u∞ · ∇u dx ≥ |∇u∞ |2 dx (4.104)
(recall that the norm in a Hilbert space is lower semi continuous for the weak topology). That is to say we have ∇u∞ · ∇(u − u∞ ) dx ≥ 0. (4.105)
From (4.15) and (4.90) we also have ∇u · ∇(u∞ − u) dx ≥ 0.
(4.106)
Thus, summing up (4.105), (4.106) it comes |∇(u − u∞ )|2 dx ≤ 0,
(4.107)
i.e., u∞ = u. Step 4: End of the proof. We have shown that un u in
H01 ().
(4.108)
To obtain the strong convergence, we go back to (4.103). Taking the lim sup we obtain 2 |∇un | dx ≤ |∇u|2 dx. lim sup n→+∞
Since – due to the lower semicontinuity of the norm for the weak topology 2 |∇un | dx ≥ |∇u|2 dx lim inf n→+∞
we have
lim
n→+∞
|∇un |2 dx =
|∇u|2 dx
and the strong convergence of un towards u follows. This completes the proof of the lemma.
56
Michel Chipot
Then we can show: Theorem 4.1. The function e = e(ζ, h) defined by (4.16) is continuous on r × R. Proof. Let (ζn , hn ) ∈ r × R be a sequence of points such that (4.74) holds. Due to Lemma 4.4 we have in H01 ().
un = u(ζn , hn ) → u = u(ζ, h) Thus, it holds 1 e(ζn , hn ) = 2
1 |∇un | dx − Ghn → e(ζ, h) = 2
2
|∇u|2 dx − Gh.
This completes the proof of the theorem. In addition to the continuity of e = e(ζ, h) we can show: Proposition 4.2. Let e be the function defined by (4.16). Let ν be a unit vector of R2 . Then, for any (ζ, h) ∈ r × R the function e is differentiable in the direction ν and we have e(ζ + λν, h) − e(ζ, h) ∂ψ ∂e (ζ, h) = lim =− (x)ψ(x) dx (4.109) λ→0 ∂ν λ ∂ν where denotes the coincidence set defined by (4.30). Proof. We give here a formal proof. Rigorous arguments can be found in [4]. Suppose that we can differentiate under the integral in (4.16), then we have: ∂u ∂e (ζ, h) = dx. (4.110) ∇u · ∇ ∂ν ∂ν For any (ζ, h) ∈ r × R it holds that u(ζ, h)(x) = u(x) = 0
on ∂.
Thus, for any ν and λ small enough such that (ζ + λh, h) ∈ r × R we have u(ζ + λν, h) − u(ζ, h) =0 λ Letting λ → 0 we obtain ∂u =0 ∂ν Integrating by parts in (4.110) we get ∂e (ζ, h) = − ∂ν
on ∂.
on ∂. u
∂u dx ∂ν
(4.111)
Elastic deformations of membranes and wires
57
when of course all the above makes sense. Applying now (4.60) it comes ∂ψ ∂e (ζ, h) = − dx ψ ∂ν ∂ν which completes the proof of the proposition. Proposition 4.2 will allow us to transform the minimization problem (4.17) to a minimization problem on a compact subset of r × R. In addition we will need Proposition 4.3. It holds that lim
|h|→+∞
e(ζ, h) = +∞
(4.112)
uniformly in ζ ∈ r . Proof. (a) When h ≤ −r it holds that e(ζ, h) = −Gh
(4.113)
and the result is clear. (b) When h → +∞, from the Poincaré inequality we derive for some constant C independent of (ζ, h) 2 |u(x)| dx − Gh ≥ C |u(x)|2 dx − Gh. (4.114) e(ζ, h) ≥ C
B(ζ,r)
Since u ∈ K(ζ, h) we have for h ≥ 0: u(x) ≥ ψ ζ h (x) ≥ h
on B(ζ, r)
and (4.114) becomes e(ζ, h) ≥ Cπ r 2 h2 − Gh.
(4.115)
Then letting h → +∞ the result follows. This completes the proof of the proposition. In order to reduce the minimization of e to a minimization on a compact subset of r × R we have to show that the ball is pushed away from the boundary of when trying to minimize its energy. Considering a direction S passing through the center of the ball it is clear with the notation of Figure 25 that if the part of the membrane below S is smaller compared to the part above, the membrane will be stiff there and thus should push the ball away. This is what we would like to express mathematically in the lemma below. Lemma 4.5. Let ζ ∈ r and ν be a unit vector. Denote by R the reflection in the axis S going through ζ and orthogonal to ν (see Figure 25). Assume that h > −r and set − = { x ∈ | (x − ζ ) · ν < 0 }, + = { x ∈ | (x − ζ ) · ν > 0 },
(4.116)
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Michel Chipot
σ γ +
+
−
ζ
ν
S
−
Figure 25. The membrane is pushing the ball inside
− = ∩ − , + = ∩ + .
(4.117)
(In (4.116) we denoted by a dot the usual scalar product in R2 .) If R(− ) +
(4.118)
+ R(− )
(4.119)
∂e (ζ, h) < 0. ∂ν
(4.120)
then it holds that
and moreover
Proof. On R(− ), if u is the solution to (4.15), we define Ru by Ru(x) = u(Rx).
(4.121)
E = R(− ) \ R(− ).
(4.122)
Consider now the open subset
Its boundary is divided in three parts that we denote by S, γ , σ – see Figure 25. We know – see (4.60) – that it holds that −u = 0
in − \ −
(4.123)
and −u ≥ 0
in .
(4.124)
Since h > −r, we cannot have u ≡ 0 and thus since u ≥ 0 (see (4.25)), by the strong maximum principle, it holds that u>0
in .
(4.125)
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Elastic deformations of membranes and wires
Since the Laplace operator is invariant by reflection we have −(Ru − u) = −Ru + u = −u(Rx) + u(x) = u(x) ≤ 0 by (4.122)–(4.124). Moreover on the boundary of E it holds that on σ, 0 − u(x) < 0 (Ru − u)(x) = u(Rx) − u(x) = ψ(x) − u(x) ≤ 0 on γ , u(x) − u(x) = 0 on S,
in E, (4.126)
(4.127)
(recall that ψ(Rx) = ψ(x)). Thus, by the strong maximum principle, it holds that Ru − u < 0
in E.
(4.128)
Let us show now that (4.119) holds. For that let us consider x ∈ + . We have u(x) = ψ(x)
⇒
u(Rx) ≥ ψ(Rx) = ψ(x) = u(x)
⇒
Ru − u(x) ≥ 0.
By (4.128) this implies that x ∈ / E. Thus x ∈ R(− ) and we have + ⊂ R(− ).
(4.129)
If we had + = R(− ) then we would have Ru − u = 0
on γ
and also ∇Ru = ∇Rψ = ∇ψ = ∇u
on
(recall that u is C 1 ). But this contradicts the Hopf maximum principle, and thus (4.119) holds. We now turn to (4.120). Using (4.109) we have ∂ψ ∂ψ ∂e (ζ, h) = − ψ dx − ψ dx ∂ν ∂ν + − ∂ν ∂ψ ∂ψ ∂ψ ψ dx − ψ dx − ψ dx =− + ∂ν R(+ ) ∂ν − \R(+ ) ∂ν (4.130) ((4.129) implies R(+ ) ⊂ − ). By (4.2) we have ν · (x − ζ ) ∂ψ (x) = ν · ∇ζ ψ(x) = ∂ν r 2 − |x − ζ |2
(4.131)
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Michel Chipot
(recall that we are differentiating in ζ ). In particular ∂ψ ν · (Rx − ζ ) ν · R(x − ζ ) ∂ψ (Rx) = (x). = =− ∂ν ∂ν r 2 − |x − ζ |2 r 2 − |x − ζ |2
(4.132)
Thus making the change of variable x → Rx in the second integral we have ∂ψ ∂ψ ∂ψ (x)ψ(x) dx = (Rx)ψ(Rx) dx = − (x)ψ(x) dx ∂ν ∂ν R(+ ) + + ∂ν (recall that ψ(Rx) = ψ(x) – see (4.51)). Thus from (4.130) we derive ∂e ∂ψ (ζ, h) = − · ψ dx. ∂ν − \R(+ ) ∂ν
(4.133)
By (4.131), (4.51) we have ∂ψ (x) < 0, ψ < 0 in − ∂ν and (4.120) follows from (4.133). This completes the proof of the lemma. We can now prove the following existence result. Theorem 4.2. Let us assume that we can find ε such that ∀ r ∈ (r, r + ε), ∀ ζ such that dist(ζ, ) = r there exists ν such that the situation of Lemma 4.5 holds.
(4.134)
Then there exists a point minimizing e on r × R and thus a solution to the problems (4.11), (4.17). Proof. For h ≤ −r we have e(ζ, h) = −Gh and thus ∂e (ζ, h) = 0. ∂ν By Proposition 4.3 to minimize e on r × R it is enough to minimize e on r × [−r, h∗ ] for some positive h∗ . Now due to (4.134) we have for h > −r ∂e (ζ, h) < 0 ∂ν for any ζ ∈ r such that dist(ζ, ) < r + ε. Clearly at such a point e cannot achieve a minimum. Thus, to minimize e on r × R, we are reduced to minimize e on { ζ ∈ | dist(ζ, ) ≥ r + ε } × [−r, h∗ ] i.e. on a compact set and the existence of a point realizing the minimum is due to the continuity of e established in Theorem 4.1. This completes the proof of the theorem.
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Elastic deformations of membranes and wires
Remark 4.2. There are many situations where (4.134) holds. For instance this is the case when is convex, smooth and r small enough. Some non-convex situations could also be handled for instance it is clear that (4.134) holds for r small enough in the case of Figure 26.
Figure 26. A non-convex case where existence holds
Remark 4.3. The property (4.120) could be used to locate the position of the equilibrium. Indeed consider for instance the convex domain described in the figure below. Then for any ν orthogonal to the axis of symmetry and as in Figure 27 we have: ∂e (ζ, h) < 0. ∂ν Thus the minimum of the energy is necessarily achieved on the axis of symmetry
Σ
ν
S
Figure 27. A convex case
of . In the case of an ellipse or a disk this shows that the minimum of the energy is necessarily achieved at the center of the ellipse or the disk. In the case of a circular membrane we can show Theorem 4.3. Suppose that = B(0, R) the open ball of center 0 and radius R. Set g = G/2π . Then the minimum of e is achieved at the single point s (4.135) ζ = 0, h = −g ln − r 2 − s 2 R
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Michel Chipot
where s is given by s 2 = {−g 2 +
4g 2 r 2 + g 4 }/2.
(4.136)
Moreover u = u(ζ, h) is given by (4.35). Proof. It follows from Remark 4.3 that ζ = 0. Next we can compute for h > ∂e (ζ, h) proceeding as we did for establishing (4.109). Indeed, differentiating −r, ∂h formally under the integral symbol we have ∂e ∂u ∇u · ∇ (ζ, h) = dx − G ∂h ∂h (this can be justified, see [4]). Since – compare to (4.111) ∂u =0 ∂h
on ,
integrating by parts we get ∂e ∂u ∂ψ (ζ, h) = − dx − G = − dx − G u ψ ∂h ∂h ∂h (4.137) ∂u ψ dx − G = − u dx − G = − dσ (x) − G = ∂n where ∂u ∂n denotes the outside normal derivative to , dσ (x) the superficial measure on . Now, clearly for any h the solution u = u(0, h) is given by (4.34), (4.35) where we have set ζ = 0. To determine h we remark that the minimum of e can be achieved only when (4.137) vanishes – that is to say when −2π Ru (R) = G Using (4.35) we derive that s2 = g r 2 − s2
−Ru (R) = g.
(4.138)
s 4 + g2 s 2 − g2 r 2 = 0
(4.139)
⇐⇒
⇐⇒
which clearly implies (4.136). To get (4.135) we remark that by (4.138) u = −g ln if ≥ s. R
(4.140)
For = s we obtain – since u = ψ ζ h : s −g ln = h + r 2 − s 2 R which completes the proof of the theorem. Remark 4.4. Note that s is independent of R and that it holds that lim h = +∞.
R→+∞
(4.141)
Elastic deformations of membranes and wires
63
Remark 4.5. One can compute the value of the energy at the minimum. Indeed we have 1 1 1 2 |∇u| dx − Gh = −uu dx − Gh = − ψψ dx − Gh. e(ζ, h) = 2 2 2 Since ψ is given by (4.2) it comes using polar coordinates 1 ψ{h + r 2 − 2 } dx − Gh e(ζ, h) = − 2 1 1 2 − 2r 2 =− u · h − Gh − d dθ 2 2 r 2 − 2 (recall (4.51)). Using the Green formula again we derive – see (4.137) s 2 ∂u − 2r 2 h d dσ (x) − Gh − π e(ζ, h) = − 2 2 2 ∂n 0 r − s 2 Gh − 2r 2 −π =− · d 2 2 2 0 r − 2 Gh π s ξ − 2r 2 =− − dξ 2 2 0 r2 − ξ 2 Gh π s r2 =− − dξ −1 − 2 2 2 0 r −ξ 2 Gh π − −ξ + r 2 ln(r 2 − ξ ) |s0 =− 2 2 Gh π 2 =− + {s − r 2 ln(1 − s 2 /r 2 )}. 2 2 It is clear from this formula and Remark 4.4 that lim e(ζ, h) = −∞.
R→+∞
(4.142)
(4.143)
4.3 Uniqueness issue In this section we would like to show that the minimum of e in r ×R is not necessarily unique. For that we consider the domain = {(−S, S) × (−1, 1)} ∪ B(−S, R) ∪ B(S, R)
(4.144)
where S is given by S =R+1 (see Figure 28). Then we can show
(4.145)
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Michel Chipot
1
−S
S
Figure 28. A domain where uniqueness fails
Theorem 4.4. Let be the domain defined by (4.144). Then, for r < 1,
R large enough,
(4.146)
the minimization problem admits at least two solutions. Proof. Suppose that (4.17) admits a unique solution. By Remark 4.3 the minimum of the energy is achieved on the axis ζ2 = 0. For symmetry reasons we have e(−ζ1 , 0, h) = e(ζ1 , 0, h) and if the minimum is unique we have necessarily ζ1 = 0. Let us denote by (uR , 0, hR ) the unique solution to (4.11). If is the square Q = (−1, 1)2 then there exists a solution (uQ , 0, hQ ) to the corresponding problem (4.11). If we extend uQ by 0 outside Q, then uQ is an admissible deformation for (4.11) associated to given by (4.144). Thus we have 1 E(uQ ; 0, hQ ) = |∇uQ |2 dx − GhQ ≥ E(uR ; 0, hR ) 2 Q (4.147) 1 1 2 2 = |∇uR | dx − GhR ≥ |∇uR | dx − GhR . 2 2 Q Since uR = 0 on (−1, 1) × {−1, 1}, by the Poincaré inequality we have for some constant c independent of R c E(uQ ; 0, hQ ) ≥ u2 dx − GhR 2 Q R (4.148) c π ≥ h2R dx − GhR = cr 2 h2R − GhR , 2 B(0,r) 2
Elastic deformations of membranes and wires
65
if hR ≥ 0 (we know by the existence part that hR ≥ −r). Thus from (4.148) we deduce that hR is bounded independently of R. But then it follows that E(uR ; 0, hR ) ≥ −GhR ≥ −M independently of R. If now we consider the deformation uB corresponding to the minimization problem (4.11) for = B(S, R + 1) and if we extend this deformation by 0 outside of B = B(S, R + 1), uB becomes an admissible deformation for the problem corresponding to defined by (4.144) and we have 1 |∇uB |2 dx − GhB ≥ −M 2 B (hB denotes the altitude of the equilibrium for = B). But since M is independent of R, by (4.143), we get a contradiction. Thus the equilibrium position of the ball cannot be unique for R large. This completes the proof of the theorem.
4.4 Some open questions It would be of course very interesting to remove the assumption (4.134) in the existence result. One could consider also a nonhomogeneous ball – compare to (2.11). Regarding uniqueness some interesting issues have yet to be addressed. In the case where is convex, we do not know if e admits at most one minimizer. For example the problem of uniqueness in the case of Figure 27 is open. Some interesting issues of non uniqueness remain also. We do not know if nonuniqueness holds in the case of Figure 26 even so it is a reasonable guess. More generally it would be interesting to find out conditions insuring non uniqueness of minimizers. Of course the above issues have to be addressed also in the case of the energy defined by (4.18) or in the case of nonhomogeneous energy of the type 1 σ (x)|∇u(x)|2 dx − Gh, 2 – see (2.10). Suppose now that = (−, ) × (−1, 1) where > 1 and that r < 1. Then, the problem of minimization (4.11) admits certainly a unique solution (for symmetry reason we must have ζ = 0, see Figure 29). Then, it would be interesting to examine what happens when → ∞. For instance does the coincidence set shrink to a curve? What is at the limit the relationship between this problem and the problem of a disk on a wire? The undeformed configuration of the wire being here (−1, 1) – cf. [10] for all these types of questions.
66
Michel Chipot 1
Figure 29. The rectangle case with one dimension large
5 The case of several balls on an elastic membrane For simplicity we restrict ourselves to the case of two balls. We denote by r1 , r2 the radii of the balls and by G1 , G2 their weights. The position of the balls will be
x2
x1
h
Figure 30. The equilibrium position of two balls rolling on a membrane
completely determined by the position of their centers C1 = (ζ1 , h1 ),
C2 = (ζ2 , h2 ),
(5.1)
R2 ,
ζi ∈ hi ∈ R, i = 1, 2. If denotes the undeformed configuration of the elastic membrane at hand we will suppose that the balls cannot touch the rim of the membrane i.e. that ζi ∈ ri = { x ∈ | dist(x, ) > ri },
i = 1, 2.
(5.2)
Moreover we will assume that these balls cannot penetrate each other in such a way that |ζ1 − ζ2 |2 + (h1 − h2 )2 ≥ (r1 + r2 )2 , | · | denotes the euclidean norm in R2 . The bottom parts of the balls will be described by the functions ψi = ψ ζi hi (x) = hi + ri2 − |x − ζi |2 , x ∈ B(ζi , ri ), i = 1, 2
(5.3)
(5.4)
67
Elastic deformations of membranes and wires
(see (4.3), (4.4) for the notation). Then, for a position ξ = (ζ1 , h1 , ζ2 , h2 ) ∈ R6
(5.5)
of the centers an admissible deformation is an element u ∈ K = K(ξ ) defined by K(ξ ) = { v ∈ H01 () | v(x) ≥ ψ ζi hi (x) a.e. x ∈ B(ζi , ri ), i = 1, 2 }.
(5.6)
Thus for ξ = (ζ1 , h1 , ζ2 , h2 ) ∈ r1 × R × r2 × R and u an admissible deformation the energy of the system is given by 1 |∇u(x)|2 dx − G1 h1 − G2 h2 . E(u; ξ ) = 2
(5.7)
(5.8)
The minimization problem that we would like to address is then to find u0 , ξ 0 such that it holds that 0 ξ = (ζ10 , h01 , ζ20 , h02 ) ∈ r1 × R × r2 × R, ξ 0 satisfies (5.3), u0 ∈ K(ξ 0 ), (5.9) E(u0 ; ξ 0 ) ≤ E(u; ξ ) ∀ ξ ∈ r1 × R × r2 × R, ξ satisfying (5.3), ∀ u ∈ K(ξ ). Arguing as in the preceding sections, we introduce u = u(ξ ) the solution to the variational inequality u ∈ K(ξ ), (5.10) ∇u · ∇(v − u) dx ≥ 0 ∀ v ∈ K(ξ ).
Then for ξ ∈ r1 × R × r2 × R, ξ satisfying (5.3), u = u(ξ ) we set 1 e(ξ ) = |∇u(x)|2 dx − G1 h1 − G2 h2 . 2
(5.11)
Then, clearly, the problem (5.9) is equivalent to minimize e on a subset of r1 × R × r2 × R, i.e. to find ξ 0 such that ξ 0 ∈ r1 × R × r2 × R, ξ 0 satisfies (5.3), (5.12) e(ξ 0 ) ≤ e(ξ ) ∀ ξ ∈ r1 × R × r2 × R, ξ satisfying (5.3). This problem is in general open. Let us collect here the results that are available with the techniques we have developed above. First let us denote by ui = u(ζi , hi ) the solution to (4.15). Then we have
(5.13)
68
Michel Chipot
Proposition 5.1. Let ξ = (ζ1 , h1 , ζ2 , h2 ) ∈ r1 × R × r2 × R and u = u(ξ ) be the solution to (5.10). Then it holds that u(ξ ) ≥ ui = u(ζi , hi ) in , i = 1, 2.
(5.14)
Proof. It is enough to show that (ui − u)+ = 0,
i = 1, 2.
(5.15)
For that remark that u + (ui − u)+ ∈ K(ξ ).
(5.16)
∇u · ∇(ui − u)+ dx ≥ 0.
(5.17)
Thus from (5.10) we deduce
Next since ui − (ui − u)+ = u or ui we have clearly ui − (ui − u)+ ∈ K(ζi , hi ) and from (4.15) we obtain
−
∇ui · ∇(ui − u)+ dx ≥ 0.
(5.18)
Adding (5.17) and (5.18) we get + |∇(ui − u) | dx = ∇(ui − u) · ∇(ui − u)+ ≤ 0
and (5.15) follows. This completes the proof of the proposition. As a consequence we can show Proposition 5.2. Let ξ = (ζ1 , h1 , ζ2 , h2 ) ∈ r1 × R × r2 × R. Let u = u(ξ ) the solution to (5.10). Then it holds that u ∈ W 1,∞ ().
(5.19)
The coincidence set = { x ∈ B(ζ1 , r1 ) ∪ B(ζ2 , r2 ) | u(x) = ψ1 (x) or u(x) = ψ2 (x) }
(5.20)
is a closed subset of such that for some ε > 0 it holds that ⊂ B(ζ1 , r1 − ε) ∪ B(ζ2 , r2 − ε).
(5.21)
Elastic deformations of membranes and wires
69
Moreover we have u ≥ 0 in , u = 0 in \ .
(5.22) (5.23)
Proof. This is an easy consequence of Proposition 4.1 and 5.1. In arguing as in Theorem 3.1 we can prove Theorem 5.1. The function e is continuous at any point of r1 × R × r2 × R satisfying (5.3). At this stage many open questions remain. We refer the reader to [16]. Indeed the system formed by two balls has not the symmetry of a single ball. For instance, if we can find an analogue of the differentiation formula (4.109), to use it as in Lemma 4.5 is a different story. So, it is not so easy to reduce the minimization problem (5.12) to a minimization on a compact subset of R6 – even in the case where is a circular membrane and the balls identical. Considering – see Lemma 4.5 – an axis S going through the center of the balls one can show that one decreases the energy when S moves toward the center of the membrane (see [16], [12]). However, we are not yet able to conclude that (5.12) achieves a minimum. The nature of this equilibrium is more mysterious. For instance do we have for small weights the two balls sitting at the same level in the middle of the membrane? Do they tilt when the weight increases (see Theorem 3.2)? Could we have the situation of Figure 17 when the balls are not identical and the small ball becomes heavy? Could we show then that the equilibrium is achieved when the two balls are centered in the middle of the membrane and that it is unique? The issue of uniqueness of a solution is, in the case of two balls, more subtle also than in the case of a single ball. For instance in the case of a circular membrane uniqueness should hold up to a rotation centered at the center of the membrane. Thus even for identical balls and simple domain the problem of existence, uniqueness and nature of the equilibrium is far from being trivial. To convince the reader further let us consider the domain introduced in the preceding section and given by = (−, ) × (−1, 1),
≥ 1.
(5.24)
Let us drop two identical balls on this membrane. What is the equilibrium position(s)? What is going to happen for = 1 i.e. for a square membrane? What will follow if we let > 1? In what direction will the balls be lined up? What will happen when → +∞. The questions are very simple. However, to be tackled they need some skills. We hope the reader will be eager to attack them and develop new elegant techniques that can be used in the wide topic of elliptic problems and calculus of variations. Acknowledgements: This work has been supported by the Swiss National Science Foundation under the contract 20-67618.02. This help is greatly acknowledged.
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Michel Chipot
References [1]
A. Aissani, Thesis, University of Metz, 2000.
[2]
A. Aissani, M. Chipot, On the equilibrium position of a square on an elastic wire, J. Convex Anal. 8 (2001), 171–195.
[3]
A. Aissani, M. Chipot, S. Fouad, On the deformation of an elastic wire by one or two heavy disks, Arch. Math. 76 (2001), 467–480.
[4]
J. Bemelmans, M. Chipot, On a variational problem for an elastic membrane supporting a heavy ball, Calc. Var. Partial Differential Equations 3 (1995), 447–473.
[5]
H. Brezis, Problèmes unilatéraux, J. Math. Pures Appl. 51 (1972), 1–168.
[6]
H. Brezis, Analyse fonctionnelle, Masson, Paris 1983.
[7]
H. Brezis, D. Kinderlehrer: The smoothness of solutions to nonlinear variational inequalities, Indiana Univ. Math. J. 23 (1974), 831–844.
[8]
M. Chipot, Variational inequalities and flows in porous media, Appl. Math. Sci. 52, Springer-Verlag, New York 1984.
[9]
M. Chipot, Elements of Nonlinear Analysis, Birkhäuser, 2000.
[10] M. Chipot, goes to plus infinity, Birkhäuser, 2001. [11] M. Chipot, On the equilibrium positions of two heavy disks supported by an elastic wire, to appear. [12] M. Chipot, in preparation. [13] R. Dautray, J. L. Lions, Mathematical analysis and numerical methods for science and technology, Springer-Verlag, 1982. [14] C. M. Elliott, A. Friedman, The contact set of a rigid body partially supported by a membrane, Nonlinear Anal. 10 (1986), 251–276. [15] A. Friedman, Variational Principles and Free-Boundary Problems, Wiley, NewYork 1982. [16] S. Fouad, Thesis, University of Metz, 1999. [17] D. Gilbarg, N. S. Trudinger, Elliptic Partial Differential Equations of Second Order, Grundlehren Math. Wiss. 224, Springer-Verlag, Berlin, Heidelberg, New York 1977. [18] D. Kinderlehrer, G. Stampacchia, An introduction to variational inequalities and their applications, Acadademic Press, 1980. [19] J. F. Rodrigues, Obstacle problems in mathematical physics, North-Holland Math. Stud. 134, North Holland, Amsterdam 1987.
Scattering, spreading, and localization of an acoustic pulse by a random medium Josselin Garnier Laboratoire de Statistique et Probabilités Université Paul Sabatier 118 Route de Narbonne, 31062 Toulouse Cedex 4, France email:
[email protected]
Abstract. Random media have material properties with such complicated spatial variations that they can only be described statistically. When looking at waves propagating in these media, we can only expect in general a statistical description of the wave. But sometimes there exists a deterministic result: the wave dynamics only depends on the statistics of the medium, and not on the particular realization of the medium. Such a phenomenon arises when the different scales present in the problem (wavelength, correlation length, and propagation distance) can be separated. In this lecture we restrict ourselves to one-dimensional wave problems that arise naturally in acoustics and geophysics.
1 Introduction Wave propagation in linear random media has been studied for a long time by perturbation techniques when the random inhomogeneities are small. One finds that the mean amplitude decreases with distance traveled, since wave energy is converted to incoherent fluctuations. The fluctuating part of the field intensity is calculated approximatively from a transport equation, a linear radiative transport equation. This theory is well-known [21], although a complete mathematical theory is still lacking (for the most recent developments, see for instance [14]). However this theory is known to be false in one-dimensional random media. This was first noted by Anderson [1], who claimed that random inhomogeneities trap wave energy in a finite region and do not allow it to spread as it would normally. This is the so-called wave localization phenomenon. It was first proved mathematically in [20]. Extensions and generalizations follow these pioneer works so that the problem is now well understood [10]. The mathematical statement is that the spectrum of the reduced wave equation is pure point with exponentially decaying eigenfunctions. However the authors did not give quantitative information associated with the wave propagation as no exact solution is
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Josselin Garnier
available. In this lecture we are not interested in the study of the strongest form of Anderson localization. We actually address the simplest form of this problem: the wave transmission through a slab of random medium. It is now well-known that the transmission of the slab tends exponentially to zero as the length of the slab tends to infinity. Furstenberg first treated discrete versions of the transmission problem [18], and finally Kotani gave a proof of this result with minimal hypotheses [25]. The connection between the exponential decay of the transmission and the Anderson localization phenomenon is clarified in [13]. Once again, these works deal with qualitative properties. Quantitative information can be obtained only for some asymptotic limits: large or small wavenumbers, large or small variances of the fluctuations of the parameters of the medium, etc. A lot of work was devoted to the quantitative analysis of the transmission problem, in particular by Rytov, Tatarski, Klyatskin [23], and by Papanicolaou and its co-workers [24]. The tools for the quantitative analysis are limit theorems for stochastic equations developed by Khasminskii [22] and by Kushner [27]. There are three basic length scales in wave propagation phenomena: the typical wavelength λ, the typical propagation distance L, and the typical size of the inhomogeneities lc . There is also a typical order of magnitude that characterizes the standard deviation of the dimensionless fluctuations of the parameters of the medium. It is not always so easy to identify the scale lc , but we may think of lc as a typical correlation length. When the standard deviation of the relative fluctuations is small ε 1, then the most effective interaction of the waves with the random medium will occur when lc ∼ λ, that is, the wavelength is comparable to the correlation length. Such an interaction will be observable when the propagation distance L is large (L ∼ λε−2 ). This is the typical configuration in optics and in optical fibers. Throughout this lecture we shall consider scalings arising in acoustics and geophysics. The main differences with optics are that the fluctuations are not small, but the typical wavelength of the pulse λ ∼ 150 m is small compared to the probing depth L ∼ 10 − 50 km, but large compared to the correlation length lc ∼ 2 − 3 m [38]. Accordingly we shall introduce a small parameter 0 < ε 1 and consider lc ∼ ε2 , λ ∼ ε, and L ∼ 1. The parameter ε is the ratio of the typical wavelength to propagation depth, as well as the ratio of correlation length to wavelength. This is a particularly interesting scaling limit mathematically because it is a high frequency limit with respect to the large scale variations of the medium, but it is a low frequency limit with respect to the fluctuations, whose effect acquires a canonical form independent of details. We shall study the asymptotic behavior of the scattered wave in the framework introduced by Papanicolaou based on the separation of these scales. The lecture is organized as follows. In Section 2 we present the method of averaging for stochastic processes that is an extension of the law of large numbers for the sums of independent random variables. These results provide the tools for the effective medium theory developed in Section 3. We give a review of the properties of Markov processes in Section 4. We focus our attention to ergodic properties, and we propose limit theorems for ordinary differential equations driven by Markov processes that
Scattering, spreading, and localization of an acoustic pulse by a random medium
73
are applied in the following sections. Section 5 is devoted to the O’Doherty–Anstey problem, that is to say the spreading of a pulse traveling through a random medium. We compute the localization length of a monochromatic pulse in Section 6. We finally study the exponential localization phenomenon for a pulse in Section 7 and show that it is a self-averaging process.
2 Averages of stochastic processes We begin by a brief review of the two main limit theorems for sums of independent random variables. • The Law of Large Numbers: If (Xi )i∈N is a sequence of independent identically distributed R-valued random variables, with E[|X1 |] < ∞, then the normalized partial sums 1 Xk n n
Sn =
k=1
converge to the statistical average X¯ = E[X1 ] with probability one (write a.s. for almost surely). • The Central Limit Theorem: If (Xi )i∈N is a sequence of independent and identically distributed R-valued random variables, with X¯ = E[X1 ] and E[X12 ] < ∞, then the normalized partial sums 1 ¯ (Xk − X) S˜n = √ n n
k=1
converge in distribution to a Gaussian random variable with mean 0 and variance ¯ 2 ]. This distribution is denoted by N (0, σ 2 ). The above assertion σ 2 = E[(X1 − X) means that, for any continuous bounded function f , 1 n→∞ E[f (S˜n )] −−−−→ √ 2π σ 2
x2 f (x) exp − 2 dx 2σ −∞ ∞
or, for any interval I ⊂ R, n→∞ 1 P S˜n ∈ I −−−−→ √ 2π σ 2
I
x2 exp − 2 dx. 2σ
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2.1 A toy model Let us consider a particle moving on the line R. Assume that it is driven by a random velocity field εF (t) where ε is a small dimensionless parameter, F is stepwise constant F (t) =
∞
Fi 1[i−1,i[ (t),
i=1
and Fi are independent and identically distributed random variables that are bounded, E[Fi ] = F¯ and E[(Fi − F¯ )2 ] = σ 2 . The position of the particle starting from 0 at time t = 0 is X(t) = ε
t
F (s) ds. 0
ε→0
Clearly X(t) −−→ 0. The problem consists in finding the adequate asymptotic, that is to say the scale of time which leads to a macroscopic motion of the particle. Regime of the Law of Large Numbers. At the scale t → t/ε, Xε (t) := X as: X ε (t) = ε
t ε
reads
t ε
F (s) ds t t ε ε Fi + ε F (s) ds =ε 0
t ε
i=1
t
ε 1 t t t × t − F t . Fi + ε = ε ε ε ε ε ε i=1 ↓ a.s. ↓ a.s. ↓ t 0 E[F ] = F¯ The convergence of
1t
ε
t
F is imposed by the Law of Large Numbers. Thus i i=1 ε
the motion of the particle is ballistic in the sense that it has constant effective velocity: ε→0 Xε (t) −−−→ F¯ t.
However, in the case F¯ = 0, the random velocity field seems to have no effect, which means that we have to consider a different scaling.
Scattering, spreading, and localization of an acoustic pulse by a random medium
75
2 ε ¯ Regime t of the Central Limit Theorem F = 0. At the scale t → t/ε , X (t) = X ε2 reads as:
X (t) = ε ε
t ε2
0
t 2
F (s) ds
ε Fi = ε +ε
i=1
t = ε ε2 ↓ √ t
t ε2 t ε2
F (s) ds
t ε2
1 t t × Fi + ε 2 − 2 F t . ε ε ε2 t i=1 a.s. ↓ ε2 0 distribution ↓ N (0, σ 2 )
The convergence of 1t
t ε2
i=1
Fi is imposed by the Central Limit Theorem. Xε (t)
ε
converges in distribution as ε → 0 to the Gaussian statistics N (0, σ 2 t). The motion of the particle in this regime is diffusive.
2.2 Stationary and ergodic processes A stochastic process (F (t))t≥0 is an application from some probability space to a functional space. This means that for any fixed time t the quantity F (t) is a random variable with values in the state space E. E is a locally compact Polish space. We shall be often concerned with the case E = Rd . Furthermore we shall only consider configurations where the functional space is either the set of the continuous functions C([0, ∞), E) equipped with the topology associated to the sup norm over the compact sets or the set of the càd-làg functions (right-continuous functions with left hand limits) equipped with the Skorohod topology. This means that the realizations of the random process are either continuous E-valued functions or càd-làg E-valued functions. (F (t))t∈R+ is a stationary stochastic process if the statistics of the process is invariant to a shift θt in the time origin: θt F
distribution
=
F
where the shift operator is θt F ( · ) := F (t + · ).
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It is a statistical steady state. Let us consider a stationary process such that E[|F (t)|] < ∞. We set F¯ = E[F (t)]. The ergodic theorem claims that the ensemble average can be substituted to the time average under the so-called ergodic hypothesis: If I is a subset of the phase space such that θt I = I a.s. for all t, then P(I ) = 0 or 1. Theorem 2.1. If F satisfies the ergodic hypothesis, then 1 T T →∞ F (t) dt −−−−→ F¯ P a.s. T 0 The ergodic hypothesis requires that the orbit (F (t))t visits all of phase space (difficult to state). This hypothesis can be made more transparent by looking at a non-ergodic process. Example 2.2. Let F1 and F2 two ergodic processes, and denote F¯j = E[Fj (t)], j = 1, 2. Now flip a coin independently of F1 and F2 , whose result is χ = 1 with probability 1/2 and 0 with probability 1/2. Let F (t) = χ F1 (t) + (1 − χ )F2 (t), which is a stationary process with mean F¯ = 21 (F¯1 + F¯2 ). The time-averaged process satisfies T T 1 T 1 1 F (t) dt = χ F1 (t) dt + (1 − χ ) F2 (t) dt T 0 T 0 T 0 T →∞
−−−−→ χ F¯1 + (1 − χ )F¯2 which is a random limit different from F¯ . This limit depends on χ because F has been trapped in a part of phase space.
2.3 Mean square theory Let (F (t))t≥0 be a stationary process, E[F 2 ] < ∞. We introduce the autocorrelation function
R(τ ) = E (F (t) − F¯ )(F (t + τ ) − F¯ ) . By stationarity, R is an even function
R(−τ ) = E (F (t) − F¯ )(F (t − τ ) − F¯ )
= E (F (t + τ ) − F¯ )(F (t ) − F¯ ) = R(τ ). By Cauchy–Schwarz inequality, R reaches its maximum at 0:
1/2
1/2 E (F (t + τ ) − F¯ )2 = R(0) = var(F ). R(τ ) ≤ E (F (t) − F¯ )2 T ∞ Proposition 2.3. Assume that 0 τ |R(τ )| dτ < ∞. Let S(T ) = T1 0 F (t) dt. Then
T →∞ E (S(T ) − F¯ )2 −−−−→ 0,
Scattering, spreading, and localization of an acoustic pulse by a random medium
more exactly
T →∞ T E (S(T ) − F¯ )2 −−−−→ 2
77
∞
R(τ ) dτ. 0
∞ One should interpret the condition 0 τ |R(τ )| dτ < ∞ as “the autocorrelation function R(τ ) decays to 0 sufficiently fast as τ → ∞.” This hypothesis is a mean square version of mixing: F (t) and F (t + τ ) are approximatively independent for long time lags τ . Mixing substitutes for independence in the law of large numbers. In the case of piecewise constant processes: F (s) =
fk 1s∈[Lk ,Lk+1 ) ,
L0 = 0,
Lk =
k∈N
k
lj
j =1
with independent and identically distributed random variables fk , and independent exponential random variables lk with mean 1 we have R(τ ) = var(f1 ) exp(−τ ). Proof. The proof consists in a straightforward calculation.
T T
1 ¯ ¯ dt dt (F (t ) − F )(F (t ) − F ) E (S(T ) − F¯ )2 = E 1 2 1 2 T2 0 0 T t1 2 = 2 dt1 dt2 R(t1 − t2 ) (by symmetry) T 0 0 T −τ T 2 = 2 dτ dhR(τ ) (with τ = t1 − t2 , h = t2 ) T 0 0 T 2 = 2 dτ (T − τ )R(τ ) T 0 T 2 T 2 = R(τ ) dτ − 2 τ R(τ ) dτ. T 0 T 0 Note that the L2 (P) convergence implies convergence in probability as the limit is deterministic. Indeed, by Chebychev inequality, for any δ > 0,
E (S(T ) − F¯ )2 T →∞ −−−−→ 0. P |S(T ) − F¯ | ≥ δ ≤ δ2
2.4 The method of averaging Let us revisit our toy model and consider a more general model for the velocity field. Let 0 < ε 1 be a small parameter and Xε satisfies dX ε t =F , Xε (0) = 0 dt ε
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where F is a stationary process with a decaying autocorrelation function such that τ |R(τ )|dτ < ∞. F (t) is a process on its own natural time scale. F (t/ε) is the t T speeded-up process. The solution is Xε (t) = 0 F (s/ε) ds = t T1 0 F (s) ds where T = t/ε → ∞ as ε → 0. So Xε (t) → F¯ t as ε → 0, or else Xε → X¯ solution of: d X¯ ¯ = 0) = 0. = F¯ , X(t dt We can generalize this result to more general configurations. Proposition 2.4 ([22, Khaminskii]). Assume that, for each fixed value of x ∈ Rd , F (t, x) is a stochastic Rd -valued process in t. Assume also that there exists a deterministic function F¯ (x) such that 1 t0 +T F¯ (x) = lim E[F (t, x)] dt T →∞ T t0 with the limit independent of t0 . Let ε > 0 and X ε be the solution of t dX ε ε =F , X , Xε (0) = 0. dt ε Define X¯ as the solution of d X¯ ¯ = F¯ (X), dt
¯ X(0) = 0.
Then under wide hypotheses on F and F¯ , we have for any T :
ε→0 ¯ sup E |X ε (t) − X(t)| −−−→ 0. t∈[0,T ]
Proof. The proof requires only elementary tools under the hypotheses: 1 T T →∞ 1) F is stationary and E F (t, x) dt − F¯ −−−−→ 0 (to check this, we T 0 can use the mean square theory since E[|Y |] ≤ E[Y 2 ]). 2) For any t F (t, · ) and F¯ ( · ) are uniformly Lipschitz with a non-random Lipschitz constant c. 3) For any compact subset K ⊂ Rd , supt∈R+ ,x∈K |F (t, x)| + |F¯ (x)| < ∞. We have t t s ε ε ¯ ¯ F , X (s) ds, X(t) = F¯ (X(s)) ds X (t) = ε 0 0 so the difference reads ¯ = X (t) − X(t) ε
t s s ε ¯ , X (s) − F , X(s) ds + g ε (t) F ε ε 0
Scattering, spreading, and localization of an acoustic pulse by a random medium
where g ε (t) :=
t 0
F
s
¯
ε , X(s)
79
¯ − F¯ (X(s)) ds. Taking the modulus,
t s s ε ¯ , X (s) − F , X(s) F ds + |g ε (t)| ε ε 0 t ¯ |Xε (s) − X(s)| ds + |g ε (t)|. ≤c
¯ ≤ |X (t) − X(t)| ε
0
Taking the expectation and applying Gronwall’s inequality,
¯ ≤ ect sup E |g ε (s)| . E |X ε (t) − X(t)| s∈[0,t]
It remains to show that the last term goes to 0 as ε → 0. Let δ > 0 [t/δ]−1 (k+1)δ
s ¯ ¯ , X(s)) − F¯ (X(s) ds F ε kδ k=0 t s ¯ ¯ , X(s)) − F¯ (X(s) ds. F + ε δ[t/δ]
g ε (t) =
¯ Denote MT = supt∈[0,T ] |X(t)|. As KT = supx∈[−MT ,MT ],t∈R+ |F (t, x)| + |F¯ (x)| is finite, the last term of the right-hand side is bounded by 2KT δ. Furthermore F is Lipschitz, so that s s ≤ cKT |s − kδ|. ¯ ¯ ¯ ¯ , X(s) −F , X(kδ) − X(kδ) ≤ c X(s) F ε ε We have similarly: F¯ (X(s)) ≤ cKT |s − kδ|. ¯ ¯ − F¯ (X(kδ)) Thus [t/δ]−1 |g (t)| ≤ ε
(k+1)δ
kδ
k=0
+ 2cKT
s ¯ ¯ ¯ F , X(kδ)) − F (X(kδ) ds ε
[t/δ]−1 (k+1)δ k=0
≤ε
[t/δ]−1 (k+1)δ/ε
k=0
kδ/ε
(s − kδ) ds + 2KT δ
kδ
¯ ¯ F (s, X(kδ)) − F¯ (X(kδ)) ds + 2KT (ct + 1)δ.
Taking the expectation and the supremum: sup E[|g ε (t)|] ≤ δ
t∈[0,T ]
[T /δ] k=0
ε E δ
(k+1)δ/ε
kδ/ε
+ 2KT (cT + 1)δ.
¯ ¯ F (s, X(kδ)) − F¯ (X(kδ)) ds
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Josselin Garnier
Taking the limit ε → 0: lim sup sup E[|g ε (t)|] ≤ 2KT (cT + 1)δ. ε→0
t∈[0,T ]
Letting δ → 0 completes the proof.
3 Effective medium theory This section is devoted to the computation of the effective velocity of an acoustic pulse traveling through a random medium. The acoustic pressure p and speed u satisfy the continuity and momentum equations ∂u ∂p + =0 ∂t ∂x ∂u ∂p +κ =0 ∂t ∂x
ρ
(3.1) (3.2)
where ρ is the material density and κ is the bulk modulus of the medium. Assume that ρ = ρ(x/ε) and κ = κ(x/ε) are stationary random functions of position on spatial scale of order ε, 0 < ε 1. They are piecewise smooth and also uniformly bounded such that ρ ∞ ≤ C, ρ −1 ∞ ≤ C, κ ∞ ≤ C, and κ −1 ∞ ≤ C a.s. Assume conditions so that the system admits a solution, for instance a Dirichlet condition at x = 0 of the type u(x = 0, t) = f (t) and p(x = 0, t) = g(t) with f, g ∈ L2 . We also assume for simplicity that the Fourier transforms fˆ and gˆ decay faster than any exponential (say for instance that f and g are Gaussian pulses). Note that the fluctuations in the sound speed are not assumed to be small. This corresponds to typical situations in acoustics and geophysics. The estimation of the vertical correlation length of the inhomogeneities in the lithosphere from well-log data is considered in [38]. They found that 2–3 m is a reasonable estimate of the correlation length of the fluctuations in sound speed. The typical pulse width is about 50 ms or, with a speed of 3 km/s, the typical wavelength is 150 m. So ε = 10−2 is our framework. We perform a Fourier analysis with respect to t. So we Fourier transform u and p ˆ ω)e−iωt dω u(x, t) = u(x, ˆ ω)e−iωt dω, p(x, t) = p(x, so that we get a system of ordinary differential equations (ODE): x dX ε =F , Xε , dx ε
81
Scattering, spreading, and localization of an acoustic pulse by a random medium
where
X = ε
pˆ uˆ
,
F (x, X) = M(x)X,
M(x) = iω
0 1 κ(x)
ρ(x) 0
.
Note first that a straightforward estimate shows that |Xε (ω, z)| ≤ |X0 (ω)| exp(Cωz) uniformly with respect to z. Apply the method of averaging. We get that X ε (x, ω) ¯ ω) solution of converges in L1 (P) to X(x,
−1 d X¯ 0 ρ¯ ¯ = M¯ X, M¯ = iω . , ρ¯ = E[ρ], κ¯ = E κ −1 1 0 dx κ¯ We now come back to the time domain. We introduce the deterministic “effective medium” with parameters ρ, ¯ κ¯ and the solution (p, ¯ u) ¯ of ∂ p¯ ∂ u¯ + =0 ∂t ∂x ∂ u¯ ∂ p¯ + κ¯ = 0. ∂t ∂x ρ¯
2
2
∂ u¯ The parameters are constant, so u¯ satisfies the closed form equation ∂∂t 2u¯ − c¯2 ∂x 2 = √ ¯ ρ. ¯ 0, which is the standard wave equation with the effective wave speed c¯ = κ/ ¯ t): We now compare uε (x, t) with u(x,
ε
−iωt ε ˆ¯ E |u (x, t) − u(x, ¯ t)| = E e (uˆ (x, ω) − u(x, ω)) dω
ˆ¯ ω)| dω. ≤ E |uˆ ε (x, ω) − u(x,
The dominated convergence theorem then gives the convergence in L1 (P) of uε to u¯ in the time domain. Thus the effective speed of the acoustic wave (p ε , uε ) as ε → 0 is c. ¯ Example 3.1 (Bubbles in water). Air and water are characterized by the following parameters: ρa = 1.2 103 g/m3 , κa = 1.4 108 g/s2 /m, ca = 340 m/s; ρw = 1.0 106 g/m3 , κw = 2.0 1018 g/s2 /m, cw = 1425 m/s. If we consider a pulse whose frequency content is in the range 10 Hz–30 kHz, then the wavelengths lie in the range 1 cm–100 m. The bubble sizes are much smaller, so the effective medium theory can be applied. Let us denote by φ the volume fraction of air. The averaged density and bulk modulus are 9.9 105 g/m3 if φ = 1% ρ¯ = E[ρ] = φρa + (1 − φ)ρw = 9 105 g/m3 if φ = 10%, −1 φ 1 − φ −1 1.4 1010 g/s2 /m if φ = 1% −1 = + = κ¯ = E[κ ] 1.4 109 g/s2 /m if φ = 10%. κa κw
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Josselin Garnier
Accordingly c¯ = 120 m/s if φ = 1% and c¯ = 37 m/s if φ = 10 %. The above example demonstrates that the average velocity may be much smaller than the minimum of the velocities of the medium components. However it cannot happen in such a configuration that the velocity be larger than the maximum (or the essential supremum) of the velocities of the medium components. Indeed,
1/2 E c−1 = E κ −1/2 ρ 1/2 ≤ E κ −1 E[ρ]1/2 = c¯−1 . −1 Thus c¯ ≤ E c−1 ≤ ess sup(c).
4 Markov processes 4.1 Definition and main properties A stochastic process (Xt )t≥0 with state space E is a Markov process if ∀ 0 ≤ s < t and B ∈ B(E) (the σ -algebra of Borel sets of E) P(Xt ∈ B|Xu , u ≤ s) = P(Xt ∈ B|Xs ) “the state Xs at time s contains all relevant information for calculating probabilities of future events”. The rigorous definition needs σ -algebras of measurable sets. This is a generalization to the stochastic case of the dynamical deterministic systems without memory of the type dx dt = f (t, x(t)). The distribution of Xt starting from x at time s is the transition probability P (s, x; t, dy). P(Xt ∈ B|Xs = x) = P (s, x; t, B) = y∈B
Definition 4.1. A transition probability P is a function from R+ × E × R+ × B(E) such that 1) P (s, x; t, A) is measurable in x for fixed s ∈ R+ ,t ∈ R+ , A ∈ B(E), 2) P (s, x; t, A) is a probability measure in A for fixed s ∈ R+ , t ∈ R+ , x ∈ E, 3) P satisfies the Chapman–Kolmogorov equation: P (s, x; τ, dz)P (τ, z; t, A) ∀ 0 ≤ s < τ < t. P (s, x; t, A) = E
Note that the Chapman–Kolmogorov equation can be deduced heuristically from a disintegration formula: P(Xt ∈ A|Xs = x, Xτ = z)P(Xτ ∈ dz|Xs = x) P(Xt ∈ A|Xs = x) = E
and the application of the Markov property.
Scattering, spreading, and localization of an acoustic pulse by a random medium
83
Note also that a Markov process is temporally homogeneous if the transition probability depends only on t − s: P (s, x; t, A) = P (0, x; t − s, A). We can now define the 2-parameter family of operators defined on the space of measurable bounded functions L∞ (E): P (s, x; t, dy)f (y). Ts,t f (x) = E[f (Xt )|Xs = x] = E
Proposition 4.2. 1) Tt,t = Id . 2) ∀s ≤ τ ≤ t Ts,t = Ts,τ Tτ,t . 3) Ts,t is a contraction Ts,t f ∞ ≤ f ∞ . Proof. The second point follows from the Chapman–Kolmogorov relations, and the third one: p(s, x; t, dy) f ∞ = f ∞ . |Ts,t f (x)| ≤ E
We shall consider Feller processes: for any continuous bounded function f such that lim|x|→∞ f (x) exists and equals 0 and for any t > 0 the function (h, x) → Ttt+h f (x) is continuous. This continuity property goes from the family operators (Ts,t ) to the process itself. The exact assertion claims that this is true for a modification of the process. A stochastic process X˜ is called a modification of a process X if P(Xt = X˜ t ) = 1 for all t. Proposition 4.3. Let (Xt ) be a Feller Markov process. Then (Xt ) has a modification whose realizations are càd-làg functions. From now on we shall always consider such a modification of the Markov process. The generator of a Markov process is the operator Qt := lim st
It is defined on a subset of
Tst − Id . t −s
L∞ (E).
Proposition 4.4. If u(s, · , · ) ∈ Dom(Qs ), then the function u(s, t, x) := Ts,t f (x) satisfies the Kolmogorov backward equation (KBE) ∂u + Qs u = 0, ∂s
s < t, u(s = t, t, x) = f (x).
Proof. Qs u = lim
s −I Ts−δ d
s Id − Ts−δ
u(s, t, x) = − lim Tst f (x) δ→0 δ δ t f Tst f − Ts−δ ∂u u(t, s, x) − u(t, s − δ, x) = − lim = − lim =− . δ→0 δ→0 δ δ ∂s δ→0
(4.1)
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Josselin Garnier
Application: ODE driven by a Feller Markov process. Let q be a Feller process with generator Qt and X be the solution of: dX = F (q(t), X(t)) dt where F is Lipschitz continuous and bounded. Then Y = (q, X) is a Markov process with generator: L = Qt +
d j =1
Fj (q, X)
∂ . ∂Xj
Proof. The fact that Y is Markov is straightforward. Let f be a smooth bounded test function. E [f (Y (t + h)) | Y (t) = (X, q)] − f (y) = Ah + Bh
(4.2)
where Ah = E [f (X(t + h), q(t + h)) − f (X(t), q(t + h)) | Y (t) = (X, q)] Bh = E [f (X(t), q(t + h)) − f (X(t), q(t)) | Y (t) = (X, q)] . On the one hand Ah = E [∇f (X(t), q(t + h)).(X(t + h) − X(t))
+O(|X(t + h) − X(t)|2 ) | Y (t) = (X, q)
= E [∇f (X(t), q(t + h)).F (X(t)) | Y (t) = (X, q)] h + O(h2 ) which implies (q is assumed to be Feller): Ah h→0 −−→ E [∇f (X(t), q(t)).F (X(t)) | Y (t) = (X, q)] = ∇f (X, q).F (X). h (4.3) On the other hand 1 Bh h→0 = E [f (X, q(t + h)) − f (X, q(t)) | q(t) = q] −−→ Qt f (X, q). (4.4) h h Substituting (4.3) and (4.4) into (4.2) yields the result.
4.2 Homogeneous Feller processes In this section we consider a temporally homogeneous Markov process. Its transition probability P (s, x; t, A) is then denoted by Pt−s (x, A), and the family of operators Ts,t is then of the form Ts,t = Tt−s : Tt f (x) = Pt (x, dy)f (y). E
Scattering, spreading, and localization of an acoustic pulse by a random medium
85
It is a semi-group as it satisfies Tt+s = Tt Ts . It is not a group as Tt does not possess an inverse. The Feller property then claims that Tt is strongly continuous. The distribution P x of the Markov process starting from x ∈ E is described by the probability transition. Indeed, by the Chapman–Kolmogorov equation, for any n ∈ N∗ , for any 0 < t1 < t2 < · · · < tn < ∞ and for any A1 , . . . , An ∈ B(E), we have: P x (Xt1 ∈ A1 , . . . , Xtn ∈ An ) = ... Pt1 (x, dx1 )Pt2 −t1 (x1 , dx2 ) . . . Ptn −tn−1 (xn−1 , dxn ). A1
An
We shall denote by E x the expectation with respect to P x . We have in particular E x [f (Xt )] = Tt f (x). The Markov processes have been extensively studied and classified. This classification is based upon the notions of recurrence and transience. From this classification simple conditions for the ergodicity of the process can be deduced. The main hypothesis that will insure most of the forthcoming results is that the transition probability has a positive, continuous density function: Hypothesis D. There exists a Borel measure µ on E supported by E, and a strictly positive function pt (x, y) continuous in (t, x, y) ∈ R+∗ × E 2 such that the transition probability Pt (x, dy) equals pt (x, y)µ(dy). Let L∞ (E) be the set of all bounded measurable functions on E and let Cb (E) be the set of all bounded continuous functions on E. For each t > 0, Tt maps L∞ (E) into Cb (E). Indeed if f ∈ L∞ (E) satisfies 0 ≤ f ≤ 1, both Tt f and Tt (1 − f ) are lower semicontinuous and their sum is constant 1. Then Tt f has to be continuous. This can be generalized to any f ∈ L∞ (E). The semigroup with this property is called a strong Feller semigroup. 4.2.1 Recurrent and transient properties. The Feller process is called recurrent (in the sense of Harris) if ∞ Px 1A (Xt ) dt = ∞ = 1 0
for every x ∈ E and A ∈ B(E) such that µ(A) > 0. This means that the time spent by the process in any subset is infinite, or else that it comes back an infinite number of times in any subset. The Feller process is called transient if
∞ sup E x 1A (Xt ) dt < ∞ x∈E
0
for every compact subset of E. This means that the process spends only a finite time in any compact subset, so that it goes to infinity. The classification result holds as follows.
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Josselin Garnier
Proposition 4.5. A Feller process that satisfies Hypothesis D is either recurrent or transient. The proof can be found in [33]. It consists in 1) introducing a suitable Markov process with discrete time parameters, 2) showing a similar transient-recurrent dichotomy for this process, and 3) applying this result to the Feller process. 4.2.2 Invariant measures. Let π be a Borel measure on E. It is called an invariant measure of the Feller semigroup (Tt ) if Tt f (x)π(dx) = f (x)π(dx) E
E
for every t ≥ 0 and for any nonnegative functions with compact support. The following proposition (whose proof can be found for instance in [26]) shows that a recurrent process possesses a unique invariant measure. Proposition 4.6. Let (Xt ) be a Feller process satisfying Hypothesis D. If it is recurrent, then there exists an invariant measure π which is unique up to a multiplicative constant and mutually absolutely continuous with respect to the reference Borel measure µ. In the case where the state space is compact, the process cannot be transient, hence it is recurrent and possesses an invariant measure, which has to be a bounded Borel measure. Corollary 4.7. Let (Xt ) be a Feller process with compact state space satisfying Hypothesis D. Then it is recurrent and has a unique invariant probability measure. 4.2.3 Ergodic properties. Let f ∈ L∞ (E). It is called Tt -invariant if Tt f (x) = f (x) for any t and x. We can characterize the ergodic hypothesis in terms of Tt -invariant functions. Proposition 4.8. A Feller process is ergodic if every bounded measurable Tt -invariant function is a constant function. Conversely if the Feller process is ergodic, then every bounded continuous Tt -invariant function is a constant function. The ergodic theorems that we are going to state are based on the following proposition. Proposition 4.9. Let (Xt ) be a recurrent Feller process satisfying Hypothesis D. For every pair of probabilities π1 and π2 , we have lim (π1 − π2 )Pt = 0
where π1 Pt ( · ) =
E
t→∞
π1 (x)Pt (x, · ) and · is here the norm of the total variations.
Scattering, spreading, and localization of an acoustic pulse by a random medium
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The proposition means that the process forgets its initial distribution as t → ∞. Applying the proposition with π1 = δx and π2 = the invariant measure of the process is the key to the proof of the ergodic theorem for Markov processes: Proposition 4.10. Let (Xt ) be a Feller process satisfying Hypothesis D. 1) If it is recurrent and has an invariant probability measure π , then for any f ∈ L∞ (E) and for any x ∈ E, t→∞ f (y)π(dy). Tt f (x) −−−−→ E
2) If it is recurrent and has an infinite invariant measure, then for any f ∈ L∞ (E) such that E |f (y)|π(dy) < ∞ and for any x ∈ E: t→∞
Tt f (x) −−−−→ 0. 3) If it is transient, then for any f ∈ L∞ (E) with compact support and for any x: t→∞
Tt f (x) −−−−→ 0. The ergodic theorem has also a pathwise version: Proposition 4.11. Let (Xt ) be a Feller process satisfying Hypothesis D. 1) If it is recurrent and has an invariant probability measure π , then for any f ∈ L∞ (E) and for any x ∈ E, 1 t t→∞ f (Xt ) −−−−→ f (y)π(dy) P x almost surely. t 0 E 2) If it is recurrent and has an infinite invariant measure, then for any f ∈ L∞ (E) such that E |f (y)|π(dy) < ∞ and for any x ∈ E: t→∞
f (Xt ) −−−−→ 0 in L1 (P x ). 3) If it is transient, then for any f ∈ L∞ (E) with compact support and for any x: t→∞
f (Xt ) −−−−→ 0 P x almost surely. 4.2.4 Resolvent equations and potential kernels. We consider a Feller process satisfying Hypothesis D. We set, for α > 0, ∞ uα (x, y) = e−αt pt (x, y) dt 0
which is a strictly lower semicontinuous function. The family of operators (Uα )α>0 ∞ Uα (x) = uα (x, y)f (y)µ(dy) = e−αt Tt f (x) dt E
0
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is called the resolvent of the semigroup (Tt ). It satisfies the resolvent equation: (α − β)Uα Uβ f + Uα f − Uβ f = 0
(4.5)
for any α, β > 0. A kernel U (i.e. a family of Borel measures {U (x, · ), x ∈ E} such that U (x, A) is measurable with respect to x for all A ∈ B(E)) is called a potential kernel if it satisfies: (I − αUα )Uf = Uα f
(4.6)
for every α > 0 and f such that Uf ∈ Cb (E). The function Uf is called a potential of f . Note that Eq. (4.6) is the limit form of Eq. (4.5) as β → 0. If the process is transient, then ∞ Pt (x, A) dt (4.7) U (x, A) := 0 ∞ L (E)
is a potential kernel. Indeed if f ∈ with compact support, Uf ∈ Cb (E) and satisfies Uf = limβ→0 Uβ f . Therefore it satisfies (4.6) since (Uα ) satisfies the resolvent equation (4.5). If the process is recurrent and has invariant measure π, then (4.7) diverges as soon as π(A) > 0. However it is possible to construct a kernel W such that Wf ∈ Cb (E) and satisfies (4.6) if f ∈ L∞ (E) with compact support satisfying E f (y)π(dy) = 0. Such a kernel is called a recurrent potential kernel. Proposition 4.12. Let (Xt ) be a recurrent Feller process satisfying Hypothesis D. There exists a recurrent potential W . Assume further that the processhas an invariant probability measure π. If f ∈ L∞ (E) with compact support satisfies E f (y)π(dy) = t 0, then 0 Ts f (x)ds is bounded in (t, x) and t t→∞ Ts f (x) ds −−−−→ Wf (x) − Wf (y)π(dy). 0
E
Note that in the recurrent case with invariant probability measure, the recurrent po Wf (y)dπ(dy) = 0 tential kernel exists and is unique if one adds the condition that E for all f ∈ L∞ (E) with compact support satisfying E f (y)π(dy) = 0. Furthermore, if Wf belongs to the domain of the infinitesimal generator of the semigroup (Tt ), then the potential kernel satisfies QWf = (α − Uα−1 )Wf = −f . We can then state the important following corollary. Corollary 4.13. Let (Xt ) be a recurrent Feller process with an invariant probability ∞ measure π satisfying Hypothesis D. If f ∈ L (E) with compact support satisfies E f (y)π(dy) = 0, and if Wf belongs to the domain of the infinitesimal generator of the semigroup (Tt ), then Wf is a solution of the Poisson equation Qu = −f .
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4.3 Diffusion Markov processes Definition 4.14. Let P be a transition probability. It is associated to a diffusion process if: 1) ∀ x ∈ Rd , ∀ ε > 0, |y−x|>ε P (s, x; t, dy) = o(t − s) uniformly over s < t, 2) ∀ x ∈ Rd , ∀ ε > 0, |y−x|≤ε (yi − xi )P (s, x; t, dy) = bi (s, x)(t − s) + o(t − s) for i = 1, . . . , d uniformly over s < t, 3) ∀ x ∈ Rd , ∀ ε > 0, |y−x|≤ε (yi − xi )(yj − xj )P (s, x; t, dy) = aij (s, x)(t − s) + o(t − s) for i, j = 1, . . . , d, uniformly over s < t. The functions bi characterize the drift of the process, while aij describe the diffusive properties of the diffusion process. In the following we shall restrict ourselves to homogeneous diffusion processes, whose diffusive matrix a and drift b do not depend on time. We are going to see that these functions completely characterize a diffusion Markov process with some more technical hypotheses. We shall assume the following hypotheses: aij are of class C 2 with bounded derivatives, 1 bi are of class C with bounded derivatives, (H) a satisfies the strong ellipticity condition: There exists some γ > 0 such that, for any (t, x), 2 a (x)ξ ξ ≥ γ ij i j ij i ξi . We introduce the second-order differential operator L: Lf (x) =
d
∂ 2 f (x) ∂f (x) + bi (x) . ∂xi ∂xj ∂xi d
aij (x)
i,j =1
i=1
Proposition 4.15. Under Hypotheses (H): 1) There exists a unique Green’s function pt (x, y) from Rd × R+ × Rd to R such that pt (x, y) > 0 ∀ t > 0, x, y ∈ Rd , p is continuous on Rd × R+∗ × Rd , p is C 2 in x and C 1 in t, as a function of t and x, p satisfies ∂p ∂t = Lp, ∀ x, Rd pt (x, y)f (y)dy → f (x) as t → 0+ for any continuous and bounded function f .
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2) There exists a unique positive function p¯ ∈ C 1 (Rd , R) such that L∗ p¯ ≡ 0 up to a multiplicative constant. If p¯ has finite mass, then we choose the normalization ¯ = 1. Rd p(y)dy
We refer to [16] for the proof. This proposition is obtained by means of PDE tools. The first point is the key for the proof of the following proposition that describes a class of Markov processes satisfying Hypothesis D.
Proposition 4.16. Under Hypotheses (H), there exists a unique diffusion Markov process with drift b and diffusive matrix a. It is Feller and it satisfies hypothesis D. It has continuous sample paths. L is its infinitesimal generator. The Green’s function p is the transition probability density pt (x, y) dy Pt (x, A) = A
which satisfies the KBE as a function of t and x: ∂p = Lp, ∂t
pt=0 (x, y) = δ(x − y),
and the Kolmogorov forward equation (KFE) as a function of t and y: ∂p = L∗ p, ∂t
pt=0 (x, y) = δ(x − y),
where L∗ is the adjoint operator of L L∗ f (x) =
d i,j =1
d ∂2 ∂ aij (x)f (x) − (bi (x)f (x)) . ∂xi ∂xj ∂xi i=1
Note that the KBE is the same as Eq. (4.1), as the probability transition reads ∂p P (s, x; t, A) = P (0, x; t − s, A) so that ∂p ∂s = − ∂t . The KFE is also known as the Fokker–Planck equation. Furthermore the KBE and KFE are not restricted to the diffusion processes. We can actually state in great generality for a Markov process (Xt ) with infinitesimal generator Q that, if dom(Q) is a subset of L∞ sufficiently large and if the probability density function p exists and is sufficiently smooth, then it satisfies both the KBE and the KFE. Indeed, from Proposition 4.4, if u ∈ dom(Q) and p is smooth enough, we have ∂u ∂p(s, x; t, y) f (y) dy = = −Qu = − Qp(s, x; t, y)f (y) dy ∂s ∂s
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which proves the KBE for p smooth enough. Second, by the Chapman–Kolmogorov equation, ∂ p(s, x; t, y) ∂τ ∂p(τ, z; t, y) ∂p(s, x; τ, z) p(τ, z; t, y) dz + p(s, x; τ, z) dz = ∂τ ∂τ ∂p(s, x; τ, z) p(τ, z; t, y) dz − p(s, x; τ, z)Qτ p(τ, z; t, y) dz = ∂τ ∂p(s, x; τ, z) p(τ, z; t, y) dz − Q∗τ p(s, x; τ, z)p(τ, z; t, y) dz = ∂τ
0=
which proves the KFE if we let τ t and use p(t, y; τ, z) → δ(y − z). A diffusion process can be either transient or recurrent. A criterion that insures recurrence is that there exists K > 0, α ≥ −1, r > 0 such that d j =1
bj (x)
xj ≤ −r|x|α |x|
∀ |x| ≥ K
(r should be large enough in the case α = −1). In such a case the drift b plays the role of a trapping force that pushes the process to the origin [37]. The following proposition expresses the Fredholm alternative for the operator L in the ergodic case. It can be seen as a consequence of Proposition 4.12 and Corollary 4.13 in the framework of diffusion processes. Proposition 4.17. Under Hypotheses ( H), if the invariant measure has finite mass, then the diffusion process is ergodic. If moreover f ∈ C 2 with compact support satisfies Rd f (y)p(y)dy ¯ = 0, then there exists a unique function χ which sat ¯ ≡ 0. It is given by isfies Lχ = f and the centering condition Rd χ (y)p(y)dy ∞ ∞ T f (x)ds = p (x, y)f (y)dyds. s 0 0 Rd s Example 4.18 (The Brownian motion). The d-dimensional Brownian motion is the Rd -valued homogeneous Markov process with infinitesimal generator: 1 2 where is the standard Laplacian operator. The probability transition density Q=
p(0, x; t, y) has the Gaussian density with mean x and covariance matrix tId : 1 |y − x|2 p(0, x; t, y) = . exp − (2π t)d/2 2t An explicit calculation demonstrates that the Brownian motion is recurrent in dimensions 1 and 2 and transient in higher dimensions. The Brownian motion possesses an
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invariant measure which is simply the Lebesgue measure over Rd which has infinite mass. Whatever the dimension is, the Brownian motion is not ergodic, it escapes to infinity. It will not be a suitable process for describing a stationary ergodic medium. However if we add a trapping potential the Brownian motion becomes ergodic, as shown in the next example. Example 4.19 (The Ornstein–Uhlenbeck process). It is an R-valued homogeneous Markov process defined as the solution of the stochastic differential equation dXt = −λXt + dWt that admits the closed form expression Xt = X0 e
−λt
+
t
e−λ(t−s) dWs ,
0
where W is a standard one-dimensional Brownian motion. It describes the evolution of the position of a diffusive particle trapped in a quadratic potential. It is the homogeneous Markov process with infinitesimal generator: Q=
∂ 1 ∂2 − λx . 2 2 ∂x ∂x
The probability transition p(0, x; t, · ) has a Gaussian density with mean xe−λt and variance σ 2 (t): (y − xe−λt )2 1 − e−2λt 1 2 , σ . exp − (t) = p(0, x; t, y) = 2σ 2 (t) 2λ 2π σ (t)2 It is a recurrent ergodic process whose invariant probability measure has a density with respect to the Lebesgue measure: λ p(y) ¯ = exp − λy 2 . (4.8) π Note that it may be more comfortable in some circumstances to deal with a process with compact state space. For instance the process Yt = arctan(Xt ) with (Xt ) an Ornstein–Uhlenbeck process is one of the models that can be used to describe the fluctuations of the parameters of a random medium.
4.4 The Poisson equation and the Fredholm alternative We consider in this section an ergodic Feller Markov process with infinitesimal generator Q. The probability transitions converge to the invariant probability measure by the ergodic theorem. The resolution of the Poisson equation Qu = f requires fast enough mixing. A set of hypotheses for rapid convergence is stated in the following proposition due to Doeblin:
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Proposition 4.20. Assume that the Markov process has a compact state space E, it is Feller, and there exists t0 > 0, c > 0 and a probability µ over E such that Pt (x, A) ≥ cµ(A) for all t ≥ t0 , x ∈ E, A ∈ B(E). Then there exists a unique invariant probability p¯ and two positive numbers c1 > 0 and δ > 0 such that ¯ ≤ c1 e−δt . sup sup |Pt (x, A) − p(A)| x∈E A∈B(E)
A Feller process with compact state space E satisfying Hypothesis D possesses nice mixing properties. Indeed, if t0 > 0, then δ := inf x,y∈E p(0, x; t0 , y) is positive as E is compact and p is continuous. Denoting by µ the uniform distribution over E we have Pt0 (x, A) ≥ δµ(A) for all A ∈ B(E). This property also holds true for any time t0 + t, t ≥ 0, as the Chapman–Kolmogorov relation implies Pt+t0 (x, A) = P (x, dz)P (z, A) ≥ δ t t 0 E E µ(A)Pt (x, dz) = δµ(A). This proves the following corollary: Corollary 4.21. If a Feller process with compact state space satisfies Hypothesis D, then it fulfills the hypotheses and conclusion of Proposition 4.20. We end up the section by revisiting the problem of the Poisson equation in terms of the Fredholm alternative. We consider an ergodic Feller Markov process satisfying the Fredholm alternative. We first investigate the null spaces of the generator Q. Considering Tt 1 = 1, we have Q1 = 0, so that 1 ∈ Null(Q). As a consequence Null(Q∗ ) is at least one-dimensional. As the process is ergodic Null(Q∗ ) is exactly one-dimensional. In other words there exists a unique invariant probability measure which satisfies Q∗ p¯ = 0. In such conditions the probability transition converges to p¯ as t → ∞. The spectrum of Q∗ gives the rate of forgetfulness, i.e. the mixing rate. For instance the existence of a spectral gap − E f Qf d p¯ >0 inf 2 f, E f d p=0 ¯ E f d p¯ insures an exponential convergence of Pt (x, · ) to p. ¯ We now investigate the solutions of the Poisson equation Qu = f . Of course Q is not invertible since it has a nontrivial null space {1}. Null(Q∗ ) has dimension 1 and is generated by the invariant probability measure p. ¯ By the Fredholm alternative, the Poisson equation admits a solution if f satisfies the orthogonality condition f ⊥ Null(Q∗ ) which means that f has zero mean E[f (X0 )] = 0 where E is the expectation with respect to the distribution of the Markov process starting with the invariant measure p: ¯ p(dx)E ¯ E[f (Xt )] = x [f (Xt )]. E
In such a case, a particular solution of the Poisson equation Qu = f is ∞ dsTs f (x). u0 (x) = − 0
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Remember that the following expressions are equivalent: Ps (x, dy)f (y) = E x [f (Xs )] = E[f (Xs )|X0 = x]. Ts f (x) = E
The convergence of the integral needs fast enough mixing. Also we have formally ∞ ∞ ∞ desQ sQ f = − esQ f dsQe f = − ds = f −E[f (X0 )] = f. Qu0 = − 0 ds 0 0 Note that another solution of the Poisson equation belongs to Null(Q) and is therefore a constant. Here we have E[u0 (X0 )] = 0 as E[f (Xs )] = E[f (X0 )] = 0, so that we can claim that −
∞
0
dsTs : D → D is the inverse of Q restricted to D = (Null(Q∗ ))⊥ .
4.5 Diffusion-approximation for Markov processes Proposition 4.22. Let us consider the system 1 t t dXε ε (t) = F X (t), q , , 2 dt ε ε ε
Xε (0) = x0 ∈ Rd .
(4.9)
Assume that 1) q is a Markov, stationary, ergodic process on a compact space with generator Q, satisfying the Fredholm alternative, 2) F satisfies the centering condition: E[F (x, q(0), τ )] = 0 for all x and τ where E[ · ] denotes the expectation with respect to the invariant probability measure of q, 3) F is of class C 2 and has bounded partial derivatives in x, 4) F is periodic with respect to τ with period T0 . If ε → 0 then the processes (X ε (t))t≥0 converge in distribution to the Markov diffusion process X with generator: ∞ Lf (x) = du E [F (x, q(0), · ).∇ (F (x, q(u), · ).∇f (x))]τ . (4.10) 0
where · τ stands for an averaging over a period in τ . A convergence in distribution means that for any continuous bounded functional from C into R we have ε→0
E[(Xε )] −−−→ E[(X)]. Remark 4.23. The infinitesimal generator also reads: L=
d i,j =1
∂2 ∂ + bi (x) ∂xi ∂xj ∂xi d
aij (x)
i=1
95
Scattering, spreading, and localization of an acoustic pulse by a random medium
with
aij (x) =
bi (x) =
∞
0
d j =1 0
!
" du E Fi (x, q(0), · )Fj (x, q(u), · ) τ ,
∞
# $ ∂Fi du E Fj (x, q(0), · ) (x, q(u), · ) . τ ∂xj
Remark 4.24. We can also consider the case when F depends continuously on the macroscopic time variable t in (4.9). We then get the same result with the limit process described as a time-inhomogeneous Markov process with generator Lt defined as above. Remark 4.25. The periodicity condition 4) can be removed if we assume instead of 4): 4bis) Assume that the limits
1 T +t0 dτ E Fi (x, q(0), τ )Fj (x, q(u), τ ) lim T →∞ T t0 ∂Fi 1 T +t0 dτ E Fj (x, q(0), τ ) (x, q(u), τ ) lim T →∞ T t0 ∂xj exist uniformly with respect to x in a compact, are independent on t0 and are integrable with respect to u. We then denote ∞
1 T du lim dτ E Fi (x, q(0), τ )Fj (x, q(u), τ ) aij (x) = T →∞ T 0 0
d ∞ ∂Fi 1 T bi (x) = du lim dτ E Fj (x, q(0), τ ) (x, q(u), τ ) T →∞ T 0 ∂xj 0 j =1
and assume that a and b are smooth enough so that there exists a unique diffusion process with generator L (assume Hypotheses H for instance). In this section we shall give a formal proof of Proposition 4.22 which contains the key points. The strategy of the complete and rigorous proof is based of the theory of martingales and is sketched out in Appendix B. It also relies on the perturbed test function method. The process X¯ ε ( · ) := (Xε ( · ), q( · /ε2 )) is Markov with generator 1 1 t ε L = 2 Q + F x, q, .∇. ε ε ε The Kolmogorov backward equation for this process can be written as follows: ∂V ε + Lε V ε = 0, ∂t
t < T,
(4.11)
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and we consider final conditions at t = T that do not depend on q: V ε (t = T , q, x) = f (x) where f is a smooth test function. We will solve (4.11) asymptotically as ε → 0 by assuming the multiple scale expansion: Vε =
∞
εn Vn (t, q, x, τ )|τ =t/ε
(4.12)
n=0
To expand (4.11) in multiple scales we must replace becomes
∂ ∂t
by
∂ ∂t
+
1 1 1 ∂V ε ∂V ε + 2 QV ε + F.∇V ε + = 0, ∂t ε ε ε ∂τ
1 ∂ ε ∂τ .
t < T,
Thus Eq. (4.11)
(4.13)
Substitution of (4.12) into (4.13) yields a hierarchy of equations: QV0 = 0 ∂V0 =0 QV1 + F.∇V0 + ∂τ ∂V0 ∂V1 + =0 QV2 + F.∇V1 + ∂τ ∂t
(4.14) (4.15) (4.16)
Taking into account the ergodicity of q, Eq. (4.14) implies that V0 does not depend 0 on q. Taking the expectation of (4.15) the equation can be reduced to ∂V ∂τ = 0 which shows that V0 does not depend on τ and V1 should satisfy: QV1 = −F (x, q, τ ).∇V0 (t, x)
(4.17)
We have assumed that q is ergodic and satisfies the Fredholm alternative, so Q has an inverse on the subspace of the functions that have mean zero with respect to the invariant probability measure P. The right-hand side of Eq. (4.17) belongs to this subspace, so we can solve this equation for V1 : V1 (t, x, q, τ ) = −Q−1 F (x, q, τ ).∇V0 (t, x)
∞
(4.18)
where −Q−1 = 0 dtetQ . We now substitute (4.18) into (4.16) and take the expectation with respect to P and the averaging over a period in τ . We then see that V0 must satisfy:
" ∂V0 ! + E F.∇(−Q−1 F.∇V0 ) τ = 0 ∂t This is the solvability condition for (4.16) and it is the limit backward Kolmogorov equation for the process Xε , which takes the form: ∂V0 + LV0 = 0 ∂t
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with the limit infinitesimal generator: ∞ !
" dt E F.∇(etQ F.∇ τ L= 0
Using the probabilistic interpretation of the semigroup etQ we can express L as (4.10).
5 Spreading of a pulse traveling through a random medium We are interested in the following question: how the shape of a pulse has been modified when it emerges from a randomly layered medium? This analysis takes place in the general framework, based on separation of scales, introduced by G. Papanicolaou and his co-authors (see for instance [7] for the one-dimensional case or [4] for the threedimensional case). We consider here the problem of acoustic propagation when the incident pulse wavelength is long compared to the correlation length of the random inhomogeneities but short compared to the size of the slab. In this framework, it has already been proved in [4] that, when the random fluctuations are weak, the O’Doherty–Anstey theory is valid, i.e. the traveling pulse retains its shape up to a low spreading; furthermore, its shape is deterministic when observed from the point of view of an observer traveling at the same random speed as the wave while it is stochastic when the observer’s speed is the mean speed of the wave. Here we do not assume the fluctuations to be small. The main result of this section consists in a complete description of the asymptotic law of the emerging pulse: we prove a limit theorem which shows that the pulse spreads in a deterministic way. For simplicity we present the proof in the one-dimensional case with no macroscopic variations of the medium and the noise only appearing in the density of the medium. We refer to [11] for the result for one-dimensional media with macroscopic variations. We consider an acoustic wave traveling in a one-dimensional random medium located in the region 0 ≤ x ≤ L, satisfying the linear conservation laws: ∂p ∂u (x, t) = 0 ρ(x) (x, t) + ∂t ∂x (5.1) ∂p (x, t) + κ(x) ∂u (x, t) = 0 ∂t ∂x here u(x, t) and p(x, t) are respectively the speed and pressure of the wave, whereas ρ(x) and κ(x) are the density and bulk modulus of the medium. In oursimplified model we suppose that κ(x) is constant equal to 1 and that ρ(x) = 1 + η εx2 where η εx2 is the rapidly varying random coefficient describing the inhomogeneities. Since these coefficients are positive we suppose that |η| is less than a constant strictly less than 1. Furthermore we assume that η(x) is stationary, centered and mixing enough. We may think for instance that η(x) = f (Xx ) where (Xx )x≥0 is the Ornstein–Uhlenbeck
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process and f is a smooth function from R to [−δ, δ], δ < 1, satisfying f (y)p(y)dy ¯ where p¯ is the invariant probability density (4.8) of the Ornstein–Uhlenbeck process. u (and p) satisfies the wave equation: x 1 + η 2 utt − uxx = 0. ε In order to precise our boundary conditions we introduce the right and left going waves A = u + p and B = u − p which satisfy the following system of equations: ∂ 1 x −1 −1 ∂ A −1 0 A = + η 2 . (5.2) 1 1 B 0 1 B ∂x 2 ε ∂t The slab of medium we are considering is located in the region 0 ≤ x ≤ L and at t = 0 an incident pulse is generated at the interface x = 0 between the random medium and the outside homogeneous medium. According to previous works [7] or [4] we choose a pulse which is broad compared to the size of the random inhomogeneities but short compared to the macroscopic scale of the medium. There is no wave entering the medium at x = L (see Fig. 1). t , B(L, t) = 0 (5.3) A(0, t) = f ε where f is a function whose Fourier transform fˆ belongs to L1 ∩ L2 . t A(0, t) = f ε B(0, t)
-
1+η
x ε2
utt − uxx = 0
0
L
-
A(L, t)
x
Figure 1. Spreading of a pulse
We are interested in the transmitted pulse A(L, t) around the arrival time t = L and in the same scale as the entering pulse A(0, t); therefore the quantity of interest is the windowed signal A(L, L + εσ )σ ∈(−∞,∞) which will be given by the following centered and rescaled quantities: Aε (x, σ ) = A(x, x + εσ ),
B ε (x, σ ) = B(x, −x + εσ ).
(5.4)
The solution of (5.2)+(5.3) takes place in an infinite-dimensional space because of the variable t. So we perform the scaled Fourier transforms: ε iωσ ε ε ˇ ˇ A (x, ω) = e A (x, σ ) dσ, B (x, ω) = eiωσ B ε (x, σ ) dσ.
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In the frequency domain, with the change of variables (5.4), Eq. (5.2) becomes: % & ε x iω x 1 d Aˇ e−2iω ε Aˇ ε η = (5.5) x Bˇ ε Bˇ ε dx 2ε ε2 −e2iω ε −1 with the boundary conditions Aˇ ε (0, ω) = fˆ(ω) and Bˇ ε (L, ω) = 0. The linearity of (5.5) enables us to replace these boundary conditions by: Aˆ ε (0, ω) = 1,
Bˆ ε (L, ω) = 0
and obtain the following representation for the transmitted pulse: 1 e−iωσ fˆ(ω)Aˆ ε (L, ω) dω A(L, L + εσ ) = Aε (L, σ ) = 2π
(5.6)
(5.7)
where (Aˆ ε , Bˆ ε ) is now the solution of problem (5.5)+(5.6). The propagator matrix Y ε (x, ω) is the solution of equation (5.5) with the initial condition Y ε (0, ω) = Id C2 . The process (Aˆ ε (x, ω), Bˆ ε (x, ω)) can be deduced from (Aˆ ε (0, ω), Bˆ ε (0, ω)) through the identity: ε ε Aˆ (0, ω) Aˆ (x, ω) ε = Y . (5.8) (x, ω) Bˆ ε (x, ω) Bˆ ε (0, ω) The structure of the propagator matrix can be exhibited. If (aˆ ε , bˆ ε )T is a solution of (5.5) with the initial condition aˆ ε (0) = 1, bˆ ε (0) = 0, then ((bˆ ε )∗ , (aˆ ε )∗ )T is another solution linearly independent from the previous one 1 and we can thus write Y ε (x, ω) as: ε aˆ (x, ω) bˆ ε (x, ω)∗ ε (5.9) Y (x, ω) = bˆ ε (x, ω) aˆ ε (x, ω)∗ The trace of the matrix appearing in the linear equation (5.5) being 0, we deduce that the determinant of Y ε (x, ω) is constant and equal to 1 which implies that |aˆ ε (x, ω)|2 − |bˆ ε (x, ω)|2 = 1 for every x. Using (5.8) and (5.9) at x = L and boundary conditions (5.6) we deduce that: Aˆ ε (L, ω) =
1 , ε aˆ (L, ω)∗
bˆ ε (L, ω) Bˆ ε (0, ω) = − ε . aˆ (L, ω)∗
(5.10)
In particular we have the following relation of conservation of energy: (5.11) |Aˆ ε (L, ω)|2 + |Bˆ ε (0, ω)|2 = 1. ε Lemma 5.1. The transmitted pulse (A (L, σ ))−∞<σ <∞ ε>0 is a tight (i.e. weakly compact) family in the space of continuous trajectories equipped with the sup norm. 1 Here the star ∗ stands for complex conjugation
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Proof. We must show that, for any δ > 0, there exists a compact subset K of the space of continuous bounded functions such that: sup P(Aε (L, · ) ∈ K) ≥ 1 − δ. ε>0
On the one hand (5.11) yields that Aε (L, σ ) is uniformly bounded by: 1 ε |A (L, σ )| ≤ |fˆ(ω)| dω. 2π On the other hand the modulus of continuity M ε (δ) = is bounded by
sup
|σ1 −σ2 |≤δ
M (δ) ≤ ε
sup
|σ1 −σ2 |≤δ
|Aε (L, σ1 ) − Aε (L, σ2 )|
|1 − exp(iω(σ1 − σ2 ))||fˆ(ω)| dω
which goes to zero as δ goes to zero uniformly with respect to ε. Moreover the finite-dimensional distributions will be characterized by the moments E[Aε (L, σ1 )p1 . . . Aε (L, σk )pk ]
(5.12)
for every real numbers σ1 < · · · < σk and every integers p1 , . . . , pk . Let us first address the first moment. Using the representation (5.7) the expectation of Aε (L, σ ) reads: 1 ε E[A (L, σ )] = e−iωσ fˆ(ω)E[Aˆ ε (L, ω)] dω. 2π We fix some ω and denote X1ε = Re(a ε ( · , ω)), X2ε = Im(a ε ( · , ω)), X3ε = Re(bε ( · , ω)) and X4ε = Im(bε ( · , ω)). The R4 -valued process X ε satisfies the linear differential equation 1 x x ε dX ε (x) = Fω η 2 , X (x), (5.13) dx ε ε ε with the initial conditions X1ε (0) = 1 and Xjε (0) = 0 if j = 2, 3, 4, where 0 −1 sin(2ωh) − cos(2ωh) ωη 1 0 cos(2ωh) sin(2ωh) Fω (η, h) = sin(2ωh) cos(2ωh) 0 1 2 − cos(2ωh) sin(2ωh) −1 0
.
Applying the approximation-diffusion Theorem 4.22, we get that Xε converges in distribution to a Markov diffusion process X characterized by an infinitesimal generator
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denoted by L: L=
4
∂2 ∂ + bi (X) ∂Xi ∂Xj ∂Xi 4
aij (X)
i,j =1
i=1
whose entries bi are all vanishing and a11
αω2 = 4
a22
αω2 = 4
% X22 %
a12 = a21 = where
α=
0
∞
X 2 + X42 + 3 2
X 2 + X42 X12 + 3 2
& , & ,
αω2 (−X1 X2 ) , 4 E[η(0)η(x)] dx.
The expectation φ(x) = E[(X1 (x) − iX2 (x))−1 ] satisfies the equation αω2 dφ = Lφ = − φ, dx 2
φ(0) = 1.
The solution of this ODE is: φ(L) = exp(−αω2 L/2). The expectation of Aˆ ε (L, ω) thus converges to φ(L). By Lebesgue’s theorem (remember |Aˆ ε | ≤ 1) the expectation of Aε (L, σ ) converges to: 1 ε→0 e−iωσ fˆ(ω) exp(−αω2 L/2) dω. E[Aε (L, σ )] −−→ 2π Let us now consider the general moment (5.12). Using the representation (5.7) for ε each factor A , these moments can be written as multiple integrals over n = jk=1 pj frequencies:
E Aε (L, σ1 )p1 . . . Aε (L, σk )pk ' 1 fˆ(ωj,l )e−iωj,l σj . . . = n (2π ) 1≤j ≤k 1 ≤ l ≤ pj
× E
' 1≤j ≤k 1 ≤ l ≤ pj
Aˆ ε (L, ωj,l )
'
dωj,l .
1≤j ≤k 1 ≤ l ≤ pj
The dependency in ε and in the randomness only appears through the quantity E[Aˆ ε (L, ω1 ) . . . Aˆ ε (L, ωn )].
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Our problem is now to find the limit, as ε goes to 0, of these moments for n distinct frequencies. In other words we want to study the limit in distribution of (Aˆ ε (L, ω1 ), . . . , Aˆ ε (L, ωn )) which results once again from the application of a diffusion-approximation theorem. We define the n-dimensional propagator Y ε (x, ω1 , ω2 , . . . , ωn ) = ⊕jN=1 Y ε (x, ωj ) which satisfies an equation similar to (5.5) with Y ε (x = 0, ω) = Id C2n . In order to be allowed to apply the diffusion-approximation theorem, we have to take care to consider separately the real and imaginary parts of each coefficient aˆ ε and bˆ ε , so that we actually deal with a system with 4n linear differential equations. ε ε ε ˆε ˆ ε ( · , ωj )), X4j ˆ ε (., ωj )), X4j Denoting X4j +1 = Re(a +2 = Im(a +3 = Re(b ( · , ωj )) ε 4n ε ˆε and X4j +4 = Im(b ( · , ωj )), j = 1, . . . , n, the R -valued process X satisfies the linear differential equation dXε (x) 1 x x ε = F η 2 , X (x), dx ε ε ε
(5.14)
ε ε
with the initial conditions X4j +j (0) = 1 if j = 1, X4j +j (0) = 0 if j = 2, 3, 4, where
F (η, h) = ⊕jn=1 Fωj (η, h). Applying the approximation-diffusion theorem 4.22, we get that Xε converges in distribution to a Markov diffusion process X characterized by an infinitesimal generator denoted by L: L=
n 4
a4i+j,4i +j (X)
i,i =1 j,j =1
a4i+1,4i+1
αωi2 = 4
a4i+2,4i+2
αωi2 = 4
% 2 X4i+2 +
% 2 X4i+1
a4i+1,4i+2 = a4i+2,4i+1 =
+
∂2 ∂X4i+j ∂X4i +j
2 2 X4i+3 + X4i+4
2 2 2 X4i+3 + X4i+4
2
& , & ,
αωi2 (−X4i+1 X4i+2 ) , 4
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and for i = i ,
a4i+2,4i +2 a4i+1,4i +2
X4i+2 X4i +2 , X4i+1 X4i +1 , αωi ωi −X4i+2 X4i +1 . = a4i +2,4i+1 = 4 αωi ωi
4 αωi ωi
= 4
a4i+1,4i +1 =
The quantity of interest E[Aˆ ε (x, ω1 ) . . . Aˆ ε (x, ωn )] is denoted by φ ε (x). An application of the infinitesimal generator to the expectation n ' (X4j +1 − iX4j +2 )−1 E j =1
gives the following equation for φ(x) = limε→0 φ ε (x): 2α k ωk2 + α k=l ωk ωl dφ(x) =− φ(x) dx 4 with the initial condition φ(0) = 1. This linear equation has a unique solution but instead of solving it and computing . explicitly
our moments one can easily see that it ˜ ˜ is also satisfied by φ(x) = E k A(x, ωk ) where √ ω2 α ω α ˜ Bx − x A(x, ω) = exp i 2 4 and (Bx ) is a standard one-dimensional Brownian motion (Bx is a Gaussian random ˜ variable with zero-mean and variance x). Therefore φ(L) = φ(L) and using (5.7) the ε (L, σ ) is equal to (2π )−1 e−iωσ fˆ(ω)A(L, ˜ ω) dω. Interpreting limit in law of A √ ω2 α ω α 2 BL as a random phase and exp − 4 L as the Fourier transform of the centered Gaussian density with variance αL 2 denoted by G αL : 2
σ2 G αL (σ ) = √ exp − 2 αL π αL 1
we get the main result of this section: Proposition 5.2. The process (Aε (L, σ ))σ ∈(−∞,∞) converges in distribution in the ¯ space of the continuous functions to (A(L, σ ))σ ∈(−∞,∞) √ α ¯ BL A(L, σ ) = f ∗ G αL σ − (5.15) 2 2 which means that the initial pulse f spreads in a deterministic way through the convolution by a Gaussian density and a random Gaussian centering appears through the Brownian motion BL .
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A¯ is what is called by physicists the “coherent part” of the wave. The energy of A¯ is non-random and given by: Ecoh = |f ∗ G αL |2 dσ. 2
If f (t) is narrowband around a high-carrier frequency: f (t) = cos(ω0 t) exp(−t 2 δω2 ),
δω ω0
then it is found that the coherent energy decays exponentially: αω02 L 1 Ecoh (L) = Ecoh (0) √ exp − 2(1 + αδω2 L) 1 + αδω2 L αω2 L . Ecoh (0) exp − 0 2 We end this section by the following remarks: • The previous analysis has been done at L fixed. It is not difficult to generalize it to the convergence in distribution of Aε (L, σ ) as a process in σ and L (see [11] for details). The limit is again given by (5.15) which means that the random centering of the spread pulse follows the trajectory of a Brownian motion as the pulse travels into the medium. • In the ε-scale, the energy entering the medium at x = 0 is equal to |f (σ )|2 dσ . The energy exiting the medium at x = L, in a coherent way around time t = L in the ε-scale, is equal to |f ∗ G αL (σ )|2 dσ which is strictly less than 2 |f (σ )|2 dσ . We may ask the following question: do we have a part of the missing energy exiting the medium in a coherent way somewhere else or at a different time? In other words what is the limit in distribution of A(L, L + t0 + εσ ) for t0 = 0 (energy exiting at x = L) or B(0, t0 + εσ ) (energy reflected at x = 0). A similar analysis shows that these two processes (in σ ) vanish as ε goes to 0 (see [11] for details). This means that there is no other coherent energy in the ε-scale exiting the slab [0, L].
6 Scattering of a monochromatic wave by a random medium This section is devoted to the propagation of monochromatic waves. This is a very natural approach since any wave can be described as the superposition of such elementary wavetrains by Fourier transform. Let uˆ ε (x) be the amplitude at x ∈ R of a monochromatic wave uε (t, x) = exp(−iωε t)uˆ ε (x) traveling in the one-dimensional
Scattering, spreading, and localization of an acoustic pulse by a random medium
105
medium described in Fig. 2. The medium is homogeneous outside the slab [0, L] and the wave uε obeys the wave equation uεtt − uεxx = 0. Accordingly uˆ ε satisfies uˆ εxx + ωε2 uˆ ε = 0 so that it has the form uˆ ε (x) = eiω
εx
+ R ε e−iω
εx
for x ≤ 0
and uˆ ε (x) = T ε eiω
εx
for x ≥ L.
The complex-valued random variables R ε and T ε are the reflection and transmission coefficients, respectively. They depend on the particular realization of ηε , the wavenumber ωε and the slab width L. Inside the slab [0, L] the perturbation is nonzero. It is the realization of a random, stationary, ergodic, and zero-mean process ηε . The dimensionless parameter ε > 0 characterizes the scale of the fluctuations of the random medium as well as the wavelength of the wave. We shall assume that: x ω ηε (x) = η 2 , ωε = ε ε which means that the correlation length of the medium ∼ ε 2 is much smaller than the wavelength ∼ ε which is much smaller than the length of the medium ∼ 1. The scalar field uˆ ε satisfies, for x ∈ [0, L]: uˆ εxx + ωε2 (1 + ηε (x))uˆ ε = 0.
(6.1)
The continuity of uˆ ε and uˆ εx at x = 0 and x = L implies that the solution uˆ ε satisfies the two point boundary conditions: iωε uˆ ε + uˆ εx = 2iωε at x = 0,
ei(ω
iωε uˆ ε − uˆ εx = 0 at x = L.
(6.2)
ε x−ωε t)
-
T ε ei(ω
ε x−ωε t)
(1 + ηε (x))uεtt − uεxx = 0
ε ε R ε e−i(ω x+ω t)
-
0
L
-
x
Figure 2. Scattering of a monochromatic pulse
The following statements hold true when the perturbation η is a stationary process that has finite moments of all orders and is rapidly mixing. We may think for instance
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that η is a Markov, stationary, ergodic process on a compact space satisfying the Fredholm alternative. Proposition 6.1. There exists a length Lεloc such that, with probability one, 1 1 ln |T ε |2 (L, ω) = − ε . L→∞ L Lloc
(6.3)
lim
This length can be expanded as powers of ε: αω2 1 + O(ε), = ε Lloc 2
α :=
∞
duE[η(0)η(u)].
(6.4)
0
Proof. The study of the exponential behavior of the transmittivity |T ε |2 can be divided into two steps. First the localization length is shown to be equal to the inverse of the Lyapunov exponent associated to the random oscillator vxx + ωε2 (1 + ηε (x))v = 0. Second the expansion of the Lyapunov exponent of the random oscillator is computed. We shall first transform the boundary value problem (6.1) and (6.2) into an initial value problem. This step is similar to the analysis carried out in Section 5. Inside the perturbed slab we expand uˆ in the form ε ε uˆ ε (x, ω) = Aˆ ε (x, ω)eiω x + Bˆ ε (x, ω)e−iω x ,
(6.5)
where Aˆ ε and Bˆ ε are respectively the forward (going to the right) and backward (going to the left) modes: iωε uˆ ε + uˆ εx −iωε x e , Aˆ ε = 2iωε
iωε uˆ ε − uˆ εx iωε x Bˆ ε = e . 2iωε
The process (Aˆ ε , Bˆ ε ) is solution of ε ε d Aˆ Aˆ ε = P (x, ω) , ε ˆ B Bˆ ε dx iωε ε 1 η (x) P ε (x, ω) = ε −e2iω x 2 % iω x 1 = η 2 x ε 2ε −e2iω ε
(6.6) e−2iω −1
εx
e−2iω ε −1
x
&
(6.7) .
The boundary conditions (6.2) read in terms of Aˆ ε and Bˆ ε : Aˆ ε (0, ω) = 1, Bˆ ε (L, ω) = 0.
(6.8)
We introduce the propagator Y ε , i.e. the fundamental matrix solution of the linear system of differential equations: Yxε = P ε Y ε , Y ε (0) = Id C2 . From symmetries in
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Eq. (6.6), Y ε is of the form Y (x, ω) = ε
aˆ ε (x, ω) bˆ ε (x, ω)
bˆ ε (x, ω)∗ aˆ ε (x, ω)∗
,
(6.9)
where (aˆ ε , bˆ ε )T is solution of (6.6) with the initial conditions aˆ ε (0, ω) = 1, bˆ ε (0, ω) = 0. The modes Aˆ ε and Bˆ ε can be expressed in terms of the propagator: ε ε Aˆ (x, ω) Aˆ (0, ω) ε = Y (x, ω) . Bˆ ε (x, ω) Bˆ ε (0, ω)
(6.10)
(6.11)
From the identity (6.11) applied for x = L and the boundary conditions (6.8) we can deduce that R ε (L, ω) = −(bˆ ε /aˆ ε )∗ (L, ω),
T ε (L, ω) = (1/aˆ ε )∗ (L, ω).
(6.12)
The transmittivity |T ε |2 is equal to 1/|a ε |2 (L, ω). We introduce the slow process ε ε v ε (x, ω) := aˆ ε (εx, ω)eiω εx + bˆ ε (εx, ω)e−iω εx . It satisfies the equation x ε vxx vε = 0 + ω2 1 + η ε with the initial condition v ε (0) = 1, vxε (0) = −iω. Let us introduce the quantity r ε (x, ω) := |v ε |2 + |vxε |2 /ω2 . A straightforward calculation shows that r ε (x, ω) = 1 + 2|aˆ ε |2 (εx, ω). By Eq. (6.12) we get a relation between r ε and T ε : r ε (x, ω) = 1 + 2|T ε (εx, ω)|−2 .
(6.13)
If γ ε (ω) denotes the Lyapunov exponent that governs the exponential growth of r ε (x, ω): 1 ln r ε (x, ω) x→∞ x
γ ε (ω) = lim
then Eq. (6.13) insures that |T ε (L, ω)|2 will decay as exp(−γ ε (ω)L/ε) as soon as γ ε (ω) > 0. In Appendix A the existence of the Lyapunov exponent γ ε (ω) is proved, and its expansion with respect to ε is derived. Note that α is a nonnegative real number since it is proportional to the power spectral density of the stationary random process η (Wiener–Khintchine theorem [29, p. 141]). The existence and positivity of the exponent 1/Lεloc can be obtained with minimal hypotheses. Kotani [25] established that a sufficient condition is that η is a stationary, ergodic process that is bounded with probability one and nondeterministic. The expansion of the localization length requires some more hypotheses about
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the mixing properties of η. A discussion and sufficient hypotheses are proposed in Appendix A. Note also that the localization length of a monochromatic wave with frequency ω0 is equal to the length that governs the exponential decay of the coherent transmitted part of a narrowband pulse with carrier frequency ω0 . Proposition 6.2. The square modulus of the transmission coefficient |T ε (L, ω)|2 converges in distribution as a continuous process in L to the Markov process τ (L, ω) whose infinitesimal generator is: Lω =
1 2 2 ∂2 1 ∂ αω τ (1 − τ ) 2 − αω2 τ 2 . 2 ∂τ 2 ∂τ
(6.14)
Proof. The square modulus of the transmission coefficient T ε can be expressed in terms of a random variable that is the solution of a Ricatti equation. Indeed, as a byproduct of the proof of Proposition 6.1 we find that |T ε |2 = 1 − | ε |2 where ε (L, ω) = bˆ ε /aˆ ε (L, ω) and (aˆ ε , bˆ ε ) are defined as the solutions of Eqs. (6.6) and (6.10). Differentiating bˆ ε /aˆ ε with respect to L yields that the coefficient ε satisfies a closed-form nonlinear equation: L L iω L d ε =− η 2 (6.15) e2iω ε + 2 ε + e−2iω ε ε2 , ε (ω, 0) = 0. dL 2ε ε One then consider the process Xε := (r ε , ψ ε ) := (| ε |2 , arg( ε )) which satisfies: 1 L L dXε (L) = F η 2 , Xε (L), , dL ε ε ε where F is defined by: ωη F (η, r, ψ, l) = 2
2 sin(ψ − 2ωl)(r 3/2 − r 1/2 ) −2 − cos(ψ − 2ωl)(r 1/2 + r −1/2 )
.
One then applies the diffusion-approximation Theorem 4.22 to the process (r ε , ψ ε ) which gives the result. In particular the expectation of the square modulus of the transmission coefficient converges to τ¯ (L, ω) := E[τ (L, ω)]: ∞ 2 x 2 e−x αω2 L 4 dx . (6.16) τ¯ (L, ω) = √ exp − 8 π 0 cosh( αω2 L/2x) This shows that L 1 αω2 1 ln τ¯ (L, ω) − . L 8
(6.17)
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109
We get actually that, for any n ∈ N∗ , L 1
E[τ (L, ω)n ]
cn (ω) αω2 L . exp − L3/2 8
By comparing Eq. (6.17) with Eqs. (6.3) and (6.4) we can see that the exponential behavior of the expectation of the transmittivity is different from the sample behavior of the transmittivity. This is a quite common phenomenon when studying systems driven by random processes. We now give some heuristic arguments to complete the discussion. Let us set τ (L) = 2/(1 + ρ(L)). ρ is a Markov process with infinitesimal generator: L=
∂2 ∂ 1 2 2 αω (ρ − 1) 2 + αω2 ρ . 2 ∂ρ ∂ρ
Applying standard tools of stochastic analysis (Itô’s formula) we can represent the process ρ as the solution of the stochastic differential equation: dρ =
√ 2 αω ρ − 1dBL + αω2 ρdL,
ρ(0) = 1.
The long-range behavior is imposed by the drift so that ρ 1 and: dρ
√ αωρdBL + αω2 ρdL
which can be solved as: ρ(L) ∼ exp
√ αωBL + αω2 L/2 .
Please note that these identities √ are just heuristic! As L 1, with probability very close to 1, we have B ∼ L L which is negligible compared to L, so ρ(L) ∼ exp αω2 L/2 and τ (L) ∼ exp −αω2 L/2 = exp(−L/Lloc ). √ But, if αωBL < −αω2 L/2, then ρ 1 and τ ∼ 1! This √ is a very rare √ event, its probability is only P( αωBL < −αω2 L/2) = P(B1 < − αω2 L/2) ∼ exp(−αω2 L/8). But this set of rare events (= realizations of the random medium) imposes the values of the moments of the transmission coefficient. Thus the expectation of the transmittivity is imposed by exceptional realizations of the medium. Apparently the “right” localization length is the “sample” one (6.4), in the sense that it is the one that will be observed for a typical realization of the medium. Actually we shall see that this holds true only for purely monochromatic waves.
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7 Scattering of a pulse by a random medium We consider an incoming wave from the left: 1 ε fˆε (ω) exp i (ωx − ωt) dω, uinc (t, x) = 2π
x ≤ 0,
(7.1)
where fˆε ∈ L2 ∩ L1 and contains frequencies of order ε−1 (i.e. wavelengths of order ε): √ 1 t . fˆε (ω) = ε fˆ(εω) ⇐⇒ f ε (t) = √ f ε ε The amplitude of the incoming pulse is normalized so that its energy is independent of ε: 1 1 ε 2 ε 2 ˆ |f (ω)| dω = |fˆ(ω)|2 dω. Einc := |uinc (t, 0)| dt = 2π 2π The total field in the region x ≤ 0 thus consists of the superposition of the incoming wave uεinc and the reflected wave: uεref (t, x)
1 = √ 2π ε
x t ε ˆ dω, f (ω)R (ω, L) exp i −ω − ω ε ε
x ≤ 0,
where R ε (ω, L) is the reflection coefficient for the frequency ω/ε. The field in the region x ≥ L consists only of the transmitted wave that is right going: t 1 x ε ε ˆ dω, x ≥ L, (7.2) utr (t, x) = f (ω)T (ω, L) exp i ω − ω √ ε ε 2π ε where T ε (ω, L) is the transmission coefficient for the frequency ω/ε. Inside the slab the wave has the general form: 1 t ε ε uˆ (ω, x) exp −iω dω, 0 ≤ x ≤ L, u (t, x) = 2π ε where uˆ ε satisfies the reduced wave equation: uˆ εxx + (1 + ηε (x))uˆ ε = 0,
0 ≤ x ≤ L.
The total transmitted energy is: 1 ε ε 2 |fˆ(ω)|2 |T ε (ω, L)|2 dω. T (L) = |utr (t, L)| dt = 2π
Scattering, spreading, and localization of an acoustic pulse by a random medium
111
uεinc (t, x) -
uεtr (t, x) (1 + ηε (x))utt − uxx = 0
uεref (t, x)
-
0
L
-
x
Figure 3. Scattering of a pulse
We first compute the two frequency correlation function. The following lemma is an extension of Proposition 6.2. Lemma 7.1. Let ω1 = ω − hε a /2 and ω2 = ω + hεa /2. 1. If a = 0, then the square moduli of the transmission coefficients (|T ε (ω1 , L)|2 , |T ε (ω2 , L)|2 ) converge in distribution to (τ (L, ω − h/2), τ (L, ω + h/2)) where τ (L, ω − h/2) and τ (L, ω + h/2) are two independent Markov processes whose infinitesimal generators are respectively Lω−h/2 and Lω+h/2 defined by (6.14). 2. If a = 1, then the square moduli of the transmission coefficients (|T ε (ω1 , L)|2 , |T ε (ω2 , L)|2 ) converge in distribution to (τ1 (L), τ2 (L)) where (τ1 (L), τ2 (L), θ (L)) is the Markov process whose infinitesimal generator is: αω2 2 ∂2 αω2 2 ∂ αω2 2 ∂2 αω2 2 ∂ + τ1 (1 − τ1 ) 2 − τ1 τ1 (1 − τ2 ) 2 − τ 2 2 ∂τ1 2 2 2 ∂τ2 ∂τ1 ∂τ2 ∂2 + αω2 cos(θ) (1 − τ1 )(1 − τ2 )τ2 τ1 ∂τ1 ∂τ2 2 αω2 (2 − τ1 )2 ∂ (2 − τ2 )2 (2 − τ1 )(2 − τ2 ) ∂ + + 2h + − 2√ cos(θ ) ∂θ 4 1 − τ1 1 − τ2 ∂θ 2 (1 − τ1 )(1 − τ2 ) √ αω2 1 − τ1 τ1 (2 − τ2 ) ∂2 + sin(θ) (7.3) √ 4 ∂τ1 ∂θ 1 − τ2 √ αω2 1 − τ2 τ2 (2 − τ1 ) ∂2 + sin(θ) √ 4 ∂τ2 ∂θ 1 − τ1
L=
starting from τ1 (0) = 1, τ2 (0) = 1, and θ(0) = 0. Proof. The most interesting case is a = 1, since this is the correct scaling that describes the correlation of the transmission coefficients at two nearby frequencies. Let us denote |T (ωj , L)|2 = 1 − |jε |2 for j = 1, 2, where jε (L) = bˆ ε /aˆ ε (ωj , L). We then
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introduce the four-dimensional process X ε := (r1ε , ψ1ε , r2ε , ψ2ε ) := (|1ε |2 , arg(1ε ), |2ε |2 , arg(2ε )) which satisfies
1 L L dXε ε (L) = F η 2 , X (L), , L , dL ε ε ε
where F is defined by
F (η, r1 , ψ1 , r2 , ψ2 , l, L) =
ωη 2
3/2
1/2
2 sin(ψ1 − 2ωl − hL)(r1 − r1 ) 1/2 −1/2 −2 − cos(ψ1 − 2ωl − hL)(r1 + r1 ) 3/2 1/2 2 sin(ψ2 − 2ωl + hL)(r2 − r2 ) 1/2 −1/2 −2 − cos(ψ2 − 2ωl + hL)(r2 + r2 )
.
Applying the diffusion-approximation theorem 4.22 to the process Xε establishes that X ε converges to a non-homogeneous Markov process X = (r1 , ψ1 , r2 , ψ2 ) whose infinitesimal generator (that depends on L) can be computed explicitly. By introducing θ := ψ1 − ψ2 − 2hL it turns out that the process (r1 , r2 , θ) is a homogeneous Markov process whose infinitesimal generator is given by (7.3). This lemma shows that the transmission coefficients corresponding to two nearby frequencies ω1 and ω2 are uncorrelated as soon as |ω1 − ω2 | ε. Once this result is known, it is easy to derive the asymptotic behavior of the transmittivity corresponding to the scattering of a pulse. Proposition 7.2. The transmittivity T ε (L) converges in probability to T (L): 1 |fˆ(ω)|2 τ¯ (L, ω)dω, T (L) = 2π where τ¯ (L, ω) is the asymptotic value (6.16) of the expectation of the square modulus of the transmission coefficient T ε (L, ω). Proof. Proposition 6.2 gives the limit value of the expectation of |T ε (L, ω)|2 for one frequency ω, so that
ε→0 1 ε |fˆ(ω)|2 τ¯ (L, ω)dω. E T (L) −−→ 2π Then one considers the second moment: 1 ε 2 ˆ(ω)|2 |fˆ(ω )|2 E |T ε (L, ω)|2 |T ε (L, ω )|2 dωdω . | f E T (L) = 4π 2 The computation of this moment requires to study the two-frequency process (|T ε (L, ω)|, |T ε (L, ω )|) for ω = ω . Applying Lemma 7.1 one finds that |T ε (L, ω)|
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and |T ε (L, ω )| are asymptotically uncorrelated as soon as ω = ω , so that
1 ε→0 E T ε (L)2 |fˆ(ω)|2 |fˆ(ω )|2 τ¯ (L, ω)τ¯ (L, ω ) −−−→ 4π 2 2 1 2 ˆ |f (ω) τ¯ (L, ω) dω , = 2π which proves the convergence of T ε (L) to T (L) in L2 (P): 2
ε→0 = E T ε (L)2 − 2E T ε (L) T (L) + T (L)2 −−−→ 0. E T ε (L) − T (L) By the Chebychev inequality this implies a convergence in probability:
ε E (T ε (L) − T (L))2 ε→0 For any δ > 0, P |T (L) − T (L)| > δ ≤ −−−→ 0. δ2 Let us assume that the incoming wave is narrowband, that is to say that the spectral content fˆ is concentrated around the carrier wavenumber ω0 and has narrow bandwidth (smaller than 1, but larger than ε). Then T (L) decays exponentially with the width of the slab as: αω2 L 1 1 ln T (L) − 0 8 L Note that this is the typical behavior of the expected value of the transmittivity of a monochromatic wave with wavenumber ω0 . In the time domain the localization process is self-averaging! This self-averaging is implied by the asymptotic decorrelation of the transmission coefficients at different frequencies. Actually T ε (ω, L) and T ε (ω , L) are correlated only if |ω − ω | ≤ ε. We can now describe the energy content of the transmitted wave. It consists of a coherent part described by the O’Doherty–Anstey theory, with coherent energy that decays quickly as exp(−αω02 L/2). There is also an incoherent part in the transmitted wave, whose energy decays as exp(−αω02 L/8). Thus the incoherent wave contains most of the energy of the total transmitted wave in the regime αω02 L ≥ 1. The spectral content of the incoherent wave is also worth studying as it contains much information [4].
Appendix A. The random harmonic oscillator The random harmonic oscillator:
t ytt + κ + σ η y=0 ε
(A.1)
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with η(t) a random process arises in many physical contexts such as solid state physics [28, 17, 31], vibrations in mechanical and electrical circuits [34, 36], and wave propagation in one-dimensional random media [23, 4]. The dimensionless parameter ε > 0 (resp. σ ) characterizes the correlation length (resp. the amplitude) of the random fluctuations. The sample Lyapunov exponent governs the exponential growth of the modulation: 1 (A.2) G := lim ln r(t), r(t) = |y(t)|2 + |yt (t)|2 . t→∞ t Note that G could be random since η is random. So it should be relevant to study the mean and fluctuations of the Lyapunov exponent. For this purpose we shall analyze the normalized Lyapunov exponent which governs the exponential growth of the p-th moment of the modulation:
1 ln E r(t)p , t→∞ pt
Gp := lim
(A.3)
where E stands for the expectation with respect to the distribution of the process η. If the process were deterministic, then we would have Gp = G for every p. But due to randomness this may not hold true since we can not invert the nonlinear power function “| · |p ” and the linear statistical averaging “E [ · ]”. The random matrix products theory applies to the problem (A.1). For instance let us assume that the random process η is piecewise constant over intervals [n, n + 1) and take random values on the successive intervals. Under appropriate assumptions on the laws of the values taken by η, it is proved in Ref. [5, Theorem 4] that there exists an analytic function g(p) such that:
1 ln E |r(t)|p = g(p), t→∞ t 1 lim ln r(t) = g (0) almost surely, t→∞ t ln r(t) − tg (0) dist. −−→ N (0, g
(0)). √ t lim
(A.4) (A.5) (A.6)
Moreover the convergence is uniform for r(t = 0) with unit modulus, and the function p → g(p)/p is monotone increasing. This proves in particular that G = g (0) is non-random. In case of non-piecewise constant processes η, various versions of the above theorem exist which yield the same conclusion [18, 35, 6, 2]. Unfortunately the expression of g(p) is very intricate, even for very simple random processes η. In the following sections we shall derive closed form expressions for the expansion of the sample Lyapunov exponent with respect to a small parameter. We shall assume in the following that the driving random process η(t) is built from a Markov process m(t) by a smooth bounded function η(t) = c(m(t)). As a consequence η may be non-Markovian itself.
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A.1 Small perturbations We shall assume here that the perturbation is slow ε = 1, but weak σ 1. There are two cases that should be distinguished: the case when κ = σ and the case when κ = 1. Introducing polar coordinates (r(t), ψ(t)) as y(t) = r(t) cos(ψ(t)) and yt (t) = r(t) sin(ψ(t)) the system (A.1) is equivalent to t r(t) = r0 exp q(ψ(s), m(s)) ds , (A.7) 0
ψt (t) = h(ψ(t), m(t)),
(A.8)
with q(ψ, m) = q0 (ψ) + σ q1 (ψ, m) and h(ψ, m) = h0 (ψ) + σ h1 (ψ, m): q0 (ψ) = 0, q1 (ψ, m) = −c(m) sin(ψ) cos(ψ), if κ = 1 h (ψ) = −1, h (ψ, m) = −c(m) cos2 (ψ) 0 1 if κ = σ
q0 (ψ) = 0,
q1 (ψ, m) = −c(m) sin(ψ) cos(ψ),
h (ψ) = 0, h (ψ, m) = −1 − c(m) cos2 (ψ). 0 1
Let us assume that the process m is an ergodic Markov process with infinitesimal generator Q on a manifold M with invariant probability π(dm). From Eq. (A.8) (ψ, m) is a Markov process on the state space S1 × M where S1 denotes the circumference of ∂ the unit circle with infinitesimal generator: L = Q + h(ψ, m) ∂ψ and with invariant measure p(ψ, ¯ m)dψπ(dm) where p¯ can be obtained as the solution of L∗ p¯ = 0. According to the theorem of Crauel [12] the long-time behavior of r(t) can be expressed in terms of the Lyapunov exponent G which is given by q(ψ, m)p(ψ, ¯ m)dψπ(dm). (A.9) G= S1 ×M
This result and the following ones hold true in particular under condition H1 [32] or H2 [3]: H1
M is a finite set and Q is a finite-dimensional matrix which generates a continuous parameter irreducible, time-reversible Markov chain.
H2
M is a compact manifold. Q is a self-adjoint elliptic diffusion operator on M with zero an isolated, simple eigenvalue.
We have considered simple situations where Q∗ = Q. Note that the result can be greatly generalized. For instance one can also work with the class of the φ-mixing processes with φ ∈ L1/2 (see [27, pp. 82–83]). The Lyapunov exponent G can be estimated in case of small noise using the technique introduced by Pinsky [32] under H1 and Arnold et al. [3] under H2. In the following we assume H2. The invariant probability measure is then simply the uniform distribution over M.
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We shall assume from now on that σ 1 and we look for an expansion of G with respect to σ 1. The strategy follows closely the one developed in Ref. [3]. We first divide the generator L into the sum L = L0 + σ L1 with L0 = Q + h0 (ψ)
∂ , ∂ψ
L1 = h1 (ψ, m)
∂ . ∂ψ
As shown in [3] the probability density p¯ can be expanded as p¯ = p¯ 0 + σ p¯ 1 + σ 2 p¯ 2 + · · · where p¯ 0 , p¯ 1 , and p¯ 2 satisfy L∗0 p¯ 0 = 0 and L∗0 p¯ 1 + L∗1 p¯ 0 = 0, L∗0 p¯ 2 + L∗1 p¯ 1 = 0, . . . For once the expansion of p is known, it can be used in (A.9) to give the expansion of G at order 2 with respect to σ : q0 p¯ 0 + σ (q1 p¯ 0 + q0 p¯ 1 ) + σ 2 (q1 p¯ 1 + q0 p¯ 2 ) (ψ, m)dψπ(dm) + O(σ 3 ). G= S1 ×M
(A.10)
If κ = σ . p¯ 0 satisfies Q∗ p¯ 0 = 0. Thus p¯ 0 is the density of the uniform density on S1 × M: p¯ 0 ≡ (2π )−1 . For p¯ 1 we have the equation Q∗ p¯ 1 = −L∗1 p¯ 0 = ∂ψ (h1 p¯ 0 ). Since Q = Q∗ this equation is of Poisson type Qp¯ 1 = ∂ψ (h1 p¯ 0 ). Note that ∂ψ (h1 p¯ 0 ) has zero-mean with respect to the invariant probability π(dm) of Q, so the Poisson equation admits a solution p¯ 1 . Let p(0, m0 ; t, m) be the transition probability density ∗ of the process m(t). It is defined by the equation ∂p ∂t = Q p, p(0, m0 ; t = 0, m) = δ(m − m0 ). In terms of p we can solve the equation for p¯ 1 to obtain ∞ 1 dt ∂ψ h1 (ψ, m)p(0, m0 ; t, m)π(dm). p¯ 1 (ψ, m0 ) = − 2π 0 M Hence
G = σ2 =−
S1 ×M 2 σ
2π
S1
q1 p¯ 1 (ψ, m0 )dψπ(dm0 ) + O(σ 3 )
dψ
M
π(dm0 )
M
∞
π(dm) 0
dtq1 (ψ, m0 )∂ψ h1 (ψ, m)p(0, m0 ; t, m)
+ O(σ 3 ). Taking into account that the autocorrelation function reads π(dm0 ) π(dm)c(m)c(m0 )p(0, m0 ; t, m), E[m(0)m(t)] = M
M
this can be simplified to give G = 0 /4 with ∞ dsE[c(m(0))c(m(s))] = α0 = σ 2α
0
∞
dsE[η(0)η(s)].
(A.11)
0
If κ = 1. Since h0 is constant = −1, p¯ 0 is the uniform density on S1 × M: p¯ 0 ≡ (2π)−1 . Further p¯ 1 satisfies L∗0 p¯ 1 = −L∗1 p¯ 0 = ∂ψ (h1 p¯ 0 ). Note that ∂ψ (h1 p¯ 0 ) has zero-mean with respect to the invariant probability p¯ 0 π(dm)dψ of L∗0 , so the Poisson
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equation admits a solution p¯ 1 . In terms of the transition probability p it is given by ∞ 1 dt ∂ψ h1 (ψ + t)p(0, m0 ; t, m)π(dm). p¯ 1 (ψ, m) = − 2π 0 M Substituting into (A.10) we obtain that G = σ 2 α1 /4 + O(σ 3 ), where α1 is nonnegative and proportional to the power spectral density of the process m evaluated at 2-frequency: ∞ ∞ ds cos(2s)E[c(m(0))c(m(s))] = ds cos(2s)E[η(0)η(s)]. (A.12) α1 = 0
0
A.2 Fast perturbations We consider the random harmonic oscillator: t 2 y=0 ytt + ω 1 + η ε
(A.13)
Proposition A.1. The Lyapunov exponent of the harmonic oscillator (A.13) can be expanded as powers of ε: G=
εω2 α0 + O(ε2 ). 4
Proof. We introduce the rescaled process y(t) ˜ := y(εt) that satisfies y˜tt + ω2 ε2 (1 + η(t))y˜ = 0. ˜ = Applying the above results establishes that the Lyapunov exponent of y˜ is G 2 2 3 ε ω α0 /4 + O(ε ) which gives the desired result.
Appendix B. Diffusion-approximation In this appendix we give the scheme for the rigorous proof of the diffusion approximation stated in Proposition 4.22. We first consider a simple case without periodic modulation. Proposition B.1. Let us consider the system 1 t dXε (t) = F Xε (t), q , dt ε ε2
Xε (0) = x0 ∈ Rd .
Assume that q is a Markov, stationary, ergodic process on a compact space with generator Q, satisfying the Fredholm alternative. F satisfies the centering condition: E[F (x, q(0))] = 0 where E[ · ] denotes the expectation with respect to the invariant probability measure of q. Instead of technical sharp conditions, assume also that F
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is smooth and has bounded partial derivatives in x. Then the continuous processes (Xε (t))t≥0 converge in distribution to the Markov diffusion process X with generator ∞ duE [F (x, q(0)).∇ (F (x, q(u)).∇f (x))] . Lf (x) = 0
Proof. For an extended version of the proof and sharp conditions we refer to [27, 19]. The process X¯ ε ( · ) := (Xε ( · ), q( · /ε2 )) is Markov with generator Lε =
1 1 Q + F (x, q).∇. 2 ε ε
This that, for any smooth function f , the process f (X¯ ε (t)) − f (X¯ ε (s)) − t εimplies ε ¯ s L f (X (u))du is a martingale. The proof consists in demonstrating the convergence of the corresponding martingale problems. It is based on the so-called perturbed test function method. Step 1. Perturbed test function method. ∀f ∈ Cb∞ , ∀ K compact subset of Rd , there exists a family f ε such that: ε→0
sup |f ε (x, q)−f (x)| −−−→ 0, x∈K,q
ε→0
sup |Lε f ε (x, q)−Lf (x)| −−−→ 0. (B.1) x∈K,q
Define f ε (x, q) = f (x) + εf1 (x, q) + ε2 f2 (x, q). Applying Lε to f ε , one gets 1 (Qf1 + F (x, q).∇f (x)) + (Qf2 + F.∇f1 (x, q)) + O(ε). ε One then defines the corrections fj as follows: Lε f ε =
1. f1 (x, q) = −Q−1 (F (x, q).∇f (x)). This function is well-defined since Q has an inverse on the subspace of centered functions (Fredholm alternative). It also admits the representation ∞ duE[F (x, q(u)).∇f (x) | q(0) = q]. f1 (x, q) = 0
2. f2 (x, q) = −Q−1 (F.∇f1 (x, q) − E[F.∇f1 (x, q(0))]) is well-defined since the argument of Q−1 has zero-mean. It thus remains: Lε f ε = E[F.∇f1 (x, q(0))] + O(ε) which proves (B.1). Step 2. Convergence of martingale problems. One first establishes the tightness of the process X ε in the space of the càd-làg functions equipped with the Skorohod topology by checking a standard criterion (see [27, Section 3.3]). Second one considers a subsequence εp → 0 such that X εp → X. One takes t1 < · · · < tn < s < t and h1 , . . . , hn ∈ Cb∞ :
s t ε ε (X (s), q − f E f ε X ε (t), q ε2 ε2 t u ε ε ε ε ε L f X (u), q 2 du h1 (X (t1 )) . . . hn (X (tn )) = 0. − ε s
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Taking the limit εp → 0 yields
t Lf (X(u))du h1 (X(t1 )) . . . hn (X(tn )) = 0 E f (X(t)) − f (X(s)) − s
which shows that X is solution of the martingale problem associated to L. This problem is well-posed (in the sense that it has a unique solution), which proves the result. Proposition B.2. Let us consider the system t t 1 dXε (t) = F Xε (t), q( 2 ), , dt ε ε ε
Xε (0) = x0 ∈ Rd .
We assume the same hypotheses as in Proposition B.1. We assume also that F (x, q, τ ) is periodic with respect to τ with period T0 and F satisfies the centering condition: E[F (x, q(0), τ )] = 0 for all x and τ . Then the continuous processes (Xε (t))t≥0 converge in distribution to the Markov diffusion process X with generator ∞ Lf (x) = du E [F (x, q(0), · ).∇ (F (x, q(u), · )∇f (x))]τ . 0
where · τ stands for an averaging over a period in τ .
Proof. The proof is similar to the one of Proposition B.1. The key consists in building a suitable family of perturbed functions from a given test function. ∀f ∈ Cb∞ , ∀ K compact subset of Rd , there exists a family f ε such that: ε→0
sup |f ε (x, q, τ )−f (x)| −−−→ 0, x∈K,q,τ
ε→0
sup |Lε f ε (x, q, τ )−Lf (x)| −−−→ 0. x∈K,q,τ
(B.2) Let us introduce τ (t) := t mod T0 . The process X¯ ε ( · ) := (Xε ( · ), q( · /ε2 ), τ ( · /ε)) is Markov with generator Lε =
1 1 1 ∂ Q + F (x, q).∇ + . 2 ε ε ε ∂τ
Let f ∈ Cb . We define f1 = f11 + f12 where f11 is the same term as in the absence of a phase: f11 (x, q, τ ) = −Q−1 (F (x, q, τ ).∇f (x)) while f12 does not depend on q so that Qf12 = 0: τ f12 (x, τ ) = − E[F (x, q(0), s).∇f11 (x, q(0), s)] − f¯1 (x) ds 0
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T where f¯1 (x) = T10 0 0 E[F (x, q(0), u).∇f11 (x, q(0), u)]du. Note that f12 is uniformly bounded because of the correction f¯1 . We finally define ∂f1 f2 (x, q, τ ) = −Q−1 F (x, q, τ ).∇f1 (x, q, τ ) + (x, q, τ ) ∂τ
∂f1 (x, q(0), τ ) − E[F (x, q(0), τ ).∇f1 (x, q(0), τ )] − E ∂τ
.
Now we set f ε (x, q, τ ) = f (x) + εf1 (x, q, τ ) + ε2 f2 (x, q, τ ). Applying the infinitesimal generator Lε we get
∂f1 L f (x, q, τ ) = E[F (x, q(0), τ ).∇f1 (x, q(0), τ )] + E (x, q(0), τ ) + O(ε) ∂τ ε
ε
which simplifies into Lε f ε (x, q, τ ) = f¯1 (x) + O(ε).
We refer to [19] for other multi-scaled versions of these propositions.
Bibliography [I]
L. Brekhovskikh, Waves in layered media, Academic Press, 1980. [ Deterministic scattering from inhomogeneous media ]
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S. N. Ethier and T. G. Kurtz, Markov processes, Wiley, New York 1986. [ Markov processes and limit theorems ]
[III] H. J. Kushner, Approximation and weak convergence methods for random processes, MIT Press, Cambridge 1984. [ Diffusion-approximation and other limit theorems ] [IV] V. I. Klyatskin, Stochastic equations and waves in random media, Nauka, Moscow 1980. [ Waves in inhomogeneous media, by a physicist ] [V]
G. Papanicolaou, Waves in one-dimensional random media, in Ecole d’été de Probabilités de Saint-Flour, P. L. Hennequin, ed., Lecture Notes in Mathematics, Springer-Verlag, 1988, pp. 205–275. [ Waves in inhomogeneous media, by a mathematician ]
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[18] H. Furstenberg, Noncommuting random products, Trans. Amer. Math. Soc. 108 (1963), 377–428. [19] J. Garnier, A multi-scaled diffusion-approximation theorem. Applications to wave propagation in random media, ESAIM Probability & Statistics 1 (1997), 183–206. [20] I. Goldsheid, S. Molchanov, and L. Pastur, A random homogeneous Schrödinger operator has a pure point spectrum, Functional Anal. Appl. 11 (1977), 1–10. [21] A. Ishimaru, Wave propagation and scattering in random media, Academic Press, San Diego 1978. [22] R. Z. Khaminskii, A limit theorem for solutions of differential equations with random right-hand side, Theory Prob. Appl. 11 (1966), 390–406. [23] V. I. Klyatskin, Stochastic equations and waves in random media, Nauka, Moscow 1980. [24] W. Kohler and G. Papanicolaou, Power statistics for wave propagation in one dimension and comparison with transport theory, J. Math. Phys. 14 (1973), 1733–1745; 15 (1974), 2186–2197. [25] S. Kotani, Lyapunov indices determine absolutely continuous spectra of stationary random one-dimensional Schrödinger operators, in Stochastic Analysis (K. Ito, ed.), North Holland, 1984, 225–247. [26] H. Kunita, Stochastic flows and stochastic differential equations, Cambridge University Press, Cambridge 1990. [27] H. J. Kushner, Approximation and weak convergence methods for random processes, MIT Press, Cambridge 1984. [28] M. Lax and J. C. Phillips, One-dimensional impurity bands, Phys. Rev. 110 (1958), 41–49. [29] D. Middleton, Introduction to statistical communication theory, Mc Graw Hill Book Co., New York 1960. [30] R. F. O’Doherty and N. A. Anstey, Reflections on amplitudes, Geophysical Prospecting 19 (1971), 430–458. [31] L. A. Pastur, Spectra of random self-adjoint operators, Russian Math. Surveys 28 (1973), 3–64. [32] M. A. Pinsky, Instability of the harmonic oscillator with small noise, SIAM J. Appl. Math. 46 (1986), 451–463. [33] D. Revuz, Markov chains, North-Holland, Amsterdam 1975. [34] R. L. Stratonovich, Topics in the theory of random noise, Gordon and Breach, New York 1963. [35] V. N. Tutubalin, On limit theorems for products of random matrices, Theory Prob. Appl. 10 (1965), 15–27. [36] N. G. Van Kampen, Stochastic processes in physics and chemistry, North Holland, Amsterdam 1981.
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Lectures on parameter identification Otared Kavian Laboratoire de Mathématiques Appliquées (UMR 7641) 45, avenue des États Unis ; 78035 Versailles cedex, France email:
[email protected]
1 Introduction and motivation The terms Inverse Problems and Parameter Identification can be considered synonymous, although the latter is probably less absconse to the non specialist. Both refer to situations in which a certain quantity, represented by a parameter or a coefficient, is to be determined via an indirect measurement, that is via the measurement or determination of a function, parameter or coefficient, which depends on the desired parameter in a more or less complicated way. A simple, elementary, example is when one determines the height of an inaccessible tower via the measurement of various angles and sides of appropriate triangulations (elementary Euclidian geometry establishes relations between the length of the sides and angles of a triangle). Another example, mathematically more involved, is when one knocks on a closed container, supposedly containing a liquid, in order to determine to what extent is it full or empty. In this case the frequencies of the resulting sound depend on the volume of the liquid or that of the air, and in a way or another by hearing the main frequencies of the resulting sound one is able to guess the height of the liquid. This is related to inverse spectral problems, where one wishes to recover some information about the coefficients of certain elliptic operators through information on the eigenvalues and some other measurements on the boundary of the eigenfunctions. Another physical example in which determination of coefficients of an elliptic operator is of importance, is the following. Consider a domain representing an inhomogeneous conducting body, with electric conductivity a(x) at x ∈ . If one assumes that this body undergoes a known electric potential ϕ(σ ) at σ ∈ ∂, the boundary of , then the electric potential u(x) in satisfies the elliptic equation: −div(a(x)∇u(x)) = 0 in (1.1) u(σ ) = ϕ(σ ) on ∂. In this setting the (local) intensity of the current is represented by a(σ )∂u(σ )/∂n(σ ),
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and the total electric power necessary to maintain the potential ϕ on the boundary ∂ will be a(x)|∇u(x)|2 dx. Qa (ϕ) :=
Now the question is the following: what can be deduced about the conductivity a(x) of the body, if one knows these physical data? (The electric conductivity is the inverse of the electric resistance). This question arises for instance when one wishes to obtain some information about the material constituents of the body, without destructive testings. Note that, as a matter of fact, in the above problem the electric power Qa (ϕ) is a redundant information, as it is entirely determined by the potential and the current on the boundary: indeed if one multiplies equation (1.1) by u, after an integration by parts (i.e. using Green’s formula) one gets: ∂u(σ ) 2 a(x)|∇u(x)| dx = a(σ ) ϕ(σ ) dσ. ∂n(σ ) ∂ Therefore we can restrict our attention to the problem in which a known potential ϕ and current a∂u/∂n on the boundary are given, and we wish to give some information on the function a. By inspection of a few physical examples, one is convinced that knowledge of a single set of data, that is one given pair of potential and current, in general would not be enough information to determine a(x) in . Therefore, we may give a new version of the above question, which is mathematically more precise: can one determine uniquely the conductivity a(x) if for all ϕ (in a certain functional space) one knows the current a∂u/∂n, where u satisfies (1.1)? Indeed, if any such space of ϕ’s exists, one may also ask what is the minimal space which permits this determination (note that since the mapping ϕ → a∂u/∂n is linear, we may restrict our attention to a basis (ϕj )j ∈J of that space). There exist numerous practical situations in which an identification problem may be reduced to the determination of the coefficients of an elliptic operator. For instance the case of a homogeneous body filling a domain in which some small inhomogeneities located in ω ⊂ are included, can be modeled by saying that the conductivity a satisfies a(x) := a0 , a known positive constant, when x is in the homogeneous part \ ω of the medium, and a(x) := a1 , an unknown positive constant, when x ∈ ω. In this case the mathematical problem would be the determination of the function a, which amounts to the determination of both the unknown positive constant a1 and the position and shape of the subdomain ω containing the inhomogeneities. These questions, aside being interesting in their own right, both from a mathematical point of view and in many physical applications, are of deep importance in many other problems (inverse scattering problems, inverse spectral problems, thermography, tomography, etc.). Later on we shall give more details in this respect. Remark 1.1. When is a bounded interval of R, say = (0, ), the above problem has a quite simple answer: it is not possible to determine the conductivity a(x) by
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measuring potentials and currents at the endpoints of the interval. From a physical point of view this is due to the fact that different conductivities in the body, represented by the interval (0, ), may lead to the same observed potential and current at the endpoints. The only information one may get is the average resistance, that is an equivalent homogeneous resistance one may put between the endpoints of the interval in order to obtain a given potential and current passing through the conductor. Indeed for j = 0 and j = 1 consider uj solution of −(a(x)uj (x)) = 0,
uj (0) = 1 − j,
uj () = j.
Then it is clear that if we denote by M(a) the harmonic mean value of a, i.e. −1 ds M(a) := , 0 a(s) then we have
u0 (x) = 1 − M(a)
x
a(s) 0
−1
ds,
u1 (x) = M(a)
x
a(s)−1 ds.
0
Now if one solves −(a(x)u (x)) = 0 with u(0) = α and u() = β, one has u(x) = αu0 (x) + βu1 (x), and the intensity of the current at the endpoints will be given by a(0)u (0) = a()u () = (β − α)M(a). From this it is clear that the only information we might obtain is the value of M(a), provided one applies a potential at the endpoints such that β − α = 0. In the following sections we address the problem of determining the coefficient a, as well as some related questions regarding elliptic operators. These notes have their origin in a series of lectures given at the Universidade Federal do Rio de Janeiro in Rio de Janeiro, Brazil and at Université Louis Pasteur in Strasbourg, France. I am indebted to my friends and colleagues Rolci Cipolatti, Flávio Dickstein and Nilson Costa Roberty in Rio de Janeiro, Bopeng Rao and Eric Sonnendrucker in Strasbourg, for both having me given the opportunity to give these lectures and having encouraged me to write these notes.
2 Some basic facts about linear elliptic equations In this section we gather a few results concerning solutions of elliptic equations which should be well known before studying the related inverse problems. In what follows we assume that ⊂ RN is a bounded Lipschitz domain, and that N ≥ 2; we shall denote the usual norm of L2 () by · 2 . For more details on any of the results sketched in this section the reader may refer to one of the classics on elliptic partial differential equations or D. Gilbarg and N. S. Trudinger [17], J. L. Lions [29], G. Stampacchia [37] or O. A. Ladyženskaja and N. N. Ural’ceva [27].
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Thanks to the trace inequality (here we denote γ0 (u) := u|∂ ): |γ0 (u)(σ )|2 dσ ≤ c( ∇u 22 + u 22 ), ∃ c > 0, ∀ u ∈ C 1 () ∂
C 1 ()
H 1 (),
is dense in the trace operator γ0 may be extended to H 1 () and since 1/2 by definition H (∂) := γ0 (H 1 ()). The Sobolev space H 1 () will be equipped with the scalar product ∇u(x) · ∇v(x) dx + u(σ )v(σ ) dσ, (u|v)H 1 :=
∂
1/2 (u|u)H 1
and, the associated norm u H 1 := is equivalent to u → ( ∇u 22 + u 22 )1/2 , 1 the usual norm of H (). In this setting H01 () is the kernel of γ0 and the norm on H01 () will be u → ∇u 2 . The orthogonal space of H01 () is the space of harmonic functions on : H0 ( ) := v ∈ H 1 (); v = 0 in H −1 () . Since by definition H 1 () = H0 ( ) ⊕ H01 (), any u ∈ H 1 () has the unique decomposition u = ψ + u0 where ψ ∈ H0 ( ) and u0 ∈ H01 (). This decomposition is characterized by the fact that ψ and u0 satisfy respectively
ψ = 0 in ψ − u = 0 on ∂, and
− u0 = − u u0 = 0
in on ∂.
In particular since ψ H 1 ≤ u H 1 , we have the following result: Lemma 2.1. For any ϕ ∈ H 1/2 (∂) there exists a unique ψ ∈ H 1 () such that
ψ = 0 in ψ =ϕ on ∂. Moreover for some positive constants c0 , c1 depending on , we have c0 ϕ H 1/2 (∂) ≤ ψ H 1 ≤ c1 ϕ H 1/2 (∂) . We shall denote by Aad the set of admissible coefficients Aad := a ∈ L∞ (); ∃ε0 > 0 such that a ≥ ε0 a.e. on .
(2.1)
For q ∈ L∞ () and a ∈ Aad we denote by La,q the elliptic operator La,q u := −div(a∇u) + qu,
(2.2)
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129
D(La,q ) := u ∈ H01 (); −div(a∇u) + qu ∈ L2 ()
(2.3)
which on the domain
is a self-adjoint operator acting on L2 () (see for instance [23], chapitre 1, § 9). The operator (La,q , D(La,q )) has a compact resolvent, more precisely there exists λ∗ ∈ R such that for all λ > λ∗ , the resolvent operator (λI + La,q )−1 : L2 () −→ L2 () is compact and, as a matter of fact, the image R((λI + La,q )−1 ) is precisely D(La,q ). As a consequence, La,q has a nondecreasing sequence of eigenvalues (λk )k≥1 , each being of finite multiplicity, and eigenfunctions (ϕk )k≥1 : j La,q ϕk = λk ϕk , ϕk ∈ D(La,q ), ϕk (x)ϕj (x) dx = δk ,
j
where δk stands for the Kronecker symbol. The first eigenvalue λ1 is simple, λ1 < λ2 , and one may choose the corresponding eigenfunction ϕ1 so that ϕ1 > 0 in . Also each eigenfunction ϕk is Lipschitz continuous in , that is it belongs to W 1,∞ (). We shall denote by N (La,q ) the kernel of (La,q , D(La,q )) and H1 := Rϕk and H2 := Rϕk , λk <0
λk >0
so that the space L2 () has the orthogonal decomposition L2 () = H1 ⊕ N(La,q ) ⊕ H2 . (Note that N (La,q ) = {0} only if zero is an eigenvalue). Any u ∈ L2 () has a unique decomposition u = u0 + u1 + u2 with uj ∈ Hj for j = 1, 2 and u0 ∈ N(La,q ), which are orthogonal projections of u on the corresponding subspaces. In particular since H1 and N (La,q ) are finite dimensional spaces spanned by eigenfunctions of La,q , one sees that H1 ⊕ N (La,q ) ⊂ D(La,q ) and so the projections u0 and u1 belong to D(La,q ). Therefore when u ∈ H01 () the above decomposition yields that we have u2 ∈ H01 (). Under the assumption that N (La,q ) = {0}, for any given f ∈ H −1 (), one can solve the problem −div(a∇u) + qu = f in H −1 () (2.4) u=0 on ∂ by a variational method. Namely, for v ∈ H01 () setting 1 1 2 2 J (v) := a(x)|∇v(x)| dx + q(x)|v(x)| dx − f (x)v(x) dx, 2 2
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there exists a unique u ∈ H01 () such that u = u1 + u2 , with uj ∈ Hj and J (u) =
min
max J (v1 + v2 ),
v2 ∈H2 ∩H01 () v1 ∈H1
(2.5)
in other words (u1 , u2 ) ∈ H1 × (H2 ∩ H01 ()) is the unique saddle point of the concave-convex function (v1 , v2 ) → J (v1 + v2 ) (see for instance [23], chapitre 3, § 4). Moreover if we denote by B := Ba,q : H −1 () −→ H01 () the mapping f → u, we have u H 1 () = Bf H 1 () ≤ 0
0
1 f H −1 () , α0
where α0 := min {|λk |; k ≥ 1} .
In the particular case of an operator La,q which is coercive, that is λ1 > 0, the functional J is strictly convex on H01 () and the above resolution of equation (2.4) is equivalent to using Lax–Milgram theorem (see for instance J. L. Lions [29], J. L. Lions and E. Magenes [30], K. Yosida [41]). When zero is an eigenvalue of (La,q , D(La,q )), that is N(La,q ) = {0}, since La,q is selfadjoint and R(La,q ) = N (La,q )⊥ = H1 ⊕ H2 , equation (2.4) has a solution if and only if f is orthogonal to N (La,q ) (this is a consequence of the Fredholm alternative; here R(La,q ) denotes the range of the operator La,q ). In this case, provided f ∈ H1 ⊕ H2 = R(La,q ), one may prove easily that (2.4) has a unique solution u ∈ H1 ⊕ (H2 ∩ H01 ()), and u is characterized by the equation (2.5). This means that u ∈ N (La,q )⊥ and that u H 1 () ≤ 0
1 f H −1 () , α0
where α0 := min {|λk |; k ≥ 1, λk = 0} .
(However note that u + ϕ0 is a solution of (2.4) for any ϕ0 ∈ N(La,q )). Closely related to equation (2.4) is the equation −div(a∇u) + qu = 0 in H −1 () u=ϕ on ∂,
(2.6)
for a given ϕ ∈ H 1/2 (∂). Choosing a ψ ∈ H 1 () such that ψ = ϕ on ∂ (for instance ψ given by Lemma 2.1), the above equation is reduced to (2.4), upon setting u =: v + ψ and seeking v ∈ H01 () which solves −div(a∇v) + qv = div(a∇ψ) − qψ in H −1 (),
v ∈ H01 ().
(2.7)
This equation has a unique solution when N (La,q ) = {0}, for any given ψ ∈ H 1 (), and since ψ H 1 () ≤ c ϕ H 1/2 (∂) , u H 1 () ≤ 0
1 c div(a∇ψ) − qψ H −1 () ≤ ϕ H 1/2 (∂) , α0 α0
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where we use the fact that qψ H −1 () ≤ c qψ 2 ≤ c ψ H 1 () ≤ c ϕ H 1/2 (∂) , and that 1 a∇ψ · ∇w dx; w ∈ H0 (), w H 1 () = 1 div(a∇ψ) H −1 () = sup − 0
≤ a ∞ ψ H 1 () ≤ c ϕ H 1/2 (∂) . When N (La,q ) = {0}, according to our above analysis, equation (2.7) has a solution if and only if the right hand side is orthogonal to the kernel N(La,q ), that is if for all ϕ0 ∈ N (La,q ) one has: ∂ϕ0 (σ ) ψ(σ ) dσ = a(σ ) a∇ψ.∇ϕ0 dx + qψϕ0 dx = 0. ∂n(σ ) ∂ Since the kernel N (La,q ) is finite dimensional, this means that equation (2.6) has a unique solution if and only if ϕ ∈ H 1/2 (∂) is orthogonal to the finite dimensional space ∂ϕ0 ; ϕ0 ∈ N(La,q ) . span a ∂n Actually this space and N (La,q ) have the same dimension, at least when the coefficient a and the boundary ∂ are smooth enough so that, by some unique continuation theorems one may assert that the conditions −div(a∇w) + qw = 0 in and w = a∂w/∂n = 0 on ∂ imply w ≡ 0 (for results concerning unique continuation theorems see for instance V. Adolfsson and L. Escauriaza [2] or N. Garofalo and F. H. Lin [15]),). In order to solve more general non homogeneous boundary value problems, that is equations such as (2.4) where the second condition is replaced with αu + ∂u/∂n = ϕ on ∂, the following result concerning existence of a normal trace is useful. Lemma 2.2. Let ⊂ RN be a bounded Lipschitz domain. There exists a constant c1 > 0 such that if F ∈ (L2 ())N satisfies div(F ) ∈ L2 (), then F · n can be defined as an element of H −1/2 (∂) and: F · n H −1/2 (∂) ≤ c1 ( F 2 + div(F ) 2 ). Proof. Assume that F ∈ W 1,∞ (). Then, using an integration by parts, for any ψ ∈ H 1 () we have ψ(σ ) F (σ ) · n(σ ) dσ = F (x) · ∇ψ(x) dx + div(F (x)) ψ(x) dx. ∂
Therefore, denoting the left hand side of the above identity as F · n, γ0 (ψ), with the brackets · , · meaning the duality between H −1/2 (∂) and H 1/2 (∂) and γ0 : H 1 () −→ H 1/2 (∂) being the trace operator, we have: |F · n, γ0 (ψ)| ≤ F 2 ∇ψ 2 + div(F ) 2 ψ 2 .
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For a given ϕ ∈ H 1/2 (∂) we may choose ψ ∈ H 1 () as in Lemma 2.1 such that ψ = ϕ on ∂, which satisfies ψ H 1 ≤ c1 ϕ H 1/2 (∂) , for a constant c1 depending only on . Hence |F · n, ϕ| ≤ c1 ( F 2 + div(F ) 2 ) ϕ H 1/2 (∂) This means that F · n H −1/2 (∂) ≤ c1 ( F 2 + div(F ) 2 ).
(2.8)
When F is no longer in W 1,∞ (), using a density argument (see the classical books on Sobolev spaces, for instance R.A. Adams [1] or V.G. Maz’ja [31]) one can find a sequence (Fk )k of W 1,∞ () such that Fk → F in (L2 ())N ,
div(Fk ) → div(F ) in L2 ().
Inequality (2.8) shows that (Fk · n)k is a Cauchy sequence in H −1/2 (∂) whose limit, which is independent of the particular choice of the sequence (Fk )k , will be denoted by F · n. The lemma is proved. One of the consequences of this lemma is that whenever u ∈ H 1 () satisfies −div(a∇u) + qu = 0 in ,
u = ϕ on ,
since a∇u ∈ (L2 ())N and div(a∇u) ∈ L2 (), we have a
∂u ∈ H −1/2 (∂). ∂n
Hence we can consider the mapping ϕ → a
∂u ∂n
as a linear continuous operator from H 1/2 (∂) into H −1/2 (∂). This operator, denoted by a,q , is the Poincaré–Steklov operator which we will study in the next sections. We recall here the following local regularity result concerning solutions of elliptic boundary value problems (see for instance D. Gilbarg and N. S. Trudinger [17], G. Stampacchia [37]). Proposition 2.3. Let be a Lipschitz domain, σ ∈ ∂ and R∗ > 0 be such that ∂ ∩ B(σ∗ , R∗ ) is of class C 1,1 , while a ∈ Aad ∩ W 1,∞ (B(σ∗ , R∗ ) ∩ ). Assume that ϕ has its support in B(σ∗ , R∗ ) ∩ ∂ and ϕ ∈ H 3/2 (B(σ∗ , R∗ ) ∩ ∂). If u ∈ H 1 () satisfies −div(a∇u) + qu = f in u=ϕ on ∂
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133
where f ∈ L2 (), then u ∈ H 2 ( ∩ B(σ∗ , R∗ )),
a
∂u ∈ H 1/2 (B(σ∗ , R∗ ) ∩ ∂). ∂n
Also there exists a positive constant c1 depending only on and R∗ and on the coefficients a, q, such that u H 2 (∩B(σ∗ ,R∗ )) ≤ c1 ( f 2 + ϕ H 3/2 (∩B(σ∗ ,R∗ )) ). This regularity result may be used to prove existence of solutions to equations such as (2.6) when ϕ is not anymore in H 1/2 (∂), for instance when ϕ ∈ H −1/2 (∂). Indeed in this case one has to interpret the equation not in H −1 (), but rather in the distributions sense, because the solution u would not be anymore in H 1 () and so the term div(a∇u) would make sense only in D (). For simplicity and clarity of exposition we shall consider only the case of two coefficients a, q such that N(La,q ) = {0}, and we shall assume global regularity assumptions on and a. However one may prove analogous local results when zero is an eigenvalue of (La,q , D(La,q )) or when and a have only local regularity properties. In the particular case a ≡ 1 and q ≡ 0, a different proof of the following lemma may be found in J.L. Lions [29], chapitre VII, § 5. Lemma 2.4. Let ∈ RN be a bounded C 1,1 domain, a ∈ Aad ∩ W 1,∞ () and q ∈ L∞ () be such that N (La,q ) = {0}. Then for a given ϕ ∈ H −1/2 (∂), there exists a unique u ∈ L2 () such that div(a∇u) ∈ L2 () and −div(a∇u) + qu = 0 in L2 () u=ϕ in H −1/2 (∂). Moreover there exist two positive constants c1 , c2 depending only on and the coefficients a, q such that div(a∇u) 2 + u 2 ≤ c1 ϕ H −1/2 (∂) , and a∂u/∂n ∈ H −3/2 (∂) satisfies
∂u
a
≤ c2 ϕ H −1/2 (∂) .
∂n −3/2 H (∂) Proof. Assume first that ϕ ∈ H 1/2 (∂). Then there exists a unique u ∈ H 1 () such that u = ϕ on the boundary ∂ and −div(a∇u) + qu = 0.
(2.9)
Now let v ∈ H01 () solve −div(a∇v) + qv = u in ,
v = 0 on ∂.
(2.10)
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According to Proposition 2.3 we know that v ∈ H 2 () and that
∂v
a
≤ c v H 2 () ≤ c u 2 .
∂n 1/2 H (∂)
(2.11)
Upon multiplying equation (2.10) by u and integrating by parts we see that since: (a∇v · ∇u + quv) dx = 0
(because v ∈ H01 () and u satisfies the homogeneous equation (2.9) in ), we have: ∂v(σ ) 2 u(σ ) dσ + u(x) dx = − a (a∇v · ∇u + quv) dx ∂n(σ ) ∂ ∂v(σ ) a ϕ(σ ) dσ =− ∂ ∂n(σ )
∂v
≤ ϕ H −1/2 (∂) a
∂n H 1/2 (∂) ≤ c ϕ H −1/2 (∂) u 2 , where in the last step we have used the estimate (2.11) . Therefore for any ϕ ∈ H 1/2 (∂), if u ∈ H 1 () solves (2.9) and u = ϕ on ∂, we have div(a∇u) 2 + u 2 ≤ c ϕ H −1/2 (∂) .
(2.12)
Also, in order to see that the normal derivative a∂u/∂n has a meaning in H −3/2 (∂) we proceed as follows. Let ψ ∈ H 3/2 (∂) be given and consider w ∈ H 2 () solution to −div(a∇w) + qw = 0 in (2.13) w=ψ on ∂. Note that such a w exists in H 1 () since N(La,q ) = {0}. By the regularity result of Proposition 2.3 we have w ∈ H 2 () and
∂w
a ≤ c w H 2 () ≤ c ψ H 3/2 (∂) .
∂n 1/2 H (∂) Multiplying equation (2.13) by u and integrating by parts twice (or alternatively multiplying also equation (2.9) by w and then subtracting the resulted identities), one sees that ∂w(σ ) a ϕ(σ ) dσ = (a∇w · ∇u + qwu) dx ∂ ∂n(σ ) ∂u(σ ) a ψ(σ ) dσ. = ∂n(σ ) ∂
Lectures on parameter identification
Therefore
∂u a , ψ ≤ ϕ H −1/2 (∂) ∂n
135
∂w
a
∂n
H 1/2 (∂)
≤ c ϕ H −1/2 (∂) w H 2 () ≤ c ϕ H −1/2 (∂) ψ H 3/2 (∂) . Finally from this we conclude that for any ϕ ∈ H 1/2 (∂) the solution u of (2.9) satisfies
∂u
a
≤ c ϕ H −1/2 (∂) . (2.14)
∂n −3/2 H
(∂)
Now if we assume only ϕ ∈ H −1/2 (∂), a standard regularization procedure allows us to see that the estimates (2.12) and (2.14) go through and the lemma is proved.
3 A. P. Calderón’s approach In this first lecture we present the results of A. P. Calderón [10] which, despite being partial as far as the identification problem is concerned, contain some very interesting ideas, in particular the use of harmonic test functions of the type u(x) := exp(iξ · x + η · x) where ξ, η ∈ RN are such that ξ · η = 0 and |ξ | = |η| (see below the proof of Proposition 3.4). We recall that by classical results on the resolution of elliptic boundary value problems (see Section 2) if a ∈ Aad , defined in (2.1), is given, then for each ϕ ∈ H 1/2 (∂) there exists a unique u ∈ H 1 () satisfying −div(a(x)∇u(x)) = 0 in (3.1) u(σ ) = ϕ(σ ) on ∂. More precisely u may be characterized by the fact that it is the unique point in the convex set (3.2) Kϕ := v ∈ H 1 (); v = ϕ on ∂ at which the strictly convex functional
a(x)|∇v(x)|2 dx
J (v) := Ja (v) :=
achieves its minimum on Kϕ . We will write u = ArgMin Ja (v) v∈Kϕ
(3.3)
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to mean that u ∈ Kϕ and for all v ∈ Kϕ we have J (u) ≤ J (v). As the mapping ϕ → u is linear continuous from H 1/2 (∂) into H 1 (), it is a matter of simple algebra to check that ϕ → Ja (u) defines a continuous quadratic form on H 1/2 (∂) with values in R+ . We will denote this quadratic form by a(x)|∇u(x)|2 dx, where u satisfies (3.1). (3.4) Qa (ϕ) := Q(ϕ) :=
The direct problem would be: knowing the coefficient a ∈ Aad study the mapping a → Qa (ϕ). It is clear that the inverse problem, namely trying to determine a assuming that the mapping a → Qa (ϕ) is known, would be better understood if one knows a good deal about the direct problem. In this regard our first observation is that, as one might expect, the mapping (a, ϕ) → u is continuous, in fact analytic, on Aad × H 1/2 (∂). Lemma 3.1. For (a, ϕ) ∈ Aad ×H 1/2 (∂) define T (a, ϕ) := u where u satisfies (3.1). Then T is analytic on Aad × H 1/2 (∂). Proof. Let (a0 , ϕ0 ) ∈ Aad × H 1/2 (∂) and assume that for some ε0 > 0 we have a0 ≥ ε0 a.e. We shall denote by B := Ba0 the continuous operator defined for f ∈ H −1 () by Bf := w where w ∈ H01 () satisfies −div(a0 ∇w) = f . Let a = a0 + b ∈ Aad where b ∈ L∞ () is such that b ∞ < ε0 . If for w ∈ H 1 () we set Ab w := −div(b∇w), then we have Ab w H −1 () ≤ b ∞ ∇w 2 . Indeed Ab w H −1 () = sup Ab w, ψ; ψ ∈ H01 (), ψ H 1 () = 1 0 1 = sup b(x)∇w(x) · ∇ψ(x) dx; ψ ∈ H0 (), ψ H 1 () = 1 0
≤ b ∞ ∇w 2 . In particular, when w ∈ H01 (), using the norm w → ∇w 2 on H01 (), we infer that BAb is a bounded operator from H01 () into itself and that BAb ≤ B b ∞ . Now denote u0 := T (a0 , ϕ0 ) and u := T (a, ϕ0 ). It is clear that v := u − u0 belongs to H01 () and satisfies (recall that div(a0 ∇u0 ) = 0) −div(a0 ∇v) − div(b∇v) = div(b∇u0 ) = −Ab u0 , which, after applying B to both sides, can be written (I + BAb )v = −BAb u0 . If we assume that a − a0 ∞ = b ∞ < min( B −1 , ε0 ) (see below to observe that min( B −1 , ε0 ) = ε0 ), then BAb < 1 and (I + BAb ) is invertible so we can write v = −(I + BAb )
−1
∞ BAb u0 = (−1)n (BAb )n u0 , n=1
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and therefore T (a, ϕ0 ) = T (a0 , ϕ0 ) +
∞ (−1)n (BAa−a0 )n T (a0 , ϕ0 ). n=1
Remarking that ϕ → T (a, ϕ) is linear, from this we conclude that T is analytic at (a0 , ϕ0 ) ∈ Aad × H 1/2 (∂). Remark 3.2. We point out the following estimates concerning the dependence of u on the coefficient a. First observe that if we denote by ε0 the essential infimum of a0 ∈ Aad , that is ε0 := sup {ε > 0; meas[a0 < ε] = 0}, then Ba0 ≤ ε0−1 . Indeed for f ∈ H −1 () and Ba0 f := w ∈ H01 () satisfying −div(a0 ∇w) = f , we have ε0 Bf 2H 1 () ≤ a0 (x)|∇w(x)|2 dx = f, w ≤ f H −1 () Bf H 1 () . 0
0
In particular we note that B −1 ≥ ε0 and therefore min( B −1 , ε0 ) = ε0 . Also if a − a0 ∞ < ε0 , then a ∈ Aad and, since BAa−a0 L(H 1 ()) ≤ a − a0 ∞ , the 0 above analysis shows that T (a, ϕ0 ) − T (a0 , ϕ0 ) H 1 () ≤ 0
a − a0 ∞ T (a0 , ϕ0 ) H 1 () . 0 ε0 − a − a0 ∞
(3.5)
This inequality may be interpreted as the expression of the fact that if we set δ :=
T (a, ϕ0 ) − T (a0 , ϕ0 ) H 1 () 0
T (a0 , ϕ0 ) H 1 ()
then
0
a − a0 ∞ ≥
ε0 δ . 1+δ
In order to obtain a stability result concerning the inverse problem, one should have a reverse inequality of the type a − a0 ∞ ≤ CF (δ) for some continuous function F such that F (0) = 0. We shall see that although such results may be proved in some cases, in general their proof is much more involved: actually this difficulty is typical of inverse and parameter identification problems (see for instance G. Alessandrini [3, 5]). Lemma 3.3. Let (a0 , ϕ) ∈ Aad × H 1/2 (∂) and u0 := T (a0 , ϕ). Then for any b0 ∈ L∞ () we have Qa0 +tb0 (ϕ) − Qa0 (ϕ) b0 (x)|∇u0 (x)|2 dx. = lim t t→0+ Proof. Setting b := tb0 and a := a0 + tb0 , for t small enough the assumptions in the proof of Lemma 3.1 are satisfied and using the notations used there, we have b0 (x)|∇u0 (x)|2 dx + 2 a0 (x)∇u0 (x) · ∇v(x) dx Qa (ϕ) = Qa0 (ϕ) + t b(x)∇u0 (x) · ∇v(x) dx + (a0 (x) + b(x))|∇v(x)|2 dx. +2
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Now, due to the fact that div(a0 ∇u0 ) = 0, multiplying this equation by v ∈ H01 () and integrating by parts, we remark that a0 (x)∇u0 (x) · ∇v(x) dx = 0. On the other hand thanks to the estimate (3.5) we point out that v H 1 ≤ c b ∞ = O(t) and 0 therefore 2 b(x)∇u0 (x) · ∇v(x) dx + (a0 (x) + b(x))|∇v(x)|2 dx = O(t 2 ).
Therefore Qa0 +tb0 (ϕ) = Qa0 (ϕ) + t the lemma is complete.
b0 (x)|∇u0 (x)|
2 dx + O(t 2 ), and the proof of
Proposition 3.4. For a ∈ Aad denote by (a) := Qa the quadratic form on H 1/2 (∂) associated to a via (3.4). Then is analytic on Aad . In particular for a0 ∈ Aad and b0 ∈ L∞ () the derivative of at a0 is given by (a0 )b0 = q where the quadratic form q is given by b0 (x)|∇u0 (x)|2 dx, q(ϕ) :=
satisfies div(a0 ∇u0 ) = 0 in and u0 = ϕ on ∂. If a0 ∈ Aad is where u0 ∈ constant, then is injective at a0 , that is if b0 ∈ L∞ () is such that (a0 )b0 = 0 then b0 ≡ 0. H 1 ()
(Here the space of quadratic forms on H 1/2 (∂) is endowed with its natural underlying norm topology: if q is a continuous quadratic form on H 1/2 (∂) then for some unique selfadjoint operator defined on H 1/2 (∂) one has q(ϕ) = (ϕ|ϕ)H 1/2 (∂) , and the norm of q is given by L(H 1/2 (∂)) ). Proof. The fact that is differentiable is a consequence of the fact that by Lemma 3.3 the mapping a0 → Qa0 is Gâteaux differentiable and its Gâteaux differential, namely b → q, is continuous. The fact that is analytic follows from Lemma 3.1 and the expression of (a0 ). Now assume that a0 is constant, so that div(a0 ∇u) = 0 means u = 0. Recall that we have denoted (see Section 2) H0 ( ) the space of harmonic functions: H0 ( ) := v ∈ H 1 (); v = 0 , which is actually the orthogonal space to H01 () in H 1 (). Note also that since a0 is constant, by Lemma 3.3 we have: b0 (x)|∇u0 (x)|2 dx, ( (a0 )b0 )(ϕ) = ∞ for u0 ∈ H0 ( ) such that u0 = ϕ on ∂. 2Therefore if b0 ∈ L () is such that for all u0 ∈ H0 ( ) one has b0 (x)|∇u0 (x)| dx = 0, then from this we conclude that b0 (x)∇u1 (x) · ∇u2 (x) dx = 0. (3.6) for all u1 , u2 ∈ H0 ( ),
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(This is the polarization of the quadratic form q). Now consider the special functions u1 , u2 defined to be u1 (x) := exp((i · ξ + η) · x),
u2 (x) := exp((iξ − η) · x),
(3.7)
where η, ξ are such that ξ, η ∈ RN ,
ξ · η = 0,
|ξ | = |η| = 0.
(3.8)
(Recall that since N ≥ 2, for any ξ = 0 there exists some η ∈ RN satisfying the above conditions). One checks easily that thanks to (3.8) the functions uj are harmonic in and thus belong to H0 ( ). However since ∇u1 (x) · ∇u2 (x) = −2|ξ |2 exp(2iξ · x), identity (3.6) implies that b0 (x)e2iξ ·x dx = 0 = (1 (x)b0 (x))e2iξ ·x dx. ∀ξ ∈ RN ,
RN
This means that the Fourier transform of 1 b0 , namely F(1 b0 ), is zero, which in turn, using the injectivity of the Fourier transform, yields b0 ≡ 0. Remark 3.5. If we could prove that (a0 ) is a homeomorphism, then the implicit functions theorem would yield that in a neighborhood B(a0 , ε1 ) of a0 the inverse problem can be solved: for a ∈ B(a0 , ε1 ) knowledge of Qa would allow us to recover a. Unfortunately at this point we are not able to prove such a result concerning (a0 )… Remark 3.6. The above analysis contains the ingredients of an approximation process to recover a conductivity a, which is a small perturbation of a constant, when the quadratic form Qa is known. Indeed let a0 ≡ 1 (for instance) and assume that a = 1 + b ∈ Aad is close to a0 . For j = 1, 2 and given ϕj ∈ H 1/2 (∂) let uj , u0j ∈ H 1 () satisfy div(a∇uj ) = div(a0 ∇u0j ) = u0j = 0 in ,
uj = u0j = ϕj on ∂.
Since Qa is known on H 1/2 (∂), so is the bilinear form 1 [Qa (ϕ1 + ϕ2 ) − Qa (ϕ1 ) − Qa (ϕ2 )] . 2 If we denote vj := uj − u0j , as we did previously, we have ∇u0j · ∇vk dx = 0 since vk ∈ H01 (), and therefore M(ϕ1 , ϕ2 ) = a(x)∇u01 (x) · ∇u02 (x) dx (3.9) + [(a − a0 )(∇u01 · ∇v2 + ∇u02 · ∇v1 ) + a∇v1 · ∇v2 ] dx. M(ϕ1 , ϕ2 ) :=
Now choose u0j as in (3.7) with ξ, η satisfying (3.8) (this means that ϕj is the trace of u0j on ∂). Then, assuming that ⊂ B(0, R) for some R > 0, we obtain from (3.9) 2|ξ |2 a (2ξ ) = −(2π )−N/2 M(ϕ1 , ϕ2 ) + h(2ξ ),
(3.10)
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with h(2ξ ) := (2π )−N/2
[(a − a0 )(∇u01 · ∇v2 + ∇u02 · ∇v1 ) + a∇v1 · ∇v2 ] dx.
(Here we take as the definition of the Fourier transform f(ξ ) := (2π )−N/2 f (x) exp(−ix · ξ ) dx. RN
Note also that as a matter of fact we should have written F(1 a)(2ξ ) instead of a (2ξ ), but this is not misleading in our analysis). The error term h(2ξ ) may be estimated by |h(2ξ )| ≤ c1 |ξ |2 b 2∞ e2R|ξ | , since ∇vj 2 ≤ c b ∞ ∇u0,j 2 = c b ∞ |ξ |, provided b ∞ is small enough (actually if b ∞ < 1) to ensure that a = a0 + b ∈ Aad (see the proof of Lemma 3.1). If F is such that −N/2 M(ϕ , ϕ ) 1 2 (2ξ ) := −(2π ) , F 2|ξ |2
(2ξ ) + (2|ξ |2 )−1 h(2ξ ) and finally then we have a(2ξ ) = F (ξ ) + 2h(ξ ) . a (ξ ) = F |ξ |2
(3.11)
Now consider ρ(x) := (2π )−N/2 exp(−|x|2 /2) and ρσ (x) := σ N ρ(σ x), so that convolution with ρσ yields an approximation of the identity when σ is large. Since (ξ/σ ), and ρ σ (x) = ρ 2h(ξ ) (ξ ) a (ξ ) ρ (ξ/σ ) = F ρ (ξ/σ ) + ρ (ξ/σ ), |ξ |2 we have a ∗ ρσ = F ∗ ρσ + r, where the remainder r is defined via r(ξ ) := 2h(ξ ) ρ (ξ/σ )/|ξ |2 . Since r ∞ ≤ −N/2 r 1 , it follows that (2π) 2 r ∞ ≤ c2 b ∞ | ρ (ξ/σ )| exp(R|ξ |)dξ = c3 b 2∞ σ N exp(σ 2 R 2 /2). RN
Finally we see that for σ > 0 large, and b ∞ small enough, then F ∗ρσ is a reasonable approximation of a ∗ ρσ , which in turn, since σ is large, is a good approximation of a.
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4 Identification on the boundary Our second lecture is devoted to the recovery of the coefficient on the boundary. To be more specific let us introduce the following notations. For a ∈ Aad (which is defined in (2.1)) and ϕ ∈ H 1/2 (∂) we denote by a (ϕ) := a
∂u ∂n
(4.1)
where u ∈ H 1 () satisfies (3.1). The operator a , called the Dirichlet to Neumann map by some authors, and the Poincaré–Steklov operator by others, is a selfadjoint operator mapping H 1/2 (∂) −→ H −1/2 (∂). Indeed if −div(a∇uj ) = 0 in and uj = ϕj on the boundary for j = 1, 2, then multiplying by u2 the equation satisfied by u1 , and vice versa, after integrating by parts we get a(x)∇u1 (x) · ∇u2 (x) dx a (ϕ1 ), ϕ2 = a(x)∇u2 (x) · ∇u1 (x) dx, a (ϕ2 ), ϕ1 =
that is a (ϕ1 ), ϕ2 = a (ϕ2 ), ϕ1 for all ϕ1 , ϕ2 ∈ H 1/2 (∂). This implies that a is a continuous selfadjoint operator mapping H 1/2 (∂) into H −1/2 (∂). Note also that Qa being defined in (3.4), we have for all ϕ ∈ H 1/2 (∂) Qa (ϕ) = a (ϕ), ϕ.
(4.2)
In this section we will show that when aj ∈ Aad is of class C 1 near the boundary ∂ for j = 0, 1, and if one has a0 = a1 ,
(4.3)
then a0 = a1 and ∇a0 = ∇a1 on the boundary. This result is due to R.V. Kohn & M. Vogelius [25] who prove actually much more: assuming that a0 and a1 are of class C k near the boundary for some k ≥ 1, then ∂ α a0 = ∂ α a1 on the boundary for |α| ≤ k, provided has enough regularity. In particular, assuming that the boundary and the coefficients a0 , a1 are real analytic, their result implies that a0 = a1 on . The idea of the proof is the following: assuming that (4.3) holds, but for instance a0 (σ0 ) > a1 (σ0 ) for some σ0 ∈ ∂, one considers appropriately chosen electric potentials ψm having an H 1/2 (∂) norm equal to 1 and small support around σ0 , and such that ψm H −1/2 = O(m−1 ). Then if −div(aj ∇uj m ) = 0 in on ∂, uj m = ψm after a minute analysis of the behavior of uj m as m → +∞, one shows that for large enough m’s one has Qa0 (ψm ) > Qa1 (ψm ). We begin by the construction of the appropriate test functions (ψm )m≥1 .
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Lemma 4.1. Let be a C 1,1 bounded domain and σ0 ∈ ∂ be given. Then there exists a sequence of functions (ψm )m≥1 such that ψm ∈ H 3/2 (∂) ∩ Cc1 (∂) and for some positive constants c1 , c2 : (i) supp(ψm+1 ) ⊂ supp(ψm ) and m≥1 supp(ψm ) = {σ0 }, (ii) ψm H 1/2 (∂) = 1, (iii)
c1 m(1+2s)/2
≤ ψm H −s (∂) ≤
c2 m(1+2s)/2
for − 1 ≤ s ≤ 1.
Proof. Via a local diffeomorphism we may assume that ∂ ⊂ RN −1 × {0} and that σ0 is the origin. Consider a function f∗ ∈ Cc∞ (R) such that f∗ is not identically zero and +1 supp(f∗ ) ⊂ [−1, +1], f∗ (t)dt = 0. (4.4) −1
Now for x := (x1 , . . . , xN−1 ) ∈ RN−1 and integers m ≥ 1 define fm (x ) :=
N−1
f∗ (mxi ).
i=1
In the following, for a function f defined on RN −1 , we denote by f(ξ ) its Fourier transform at ξ ∈ RN−1 . We see that supp(fm ) is contained in [−1/m, 1/m]N −1 , and since fm (x ) = f1 (mx ), we have fm (ξ ) = m−(N−1) f1 (ξ /m) and therefore for any s ≥ 0 (1 + |ξ |2 )−s |fm (ξ )|2 dξ fm 2H −s (RN −1 ) = N −1 R 1 = N−1 (1 + m2 |ξ |2 )−s |f1 (ξ )|2 dξ . (4.5) N −1 m R Note that since (1 + m2 θ 2 )−1 ≥ m−2 (1 + θ 2 )−1 for θ ≥ 0, we have c3 (1 + m2 |ξ |2 )−s |f1 (ξ )|2 dθ ≥ 2s . N −1 m R
(4.6)
On the other hand the mean value of f∗ being zero, we have f1 (0) = 0. Whence using the fact that f1 is analytic we have | · |−1 f1 ∈ S(RN −1 ) (remember that f1 has a compact support), and recalling the inequality (1 + m2 θ 2 )−1 ≤ θ −2 m−2 for θ > 0, we obtain the upper bound for 0 ≤ s ≤ 1 1 c4 |f1 (ξ )|2 dξ (1 + m2 |ξ |2 )−s |f1 (ξ )|2 dξ ≤ 2s = 2s . (4.7) 2s m |ξ | m RN −1 RN−1 This and (4.6) plugged into (4.5) yield that for 0 ≤ s ≤ 1 c5 (s) m
−(N−1) −s 2
≤ fm H −s (RN−1 ) ≤ c6 (s) m
−(N−1) −s 2
.
(4.8)
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The same line of arguments shows that for any s ≥ 0 we have c1 (s)m
−(N−1) +s 2
≤ fm H s (RN−1 ) ≤ c2 (s)m
−(N−1) +s 2
.
(4.9)
Finally from (4.9) and (4.8) we obtain for s ≥ −1: c7 (s) m
−(N−1) +s 2
≤ fm H s (RN−1 ) ≤ c8 (s) m
−(N−1) +s 2
,
and upon setting ψm (x ) :=
fm (x ) , fm H 1/2 (RN−1 )
we conclude that for −1 ≤ s ≤ 1 we have c9 (s) m−(1−2s)/2 ≤ ψm H s (RN−1 ) ≤ c10 (s) m−(1−2s)/2 , and that the sequence (ψm )m≥1 satisfies the conditions of the lemma, at least when the boundary ∂ is a portion of RN−1 × {0}. The general case is treated via a local diffeomorphism and the resulting sequence (ψm )m≥1 will inherit the regularity of that is, with our assumptions, each ψm will be in C 1,1 (∂) ∩ H 3/2 (∂). In what follows, σ0 ∈ ∂ and (ψm )m≥1 being as in Lemma 4.1 and a ∈ L∞ () so that a ≥ ε0 for some ε0 > 0, we shall denote by um ∈ H 1 () the solution of −div(a∇um ) = 0 in (4.10) on ∂. um = ψm Our purpose in the following lemma is to show that, in some sense, um concentrates around σ0 , more precisely that, far from σ0 , the H 1 -norm of um is small. Lemma 4.2. Assume that is a bounded C 1,1 domain and that the coefficient a in (4.10) is in W 1,∞ (B(σ0 , R) ∩ ). Then if ω := \ B(σ0 , R), there exist m0 ≥ 1 and a constant c > 0 such that for all m ≥ m0 one has um H 1 (ω) ≤
c . m
(4.11)
Proof. First we may take m0 ≥ 1 large enough so that the support of ψm is contained in B(σ0 , R/8) ∩ ∂ for all m ≥ m0 . Then we set ω0 := \ B(σ0 , R/4) and choose a cut-off function ζ ∈ Cc∞ (RN ) such that 0 ≤ ζ ≤ 1, and ζ ≡ 1 in ω,
ζ ≡ 0 in B(σ0 , R/2).
Now consider vm := ζ um and observe that vm = 0 on ∂ω0 and, since div(a∇um ) = 0, −div(a∇vm ) = −a∇ζ · ∇um − div(aum ∇ζ ).
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Multiplying this equation by vm = ζ um , and integrating by parts over ω0 , we get 2 a(x)|∇vm (x)| dx = − a(∇ζ · ∇um )um ζ dx + aum ∇ζ · ∇(ζ um ) dx ω0 ω0 ω0 = a(x)um (x)2 |∇ζ (x)|2 dx, ω0
and finally, observing that um = vm on ω ⊂ ω0 , 2 |∇um (x)| dx ≤ c |um (x)|2 |∇ζ (x)|2 dx. ω
(4.12)
Now we need to estimate the right hand side of (4.12) in terms of ψm H −1/2 . To this end consider ∈ H01 () solution to −div(a∇) = um |∇ζ |2 in ,
= 0 on ∂.
(4.13)
Since a ∈ W 1,∞ (B(σ0 , R) ∩ ), and ψm ∈ H 3/2 (B(σ0 , R) ∩ ∂)), by the local regularity results of solutions to elliptic equations (see Section 2) we know that is in H 2 (B(σ0 , R) ∩ ) and therefore ∂/∂n is in H 1/2 (B(σ0 , R) ∩ ∂). Moreover we have
∂
a ≤ c H 2 (B(σ0 ,R)∩) ≤ c um |∇ζ |2 L2 () . (4.14)
∂n 1/2 H
(B(σ0 ,R)∩∂)
Multiplying equation (4.13) by um and integrating by parts we have: ∂(σ ) 2 2 ψm (σ ) dσ um (x) |∇ζ (x)| dx = a∇ · ∇um dx − a ∂ ∂n(σ ) ∂(σ ) a ψm (σ ) dσ, =− ∂ ∂n(σ ) where we have used the fact that since um satisfies (4.10) and ∈ H01 () we have a∇ · ∇um dx = 0. It follows that
∂
um (x)2 |∇ζ (x)|2 dx ≤ c
a ψm H −1/2 (∂) ,
∂n 1/2 H (B(σ0 ,R)∩∂) which, using (4.14) and the fact that |∇ζ |2 ≤ c|∇ζ |, yields um ∇ζ 2L2 ≤ c um |∇ζ |2 L2 ψm H −1/2 (∂) ≤ c um ∇ζ L2 ψm H −1/2 (∂) . Finally the estimate on ψm given by Lemma 4.1 together with inequality (4.12) prove that ∇um L2 (ω) ≤ c/m for m ≥ m0 . Since um = 0 on ∂ω ∩ ∂, by Poincaré inequality we have um L2 (ω) ≤ c(ω, ) ∇um L2 (ω) , and the lemma is proved. Our next step will be to show that in the ball B(σ0 , R) the solution um is not too small:
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Lemma 4.3. With the notations and assumptions of Lemma 4.2 there exist m1 ≥ m0 and a constant c0 > 0 depending on R such that for all m ≥ m1 |∇um (x)|2 dx ≥ c0 . B(σ0 ,R)∩
Proof. Recall that if we denote by γ0 : H 1 () −→ H 1/2 (∂) the trace operator u → u|∂ , there exists a positive constant c such that for all u ∈ H 1 () we have (see Section 2) γ0 (u) 2H 1/2 (∂) ≤ c ∇u 22 + γ0 (u) 2L2 (∂) . Applying this to um for m ≥ m0 , one sees that since γ0 (um ) = ψm and ψm L2 (∂) ≤ cm−1/2 (cf. Lemma 4.1): c 1 = γ0 (um ) 2H 1/2 (∂) ≤ c ∇um 22 + m c c |∇um (x)|2 dx + 2 + . ≤c m m B(σ0 ,R)∩ Clearly, upon taking m1 ≥ m0 large enough and m ≥ m1 , this yields the claimed result of the lemma. We can now state and prove the first identification result we had in mind: namely that if a0 = a1 then a0 = a1 on the boundary. Theorem 4.4. Let be a bounded domain and σ∗ ∈ ∂ such that is C 1,1 in a neighborhood B(σ∗ , R∗ ) of σ∗ . Assume that aj ∈ L∞ () ∩ W 1,∞ ( ∩ B(σ∗ , R∗ )) with aj ≥ ε0 > 0 in . Let aj be as in (4.1) and assume that a0 = a1 . Then one has a0 = a1 on B(σ∗ , R∗ ) ∩ ∂. In particular if is C 1,1 and aj ∈ W 1,∞ (), the equality a0 = a1 implies that a0 = a1 on ∂. Proof. If there exists σ0 ∈ B(σ∗ , R∗ ) ∩ ∂ such that (for instance) a0 (σ0 ) > a1 (σ0 ), then the aj ’s being continuous in a neighborhood of σ0 , there exist β > 0 and R > 0 such that B(σ0 , R) ⊂ B(σ∗ , R∗ ) and ∀x ∈ B(σ0 , R) ∩ ,
a0 (x) ≥ a1 (x) + β.
Now denote by uj m ∈ H 1 () the solution of −div(aj ∇uj m ) = 0 uj m = ψm
in on ∂.
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Denoting B := B(σ0 , R) ∩ we have a0 (x)|∇u0m |2 dx ≥ a0 (x)|∇u0m |2 dx Qa0 (ψm ) = B a1 (x)|∇u0m |2 dx + β |∇u0m |2 dx ≥ B B 2 a1 (x)|∇u0m | dx + βc0 ≥ B c ≥ a1 (x)|∇u0m |2 dx − 2 + βc0 , m provided m ≥ m1 , where m1 is given by Lemma 4.3 (in the last estimate we have used Lemma 4.2, that is \B a1 (x)|∇u0m (x)|2 dx ≤ cm−2 ). It follows that since by definition of u1m (we denote by Kψm the convex set Kψm := ψm + H01 () := {v ∈ H 1 (); v = ψm on ∂}) a1 (x)|∇u0m |2 dx ≥ min a1 (x)|∇v|2 dx v∈Kψm = a1 (x)|∇u1m |2 dx = Qa1 (ψm ),
if m2 ≥ m1 is large enough so that for m ≥ m2 we have cm−2 ≤ βc0 /2, we have 1 1 Qa0 (ψm ) ≥ a1 (x)|∇u0m |2 dx + βc0 ≥ Qa1 (ψm ) + βc0 . (4.15) 2 2 Since a0 = a1 , we have Qa0 (ψm ) = Qa1 (ψm ), which is in contradiction with the above inequality. Therefore we have a0 = a1 on B(σ∗ , R∗ ) ∩ ∂ and the proposition is proved. In order to prove that the assumption a0 = a1 implies that a0 = a1 at order one on the boundary, that is a0 = a1 and ∇a0 = ∇a1 on ∂, we need to show a little bit more than the estimate of Lemma 4.3. Indeed, since we know that a0 = a1 on the boundary, the tangential derivatives of a0 and a1 coincide and so it remains to show that ∂a0 /∂n = ∂a1 /∂n. If this were not so, we would find β > 0 such that in a neighborhood B := B(σ0 , R) ∩ of some σ0 ∈ ∂: ∀x ∈ B,
a0 (x) ≥ a1 (x) + βdist(x, ∂).
(Here dist(x, ∂) denotes the distance of x to the boundary). After a detailed inspection that if we knew of the above proof of Proposition 4.4, one is convinced that B dist(x, ∂)|∇u0m |2 dx has a greater magnitude than \B a1 (x)|∇u0m |2 dx (which is O(m−2 )), then the whole line of the above argument might be carried out, and we would obtain a contradiction to the fact that a0 = a1 , or equivalently Qa0 = Qa1 .
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Remark 4.5. One can see through the above proof that the identification result is in some sense local. This means that if for all ϕ ∈ H 1/2 (∂) with supp(ϕ) ⊂ B(σ, R∗ ) ∩ ∂ one has a0 (ϕ) = a1 (ϕ), then a0 = a1 on B(σ∗ , R∗ ) ∩ ∂. In order to obtain the estimate we are lacking for now, we shall use a special function equivalent to dist(x, ∂). Indeed let us denote by ϕ1 ∈ H01 () the first eigenfunction of the operator u → −div(a∇u) with Dirichlet boundary condition, that is ϕ1 (x)2 dx = 1. (4.16) −div(a∇ϕ1 ) = λ1 ϕ1 > 0 in , ϕ1 ∈ H01 (),
It is a classical consequence of the Hopf maximum principle (see for instance D. Gilbarg and N. S. Trudinger [17]) that on a part B(σ0 , R) ∩ ∂ of the boundary which is C 1,1 one has ∂ϕ1 /∂n ≤ −c∗ (R, ) < 0 and ∀x ∈ B(σ0 , R) ∩ ,
c0 dist(x, ∂) ≤ ϕ1 (x) ≤ c1 dist(x, ∂).
Lemma 4.6. With the notations and assumptions of Lemma 4.2 there exist m1 ≥ m0 and a constant c0 > 0 depending on R such that for all m ≥ m1 c0 . dist(x, ∂)|∇um (x)|2 dx ≥ m B(σ0 ,R)∩ Proof. Let ϕ1 be as in (4.16). Then we know that for some positive constants c∗ , c1∗ , c2∗ we have (here again B := B(σ0 , R) ∩ ) 1 ∀σ ∈ B(σ0 , R) ∩ ∂, − ∂ϕ ∂n ≥ c∗ > 0 (4.17) ∀x ∈ B, c1∗ dist(x, ∂) ≤ ϕ1 (x) ≤ c2∗ dist(x, ∂). Assuming that m ≥ m0 is large enough so that supp(ψm ) ⊂ B(σ0 , R) ∩ ∂, and multiplying (4.10) by um ϕ1 , which belongs to H01 () since ϕ1 is in W 1,∞ () and vanishes on ∂, we obtain 2 a(x)|∇um | ϕ1 (x) dx + a(x)um (x)∇um (x) · ∇ϕ1 (x) dx = 0. (4.18)
But noting that um ∇um = ∇(u2m )/2, we may integrate by parts the second term of the above equality to see that (we use also the equation satisfied by ϕ1 ): 1 ∂ϕ1 (σ ) dσ a(x)um (x)∇um (x) · ∇ϕ1 (x) dx = ψm (σ )2 a(σ ) 2 ∂n(σ ) ∂ λ1 + um (x)2 ϕ1 (x) dx. 2
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Inserting this into (4.18) yields: λ1 2 a(x)|∇um | ϕ1 (x) dx = − um (x)2 ϕ1 (x) dx 2 ∂ϕ1 1 dσ. ψm (σ )2 a(σ ) − 2 ∂ ∂n(σ )
(4.19)
Now we may use the Hopf maximum principle expressed in the first inequality of (4.17) to see that for some positive constant c, due to property (iii) of Lemma 4.1 with s = 0, we have ∂ϕ1 c 2 − dσ ≥ c∗ (4.20) ψm (σ ) a(σ ) a(σ )ψm (σ )2 dσ ≥ . ∂n(σ ) m ∂ ∂ As far as the first term in the right hand side of (4.19) is concerned, observe that if ∈ H01 () solves −div(a∇) = um ϕ1 in ,
= 0 on ∂,
after multiplying this equation by um and integrating by parts, we obtain (because um satisfies (4.10) and so we have a∇um · ∇ dx = 0) ∂(σ ) 2 um (x) ϕ1 (x) dx = − ψm (σ )a(σ ) dσ ∂n(σ ) ∂
∂
(4.21)
a ≤ c ψm H −1/2 (∂)
∂n 1/2 H (B(σ0 ,R)∩∂) 1/2
≤ c ψm H −1/2 (∂) um ϕ1 2 , where we have used the fact that ∈ H 2 (B) with B := B(σ0 , R) ∩ , and
∂
1/2
a
≤ c H 2 (B) ≤ c um ϕ1 2 ≤ c um ϕ1 2 ,
∂n 1/2 H
(B(σ0 ,R)∩∂)
thanks to classical regularity theorems (see Proposition 2.3) and the fact that ϕ1 ≤ 1/2 1/2 cϕ1 . Finally from (4.21) we obtain that um ϕ1 2 ≤ c ψm H −1/2 (∂) and using Lemma 4.1 with s = −1/2, the latter estimate shows that c um (x)2 ϕ1 (x) dx ≤ c ψm 2H −1/2 (∂) ≤ 2 . (4.22) m It is now plain that plugging estimates (4.20) and (4.22) into (4.19) finishes the proof of the lemma. At this point we can prove the following identification result. Theorem 4.7. Let be a bounded domain and σ∗ ∈ ∂ such that is C 1,1 in a neighborhood B(σ∗ , R∗ ) of σ∗ . Assume that aj ∈ L∞ () ∩ C 1 ( ∩ B(σ∗ , R∗ )) with aj ≥ ε0 > 0 in . Let aj be as in (4.1) and assume that a0 = a1 . Then one has
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a0 = a1 and ∇a0 = ∇a1 on B(σ∗ , R∗ ) ∩ ∂. In particular if is C 1,1 and aj ∈ C 1 (), the equality a0 = a1 implies that a0 = a1 and ∇a0 = ∇a1 on ∂. Proof. By Proposition 4.7 we know that a0 = a1 on B(σ∗ , R∗ )∩∂, therefore in order to prove that ∇a0 = ∇a1 on this portion of the boundary, it is enough to show that ∂a0 /∂n = ∂a1 /∂n. If there exists σ0 ∈ B(σ∗ , R∗ ) ∩ ∂ such that ∂a0 (σ0 )/∂n(σ0 ) = ∂a1 (σ0 )/∂n(σ0 ), then the aj ’s being C 1 and equal on B(σ∗ , R∗ ) ∩ ∂, there exist β > 0 and R > 0 such that B(σ0 , R) ⊂ B(σ∗ , R∗ ) and (for instance) ∀x ∈ B(σ0 , R) ∩ ,
a0 (x) ≥ a1 (x) + β dist(x, ∂).
We denote again by uj m the solution of div(aj ∇uj m ) = 0 and uj m = ψm on ∂, and by ϕ1 ∈ H01 () the eigenfunction considered in (4.16) with a := a0 . If B := B(σ0 , R) ∩ we have (here we use Lemma 4.6) 2 a0 (x)|∇u0m | dx ≥ a0 (x)|∇u0m |2 dx B 2 a1 (x)|∇u0m | dx + β dist(x, ∂)|∇u0m |2 dx ≥ B B βc0 a1 (x)|∇u0m |2 dx + ≥ m B c βc0 a1 (x)|∇u0m |2 dx − 2 + , ≥ m m provided m ≥ m1 , given by Lemma 4.6 (in the last estimate we have used Lemma 4.2). It follows that since by definition of u1m a1 (x)|∇u0m |2 dx ≥ Qa1 (ψm ),
for m2 ≥ m1 large enough so that for m ≥ m2 we have cm−1 ≤ βc0 /2, βc0 βc0 ≥ Qa1 (ψm ) + . a1 (x)|∇u0m |2 dx + Qa0 (ψm ) ≥ 2m 2m This is a contradiction with the fact that Qa0 (ψm ) = Qa1 (ψm ). Therefore we have ∂a0 /∂n = ∂a1 /∂n on B(σ∗ , R∗ ) ∩ ∂ and the theorem is proved. Remark 4.8. It is interesting to note that the arguments used to prove Theorem 4.4 may be used to give a stability estimate of the type a0 − a1 L∞ (∂) ≤ c(M) a0 − a1 L(H 1/2 (∂),H −1/2 (∂)) ,
(4.23)
for all aj ∈ W 1,∞ () such that aj W 1,∞ () ≤ M, where c(M) is a constant depending only on and M. Indeed assume that a0 , a1 ∈ W 1,∞ () are such that a0 −a1 ≡ 0 on the boundary and that if 2β := a0 − a1 L∞ (∂) , for some σ0 ∈ ∂ and R > 0
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we have ∀ x ∈ B(σ0 , R) ∩ ,
a0 (x) ≥ a1 (x) + β.
Then according to (4.15) for some constant c0 > 0 we have that 1 βc0 ≤ Qa0 (ψm ) − Qa1 (ψm ) = (a0 − a1 )ψm , ψm 2 ≤ a0 − a1 L(H 1/2 (∂),H −1/2 (∂)) since ψm H 1/2 (∂) = 1. It follows that (4.23) holds with c(M) := 4/c0 .
5 Identification of potentials in Schrödinger operators Our third lecture will treat a few more general identification problems for elliptic operators, through appropriate boundary measurements. For a ∈ Aad and q ∈ L∞ () denote by L := La,q the (formal) elliptic operator Lu := La,q u := −div(a∇u) + qu. It is known that (see Section 2) the subspace N(L) or N(La,q ) defined to be N (La,q ) := ϕ ∈ H01 (); La,q ϕ = 0 in H −1 () ,
(5.1)
(5.2)
is finite dimensional. Then we introduce the following way of saying: Definition 5.1. We shall say that zero is not an eigenvalue of La,q if N(La,q ) = {0}. If zero is not an eigenvalue of La,q then for any given ϕ ∈ H 1/2 (∂) there exists a unique u ∈ H 1 () such that −div(a∇u) + qu = 0 in (5.3) u=ϕ on ∂, and we may define the corresponding Poincaré–Steklov operator a,q ∂u (5.4) , a,q : H 1/2 (∂) −→ H −1/2 (∂). ∂n Now the question we may ask is the following: if for two sets of coefficients aj , qj (where j = 0, 1) we have a0 ,q0 = a1 ,q1 can we conclude that a0 = a1 and q0 = q1 ? Before trying to answer this question let us show that a,q is a continuous selfadjoint operator on H 1/2 (∂). a,q (ϕ) = a
Lemma 5.2. For given a ∈ Aad and q ∈ L∞ () assume that zero is not an eigenvalue of a,q defined by (5.4). Then a,q is a selfadjoint operator mapping H 1/2 (∂) into H −1/2 (∂).
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Proof. The fact that a,q is continuous from H 1/2 (∂) into H −1/2 (∂) is a consequence of Lemma 2.2. In order to see that a,q is selfadjoint, let u solve (5.3) and for ψ ∈ H 1/2 (∂) let v ∈ H 1 () solve −div(a∇v) + qv = 0 in ,
v = ψ on ∂.
Upon multiplying this equation by u, equation (5.3) by v and integrating by parts, one sees that: ∂v(σ ) ϕ(σ ) dσ a(x)∇v(x) · ∇u(x) dx + q(x)v(x)u(x) dx = a(σ ) ∂n(σ ) ∂ ∂u(σ ) ψ(σ ) dσ, a(x)∇u(x) · ∇v(x) dx + q(x)u(x)v(x) dx = a(σ ) ∂n(σ ) ∂ which yields that a,q ψ, ϕ = a,q ϕ, ψ, and hence ∗a,q = a,q . This gives the following important identity: Corollary 5.3. If aj ∈ Aad and qj ∈ L∞ () (for j = 0, 1) are such that zero is not an eigenvalue of aj ,qj , then for all ϕj ∈ H 1/2 (∂) if uj ∈ H 1 () solves (5.3) with a := aj and q := qj and ϕ := ϕj , we have: (a0 ,q0 − a1 ,q1 )ϕ0 , ϕ1 = (a0 (x) − a1 (x))∇u0 (x) · ∇u1 (x) dx (5.5) + (q0 (x) − q1 (x))u0 (x)u1 (x) dx.
Proof. Multiplying by u1 the equation satisfied by u0 , and vice versa, integrating by parts gives: a0 (x)∇u0 (x) · ∇u1 (x) dx + q0 (x)u0 (x)u1 (x) dx a0 ,q0 ϕ0 , ϕ1 = a1 (x)∇u1 (x) · ∇u0 (x) dx + q1 (x)u1 (x)u0 (x) dx. a1 ,q1 ϕ1 , ϕ0 =
Subtracting the second equality from the first, and using the fact that according to Lemma 5.2 one has a1 ,q1 ϕ1 , ϕ0 = a1 ,q1 ϕ0 , ϕ1 , we see that identity (5.5) is proved. This corollary, although quite simple to prove, is of the utmost importance in the study of inverse or identification problems. Indeed it shows that identification of the coefficients aj , qj boils down to the proof of an abstract density result. Indeed denote
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by E0 := E0 (a0 , q0 , a1 , q1 ) the space E0 := span (∇u0 · ∇u1 , u0 u1 ); uj ∈ H 1 (), Laj ,qj uj = 0 ,
(5.6)
in L1 () × L1 (). Then if one could prove that E0 is dense in L1 () × L1 (), one could conclude that if a0 ,q0 = a1 ,q1 then necessarily a0 = a1 and q0 = q1 : indeed by identity (5.5) we would infer that (a0 − a1 , q0 − q1 ) is orthogonal to E0 , and the density of this space in L1 () × L1 () would imply that (a0 − a1 , q0 − q1 ) = (0, 0) in L∞ () × L∞ (). More precisely, rather than the density of E0 in L1 () × L1 (), since E0 depends on a0 , q0 , a1 , q1 , we need to know that if (a0 − a1 , q0 − q1 ) ∈ E0 (a0 , a1 , q0 , q1 )⊥ , then (a0 , q0 ) = (a1 , q1 ). At this point we are not able to show such a general result for any two pairs of coefficients (aj , qj ). However we may state the following: Theorem 5.4. For aj ∈ Aad and qj ∈ L∞ () so that zero is not an eigenvalue of Laj ,qj , define the spaces V0 (q0 , q1 ) := span u0 u1 ; uj ∈ H 1 (), Laj ,qj uj = 0 V1 (a0 , a1 ) := span ∇u0 · ∇u1 ; uj ∈ H 1 (), Laj ,qj uj = 0 . Then if a0 = a1 and if V0 (q0 , q1 ) is dense in L1 (), the equality a0 ,q0 = a0 ,q1 implies that q0 = q1 in . Analogously if q0 = q1 and if V1 (a0 , a1 ) is dense in L1 (), the equality a0 ,q0 = a1 ,q0 implies that a0 = a1 in . Proof. Indeed if for instance a0 = a1 and a0 ,q0 = a0 ,q1 then thanks to identity (5.5) for any ϕj ∈ H 1/2 (∂) we have (q0 (x) − q1 (x))u0 (x)u1 (x) dx = 0.
This means that q0 − q1 is orthogonal to V0 (q0 , q1 ), and since the latter is dense in L1 () we conclude that q0 = q1 . The other claim of the theorem is seen to be true in an analogous manner. We observe now that a practical way to prove that some coefficients may be identified in inverse problems, would be to prove that the spaces V0 or V1 are dense in L1 (). In these lectures, due to lack of time we shall not consider this general problem (see [22] for a detailed treatment), but rather consider the simpler case a0 = a1 ≡ 1. The following results are essentially due to J. Sylvester and G. Uhlmann [38], although in part we shall use a simpler proof of one of their arguments, thanks to a simplification given by P. Hähner [18]. We begin with a result which says that there is a direct connection between the study of the general case and the special case a0 = a1 ≡ 1 and q0 , q1 ∈ L∞ ().
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Proposition 5.5. Let aj ∈ Aad ∩ W 2,∞ () and qj ∈ L∞ () (for j = 0, 1) be such that zero is not an eigenvalue of Laj ,qj . Define 1/2
aj qj + 1/2 qj := aj aj
and assume that a0 = a1 and ∇a0 = ∇a1 on the boundary ∂. Then a0 ,q0 = a1 ,q1 ⇐⇒ 1, q0 = 1, q1 . Proof. For a given ϕ ∈ H 1/2 (∂), if uj ∈ H 1 () satisfies −div(aj ∇uj ) + qj uj = 0 in ,
uj = ϕ on ∂ , 1/2
then using the Liouville’s transform, that is setting vj := aj uj , or equivalently −1/2
uj = aj
v, one checks that ∇aj 1/2 1/2 1/2 div(aj ∇uj ) = div aj ∇vj − 1/2 vj = div aj ∇vj − vj ∇(aj ) 2aj 1/2
aj 1/2
vj − 1/2 vj . = aj aj
Hence vj satisfies − vj +
1/2
aj qj + 1/2 aj aj
vj = 0 in ,
1/2
vj = aj ϕ on ∂ .
Observe also that since vj satisfies the above equation, it follows that 1/2
1, qj (aj ϕ) =
∂vj 1 −1/2 ∂aj 1/2 ∂uj = aj + aj ϕ 2 ∂n ∂n ∂n 1 −1/2 ∂aj −1/2 aj ,qj (ϕ) + aj = aj ϕ 2 ∂n
(5.7)
1/2
Since a0 = a1 and ∇a0 = ∇a1 on ∂, and due to the fact that ϕ → aj ϕ is a homeomorphism on H 1/2 (∂), we conclude that saying a0 ,q0 = a1 ,q1 is equivalent to say that 1, q0 = 1, q1 . Corollary 5.6. Let be a bounded C 1,1 domain. Assume that for qj ∈ L∞ () (with j = 0, 1) such that zero is not an eigenvalue of L1,qj we have the following property: 1,q0 = 1,q1 ⇒ q0 = q1 in .
(5.8)
Then if aj ∈ Aad ∩ W 2,∞ () are such that a0 ,0 = a1 ,0 , we have a0 = a1 in .
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Proof. First according to Theorem 4.4 and Theorem 4.7, we know that whenever a0 ,0 = a1 ,0 , then a0 = a1 and ∇a0 = ∇a1 on the boundary ∂. Next setting 1/2
qj :=
aj
1/2
,
aj
1 then clearly zero is not an eigenvalue of L1, qj , because saying that v ∈ H0 () satisfies − v + qj v = 0 is equivalent to say that div(aj ∇u) = 0 with u := a −1/2 v ∈ H01 (), and the latter equality implies u = 0 = v. Now Proposition 5.5 shows that if a0 ,0 = a1 ,0 , since a0 and a1 , as well as ∇a0 and ∇a1 , are equal on the boundary, we have q0 = q1 in , but it is clear that if we set 1, q0 = 1, q1 . Property (5.8) implies that 1/2 1 q := qj , the functions aj (which are also in H ()) satisfy the elliptic equation
1/2
1/2
− aj + qaj aj = a0
=0
in on ∂.
Since zero is not an eigenvalue of L1,q it follows that the above equation has a unique solution, that is a0 = a1 in , and the corollary is proved. Now our objective will be to prove Property (5.8), which is true when the dimension N ≥ 3, and its proof is due to J. Sylvester and G. Uhlmann [38]. We state the following density theorem which will be proved in the next section. Theorem 5.7. Assume that ⊂ RN is a bounded Lipschitz domain with N ≥ 3. If qj ∈ L∞ () are such that zero is not an eigenvalue of L1,qj , then the space V0 (q0 , q1 ) := span u0 u1 ; uj ∈ H 1 () and − uj + qj uj = 0 in
(5.9)
is dense in L1 (). As a consequence if the Poincaré–Steklov operators 1,q0 and 1,q1 are equal, then q0 = q1 in . Remark 5.8. The case of dimension N = 2 is particular in two respects. First by a result of A.I. Nachman [33], which uses a completely different approach, one knows that if a0 and a1 are smooth enough then the equality of the Poincaré–Steklov operators a0 ,0 = a1 ,0 implies that a0 = a1 in . Using this result one may show that if q0 , q1 are such that the first eigenvalue of the operator (L1,qj , D(L1,qj )) is positive, then the fact that 1,q0 = 1,q1 implies q0 = q1 . However the case of dimension two is not thoroughly well understood: it is not known whether for a general pair q0 , q1 ∈ L∞ () the space V0 (q0 , q1 ) is dense in L1 (), nor is it known whether the functions q0 , q1 must be equal whenever 1,q0 = 1,q1 . Due to lack of time unfortunately we will not treat the case N = 2 in these lectures.
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6 Density of products of solutions In our fourth and last lecture we prove the density result invoked at the end of the previous lecture. In order to establish the density Theorem 5.7 the main idea, as pointed out in the original paper of J. Sylvester and G. Uhlmann [38], is the following. Since the exponential functions exp((iξ ±η)·x) used byA.P. Calderón (see Section 3, in particular (3.7) and (3.8)) are such that the space spanned by their product is dense, each of them being a harmonic function, one may consider the operators u → − u + qj u as perturbations of the Laplacian and prove that the space E(q0 , q1 ) defined in (5.9) is dense. The point is that this clever idea works, at least when N ≥ 3. However here we will follow the point of view presented in a later work by P. Hähner [18]. In what follows we consider a potential q ∈ L∞ () such that zero is not an eigenvalue of the operator L1,q . We wish to construct solutions of u ∈ H 1 (),
− u + qu = 0
(6.1)
such that u(x) = eζ ·x (1 + ψ(x)),
with ζ = iξ + η,
and ξ, η ∈ RN ,
|ξ | = |η|,
ξ · η = 0.
(6.2)
Since L1,q u = eζ ·x [− ψ − 2ζ · ∇ψ + q(1 + ψ)], in order for u to be a solution of (6.1), the new unknown function ψ should be a solution of ψ ∈ H 1 (),
− ψ − 2ζ · ∇ψ + qψ = −q.
(6.3)
Moreover it is desirable to have ψ small in some appropriate norm, at least when |ζ | → +∞, so that u is close to e(iξ +η)·x . We are going to solve (6.3) when ⊂ (−R, R)N for some R > 0 and in the special case ζ = se1 + iξ for some s ∈ R, s = 0, and ξ ∈ RN such that e1 · ξ = 0. Note that this is not any restriction since the operators are essentially invariant under translations and rotations of RN ; also we should say that we denote e1 := (1, 0, . . . , 0), and analogously ek will be the k-th vector of the standard basis of RN . We denote by Q the cube (−R, R)N and we introduce αj R 1 α1 R − ∈ Z, ∈ Z for j ≥ 2 . (6.4) Z0 := α ∈ RN ; π 2 π Observe that when α ∈ Z0 then α1 = 0, more precisely |α1 | ≥ π/2R and, as we shall see below, this is the reason for the choice of a shifted grid. Also note that we may write π π (6.5) e1 + ZN . Z0 = R 2R We recall here the classical definition of Q-periodic functions:
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1 (RN ) is said to be Q-periodic if for all 1 ≤ k ≤ N Definition 6.1. A function u ∈ Hloc it satisfies u(· + 2Rek ) = u(·). Such a function is completely determined by its values 1 (RN ) is denoted by H 1 (Q). on Q and the subspace of Q-periodic functions of Hloc per 1 (Q). A function u ∈ H 1 (Q) Equivalently one can give a local definition of Hper 1 is said to belong to Hper (Q), or to be Q-periodic, if for all 1 ≤ k ≤ N we have u(σ + 2Rek ) = u(σ ) for σ ∈ ∂Q (since the trace u|∂Q is defined, the condition makes sense). In this case, a Q-periodic function in the latter sense can be extended by 1 (Q) coincide. periodicity to all RN , and one may check that the two definitions of Hper 1 1 The space Hper (Q) is a closed subspace of H (Q) and it is endowed with the 2 (Q) is defined to be: natural topology of H 1 (Q). Analogously the space Hper 2 1 1 (Q) := u ∈ Hper (Q); ∂k u ∈ Hper (Q) for 1 ≤ k ≤ N . Hper
Next we define the notion of Z0 -quasiperiodic functions. 1 (RN ) is said to be Z -quasiperiodic if the function Definition 6.2. A function u ∈ Hloc 0 x → exp(−iπ x1 /2R)u(x) is Q-periodic. The space of Z0 -periodic functions which 1 (RN ) will be denoted by H 1 (Q). This is a closed subspace of H 1 (Q). are in Hloc Z0
One may check that this is a closed subspace of H 1 (Q). Analogously the space 2 (RN ). Equivis the space of Z0 -quasiperiodic functions which are in Hloc 1 2 alently, the space HZ0 (Q) (resp. HZ0 (Q)) can be defined as the space generated by 1 (Q) (resp. ϕ ∈ H 2 (Q)). exp(iπx1 /2R)ϕ for ϕ ∈ Hper per For α ∈ Z0 we denote by ϕα the function HZ20 (Q)
ϕα (x) := (2R)−N/2 eiα·x .
(6.6)
One checks easily that each ϕα is Z0 -quasiperiodic, that (ϕα )α∈Z0 is a Hilbert basis of L2 (Q) and that for any α ∈ ZN ∇ϕα = iαϕα ,
ϕα = −|α|2 ϕα .
A useful observation about Q-periodic functions is that when u, v ∈ C 1 (Q) ∩ then we have ∂Q u(σ )v(σ )nk (σ ) dσ = 0 (because σ → nk (σ ) is anti periodic on ∂Q) and therefore Q ∇u(x) v(x) dx = − Q u(x) ∇v(x) dx. Also for 2 (Q), since σ → ∇u(σ ) · n(σ ) is anti-periodic, we have u, v ∈ C 2 (Q) ∩ Hper ∂v(σ ) ∂u(σ ) ∂u(σ ) v(σ ) dσ = 0, and v(σ ) − u(σ ) dσ = 0, ∂n(σ ) ∂Q ∂n(σ ) ∂Q ∂n(σ )
1 (Q) Hper
and hence
u(x) v(x) dx =
Q
u(x) v(x) dx. Q
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For our study of equation (6.3) it is useful to point out that if u, v ∈ C 2 (Q) are Z0 -quasiperiodic, then setting ϕ(x) := exp(−iπ x1 /2R)u(x),
ψ(x) := exp(−iπ x1 /2R)v(x),
we have two Q-periodic functions and iπ exp(iπ x1 /2R)ϕ(x)e1 2R iπ exp(iπ x1 /2R)ψ(x)e1 . ∇v = exp(iπ x1 /2R)∇ψ(x) + 2R
∇u = exp(iπ x1 /2R)∇ϕ(x) +
Hence one can easily verify that ∂v(σ ) ∂ψ(σ ) ∂u(σ ) ∂ϕ(σ ) dσ = dσ v(σ ) − u(σ ) ψ(σ ) − ϕ(σ ) ∂n(σ ) ∂n(σ ) ∂Q ∂n(σ ) ∂Q ∂n(σ ) iπ ϕ(σ )ψ(σ ) e1 · n(σ ) dσ + R ∂ = 0. Therefore for any u, v ∈ C 2 (Q) which are Z0 -quasiperiodic we have ∇u(x) v(x) dx = − u(x) ∇v(x) dx Q Q
u(x) v(x) dx = u(x) v(x) dx. Q
(6.7)
Q
Invoking a density argument, one is convinced that the above relations are still valid when the derivatives exist in the weak sense (the first equality in (6.7) holds for u, v ∈ HZ10 (Q), while the second one is valid for u, v ∈ HZ20 (Q) which may be defined in much the same way as HZ10 (Q)). We are now in a position to solve an elliptic equation which corresponds to the differential part of equation (6.3). Proposition 6.3. Let s ∈ R, s = 0 and ξ ∈ RN such that ξ · e1 = 0 and set ζ := iξ + se1 . Then for any f ∈ L2 (Q) there exists a unique ψ ∈ HZ10 (Q) ∩ H 2 (Q) such that − ψ − 2ζ · ∇ψ = f.
(6.8)
Moreover we have ψ L2 (Q) ≤
R f L2 (Q) . π |s|
Proof. Using the Hilbert basis (ϕα )α∈Z0 defined in (6.6), one sees that equation (6.8) is equivalent to: ∀α ∈ Z0 ,
(− ψ − 2ζ · ∇ψ|ϕα ) = (f |ϕα ),
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where ( ·| · ) denotes the scalar product of L2 (Q). Using the above remarks summarized in (6.7), we have (− ψ − 2ζ · ∇ψ|ϕα ) = (α · α − 2iζ.α) (ψ|ϕα ). It follows that, provided α · α − 2iζ · α = 0, we can express (ψ|ϕα ) in terms of (f |ϕα ). However this is the case, since iζ · α = isα1 − ξ · α and therefore |α · α − 2iζ · α| ≥ |Im(α · α − 2iζ · α)| = |2sα1 | ≥
|s|π . R
Therefore we have (ψ|ϕα ) = (f |ϕα )/(α · α − 2iζ · α) and (f |ϕα ) ϕα . ψ= α · α − 2iζ · α α∈Z0
This, since (ϕα )α is a Hilbert basis of L2 (Q) , yields the estimate ψ 2L2 (Q) =
α∈Z0
R2 |(f |ϕα )|2 ≤ f 2L2 (Q) , |α · α − 2iζ · α|2 |s|2 π 2
and, using the regularity results for elliptic equations, we have also ψ ∈ H 2 (Q). Corollary 6.4. Let ζ := iξ + η where ξ, η ∈ RN are such that ξ · η = 0 and η = 0. Then if ⊂ RN with N ≥ 2 is a bounded domain, there exists a linear continuous operator Bζ : L2 () −→ H 2 () such that for any f ∈ L2 (), the function ψ := Bζ (f ) satisfies ψ ∈ H 2 (),
− ψ − 2ζ · ∇ψ = f.
(6.9)
Moreover for some constant c > 0, independent of ξ, η and f , we have c f 2 . Bζ (f ) 2 ≤ |η| Proof. Without loss of generality we may assume that 0 ∈ , and we may fix R > 0 such that for any rotation S of RN around the origin we have S() ⊂ (−R, R)N . For a given η ∈ RN \ {0}, we may set s := |η| and find a rotation S in RN such that Se1 = η/s. If f : −→ C is a function in L2 (), first we may extend it to all the space RN by setting f(x) := 0 whenever x ∈ / and f(x) = f (x) for 1 (RN ), upon setting N x ∈ . Next one checks that if u : R −→ R is a function in Hloc ∗ v(x) := u(Sx) one has ∇v(x) = S (∇u)(Sx) and ζ0 · ∇v(x) = (Sζ0 ) · (∇u)(Sx),
v(x) = u(Sx).
It follows that if we set ζ0 := iS ∗ ξ + S ∗ η, then in order to solve (6.9) in it is enough to set g(x) := f(Sx) and solve v ∈ H 2 (Q) ∩ HZ10 (Q),
− v − 2ζ0 · ∇v = g.
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Since this has been done in Proposition 6.3 with v L2 (Q) ≤ c g L2 (Q) /|η|, one sees that if we set Bζ (f ) := v| , all the requirements of the corollary are satisfied. Now we can solve equation (6.3) provided |ζ | is large enough. Proposition 6.5. Let ζ := iξ + η where ξ, η ∈ RN are such that ξ · η = 0 and η = 0. Then if ⊂ RN (with N ≥ 2) is a bounded domain and q ∈ L∞ (), there exist a constant c0 > 0 and for |η| ≥ c0 a linear continuous operator Mζ : L2 () −→ H 2 () such that for any f ∈ L2 (), the function ψ := Mζ (f ) satisfies ψ ∈ H 2 (),
− ψ − 2ζ · ∇ψ + qψ = f.
(6.10)
Moreover for some constant c > 0, independent of ξ, η and f , we have c f 2 . ψ 2 = Mζ (f ) 2 ≤ |η| Proof. The operator Bζ being as in Corollary 6.4, we note that solving (6.10) is equivalent to solve − ψ − 2ζ · ∇ψ = f − qψ and hence ψ = Bζ (f ) − Bζ (qψ).
(6.11)
If we show that for |η| large enough ψ → Bζ (qψ) is a contraction from L2 () into itself, then the mapping ψ → Bζ (f ) − Bζ (qψ) has a unique fixed point and equation (6.11), and thus (6.10), has a unique solution. But Bζ (qψ) 2 ≤ c|η|−1 qψ 2 ≤ c|η|−1 q ∞ ψ 2 , and therefore when, for instance |η| ≥ 2c q ∞ , we have Bζ (qψ) 2 ≤ 2−1 ψ 2 and thus equation (6.11) has a unique solution. Also if we set Aζ (ψ) := Bζ (qψ) we have Aζ ≤ 1/2 for |η| ≥ 2c q ∞ and the solution ψ of (6.11) is given by ψ = (I + Aζ )−1 Bζ (f ). Whence if we set Mζ := (I + Aζ )−1 Bζ , we have ψ 2 = Mζ (f ) 2 ≤ (I + Aζ )−1 Bζ (f ) 2 ≤
2c f 2 , |η|
since (I + Aζ )−1 ≤ 2. Finally using classical regularity results in equation (6.10) one sees that ψ ∈ H 2 (), and the proposition is proved. We are now in a position to prove the result announced in Theorem 5.7. Theorem 6.6. Theorem. Assume that ⊂ RN is a bounded domain and that N ≥ 3. For qj ∈ L∞ () (with j = 0, 1) let V0 (q0 , q1 ) be the space V0 (q0 , q1 ) := u0 u1 ; uj ∈ H 1 (), and − uj + qj uj = 0 . Then V0 (q0 , q1 ) is dense in L1 (). Proof. Let ξ ∈ RN with ξ = 0 be given. It is interesting to emphasize that the assumption N ≥ 3 comes in at this point: when the dimension N ≥ 3, there exist
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at least two other directions η, σ orthogonal to ξ and such that η is orthogonal to σ . That is we may fix η and σ ∈ RN such that for a given t > 0 we have: η · ξ = ξ · σ = 0,
|σ | = 1,
|η|2 = |ξ + tσ |2 = |ξ |2 + t 2 .
Then consider ζ0 := η + i(ξ + tσ ),
ζ1 := −η + i(ξ − tσ ),
so that x → exp(ζj · x) is harmonic in RN , and let ψj be the solution given by Proposition 6.5 solving (6.3). Or rather more precisely let ψj satisfy: ψj ∈ H 2 (),
− ψj − 2ζj · ∇ψj + qj ψj = −qj ,
in such a way that uj := (1 + ψj ) exp(ζj · x) solves uj ∈ H 2 (),
− uj + qj uj = 0,
and u0 u1 ∈ V0 (q0 , q1 ). In order to show that V0 (q0 , q1 ) is dense in L1 (), we ∞ have to prove that if for some h ∈ L () and for all w ∈ V0 (q0 , q1 ) we have h(x)w(x) dx = 0, then necessarily h = 0. Now choosing w := u0 u1 , we have h(x)u0 (x)u1 (x) dx = h(x)(1 + ψ0 (x))(1 + ψ1 (x)) exp((ζ0 + ζ1 ) · x) dx = 0,
and since ζ0 + ζ1 = 2iξ , we conclude that h(x)e2iξ ·x dx = − h(x)e2iξ ·x (ψ0 (x) + ψ1 (x) + ψ0 (x)ψ1 (x)) dx.
Therefore
h(x)e2iξ ·x dx ≤ h ∞ ( ψ0 2 + ψ1 2 + ψ0 2 ψ1 2 ) ,
and using the estimates ψj 2 ≤ c|η|−1 and the fact that |η| ≥ t we conclude, as 2iξ t → +∞, that h(x)e ·x dx = 0, and hence h = 0.
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